1. Electronica De Potencia Rashid

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Electronica De Potencia Rashid

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• 1. www.elsolucionario.net

2. Chapter 1 Exercise Problems EX1.1 Eg ni = BT 3 / 2 exp 2kT GaAs: ni = ( 2.1 1014 ) ( 300 ) Ge: ni = (1.66 1013 ) ( 300 )3/ 23/ 2 1.4 or ni = 1.8 106 cm 3 exp 2 ( 86 106 ) ( 300 ) 0.66 or ni = 2.40 1013 cm 3 exp 2 ( 86 106 ) ( 300 ) EX1.2 (a) majority carrier: holes, po = 1017 cm 3 minority carrier: electrons, n 2 (1.5 10 no = i = 1017 po)10 2= 2.25 103 cm 3(b) majority carrier: electrons, no = 5 1015 cm 3 minority carrier: holes, n 2 (1.5 10 ) = 4.5 104 cm 3 po = i = 5 1015 no 10 2EX1.3 For n-type, drift current density J en nE or 200 = (1.6 1019 ) ( 7000 ) (1016 ) E which yields E = 17.9 V / cmEX1.4 Diffusion current density due to holes: dp J p = eD p dx 1 x = eD p (1016 ) exp L L p p (a) At x = 0(1.6 10 ) (10 ) (10 ) = 16 A / cm = 19Jp162103 3 (b) At x = 10 cm 103 J p = 16 exp 3 = 5.89 A / cm 2 10 EX1.5 N N Vbi = VT ln a 2 d ni (1016 )(1017 ) or Vbi = 1.23 V = ( 0.026 ) ln (1.8 106 )2 EX1.6 V C j = C jo 1 + R Vbi and1/ 2www.elsolucionario.net 3. N N Vbi = VT ln a 2 d ni (1017 )(1016 ) = 0.757 V = ( 0.026 ) ln (1.5 1010 )2 5 Then 0.8 = C jo 1 + 0.757 or C jo = 2.21 pF1/ 2= C jo ( 7.61)1/ 2EX1.7 v iD = I S exp D 1 VT v so 103 = (1013 ) exp D 1 0.026 103 Solving for the diode voltage, we find vD = ( 0.026 ) ln 13 + 1 10 or vD ( 0.026 ) ln (1010 )which yields vD = 0.599 V EX1.8 V VPS = I D R + VD and I D I S exp D VT ( 4 VD ) so 4 = I D ( 4 103 ) + VD I D = 4 103 and V I D = (10 12 ) exp D 0.026 By trial and error, we find I D 0.864 mA and VD 0.535 VEX1.9(a)ID =(b)ID =Then R =VPS V R VPS V R5 0.7 I D = 1.08 mA 4 VPS V R= ID =8 0.7 = 6.79 k 1.075(c)www.elsolucionario.net 4. ID(mA) Diode curve 1.25 1.08Load lines (b) (a)00.724 VD(v)68EX1.10 PSpice analysis EX1.11Quiescent diode current I DQ =VPS V=10 0.7 = 0.465 mA 20R Time-varying diode current: V 0.026 We find that rd = T = = 0.0559 k I DQ 0.465 Then id =vI 0.2sin t (V ) = or id = 9.97sin t ( A) rd + R 0.0559 + 20 ( k )EX1.12 I 1.2 103 or VD = 0.6871 V For the pn junction diode, VD VT ln D = ( 0.026 ) ln 15 4 10 IS The Schottky diode voltage will be smaller, so VD = 0.6871 0.265 = 0.4221 V V Now I D I S exp D VT or 1.2 103 IS = I S = 1.07 1010 A 0.4221 exp 0.026 EX1.13 P = I VZ 10 = I ( 5.6 ) I = 1.79 mAAlso I =10 5.6 = 1.79 R = 2.46 k RTest Your Understanding Exercises TYU1.1 (a) T = 400K Eg Si: ni = BT 3 / 2 exp 2kT ni = ( 5.23 1015 ) ( 400 )3/ 2 1.1 exp 6 2 ( 86 10 ) ( 400 ) or ni = 4.76 1012 cm 3www.elsolucionario.net 5. Ge: ni = (1.66 1015 ) ( 400 )3/ 2 0.66 exp 6 2 ( 86 10 ) ( 400 ) or ni = 9.06 1014 cm 3 GaAs: ni = ( 2.1 1014 ) ( 400 )3/ 2 1.4 exp 6 2 ( 86 10 ) ( 400 ) or ni = 2.44 109 cm 3 (b) T = 250 K Si: ni = ( 5.23 1015 ) ( 250 )3/ 2 1.1 exp 2 ( 86 106 ) ( 250 ) or ni = 1.61 108 cm 3 Ge: ni = (1.66 1015 ) ( 250 )3/ 2 0.66 exp 6 2 ( 86 10 ) ( 250 ) or ni = 1.42 1012 cm 3 GaAs: ni = ( 2.10 1014 ) ( 250 )3/ 2 1.4 exp 2 ( 86 106 ) ( 250 ) or ni = 6.02 103 cm 3 TYU1.2 (a) n = 5 1016 cm 3 , p 0 V = 0 Voltage across RL + R1 = vi RL 1 Voltage Divider v0 = vi = vi 2 RL + R1 www.elsolucionario.net 38. 0 202.13 For vi > 0, (V = 0 )iR1 0 R2RLa. R2 RL v0 = vi R2 RL + R1 R2 RL = 2.2 6.8 = 1.66 k0 1.66 v0 = vi = 0.43 vi 1.66 + 2.2 4.3v0 ( rms ) =b.v0 ( max ) 2 v0 ( rms ) = 3.04 V2.14 3.9 I 2 = 0.975 mA 4 20 3.9 IR = = 1.3417 mA 12 I Z = 1.3417 0.975 I Z = 0.367 mA IL =PT = I Z VZ = ( 0.367 )( 3.9 ) PT = 1.43 mW2.15 (a) 40 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 WIZ =(b)IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A 12 So 0.21 = RL = 57.1 RL P = ( 0.1)( 0.233)(12 ) P = 0.28 W(c) 2.16Ri VI II IZ VZ V0 RL ILVI = 6.3 V, Ri = 12, VZ = 4.8www.elsolucionario.net 39. a. 6.3 4.8 125 mA 12 I L = I I I Z = 125 I Z II =25 I L 120 mA 40 RL 192b. PZ = I Z VZ = (100 )( 4.8 ) PZ = 480 mW PL = I LV0 = (120 )( 4.8 ) PL = 576 mW2.17 a. 20 10 I I = 45.0 mA 222 10 IL = I L = 26.3 mA 380 I Z = I I I L I Z = 18.7 mA II =b. PZ ( max ) = 400 mW I Z ( max ) = I L ( min ) = I I I Z ( max ) = 45 40 I L ( min ) = 5 mA =400 = 40 mA 1010 RL RL = 2 k(c)For Ri = 175 I I = 57.1 mAI L = 26.3 mAI Z = 30.8 mAI Z ( max ) = 40 mA I L ( min ) = 57.1 40 = 17.1 mA RL =10 RL = 585 17.12.18 a.From Eq. (2-31) 500 [ 20 10] 50 [15 10] I Z ( max ) = 15 ( 0.9 )(10 ) ( 0.1)( 20 ) 5000 250 4 I Z ( max ) = 1.1875 A I Z ( min ) = 0.11875 A =From Eq. (2-29(b)) Ri =20 10 Ri = 8.08 1187.5 + 50b. PZ = (1.1875 )(10 ) PZ = 11.9 WPL = I L ( max ) V0 = ( 0.5 )(10 ) PL = 5 W2.19 (a)As approximation, assume I Z ( max ) and I Z ( min ) are the same as in problem 2.18.V0 ( max ) = V0 ( nom ) + I Z ( max ) rZ = 10 + (1.1875)(2) = 12.375 V V0 ( min ) = V0 ( nom ) + I Z ( min ) rZ = 10 + (0.11875)(2) = 10.2375 Vwww.elsolucionario.net 40. % Reg =b. 2.20 % Reg = =12.375 10.2375 100% % Reg = 21.4% 10VL ( max ) VL ( min ) VL ( nom ) 100%VL ( nom ) + I Z ( max ) rz (VL ( nom ) + I Z ( min ) rz ) VL ( nom ) I Z ( max ) I Z ( min ) ( 3) = = 0.05 6 So I Z ( max ) I Z ( min ) = 0.1 A6 6 = 0.012 A, I L ( min ) = = 0.006 A 500 1000 VPS ( min ) VZNow I L ( max ) = Now Ri = or 280 =I Z ( min ) + I L ( max )15 6 I Z ( min ) = 0.020 A I Z ( min ) + 0.012Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri = or 280 =VPS ( max ) 6 0.12 + 0.006VPS ( max ) VZI Z ( max ) + I L ( min ) VPS ( max ) = 41.3 V2.21 Using Figure 2.21 a. VPS = 20 25% 15 VPS 25 V For VPS ( min ) :I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mARi =b.VPS ( min ) VZ II=15 10 Ri = 200 25For VPS ( max ) I I ( max ) =25 10 I I ( max ) = 75 mA RiFor I L ( min ) = 0 I Z ( max ) = 75 mAVZ 0 = VZ I Z rZ = 10 ( 0.025 )( 5 ) = 9.875 V V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25 V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90V0 = 0.35 Vc.% Reg =V0 100% % Reg = 3.5% V0 ( nom )2.22 From Equation (2.29(a)) VPS ( min ) VZ 24 16 or Ri = 18.2 Ri = = I Z ( min ) + I L ( max ) 40 + 400 Also Vr =VM VM C = 2 fRC 2 fRVrR Ri + rz = 18.2 + 2 = 20.2Thenwww.elsolucionario.net 41. C=24 C = 9901 F 2 ( 60 )(1)( 20.2 )2.23 VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V 8 = VZ 0 + ( 0.1)( 0.5 ) VZ 0 = 7.95 V Ii =VS ( max ) VZ ( nom ) Ri=12 8 = 1.333 A 3For I L = 0.2 A I Z = 1.133 A For I L = 1 A I Z = 0.333 A VL ( max ) = VZ 0 + I Z ( max ) rZ= 7.95 + (1.133)( 0.5 ) = 8.5165VL ( min ) = VZ 0 + I Z ( min ) rZ= 7.95 + ( 0.333)( 0.5 ) = 8.1165VL = 0.4 V VL 0.4 = % Reg = 5.0% % Reg = V0 ( nom ) 8 Vr =VM VM C = 2 fRC 2 fRVrR = Ri + rz = 3 + 0.5 = 3.5Then C =12 C = 0.0357 F 2 ( 60 )( 3.5 )( 0.8 )2.24 (a) For 10 vI 0, both diodes are conducting vO = 0 For 0 vI 3, Zener not in breakdown, so i1 = 0, vO = 0 vI 3 mA 20 1 v 3 vo = I (10 ) = vI 1.5 20 2 At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA For vI > 3 i1 =O(V) 4 3.510(b)3.010I(V)For vI < 0, both diodes forward biased0 vI . At vI = 10 V , i1 = 1 mA 10 v 3 For vI > 3, i1 = I . At vI = 10 V , i1 = 0.35 mA 20 i1 =www.elsolucionario.net 42. i1(mA)0.351010 I(V)312.25 (a) 1K I01K2K 15V11 V1 = 15 = 5 V for vI 5.7, v0 = vI 3 For vI > 5.7 V vI (V1 + 0.7 ) 15 V1 V1 + = , v0 = V1 + 0.7 1 2 1 15 ( v0 0.7 ) v0 0.7 vI v0 + = 1 2 1 vI 15.7 0.7 1 1 1 + + = v0 + + = v0 ( 2.5 ) 1 2 1 1 2 1 1 vI + 8.55 = v0 ( 2.5 ) v0 = vI + 3.42 2.5 vI = 5.7 v0 = 5.7 vI = 15 v0 = 9.42 0(V) 9.425.705.715I (V)(b) iD = 0 for 0 vI 5.7 Then for vI > 5.7 V v vI I + 3.42 vI vO 2.5 or i = 0.6vI 3.42 For v = 15, i = 5.58 mA = iD = I D D 1 1 1www.elsolucionario.net 43. iD(mA) 5.585.715I (V)2.26 20 For D off, vo = (20) 10 = 3.33 V 30 Then for vI 3.33 + 0.7 = 4.03 V vo = 3.33 V For vI > 4.03, vo = vl 0.7; For vI = 10, vo = 9.3(a)O(V) 9.33.330(b)4.0310 I (V)For vI 4.03 V , iD = 010 vo vo ( 10 ) = 10 20 3 Which yields iD = vI 0.605 20 For vI = 10, iD = 0.895 mAFor vI > 4.03, iD +iD(mA) 0.89502.27304.0310 I(V)O 12.5 10.7 10.730I3030 10.7 = 0.175 A 100 + 10 v0 = i(10) + 10.7 = 12.5 VFor vI = 30 V, i =www.elsolucionario.net 44. b.O12.5 10.7 0302.28 5IO R 6.8 KV = 0.6 V vI = 15sin t OO4.4 19.42.29 a. V = 0 0 3 VV = 0.6 0 2.4b. V = 0 205V = 0.6 19.452.30www.elsolucionario.net 45. 10 6.7 0 4.7 102.31 One possible example is shown. Ri Ii DZVignLD VZ 14 V RADIO VRADIOL will tend to block the transient signals Dz will limit the voltage to +14 V and 0.7 V. Power ratings depends on number of pulses per second and duration of pulse. 2.32 O(V) 40 (a)0 O(V) 35(b)0 52.33 C IO Vx a.For V = 0 Vx = 2.7 Vb.For V = 0.7 V Vx = 2.0 V2.34 C I 10 V O 2.35www.elsolucionario.net 46. 20O10 VB 00 I1020O13 10 3 0VB 3 VI VB720O10 7 VB 3 V0 3 I VB 132.36 For Figure P2.32(a) 10 I 0 10O202.37 a. 10 0.6 I D1 = 0.94 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) V0 = 8.93 VI D1 =ID2 = 0b. 5 0.6 I D1 = 0.44 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) V0 = 4.18 VI D1 =c. d. 10 =ID2 = 0Same as (a)(I ) 2( 0.5 ) + 0.6 + I ( 9.5 ) I = 0.964 mAV0 = I ( 9.5 ) V0 = 9.16 V I D1 = I D 2 =2.38 a.I I D1 = I D 2 = 0.482 mA 2I = I D1 = I D 2 = 0 V0 = 10b.www.elsolucionario.net 47. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) I = I D 2 = 0.94 mAI D1 = 0V0 = 10 I ( 9.5 ) V0 = 1.07 Vc. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 I = I D 2 = 0.44 mAI D1 = 0V0 = 10 I ( 9.5 ) V0 = 5.82 Vd. 10 = I ( 9.5 ) + 0.6 +I ( 0.5) I = 0.964 mA 2I I D1 = I D 2 = 0.482 mA 2 V0 = 10 I ( 9.5 ) V0 = 0.842 VI D1 = I D 2 =2.39 a. V1 = V2 = 0 D1 , D2 , D3 , on V0 = 4.4 V 10 4.4 I = 0.589 mA 9.5 4.4 0.6 = ID2 = I D1 = I D 2 = 7.6 mA 0.5 = I D1 + I D 2 I = 2 ( 7.6 ) 0.589 I D 3 = 14.6 mAI= I D1 I D3b. V1 = V2 = 5 V D1 and D2 on, D3 off I 10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 I = 0.451 mA 2 I I D1 = I D 2 = I D1 = I D 2 = 0.226 mA 2 I D3 = 0 V0 = 10 I ( 9.5 ) = 10 ( 0.451)( 9.5 ) V0 = 5.72 VV1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 Vc.10 4.4 9.5 4.4 0.6 = 0.5I= I D2I = 0.589 mA I D 2 = 7.6 mA I D1 = 0I D 3 = I D 2 I = 7.6 0.589 I D 3 = 7.01 mAV1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 Vd.10 4.4 9.5 4.4 0.6 2 = 0.5I= I D2I = 0.589 mA I D 2 = 3.6 mA I D1 = 0I D 3 = I D 2 I = 3.6 0.589 I D 3 = 3.01 mA2.40 (a) D1 on, D2 off, D3 on So I D 2 = 0 Now V2 = 0.6V , I D1 =10 0.6 ( 0.6 ) R1 + R2=10 I D1 = 1.25 mA 2+6www.elsolucionario.net 48. V1 = 10 0.6 (1.25 )( 2 ) V1 = 6.9 V I R3 = I D30.6 ( 5 )= 2.2 mA 2 = I R 3 I D1 = 2.2 1.25 I D 3 = 0.95 mA(b) D1 on, D2 on, D3 off So I D 3 = 0 V1 = 4.4 V , I D1 =10 0.6 4.4 5 = 6 R1or I D1 = 0.833 mA I R2 =4.4 ( 5 ) R2 + R3=9.4 = 0.94 mA 10I D 2 = I R 2 I D1 = 0.94 0.833 I D 2 = 0.107 mA V2 = I R 2 R3 5 = ( 0.94 )( 5 ) 5 V2 = 0.3 VAll diodes are on V1 = 4.4V , V2 = 0.6 V(c)I D1 = 0.5 mA =10 0.6 4.4 R1 = 10 k R1I R 2 = 0.5 + 0.5 = 1 mA = I R 3 = 1.5 mA =4.4 ( 0.6 )0.6 ( 5 ) R3R2 R2 = 5 k R3 = 2.93 k 2.41 0.5 For vI small, both diodes off vO = vI = 0.0909vI 0.5 + 5 When vI vO = 0.6, D1 turns on. So we have vI 0.0909vI = 0.6 vI = 0.66, vO = 0.06vI 0.6 vO vI vO vO 2v 0.6 + = which yields vO = I 5 5 0.5 12 2vI 0.6 When vO = 0.6, D2 turns on. Then 0.6 = vI = 3.9 V 12 v 0.6 vO vI vO vO vO 0.6 + = + Now for vI > 3.9 I 5 5 0.5 0.5 2vI + 5.4 ; For vI = 10 vO = 1.15 V Which yields vO = 22 For D1 on2.4210 V10 KD2D1 I0 D3D410 K10 K10 Vwww.elsolucionario.net 49. For vI > 0. when D1 and D4 turn off 10 0.7 = 0.465 mA 20 v0 = I (10 k ) = 4.65 V I=04.65 104.65 4.6510I4.65v0 = vI for 4.65 vI 4.652.43 a. R1D210 VV0 D1 ID1R1 = 5 k, R2 = 10 k D1 and D2 on V0 = 0R210 V10 0.7 0 ( 10 ) = 1.86 1.0 5 10 = 0.86 mAI D1 = I D1b. R1 = 10 k, R2 = 5 k, D1 off, D2 on I D1 = 0 I=10 0.7 ( 10 )= 1.287 15 V0 = IR2 10 V0 = 3.57 V2.44 If both diodes on (a) VA = 0.7 V, VO = 1.4 V I R1 = IR2 I R1 + I D110 ( 0.7 )= 1.07 mA 10 1.4 ( 15 ) = = 2.72 mA 5 = I R 2 I D1 = 2.72 1.07I D1 = 1.65 mA D1 off, D2 on 10 0.7 ( 15 ) I R1 = I R 2 = = 1.62 mA 5 + 10 VO = I R 2 R2 15 = (1.62 )(10 ) 15 VO = 1.2 V(b)VA = 1.2 + 0.7 = 1.9 V D1 off ,I D1 = 02.45www.elsolucionario.net 50. (a)D1 on, D2 off 10 0.7 I D1 = = 0.93 mA 10 VO = 15 V (b) D1 on, D2 off 10 0.7 I D1 = = 1.86 mA 5 VO = 15 V2.46 15 (V0 + 0.7 )V0 + 0.7 V0 + 10 20 20 15 0.7 0.7 1 1 1 4.0 = V0 + + = V0 10 10 20 10 20 20 20 V0 = 6.975 V ID V0 I D = 0.349 mA 202.47 10 K V1Va VD 10 KVbV2ID 10 K10 Ka. V1 = 15 V, V2 = 10 V Diode off Va = 7.5 V, Vb = 5 V VD = 2.5 V ID = 0b.V1 = 10 V, V2 = 15 V Diode onV2 Vb Vb Va Va V1 = + + Va = Vb 0.6 10 10 10 10 15 10 1 1 1 1 1 1 + = Vb + + Vb + 0.6 + 10 10 10 10 10 10 10 10 4 2.62 = Vb Vb = 6.55 V 10 15 6.55 6.55 ID = I D = 0.19 mA 10 10 VD = 0.6 V2.48vI = 0, D1 off, D2 on10 2.5 = 0.5 mA 15 vo = 10 ( 0.5 )( 5 ) vo = 7.5 V for 0 vI 7.5 V I=For vI > 7.5 V , Both D1 and D2 on vI vo vo 2.5 vo 10 = + or vI = vo ( 5.5 ) 33.75 15 10 5 When vo = 10 V, D2 turns off vI = (10 )( 5.5 ) 33.75 = 21.25 VFor vI > 21.25 V, vo = 10 Vwww.elsolucionario.net 51. 2.49 a.V01 = V02 = 0b.V01 = 4.4 V, V02 = 3.8 Vc.V01 = 4.4 V, V02 = 3.8 VLogic 1 level degrades as it goes through additional logic gates. 2.50 a.V01 = V02 = 5 Vb.V01 = 0.6 V, V02 = 1.2 Vc.V01 = 0.6 V, V02 = 1.2 VLogic 0 signal degrades as it goes through additional logic gates. 2.51 (V1 AND V2 ) OR (V3 AND V4 ) 2.52 10 1.5 0.2 I= = 12 mA = 0.012 R + 10 8.3 R + 10 = = 691.7 0.012 R = 681.7 2.53 10 1.7 VI =8 0.75 VI = 10 1.7 8 ( 0.75 ) VI = 2.3 V I=2.54 VR VPS R 2 KVR = 1 V, I = 0.8 mA VPS = 1 + ( 0.8 )( 2 ) VPS = 2.6 V2.55 I Ph = eA0.6 103 = (1) (1.6 1019 )(1017 ) A A = 3.75 102 cm 2www.elsolucionario.net 52. Chapter 3 Exercise Solutions EX3.1 VTN = 1 V , VGS = 3 V , VDS = 4.5 V VDS = 4.5 > VDS ( sat ) = VGS VTN = 3 1 = 2 V Transistor biased in the saturation region I D = K n (VGS VTN ) 0.8 = K n ( 3 1) K n = 0.2 mA / V 2 22(a) VGS = 2 V, VDS = 4.5 V Saturation region: I D = ( 0.2 )( 2 1) I D = 0.2 mA 2(b) VGS = 3 V, VDS = 1 V Nonsaturation region: 2 I D = ( 0.2 ) 2 ( 3 1)(1) (1) I D = 0.6 mA EX3.2 VTP = 2 V , VSG = 3 VVSD ( sat ) = VSG + VTP = 3 2 = 1 V(a) (b) (c)VSD = 0.5 V Nonsaturation VSD = 2 V Saturation VSD = 5 V SaturationEX3.3 R2 160 VG = (VDD ) = (10 ) = 3.636 V = VGS 160 + 280 R1 + R2 I D = 0.25 ( 3.636 2 ) = 0.669 mA 2VDS = 10 ( 0.669 )(10 ) = 3.31 V P = I DVDS = ( 0.669 )( 3.31) = 2.21 mWEX3.4 I DQ = K P (VSG + VTP )21.2 = 0.4 (VSG 1.2 ) VSG = 2.932 V 2 R1 1 VSG = VTTN VDD VDD = R2 R1 + R2 Note K = k 1 2.932 = ( 200 )(10 ) R2 = 682 K R2682 R1 = 200 R1 = 283 K 682 + R1 RD =10 4 =5K 1.2EX3.5 R2 40 VG = (10 ) 5 = (10 ) 5 = 1 V R1 + R2 40 + 60 V ( 5 ) 2 ID = S = K n (VGS VTN ) RS(a)VS = VG VGSwww.elsolucionario.net 53. 2 ( 5 1) VGS = ( 0.5 )(1) (VGS 2VGS + 1) 2 2 0.5VGS 3.5 = 0 VGS = 7 VGS = 2.646 VI D = ( 0.5 )( 2.646 1) I D = 1.354 mA VDS = 10 (1.354 )( 3) = 5.937 V 24 VGS = K n (1)(VGS VTN )(b)2(1) K n = (1.05 )( 0.5 ) = 0.525(2) K n = ( 0.95)( 0.5 ) = 0.475 (3) VTN = (1.05 )(1) = 1.05 V(4) VTN = ( 0.95 )(1) = 0.95 V(1)-(3)2 4 VGS = 0.525 (VGS 2.1VGS + 1.1025 )2 0.525VGS 0.1025VGS 3.421 = 0VGS =0.1025 0.010506 + 7.1841 = 2.652 V 2 ( 0.525 )I D = 0.525 ( 2.652 1.05 ) = 1.348 mA VDS = 10 (1.348 )( 3) = 5.957 V (2)-(4) 2 4 VGS = 0.475 (VGS 1.9VGS + 0.9025 ) 22 0.475VGS + 0.0975VGS 3.5713 = 0VGS =0.0975 0.00950625 + 6.78547 2 ( 0.475 )VGS = 2.641 V I D = 0.475 ( 2.641 0.95 ) = 1.359 mA VDS = 10 (1.359 )( 3) = 5.924 V (1)-(4) 2 4 VGS = ( 0.525) (VGS 1.9VGS + 0.9025) 22 0.525 VGS + 0.0025VGS 3.5262 = 0VGS =0.0025 0.00000625 + 7.40502 2 ( 0.525)= 2.5893 V I D = ( 0.525)( 2.5893 0.95) = 1.411 2VDS = 10 I D ( 3) = 5.7678 V (2)-(3) 2 4 VGS = 0.475 (VGS 2.1VGS +1.1025 ) 2 0.475VGS + 0.0025VGS 3.4763 = 00.0025 0.00000625 + 6.60499 2(0.475) = 2.7027VGS = VGSI D = (0.475)(2.7027 1.05) 2 = 1.2973 mA VDS = 10 I D (3) = 6.108 V1.297 I DQ 1.411 mA 5.768 VDS 6.108 V EX3.6www.elsolucionario.net 54. R2 VG = (10 ) 5 R1 + R2 200 = (10 ) 5 = 0.714 V 350 VS = 5 I D RS = 5 (1.2 ) I D So VSG = VS VG = 5 (1.2 ) I D 0.714= 4.286 (1.2 ) I DID =4.286 VSG 1.2I D = K p (VSG + VTP )2(2 4.286 VSG = (1.2 )( 0.25 ) VSG 2VSG ( 1) + ( 1)24.286 VSG = ( 0.3) V 0.6VSG + 0.3)2 SG2 0.3VSG + 0.4VSG 3.986 = 0VSG =( 0.4 )0.4 + 4 ( 0.3)( 3.986 )22 ( 0.3)Must use + sign VSG = 3.04 V I D = ( 0.25 )( 3.04 1) I D = 1.04 mA 2VSD = 10 I D ( RS + RD ) = 10 (1.04 )(1.2 + 4 ) VSD = 4.59 V VSD > VSD ( sat ) , YesEX3.7 VSD = 10 I DQ ( RS + RP ) VSD = 10 K P (VSG + VTP ) ( RS + RP ) 2Set VSD = VSG + VTP VSG + VTP = 10 ( 0.25 )(VSG + VTP ) ( 5.2 ) 21.3 (VSG + VTP ) + (VSG + VTP ) 10 = 0 2(VSG + VTP ) =1 1 + 4 (1.3)(10 ) 2 (1.3)= 2.415 V( 3.42 V ) VSD = 2.415 V ( 2.42 V ) 2 I D = ( 0.25 )( 2.415 ) = 1.46 mA VSG = 3.415 VEX3.8 R2 240 VG = (10 ) 5 = (10 ) 5 R1 + R2 240 + 270 VG = 0.294 V ID =VS ( 5 ) RS=VG VGS + 5 2 = K n (VGS VTN ) RS 0.08 2 4.706 VGS = ( 4 )( 3.9 ) (VGS 2.4VGS + 1.44 ) 2 2 0.624VGS 0.4976VGS 3.80744 = 0www.elsolucionario.net 55. VGS =0.4976 0.2476 + 9.50337 2 ( 0.624 )VGS = 2.90 V 2 0.08 ID = ( 4 )( 2.90 1.2 ) I D = 0.463 mA 2 VDS = 10 I D ( 3.9 + 10 ) VDS = 3.57 VEX3.9 10 VSG 2 ID = and I D = K p (VSG + VTP ) RS 0.12 = ( 0.050 )(VSG 0.8 ) VSG = 2.35 V210 2.35 RS = 63.75 k 0.12 VSD = 8 = 20 I D ( RS + RD ) 8 = 20 ( 0.12 )( 63.75 ) ( 0.12 ) RD RS =RD = (1) (2) (3) (4)20 ( 0.12 )( 63.75 ) 8KP KP VTP VTP ID RD = 36.25 k 0.12 = ( 0.05 )(1.05 ) = 0.0525 = ( 0.05 )( 0.95 ) = 0.0475 = 0.8 (1.05 ) = 0.84 V = 0.8 ( 0.95 ) = 0.76 V 10 VSG 2 = = K P (VSG + VTP ) RS(1)-(3) 2 10 VSG = ( 63.75 )( 0.0525 ) VSG 1.68VSG + 0.7056 2 3.347VSG 4.623VSG 7.6384 = 0VSG =4.623 21.372 + 102.263 2 ( 3.347 )VSG = 2.352 V I D 0.120 mA VSD 8.0 V (2)-(4) 2 10 VSG = ( 63.75 )( 0.0475 ) VSG 1.52VSG + 0.5776 2 3.028VSG 3.603VSG 8.251 = 0VSG =3.603 12.9816 + 99.936 2 ( 3.028 )VSG = 2.35 V I D 0.120 VSD 8.0 (1)-(4) 2 10 VSG = ( 63.75 )( 0.0525 ) VSG 1.52VSG + 0.5776 2 3.347VSG 4.087VSG 8.06685 = 0www.elsolucionario.net 56. VSG =4.087 16.7036 + 107.999 2 ( 3.347 )VSG = 2.279 V I D = 0.121 mA VSD = 7.89 V (2)-(3) 2 10 VSG = ( 63.75 )( 0.0475 ) VSG 1.68VSG + 0.7056 2 3.028VSG 4.0873VSG 7.8634 = 0VSG =4.0873 16.706 + 95.242 2 ( 3.028 )VSG = 2.422 V I D = 0.119 mA VSD = 8.11 VSummary 0.119 I D 0.121 mA 7.89 VSD 8.11 V EX3.10 V VGS , I D = DD RSI D = K n (VGS VTN )22 2 10 VGS = (10 )( 0.2 ) (VGS 2VGSVTN + VTN ) 2 10 VGS = 2VGS 8VGS + 8 2 2VGS 7VGS 2 = 0VGS =7(7) + 4 ( 2) 2 2 ( 2) 2Use + sign: VGS = VDS = 3.77 V 10 3.77 I D = 0.623 mA 10 Power = I DVDS = ( 0.623)( 3.77 ) Power = 2.35 mW ID =EX3.11 (a) VI = 4 V, Driver in Non Sat. K nD 2 (VI VTND ) VO VO2 = K nL [VDD VO VTNL ] 22 5 2 ( 4 1) VD VD = ( 5 VD 1) = ( 4 VO ) = 16 8VO + VO2 222 6VD 38VO + 16 = 0VD =38 1444 384 2 ( 6)VD = 0.454 V (b) VI = 2 V Driver: SatK nD [VI VTND ] = K nL [VDD VO VTNL ] 225 [ 2 1] = [5 VO 1] 225 = 4 VO VO = 1.76 VEX3.12 If the transistor is biased in the saturation regionwww.elsolucionario.net 57. I D = K n (VGS VTN ) = K n ( VTN ) 22I D = ( 0.25 )( 2.5 ) I D = 1.56 mA 2VDS = VDD I D RS = 10 (1.56 )( 4 ) VDS = 3.75 VDS > VGS VTN = VTN 3.75 > ( 2.5 )Yes biased in the saturation region Power = I DVDS = (1.56 )( 3.75 ) Power = 5.85 mW EX3.13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation. I DD = I DL K nD 2 (VI VTND ) VO VO2 = K nL ( VTNL ) 2K nD K 2 2 5 1)( 0.25 ) ( 0.25 ) = 4 nD = 2.06 ( K nL K nL (b) I DL = K nL ( VTNL ) 0.2 = K nL ( 2 ) 22K nL = 50 A / V 2 and K nD = 103 A / V 2EX3.14 For M N I DN = I DP K n (VGSN VTN ) = K p (Vscop + VTP ) 22VGSN = 1 + ( 5 3.25 1) = 1.75 V = VI Vo = VDSN ( sat ) = 1.75 1 Vo = 0.75 V For M P : VI = 1.75 VVDD VO = VSD ( sat ) = VSGP + VTP = ( 5 3.25 ) 1 = 0.75 V So Vot = 5 0.75 Vot = 4.25 VEX3.15 For RD = 10 k , VDD = 5 V, and Vo = 1 V 5 1 = 0.4 mA 10 2 I D = K n 2 (VGS VTN ) VDS VDS ID =2 I D = 0.4 = K n 2 ( 5 1)(1) (1) K n = 0.057 mA / V 2 P = I D VDS = ( 0.4 )(1) P = 0.4 mWEX3.16 a. V1 = 5 V, V2 = 0, M 2 cutoff I D 2 = 0 I D = K n 2 (VI VTN ) VO VO2 = 5 VO RD( 0.05 )( 30 ) 2 ( 5 1)V0 V02 = 5 V0 www.elsolucionario.net 58. 1.5V02 13V0 + 5 = 0 V0 =13 (13) 4 (1.5 )( 5) V0 = 0.40 V 2 (1.5 ) 25 0.40 I R = I D1 = 0.153 mA 30 V1 = V2 = 5 VI R = I D1 =b.5 VO = 2 K n 2 (VI VTN ) VO VO2 RD{}5 V0 = 2 ( 0.05 )( 30 ) 2 ( 5 1)V0 V02 3V02 25V0 + 5 = 0 V0 =25 ( 25) 4 ( 3)( 5 ) V0 = 0.205 V 2 ( 3) 25 0.205 I R = 0.160 mA 30 = I D 2 = 0.080 mAIR = I D1EX3.17 M 2 & M 3 watched I Q1 = I REF 1 = 0.4 mA 0.4 = 0.3 (VGS 3 1) VGS 3 = VGS 2 = 2.15 V 20.4 = 0.6 (VGS 1 1) VGS1 = 1.82 V 2EX3.18 2 0.04 0.1 = (15 )(VSGC 0.6 ) 2 VSGC = 1.177 V = VSGB 2 0.04 W 0.2 = (1.177 0.6 ) 2 L B W = 30 L B 2 0.04 0.2 = ( 25 )(VSGA 0.6 ) 2 VSGA = 1.23 VEX3.19(a)I REF = K n 3 (VGS 3 VTN ) = K n 4 (VGS 4 VTN ) VGS 3 = 2 V VGS 4 = 3 V K K 1 2 2 ( 2 1) = n 4 ( 3 1) n 4 = K n3 K n3 4(b)I Q = K n 2 (VGS 2 VTN ) But VGS 2 = VGS 3 = 2 V20.1 = K n 2 ( 2 1) 2(c)22K n 2 = 0.1 mA / V 20.2 = K n 3 ( 2 1) K n 3 = 0.2 mA / V 2 20.2 = K n 4 ( 3 1) K n 4 = 0.05 mA / V 2 2EX3.20www.elsolucionario.net 59. VS 2 = 5 5 = 0RS 2 =I D 2 = K n 2 (VGS 2 VTN 2 )5 = 16.7 K 0.320.3 = 0.2 (VGS 2 1.2 ) VGS 2 = 2.425 V VG 2 = VGS 2 + VS = 2.425 V 25 2.425 = 25.8 K 0.1 VS 1 = VG 2 VDSQ1 = 2.425 5 = 2.575 V RD1 =RS 1 =2.575 ( 5 ) 0.1 RS 1 = 24.3 KI D1 = K n1 (VGS 1 VTN 1 )20.1 = 0.5 (VGS1 1.2 ) VGS 1 = 1.647 V VG1 = VGS 1 + VS 1 = 1.647 + ( 2.575 ) VG1 = 0.928 V 2 R2 1 VG1 = (10 ) 5 = RTN (10 ) 5 R1 R1 + R2 1 0.928 = ( 200 )(10 ) 5 R1 = 491 K R1 491 R2 = 200 R2 = 337 K 491 + R2EX3.21 VS1 = I D RS 5 = (0.25)(16) 5 = 1 V I DQ = K n (VGS1 VTN ) 2 0.25 = 0.5(VGS 1 0.8) 2 VGS 1 = 1.507 V VG1 = VGS 1 + VS 1 = 1.507 1 = 0.507 V R3 R3 (5) R3 = 50.7 K VG1 = (5) 0.507 = 500 R1 + R2 + R3 VS 2 = VS 1 + VDS1 = 1 + 2.5 = 1.5 V VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V R2 + R3 R2 + R3 VG 2 = (5) 3.007 = (5) R1 + R2 + R3 500 R2 + R3 = 300.7 R2 = 300.7 50.7 R2 = 250 K R1 = 500 250 50.7 R1 = 199.3 K VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V 54 RD = 4 K RD = 0.25EX3.22 VDS ( sat ) = VGS VP = 1.2 ( 4.5 ) VDS ( sat ) = 3.3 V ( 1.2 ) V I D = I DSS 1 GS = 12 1 I D = 6.45 mA ( 4.5 ) VP 22EX3.23 Assume the transistor is biased in the saturation region.www.elsolucionario.net 60. V I D = I DSS 1 GS VP 22 V 8 = 18 1 GS VGS = 1.17 V VS = VGS = 1.17 ( 3.5 ) VD = 15 ( 8 )( 0.8 ) = 8.6 VDS = 8.6 (1.17 ) = 7.43 V VDS = 7.43 > VGS VP = 1.17 ( 3.5 ) = 2.33Yes, the transistor is biased in the saturation region. EX3.24 I D = 2.5 mA V I D = I DSS 1 GS VP 22 V 2.5 = 6 1 GS VGS = 1.42 V ( 4 ) VS = I D RS 5 = ( 2.5 )( 0.25 ) 5 VS = 4.375 VDS = 6 VD = 6 4.375 = 1.625 5 1625 RD = RD = 1.35 k 2.5( 20 )2R1 + R2= 2 R1 + R2 = 200 kVG = VGS + VS = 1.42 4.375 = 5.795 R2 VG = ( 20 ) 10 R1 + R2 R 5.795 = 2 ( 20 ) 10 R2 = 42.05 k 42 k 200 R1 = 157.95 k 158 kEX3.25 VS = VGS . I D =0 VS VGS = RS RS V I D = I DSS 1 GS VP 2 V VGS V2 V = 6 1 GS = 6 1 GS + GS 1 4 2 16 2 0.375VGS 4VGS + 6 = 0 2VGS =4 16 4 ( 0.375 )( 6 ) 2 ( 0.375 )VGS = 8.86 or VGS = 1.806 V impossibleID =VGS = 1.806 mA RSwww.elsolucionario.net 61. VD = I D RD 5 = (1.81)( 0.4 ) 5 = 4.278VSD = VS V0 = 1.81 ( 4.276 ) VSD = 2.47 V VSD ( sat ) = VP VGS = 4 1.81 = 2.19 So VSD > VSD ( sat )EX3.26 Rin = R1 R2 =R1 R2 = 100 k R1 + R2I DQ = 5 mA, VS = I DQ RS = ( 5 )(1.2 ) = 6 VVSDQ = 12 V VD = VS VSDQ= 6 12 = 18 VRD =18 ( 20 ) 5 RD = 0.4 k 2 V V I DQ = I DSS 1 GS 5 = 8 1 GS VP 4 VGS = 0.838 V2VG = VGS + VS = 0.838 6 = 5.162 R2 VG = ( 20 ) R1 + R2 1 5.162 = (100 )( 20 ) R1 = 387 k R1 R1 R2 = 100 ( 387 ) R2 = 100 ( 387 ) + 100 R2 R1 + R2( 387 100 ) R2 = (100 )( 387 ) R2 = 135 k TYU3.1 VTN = 1.2 V , VGS = 2 V (a) V DS ( sat ) = VGS VTN = 2 1.2 = 0.8 V(i) VDS = 0.4 Nonsaturation (ii) VDS = 1 Saturation (iii) VDS = 5 Saturation (b) VTN = 1.2 V , VGS = 2 V V DS ( sat ) = VGS VTN = 2 ( 1.2 ) = 3.2 V (i) VDS = 0.4 Nonsaturation (ii) VDS = 1 Nonsaturation (iii) VDS = 5 Saturation TYU3.2(a)W n Cox 2L 14 ox ( 3.9 ) ( 8.85 10 ) Cox = = = 7.67 108 F / cm 8 tox 450 10 Kn =Kn =(100 )( 500 ) ( 7.67 108 ) K n = 0.274 mA / V 2 2 (7)(b) VTN = 1.2 V, VGS = 2 Vwww.elsolucionario.net 62. (i)VDS = 0.4 V Nonsaturation(ii)2 I D = ( 0.274 ) 2 ( 2 1.2 )( 0.4 ) ( 0.4 ) I D = 0.132 mA VDS = 1 V SaturationI D = ( 0.274 )( 2 1.2 ) I D = 0.175 mA 2(iii) VDS = 5 V Saturation I D = ( 0.274 )( 2 1.2 ) I D = 0.175 mA 2VTN = 1.2 V , VGS = 2 V(i)VDS = 0.4 V Nonsaturation(ii)2 I D = ( 0.274 ) 2 ( 2 + 1.2 )( 0.4 ) ( 0.4 ) I D = 0.658 mA VDS = 1 V Nonsaturation2 I D = ( 0.274 ) 2 ( 2 + 1.2 )(1) (1) I D = 1.48 mA (iii) VDS = 5 V SaturationI D = ( 0.274 )( 2 + 1.2 ) I D = 2.81 mA 2TYU3.3 (a) VSD (sat) = VSG + VTP = 2 1.2 = 0.8 V (i) Non Sat (ii) Sat (iii) Sat (b) VSD (sat) = 2 + 1.2 = 3.2 V (i) Non Sat (ii) Non Sat (iii) Sat TYU3.4 (a) (3.9)(8.85 1014 ) W p Cox KP = Cox = 350 108 L Z = 9.861 108 KP =(40) ( 300 ) ( 9.861 10 (2) 2 K P = 0.296 mA / V(b) (i)8) 2I D = (0.296) 2(2 1.2)(0.4) (0.4) 2 = 0.142 mA(ii)I D = (0.296) [ 2 1.2] I D = 0.189 mA 2(iii) ID = 0.189 mA 2 (i) I D = (0.296) 2 ( 2 + 1.2 )( 0.4 ) ( 0.4 ) = 0.710 mA (ii)2 I D = (0.296) 2 ( 2 + 1.2 )(1) (1) =1.60 mA(iii) I D = ( 0.296 )( 2 + 1.2 ) = 3.03 mA2TYU3.5www.elsolucionario.net 63. (a) = 0, VDS ( sat ) = 2.5 0.8 = 1.7 VFor VDS = 2 V , VDS = 10 V Saturation Region I D = ( 0.1)( 2.5 0.8 ) I D = 0.289 mA 2(b) = 0.02 V 1 I D = K n (VGS VTN ) (1 + VDS ) For VDS = 2 V 2I D = ( 0.1)( 2.5 0.8 ) 1 + ( 0.02 )( 2 ) I D = 0.300 mA VDS = 10 V 2(c)2 I D = ( 0.1) ( 2.5 0.8 ) (1 + ( 0.02 )(10 ) ) I D = 0.347 mA For part (a), = 0 ro = For part (b), = 0.02 V 1 , 12 2 ro = K n (VGS VTN ) = ( 0.02 )( 0.1)( 2.5 0.8 ) 1or ro = 173 k TYU3.6 VTN = VTNO + 2 f + VSB 2 f 2 f = 0.70 V , VTNO = 1 V(a)VSB = 0 , VTN = 1 V(b) VSB = 1 V , VTN = 1 + ( 0.35 ) 0.7 + 1 0.7 VTN = 1.16 V VSB = 4 V , VTN = 1 + ( 0.35 ) 0.7 + 4 0.7 VTN = 1.47 V (c)TYU3.7 I D = K n (VGS VTN )20.4 = 0.25 (VGS 0.8 ) VGS = 2.06 V 2 R2 VGS = VDD R1 + R2 R 2.06 = 2 ( 7.5 ) R2 = 68.8 k 250 R1 = 181.2 k VDS = 4 = VDD I D RD 7.5 4 RD = RD = 8.75 k 0.4 VDS > VDS ( sat ) , YesTYU3.8www.elsolucionario.net 64. VS ( 5 )ID =and VS = VGSRSSo RS =5 VGS 0.1I D = K n (VGS VTN )20.1 = ( 0.080 )(VGS 1.2 ) VGS = 2.32 V 5 2.32 So RS = RS = 26.8 k 0.1 VDS = VD VS VD = VDS + VS = 4.5 2.32 2VD = 2.18 5 VD 5 2.18 RD = = RD = 28.2 k ID 0.1VDS > VDS ( sat ) , YesTYU3.9 For VDS = 2.2 V 5 2.2 I D = 0.56 mA ID = 5 I D = K n (VGS VTN ) 0.56 = K n ( 2.2 1)22K n = 0.389 mA / V =W n Cox 2 LW ( 389 )( 2 ) W = = 19.4 L L ( 40 )TYU3.10 (a) The transition point is(VDD VTNL + VTND 1 + K nD / K nLVIt = =)1 + K nD /K nL(5 1 + 1 1 + 0.05/ 0.01)1 + 0.05/ 0.01 7.236 = VIt = 2.236 V 3.236 VOt = VIt VTND = 2.24 1 VOt = 1.24 V(b) We may write I D = K n D (VGSD VTND ) = ( 0.05 )( 2.236 1) I D = 76.4 A 2TYU3.112(VDD VTNL + VTND 1 + K nD /K nLVIt =)1 + K nD /K nL2.5 =(5 1 + 1 1 + K nD /K nL)1 + K nD /K nL2.5 + 2.5 K nD /K nL = 5 + K nD /K nL K nD /K nL =b.5 2.5 = 1.67 K nD /K nL = 2.78 1.5For VI = 5, driver in nonsaturated region.www.elsolucionario.net 65. I DD = I DL K nD 2 (VI VTND ) VO VO2 = K nL (VGSL VTNL ) 2K nD 2 2 (VI VTND ) VO VO2 = [VDD VO VTNL ] K nL 2.78 2 ( 5 1) V0 V02 = [5 V0 1] 222.24V0 2.78V02 = ( 4 V0 )2= 16 8V0 + V02 3.78V02 30.24V0 + 16 = 0 V0 =30.24 ( 30.24 ) 4 ( 3.78 )(16 ) V0 = 0.57 V 2 ( 3.78 ) 2TYU3.12 We have VDS = 1.2 V < VGS VTN = VTN = 1.8 V Transistor is biased in the nonsaturation region. V VDS 5 1.2 2 = I D = 0.475 mA I D = K n 2 (VGS VTN ) VDS VDS and I D = DD 8 RS 0.475 = K n 2 ( 0 ( 1.8 ) ) (1.2 ) (1.2 ) 0.475 = K n ( 2.88 ) K n = 0.165 mA/V 2 2W n Cox 2 L 165 )( 2 ) W ( W = = 9.43 35 L L Kn =TYU3.13 (a) Transition point for the load transistor Driver is in the saturation region. I DD = I DL K nD (VGSD VTND ) = K nL (VGSL VTNL ) 22VDSL ( sat ) = VGSL VTNL = VTNL VDSL = VDD VOt = 2 V Then VOt = 5 2 = 3 V , VOt = 3 V K nD (VIt 1) = ( VTNL ) K nL 0.08 (VIt 1) = 2 VIt = 1.89 V 0.01 (b) For the driver: VOt = VIt VTND VIt = 1.89 V , VOt = 0.89 VTYU3.14 2 I D = K n 2 (VGS VTN ) VDS VDS 2 = ( 0.050 ) 2 (10 0.7 )( 0.35 ) ( 0.35 ) I D = 0.319 mARD =VDD Vo 10 0.35 = RD = 30.3 k ID 0.319TYU3.15 (a) Transistor biased in the nonsaturation regionwww.elsolucionario.net 66. 5 1.5 VDS = 12 R 2 I D = K n 2 (VGS VTN ) VDS VDS ID =2 12 = 4 2 ( 5 0.8 ) VDS VDS 2 4VDS 33.6VDS + 12 = 0 VDS = 0.374 VThen R =5 1.5 0.374 R = 261 12TYU3.16a.ID =5 VO = K n 2 (V2 VTN ) VO VO2 RD5 ( 0.10 )b.2 = K n 2 ( 5 1)( 0.10 ) ( 0.10 ) K n = 0.248 mA / V 2 25 5 V0 = 2 ( 0.248 ) 2 ( 5 1) V0 V02 25 2 5 V0 = 12.4 8V0 V0 12.4V02 100.2V0 + 5 = 0 V0 =100.2 (100.2 ) 4 (12.4 )( 5 ) V0 = 0.0502 V 2 (12.4 ) 2TYU3.17 2 2 I DQ = K (VGS VTN ) 5 = 50 (VGS 0.15 ) VGS = 0.466 V VS = ( 0.005 )(10 ) = 0.050 V VGG = VGS + VS = 0.466 + 0.050 VGG = 0.516 V VD = 5 ( 0.005 )(100 ) VD = 4.5 VVDS = VD VS = 4.5 0.050 VDS = 4.45 VTYU3.18 2 I D = K 2 (VGS VTN ) VDS VDS 2 = 100 2 ( 0.7 0.2 )( 0.1) ( 0.1) ID = 9 ARD =2.5 0.1 RD = 267 k 0.009www.elsolucionario.net 67. Chapter 3 Problem Solutions 3.1 W k 10 0.08 2 Kn = n = = 0.333 mA/V L 2 1.2 2 For VDS = 0.1 V Non Sat Bias Region(a)VGS = 0 I D = 0(b)2 VGS = 1 V I D = 0.333 2 (1 0.8 )( 0.1) ( 0.1) = 0.01 mA (c)VGS = 2 V2 I D = 0.333 2 ( 2 0.8 )( 0.1) ( 0.1) = 0.0767 mA (d)VGS = 3 V2 I D = 0.333 2 ( 3 0.8 )( 0.1) ( 0.1) = 0.143 mA 3.2 All in Sat region 10 0.08 2 Kn = = 0.333 mA/V 1.2 2 (a) ID = 0 (b)I D = 0.333[1 0.8] = 0.0133 mA(c)I D = 0.333[ 2 0.8] = 0.480 mA(d)I D = 0.333[3 0.8] = 1.61 mA2223.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now 2 0.03 = K n ( 2 1.5 ) = 0.25 K n K n = 0.12 0.15 = K n ( 3 1.5 ) = 2.25 K nK n = 0.06660.39 = K n ( 4 1.5 ) = 6.25 K nK n = 0.06240.77 = K n ( 5 1.5 ) = 12.25 K nK n = 0.0629222From last three, K n (Avg) = 0.0640 mA/V 2 (c)3.4 a.iD (sat) = 0.0640(3.5 1.5) 2 iD (sat) = 0.256 mA for VGS = 3.5 V iD (sat) = 0.0640(4.5 1.5) 2 iD (sat) = 0.576 mA for VGS = 4.5 V VGS = 0VDS ( sat ) = VGS VTN = 0 ( 2.5 ) = 2.5 V i.VDS = 0.5 V Biased in nonsaturationii.2 I D = (1.1) 2 ( 0 (2.5) )( 0.5 ) ( 0.5 ) I D = 2.48 mA VDS = 2.5 V Biased in saturationI D = (1.1) ( 0 ( 2.5 ) ) I D = 6.88 mA 2iii.VDS = 5 V Same as (ii) I D = 6.88 mAb. VGS = 2 V VDS ( sat ) = 2 ( 2.5 ) = 4.5 V i.VDS = 0.5 V Nonsaturation I D = (1.1) 2(2 (2.5))(0.5) (0.5) 2 I D = 4.68 mA www.elsolucionario.net 68. VDS = 2.5 V Nonsaturationii.I D = (1.1) 2(2 (2.5))(2.5) (2.5) 2 I D = 17.9 mA VDS = 5 V Saturationiii.I D = (1.1) ( 2 ( 2.5 ) ) I D = 22.3 mA 23.5 VDS > VGS VTN = 0 ( 2 ) = 2 V Biased in the saturation region k W 2 I D = n (VGS VTN ) 2 L 2 W 0.080 W = 9.375 1.5 = 0 ( 2 ) L 2 L 3.6 kn = n Cox = n ox tox=( 600 )( 3.9 ) (8.85 1014 ) tox(a)500 A 250 kn = 82.8 A/V 2(c)100 kn = 207 A/V 2(d)50 kn = 414 A/V 2(e)252.071 1010 tox kn = 41.4 A/V 2(b)= kn = 828 A/V 23.7 a. Cox = Kn =ox ( 3.9 ) ( 8.85 10 = t0 x 450 10814)oxt0 x= 7.67 108 F/cm 2 n Cox W 2L1 64 ( 650 ) ( 7.67 108 ) 2 4 K n = 0.399 mA / V 2 =b.VGS = VDS = 3 V Saturation I D = K n (VGS VTN ) = ( 0.399 )( 3 0.8 ) I D = 1.93 mA 223.8 2 k I D = n (VGS VTN ) 2 2 2 0.08 1.25 ( 2.5 1.2 ) = 23.1 m 1.25 2 3.9 Cox = ox t0 x=( 3.9 ) (8.85 1014 ) 400 108= 8.63 108 F/cm 2www.elsolucionario.net 69. Kn = n Cox W L 1 W = ( 600 ) ( 8.63 108 ) 2 2.5 2K n = (1.036 105 ) W I D = K n (VGS VTN )21.2 10 3 = (1.036 10 5 ) W ( 5 1) W = 7.24 m 23.10 Biased in the saturation region in both cases. kp W 2 I D = (VSG + VTP ) 2 L 2 0.040 W (1) 0.225 = ( 3 + VTP ) 2 L 2 0.040 W 1.40 = (2) ( 4 + VTP ) 2 L Take ratio of (2) to (1): (4 + VTP ) 2 1.40 = 6.222 = 0.225 (3 + VTP ) 2 6.222 = 2.49 =4 + VTP VTP = 2.33 V 3 + VTPW 2 0.040 W Then 0.225 = = 25.1 ( 3 2.33) L 2 L 3.11 VS = 5 V, VG = 0 VSG = 5 VVTP = 0.5 V VSD ( sat ) = VSG + VTP = 5 0.5 = 4.5 Va.VD = 0 VSD = 5 V Biased in saturation I D = 2 ( 5 0.5 ) I D = 40.5 mA 2b.VD = 2 V VSD = 3 V Nonsaturationc.2 I D = 2 2 ( 5 0.5 )( 3) ( 3) I D = 36 mA VD = 4 V VSD = 1 V Nonsaturationd.2 I D = 2 2 ( 5 0.5 )(1) (1) I D = 16 mA VD = 5 V VSD = 0 I D = 03.12 (a) (b)Enhancement-mode From Graph VTP = + 0.5 V0.45 = k p ( 2 0.5 ) = 2.25 K p K p =0.201.25 = k p ( 3 0.5 ) = 6.25 K p0.202.45 = k p ( 4 0.5 ) = 12.25 K p0.202224.10 = k p ( 5 0.5 ) = 20.25 K p 2(c)0.202 Avg K p = 0.20 mA/V 2iD (sat) = 0.20 (3.5 0.5) 2 = 1.8 mA iD (sat) = 0.20 (4.5 0.5) 2 = 3.2 mAwww.elsolucionario.net 70. 3.13 VSD ( sat ) = VSG + VTP (a)VSD ( sat ) = 1 + 2 VSD ( sat ) = 1 V(b)VSD ( sat ) = 0 + 2 VSD ( sat ) = 2 V(c)VSD ( sat ) = 1 + 2 VSD ( sat ) = 3 VID =(a) (b) (c) kp W k W 2 2 p (VSG + VTP ) = VSD ( sat ) 2 L 2 L 2 0.040 ID = ( 6 )(1) I D = 0.12 mA 2 2 0.040 ID = ( 6 )( 2 ) I D = 0.48 mA 2 2 0.040 ID = ( 6 )( 3) I D = 1.08 mA 2 3.14 VSD (sat) = VSG + VTP = 3 0.8 = 2.2 V 15 0.04 2 KP = = 0.25 mA/V 1.2 2 a) b) c) d) e)2 VSD = 0.2 Non Sat I D = 0.25 2 ( 3 0.8 )( 0.2 ) ( 0.2 ) = 0.21 mA 2 VSD = 1.2 V Non Sat I D = 0.25 2 ( 3 0.8 )(1.2 ) (1.2 ) = 0.96 mA 2 VSD = 2.2 V Sat I D = 0.25(3 0.8) = 1.21 mA VSD = 3.2 V Sat ID = 1.21 mA VSD = 4.2 V Sat ID = 1.21 mA3.15 k p = p Cox = p ox t0 x=( 250 )( 3.9 ) (8.85 1014 ) t0 x(a)250 k = 34.5 A/V 2 p(c)100 k = 86.3 A/V 2 p(d) 50 k p = 173 A/V 2(e)8.629 1011 t0 xtox = 500 k = 17.3 A/V 2 p(b)=25 k = 345 A/V 2 p3.1614 ox ( 3.9 ) ( 8.85 10 ) = = 6.90 108 F/cm 2 Cox = 8 t0 x 500 10 kn = ( n Cox ) = ( 675 ) ( 6.90 108 ) 46.6 A/V 2k = ( p Cox ) = ( 375 ) ( 6.90 108 ) 25.9 A/V 2 pPMOS:www.elsolucionario.net 71. ID =k W 2 p (VSG + VTP ) 2 L p2 0.0259 W W 0.8 = ( 5 0.6 ) = 3.19 2 L p L pL = 4 m W p = 12.8 m 0.0259 2 Kp = ( 3.19 ) K p = 41.3 A/V = K n 2 Want Kn = Kp k W kn W p = = 41.3 2 L N 2 L p 46.6 W W = 41.3 = 1.77 2 L N L N L = 4 m WN = 7.09 m3.17 VGS = 2 V, I D = ( 0.2 )( 2 1.2 ) = 0.128 mA 1 1 r0 = = r0 = 781 k I D ( 0.01)( 0.128 ) 2VGS = 4 V, I D = ( 0.2 )( 4 1.2 ) = 1.57 mA 1 r0 = r = 63.7 k ( 0.01)(1.57 ) 0 2VA =1=1 VA = 100 V ( 0.01)3.18 2 2 0.080 ID = ( 4 )( 3 0.8 ) = ( 0.16 )( 3 0.8 ) I D = 0.774 mA 2 1 1 1 = = (max) = 0.00646 V 1 r0 = ID r0 I D ( 200 )( 0.774 )VA ( min ) =1 ( max )=1 VA ( min ) = 155 V 0.006463.19 VTN = VTNO + 2 f + VSB 2 f VTN = 2 = ( 0.8 ) 2 f + VSB 2 ( 0.35 ) 2.5 + 0.837 = 2 ( 0.35 ) + VSB VSB = 10.4 V3.20 VTN = VTNo + r 2 f + VSB 2 f 2 ( 0.37 ) + 3 2 ( 0.37 ) = 0.75 + 0.6 = 0.75 + 0.6 [1.934 0.860] VTN = 1.39 VVDS (sat) = 2.5 1.39 = 1.11 Vwww.elsolucionario.net 72. 2 0.08 Sat Region I D = (15 ) ( 2.5 1.39 ) 2 I D = 0.739 mA(a)2 0.08 Non-Sat I D = (15 ) 2 ( 2.5 1.39 )( 0.25 ) ( 0.25 ) 2 I D = 0.296 mA(b)3.21 a. VG = %ox t0 x = ( 6 106 )( 275 108 ) VG = 16.5 V VG =b.16.5 VG = 5.5 V 33.22 Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 106 ) t0 x t0 x = 1.2 105 cm = 1200 Angstroms3.23 R2 18 VG = VDD = (10 ) = 3.6 V 18 + 32 R1 + R2 Assume transistor biased in saturation region V V VGS 2 = K n (VGS VTN ) ID = S = G RS RS 3.6 VGS = ( 0.5 )( 2 )(VGS 0.8 ) 2 = VGS 1.6VGS + 0.6422 VGS 0.6VGS 2.96 = 0VGS = ID =0.6 VG VGS RS( 0.6 )2+ 4 ( 2.96 ) VGS = 2.046 V 2 3.6 2.046 = I D = 0.777 mA 2VDS = VDD I D ( RD + RS )= 10 ( 0.777 )( 4 + 2 ) VDS = 5.34 VVDS > VDS ( sat )3.24www.elsolucionario.net 73. ID(mA) 4 (a)Q-pt Q-pt1.67(b) 45 V (V) DSVGS = 4 V VDS (sat) = 4 0.8 = 3.2 V(a)If Sat I D = 0.25 ( 4 0.8 ) = 2.56 2VDS = 1.44 Non-Sat 2 4 = I D RD + VDS = K n RD 2 (VGS VT ) VDS VDS + VDS 2 4 = ( 0.25 )(1) 2 ( 4 0.8 ) VDS VDS + VDS 2 4 = 2.6VDS 0.25VDS 2 0.25VDS 2.6VDS + 4 = 02.6 6.76 4 = 1.88 V 2 ( 0.25 )VDS =4 1.88 = 2.12 mA 1 (b) Non-Sat region 2 5 = I D RD + VDS = K n RD 2 (VGS VT )VDS VDS + VDS ID =2 5 = ( 0.25 )( 3) 2 ( 5 0.8 ) VDS VDS + VDS 2 5 = 7.3VDS 0.75VDS 2 0.75 VDS 7.3VDS + 5 = 0VDS =7.3 53.29 15 2 ( 0.75 )VDS = 0.741 V 5 0.741 ID = = 1.42 mA 33.25 ID(mA) 2.92 (a)Q-pt 1.25 (b) 3.55 V (V) SDwww.elsolucionario.net 74. VSG = VDD = 3.5(a)VSD ( sat ) = 3.5 0.8 = 2.7 VIf biased in Sat region, I D = ( 0.2 )( 3.5 0.8 ) = 1.46 mA VSD = 3.5 (1.46 )(1.2 ) = 1.75 V2Biased in Non-Sat Region. 2 3.5 = VSD + I D RD = VSD + K p RD 2 (VSG + VTP ) VSD VSD 2 3.5 = VSD + ( 0.2 )(1.2 ) 2 ( 3.5 0.8 ) VSD VSD 2 3.5 = VSD + 1.296 VSD 0.24 VSD 2 0.24 VSD 2.296 VSD + 3.5 = 0VSD =+2.296 5.272 3.36 use sign VSD = 1.90 V 2 ( 0.24 )2 I D = ( 0.2 ) 2 ( 3.5 0.8 )(1.9 ) (1.9 ) = 0.2 [10.26 3.61] 3.5 1.90 ID = = 1.33 mA 1.2 I D = 1.33 mAVSG = VDD = 5 V VSD ( sat ) = 5 0.8 = 4.2 V(b)If Sat Region I D = ( 0.2 )( 5 0.8 ) = 3.53 mA, VSD < 0 2Non-Sat Region. 2 5 = VSD + K p RD 2 (VSG + VTP ) VSD VSD 2 5 = VSD + ( 0.2 )( 4 ) 2 ( 5 0.8 ) VSD VSD 2 5 = VSD + 6.72 VSD 0.8 VSD 2 0.8 VSD 7.72 VSD + 5 = 0VSD = ID =7.72 59.598 16 use sign VSD = 0.698 V 2 ( 0.8 )5 0.698 I D = 1.08 mA 43.26 10 VS 2 = K p (VSG + VTP ) RS Assume transistor biased in saturation region R2 VG = ( 20 ) 10 R1 + R2 ID = 22 = ( 20 ) 10 VG = 4.67 V 8 + 22 VS = VG + VSG 10 ( 4.67 + VSG ) = (1)( 0.5 )(VSG 2 )22 5.33 VSG = 0.5 (VSG 4VSG + 4 )2 0.5VSG VSG 3.33 = 0VSG =1(1)2+ 4 ( 0.5 )( 3.33)2 ( 0.5 ) VSG = 3.77 Vwww.elsolucionario.net 75. VSD10 ( 4.67 + 3.77 ) I D = 3.12 mA 0.5 = 20 I D ( RS + RD ) = 20 ( 3.12 )( 0.5 + 2 ) VSD = 12.2 VID =VSD > VSD ( sat )3.27 VG = 0, VSG = VS Assume saturation region 2 I D = 0.4 = K p (VSG + VTP ) 0.4 = ( 0.2 )(VS 0.8 )20.4 + 0.8 VS = 2.21 V 0.2 VD = I D RD 5 = ( 0.4 )( 5 ) 5 = 3 V VSD = VS VD = 2.21 ( 3) VSD = 5.21 V VS =VSD > VSD ( sat )3.28 VDD = I DQ RD + VDSQ + I DQ RS 2 k W (1) 10 = I DQ ( 5 ) + 5 + VGS and I DQ = n (VGS VTN ) 2 L 2 0.060 W or (2) I DQ = (VGS 1.2 ) 2 L Let VGS = 2.5 VThen from (1), 10 = I DQ ( 5 ) + 5 + 2.5 I D = 0.5 mA W 2 0.060 W Then from (2), 0.5 = = 9.86 ( 2.5 1.2 ) 2 L L V 2.5 I DQ RS = VGS RS = GS = RS = 5 k I DQ 0.5 IR =10 = ( 0.5 )( 0.05 ) = 0.025 mA R1 + R2Then R1 + R2 =10 = 400 k 0.025 R2 R2 (VDD ) = 2VGS (10 ) = 2 ( 2.5 ) R1 = R2 = 200 k R1 + R2 400 3.29 75 K n = ( 25 ) 0.9375 mA/V 2 2 6 VG = (10 ) 5 = 2 V 6 + 14 (VG VGS ) ( 5) 2 = I D = K n (VGS VTN ) RS 2 VGS + 5 = ( 0.9375 )( 0.5 )(VGS 1)22 3 VGS = 0.469 (VGS 2VGS + 1)www.elsolucionario.net 76. 2 0.469 VGS + 0.0625 VGS 2.53 = 0VGS =0.0625 0.003906 + 4.746 VGS = 2.26 V 2 ( 0.469 )I D = 0.9375 ( 2.26 1) I D = 1.49 mA 2VDS = 10 (1.49 )(1.7 ) VDS = 7.47 V3.30 20 = I DQ RS + VSDQ + I DQ RD (1) 20 = VSG + 10 + I DQ RD k W 2 p I DQ = (VSG + VTP ) 2 L 2 0.040 W (2) I DQ = (VSG 2 ) 2 L For example, let I DQ = 0.8 mA and VSG = 4 V 0.040 W Then 0.8 = 2 L I DQ RS = VSG ( 0.8 ) RSW 2 = 10 ( 4 2) L = 4 RS = 5 k From (1) 20 = 4 + 10 + ( 0.8 ) RD RD = 7.5 k IR =20 = ( 0.8 )( 0.1) R1 + R2 = 250 k R1 + R2 R1 ( 20 ) = 2VSG = ( 2 )( 4 ) R1 + R2 R1 ( 20 ) = 8 R1 = 100 k , R2 = 150 k 2503.31 (a)(i)I Q = 50 = 500 (VGS 1.2 ) VGS = 1.516 V 2VDS = 5 ( 1.516 ) = VDS = 6.516 V(ii)I Q = 1 = ( 0.5 )(VGS 1.2 ) VGS = 2.61 V 2VDS = 5 ( 2.61) VDS = 7.61 V (b)(i) Same as (a) VGS = VDS = 1.516 V(ii)VGS = VDS = 2.61 V3.32 I D = K n (VGS VTN )20.25 = ( 0.2 )(VGS 0.6 )20.25 + 0.6 VGS = 1.72 V VS = 1.72 V 0.2 VD = 9 ( 0.25 )( 24 ) VD = 3 V VGS =3.33 (a)www.elsolucionario.net 77. ID(mA)1.0 0.808 Q-pt 0.53.81RD =10 V (V) DS5 1 RD = 8 K 0.5I DQ = 0.5 = 0.25 (VGS 1.4 ) VGS = 2.81 V 2RS =2.81 ( 5 ) RS = 4.38 K 0.5 Let RD = 8.2 K, RS = 4.3 K(b) NowVGS ( 5 )5 VGS= I D = 0.25 (VGS 1.4 ) 4.3 2 = 1.075 (VGS 2.8 VGS + 1.96 )22 1.075 VGS 2.01 VGS 2.89 = 0VGS =2.01 4.04 + 12.427 VGS = 2.82 V 2 (1.075 )I D = 0.25 ( 2.82 1.4 ) I D = 0.504 mA 2VDS = 10 ( 0.504 )( 8.2 + 4.3) VDS = 3.70 V(c)If RS = 4.3 + 10% = 4.73 K2 5 VGS = 1.18 (VGS 2.8VGS + 1.96 ) 2 1.18 VGS 2.31 VGS 2.68 = 0VGS =2.31 5.336 + 12.65 = 2.78 V 2 (1.18 )I D = ( 0.25 )( 2.78 1.4 ) I D = 0.476 mA 2If Rs = 4.3 10% = 3.87 K2 5 VGS = ( 0.9675 ) (VGS 2.8VGS + 1.96 ) 2 0.9675VGS 1.71VGS 3.10 = 0VGS =1.71 2.924 + 12.0 = 2.88 V 2 ( 0.9675 )I D = ( 0.25 )( 2.88 1.4 ) = 0.548 mA 23.34 VDD = VSD + I DQ R 9 = 2.5 + ( 0.1) R R = 65 k k W 2 p I DQ = (VSG + VTP ) 2 L W 2 0.025 W =8 ( 0.1) = ( 2.5 1.5 ) L L 2 Then for L = 4 m, W = 32 mwww.elsolucionario.net 78. 3.35 5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5 I DQ = 1.25 mA IR =10 = (1.25 )( 0.1) R1 + R2 = 80 k R1 + R2I DQ = K p (VSG + VTP )21.25 = 0.5 (VSG + 1.5 ) 21.25 1.5 = VSG 0.5VSG = 0.0811 V VG = VS VSG = 2.5 0.0811 = 2.42 V R2 VG = (10 ) 5 R1 + R2 R 2.42 = 2 (10 ) 5 R2 = 59.4 k , R1 = 20.6 k 80 3.36 (a) ID(mA)0.429 Q-ptRD =VD ( 5 ) I DQ5 V (V) SD=52 RD = 12 K 0.252 k p (VSG + VTP ) 2 2 0.035 0.25 = (15 ) (VSG 1.2 ) VSG = 2.18 V 2 5 2.18 RS = RS = 11.3 K 0.25 VSD = 2.18 ( 2 ) = 4.18 V (b) k = 35 + 5% = 36.75 A/V 2 pW ID = L5 VSG 2 0.03675 I D = (15 ) (VSG 1.2 ) = 2 11.3 2 3.11(VSG 2.4VSG + 1.44 ) = 5 VSG 2 3.11VSG 6.46VSG 0.522 = 0www.elsolucionario.net 79. VSG =6.46 41.73 + 6.49 = 2.155 V 2 ( 3.11)5 2.155 = 0.252 mA 11.3 = 10 ( 0.252 )(12 + 11.3) = 4.13 VID = VSDk = 35 5% = 33.25 A/V 2 p 5 VSG 2 0.03325 I D = (15 ) (VSG 1.2 ) = 2 11.3 2 2.82 (VSG 2.4VSG + 1.44 ) = 5 VSG 2 2.82VSG 5.77VSG 0.939 = 0VSG =5.77 33.29 + 10.59 = 2.198 V 2 ( 2.82 )5 2.198 = 0.248 mA 11.3 = 10 ( 0.248 )(12 + 11.3) = 4.22 VID = VSD3.37 ID =VSD ( 10 ) RD5=6 + 10 RD = 0.8 k RDI D = K P (VSG + VTP ) 5 = 3 (VSG 1.75 ) 2VSG =25 + 1.75 = 3.04 V VG = 3.04 3 R2 VG = (10 ) 5 = 3.04 R1 + R2 Rin = R1 R2 = 80 k 1 ( 80 )(10 ) = 5 3.04 R1 = 408 k R1 408 R2 = 80 R2 = 99.5 k 408 + R23.38 60 K n1 = ( 4 ) = 120 A/V 2 2 60 K n 2 = (1) = 30 A/V 2 2 For vI = 1 V , M1 Sat. region, M2 Non-sat region.(a)I D 2 = I D1 30 2 ( VTNL )( 5 vO ) ( 5 vO ) = 120 (1 0.8 ) 2 We find vO 6.4vO + 7.16 = 0 vO = 4.955 V 2(b)2For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D12 30 ( 1.8 ) = 120 2 ( 3 0.8 ) vO vO 22 We find 4vO 17.6vO + 3.24 = 0 vO = 0.193 V(c)For vI = 5 V , biasing same as (b)2 30 ( 1.8 ) = 120 2 ( 5 0.8 ) vO vO 2www.elsolucionario.net 80. 2 We find 4vO 33.6vO + 3.24 = 0 vO = 0.0976 V3.39 For vI = 5 V , M1 Non-sat region, M2 Sat. region. I D1 = I D 2 2 kn W kn W 2 2 L 2 (VGS1 VTN 1 ) VDS 1 VDS 1 = 2 L (VGS 2 VTN 2 ) 1 2 2 W 2 2 ( 5 0.8 )( 0.15 ) ( 0.15 ) = (1) 0 ( 2 ) L 1 W which yields = 3.23 L 13.40 a.M1 and M2 in saturationK n1 (VGS 1 VTN 1 ) = K n 2 (VGS 2 VTN 2 ) K n1 = K n 2 , VTN 1 = VTN 2 VGS1 = VGS 2 = 2.5 V, V0 = 2.5 V 22I D = (15 )( 40 )( 2.5 0.8 ) I D = 1.73 mA 2b. W W > VGS1 < VGS 2 L 1 L 2 40 (VGS1 0.8 ) = (15 )(VGS 2 0.8 ) 22VGS 2 = 5 VGS 1 1.633 (VGS 1 0.8 ) = ( 5 VGS1 0.8 )2.633VGS 1 = 5.506 VGS 1 = 2.09 VVGS 2 = 2.91 V, V0 = VGS1 = 2.91 VI D = (15 )(15 )( 2.91 0.8 ) I D = 1.0 mA 23.41 (a) V1 = VGS 3 = 2.5 V 2 W 0.06 I D = 0.5 = ( 2.5 1.2 ) L 3 2 W = 9.86 L 3V2 = 6 V VGS 2 = V2 V1 = 6 2.5 = 3.5 V 2 W 0.06 W 0.5 = ( 3.5 1.2 ) = 3.15 L 2 2 L 2 VGS1 = 10 V2 = 10 6 = 4 V 2 W 0.06 W 0.5 = ( 4 1.2 ) = 2.13 L 1 2 L 1 (b) kn1 = 0.06 + 5% = 0.063 mA/V 2 kn 2 = k n3 = 0.6 5% = 0.057 mA/V 2 2 0.057 For M3: I D = ( 9.86 ) (V1 1.2 ) 2 2 0.057 For M2: I D = ( 3.15 ) (V2 V1 1.2 ) 2 www.elsolucionario.net 81. 2 0.063 For M1: I D = ( 2.13) (10 V2 1.2 ) 2 0.281(V1 1.2 ) = 0.0898 (V2 V1 1.2 ) = 0.0671( 8.8 V2 ) 222Take square root. 0.530 (V1 1.2 ) = 0.300 (V2 V1 1.2 ) = 0.259 ( 8.8 V2 ) (1) 0.830V1 = 0.300V2 + 0.276(2) 0.559V2 = 0.300V1 + 2.64From (2) V2 = 0.537V1 + 4.72 Substitute into (1) 0.830V1 = 0.300 [ 0.537V1 + 4.72] + 0.276 = 0.161V1 + 1.69 V1 = 2.53 V Then V2 = 0.537 ( 2.53) + 4.72 V2 = 6.08 V3.42 ML in saturation MD in nonsaturation 2 W W 2 (VGSL VTNL ) = 2 (VGSD VTND )VDSD VDSD L L L D W 2 2 (1)( 5 0.1 0.8) = 2 ( 5 0.8)( 0.1) ( 0.1) L D W 16.81 = [ 0.83] L D W = 20.3 L D3.43 ML in saturation MD in nonsaturation 2 W W 2 (VGSL VTNL ) = 2 (VGSD VTND )VDSD VDSD L L L D W 2 2 (1)(1.8 ) = 2 ( 5 0.8 )( 0.05) ( 0.05) L D W 3.24 = [ 0.4175] L D W = 7.76 L D3.44 VDD V0 5 0.1 = = 0.49 mA 10 RD Transistor biased in nonsaturation I D = 0.49 ID =2 W = ( 0.015 ) 2 ( 4.2 0.8 )( 0.1) ( 0.1) L W W 0.49 = 0.01005 = 48.8 L L3.45www.elsolucionario.net 82. 5 = I D RD + V + VDS5 = (12 ) RD + 1.6 + 0.2 RD = 267 2 k W I D = n (VGS VTN ) 2 L W 2 0.040 W = 34 12 = ( 5 0.8 ) 2 L L 3.46 5 = VSD + I D RD + V 5 = 0.15 + (15 ) RD + 1.6 RD = 217 k W 2 p I D = (VSG + VTP ) 2 L W 2 0.020 W = 85 15 = ( 5 0.8 ) L 2 L 3.47 (a) VDD VO W = 2 RD L 5 0.2 W = 2 20 L 0.060 2 ( 2 )(VGS VTN ) VO VO 2 2 ( 0.030 ) 2 ( 5 0.8 )( 0.2 ) ( 0.2 ) W W W 0.24 = 0.0984 = = 2.44 L L 1 L 2(b) 5 VO 0.06 2 = ( 2.44 ) 2 ( 5 0.8 ) VO VO 20 2 5 VO = 12.30VO 1.464VO2 1.464VO2 13.30VO + 5 = 0 VO =13.30 176.89 29.28 2 (1.464 )VO = 0.393 V3.48 2 kn (VGS 1 VTN ) 2 1 2 kn (VDS 2 ( sat ) ) 2 2 2 W 0.08 W W 0.1 = ( 0.5 ) = 10 = L 2 2 L 2 L 1 W 200 W = = 20 L 3 100 L 2 M1 & M2 matched. 2 0.08 Then 0.1 = (10 ) (VGS 1 0.25 ) 2 VGS1 = 0.75 V W I Q1 = L W I Q1 = LVD1 = 0.75 + 2 = 1.25 V RD =2.5 1.25 RD = 12.5 K 0.1www.elsolucionario.net 83. 3.49 (a) 2 W k p I Q 2 = (VSDB ( sat ) ) L B 2 2 W 0.04 0.25 = ( 0.8 ) L B 2 W W = 19.5 = L B L A I KQ 2 W W = IQ 2 L B L C 100 = (19.5 ) = 7.81 250 2 W kp I Q 2 = (VSGA + VTP ) L A 2 2 0.04 0.25 = (19.5 ) (VSGA 0.5 ) 2 VSGA = 1.30 V (b) VDA = 1.3 4 = 2.7 VRD =2.7 ( 5 ) 0.25 RD = 9.2 K3.50 2 kn (VDS 2 ( sat ) ) 2 2 2 0.06 W W ( 0.5 ) = 53.3 = 2 2 L 2 L 1 2 W k I Q = n (VGS 1 VTN ) L 1 2 2 0.06 0.4 = ( 53.3) (VGS 1 0.75 ) 2 VGS1 = 1.25 V VD1 = 1.25 + 4 = 2.75 V 5 2.75 RD = RD = 5.625 K 0.4 W IQ = L W 0.4 = L3.51 VDS ( sat ) = VGS VP So VDS > VDS ( sat ) = VP , I D = I DSS 3.52 VDS ( sat ) = VGS VP = VGS + 3 = VDS ( sat ) a.VGS = 0 I D = I DSS = 6 mAb. V 1 I D = I DSS 1 GS = 6 1 I D = 2.67 mA VP 3 222c. d. 2 I D = 6 1 I D = 0.667 mA 3 ID = 0www.elsolucionario.net 84. 3.53 V I D = I DSS 1 GS VP 1 2.8 = I DSS 1 VP 22 3 0.30 = I DSS 1 VP 1 1 2.8 VP = 0.30 3 1 VP 22 = 9.33 2 1 1 VP 3 1 VP = 3.055 1 9.165 = 3.055 1 VP VP 8.165 = 2.055 VP = 3.97 V VP 21 2.8 = I DSS 1 = I DSS ( 0.560 ) I DSS = 5.0 mA 3.97 3.54 VS = VGS , VSD = VS VDD Want VSD VSD ( sat ) = VP VGS VS VDD VP VGS VGS VDD VP VGS VDD VPSo VDD 2.5 V V I D = 2 = I DSS 1 GS VP 22 V 2 = 6 1 GS VGS = 1.06 V VS = 1.06 V 2.5 3.55 I D = K n (VGS VTN )218.5 = K n ( 0.35 VTN ) 86.2 = K n ( 0.5 VTN )22Then( 0.35 VTN ) 18.5 = 0.2146 = VTN = 0.221 V 2 86.2 ( 0.50 VTN ) 218.5 = K n ( 0.35 0.221) K n = 1.11 mA / V 2 23.56 I D = K (VGS VTN )2250 = K ( 0.75 0.24 ) K = 0.961 mA / V 2 2www.elsolucionario.net 85. 3.57 2 V V V I D = I DSS 1 GS = S = GS VP RS RS 2V V 10 1 GS = GS 0.2 5 2 2V V 2 1 + GS + GS = VGS 5 25 2 2 9 VGS + VGS + 2 = 0 25 5 2 2VGS + 45VGS + 50 = 0 VGS =45 ID = ( 45 ) 4 ( 2 )( 50 ) VGS 2 ( 2) 2= 1.17 VVGS 1.17 = I D = 5.85 mA RS 0.2VD = 20 ( 5.85 )( 2 ) = 8.3 V VDS = VD VS = 8.3 1.17 VDS = 7.13 V3.58 VDS = VDD VS 8 = 10 VS VS = 2 V = I D RS = ( 5 ) RS RS = 0.4 k V I D = I DSS 1 GS VP 22 1 5 = I DSS 1 Let I DSS = 10 mA VP 2 1 5 = 10 1 VP = 3.41 V VP VG = VGS + VS = 1 + 2 = 1 V R2 1 VG = VDD = Rin VDD R1 R1 + R2 1 1 = ( 500 )(10 ) R1 = 5 M R1 5R2 = 0.5 R2 = 0.556 M 5 + R23.59 V I D = I DSS 1 GS VP 22 V 5 = 8 1 GS VGS = 0.838 V 4 VSD = VDD I D ( RS + RD ) = 20 ( 5 )( 0.5 + 2 ) VSD = 7.5 Vwww.elsolucionario.net 86. VS = 20 ( 5 )( 0.5 ) = 17.5 V VG = VS + VGS = 17.5 + 0.838 = 18.3 V R2 1 VG = VDD = Rin VDD R1 R1 + R2 1 18.3 = (100 ) ( 20 ) R1 = 109 k R1 109 R2 = 100 R2 = 1.21 M 109 + R23.60 V I D = I DSS 1 GS VP 22 V 5 = 7 1 GS VGS = 0.465 V 3 VSD = VDD I D ( RS + RD )6 = 12 ( 5 )( 0.3 + RD ) RD = 0.9 kVS = 12 ( 5 )( 0.3) = 10.5 V VG = VS + VGS = 10.5 + 0.465 = 10.965 V R2 VG = VDD R1 + R2 R 10.965 = 2 (12 ) R2 = 91.4 k R1 = 8.6 k 100 3.61 R2 60 VG = VDD = ( 20 ) VG = 6 V 140 + 60 R1 + R2 2 V V V VGS I D = I DSS 1 GS = S = G VP RS RS (8 )( 2 ) 1 2VGS = 6 VGS ( 4 ) V V2 16 1 + GS + GS = 6 VGS 2 16 2 VGS + 9VGS + 10 = 0 VGS =9 (9)2 4 (10 )2 VGS = 1.30 ( 1.30 ) I D = 8 1 I D = 3.65 mA ( 4 ) VDS = VDD I D ( RS + RD ) 2= 20 ( 3.65 )( 2 + 2.7 )VDS = 2.85 V VDS > VDS ( sat ) = VGS VP = 1.30 ( 4 ) = 2.7 V (Yes)3.62www.elsolucionario.net 87. VDS = VDD I D ( RS + RD )5 = 12 I D ( 0.5 + 1) I D = 4.67 mA VS = I D RS = ( 4.67 ) ( 0.5 ) VS = 2.33 V R2 20 VG = VDD = (12 ) VG = 0.511 V R1 + R2 450 + 20 VGS = VG VS = 0.511 2.33 VGS = 1.82 V V I D = I DSS 1 GS VP 2 ( 1.82 ) 4.67 = 10 1 VP = 5.75 V VP 23.63 2 V I D = I DSS 1 GS , VGS = 0 VP I D = I DSS = 4 mA RD =VDD VDS 10 3 RD = 1.75 k = 4 ID3.64 VSD = VDD I D RS 10 = 20 (1) RS RS = 10 k R1 + R2 =VDD 20 = = 200 k 0.1 I V I D = I DSS 1 GS VP 22 V 1 = 2 1 GS VGS = 0.586 V 2 VG = VS + VGS = 10 + 0.586 = 10.586 R2 VG = VDD R1 + R2 R 10.586 = 2 ( 20 ) R2 = 106 k 200 R1 = 94 k3.65www.elsolucionario.net 88. VDS = VDD I D ( RS + RD ) 2 = 3 ( 0.040 )(10 + RD ) RD = 15 k I D = K (VGS VTN )240 = 250 (VGS 0.20 ) VGS = 0.60 V 2VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V R2 VG = VDD R1 + R2 R 1 = 2 ( 3) R2 = 50 k 150 R1 = 100 k3.66 For VO = 0.70 V VDS = 0.70 > VDS ( sat ) = VGS VTN 0.75 0.15 = 0.6 Biased in the saturation region V VDS 3 0.7 I D = 46 A I D = DD = 50 RD I D = K (VGS VTN ) 46 = K ( 0.75 0.15 ) K = 128 A / V 2 22www.elsolucionario.net 89. Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS VTN ) and I D = K n (VGS VTN )20.75 = 0.5 (VGS 0.8 ) VGS = 2.025 V g m = 2 ( 0.5 )( 2.025 0.8 ) g m = 1.22 mA / V 2ro =1 I DQ1 (0.01)(0.75) = 133 k ro = 133 k =EX4.2 Av = g m RD g m = 2 K n I DQ = 2( 0.5)( 0.4 ) = 0.8944 mA/VAv = ( 0.8944 )(10 ) = 8.94EX4.3 (a) R2 320 VGS = VDD = ( 5 ) = 1.905 V R1 + R2 520 + 320 I DQ = 0.20 (1.905 0.8 ) = 0.244 mA 2g m = 2 K n I DQ = 2( 0.2 )( 0.244 ) = 0.442 mA/Vro = (b)Av = g m RD = ( 0.422 )(10 ) = 4.22(c) (d)Ri = R1 R2 = 520 320 = 198 K RO = RD = 10 KEX4.4 At transition point, I D = 1 mA I D = K n (VGSt VTN ) = K n (VDS ( sat ) ) 221 = 0.2 (VDS ( sat ) ) VDS ( sat ) = 2.236 V 25 2.236 + 2.236 = 3.62 V 2 5 3.62 RD = = 2.76 K 0.5Want VDSQ =0.5 = 0.2 (VGSQ 0.8 ) VGSQ = 2.38 V 2 R2 1 VGSQ = VDD = ( R1 R2 ) VDD R1 R1 + R2 1 So 2.38 = ( 200 )( 5 ) R1 = 420 K and R2 = 382 K R1www.elsolucionario.net 90. Av = g m RD( 0.2 )( 0.5 ) = 0.6325 mA/V Av = ( 0.6325 )( 2.76 )g m = 2 K n I DQ = 2 = 1.75EX4.5 (a) R2 250 VG = (10 ) 5 = (10 ) 5 = 3 V R1 + R2 250 + 1000 (V VGS ) ( 5 ) 2 = K n (VGS VTN ) ID = G 2 3 VGS + 5 = 2 ( 0.5 )(VGS 0.6 ) 2 2 VGS = VGS 1.2VGS + 0.36 2 VGS 0.2VGS 1.64 = 0 VGS =0.2 ( 0.04 ) + 4 (1.64 ) 22= 1.385 VI DQ = ( 0.5 )(1.385 0.6 ) I DQ = 0.308 mA VDSQ = 10 ( 0.308 )(10 + 2 ) VDSQ = 6.30 V 2(b) Av = g m RD 1 + g m RSg m = 2 K n I DQ = 2( 0.5 )( 0.308 )g m = 0.7849 mA/V Av = ( 0.7849 )(10 )1 + ( 0.7849 )( 2 )Av = 3.05EX4.6 VSDQ = 3 V and I DQ = 0.5 mA RD = I DQ = K P (VSG VTP)53 RD = 4 k 0.520.5 = 1(VSG 1) VSG = 1.71 V VGG = 5 1.71 VGG = 3.29 V 2Av = g m RD g m = 2 K P I DQ = 2 (1)( 0.5 ) g m = 1.414 mA/VAv = (1.414 )( 4 ) Av = 5.66 Av =v0 vsd 0.46sin t = = = 5.66 vi = 0.0813sin t vi vi viEX4.7 a. VSG = 9 I DQ RS , I DQ = K P (VSG VTP VSG = 9 ( 2 )(1.2 )(VSG 2 ))222 = 9 2.4 (VSG 4VSG + 4 )2 2.4VSG 8.6VSG + 0.6 = 0www.elsolucionario.net 91. (8.6 ) 4 ( 2.4 )( 0.6 ) 2 ( 2.4 ) 2 VSG = 3.51 V, I DQ = 2 ( 3.51 2 ) I DQ = 4.57 mA VSDQ = 9 + 9 I DQ (1.2 + 1) = 18 ( 4.57 )( 2.2 ) VSDQ = 7.95 V VSG =28.6 b. V0 VSG VigmVSG RDRSCS( 2 )( 4.57 ) = 6.046 mA/Vg m = 2 K P I DQ = 2V0 = g mVSG RD Av = g m RD = ( 6.046 )(1) Av = 6.05EX4.8 VDSQ = VDD I DQ RS 5 = 10 (1.5 ) RS RS = 3.33 k I DQ = K n (VGS VTN ) 1.5 = (1)(VGS 0.8 ) 22 R2 VGS = 2.025 V = VG VS = VG 5 VG = 7.025 V = R1 + R2 So R2 = 281 k, R1 = 119 kNeglecting RSi , Av =g m ( RS1 + g m ( RS1 R2 10 VDD = 400 r0 )r0 )1r0 = I DQ = ( 0.015 )(1.5 ) = 44.4 k RS r0 = 3.33 44.4 = 3.1 k g m = 2 K n I DQ = 2 (1)(1.5 ) = 2.45 mA / V Av =( 2.45)( 3.1) Av = 0.884 1 + ( 2.45 )( 3.1)EX4.9 I DQ = K P (VSG VTP)23 = 2 (VSG 2 ) VSG = 3.22 V 5 VSG 5 3.22 I DQ = 3= RS = 0.593 k RS RS 211r0 = I DQ = ( 0.02 )( 3) = 16.7 k ( 2 )( 3) = 4.9 mA / V g m ( r0 RS ) RL = , Av = 1 + g m ( r0 RS )g m = 2 K P I DQ = 2Forr0 RS = 16.7 0.593 = 0.573 kwww.elsolucionario.net 92. Av =( 4.9 )( 0.573) Av = 0.737 1 + ( 4.9 )( 0.573)If Av is reduced by 10% Av = 0.737 0.0737 = 0.663 g m ( r0 RS RL )Av =Let r0 RS 0.663 =1 + g m ( r0 RS RL )RL = x( 4.9 ) x 0.663 = 4.9 x (1 0.663) 1 + ( 4.9 ) xx = 0.402 = 0.573 RL 0.573RL = 0.402 ( 0.573 0.402 ) RL = ( 0.402 )( 0.573) RL = 1.35 k RL + 0.573EX4.10 R2 9.3 VG = VDD = (5) 70.7 + 9.3 R1 + R2 = 0.581 V I DQ = K p (VSG VTP)2= K P (VS VG VTP =)25 VS RSThen ( 0.4 )( 5 )(VS 0.581 0.8 ) = 5 VS 22 (VS 1.381) = 5 VS 22 (VS2 2.762VS + 1.907 ) = 5 VS 2VS2 4.52VS 1.19 = 0VS =4.52 ( 4.52 ) + 4 ( 2 )(1.19 ) 2 ( 2) 2VS = 2.50 V I DQ = g m = 2 K P I DQ = 2 Av = =5 2.5 = 0.5 mA 5( 0.4 )( 0.5 ) = 0.894 mA / Vg m RS R1 R2 1 + g m RS R1 R2 + RSi( 0.894 )( 5) 70.7 9.3 Av = 0.770 1 + ( 0.894 )( 5 ) 70.7 9.3 + 0.5Neglecting RSi , Av = 0.817 R0 = RS1 1 =5 = 5 1.12 R0 = 0.915 k 0.894 gmEX4.11 gmVsg V0 Vi RSVsgRDRLwww.elsolucionario.net 93. V0 = g mVsg ( RD RL ) and Vsg = Vi Av = g m ( RD RL ) I DQ =5 VSG = K p (VSG VTP RS5 VSG = (1)( 4 )(VSG 0.8 ))222 5 VSG = 4 (VSG 1.6VSG + 0.64 ) 2 4VSG 5.4VSG 2.44 = 0VSG =5.4 (5.4) 2 + ( 4 )( 4 )( 2.44 ) 2 ( 4)VSG = 1.71 V 5 1.71 I DQ = = 0.822 mA 4 g m = 2 K p I DQ = 2 (1)( 0.822 ) = 1.81 mA / V Av = (1.81)( 2 4 ) = (1.81)(1.33) Av = 2.41 Rin = RS1 1 =4 = 4 0.552 Rin = 0.485 k gm 1.81EX4.12 Kn2 = n Cox W Av = 2 = ( 0.015 )( 2 ) = 0.030 mA / V 2 L 2K n1 K = 6 n1 = 36 Kn2 Kn2K n1 = ( 36 )( 0.030 ) = 1.08 mA / V 2 W W 1.08 = ( 0.015 ) = 72 L 1 L 1 The transition point is found from vGSt 1 = (10 1) ( 6 )( vGSt 1) 10 1 + 6 + 1 = 2.29 V 1+ 6 2.29 1 For Q-point in middle of saturation region VGS = + 1 VGS = 1.645 V 2 vGSt =EX4.13 (a) Transition points: For M 2 : vOtB = VDD VTNL = 5 1.2 = 3.8 V 2 2 For M 1 : K n1 ( vOtA ) (1 + vOtA ) = K n 2 (VTNL ) (1 + 2 [VDD vOtA ]) 2 2 3 250 vOtA + ( 0.01) vOtA = 25 (1.2 ) (1 + ( 0.01)( 5 ) ( 0.01) vOtA ) 2 3 3 2 10 vOtA + ( 0.01) vOtA = 1.512 0.0144vOtA ( 0.01) vOtA + vOtA + 0.00144vOtA 0.512 = 0 which yields vOtA 0.388 Vwww.elsolucionario.net 94. 3.8 0.388 + 0.388 VDSQ1 = 2.094 V 2 2 2 K n1 (VGS 1 VTND ) (1 + 1vO ) = K n 2 (VTNL ) (1 + 2 [VDD vO ]) Then middle of saturation region V0Q =250 (VGS 1 0.8 ) (1 + [ 0.01][ 2.094]) = 25 (1.2 ) (1 + [ 0.01][5 2.094]) 222 10 (VGS1 0.8 ) = (1.0209 ) = 1.482 (VGS1 0.8 )2= 0.145 VGS1 = 1.18 VI DQ = K n1 (VGS1 0.8 ) (1 + ( 0.01)( 2.094 ) ) 2b.2 I DQ = ( 0.25 ) ( 0.145 ) (1.02094 ) I DQ = 37.0 A g m1 Av = = g m1 ( r01 r02 ) I DQ ( 1 + 2 )c.g m1 = 2 K n1 (VGS 1 VTND ) = 2 ( 0.25 )(1.18 0.8 ) = 0.19 mA/V Av =0.19 Av = 257 ( 0.037 )( 0.01 + 0.01)EX4.14 Av = g m ( ron rop ) ron = rop =1 = 666.7 K ( 0.015)( 0.1)250 = g m ( 666.7 666.7 )g m = 0.75 mA/V = 2 K n I DQ = 2 K n ( 0.1) K n = 1.406 mA/V 2 = kn W 2L 0.080 W = 2 L W = 35.2 L 1EX4.15 (a) RO = So g m1 =1 1 ro 2 ro1 g m1 g m1 1 1 = = 0.5 mA/V R0 2g m1 = 2 K n I D( 0.2 ) I D0.5 = 2(b) Av =g m1 ( ro1 ro 2 )1 + g m1 ( ro1 ro 2 )ro1 = ro 2 = Av = I D = 0.3125 mA1 = 320 K ( 0.01)( 0.3125 )0.5 ( 320 320 )1 + ( 0.5 )( 320 320 )Av = 0.988EX4.16www.elsolucionario.net 95. (a) 1 ro1 2 K n I D + 1 I D Av = = 1 1 2 I D + 1 I D + ro 2 ro1 g m1 +120 =2 0.2 I D + 0.01I D 0.01I D + 0.01I D2.4 I D 0.01I D = 2 0.2 I D 2.39 I D = 2 0.2 I D = 0.140 mA g m1 = 2( 0.2 )( 0.14 ) g m1 = 0.335 mA/V(b) Ro = ro1 ro 2 ro1 = ro 2 =1 = 714 K 0.01)( 0.14 ) (Ro = 357 KEX4.17 R0 = RS 21 gm2g m 2 = 0.632 mA/V, RS 2 = 8 k R0 = 81 = 8 1.58 R0 = 1.32 k 0.632EX4.18 a.I DQ1 = K n1 (VGS 1 VTN 1 )21 = 1.2 (VGS 1 2 ) VGS 1 = VGS 2 = 2.91 V 2RS = 10 k VS1 = I DQ RS 10 = (1)(10 ) 10 = 0 R3 VG1 = 2.91 = (10 ) R1 + R2 + R3 R = 3 (10 ) R3 = 145.5 k 500 VDSQ1 = 3.5 VS 2 = 3.5 V VG 2 = 3.5 + 2.91 VG 2 = 6.41 R2 + R3 VG 2 = (10 ) = 6.41 R1 + R2 + R3 R + R3 = 2 (10 ) 500 R2 + R3 = 320.5 = R2 + 145.5 R2 = 175 kThen R1 + R2 + R3 = 500 = R1 + 175 + 145.5 R1 = 179.5 k Now VS 2 = 3.5 VD 2 = VS 2 + VSDQ 2 = 3.5 + 3.5 = 7 V So RD = b.10 7 RD = 3 k 1 From Example 6-18:www.elsolucionario.net 96. Av = g m1 RD g m1 = 2 K n1 I DQ = 2 (1.2 )(1) = 2.19 mA / V Av = ( 2.19 )( 3) Av = 6.57EX4.19 VS = I DQ RS = (1.2 )( 2.7 ) = 3.24 V VD = VS + VDSQ = 3.24 + 12 = 15.24 20 15.24 RD = RD = 3.97 k 1.2 V I D = I DSS 1 GS VP 22 V V 1.2 = 4 1 GS GS = 0.4523 VP VP VGS = ( 0.4523)( 3) = 1.357 VG = VS + VGS = 3.24 1.357 = 1.883 V R2 R2 VG = ( 20 ) = ( 20 ) = 1.88 R2 = 47 k. R1 = 453 k 500 R1 + R2 1 1 = = 167 k r0 = I DQ ( 0.005 )(1.2 ) 2 I DSS VGS 2 ( 4 ) 1.357 = 1 1 = 1.46 mA/V 3 ( VP ) VP 3 Av = g m ( r0 RD RL ) = (1.46 )(167 3.97 4 ) Av = 2.87gm =EX4.20 a. 22 V V V I DQ = I DSS 1 GS 2 = 8 1 GS GS = 0.5 VP VP VP VGS = ( 0.5 )( 3.5 ) VGS = 1.75Also I DQ =VGS ( 10 ) RS2=1.75 + 10 RS = 5.88 k RSb. Vi gmVgs Vgs V0 RSwww.elsolucionario.net 97. gm = r0 =2 I DSS VP VGS 1 VP 2 ( 8 ) 1.75 = 1 = 2.29 mA/V 3.5 3.5 1 = 50 k ( 0.01)( 2 )Vi = Vgs + g m RS Vgs Vgs = Av =c. Av =Vi 1 + g m RS( 2.29 ) [5.88 50] V0 g m RS r0 = = Av = 0.9234 V1 1 + g m RS r0 1 + ( 2.29 ) [5.88 50] g m ( RS1 + g m ( RS( 2.29 )( RS 1 + ( 2.29 )( RSRL ) roRL ro )RL ro )RL ro )= ( 0.80 )( 0.9234 ) = 0.7387= 0.7387 RSRL ro = 1.235 kRS ro = 5.261 k 5.261RL = 1.235 k RL = 1.61 k 5.261 + RLTYU4.1 g m = 2 K n (VGS VTN ) and I D = K n (VGS VTN ) VGS VTN = 2and g m = 2 K nI DQ KnI DQ Kn= 2 K n I DQ( 3.4 ) g2 = 1.445 mA / V Kn = m = 4 I DQ 4 ( 2) 2Kn= n Cox W 2LW 1.445 = ( 0.018 ) L W = 80.3 L 12 ro = K n (VGS VTN ) = I DQ 11ro = ( 0.015 )( 2 ) ro = 33.3 k TYU4.2a.I DQ = K n (VGS VTN )20.4 = 0.5 (VGS 2 ) VGS = 2.894 V 2VDSQ = VDD I DQ RD = 10 ( 0.4 )(10 ) VDSQ = 6 Vb.g m = 2 K n (VGS VTN ) = 2 ( 0.5 )( 2.894 2 ) g m = 0.894 mA/V 1r0 = I DQ , = 0 r0 = v0 Av = = g m RD = ( 0.894 )(10 ) Av = 8.94 vic.vi = 0.4sin t vds = ( 8.94 )( 0.4 ) sin t vds = 3.58sin twww.elsolucionario.net 98. At VDS 1 = 6 3.58 = 2.42 VGS1 = 2.89 + 0.4 = 3.29VGS1 VTN = 3.29 2 = 1.29 = VDS ( sat )So VDS 1 > VGS 1 VTN Biased in saturation regionTYU4.3 I DQ = K n (VGS VTN ) = ( 0.25 )( 2 0.8 ) I DQ = 0.36 mA 2a.2VDSQ = VDD I DQ RD = 5 ( 0.36 )( 5 ) VDSQ = 3.2 Vb.g m = 2 K n (VGS VTN ) = 2 ( 0.25 )( 2 0.8 ) g m = 0.6 mA/V,c.Av =r0 = v0 = g m RD = ( 0.6 )( 5 ) Av = 3.0 viTYU4.4 vi = vgs = 0.1sin t id = g m vgs = ( 0.6 )( 0.1) sin tid = 0.06sin t mA vds = ( 3)( 0.1) sin t = 0.3sin t ( V ) Then iD = I DQ + id = 0.36 + 0.06sin t = iD mA vDS = VDSQ + vds = 3.2 0.3sin t = vDSTYU4.5 a. VSDQ = VDD I DQ RD 7 = 12 I DQ ( 6 ) I DQ = 0.833 mA I DQ = K P (VSG VTP)20.833 = 2 (VSG 1) VSG = 1.65 V 2g m = 2 K P (VSG VTP ) = 2 ( 2 )(1.65 1) g m = 2.58 mA/V, r0 = b.Av =v0 = g m RD = ( 2.58 )( 6 ) Av = 15.5 vi V0 Vi gmVsgVSGRDTYU4.6 I DQ = K n (VGS VTN ) VGS VTN = 2g m = 2 K n (VGS VTN ) = 2 K nI DQ KnI DQ KnSo g m = 2 K n I DQTYU4.7= 2 2 f + vSBwww.elsolucionario.net 99. =(a)0.40 2 2 ( 0.35 ) + 1=(b) = 0.1530.40 2 2 ( 0.35 ) + 3 = 0.104( 0.5 )( 0.75) = 1.22 mA / V g mb = g m = (1.22 )( 0.153) g mb = 0.187 mA / V g mb = (1.22 )( 0.104 ) g mb = 0.127 mA / V g m = 2 K n I DQ = 2For (a), For (b),TYU4.8 I DQ = I Q = 0.5 mA W Let = 25 L K n = ( 20 )( 25 ) = 500 A / V 2 0.5 + 1.5 = 2.5 V VS = 2.5 V 0.5 Av = g m RDVGS =g m = 2 ( 0.5 )( 2.5 1.5 ) = 1 mA/VFor Av = 4.0 RD = 4 kVD = 5 ( 0.5 )( 4 ) = 3 V VDSQ = 3 ( 2.5 ) = 5.5 VTYU4.9 a. With RG VGS = VDS transistor biased in sat. region I D = K n (VGS VTN ) = K n (VDS VTN ) 22VDS = VDD I D RD = VDD K n RD (VDS VTN ) VDS = 15 ( 0.15 )(10 )(VDS 1.8 )222 = 15 1.5 (VDS 3.6VDS + 3.24 )2 1.5VDS 4.4VDS 10.14 = 0VDS =4.4 ( 4.4 )2+ ( 4 )(1.5 )(10.14 )2 (1.5 ) VDSQ = 4.45 VI DQ = ( 0.15 )( 4.45 1.8 ) I DQ = 1.05 mA 2b. Neglecting effect of RG: Av = g m ( RD RL ) g m = 2 K n (VGS VTN ) = 2 ( 0.15 )( 4.45 1.8 ) g m = 0.795 mA/V Av = ( 0.795 )(10 5 ) Av = 2.65 c.RG establishes VGS = VDS essentially no effect on small-signal voltage gain.TYU4.10 a. 2 I DQ = K n (VGS VTN ) I DQ = 0.8 ( 2 VSG ) = 2VSG VSG = 4 RS2 3.2 ( 4 4VSG + VSG ) = VSG2 3.2VSG 13.8VSG + 12.8 = 0www.elsolucionario.net 100. (13.8) 4 ( 3.2 )(12.8 ) 2 ( 3.2 ) 2 = 1.35 V I DQ = 0.8 ( 2 1.35 ) I DQ = 0.338 mAVSG = VSG213.8 b. VDSQ = VDD I DQ ( RD + RS ) 6 = 10 ( 0.338 )( RD + 4 ) RD =10 ( 0.338 )( 4 ) 6 0.338 RD = 7.83 kc. V0 Vgs Vi gmVgs RD RSVi = Vgs + g mVgs RS Vgs =Vi 1 + g m RSV0 = g mVgs RD g m = 2 K n (VGS VTN ) = 2 ( 0.8 )( 1.35 + 2 ) = 1.04 mA/V Av = (1.04 )( 7.83) V0 g m RD = = Av = 1.58 1 + (1.04 )( 4 ) Vi 1 + g m RSTYU4.11 a. 5 = I DQ RS + VSG and I DQ = K p (VSG + VTP ) 2 0.8 = 0.5(VSG + 0.8) 2 VSG = 0.465 V 5 = ( 0.8 ) RS + 0.465 RS = 5.67 k VSDQ = 10 I DQ ( RS + RD ) 3 = 10 ( 0.8 )( 5.67 + RD ) RD =10 ( 0.8 )( 5.67 ) 3 0.8 RD = 3.08 kb. V0 Vi VSGgmVsgr0RDV0 = g mVsg ( RD r0 ) = g mVi ( RD r0 ) V Av = 0 = g m ( RD r0 ) Vi g m = 2 K p (VSG + VTP ) = 2(0.5)(0.465 + 0.8) = 1.265 mA/V 1 1 = = 62.5 k r0 = I 0 ( 0.02 )( 0.8 ) Av = (1.265 )( 3.08 62.5 ) Av = 3.71www.elsolucionario.net 101. TYU4.12 (a) V0 = g mVgs r0 Vi = Vgs + V0 Vgs = Vi V0 So V0 = g m r0 (Vi V0 ) Av =( 4 )( 50 ) V0 g m r0 = = Av = 0.995 Vi 1 + g m r0 1 + ( 4 )( 50 ) Vgs IxI x + g mVgs = I x = g mVx +VxVx and Vgs = Vx r0Vx 1 1 R0 = r0 = 50 R0 0.25 k 4 r0 gmWith RS = 4 k Av =(b) r0gmVgsr0 Rs = 50 4 = 3.7 k Av =g m ( r0 RS )1 + g m ( r0 RS )( 4 )( 3.7 )1 + ( 4 )( 3.7 ) Av = 0.937TYU4.13 (a) g m = 2 K n I DQ 2 = 2 K n ( 0.8 ) K n = 1.25 mA / V 2 Kn = So n Cox W 2W 1.25 = ( 0.020 ) L LW = 62.5 LI DQ = K n (VGS VTN ) 0.8 = 1.25 (VGS 2 ) VGS = 2.8 V 22b. 11 r0 = I DQ = ( 0.01)( 0.8 ) = 125 k g m ( r0 RL ) Av = 1 + g m ( r0 RL )r0 RL = 125 4 = 3.88 Av =( 2 )( 3.88 ) Av = 0.886 1 + ( 2 )( 3.88 )R0 =1 1 r0 = 125 R0 0.5 k gm 2TYU4.14www.elsolucionario.net 102. Rin =1 = 0.35 k g m = 2.86 mA/V gmV0 4 RD = RD RL = 2.4 = RD 4 = Ii 4 + RD( 4 2.4 ) RD = ( 2.4 )( 4 ) RD = 6 k g m = 2 K n I DQ 2.86 = 2 K n ( 0.5 ) K n = 4.09 mA / V 2 I DQ = K n (VGS VTN )20.5 = 4.09 (VGS 1) VGS = 1.35 V VS = 1.35 V, VD = 5 ( 0.5 )( 6 ) = 2 V 2VDS = VD VS = 2 ( 1.35 ) = 3.35 VWe have VDS = 3.35 > VGS VTN = 1.35 1 = 0.35 V Biased in the saturation region TYU4.15 C K n1 = n ox 2 Kn2W L C W = n ox 2 LAv = 2 = ( 0.020 )( 80 ) = 1.6 mA / V 1 2 = ( 0.020 )(1) = 0.020 mA / V 2K n1 1.6 = Av = 8.94 Kn2 0.020The transition point is determined from vGSt VTND = VDD VTNL K n1 ( vGSt VTND ) Kn2vGSt 0.8 = ( 5 0.8 ) ( 8.94 )( vGSt 0.8 ) vGSt =( 5 0.8) + ( 8.94 )( 0.8) + 0.8 1 + 8.94vGSt = 1.22 VFor Q-point in middle of saturation region VGS =1.22 0.8 + 0.8 VGS = 1.01 V 2TYU4.16 a. I DQ 2 = 2mA, VDSQ 2 = 10 V I DQ 2 RS 2 = 10 = 2 RS 2 RS 2 = 5 k I DQ 2 = K n 2 (VGS 2 VTN 2 )22 = 1(VGS 2 2 ) VSG 2 = 3.41 V VG 2 = 3.41 V 210 3.41 RD1 = 3.29 k 2 = 10 V VS 1 = 3.41 10 = 6.59 VThen RD1 = For VDSQ1Then RS 1 =6.59 ( 10 ) 2I D1 = K n1 (VGS 1 VTN 1 ) RS1 = 1.71 k22 = 1(VGS 1 2 ) VGS 1 = 3.41 V 2 R2 R2 1 VGS1 = = Rin ( 20 ) I DQ1 RS 1 R1 + R2 R1 + R2 R1 1 3.41 = ( 200 )( 20 ) ( 2 )(1.71) R1 = 586 k R1www.elsolucionario.net 103. 586 R2 = 200 ( 586 200 ) R2 = ( 200 )( 586 ) R2 = 304 k 586 + R2 g m1 = 2 K n1 I DQ1 = 2 (1)( 2 ) g m1 = g m 2 = 2.828 mA/Vb.From Example 6.17 g g R ( R RL ) Av = m1 m 2 D1 S 2 1 + g m 2 ( RS 2 RL ) RS 2 RL = 5 4 = 2.222 k ( 2.828 )( 2.828 )( 3.29 )( 2.222 )Av =1 + ( 2.828 )( 2.222 ) Av = 8.031 1 RS 2 = 5 = 0.3536 5 R0 = 0.330 k gm2 2.828R0 =TYU4.17 From Example 6.19 g m = 3.0 mA/V, r0 = 41.7 k R1 R2 = 420 180 = 126 k R1 R2 Vgs = Vi R1 R2 + Ri 126 = Vi = 0.863Vi 126 + 20 g mVgs ( r0 RD RL )Av =Vi= ( 3.0 )( 0.863)( 41.7 2.7 4 ) = ( 2.589 )( 41.7 1.61) = ( 2.589 )(1.55 ) Av = 4.01TYU4.18 a. R2 VG1 = (VDD ) R1 + R2 430 VG1 = ( 20 ) = 17.2 V 430 + 70 2 V V V I DQ1 = I DSS 1 GS = 6 1 G1 S1 VP 2 22220 VS1 17.2 VS 1 V = 6 1 + = 6 S 1 7.6 and I DQ1 = 2 2 2 4 220 VS 1 V = 6 S1 7.6 4 2 2 V 20 VS1 = 24 S 1 7.6VS 1 + 57.76 4 Then= 6VS21 182.4VS 1 + 1386.24 6VS21 181.4VS 1 + 1366.24 = 0 VS 1 =181.4 (181.4 ) 4 ( 6 )(1366.24 ) 2 ( 6) 2VS 1 = 14.2 V VGS 1 = 17.2 14.2 = 3 V > VPwww.elsolucionario.net 104. So VS1 = 16.0 VGS1 = 17.2 16 = 1.2 < VP Q 20 16 I DQ1 = 1 mA 4 = 20 I DQ1 ( RS 1 + RD1 ) = 20 (1)( 8 ) VSDQ1 = 12 Von I DQ1 = VSDQ1VD1 = I DQ1 RD1 = (1)( 4 ) = 4 V = VG 2 2 V VS 2 V I DQ 2 = I DSS 1 GS = 6 1 G 2 VP ( 2 ) 222V V 4 V V = 6 1 + S 2 = 6 3 S 2 and I DQ 2 = S 2 = S 2 RS 2 2 4 2 2 ThenVS 2 V = 63 S2 4 2 2 V2 VS 2 = 24 9 3VS 2 + S 2 4 2 6VS 2 73VS 2 + 216 = 0 VS 2 =( 73)73 2 4 ( 6 )( 216 )2 ( 6) VS 2 = 7.09 V or = 5.08 VFor VS 2 = 5.08 V VGS 2 = 4 5.08 = 1.08 transistor biased ON 5.08 I DQ 2 = 1.27 mA 4 = 20 VS 2 = 20 5.08 VDS 2 = 14.9 VI DQ 2 = VDS 2b. Vg2 gm2Vgs2 Vgs2Vi RD1V0Vsg1 gm1Vsg1RS2RLVg 2 = g m1Vsg1 RD1 = g m1V1 RD1 V0 = g m 2Vgs 2 ( RS 2 RL )Vg 2 = Vgs 2 + V0 Vgs 2 =Vg 2 1 + g m 2 ( RS 2 RL )Av =V0 g m1 RD1 = Vi 1 + g m 2 ( RS 2 RL )g m1 =2 I DSS VP VGS 1 VP 2 ( 6 ) 1.2 1 = 2.4 mA/V 2 2 2 ( 6 ) 1.08 gm2 = 1 = 2.76 mA/V 2 2 ( 2.4 )( 4 ) = 2.05 = Av Then Av = 1 + ( 2.76 )( 4 )( 2 ) =www.elsolucionario.net 105. Chapter 4 Problem Solutions 4.1 (a) C W g m = 2 K n I D = 2 n ox I D 2 L 0.5 = 2 ( 0.040 ) W = 3.125 ( 0.5 ) L L(b) 2 C W I D = n ox (VGS VTN ) 2 L 0.5 = ( 0.04 )( 3.125 )(VGS 0.8 ) VGS = 2.80 V 24.2 p Cox W gm = 2 K p I D = 2 ID 2 L (a)2W 50 W = 0.3125 = ( 20 ) (100 ) L 2 L p Cox W 2 ID = (VSG + VTP ) 2 L (b)100 = ( 20 )( 0.3125 )(VSG 1.2 ) VSG = 5.2 V 24.3 I D = K n (VGS VTN ) (1 + VDS ) 2I D1 1 + VDS1 3.4 1 + (10 ) = = I D 2 1 + VDS 2 3.0 1 + ( 5 )3.4 [1 + 5 ] = 3.0 [1 + 10 ]3.4 3.0 = ( 3 10 ( 3.4 ) 5 ) = 0.0308 ro =VDS 5 = = 12.5 k I D 0.44.4 r0 = ID =1 ID 1 r0=1( 0.012 )(100 ) I D = 0.833 mA4.5 2 I D = K n (VGS VTN ) (1 + VDS ) I D1 1 + VDS 1 = I D 2 1 + VDS 2 0.20 1 + ( 2 ) = 0.22 1 + ( 4 )0.20 (1 + 4 ) = 0.22 (1 + 2 )( 0.8 0.44 ) = 0.22 0.20 = 0.0556 V 1www.elsolucionario.net 106. ro =VDS 2 = ro = 100 K 0.02 I D4.6 (a) (i)ro =1 1 = = 1000 K I D ( 0.02 )( 0.05 )(ii)ro =1 = 100 K ( 0.02 )( 0.5)(b) I D =(i)VDS 1 = = 0.001 mA = 1.0 A 1000 roI D 1.0 2% = 50 ID I D =(ii)VDS 1 = = 0.01 10 A ro 100I D 10 = 2% ID 5004.7 I D = 1.0 mA 1 1 ro = = = 100 K I D ( 0.01)(1) 4.8 Av = g m ( r0 RD ) = (1)( 50 10 ) Av = 8.33 4.9 RD =a.VDD VD SQ I DQ=10 6 RD = 8 k 0.5For VGSQ = 2 V, for example, 2 C W I DQ = n ox (VGSQ VTN ) 2 L W 2 W 0.5 = ( 0.030 ) ( 2 0.8 ) = 11.6 L L b.g m = 2 K n I DQ = 2 K n (VGSQ VTN )g m = 2 ( 0.030 )(11.6 )( 2 0.8 ) g m = 0.835 mA/V ro =1c.=1 r = 133 k( 0.015)( 0.50 ) o Av = g m ( r0 RD ) = ( 0.835 )(133 8 ) Av = 6.30 I DQ4.10 22 2 K n vgs = K n Vgs sin t = K nVgs sin 2 t 1 [1 cos 2 t ] 2 2 K nVgs 2 So K n vgs = [1 cos 2 t ] 2 sin 2 t =www.elsolucionario.net 107. 2 K nVgsRatio of signal at 2 to that at :2 cos 2 t2 K n (VGSQ VTN ) Vgs sin tThe coefficient of this expression is then:Vgs4 (VGSQ VTN )4.11 0.01 =Vgs4 (VGSQ VTN )So Vgs = ( 0.01)( 4 )( 3 1) Vgs = 0.08 V4.12 Vo = g mVgs ( ro RD ) 50 Vi = Vi = ( 0.9615 ) Vi R1 R2 + RSi 50 + 2 Av = g m ( 0.9615 ) ( ro RD ) = (1)( 0.9615 ) ( 50 10 ) Av = 8.01 R1 R2Vgs =4.13 Av = g m ( r0 RD ) 10 = g m (100 5 ) g m = 2.1 mA/V 4.14 a. R2 VG = VDD R1 + R2 200 VG = (12 ) = 4.8 V 200 + 300 V VGS 2 = K n (VGS VTN ) ID = G RS2 4.8 VGS = (1)( 2 ) (VGS 4VGS + 4 ) 2 2VGS 7VGS + 3.2 = 0VGS =7(7)2 4 ( 2 )( 3.2 )2 ( 2)= 2.96 VI D = (1)( 2.96 2 ) I D = 0.920 mA 2VDS = VDD I D ( RD + RS )= 12 ( 0.92 )( 3 + 2 ) VDS = 7.4 V(b) Vo = g mVg ( RD RL ) 1 + g m RSwhere Vg =R1 R2 R1 R2 + RSi Vi =300 200 300 200 + 2 Vi =120 Vi = ( 0.9836 ) Vi 120 + 2Thenwww.elsolucionario.net 108. Av = g m ( 0.9836 )( RD RL ) 1 + g m RSg m = 2 (1)( 2.96 2 ) = 1.92 mA / V So Av = (1.92 )( 0.9836 )( 3 3) 1 + (1.92 )( 2 ) Av = 0.585c.AC load line 1 1 Slope 3.5 K 33 2 0.92127.41 i D = vds 3.5 k i D = 0.92 mA vds = ( 0.92 )( 3.5 ) = 3.22 6.44 V peak-to-peak4.15 a. I D Q = 3 mA VS = I DQ RS = ( 3)( 0.5 ) = 1.5 V I DQ = K n (VGS VTN )23 = ( 2 )(VGS 2 ) VGS = 3.225 V 2VG = VGS + VS = 3.225 + 1.5 = 4.725 V R2 1 VG = VDD = Rin VDD R1 R1 + R2 1 4.725 = ( 200 )(15 ) R1 = 635 k R1 635R2 = 200 R2 = 292 k 635 + R2 b. g m ( RD RL ) Av = 1 + g m RSg m = 2 ( 2 )( 3.225 2 ) = 4.90 mA / V Av = ( 4.90 )( 2 5 )1 + ( 4.90 )( 0.5 ) Av = 2.034.16 From Problem 4.14: I D = 0.920 mA VDS = 7.4 V g m = 1.92 mA/V R1 R2 Av = g m ( RD RL ) R1 R2 + RSi 200 300 = (1.92 )( 3 3) = ( 2.88 )( 0.9836 ) 200 300 + 2 Av = 2.83www.elsolucionario.net 109. AC load line 1 1 Slope 33 1.5 K 0.92127.41 vDS , iD = 0.92 mA vDS = ( 0.92 ) (1.5 ) = 1.38 1.5 k 2.76 V peak-to-peak output voltage swing iD =4.17 (a) Av = g m RD 15 = 2 RD RD = 7.5 K (b) Av = 5 = g m RD 1 + g m RS ( 2 )( 7.5 ) 1 + ( 2 ) RS RS = 1 K4.18 (a)Av = g m RD 1 + g m RS(1)8 = g m RD 1 + g m (1)(2)16 = g m RDThen 8 =16 1 + g m (1)g m = 1 mA/V RD = 16 K(b)Av = 10 = (1)(16 )1 + (1) RSRS = 0.6 K4.19 a.AC load line 1 Slope 5K IDQVDS(sat)VDSQwww.elsolucionario.net 110. VDSQ = V + I DQ RD (VGSQ ) Output Voltage Swing = VDSQ VDS ( sat ) = V + I DQ RD + VGSQ (VGSQ VTN ) = V + I DQ RD + VTN So I D = I DQ =1 1 V + I DQ RD + VTN VDS = 5 k 5 k 1 5 I DQ (10) + 1 5 k = 1.2 2 I DQ = I DQ I DQ = 0.4 mA = I QI DQ =b.( 0.5 )( 0.4 ) = 0.8944 mA / Vg m = 2 K n I DQ = 2 r0 =1 1 = = 250 k I DQ ( 0.01)( 0.4 )Av = g m ( RD RL r0 ) = ( 0.8944 )(10 10 250 ) Av = 4.384.20 (a) I DQ = K n (VGSQ VTN )20.1 = 0.85 (VGSQ 0.8 )2VGSQ = 1.143 V RS =1.143 ( 5 ) RS = 38.6 K0.1V = I ( RD RL ) roro =1 = 0.1( RD RL ) ro RD RL ro = 10 K =1 = 500 K ( 0.02 )( 0.1)40 500 RD 40 500 + RD37.04 RD = 10 37.04 + RDRD = 13.7 K(b) gm = 2 Kn I D = 2( 0.85)( 0.1)g m = 0.583 mA/V ro = 500 K (c) Av = g m ( RD RL ro )= ( 0.583) (13.7 40 500 ) = ( 0.583)(10 )Av = 5.834.21 a. 2 I D = K n (VGS VTN ) 2 = 4 (VGS ( 1) )2VGS = 0.293 V VS = 0.293 V = I D RS = (2) RS RS = 0.146 kwww.elsolucionario.net 111. VD = VDS + VS = 6 + 0.293 = 6.293 RD =b. Av =10 6.293 RD = 1.85 k 2 g m ( RD RL ) 1 + g m RSg m = 2 K n (VGS VTN ) g m = 2 ( 4 )( 0.293 + 1) = 5.66 mA/V Av = ( 5.66 ) (1.85 2 )1 + ( 5.66 )( 0.146 ) Av = 2.98c.AC load line1 1.852 0.146 1 1.107 KSlope 2 mA106v0 = ( iD )(1.85 2 ) = ( 2 )(1.85 2 ) = 1.92 V vi =1.92 = 0.645 Vi = 0.645 V 2.984.22 a. VDSQ = VDD I DQ ( RD + RS ) 2.5 = 5 I DQ ( 2 + RS ) 2.5 I DQ = 2 + RS I DQ = K n (VGS VTN ) = 2VGS 2.5 RS VGS = I DQ RS = 2 + RS RS2 2.5RS 2.5 Kn VTN = 2 + RS 2 + RS 2 2.5 RS 2.5 4 1 = 2 + RS 2 + RS 2 2 + RS 2.5 RS 2.5 4 = 2 + RS 2 + RS 4( 2 1.5RS ) 2 + RS2= 2.52 4 ( 4 6 RS + 2.25 RS ) = 2.5 ( 2 + RS ) 2 9 RS 26.5RS + 11 = 0RS =26.5 ( 26.5) 4 ( 9 )(11) 2 (9) 2RS = 0.5 k or 2.44 kBut RS = 2.44 k VGS = 1.37 Cut off. RS = 0.5 k,I DQ = 1.0 mAwww.elsolucionario.net 112. b. Av = g m ( RD RL ) 1 + g m RS( 4 )(1) = 4 mA / Vg m = 2 K n I DQ = 2 Av = ( 4 )( 2 2 )1 + ( 4 )( 0.5 ) Av = 1.334.23 a. 5 = I DQ RS + VSDQ + I DQ RD 5 5 = I DQ RS + 6 + I DQ (10 ) 5 1.I DQ =4 RS + 10VS = VSDQ + I DQ RD 5 = VSGQ2.1 + I DQ (10 ) = VSGQ3.I DQ = K p (VSGQ 2 )2Choose RS = 10 k I DQ =4 = 0.20 mA 20VSGQ = 1 + (0.2)(10) = 3 V 0.20 = K P (3 2) 2 K P = 0.20 mA / V 2b. I DQ = ( 0.20 )( 3 2 ) = 0.20 mA 2( 0.2 )( 0.2 ) = 0.4 mA / V Av = g m ( RD RL ) = ( 0.4 )(10 10 ) Av = 2.0g m = 2 K P I DQ = 2c. 4 = 0.133 mA 30 = 1 + (0.133)(10) = 2.33 VChoose RS = 20 k I DQ = VSGQ0.133 = K p (2.33 2) 2 K p = 1.22 mA / V 2 g m = 2 (1.22)(0.133) = 0.806 mA/V Av = (0.806)(10 10) Av = 4.03A larger gain can be achieved. 4.24 (a) I DQ = K p (VSGQ + VTP )20.25 = 0.8 (VSGQ 0.5 )2VSGQ = 1.059 V 3 1.059 RS = RS = 7.76 K 0.25 VD = VS VSDQ = 1.059 1.5 = 0.441 V RD =0.441 ( 3) 0.25 RD = 10.2 K(b)www.elsolucionario.net 113. Av = g m ( RD RL )( 0.8)( 0.25 ) = 0.8944 mA/V Av = ( 0.8944 )(10.2 2 )g m = 2 K p I DQ = 2Av = 1.50 (c) VO = I ( RD RL ) = 0.25 (10.2 2 ) = 0.418 So VO = 0.836 peak-to-peak4.25 I DQ = K n (VGSQ VTN )2g m = 2 K n I DQ 2.2 = 2 K n ( 6 ) K n = 0.202 mA / V 2 6 = 0.202 ( 2.8 VTN ) VTN = 2.65 V 2VDSQ = 18 I DQ ( RS + RD ) 18 10 = 1.33 k RS = 1.33 RD RS + RD = 6 g m ( RD RL ) Av = 1 + g m RS R 1 2.2 D RD + 1 1 = 1 + ( 2.2 )(1.33 RD ) 1 + 2.93 2.2 RD =2.2 RD 1 + RD( 3.93 2.2 RD )(1 + RD ) = 2.2 RD 2 3.93 + 1.73RD 2.2 RD = 2.2 RD 2 2.2 RD + 0.47 RD 3.93 = 0RD =0.47 +( 0.47 ) + 4 ( 2.2 )( 3.93) RD = 1.23 k, 2 ( 2.2 ) = 2.8 + ( 6 )( 0.1) = 3.4 V 2RS = 0.10 kVG = VGS + VS 1 1 VG = Rin VDD = (100 )(18 ) = 3.4 R1 = 529 k R1 R1 529 R2 = 100 R2 = 123 k 529 + R24.26 a.VSD(sat)VSDQV1www.elsolucionario.net 114. V1 = 9 + VSG , VSD ( sat ) = VSG + VTP VSDQ = =V1 VSD ( sat ) 2+ VSD ( sat )( 9 + VSG ) (VSG + VTP ) 2+ (VSG + VTP )9 + 1.5 + VSG 1.5 2 = 3.75 + VSG = 9 + VSG I DQ RD =VSDQI DQ = K p (VSG + VTP ) Set RD = 0.1RL = 2 k 3.75 = 9 I DQ ( 2 ) I DQ = 2.625 mA 2b. g m = 2 K p I DQ = 2 r0 =( 2 )( 2.625) = 4.58 mA / V1 1 = = 38.1 k I DQ ( 0.01)( 2.625 )Open circuit. Av = g m ( RD r0 ) Av = 4.58 ( 2 38.1) Av = 8.70 c. With RL Av = 4.58 ( 2 20 38.1) Av = 7.947 Change = 8.66% 4.27 (a) I DQ = K p (VSGQ + VTP )20.5 = 0.25 (VSGQ + 0.8 )2VSGQ = 0.614 V = VS 10 0.614 RS = RS = 18.8 K 0.5 VD = VS VSDQ = 0.614 3 = 2.386 V RD =2.386 ( 10 ) 0.5 RD = 15.2 K(b) Av = g m RD( 0.25)( 0.5 ) = 0.7071 mA/V Av = ( 0.7071)(15.2 )g m = 2 K p I DQ = 2 Av = 10.74.28 Av = g m ( RD RL )VDSQ = VDD I DQ ( RS + RD )10 = 20 (1)( RS + RD ) RS + RD = 10 kLet RD = 8 k, RS = 2 k Av = 10 = g m ( 8 20 ) g m = 1.75 mA / V = 2 K n I DQ = 2 K n (1) K n = 0.766 mA / V 2www.elsolucionario.net 115. VS = I DQ RS = (1)( 2 ) = 2 V I DQ = K n (VGS VTN ) 1 = 0.766 (VGS 2 ) VGS = 3.14 V 22VG = VGS + VS = 3.14 + 2 = 5.14 VG =1 1 Rin VDD ( 200 )( 20 ) = 5.14 R1 = 778 k R1 R1778R2 = 200 R2 = 269 k 778 + R24.29 Av =( 4 )( 50 ) g m r0 = Av = 0.995 1 + g m r0 1 + ( 4 )( 50 )Ro =1 1 r0 = 50 R0 = 0.249 k 4 gmAv = = R0 =g m ( r0 RS )1 + g m ( r0 RS ) 4 ( 2.38 )1 + 4 ( 2.38 )=4 ( 50 2.5 )1 + 4 ( 50 2.5 ) Av = 0.9051 1 r0 RS = 50 2.5 R0 = 0.226 k 4 gm4.30 Av = 0.98 =g m ( RL ro )1 + g m ( RL ro ) g m ro g m ro = 49 1 + g m roAlso 0.49 =0.49 = 0.49 =g m ( RL ro )1 + g m ( RL ro ) Rr gm L o RL + ro = Rr 1 + gm L o RL + ro g m ( RL ro )RL + ro + g m ( RL ro )( 49 )(1) 49 = 1 + ro + ( 49 )(1) 50 + roro = 50 K g m = 0.98 mA/V4.31 (a) Av =( 2 )( 25) g m ro = 1 + g m ro 1 + ( 2 )( 25 )Av = 0.98 Ro =1 1 ro = 25 = 0.5 25 gm 2Ro = 0.49 K(b)www.elsolucionario.net 116. Av =g m ( ro RL )1 + g m ( ro RL )=2 ( 25 2 )1 + 2 ( 25 2 )=2 (1.852 )1 + 2 (1.852 )Av = 0.7874.32 Av =g m ( RL ro )1 + g m ( RL ro )=5 (10 100 )1 + ( 5 )(10 100 )=5 ( 9.09 )1 + 5 ( 9.09 )Av = 0.978 Ro =1 1 RL ro = 10 100 = 0.2 9.0909 gm 5Ro = 0.196 K4.33 a. R2 396 VG = VDD = (10 ) VG = 2.42 V 396 + 1240 R1 + R2 10 (VSG + 2.42 ) 2 = K p (VSG + VTP ) I DQ = RS 2 7.58 VSG = ( 2 )( 4 ) (VSG 4VSG + 4 )2 8VSG 31VSG + 24.4 = 0VSG =31 ( 31)2 4 ( 8 )( 24.4 )2 (8) VSG = 2.78 VI DQ = ( 2 )( 2.78 2 ) I DQ = 1.21 mA 2VSDQ = 10 I DQ RS = 10 (1.21)( 4 ) VSDQ = 5.16 Vb. g m = 2 K p I DQ = 2 r0 = Av = =( 2 )(1.21) = 3.11 mA / V1 1 = = 41.3 k I DQ ( 0.02 )(1.21) g m ( RS RL r0 )1 + g m ( RS RL r0 )3.11( 4 4 41.3)1 + ( 3.11) ( 4 4 41.3) Av = 0.856 Ii Vsg Vi R1R2r0gmVsg RSRL I0www.elsolucionario.net 117. RS r0 I 0 = ( g mVsg ) R r +R L S 0 Vi Vsg = 1 + g m ( RS RL r0 ) Vi = I i ( R1 R2 ) Ai =g m ( R1 R2 ) RS r0 I0 = I i 1 + g m ( RS RL r0 ) RS r0 + RL( 3.11) ( 396 1240 ) 4 41.3 1 + ( 3.11) ( 4 4 41.3) 4 41.3 + 4 ( 3.11)( 300 ) 3.647 = Av = 64.2 1 + ( 3.11)(1.908 ) 3.647 + 4 =R0 =1 1 4 41.3 R0 = 0.295 k RS r0 = 3.11 gm4.34 g m = 2 K n I Q = 2 (1)(1) = 2 mA / V 1 1 = = 100 k I Q ( 0.01)(1)r0 =g m ( r0 RL ) Av = 0.8851 + g m ( r0 RL )R0 =1 1 r0 = 100 R0 = 0.498 k 2 gm4.35 a. Av 2 (100 4 )Av =1 + 2 (100 4 )gm ( 4) g m RL 0.95 = 1 + g m RL 1 + gm ( 4)0.95 = 4 (1 0.95 ) g m g m = 4.75 mA/V1 W g m = 2 n Cox 2 L IQ 4.75 = 2 ( 0.030 ) W W = 47.0 ( 4) L Lb.1 W g m = 2 n Cox 2 L4.75 = 2 IQ ( 0.030 )( 60 ) I Q I Q = 3.13 mA4.36 2 I DQ = K n (VGS VTN ) 5 = 5 (VGS + 2 ) VGS = 1 V VS = VGS = 1 V 2I DQ =VS ( 5 ) RS RS =g m = 2 K n I DQ = 2 r0 =1 I DQ=1+ 5 RS = 1.2 k 5( 5 )( 5) = 10 mA / V1 = 20 k 0.01)( 5 ) (www.elsolucionario.net 118. Av = = R0 =g m ( r0 RS RL )1 + g m ( r0 RS RL )(10 ) ( 20 1.2 2 ) Av = 0.878 1 + (10 ) ( 20 1.2 2 ) 1 1 r0 RS = 20 1.2 Ro = 91.9 10 gm4.37 Av =g m (10 ) g m RS 0.90 = 1 + g m RS 1 + g m (10 )0.90 = 10 (1 0.90 ) g m g m = 0.90 mA/V 1 1 RS = 10 R0 = 1 k gm 0.90 With RL : R0 =Av =g m ( RS RL )( 0.90 )(10 2 ) Av = 0.60 1 + g m ( RS RL ) 1 + ( 0.90 )(10 2 ) =4.38 R0 =1 RS gmOutput resistance determined primarily by gm 1 = 0.2 k g m = 5 mA/V Set gm g m = 2 K n I DQ 5 = 2 I DQ = K n (VGS VTN )( 4 ) I DQ I DQ = 1.56 mA21.56 = 4 (VGS + 2 ) VGS = 1.38 V, VS = VGS = 1.38 V 21.38 ( 5 ) RS = 4.09 k 1.56 5 ( 4.09 2 ) g m ( RS RL ) Av = = Av = 0.870 1 + g m ( RS RL ) 1 + 5 ( 4.09 2 )RS =4.39 a.g m = 2 K p I DQ = 2( 5)( 5 ) = 10 mA / V1 1 = Ro = 100 g m 10 b. Open-circuit gain g m r0 But r0 = so Av = 1.0 Av = 1 + g m r0 With RL: g m RL Av = 1 + g m RL Ro =0.50 =10 RL 0.50 = 10 (1 0.5 ) RL RL = 0.1 k 1 + 10 RL4.40www.elsolucionario.net 119. 1 v0 RS RLiD = I DQ =v0 = I DQ RS RL = v0 ( min ) = Av = vi =I DQ RS RL RS + RLI DQ RS R 1+ S RL g m ( RS RL )1 + g m ( RS RL )=v0 vi I DQ ( RS RL ) 1 + g m ( RS RL ) g m ( RS RL )vi ( min ) = I DQ1 + g m ( RS RL ) gm 4.41 (a) VDD = VDSQ + I DQ RS 3 = 1.5 + ( 0.25 ) RS RS = 6 K VS = 1.5 V I DQ = K n (VGSQ VTN ) 0.25 = 0.5 (VGSQ 0.4 )22VGSQ = 1.107 V VG = VGSQ + VS + 1.107 + 1.5 = 2.607 V R2 1 VG = VDD = RL VDD R1 + R2 R1 1 2.607 = ( 300 )( 3) R1 = 345.2 K R2 = 2291 K R1 (b) g m RS Av = g m = 2 K n I DQ = 2 ( 0.5 )( 0.25 ) = 0.7071 mA/V 1 + g m RS Av =( 0.7071)( 6 ) Av = 0.809 1 + ( 0.7071)( 6 )Ro =1 1 RS = 6 = 1.414 6 gm ( 0.7071)Ro = 1.14 K4.42 Av = + g m RD = ( 5 )( 4 ) Av = 20 Ri =1 1 10 = 10 = 0.2 10 5 gmRi = 0.196 K4.43 a.www.elsolucionario.net 120. VGS + I DQ RS = 5 5 VGS 2 = K n (VGS VTN ) I DQ = RS2 5 VGS = (10 )( 3) (VGS 2VGS + 1) 2 30VGS 59VGS + 25 = 0VGS =59 ( 59 )2 4 ( 30 )( 25 )2 ( 30 ) VGS = 1.35 VI DQ = ( 3)(1.35 1) I DQ = 0.365 mA 2VDSQ = 10 ( 0.365 )( 5 + 10 ) VDSQ = 4.53 Vb. g m = 2 K n I DQ = 2 r0 =c.1( 3)( 0.365 ) g m = 2.093 mA / V1=r =( 0 )( 0.365 ) 0 Av = g m ( RD RL ) = ( 2.093) ( 5 4 ) Av = 4.65 I DQ4.44 a. I DQ = K p (VSG + VTP )20.75 = ( 0.5 )(VSG 1) VSG = 2.225 V 5 2.225 5 = I DQ RS + VSG RS = RS = 3.70 k 0.75 VSDQ = 10 I DQ ( RS + RD ) 6 = 10 ( 0.75 )( 3.70 + RD ) RD = 1.63 k 2b. Ri =1 gmg m = 2 K p I DQ = 2( 0.5 )( 0.75) = 1.225 mA / V1 Ri = 0.816 k 1.225 Ro = RD Ro = 1.63 k Ri =c. RS i R + [1/ g ] i S m 3.70 1.63 i0 = ii 1.63 + 2 3.70 + 0.816 i0 = 0.368ii = i0 = 1.84sin t ( A ) RD i0 = RD + RLv0 = i0 RL = (1.84 )( 2 ) sin t v0 = 3.68sin t ( mV )4.45 a. 2 I DQ = K n (VGS VTN ) 5 = 4 (VGS 2 ) VGS = 3.12 V V + = I DQ RD + VDSQ VGS 210 = ( 5 ) RD + 12 3.12 RD = 0.224 kb.www.elsolucionario.net 121. g m = 2 K n I DQ = 2 Ri =( 4 )( 5 ) g m = 8.944 mA / V1 1 = Ri = 0.112 k g m 8.94Av = g m ( RD RL ) = ( 8.944) ( 0.224 2 ) Av = 1.80c.4.46 a. 2 I DQ = K p (VSG + VTP ) 3 = 2 (VSG 2 ) VSG = 3.225 V V + = I DQ RS + VSG 10 3.225 RS = RS = 2.26 k 3 VSDQ = 20 I DQ ( RS + RD ) 10 = 20 ( 3)( 2.26 + RD ) RD = 1.07 k 2b. Av = g m ( RD RL )( 2 )( 3) = 4.90 mA / V Av = ( 4.90 )(1.07 10 ) Av = 4.74g m = 2 K p I DQ = 24.47 a. Av =(W / L )D (W / L )L= 5 (W / L ) D = 25 ( 0.5 ) (W / L ) D = 12.51 W n Cox = ( 30 )(12.5 ) = 375 A/V 2 2 L D K L = ( 30 )( 0.5 ) = 15 A / V 2KD =Transition point: vGSD VTND = (VDD VTNL ) KD ( vGSD VTND ) KL375 ( vGSD 2 ) 15 vGSD (1 + 5 ) = (10 2 ) + 2 + 5 ( 2 ) vGSD = 3.33 V and vDSD = 1.33 V 3.33 2 + 2 VGSQ = 2.67 V VGSQ = 2 vGSD 2 = (10 2 ) O8 Q-pointVDSQ 1.333.332GSDVGSQb.www.elsolucionario.net 122. I DQ = K D (VGSQ VTND )2I DQ = 0.375 ( 2.67 2 ) I DQ = 0.166 mA 2VDSQ =4.48 (a)8 1.33 + 1.33 VDSQ = 4.67 V 2 Transition point: Load: VOtB = VDD VTNL = 10 2 = 8 VDriver: 2 2 K D ( vOt A ) (1 + D vOt A ) = K L ( VTNL ) 1 + L (VDD vOt A ) 2 3 0.5 v0t A + ( 0.02 ) v0t A = 0.1 ( 4 )(1 + 0.02 ) (10 v0t A ) 3 2 We have 0.01v0t A + 0.5v0t A + 0.008v0t A 0.48 = 0()Therefore v0t A = 0.963 V 8 0.963 + 0.963 = 4.48 V = VDSQ 2 2 2 Then K D (VGSD VTND ) (1 + D vOQ ) = K L ( VTNL ) 1 + L (VDD vOQ ) (VGSD 1.2 )2 (1 + ( 0.02 )( 4.48 ) ) = 0.1 ( 4 ) (1 + 0.02 (10 4.48 ) ) which yields VGSD = 2.103 V 0.5 b. 2 I DQ = K D (VGSD VTND ) (1 + D vOQ ) 2 I DQ = 0.5 ( 2.103 1.2 ) (1 + ( 0.02 )( 4.48 ) ) So I DQ = 0.444 mANow v0 Q =()c. r0 D = r0 L =1 1 = = 112.6 k I DQ ( 0.02 )( 0.444 )g mD = 2 K D (VGSD VTND ) = 2 ( 0.5 )( 2.103 1.2 ) g mD = 0.903 mA / V Then Av = g mD ( r0 D r0 L ) = ( 0.903) (112.6 112.6 ) or Av = 50.84.49 2 I D = K n (VGS VTN ) , VDS = VGS I D = 0 when VDS = 1.5 V VTN = 1.5 V 0.8 = K n ( 3 1.5 ) K n = 0.356 mA / V 2 dI I go = D = = 2 K n (VDS VTN ) = 2 ( 0.356 )( 3 1.5 ) Ro = 0.936 k dVDS Ro 24.50 a. I DQ = K nD (VGS VTND ) = ( 0.5 ) ( 0 ( 1) ) 22I DQ = 0.5 mA I DQ = K nL (VGSL VTNL ) = K nL (VDD VO VTNL ) 20.5 = 0.030 (10 V0 1)220.5 = 9 V0 V0 = 4.92 V 0.030b.www.elsolucionario.net 123. I DD = I DL K nD (Vi VTND ) = K nL (VDD Vo VTNL ) 22K nD (Vi VTND ) = VDD Vo VTNL K nL K nD (Vi VTND ) K nLVo = VDD VTNL Av =(W / L )D (W / L )LdVo K nD = = dVi K nLAv = 500 Av = 4.08 304.51 (a) I DQ = K L (VGSL VTNL ) = K L (VDSL VTNL ) 22I D = ( 0.1)( 4 1) = 0.9 mA 2I DQ = K D (VGSD VTND )20.9 = (1)(VGSD 1) VGSD = 1.95 V VGG = VGSD + VDSL = 1.95 + 4 VGG = 5.95 V 2b. I DD = I DL K D (VGSD VTND ) = K L (VGSL VTNL ) 22KD (VGG + Vi Vo VTND ) = Vo VTNL KL KD Vo 1 + = KL Av =(c) RLDKD (VGG + Vi VTND ) + VTNL KLKD / KL dVo = dVi 1 + K D / K LAv =1 1+ KL / KDFrom Problem 4.49. 1 = 2 K L (VDSL VTNL ) =1 = 1.67 k 2 ( 0.1)( 4 1)g m = 2 K D I DQ = 2 (1)( 0.9 ) = 1.90 mA / V Av =g m ( RLD RL )1 + g m ( RLD RL )=(1.90 )(1.67 4 )1 + (1.90 )(1.67 4 ) Av = 0.6914.52 a.From Problem 4.51. g m ( RLD RL ) (1.90 )(1.67 10 ) Av = = 1 + g m ( RLD RL ) 1 + (1.90 )(1.67 10 ) Av = 0.731b.www.elsolucionario.net 124. R0 =1 1 RLD = 1.67 = 0.526 1.67 gm 1.90R0 = 0.40 k4.53 85 K n1 = ( 50 ) 2.125 mA/V 2 2g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 =11 I D1=( 2.125 )( 0.1) = 0.92201 = 200 K 0.05 )( 0.1) (1 1 = = 133.3 K 2 I D 2 ( 0.075 )( 0.1)Av = g m1 ( ro1 ro 2 ) = ( 0.922 )( 200 133.3) Av = 73.74.54 K p1 =k w 40 p 2 = ( 50 ) 1.0 mA/V 2 L 2 g m1 = 2 K p1 I D1 = 2 (1)( 0.1) = 0.6325 mA/Vro1 = ro 2 =11 I D1=1 = 133.3 K 0.075 )( 0.1) (1 1 = = 200 K 2 I D 2 ( 0.05 )( 0.1)Av = g m1 ( ro1 ro 2 ) = ( 0.6325 )(133.3 200 ) Av = 50.64.55 (a) 85 K n1 = ( 50 ) 2.125 mA/V 2 2g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 =11 I D1=( 2.125 )( 0.1) = 0.922 mA/V1( 0.05 )( 0.1)= 200 K1 1 = = 133.3 K 2 I D 2 ( 0.075 )( 0.1)(b) Ri1 =1 1 = = 1.085 K g m1 0.922 Ri1 1.085 Vgs1 = Vi = Vi = 0.956Vi Ri1 + 0.050 1.085 + 0.050 Vgs1 = + ( 0.956 )( 0.922 )( 200 )(133.3) Av = g m1 ( ro1 ro 2 ) ViAv = 70.5 1 1 = 0.05 + Ri = 1.135 K 0.922 g m1(c)Ri = 0.05 +(d)Ro ro1 ro 2 = 200 133.7 Ro 80 Kwww.elsolucionario.net 125. 4.56 (a) g m1 = 2 K n I D1 = 2( 2 )( 0.1) = 0.8944 mA/Vgm2 = 2 K p I D 2 = 2( 2 )( 0.1) = 0.8944 mA/Vro1 = ro 2 =1=1= 100 K I D ( 0.1)( 0.1) (b) The small-signal equivalent circuit gm2Vsg2VorO2 gm1ViViVsg2 rO1(1) g m1Vi +Vsg 2 ro1+ g m 2Vsg 2 +Vsg 2 Vo ro 2=0(2) Vo Vo Vsg 2 + = g m 2Vsg 2 ro ro 2 1 1 1 + gm2 Vo + = Vsg 2 ro ro 2 ro 2 1 1 1 Vo + + 0.8944 Vsg 2 = Vo ( 0.03317 ) = Vsg 2 50 100 100 (1) 1 1 V g m1Vi + Vsg 2 + g m 2 + = o ro1 ro 2 ro 2 1 Vo 1 0.8944 Vi + Vo ( 0.03317 ) + 0.8944 + = 100 100 100 0.8944 Vi = Vo ( 0.01 0.03033) Vo = 44 Vi(c) For output resistance, set Vi = 0. gm2Vsg2 ROIxrO2 Vsg2rO1rO Vxwww.elsolucionario.net 126. (1) (2)g m 2Vsg 2 + I x =Vsg 2 ro1Vx Vx Vsg 2 + ro ro 2+ g m 2Vsg 2 +Vsg 2 Vx ro 2=0(2) 1 1 V Vsg 2 + g m 2 + = x ro 2 ro 2 ro1 1 Vx 1 + 0.8944 + Vsg 2 = 100 100 100 Vsg 2 = Vx ( 0.010936 )(1) 1 1 1 I x = Vx + Vsg 2 + gm2 ro ro 2 ro 2 1 1 1 + 0.8944 I x = Vx + Vx ( 0.010936 ) 50 100 100 I x = Vx ( 0.03 0.0098905 ) V Ro = x = 49.7 K Ix4.57 a. 2 I D1 = K1 (VGS 1 VTN 1 ) 0.4 = 0.1(VGS 1 2 ) VGS1 = 4 V VS 1 = I D1 RS 1 = ( 0.4 )( 4 ) = 1.6 V VG1 = VS 1 + VGS 1 = 1.6 + 4 = 5.6 V 2 R2 1 VG1 = VDD = Rin VDD R1 R1 + R2 1 5.6 = ( 200 )(10 ) R1 = 357 k R1357 R2 = 200 R2 = 455 k 357 + R2VDS1 = VDD I D1 ( RS1 + RD1 ) 4 = 10 ( 0.4 )( 4 + RD1 ) RD1 = 11 k VD1 = 10 ( 0.4 )(11) = 5.6 V I D 2 = K 2 (VSG 2 + VTP 2 )22 = 1(VSG 2 2 ) VSG 2 = 3.41 V VS 2 = VD1 + VSG 2 = 5.6 + 3.41 = 9.01 10 9.01 RS 2 = RS 2 = 0.495 k 2 VSD 2 = VDD I D 2 ( RS 2 + RD 2 ) 5 = 10 ( 2 )( 0.495 + RD 2 ) RD 2 = 2.01 k 2b.www.elsolucionario.net 127. ViVgs1RD1R1R2 V0gm1Vgs1Vgs2RD2 gm2Vsg2V0 = g m 2Vsg 2 RD 2 = ( g m 2 RD 2 ) ( g m1Vgs1 RD1 ) Vgs1 = Vi V Av = 0 = g m1 g m 2 RD1 RD 2 Vi g m1 = 2 K1 I D1 = 2( 0.1)( 0.4 ) = 0.4 mA / Vg m 2 = 2 K 2 I D 2 = 2 (1)( 2 ) = 2.828 mA / V Av = ( 0.4 )( 2.828 )(11)( 2.01) Av = 25.0c.42VSD(sat)510VSD ( sat ) = VSG + VTh = VDD I D 2 ( RD 2 + RS 2 ) 2 = VDD K p 2 ( RD 2 + RS 2 )VSD ( sat )So 2 (1)( 2.01 + 0.495 )VSD ( sat ) + VSD ( sat ) VDD = 0 2 2.50VSD ( sat ) + VSD ( sat ) 10 = 0 VSD ( sat ) =1 1 + 4 ( 2.501)(10 ) 2 ( 2.501)VSD ( sat ) = 1.81 V 5 1.81 = 3.19 Max. output swing = 6.38 V peak-to-peak4.58 a. 10 4 mA 2.5VSD2(sat)10www.elsolucionario.net 128. 2 VSD 2 ( sat ) = VDD I D 2 ( RD 2 + RS 2 ) = VDD K p 2 ( RD 2 + RS 2 ) VSD 2 ( sat ) 2 (1)( 2 + 0.5)VSD 2 ( sat ) + VSD 2 ( sat ) 10 = 0 2 2.5VSD 2 ( sat ) + VSD 2 ( sat ) 10 = 0VSD 2 ( sat ) =1 1 + 4 ( 2.5 )(10 ) 2 ( 2.5 )= 1.81 V10 1.81 + 1.81 VSDQ 2 = 5.91 V 2 VSDQ 2 = VDD I DQ 2 ( RD 2 + RS 2 ) 5.91 = 10 I DQ 2 ( 2 + 0.5 ) I DQ 2 = 1.64 mA VS 2 = 10 (1.64 )( 0.5 ) = 9.18 V VSDQ 2 =I DQ 2 = K p 2 (VSG 2 + VTP 2 )21.64 = (1)(VSG 2 2 ) VSG 2 = 3.28 V VD1 = VS 2 VSG 2 = 9.18 3.28 = 5.90 V 10 5.90 RD1 = RD1 = 10.25 k 0.4 2I DQ1 = K n1 (VGS1 VTN 1 )20.4 = ( 0.1)(VGS 1 2 ) VGS1 = 4 V VS 1 = ( 0.4 )(1) = 0.4 V VG1 = VS 1 + VGS 1 = 0.4 + 4 = 4.4 V 2 R2 1 VG1 = VDD = Rm VDD R1 + R2 R1 1 4.4 = ( 200 )(10 ) R1 = 455 k R1455 R2 = 200 R2 = 357 k 455 + R2 b. I DQ 2 = 1.64 mA VSDQ 2 = 10 (1.64 )( 2 + 0.5 ) VSDQ 2 = 5.90 V VDSQ1 = 10 ( 0.4 )(10.25 + 1) VDSQ1 = 5.50 V(c) g m1 = 2 K n1 I DQ1 = 2( 0.1)( 0.4 ) = 0.4 mA / Vg m 2 = 2 K p 2 I DQ 2 = 2 (1)(1.64 ) = 2.56 mA / V Av = g m1 g m 2 RD1 RD 2 = ( 0.4 )( 2.56 )(10.25 )( 2 ) Av = 21.04.59 a. VDSQ 2 = 7 = VDD I DQ 2 RS 2 = 10 I DQ 2 ( 6 ) I DQ 2 = 0.5 mA I DQ 2 = K 2 (VGS 2 VTN 2 )20.5 = ( 0.2 )(VGS 2 0.8 ) VGS 2 = 2.38 V VS 1 = VS 2 + VGS 2 = 3 + 2.38 = 5.38 2www.elsolucionario.net 129. I DQ1 =VS 1 5.38 = = 0.269 mA 20 RS1I DQ1 = K1 (VGS 1 VTN 1 )20.269 = ( 0.2 )(VGS 1 0.8 ) VGS 1 = 1.96 V VG1 = VS 1 + VGS 1 = 5.38 + 1.96 = 7.34 V 2 R2 1 VG1 = VDD = Rin VDD R1 + R2 R1 1 7.34 = ( 400 )(10 ) R1 = 545 k R1 545 R2 = 400 R2 = 1.50 M 545 + R2 b. I DQ1 = 0.269 mA, I DQ 2 = 0.5 mAVDSQ1 = 10 ( 0.269 )( 20 ) VDSQ1 = 4.62 Vc. Av =g m1 RS 1 g R m2 S 2 1 + g m1 RS1 1 + g m 2 RS 2g m1 = 2 K1 I DQ1 = 2( 0.2 )( 0.269 ) = 0.464 mA / V( 0.2 )( 0.5) = 0.6325 mA / V ( 0.464 )( 20 ) ( 0.6325)( 6 ) Av = 1 + ( 0.464 )( 20 ) 1 + ( 0.6325 )( 6 ) = ( 0.9027 )( 0.7915 )g m 2 = 2 K 2 I DQ 2 = 2= Av = 0.714gm1Vgs1 Vgs1 Vgs2 Ix RS2RS1R0 =gm2Vgs2 Vx1 1 RS 2 = 6 = 1.581 6 Ro = 1.25 k gm2 0.63254.60 (a) I DQ1 =10 VGS 1 2 = K n1 (VGS 1 VTN 1 ) RS 22 10 VGS 1 = ( 4 )(10 ) (VGS 1 4VGS 1 + 4 ) 2 40VGS 1 159VGS 1 + 150 = 0VGS1 =159 (159 ) 4 ( 40 )(150 ) VGS 1 = 2.435 V 2 ( 40 ) 2www.elsolucionario.net 130. I DQ1 = ( 4 )( 2.435 2 ) I DQ1 = 0.757 mA 2VDSQ1 = 20 ( 0.757 )(10 ) VDSQ1 = 12.4 V Also I DQ 2 = 0.757 mA VDSQ 2 = 20 ( 0.757 )(10 + 5 ) VDSQ 2 = 8.65 V g m1 = g m 2 = 2 KI DQ = 2(b)( 4 )( 0.757 ) g m1 = g m 2 = 3.48 mA / Vc. Vgs1 Vi gm1Vgs1 gm2Vgs2RGV0 RS1Vgs2 RS2RDRLV0 = ( g m 2Vgs 2 ) ( RD RL )Vgs 2 = ( g m1Vgs1 g m 2Vgs 2 ) ( RS1 RS 2 ) Vi = Vgs1 Vgs 2 Vgs1 = Vi + Vgs 2Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) = g m1 (Vi + Vgs 2 ) ( RS 1 RS 2 )Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) + g m1Vgs 2 ( RS 1 Rs 2 ) = g m1Vi ( RS1 RS 2 ) Vgs 2 = Av = g m1Vi ( RS 1 RS 2 )1 + g m 2 ( RS 1 RS 2 ) + g m1 ( RS 1 RS 2 )V0 g m1 g m 2 ( RS 1 RS 2 )( RD RL ) = Vi 1 + ( g m1 + g m 2 ) ( RS1 RS 2 )( 3.48 ) (10 10 )( 5 2 ) Av = Av = 2.42 1 + ( 3.48 + 3.48 ) (10 10 ) 24.61 a. I DQ = 3 mA VS 1 = I DQ RS 5 = ( 3)(1.2 ) 5 = 1.4 V I DQ = K1 (VGS VTN )23 = 2 (VGS 1) VGS = 2.225 V 2VG1 = VGS + VS 1 = 2.225 1.4 = 0.825 V R3 R3 VG1 = ( 5 ) 0.825 = ( 5 ) R3 = 82.5 k 500 R1 + R2 + R3 VD1 = VS1 + VDSQ1 = 1.4 + 2.5 = 1.1 V VG 2 = VD1 + VGS = 1.1 + 2.225 = 3.325 V R2 + R3 R2 + R3 VG 2 = ( 5 ) 3.325 = ( 5) 500 R1 + R2 + R3 R2 + R3 = 332.5 R2 = 250 kwww.elsolucionario.net 131. R1 = 500 250 82.5 R1 = 167.5 k VD 2 = VD1 + VDSQ 2 = 1.1 + 2.5 = 3.6 V RD =5 3.6 RD = 0.467 k 3b. Av = g m1 RD( 2 )( 3) = 4.90 mA / Vg m1 = 2 K n I DQ = 2Av = ( 4.90 )( 0.467 ) Av = 2.294.62 a. VS 1 = I DQ RS 10 = ( 5 )( 2 ) 10 VS 1 = 0 I DQ = K1 (VGS 1 VTN )25 = 4 (VGS1 1.5 ) VGS 1 = 2.618 V VG1 = VGS 1 + VS1 = 2.618 V = IR3 = ( 0.1) R3 R3 = 26.2 k 2VD1 = VS1 + VDSQ1 = 0 + 3.5 = 3.5 V VG 2 = VD1 + VGS = 3.5 + 2.62 = 6.12 V = ( 0.1)( R2 + R3 ) R2 + R3 = 61.2 k R2 = 35 k VD 2 = VD1 + VDSQ 2 = 3.5 + 3.5 = 7.0 V 10 7 RD = RD = 0.6 k 5 10 6.12 R1 = R1 = 38.8 k 0.1 b. Av = g m1 RD( 4 )( 5 ) = 8.944 mA / Vg m1 = 2 K n I DQ = 2Av = ( 8.944 )( 0.6 ) Av = 5.374.63 a. V I DQ = I DSS 1 GS VP V 4 = 6 1 GS ( 3) 22 4 VGS = ( 3) 1 VGS = 0.551 V 6 VDSQ = VDD I DQ RD 6 = 10 ( 4 ) RD RD = 1 kb.2 I DSS VGS 2 ( 6 ) 0.551 = 1 1 g m = 3.265 mA/V 3 ( VP ) VP 3 1 1 = r0 = 25 k r0 = I DQ ( 0.01)( 4 ) gm =c.Av = g m ( r0 RD ) = ( 3.265 )( 25 1) Av = 3.14www.elsolucionario.net 132. 4.64 VGS + I DQ ( RS1 + RS 2 ) = 0 V I DQ = I DSS 1 GS VP 22VGS V + I DSS ( RS1 + RS 2 ) 1 GS = 0 VP 2 V VGS + ( 2 )( 0.1 + 0.25 ) 1 GS = 0 VP 2 V2 VGS + 0.7 1 VGS + GS 2 = 0 ( 2 ) ( 2 ) 2 0.175VGS + 1.7VGS + 0.7 = 0 VGS =1.7 (1.7 ) 4 ( 0.175)( 0.7 ) VGS 2 ( 0.175 ) VGS 2 ( 2 ) 0.431 2= 0.4314 V2 I DSS 1 = 1 g m = 1.569 mA/V VP 2 VP 2 g m ( RD RL ) (1.569 ) ( 8 4 ) = Av = 3.62 Av = 1 + g m RS1 1 + (1.569 )( 0.1)gm =Ai =i0 ( v0 / RL ) v0 RG 50 = = = ( 3.62 ) Ai = 45.2 ii ( vi / RG ) vi RL 44.65 I DSS = 4 mA 2 V = DD = 10 V 2 = VDD I DQ ( RS + RD )I DQ = VDSQ VDSQ10 = 20 ( 4 )( RS + RD ) RS + RD = 2.5 kVS = 2 V = I DQ RS = 4 RS RS = 0.5 k, RD = 2.0 k I DQ V = I DSS 1 GS VP 22 V 4 4 = 8 1 GS VGS = ( 4.2 ) 1 VGS = 1.23 V ( 4.2 ) 8 VG = VS + VGS = 2 1.23 R2 R2 VG = 0.77 V = ( 20 ) = ( 20 ) R2 = 3.85 k, R1 = 96.2 K 100 R1 + R2 4.66 a.www.elsolucionario.net 133. I DSS = 5 mA 2 V 12 VDSQ = DD = =6V 2 2 12 6 RS = RS = 1.2 k 5 I DQ =I DQ V = I DSS 1 GS VP 22 V 5 5 = 10 1 GS VGS = ( 5 ) 1 VGS = 1.464 V ( 5 ) 10 VG = VS + VGS = 6 1.464 = 4.536 V R2 1 VG = VDD = Rin VDD R1 R1 + R2 1 4.536 = (100 )(12 ) R1 = 265 k R1 265 R2 = 100 R2 = 161 k 265 + R2 b. V 2 (10 ) 1.46 2I g m = DSS 1 GS = 1 g m = 2.83 mA/V VP ) 5 5 VP ( r0 =1 1 = = 20 k I DQ ( 0.01)( 5 )Av =(g m r0 Rs RL()1 + g m r0 RS RL)Av =( 2.83) ( 20 1.2 0.5 ) Av = 0.495 1 + ( 2.83) ( 20 1.2 0.5 )R0 =1 1 1.2 = 0.353 1.2 R0 = 0.273 k RS = 2.83 gm4.67 a. R2 110 VG = VDD = (10 ) = 5.5 V 110 + 90 R1 + R2 I DQ =10 (VG VGS ) RS V = I DSS 1 GS VP V 10 5.5 + VGS = ( 2 )( 5 ) 1 GS 1.75 222 4.5 + VGS = 10 (1 1.143VGS + 0.3265VGS )2 3.265VGs 12.43VGS + 5.5 = 0VGS =12.43 (12.43) 4 ( 3.265 )( 5.5) VGS 2 ( 3.265 ) 2= 0.511 V2 0.511 I DQ = ( 2 ) 1 I DQ = 1.00 mA 1.75 VSDQ = 10 (1.00 )( 5 ) VSDQ = 5.0 Vwww.elsolucionario.net 134. b.2 I DSS VGS 2 ( 2 ) 0.511 1 = 1 g m = 1.618 mA/V VP VP 1.75 1.75 g m ( RS RL ) (1.618 ) ( 5 10 ) Av = = Av = 0.844 1 + g m ( RS RL ) 1 + (1.618 ) ( 5 10 )gm =R i0 ( v0 / RL ) = = Av i ii ( vi / Ri ) RL Ri = R1 R2 = 90 110 = 49.5 k Ai = 49.5 Ai = ( 0.844 ) Ai = 4.18 10 c.AC load line 1 Slope 510 1 3.33 K1.05.010id = 1.0 mAvsd = ( 3.33)(1.0 ) = 3.33 V Maximum swing in output voltage = 6.66 V peak-to-peak4.68 V I DQ = I DSS 1 GS VP 22 4 V 4 = 8 1 GS VGS = 4 1 VGS = 1.17 V 4 8 VSDQ = VDD I DQ ( RS + RD ) 7.5 = 20 4 ( RS + RD ) RS + RD = 3.125 k VGS 1 VP RS = 3.125 RD g m RD Av = 1 + g m RS gm =2 I DSS VP 2 ( 8 ) 1.17 = 1 g m = 2.83 mA/V 4 4 3 (1 + g m RS ) = g m RD 3 1 + ( 2.83)( 3.125 RD ) = ( 2.83) RD 9.844 2.83RD = 0.9433RD RD = 2.61 k RS = 0.516 k VS = 20 ( 4 )( 0.516 ) VS = 17.94 V VG = VS VGS = 17.94 1.17 = 16.77 V R2 R2 VG = VDD = ( 20 ) R2 = 335 k. R1 = 65 k R1 + R2 400 www.elsolucionario.net 135. Chapter 5 Exercise Problems EX5.1= 10.980 = 49 1 0.980 0.995 = 199 For = 0.995, = 1 0.995 So 49 199For = 0.980, =EX5.2 BVCEO =BVCBO n=200 3120or BVCEO = 40.5 VEX5.3 V VBE ( on ) 2 0.7 I B = BB = 6.5 A RB 200I C = I = (120 )( 6.5 A ) I C = 0.78 mAVCE = VCC I C RC = 5 ( 0.78 )( 4 ) or VCE = 1.88 VP = I BVBE ( on ) + I CVCE = ( 0.0065 )( 0.7 ) + ( 0.78 )(1.88 ) 1.47 mWEX5.4 V + VEB ( on ) VBB 5 0.7 2.8 IB = = or I B = 4.62 A RB 325 I C = I B = ( 80 )( 4.615 ) I C = 0.369 mAVEC = 2 = 5 ( 0.369 ) RC which yields RC = 8.13 k EX5.5(a)IB =VBB VBE ( on ) RB=2 0.7 5.91 A 220I C = I B = (100 )( 5.91 A ) 0.591 mA I E = (1 + ) I B = 0.597 mAVCE = VCC I C RC = 10 ( 0.591)( 4 ) or VCE = 7.64 V6.5 0.7 26.4 A 220 Transistor is biased in saturation mode, so(b)IB =VCE = VCE ( sat ) = 0.2 V IC =VCC VCE ( sat ) RC=10 0.2 or I C = 2.45 mA 4I E = I C + I B = 2.45 + 0.0264 2.48 mAEX5.6 For 0 VI < 0.7 V , Qn is cutoff, VO = 9 VWhen Qn is biased in saturation, we have 0.2 = 9 (100 )(VI 0.7 )( 4 ) 200So, for VI 5.1 V , VO = 0.2 V EX5.7www.elsolucionario.net VI = 5.1 V 136. IB =VBB VBE ( on )RB + (1 + ) RE=8 0.7 30 + ( 76 )(1.2 )or I B = 60.2 A I C = I B = ( 75 )( 60.23 A ) 4.52 mAI E = (1 + ) I B = 4.58 mA VCE = VCC I C RC I E RE= 12 ( 4.517 )( 0.4 ) ( 4.577 )(1.2 )or VCE = 4.70 VEX5.8 For VC = 4 V and ICQ = 1.5 mA, V + VC 10 4 RC = = RC = 4 k I CQ 1.5 1+ 101 IE = IC = (1.5 ) = 1.515 mA 100 V ( on ) V also I E = BE RE Then RE =0.7 ( 10 ) 1.515 RE = 6.14 k EX5.9 3 0.7 = 0.25 I EQ = RE RE = 9.2 k 75 I CQ = ( 0.25 ) = 0.2467 mA 76 0.7 + VCEQ + I CQ RC 3 = 0 RC =3 + 0.7 2 RC = 6.89 k 0.2467EX5.10 5 = I E RE + VEB ( on ) + I B RB 2(a)(b)(c)(d)180 5 + 2 0.7 = I E 2 + I E = 0.9859 mA 41 I C = 0.962 mA 180 6.3 = I E 2 + I E = 1.2725 mA 61 I C = 1.25 mA 180 6.3 = I E 2 + I E = 1.6657 mA 101 I C = 1.64 mA 180 6.3 = I E 2 + I E = 1.97365 mA 151 I C = 1.94 mAEX5.11www.elsolucionario.net 137. IE =VBB VEB ( on ) RE RE =4 0.7 1.0or RE = 3.3 k I C = I E = ( 0.992 )(1) = 0.992 mAI B = I E I C = 1.0 0.992 or I B = 8 AVCB = I C RC VCC = ( 0.992 )(1) 5 or VCB = 4.01 VEX5.12 V + (V + VCE ( sat ) ) 5 (1.5 + 0.2 ) R= = IC 15 or R = 220 15 IB = = 1 mA 15 v VBE ( on ) 5 0.7 = = 4.3 k RB = I 1 IB P = I BVBE (on) + I CVCE= (1)( 0.7 ) + (15 )( 0.2 ) = 3.7 mWEX5.13 (a) For V1 = V2 = 0, All currents are zero and VO = 5 V. (b) For V1 = 5 V, V2 = 0; IB2 = IC2 = 0, 5 0.7 I B1 = = 4.53 mA 0.95 5 0.2 I C1 = I C1 = I R = 8 mA 0.6 VO = 0.2 V (c) For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V EX5.14 vO = 5 iC RC = 5 iB RC vO = iB RC iB =VBB + vI VBE ( on ) RBiB =vI RBThenvO RC = vI RBLet = 100, RC = 5 k , RB = 100 k Sovo (100 )( 5 ) = = 5 vI 100Want Q-point to be vo ( Q pt ) = 2.5 = 5 (100 ) I BQ ( 5 ) Then I BQ = 0.005 mA, I BQ = 0.005 =VBB 0.7 100so VBB = 1.2 V Also, I CQ = I BQ = (100 )( 0.005 ) = 0.5 mAEX5.15www.elsolucionario.net 138. VCEQ = 2.5 = 5 I CQ RC5 2.5 = 10 k 0.25 I CQ 0.25 = = = 0.002083 mA 120 or RC = I BQThen RB =5 0.7 RB = 2.06 M 0.002083EX5.16 (a) RTH = R1 R2 = 9 2.25 = 1.8 k or VTH R2 2.25 VTH = VCC = ( 5) R1 + R2 9 + 2.25 = 1.0 VI BQ =(b)VTH VBE ( on )RTH + (1 + ) RE=1 0.7 1.8 + (151)( 0.2 )or I BQ = 9.375 A I CQ = I BQ = (150 )( 9.375 A ) or I CQ = 1.41 mAI EQ = (1 + ) I BQ I EQ = 1.42 mAVCEQ = 5 I CQ RC I EQ RE= 5 (1.41)(1) (1.42 )( 0.2 )or VCEQ = 3.31 VFor = 75(c) I BQ =1 0.7 = 17.6 A 1.8 + ( 76 )( 0.2 )I CQ = I BQ = ( 75 )(17.6 A ) or I CQ = 1.32 mAI EQ = (1 + ) I BQ = ( 76 )(17.6 A )or I EQ = 1.34 mAVCEQ = 5 (1.32 )(1) (1.34 )( 0.2 )or VCEQ = 3.41 V EX5.17 VCEQ VCC I CQ ( RC + RE ) or 2.5 5 I CQ (1 + 0.2 ) which yields I CQ = 2.08 mA, I CQ 2.08 I BQ = = = 0.0139 mA 150 RTH = ( 0.1)(1 + ) RE = ( 0.1)(151)( 0.2 ) or RTH = 3.02 k R2 1 Now VTH = VCC = RTH VCC R1 + R2 R1 1 so VTH = ( 3.02 )( 5 ) R1We can write VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ REwww.elsolucionario.net 139. or1 ( 3.02 )( 5 ) = ( 0.0139 )( 3.02 ) + 0.7 + (151)( 0.0139 )( 0.2 ) R1We obtain R1 = 13 k and then R2 = 3.93 k EX5.18 = 150, RTH = R1 R2 50 I CQ = = 5 mA 1 I CQ 5 = = 0.0333 mA I BQ = 150 I BQ =VTH VBE ( on ) ( 5 ) RTH + (1 + ) RESet RTH = ( 0.1)(1 + ) RE R2 We have VTH = (10 ) 5 R1 + R2 Then I BQ R2 (10 ) 0.7 R + R2 = 0.0333 = 1 (1.1)(151)( 0.2 ) R2 which yields = 0.1806 R1 + R2 RR Now RTH = 1 2 = ( 0.1)(151)( 0.2 ) = 3.02 k R1 + R2 so R1 ( 0.1806 ) = 3.02 k We obtain R1 = 16.7 k and R2 = 3.69 k EX5.19 V + V VECQ 5 ( 5 ) 5 I CQ = 4.5 + 0.5 RC + REso I CQ = 1 mA, and I BQ =I CQ=1 = 0.00833 mA 120RTH = ( 0.1)(1 + ) RE = ( 0.1)(121)( 0.5 )or RTH = 6.05 k We can write V + = I EQ RE + VEB ( on ) + I BQ RTH + VTH We have VTH =1 RTH (10 ) 5 and if we let I EQ I CQ = 1 mA, R1then we have 5 = (1)( 0.5 ) + 0.7 + ( 0.00833)( 6.05 ) +1 ( 6.05 )(10 ) 5 R1which yields R1 = 6.91 k and R2 = 48.6 k EX5.20 2 2 I1 = I Q 1 + = ( 0.25 ) 1 + = 0.2625 mA 40 0 VBE ( on ) V 0 0.7 ( 5 ) = R1 = 0.2625 I1 or R1 = 16.38 k www.elsolucionario.net 140. For VCEO = 3 V , then VCO = 2.3 V 40 I CO = I Q = ( 0.25 ) = 0.2439 mA 41 1+ + V VCO 5 2.3 RC = = = 11.07 k I CO 0.2439EX5.21 RTH 50 100 = 33.3 k 50 VTH = VTH = (10 ) 5 = 1.67 V 50 + 100 1.67 0.7 ( 5 ) I B1 = 11.2 A 33.3 + (101)( 2 ) I C1 = 1.12 mA, I E1 = 1.13 mAVE1 = I E1 RE1 5 = (1.13)( 2 ) 5 = 2.74 V VCE1 = 3.25 V VC1 = 0.51 V Now VE 2 = 0.51 + 0.7 = 1.21 V 5 1.21 = 1.90 mA I B 2 = 18.8 A 2 = 1.88 mAIE2 = IC 2I R1 = I C1 I B 2 = 1.12 0.0188 = 1.10 mA 5 0.51 = 4.08 k 1.10 = 2.5 VC 2 = VE 2 VEC 2RC1 = VEC 2= 1.21 2.5 = 1.29 V RC 2 =1.29 ( 5 ) 1.88= 1.97 k EX5.22 12 = 240 k = R1 + R2 + R3 0.05 Then VB1 = ( 0.5)( 2 ) + 0.7 = 1.7 V We find1.7 = 34 k 0.05 = ( 0.5 )( 2 ) + 4 + 0.7 = 5.7 VR3 = Also VB 2VR 2 = 5.7 1.7 = 4 V 4 = 80 k 0.05 and R1 = 240 80 34 = 126 k so R2 =VC 2 = 1 + 4 + 4 = 9 V Then RC =V + VC 2 12 9 = = 6 k I CQ 0.5Test Your Understanding Exercises TYU5.1= 1+ For = 75, =75 = 0.9868 76www.elsolucionario.net 141. For = 125, =125 = 0.9921 126TYU5.2 I E = (1 + ) I B IE 0.780 = = 81.25 then = 80.3 I B 0.00960so 1 + = Now1+ 80.25 = 0.9877 81.25I C = I E = ( 0.9877 )( 0.78 ) = 0.770 mATYU5.3= 0.990 = = 99 1 1 0.990Now I B =IE 2.15 = 21.5 A and I C = I E = ( 0.990 )( 2.15 ) = 2.13 mA 1 + 100TYU5.4 V 150 ro = A = IC ICFor I C = 0.1 mA ro = 1.5 M For I C = 1.0 mA ro = 150 k For I C = 10 mA ro = 15 k TYU5.5 V I C = I O 1 + CE VA At VCE = 1 V , I C = 1 mA 1 For VA = 75 V , I C = 1 = I O 1 + I O = 0.9868 mA 75 Then, at VCE = 10 V(a) 10 I C = ( 0.9868 ) 1 + = 1.12 mA 75 (b) At VCE1 For VA = 150 V , I C = 1 = I O 1 + I O = 0.9934 mA 150 10 = 10 V , I C = ( 0.9934 ) 1 + = 1.06 mA 150 TYU5.6BVCEO =BVCBO n so BVCBO = 3 100 ( 30 ) = 139 VTYU5.7 (a) For VI = 0.2 V < VBE ( on ) I B = I C = 0, VO = 5 V and P = 0(b) IB =For VI = 3.6 V , transistor is driven into saturation, so VI VBE ( on ) RB=V + VCE ( sat ) 5 0.2 3.6 0.7 = 4.53 mA and I C = = = 10.9 mA RC 0.440 0.64www.elsolucionario.net 142. I C 10.9 = = 2.41 < which shows that the transistor is indeed driven into saturation. Now, I B 4.53Note thatP = I BVBE ( on ) + I CVCE ( sat )= ( 4.53)( 0.7 ) + (10.9 )( 0.2 ) = 5.35 mWTYU5.8 For VBC = 0 VO = 0.7 VThen I C =I 5 0.7 9.77 = 9.77 mA and I B = C = = 0.195 mA 0.44 50Now VI = I B RB + VBE ( on ) = ( 0.195 )( 0.64 ) + 0.7 or VI = 0.825 V Also P = I BVBE ( on ) + I CVCE= ( 0.195 )( 0.7 ) + ( 9.77 )( 0.7 ) = 6.98 mWTYU5.9 V + VC 10 6.34 IC = = = 0.915 mA RC 4 VBE ( on ) V And I E = Now =RE=0.7 ( 10 ) 10or IE = 0.930I C 0.915 0.9839 = = 0.9839 and = = = 61 I E 0.930 1 1 0.9839Also I B = I E I C = 0.930 0.915 15 A and VCE = VC VE = 6.34 ( 0.7 ) = 7.04 V TYU5.10 10 0.7 IE = = 1.16 mA 8 1.1625 IB = = 22.8 A 51 I C = ( 50 )( 22.8 A ) = 1.14 mA VE = V + I E RE = 10 (1.1625 )( 8 ) = 0.7 V VC = I C RC 10 = (1.1397 )( 4 ) 10 = 5.44 VEC = 0.7 ( 5.44 ) VEC = 6.14 V TYU5.11 VBB = I B RB + VBE ( on ) + I E REor VBB = I B RB + VBE ( on ) + (1 + ) I B RE Then I B =VBB VBE ( on )RB + (1 + ) RE=2 0.7 10 + ( 76 )(1)or I B = 15.1 A Also I C = ( 75 )(15.1 A ) = 1.13 mA and I E = ( 76 )(15.1 A ) = 1.15 mA Now VCE = VCC + VBB I C RC I E RE= 8 + 2 (1.13)( 2.5 ) (1.15 )(1) = 6.03 VTYU5.12 VCE = 2.5 V VE = 2.5 V = I E REWe have VBB = I B RB + VBE ( on ) + VEwww.elsolucionario.net 143. so I B =VBB VBE ( on ) VE RB=5 0.7 2.5 10or I B = 0.18 mA Then I E = (101)( 0.18 ) = 18.2 mA And RE =2.5 = 0.138 k 138 18.18TYU5.13 VBB = I E RE + VEB ( on ) + I B RB 2.2 = 0.0431 mA 51 50 and I C = I E = ( 2.2 ) = 2.16 mA 51 1+ Then VBB = ( 2.2 )(1) + 0.7 + ( 0.0431)( 50 ) I E = 2.2 mA I B =or VBB = 5.06 V Now VEC = 5 I E RE = 5 ( 2.2 )(1) = 2.8 V TYU5.14 (a) For vI = 0, iB = iC = 0, vO = 12 V , P = 0For vI = 12 V , iB =(b) iC =VCC VCE ( sat ) RC=vI VBE ( on ) RB=12 0.7 = 47.1 mA 0.2412 0.1 = 2.38 A 5vO = 0.1 V and P = iBVBE ( on ) + iCVCE ( sat ) = ( 0.0471)( 0.7 ) + ( 2.38 )( 0.1) = 0.271 WTYU5.15For VCEQ = 2.5 V , I CQ =(a) I BQ =I CQ=5 2.5 = 1.25 mA 21.25 I BQ = 12.5 A 1005 0.7 = 344 k 0.0125 IBQ is independent of .Then RB = (b)For VCEQ = 1 V , I C 5 1 = 2 mA 2IC 2 = = 160 I B 0.0125For VCEQ = 4 V , I C =54 = 0.5 mA 2IC 0.5 = = 40 I B 0.0125 So 40 160=TYU5.16 5 0.7 I BQ = = 0.005375 mA 800www.elsolucionario.net 144. For = 75, I CQ = I BQ = ( 75 )( 0.005375 ) Or I CQ = 0.403 mA For = 150, I CQ = (150 )( 0.005375 ) Or I CQ = 0.806 mA Largest I CQ Smallest VCEQ 5 1 = 4.96 k 0.806 54 For = 75, RC = = 2.48 k 0.403 For = 150, RC =5 2.5 = 4.14 k 0.604 = 5 ( 0.403)( 4.14 ) = 3.33 VFor a nominal I CQ = 0.604 mA and VCEQ = 2.5 V , RC = Now for I CQ = 0.403 mA, VCEQFor I CQ = 0.806 mA, VCEQ = 5 ( 0.806 )( 4.14 ) = 1.66 V So, for RC = 4.14 k , 1.66 VCEQ 3.33 V TYU5.17 (a) 100 I CQ = I EQ = (1) = 0.99 mA 101 1+ I EQ 1 = 9.90 A I BQ = 1 + 101 VB = I BQ RB = ( 0.0099 )( 50 ) or VB = 0.495 V I CQ 0.99 103 VBE = VT ln = ( 0.026 ) ln 14 3 10 IS or VBE = 0.630 V Then VE = VB VBE = 0.495 0.630 = 1.13 V VC = 10 ( 0.99 )( 5 ) = 5.05 VThen VCEQ = VC VE = 5.05 ( 1.13) = 6.18 V(b) I EQ1 = 0.0196 mA 1 + 51 VB = ( 0.0196 )( 50 ) = 0.98 VI EQ = 1 mA, I BQ = I CQ = 1+ = 50 I EQ = (1) = 0.98 mA 51 0.98 103 VBE = ( 0.026 ) ln = 0.629 V 14 3 10 VE = 0.98 0.629 = 1.61 V VC = 10 ( 0.98 )( 5 ) = 5.1 VVCEQ = VC VE = 5.1 ( 1.61) = 6.71 VTYU5.18 IQ IQ IB = = 1 + 121 andwww.elsolucionario.net 145. IQ VB = ( 20 ) = I Q ( 0.165 ) 121 VE = I Q ( 0.165 ) + 0.7 120 I CQ = I EQ = I Q = ( 0.992 ) I Q 121 1+ VC = I CQ RC 5 = ( 0.992 ) I Q ( 4 ) 5 = 3.97 I Q 5 VECQ = VE VC = I Q ( 0.165 ) + 0.7 I Q ( 3.97 ) 5 = 3.805 I Q + 5.7Then 3 = 5.7 3.805I Q which yields I Q = 0.710 mAwww.elsolucionario.net 146. Chapter 5 Problem Solutions 5.1 (a)=iC 510 = = 85 6 iB85 = 0.9884 1 + 86 iE = (1 + ) iB = ( 86 )( 6 ) iE = 516 A=(b)=2.65 = 53 0.050 53 = = 0.9815 54 iE = (1 + ) iB = ( 54 )( 0.050 ) iE = 2.70 mA=5.2 (a)For = 110: = 1+ =110 = 0.99099 111180 = 0.99448 181 0.99099 0.99448For = 180: =(b)I C = I B = 110 ( 50 A ) I C = 5.50 mAor I C = 180 ( 50 A ) I C = 9.00 mA so 5.50 I C 9.0 mA 5.3 (a)iE (b)1.12 9.33 A 120 121 = (1.12 ) = 1.13 mA 120 120 = = 0.9917 121 50 = = 2.5 mA 20 21 = ( 50 ) = 52.5 mA 20 20 = = 0.9524 21iB =iB iE 5.4 (a)=0.9 0.95 0.98 0.99 0.995 0.9999 19 49 99 199 9991(b)www.elsolucionario.net 147. =1+ 0.9524 0.9804 0.9901 0.9934 0.9955 0.997520 50 100 150 220 400 5.51.2 14.8 A 81 80 I C = (1.2 ) = 1.185 mA 81 80 = = 0.9877 81 VC = 5 (1.185 )( 2 ) = 2.63 V IB =(a)0.80 9.88 A 81 80 I C = ( 0.80 ) = 0.790 mA 81 80 = = 0.9877 81 VC = 5 ( 0.790 )( 2 ) = 3.42 V IB =(b)Yes, VC > VB so B-C junction is reverse biased in both areas.(c) 5.6For VC = 0, I C = IE =IC=5 = 2.5 mA 22.5 I E = 2.546 mA 0.9825.7 (a)IC VC VC(b)(c)0.75 12.3 A 61 60 = ( 0.75 ) = 0.738 mA 61 60 = = 0.9836 61 = I C RC 10 = ( 0.738 )( 5 ) 10 = 6.31 VIB =1.5 24.6 A 61 60 I C = (1.5 ) = 1.475 mA 61 60 = = 0.9836 61 VC = (1.475 )( 5 ) 10 VC = 2.625 V IB =Yes, VC < 0 in both cases so that B-C junction is reverse biased.www.elsolucionario.net 148. 5.8 IC = IE =VC ( 10 ) RC IC10 1.2 = 1.76 mA 51.76 I E = 1.774 mA 0.9925.9 I C = I S eVRE / VT 0.685 = 1013 exp I C = 27.67 mA 0.026 1+ 91 IE = I C = ( 27.67 ) 90 I D = 27.98 mA I 27.67 IB = C = I B = 0.307 mA 905.10 Device 1: iE = I Eo1evEB / VT 0.5 103 = I Eo1e0.650 / 0.026 So that I EO1 = 6.94 1015 A Device 2: 12.2 103 = I Eo 2 e0.650 / 0.026 Or I Eo 2 = 1.69 1013 A Ratio of areas =I Eo 2 1.69 1013 = Ratio = 24.4 I Eo1 6.94 10155.11 (a)ro =VA 250 = ro = 250 k 1 IC(b)ro =VA 250 = ro = 2.50 M IC 0.15.12 BVC E 0 =BVC B 0 3=60 3100BVC E 0 = 12.9 V5.13 BVC E 0 = 56 =220 3BVC B 0 33 =220 = 3.93 56 = 60.6 5.14www.elsolucionario.net 149. BVC E 0 =BVC B 0 3BVC B 0 = ( BVC E 0 ) 3 = ( 50 ) 3 50 BVC B 0 = 184 V5.15 (a)IE =0.7 ( 10 )= 1.86 mA 5 75 I C = (1.86 ) = 1.836 mA 76 VC = 0.7 + 4 = 3.3 V 10 3.3 RC = 3.65 K 1.836 0.5 IB = = 0.00658 mA 76 VB = I B RB = ( 0.00658 )( 25 ) VB = 0.164 VRC =(b)(c) 75 I C = ( 0.5 ) = 0.493 mA 76 1 ( 5 ) RC = RC = 8.11 K 0.493 I O = E (10 ) + 0.7 + I E ( 4 ) 8 76 7.3 = I E ( 4 + 0.132 ) I E = 1.767 mA 75 I C = (1.767 ) = 1.744 mA 76 VCE = 8 (1.744 )( 4 ) (1.767 )( 4 ) 8 = 16 6.972 7.068 VCE = 1.96 V(d)I = I E (10 + 0.263 + 2 ) + 0.7 5 = I E (10 ) + E ( 20 ) + 0.7 + I E ( 2 ) 76 I E = 0.3506 mA I B = 4.61 A VC = 5 ( 0.3506 )(10 ) VC = 1.49 V5.16 For Fig. 5.15 (a) RE = 5 + 5% = 5.25 K IE =0.7 ( 10 )= 1.77 mA 5.25 I C = 1.75 mA 10 3.3 = 3.83 K RC = 1.75 RE = 5 5% = 4.75 K IE =0.7 ( 10 )= 1.96 mA 4.75 I C = 1.93 mA 10 3.3 = 3.47 K RC = 1.93 So 1.75 I C 1.93 mA 3.47 RC 3.83 K For Fig. 5.15(c) RE = 4 + 5% = 4.2 Kwww.elsolucionario.net 150. IB =8 0.7 = 0.0222 mA 10 + ( 76 )( 4.2 )I C = 1.66 mAI E = 1.69 mA VCE = 16 (1.66 )( 4 ) (1.69 )( 4.2 ) = 16 6.64 7.098 VCE = 2.26 V RE = 4 5% = 3.8 K 8 0.7 = 0.0244 I C = 1.83 mA IB = 10 + ( 76 )( 3.8 ) I E = 1.86 mA VCE = 16 (1.83)( 4 ) (1.86 )( 3.8 ) = 16 7.32 7.068 VCE = 1.61 V So 1.66 I C 1.83 mA 1.61 VCE 2.26 V5.17 RB =VBB VEB 2.5 0.7 = RB = 120 k IB 0.015I CQ = ( 70 )(15 A ) 1.05 mA RC =VCC VECQ I CQ=5 2.5 1.055.18 (a) VB = I B RB I B =RC = 2.38 k VB ( 1) = 500 RBI B = 2.0 A VE = 1 0.7 = 1.7 V IE =VE ( 3) RE=1.7 + 3 = 0.2708 mA 4.8IE 0.2708 = (1 + ) = = 135.4 = 134.4 0.002 IB = 0.9926 1+ I C = I B I C = 0.269 mA=VCE = 3 VE = 3 ( 1.7 ) VCE = 4.7 V(b) 54 I E = 0.5 mA 2 4 = 0.7 + I B RB + ( I B + I C ) RC 5 IE =I B + IC = I E I B + IC = I E4 = 0.7 + I B (100 ) + ( 0.5 )( 8 ) 5 I B = 0.043 IE 0.5 = (1 + ) = = 11.63 IB 0.043 = 10.63, = 1+ = 0.91405.19www.elsolucionario.net 151. VB 0.7 ( 5 )VB + 4.3 3 3 VB + 4.3 50 50 IC = I E = 3 51 51 V + 4.3 50 VC = 5 I C RC = 5 B (10 ) 3 51 IE Now VB = VC , so VB [1 + 3.27 ] = 5 14.1 = 9.05 VB = 2.12 V IE =2.12 + 4.3 I E = 0.727 mA 35.20 10 VE 10 2 = I E = 0.80 mA 10 10 VB = VE 0.7 = 2 0.7 = 1.3 V IE =IB =VB 1.3 = I B = 0.026 mA RB 50I C = I E I B = 0.80 0.026 I C = 0.774 mA= =I C 0.774 = = 29.77 I B 0.026 1+ =29.77 = 0.9675 30.77VEC = VE VC = VE ( I C RC 10 ) = 2 ( 0.774 )(10 ) 10 VEC = 4.26 VLoad line developed assuming the VB voltage can change and the RB resistor is removed.1 mA 0.774Q-pointIC4.2620VEC5.21 5 0.7 17.2 A 250 I C = (120 )( 0.0172 ) = 2.064 mA IB =VC = ( 2.064 )(1.5 ) 5 = 1.90 VVEC = 5 ( 1.90 ) VEC = 6.90 Vwww.elsolucionario.net 152. IC (mA) 6.67Q-point 2.066.910 VEC (V)5.22 50 I C = (1) = 0.98 mA 51 VC = I C RC 9 = ( 0.98 )( 4.7 ) 9 or VC = 4.39 V 1 = 0.0196 mA 51 VE = I B RB + VEB ( on ) = ( 0.0196 )( 50 ) + 0.7 or VE = 1.68 V IB =5.23 0.5 50 I C = ( 0.5 ) = 0.49 mA, I B = = 0.0098 mA 51 51 VE = I B RB + VEB ( on ) = ( 0.0098 )( 50 ) + 0.7 or VE = 1.19 V VC = I C RC 9 = ( 0.49 )( 4.7 ) 9 = 6.70 V Then VEC = VE VC = 1.19 ( 6.7 ) 7.89 V PQ = I CVEC + I BVEB = ( 0.49 )( 7.89 ) + ( 0.0098 )( 0.7 ) or PQ = 3.87 mW Power Dissipated = PS = I Q ( 9 VE ) = ( 0.5 )( 9 1.19 )Or PS = 3.91 mW 5.24 I I E1 = I E 2 = 0.5 mA 2 0.5 mA = 5 ( 0.5 )( 4 ) VC1 = VC 2 = 3 VI E1 = I E 2 = I C1 = I C 2 VC1 = VC 25.25 (a)RE = 0 I B =2 0.7 1.3 = RB RB 1.3 5 2 = 0.8 RC = 3.75 K I C = ( 80 ) = RC RB RB = 130 K(b)0.8 81 = 0.010 mA I E = 0.8 = 0.81 mA 80 80 2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) RB = 49 K RE = 1 KIB =5 = ( 0.8 ) RC + 2 + ( 0.81)(1) RC = 2.74 K(c)For part (a)IB =2 0.7 = 0.01 mA 130www.elsolucionario.net 153. I C = (120 )( 0.01) I C = 1.20 mA VCE = 5 (1.2 )( 3.75 ) VCE = 0.5 V 2 = I B ( 49 ) + 0.7 + (121) I B (1)For part (b)I B = 0.00765 mA, I E = 0.925 mA, I C = 0.918 mA VCE = 5 ( 0.918 )( 2.74 ) ( 0.925 )(1) VCE = 1.56 V Including RE result in smaller changes in Q-point values. 5.26 I BQ =a.VCC VBE ( on ) RB2 = 0.0333 mA 60 24 0.7 RB = RB = 699 k 0.0333 VCC VCEQ 24 12 I CQ = RC = RC = 6 k RC 2 I BQ =I CQI BQ =b.=VCC VBE ( on ) RB24 0.7 699 0.0333 mA ( Unchanged )I CQ = I BQ = (100 )( 0.0333) I CQ = 3.33 mA VCEQ = VCC I CQ RC = 24 ( 3.33)( 6 ) VCEQ = 4.02 V VCE = VCC I C RC = 24 I C ( 6 )(c) IC (mA) 4 3.33Q-pt ( 100)Q-pt ( 60)24.0212245.27 a.VB = 0 Cutoff I E = 0, VC = 6 Vb.VB = 1 V, I E =c.VCEVB = 2 V. Assume active-mode1 0.7 I E = 0.3 mA 1 I C I E VC = 6 ( 0.3)(10 ) VC = 3 V2 0.7 = I E = 1.3 mA I C 1 VC = 6 (1.3)(10 ) = 7 V! IE =www.elsolucionario.net 154. Transistor in saturation 2 0.7 I E = 1.3 mA 1 VE = 1.3 V, VCE ( sat ) = 0.2 V IE =VC = VE + VCE ( sat ) = 1.3 + 0.2 VC = 1.5 V5.28 a.VBB = 0. RL Cutoff V0 = RC + RL V0 = 3.33 V 10 VCC = ( 5) 10 + 5 VBB = 1 Vb.1 0.7 6 A 50 I C = I B = ( 75 )( 6 ) I C = 0.45 mA IB =5 V0 V = IC + 0 5 10 1 1 1 0.45 = V0 + V0 = 1.83 V 5 10 Transistor in saturation V0 = VCE ( sat ) = 0.2 Vc.5.29 (a) = 100 100 (i) I Q = 0.1 mA I C = ( 0.1) = 0.0990 mA 101 VO = 5 ( 0.099 )( 5 ) VO = 4.505 V 100 (ii) I Q = 0.5 mA I C = ( 0.5 ) = 0.495 mA 101 VO = 5 ( 0.495 )( 5 ) VO = 2.525 V(iii) I Q = 2 mA Transistor is in saturation VO = VBE ( sat ) + VCE ( sat ) = 0.7 + 0.2 VO = 0.5 V (b) = 150 150 (i) I Q = 0.1 mA I C = ( 0.1) = 0.09934 mA 151 VO = 5 ( 0.09934 )( 5 ) VO = 4.503 V 4.503 4.505 100% = 0.044% 4.503 150 (ii) I Q = 0.5 mA I C = ( 0.5 ) = 0.4967 mA 151 VO = 5 ( 0.4967 )( 5 ) VO = 2.517 V % change =2.517 2.525 100% = 0.32% 2.525 (iii) I Q = 2 mA Transistor in saturation Vo = 8.5 V No change % change =5.30www.elsolucionario.net 155. VCB = 0.5 V VO = 0.5 V , I C =5 0.5 = 0.90 mA 5 101 IQ = ( 0.90 ) I Q = 0.909 mA 100 5.31 For I Q = 0, then PQ = 0 50 For I Q = 0.5 mA, I C = ( 0.5 ) = 0.49 mA 51 0.5 IB = = 0.0098 mA, VB = 0.490 V , VE = 1.19 V 51 VC = ( 0.49 )( 4.7 ) 9 = 6.70 V VEC = 7.89 V P I CVEC = ( 0.49 )( 7.89 ) P = 3.87 mWFor I Q = 1.0 mA, Using the same calculations as above, we find P = 5.95 mW For I Q = 1.5 mA, P = 6.26 mW For I Q = 2 mA, P = 4.80 mW For I Q = 2.5 mA, P = 1.57 mW For I Q = 3 mA, Transistor is in saturation. 0.7 + I B ( 50 ) = 0.2 + I C ( 4.7 ) 9 I E = IQ = I B + IC I B = 3 IC Then, 0.7 + ( 3 I C ) ( 50 ) = 0.2 + I C ( 4.7 ) 9Which yields I C = 2.916 mA and I B = 0.084 mA P = I BVEB + I CVEC = ( 0.084 )( 0.7 ) + ( 2.916 )( 0.2 ) or P = 0.642 mW5.32 IE =VEE VEB ( on ) RE=9 0.7 I E = 2.075 mA 4I C = I E = ( 0.9920 ) ( 2.075 ) I C = 2.06 mA VBC + I C RC = VCC VBC = 9 ( 2.06 ) ( 2.2 ) VBC = 4.47 V5.33 I CQ = I BQ = IR2 =VCC VCEQ RC I CQ12 6 = 2.73 mA 2.22.73 I BQ = 0.091 mA 300.7 ( 12 )= 0.127 mA 100 I R1 = I R 2 + I BQ = 0.127 + 0.091 = 0.218 mA V1 = I R1 R1 + 0.7 = ( 0.218 )(15 ) + 0.7 V1 = 3.97 V5.34www.elsolucionario.net 156. For VCE = 4.5 5 4.5 = 0.5 mA I CQ = 1 0.5 = 0.02 mA I BQ = 25 0.7 ( 5 ) = 0.057 mA I R2 = 100 I R1 = I R 2 + I BQ = 0.057 + 0.02 = 0.077 mA V1 = I R1 R1 + VBE ( on ) = ( 0.077 )(15 ) + 0.7 = 1.86 V For VCE = 1.0 5 1 = 4 mA 1 4 = = 0.16 mA 25 = 0.057 mAI CQ = I BQ I R2I R1 = I R 2 + I BQ = 0.057 + 0.16 = 0.217 mAV1 = ( 0.217 )(15 ) + 0.7 3.96 VSo 1.86 V1 3.96 V IC 5 4 Range of Q-pt values0.5 014.555.35 5 2.5 =5K 0.5 0.5 IB = = 0.00417 mA 120 5 0.7 RB = = 1032 K 0.00417RC =(a)IC (mA) 1.0Q-point 0.52.5(b)5 V (V) CEChoose RC = 5.1 K RB = 1 Mwww.elsolucionario.net 157. For RB = 1 M + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K 5 0.7 I BQ = = 3.91 A I CQ = 0.469 mA 1.1 VCEQ = 2.37 V RB = 1 M + 10% = 1.1M, RC = 5.1 K 10% = 4.59 K I BQ = 3.91 A I CQ = 0.469 mA VCEQ = 2.85 V RB = 1 M 10% = 0.90 M RC = 5.1 k + 10% = 5.61 K 5 0.7 I BQ = = 4.78 A I C = 0.573 mA 0.90 VCEQ = 1.78 V RB = 1 M 10% = 0.90 M RC = 5.1 k 10% = 4.59 K I BQ = 4.78 A I C = 0.573 mA VCEQ = 2.37 V IC (mA) 1.09 0.891 0.573 0.4691.78 2.37 2.855 V (V) CE5.36 VE 2 = 5 VBE 2 VE1 = 5 VBE1 VO = VE 2 VE1 = ( 5 VBE 2 ) ( 5 VBE1 ) VO = VBE1 VBE 2 I We have VBE1 = VE ln E1 I EO I VBE 2 = VT ln E 2 I EO I I VO = VT ln E1 ln E 2 I EO I EO I 10 VO = VT ln E1 = VT ln I I IE2 VO =5.37 (a)kT ln (10 ) eRE = 0(120 )( 4 ) VI 0.7 IC = I B VO = 5 I C ( 4 ) = 5 (VI 0.7 ) 200 200 When VO = 0.2, 0.2 = 5 2.4 VI + 1.68 VI = 2.7 IB =www.elsolucionario.net 158. VO(V) 50.2 0.72.75VI (V)RE = 1 K VI 0.7 V 0.7 IB = = I I C = IB 200 + (121)(1) 321(b)(120 )( 4 )(VI 0.7 ) 321 When VO = 0.2 = 5 1.495VI + 1.047VO = 5 VI = 3.91 V VO(V) 50.2 0.73.915VI (V)5.38 For 4.3 VI 5 Q is cutoff I C = 0 VO = 0 If Q reaches saturation, VO = 4.8 4.8 IC = = 1.2 mA 4 5 0.7 VI 1.2 IB = = 0.015 = VI = 1.6 80 180 So VI 1.6, VO = 4.8www.elsolucionario.net 159. VO(V)4.81.65.39 (a)4.35VI (V)For VI 4.3, Q is off and VO = 0 101 When transistor enters saturation, 5 = I C (1) + 0.2 + I C ( 4 ) I C = 0.958 mA 100 VO = 3.832 V I B = 0.00958 mA 101 5= ( 0.958 )(1) + 0.7 + ( 0.00958 )(180 ) + VI 100 VI = 5 0.7 0.9676 1.7244 VI = 1.61 V For VI = 0, transistor in saturation5 = I E (1) + 0.2 + I C ( 4 ) 5 = I C (1) + I B (1) + 0.2 + I C ( 4 ) 5 = I E (1) + 0.7 + I B (180 ) 5 = I C (1) + I B (1) + 0.7 + I B (180 ) I E = IC + I B4.8 = 5 I C + I B (1) 4.3 = 1I C + 181I B I B = 4.8 5 I C4.3 = I C + (181)( 4.8 5 I C ) 904 I C = 864.5 I C = 0.956 mA VO = 3.825 Vwww.elsolucionario.net 160. VO(V) 3.832 3.8251.614.35VI (V)5.40 RTH = R1 R2 = 33 10 = 7.67 k R2 10 VTH = VCC = (18 ) = 4.186 V R1 + R2 10 + 33 V VBE ( on ) 4.186 0.7 I BQ = TH = RTH + (1 + ) RE 7.67 + ( 51)(1) I BQ = 0.0594 mA I CQ = I BQ I CQ = 2.97 mA I EQ = 3.03 mA VCEQ = VCC I CQ RC I EQ RE= 18 ( 2.97 )( 2.2 ) ( 3.03)(1) VCEQ = 8.44 V5.41 I CQ = 1.2 mA, VCEQ = 9 V , RTH = 50 k 1.2 = 0.015 mA Also I B = 80 VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE R2 1 1 VTH = (VCC ) = RTH VCC = ( 50 )(18 ) R1 R1 R1 + R2 1 Then ( 50 )(18) = ( 0.015)( 50 ) + 0.7 + ( 81)( 0.015)(1) or R1 = 338 k . R1 Then338R2 = 50 R2 = 58.7 k 338 + R2 81 I EQ = (1.2 ) = 1.215 mA 80 18 = I CQ RC + VCEQ + I EQ RE18 = (1.2 ) RC + 9 + (1.215 )(1) RC = 6.49 k 5.42www.elsolucionario.net 161. RTH = R1 R2 = 20 15 = 8.57 k R2 15 VTH = (VCC ) = (10 ) = 4.29 V 15 + 20 R1 + R2 I EQ VCC = I EQ RE + VEB ( on ) + RTH + VTH 1+ 8.57 10 = I EQ (1) + 0.7 + I EQ + 4.29 101 10 0.7 4.29 5.01 Then I EQ = = I EQ = 4.62 mA 8.57 1.085 1+ 101 I EQ 4.62 VB = RTH + VTH = ( 8.57 ) + 4.29 or VB = 4.68 V 1+ 101 5.43 (a) RTH = 42 58 = 24.36 K 42 VTH = ( 24 ) = 10.08 V 100 10.08 0.7 9.38 I BQ = = 7.30 A 24.36 + (126 )(10 ) 1284.36 I CQ = 0.913 mA VCEQ = 14.8 VI EQ = 0.9202IC (mA)2.38Q-point 0.91314.824 V(b) R1 + 5% = 60.9, R2 + 5% = 44.1 RTH = 25.58 K 10.08 0.7 9.38 I BQ = = 7.30 A 25.58 + 126 (10 ) 1285.58 I CQ = 0.912 mA VCEQ = 14.81CE (V)VTH = 10.08I EQ = 0.919www.elsolucionario.net 162. R1 + 5% = 60.9, R2 5% = 39.90 RTH = 24.11 K 9.50 0.7 8.8 I BQ = = = 6.85 A 24.11 + (126 )(10 ) 1284.11 I CQ = 0.857 mA VCEQ = 15.37 VVTH = 9.50I EQ = 0.8635 mAR1 5% = 55.1 K R2 + 5% = 44.1 K 10.67 0.7 9.97 I BQ = = = 7.76 A 24.50 + 1260 1284.5 I CQ = 0.970 mA I EQ = 0.978 mA VCEQ = 14.22 V R1 5% = 55.1 K R2 5% = 39.90 10.08 0.7 9.38 I BQ = = = 7.31 A 23.14 + 1260 1283.14 I CQ = 0.914 mA I EQ = 0.9211 mA VCEQ = 14.79 VRTH = 24.50 KRTH = 23.14 KVTH = 10.67 VVTH = 10.08So we have 0.857 I CQ 0.970 mA 14.22 VCEQ 15.37 V 5.44 a. RTH = R1 R2 = 25 8 = 6.06 k R2 8 VTH = VCC = ( 24 ) 8 + 25 R1 + R2 = 5.82 V V VBE (on) 5.82 0.7 I BQ = TH = RTH + (1 + ) RE 6.06 + ( 76 )(1)I BQ = 0.0624 mA, I CQ = 4.68 mA I EQ = 4.74 VCEQ = VCC I CQ RC I EQ RE = 24 ( 4.68 )( 3) ( 4.74 )(1) VCEQ = 5.22 Vb. I BQ =5.82 0.7 I BQ = 0.0326 mA 6.06 + (151)(1)I CQ = 4.89 mA I EQ = 4.92VCEQ = 24 ( 4.89 )( 3) ( 4.92 )(1) VCEQ = 4.41 V5.45 (a)www.elsolucionario.net 163. I CQ I EQ = 0.4 mA 3 3 RC = 7.5 k ; RE = RE = 7.5 k 0.4 0.4 9 R1 + R2 = 112.5 k ( 0.2 )( 0.4 ) RC = R2 VTH = (VCC ) = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE R1 + R2 (112.5 R2 ) R2 RR 0.4 , I BQ = = 0.004 mA RTH = 1 2 = 112.5 100 R1 + R2 (112.5 R2 ) R2 9 R2 + 0.7 + (101)( 0.004 )( 7.5 ) = ( 0.004 ) 112.5 112.5 2 We obtain R2 ( 0.08 ) = 0.004 R2 3.56 105 R2 + 3.73From this quadratic, we find R2 = 48 k R1 = 64.5 k (b) Standard resistor values: Set RE = RC = 7.5 k and R1 = 62 k , R2 = 47 k Now RTH = R1 R2 = 62 47 = 26.7 k R2 47 VTH = (VCC ) = ( 9 ) = 3.88 V 47 + 62 R1 + R2 VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE So I BQ =3.88 0.7 = 0.00406 mA 26.7 + (101)( 7.5 )Then I CQ = 0.406 mA VRC = VRE = ( 0.406 )( 7.5 ) = 3.05 V5.46 (a) RTH = R1 R2 = 12 2 = 1.714 K R2 2 VTH = (10 ) 5 = (10 ) 5 VTH = 3.571 V R1 + R2 14 (b) VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 5 3.57 = I BQ (1.714 ) + 0.7 + (101) I BQ ( 0.5 ) 5 I BQ =5 0.7 3.571 0.729 = 13.96 A 1.714 + (101)( 0.5 ) 52.21I CQ = 1.396 mA, I EQ = 1.410 mAVCEQ = 10 (1.396 )( 5 ) (1.41)( 0.5 ) VCEQ = 2.32 V(d)www.elsolucionario.net 164. RE = 0.5 + 5% = 0.525 K I BQ =RC = 5 + 5% = 5.25 K0.729 13.32 A 1.714 + (101)( 0.525 )I CQ = 1.332 mAI EQ = 1.345 mAVCEQ = 10 (1.332 )( 5.25 ) (1.345 )( 0.525 )= 10 6.993 0.7061 VCEQ = 2.30 VRE = 0.5 + 5% = 0.525 K I CQ = 1.332 mARC = 5 5% = 4.75 KI EQ = 1.345 mAVCEQ = 10 (1.332 )( 4.75 ) (1.345 )( 0.525 )= 10 6.327 0.7061 VCEQ = 2.97 VRE = 0.5 5% = 0.475 K I BQ =RC = 5 + 5% = 5.25 K0.729 14.67 A 1.714 + (101)( 0.475 )I CQ = 1.467 mAI EQ = 1.482 mAVCEQ = 10 (1.467 )( 5.25 ) (1.482 )( 0.475 )= 10 7.70175 0.70395 VCEQ = 1.59 VRE = 0.5 5% = 0.475 K I CQ = 1.467 mARC = 5 5% = 4.75 KI EQ = 1.482 mAVCEQ = 10 (1.467 )( 4.75 ) (1.482 )( 0.475 )= 10 6.96825 0.70395 VCEQ = 2.33 V5.47 RTH = R1 R2 = 9 1 = 0.91 k R2 1 VTH = ( 12 ) = ( 12 ) = 1.2 V R1 + R2 1+ 9 I EQ RE + VEB ( on ) + I BQ RTH + VTH = 0 I BQ =VTH VEB (on) 1.2 0.7 = RTH + (1 + ) RE 0.90 + ( 76 )( 0.1)I BQ = 0.0588,I CQ = 4.41 mAI EQ = 4.47 mA Center of load line VECQ = 6 V I EQ RE + VECQ + I CQ RC 12 = 0( 4.47 ) ( 0.1) + 6 + ( 4.41) RC= 12 RC = 1.26 k5.48 (a) RTH = 36 68 = 23.5 K 36 VTH = (10 ) = 3.46 V 36 + 68 3.46 0.7 I BQ = = 0.00178 mA 23.5 + ( 51)( 30 ) I CQ = 0.0888 mAI EQ = 0.0906 mAVCE = 10 ( 0.0888 )( 42 ) ( 0.0906 )( 30 ) = 10 3.73 2.72 VCE = 3.55 Vwww.elsolucionario.net 165. (b) R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K RTH = 7.85 k VTH = 3.46 3.46 0.7 I BQ = = 0.00533 mA 7.85 + ( 51)(10 ) I CQ = 0.266 mAI EQ = 0.272 mAVCE = 10 ( 0.266 )(14 ) ( 0.272 )(10 ) VCE = 3.56 V5.49 (a) R2 68 RTH = 36 68 = 23.5 K VTH = (10 ) 5 = (10 ) 5 = 1.54 V R1 + R2 36 + 68 5 = ( 51) I BQ ( 30 ) + 0.7 + I B ( 23.5 ) + 1.54 I BQ =2.76 = 1.78 A I CQ = 0.0888 mA 1553.5 I EQ = 0.0906 mAVECQ = 10 ( 0.0906 )( 30 ) ( 0.0888 )( 42 )= 10 2.718 3.7296 VECQ = 3.55 V(b) RTH = 12 22.7 = 7.85 K VTH = 1.54 RE = 10 KRC = 14 K5 = ( 51) I BQ (10 ) + 0.7 + I B ( 7.85 ) + 1.54 I BQ =2.76 = 5.33 A I CQ = 0.266 mA 517.85 I EQ = 0.272 mAVECQ = 10 ( 0.272 )(10 ) ( 0.266 )(14 )= 10 2.72 3.724 VECQ = 3.56 V5.50 (a) RTH = ( 0.1)(1 + ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k 1 VTH = RTH VCC = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE R1 I BQ =I CQ=0.8 = 0.008 mA 1001 Then ( 5.05)(10 ) = ( 0.008)( 5.05) + 0.7 + (101)( 0.008)( 0.5) R1 or R1 = 44.1 k ,44.1R2 = 5.05 R2 = 5.70 k 44.1 + R2 101 Now I EQ = ( 0.8 ) = 0.808 mA 100 VCC = I CQ RC + VCEQ + I EQ RE10 = ( 0.8 ) RC + 5 + ( 0.808 )( 0.5 ) RC = 5.75 k (b)For 75 150www.elsolucionario.net 166. R2 5.7 VTH = (VCC ) = (10 ) = 1.145 V R1 + R2 5.7 + 44.1 VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ REFor = 75, I BQ =1.145 0.7 = 0.0103 mA 5.05 + ( 76 )( 0.5 )Then I CQ = ( 75 )( 0.0103) = 0.775 mA For = 150, I BQ =1.145 0.7 = 0.00552 mA 5.05 + (151)( 0.5 )Then I CQ = 0.829 mA % Change =I CQ=I CQ0.829 0.775 100% % Change = 6.75% 0.80For RE = 1 k (c)RTH = ( 0.1)(101)(1) = 10.1 k VTH =1 1 RTH VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1) R1 R1which yields R1 = 63.6 k And63.6 R2 = 10.1 R2 = 12.0 k 63.6 + R2 R2 12 Now VTH = (VCC ) = (10 ) = 1.587 V R1 + R2 12 + 63.6 1.587 0.7 = 0.0103 mA For = 75, I BQ = 10.1 + ( 76 )(1)So I CQ = 0.773 mA For = 150, I BQ =1.587 0.7 = 0.00551 mA 10.1 + (151)(1)Then I CQ = 0.826 mA % Change =I CQ I CQ=0.826 0.773 100% % Change = 6.63% 0.85.51 VCC I CQ ( RC + RE ) + VCEQ 10 = ( 0.8 )( RC + RE ) + 5 RC + RE = 6.25 k Let RE = 0.875 k Then, for bias stable RTH = ( 0.1)(121)( 0.875 ) = 10.6 k I BQ =0.8 = 0.00667 mA 1201 (10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 ) R1So R1 = 71.8 k and71.8R2 = 10.6 R2 = 12.4 k 71.8 + R210 = 0.119 mA 71.8 + 12.4 This is close to the design specification.Then I R 5.52www.elsolucionario.net 167. = VCC I CQ ( RC + RE )I CQ I EQ VCEQ6 = 12 I CQ ( 2 + 0.2 )I CQ = 2.73 mA,I BQ = 0.0218 mAVCEQ = 6 V VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 6 R2 VTH = (12 ) 6, R1 + R2 RTH = R 1 R2Bias stable RTH = ( 0.1)(1 + ) RE = ( 0.1)(126 )( 0.2 ) = 2.52 k 1 VTH = ( RTH )(12 ) 6 R1 1 ( 2.52 )(12 ) 6 = ( 0.0218)( 2.52 ) + 0.7 + (126 )( 0.0218)( 0.2 ) 6 R1 1 ( 30.24 ) = 0.7549 + 0.5494 R1 R1 = 23.2 k,23.2R 2 = 2.52 23.2 + R 2R2 = 2.83 k5.53 a. 81 I CQ = 1 mA. I EQ = (1) = 1.01 mA 80 VCEQ = 12 (1)( 2 ) (1.01)( 0.2 ) VCEQ = 9.80 V 1 = 0.0125 mA 80 = + ( 0.1) (1 + ) RE = ( 0.1)( 81)( 0.2 ) = 1.62 k I BQ =RTH R2 1 1 VTH = ( RTH )(12 ) 6 = (19.44 ) 6 (12 ) 6 = R1 + R2 R1 R1 VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 6 1 (19.44 ) 6 = ( 0.0125 )(1.62 ) + 0.7 + ( 81)( 0.0125)( 0.2 ) 6 R1 1 (19.44 ) = 0.923 R1 R1 = 21.1 k,21.1R2 = 1.62 21.1 + R2R2 = 1.75 kb. R1 = 22.2 k or R1 = 20.0 k R2 = 1.84 k or R2 = 1.66 k R2 ( max ) , R1 ( min )RTH = (1.84 )( 20.0 ) = 1.685 k 1.84 VTH = (12 ) 6 = 4.99 V 1.84 + 20.0 4.99 0.7 ( 6 ) 0.31 = = 0.0173 mA I BQ = 1.685 + ( 81)( 0.2 ) 17.89 I CQ = 1.39 mA, I EQ = 1.40 mAwww.elsolucionario.net 168. For max, RC VCE = 12 (1.39 )( 2 ) (1.40 )( 0.2 ) VCE = 8.94 V R2 ( min ) , R1 ( max ) RTH = (1.66 )( 22.2 ) = 1.545 k 1.66 VTH = (12 ) 6 = 5.165 V 1.66 + 22.2 5.165 0.7 + 6 0.135 = = 0.00761 mA I CQ = 0.609 mA, I E = 0.616 I BQ = 1.545 + ( 81)( 0.20 ) 17.745 VCEQ = 12 ( 0.609 )( 2 ) ( 0.616 )( 0.2 ) VCEQ = V 10.7 VSo 0.609 I C 1.39 mA 8.94 VCEQ 10.7 V 5.54 VCEQ VCC I CQ ( RC + RE )5 = 12 3 ( RC + RE ) RC + RE = 2.33 k Let RE = 0.333 k and RC = 2 k Nominal value of = 100 RTH = ( 0.1)(1 + ) RE = ( 0.1)(101)( 0.333) = 3.36 k 3 = 0.03 mA 100 1 1 = RTH (12 ) 6 = ( 3.36 )(12 ) 6 R1 R1I BQ = VTHThen VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 6 1 ( 3.36 )(12 ) 6 = ( 0.03)( 3.36 ) + 0.7 + (101)( 0.03)( 0.333) 6 R1 which yields R1 = 22.3 k and R2 = 3.96 k R2 3.96 Now VTH = (12 ) 6 = (12 ) 6 or VTH = 4.19 V 3.96 + 22.3 R1 + R2 For = 75, VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 6 I BQ =VTH + 6 0.7 4.19 + 6 0.7 = = 0.0387 mA I C = 2.90 mA RTH + (1 + ) RE 3.36 + ( 76 )( 0.333)For = 150, I BQ =4.19 + 6 0.7 = 0.0207 mA 3.36 + (151)( 0.333)Then I C = 3.10 mA Specifications are met. 5.55 RTH = R1 R2 = 3 12 = 2.4 k R2 12 VTH = VCC = ( 20 ) = 16 V 12 + 3 R1 + R2 (a) For = 7520 = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 20 0.7 16 = I BQ ( 76 )( 2 ) + 2.4 So I BQ = 0.0214 mA, I CQ = 1.60 mA, I EQ = 1.62 EVECQ = 20 (1.6 )(1) (1.62 )( 2 ) or VECQ = 15.16 Vwww.elsolucionario.net 169. (b) For = 100, we find I BQ = 0.0161 mA, I CQ = 1.61 mA, VECQ = 15.13 V , I EQ = 1.63 mA 5.56 I CQ = 4.8 mA I EQ = 4.84 mA VCEQ = VCC I CQ RC I EQ RE 6 = 18 ( 4.8 )( 2 ) ( 4.84 ) RE RE = 0.496 k RTH = ( 0.1)(1 + ) RE = ( 0.1)(121)( 0.496 ) = 6.0 kVTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE I BQ = 0.040 mA VTH =1 1 RTH VCC = ( 6.0 )(18 ) R1 R11 ( 6.0 )(18) = ( 0.04 )( 6.0 ) + 0.70 + (121)( 0.04 )( 0.496 ) R1 1 (108 ) = 3.34 R1 R1 = 32.3 k,32.3 R2 = 6.0 32.3 + R2R2 = 7.37 k5.57 For nominal = 70 2 = 0.0286 mA I EQ = 2.03 mA 70 = VCC I CQ RC I EQ REI BQ = VCEQ10 = 20 ( 2 )( 4 ) ( 2.03) RE RE = 0.985 K RTH = ( 0.1)(1 + ) RE = ( 0.1)( 71)( 0.985 ) = 6.99 KVTH = I BQ RTH + VBE ( on ) + I EQ RE1 RTH VCC = I BQ RTH + VBE ( on ) + I EQ RE R1 1 ( 6.99 )( 20 ) = ( 0.0286 )( 6.99 ) + 0.70 + ( 2.03)( 0.985 ) R1 1 (139.8 ) = 2.90 R1 R1 = 48.2 K,48.2 R2 = 6.99 48.2 + R2R2 = 8.18 KCheck: For = 50 8.18 VTH = ( 20 ) = 2.90 8.18 + 48.2 V VBE ( on ) 2.90 0.7 I BQ = TH = = 0.0384 mA RTH + (1 + ) RE 6.99 + ( 51)( 0.985 ) I CQ = 1.92 mAFor = 90 I BQ =2.90 0.7 = 0.0228 mA 6.99 + ( 91)( 0.985 )I CQ = 2.05 mADesign criterion is satisfied.www.elsolucionario.net 170. 5.58 I CQ = 1 mA I EQ = 1.017 mA VCEQ = VCC I CQ RC I EQ RE 5 = 15 (1)( 5 ) (1.017 ) RE RE = 4.92 k Bias stable: RTH = ( 0.1)(1 + ) RE = ( 0.1)( 61)( 4.92 ) = 30.0 k 1 = 0.0167 mA I BQ = 60 1 VTH = RTH VCC = I BQ RTH + VBE ( on ) + I EQ RE R1 1 ( 30.0 )(15 ) = ( 0.0167 )( 30.0 ) + 0.70 + (1.017 )( 4.92 ) R1 1 ( 448.5) = 6.197 R1 R1 = 72.5 k,72.5 R2 = 30.0 72.5 + R2R2 = 51.2 kCheck: For = 45 51.2 VTH = (15 ) = 6.21V 51.2 + 72.5 V VBE ( on ) 6.21 0.7 I BQ = TH = = 0.0215 mA RTH + (1 + ) RE 30 + ( 46 )( 4.92 )I CQ = 0.967 mA,I C = 3.27% ICCheck: For = 75 6.21 0.7 I BQ = = 0.0136 mA 30.0 + ( 76 )( 4.92 ) I C = 2.31% IC Design criterion is satisfied. I CQ = 1.023 mA,5.59 (a) VCC I CQ ( RC + RE ) + VCEQ 3 = ( 0.1)( 5 RE + RE ) + 1.4 RE = 2.67 k 100 = 0.833 A 120 = ( 0.1)(1 + ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k RC = 13.3 k , I BQ = RTHVTH =1 1 RTH VCC = ( 32.3)( 3) R1 R1= I BQ RTH + VBE ( on ) + (1 + ) I BQ RE= ( 0.000833)( 32.3) + 0.7 + (121)( 0.000833)( 2.67 ) which gives R1 = 97.3 k , and R2 = 48.4 k (b)www.elsolucionario.net 171. IR 3 3 = 20.6 A R1 + R2 97.3 + 48.4I CQ = 100 AP = ( I CQ + I R )VCC = (100 + 20.6 )( 3) or P = 362 W5.60 IE =5 VE 5 = = 1.67 mA RE 3RTH = R1 R2 = ( 0.1)(1 + ) RE = ( 0.1)(101)( 3) = 30.3 k R2 1 VTH = ( 4 ) 2 = RTH ( 4 ) 2 R1 + R2 R1 I EQ = 0.0165 mA I BQ = 1+ 5 = I EQ RE + VEB ( on ) + I B RTH + VTH 1 5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) + ( 30.3)( 4 ) 2 R1 0.80 =1 ( 30.3)( 4 ) R1 = 152 k R1152 R2 = 30.3 R2 = 37.8 k 152 + R25.61 a.RTH = R1 R2 = 10 20 RTH = 6.67 k R2 20 VTH = (10 ) 5 = (10 ) 5 VTH = 1.67 V 20 + 10 R1 + R2 b. 10 = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 0.7 1.67 7.63 I BQ = = I BQ = 0.0593 mA 6.67 + ( 61)( 2 ) 128.7 I CQ = 3.56 mA, I EQ =3.62 mA VE = 10 I EQ RE = 10 ( 3.62 )( 2 ) VE = 2.76 V VC = I CQ RC 10 = ( 3.56 )( 2.2 ) 10 VC = 2.17 V5.62 V + V I CQ ( RC + RE ) + VECQ 20 = ( 0.5 )( RC + RE ) + 8 ( RC + RE ) = 24 k Let RE = 10 k then RC = 14 k Let = 60 from previous problem. RTH = ( 0.1)(1 + ) RE = ( 0.1)( 61)(10 )Or RTH = 61 k www.elsolucionario.net 172. 0.5 = 0.00833 mA 60 R2 1 = (10 ) 5 = RTH 10 5 R1 R1 + R2 I BQ = VTHNow 10 = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 = ( 61)( 0.00833)(10 ) + 0.7 + ( 0.00833) ( 61) +1 ( 61) (10 ) 5 R1Then R1 = 70.0 k and R2 = 474 k 10 10 = 18.4 A R1 + R2 70 + 474 So the 40 A current limit is met. IR 5.63 a.RTH = R1 R2 = 35 20 RTH = 12.7 k R2 20 VTH = (7) 5 = ( 7 ) 5 VTH = 2.45 V R1 + R2 20 + 35 b. I BQ VTH VBE ( on ) ( 10 ) RTH + (1 + ) RE2.45 0.7 + 10 I BQ = 0.136 mA 12.7 + ( 76 )( 0.5 )I CQ = 10.2 mA, I EQ = 10.4 mA VCEQ = 20 I CQ RC I EQ RE= 20 (10.2 )( 0.8 ) (10.4 )( 0.5 )VCEQ = 6.64 Vc. R2 = 20 + 5% = 21 k R1 = 35 5% = 33.25 k RE = 0.5 5% = 0.475 k RTH = R1 R2 = 21 33.25 = 12.9 k R2 VTH = (7) 5 R1 + R2 21 = ( 7 ) 5 = 2.29 V 21 + 33.25 I BQ =2.29 0.7 ( 10 ) 12.9 + ( 76 )( 0.475 )= 0.143 mAI CQ = 10.7 mA, I EQ = 10.9 mAFor RC = 0.8 + 5% = 0.84 k VCEQ = 20 (10.7 )( 0.84 ) (10.9 )( 0.475 ) VCEQ = 5.83 VFor RC = 0.8 5% = 0.76 kwww.elsolucionario.net 173. VCEQ = 20 (10.7 )( 0.76 ) (10.9 )( 0.475 ) VCEQ = 6.69 V R2 = 20 5% = 19 k R1 = 35 + 5% = 36.75 k RE = 0.5 + 5% = 0.525 k RTH = R1 R2 = 19 36.75 = 12.5 k19 VTH = ( 7 ) 5 = 2.61 V 19 + 36.75 2.61 0.7 ( 10 ) = 0.128 mA I BQ = 12.5 + ( 76 )( 0.525 ) I CQ = 9.58 mA, I EQ = 9.70 mAFor RC = 0.84 k VCEQ = 20 ( 9.58 )( 0.84 ) ( 9.70 )( 0.525 ) VCEQ = 6.86 VFor RC = 0.76 k VCEQ = 20 ( 9.58 )( 0.76 ) ( 9.70 )( 0.525 ) VCEQ = 7.63 VSo 9.58 I CQ 10.7 mA and 5.83 VCEQ 7.63 V 5.64 a. RTH = 500 k 500 k 70 k = 250 k 70 k RTH = 54.7 k 5 VTH 3 VTH VTH ( 5 ) + = 500 500 70 5 3 5 1 1 1 + = VTH + + 0.0554 = VTH ( 0.0183) 500 500 70 500 500 70 VTH = 3.03 Vb. I BQ VTH VBE ( on ) ( 5 ) RTH + (1 + ) RE3.03 0.7 + 5 54.7 + (101)( 5 )I BQ = 0.00227 mA I CQ = 0.227 mA, I EQ = 0.229 VCEQ = 20 ( 0.227 )( 50 ) ( 0.229 )( 5 ) VCEQ = 7.51 V5.65 RTH = 30 60 20 RTH = 10 k 5 VTH 5 VTH VTH + = 30 60 20 5 1 1 5 1 + = VTH + + 30 60 30 60 20 VTH = 2.5 VFor = 100www.elsolucionario.net 174. I BQ = =VTH VBE ( on ) ( 5 ) RTH + (1 + ) RE2.5 0.7 + 5 10 + (101)( 0.2 )I BQ = 0.225 mA I CQ = 22.5 mA, I EQ = 22.7 mA VCEQ = 15 ( 22.5 )( 0.5 ) ( 22.7 ) ( 0.2 )VCEQ = 0.79! In saturation VCEQ = 0.2 V VTH = I BQ RTH + VBE + I EQ RE 5 2.5 + 5 0.7 = I BQ (10 ) + I EQ ( 0.2 ) 6.8 = I BQ (10 ) + I EQ ( 0.2 ) 14.8 = I CQ ( 0.5 ) + I EQ ( 0.2 ) Transistor in saturation, I EQ = I BQ + I CQ 6.8 = I BQ (10 ) + I BQ ( 0.2 ) + I CQ ( 0.2 ) 6.8 = I BQ (10.2 ) + I CQ ( 0.2 )51 14.8 = I BQ ( 0.2 ) + I CQ ( 0.7 ) 754.8 = I BQ (10.2 ) + I CQ ( 35.7 ) 748 = I CQ ( 35.5 ) I CQ = 21.1 mA VCEQ = 0.2 V5.66 I CQ = 50 A, I BQ = 0.625 A, I EQ = 50.6 A (a) 1 = 19.8 K 0.0506 5 = ( 0.050 ) RC + 5 + ( 0.0506 )(19.8 ) 5 RE =RC = 80 KRTH = R1 R2 Design bias stable circuit. RTH = ( 0.1)( 51)(19.8 ) = 101 K R2 1 VTH = (10 ) 5 = RTH (10 ) 5 R1 R1 + R2 1 So (101)(10 ) 5 = I BQ (101) + 0.7 + ( 0.0506 )(19.8 ) 5 R1 1 (1010 ) = 0.0631 + 0.7 + 1 R1 R1 = 573 K573 R2 = 101 573 + R2R2 = 123 K (b)www.elsolucionario.net 175. RTH = 101 K, VTH = 3.23 VVTH = I BQ RTH + 0.7 + (121)(19.8 ) I BQ 51.07 = I BQ (101 + 2395.8 ) I BQ = 0.429 A I CQ = 0.0514 mA, I EQ = 0.0519 mAVCEQ = 10 ( 0.0514 )( 80 ) ( 0.0519 )(19.8 ) = 10 4.11 1.03 VCEQ = 4.86 V5.67 (a) 2 = 2.5 K 0.8 + I EQ REI EQ = 0.80 mA, RE = 12 = I CQ RC + VCEQ12 = ( 0.8 ) RC + 7 + 2 RC = 3.75 K VTH = I BQ RTH + VBE + I EQ RE For a bias stable circuit RTH = ( 0.1)(12.1)( 2.5 ) = 30.25 K 1 1 RTH VCC = ( 30.25 )(12 ) = ( 0.00667 )( 30.25 ) + 0.7 + 2 R1 R1 1 ( 363) = 2.90 R1 = 125 K R1 125 R2 = 30.25 R2 = 39.9 K 125 + R2(b) Let RE = 2.4 K, RC = 3.9 K R1 = 120 K R2 = 39 K Then RTH = R1 R2 = 120 39 = 29.4 K 39 VTH = (12 ) = 2.94 V 120 + 39 2.94 0.7 2.24 = I BQ = 7.00 A 29.4 + (121)( 2.4 ) 319.8 I CQ = 0.841 m I EQ = 0.848 mA VCEQ = 12 ( 0.841)( 3.9 ) ( 0.848 )( 2.4 ) = 12 3.28 2.04 VCEQ = 6.68 V5.68 (a)I CQ = 100 A, I BQ = 1.18 A, I EQ = 101 A 2 RE = 19.8 K 0.101 9 = ( 0.101)(19.8 ) + 6 + ( 0.1) RC 9RE =RC = 100 Kwww.elsolucionario.net 176. Design a bias stable circuit. BTH = ( 0.1)( 86 )(19.8 ) = 170 K R2 1 VTH = (18 ) 9 = (170 )(18 ) 9 R1 + R2 R1 9 = ( 0.101)(19.8 ) + 0.7 + ( 0.00118 )(170 ) +1 (170 )(18 ) 9 R1203R2 = 170 203 + R2R1 = 203 KR2 = 1046 K (b) = 125 9 = (126 ) I BQ (19.8 ) + 0.7 + I BQ (170 ) + (15.07 9 ) 2.23 = 0.837 A 2664.8 = 0.1054 mAI BQ = I EQI CQ = 0.1046 mAVECQ = 18 ( 0.1046 )(100 ) ( 0.1054 )(19.8 ) VECQ = 5.45 V5.69 (a) 3 51 I EQ ( 20 ) = 20.4 mA RE = = 0.147 K 20.4 50 6 = 0.3 K VRC = 18 9 3 = 6 V RC = 20 For a bias stable circuit. RTH = ( 0.1)( 51)( 0.147 ) = 0.750 K V + = I EQ RE + VEB + I BQ RTH + VTH 18 = 3 + 0.7 + ( 0.4 )( 0.75 ) + 14 =1 ( 0.75 )(18) R11 (13.5) R1 = 0.964 K R1( 0.964 ) R2 0.964 + R2= 0.75 R2 = 3.38 K(b) Let RE = 0.15 K, RC = 0.3 K R1 = 1.0 K, R2 = 3.3 K 3.38 RTH = 1/13.3 = 0.767 K, VTH = (18 ) = 13.8 V 1 + 3.38 18 13.8 0.7 3.5 = = 0.416 mA I BQ = 0.767 + ( 51)( 0.15 ) 8.417 I CQ = 20.8 mA, I EQ = 21.2 mAVECQ = 18 ( 20.8 )( 0.3) ( 21.2 )( 0.15 ) = 18 6.24 3.18 VECQ = 8.58 V5.70www.elsolucionario.net 177. RTH = R1 R2 = 100 40 = 28.6 k R2 40 VTH = (10 ) = (10 ) = 2.86 V 40 + 100 R1 + R2 VTH VBE ( on ) 2.86 0.7 I B1 = = RTH + (1 + ) RE1 28.6 + (121) (1) I B1 = 0.0144 mA I C1 = 1.73 mA,I E1 = 1.75 mA10 VB 2 = I C1 + I B 2 3 VB 2 VBE ( on ) ( 10 ) IE2 = 5 10 VB 2 VB 2 0.7 + 10 = I C1 + 3 (121) ( 5 )1 10 9.3 1 1.73 = VB 2 + 3 (121) ( 5 ) 3 605 1.588 = VB 2 ( 0.335 ) VB 2 = 4.74 V IE2 = I B24.74 0.7 ( 10 )5 = 0.0232 mA I E 2 = 2.808 mAI C 2 = 2.785 mAVCEQ1 = 4.74 (1.75 ) (1) VCEQ1 = 2.99 V VCEQ 2 = 10 ( 4.74 0.7 ) VCEQ 2 = 5.96 V5.71 VE1 = 0.7 I R1 = VE 20.7 ( 5 )= 0.215 mA 20 = 0.7 0.7 = 1.4IE2 =1.4 ( 5 ) 1 I E 2 = 3.6 mA I B 2 = 0.0444 mA I C 2 = 3.56 mAI E1 = I R1 + I B 2 = 0.215 + 0.0444 I E1 = 0.259 mA I B1 = 0.00320 mA I C1 = 0.256 mA5.72 Current through V source = I E1 + I E 2 and I E1 = I E 2 = (1 + ) I B1 = ( 51)( 8.26 ) A So total current = 2 ( 51)( 8.26 ) A=843 AP = I V = ( 0.843)( 5 ) P = 4.22 mW (From V source) From Example 5.19, I Q = 0.413 mA 50 So I C 0 = ( 0.413) = 0.405 mA 51 www.elsolucionario.net 178. P + = I V + = ( 0.405 )( 5 ) P + = 2.03 mW(From V + source) 5.73 RTH = R1 R2 = 50 100 = 33.3 k R2 VTH = (10 ) 5 R1 + R2 100 = (10 ) 5 = 1.67 V 100 + 50 5 = I E1 RE1 + VEB ( on ) + I B1 RTH + VTH 101 I E1 = ( 0.8 ) = 0.808 mA 100 I B1 = 0.008 mA 5 = ( 0.808 ) RE1 + 0.7 + ( 0.008 )( 33.3) + 1.67 RE1 = 2.93 k VE1 = 5 ( 0.808 ) ( 2.93) = 2.63 V VC1 = VE1 VECQ1 = 2.63 3.5 = 0.87 V VE 2 = 0.87 0.70 = 1.57 V IE2 =1.57 ( 5 ) RE 2= 0.808 RE 2 = 4.25 kVCEQ 2 = 4 VC 2 = 1.57 + 4 = 2.43 V 5 2.43 RC 2 = RC 2 = 3.21 k 0.8 I RC1 = I C1 I B 2 = 0.8 0.008 = 0.792 mA RC1 =0.87 ( 5 ) 0.792 RC1 = 5.21 kwww.elsolucionario.net 179. Chapter 6 Exercise Problems EX6.1I BQ =(a)VBB VBE ( on ) RB=2 0.7 2 A 650Then I CQ = I BQ = (100 )( 2 A ) = 0.20 mA VCEQ = VCC I CQ RC = 5 ( 0.2 )(15 ) = 2 V (b)Now I CQ 0.20 gm = = = 7.69 mA / V VT 0.026 and VT (100 )( 0.026 ) r = = = 13 k 0.20 I CQ (c) We find r Av = g m RC r + RB 13 = ( 7.69 )(15 ) = 2.26 13 + 650 EX6.2 V VBE ( on ) 0.92 0.7 I BQ = BB = 100 RBor I BQ = 0.0022 mA and I CQ = I BQ = (150 )( 0.0022 ) = 0.33 mA (a) gm = r = ro =I CQ VT VT I CQ=0.33 = 12.7 mA / V 0.026=(150 )( 0.026 ) 0.33= 11.8 k VA 200 = = 606 k I CQ 0.33(b) vo = g m v ( ro r RC ) and v = vs r + RB so r vo = gm ( ro RC ) v r + RB 11.8 = (12.7 ) ( 606 15 ) 11.8 + 100 which yields Av = 19.6 EX6.3 (a) V VEB ( on ) 1.145 0.7 I BQ = BB = 50 RB Av =or I BQ = 0.0089 mA Then I CQ = I BQ = ( 90 )( 0.0089 ) = 0.801 mAwww.elsolucionario.net 180. Now gm = r = ro =I CQ VT VT I CQ=0.801 = 30.8 mA / V 0.026=( 90 )( 0.026 ) 0.801= 2.92 k VA 120 = = 150 k I CQ 0.801We have Vo = g mV ( ro RC )(b) r and V = Vs r + RB so r V Av = o = g m ( ro RC ) Vs r + RB 2.92 = ( 30.8 ) (150 2.5 ) 2.92 + 50 which yields Av = 4.18 EX6.4 Using Figure 6.23 (a) For I CQ = 0.2 mA, 7.8 < hie < 15 k , 60 < h fe < 125, 6.2 104 < hre < 50 104 , 5 < hoe < 13 mhosFor I CQ = 5 mA, 0.7 < hie < 1.1 k , 140 < h fe < 210, 1.05 104 < hre < 1.6 104 ,(b)22 < hoe < 35 mhos EX6.5 RTH = R1 R2 = 35.2 5.83 = 5 k R2 5.83 VTH = VCC = (5) R1 + R2 5.83 + 35.2 or VTH = 0.7105 V Then V VBE ( on ) 0.7105 0.7 I BQ = TH = RTH 5 or I BQ = 2.1 Aand I CQ = I BQ = (100 )( 2.1 A ) = 0.21 mA VCEQ = VCC I CQ RC = 5 ( 0.21)(10 ) and VCEQ = 2.9 V Now gm = ro =I CQ VT=0.21 = 8.08 mA 0.026VA 100 = = 476 k I CQ 0.21And Av = g m ( roRC ) = ( 8.08 ) ( 476 10 )so Av = 79.1 EX6.6www.elsolucionario.net 181. RTH = R1 R2 = 250 75 = 57.7 k R2 75 VTH = (VCC ) = (5) 75 + 250 R1 + R2 or VTH = 1.154 V I BQ =VTH VBE ( on )RTH + (1 + ) RE=1.154 0.7 57.7 + (121)( 0.6 )or I BQ = 3.48 A I CQ = I BQ = (120 )( 3.38 A ) = 0.418 mA (a) Now gm = r =I CQ VT=0.418 = 16.08 mA / V 0.026VT (120 )( 0.026 ) = = 7.46 k 0.418 I CQWe have Vo = g mV RC We find Rib = r + (1 + ) RE = 7.46 + (121)( 0.6 ) or Rib = 80.1 k Also R1 R2 = 250 75 = 57.7 k R1 R2 Rib = 57.7 80.1 = 33.54 k We find R1 R2 Rib 33.54 Vs = Vs = Vs 33.54 + 0.5 R1 R2 Rib + RS or Vs = ( 0.985 ) Vs Now 1+ 121 Vs = V 1 + RE = V 1 + ( 0.6 ) r 7.46 or V = ( 0.0932 ) Vs = ( 0.0932 )( 0.985 ) VsSo Av =Vo = (16.08 )( 0.0932 )( 0.985 )( 5.6 ) Vsor Av = 8.27 EX6.7 RTH = R1 R2 = 15 85 = 12.75 k R2 85 VTH = VCC = (12 ) R1 + R2 15 + 85 or VTH = 10.2 V Now VCC = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTHwww.elsolucionario.net 182. so 12 = (101) I BQ ( 0.5 ) + 0.7 + I BQ (12.75 ) + 10.2 which yields I BQ = 0.0174 mA and I CQ = I BQ = (100 )( 0.0174 ) = 1.74 mA I EQ = (101)( 0.0174 ) = 1.76 mA VECQ = VCC I EQ RE I CQ RC = 12 (1.76 )( 0.5 ) (1.74 )( 4 ) or VECQ = 4.16 V Now r = VT I CQ=(100 )( 0.026 ) 1.74= 1.49 k We have Vo = g mV ( RC RL ) V Vs = V g mV + RE r Solving for V and noting that = g m r , we find Av =Vo Vs= = ( RCRL )r + (1 + ) RE (100 )( 4 2 )1.49 + (101)( 0.5 )or Av = 2.56 EX6.8 Dc analysis 10 0.7 I BQ = = 0.00439 mA 100 + (101)( 20 ) I CQ = 0.439 mA, I EQ = 0.443 mANow r =(100 )( 0.026 )= 5.92 k 0.439 0.439 gm = = 16.88 mA / V 0.026 100 ro = = 228 k 0.439 (a) Now Vo = g mV ( ro RC ) RB r V = Vs RB r + RS RB r = 100 5.92 = 5.59 k 5.59 Then V = Vs = ( 0.918 ) Vs 5.59 + 0.5 V Then Av = o = (16.88 ) ( 0.918 ) ( 228 10 ) Vs or Av = 148 (b) Rin = RS + RB r = 0.5 + 100 5.92 = 6.09 k www.elsolucionario.net 183. and Ro = RC ro = 10 228 = 9.58 k EX6.9 (a) I CQ 0.25 gm = = = 9.615 mA / V 0.026 VT ro =VA 100 = = 400 k I CQ 0.25Av = g m ( ro rc ) = ( 9.615 )( 400 100 ) = 769 (b) Av = g m ( ro rc rL ) = ( 9.615 )( 400 100 100 ) Av = 427 EX6.10 5 0.7 I BQ = = 0.00672 mA 10 + (126 )( 5 ) I CQ = 0.84 mA, I EQ = 0.847 mAVCEQ = 10 ( 0.84 )( 2.3) ( 0.847 )( 5 )or VCEQ = 3.83 V dc load line VCE (V + V ) I C ( RC + RE ) or VCE = 10 I C ( 7.3) ac load line (neglecting ro) vce = ic ( RC RL ) = ic ( 2.3 5 ) = ic (1.58 ) IC (mA) 1.370.84AC load lineQ-pointDC load line3.8310 VCE (V)EX6.11 (a) dc load line VEC VCC I C ( RC + RE ) = 12 I C ( 4.5 )ac load line vec ic ( RE + RCRL ) or vec = ic ( 0.5 + 4 2 ) = ic (1.83)Q-point values 12 0.7 10.2 = 0.0150 mA I BQ = 12.75 + (121)( 0.5 ) I CQ = 1.80 mA, I EQ = 1.82 mA VECQ = 3.89 Vwww.elsolucionario.net 184. IC (mA) AC load line2.67Q-point1.80DC load line3.8912 V (V) ECiC = I CQ = 1.80 mA vEC = (1.8 )(1.83) = 3.29 V(b)vEC ( min ) = 3.89 3.29 = 0.6 VSo maximum symmetrical swing = 2 3.29 = 6.58 V peak-to-peak EX6.12 R2 1 VTH = (a) VCC = RTH VCC R1 R1 + R2 RTH = ( 0.1)(1 + ) RE = ( 0.1)(121)(1)so RTH = 12.1 k , VTH =1 (12.1)(12 ) R1We can write VCC = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTH We have I CQ = 1.6 mA, I BQ =1.6 = 0.0133 mA 120Then 1 (12.1)(12 ) R1 12.1 + (121)(1)12 0.7 I BQ = 0.0133 =which yields R1 = 15.24 k Since RTH = R1 R2 = 12.1 k , we find R2 = 58.7 k Also VECQ = 12 (1.6 )( 4 ) (1.61)(1) = 3.99 V (b)ac load line vec = ic ( RCRL )Want ic = I CQ 0.1 = 1.6 0.1 = 1.5 mA Also vec = 3.99 0.5 = 3.49 V Nowvec 3.49 = = 2.327 k = RC ic 1.5RLSo 4 RL = 2.327 k which yields RL = 5.56 k EX6.13www.elsolucionario.net 185. RTH = R1 R2 = 25 50 = 16.7 k R2 50 VTH = VCC = (5) 25 + 50 R1 + R2 or VTH = 3.33 VI BQ =VTH VBE ( on )RTH + (1 + ) RE=3.33 0.7 16.7 + (121)(1)or I BQ = 0.0191 mA Also I CQ = (120 )( 0.0191) = 2.29 mANow gm = r = ro =I CQ VT= VT I CQ2.29 = 88.1 mA / V 0.026=(120 )( 0.026 ) 2.29= 1.36 k VA 100 = = 43.7 k I CQ 2.29(a) R1 R2 Rib Vs = Vs R1 R2 Rib + RS Rib = r + (1 + ) ( RE ro ) = 1.36 + (121)(1 43.7 )or Rib = 120 k and R1 R2 = 16.7 k Then R1 R2 Rib = 16.7 120 = 14.7 k Now 14.7 Vs = Vs = ( 0.967 ) Vs 14.7 + 0.5 and V 1+ Vo = + g mV ( RE ro ) = V r r RE ro We have Vs = V + Vo thenVs ( 0.967 )Vs = 1+ 1+ 1+ RE ro 1 + RE ro r r We then obtainV =V Av = o = Vs=1+ r( 0.967 ) RE ro1+ 1+ RE ro r ( 0.967 )(1 + ) RE ror + (1 + ) REroNow RE ro = 1 43.7 = 0.978 k www.elsolucionario.net 186. Then Av =( 0.967 )(121)( 0.978 ) = 0.956 1.36 + (121)( 0.978 )(b) Rib = r + (1 + )( RE ro ) or Rib = 1.36 + (121)( 0.978 ) = 120 k EX6.14 For RS = 0, then r Ro = RE ro 1+ Using the parameters from Exercise 4.13, we obtain 1.36 Ro = 1 43.7 121 or Ro = 11.1 EX6.15 (a) For I CQ = 1.25 mA and = 100, we find I EQ = 1.26 mA and I BQ = 0.0125 mANow VCEQ = 10 I EQ RE Or 4 = 10 (1.26 ) RE which yields RE = 4.76 k Then RTH = ( 0.1)(1 + ) RE = ( 0.1)(101)( 4.76 ) or RTH = 48.1 k We have R2 1 VTH = (10 ) 5 = RTH (10 ) 5 R1 R1 + R2 or VTH =1 ( 481) 5 R1We can write I BQ =VTH 0.7 ( 5 ) RTH + (1 + ) REOr 1 ( 481) 5 0.7 + 5 R1 0.0125 = 48.1 + (101)( 4.76 )which yields R1 = 65.8 k Since R1 R2 = 48.1 k , we obtain R2 = 178.8 k (b)www.elsolucionario.net 187. r = ro = VT I CQ=(100 )( 0.026 ) 1.25= 2.08 k VA 125 = = 100 k I CQ 1.25We may note that g mV = g m ( I b r ) = I b Also Rib = r + (1 + )( RE RL ro ) = 2.08 + (101) ( 4.76 1 100 ) or Rib = 84.9 k Now RE ro Io = 1 + ) Ib R r + R ( L E o where R1 R2 Ib = I R R +R s ib 1 2 We can then write RE ro R1 R2 I AI = o = (1 + ) R R +R I s RE ro + RL ib 1 2 We have RE ro = 4.76 100 = 4.54 k so 48.1 4.54 AI = (101) 4.54 + 1 48.1 + 84.9 or AI = 29.9 r 2.08 (c) Ro = RE ro = 4.76 100 1+ 101or Ro = 20.5 EX6.16 (a) RTH = R1 R2 = 70 6 = 5.53 k R2 6 VTH = (10 ) 5 = (10 ) 5 70 + 6 R1 + R2 or VTH = 4.2105 V We find 4.2105 0.7 ( 5 ) I BQ1 = 2.91 A 5.53 + (126 )( 0.2 )and I CQ1 = I BQ1 = (125 )( 2.91 A ) = 0.364 mA I EQ1 = (1 + ) I BQ1 = 0.368 mAAt the collector of Q1, V 0.7 ( 5 ) 5 VC1 = I CQ1 + C1 RC1 (1 + )( RE 2 ) orwww.elsolucionario.net 188. V 0.7 ( 5 ) 5 VC1 = 0.364 + C1 5 (126 )(1.5)which yields VC1 = 2.99 V also VE1 = I EQ1 RE1 5 = ( 0.368 )( 0.2 ) 5 or VE1 = 4.93 V Then VCEQ1 = VC1 VE1 = 2.99 ( 4.93) = 7.92 V We find V 0.7 ( 5 ) I EQ 2 = C1 = 4.86 mA 1.5 and 125 I CQ 2 = I EQ1 = ( 4.86 ) = 4.82 mA 1+ 126 We find VE 2 = VC1 0.7 = 2.99 0.7 = 2.29 V and VCEQ 2 = 5 VE 2 = 5 2.29 = 2.71 V (b) The small-signal transistor parameters are: r 1 = VT I CQ1g m1 = r 2 = gm2 I CQ1 VT VT I CQ 2 I CQ 2 VT(125 )( 0.026 ) 0.3640.364 = 14.0 mA / V 0.026= 8.93 k (125 )( 0.026 )=4.82= 0.674 k 4.82 = 185 mA / V 0.026We find Rib1 = r 1 + (1 + ) RE1 = 8.93 + (126 )( 0.2 ) Or Rib1 = 34.1 k and Rib 2 = r 2 + (1 + )( RE 2 RL ) = 0.674 + (126 )(1.5 10 ) = 165 k The small-signal equivalent circuit is:www.elsolucionario.net 189. V1r1gm1V1V2 R1R2RC1RE1We can write Vo = (1 + ) I b 2 ( RE 2 RL ) where RC1 Ib2 = ( g m1V 1 ) RC1 + Rib 2 V V 1 = s r 1 Rib1 Then V Av = o Vs RC1 g m1r 1 = (1 + )( RE 2 RL ) RC1 + Rib 2 Rib1 so 5 125 Av = (126 )(1.5 10 ) 5 + 165 34.1 or Av = 17.7 (c) Ri = R1 R2 Rib1 = 70 6 34.1 = 4.76 k r + RC1 0.676 + 5 and Ro = RE 2 2 = 1.5 126 1+ or Ro = 43.7 Test Your Understanding Exercises TYU6.1 I CQ 0.25 = = 9.62 mA / V gm = 0.026 VTr = ro = VT I CQ=(120 )( 0.026 ) 0.25gm2V2 Vsr2= 12.5 k VA 150 = = 600 k I CQ 0.25TYU6.2 V V 75 ro = A I CQ = A = I CQ ro 200 k or ICQ = 0.375 mAwww.elsolucionario.netVoRE2RL 190. TYU6.3 RC RE Resulting gain is always smaller than this value. The effect of RS is very small. Set RC = 10 RE Now 5 I C ( RC + RE ) + VCEQAs a first approximation, Av or 5 = ( 0.5 )( RC + RE ) + 2.5 which yields RC + RE = 5 k So 10 RE + RE = 5 or RE = 0.454 k and RC = 4.54 k We have I CQ 0.5 I BQ = = = 0.005 mA 100 and RTH = ( 0.1)(1 + ) RE = ( 0.1)(101)( 0.454 ) or RTH = 4.59 k also R2 1 VTH = VCC = RTH VCC R1 + R2 R1 or 1 23 ( 4.59 )( 5 ) = R1 R1 We can write VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE VTH =or 23 = ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 ) R1 which yields R1 = 24.1 k and since R1 R2 = 4.59 k we find R2 = 5.67 k TYU6.4 dc analysis RTH = R1 R2 = 15 85 = 12.75 k R2 85 VTH = VCC = (12 ) 15 + 85 R1 + R2 or VTH = 10.2 V We can write 12 0.7 VTH 12 0.7 10.2 I BQ = = RTH + (1 + ) RE 12.75 + (101)( 0.5 )or IBQ = 0.0174 mAwww.elsolucionario.net 191. and I CQ = I BQ = 1.74 mA ac analysis Vo = h fe I b ( RCRL )and Ib =so Av =Vshie + (1 + h fe ) RE h fe ( RC RL ) Vo = Vs hie + (1 + h fe ) REFor ICQ = 1.74 mA, we find h fe ( max ) = 110, h fe ( min ) = 70 hie ( max ) = 2 k , hie ( min ) = 1.1 k We obtain 110 ( 4 2 ) Av ( max ) = = 2.59 1.1 + (111)( 0.5 ) and Av ( min ) =70 ( 4 2 )2 + ( 71)( 0.5 )= 2.49TYU6.5First approximation, Av RC REThis predicts a low value, so setRC =9 RENow VCC I CQ ( RC + RE ) + VECQ or 7.5 = ( 0.6 )( 9 RE + RE ) + 3.75 which yields RE = 0.625 k and RC = 5.62 k We have RTH = ( 0.1)(1 + ) RE = ( 0.1)(101)( 0.625 ) or RTH = 6.31 k Also 1 1 VTH = RTH VCC = ( 6.31)( 7.5 ) R1 R1 We have I CQ 0.6 I BQ = = = 0.006 mA 100 The KVL equation around the E-B loop gives VCC = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTH or 7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) +1 ( 6.31)( 7.5) R1which yields R1 = 7.41 k Since RTH = R1 R2 = 6.31 k , then R2 = 42.5 k TYU6.6www.elsolucionario.net 192. R RC = ( 0.95 ) C r + (1 + ) RE RE 2 or Av = ( 0.95 ) = 4.75 0.4 Assume r = 1.2 k from Example 4.6. Then ( 2) = 4.75 1.2 + (1 + )( 0.4 )We have Av =or = 76 TYU6.7 dc analysis: By symmetry, VTH = 0 RTH = R1 R2 = 20 20 = 10 k We can write 0 0.7 ( 5 ) I BQ = = 0.00672 mA 10 + (126 )( 5 )I CQ = I BQ = (125 )( 0.00672 ) = 0.84 mASmall-signal transistor parameters: VT (125 )( 0.026 ) r = = = 3.87 k I CQ 0.84 gm = ro =I CQ VT=0.84 = 32.3 mA / V 0.026VA 200 = = 238 k I CQ 0.84(a) We have Vo = g mV ( ro RCRL ) and V = Vsso Av =Vo Vs= g m ( ro RCRL )= ( 32.3)( 238 2.3 5 ) or Av = 50.5 (b) Ro = ro RC = 238 2.3 = 2.28 k TYU6.8 We find I CQ = 0.418 mA, then 121 VCEQ = 5 ( 0.418 )( 5.6 ) ( 0.418 )( 0.6 ) 120 or VCEQ = 2.41 V So vCE = ( 2.41 0.5 ) 2or vCE = 3.82 V peak-to-peak TYU6.9 dc load line VCE (10 + 10 ) I CQ ( RC + RE ) orwww.elsolucionario.net 193. VCE = 20 I C (10 + RE )ac load line vce = ic RC = ic (10 ) Now vCE = VCEQ 0.7 = iC (10 ) = I CQ (10 ) so VCEQ 0.7 = I CQ (10 ) We have that VCEQ = 20 I CQ (10 + RE ) then I CQ (10 ) + 0.7 = 20 I CQ (10 + RE ) or I CQ ( 20 + RE ) = 19.3(1)The base current is found from 10 0.7 I BQ = 100 + (101) RE so I CQ =(100 )( 9.3)100 + (101) RESubstituting into Equation (1), (100 )( 9.3) ( 20 + RE ) = 19.3 100 + (101) RE which yields RE = 16.35 k Then (100 )( 9.3) I CQ = = 0.531 mA 100 + (101)(16.35 ) and VCEQ 20 ( 0.531)(10 + 16.35 ) = 6.0 V Now vCE = VCEQ 0.7 = 6 0.7 = 5.3 V or, maximum symmetrical swing = 2 5.3 = 10.6 V peak-to-peak TYU6.10 We can write 0 0.7 ( 10 ) I BQ = 6.60 A 100 + (131)(10 ) and I CQ = (130 )( 6.60 A ) = 0.857 mA Assume nominal small-signal parameters of: hie = 4 k , h fe = 134 hre = 0, hoe = 12 S 1 = 83.3 k hoeWe find 1 Rib = hie + (1 + h fe ) RE RL hoe = 4 + (135 )(10 10 83.3) = 641 k To find the voltage gain:www.elsolucionario.net 194. RB Rib 100 641 Vs = Vs 100 641 + 10 RB Rib + RSVs =or Vs = ( 0.896 ) Vs Also(1 + h fe ) R Vo = Vs hie + (1 + h fe ) Rwhere R = REThen Av =RL1 = 10 10 83.3 = 4.72 k hoeVo ( 0.896 )(135 )( 4.72 ) = = 0.891 Vs 4 + (135 )( 4.72 )To find the current gain: 1 RE hoe RB Io 1 + h fe ) Ai = = ( Ii 1 RB + Rib + RL RE hoe 10 83.3 100 = (135 ) 10 83.3 + 10 100 + 641 or Ai = 8.59 To find the output resistance: 1 hie + RS RB Ro = RE 1 + h fe hoe = 10 83.34 + 10 100 96.0 135TYU6.11 We find RTH = R1 R2 = 50 50 = 25 k R2 1 VTH = VCC = ( 5 ) = 2.5 V 2 R1 + R2 Now V VEB ( on ) VTH I BQ = CC RTH + (1 + ) RE =5 0.7 2.5 = 0.00793 mA 25 + (101)( 2 )and I CQ = (100 )( 0.00793) = 0.793 mAThe small-signal transistor parameters: I CQ 0.793 = = 30.5 mA / V gm = 0.026 VT r = ro = VT I CQ=(100 )( 0.026 ) 0.793= 3.28 k VA 125 = = 158 k I CQ 0.793(a)www.elsolucionario.net 195. Define R = RE RL ro = 2 0.5 158 0.40 k Now Av =(1 + ) R (101)( 0.4 ) = r + (1 + ) R 3.28 + (101)( 0.4 )or Av = 0.925 (b) Rib = r + (1 + ) R = 3.28 + (101)( 0.4 ) or Rib = 43.7 k Also Ro = RE ror 3.28 = 2 158 1+ 101or Ro = 32.0 TYU6.12 For VECQ = 2.5, I EQ =VCC VECQ RE=5 2.5 = 5 mA 0.5then 75 I CQ = ( 5 ) = 4.93 mA 76 5 I BQ = = 0.0658 mA 76 Small-signal transistor parmaters: VT ( 75 )( 0.026 ) = = 0.396 k r = I CQ 4.93 ro =VA 75 = = 15.2 k I CQ 4.93Define the small-signal base current into the base, then g mV = I b Now, RE ro Io = (1 + ) I b RE ro + RL and R1 R2 Ib = Ii R1 R2 + Rib The current gain is RE ro R1 R2 I AI = o = (1 + ) I i RE ro + RL R1 R2 + Rib We have RE = RL = 0.5 k Rib = r + (1 + )( RE RL ro ) = 0.396 + ( 76 )( 0.5 0.5 15.2 ) = 19.1 k and RE ro = 0.5 15.2 = 0.484 k Then R1 R2 0.484 AI = 10 = ( 76 ) 0.484 + 0.5 R1 R2 + 19.1 which yields R1 R2 = 6.975 k Nowwww.elsolucionario.net 196. R2 1 VTH = VCC = RTH VCC R1 + R2 R1 or 1 VTH = ( 6.975 )( 5 ) R1 We can write V VEB ( on ) VTH I BQ = CC RTH + (1 + ) REor 0.0658 =5 0.7 VTH 6.975 + ( 76 )( 0.5 )which yields VTH = 1.34 =1 ( 6.975)( 5 ) R1or R1 = 26.0 k and R2 = 9.53 k TYU6.13 (a) dc analysis: V VEB ( on ) 10 0.7 I EQ = EE = = 0.93 mA 10 RE 100 I CQ = I EQ = ( 0.93) = 0.921 mA 101 1+ VECQ = VEE I EQ RE I CQ RC ( VCC )= 10 ( 0.93)(10 ) ( 0.921)( 5 ) ( 10 )or VECQ = 6.1 V (b) Small-signal transistor parameters: VT (100 )( 0.026 ) r = = = 2.82 k I CQ 0.921 I CQ0.921 = 35.42 mA / V 0.026 VT Small-signal current gain: I o = g mV and V = Vs also 1 Vs + g mV = Vs Ii = R r + gm RE r E Then g m ( RE r ) I g mV AI = o = = Ii 1 1 + g m ( RE r ) + gm V RE r gm ( 35.42 ) (10 2.82 ) 1 + ( 35.42 ) (10 2.82 )or AI = 0.987 (c) Small-signal voltage gain:www.elsolucionario.net 197. Vo = g mV RC = g mVs RC or V Av = o = g m RC = ( 35.42 )( 5 ) Vs or Av = 177 TYU6.14 (a) V VBE ( on ) 10 0.7 I BQ = EE = RB + (1 + ) RE 100 + (101)(10 )or I BQ = 8.38 A and I CQ = 0.838 mA r = gm = ro = VT=I CQ I CQ VT=(100 )( 0.026 ) 0.838= 3.10 k 0.838 = 32.23 mA / V 0.026VA = = I CQ 0.838(b) Summing currents, we have V V Vs V g mV + = + r RE RS or 1 + 1 V 1 V + = s + r RE RS RS We can then write V r V = s RE RS RS 1 + Now Vo = g mV ( RC RL ) So( RC RL ) r R R Vo = gm E S Vs RS 1 + 32.23) (10 1) 3.10 ( = 101 10 1 1) ( Av =or Av = 0.870 Now RE Ie = RE + r Ic = 1+ 10 Ii = I i = ( 0.763) I i 10 + 3.10 100 Ie = Ie 101 RC 10 Io = Ic = I c = ( 0.909 ) I c RC + RL 10 + 1 So we have I 100 AI = o = ( 0.909 ) ( 0.763) Ii 101 www.elsolucionario.net 198. or AI = 0.687 (c) We have r 3.10 Ri = RE = 10 1+ 101 or Ri = 30.6 Also Ro = RC = 10 k TYU6.15 dc analysis: We can write 5 = I BQ RB + VBE ( on ) + I EQ RE or I BQ = I CQ =5 0.7 4.3 = RB + (101) RE RB + (101) RE(100 )( 4.3) RB + (101) REAlso 5 = I CQ RC + VCEQ + I EQ RE 5 or 101 VCEQ = 10 I CQ RC + RE 100 ac analysis: Vo = g mV ( RC RL )and Vs = V V RB = V r RB 1 + r or r V = Vs r + RB Then V Av = o = ( RC RL ) Vs r + RBwhere = g m r For I CQ = 1 mA, r = Now Av = 20 = VT I CQ=(100 )( 0.026 ) 1= 2.6 k (100 ) ( 2 2 )2.6 + RB which yields RB = 2.4 k Also (100 )( 4.3) I CQ = 1 = 2.4 + (101) REwhich yields RE = 4.23 k TYU6.16 (a)www.elsolucionario.net 199. dc analysis: For I EQ 2 = 1 mA, 100 I CQ 2 = (1) = 0.990 mA 101 I EQ 2 1 I EQ1 = = = 0.0099 mA 1 + 101 and I EQ1 0.0099 I BQ1 = = = 0.000098 mA 1+ 101 and I CQ1 = (100 )( 0.000098 ) = 0.0098 mAand VB1 = I BQ1 RB = ( 0.000098 )(10 ) or VB1 = 0.00098 0 so VE1 = 0.7 V and VE 2 = 1.4 V Now I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 1 mA then VO = 5 (1)( 4 ) = 1 V and VCEQ 2 = 1 ( 1.4 ) = 2.4 V VCEQ1 = 1 ( 0.7 ) = 1.7 V(b) Small-signal transistor parameters: VT (100 )( 0.026 ) = = 265 k r 1 = 0.0098 I CQ1 g m1 = r 2 = gm2 =I CQ1 VT VT I CQ 2 I CQ 2 VT0.0098 = 0.377 mA 0.026= =(100 )( 0.026 )=0.990= 2.63 k 0.990 = 38.1 mA / V 0.026ro1 = ro 2 = (c) Small-signal voltage gain: From Figure 4.73 (b) Vo = ( g m1V 1 + g m 2V 2 ) RC Vs = V 1 + V 2 and V 1+ V 2 = 1 + g m1V 1 r 2 = V 1r 2 r 1 r 1 Then 1+ Vo = g m1V 1 + g m 2 r 1 Also r 2V 1 RC www.elsolucionario.net 200. 1+ Vs = V 1 + r 1 r 2V 1 r = V 1 1 + (1 + ) 2 r 1 or V 1 =Vs r 1 + (1 + ) 2 r 1 Now r 2 g m1 + g m 2 (1 + ) RC V r 1 Av = o = Vs r 1 + (1 + ) 2 r 1 so 2.63 0.377 + ( 38.1)(101) 265 ( 4 ) Av = 2.63 1 + (101) 265 or Av = 77.0 (d) Ri = r 1 + (1 + ) r 2 = 265 + (101)( 2.63)or Ri = 531 k TYU6.17 (a) dc analysis: For R1 + R2 + R3 = 100 k I1 =VCC 12 = = 0.12 mA R1 + R2 + R3 100Now VE1 = I CQ 2 RE = ( 0.5 )( 0.5 ) = 0.25 V VC1 = VE1 + VCEQ1 = 0.25 + 4 = 4.25 V and VC 2 = VC1 + VCEQ 2 = 4.25 + 4 = 8.25 V So RC =VCC VC 2 12 8.25 = = 7.5 k 0.5 I CQAlso VB1 = VE1 + VBE ( on ) = 0.25 + 0.7 = 0.95 V And R3 R3 VB1 = (12 ) VCC 0.95 = 100 R1 + R2 + R3 which yields R3 = 7.92 k We have VB 2 = VC1 + VBE ( on ) = 4.25 + 0.7 = 4.95 Vandwww.elsolucionario.net 201. R2 + R3 VB 2 = VCC R1 + R2 + R3 or R + 7.92 4.95 = 2 (12 ) 100 which yields R2 = 33.3 k Then R1 = 100 33.3 7.92 = 58.8 k (b) Small-signal transistor parameters: (100 )( 0.026 ) V r 1 = r 2 = T = I CQ 0.5or r 1 = r 2 = 5.2 k and I CQ 0.5 g m1 = g m 2 = = 0.026 VT or g m1 = g m 2 = 19.23 mA/V and ro1 = ro 2 = (c) Small-signal voltage gain: We have Vo = g m 2V 2 ( RC RL ) Also V 2 + g m 2V 2 = g m1V 1 r 2 or r V 2 = g m1V 1 2 and V 1 = Vs 1+ We find r V Av = o = g m 2 ( RC RL ) g m1 2 Vs 1+ = g m 2 ( RC RL ) 1+ We obtain 100 Av = (19.23) ( 7.5 2 ) 101 or Av = 30.1 TYU6.18 (a) dc analysis: with vs = 0 RTH = R1 R2 = 125 30 = 24.2 k R2 30 VTH = VCC = (12 ) 125 + 30 R1 + R2 or VTH = 2.32 Vwww.elsolucionario.net 202. Now VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE orI BQ =2.32 0.7 = 0.0250 mA 24.2 + ( 81)( 0.5 )and I CQ = ( 80 )( 0.025 ) = 2.00 mA Also 1+ VCEQ = VCC I CQ RC + RE 81 = 12 ( 2 ) 2 + ( 0.5 ) 80 or VCEQ = 6.99 VPower dissipated in RC: 2 PRC = I CQ RC = ( 2.0 ) ( 2 ) = 8.0 mW 2Power dissipated in RL: I LQ = 0 PRL = 0 Power dissipated in transistor: PQ = I BQVBEQ + I CQVCEQ = ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW (b) With vs = 18cos t ( mV ) VTr =I CQ=(80 )( 0.026 ) 2.0= 1.04 k We can write( RC RL )VP cos t r Power dissipated in RL: vce =pRL = =vce ( rms ) RL 1 1 2 2 1032 1 1 = ( RC RL ) VP 2 RL r 80 ( 2 2 ) ( 0.018) 1.04 22or pRL = 0.479 mW Power dissipated in RC: Since RC = RL = 2 k , we have pRC = 8.0 + 0.479 = 8.48 mW Power dissipated in transistor: From the text, we have 2 V pQ I CQVCEQ P ( RC RL ) r 2 22280 0.018 3 = ( 2 103 ) ( 6.99 ) ( 2 10 2 3 1.04 10 2 3)or pQ = 13.0 mW TYU6.19 (a) dc analysis:www.elsolucionario.net 203. RTH = R1 R2 = 53.8 10 = 8.43 k R2 10 VTH = VCC = (5) R1 + R2 53.8 + 10 or VTH = 0.7837 V Now 0.7837 0.7 I BQ = = 0.00993 mA 8.43 and I CQ = (100 )( 0.00993) = 0.993 mAWe have VCEQ = VCC I CQ RC or 2.5 = 5 ( 0.993) RC which yields RC = 2.52 k (b) Power dissipated in RC: 2 PRC = I CQ RC = ( 0.993) ( 2.52 ) 2or PRC = 2.48 mW Power dissipated in transistor: PQ I CQVCEQ = ( 0.993)( 2.5 ) or PQ = 2.48 mW (c) ac analysis: Maximum ac collector current: ic = ( 0.993) cos t ( mA ) average ac power dissipated in RC: 1 1 2 2 pRC = ( 0.993) RC = ( 0.993) ( 2.52 ) 2 2 or pRC = 1.24 mW Now pRC 1.24 Fraction = = = 0.25 PRC + PQ 2.48 + 2.48www.elsolucionario.net 204. Chapter 6 Problem Solutions 6.1 a. gm = r = r0 =I CQ VT VT I CQ=2 g m = 76.9 mA/V 0.026=(180 )( 0.026 ) 2 r = 2.34 kVA 150 = r0 = 75 k I CQ 2b. 0.5 g m = 19.2 mA/V 0.026 (180 )( 0.026 ) r = r = 9.36 k 0.5 150 r0 = r0 = 300 k 0.5gm =6.2 (a) gm = r = ro =I CQ VT VT I CQ=0.8 = 30.8 mA/V 0.026=(120 )( 0.026 ) 0.8= 3.9 KVA 120 = = 150 K I CQ 0.8(b) 0.08 = 3.08 mA/V 0.026 (120 )( 0.026 ) r = = 39 K 0.08 120 ro = = 1500 K 0.08gm =6.3 gm = r = r0 =I CQ VT VT I CQ 200 = =I CQ 0.026 I CQ = 5.2 mA(125)( 0.026 ) 5.2 r = 0.625 kVA 200 = r0 = 38.5 k I CQ 5.26.4www.elsolucionario.net 205. gm = r =I CQ 80 =VT VTI CQ 0.026 1.20 =I CQ I CQ = 2.08 mA ( 0.026 ) 2.08 = 966.5 (a) 2 0.7 = 0.0052 mA 250 I C = (120 )( 0.0052 ) = 0.624 mA I BQ =0.624 g m = 24 mA / V 0.026 (120 )( 0.026 ) r = r = 5 k 0.624 ro = gm = r 5 Av = g m RC = ( 24 )( 4 ) Av = 1.88 r + RB 5 + 250 v v vS = O = O vS = 0.426sin100t V Av 1.88(b) (c) 6.6 gm =I CQ VT, 1.08 I CQ 1.32 mA1.08 1.32 gm 41.5 g m 50.8 mA/V 0.026 0.026 (120 )( 0.026 ) VT r = ; r ( max ) = = 2.89 k I CQ 1.08 r ( min ) =(80 )( 0.026 ) 1.32= 1.58 k1.58 r 2.89 k6.7 a. r = 5.4 = VT I CQ=(120 )( 0.026 ) I CQ I CQ = 0.578 mA1 1 VCEQ = VCC = ( 5 ) = 2.5 V 2 2 VCEQ = VCC I CQ RC 2.5 = 5.0 ( 0.578 ) RC RC = 4.33 k I CQ0.578 = 0.00482 mA 120 VBB = I BQ RB + VBE ( on ) = ( 0.00482 )( 25 ) + 0.70 VBB = 0.820 V I BQ b.www.elsolucionario.net 206. r = gm = r0 = VT I CQ I CQ=0.578= 5.40 k0.578 = 22.2 mA/V 0.026=VT(120 )( 0.026 )VA 100 = = 173 k I CQ 0.578 r V0 = g m ( r0 RC ) V , V = VS r + RB ( r0 RC ) r Av = g m ( r0 RC ) = r + RB r + RB Av = (120 ) 173 4.33 5.40 + 25=(120 )( 4.22 ) 30.4 Av = 16.76.8 a. 1 VECQ = VCC = 5 V 2 VECQ = 10 I CQ RC 5 = 10 ( 0.5 ) RC RC = 10 k I CQ0.5 = 0.005 100 VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) VBB = 0.95 V I BQ b. gm = r = r0 =c.I CQ=VT VT I CQ0.5 g m = 19.2 mA/V 0.026=(100 )( 0.026 ) 0.5 r = 5.2 kVA = r0 = I CQ 0.5 Av = (100 )(10 ) A = 18.1 RC = v 5.2 + 50 r + RB6.9 10 4 = 1.5 mA 4 1.5 I BQ = = 0.015 mA 100 (100 )( 0.026 ) r = = 1.73 K 1.5 5sin t ( mV ) v ib = be = = 2.89sin t ( A ) r 1.73 k So I CQ =www.elsolucionario.net 207. iB ( t ) = I BQ + iEb = 15 + 2.89sin t ( A )iC1 ( t ) = iB iC1 ( t ) = 1.5 + 0.289sin t ( mA )vC ( t ) = 10 iC1 ( t ) RC = 10 [1.5 + 0.289sin t ] ( )vC1 ( t ) = 4 1.156sin t ( v ) Av =vC ( t )vbe ( t )=1.156 Av = 231 0.0056.10 vo = 1.2sin t ( V ) iC ( t ) RC + vo = 0 iC ( t ) = iC ( t ) = 0.60sin t ( mA ) ib ( t ) =iC ( t )1.2sin t 2= 6sin t ( A )vbe ( t ) = ib ( t ) rg m r = 100 =2K 50 vbe ( t ) = 12sin t ( mV ) r =6.11 a. I CQ I EQ VCEQ = 5 = 10 I CQ ( RC + RE ) = 10 I CQ (1.2 + 0.2) I CQ = 3.57 mA 3.57 = 0.0238 mA 150 R2 = RTH = ( 0.1)(1 + ) REI BQ = R1VTH ( 0.1)(151)( 0.2 ) = 3.02 k1 RTH (10) 5 R1VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 5 1 (3.02)(10) 5 = ( 0.0238)(3.02) + 0.7 + (151)( 0.0238)( 0.2) 5 R1 1 ( 30.2 ) = 1.50 R1 = 20.1 k R1 20.1R2 = 3.02 R2 = 3.55 k 20.1 + R2 b. (150 )( 0.026 ) rp = = 1.09 k 3.57 3.57 gm = = 137 mA/V 0.026www.elsolucionario.net 208. V0 V VSrgmV R1R2RC REAv =(150 )(1.2 ) 2 RC =2 Av = 2 5.75 rp + (1 + ) RE 1.09 + (151)( 0.2 )6.12 a. R2 50 VTH = VCC = (12 ) = 10 V R1 + R2 50 + 10 RTH = R1 R2 = 50 10 = 8.33 k I BQ =12 0.7 10 = 0.0119 mA 8.33 + (101)(1)I CQ = 1.19 mA, I EQ = 1.20 mA VECQ = 12 (1.20 )(1) (1.19 )( 2 ) VECQ = 8.42 V iC 41.198.4212ECb. V0 V VS rgmV R1R2RC REwww.elsolucionario.net 209. rp =(100 )( 0.026 )1.19 V0 = g mVp RC= 2.18 kV VS = 2 Vp p + g mVp RE rp r + (1 + ) RE = Vp rp 2 (100 )( 2 ) 2 RC = Av = 2 1.94 Av = rp + (1 + ) RE 2.18 + (101)(1)c.Approximation: Assume rp does not vary significantly.RC = 2 k 5% = 2.1 k or 1.9 k RE = 1 k 5% = 1.05 k or 0.95 kFor RC ( max ) = 2.1 k and RE ( min ) Av = (100 )( 2.1)2.18 + (101)( 0.95 )= 2.14For RC ( min ) = 1.9 k and RE ( max ) = 1.05 k Av = (100 )(1.9 )2.18 + (101)(1.05 )= 1.76So 1.76 Av 2.146.13 (a) VCC = 1+ I CQ RE + VECQ + I CQ RC 101 12 = I CQ (1) + 6 + I CQ ( 2 ) 100 so that I CQ = 1.99 mA 1.99 = 0.0199 mA 100 = ( 0.1)(1 + ) RE = ( 0.1)(101)(1) = 10.1 k I BQ = RTH R2 1 1 VTH = V = R V = (10.1)(12 ) R1 + R2 CC R1 TH CC R1 VCC = (1 + ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 121.2 12 = (101)( 0.0199)(1) + 0.7 + ( 0.0199)(10.1) + R1 which yields R1 = 13.3 k and R2 = 41.6 k (b)Av =2 (100 )( 2 ) 2 RC = Av = 2 1.95 rp + (1 + ) RE 1.31 + (101)(1)6.14www.elsolucionario.net 210. I CQ = 0.25 mA, I EQ = 0.2525 mA I BQ = 0.0025 mAI BQ RB + VBE ( on ) + I EQ ( RS + RE ) 5 = 0 ( 0.0025)( 50 ) + 0.7 + ( 0.2525)( 0.1 + RE ) = 5 RE = 16.4 kVE = ( 0.0025 )( 50 ) 0.7 = 0.825 V VC = VCEQ + VE = 3 0.825 = 2.175 V 5 2.175 RC = RC = 11.3 k 0.25 RC Av = r + (1 + ) RS r =(100 )( 0.026 )= 10.4 k 0.25 (100 )(11.3) Av = Av = 55.1 10.4 + (101)( 0.1)Ri = RB r + (1 + ) RS = 50 10.4 + (101)( 0.1) Ri = 50 20.5 Ri = 14.5 k6.15 (a) VCC > I CQ ( RC + RE ) + VCEQ9 = I CQ ( 2.2 + 2 ) + 3.75 So that I CQ = 1.25 mAAssume circuit is to be designed to be bias stable. RTH = R1 R2 = ( 0.1)(1 + ) RE = ( 0.1)(121)( 2 ) = 24.2 1.25 I BQ = = 0.01042 mA 120 1 VTH = RTH VCC = I BQ RTH + VBE ( on ) + I BQ (121)( RE ) R1 1 ( 24.2 )( 9 ) = ( 0.01042 )( 24.2 ) + 0.7 + ( 0.01042 )(121)( 2 ) R1 = 0.2522 + 0.7 + 2.5216 = 3.474 R1 = 62.7 K62.7 R2 = 24.2 62.7 + R2R2 = 39.4 K(b)www.elsolucionario.net 211. 1.25 = 48.08 mA / V 0.026 (120 )( 0.026 ) rp = = 2.50 k 1.25 100 ro = = 80 k 1.25gm =Vo ISVrR1R2roRCRLgmVVo = 2 g mVp ( ro RC RL )Vp = I S ( R1 R2 rp )Then Rm =Vo = 2 g m ( R1 R2 rp )( ro RC RL ) IsRm = 2 48.08 ( 24.2 2.5 )( 80 2.2 1) = 2 48.08 ( 2.266 )( 0.6816 )or Rm =Vo = 74.3 k = 74.3 V / mA Is6.16 a. I EQ = 0.80 mA, I BQ =0.80 = 0.0121 mA 66I CQ = 0.788 mA 0.3 RB = 24.8 k 0.0121 V ( 5 ) 5 3 = RC = 2.54 k RC = C I CQ 0.788 VB = I BQ RB RB =b. 0.788 = 30.3 mA / V 0.026 ( 65)( 0.026) r = = 2.14 k 0.788 75 r0 = = 95.2 k 0.788 RC r0 i0 = g V , V = vS R r +R m L C 0 gm =Gf = RC r0 i0 = gm R r +R vS L C 0 2.54 95.2 = ( 30.3) 2.54 95.2 + 4 G f = 11.6 mA/Vwww.elsolucionario.net 212. 6.17 (a) I CQ = 0.8 mA I BQ = 0.00667 mA I BQ RS + 0.7 + (121) I BQ RE 15 = 0 ( 0.01667 )( 2.5) + 0.7 + (121)( 0.00667 ) RE = 15 RE = 17.7 KVE = ( 0.00667 )( 2.5 ) 0.7 = 0.717 V VC = 0.717 + 7 = 6.283 V 15 6.283 RC = = 10.9 K 0.8 (120 )( 0.026 ) 0.8 gm = = 30.77 mA/V r = = 3.9 K 0.026 0.8 r vo = g m ( RC RL ) v v = vS r + RS Av = ( RC RL ) r + RS= (120 ) (10.9 5 ) 3.9 + 2.5Av = 64.3 (b) For RS = 0 0.7 + (121)( 0.00667 ) RE = 15 RE = 17.7 K VE = 0.7 VC = 0.7 + 7 = 6.3 15 6.3 RC = 10.9 K 0.8 ( RC RL ) Av = = 30.77 (10.9 5 ) rRC =Av = 1056.18 (a) 15 = ( 81) I BQ (10 ) + 0.7 + I BQ ( 2.5 ) 15 0.7 I BQ = = 0.0176 mA 2.5 + ( 81)(10 ) I CQ = 1.408 mA(80 )( 0.026 ) 1.408 = 54.15 mA/V r = 0.026 1.408 r = 1.48 K gm =RS V0 IS VS V rgmVRCIoRLwww.elsolucionario.net 213. Vo = g mV ( RC RL ) Av = i AI = o = iS ( RC RL ) r + RS= ( 80 ) ( 5 5 ) 1.48 + 2.5 Av = 50.3 RC g mV RC + RL = RC V RC + RL rAI = 40vo ( t ) = ( 50.3)( 4sin t ) vo ( t ) = 0.201sin t ( V ) 4 sin t ( mV ) = 1.005sin t ( A ) 2.5 + 1.48 io = 40.2 sin t ( A ) is =(b) 15 0.7 = 1.43 mA 10 80 = (1.43) = 1.412 mA 81 I EQ = I CQgm =(80 )( 0.026 ) 1.412 = 54.3 mA/V r = = 1.47 K 0.026 1.412Av = g m ( RC RL ) = ( 54.3) ( 5 5 ) Av = 136 RC 5 AI = = 80 AI = 40 5+5 RC + RL vo ( t ) = ( 136 )( 4sin t ) vo ( t ) = 544sin t vo ( t ) = 0.544sin t ( V ) is ( t ) =4sin t ( mV )= 2.72sin t ( A ) 1.47 k io ( t ) = ( 40 )( 2.72sin t ) io ( t ) = 109sin t ( A )6.19 RTH = R1 R2 = 27 15 = 9.64 K R2 15 VTH = VCC = ( 9 ) = 3.214 V R1 + R2 15 + 27 V VBE ( on ) 3.214 0.7 2.514 I BQ = TH = = RTH + (1 + ) RE 9.64 + (101)(1.2 ) 130.84I BQ = 0.0192 mA I CQ = 1.9214 mA gm =(100 )( 0.026 ) 1.92 = 73.9 mA/V r = = 1.35 K 0.026 1.92www.elsolucionario.net 214. RS V0 IS VSro =VRTHrgmVr0100 = 52.1 K 1.92 r R V = TH r R +R S TH = 1.35 9.64 = 1.184 K(Vo = g mV r0 RC RL r RTH 1.184 V = VS 1.184 + 10 = 0.1059VS)(Av = ( 73.9 ) ( 0.1059 ) 52.1 2.2 2 VS )= ( 73.9 ) ( 0.1059 ) ( 52.1 1.0476 )= ( 73.9 ) ( 0.1059 ) (1.027 ) Av = 8.04 ro RC g mV r R +R I L o C AI = o = V IS RTH r ro RC AI = g m ( RTH r ) r R +R L o C ro RC = 52.1 2.2 = 2.11 K RTH r = 9.64 1.35 = 1.184 K 2.11 AI = ( 73.9 ) (1.184 ) 2.11 + 2 AI = 44.9 Ri = RTH r = 9.64 1.35 Ri = 1.184 K6.20 a. 0.35 = 0.00347 mA 101 VB = 2 I B RB = 2 ( 0.00347 )(10 ) VB = 2 0.0347 V I E = 0.35 mA, I B =VE = VB VBE ( on ) VE = 2 0.735 Vb.www.elsolucionario.netRCI0RL 215. VC = VCEQ + VE = 3.5 0.735 = 2.77 V b 100 IC = IE = ( 0.35 ) = 0.347 mA 101 1+ b V 1 VC 5 2.77 RC = = RC = 6.43 k IC 0.347 (c) RB rp Av = 2 g m R r R r + R ( C o ) S B 0.347 100 gm = = 13.3 mA/V , ro = = 288 k 0.026 0.347 (100 )( 0.026 ) = 7.49 k rp = 0.347 RB rp = 10 7.49 = 4.28 k 4.28 Av = 2 (13.3) ( 6.43 288 ) Av = 2 81.7 4.28 + 0.1 d. RB rp Av = 2 g m R r R r + R ( C 0 ) S B p RB rp = 10 7.49 = 4.28 k 4.28 Av = 2 (13.3) ( 6.43 288 ) Av = 2 74.9 4.28 + 0.5 6.21 a. RTH = R1 R2 = 6 1.5 = 1.2 k R2 + 1.5 VTH = V = ( 5 ) = 1.0 V 1.5 + 6 R1 + R2 V VBE ( on ) 1.0 0.7 I BQ = TH = = 0.0155 mA RTH + (1 + ) RE 1.2 + (181)( 0.1)I CQ = 2.80 mA, I EQ = 2.81 VCEQ = V + I CQ RC I EQ RE= 5 ( 2.8 )(1) ( 2.81)( 0.1) VCEQ = 1.92 Vb.(180 )( 0.026 ) rp = 1.67 k 2.80 2.80 gm = g m = 108 mA/V, r0 =` 0.026 (c) R1 R2 rp Av = 2 g m ( RC RL ) R1 R2 rp + RS R1 R2 rp = 6 1.5 1.67 = 0.698 k V rp = 0.698 Av = 2 (108 ) (1 1.2 ) Av = 2 45.8 0.698 + 0.2 www.elsolucionario.net 216. 6.22 a. 9 = I EQ RE + VEB ( on ) + I BQ RS 0.75 = 0.00926 mA I EQ = 0.75 mA, I BQ = 81 I CQ = 0.741 mA 9 = ( 0.75 ) RE + 0.7 + ( 0.00926 )( 2 ) RE = 11.0 k b. VE = 9 ( 0.75 )(11) = 0.75 V VC = VE VECQ = 0.75 7 = 6.25 V RC =VC ( 9 ) I CQ=9 6.25 RC = 3.71 k 0.741c. rp Av = 2 g m ( RC RL r0 ) rp + RS (80 )( 0.026 ) rp = = 2.81 k 0.741 80 r0 = = 108 kV 0.741 2 80 Av = ( 3.71 10 108 ) 2.81 + 2 Av = 2 43.9d. Ri = RS + rp = 2 + 2.81 Ri = 4.81 k 6.23 I BQ =4 0.7 = 0.00647 5 + (101)( 5 )I CQ = 0.647 mA80 h fe 120, 10 h0e 20 mSa.2.45 k hie 3.7 k low gainhigh gainRS V0 IS VS V rgmVRCIoRLwww.elsolucionario.net 217. 1 V0 = 2 h fe I b RC RL hoe RB ?S V R + RS Ib = B RTH + hie RTH = RB RS = 5 1 = 0.833 kHigh-gain 5 VS 5 +1 Ib = = 0.1838VS 0.833 + 3.7 Low-gain 5 VS 5 +1 Ib = = 0.2538VS 0.833 + 2.45 1 1 For hoe = 10 Rc RL = 4 4 0.010 hoe = 100 2 = 1.96 k1 4 4 = 50 2 = 1.92 k 0.020 = (120 )( 0.1838 )(1.96 ) = 43.2 = ( 80 )( 0.2538 )(1.92 ) = 39.0For hoe = 20 Av Avmax min39.0 Av 43.2b. Ri = RB hie = 5 3.7 = 2.13 kV or Ri = 5 2.45 = 1.64 k 1.64 Ri 2.13 k R0 =1 1 4 = 100 4 = 3.85 k RC = 0.010 hoe1 4 = 50 4 = 3.70 k 0.020 3.70 R0 3.85 k or R0 =6.24 VCC 10 VRC R1 o RS 1 k CC s R2 RECEwww.elsolucionario.net 218. Assume an npn transistor with b = 100 and VA = . Let VCC = 10 V . 0.5 = 50 0.01 Bias at I CQ = 1 mA and let RE = 1 k Av =For a bias stable circuit RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)(1) = 10.1 k 1 1 101 VTH = RTH VCC = (10.1)(10 ) = R1 R1 R1 1 I BQ = = 0.01 mA 100 VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 101 = ( 0.01)(10.1) + 0.7 + (101)( 0.01)(1) R1 which yields R1 = 55.8 k and R2 = 12.3 k Now rp =(100 )( 0.026 ) 1= 2.6 k 1 = 38.46 mA/V 0.026 Vo = g mVp RCgm = R1} R2 } rp 10.1} 2.6 where Vp = Vs = .Vs 10.1} 2.6 + 1 R1} R2 } rp + RS or Vp = 0.674 Vs V Then Av = o = ( 0.674 ) g m RC = ( 0.674 )( 38.46 ) RC = 50 Vs which yields RC = 1.93 k With this RC, the dc bias is OK. Finish Design, Set RC = 2 KRE = 1 KR1 = 56 K R2 = 12 KRTH = R1 R2 = 9.88 K R2 12 VTH = VCC = (10 ) = 1.765 V 12 + 56 R1 + R2 1.765 0.7 I BQ = = 9.60 A 9.88 + (101)(1) I CQ = 0.9605 mA r = RTH(100 )( 0.026 ) 0.9605 r = 2.125 K= 2.707 Kgm =0.9605 = 36.94 0.026 RTH r 2.125 V = Vi = Vi = ( 0.680 ) Vi 2.125 + 1 RTH r + RS Av = ( 0.680 ) g m RC = ( 0.680 )( 36.94 )( 2 ) = 50.2www.elsolucionario.net 219. Design specification met. 6.25 a. I BQ =6 0.7 = 0.0169 mA 10 + (101)( 3)I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) (1.69 )( 6.8 ) (1.71)( 3) VCEQ = 5.38 Vb. 1.69 g m = 65 mA/V 0.026 (100 )( 0.026 ) rp = rp = 1.54 kV , 1.69 (c) ( RC RL ) RB Rib Av = r + (1 + ) RE RB Rib + RS gm =r0 = Rib = r + (1 + ) RE = 1.54 + (101)(3) = 304.5 k RB Rib = 10 304.5 = 9.68 k Then Av = (100 )( 6.8 6.8 ) 9.68 Av = 1.06 1.54 + (101)( 3) 9.68 + 0.5 RC i0 = ( ib ) RC + RL RB ib = RB + r + (1 + ) RE iS RB R + r + (1 + ) R E B 10 6.8 = (100 ) 10 + 1.54 + (101)( 3) Ai = 1.59 6.8 + 6.8 (d) Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k RC Ai = ( ) RC + RL(e) Av = Av =2 b ( RC RL )rp + (1 + b ) RE2 (100 ) ( 6.8} 6.8 ) 1.54 + (101)( 3) Av = 2 1.12Ai = same as ( c ) Ai = 2 1.596.26www.elsolucionario.net 220. ie vCE vbe vCEgmvriso vCEgmvr=vCe 1 = g m vCe g m 1 So re = rp r0 gm 6.27 Let b = 100, VA = VCCRC R1 o RS 100 CC s R2 RELet VCC = 2.5 VP = ( I R + I C ) VCC 0.12 = ( I R + I C )( 2.5 ) I R + I C = 48 mA, Let I R = 8mA, I C = 40 mAR1 + R2 > I BQ =VCC 2.5 312.5 k = IR 840 = 0.4 mA 100www.elsolucionario.net 221. Let RE = 2 k . For a bias stable circuit RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)( 2 ) = 20.2 k 1 VTH = RTH VCC = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE R1 1 ( 20.2 )( 2.5 ) = ( 0.0004 )( 20.2 ) + 0.7 + (101)( 0.0004 )( 2 ) R1 which yields R1 = 64 k V and R2 = 29.5 k (100 )( 0.026 )r =0.04Av 65 k Neglect RSVo 2 b RC > Vs r + (1 + b ) RE10 =2 100 RC RC = 26.7 k 65 + (101)( 2 )With this RC , dc biasing is OK. 6.28 100 = 20. 5 Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31. Let RS = 0. Need an input resistance ofNeed a voltage gain of5 102 3 = 25 103 = 25 k 0.2 102 6 Ri = RTH Rib . Let RTH = 50 k , Rib = 50 k Ri =Rib = rp + (1 + b ) RE > (1 + b ) RERib 50 = = 0.495 k 1 + b 101 Let RE = 0.5 k V , VCC = 10 V , I CQ = 0.2 mA For b = 100, RE =0.2 = 0.002 mA 100 = I BQ RTH + VBE ( on ) + (1 + ) I BQ REThen I BQ = VTH1 1 RTH VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002 )( 0.5) R1 R1 which yields R1 = 555 k and R2 = 55 k Now Av =(100)( 0.026) RC = 13 k , r = r + (1 + ) RE 0.2So 20 = (100 ) RC13 + (101)( 0.5) RC = 12.7 k [Note: I CQ RC = ( 0.2 )(12.7 ) = 2.54 V. So dc biasing is OK.] 6.29www.elsolucionario.net 222. VCC 10 VR1RECCso R2 RCb = 80, Av =2 b RC rp + (1 + b ) REFirst approximation: R ( Av ) C = 10 RC = 10 RE RE Set RC = 12 REVEC VCC I C ( RC + RE ) = 10 I C (13RE ) 1 For VEC = VCC = 5 2 5 = 10 I C (13RE ) For I C = 0.7 mA I E = 0.709, I B = 0.00875 mA RE = 0.55 k RC = 6.6 k Bias stable R1 R2 = RTH = ( 0.1)(1 + ) RE = ( 0.1)(81)( 0.55 ) = 4.46 k 1 10 = ( 0.709 )( 0.55 ) + 0.7 + ( 0.00875 )( 4.46 ) + ( 4.46 )(10 ) R1 8.87 =1 ( 4.46 ) R1 = 5.03 k R15.03R2 = 4.46 R2 = 39.4 k 5.03 + R2 10 10 = = 0.225 mA R1 + R2 5.03 + 39.4 0.7 + 0.225 0.925 mA from VCC source. Now r = Av =(80 ) ( 0.026 ) 0.7 (80 )( 6.6 )= 2.97 k2.97 + ( 81)( 0.55 )= 11.16.30www.elsolucionario.net 223. 5VRCR1CC2 oCC1RL 10 K s R2CE RE5V = 120Let I CQ = 0.35 mA, I EQ = 0.353 mA I BQ = 0.00292 mA Let RE = 2 k. For VCEQ = 4 V 10 = 4 + ( 0.35) RC + ( 0.353)( 2) RC = 15.1 k, r = Av = ( RC RL ) r(120 )( 0.026 ) 8.91 k 0.35 (120 ) (15.1 10 ) 8.91Av = 81.0For bias stable circuit: R1 R2 = RTH = ( 0.1)(1 + ) RE = ( 0.1)(121)( 2 ) = 24.2 k R2 1 VTH = (10) 5 = RTH (10 ) 5 R1 R1 + R2 VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 5 1 ( 24.2 )(10 ) 5 = ( 0.00292 )( 24.2 ) + 0.7 + (121)( 0.00292 )( 2 ) 5 R1 1 ( 242 ) = 1.477, R1 = 164 k R1 164 R2 = 24.2 R2 = 28.4 k 164 + R2 10 = 0.052, 0.35 + 0.052 = 0.402 mA 164 + 28.4 So bias current specification is met.6.31 From Prob. 6.12,www.elsolucionario.net 224. RTH = R1} R2 = 10}50 = 8.33 k R2 50 VTH = (12 ) = (12 ) = 10 V 50 + 10 R1 + R2 12 0.7 10 = 0.0119 mA I BQ = 8.33 + (101)(1) I CQ = 1.19 mA, I EQ = 1.20 mA VECQ = 12 (1.19 )( 2 ) (120 )(1) = 8.42 V1.1918.4211 12For 1 vEC 11 DvEC = 11 8.42 = 2.58 Output voltage swing = 5.16 V (peak-to-peak) 6.32 I BQ =5 0.7 = 0.00315 mA 50 + (101)( 0.1 + 12.9)I CQ = 0.315 mA, I EQ = 0.319 mA VCEQ = ( 5 + 5 ) ( 0.315 )( 6 ) ( 0.319 )(13) VCEQ = 3.96 VAC load line 1 Slope 6.1 K 0.3153.96101 v 6.1 eC For iC = 0.315 0.05 = 0.265 vEC = 1.62iC = vEC ( min ) = 3.96 1.62 = 2.34Output signal swing determined by current:www.elsolucionario.net 225. Max. output swing = 3.24 V peak-to-peak 6.33 From Problem 4.18, I CQ = 1.408 mA, I EQ = 1.426 mA (a) VECQ = 30 (1.408 )( 5 ) (1.426 )(10 ) = 8.7 V IC (mA) AC load line 1 Slope RC RL 1 2.5 k1.4088.7EC (V)vEC ( max ) = 8.7 + I C ( 2.5 ) = 8.7 + (1.408 )( 2.5 ) = 12.22Set vEC ( max ) = 12 = 8.7 + I C ( 2.5 ) I C = 1.32 mA So vEC (peak-to-peak) = 2(12 8.7) = 6.6 V (b) iC (peak-to-peak) = 2(1.32) = 2.64 mA 6.34 I EQ I BQ VE VC VECQ= 0.80 mA, I CQ = 0.792 mA = 0.00792 mA = 0.7 + ( 0.00792 )(10 ) = 0.779 V = I CQ RC 5 = ( 0.792 )( 4 ) 5 = 2 1.83 V = 0.779 ( 1.83) = 2.61 VLoad line: Assume VE remains constant at 0.78 V IC (mA) AC load line 1 Slope RC RL 1 2.5 k1.4088.7EC (V)21 ? ec v 2 kV Collector current swing = 0.792 0.08 = 0.712 mA Dvec = ( 0.712 )( 2 ) = 1.424 V DiC =Output swing determined by current. Max. output swing = 2.85 V peak-to-peak 2.85 4 = 0.712 mA peak-to-peakSwing in i0 current =www.elsolucionario.net 226. 6.35 I BQ =6 0.7 = 0.0169 mA 10 + (101)( 3)I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) (1.69 )( 6.8 ) (1.71)( 3) VCEQ = 5.38 VAC load line 1 Slope 3.4 3 1 6.4 K1.695.38221 DiC = 2 Dvce 6.4For vce ( min ) = 1 V, Dvce = 5.38 1 = 4.38 V DiC =4.38 = 0.684 mA 6.4Output swing limited by voltage: vce = Max. swing in output voltage = 8.76 V peak-to-peak 1 iC i0 = 0.342 mA 2 or i0 = 0.684 mA (peak-to-peak) i0 =6.36 AC load line 1 Slope 1.05 K2.65Q-pointICQVCEQro =9100 I CQNeglect ro as (E) approx. dc load line VCE = 9 I C ( 3.4 )www.elsolucionario.net 227. I C = I CQ 0.1 VCE = VCEQ 1 Also VCE = I C ( RC RL ) = I C (1.05 ) Or VCEQ 1 = ( I CQ 0.1) (1.05 )Substituting the expression for the dc load line. 9 I CQ ( 3.4 ) 1 = ( I CQ 0.1) (1.05 ) 8.105 = I CQ ( 4.45 ) I CQ = 1.821 mA VCEQ = 2.81 V 1.821 I BQ = = 0.01821 100 RTH = ( 0.1)(101)(1.2 ) = 12.12 K 1 1 VTH = RTH VCC = (12.12 ) ( 9 ) = ( 0.01821) (12.12 ) + 0.7 + (101)( 0.01821)(1.2 ) R1 R1 = 0.2207 + 0.7 + 2.20705 R1 = 34.9 K R2 = 18.6 K 34.9 R2 = 12.12 34.9 + R26.37 dc load line 5 4.55 mA 1 0.1 AC load line 1 Slope 11.2 1 0.545 KICQVCEQ5For maximum symmetrical swing iC = I CQ 0.25 vCE = VCEQ 0.5 and iC = I CQ 0.25 =VCEQ 0.51 vCE 0.545 k0.545 VCEQ = 5 I CQ (1.1)0.545 ( I CQ 0.25 ) = 5 I CQ (1.1) 0.5 ( 0.545 + 1.1) I CQ = 5 0.5 + 0.136 I CQ = 2.82 mA,I BQ = 0.0157 mARTH = R1 R2 = ( 0.1)(1 + ) RE= ( 0.1)(181)( 0.1) = 1.81 kVTH =1 RTH V + = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE R1www.elsolucionario.net 228. 1 (1.81)( 5 ) = ( 0.0157 )(1.81) + 0.7 + (181)( 0.0157 )( 0.1) R1 1 ( 9.05 ) = 1.013 R1 = 8.93 k R1 8.93R2 = 1.81 R2 =2.27 k 8.93 + R26.38 I CQ = 0.647 mA , VCEQ > 10 ( 0.647 )( 9 ) = 4.18 V DiC = I CQ = 0.647 mA So DvCE = DiC ( 4} 4 ) = ( 0.647 )( 2 ) = 1.294 VVoltage swing is well within the voltage specification. Then DvCE = 2 (1.294 ) = 2.59 V peak-to-peak 6.39 a. RTH = R1} R2 = 10}10 = 5 k R2 10 VTH = (18 ) 9 = (18 ) 9 = 0 10 + 10 R1 + R2 0 0.7 ( 9 ) = 0.0869 mA I BQ = 5 + (181)( 0.5 )I CQ = 15.6 mA, I EQ = 15.7 mAVCEQ = 18 (15.7 )( 0.5 ) VCEQ = 10.1 Vb.AC load line 1 Slope 0.50.3 1 0.188 K15.610.1c. r =(180 )( 0.026 ) 15.6 (1 + )( RE18= 0.30 kRL ) R1 R2 Rib r + (1 + ) ( RE RL ) R1 R2 Rib + RS Rib = r + (1 + )( RE RL ) = 0.30 + (181)( 0.5 Av = 0.3) or Rib = 34.2 k R1 R2 Rib = 5 34.2 = 4.36 k Av =(181)( 0.5 0.3) 4.36 Av 0.3 + (181)( 0.5 0.3) 4.36 + 1 = 0.806d.www.elsolucionario.net 229. Rib = rp + (1 + b ) ( RE } RL )Rib = 0.30 + (181)( 0.188 ) Rib = 34.3 k Ro = RErp + R1} R2 } RS 0.3 + 5}1 Ro = 6.18 = 0.5 1+ b 1816.40 a. RTH = R1} R2 = 10}10 = 5 k R2 VTH = ( 10 ) = 2 5 V R1 + R2 VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 10 I BQ =2 5 0.7 ( 10 )5 + (121)( 2 )= 0.0174 mAI CQ = 2.09 mA, I EQ = 2.11 mA VCEQ = 10 ( 2.09 )(1) ( 2.11)( 2 ) VCEQ = 3.69 Vb. AC load line 1 Slope 22 1 1K2.093.69c. r =10(120 )( 0.026 )= 1.49 k 2.09 (1 + ) ( RE RL ) R1 R2 Rib Av = r + (1 + ) ( RE RL ) R1 R2 Rib + RS Rib = r + (1 + ) ( RE RL ) = 1.49 + (121) ( 2 2) Rib = 122.5 k , Av =R1 R2 Rib = 5 122.5 = 4.80 k (121) ( 2 2) 4.80 Av 1.49 + (121) ( 2 2) 4.80 + 5 = 0.484d. Rib = r + (1 + b ) ( RE } RL ) Rib = 1.49 + (121) ( 2} 2 ) 1 Rib = 122 k Ro = REr + R1} R2 } RS 1.49 + 5}5 1 Ro = 32.4 =2 1+ b 1216.41 a.www.elsolucionario.net 230. RTH = R1 R2 = 60 40 = 24 k R2 40 VTH = VCC = ( 5) = 2 V 40 + 60 R1 + R2 5 0.7 2 I BQ = = 0.0130 mA 24 + ( 51)( 3) I CQ = 0.650 mA, I EQ = 0.663 mA VECQ = 5 I EQ RE = 5 ( 0.663)(3) VECQ = 3.01 Vb.1.63 AC load line 1 Slope 51 50 34 1 1.75 K0.6553.01c.( 50 )( 0.026 )80 = 2 k, r0 = = 123 k 0.650 0.65 Define RL = RE RL r0 = 3 4 123 = 1.69 kr =Av = (1 + ) RL r (1 + ) RL=( 51)(1.69 )2 + ( 51)(1.69 ) RE r0 Ai = (1 + ) I b R r +R L E 0 Av = 0.977 RTH Ib = I S RTH + Rib Rib = r + (1 + ) RL = 2 + ( 51)(1.69 ) = 88.2 RE r0 = 3 r0 = 3 123 = 2.93 24 2.93 Ai = ( 51) Ai = 4.61 2.93 + 4 24 + 88.2 d. Rib = r + (1 + ) RE RL r0 = 2 + ( 51)(1.69 ) Rib = 88.2 k r 2 RE = 3 = 0.0392 3 1+ 51 R0 = 38.7 e. Assume variations in r and r0 have negligible effects R1 = 60 5% R1 = 63 k, R1 = 57 k R0 =R2 = 40 5% R2 = 42 k,R2 = 38 kRE = 3 5%RE = 3.15 k, RE = 2.85 kRL = 4 5%RL = 4.2 k,RL = 3.8 kwww.elsolucionario.net 231. RE r0 RTH Ai = (1 + ) R r + R R + R L TH ib E 0 Rib = r + (1 + ) ( RE RL r0 )RTH ( max ) = 25.2 k, RTH ( min ) = 22.8 k Rib ( max ) = 92.5 k, Rib ( min ) = 84.0 k RE ( max ) , RL ( min ) , Rib = 88.6 k RE ( min ) , RL ( max ) , Rib = 87.4 kRE ( max ) r0 = 3.07 kRE ( min ) r0 = 2.79 kFor RE ( min ) , RL ( max ) , RTH ( min ) 22.8 2.79 Ai = ( 51) Ai = 4.21 2.79 + 4.2 22.8 + 87.4 For RE ( max ) , RL ( min ) , RTH ( max ) 25.2 3.07 Ai = ( 51) Ai = 5.05 3.07 + 3.8 25.2 + 88.6 6.42 (a) 0.5 = 0.00617 mA 81 VB = I BQ RB = ( 0.00617 )(10 ) VB = 0.0617 VI BQ =VE = VB + 0.7 VE = 0.7617 V(b) 80 I CQ = ( 0.5 ) = 0.494 mA 81 I CQ 0.494 gm = = g m = 19 mA / V VT 0.026r = ro = VT I CQ=(80 )( 0.026 ) 0.494 r = 4.21 k VA 150 ro = 304 k = I CQ 0.494(c) RSVs ISV S RB Vrro gmVVo RL IoFor RS = 0 V Vo = + g mV ( RL ro ) r www.elsolucionario.net 232. Vo 1+ ( RL ro ) r Now Vs + V = Voso that V =or Vs = Vo V = Vo +Vo 1+ ( RL ro ) r We find(1 + )( RL ro ) (81)( 0.5 304 ) = r + (1 + )( RL ro ) 4.21 + ( 81)( 0.5 304 ) (81)( 0.5 ) Av = 0.906 4.21 + ( 81)( 0.5 ) Rib = r + (1 + )( RL ro ) 4.21 + ( 81)( 0.5 ) = 44.7 k Av =Vo Vs= RB ro Ib = I s and I o = (1 + ) I b RB + Rib ro + RL Then RB ro Io = (1 + ) Is RB + Rib ro + RL 10 Ai ( 81) (1) Ai = 14.8 10 + 44.7 (d) RB + Rib 10 44.7 Vs = V = V = 0.803) Vs R R + R s 10 44.7 + 2 s ( s B ib Then Av = ( 0.803)( 0.906 ) Av = 0.728 Ai =Ai = 14.8 (Unchanged)6.43 (a) I CQ = 1.98 mA ro =r =(100 )( 0.026 ) 1.98= 1.313 KVA 100 = I CQ 1.98= 50.5 K r + RS 1.31 + 10 ro = 50.5 Ro = 112 1+ 101 0.112 50.5 Ro 112 (b) From Equation 4.68 (1 + ) ( ro RL ) 100 Av = ro = = 50.5 K 1.98 r + (1 + ) ( ro RL ) Ro =(i)www.elsolucionario.net 233. RL = 0.5 K(101) ( 50.5 0.5) 1.31 + (101) ( 50.5 0.5 ) (101)( 0.4951) Av = 0.974 Av = 1.31 + (101)( 0.4951) Av =(ii) RL = 5 K Av =ro RL = 50.5 5 = 4.5495(101)( 4.55) Av = 0.997 1.31 + (101)( 4.55 )6.44 5 0.7 I CQ = 1.293 mA = 1.303 3.3 (125 )( 0.026 ) r = = 2.51 K 1.293 1.293 gm = = 49.73 mA/V 0.026 (a) Rib = r + (1 + ) ( RE RL ) = 2.51 + (126 ) ( 3.3 1) I EQ =Rib = 99.2 K Ro = REr 2.51 = 3.3 = 3.3 0.01992 1+ 126Ro = 19.8 (b) v 2sin t is ( t ) = 20.2sin t ( A ) is = s = Rib 99.2 veb ( t ) = is ( t ) r = ( 20.2 )( 2.51) sin t veb ( t ) = 50.6sin t ( mV )(126 ) ( 3.3 1) (1 + ) ( RE RL ) (126 )( 0.7674 ) = = r + (1 + ) ( RE RL ) 2.51 + (126 ) ( 3.3 1) 2.51 + (126 )( 0.7674 ) Av = 0.9747 vo ( t ) = 1.95sin t ( V ) v (t ) io ( t ) = 1.95sin t ( mA ) io ( t ) = o Av =RL6.45 a. I EQ = 1 mA , VCEQ = VCC I EQ RE5 = 10 (1)( RE ) RE = 5 k1 = 0.0099 mA 101 10 = I BQ RB + VBE ( on ) + I EQ RE 10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) RB = 434 kI BQ =b.www.elsolucionario.net 234. b 0RBr =RE(100 )( 0.026 ) 0.99 (1 + ) RE= 2.63 k(101)( 5 ) v0 = = = 0.995 vb r + (1 + ) RE 2.63 + (101)( 5 ) vb =v0 4 = vb = 4.02 V peak-to-peak at base 0.995 0.995 RS bS RBRibRib = r + (1 + ) RE = 508 k RB Rib = 434 508 = 234 k vb =RB Rib RB Rib + RS vS =vb = 0.997vS vS =234vS 234 = vS 234 + 0.7 234.74.02 vS = 4.03 V peak-to-peak 0.997c. Rib = r + (1 + ) ( RE RL )Rib = 2.63 + (101) ( 5 1) = 86.8 k RB Rib = 434 86.8 = 72.3 k 72.3 vb = vS = 0.99vS = ( 0.99 )( 4.03) 72.3 + 0.7 vb = 3.99 V peak-to-peak(1 + )( RE RL ) vb r + (1 + )( RE RL ) (101)( 0.833) = ( 3.99 ) 2.63 + (101)( 0.833)v0 =v0 = 3.87 V peak-to-peak6.46www.elsolucionario.net 235. RTH = R1 R2 = 40 60 = 24 k 60 VTH = (10 ) = 6 V 60 + 40 6 0.7 = 75 I BQ = = 0.0131 mA 24 + ( 76 )( 5 ) I CQ = 0.984 mA = 150 I BQ =6 0.7 = 0.00680 mA 24 + (151)( 5 )I CQ = 1.02 mA = 75 r =( 75 )( 0.026 )0.984 = 150 r = 3.82 k= 1.98 k = 75 Rib = r + (1 + )( RE RL ) = 65.3 k = 150 Rib = 130 k (1 + )( RE RL ) R1 R2 Rib Av = r + (1 + )( RE RL ) R1 R2 Rib + RS For = 75, R1 R2 Rib = 40 60 65.3 = 17.5 k Av =( 76 )( 0.833) 17.5 Av = 0.789 1.98 + ( 76 )( 0.833) 17.5 + 4For = 150, R1 R2 Rib = 40 60 130 = 20.3 k Av =(151)( 0.833) 20.3 Av = 0.811 3.82 + (151)( 0.833) 20.3 + 4So 0.789 Av 0.811 RE RTH Ai = (1 + ) RE + RL RTH + Rib = 75 24 5 Ai = ( 76 ) Ai = 17.0 5 + 1 24 + 65.3 = 150 5 24 Ai = (151) Ai = 19.6 6 24 + 130 17.0 Ai 19.66.47 (a)www.elsolucionario.net 236. I 9 = E (100 ) + VBE ( on ) + I E RE 1+ 9 0.7 IE = 100 + RE 1+ 8.3 = 2.803 mA 100 +1 51 8.3 = 200 I E = = 5.543 mA 100 +1 201 2.80 I E 5.54 mA = 50 I E =VE = I E RE , = 50, VE = 2.80 V = 200, VE = 5.54 V = 50, I CQ = 2.748 mA, r = 0.473 K (b) = 200, I CQ = 5.515 mA, r = 0.943 K Ri = RB r + (1 + ) RE RL = 50 Ri = 100 0.473 + ( 51)(1 1) = 100 25.97 = 20.6 K = 200 Ri = 100 0.943 + ( 201)(1 1) = 100 101.4 = 50.3 K From Fig. (4.68) (1 + ) ( RE RL ) Ri Av = r + (1 + ) ( RE RL ) Ri + RS =( 51) (1 1) 20.6 0.473 + ( 51) (1 1) 20.6 + 10 = 50 Av = 0.661 = 200 Av =( 201) (1 1) 50.3 0.943 + ( 201) (1 1) 50.3 + 10 Av = 0.8266.48 Vo = (1 + ) I b RL Vs Ib = r + (1 + ) RL so Av =(1 + ) RL r + (1 + ) RLFor = 100, RL = 0.5 k r =(100 )( 0.026 ) 0.5= 5.2 k www.elsolucionario.net 237. (101)( 0.5 ) = 0.9066 5.2 + (101)( 0.5 )Then Av ( min ) =Then = 180, RL = 500 k r =(180 )( 0.026 ) 0.5Then Av ( max ) 9.36 k (181)( 500 ) = 0.9999 9.36 + (181)( 500 )6.49 RibISIbV S gmV Ibr R1R2 RERL I0 RE I 0 = (1+ ) I b RE + RL R1 R2 Ib = I S R1 R2 + Rib Rib = r + (1 + )( RE RL ) VCC = 10 V, For VCEQ = 5 V1+ 5 = 10 I CQ RE = 80, For RE = 0.5 k I CQ = 9.88 mA, I EQ = 10 mA, I BQ = 0.123 mA r =(80 )( 0.026 )= 0.211 k 9.88 Rib = 0.211 + ( 81)( 0.5 0.5 ) Rib = 20.46 k Ai = RE R1 R2 I0 = (1 + ) IS RE + RL R1 R2 + Rib R1 R2 1 8 = ( 81) 2 R1 R2 + 20.46 0.1975 R1} R2 + 20.46 = R1} R2 R1} R2 5.04 kwww.elsolucionario.net 238. VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 1 ( 5.04 )(10 ) = ( 0.123)( 5.04 ) + 0.7 + (10 )( 0.5 ) R1 = 7.97 k R1 7.97 R2 = 5.04 R2 = 13.7 k 7.97 + R2 Now Ro = REr 0.211 = 0.5 or Ro = 2.59 1+ b 81(b) Rib = 0.211 + (81) ( 0.5 2) = 32.6 k 0.5 5.04 Ai = ( 81) = ( 81)( 0.2 )( 0.134 ) 0.5 + 2 5.04 + 32.6 Ai = 2.176.50 Ri = RTH Rib where Rib = r + (1 + ) RE 5 3.5 VCEQ = 3.5, I CQ = 0.75 mA 2 (120 )( 0.026 ) r = = 4.16 k 0.75 Rib = 4.16 + (121) ( 2 ) = 246 k Then Ri = 120 = RTH 246 RTH = 234 k 0.75 = 0.00625 mA I BQ = 120 VTH = I BQ RTH + VBE ( on ) + (1+ ) I BQ RE 1 1 RTH VCC = ( 234)(5) = ( 0.00625) ( 234) + 0.7 + (121)( 0.00625 )( 2 ) R1 R1 which yields R1 = 318 k and R2 = 886 k 6.51 a. Let RE = 24 and VCEQ = 1 VCC = 12 V I EQ = 212 = 0.5 A 24I CQ = 0.493 A, I BQ = 6.58 mA r =( 75)( 0.026 ) 0.493= 3.96 RebIsIb VVS gmV Ibr R1 R2 Rrn REIoRLwww.elsolucionario.net 239. RE I 0 = (1 + ) I b RE + RL RTH Ib = I S RTH + Rib Rib = r + (1 + ) ( RERL )= 3.96 + ( 76 )( 24 8 ) Rib = 460 Ai = RE RTH I0 = (1 + ) IS RE + RL RTH + Rib 24 RTH 8 = ( 76) 24 + 8 RTH + 460 RTH 0.140 = RTH = 74.9 (Minimum value) RTH + 460 dc analysis: 1 VTH = RTH VCC R1 = I BQ RTH + VBE ( on ) + I EQ RE1 ( 74.9 )( 24 ) = ( 0.00658)( 74.9 ) + 0.70 + ( 0.5 )( 24 ) R1 = 13.19 R1 = 136 ,136 R2 = 74.9 R2 = 167 136 + R2b.0.493AC load line 1 Slope 248 1 612241 iC = vce 6 For iC = 0.493 vce = ( 0.493)( 6 ) Max. swing in output voltage for this design = 5.92 V peak-to-peakc. R0 =r 3.96 RE = 24 = 0.0521 24 R0 = 52 m 1+ 766.52 The output of the emitter follower is RL vo = vTH RL + Ro www.elsolucionario.net 240. Ro THORLFor vO to be within 5% for a range of RL , we have RL ( min )RL ( min ) + Ro= ( 0.95 )RL ( max )RL ( max ) + Ro4 10 = ( 0.95 ) which yields Ro = 0.364 k 4 + Ro 10 + Ro r + R1 R2 RS We have Ro = RE ro 1+ The first term dominates Let R1 R2 RS RS , then r + RS r +4 0.364 = 1+ 1+ Ro or 0.364 =0.364 r VT 4 4 + = + 1 + 1 + I CQ (1 + ) 1 + VT 4 + I CQ 1 + The factorV 4 4 4 = 0.044 to = 0.0305. We can set Ro 0.32 = T is in the range of I CQ 91 131 1+ Or I CQ = 0.08125 mA. To take into account other factors, set I CQ = 0.15 mA, I BQ =0.15 = 0.00136 mA 1105 = 33.3 k 0.15 Design a bias stable circuit. R2 1 VTH = (10) 5 = ( RTH )(10) 5 R1 R1 + R2 For VCEQ 5 V , set RE =RTH = ( 0.1)(1 + ) RE = ( 0.1)(111)(33.3) = 370 k VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE 51 ( 370 )(10 ) 5 = ( 0.00136 )( 370 ) + 0.7 + (111)( 0.00136 )( 33.3) 5 R 1 which yields R1 = 594 k and R2 = 981 k SoNow Av =(1 + ) ( RE RL ) RTH Rib r + (1 + ) ( RE RL ) RTH Rib + RS Rib = r + (1 + ) ( RE RL )and r= VT I CQFor = 90, RL = 4 k , r = 15.6 k , Rib = 340.6 k Av =( 91)( 33.3 4 ) 370 340.6 Av = 0.9332 15.6 + ( 91)( 33.3 4 ) 370 340.6 + 4www.elsolucionario.net 241. For = 90, RL = 10 k Rib = 715.4 k Av =( 91)( 33.3 10 ) 370 715.4 Av = 0.9625 15.6 + ( 91)( 33.3 10 ) 370 715.4 + 4For = 130, RL = 4 k r = 22.5 k , Rib = 490 k Av =(131)( 33.3 4 ) 370 490 Av = 0.9360 22.5 + (131)( 33.3 4 ) 370 490 + 4For = 130, RL = 10 k Rib = 1030 k (131)( 33.3 10 ) 370 1030 Av = 0.9645 22.5 + (131)( 33.3 10 ) 370 1030 + 4 Now vO ( min ) = Av ( min ) .vS = 3.73sin t vO ( max ) = Av ( max ) .vS = 3.86sin t Av =vO = 3.5% vO6.53 2 2 PAVG = iL ( rms ) RL 1 = iL ( rms )(12 ) so iL ( rms ) = 0.289 A iL ( peak ) = 2 ( 0.289 ) iL ( peak ) = 0.409 AvL ( peak ) = iL ( peak ) RL = ( 0.409 )(12 ) = 4.91 V4.91 = 0.982 5 With RS = 10 k , we will not be able to meet this voltage gain requirement. Need to insert a buffer or an Need a gain ofop-amp voltage follower (see Chapter 9) between RS and CC1 . 1 Set I EQ = 0.5 A, VCEQ = (12 ( 12 ) ) = 8 V 3 24 = I EQ RE + VCEQ = ( 0.5 ) RE + 8 RE = 32 50 ( 0.5 ) = 0.49 A 51 VT ( 50 )( 0.026 ) r = = = 2.65 I CQ 0.49 Let = 50, I CQ =Rib = r + (1 + ) ( RE RL ) = 2.65 + ( 51) ( 32 12 ) Rib = 448 Av =(1 + ) ( RE RL ) ( 51) ( 32 12 ) = = 0.994 r + (1 + ) ( RE RL ) 2.65 + ( 51) ( 32 12 )So gain requirement has been met.www.elsolucionario.net 242. 0.49 = 0.0098 A = 9.8 mA 50 24 10 I B = 98 mA Let I R R1 + R2 I BQ =So that R1 + R2 = 245 VTH =R2 ( 24 ) 12 = I BQ RTH + VBE ( on ) + I EQ RE 12 R1 + R2( 0.0098) R1 R2 R2 + 0.7 + ( 0.5 )( 32 ) 245 ( 24 ) = 245 Now R1 = 245 R2 So we obtain 2 4 105 R2 + 0.0882 R2 16.7 = 0 which yields R2 = 175 and R1 = 70 6.54 (a) RTH = R1 R2 = 25.6 10.4 = 7.40 k R2 10.4 VTH = (VCC ) = (18 ) = 5.2 V 10.4 + 25.6 R1 + R2 VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ REI BQ =5.2 0.7 = 0.0117 mA 7.40 + (126 )( 3)Then I CQ = 1.46 mA and I EQ = 1.47 mA VCEQ = VCC I CQ RC I EQ REVCEQ = 18 (1.46 )( 4 ) (1.47 )( 3) VCEQ = 7.75 V(b) r =(125) ( 0.026 )= 2.23 k 1.46 1.46 gm = = 56.2 mA / V 0.026 Re Vo Ie IsRSREr IbIbRCRLRTHwww.elsolucionario.net 243. r + RTH 2.23 + 7.40 = = 0.0764 k 1+ 126Re = Ie = ( RS RE )(RSRE ) + Re Is = (100 3)(1003) + 0.0764 Isor I e = ( 0.974 ) I s Vo = I c ( RC RL ) = 1+ I e ( RC RL ) Vo 125 = ( 0.974 )( RC RL ) = ( 0.974 )( 4 4 ) Is 1+ 126 V Then Rm = o = 1.93 k = 1.93 V / mA IsThen(c)() ()Vs = I s RS R E Re = I s 100 3 0.0764 = I s ( 0.0744 ) Vs 0.0744 V V which yields o = o ( 0.0744 ) = 1.93 I s Vsor I s =or Av =6.55 (a)Vo = 25.9 Vs ( RC RL )Av =, RL = 12 k , = 100r + R1 R2Let R1 R2 = 50 k , I CQ = 0.5 mA VTH = I BQ RTH + VBE ( on ) + (1+ ) I BQ RE(100 )( 0.026 ) 0.5 = 0.005 mA, r = = 5.2 k 100 0.5 1 1 RTH VCC = ( 50 )(12 ) = ( 0.005 )( 50 ) + 0.7 + (101)( 0.005 )( 0.5 ) R1 R1I BQ =which yields R1 = 500 k and R2 = 55.6 k Av =(b) I CQ(100 )(1212 )5.2 + 50= 10.9, Design criterion is met.= 0.5 mA, I EQ = 0.505 mAVCEQ = 12 ( 0.5)(12) ( 0.505)( 0.5) VCEQ = 5.75 V 0.5 = 19.23 mA / V 0.026 Av = (19.23) (12 12 ) Av = 115 Av = g m ( RC RL ) , g m =6.56 a. Emitter currentwww.elsolucionario.net 244. I EQ = I CC = 0.5 mA 0.5 = 0.00495 mA 101 VE = I EQ RE = ( 0.5 )(1) VE = 0.5 VI BQ =VB = VE + VBE ( on ) = 0.5 + 0.7 VB = 1.20 V VC = VB + I BQ RB = 1.20 + ( 0.00495 )(100 ) VC = 1.7 Vb. r =(100 )( 0.026 ) = 5.25 k (100 )( 0.00495 ) (100 )( 0.00495 )gm 19.0 mA/V0.026 RigmVVSRS RE VrV0 RBRLgmVRE V RE RS S RS RE V Vo = g mV ( RB V = RE RErRL )Rie VS = ( 0.4971) VS Rie + RSVo = (19 )( 0.4971) VS (100 1) Av = 9.37c.VX IXgmV RE VrIX =VX VX + g mV , V = VX RE rIX 1 1 1 = = + + gm VX Ri RE r or Ri = RE r1 1 = 1 5.253 gm 19Ri = 0.84 0.05252 Ri = 49.4 6.57 (a)I EQ = 1 mA, I CQ = 0.9917 mAwww.elsolucionario.net 245. VC = 5 ( 0.9917 )( 2 ) = 3.017 V VE = 0.7 V VCEQ = 3.72 V(b) Av = g m ( RC RL ) 0.9917 = 38.14 mA/V 0.026 Av = ( 38.14 ) ( 2 10 ) Av = 63.6gm =6.58 (a) 10 0.7 = 0.93 mA 10 = 0.921 mAI EQ = I CQVECQ = 20 ( 0.93)(10 ) ( 0.921)( 5 ) VECQ = 6.10 V(b) 0.921 = 35.42 mA/V 0.026 Av = g m ( RC RL ) = ( 35.42 ) ( 5 50 )gm =Av = 1616.59 (a)I EQ = 0.93 mA, I CQ = 0.921 mAVECQ = 6.10 V (b)gm =0.921 = 35.42 mA/V r = 2.82 K 0.026From Eq. 6.90 ( RC RL ) r R R Av = g m 1 + E S RS ( 35.42 ) ( 50 5 ) 2.82 = 101 10 0.1 0.1 ( 35.42 )( 4.545 ) Av = [0.0218] 0.1 Av = 35.1 6.60 (a) 60 I CQ = (1) I CQ = 0.984 mA 61 1 VCEQ = I BQ RB + VBE ( on ) = (100 ) + 0.7 61 VCEQ = 2.34 V(b)www.elsolucionario.net 246. Av = g m(RRL ) r RS RS 1 + B0.984 = 37.85 mA/V 0.026 r = 1.59 Kgm =( 37.85) (100 2 ) 1.59 61 0.05 0.05 = 1484 0.0261 0.05 Av = 25.4Av =6.61is ( peak ) = 2.5 mA, Vo ( peak ) = 5 mVvo 5 102 3 = = 2 103 = 2 k is 2.5 102 6 From Problem 4.54 RS RE Vo = ( RC RL ) R R +R Is 1+ ie S E Let RC = 4 k , RL = 5 k , RE = 2 k So we need Rm =Now = 120, so we have RS RE 120 2= ( 4 5) R R +R 121 ie S E RS RE = 0.9075 Then RS RE + Re RS RE = 2.204 R R +R ie S ERS RE = 50 2 = 1.923 k , so that Rie = 0.196 k Assume VCEQ = 3 VVCC I CQ ( RC + RE ) + VCEQ5 = I CQ ( 4 + 2 ) + 3 I CQ = 0.333 mA r =(120 )( 0.026 )= 9.37 k 0.333 r + RTH 9.37 + RTH 0.196 = Rie = 1+ 121 which yields RTH = 14.35 k Now VTH = I BQ RTH + VBE ( on ) + I EQ RE 1 121 = 0.00833 mA, I EQ = (1) = 1.008 mA 120 120 1 1 = RTH VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 ) R1 R1I BQ = VTHwhich yields R1 = 25.3 k and R2 = 33.2 k 6.62 a.www.elsolucionario.net 247. 20 0.7 = 1.93 mA 10 = 1.91 mAI EQ = I CQVECQ = VCC + VEB ( on ) I C RC= 25 + 0.7 (1.91)( 6.5 ) VECQ = 13.3 Vb. RSVS RieISh fe IbV0Ie RERChieRLIbNeglect effect hoe From Problem 6-16, assume 2.45 hie 3.7 k 80 h fe 120 Vo = ( h fe I b ) ( RC Rie =RL ) RE hie , Ie = IS 1 + h fe RE + Rie I VS Ib = e , I S = 1+ h RS + RE fe Rie h fe RE 1 Av = R RL ) 1+ h ( C fe RE + Rie RS + RE High gain device: hie = 3.7 k, h fe = 120Rie = RE Rie 3.7 = 0.0306 k 121 Rie = 10 0.0306 = 0.030510 1 120 Av = ( 6.5 5 ) Av = 2.711 121 10 + 0.0306 1 + 0.0305 Low gain device: hie = 2.45 k, h fe = 80 Rie = RE2.45 = 0.03025 k 81 Rie = 10 0.03025 = 0.030210 1 80 Av = ( 6.5 5 ) Av = 2.70 So Av constant 81 10 + 0.03025 1 + 0.0302 2.70 Av 2.71c. Ri = RE Rie We found 0.0302 Ri 0.0305 k Neglecting hoe , Ro = RC = 6.5 k 6.63 a.Small-signal voltage gainwww.elsolucionario.net 248. Av = g m ( RCRL ) 25 = g m ( RC 1)For VECQ = 3 V VC = VECQ + VEB ( on ) = 3 + 0.7 VC = 2.3 VCC I CQ RC + VC = 0 I CQ =5 2.3 2.7 = = I CQ RC RCFor I CQ = 1 mA, RC = 2.7 k 1 = 38.5 mA/V 0.026 Av = ( 38.5 )( 2.7 1) = 28.1gm =Design criterion satisfied and VECQ satisfied. 101 IE = (1) = 1.01 mA 100 VEE = I E RE + VEB ( on ) RE =b. r = VT I CQ=(100)( 0.026) 15 0.7 RE = 4.26 k 1.01 r = 2.6 k, g m = 38.5 mA/V, ro = 6.64 a. R2 20 VTH 1 = VCC = (10 ) VTH 1 = 2.0 V 20 + 80 R1 + R2 RTH 1 = R1 R2 = 20 80 = 16 k I B1 =2 0.7 = 0.0111 mA 16 + (101)(1)I C1 = 1.11 mA g m1 = r 1 =(100)( 0.026) 1.111.11 g m1 = 42.74 mA/V 0.026 r 1 = 2.34 k r01 = 1.11 R4 15 VTH 2 = VCC = (10 ) = 1.50 V 15 + 85 R3 + R4 RTH 2 = R3 R4 = 15 85 = 12.75 kr01 =IB2 =1.50 0.70 = 0.01265 mA 12.75 + (101)( 0.5 )I C 2 = 1.265 mA g m 2 = r 2 =(100 )( 0.026 ) 1.261.265 g m2 = 48.65 mA/V 0.026 r 2 = 2.06 k r02 = b. Av1 = g m1 RC1 = ( 42.7 )( 2 ) Av1 = 85.48 Av 2 = g m 2 ( RC 2 RL ) = ( 48.5) ( 4 4) Av 2 = 97.3 c.Input resistance of 2nd stagewww.elsolucionario.net 249. Ri 2 = R3 R4 r 2 = 15 85 2.06 = 12.75 2.06 Ri 2 = 1.773 k 1 = g m1 ( RC1 Ri 2 ) = ( 42.7 ) ( 2 1.77B) Av Av1 = 40.17 Overall gain: Av = ( 40.17 )( 97.3) Av = 3909 If we had Av1 Av 2 = ( 85.48)( 97.3) = 8317 Loading effect reduces overall gain6.65 a. R2 12.7 VTH 1 = VCC = (12) VTH 1 = 1.905 V 12.7 + 67.3 R1 + R2 RTH 1 = R1 R2 = 12.7 67.3 = 10.68 k 1.905 0.70 I B1 = = 0.00477 mA 10.68 + (121)( 2 ) I C1 = 0.572 mA 0.572 g m1 = g m1 = 22 mA/V 0.026 (120 )( 0.026 ) r 1 = r 1 = 5.45 k 0.572 r01 = r01 = 0.572 R4 45 VTH 2 = VCC = (12) VTH 2 = 9.0 V 45 + 15 R3 + R4 RTH 2 = R3 R4 = 15 45 = 11.25 k 9.0 0.70 I B2 = = 0.0405 mA 11.25 + (121)(1.6) I C2 = 4.86 mA 4.86 gm2 = g m 2 = 187 mA/V 0.026 (120 )( 0.026 ) r 2 = r 2 = 0.642 k 4.86 r02 = b. I E1 = 0.577 mA VCEQ1 = 12 ( 0.572 ) (10 ) ( 0.577 ) ( 2 ) VCEQ1 = 5.13 V I E 2 = 4.90 VCEQ 2 = 12 ( 4.90 )(1.6 ) VCEQ 2 = 4.16 Vwww.elsolucionario.net 250. Q1 AC load line 1 107.92 1 4.42 KSlope 0.5725.1312Q24.86AC load line 1 Slope 1.60.25 1 0.216 K4.1612Ri 2 = R3 R4 Rib Rib = r 2 + (1 + ) ( RE 2 RL ) = 0.642 + (121) (1.6 0.25 ) Rib = 26.8 Ri 2 = 15 45 26.8 Ri 2 = 7.92 k c. Av1 = g m1 ( RC1 Ri 2 ) = ( 22 )(10 7.92 ) Av 2 = 97.2(1 + )( RE 2 RL ) r 2 + (1 + )( RE 2 RL ) (121)( 0.216 ) = = 0.976 0.642 + (121)( 0.216 ) Overall gain = ( 97.2 )( 0.976 ) = 94.9 Av 2 =d. RiS = R1 R2 r 1 = 67.3 12.7 5.45 RiS = 3.61 k Ro =r 2 + RS RE 2 where 1+ RS = R3 R4 RC1 = 15 45 10 RS = 5.29 k Ro =0.642 + 5.29 1.6 0.049 1.6 Ro = 47.6 121e. 1 vce , iC = 4.86 0.216 k vce = ( 4.86 )( 0.216 ) = 1.05 V Max. output voltage swing = 2.10 V peak-to-peakiC =6.66 (a)www.elsolucionario.net 251. I R1 = IR2 IC 25 2 ( 0.7 )= 72 mA 0.050 0.7 = = 1.4 mA 0.5 = ( 72 1.4 ) I C 2 = 69.9 mA 1+ 69.9 = 0.699 mA 100 = (1.4 + 0.699 ) I C1 = 2.08 mA 1+ IB2 = I C1(b)Vs V1r1gm1V1r2 0.5 kgm2V2 V2 Vo 50 Vs = V 1 + V 2 + Vo (1) V V Vo = 2 + 2 + g m 2V 2 ( 0.05 ) 0.5 r 2 r 2 =(100 )( 0.026 )= 0.0372 k 69.9 69.9 gm2 = = 2688 mA / V 0.026 V 1 1 Vo = V 2 + + 2688 ( 0.05 ) so that (1) V 2 = o 135.8 0.5 0.0372 (2)www.elsolucionario.net 252. V 1 V V + g m1V 1 = 2 + 2 r 1 0.5 r 2 r 1 =(100 )( 0.026 )= 1.25 k 2.08 2.08 g m1 = = 80 mA / V 0.026 1 1 1 V 1 + 80 = V 2 + 1.25 0.5 0.0372 V V 1 ( 80.8 ) = V 2 ( 28.88 ) = o ( 28.88 ) or (2) V 1 = Vo ( 0.00261) 136.7 V V Then Vs = Vo ( 0.00261) + o + Vo = Vo (1.00993) or Av = o = 0.990 Vs 136.7(c) Rib = r 1 (1 + ) [ Rx ] Ix Vx 0.5 k V2r2gm2V2Vo 50 1 V 2 V 2 1 + = V 2 + 0.5 r 2 0.5 r 2 Vo V V 2 = x = I x + g m 2V 2 0.05 0.05Ix = 1 Ix + gm2 Vx 1 0.05 I x = V 2 + gm2 = 0.05 0.05 1 1 + 0.5 r 2 We findVx = Rx = 4.74 k IxThen Rib = 1.25 + (101) ( 2.89 ) Rib = 480 k www.elsolucionario.net 253. V1r1gm1V1r2 0.5 kgm2V2 V2 Ix 50 VxTo find Ro: (1) (2) (3)Ix =Vx V g m 2V 2 2 0.05 0.5 r 2V 1 V 2 = 1 + g m1V 1 ( 0.5 r 2 ) = V 1 + 80 ( 0.5 0.0372 ) or V 2 = ( 2.77 ) V 1 1.25 r 1 V 1 + V 2 + Vx = 0 V 1 + ( 2.77 ) V 1 + Vx = 0so that V 1 = ( 0.2653) Vxand V 2 = ( 2.77 ) ( 0.2653) Vx = ( 0.735 ) Vx Now I x = Vx 1 V 2 g m 2 + 0.05 0.5 r 2 So that I x = Vx Vx 1 + ( 0.735 ) Vx 2688 + = 0.496 which yields Ro = Ix 0.05 0.5 0.0372 6.67 a. RTH = R1 R2 = 335 125 = 91.0 k R2 VTH = VCC R1 + R2 125 = (10 ) = 2.717 V 125 + 335 VTH = I B1 RTH + VBE1 + VBE 2 + I E 2 RE 2I E 2 = (1 + ) I E1 = (1 + ) I B1 2I B1 =2.717 1.40 91.0 + (101) (1) 2 I B1 = 0.128 I C1 = 12.8 I C 2 = I E1 = (1 + ) I B1 = (100 )(101)( 0.128 ) I C 2 = 1.29 m, I E 2 = 1.31 m I RC = I C 2 + I C1 = 1.29 + 0.0128 = 1.30 mwww.elsolucionario.net 254. VC = 10 I RC RC = 10 (1.30 )( 2.2 ) = 7.14 V VE = I E 2 RE 2 = (1.30 )(1) = 1.30 V VCE 2 = 7.14 1.30 = 5.84 V VCE1 = VCE 2 VBE 2 = 5.84 0.7 VCE1 = 5.14 V Summary: I C1 = 12.8 I C 2 = 1.29 m VCE1 = 5.14 V VCE 2 = 5.84 V b. 0.0128 g m1 = = 0.492 m / V 0.026 1.292 gm2 = = 49.7 m / V 0.026 Rib Ib VSV0 V1 R1 R2r1gm1V1 RC V2r2gm2V2V0 = ( g m1V 1 + g m 2V 2 ) RC VS = V 1 + V 2 , V 1 = VS V 2 V V 2 = 1 + g m1V 1 r 2 r 1 1+ V 2 = V 1 r 2 r 1 V0 = g m1 (VS V 2 ) + g m 2V 2 RC V0 = g m1VS + ( g m 2 g m1 ) V 2 RC r V 2 = (VS V 2 )(1 + ) 2 r 1 r r V 2 1 + (1 + ) 2 = VS (1 + ) 2 r 1 r 1 www.elsolucionario.net 255. r VS (1 + ) 2 r 1 R V0 = g m1VS + ( g m 2 g m1 ) C r 1 + (1 + ) 2 r 1 V0 Av = VS 2.01 ( 49.7 0.492 )(101) 203 2.2 = ( 0.492 ) + 2.01 1 + (101) 203 Av = 55.2 c. Ris = R1 R2 Rib Rib = r 1 + (1 + ) r 2 = 203 + (101)( 2.01) = 406 k Ris = 91 406 = 74.3 k = Ris R0 = RC = 2.2 k6.68 R0 Ix V1r1 ro1 gm1V1VxVA V2r2gm2V2ro2Vx Vx VA + + g m1V 1 ro 2 ro1(1)I x = g m 2V 2 +(2)Vx VA VA + g m1V 1 = ro1 r 1 r 2V 2 = VA = V 1 (3) Then from (2) 1 Vx 1 = VA + g m1 + r ro1 r 1 r 2 o1 1 Vx Vx VA 1 1 + g m1VA or I x = Vx + + VA g m 2 g m1 ro 2 ro1 ro1 ro1 ro1 ro 2 Solving for VA from Equation (2) and substituting into Equation (1), we find (1) I x = g m 2VA +www.elsolucionario.net 256. 1 1 + g m1 + V ro1 r 1 r 2 Ro = x = Ix 1 1 1 1 1 + gm2 + g m1 + + ro 2 ro1 r 1 r 2 ro1 r 1 r 2 For = 100, VA = 100 V , I C1 = I Bias = 1 mA ro1 = ro 2 = r 1 = r 2 =100 = 100 k 1 (100 )( 0.026 ) 1g m1 = g m 2 =Then Ro 2.6 k 1 = 38.46 mA/V 0.026 1 1 + 38.46 + 100 2.6 2.6 1 1 1 1 1 + 38.46 + + 38.46 + 100 100 2.6 2.6 100 2.6 2.6 or Ro = 50.0 k Now I C 2 = 1 mA, I Bias = 0 Replace I Bias byI = C 2 , I C1 0.01 mA 1+ 1+ IC 2100 100 = 100 k , ro1 = = 10, 000 k 1 0.01 1 = 38.46 mA/V , g m1 = 0.3846 mA/V gm2 = 0.026 (100 )( 0.026 ) = 2.6 k , r 1 = 260 k r 2 = 1 Then Ro = 66.4 k ro 2 =6.69 a. RTH = R1 R2 = 93.7 6.3 = 5.90 k R2 VTH = VCC R1 + R2 6.3 = (12 ) = 0.756 V 6.3 + 93.7 0.756 0.70 I BQ = = 0.00949 mA 5.90 I CQ = 0.949 mA VCEQ = 12 ( 0.949 )( 6 ) VCEQ = 6.305 VTransistor: PQ I CQVCEQ = ( 0.949 )( 6.305 ) PQ = 5.98 mW 2 RC : PR = I CQ RC = ( 0.949 ) ( 6 ) PR = 5.40 mW 2b.www.elsolucionario.net 257. 2AC load line 1 Slope 6105 1 5.68 K0.9496.3112100 r0 = = 105 k 0.949 Peak signal current = 0.949 mAV0 ( max ) = ( 5.68 )( 0.949 ) = 5.39 VPRC =2 2 1 V0 ( max ) 1 ( 5.39 ) = PRC = 2.42 mW RC 2 2 6 6.70 (a) 10 = I BQ RB + VBE ( on ) + (1 + ) I BQ RE 10 0.7 I BQ = = 0.00369 mA 100 + (121)( 20 ) I CQ = 0.443 mA, I EQ = 0.447 mA For RC : PRC = ( 0.443) (10 ) PRC = 1.96 mW 2For RE : PRE = ( 0.447 ) ( 20 ) PRE = 4.0 mW 2(b) iC = 0.667 0.443 = 0.224 mA 1 1 2 2 ( iC ) RC = ( 0.224 ) (10 ) 2 2 = 0.251 mWThen P RC = P RC6.71 a. I BQ =10 0.7 = 0.00596 mA 50 + (151)(10 )I CQ = 0.894 mA, I EQ = 0.90 mAVECQ = 20 ( 0.894 )( 5 ) ( 0.90 )(10 ) VECQ = 6.53 V PQ I CQVECQ = ( 0.894 )( 6.53) PQ = 5.84 mW2 PRC I CQ RC = ( 0.894 ) ( 5 ) PRC = 4.0 mW 22 PRE I EQ RE = ( 0.90 ) (10 ) PRE = 8.1 mW 2b.www.elsolucionario.net 258. AC load line 1 52 1 1.43 KSlope 0.8946.53201 iC = vec 1.43 k iC = 0.894 vec = ( 0.894 )(1.43) = 1.28 V 5 i0 = iC = 0.639 mA 5+2 1 2 PRL = ( 0.639 ) ( 2 ) PRL = 0.408 mW 2 1 2 PRC = ( 0.894 0.639 ) ( 5 ) PRC = 0.163 mW 2 PRE = 0 PQ = 5.84 0.408 0.163 PQ = 5.27 mW6.72 I BQ =10 0.70 = 0.00838 mA 100 + (101)(10 )I CQ = 0.838 mA, I EQ = 0.846 mAVCEQ = 20 ( 0.838 )(10 ) ( 0.846 )(10 ) VCEQ = 3.16 V AC load line 1 RE RLr0Slope 0.8383.1620100 r0 = = 119 k 0.838 Neglecting base currents: a. RL = 1 k slope =1 1 = 10 1 119 0.902 k1 Vce 0.902 k iC = 0.838 Vce = ( 0.902 )( 0.838 ) = 0.756 V iC =1 ( 0.756 ) PRL = 0.286 mW 2 1 2PRL =b.www.elsolucionario.net 259. RL = 10 k slope =1 1 = 10 10 119 4.80For iC = 0.838 vce = ( 0.838 )( 4.80 ) = 4.02 1 ( 3.16 ) PRL = 0.499 mW 2 10 2Max. swing determined by voltage PRL =6.73 a. I BQ =10 0.7 = 0.00838 mA 100 + (101)(10 )I CQ = 0.838 mA, I EQ = 0.846 mAVCEQ = 20 ( 0.838 )(10 ) ( 0.846 )(10 ) VCEQ = 3.16 V PQ I CQVCEQ = ( 0.838 )( 3.16 ) PQ = 2.65 mW 2 PRC I CQ RC = ( 0.838 ) (10 ) PRC = 7.02 mW 2b. AC load line 1 RC RL 1 1 101 0.909 KSlope 0.8383.16201 iC = vce 0.909 k For iC = 0.838 vce = ( 0.909 )( 0.838 ) = 0.762 V RC 10 i0 = iC = iC = 0.762 mA 10 + 1 RC + RL 1 2 PRL = ( 0.762 ) (1) PRL = 0.290 mW 2 1 2 PRC = ( 0.838 0.762 ) (10 ) PRC = 0.0289 mW 2 PQ = 2.65 0.290 0.0289 PQ = 2.33 mWwww.elsolucionario.net 260. Chapter 7 Exercise Solutions EX7.1 (a) (i) RS = RP = 4 k 1 1 = = rS ( RS + RP ) CS CS =1 2 f ( RS + RP )=1 2 ( 20 )( 4 + 4 ) 10 3CS = 0.995 F(ii) RP rS T ( j ) = RS + RP 1 + 2 rS2 rS = ( RS + RP ) CS = 7.96 10 3 RP 4 = = 0.5 RS + RP 4 + 4 f = 40 Hz T ( j ) =( 0.5)( 2 )( 40 ) ( 7.96 10 3 )()1 + 2 ( 40 ) 7.96 10 3 T ( j ) = 0.447f = 80 Hz T ( j ) =2( 0.5)( 2 )(80 ) ( 7.96 10 3 )()1 + 2 ( 80 ) 7.96 10 3 T ( j ) = 0.4852f = 200 Hz T ( j ) =( 0.5 )( 2 )( 200 ) ( 7.96 10 3 )()1 + 2 ( 200 ) 7.96 10 3 T ( j ) = 0.4982(b)=1 1 = rP ( RS RP ) CPCP = =1 2 f ( RS(RP )2 500 1013) (1010 ) 10 3CP = 63.7 pFEX7.2 a.www.elsolucionario.net 261. RP RP RP = 0.891 = (1 0.891) RP = 0.891 RP = 8.20 k 20 log10 = 1 RP + RS RP + RS RP + 1 1 1 CS = fL = 2 ( RS + RP ) CS 2 (100 )(1 + 8.20 ) 10 3 CS = 0.173 F fH =1 2 ( RSRP ) CP1 CP =( ) (1 8.20 ) 102 1063CP = 179 pFb. rS = ( RS + RP ) CS()(rS = 1 10 3 + 8.20 10 3 0.173 10 6)rS = 1.59 ms open-circuit time-constant rP = ( RSRP ) CP(rP = (1 8.20 ) 10 3 179 10 12)rP = 0.160 s short-circuit time-constantEX7.3 a.rS = ( Ri + RS i ) CCb. f =1 2 rSRTH = R1 R2 = 2.2 20 = 1.98 k R2 2.2 VTH = VCC = (10 ) = 0.991 V 2.2 + 20 R1 + R2 VTH VBE ( on ) 0.991 0.7 I BQ = = = 0.0132 mA RTH + (1 + ) RE 1.98 + ( 201)( 0.1) I CQ = 2.636 mA r = gm = VT I CQ ICQ VT= =( 200 )( 0.026 ) 2.636= 1.97 k2.636 = 101.4 mA/V 0.026Rib = + (1 + ) RE = 1.97 + ( 201)( 0.1) = 22.1 k RB = R1 R2 = 1.98 k Ri = RB Rib = 1.98 22.1 = 1.817 k rS = ( Ri + RSi ) CC(= (1.817 + 0.1) 10 3)( 47 10 ) 6= 90.1 ms 1 f = f = 1.77 Hz 2 90.1 10 3()Midband Gainwww.elsolucionario.net 262. A = = RC Ri r + (1 + ) RE Ri + RSi ( 200 )( 2 )1.82 1.97 + ( 201)( 0.1) 1.82 + 0.1 A = 17.2EX7.4 I DQ = K n (VGS VTN )2a. 0.8 + 2 = VGS VGS = 3.265 V 0.5 VS ( 5 ) VS = 3.265 I DQ = RS3.265 + 5 RS = 2.17 k 0.8 5 RD = 6.25 k VD = 0 RD = 0.8 b. rS = ( RD + RL ) CC = (10 + 6.25 ) 10 3 CC 1 1 f = CC = 2 rS 2 f 16.25 10 3 RS =(CC =(12 ( 20 ) 16.25 10 3)) CC = 0.49 FEX7.5 rS = ( RL + Ro ) CC 2 1 1 f = CC 2 = 2 rS 2 f ( RL + Ro ) r + ( RS RB ) R0 = RE r0 1+ From Example 7-5, R0 = 35.5 CC 2 =1 2 (10 ) 10 10 3 + 35.5 CC 2 = 1.59 FEX7.6 a.I DQ = K P (VSG + VTP )21 ( 2 ) = VSG VSG = 3.41 V 0.5 VS = 3.41 5 3.41 RS = 1.59 k RS = 1 5 For VSDG = VSGQ VD = 0 RD = RD = 5 k 1www.elsolucionario.net 263. rP = ( RD RL ) CLb.f =1 1 CL = 2 rP 2 f ( RD RL )CL =( )2 10 61( 5 10 ) 10 3 CL = 47.7 pFEX7.7 (a) RTH = 5 K VTH = 3.7527 I BQ =3.7527 0.7 ( 5) 0.54726 = 5 + (101)( 0.5) 55.5= 0.00986 I CQ = 0.986 mA gm = 37.925 r = 2.637 K 5 Vo Vo = 1 Vo ( 0.4 ) 5 5 Vo = 0.035 (b) 0.986 =Rib RS 0.1 k V0 Vi RTH 5 kVrgmVRE 0.5 kVo = gmV ( RCRL )Rib = r + (1 + ) RE = 2.64 + (101)( 0.5) = 53.14 K V =Vb Vb Vb = = 101 14.885 1+ 1+ RE 1 + 2.637 ( 0.5) r RTH Rib 5 53.14 Vb = Vi = Vi RTH Rib + RS 5 53.14 + 0.1 4.57 = Vi = 0.9786 4.57 + 0.1 Av = (37.925)( 2.5) 14.885( 0.9786 ) = Av = 6.23(c) EX7.8 a.www.elsolucionario.netRC RL 2.5 k 264. I BQ =0 0.7 ( 10 )= 0.0230 mA0.5 + (101)( 4 )I CQ = 2.30 mA VTr =I CQ I CQgm =rB =VT= =(100 )( 0.026 ) 2.30= 1.13 k2.30 = 88.46 mA / V 0.026RE ( RS + r ) CERS + r + (1 + ) RE( 4 10 ) ( 0.5 + 1.13) C = 3E0.5 + 1.13 + (101)( 4 )rB =1 1 = = 0.7958 ms 2 f B 2 ( 200 )rB = 16.07CE CE =b. rA = RE CE = 4 10 3(fA =f = =)( 49.5 10 ) r 6A= 0.198 s1 1 = f A = 0.804 H Z 2 rA 2 ( 0.198 )EX7.9 r =0.796 10 3 CE = 49.5 F 16.07 0VT I CQ=(150 )( 0.026 ) 0.5= 7.8 k1 2 r ( C + C ) 1(2 7.8 103) ( 2 + 0.3) 1012 f = 8.87 MH ZEX7.10 r=f = = 0VT I CQ=(150)(0.026) r = 3.9 k 11 2 r ( C + C )(2 3.9 1013) ( 4 + 0.5) (10 ) 12 f = 9.07 MHzfT = 0 f = (150 )( 9.07 ) fT = 1.36 GHzEX7.11 RTH = R1 R2 = 200 220 = 104.8 k R2 220 VTH = VCC = (5) = 2.619 V 200 + 220 R1 + R2 2.62 0.7 = 0.009316 mA I BQ = 105 + (101)(1) I CQ = 0.9316 mAwww.elsolucionario.net 265. gm = r =I CQ=VT VT I CQ0.9316 gm = 35.83 mA/V 0.026=(100)(0.026) r = 2.79 k 0.932a. C M = C 1 + gm ( RC RL ) C M = ( 2 ) 1 + ( 35.83 )( 2.2 4.7 ) C M = 109 pF b. RB = rS f3 dB = =R2 = 100 200 220 = 51.17 kR11 2 ( RB r) (C+ C )1 f3dB = 0.506 MHz 2 [ 51.17 2.79] 10 3 (10 + 109 ) 10 12EX7.12 fT =gm 2 ( Cgs + Cgd )gm = 2 K n I DQ =2( 0.2 )( 0.4 )= 0.5657 mA/V fT =0.5657 10 3 fT = 333 MHz 2 ( 0.25 + 0.02 ) 10 12EX7.13 dc analysis R2 166 VG = VDD = (10 ) = 4.15 V R1 + R2 166 + 234 V I D = S and VS = VG VGS RS K n (VGS VTN ) = 2VG VGS RS2 ( 0.5)( 0.5) (VGS 4VGS + 4 ) = 4.15 VGS 2 0.25VGS 3.15 = 0 VGS = 3.55 Vgm = 2K n (VGS VTN ) = 2 ( 0.5 )( 3.55 2 ) = 1.55 mA/V Small-signal equivalent circuit. Ri 10 kVi RG R1R2CgsV0 Cgd VgsRDRLgmVgs a.C M = Cgd (1 + gm ( RD RL ) )C M = ( 0.1) 1 + (1.55) ( 4 20 ) C M = 0.617 pF b.www.elsolucionario.net 266. 1fH =2 P P = ( RG Ri ) ( Cgs + C M ) RG = R1 R2 = 234 166 = 97.1 k RG Ri = 97.1 10 = 9.07 k()rP = 9.07 10 3 (1 + 0.617 ) 10 12 = 14.7 ns fH =1(2 14.7 10 9) f H = 10.9 MHzEX7.14 dc analysis VTH = 0, RTH = 10 k I BQ =0 0.7 ( 5 )10 + (126 )( 5 )= 0.00672 mAI CQ = 0.840 mA r = gm = r0 = VT I CQ I CQ VT= =(125)( 0.026 ) 0.840= 3.87 k0.840 = 32.3 mA/V 0.026VA 200 = = 238 k I CQ 0.84High-frequency equivalent circuit CRS 1 kV0 ViRB V R1R2 rr0 CgmVRCa. Miller Capacitance C M = C (1 + gm RL ) RL = r0 RC RL RL = 238 2.3 5 = 1.565 kC M = ( 3 ) 1 + ( 32.3 )(1.57 ) C = 155 pF b. Req = RS RB r = RS R1 R2 r Req = 1 20 20 3.87 = 0.736 k rP = Re q ( C + C M )()= 0.736 10 3 ( 24 + 155 ) 10 12 = 1.314 10fH =(712 1.314 10 7) f H = 1.21 MHzc.www.elsolucionario.netRL 267. ( Av )M ( Av )M RB = gm RL RB + RS 10 3.87 = ( 32.3 )(1.565 ) ( Av ) M = 37.2 10 3.87 + 1 EX7.15 The dc analysis 10 0.7 = 0.00838 mA I BQ = 100 + (101)(10 ) I CQ = 0.838 mA r = gm = VT I CQ=(100 )( 0.026 ) 0.838= 3.10 kI CQ= 32.22 mA/V VT For the input r rP = RE RS C 1 + 3.10 10 1 10 3 24 10 12 = 101 = 7.13 10 10 s f H =1 1 = f H = 223 MHz 2 rP 2 7.13 10 10()For the output rP = [ RC RL ] C = (10 1) 10 3 3 10 12 = 2.73 10 9 1 1 fH = = f H = 58.4 MHz 2 rP 2 2.73 10 9(( A )M= gm ( RC) r RE 1+ RL ) r RE + RS 1+ 3.1 10 101 A = ( 32.22 )(10 1) ( )M = 0.870 3.1 10 101 + 1 EX7.16 R3 7.92 VB1 = (12 ) = 0.9502 V (12 ) = R1 + R2 + R3 58.8 + 33.3 + 7.92 Neglecting lose currents 0.9502 0.7 IC = = 0.50 mA 0.5 VT (100 )( 0.026 ) r = = = 5.2 K IC 0.5 IC 0.5 = = 19.23 mA/V VT 0.026 From Eq (7.127(a)), gm =www.elsolucionario.net 268. P= [ RSRB1 r ][C 1 + C M 1 ]RB1 = R2 R3 = 33.3 7.92 = 6.398 K C M 1 = 2C 1 = 6 pFThen 3 12 P = [1 6.398 5.2] 10 [24 + 6] 10 22.24 ns 1 1 = 7.15 MHz f H = 2 P 2 ( 22.24 10 9 ) From Eq (7.128(a)) 3 12 P = [ RC RL ] C 2 = ( 7.5 2 ) 10 3 10 4.737 ns 1 1 fH = = = 33.6 MHz 2 P 2 ( 4.737 10 9 ) From Eq. 7.133 AvAvMM RB1 r 1 RC ) RB1 r 1 + RS 6.40 5.2 = (19.23)( 7.5 2 ) 6.40 5.2 + 1 = gm 2 ( RC 2.869 = (19.23)(1.579 ) 2.869 + 1 = 22.5TYU7.1 a. V0 = ( gmV ) RL r V = Vi 1 r + + RS sCC T (s) = =V0 ( s ) Vi ( s )= gm r RL r + RS + (1 / sCC ) gm r RL ( sCC )1 + s ( r + RS ) CCgm r = T (s) = RL r + RS s ( r + RS ) CC 1 + s (r + R ) C S C Then = ( r + RS ) CC b. 1 f3 dB = 2 ( r + RS ) CC f3 dB =12 2 10 + 1 10 3 10 6 ( 2 )( 50 )( 4 ) rg R T ( j ) max = m L = 2 +1 r + RS 3 f3 dB = 53.1 HzT ( j ) max = 133c.www.elsolucionario.net 269. T( j) 133f53.1 HzTYU7.2 a. V0 = g mV RL 1 sCL r V = Vi r + RS 1 RL sC V0 ( s ) r L T (s) = = gm 1 Vi ( s ) r + RS RL + sC L T (s) = RL 1 r + RS 1 + sRL CL Then = RL CL b. 1 1 f3 dB = = f3dB = 3.18 MH Z 3 2 RL CL 2 5 10 10 10 12(T ( j ) =)()gm r RL ( 75)(1.5)( 5) = 1.5 + 0.5 r + RST ( j ) max = 281c.281T( j)3.18 MHzfTYU7.3 a. Open-circuit time constant ( CL open ) rS = ( RS + r ) CC()= ( 0.25 + 2 ) 10 3 2 10 6 = 4.5 msShort-circuit time constant ( CC short )()(rP = RL CL = 4 10 3 50 10 12)rP = 0.2 s b.Midband gainwww.elsolucionario.net 270. r V0 = gmV RL , V = Vi + RS V g r R Av = 0 = m L Vi r + RS = ( 65 )( 2 )( 4 )2 + 0.25 Av = 231c. fL =1 1 = f L = 35.4 Hz 2 rS 2 4.5 10 3fH =1 1 f H = 0.796 MHz = 2 rP 2 0.2 10 6()()TYU7.4 Computer Analysis TYU7.5 Computer Analysis TYU7.6 (100 )( 0.026 ) V = 10.4 k r = 0 T = 0.25 I CQ f =1 2 r ( C + C )C + C =1 1 = 2 f r 2 11.5 10 6 10.4 10 3()()C + C = 1.33 pF C = 0.1 pF C = 1.23 pFTYU7.7 h fe =01 + j ( f / f )f = 5 MH Z , 0 = 100 At f = 50 MH Z h fe =100 50 1+ 5 2 h fe = 9.95 50 Phase = tan 1 Phase = 84.3 5 TYU7.8 f 500 f = 4.17 MHz f = T = 0 120 f =1 2 r (C + C )C + C =1 1 = 2 f r 2 (4.167 10 6 )(5 10 3 )C + C = 7.639 pF C = 0.2 pF C = 7.44 pFwww.elsolucionario.net 271. TYU7.9 (a) gm = 2K n (VGS VTN ) = 2 ( 0.4 )( 3 1) gm = 1.6 mA/V gm = 80% of gm = 1.28 mA/V gm gm = 1 + gm rS 1 + gm rS =gm gm gm 1 1.6 1 = 1 gm 1.6 1.28 rS = 0.156 k rS = 156 ohms 1 gmrS =(b) gm = 2K n (VGS VTN ) = 2 ( 0.4 )( 5 1) gm = 3.2 mA/V gm 3.2 gm = = = 2.134 1 + gm rS 1 + ( 3.2 )( 0.156 ) gm 3.2 2.134 = A 33.3% reduction 3.2 gmTYU7.10 fT = = Cgs = =gm2 ( CgsT + C gdT ) gm2 ( Cgs + Cgsp + Cgdp ) gm Cgsp C gdp 2 fT 0.5 10 3(2 500 106) ( 0.01 + 0.01) 10 12 Cgs = 0.139 pFTYU7.11 fT =gm 2 (Cgs + Cgsp + Cgdp )Cgsp = Cgdp 2Cgsp =gm 1 103 C gs = 0.4 1012 2 fT 2 (350 106 )2Cgsp = 0.0547 pF Cgsp = Cgdp 0.0274 pFTYU7.12 dc analysis 50 VG = (10 ) 5 = 2.5 50 + 150 VS = VG VGS . I D = K n (VGS VTN ) = 2VS ( 5 ) RSVG VGS + 5 RSwww.elsolucionario.net 272. 2 (1)( 2 ) VGS 1.6VGS + 0.64 = 2.5 VGS + 5 2 2VGS 2.2VGS 1.22 = 02.2 VGS =( 2.2 )2+ 4 ( 2 )(1.22 )2 (2) VGS = 1.505 Vgm = 2 K n (VGS VTN ) = 2 (1)(1.505 0.8 ) = 1.41 mA/VEquivalent circuit CgdRiV0 ViRG Vgs R1R2 CgsgmVgsRDC M = Cgd (1 + gm RD ) = ( 0.2 ) 1 + (1.42 )( 5 ) C M = 1.61 pF (a)(b) P = ( Ri RG ) ( Cgs + CM )rP = [ 20 50 150 ] 10 3 ( 2 + 1.61) 10 12()()= 13 10 3 3.62 10 12 = 4.71 10 8 s fH =c.1 1 = f H = 3.38 MHz 2 rP 2 4.71 10 8(( Av )M ( Av )M) RG = gm RD RG + RS 37.5 = (1.41)( 5 ) ( Av ) M = 4.60 37.5 + 20 TYU7.13 Computer Analysiswww.elsolucionario.net 273. Chapter 7 Problem Solutions 7.1 a. T (s) = T (s) =V0 ( s ) Vi ( s )=1/ ( sC1 )1/ ( sC1 ) + R1 1 1 + sR1C1b.1T fH 159 HzffH =1 1 = f H = 159 Hz 2 R1C1 2 (103 )(106 )c. V0 ( s ) = Vi ( s ) 1 1 + sR1C1For a step function Vi ( s ) =1 sK K2 1 1 = 1+ V0 ( s ) = s 1 + sR1C1 s 1 + sR1C1 = =K1 (1 + sR1C1 ) + K 2 s s (1 + sR1C1 )K1 + s ( K1 R1C1 + K 2 ) s (1 + sR1C1 )K 2 = K1 R1C1 and K1 = 1 V0 ( s ) = R1C1 1 + s 1 + sR1C1=1 1 1 s +s R1C1v0 ( t ) = 1 e t / R1C17.2 a. T (s) = T (s) =V0 ( s ) Vi ( s )=R2 R2 + 1/ ( sC2 ) sR2 C2 1 + sR2 C2b.www.elsolucionario.net 274. 1T fL 1.59 Hz1 1 fL = = f L = 1.59 Hz 4 2 R2 C2 2 (10 )(10 106 )c. V0 ( s ) = Vi ( s ) sR2 C2 1 + sR2 C2Vi ( s ) =1 sV0 ( s ) =R2 C2 1 = 1 + sR2 C2 s + 1 R2 C2v0 ( t ) = e t / R2C27.3 a. T (s) =RPV0 ( s ) Vi ( s )1 sCP 1 1 + RS + sCP sCS RP= RP1 RP sCP RP 1 = = sCP R + 1 1 + sRP CP P sCPThen RP 1 RP + RS + (1 + sRP CP ) sCS RP = RP CP 1 RP + RS + + + sRS RP CP CS sCST (s) = RP T (s) = RP + RS b. sRP RS RP C 1 P+ + CP 1/ 1 + RP + RS CS s ( RS + RP ) CS RS + RP www.elsolucionario.net 275. 11 1 10 10 10 T (s) = 1/ 1 + 6 + + s ( 5 103 ) 1011 s ( 2 104 ) 106 10 + 10 20 10 1 1 1 2 1+ + s ( 5 108 ) s ( 0.02 )s = j T ( j ) =1 2T ( j ) =11 2 1 1 + j ( 5 108 ) ( 0.02 ) 1 1 For L = = = 50 4 ( RS + RR ) CS ( 2 10 )(106 ) 1 1 1 + j ( 50 ) ( 5 108 ) ( 50 )( 0.02 ) 1 1 1 1 T ( j ) = 2 1 j 2 2ForH =( RS1 1 = = 2 107 RP ) CP ( 5 103 )(1011 )1 1 1 + j ( 2 107 )( 5 108 ) ( 2 107 ) ( 0.02 ) 1 1 1 1 T ( j ) = T ( j ) 2 1+ j 2 2 T ( j ) =1 2In each case, T ( j ) =c.RP 2 RP + RS1RS = RP = 10 k, CS = CP = 0.1 FT (s) = 1 1 1 1 1/ 1 + + + s ( 5 103 )(107 ) 2 2 1 s 2 104 (107 ) ()s = j T ( j ) =1 21 1 1 1 + + j ( 5 104 ) 2 ( 2 103 ) 1 = 500 For = ( 2 104 )(107 )T ( j ) =1 21 1 1.5 + j ( 500 ) ( 5 104 ) ( 500 ) ( 2 103 ) 1 1 = T ( j ) = 0.298 2 1.5 j ( 0.75 )www.elsolucionario.net 276. For =1( 5 10 )(10 ) 73= 2 103 1 1 1/ 1.5 + j ( 2 103 )( 5 104 ) 2 ( 2 103 )( 2 103 ) 1 1 = T ( j ) = 0.298 2 1.5 + j ( 0.75 )T ( j ) =In each case, T ( j ) C and f = C1C2f 3 dB ( CC 2 ) = 25 Hz =1 2 2, so that 2 =1 = 0.006366 s = Req CC 2 2 ( 25 )where r + R1 R2 RS Req = RL + RE 1+ 81.5 + 600 300 = 10 + 50 = 10 + 50 2.787 101 0.00637 Req = 12.64 CC 2 = CC 2 = 504 F 12.6 Rib = r + (1 + ) RE Assume CC 2 an open Rib = 81.5 + (101)( 50 ) = 5132 1 = (100 ) 2 = (100 )( 0.006366 ) = 0.6366 s = Req1CC1 Req1 = RS + RTH Rib = 300 + 600 5132 = 837.2 So CC1 =0.6366 CC1 = 760 F 837.27.26 From Problem 7.25 RTH = 0.6 K, I CQ = 31.9 mA, r = 81.5 1 so f 3 dB ( CC 2 ) f 3 dB ( CC1 ) 2 Then f 3 dB ( CC 2 ) CC1 acts as an open circuit and for f 3 dB ( CC1 ) CC 2 acts as a short circuit.C2 C1 and f =f 3 dB ( CC1 ) = 20 Hz =1 2 C1 C1 = 0.007958 sRib = r + (1 + ) ( RE RL ) = 81.5 + (101) ( 50 10 ) = 923.2 C1 Req1 = RS + RTH Rib = 300 + 600 923.2 = 663.7 0.007958 CC1 = 12 F 663.7 = 100 C1 = 0.7958 sCC1 =C2 r + RTH Req 2 = RL + RE = 10 + 50 1+ Req 2 = 10 + 50 6.748 = 15.95 0.7958 CC 2 = CC 2 = 0.050 F 15.95 81.5 + 600 101 7.27 a.www.elsolucionario.net 288. I D = K n (VGS VTN ) VGS =2ID 0.5 + VTN = + 0.8 = 1.8 V Kn 0.5 VGS ( 5 )5 1.8 RS = 6.4 k 0.5 0.5 VD = VDSQ + VS = 4 1.8 = 2.2 V 5 2.2 RD = RD = 5.6 k 0.5 (b) RS =gm = 2 Kn I D = 2=( 0.5 )( 0.5 ) = 1 mA / VrA = RS CS = ( 6.4 103 )( 5 106 )= 3.2 102 s 1 1 = f A = 4.97 Hz fA = 2 rA 2 ( 3.2 102 ) 6.4 103 6 ( 5 10 ) CS = 1 + (1)( 6.4 ) = 4.32 103 s 1 1 = f B = 36.8 Hz fB = 2 rB 2 ( 4.32 103 ) RS rB = 1 + g m RSc.g m RD (1 + sRS CS )Av =RS 1 + g m RS (1 + g m RS ) 1 + s CS As RS becomes large g m RD ( sRS CS ) Av R ( g m RS ) 1 + s S CS g m RS 1 CS gm ( g m RD ) s Av = 1 1+ s gm CS The corner frequency f B = gm = 2 Kn I D = 2 fB =( 0.5 )( 0.5 ) = 1 mA / V1 f B = 31.8 Hz 1 6 2 3 ( 5 10 ) 10 7.28 a.1 and the corresponding f A 0 2 (1/ g m ) CSfB =RE ( RS + r ) CE 1 and rB = 2 rB RS + r + (1 + ) REwww.elsolucionario.net 289. For RS = 0 rB = I EQ =RE r CE r + (1 + ) RE0.7 ( 10 ) 5= 1.86 mA = 75 I CQ = 1.84 mA = 125 I CQ = 1.85 mA For f B 200 Hz rB 1 = 0.796 ms 2 ( 200 )r so smallest rB will occur for smallest . = 75; r =( 75)( 0.026 )= 1.06 k 1.84 ( 5 103 ) (1.06 ) CE C = 57.2 F 0.796 103 = E 1.06 + ( 76 )( 5 ) b. (125 )( 0.026 ) For = 125; r = = 1.76 k 1.85 ( 5 103 ) (1.76 ) ( 57.2 106 ) = 0.797 ms rB = 1.76 + (126 )( 5 ) fB =1 1 = f B = 199.7 Hz Essentially independent of . 2 rB 2 ( 0.797 103 )rA = RE CE = ( 5 103 )( 57.2 106 ) = 0.286 sec fA =1 1 = f A = 0.556 Hz Independent of . 2 rA 2 ( 0.286 )7.29 a. Expression for the voltage gain is the same as Equation (7.66) with RS = 0. b. rA = RE CE RE r CE rB = r + (1 + ) RE 7.30 5 0.7 I C = 0.99 mA = 1 mA = I E 4.3 VT (100 )( 0.026 ) r = = = 2.626 K 0.99 I CQ rA = RE CE = ( 4.3 103 )( 5 106 ) = 2.15 10 2 s rB =( 4.3 103 )( 2.626 103 )( 5 106 ) = 1.292 104 s RE r CE = r + (1 + ) RE 2.626 103 + (101) ( 4.3 103 )fA =1 1 = = 7.40 Hz 2 rA 2 ( 2.15 102 )fB =1 1 = = 1.23 103 = 1.23 kHz 2 rB 2 (1.292 104 )www.elsolucionario.net 290. Avw0=Avw =(100 )( 2 ) RC = = 0.458 r + (1 + ) RE 2.626 + (101)( 4.3) RC r=(100 )( 2 ) 2.626= 76.2A 76.20.4587.40 Hz1.23 kHzf7.31 rH = ( RL RC ) CL = (10 5 ) 103 15 1012 rH = 5 108 s 1 1 = f H = 3.18 MHz 2 rH 2 ( 5 108 )fH =10 0.7 = 0.93 mA, I CQ = 0.921 mA 10 0.921 gm = = 35.4 mA/V 0.026 Av = g m ( RC RL ) = 35.4 ( 5 10 ) Av = 118I EQ =7.32 R2 VG = VDD R1 + R2 ID = 166 = (10 ) 166 + 234 = 4.15 VVG VGS 2 = K n (VGS VTN ) RS2 4.15 VGS = ( 0.5 )( 0.5 ) (VGS 4VGS + 4 ) 2 0.25VGS 3.15 = 0 VGS = 3.55 Vg m = 2 K n (VGS VTN ) = 2 ( 0.5 )( 3.55 2 ) g m = 1.55 mA / V 1 1 = 0.5 = 0.5 0.645 1.55 gmR0 = RSR0 = 0.282 k r = ( R0 RL ) CL and f H = f H = 5 MHz r = CL =r R0 RL=1 2 r 12 ( 5 106 )= 3.18 108 s3.18 108 CL = 121 pF ( 0.282 4 ) 1037.33www.elsolucionario.net 291. (a) Low-frequency RSCC Vo VsRBVRCrRLgmVMid-Band RS Vo VsRBVrRCRLgmVHigh-frequency RS Vo VsRBV rRCRLCLgmV(b) AmfLfHf(c) 12 0.7 = 11.3 A 1 M = 1.13 mAI BQ = I CQr =(100 )( 0.026 )= 2.3 k 1.13 1.13 gm = = 43.46 mA / V 0.026 Am = R R V0 ( midband ) = g m ( RC RL ) B R r +R Vs S B 1000 2.3 = ( 43.46 ) ( 5.1 500 ) 1000 2.3 + 1 2.29 = ( 43.46 )( 5.05 ) Am = 153 2.29 + 1 Am dB = 43.7 dBwww.elsolucionario.net 292. fL =1 2 L, L = ( RS + RB r ) CC or L = (1 + 1000 2.3) 103 (10 106 ) L = 3.29 102 s f L = 4.83 Hz fH =1 , H = ( RC RL ) CL L = ( 5.1 500 ) 103 (10 1012 ) 2 H= 5.05 108 s f H = 3.15 MHz7.34 a. V0 Vi VsgRDgmVsgRLCL 1 V0 = ( g mVsg ) R0 RL sCL Vsg = Vi Av ( s ) =V0 ( s ) 1 = g m RD RL Vi ( s ) sCL 1 RD RL sCL = gm 1 RD RL + sCL Av ( s ) = g m ( RD RL ) b. c. 1 1 + s ( RD RL ) CLr = ( RD RL ) CLr = (10 20 ) 103 10 1012 r = 6.67 10 8 s fH =1 1 = f H = 2.39 MHz 2 r 2 ( 6.67 108 )From Example 7.6, gm = 0.705 mA/V Av = g m ( RD RL ) = ( 0.705 ) (10 20 ) Av = 4.7 7.35 Computer Analysis 7.36 Computer Analysis 7.37 Computer Analysis 7.38www.elsolucionario.net 293. fT =gm2 ( C + C ) I CQgm = fT =1 = 38.46 mA/V 0.026=VT38.46 103 2 (10 + 2 ) 1012fT = 510 MHz fTf 510 f = 4.25 MHz 1205000 MHz f = 33.3 MHz 150 gm7.39 fTf = fT =2 ( C + C )0.5 = 19.23 mA/V 0.026 19.2 10 3 5 109 = 2 ( C + 0.15 ) 1012 gm =C + 0.15 =19.2 1032 (1012 )( 5 109 )= 0.612 pFC = 0.462 pF7.40 f =a.fT=2000 MHz = 13.3 MHz = f 150b. h fe =150 1 + j ( f / f )h fe =150 1 + ( f / f )2= 102 f 150 2 1+ = = 225 f 10 f = f 224 = (13.33) 224 f = 199.6 MHz7.41 (a) V0 = g mV RL where 1 sC1r 1 + sr C1 V = Vi = Vi r 1 + rb r + rb 1 + sr C1 sC1 r= r r 1 Vi = Vi r + rb + srb r C1 r + rb 1 + s ( rb r ) C1 www.elsolucionario.net 294. So Av ( s ) =V0 ( s ) r 1 = g m RL 1 + s ( rb r ) C1 Vi ( s ) r + rb (100 )( 0.026 )(b) Midband gain: r =1= 2.6 k , g m =1 = 38.46 mA / V 0.026(i)For rb = 100 (ii) 2.6 Av1 = ( 38.46 )( 4 ) Av1 = 148.1 2.6 + 0.1 For rb = 500 (c) (i) 2.6 Av 2 = ( 38.46 )( 4 ) Av 2 = 129.0 2.6 + 0.5 1 , = ( rb r ) C1 f 3 dB = 2 For rb = 100 1 = ( 0.1 2.6 ) 103 ( 2.2 1012 ) = 2.12 1010 s f 3 db = 751 MHz(ii)For rb = 500 2 = ( 0.5 2.6 ) 103 ( 2.2 1012 ) = 9.23 1010 s f 3 dB = 173 MHz7.42 (b)f = 10 kHz = 104 Z i = 200 +(2500 1 j (104 )(1.333 106 ) 1 + (10) (1.333 10 )4 2)6 2= 200 + 2500 j 33.3 = 2700 j 33.3 f = 100 kHz = 105(c)Z i = 200 +(2500 1 j (105 )(1.333 106 ) 1 + (10) (1.333 10 )5 2)6 2Z i = 200 + 2456 j 327 = 2656 j 327 f = 1 MHz = 106(d)Z i = 200 +(2500 1 j (106 )(1.333 106 ) 1 + (10) (1.333 10 )6 2)6 2Z i = 200 + 900 j1200 = 1100 j12007.43 a.CM = C (1 + g m RL )b. RBRSrb V0 RB Vi RB RS V rCCMgmVRLwww.elsolucionario.net 295. V0 = g mV RL rV =Let C + CM = Ci 1 sCi RB Vi R + RS 1 r + RB RS + rb B sC1Av ( s ) =Vo ( s ) Vi ( s )1 r sCi 1 r + RB sCi = g m RL RB + RS r 1 sCi + RB RS + rb 1 r + sCi RB r = g m RL RB + RS r + (1 + sr Ci ) ( RB RS + rb ) Let Req = ( RB RS + rb ) RB Av ( s ) = RL RB + RS Av ( s ) =c. 1 1 + s r Req Ci ( r + Req ) () RL RB 1 r + Req RB + RS 1 + s r Req Ci(fH =(1))2 r Req Ci7.44 High Freq. CC1 , CC 2 , CE short circuits C V0 ISR1R2 VrCgmVRCRL I0www.elsolucionario.net 296. gm = fT =I CQ VT=5 = 192.3 mA/V 0.026gm2 ( C + C ) 250 106 =192 103 2 ( C + C )C + C = 122.4 pF C = 5 pF, C = 117.4 pF(CM = C 1 + g m ( RC RL ))= 5 1 + (192.3) (1 1) CM = 485.8 pF Ci = C + CM = 117 + 485 = 603 pF r =( 200 )( 0.026 )= 1.04 k5Req = R1 R2 r = 5 1.04 = 0.861 k r = Req Ci= ( 0.861 103 )( 603 1012 ) = 5.19 10 7 s1f =2 r=12 ( 5.19 107 ) f = 307 kHz7.45 RTH = R1 R2 = 60 5.5 = 5.04 k R2 5.5 VTH = VCC = (15 ) = 1.26 V 5.5 + 60 R1 + R2 1.26 0.7 I BQ = = 0.0222 mA 5.04 + (101)( 0.2 ) I CQ = 2.22 mA r =(100 )( 0.026 )= 1.17 k 2.22 2.22 gm = = 85.4 mA/V 0.026 Lower 3 dB frequency: rL = Req CC1Req = RS + R1 R2 r = 2 + 60 5.5 1.17 = 2.95 k rL = ( 2.95 103 )( 0.1 106 ) = 2.95 104 s fL =1 1 = f L = 540 Hz 2 rL 2 ( 2.95 104 )Upper 3 dB frequency:www.elsolucionario.net 297. gmfT =2 ( C + C ) 400 106 =85.4 103 2 ( C + C )C + C = 34 pF; C = 2 pF C = 32 pF CM = C (1 + g m RC ) = 2 (1 + ( 85.4 )( 4 ) ) CM = 685 pF Ci = C + CM = 32 + 685 = 717 pF Req = RS R1 R2 r= 2 60 5.5 1.17 = 0.644 k= ( 0.644 103 )( 717 1012 )r = Req Ci= 4.62 107 s fH =1 2 r f H = 344 kHz7.46 RTH = R1 R2 = 600 55 = 50.38 K R2 55 VTH = (15 ) = (15 ) = 1.2595 V 600 + 55 R1 + R2 1.26 0.7 I BQ = = 0.00222 mA 50.4 + (101)( 2 ) I CQ = 0.2217 mA r =(100 )( 0.026 )= 11.73 K 0.222 0.2217 gm = = 8.527 mA/V 0.026 Lower 3dB Fig. L = R e q1 Cc1 ; Req1 = RS + RTH r = 0.50 + 50.38 11.73 = 10.0 K L = (10 103)( 0.110 ) = 10 63s L = R e q1 Cc1 ; Req1 = RS + RTH r = 0.50 + 50.38 11.73 = 10.0 K L = (10 103 )( 0.1106 ) = 103 s 1fL =2 2=12 (103 ) f L = 159 HzUpper 3dB Fig. gm 8.527 103 = = 400 106 fT = 12 2 ( C + C ) 2 ( C + 2 ) 10 C + C = 3.393 pF C = 1.393 pF CM = C (1 + g m RC ) = 2 1 + ( 8.527 )( 40 ) = 684 pF CT = C + CM = 1.393 + 684 = 685.4 pF Req 2 = RS RTH r = 0.5 50.38 11.73 = 50.38 0.480 = 0.4750 K H = R eq 2 .CT fH ( 0.4750 103 ) ( 685.4 1012 ) = 3.256 107 s 11 = f H = 489 KHz 2 H 2 ( 3.256 107 )www.elsolucionario.net 298. 7.47 fT =gm2 ( Cgs + Cgd ) 40 g m = 2 K n I D , K n = (15 ) = 60 A / V 2 10 gm = 2 fT =( 60 )(100 ) = 154.9 A / V154.9 106 fT = 44.8 MHz 2 ( 0.5 + 0.05 ) 10127.48 fT =gm2 ( Cgs + Cgd )g m = 2 K n (VGs VT )ID then g m = 2 K n I D KnI D = K n (VGs VTN ) or VGs VTN = 2So fT =2 Kn I D2 ( Cgs + Cgd )=Kn I D ( Cgs + Cgd )( 0.2 10 )( 20 10 ) 3(a)fT =(b)fT =(c)10 =6 ( 0.5 + 0.1) 1012 fT = 33.6 MHz( 0.2 10 )( 250 10 ) 36 ( 0.5 + 0.1) 1012( 0.2 10 ) I fT = 118.6 MHz39D ( 0.5 + 0.1) 1012 I D = 17.8 mA7.49 fT =gm2 ( Cgs + Cgd )Cgs + Cgd = WLCox W C g m = 2 K n (VGS VTN ) = 2 n ox (VGS VTN ) L 2 W ( n Cox )(VGS VTN ) L Then fT = 2 WLCox fT = n (VGS VTN ) 2 L2(a)fT =(b)fT =450 ( 0.5 )2 (1.2 104 )450 ( 0.5 )22 ( 0.18 104 ) fT = 2.49 GHz fT = 111 GHz7.50 a.www.elsolucionario.net 299. (CM = Cgd 1 + g m ( ro RD ))CM = 5 1 + ( 3) (15 10 ) CM = 95 pF b. r = ( ri ) ( Cgs + CM )r = (10 103 ) ( 50 + 95 ) 1012 = 1.45 106 sf =1 1 = f = 110 kHz 2 r 2 (1.45 106 )7.51 fT =gm2 ( CgsT + CgdT )( Eq. 7.104 ) 2 Let CgdT = 0 and CgsT = (WLCox ) 3 C g m = 2 K n I D = 2 n ox 2 W L ID 1 W 2 n Cox I D 2 L So fT = 2 2 (W LCox ) 3 1 W n Cox I D 2 3 L = 2 L W Cox fT =3 2 Ln I D 2W Cox L7.52 (a) gm =gm 1 + g m rSg m = 2 K n (VGS VTN )8 C W ( 400 ) ( 7.25 10 ) (10 ) K n = n ox = 2 2 L K n = 1.45 104 A / V 2For VGS = 5 Vg m = 2 (1.45 10 4 ) ( 5 0.65 ) = 1.26 103 A/V g m = ( 0.80 ) g m = 1.01 103 A/V1.01 103 =1.26 1031 + (1.26 103 ) rS1 + (1.26 103 ) rS = 1.25 rS = 196 b.www.elsolucionario.net 300. For VGS = 3 Vg m = 2 (1.45 104 ) ( 3 0.65 ) = 0.6815 103 A/V gm =0.6815 1031 + ( 0.6815 103 ) (196 ) g m = 0.60 103 A/V Re duced by 12%7.53 a. Vgs IiRigmVgsRLI0rSI 0 = g mVgs and Vgs = I i Ri g mVgs rS so Vgs = Then Ai = b.I i Ri 1 + g m rSI0 g R = m i I i 1 + g m rSAs an approximation, consider IiVgsRiCgsTI0CMRLg Vgs mIn this case I gm 1 Ai = 0 = g m Ri where CM = C gdT (1 + g m RL ) and g m = Ii 1 + g m rs 1 + sRi ( C gsT + CM ) c. As rS increases, CM decreases, so the bandwidth increases, but the current gain magnitude decreases. 7.54 R2 225 VGS = VDD = (10 ) 225 + 500 R1 + R2 VGS = 3.10 Vg m = 2 K n (VGS VTN ) = 2 (1)( 3.10 2 ) g m = 2.207 mA / V a.CgdTRiV0Vi R1R2VgsCgsTgmVgsRDb.www.elsolucionario.net 301. CM = C gdT (1 + g m RD ) = (1) 1 + ( 2.207 )( 5 ) CM = 12 pFc. r = ( Ri R1 R2 ) ( CgsT + CM ) Ri R1 R2 = 1 500 225 = 1 155 = 0.9936 kr = ( 0.9936 103 ) ( 5 + 12 ) 1012 = 1.69 10 8 s fH = Av =1 2 r=12 (1.69 108 ) g mVgs RD Viand Vgs = f H = 9.42 MHz R1 R2R1 R2 + Ri Vi =155 Vi = 0.994 Vi 155 + 1Av = ( 2.2 )( 5 )( 0.994 ) Av = 10.97.55 RTH = R1 R2 = 33 22 = 13.2 k R2 22 VTH = ( 5) = ( 5) = 2 V 22 + 33 R1 + R2 2 0.7 I BQ = = 0.00261 mA 13.2 + (121)( 4 ) I CQ = 0.3138 r =(120 )( 0.026 )= 9.94 k 0.3138 0.3138 gm = = 12.07 mA/V 0.026 100 r0 = = 318 k 0.3138 a. gm fT = 2 ( C + C )C + C =gm 12.07 103 = 2 fT 2 ( 600 106 )C + C = 3.20 pF; C = 1 pF C = 2.20 pF()CM = C 1 + g m ro RC RL = (1) 1 + (12.07 ) 318 4 5 CM = 27.6 pF b.()www.elsolucionario.net 302. r = Req ( C + CM ) Req = R1 R2 Rs r = 33 22 2 r = 1.74 9.94 k Req = 1.48 k r = (1.48 103 ) ( 2.20 + 27.6 ) 1012 r = 4.41 10 8 s fH =1 2 r1=(2 ( 4.41 108 )V0 = g mV ro RC RL f H = 3.61 MHz)x R1 R2 r V = V R1 R2 r + RS i R1 R2 r = 33 22 9.94 = 5.67 k 5.67 v = Vi = 0.739Vi 5.67 + 2 r0 RC RL = 318 4 5 = 2.18 k Av = (12.07 )( 0.739 )( 2.18 ) Av = 19.77.56 RTH = R1 R2 = 40 5 = 4.44 k R2 5 VTH = VCC = (10 ) = 1.111 V 5 + 40 R1 + R2 1.111 0.7 = 0.00633 mA I BQ = 4.44 + (121)( 0.5 ) I CQ = 0.760 mA r =(120 )( 0.026 )= 4.11 k 0.760 0.760 = 29.23 mA/V gm = 0.026 r0 = fT =gm2 ( C + C )C + C =gm 29.23 103 = 2 fT 2 ( 250 106 )C + C = 18.6 pF; C = 3 pF C = 15.6 pFa. CM = C 1 + g m ( RC RL ) 1 + ( 29.2 ) ( 5 2.5 ) CM = 149 pF CM = 3 For upper frequency:www.elsolucionario.net 303. rH = Req ( C + CM ) Req = r R1 R2 RS = 4.11 40 5 0.5 Req = 0.405 k rH = ( 0.405 103 ) (15.6 + 149 ) 1012 = 6.67 10 8 s 1 fH = f H = 2.39 MHz 2 rH For lower frequency: rL = Req CC1 Req = RS + R1 R2 r = 0.5 + 40 5 4.11 Req = 2.64 krL = ( 2.64 103 )( 4.7 106 ) = 1.24 102 sfL =1 f L = 12.8 Hz 2 rLb. A 39.5fLfHV0 = g mV ( RC RL )f R1 R2 r V = V R1 R2 r + RS i 2.135 V = Vi = 0.8102Vi 2.135 + 0.5 AV = ( 29.23)( 0.8102 ) ( 5 2.5 ) AV = 39.57.57 I D = K P (VSG + VTP ) = 29 VSG RS2 ( 2 )(1.2 ) (VSG 4VSG + 4 ) = 9 VSG 2 2.4VSG 8.6VSG + 0.6 = 0VSG =8.6 (8.6 ) 4 ( 2.4 )( 0.6 ) 2 ( 2.4 ) 2VSG = 3.512 V g m = 2 K P (VSG + VTP ) = 2 ( 2 )( 3.512 2 ) g m = 6.049 mA / V I D = ( 2 )( 3.512 2 ) = 4.572 mA 1 1 r0 = 21.9 k r0 = = I o ( 0.01)( 4.56 ) 2www.elsolucionario.net 304. (CM = CgdT 1 + g m ( ro RD )a.)CM = (1) 1 + ( 6.04 ) ( 21.9 1) CM = 6.785 pF b. rH = ( Ri RG ) ( CgsT + CM )rH = ( 2 100 ) 103 (10 + 6.78 ) 1012 rH = 3.29 108 s 1fH = f H = 4.84 MHz2 HV0 = g m ( ro RD ) Vgs RG 100 Vgs = Vi = Vi 102 RG + Ri 100 Av = ( 6.04 ) ( 21.9 1) 102 Av = 5.677.58 R2 22 VG = ( 20 ) 10 = ( 20 ) 10 R1 + R2 22 + 8 VG = 4.67 VID =10 VSG 4.67 2 = K P (VSG + VTP ) RS2 5.33 VSG = (1)( 0.5 ) (VSG 4VSG + 4 ) 2 0.5VSG VSG 3.33 = 0VSG =1 1 + 4 ( 0.5 )( 3.33) 2 ( 0.5 ) VSG = 3.77 Vg m = 2 K p (VSG + VTP ) = 2 (1)( 3.77 2 ) g m = 3.54 mA / V b.(CM = CgdT 1 + g m ( RD RL ))CM = ( 3) 1 + ( 3.54 ) ( 2 5 ) CM = 18.2 pF a. r = Req ( CgsT + CM ) Req = Ri R1 R2 = 0.5 8 22 = 0.461 k r = ( 0.461 103 ) (15 + 18.2 ) 1012 = 1.53 108 s fH =1 2 r f H = 10.4 MHzc. V0 = g mVgs ( RD RL ) R R 5.87 Vgs = 1 2 Vi = Vi Vgs = ( 0.9215 ) Vi R1 R2 Ri 5.87 + 0.5 Av = ( 3.54 )( 0.9215 ) ( 2 5 ) Av = 4.66www.elsolucionario.net 305. 7.59 100 I E = 0.5 mA I CQ = ( 0.5 ) = 0.495 mA 101 0.495 = 19.0 mA/V gm = 0.026 (100 )( 0.026 ) = 5.25 k r = 0.495 a. Input: From Eq. 7.107b r rP = RE RS C 1 + 5.25 0.5 0.05 103 10 1012 = 101 = 2.43 1010 s1 f H = 656 MHz 2 rp Output: From Eq. 7.108b rP = ( RB RL ) C = (100 1) 103 1012 f H 9.90 1010 s fH =1 f H = 161 MHz 2 rPb. RSVigmVV0 REVrRBRLV0 = g m V ( RB RL ) g mV +V V Vi ( V ) + + =0 r RE RS 1 1 1 V + = i V g m + + r RE RS RS 1 1 1 Vi + + = V 19 + 5.25 0.5 0.05 0.05 V ( 41.19 ) = Vi ( 20 )V = ( 0.4856 ) ViV0 = (19 )( 0.4856 )(100 1) Vi Av = 9.14c. r = CL ( RL RB ) = (15 1012 ) (1 100 ) 103 r = 1.485 108 s 1 f = 10.7 MHz 2 r Since f < f H 3d B freq. dominated by CL . f =7.60www.elsolucionario.net 306. 20 0.7 = 1.93 mA 10 100 I CQ = (1.93) = 1.91 mA 101 1.91 gm = = 73.5 mA/V 0.026 (100 )( 0.026 ) r = = 1.36 k 1.91 a. Input: r rP = RE RS C 1 + I EQ =rP1.36 10 1 103 10 1012 = 101 = 1.327 1010 s1 f P = 1.20 GHz 2 rP Output: rP = ( RC RL ) C = ( 6.5 5 ) 103 1012 f P =rP = 2.826 109 s 1 f P = f P = 56.3 MHz 2 rP b. V0 = g m V ( RC RL ) g mV +V V Vi ( V ) + + =0 r RE RS Vi 1 1 1 + V g m + + = r RE RS RS 1 1 1 V + + = i V 73.5 + 1.36 10 1 (1) V ( 75.34 ) = Vi V = ( 0.01327 ) Vi V0 = ( 73.5 )( 0.01327 ) ( 6.5 5 ) Vi Av = 2.76 c.r = CL ( RL Rc ) = (15 1012 ) ( 6.5 5 ) 103 r = 4.24 108 sf =1 f = 3.75 MHz 2 r Since f < fp , 3dB frequency is dominated by CL.7.61www.elsolucionario.net 307. VGS + I D RS = 5 5 VGS 2 ID = = K n (VGS VTN ) RS5 VGS = ( 3)(10 ) (V 2GS 2VGS + 1) 30V 2GS 59VGS + 25 = 0VGS =59 ( 59 ) 4 ( 30 )( 25 )22 ( 30 ) VGS = 1.349 Vg m = 2 K n (VGS VTN ) = 2 ( 3)(1.35 1) g m = 2.093 mA / V On the output: rP = ( RD RL ) Cgd T = ( 5 4 ) 103 4 1012 rP = 8.89 109 s f P =1 f P = 17.9 MHz 2 rP RiVigmVgsV0 RSVgsRDRLV0 = g mVgs ( RD RL ) g mVgs +Vgs RS+Vi ( Vgs ) RS=0 V 1 1 + = i Vgs g m + RS Ri Ri V 1 1 Vgs 2.093 + + = i 10 2 2 Vgs = ( 0.1857 )ViAv =V0 = ( 2.093)( 0.1857 ) ( 5 4 ) ViAv = 0.8647.62 dc analysis ID =V + VSG 2 = K P (VSG + VTP ) RS5 VSG = (1)( 4 )(VSG 0.8 )22 = 4 (VSG 1.6VSG + 0.64 )2 4VSG 5.4VSG 2.44 = 0VSG =5.4 ( 5.4 )2+ 4 ( 4 )( 2.44 )2 ( 4)= 1.707g m = 2 K P (VSG + VTP ) = 2 (1)(1.707 0.8 ) g m = 1.81 mA / Vwww.elsolucionario.net 308. RigmVgs RSVgsCgsTCgdTRDRL3 dB frequency due to CgsT : Req =1 RS Ri gm1 2 Req CgsTfA =1 4 0.5 = 0.246 k 1.81 1 fA = = 162 MHz 2 ( 246 ) ( 4 1012 )Req =3 dB frequency due to CgdT 1 2 ( RD RL ) CgdTfB = =1 2 ( 2 4 ) 103 1012f = 119 MHzMidband gain RiVigmVgsV0 VgsRSRDRL1 RS gm1 4 1.81 Vgs = Vi = Vi 1 1 RS + R i 4 + 0.5 gm 1.81 = 0.492ViV0 = g mVgs ( RD RL )Av = ( 0.492 )(1.81) ( 4 2 ) Av = 1.197.63 r =(120 )( 0.026 )1.02 g m = 39.23 mA/V a.= 3.059 kwww.elsolucionario.net 309. Input: f H =1 2 rr = Rs R2 R3 r ( C + 2C ) Req = 0.1 20.5 28.3 3.06 = 0.096 k r = ( 96 ) (12 + 2 ( 2 ) ) 1012 = 1.537 109 s f H =12 (1.536 109 )Output: f H 103.6 MHz1 2 rr = ( RC RL ) C= (15 10 ) 103 2 1012 = 6.67 109 fH =12 ( 6.67 109 )= 23.9 MHzb. A = g m ( RC R2 R3 r RL ) R2 R3 r + RS R2 R3 r = 20.5 28.3 3.059 = 2.433 k 2.433 A = ( 39.23) ( 5 10 ) A = 125.6 2.433 + 0.1 c. CL = 15 pF > C CL dominates frequency response.www.elsolucionario.net 310. Chapter 8 Exercise Solutions EX8.1VCC = 30 V, VCE = 30 I C RC , I CVCE = 10 1 Maximum power at VCE = VCC = 15 2 10 10 2 IC = = = VCE 15 3a.2 So 15 = 30 RL RL = 22.5 Maximum Power = 10 W 3 b. VCC = 15 V, I C ,max = 2 A VCE = 15 I C RL0 = 15 2 RL RL = 7.5 Maximum Power = (1)( 7.5 ) = 7.5 WEX8.2 (a)(b)T = P = ( 8 )( 2.4 ) T = 19.2 C T = P T 85 = P= 3.7 P = 23.0 WEX8.3 Power = iD vDS = (1)(12 ) = 12 wattsc.Tsink = Tamb + P snk ambTsink = 25 + (12 )( 4 ) Tsink = 73Cb.Tcase = Tsink + P case snk Tcase = 73 + (12 )(1) Tcase = 85Ca.Tdev = Tcase + P dev caseTdev = 85 + (12 )( 3) Tdev = 121CEX8.4www.elsolucionario.net 311. dev case = PD ,max =TJ ,max Tamb PD,rated=200 25 = 3.5C/W 50TJ ,max Tamb dev case + case snk + snk amb200 25 PD ,max = 29.2 W 3.5 + 0.5 + 2 = Tamb + PD ,max ( case snk + snk amb ) = 25 + ( 29.2 )( 0.5 + 2 ) Tcase = 98C= TcaseEX8.5 I DQ =a.10 4 I DQ = 60 mA 0.1b.9 vds = ( 60 )( 0.050 ) = 2.7 V vDS ( min ) = 4 2.7 = 1.3 V 10 So maximum swing is determined by drain-to-source voltage. VPP = 2 ( 2.5 ) = 5.0 Vc. 1 VP2 1 ( 2.5 ) = PL = 31.25 mW 2 RL 2 0.1 2PL =PS = VDD I DQ = (10 )( 60 ) = 600 mW=PL PS=31.25 = 5.2% 600EX8.6 Computer Analysis EX8.7 No Exercise Problem EX8.8 a. vI = v0 + vGSn VBB 2dv dvI = 1 + GSn dv0 dv0 iDn = K n ( vGSn VTN ) vGSn =2iDn + VTN KndvGSn dvGSn diDn = dvo diDn dvowww.elsolucionario.net 312. SodvGSn = diDn11 1 K n 2 iDnAt v0 = 0, iDn = 0.050 A SoiDndvGSn 1 1 1 = =5 diDn 2 0.050 0.2 = iL + iDpFor a small change in v0 iL = iDn ( iDp ) 1 iL 2 di 1 di 1 1 1 1 = = 0.025 or Dn = L = dv0 2 dv0 2 RL 2 20So iDn =ThendvGSn = ( 5 )( 0.025 ) = 0.125 dv0Thendv dvI 1 = 1 + 0.125 = 1.125 and Av = 0 = Av = 0.889 dv0 dvI 1.125For v0 = 5 V, iL = 0.25 A = iDn , and iDp = 0b.dvGSn dvGSn diDn = dv0 diDn dv0 dvGSn = diDn11 1 1 1 1 = = 2.24 0.2 2 0.25 K n 2 iDndiDn diL 1 = = = 0.05 dv0 dv0 20 dvGSn = ( 2.24 )( 0.05 ) = 0.112 dv0 dvI = 1 + 0.112 = 1.112 dv0 Av =dv0 1 = Av = 0.899 dvI 1.112EX8.9 a. 2 Rb = r + (1 + ) RE and RE = a 2 RL = (10 ) ( 8 ) = 800 Ri = 1.5 k = RTH Rb V 18 I Q = 2CC = = 22.5 mA a RL (10 )2 ( 8 ) r =(100 )( 0.026 )= 0.116 k 22.5 Rb = 0.116 + (101)( 0.8 ) = 80.9 k1.5 = RTH 80.9 =RTH ( 80.9 )RTH + ( 80.9 ) ( 80.9 1.5 ) RTH = (1.5 )( 80.9 ) RTH = 1.53 k R2 1 VTH = VCC = RTH VCC R1 + R2 R1 www.elsolucionario.net 313. 22.5 = 0.225 mA 100 V 0.7 1 = TH (1.53)(18 ) = ( 0.225 )(1.53) + 0.7 R1 = 26.4 k RTH R1I BQ = I BQIQ=26.4 R2 = 1.53 26.4 + R2( 26.4 1.53) R2 = (1.53)( 26.4 ) R2 = 1.62 k b. vE = 0.9VCC = ( 0.9 )(18 ) = 16.2 V iE = 0.9 I CQ = ( 0.9 )( 22.5 ) = 20.25 mA v 16.2 v0 = E = VP = 1.62 V a 10 i0 = aiE = (10 )( 20.25 ) I P = 203 mA PL =1 (1.62 )( 0.203) PL = 0.164 W 2EX8.10 a. IC VBE = VT ln I SQ 5 103 = ( 0.026 ) ln = 0.6225 V VD1 = VD 2 = 0.6225 13 2 10 V I C = I SQ exp BE VT VBE 0.6225 I Bias = I D = I SD exp 0.026 0.6225 = 5 1013 exp 0.026 I Bias = 12.5 mAb.V0 = 2 V, iL =2 = 26.7 mA 0.0751st approximation: iCn 26.7 mA, iBn = 0.444 mA 26.7 103 VBE = ( 0.026 ) ln = 0.6661 13 2 10 I D = 12.5 0.444 = 12.056 mA 12.056 103 VD = ( 0.026 ) ln = 0.6216 13 5 10 2VD = 1.243 V VEB = 2VD VBE = 0.5769 0.5769 icP = 2 1013 exp = 0.866 mA 0.026 2nd approximation: iEn = iL + iCP = 26.7 + 0.866 27.6 mA = iEn 60 iCn = ( 27.6 ) iCn = 27.1 mA 61 iBn = 0.452 mA I D = 12.5 0.452 I D = 12.05 mAwww.elsolucionario.net 314. 27.1 103 VBEn = 0.6664 V VBEn = ( 0.026 ) ln 13 2 10 12.05 103 VD = ( 0.026 ) ln = 0.6215 V 13 5 10 2VDD = 1.243 V VEB = 1.243 0.6664 VEBp = 0.5766 V 0.5766 iCP = 2 1013 exp iCP = 0.856 mA 0.026 c. 10 = 133 mA 0.075 iL = 133 mA iCN = 131 mAV0 = 10 V, iL = iEniBn = 2.18 mA I D = 12.5 2.18 I D = 10.3 mA 10.3 103 VD = ( 0.026 ) ln = 0.6175 13 5 10 2VDD = 1.235 V 131 103 VBEn = 0.7074 V VBEn = ( 0.026 ) ln 13 2 10 VEBp = 1.235 0.7074 VEBp = 0.5276 V 0.5276 iCP = 2 1013 exp iCP = 0.130 mA 0.026 EX8.11 No Exercise Problem EX8.12 a. vI = 0 = v0 , vB 3 = 0.7 V 12 0.7 11.3 I R1 = = I R1 = 45.2 mA R1 0.25 If transistors are matched, then iE1 = iE 3 i iR1 = iE1 + iB 3 = iE1 + E 3 1+ 1 1 iR1 = iE1 1 + = iE1 1 + 1+ 41 45.2 iE1 = iE1 = iE 2 = 44.1 mA 1.024 i 44.1 iB1 = iB 2 = E1 = iB1 = iB 2 = 1.08 mA 1+ 41 b.www.elsolucionario.net 315. For vI = 5 V v0 = 5 V 5 i0 = 0.625 A 8 0.625 iE 3 0.625 A, iB 3 = iB 3 = 15.2 mA 41 12 5.7 = 25.2 mA vB 3 = 5.7 V iR1 = 0.25 10 = 0.244 mA iE1 = 25.2 15.2 iE1 = 10.0 mA iB1 = 41 vB 4 = 5 0.7 = 4.3 V i0 =I R2 =4.3 ( 12 ) 0.25= 65.2 mA iE 265.2 = 1.59 mA 41 iI = iB 2 iB1 = 1.59 0.244 iI = 1.35 mA iB 2 =i0 625 = AI = 463 iI 1.35 From Equation (8.54) (1 + ) R ( 41)( 250 ) AI = = = 641 2 RL 2 (8) AI =c.TYU8.124 = 1.2 A = I D ( max ) 20 For I D = 0 VDS ( max ) = 24 V For VDS = 0, I D ( max ) =Maximum power when VDS = ID =VDS ( max ) 2 I D ( max ) 2= 12 V and= 0.6 A PD ( max ) = (12 )( 0.6 ) = 7.2 WattsTYU8.2Maximum power at center of load line Pmax = ( 0.05 )(10 ) Pmax = 0.5 W TYU8.3www.elsolucionario.net 316. a.PQ = VCEQ I CQ = ( 7.5 )( 7.5 ) PQ = 56.3 mW1 VP2 1 ( 6.5 ) = PL = 21.1 mW 2 RL 2 1 2b.PL =PS = (15 )( 7.5 ) PS = 113 mW=PL PS=21.1 = 18.7% 113PQ = 56.3 21.1 = 35.2 mWTYU8.4a.PL =1 VP2 20 VP = 2 RL PL = 2 ( 8 )( 25 ) VP = 20 V VCC = VCC = 25 V 2 RL 0.8b.IP =VP 20 = I P = 2.5 A 8 RLc.PQ =VCCVP VP2 RL 4 RL( 25 )( 20 ) ( 20 ) PQ = (8) 4 (8)2d. 19.9 12.5 PQ = 7.4 W VP 20 = = 62.8% 4VCC 4 25TYU8.5a.PL =( 4) 1 VP2 = PL = 80 mW 2 RL 2 ( 0.1)b.IP =VP 4 I P = 40 mA = RL 0.1c.PQ =VCCVP VP2 RL 4 RLPQ =( 5 )( 4 ) ( 4 ) = 63.7 40 PQ = 23.7 mW ( 0.1) 4 ( 0.1)22d.= VP 4VCCTYU8.6 a. 1 2V I CQ CC 2 RL= 4 = 62.8% 4 5 VCC 12 = = 8 mA = RL 1.5RTH = R1 R2 R2 1 VTH = VCC = RTH VCC R1 R1 + R2 I CQ VTH VBE 8 = = 0.107 mA = I BQ = 75 RTH + (1 + ) REwww.elsolucionario.net 317. Let RTH = (1 + ) RE = ( 76 )( 0.1) = 7.6 k 1 ( 7.6 )(12 ) 0.7 R1 0.107 = 7.6 + 7.6 1 ( 91.2 ) = 2.33 R1 = 39.1 k R1 39.1R2 = 7.6 ( 39.1 7.6 ) R2 = ( 7.6 )( 39.1) R2 = 9.43 k 39.1 + R2 b. 2 2 1 1 PL = ( 0.9 I CQ ) RL = ( 0.9 )( 8 ) (1.5 ) PL = 38.9 mW 2 2 PS = VCC I CQ = (12 )( 8 ) = 96 mWPQ = PS PL = 96 38.9 PQ = 57.1 mW=PL PS=38.9 = 40.5% 96TYU8.7I E = I E 3 + IC 4 + IC 5 = I E 3 + IC 4 + 5 I B5= I E 3 + 4 I B 4 + 5 (1 + 4 ) I B 4I E = (1 + 3 ) I B 3 + 4 3 I B 3 + 5 (1 + 4 ) 3 I B 3If 4 and 5 are large, then I E 3 4 5 I B 3 So that composite current gain is 3 4 5www.elsolucionario.net 318. Chapter 8 Problem Solutions 8.1 a.b.i.VD D = 80 VMaximum power at VDS = ID = RD =ii.VD D 2= 40 VPT 25 = = 0.625 A VD S 40 80 40 RD = 64 0.625 VD D = 50 VMaximum power at VD S =VD DID 25 VPT 25 = =1 A VDS 25RD =250 25 RD = 25 18.2 a.www.elsolucionario.net 319. PQ (max) = I C Q So I C Q = RL =VC C2 2 PQ (max) VCC=2(20) = 1.67 A 24VCC (VCC / 2) 24 12 = RL = 7.2 I CQ 1.671.67 20.8 mA 80 24 0.7 RB = RB = 1.12 k 20.8 b. I CQ RL (1.67 )( 7.2 ) = = 462 Av = g m RL = VT 0.026 IB =I CQ=V0 (max) = 12 V VP =V0 (max) 12 = VP 26 mV Av 4628.3 a.For maximum power delivered to the load, set VC EQ =VC C 2Set VC C = 25 V = VCE ( sus )Then I Cm =VCC 25 = RL 0.1I Cm = 250 mA < I C ,max 25 12.5 = 125 mA 0.1 V PQ ( max ) = I CQ CC = ( 0.125 )(12.5 ) 2 = 1.56 W < PD,max I CQ =125 = 1.25 mA 100 25 0.7 RB = RB = 19.4 k 1.25 1 2 1 2 b. PL ( max ) = I CQ RL = ( 0.125 ) (100 ) PL ( max ) = 0.781 W ( rms ) 2 2 I BQ =8.4Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output.www.elsolucionario.net 320. 8.5 a.b. VGG = 5 V, I D = 0.25 ( 5 4 ) = 0.25 A, VD S = 37.5 V, P = 9.375 W 2VGG = 6 V, I D = 0.25 ( 6 4 ) = 1.0 A, VD S = 30 V, P = 30 W 2VGG = 7 V, I D = 0.25 ( 7 4 ) = 2.25 A, VD S = 17.5 V, P = 39.375 W 2VGG = 8 V, I D2 = 0.25 2 ( 8 4 )VD S VD S =40 VD S10 I D = 3.71 A, P = 10.8 W VGG = 9 V, I D VD S = 2.922 = 0.25 2 ( 9 4 ) VD S VD S 40 VD S VD S = 1.88 V 10 I D = 3.81 A, P = 7.16 W c. Yes, at VGG = 7 V, P = 39.375 W > PD ,max = 35 W =8.6 a. VDD = 25 V 2 50 25 = = 1.25 A 20Set VDSQ = I DQI DQ = K n (VGS VTN )21.25 + 4 = VGS = 6.5 V 0.2 R2 VGS = VDD R1 + R2 Let R1 + R2 = 100 k R 6.5 = 2 ( 50 ) R2 = 13 k 100 R1 = 87 kb.PD = I DQVDSQ = (1.25 )( 25 ) PD = 31.25 Wc. I D ,max = 2 I DQ I D ,max = 2.5 A VDS ,max = VDD VDS ,max = 50 V PD ,max = 31.25 Wd.www.elsolucionario.net 321. V0 = g m RL Vi g m = 2 K n I DQ = 2( 0.2 )(1.25) = 1 A / VV0 = (1)( 20 )( 0.5 ) = 10 V 1 V02 1 (10 ) = PL = 2.5 W 2 RL 2 20 2PL =PQ = 31.25 2.5 PQ = 28.75 W8.7 (a)PD = PD ,max ( Slope ) (T j 25 )(b)60 + 25 T j ,max = 145C 0.5 T j ,max Tcase 145 25 or dev amb = dev amb = 2C/W = 60 dev ambAt PD = 0, T j ,max = (c)PD ,max8.8 PD ,rated =T j ,max Tamb dev caseor dev case =T j ,max Tamb PD ,rated150 25 = 2.5C/W 50 Tamb = PD ( dev case + case amb ) =Then Tdev150 25 = PD ( 2.5 + case amb ) 125 = PD ( 2.5 + case amb )8.9 PD = I D VDS = ( 4 )( 5 ) = 20 W Tdev Tamb = PD ( dev case + case snk + snk amb ) Tdev 25 = 20 (1.75 + 0.8 + 3) = 111 Tdev = 136C Tdev Tcase = PD dev case = ( 20 )(1.75 ) = 35Tcase = Tdev 35 = 136 35 Tcase = 101C Tcase Tsink = PD case snk = ( 20 )( 0.8 ) = 16C Tsink = Tcase 16 = 101 16 Tsink = 85C8.10www.elsolucionario.net 322. Tdev Tamb = PD ( dev case + case amb )200 25 = 25 ( 3 + case amb ) case amb = 4C/W8.11 dev case = PD = =T j ,max Tamb PD ,rated=175 25 = 10C/W 15T j ,max Tamb dev case + case snk + snk amb 175 25 PD = 10 W 10 + 1 + 48.12=PL PSPS = VCC I Q V PL = VP I P = CC ( I Q ) 2 1 VCC I Q = 50% = 2 VCC I Q8.13 vo ( max ) = 4.8 V iC 3 = iC 2 = vI = vo + 0.70.7 ( 5 ) 1iL ( max ) = 4.3 mA =so 3.6 vI 5.5 V vo ( min ) = 4.3 V= 4.3 mA vS ( min ) 18.14www.elsolucionario.net 323. I D 3 = K (VGS 3 VTN ) = 20 VGS 3 ( 5 ) R12 (VGS 3 0.5 ) = 5 VGS 3 2 2VGS 3 11VGS 2 = 0 2VGS 3 =11 (11)2+ 4 (12 )( 2 )2 (12 )VGS 3 = VGS 2 = 1.072 V I D 3 = I D 2 = 12 (1.072 0.5 ) = 3.93 mA VDS 2 ( sat ) = VGS 2 VTN = 1.072 0.5 = 0.572 V 2vo ( min ) : i2 ( max ) = 3.93 = vI ( min ) = vo ( min ) + VTN vI ( min ) = 3.43 VV0 ( min )1 = 3.93 + 0.5 V0 ( min ) = 3.93 Vvo ( max ) = 5 VDS ( sat ) = 5 0.572 vo ( max ) = 4.43 V 4.43 I D1 ( max ) = 3.93 + = 8.36 mA 1 I D1 = 8.36 = 12 (VGS 1 0.5 ) VGS 1 = 1.33 V vI ( max ) = vo + VGS1 = 4.43 + 1.33 vI ( max ) = 5.76 V 28.15 a. Neglect base currents. v0 ( max ) = V + VCE (sat) = 10 0.2 = 9.8 V 9.8 9.8 iL (max) = I Q = = I Q = 9.8 mA RL 1 0 0.7 ( 10 ) R = 949 9.8 iE1 ( max ) = 2 I Q iE1 ( max ) = 19.6 mA R=iE1 ( min ) = 0 iL ( max ) = I Q = 9.8 mA iL ( min ) = I Q = 9.8 mAb. 2 1 1 ( iL ( max ) ) RL = 2 ( 9.8)2 (1) PL = 48.02 mW 2 PS = I Q (V + V ) + I Q ( 0 V )PL 9.8 ( 20 ) + 9.8 (10 ) PS = 294 mW=PL PS=8.16 a. I Q ( min ) = R=48.02 = 16.3% 294v0 ( max ) RL0 0.7 ( 12 ) 100=10 I Q ( min ) = 100 mA 0.1 R = 113 b.www.elsolucionario.net 324. PQ1 = I Q VCE1 = (100 )(12 ) PQ1 = 1.2 W P (source) = 2 I Q (12 ) = 2.4 Wc.(10 ) 1 VP2 = = 0.5 W 2 RL 2 (100 ) 2PL =PS = 1.2 + 2.4 = 3.6 W=PL PS=0.5 = 13.9% 3.68.17 I D1 = K n (VGS VTN ) = 12 ( 0 ( 1.8 ) ) 22I D1 = 38.9 mA (a) For RL = vo ( max ) = 4.8 VVDS ( sat ) = VGS VTN = 1.8 V vo ( min ) = 5 + 1.8 = 3.2 VvI = vo + 0.7 2.5 vI 5.5 VFor RL = 500 vo ( max ) = 4.8 V(b)For vo < 0, vo ( min ) = 3.2 V I2 =vo 3.2 = = 6.4 mA RL 0.52.5 vI 5.5 V (c) For vo = 2V , I 2 ( max ) = 38.9 mA 2 R2 ( min ) = RL ( min ) = 51.4 38.91 v2 1 ( 2) PL = o = PL = 38.9 mW 2 RL 2 51.4 2PL = 10 ( 38.9 ) = 389 mW % =38.9 = 10% 3898.18 + V 2 (V ) PL = P = RL RL21 (V ) 1 (V ) + , V = V + PS = 2 RL 2 RL + 2 2(V ) =+ 2So PS=PL PSRL = 100%8.19 (a)www.elsolucionario.net 325. As maximum conversion efficiency V = , P = 0.785 4 VCC 4 So V p ( max ) = ( 0.785 )( 5 ) V p ( max ) = 5 V(b)Maximum power dissipation occurs when V p =(c)P ( max ) = 2=2VCC=2 ( 5)2 VCC 2 RL( 5)2 2 RL RL = 1.27 8.20 P=(a)2 1 Vp 2 RL2 1 Vp V p = 49 V V + = 52 V, V = 52 V 2 24 V 49 IP = P = = 2.04 A RL 2450 =(b)=(c) VP =4 VCC 49 4 52 = 74.0% 8.21 (a) VDS VDS ( sat ) = VGS VTN = VGS VDS = 10 Vo ( max ) and I D = I L = K n (VGS ) Vo ( max ) RL VGS K n (VGS )2Vo ( max ) RL K nSo 10 Vo ( max ) = 210 V0 ( max ) = Vo ( max ) RL K n=2VGS20.5 ( 5 )( 0.4 )V0 ( max )V02 ( max ) 20.5V0 ( max ) + 100 = 0 V0 ( max ) =Vo ( max )V0 ( max )100 20V0 ( max ) + V02 ( max ) =iL =2( 20.5 ) 222 4 (100 ) V0 ( max ) = 8 V8 iL = 1.6 mA 5 i 1.6 = L = = 2 V VI = 10 V Kn 0.4b.www.elsolucionario.net= 3.183 V 326. 1 (8) PL = = 6.4 mW 2 5 20 (1.6 ) PS = = 10.2 mW 2=PL PS=6.4 = 62.7% 10.28.22vO = iL RL and iL = iD = K n ( vGS VTN ) or iL = K n ( vGS ) and vGS = vI vO Then 2vO = K n RL ( vI vO ) or vO = 2 ( vI vO ) 222 dv dv0 = ( 2 )( 2 )( vI v0 ) 1 0 dvI dvI dv0 1 + 4 ( vI v0 ) = 4 ( vI v0 ) dvI or4 ( vI v0 ) dv0 = dvI 1 + 4 ( vI v0 )For vI = 10 V, v0 = 8 V At vI = 0, v0 = 0 4 (10 8 ) dv0 dv = 0 = 0.889 dvI 1 + 4 (10 8 ) dvIdv0 =0 dvIAt vI = 1, v0 = 0.5 dv0 = 0.667 dvI8.23 a. i 5 103 VBE = VT ln C = ( 0.026 ) ln 13 5 10 IS V VBE = BB = 0.5987 V VBB = 1.1973 V 2 PQ = iC vCE = ( 5 )(10 ) PQ = 50 mWb.www.elsolucionario.net 327. v0 = 8 V iL =8 iL = 80 mA 0.1 iCp 80 mA iCp 80 103 vEB = VT ln = ( 0.026 ) ln 13 5 10 IS vEB = 0.6708 V VBB vEB + v0 = 0.5987 0.6708 8 vI = 8.072 V 2 = VBB vEB = 1.1973 0.6708 = 0.5265 VvI = VBEv 0.5265 iCn = I S exp BE = 5 1013 exp iCn = 0.311 mA 0.026 VT 2 PL = iL RL = ( 80 ) ( 0.1) PL = 640 mW 2PQn = iCn vCE = ( 0.311) (10 ( 8 ) ) PQn = 5.60 mW PQp = iCp vEC = ( 80 )( 2 ) PQp = 160 mW8.24 (a)iDn = K n ( vGSn VTN )2V 0.5 + 2 = vGSn = 2.5 V = BB VBB = 5.0 V 2 2 Pn = ( 0.5 )(10 ) Pn = Pp = 5 mW(b)VDS = VGS VTN VDS = VGS 2 VDS = 10 vo ( max ) and v ( max ) v ( max ) iL + VTN = O +2 = O +2 Kn RL K n ( 2 )(1)VGS =so 10 v0 ( max ) =v0 ( max ) 2+22 =v0 ( max ) 2so v0 ( max ) = 8 V iDn = iL =8 iDn = iL = 8 mA 18 + 2 VGS = 4 V 2 V Then vI = vo + VGS BB = 8 + 4 2.5 vI = 9.5 V 2 VBB vSGp = vo vI = 8 ( 9.5 2.5 ) 2 vSGp = 1 V M p cutoff iDp = 0VGS =2 PL = iL RL = ( 8 ) (1) PL = 64 mW 2PMn = iDn vDS = ( 8 )(10 8 ) PMn = 16 mW PMp = iDp vSD PMp = 08.25 a.www.elsolucionario.net 328. v0 = 24 V iL =24 iL iN = 3 A 83 iBn = 73.2 mA 41 For iD = 25 mA iR1 = 25 + 73.2 = 98.2 mA iBn =i VBE = VT ln N IS Then 98.2 = 3 = ( 0.026 ) ln 12 6 10 = 0.7004 V30 ( 24 + 0.7 ) R1 R1 =5.3 R1 = 53.97 98.2 25 103 = 0.5759 V VD = ( 0.026 ) ln 12 6 10 VEB = 2VD VBE = 2 ( 0.5759 ) 0.7004 = 0.4514 V V 0.4514 iP = I S exp EB = ( 6 1012 ) exp iP = 0.208 mA VT 0.026 b. Neglecting base current 30 0.6 30 0.6 iD = iD 545 mA R1 53.97 0.545 VD = ( 0.026 ) ln = 0.656 V 12 6 10 Approximation for iD is okay. Diodes and transistors matched iN = iP = 545 mA8.26 (a) I D1 = K1 (VGS 1 VTN ) VGS1 =25 +1 = 2 V 5I D 3 = K 3 (VGS 3 VTN )2200 = K 3 ( 2 1) K n3 = K p 4 = 200 A / V 2 2(b) vI + VSG 4 + VGS 3 VGS1 = vO For vo large, iL = i1 = K n1 (VGS1 VTN ) VGS1 =iL + VTN = K n1 So vI + 2 + 2 2vo + VTN RL K n1 vo + 1 = v0 ( 0.5)( 5) v0 3 2.5 dv 1 dv dvI 1 =1= 0 + 0 dvI dvI 2 2.5v0 dvIvI = v0 + 1 1 + 2 2.5v0 For vO = 5 V : 1=dv0 dvIwww.elsolucionario.net 329. 1= dv dv0 dv 1 1 + = 0 (1.1414 ) 0 = 0.876 dvI 2 2.5 ( 5 ) dvI dvI 8.27 vO = vI +I Dn + VTN KnVBB VGS and VGS = 2For vO 0, I Dn = I DQ + iL = I DQ + Then vO = vI +I DQ + ( vO / RL ) I DQ v VBB V or vO = vI + BB VTN 1+ O VTN 2 Kn I DQ RL 2 KnFor vO small, vO vI + 1 I DQ vO 1 + 2 Kn Now dvO = dvI 1 1 + 2 SoI DQ VBB 1 v VTN 1+ O Kn 2 2 I DQ RLI DQ V 1 = vI + BB VTN 2 I DQ RL Kn 11 K n I DQ RL I DQ= 0.951 I DQ 1 1 = 1 = 0.0526 2 K n I DQ RL 0.95For RL = 0.1 k , then OrvO RL1 K n I DQ= 0.01052K n I DQ = 95.1We can write g m = 2 K n I DQ = 190 mA/V This is the required transconductance for the output transistor. This implies a very large transistor. 8.28 Av = g m RLI CQSo 12 = g m ( 2 ) g m = 6 mA/V=VTI CQ = ( 6 )( 0.026 ) I CQ = 0.156 mABut for maximum symmetrical swing, set I CQ =VCC 10 = = 5 mA Av > 12 2 RLMaximum power to the load: 2 1 VCC (10 ) = PL ( max ) = 25 mW 2 RL 2 ( 2) 2PL ( max ) =PS = VCC I CQ = (10 )( 5 ) = 50 mWSo = 50%www.elsolucionario.net 330. 5 = 0.0278 mA 180 R1 = RTH = 6 k I CQI BQ VTH = I BQ RTH + VBE ( on ) + (1 + ) I BQ RE Set RE = 20 VTH = ( 0.0278 )( 6 ) + 0.7 + (181)( 0.0278 )( 0.020 ) VTH = 0.967 V 1 RTH VCC R1VTH =0.967 =1 ( 6 )(10 ) R1 = 62.0 k R1 R2 = 6.64 k8.29 I CQ =VCC 15 = = 15 mA 1 RLI BQ =15 = 0.15 mA 100(15) 1 V2 PL ( max ) = 112.5 mW PL ( max ) = CC = 2 RL 2 (1) 2Let RTH = 10 k VTH = I BQ RTH + VBE + (1 + ) I BQ RE= ( 0.15 )(10 ) + 0.7 + (101)( 0.15 )( 0.1)VTH = 3.715 =1 1 RTH VCC = (10 )(15 ) R1 R1R1 = 40.4 k R2 = 13.3 k8.30 R2 1.55 VTH = VCC = (10 ) R1 + R2 1.55 + 0.73 = 6.80 V RTH = R1 R2 = 0.73 1.55 = 0.496 k I BQ =VTH VBE 6.80 0.70 = RTH + (1 + ) RE 0.496 + ( 26 )( 0.02 )I BQ = 6.0 mA, I CQ = 150 mA Av = g m RL and RL = a 2 RL = ( 3) ( 8 ) = 72 2gm =I CQ VT=150 5.77 A/V 0.026www.elsolucionario.net 331. Av = ( 5.77 )( 72 ) = 415 Vo = Av Vi = ( 415 )( 0.017 ) = 7.06 V 7.06 = 2.35 V 3 7.06 Vo = = 2.35 V 3 7.06 Vo = = 2.35 V 3 PS = I CQ VCC = ( 0.15 )(10 ) = 1.5 W Vo PL PS=0.345 = 23% 1.58.31 a.Assuming the maximum power is being delivered, then 36 9 Vo ( peak ) = 36 V Vo = = 9 V Vrms = Vrms = 6.36 V 4 2 36 Vo = Vo = 25.5 V b. 2c.Secondary I rms =Primary I P =PL 2 = I rms = 0.314 A Vrms 6.360.314 I P = 78.6 mA 4d. PS = I CQ .VCC = ( 0.15 )( 36 ) = 5.4 W 2 = 37% = 5.4 8.32 a.www.elsolucionario.net 332. V ve = + g mV r RE = V 1 + g m RE r 1+ = V r vi = V + ve V = vi ve RE 1+ ve = (vi ve ) RE r 1+ RE 2 n (1 + ) RE ve r v = = = e where RE = 1 RL vi 1 + 1 + R r + (1 + ) RE vi n2 E r n ve so ve v0 1 n1 n2 n2 (1 + ) RE v 1 so 0 = vi n1 r + (1 + ) RE n2 v0 =b. I n1 1 2 1 2 I P RL , a = , I CQ = P so PL = .a 2 I CQ RL n2 a 2 2 PS = I CQ .VCC For = 50% : PL =1 2 2 a I CQ RL a 2 I R VCC VCC V PL CQ L so a 2 = = 0.5 = 2 = = a 2 = CC I CQ VCC I CQ RL ( 0.1)( 50 ) 2VCC 5 PS c. 49 ( 0.026 ) r VT R0 = = = R0 = 0.255 1 + (1 + ) I CQ ( 50 )( 0.1)8.33 a.With a 10:1 transformer ratio, we need a current gain of 8 through the transistor. R1 R2 R1 R2 ie ie = (1 + ) ib and ib = ii so we need = 8 = (1 + ) R R +R R R +R ii ib ib 1 2 1 2 Rib = r + (1 + ) RL (1 + ) RL = (101)( 0.8 ) = 80.8www.elsolucionario.net where 333. R1 R2 Then 8 = (101) R R + 80.8 1 2 R1 R2 = 0.0792 or R1 R2 = 6.95 k R1 R2 + 80.8 2VCC V 12 = RL I CQ = CC = = 15 mA 2 I CQ RL 0.8Set15 = 0.15 mA 100 = I BQ RTH + VBEI BQ = VTH1 RTH VCC = I BQ RTH + VBE R1 1 ( 6.95)(12 ) = ( 0.15)( 6.95) + 0.7 R1 = 47.9 k then R2 = 8.13 k R1 b. I I e = 0.9 I CQ = 13.5 mA = L I L = 135 mA a 1 2 PL = ( 0.135 ) ( 8 ) PL = 72.9 mW 2 PS = VCC I CQ = (12 )(15 ) PS = 180 mW=PL PS = 40.5%8.34 a. VP = 2 RL PL VP = 2 ( 8 )( 2 ) = 5.66 V = peak output voltage IP =VP 5.66 = = 0.708 A = peak output current RL 8Set Ve = 0.9VCC = aVP to minimize distortion Then a =( 0.9 )(18) 5.66 a = 2.86b. 1 I P 1 0.708 = I CQ = 0.275 A 0.9 a 0.9 2.86 Then PQ = VCC I Q = (18 )( 0.275 ) PQ = 4.95 W Power rating of transistor Now I CQ =8.35 a.Need a current gain of 8 through the transistor.www.elsolucionario.net 334. R1 R2 ib = 8 = (1 + ) R R +R ii ib 1 2 where Rib (1 + )( 0.9 ) = 90.9 k R1 R2 8 = = 0.0792 or R1 R2 = 7.82 k R R + 90.9 101 1 2 2VCC 12 = 0.9 k I CQ = = 13.3 mA 2 I CQ 0.9Set13.3 = 0.133 mA 100 1 Then ( 7.82 )(12 ) = ( 0.133)( 7.82 ) + 0.7 R1 = 53.9 k and R 2 = 9.15 k R1 b. I I e = ( 0.9 ) I CQ = 12 mA = L I L = 120 mA a 1 2 PL = ( 0.12 ) ( 8 ) PL = 57.6 mW 2 PS = VCC I CQ = (12 )(13.3) PS = 159.6 mW I BQ PL 57.6 = = 36.1% PS 159.68.36 a.All transistors are matched. 1+ iC 3 mA = iE1 + iB 3 = iC + 61 1 3 = + iC iC = 2.90 mA 60 60 b. For vo = 6 V , let RL = 200 . 6 = 0.03 A = 30 mA iE 3 200 30 iB 3 = = 0.492 mA 61 iE1 = 3 0.492 = 2.508 mA io =2.508 iB1 = 41.11 A 61 3 iE 2 3 mA iB 2 = 49.18 A 61 iI = iB 2 iB1 = 49.18 41.11 iI = 8.07 A iB1 =Current gain Ai = VBE 330 103 Ai = 3.72 103 8.07 106 i 30 103 = VT ln E 3 = ( 0.026 ) ln 13 5 10 IS VBE 3 = 0.6453 V i 2.508 103 VEB1 = VT ln E1 = ( 0.026 ) ln 13 5 10 IS VEB1 = 0.5807 Vwww.elsolucionario.net 335. vI = v0 + VBE 3 VEB1 = 6 + 0.6453 0.5807 vI = 6.0646 V Voltage gain Av =v0 6 = Av = 0.989 vI 6.06468.37 a.For i0 = 1 A, I B 3 1 20 mA 5010 ( v0,max + VBE 3 ) 10 VEB1 = 2 20 R1 R1 10 VBE 2vo,max = + 40 If, for simplicity, we assume VEB1 = VBE 3 = 0.7 V, then R1 R1 We can then writeIf we assume v0,max = 4 V, then b.9.3 2 ( 4 ) = + 40 which yields R1 = R2 = 32.5 R1 R19.3 I E1 = 0.286 A = I E 2 32.5 = 10 I S1,2 , then I E 3 = I E 4 = 2.86 AFor vI = 0, I E1 =Since I S 3,4c. We can write r 1 r 3 + R1 1 + 1 1 R0 = 2 1 + 3 ( 50 )( 0.026 ) V = 0.4545 Now r 3 = 3 T = 2.86 IC 3 r 1 =1VT I C1=(120 )( 0.026 ) 0.286= 10.91 So 10.91 0.4545 + 32.5 1 121 R0 = 2 51 10.91 = 32.5 0.0902 = 0.0900 32.5 121 Then R0 =1 0.4545 + 0.0900 or R 0 = 0.00534 2 51 8.38www.elsolucionario.net 336. {}1 r 1 + (1 + ) R1 ( r 3 + (1 + ) 2 RL ) 2 iC1 7.2 mA and iC 3 7.2 mA Ri =Then r =( 60 )( 0.026 ) 7.2= 0.217 k{}1 0.217 + ( 61) 2 ( 0.217 + ( 61)( 0.2 ) ) 2 1 = 0.217 + 61 2 12.4 or Ri = 52.6 k 2So Ri ={}8.39 a.b. I1 = K1 (VSG + VTP ) = 2V + VSG R15 = 10 (VSG 2 ) VSG = 2.707 V 210 2.707 R1 = R2 = 1.46 k R1 RL = 100 For a sinusoidal output signal: c. 5=www.elsolucionario.net 337. 1 (v ) 1 ( 5) PL = o = PL = 125 mW 2 RL 2 0.1 2iD 3 2( vo ) ( 5 ) RL=0.1 iD 3 = 50 mA50 + 2 = 4.236 V 10 10 ( 4.236 + 5 ) I1 = I D1 = 0.523 mA 1.46 0.523 VSG1 = + 2 = 2.229 V 10 vI = 5 + 4.236 2.229 vI = 7.007 VVGS 3 =I D2 =(VI VGS ) ( 10 )= 10 (VGS 2 )21.46 17.007 VGS 2 = 10 (VGS 4VGS + 4 ) 1.46 2 14.6VGS 57.4VGS + 41.4 = 0 VGS =57.4 ( 57.4 ) 4 (14.6 )( 41.4 ) 2 (14.6 ) 2VGS 2 = 2.98 V I D 2 = 10 ( 2.98 2 ) I D 2 = 9.60 mA 2VG 4 = vI VGS 2 = 7 2.98 = 4.02 V VSG 4 = 5 4.02 = 0.98 V I D 4 = 08.40 For v0 = 0 I Q = I C 3 + I C 2 + I E1 1+ n IC 3 I B3 = I E 2 = IC 2 = n n I C 3 = (1 + n ) I C 2 I I B 2 = I C 1 = P I E1 = C 2 n 1+ P I C 2 = n P I E1 1+ P I C 3 = (1 + n ) n P I E1 1+ p I Q = (1 + n ) n P I E1 + n P I E1 + I E1 1+ P 1+ P 10 10 = ( 51)( 50 ) I E1 + ( 50 ) I E1 + I E1 11 11 I Q = 2318.18I E1 + 45.45 I E1 + I E1 I E1 = 1.692 A I C1 = 1.534 A 10 I C 2 = ( 50 ) (1.692 ) I C 2 = 76.9 A 11 10 I C 3 = ( 51)( 50 ) (1.692 ) I C 3 = 3.92 mA 11 www.elsolucionario.net 338. Because of r 1 and Z, neglect effect of r0. Then neglecting r01, r02 and r03, we find I X = g m 3V 3 + g m 2V 2 + g m1V 1 +VX r 1 + ZNow r V 1 = 1 VX , V 2 g m1V 1r 2 r 1 + Z and V 3 = ( g m1V 1 + g m 2V 2 ) r 3 = g m1V 1 + g m 2 ( g m1V 1r 2 ) r 3 r V 3 = 1 [ g m1 + g m1 g m 2 r 2 ] r 3 VX r 1 + Z ( + 1 2 ) r 3 VX V 3 = 1 r 1 + Z r r and V 2 = g m1 1 r 2VX = 1 2 VX r 1 + Z r 1 + Z ( + 1 2 ) 3 VX 1 Then I X = 1 V X + 1 2 VX + VX + r 1 + Z r 1 + Z r 1 + Z r 1 + Z Then R0 = r 1 =r 1 + Z VX = I X 1 + 1 + 1 2 + ( 1 + 1 2 ) 3(10 )( 0.026 )1.534 Z = 25 k Then= 0.169 MR0 =169 + 25 1 + (10 ) + (10 )( 50 ) + 10 + (10 )( 50 ) ( 50 ) R0 =194 = 0.00746 k or Ro = 7.46 26, 0118.41 a VBBNeglect base currents. I = 2VD = 2VT ln Bias IS 5 103 = 2 ( 0.026 ) ln VBB = 1.281 V 13 10 www.elsolucionario.net 339. VBE1 + VEB 3 = VBB I E1 = I E 3 + I C 2 I B2 = IC 3 = P I E 3 1+ P IC 2 = n I B 2 = n P I E 3 1+ P I E1 = I E 3 + n P 1+ P IE3 I E1 = I E 3 1 + n P 1+ P 1+ n 1+ P P I C1 = I C 3 1 + n P 1+ P n I I VBE1 = VT ln C1 , VEB 3 = VT ln C 3 IS IS (1.01) I C1 = 21 IC 3 20 I C1 20 1 + (100 ) 21 21 = I C 3 + 100 = 101.05 I C 3 20 = 100.05 I C 3 100.05 I C 3 IC 3 VT ln + VT ln = VBB IS IS 2 100.05 I C 3 VT ln = VBB 2 IS 2 100.05 I C 3I2 SV = exp BB VT V exp BB = 0.4995 mA = I C3 100.05 VT Then I E 3 = 0.5245 mA Now I C1 = 100.05 I C 3 = 49.97 mA = I C1 IC 3 =IS 20 I C 2 = (100 ) ( 0.5245 ) = 49.95 mA = I C 2 21 I 49.97 103 VBE1 = VT ln C1 = 0.026 ln 13 10 IS = 0.70037 I 0.4995 103 VEB 3 = VT ln C 3 = 0.026 ln 1013 IS = 0.58062 Note: VBE1 + VEB 3 = 0.70037 + 0.58062 = 1.28099 = VBBb.www.elsolucionario.net 340. v0 = 10 V iE1 10 = 0.10 A = iC1 100100 = 1 mA 100 4 103 = 2 ( 0.026 ) ln = 1.2694 V 13 10 iB1 = VBB 0.1 VBE1 = ( 0.026 ) ln 13 = 0.7184 10 VEB 3 = 1.2694 0.7184 = 0.55099 V 0.55099 I C 3 = 1013 exp = 0.1598 mA 0.026 V 2 (10 ) PL = 0 = PL = 1 W RL 100 2PQ1 = iC1 vCE1 = ( 0.1)(12 10 ) PQ1 = 0.2 W PQ 3 = iC 3 vEC 3 = ( 0.1598 ) (10 [ 0.7 12]) PQ 3 = 3.40 mW iC 2 = (100 )( iC 3 ) = (100 )( 0.1598 ) = 15.98 mAPQ 2 = iC 2 vCE 2 = (15.98 ) (10 [ 12]) PQ 2 = 0.352 W8.42 a. 10 103 VBB = 1.74195 V VBB = 3 ( 0.026 ) ln 12 2 10 VBE1 + VBE 2 + VEB 3 = VBB I C1 IC 2n, IC 3 IC 2 n2I I I VT ln C1 + VT ln C 2 + VT ln C 3 = VBB IS IS IS 3 IC VT ln 3 2 3 = VBB n IS V I C 2 = n I S 3 exp BB VT 1.74195 = ( 20 ) ( 20 1012 ) 3 exp 0.026 I C 2 = 0.20 A, I C1 10 mA, I C 3 0.5 mA 10 103 VBE1 = ( 0.026 ) ln VBE1 = 0.58065 V 12 2 10 0.2 VBE 2 = ( 0.026 ) ln VBE 2 = 0.6585 V 12 2 10 0.5 103 VEB 3 = ( 0.026 ) ln VEB 3 = 0.50276 V 12 2 10 b. 1 V2 1 V2 PL = 10 W= 0 = 0 V0 ( max ) = 20 V 2 RL 2 20For v0 ( max ) :www.elsolucionario.net 341. 2 v0 ( 20 ) = PL = 20 W 20 RL 2PL =20 = 1 A 20 iC 5 + iC 4 + iE 3 = io ( max ) = 1 A i0 ( max ) = =1 i i 1 1+ p iC 5 + C 5 n + C 5 n n 1 + n n 1 + n n p iC 5 +iC 5 n iC 4 1 + p + n 1 + n n p = 1 1 20 1 1 6 iC 5 1 + + = 1 20 21 21 20 5 iC 5 (1.05048 ) = 1 iC 5 = 0.952 A iC 4 = 0.0453 A iE 3 = 0.00272 A5 iC 3 = 0.00272 6 = 0.002267 A 2.267 103 VEB 3 = ( 0.026 ) ln = 0.54206 V 12 2 10 VBE1 + VBE 2 = 1.74195 0.54206 = 1.19989 I I VT ln C 2 + VT ln C 2 = 1.19989 n IS IS 1.19989 iC 2 = n I S exp 0.026 = 20 (18.83) mA iC 2 = 93.9 mA iC 2 n 93.9 = 4.47 mA = n 1 + n 21 PQ 2 = I C 2 ( 24 ( 20 ) ) = ( 0.0939 ) ( 44 ) = 4.13 WiC1 =PQ 5 = ( 0.952 ) ( 10 ( 24 ) ) = 13.3 Wwww.elsolucionario.net 342. Chapter 9 Exercise Solutions EX9.1 Av = 15 = R2 R1R1 = Ri = 20 K R2 = 15 R1 = 15 ( 20 ) = 300 KEX9.2 We can write ACL = R2 R3 R3 1 + R1 R4 R1R1 = R1 = 10 k Want ACL = 50 Set R2 = R3 = 50 k R ACL = 50 = 5 1 + 3 5 R4 R R 50 1+ 3 = 9 3 = 8 = R4 R4 R4 R4 = 6.25 kEX9.3 R2 1 R1 R2 1 1 + 1 + Ad R1 R Ri = R1 = 25 k Let 2 = x R1We have ACL = x 1 1+ (1 + x ) 5 103 x = x 1.0002 + 5 103 x 12 1.0002 + =x 5 103 12 = 12.0024 = x ( 2.4 10 3 ) x R2 12.0024 = 12.0313 = 0.9976 25 k R2 = 300.78 k x=EX9.4 R R R R v0 = F vI 1 + F vI 2 + F vI 3 + F VI 4 R2 R3 R4 R1 www.elsolucionario.net 343. We needRF R R R = 7, F = 14, F = 3.5, F = 10 R1 R2 R3 R4Set RF = 280 k 280 7 280 R2 = 14 280 R3 = 3.5 280 R4 = 10Then R1 40 k = 20 k = 80 k = 28 kEX9.5 We may note thatR3 R R R 3 20 = = 2 and F = = 2 so that 3 = F R2 1.5 R1 10 R2 R1Then iL = ( 3) vI iL = 2 mA = R2 1.5 kvL = iL Z L = ( 2 103 ) ( 200 ) = 0.4 V i4 =vL 0.4 = = 0.267 mA R2 1.5 ki3 = i4 + iL = 0.267 + 2 = 2.267 mAv0 = i3 R3 + vL = ( 2.267 103 )( 3 103 ) 0.4 v0 = 7.2 VEX9.6 Refer to Fig. 9.24 R1 = 2 R1 = 5 k Let R1 = R3 = 2.5 k Set R2 = R4 Differential Gain =v0 R2 R2 = = 100 = R2 = R4 = 250 k v1 R1 2.5 kEX9.7www.elsolucionario.net 344. We have the general relation that R [ R4 / R3 ] R v0 = 1 + 2 v 2v 1 + [ R / R ] I 2 R I1 R1 4 3 1 R1 = R3 = 10 k, R2 = 20 k, R4 = 21 k 20 [ 21/10] 20 v0 = 1 + vI 2 vI 1 10 1 + [ 21/10] 10 v0 = 2.0323vI 2 2.0vI a. vI 1 = 1, vI 2 = 1 v0 = 2.0323 2.0 v0 = 4.032 Vb.vI 1 = vI 2 = 1 V v0 = 2.0323 2.0 v0 = 0.0323 Vc.vcm = vI 1 = vI 2 so common-mode gain v0 = 0.0323 vcmAcm =d. A C M R RdB = 20 log10 d Acm 2.0323 1 Ad = ( 2.0 ) = 2.016 2 2 2.016 C M R RdB = 20 log10 = 35.9 d B 0.0323 EX9.8 v0 = R4 2 R2 1 + ( vI 1 vI 2 ) R3 R1 Differential gain (magnitude) =R4 2 R2 1 + R3 R1 Minimum Gain Maximum R1 = 1 + 50 = 51 k So Ad =20 2 (100 ) 1 + Ad = 4.92 20 51 Maximum Gain Minimum R1 = 1 k Ad =20 2 (100 ) 1 + Ad = 201 20 1 Range of Differential Gain = 4.92 201 EX9.9 Time constant = r = R1C2 = (104 )( 0.1106 ) = 1 m sec 1 t 0 t 1 v0 = R1 C2 At t = 1 m sec v0 = 1 V 0 t 2 v0 = 1 +1 ( t 1) R1 C2At t = 2 m sec v0 = 1 +( 2 1) 1=0www.elsolucionario.net 345. EX9.10 v0 = vI 1 + 10vI 2 25vI 3 80vI 4 From Figure 9.40, v13 input to R1, vI4 input to R2, vI1 input to RA, and vI2 input to RB. From Equation (9.94) RF R = 25 and F = 80 R1 R2 Set RF = 500 k, then R1 = 20 k, and R2 = 6.25 k. R R R R Also 1 + F P = 1 and 1 + F P = 10 RN RA RN RB where RN = R1 R2 = 20 6.25 = 4.76 k and RP = RA RB RCWe find thatRA = 10 RBLet RA = 200 k, RB = 20 k 500 RP RP Now 1 + = 1 = (106 ) 4.76 RA 200 Then RP = 1.89 k RA RB = 200 20 = 18.2 kSo RP = 1.89 =18.2 RC RC = 2.11 k 18.2 + RCEX9.11 Computer Analysis TYU9.1 ACL = R2 100 k = ACL = 10 10 k R1vI = 0.25 V vo = 2.5 V i1 =vI 0.25 = = 0.025 mA i1 = 25 A R1 10 ki2 = i1 = 25 A Ri = R1 = 10 kTYU9.2 (a) R2 Av = R1 + RS 100 = 4.926 19 + 1.3 100 = 5.076 Av ( max ) = 19 + 0.7 so 4.926 Av 5.076 Av ( min ) =(b)www.elsolucionario.net 346. 0.1 = 5.076 A 19 + 0.7 0.1 = 4.926 A i1 ( min ) = 19 + 1.3 so 4.926 i1 5.076 A i1 ( max ) =(c)Maximum current specification is violated.TYU9.3 v0 = Ad ( v2 v1 )Ad = 103a. v2 = 0, v0 = 5 v 5 v1 = 0 = 3 v1 = 5 mV Ad 10 b. v1 = 5, v0 = 10 v0 = v2 v1 Ad 10 = v2 5 v2 = 4.99 V 103 c. v1 = 0.001, v2 = 0.001 v0 = 103 ( 0.001 0.001) v0 = 2 Vd. v2 = 3, v0 = 3v0 = Ad ( v2 v1 )v0 = v2 v1 Ad 3 = 3 v1 v1 = 2.997 V 103TYU9.4 R R R v0 = 4 vI 1 + 4 vI 2 + 4 vI 3 R2 R3 R1 40 40 40 v0 = ( 250 ) + ( 200 ) + ( 75 ) 20 30 10 v0 = [1000 + 400 + 100]v0 = 1500 V = 1.5 mVTYU9.5www.elsolucionario.net 347. vI 1 + vI 2 + vI 3 RF = ( vI 1 + vI 2 + vI 3 ) R 3 RF 1 = R1 = R2 = R3 R = 1 M R 3 1 Then RF = M = 333 k 3 vO =TYU9.6 v R R Av = 0 = 1 + 2 = 5 so that 2 = 4 vI R1 R1 For v0 = 10 V, vI = 2 V Then i1 =2 = 50 A R1 = 40 k R1Then R2 = 160 k we find i2 =v0 vI 10 2 = = 50 A 160 R2TYU9.7v0 = Aod ( v2 v1 ) = Aod ( vI v1 ) v0 v vI = v1 or v1 = vI 0 Aod Aod i1 =v v v1 = i2 and i2 = 0 1 R1 R2 1 v v Then v1 = 0 1 R2 R1 1 1 v v1 + = 0 R1 R2 R2 R R v v0 1 + 2 v1 = 1 + 2 vI 0 R1 R1 Aod R2 1 + R1 v0 So Av = = vI 1 R2 1+ 1 + Aod R1 TYU9.8www.elsolucionario.net 348. Rb For vI 2 = 0, v2 = Rb + Ra vI 1 and R Rb v0 ( vI 1 ) = 1 + 2 vI 1 R1 Rb + Ra 70 50 = 1 + vI 1 5 50 + 25 = 10vI 1 Ra For vI 1 = 0, v2 = vI 2 Rb + Ra R Ra v0 ( vI 2 ) = 1 + 2 vI 2 R1 Rb + Ra 70 25 = 1 + vI 2 5 25 + 50 = 5vI 2 Then v0 = v0 ( vI 1 ) + v0 ( vI 2 ) v0 = 10vI 1 + 5vI 2TYU9.9Ri so i1 = i2 = iS = 100 ARSv0 = iS RFWe want 10 = (100 106 ) RF RF = 100 k TYU9.10 We want iL = 1 mA when vI = 5 V iL = ( 5 ) VI v R2 = I = R2 = 5 k R2 i2 103vL = iL Z L = (103 ) ( 500 ) vL = 0.5 V i4 =vL 0.5 = i4 = 0.1 mA R2 5 ki3 = i4 + iL = 0.1 + 1 = 1.1 mA If op-amp is biased at 10 V, output must be limited to 8 V. So v0 = i3 R3 + vL 8 = (1.1 103 ) R3 + 0.5 R3 = 6.82 kLet R3 = 7.0 kwww.elsolucionario.net 349. Then we wantR3 RF 7 = = = 1.4 R2 R1 5Can choose R1 = 10 k and RF = 14 k TYU9.11 a. v v i1 = I 1 I 2 R1 v01 = vI 1 + i1 R2 , v02 = vI 2 i1 R2 and v0 = v0 =R4 [vI 2 i1 R2 vI 1 i1 R2 ] R3v0 =R4 ( v02 v01 ) R3R4 ( vI 2 vI 1 ) i1 ( R2 + R2 ) R3 vI 2 vI 1 R4 ( vI 2 vI 1 ) ( R2 + R2 ) R3 R1 For common-mode input vI 2 = vI 1 v0 = 0 Common Gain = 0, C M R R = v0 =b. Ad ( min ) R2 min, R1 max 20 100 + 95 Ad = 1 + = 4.82 51 20 20 100 + 105 Ad ( max ) = 1 + = 206 1 20 c. A CM RR = d AcmAcm = 0 C M R R = TYU9.12 Differential Gain =R4 2 R2 1 + R3 R1 Let R3 = R4 so the difference amplifier gain is unity. Minimum Gain Maximum R1 2 R2 So 1 + R ( max ) = 2 1 We want 2 R2 = R1 ( max )Maximum Gain Minimum R1 2 R2 So 1 + R ( min ) = 1000 or 2 R2 = 999 R1 ( min ) 1 If R2 = 50 k, let R1 ( min ) = 100 fixed resistor and let R1 ( max ) = 100 k + 100 = 100.1 potThen actual differential gain is in the range of 1.999 1001 TYU9.131 10 sec 10 106 t 0 = r r (10 ) (10 106 ) After 10 pulses: v0 = 5 = rEnd of 1st pulse: v0 =www.elsolucionario.net 350. 100 106 = 20 sec = r 5 r = R1C2 = 20 sec = 20 106So r =For example, C2 = 0.01 106 = 0.01 F R1 = 2 k TYU9.14 + R R R + R v01 = V ( R R ) + ( R + R ) ( R + R ) + ( R R ) R R R + R + V = 2R 2R R R R R + = V 2R R + v01 = V R For V + = 3.5 V, R = 50, R = 10 103 50 v01 = 4 ( 3.5 ) = 1.75 102 10 Need an amplifier with a gain of Ad =v0 5 = Ad = 285.7 vi 1.75 102Use instrumentation amplifier, Fig. 9-25. Connect v01 to vI1 and ( v01 ) to vI2. R 2R Ad = 4 1 + 2 = 285.7 R1 R3 Let R4 = 150 k, R3 = 10 k ThenR2 = 9.02 R1Let R2 = 100 k, R1 = 11.1 k TYU9.15 1 + R v01 = V 2 R + R (1 + ) R + R (1 + ) 2 R + = V 2 ( R + R (1 + ) ) =R V + 2R ( 2 + ) v01 V + 4 + V = 5 For = 0.01 0.01 v01 = ( 5 ) = 0.0125 4 v 5 = 400 Need a gain Ad = 0 = v01 0.0125 R 2R Use an instrumentation amplifier Ad = 400 = 4 1 + 2 R1 R3 R Let R4 = 150 k, R3 = 10 k then 2 = 12.8 R1Let R2 = 150 k, R1 = 11.7 kwww.elsolucionario.net 351. Chapter 9 Problem Solutions 9.1 (a) vO = Ad ( v2 v1 )()1 = Ad 103 ( 103 ) Ad = 500(b) 1 = 500 ( v2 103 ) = 1 + 0.5 = 500v2v2 = 3 mV (c) 5 = 500 (1 v1 ) 500v1 = 495 v1 = 0.990 V (d) (e)vO = 0 3 = 500 ( v2 ( 0.5 ) )250 3 = 500v2 v2 = 0.506 V9.2 (a) 1 3 v2 = vI = ( 0.49975 10 ) ( 3) 1 + 2000 v2 = 1.49925 103vO = Aod ( v2 v1 ) = ( 5 103 )(1.49925 103 0 ) vO = 7.49625 V(b) vO = Aod ( v2 v1 )3 = Aod (1.49925 103 0 ) Aod = 2 1039.3 Av = R2 = 12 R2 = 12 R1 R1Ri = R1 = 25 k R2 = (12 )( 25 ) = 300 k9.3 (a) (b)v2 = 3.00 VvO = Aod ( v2 v1 ) 2.500 = Aod ( 3.010 3.00 ) Aod = 2509.4www.elsolucionario.net 352. Ri vid = vI Ri + 25 Ri 0.790 = ( 0.80 ) Ri + 25 0.9875 ( Ri + 25 ) = Ri 24.6875 = 0.0125 Ri Ri = 1975 K9.5 200 = 10 20 and for each case Ri = 20 k Av = 9.6 a. 100 = 10 10 Ri = R1 = 10 k b. 100 100 Av = = 5 10 Ri = R1 = 10 k c. 100 Av = = 5 10 + 10 Ri = 10 + 10 = 20 K Av = 9.7www.elsolucionario.net 353. I1 =vI 0.5 R1 = R1 = 5 K R1 0.1R2 = 15 R2 = 75 K R19.8 Av = R2 R1 Av = 10(a) (b) (c) (d) (e) (f)Av = 1 Av = 0.20 Av = 10 Av = 2 Av = 19.9 Av = R2 R1 R1 = 20 K, R2 = 40 K(a) (b) (c) (d)R1 = 20 K, R2 = 200 K R1 = 20 K, R2 = 1000 K R1 = 80 K, R2 = 20 K9.10 Av = R2 = 8 R2 = 8 R1 R1For vI = 1, i1 =1 = 15 A R1 = 66.7 k R2 = 533.3 k R19.11 Av = R2 = 30 R2 = 30 R1 R1Set R2 = 1 M R1 = 33.3 k9.12 a. Av =R R2 1.05R2 = 1.105 2 R1 0.95 R1 R1 R 0.95R2 = 0.905 2 1.05R1 R1 Deviation in gain is +10.5% and 9.5% b. R R 1.01R2 0.99 R2 = 1.02 2 = 0.98 2 Av 0.99 R1 1.01R1 R1 R1 Deviation in gain = 2%9.13 (a)www.elsolucionario.net 354. Av =vO 15 = = 15 vl 1vO = 15vl vO = 150sin t ( mV )(b) i2 = i1 =vI = 10sin t ( A ) R1vO iL = 37.5sin t ( A ) RLiL =iO = iL i2iO = 47.5sin t ( A )9.14 Av = R2 R1 + R5Av = 30 2.5% 29.25 Av 30.75 SoR2 R2 = 29.25 and = 30.75 R1 + 2 R1 + 1We have 29.25 ( R1 + 2 ) = 30.75 ( R1 + 1) Which yields R1 = 18.5 k and R2 = 599.6 k For vI = 25 mV , then 0.731 vO 0.769 V9.15 vO1 = R2 120 , vI = ( 0.2 ) vO1 = 1.2 V 20 R1R4 75 , vO1 = ( 1.2 ) vO = +6 V R3 15 0.2 i1 = i2 = 10 A i1 = i2 = 20 v 1.2 i3 = i4 = 80 A i3 = i4 = O1 = 15 R3 1st op-amp: 90 A into output terminal 2nd op-amp: 80 A out of output terminal. vO = 9.16 (a) R2 22 = Av = 22 1 R1 (b) From Eq. (9.23) R2 1 1 Av = = 22 1 R1 1 R2 1 + 1 + 104 ( 23) 1 + R1 Aod Av = Av = 21.95 (c)www.elsolucionario.net 355. Want Av = 22 ( 0.98 ) = 21.56 So 21.56 =1+22 1 1+ ( 23) Aod1 22 ( 23) = Aod 21.56 1 ( 23) = 0.020408 Aod = 1127 Aod9.17 (a) R2 1 R1 1 R2 1 + 1 + R1 Aod 100 1 = 1 25 1 + 5 103 ( 5 ) Av = 3.9960Av = (b)vO = 3.9960 (1.00 ) vO = 3.9960 V(c)4 3.9960 100% = 0.10% 4(d) vO = Aod ( v2 v1 ) = Aod v1 ( 3.9960 ) vO = Aod 5 10+3v1 = v1 = 0.7992 mV9.18 vO = Aod ( v2 v1 ) = Aod v1 v 5 v1 = O = Aod 5 10+3 v1 = 1 mV9.19 Av = R2 R3 R3 + 1 + R1 R4 R2 a. 10 = 10 =b.R2 100 100 + 1 + 100 100 R2 2 R2 + 1 R2 = 450 k 100 2R 100 = 2 + 1 R2 = 4.95 M 1009.20 a.www.elsolucionario.net 356. R2 R3 R3 + 1 + R1 R4 R2 R1 = 500 k Av = R2 R3 R3 + 1 + 500 R4 R2 Set R2 = R3 = 500 k 80 = 500 500 80 = 1 1 + + 1 = 2 + R4 = 6.41 k R4 R4 b. For vI = 0.05 V 0.05 i1 = i2 = 0.1 A i1 = i2 = 500 k v X = i2 R2 = ( 0.1 106 )( 500 103 ) = 0.05 i4 = vX 0.05 = i4 = 7.80 A 6.41 R4i3 = i2 + i4 = 0.1 7.80 i3 = 7.90 A9.21 (a) Av = 1000 = R2 500 = R1 R1R1 = 0.5 K(b) R2 R3 R3 + 1 + R1 R4 R2 250 500 500 1250 1000 = + 1 + = R1 250 250 R1 R1 = 1.25 K Av =9.22i1 =vI = i2 Rv v A = i2 R = I R = vI R v v i3 = A = I R Rwww.elsolucionario.net 357. vA vA 2v 2v = A = I R R R R 2vI vB = v A i4 R = vI ( R ) = 3vI R i4 = i2 + i3 = ( 3vI ) 3vI vB = = R R R 2vI 3vI 5vI i6 = i4 + i5 = + = R R R v0 5vI v0 = vB i6 R = 3vI R v = 8 R I i5 = From Figure 9.12 Av = 39.23 (a) R2 1 R1 1 R2 1 + 1 + R1 Aod 50 1 = Av = 4.99985 10 1 50 1+ 1 + 2 105 10 Av = (b)vO = ( 4.99985 ) (100 103 ) vO = 499.985 mV(c)Error =0.5 0.499985 100% 0.003% 0.59.24 a.From Equation (9.23) R2 1 Av = R1 1 R2 1 + 1 + R1 Aod 100 1 = = 0.9980 100 1 100 1 + 3 1 + 10 100 Then v0 = Av vI = ( 0.9980 )( 2 ) v0 = 1.9960 Vb.v0 = Aod ( v A vB ) 1 vB v0 vB 1 v = vB + = 0 R1 R2 R1 R2 R2 v0 vB = R2 1 + R1 www.elsolucionario.net 358. Then v0 = Aod v A Aod v0 R2 1 + R1 Aod v0 1 + = Aod v A R 2 1 + R1 R2 1 + + Aod R1 =A v v0 od A R 1 + 2 R1 R2 Aod 1 + v A R1 v0 = R Aod + 1 + 2 R1 R2 1 + vA R1 v0 = 1 R2 1+ 1 + Aod R1 10 vI 1 + 10 2 So v0 = = 0.9980vI 1 10 1 + 3 1 + 10 10 For vI = 2 V v0 = 1.9960 V9.25 ii =(a)vl v v R = i2 = O O = 2 R1 R2 vl R1(b) vl v 1 R2 = i3 + O = i3 + vl R1 RL RL R1 v R Then i3 = l 1 + 2 R1 RL i2 = i1 =9.26 R3 R1 + 0.1 1 VX .max = 10 VX .max = 0.09008 V V = R R +R 0.1 1 + 10 ( ) 4 3 1 R vO = 2 VX .max R1 10 =R2 R ( 0.09008 ) 2 = 111 R1 R1So R2 = 111 k 9.27 (a)www.elsolucionario.net 359. R R R vO = F vI 1 + F vI 2 + F vI 3 R2 R3 R1 100 100 100 = ( 0.5 ) + ( 0.75 ) + ( 2.5 ) 50 20 100 = [1 + 3.75 + 2.5]vO = 7.25 V(b) 100 100 100 2 = (1) + ( 0.8 ) + vI 3 20 100 50 2 = 2 + 4 + vI 3vI 3 = 4 V9.28 vo =R R RF vI 1 F vI 2 F vI 3 R1 R2 R3= 4vI 1 8vI 2 2vI 3 RF =4 R1RF =8 R2RF =2 R3Largest resistance = RF = 250 K R1 = 62.5 KR2 = 31.25 KR3 = 125 K9.29 v0 = 4vI 1 0.5vI 2 = RF =4 R1RF R vI 1 F vI 2 R1 R2RF = 0.5 R1 is the smallest resistor R2i = 100 A =vI 2 = R1 R1 R1 = 20 k RF = 80 k R2 = 160 k9.30 vI 1 = ( 0.05 ) 2 sin ( 2 ft ) = 0.0707 sin ( 2 ft ) 1 1 10 ms f = 1 kHz T = 3 1 ms vI 2 T2 = 100 10 R R 10 10 vO = F vI 1 F vI 2 = vI 1 vI 2 R1 R2 1 5 vO = (10 ) ( 0.0707 sin ( 2 ft ) ) ( 2 )( 1 V ) vO = 0.707 sin ( 2 ft ) ( 2 V )www.elsolucionario.net 360. 9.31 vO = RF R R vI 1 F vI 2 F vI 3 R1 R2 R320 20 20 vI 1 vI 2 vI 3 10 5 2 K sin t = 2vI 1 4 [ 2 + 100sin t ] 0vO = Set vI 1 = 4 mV9.32 Only two inputs. R R vO = F vI 1 + F vI 2 R2 R1 1 = 3vI 1 + vI 2 4 RF RF 1 =3 = R1 R2 4 Smallest resistor = 10 K = R1 RF = 30 KR2 = 120 K9.33 R R vO = F vI 1 + F vI 2 R1 R2 R RF R 5 5sin t = ( 2.5sin t ) F ( 2 ) F = 2 R1 R2 R1RF = 2.5 R2RF = largest resistor RF = 200 K R1 = 100 KR2 = 80 K9.34 a. v0 = RF R R R a3 ( 5 ) F a2 ( 5 ) F a1 ( 5 ) F a0 ( 5 ) R3 R2 R1 R0So v0 =b.RF a3 a2 a1 a0 + + + ( 5) 10 2 4 8 16 R 1 v0 = 2.5 = F 5 RF = 10 k 10 2c. i.v0 =10 1 5 v0 = 0.3125 V 10 16www.elsolucionario.net 361. v0 =ii.10 1 1 1 1 + + + ( 5 ) v0 = 4.6875 V 10 2 4 8 16 9.35 (a) 10 vI 1 1 20 20 vO = vO1 vI 2 = ( 20 )( 10 ) vI 1 ( 20 ) vI 2 1 1 vO = 200vI 1 20vI 2 vO1 = (b) vI 1 = 1 + 2sin t ( mV ) vI 2 = 10 mV Then vO = 200 (1 + 2sin t ) 20 ( 10 ) So vO = 0.4 + 0.4sin t (V ) 9.36 For one-inputv1 = v0 Aodv v vI 1 v1 v1 = + 1 0 R1 R2 R3 RF 1 VI 1 1 1 v0 = v1 + + R1 R1 R2 R3 RF RF =v0 Aod1 1 1 v0 + + R1 R2 R3 RF RF 1 1 1 1 1 = v0 + + + Aod RF RF Aod R1 R2 R3 = v0 1 RF 1 +1+ RF Aod Aod R1 R2 R3 R 1 v0 = F vI 1 where RP = R1 R2 R3 R1 1 RF 1 + 1+ Aod RP Therefore, for three-inputs v0 =R R R 1 F vI 1 + F vI 2 + F vI 3 R2 R3 1 RF R1 1+ 1 + Aod RP 9.37www.elsolucionario.net 362. R R Av = 12 = 1 + 2 2 = 11 R1 R1 v v 0.5 i1 = I R1 = I = R1 i1 0.15R1 = 3.33 K R2 = 36.7 K9.38 1 vB = vI 1 + 500 a. b. 1 v0 = Aod vi 501 1 2.5 = Aod ( 5 ) Aod = 250.5 501 1 v0 = 5000 ( 5 ) v0 = 49.9 V 501 9.39 R Av = 1 + 2 R1 Av = 11 (a)(b) (c) (d) (e) (f)Av = 2 Av = 1.2 Av = 11 Av = 3 Av = 29.40 (a)R2 = 1 R1 = R2 = 20 K R1(b)R2 = 9 R1 = 20 K, R2 = 180 K R1(c)R2 = 49 R1 = 20 K, R2 = 980 K R1(d)R2 = 0 can set R2 = 20 K, R1 = (open circuit) R19.41 50 20 40 v0 = 1 + vI 2 + vI 1 50 20 + 40 20 + 40 v0 = 1.33vI 1 + 0.667vI 29.42 (a)www.elsolucionario.net 363. vI 1 v2 vI 2 v2 v2 + = 20 40 10 100 vO = 1 + v2 = 3v2 50 Now 2vI 1 2v2 + vI 2 v2 = 4v2 v 2vI 1 + vI 2 = 7v2 = 7 o 3 6 3 So vO = vI 1 + vI 2 7 7(b) (c)6 3 ( 0.2 ) + ( 0.3) vO = 0.3 V 7 7 6 3 vO = ( 0.25 ) + ( 0.4 ) vO = 42.86 mV 7 7 vO =9.43 R4 v2 = vI R3 + R4 R R R4 vO = 1 + 2 v2 = 1 + 2 vI R1 R1 R3 + R4 Av =vO R2 R4 = 1 + vI R1 R3 + R4 9.44 (a) vO 50 x = 1 + (1 x ) 50 vI vO x 1 x + x = 1 + = vI 1 x 1 x v 1 Av = O = vI 1 x (b) 1 Av (c) If x = 1, gain goes to infinity.9.45 Change resister values as shown.www.elsolucionario.net 364. i1 =vI = i2 Rv vx = i2 2 R + vI = I 2 R + vI = 3vI R v x 3I i3 = = R R 4v v 3v i4 = i2 + i3 = I + I = I R R R 4vI v0 = i4 2 R + vx = 2 R + 3vI R v0 = 11 vI9.46 (a) (b)vO =1 vI From Exercise TYU9.7 R2 1 + R1 vO = vI 1 R2 1 + 1 + R1 Aod But R2 = 0, R1 = vO v 1 1 = = O = 0.999993 1 1 vI 1 + vI 1+ Aod 1.5 105(b)WantvO 1 Aod = 99 = 0.990 = 1 vI 1+ Aod9.47www.elsolucionario.net 365. v0 = Aod ( vI v0 ) 1 + 1 v0 = vI Aod v0 1 = vI 1 1 + Aod v Aod = 104 ; 0 = 0.99990 vI Aod = 103 ;v0 = 0.9990 vIAod = 102 ;v0 = 0.990 vIAod = 10;v0 = 0.909 vI9.48 R v0 A = 1 + 2 vI R1 R R v01 = 1 + 2 vI , v02 = 1 + 2 vI R1 R1 So v01 = v029.49 (a)iL =vI R1(b) vO1 = iL RL + vI = iL RL + iL R1 vOI ( max ) 10 V = iL (1 + 9 ) = 10iL So iL ( max ) 1 mA Then vI ( max ) iL R1 = (1)( 9 ) vI ( max ) 9 V9.50 (a) 20 20 vX = vI = ( 6 ) = 2 20 + 40 60 vO = 2 V(b) (c)Same as (a) 6 vX = ( 6 ) = 0.666 V 6 + 48 10 vO = 1 + v X vO = 1.33 V 10 9.51 a.www.elsolucionario.net 366. Rin =v v v1 and 1 0 = i1 and v0 = Aod v1 i1 RFSo i1 =v1 ( Aod v1 )Then Rin =RF=v1 (1 + Aod ) RFv1 RF = i1 1 + Aodb. RS RF i1 i1 = iS and v0 = Aod 1 + Aod RS + Rin A RS So v0 = RF od iS 1 + Aod RS + Rin Rin =RF 10 = = 0.009990 1 + Aod 1001 RS 1000 v0 = RF iS 1001 RS + 0.009990 RS 1000 Want 0.990 1001 RS + 0.009990 which yields RS 1.099 k9.52vO = iC RF , 0 iC 8 mA For vO ( max ) = 8 V, Then RF = 1 k 9.53 v 10 i = I so 1 = R = 10 k R R In the ideal op-amp, R1 has no influence. R Output voltage: v0 = 1 + 2 vI R v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input voltage vI in which the output is valid. 9.54 (a)www.elsolucionario.net 367. iL = vI R210mA = ( 10V ) R2R2 = 1 K R 1 = F R2 R1 R3AlsovL = (10mA )( 0.05k ) = 0.5 V 0.5 = 0.5 mA i2 = 1 iR 3 = 10 + 0.5 = 10.5 mA v v 13 0.5 Limit vo to 13V R3 = O L = R3 = 1.19 K 10.5 iR 3 ThenRF R3 1.19 R = = = 1.19 = F 1 R1 R2 R1For example, RF = 119 K, R1 = 100 K(b)From part (a), vO = 13 V when vI = 10 V9.55 (a) i1 = i2 and i2 =vx + iD , vx = i2 RF R2R Then i1 = i1 F R2 R Or iD = i1 1 + F R2 + iD (b) R1 =vI 5 = R1 = 5 k i1 1 R R 12 = (1) 1 + F F = 11 R2 R2 For example, R2 = 5 k , RF = 55 k 9.56 VX VX vO + R2 R3(1)IX =(2)VX VX vO + =0 R1 RFwww.elsolucionario.net 368. R From (2) vO = VX 1 + F R1 1 R 1 1 Then (1) I X = VX + VX 1 + F R1 R2 R3 R3 IX R R 1 1 1 1 1 = = + F = F VX R0 R2 R3 R3 R1 R3 R2 R1 R3 =R1 R3 R2 RF R1 R2 R3or Ro =R1 R2 R3 R1 R3 R2 RFNote: IfRF 1 = R2 RF = R1 R3 then Ro = , which corresponds to an ideal current source. R1 R3 R29.57 R2 R4 = =5 R1 R3 Minimum resistance seen by vI1 is R1. Set R1 = R3 = 25 k Then R2 = R4 = 125 k Ad =iL =v0 v0 = iL RL = ( 0.5 )( 5 ) = 2.5 V RLv0 = 5 ( vI 2 vI 1 )2.5 = 5 ( vI 2 2 ) vI 2 = 2.5 V9.58 vO =R2 ( vI 2 vI 1 ) R1R2 R R and 2 = 4 with R2 = R4 and R1 = R3 R1 R1 R3 Differential input resistance R 20 Ri = 2 R1 R1 = i = = 10 K 2 2 R2 (a) = 50 R2 = R4 = 500 K R1 Ad =R1 = R3 = 10 K(b)R2 = 20 R2 = R4 = 200 K R1 R1 = R3 = 10 K(c)R2 = 2 R2 = R4 = 20 K R1 R1 = R3 = 10 K(d)R2 = 0.5 R2 = R4 = 5 K R1 R1 = R3 = 10 K9.59www.elsolucionario.net 369. We have R R / R R R R2 1 vO = 1 + 2 4 3 vI 2 2 vI 1 or vO = 1 + 2 vI 2 vI 1 R1 1 + R4 / R3 R1 1 + R3 / R4 R1 R1 Set R2 = 50 (1 + x ) , R1 = 50 (1 x ) R3 = 50 (1 x ) , R4 = 50 (1 + x ) 1 + x 1+ x 1 vI 2 vO = 1 + vI 1 1 x 1 + 1 x 1 x 1+ x 1 x + (1 + x ) 1+ x 1+ x vO = vI 2 vI 1 1 x 1 x 1 + x + (1 x ) 1+ x 1+ x = vI 2 vI 1 1 x 1 x For vI 1 = vI 2 vO = 0 SetR2 = 50 (1 + x )R1 = 50 (1 x )R3 = 50 (1 + x ) R4 = 50 (1 x ) 1 1+ x 1+ x vO = 1 + 1 + x vI 2 vI 1 1 x 1+ 1 x 1 x 1+ x = vI 2 vI 1 1 x vI 1 = vI 2 = vcm vO 1 + x 1 x (1 + x ) 2 x = 1 = = vcm 1 x 1 x 1 x SetR2 = 50 (1 x )R1 = 50 (1 + x )R3 = 50 (1 x ) R4 = 50 (1 + x ) 1 1 x 1 x vO = 1 + 1 x vI 2 vI 1 1+ x 1+ 1+ x 1+ x 1 x = 1 vcm 1+ x 1 + x (1 x )2x 1+ x 1+ x Worst common-mode gain Acm =Acm 2 x 1 x(b)www.elsolucionario.net 370. 2 x 2 ( 0.01) = = 0.0202 1 x 1 0.01 2 ( 0.02 ) For x = 0.02, Acm = = 0.04082 1 0.02 2 ( 0.05 ) For x = 0.05, Acm = = 0.1053 1 0.05 1 1 For this condition, set vI 2 = + , vI 1 = vd = 1 V 2 2 1 1 1 + x 1 1 x + (1 + x ) 1 2 = Ad = 1 + = = 2 1 x 2 1 x 2 1 x 1 x For x = 0.01,Acm =For x = 0.01 Ad = 1.010C M R RdB = 20 log101.010 = 33.98 dB 0.0202C M R RdB = 20 log101.020 = 27.96 dB 0.04082For x = 0.02, Ad =1 = 1.020 0.98For x = 0.05 Ad =1 1.0526 = 1.0526 C M R RdB = 20 log10 20 dB 0.95 0.10539.60 10R 10 vy = v2 = ( 2.65 ) v y = vx = 2.40909 V 10R+R 11 v2 v y 2.65 2.40909 i3 = i4 = = = 0.0120 mA 20 R v v 2.50 2.40909 i1 = i2 = 1 x = = 0.0045455 mA 20 R vO = vx i2 (10R ) = ( 2.40909 ) ( 0.0045455 )( 200 ) vO = 1.50 V9.61 iE = (1 + )( iB ) = ( 81)( 2 ) = 162 mA =10 RR = 61.73 9.62 a.From superposition: R2 v01 = vI 1 R1 R R1 v02 = 1 + 2 vI 2 R1 R3 + R4 Setting vI 1 = vI 2 = vcm R 1 v0 = v01 + v02 = 1 + 2 R3 R1 1+ R 4 R 2 vcm R1 www.elsolucionario.net 371. v R R 1 Acm = 0 = 4 1 + 2 R4 vcm R3 R1 1+ R 3 R4 R2 R2 R4 1 + 1 + R R1 R1 R3 = 3 R4 1 + R3 R 2 R1 R4 R2 R R1 Acm = 3 R4 1 + R3 b.Max. AcmMax. Acm Min.R4 R and Max. 2 R3 R147.5 52.5 10.5 9.5 = 4.5238 5.5263 A = cm 47.5 1 + 4.5238 1+ 10.5max= 0.18159.63vI 1 v A v A vB vA v0 = + R1 + R2 Rv R2(1)vI 2 vB vB v A vB = + R1 + R2 Rv R2(2) R1 R2 v = vA + vI 1 R1 + R2 R1 + R2 R1 R2 v+ = vB + vI 2 R1 + R2 R1 + R2 (3) (4)www.elsolucionario.net 372. Now v = v+ R1vA + R2 vI 1 = R1vB + R2 vI 2 R So that v A = vB + 2 ( vI 2 vI 1 ) R1 1 v vI 1 1 1 v = vA + + B 0 R1 + R2 R1 + R2 RV R2 RV R2 1 vI 2 1 1 v = vB + + A R1 + R2 R1 + R2 RV R2 RV (1) ( 2)Then 1 v R 1 vI 1 1 1 v 1 1 + ( vI 2 vI 1 ) = vB + + B 0 + 2 + R1 + R2 R1 + R2 RV R2 RV R2 R1 R1 + R2 RV R2 1 vI 2 R2 1 1 1 = vB + + vB + ( vI 2 vI 1 ) R1 + R2 R1 + R2 RV R2 RV R1 Subtract (2) from (1) R 1 v 1 1 1 1 R2 + + ( vI 2 vI 1 ) 0 + ( vI 2 vI 1 ) ( vI 1 vI 2 ) = 2 R1 + R2 R1 R1 + R2 RV R2 R2 RV R1 v0 1 1 1 1 R2 R 1 = ( vI 2 vI 1 ) 2 + + + + R2 R1 R1 + R2 RV R2 R1 + R2 RV R1 R R2 R R1 R v0 = ( vI 2 vI 1 ) 2 + 2 +1+ + 2 R1 + R2 RV R1 R1 + R2 RV v0 =R2 2 R2 1 + ( vI 2 vI 1 ) R1 RV 9.64www.elsolucionario.net(1) (2) 373. i1 =vI 1 vI 2 ( 0.50 0.030sin t ) ( 0.50 + 0.030sin t ) = 20 R10.060sin t 20 i1 = 3sin t ( A ) vO1 = i1 R2 + vI 1 = ( 0.0030sin t )(115 ) + 0.50 0.030sin t vO1 = 0.50 0.375sin t =vO 2 = vI 2 i1 R2 = 0.50 + 0.030sin t ( 0.003sin t )(115 ) vO 2 = 0.50 + 0.375sin t R4 200 0.50 + 0.375sin t ( 0.50 0.375sin t ) ( vO 2 vO1 ) = 50 R3vO =vO = 3sin t ( V ) i3 =vO 2 0.50 + 0.375sin t = 50 + 200 R3 + R4i3 = 2 + 1.5sin t ( A ) i2 =vO1 vO ( 0.5 0.375sin t ) ( 3sin t ) = 250 R3 + R4i2 = 2 13.5sin t ( A )9.65 40 vOB = 1 + vI = 2.1667 sin t 12 30 vOC = vI = 1.25sin t 12(a) (b)(c) vO = vOB vOC = 2.1667 sin t ( 1.25sin t ) vO = 3.417 sin tvO 3.417 = = 6.83 0.5 vI(d) 9.66 iO =vI R9.67 vO R 2R = 4 1 + 2 vI 2 vI 1 R3 R1 200 2 (115 ) vO = 1 + ( 0.06sin t ) R1 50 230 For vO = 0.5 = 1.0833 R1 = 212.3 K R1 Ad =vO = 8 V230 = 32.33 R1 = 7.11 K R1 f = 7.11 K, R1 (potentiometer) = 205.2 K R19.68www.elsolucionario.net 374. 2 R2 1 + ( vI 2 vI 1 ) R1 Set R2 = 15 K, Set R1 = 2 K + 100 k ( Rot ) R4 R3vO =WantR4 8 R3Set R3 = 10 K R 4 = 75 KNow Gain (min) =75 2 (15 ) 1 + = 9.71 10 102 Gain ( max ) =75 2 (15 ) 1 + = 120 10 2 9.69 For a common-mode gain, vcm = vI 1 = vI 2 Then R R v01 = 1 + 2 vcm 2 vcm = vcm R1 R1 R R v02 = 1 + 2 vcm 2 vcm = vcm R1 R1 From Problem 9.62 we can write R4 R4 R3 R3 Acm = R4 1 + R3 R3 = R4 = 20 k, R3 = 20 k 5% 20 1 R3 1 20 Acm = = 1 R3 1 + 1) 2 ( For R3 = 20 k 5% = 19 k 1 20 1 = 0.0263 2 19 For R3 = 20 k + 5% = 21 k Acm =1 20 1 = 0.0238 2 21 So Acm max = 0.0263 Acm =9.70 a. v0 =1 vI ( t ) dt R1C2 0.5 0.5sin t dt = v0 = 0.5 = f =cos t1 ( 0.5 ) 0.5 = R1C2 2 R1C2 f1 1 = f = 31.8 Hz 2 R1C2 2 ( 50 103 )( 0.1 106 )Output signal lags input signal by 90www.elsolucionario.net 375. b. i.f =ii.f =0.52 ( 50 103 )( 0.1 106 ) f = 15.9 Hz0.5( 0.1)( 2 ) ( 50 103 )( 0.1106 ) f = 159 Hz9.71 vI t 1 vI ( t ) dt = RC RC vI = 0.2 vO = Now 8= ( 0.2 )( 2 )(a)RC RC = 0.05 s(b)14 =( 0.2 ) t 0.05 t = 3.5 s9.72 a. v0 = vI R21 j C2 R1R2 =1 j C2 1 R1 R2 + j C2 v0 R 1 = 2 vI R1 1 + j R2 C2b.v0 R = 2 vI R1c.f =1 2 R2 C29.73 a. R ( j C1 ) v0 R2 = = 2 vI R + 1 1 + j R1C1 1 j C1 v0 R j R1C1 = 2 vI R1 1 + j R1C1b.v0 R = 2 vI R1c.f =1 2 R1C19.74 Assuming the Zener diode is in breakdown,www.elsolucionario.net 376. vO = i2 =R2 1 Vz = ( 6.8 ) vO = 6.8 V R1 10 vO 0 ( 6.8 ) = i2 = 6.8 mA R2 110 Vz 10 6.8 i2 = 6.8 iz = 6.2 mA!!! Rs 5.6 Circuit is not in breakdown. Now 10 0 10 = i2 = i2 = 1.52 mA 5.6 + 1 Rs + R1 iz =vO = i2 R2 = (1.52 )(1) vO = 1.52 Viz = 09.75 v v vI vO = VT ln I = ( 0.026 ) ln 14 I 4 vO = 0.026 ln 10 I s R1 (10 )(10 ) 10 For vI = 20 mV , vO = 0.497 V For vI = 2 V , vO = 0.617 V9.76www.elsolucionario.net 377. 333 v0 = ( v01 v02 ) = 16.65 ( v01 v02 ) 20 i v01 = vBE1 = VT ln C1 IS i v02 = vBE 2 = VT ln C 2 IS i i v01 v02 = VT ln C1 = VT ln C 2 iC 2 iC1 v v iC 2 = 2 , iC1 = 1 R2 R1 v R So v01 v02 = VT ln 2 1 R2 v1 Then v R v0 = (16.65 )( 0.026 ) ln 2 1 v1 R2 v R v0 = 0.4329 ln 2 1 v1 R2 ln ( x ) = log e ( x ) = log10 ( x ) log e (10 ) = 2.3026 log10 ( x )v R Then v0 (1.0 ) log10 2 1 v1 R2 9.77 vO = I s R evI / VT = (1014 )(104 ) evI / VT vO = (10(10)e)vI / 0.026For vI = 0.30 V ,vo = 1.03 105 VFor vI = 0.60 V ,vo = 1.05 Vwww.elsolucionario.net 378. Chapter 10 Exercise Solutions EX10.1 V + VBE ( on ) 10 0.7 = I REF = 15 R1 I REF = 0.62 mA I REF 0.62 = 2 2 1+ 1+ 75 I 0 = 0.604 mA I0 =EX10.2 V + VBE ( on ) V 5 0.7 ( 5 ) I REF = = R1 12 I REF = 0.775 mA I 0.775 I 0 = REF = = 0.7549 mA 2 2 1+ 1+ 75 I 0 = ( 0.02 )( 0.7549 ) = 0.0151 mA and I 0 = r0 =V 1 VCE 2 r0 = CE 2 I 0 r0V 4 = 265 k = A VA = ( 265 )( 0.7549 ) VA 200 V 0.0151 I0EX10.3 V + 2VBE ( on ) 9 2 ( 0.7 ) I REF = = R1 12 I REF = 0.6333 mA I0 =I REF 0.6333 = = 0.6331 mA 2 2 1+ 1+ (1 + ) 75 ( 76 )I 0 = 0.6331 mA = I C1 I B1 = I B 2 =I0 I B1 = I B 2 = 8.44 AI E 3 = I B1 + I B 2 I E 3 = 16.88 A I B3 =I E3 I B 3 = 0.222 A 1+ EX10.4 I I 0 RE = VT ln REF I0 I 0.026 0.75 V ln RE = T ln REF = RE = 3.54 k I 0 I 0 0.025 0.025 5 0.7 R1 = R1 = 5.73 k 0.75 VBE1 VBE 2 = I 0 RE = ( 0.025 )( 3.54 ) VBE1 VBE 2 = 88.5 mVEX10.5www.elsolucionario.net 379. 5 0.7 ( 5 ) I REF = 0.775 mA 12 I I 0 RE = VT ln REF I0 0.775 I 0 ( 6 ) = ( 0.026 ) ln I 0 16.6 A I0 I REF =EX10.6 I I 0 RE = VT ln REF I0 0.026 0.70 RE = ln RE = 3.465 k 0.025 0.025 I 0.025 gm2 = 0 = g m 2 = 0.9615 mA/V VT 0.026 VTr 2 = r02 =I0=(150 )( 0.026 ) 0.025= 156 kVA 100 = = 4000 k I 0 0.025 RE = RE r 2 = 3.47 156 = 3.39 k R0 = r02 (1 + g m 2 RE ) = 4000 1 + ( 0.962 )( 3.39 ) R0 = 17.04 M 1 3 dI 0 = dVC 2 = dI 0 = 0.176 A R0 17, 040EX10.7 I REF = I R + I BR + I B1 + ... + I BN I R = I 01 = I 02 = ... = I 0 N and I BR = I B1 = I B 2 = ... = I BN =I 01I N +1 I REF = I 01 + ( N + 1) 01 = I 01 1 + I REF So I 01 = I 02 = ... = I 0 N = N +1 1+I 01 1 = 0.90 = N +1 I REF 1+ 50 N +1 1 1+ = 50 0.9 1 1 ( 50 ) N +1 = 0.9 1 1 ( 50 ) 1 N = 0.9 N = 4.55 N = 4EX10.8www.elsolucionario.net 380. VDS ( sat ) = 1 V = VGS 2 VTN = VGS 2 2 VGS 2 = 3 V 2 2 C W I O = K n 2 (VGS 2 VTN ) = n ox (VGS 2 VTN ) 2 L 2 2 W W 0.20 = ( 0.020 ) ( 3 2 ) = 10 L 2 L 2 2 C W I REF = n ox (VGS 1 VTN ) 2 L 1 VGS 1 = VGS 2 2 W W 0.5 = ( 0.020 ) ( 3 2 ) = 25 L 1 L 1VGS 3 = V + VGS 1 = 10 3 = 7 V C W 2 I REF = n ox (VGS 3 VTN ) 2 L 3 2 W W 0.5 = ( 0.020 ) ( 7 2 ) = 1 L 3 L 3EX10.9 I REF = K n (VGS VTN )20.020 = 0.080 (VGS 1)a.2VGS = 1.5 V all transistorsb. VG 4 = VGS 3 + VGS1 + V = 1.5 + 1.5 5 = 2 V VS 4 = VG 4 VGS 4 = 2 1.5 = 3.5 VVD 4 ( min ) = VS 4 + VDS 4 ( sat ) and VDS 4 ( sat ) = VGS 4 VTN = 1.5 1 = 0.5 V So VD 4 ( min ) = 3.5 + 0.5 VD 4 ( min ) = 3.0 Vc. R0 = r04 + r02 (1 + g m r04 ) 1 1 r02 = r04 = = = 2500 k I 0 ( 0.02 )( 0.020 )g m = 2 K n (VGS VTN ) = 2 ( 0.080 )(1.5 1) g m = 0.080 mA / V R0 = 2500 + 2500 (1 + ( 0.080 )( 2500 ) ) R0 = 505 MEX10.10 For Q2 : vDS ( min ) = VP = 2 V VS ( min ) = vDS ( min ) 5 = 2 5 VS ( min ) = 3 V I 0 = I DSS 2 (1 + vDS 2 ) = 0.5 (1 + ( 0.15 )( 2 ) ) I 0 = 0.65 mA v I 0 = I DSS1 1 GS 1 VP1 22 v 0.65 = 0.80 1 GS1 2 vGS 1 = 0.0986 vGS 1 = 0.197 V 2 vGS 1 = VI VS 0.197 = VI ( 3) VI ( min ) = 3.2 Vwww.elsolucionario.net 381. Vgs 2 = 0, Vgs1 = VXIX =VX VX V1 + + g m1VX r02 r01V1 V1 VX + = g m1VX RD r01(1) (2) 1 VX + g m1 r01 V1 = 1 1 + RD r01 1 1 + g m1 r01 r01 IX 1 1 1 = = + + g m1 1 1 VX R0 r02 r01 + RD r01 1 r01 1 1 = + + g m1 1 1 1 r02 r01 + RD r01 1 RD 1 1 = + + g m1 r02 r01 1 + 1 R D r01 www.elsolucionario.net 382. r01 For RD 1 1 1 + + g m1 R0 r02 r01 For Q1: 2I g m1 = DSS 1 VP VGS 1 2 ( 0.8 ) 0.197 1 = 1 2 2 VP g m1 = 0.721 mA/V 1 1 = = 10.3 k r0 = I 0 ( 0.15 )( 0.65 )1 1 1 = + + 0.721 = 0.915 R0 = 1.09 k R0 10.3 10.3EX10.11a.V I REF = I S exp EB 2 VT I 0.5 103 VEB 2 = VT ln REF = ( 0.026 ) ln VEB 2 = 0.521 V 12 10 IS 5 0.521 R1 = R1 = 8.96 k b. 0.5 c. Combining Equations (10.79), (10.80), and (10.81), we find VEC 2 1 + VAP V V I S 0 exp I 1 + CEo = I REF VEB 2 VT VAN 1 + VAP 2.5 1 + V 2.5 100 = ( 0.5 103 ) 1012 exp I 1 + 0.521 VT 100 1 + 100 V 1.025 10 12 exp I VTd. VI 4 = 5.098 10 exp VT 8 = 4.974 10 VI = 0.521 V 1 1 VT = 0.026 = 38.46 A = 1923 Av = v 1 1 1 1 0.01 + 0.01 + + VAN VAP 100 100EX10.12 I CQ 0.8 gm = = = 30.77 mA/V 0.026 VT r0 = r02 =VA 80 = = 100 k I CQ 0.8a.V0 = g mV 1 ( r0 r02 ) , V 1 = Vi Av = g m ( r0 r02 ) = ( 30.77 ) [100 100] Av = 1538b.Av = g m ( r0 r02 RL ) Av = 1540 = 770 770 = ( 30.77 )( 50 RL ) ( 50 RL ) = 25 RL = 50 k 2EX10.13 (a) Neglecting effect of and RLwww.elsolucionario.net 383. I O = I REF = K n (VIQ VTN ) 0.40 = 0.25 (VIQ 1)22Then VIQ = 2.265 Vr0 = r02 =b.1 1 = = 125 k I 0 ( 0.02 )( 0.4 )g m = 2 K n (VIQ VTN ) = 2 ( 0.25 )( 2.26 1) = 0.632 mA/V Av = g m ( r0 r02 ) = ( 0.632 )(125 125 ) Av = 39.5Av = g m ( r0 r02 RL ) c.39.4 = ( 0.632 )( 62.5 RL ) 62.5 RL = 31.25 RL = 62.5 k 2TYU10.1 For I 0 = 0.75 mA 2 2 I REF = I 0 1 + = ( 0.75 ) 1 + 100 I REF = 0.765 mA I REF = R1 =V + VBE ( on ) V R15 0.7 ( 5 )0.765 R1 = 12.2 kTYU10.2 10 ( 0.7 )( 2 ) I REF = = 0.717 mA 12 I 0 I REF = 0.717 V 100 r0 = A = r0 = 139.5 k I 0 0.717 I 0 =1 4 VCE 2 = I 0 = 0.0287 mA r0 139.5TYU10.3 I 0 = I REF I B3 =I01 0.50 = I 0 = 0.4996 mA 2 2 1 + (1 + ) 1 + 50 ( 51) I B 3 = 9.99 A1+ I E3 = I C 3 = I E 3 = 0.5096 mA IE3 0.5096 IC 2 = = I C 2 = 0.490 mA = I C1 2 2 1 + 1 + 50 I B1 = I B 2 =IC 2 I B1 = I B 2 = 9.80 ATYU10.4www.elsolucionario.net 384. I REF = k n1 W kn 3 W 2 2 (VGS 1 VTN 1 ) = ( 5 VGS 1 VTN 3 ) 2 L 1 2 L 3 1 40 17.3 38 2.70 (VGS1 0.98 ) = ( 3.98 VGS 1 ) VGS 1 = 1.814 V 2 0.040 I REF = (17.3)(1.814 0.98 ) 2 I REF = 0.241 mA 2 kn 2 W 2 (VGS 1 VTN 2 ) 2 L 2 2 0.042 = ( 6.92 )(1.814 1.0 ) 2 I O = 0.0963 mA IO =TYU10.5 a. From Equation (10.52), 3 3 1 12 10 + 12 1.8 VGS 1 = ( ) 3 3 1+ 1+ 12 12 0.5 1 0.5 VGS 1 = (10 ) + (1.8 ) 1 + 0.5 1 + 0.5 VGS 1 = 3.93 V also VDS1 = 3.93 V I REF = (12 )( 0.020 ) [3.93 1.8] 1 + ( 0.01)( 3.93) I REF = 1.13 mA W / L )2 (1 + VDS 2 ) ( b. I 0 = I REF (W / L )1 (1 + VDS1 ) 2 6 1 + ( 0.01)( 2 ) I = 0.555 mA I 0 = (1.13) 0 12 1 + ( 0.01)( 3.93) c.For VDS 2 = 6 V I o = 0.576 mATYU10.6 K n1 (VGS 1 VTN ) = K n3 (VGS 3 VTN ) 22 0.10 VGS 1 2 = 0.25 (VGS 3 2 ) 2 = ( 0.6325 )(VGS 3 2 ) VGS 1 VGS 3 = 10 VGS 1 VGS 1 2 = ( 0.6325 )(10 VGS 1 ) ( 0.6325 )( 2 )1.6325VGS1 = 7.06 VGS 1 = 4.325 VI REF = K n1 (VGS 1 VTN ) = ( 0.25 ) ( 4.325 2 ) I REF = 1.35 mA 22I O = 3K n 2 (VGS 1 VTN ) = 3 ( 0.25 )( 4.325 2 ) I O = 3I REF I O = 4.05 mA 22TYU10.7www.elsolucionario.net 385. I REF = 0.20 = K n1 (VGS 1 VTN ) = 0.15 (VGS 1 1) VGS 1 = VGS 2 = 2.15 V 2I O = K n 2 (VGS 2 VTN ) = 2I O = K n 3 (VGS 3 VTN )20.15 2 ( 2.15 1) I O = 0.10 mA 220.10 = 0.15 (VGS 3 1) VGS 3 = 1.82 V 2TYU10.8 All transistors are identical I 0 = I REF = 250 A I REF = K n (VGS VTN )20.25 = 0.20 (VGS 1) VGS = 2.12 V 2TYU10.9 For Q1 : iD = I DSS 1 (1 + vDS1 ) 2 v For Q2 : iD = I DSS 2 1 GS 2 (1 + vDS 2 ) VP vGS 2 = vDS1 and vDS 2 = VDS vDS 1 So 2 v I DSS 1 (1 + vDS1 ) = I DSS 2 1 DS 1 1 + (VDS vDS1 ) VP I DSS 1 = I DSS 2 2 v 1 + ( 0.1) vDS1 = 1 DS1 1 + ( 0.1) ( 3) ( 0.1) vDS 1 2 2 1 + 0.1vDS 1 = (1 vDS 1 + 0.25vDS 1 ) (1.3 0.1vDS 1 )3 2 This becomes 0.025vDS 1 0.425vDS 1 + 1.5vDS 1 0.3 = 0 We find vDS 1 = 0.2127 V, vDS 2 = 2.787 V, vGS 2 = 0.2127 ViD = I DSS 1 (1 + vDS 1 ) = 2 1 + ( 0.1)( 0.2127 ) iD = 2.04 mA R0 = r02 + r01 (1 + g m 2 r02 )2 I DSS vGS 2 2 ( 2 ) 0.2127 1 = 1 VP VP 2 2 g m = 1.787 mA/V 1 1 r02 = r04 = = = 5 k I DSS ( 0.1)( 2 ) gm =R0 = 5 + 5 1 + (1.787 )( 5 ) R0 = 54.7 k TYU10.10a. b. 0.1 103 VEB 2 = 0.557 V VEB 2 = ( 0.026 ) ln 14 5 10 5 0.557 R1 = R1 = 44.4 k 0.1c.www.elsolucionario.net 386. VEC 2 1+ V VCE 0 AP = I REF 1+ VAN 1 + VEB 2 VAP 2.5 1 + 100 VI 2.5 14 3 5 10 exp 1 + = ( 0.1 10 ) 0.557 VT 100 1+ 100 V ( 5.125 1014 ) exp VI = 1.019 104 T V I S 0 exp I VT V exp I VT 1 0.026 Av = 1 1 + 100 100d. 9 = 1.988 10 VI = 0.557 V Av = 1923TYU10.11 I REF = K p1 (VSG + VTP )a.20.25 = 0.20 (VSG 1) VSG = 2.12 V 2b. VDSOFrom Equation (10.89) 1 + P (V + VSG ) K (V V )2 TN n I = Vo = I REF ( n + P ) n + P5=1 + ( 0.015 )(10 2.12 ) 0.030( 0.2 )(VI 1) 0.25 ( 0.030 )20.15 = 1.12 0.8 (VI 1) VI = 2.10 V 2c.Av =2 K n (VI VTN ) I REF ( n + P )Av = 2 ( 0.2 )( 2.10 1.0 ) 0.25 ( 0.030 ) Av = 58.7TYU10.12(a)I REF = K p1 (VSG + VTP )280 = 50 (VSG 1) VSG = 2.26 V 2(b)VDSo1 + p (V + VSG ) K (V V )2 TN n I = Vo = n + p I REF ( n + p )1 + ( 0.015 )(10 2.26 ) ( 50 )(VI 1) 5= 0.030 (80 )( 0.030 )220.83 (VI 1) = 32.2 VI = 2.243 V 2(c)Av =2 K n (VI VTN ) I REF ( n + p )=2 ( 50 )( 2.243 1)(80 )( 0.030 ) Av = 51.8TYU10.13 a.www.elsolucionario.net 387. gm =IC 0 0.5 = g m = 19.2 mA/V 0.026 VTr0 =VAN 120 = r0 = 240 k I CQ 0.5r02 =VAP 80 = r02 = 160 k I CQ 0.5 Av = g m ( r0 r02 RL ) = (19.2 ) [ 240 160 50] Av = 631b.TYU10.14 1 = 38.46 mA/V 0.026 (100 )( 0.026 ) r 1 = r 2 = = 2.6 K 1 80 rO1 = rO 2 = = 80 K 1 120 rO = = 120 K 1 1 80 = 0.0257 K RO1 = 2.6 38.46 For R1 = 9.3 K I C = 1mA, g m =( RO1 + RE ) = 9.3 ( 0.0257 + 1) = 0.924 K RE = 1 [ 2.6 + 0.924] = 0.779 K RO 2 = 80 1 + ( 38.46 )( 0.779 ) = 2476.7 K Av = g m ( rO RO 2 ) = ( 38.46 )(120 2476.7 ) = ( 38.46 )(114.5 ) R = R1Av = 4404 For RL = 100 K Av = 38.46 114.5 100 = 2053 For RL = 10 KAv = 38.46 [114.5 10] = 354TYU10.15 M 1 and M 2 identical I o = I REF a. I O = K n (VI VYN )20.25 = 0.2 (VI 1) VI = 2.12 V g m = 2 K n (VI VTN ) = 2 ( 0.2 )( 2.12 1) g m = 0.447 mA/V 2r0 n = r0 p =b.1n I 0=1( 0.01)( 0.25 ) r0 n = 400 k1 1 = r0 p = 200 k p I 0 ( 0.02 )( 0.25 ) Av = g m ( r0 r02 RL ) Av = ( 0.447 ) [ 400 200 100] Av = 25.5www.elsolucionario.net 388. Chapter 10 Problem Solutions 10.1 a.I1 = I 2 =0 2V V R1 + R22V + I 2 R2 = VBE + I C R3 R2 2V + ( 2V V ) = VBE + IC R3 R1 + R2 R2 2V ( 2V + V ) VBE R1 + R2 V = VBE and R1 = R2 IC =IC =b.1 R31 1 2V ( 2V + V ) VBE R3 2 or I C =c.V 2 R3I C = 2 mA = ( 10 ) 2 R3I1 = I 2 = 2 mA = R3 = 2.5 k2 ( 0.7 ) ( 10 ) R1 + R2 R1 + R2 = 4.3 k R1 = R2 = 2.15 k10.2 (a)I VBE1 = VT ln C1 IS (i)I REF = I C1 = 10 A,(ii)I REF = I C1 = 100 A,(iii)(b) 10 106 VBE1 = ( 0.026 ) ln = 0.5388 V 14 10 I O = 10 A 100 106 VBE1 = ( 0.026 ) ln = 0.5987 V 14 10 I O = 100 A 103 I REF = I C1 = 1 mA, VBE1 = ( 0.026 ) ln 14 = 0.6585 V 10 I O = 1 mA IO =I REF 2 1+(i)IO =10 I O = 9.615 A VBE1 = VBE 2 2 1+ 50I = VT ln O IS 9.615 106 = ( 0.026 ) ln 14 10 = 0.5378 V(ii)IO = 96.15 106 100 I O = 96.15 A VBE1 = ( 0.026 ) ln 14 2 10 1+ 50 = 0.5977 Vwww.elsolucionario.net 389. IO =(iii)10.3 I REF = 0.9615 103 I O = 0.9615 mA VBE1 = ( 0.026 ) ln 2 1014 1+ 50 = 0.6575 V 1V + VBE ( on ) V R1 0.250 =3 0.7 ( 3) R1R1 = 21.2 K I REF 0.250 = I C1 = I C 2 = 0.2419 mA 2 2 1+ 1+ 60 = 4.03 AI C1 = I C 2 = I B1 = I B 210.4 I REF =V + VBE ( on ) V R1=5 0.7 ( 5 ) 18.3I REF = 0.5082 mA I REF 0.5082 = I C1 = I C 2 = 0.4958 mA 2 2 1+ 1+ 80 = ( 6.198 A )I C1 = I C 2 = I B1 = I B 210.5 (a)I REF =(b)R1 =V + VBE ( on ) V R1V + VBE ( on ) V =or R1 =15 0.7 ( 15 )0 0.7 ( 15 )I REF 0.5 Advantage: Requires smaller resistance. (c) For part (a): 29.3 = 0.526 mA I O ( max ) = ( 58.6 )( 0.95 ) I O ( min ) =0.5 R1 = 58.6 k R1 = 28.6 k 29.3 = 0.476 mA ( 58.6 )(1.05 )I O = 0.526 0.476 = 0.05 mA 5% For part (b): 14.3 = 0.526 mA I O ( max ) = ( 28.6 )( 0.95)I O ( min ) =14.3 = 0.476 mA ( 28.6 )(1.05 )I O = 0.05 mA 5%10.6 a. 2 2 I REF = I 0 1 + = 2 1 + or I REF = 2.04 mA 100 15 0.7 R1 = R1 = 7.01 k 2.04www.elsolucionario.net 390. r0 =b.VA 80 = = 40 k 2 I0I 0 1 1 = I 0 = ( 9.3) = 0.2325 mA VCE r0 40 I 0 0.2325 I = 0 = 11.6% 2 I0 I010.7 I 0 = nI C1 I REF = I C1 + I B1 + I B 2 = I C1 +I C1+I0 1+ n 1 n I REF = I C1 1 + + = I C1 1 + =I0 1 + n nI REF 1 + or I 0 = n 1+ n 1 + 10.8 IO =I REF 2 I REF = ( 0.20 ) 1 + = 0.210 mA 2 40 1+5 0.7 4.3 = R1 = 20.5 K R1 = I REF 0.2110.9 a.b.5 0.7 = 0.239 mA 18 0.239 I0 = I 0 = 0.230 mA 2 1+ 50 VA 50 r0 = = = 218 k I 0 0.230I REF =1 1 VEC = (1.3) = 0.00597 mA I 0 = 0.236 mA r0 217 1 I 0 = ( 3.3) = 0.01516 mA I 0 = 0.245 mA 217 I 0 =c. 10.105 0.7 ( 5 ) R1 = 9.3 ka.I REF = 1 =b.I 0 = 2 I REF I 0 = 2 mAc.For VEC 2 ( min ) = 0.7 RC 2 =R15 0.7 RC 2 = 2.15 k 210.11www.elsolucionario.net 391. I O = 0.50 mA I OA = I OB = 0.25 mA 3 3 I REF = I OA 1 + = 0.25 1 + 60 I REF = 0.2625 mA R1 =2.5 0.7 R1 = 6.86 K 0.262510.12R1 =10 0.7 = 37.2 K 0.2510.13 I 2 = 2 I1 and I3 = 3I1 (a) I 2 = 1.0 mA, I 3 = 1.5 mA (b) I1 = 0.25 mA, I 3 = 0.75 mA (c) I1 = 0.167 mA, I 2 = 0.333 mA 10.14 a.www.elsolucionario.net 392. I 0 = I C1 and I REF = I C1 + I B 3 = I C1 + VBE 2 I C1 VBE = + R2 R2 I E 3 = I B1 + I B 2 + I REF = I C1 + I REF 2 I C1 (1 + )+VBE 1 + ) R2 ( VBE 2 = I 0 1 + (1 + ) (1 + ) R2 I REF I0 =IE3 1+ VBE 1 + ) R2 ( 2 1 + (1 + ) 2 0.7 I REF = ( 0.70 ) 1 + ( 80 )( 81) + ( 81)(10 ) I REF = 0.700216 + 0.000864b.I REF = 0.7011 mA =10 2 ( 0.7 ) R1 R1 = 12.27 k10.15 a. I ES 1+ = (1 + N ) I BRI 0i = I CR and I REF = I CR + I BS = I CR + I ES = I BR + I B1 + I B 2 + ... + I BN =(1 + N ) I CR Then I REF = I CR + or I 0i =b.(1 + N ) I CR (1 + )I REF (1 + N ) 1 + (1 + ) 6 I REF = ( 0.5 ) 1 + = 0.5012 mA ( 50 )( 51) R1 =5 2 ( 0.7 ) ( 5 ) 0.5012 R1 = 17.16 k10.16www.elsolucionario.net 393. 2 2 I REF = I 0 1 + (1 + ) = ( 0.5 ) 1 + ( 50 )( 51) I REF = 0.5004 mA 5 2 ( 0.7 ) ( 5 ) R1 = R1 = 17.19 k 0.500410.17I 0 = I REF 1 2 1 + (2 + ) For I 0 = 0.8 mA 2 I REF = ( 0.8 ) 1 + 25 ( 27 ) I REF = 0.8024 mA 18 2 ( 0.7 ) R1 = 20.69 k R1 = 0.802410.18www.elsolucionario.net 394. The analysis is exactly the same as in the text. We have 1 I 0 = I REF 2 1 + (2 + ) 10.19 2 = 0.0267 mA 75 1 I C1 = 1 mA, I B1 = = 0.0133 mA 75 I E 3 = I B1 + I B 2 = 0.0133 + 0.0267 = 0.04 mA I 0 = 2 mA, I B 2 =I E3 0.04 = = 0.000526 mA 1+ 76 = I C1 + I B 3 I REF = 1.000526 1 mAI B3 = I REF R1 =10 2 ( 0.7 ) I REF=8.6 R1 = 8.6 k 110.20 (a) Assuming RO rO 3 = RO =VA V = A I O I REF ro3 2 100 = = 400 K 0.25(100 )( 400 ) 2 RO = 20 M(b) RO =V V 5 I O = = I O 20 M 20 MI O = 0.25 A10.21 I REF =V + VBE1 V 5 0.7 = 9.3 R1I REF = 0.4624 mA VT I REF 0.026 0.4624 ln ln = 1.5 RE I O IO 0.4624 I O = 0.01733ln IO IO =By trial and error I O 41.7 A VBE 2 = 0.7 I O RE VBE 2 = 0.7 ( 0.0417 )(1.5 ) VBE 2 = 0.6375 V 10.22 (a)www.elsolucionario.net 395. I REF =V + VBE1 V 5 0.7 ( 5 ) = I REF = 93 A 100 R1I 93 103 mA I O RE = VT ln REF I O (10 ) = 0.026 ln IO IO By trial and error, I O 6.8 A Ro = ro 2 (1 + g m 2 RE )Now 30 = 4.41 M 6.8 0.0068 = 0.2615 mA / V gm2 = 0.026 (100 )( 0.026 ) = 382.4 k r 2 = 0.0068 So RE = r 2 RE = 382 10 = 9.74 k Then Ro = 4.41 1 + ( 0.262 )( 9.74 ) Ro = 15.6 M ro 2 =VBE1 VBE 2 = I o RE = ( 0.0068 )(10 ) VBE1 VBE 2 = 0.068 V(b) 10.23I VC R0I 0 = R0 = r02 (1 + g m 2 RE ) V 80 r02 = A = = 11.76 M I 0 6.8 gm2 = r 2 =I 0 0.0068 = = 0.2615 mA/V VT 0.026(80 )( 0.026 )= 306 k 0.0068 RE = RE r 2 = 10 306 = 9.68 K R0 = (11.76 ) 1 + ( 0.2615 )( 9.68 ) = 41.54 M Now 1 I 0 = ( 5 ) I 0 = 0.120 A 41.54 10.24 (a)I REF =5 0.7 ( 5 ) R1= 0.50R1 = 18.6 K I I O RE = VT ln REF IO 0.026 0.50 RE = ln 0.050 0.050 RE = 1.20 Kwww.elsolucionario.net 396. (b) RO = rc 2 [1 + RE g m 2 ] RE = RE r 2 r 2 =( 75 )( 0.026 )= 39 K0.050 VA 100 ro 2 = = 2 M I O 0.05(c)gm2 =0.050 = 1.923 mA/V 0.026 RE = 1.20 39 = 1.164 KRO = 2 1 + (1.164 )(1.923) RO = ( 6.477 ) M V 5 I O = = = 0.772 A RO 6.477 I O 0.772 100% = 100 = 1.54% IO 5010.25 Let R1 = 5 k , Then I REF =12 0.7 ( 12 ) 5 I REF = 4.66 mANow I 0.026 4.66 I O RE = VT ln REF RE = ln RE 1 k 0.10 0.10 IO 10.26 I VBE = VT ln REF IS 103 15 0.7 = ( 0.026 ) ln I S = 2.03 10 A IS 2 103 At 2 mA, VBE = ( 0.026 ) ln 15 2.03 10 = 0.718 V 15 0.718 R1 = 7.14 k 2 I 0.026 2 V RE = T ln REF = ln RE = 1.92 k I 0 I 0 0.050 0.050 R1 =10.27 a. 10 0.7 = 0.465 mA 20 Let V = 0 I REF I VBE VT ln REF IS 103 15 0.7 = ( 0.026 ) ln I S = 2.03 10 A IS Then 0.465 103 = 0.680 V VBE ( 0.026 ) ln 15 2.03 10 Thenwww.elsolucionario.net 397. I REF b. 10.28 I REF 10 0.680 I REF = 0.466 mA 20 RE =VT I REF ln I0 I010 0.7 ( 10 ) 40 0.026 0.466 ln = RE = 400 0.10 0.10 = 0.4825 mAI VBE VT ln REF IS 103 15 0.7 = ( 0.026 ) ln I S = 2.03 10 A IS Now 0.4825 103 VBE = ( 0.026 ) ln = 0.681 V 15 2.03 10 VBE1 = 0.681 V So 10 0.681 ( 10 ) I REF = 0.483 mA 40 I I 0 RE = VT ln REF I0 0.483 I 0 (12 ) = ( 0.026 ) ln I0 I REF By trial and error. I 0 8.7 A VBE 2 = VBE1 I 0 RE = 0.681 ( 0.0087 )(12 ) VBE 2 = 0.5766 V10.29 VBE1 + I REF RE1 = VBE 2 + I 0 RE 2 VBE1 VBE 2 = I 0 RE 2 I REF RE1 For matched transistors I VBE1 = VT ln REF IS I VBE 2 = VT ln 0 IS I Then VT ln REF = I 0 RE 2 I REF RE1 I0 Output resistance looking into the collector of Q2 is increased.10.30V + VBE1 V 5 0.7 ( 5 ) = = 0.3174 mA R1 + RE1 27.3 + 2(a)I REF =(b)I O = I REF = 0.3174 mA Using the same relation as for the widlar current source.www.elsolucionario.net 398. RO = ro 2 1 + g m 2 ( RE r 2 ) ro 2 = r 2 =VA 80 = = 252 K I O 0.3174gm2 =0.3174 = 12.21 mA/V 0.026(100 )( 0.026 )= 8.192 K RE r 2 = 2 8.192 = 1.608 K 0.3174 RO = 252 1 + (12.21)(1.608 ) RO = 5.2 M (c) I O = I REF =5 0.7 ( 5 )= 0.3407 mA 27.3 V 80 RO = ro 2 = A = RO = 235 K I O 0.340710.31 Assume all transistors are matched. a. 2VBE1 = VBE 3 + I 0 RE I VBE1 = VT ln REF IS I VBE 3 = VT ln 0 IS I I 2VT ln REF VT ln 0 = I 0 RE IS IS I I VT ln REF ln 0 IS IS 2 = I 0 RE I 2 REF VT ln = I 0 RE I0 I S b. 0.7 15 VBE = 0.7 V at 1 mA 103 = I S exp or I S = 2.03 10 A 0.026 0.1 103 VBE at 0.1 mA VBE = ( 0.026 ) ln = 0.640 V 15 2.03 10 0.640 Since I 0 = I REF , then VBE = I 0 RE RE = or RE = 6.4 k 0.110.32 (a) I REF =5 0.7 ( 5 ) R1= 0.80 mAR1 = 11.6 K RE 2 =0.026 0.80 ln RE 2 = 1.44 K 0.050 0.050 RE 3 =0.026 0.80 ln RE 2 = 4.80 K 0.020 0.020 (b) VBE 2 = 0.7 ( 0.05 )(1.44 ) VBE 2 = 0.628 V VBE 3 = 0.7 ( 0.02 )( 4.80 ) VBE 3 = 0.604 Vwww.elsolucionario.net 399. 10.33 (a) VBE1 = VBE 2 I REF =V + 2VBE1 V R1 + R2Now 2VBE1 + I REF R2 = VBE 3 + I O RE or I O RE = 2VBE1 VBE 3 + I REF R2 We have I I VBE1 = VT ln REF and VBE 3 = VT ln O IS IS (b) Let R1 = R2 and I O = I REF VBE1 = VBE 3 VBE Then VBE = I O RE I REF R2 = I O ( RE R2 ) so I REF = I O = =V + V 2 I O ( RE R2 ) 2 R2R V V IO E + IO 2 R2 R2 +Then IO =V + V 2 R(c) Want I O = 0.5 mA So RE = 2 R2 =5 ( 5 ) 2 ( 0.5 ) RE = 10 k 5 2 ( 0.7 ) ( 5 )0.5 Then R1 = R2 = 8.6 k = 17.2 k 10.34 a. 20 0.7 0.7 = 1.55 mA 12 I 01 = 2 I REF = 3.1 mA I 02 = I REF = 1.55 mA I 03 = 3I REF = 4.65 mA b. VCE1 = I 01 RC1 ( 10 ) = ( 3.1)( 2 ) + 10 VCE1 = 3.8 V I REF =VEC 2 = 10 I 02 RC 2 = 10 (1.55 )( 3) VEC 2 = 5.35 V VEC 3 = 10 I 03 RC 3 = 10 ( 4.65 )(1) VEC 3 = 5.35 V10.35 a.Ist approximationwww.elsolucionario.net 400. 20 1.4 = 2.325 mA 8 2.32 Now VBE 0.7 = ( 0.026 ) ln VBE = VEB = 0.722 V 1 Then 2nd approximation 20 2 ( 0.722 ) I REF = 2.32 mA 8 I 01 = 2 I REF = 4.64 mA I 02 = I REF = 2.32 mA I 03 = 3I REF = 6.96 mA b. At the edge of saturation, VCE = VBE = 0.722 V I REF RC 2 RC 30 0.722 ( 10 ) RC1 = 2.0 k 4.64 10 0.722 RC 2 = 4.0 k = 2.32 10 0.722 RC 3 = 1.33 k = 6.96RC1 =10.36 I C1 = I C 2 =10 0.7 0.7 ( 10 ) 10= 1.86 mAI C 3 = I C 4 = 1.86 mA 1.86 I C 5 ( 0.5 ) = 0.026 ln IC 5 By Trial and error. I C 5 = 0.136 mA = I C 6 = I C 72 I C 3 ( 0.8 ) + VCE 3 = 10 VCE 3 = 10 2 (1.86 )( 0.8 ) VCE 3 = 7.02 V 5 = VEB 6 + VCE 5 + I C 5 ( 0.5 ) 10VCE 5 = 5 + 10 0.7 ( 0.136 )( 0.5 ) VCE 5 = 14.2 V 5 = VEC 7 + I C 7 ( 0.8 )VEC 7 = 5 ( 0.136 )( 0.8 ) VEC 7 = 4.89 V10.37 I C1 = I C 2 =10 0.7 0.7 ( 10 ) 10 I C1 = I C 2 = 1.86 mAI C 4 = I C 5 = 1.86 mAI 1.86 I C 3 RE1 = VT ln C1 I C 3 ( 0.3) = 0.026 ln IC 3 IC 3 By trial and error I C 3 = 0.195 mA I 1.86 I C 6 RE 2 = VT ln C 5 I C 6 ( 0.5 ) = 0.026 ln IC 6 IC 6 By trial and error I C 6 = 0.136 mAwww.elsolucionario.net 401. 10.38 10 0.7 = 1 mA 6.3 + 3 VBE ( QR ) = 0.7 V as assumed VRER = I REF RER = (1)( 3) = 3 V I REF =VRE1 = 3 V RE1 =VRE1 3 = RE1 = 3 k I 01 1VRE 2 = 3 V RE 2 =VRE 2 3 = RE 2 = 1.5 k I 02 2VRE 3 = 3 V RE 3 =VRE 3 3 = RE 3 = 0.75 k I 03 4I 01 = 1 mA I 02 = 2 mA I 03 = 4 mA10.38 VDS 2 ( sat ) = 2 V = VGS 2 VTN 2 = VGS 2 1.5 VGS 2 = 3.5 V 2 1 W I O = n Cox (VGS 2 VTN 2 ) 2 L 2 2 W W 250 = ( 20 ) ( 3.5 1.5 ) = 3.125 L 2 L 21 W I REF = n Cox (VGS 2 VTN 1 ) 2 L 1 2 W W 100 = ( 20 ) ( 3.5 1.5 ) = 1.25 L 2 L 1 Now VGS 3 = 10 VGS 2 = 10 3.5 = 6.5 V 2 W W So 100 = ( 20 ) ( 6.5 1.5 ) = 0.2 L 3 L 310.39 I REF =2.5 VGS 0.08 2 = ( 6 )(VGS 0.5 ) 15 2 2 2.5 VGS = 3.6 (VGS VGS + 0.25 ) 2 3.6VGS 2.6VGS 1.6 = 0VGS =2.6 6.76 + 23.04 2 ( 3.6 )VGS = 1.12 V (1.1193) 2.5 1.1193 I REF = 92.0 A ( 92.05 ) I REF = 15 I o = 92.0 A VDS 2 ( sat ) = VGS VTN = 1.1193 0.5 VDS 2 ( sat ) = 0.619 V10.39 a.From Equation (10.50),www.elsolucionario.net 402. VGS1 = VGS 2VGS 1 = VGS 2 5 5 1 25 5 + 25 0.5 = ( ) ( ) 5 5 1+ 1+ 25 25 0.447 1 0.447 = (5) + ( 0.5 ) 1 + 0.447 1 + 0.447 = 1.74 VI REF K n1 (VGS 1 VTN ) = (18 )( 25 )(1.74 0.5 ) I REF = 0.692 mA 222 1 W I O = n Cox (VGS 2 VTN ) (1 + VDS 2 ) 2 L 2 b.I 0 = (18 )(15 )(1.74 0.5 ) 1 + ( 0.02 )( 2 ) = ( 415 )(104 ) I 0 = 0.432 mA 2I 0 = ( 415 ) 1 + ( 0.02 )( 4 ) I 0 = 0.448 mA c.10.40 (a) 2 80 W I REF = 50 = (VGS 0.5 ) 2 L 1 2.0 VGS I REF = 0.050 = R Design such that VDS 2 ( sat ) = 0.25 = VGS 0.5VGS = 0.75 V 2 0.75 R = 25 K R 2 80 W W 50 = ( 0.75 0.5 ) = 20 2 L 1 L 1So 0.050 =W 20 50 W L 1 I REF = = = 40 100 L 2 IO W W L 2 L 2 1 1 RO = = RO = 667 K (b) I O ( 0.015 )( 0.1)(c)I O =V 1 = 1.5 A RO 666I O 1.5 100% = 100% 1.5% IO 100 10.41 (a)2 80 I REF = 250 = ( 3)(VGS 1) 2 VGS = 2.44 V I O = 250 A at VDS 2 = VGS = 2.44 V 1 1 = = 200 K RO = I O ( 0.02 )( 0.25 )www.elsolucionario.net 403. I O =(i)V 3 2.44 = 2.8 A RO 200I O = 252.8 A I O =(ii)V 4.5 2.44 = 10.3 A RO 200I O = 260.3 A I O =(iii)V 6 2.44 = 17.8 A RO 200I O = 267.8 A 4.5 ( 250 ) = 375 A at VDS = 2.44 V 3 1 1 RO = = = 133.3 K I O ( 0.02 )( 0.375 ) IO =(b)I O =(i)V 3 2.44 = 4.20 A RO 133.3I O = 379.2 A I O =(ii)V 4.5 2.44 = 15.5 A RO 133.3I O = 390.5 A I O =(iii)V 6 2.44 = 26.7 A RO 133.3I O = 401.7 A10.41 VSD 2 ( sat ) = 0.25 = VSG + VTP = VSG 0.4 VSG 2 = 0.65 V k W 2 p I O = (VSG 2 + VTP ) 2 L 2 40 W 2 W 25 = ( 0.65 0.4 ) = 20 2 L 2 L 2 I REF = 75 A =(W /L )1 W I O = 60 (W /L )2 L 1k W 2 p (VSG 3 + VTP ) 2 L 3 = 3 0.65 = 2.35 VI REF = VSG 3Then 75 =40 W 2 W ( 2.35 0.4 ) = 0.986 2 L 3 L 310.42 (a) VGS = VTN 1 +I REF 0.5 = 1+ =2V K n1 0.5 I I O = K n 2 REF K n1 2 I REF = Kn2 K n1 www.elsolucionario.net 404. 0.5 I 0 ( max ) = ( 0.5 )(1.05 ) I 0 ( max ) = 0.525 mA 0.5 0.5 I 0 ( min ) = ( 0.5 )( 0.95 ) I 0 ( min ) = 0.475 mA 0.5 So 0.475 I 0 0.525 mA (b) I I O = K n 2 REF + VTN 1 VTN 2 K n1 22 0.5 I 0 ( min ) = ( 0.5 ) + 1 1.05 I 0 (min) = 0.451 mA 0.5 2 0.5 I 0 ( max ) = ( 0.5 ) + 1 0.95 I 0 (max) = 0.551 mA 0.5 So 0.451 I 0 0.551 mA10.43(1) (2)Ix =Vx VA + g mVgs 2 roIx =VA + g mVgs1 roVgs1 = Vx , Vgs 2 = VASo (1)Ix =1 Vx VA + g m ro ro (2)Ix =VA + g mVx VA = ro [ I x g mVx ] rowww.elsolucionario.net 405. Then Ix =1 Vx ro ( I x g mVx ) + g m ro ro Ix =I Vx g 2 ro x + g m I x m Vx g mVx ro ro ro Ix =Vx 2 I x g m ro I x + g mVx + g m roVx ro1 2 I x [ 2 + g m ro ] = Vx + g m + g m ro ro 1 Since g m >> ro I x [ 2 + g m ro ] Vx ( g m )(1 + g m ro ) ThenVx 2 + g m ro = Ro = Ix g m (1 + g m ro )Usually, g m ro >> 2, so that Ro 1 gm10.44 VDS 2 (sat) = 2 = VGS 2 0.8 VGS 2 = 2.8 V I O = 200 =60 W 2 W ( 2.8 0.8 ) = 1.67 2 L 2 L 2W W I REF L 1 0.4 L 1 W = = = 3.33 IO 0.2 (1.67 ) L 1 W L 2 VGS 3 = 6 2.8 = 3.2 V 2 60 W W I REF = 400 = ( 3.2 0.8 ) = 2.31 2 L 3 L 310.45 (a) 2 2 60 60 I REF = ( 20 )(VGS 1 0.7 ) = ( 3)(VGS 3 0.7 ) 2 2 VGS1 + VGS 3 = 520 (VGS1 0.7 ) = 5 VGS1 0.7 3 3.582VGS 1 = 6.107 VGS1 = VGS 2 = 1.705 V 2 60 I O = (12 )(1.705 0.7 ) = 363.6 A at VDS 2 = 1.705 V 2 2 60 I REF = ( 20 )(1.705 0.7 ) = 606 A 2 (b)RO = I O =1 IO=1 = 183.4 K 0.015 )( 0.3636 ) (V 1.5 1.705 = 1.12 A RO 183.4I O = 362.5 Awww.elsolucionario.net 406. (c)I O =V 3 1.705 = 7.06 A RO 183.4I O = 370.7 A10.46 2 2 50 50 I REF = (15 )(VSG1 0.5 ) = ( 3)(VSG 3 0.5 ) 2 2 VSG1 + VSG 3 = 10 VSG 3 = 10 VSG115 (VSG1 0.5) = 10 VSG1 0.5 3 3.236VSG1 = 10.618 VSG1 = 3.28 V 2 50 I REF = (15 )( 3.28 0.5 ) I REF = 2.90 mA 2 I O = I REF = 2.90 mAVSD 2 (sat) = VSG 2 + VTP = 3.28 0.5 VSD 2 (sat) = 2.78 V10.47 VSD 2 (sat) = 1.2 = VSG 2 0.35 VSG 2 = 1.55 V 2 50 W W I O = 100 = (1.55 0.35 ) = 2.78 2 L 2 L 2 W W I REF 200 W L1 L1 = = = 5.56 W IO 100 2.78 L 1 L 2 VSG1 + VSG 3 = 4 VSG 3 = 2.45 V( ) ( )( )2 50 W W I REF = 200 = ( 2.45 0.35 ) = 1.81 2 L 3 L 310.48 2 2 80 80 I REF = ( 25 )(VSG1 1.2 ) = ( 4 )(VSG 3 1.2 ) 2 2 10 VSG1 VSG1 + 2VSG 3 = 10 VSG 3 = 2 10 VSG1 25 1.2 Then (VSG1 1.2 ) = 4 2 3VSG1 = 6.8 VSG1 = 2.27 V 2 80 I REF = ( 25 )( 2.267 1.2 ) I REF = I O = 1.14 mA 2 VSD 2 (sat) = VSG 2 + VTP = 2.27 1.2 VSD 2 ( sat ) = 1.07 V10.49 VSD 2 (sat) = 1.8 = VSG 2 1.4 VSG 2 = 3.2 V 2 80 W W I O = ( 3.2 1.4 ) = 100 = 0.772 2 L 2 L 2 W I REF L1 = W IO L 2 W 200 W L1 = = 1.54 100 0.772 L 1( ) ( ) ( )www.elsolucionario.net 407. Assume M3 and M4 are matched. 10 3.2 = 3.4 V 2VSG 3 + VSG1 = 10 VSG 3 = 2 2 80 W I REF = 200 = ( 3.4 1.4 ) 2 L 3,4 W = 1.25 L 3,410.50 (a) k W 2 p I REF = (VSG1 + VTP ) 2 L 1 k W 2 p = (VSG 3 + VTP ) 2 L 3 But VSG 3 = 3 VSG1 So 25 (VSG1 0.4 ) = 5 ( 3 VSG1 0.4 ) 22which yields VSG1 = 1.08 V and VSG 3 = 1.92 V I REF = 20 ( 25 )(1.08 0.4 ) I REF = 231 A 2(W / L )2 15 = = 0.6 I REF (W / L )1 25 Then I O = ( 0.6 )( 231) = 139 A IO=(b) VDS 2 ( sat ) = 1.08 0.4 = 0.68 V VR = 3 0.68 = 2.32 = I O R then R=2.32 R = 16.7 k 0.13910.51 VSD 2 (sat) = 0.35 = VSG 2 0.4 VSG 2 = 0.75 V 2 W W I O = 80 = ( 20 ) ( 0.75 0.4 ) = 32.7 L 2 L 2( ) ( )( )W W I REF 50 L1 L1 = = W = 20.4 L1 W 80 32.7 IO L 2 VSG 3 = 3 0.75 = 2.25( )2 W W I REF = 50 = ( 20 ) ( 2.25 0.4 ) = 0.730 L 3 L 310.52 a.I REF = K n (VGS VTN )2100 = 100 (VGS 2 ) VGS = 3 V 2For VD 4 = 3 V, I 0 = 100 Awww.elsolucionario.net 408. R0 = r04 + r02 (1 + g m r04 )b.r02 = r04 =1 1 = = 500 k I 0 ( 0.02 )( 0.1)g m = 2 K n (VGS VTN ) = 2 ( 0.1)( 3 2 ) = 0.2 mA / VR0 = 500 + 500 1 + ( 0.2 )( 500 ) R0 = 51 M I 0 =1 6 VD 4 = I 0 = 0.118 A 51 R010.53Vgs 4 = I X r02VS 6 = ( I X g mVgs 4 ) r04 + I X r02= ( I X + g m I X r02 ) r04 + I X r02VS 6 = I X r02 + (1 + g m r02 ) r04 = Vgs 6 V VS 6 VX 1 I X = g mVgs 6 + X = VS 6 g m + r06 r06 r06 1 g m + r02 + (1 + g m r02 ) r04 r06 V 1 I X 1 + g m + r02 + (1 + g m r02 ) r04 = X r06 r06 VX = R0 = r06 + (1 + g m r06 ) r02 + (1 + g m r02 ) r04 IX IX =VX IX r06I 0 I REF = 0.2 mA = 0.2 (VGS 1)2VGS = 2 Vg m = 2 K n (VGS VTN ) = 2 ( 0.2 )( 2 1) = 0.4 mA / Vr02 = r04 = r06 =1=1= 250 k( 0.02 )( 0.2 ) R0 = 250 + 1 + ( 0.4 )( 250 ) {250 + 1 + ( 0.4 )( 250 ) ( 250 )} I0R0 = 2575750 k R0 = 2.58 109 www.elsolucionario.net 409. 10.54 kn W kn W 2 2 (VGS1 VTN ) = (VGS 3 VTN ) 2 L 1 2 L 3 k W 2 p = (VGS 4 + VTP ) 2 L 4 (1) 50 ( 20 )(VGS 1 0.5 ) = 50 ( 5 )(VGS 3 0.5 ) 22(2) 50 ( 20 )(VGS 1 0.5 ) = 20 (10 )(VGS 4 0.5 ) 22(3) VSG 4 + VGS 3 + VGS1 = 6 From (1) 4 (VGS 1 0.5 ) = (VGS 3 0.5 ) VGS 3 = 2 (VGS 1 0.5 ) + 0.5 22From (2) 5 (VGS1 0.5 ) = (VGS 4 0.5 ) VSG 4 = 5 (VGS1 0.5 ) + 0.5 22Then (3) becomes 5 (VGS1 0.5 ) + 0.5 + 2 (VGS 1 0.5 ) + 0.5 + VGS1 = 6 which yields VGS 1 = 1.36 V and VGS 3 = 2.22 V , VSG 4 = 2.42 V k W 2 2 Then I REF = n (VGS1 VTN ) = 50 ( 20 )(1.36 0.5 ) or I REF = I O = 0.740 mA 2 L 1 VGS 1 = VGS 2 = 1.36 V VDS 2 ( sat ) = VGS 2 VTN = 1.36 0.5 VDS 2 ( sat ) = 0.86 V10.55 VDS 2 ( sat ) = 0.5 V = VGS 2 VTN = VGS 2 0.5 VGS 2 = 1 V kn W 2 (VGS 2 VTN ) 2 L 2 2 W W = 50 (1 0.5 ) = 4 L 2 L 2I O = 50 A =VGS1 = VGS 2 = 1 V I REF = 150 = kn W 2 2 W W (VGS 1 VTN ) = 50 (1 0.5 ) = 12 2 L 1 L 1 L 1VGS 3 + VSG 4 + VGS 1 = 6 2VGS 3 = 6 1 = 5 V VGS 3 = 2.5 V 2 W W I REF = 150 = 50 ( 2.5 0.5 ) = 0.75 L 3 L 3k W 2 p (VSG 4 + VTP ) 2 L 4 2 W W 150 = 20 ( 2.5 0.5 ) 1.88 L 4 L 4 I REF =10.56 a. As a first approximation 2 I REF = 80 = 80 (VGS 1 1) VGS 1 = 2 V Then VDS 1 2 ( 2 ) = 4 V The second approximation 80 = 80 (VGS1 1) 1 + ( 0.02 )( 4 ) 80 2 Or = (VGS 1 1) VGS 1 = 1.962 86.4 2www.elsolucionario.net 410. Then I O = K n (VGS1 VTN ) (1 + nVGS 1 ) 2= 80 (1.962 1) 1 + ( 0.02 )(1.962 ) Or I 0 = 76.94 A 2b. From a PSpice analysis, I 0 = 77.09 A for VD 3 = 1 V and I 0 = 77.14 A for VD 3 = 3 V. The change is I 0 0.05 A or 0.065%. 10.57 a.For a first approximation,I REF = 80 = 80 (VGS 4 1) VGS 4 = 2 V 2As a second approximation I REF = 80 = 80 (VGS 4 1) 1 + ( 0.02 )( 2 ) Or VGS 4 = 1.98 V = VGS 1 2I O = K n (VGS 2 VTN ) (1 + VGS 2 ) 2To a very good approximation I 0 = 80 A b. From a PSpice analysis, I 0 = 80.00 A for VD 3 = 1 V and the output resistance is R0 = 76.9 M. Then For VD = +3 V 1 4 I 0 = VD 3 = = 0.052 A R0 76.9 I 0 = 80.05 A10.58 (a)VDS 3 ( sat ) = VGS 3 VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0 k W 2 I D = n (VGS 3 VTN ) 2L 2 W W 50 = 48 ( 0.2 ) = 26 L L 3(b)VGS 5 VTN = 2 (VGS 3 VTN ) VGS 5 = 0.8 + 2 ( 0.2 ) VGS 5 = 1.2 V(c)VD1 ( min ) = 2VDS ( sat ) = 2 ( 0.2 ) VD1 ( min ) = 0.4 V10.59 (a) K n1 = R= = kn W 2 = 50 ( 5 ) = 250 A / V 2 L 1 1 K n1 I D1 1 (W / L )1 (W / L )2 5 1 = ( 8.944 )( 0.6838 ) 50 ( 0.25 )( 0.05 ) 1R = 6.12 k (b)www.elsolucionario.net 411. V + V = VSD 3 ( sat ) + VGS 1 VSD 3 ( sat ) = VSG 3 + VTP I D1 = 50 = 20 ( 5 )(VSG 3 0.5 ) VSG 3 = 1.207 V Then VSD 3 ( sat ) = 1.21 0.5 = 0.707 V 2Also I D1 = 50 = 50 ( 5 )(VGS1 0.5 ) VGS 1 = 0.9472 V 2Then (V + V )min= 0.71 + 0.947 = 1.66 V(c) 2 W W I O1 = 25 = 50 ( 0.947 0.5 ) = 2.5 L 5 L 5 2 W W I O 2 = 75 = 20 (1.207 0.5 ) = 7.5 L 6 L 610.60 1 ( 5 ) = 1.667 V 3 2 1 W I REF = n Cox (VGS 3 VTN ) 2 L 3 2 W W W W 100 = ( 20 ) (1.667 1) = = = 11.25 L 3 L 3 L 4 L 5 VGS 3 =2 1 W I O1 = n Cox (VGS 3 VTN ) 2 L 1 W I REF L 3 Or = I 01 W L 1 W I 01 = L 1 I REF W 0.2 W = (11.25 ) = 22.5 L 1 L 3 0.1 W I W 0.3 W And = 02 = (11.25 ) = 33.75 L 2 I REF L 3 0.1 L 210.61 I REF =24 VSGP VGSN RAlso I REF = 40 (1)(VGSN 1.2 ) I REF = 18 (1)(VSGP 1.2 )22Then 40 (VGSN 1.2 ) = 18 (VSGP 1.2 ) which yields VSGP =6.325 (VGSN 1.2 ) + 1.2 4.243www.elsolucionario.net 412. 2 Then 0.040 (VGSN 1.2 ) R = 24 VGSN 1.49 (VGSN 1.2 ) 1.2 which yields VGSN = 2.69 Vand VSGP = 3.43 V 24 3.42 2.69 I REF = 89.4 A 200Now I REF =89.4 = 17.9 A 5 I 2 = (1.25 )( 89.4 ) = 112 A I 3 = ( 0.8 )( 89.4 ) = 71.5 A I 4 = 4 ( 89.4 ) = 358 A I1 =10.61 a.g m ( M 0 ) = 2 K n I REF gm ( M 0 ) = 2( 0.25)( 0.2 ) g m ( M 0 ) = 0.447 mA/Vr0 n =1 1 = r0 n = 250 k n I REF ( 0.02 )( 0.2 )r0 p =1 1 = r0 p = 167 k p I REF ( 0.03)( 0.2 )b.Av = g m ( r0 n r0 p ) = ( 0.447 )( 250 167 ) Av = 44.8c.RL = 205 167 = r0 n r0 p or RL = 100 k10.62 We have VGSN = 2.69 V and VSGP = 3.43 V 10 2.69 3.43 3.88 So I REF = = I REF = 19.4 A R 200 Then I1 = ( 0.2 )(19.4 ) = 3.88 A I 2 = (1.25 )(19.4 ) = 24.3 A I 3 = ( 0.8 )(19.4 ) = 15.5 A I 4 = 4 (19.4 ) = 77.6 A 10.63 I D2(W L ) = (W L ) (W L ) = (W L )2 I REF =9 ( 200 ) I D 2 = 120 A 151IO4 20 I D 2 = (120 ) I O = 267 A 9 32 40 I O = 266.7 = ( 20 )(VSG 4 0.6 ) 2 VSG 4 = 1.416 V VSD 4 (sat) = 1.416 0.6 VSD 4 ( sat ) = 0.816 V10.64 2 40 W I REF = 50 = (VSG1 0.6 ) 2 L 1 1.75 VSG1 I REF = = 50 Rwww.elsolucionario.net 413. VSD 2 (sat) = 0.35 = VSG 2 0.6 VSG 2 = 0.95 V 1.75 0.95 R= R = 16 K 0.05 2 40 W W 50 = ( 0.95 0.6 ) = 20.4 2 L 1 L 1( )W I O1 120 W L 2 = = = 49 50 I REF ( 20.4 ) L 2 ( )W I D 3 25 W L 3 = = = 10.2 I REF 50 ( 20.4 ) L 3 VDS 5 (sat) = 0.35 = VGS 5 0.4 VGS 5 = 0.75 V 2 100 W W IO 2 = ( 0.75 0.4 ) = 150 = 24.5 2 L 5 L 5 W I D4 I D3 25 W L 4 = = = = 4.08 I O 2 I O 2 150 24.5 L 4( )10.65 For vGS = 0, iD = I DSS (1 + vDS ) a. b. c.VD = 5 V, vDS = 5 iD = ( 2 ) 1 + ( 0.05 )( 5 ) iD = 2.5 mA VD = 0, vDS = 10 iD = ( 2 ) 1 + ( 0.05 )(10 ) iD = 3 mA VD = 5 V, vDS = 15 V iD = ( 2 ) 1 + ( 0.05 )(15 ) iD = 3.5 mA 10.66 V I 0 = I DSS 1 GS VP V 2 = 4 1 GS VP 22VGS 2 = 1 = 0.293 4 VPSo VGS = ( 0.293)( 4 ) = 1.17 V VS and VS = VGS R ( 1.17 ) V R = 0.586 k R = GS = 2 I0 Finish solution: See solution Then I 0 =10.66 Completion of solution Need vDS vDS ( sat ) = vGS VP = 1.17 ( 4 ) vDS 2.83 V So VD vDS ( sat ) + VS = 2.83 + 1.17 VD 4 Vwww.elsolucionario.net 414. 10.67 V I REF = I S 1 exp EB1 VT a.I 1 103 or VEB1 = VT ln REF = ( 0.026 ) ln VEB1 = 0.5568 13 5 10 I S1 5 0.5568 R1 = R1 = 4.44 k 1 From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 Vb. c. 10122.5 1 + 80 VI 2.5 3 exp 1 + = 10 0.5568 VT 120 1+ 80 V 1.03125 1.0208333 1012 exp I = (103 ) 1.00696 VT Then VI = 0.026 ln (1.003222 109 )So VI = 0.5389 Vd.Av = (1/ VT )(1/ VAN ) + (1/ VAP )1 38.46 Av = 0.026 = 1 1 0.00833 + 0.0125 + 120 80 Av = 1846 10.68 a. b. c. other.I 0.5 103 VBE = VT ln REF = ( 0.026 ) ln VBE = 0.5208 12 10 I S1 5 0.5208 R1 = R1 = 8.96 k 0.5 Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each V V I CO = I SO exp EBO 1 + ECO VAP VT VBE VCE 2 = I C 2 = I S 2 exp 1 + VT VAN V 2.5 VBE 2.5 12 5 1013 exp EBO 1 + 1 + = 10 exp 80 VT VT 120 V V 5.15625 1013 exp EBO = 1.020833 1012 exp BE VT VT V exp EBO VT = 1.9798 = exp VEBO VBE VT V exp BE VT VEBO = VBE + VT ln (1.9798 ) = 0.5208 + ( 0.026 ) ln (1.9798 ) VEBO = 0.5386 VI = 5 0.5386 VI = 4.461 Vwww.elsolucionario.net 415. Av =d. (1/ VT )(1/ VAN ) + (1/ VAP )1 38.46 0.026 = Av = 1 1 0.00833 + 0.0125 + 120 80 Av = 1846 10.69 a. M1 and M2 matched. For I REF = I 0 , we have VSD 2 = VSG = VSG 3 = VDS 0 = 2.5 V For M1 and M3: 2 1 W I REF = p Cox (VSG + VTP ) (1 + PVSD ) 2 L 1 2 W W W W 100 = 10 ( 2.5 1) 1 + ( 0.02 )( 2.5 ) = 4.23 = = L 1 L 1 L 3 L 2 For M0: 2 1 W I O = n Cox (VGS VTN ) (1 + nVDS ) 2 L 0 2 W W 100 = 20 ( 2 1) 1 + ( 0.02 )( 2.5 ) = 4.76 L 0 L 0 b. r0 n = r0 p =1 I0=1( 0.02 )( 0.1)= 500 k1 W g m = 2 K n I O = 2 n Cox I O 2 L o( 0.02 )( 4.76 )( 0.1)=2g m = 0.195 mA/VAv = g m ( r0 n r0 p ) = ( 0.195 )( 500 500 ) Av = 48.810.70 I REF = K p1 (VSG + VTP )a.2100 = 100 (VSG 1) VSG = 2 V 2b.From Eq. 10.89 1 + p (V + VSG ) K (V V )2 TN n I VO = n + p I REF ( n + p ) 1 + ( 0.02 )(10 2 ) 100 (VI 1) 5= 0.02 + 0.02 100 ( 0.02 + 0.02 ) 25 = 29 (VI 1)2(VI 1)20.04 = 0.96VI = 1.98 Vwww.elsolucionario.net 416. Av = g m ( r0 n r0 p )c.r0 n = r0 p =1 I REF=1 = 500 k 0.02 )( 0.1) (( 0.1)( 0.1) = 0.2 mA / V Av = ( 0.2 )( 500 500 ) Av = 50g m = 2 K n I REF = 210.71 5 0.6 5 0.6 = = 0.22 mA R1 20 From Eq. 10.96 I 0.22 C VT 0.026 Av = = IC I C 0.22 + 1 + 0.22 1 + + 90 VAN RL VAP 140 RL I REF 8.4615 8.4615 = 1 1 0.0015714 + + 0.002444 0.004016 + RL RL RL = , Av = 2107(a) (b) (c)RL = 250 K, Av = 1056 RL = 100 K, Av = 60410.72 I REF =5 0.6 = 0.1257 mA 35Then I CO = 2 I REF = 0.2514 mA From Eq. 10.96 0.2514 9.6692 0.026 Av = = 0.2514 0.2514 1 1 0.002095 + 0.0031425 + + + 120 80 RL RL Av =(a) (b)9.6692 1 RL RL = Av = 18460.0052375 +RL = 250 K, Av = 104710.73 (a) To a good approximation, output resistance is the same as the widlar current source. R0 = r02 1 + g m 2 ( r 2 RE ) (b)Av = g m 0 ( r0 RL R0 )10.74 Output resistance of Wilson sourcewww.elsolucionario.net 417. r03R0 2 Then Av = g m ( r0 R0 ) V 80 r03 = AP = = 400 k I REF 0.2 r0 =VAN 120 = = 600 k I REF 0.2gm =I REF 0.2 = = 7.692 mA/V VT 0.026 (80 )( 400 ) Av = 7.69 600 = 7.69 [ 600 16, 000] Av = 4448 2 10.75 (a)I D 2 = I D 0 = I REF = 200 AFor M 2 ; ro 2 =1P I D 2=1 = 250 K 0.02 )( 0.2 ) ( 0.04 gm2 = 2 K P I D2 = 2 ( 35 )( 0.2 ) 2 g m 2 = 0.748 mA/V 1 1 For M 0 ; r = = = 333 K n I Do ( 0.015 )( 0.2 ) 0.08 g mo = 2 ( 20 )( 0.2 ) g mo = 0.80 mA/V 2 Av = g mo ( ro 2 roo ) = ( 0.80 )( 250 333)(b)Av = 114.3 Want Av = 57.15 = 0.80 (142.8 RL ) 142.8 RL 142.8 RL = 71.375 = RL = 143 K 142.8 + RL(c)10.76 Assume M1, M2 matched I REF = I D 2 = I Do = 200 A 1 1 ro 2 = = = 250 K p I D 2 ( 0.02 )( 0.2 ) roo =1 1 = = 333 K n I D 0 ( 0.015 )( 0.2 )Av = g mo ( ro 2 roo )100 = g mo ( 250 333) g mo = 0.70 mA/V 0.08 W g mo = 2 ( 0.2 ) = 0.70 2 L 0 W = 15.3 L 0www.elsolucionario.net 418. k W k W p Now n = 2 L 0 2 L 2 80 40 W (15.3) = 2 2 L 2 W W = = 30.6 L 2 L 110.77Since Vsg 3 = 0, the circuit becomeswww.elsolucionario.net 419. I x = g mVsg 2 +Vx Vsg 2 r02and Vsg 2 = I x ro 3Then r V I x 1 + g m ro3 + o3 = x ro 2 ro 2 so that Vx r = Ro = ro 2 1 + g m ro 3 + o 3 Ix ro 2 or Ro = ro 2 + ro 3 (1 + g m ro 2 ) vo = g m1 ( ro1 Ro ) viAv = Nowg m1 = 2 ro1 =( 0.050 )( 20 )( 0.10 ) = 0.632 mA / V1 1 = = 500 k n I DQ ( 0.02 )( 0.10 ) g m = 2 K p I DQ = 2 ro 2 = ro 3 =( 0.020 )(80 )( 0.1) = 0.80 mA / V1 1 = = 500 k p I DQ ( 0.020 )( 0.1)Then Ro = 500 + 500 1 + ( 0.8 )( 500 ) 201 M Av = ( 0.632 ) ( 500 201000 ) Av = 31510.78 From Eq. 10.105 2 gm Av = 1 1 + ro3 ro 4 ro1 ro 2 k W g m = 2 n I D1 2 L =2( 0.050 ) ( 20 ) ( 0.08 )g m = 0.5657 mA / Vro =1 1 = = 625 K I D ( 0.02 )( 0.08 ) ( 0.5657 ) 0.3200 Av = = 1 1 2 ( 0.00000256 ) + 2 2 ( 625) ( 625) 2Av = 62,50010.79www.elsolucionario.net 420. V ( V 2 ) V 2 V 2 + + g m 2V 2 + O r 2 ro1 ro 2(1)g m1Vi =(2)VO VO ( V 2 ) + + g m 2V 2 = 0 RO 3 ro 2(1) 1 1 1 V + + gm2 + + O g m1Vi = V 2 ro 2 ro 2 r 2 ro1(2) 1 1 1 VO + + V 2 + gm2 = 0 RO 3 ro 2 ro 2 g m >>1 ro(1) 1 + VO g m1Vi = V 2 + r 2 ro 2(2) 1 1 VO + + V 2 g m 2 = 0 RO 3 ro 2 (3)V 2 = VO gm2 1 1 + RO 3 ro 2 Then 1 1 1 + VO + + RO 3 ro 2 r 2 ro 2 1 1 1 + VO = VO + + RO 3 ro 2 ro 2 (1) g m1Vi = VO gm2VO 1 + RO 3 VO = g m1 RO 3 Vi 1+ From Equation (10.20) RO 3 rO 3 So www.elsolucionario.net 421. VO g m1ro3 2 = 1+ ViAv =gm =0.25 = 9.615 mA/V 0.026ro3 =Av =80 = 320 K 0.25 ( 9.615 )( 320 )(120 ) 1212= 366,165www.elsolucionario.net 422. Chapter 11 Exercise Solutions EX11.1 vE = VBE ( on ) vE = 0.7 V I C1 = I C 2 = 0.5 mA vC1 = vC 2 = 10 ( 0.5 )(10 ) vC1 = vC 2 = 5 VEX11.2 iC 2 1 = = 0.99 IQ vd 1 + exp VT v 1 1 + exp d = VT 0.99 v exp d VT 1 1 = 0.99 1 1 vd = 119.5 mV vd = VT ln 0.99 EX11.3 a.v1 = v2 = 0 vE = 0.7 V VRC = ( 0.25 )( 8 ) = 2 V vC1 = vC 2 = 3 V vEC1 = 3.7 Vb.v1 = v2 = 2.5 V vE = 3.2 V vEC1 = 6.2 Vc.v1 = v2 = 2.5 V vE = 1.8 V vEC1 = 1.2 VEX11.4 Let I Q = 1 mA, then I CQ1 = I CQ 2 = 0.5 mA 0.5 = 19.23 mA / V 0.026 v 1 At vC 2 , Ad = c 2 = g m RC 2 vd 2 g m1 = g m 2 =1 (19.23) RC 2 RC 2 = 15.6 k 2 v 1 At vC1 , Ad = c1 = g m RC1 vd 2So, 150 =1 (19.23) RC1 RC1 = 10.4 k 2 If V + = +10 V and V = 10 V , dc biasing is OK.So, 100 = EX11.5Acm = 1+ I Q RC 2VT = (1 + ) IQ RO VT ( 0.8 )(12 ) 2 ( 0.026 ) = 0.0594 (101)( 0.8 )(100 ) 1+ ( 0.026 )(100 )vo = Acm vcm = ( 0.0594 )(1sin t )( mV ) = 59.4sin t ( V )EX11.6www.elsolucionario.net 423. vd = 40 V , vcm = 0 vo = ( 50 )( 40 ) 2.0 mV(a)(b) vd = 40 V vcm = 200 V Assuming Acm > 0 50 Acm60 = 20 log10Acm = 0.05 vo = ( 50 )( 40 ) + ( 0.05 )( 200 ) vo = 2.010 mVEX11.7Diff. Gain Ad =a.I Q RC 4VTFor v1 = v2 = 5 V Minimum collector voltagevC 2 = 5 V IQ RC = 15 5 = 10 V or I Q RC = 20 V for max. Ad 2 20 Then Ad = Ad ( max ) = 192 2 ( 0.026 ) If I Q = 0.5 mA, RC = 40 kb. I Q RC 2VT Acm = (1 + ) I Q R0 1 + VT Then 20 2 ( 0.026 ) Acm = Acm = 0.199 ( 201)( 0.5 )(100 ) 1 + ( 0.026 )( 200 ) 192 and C M RRdB = 20 log10 C M RRdB = 59.7 dB 0.199 EX11.8 Rid = 2 r + (1 + ) RE r = VT I CQ=(100 )( 0.026 ) 0.25= 10.4 KRid = 2 10.4 + (101)( 0.5 ) = 122 K EX11.9 Ad =g m RC 2 (1 + g m RE )( 9.62 )(10 ) 2 1 + ( 9.62 ) RE 1 + ( 9.62 ) RE = 4.81 RE = 0.396 K Rid = 2 r + (1 + ) RE = 2 10.4 + (101)( 0.396 ) 10 =Rid = 100.8 Kwww.elsolucionario.net 424. EX11.10 10 VGS 4 2 = K n 3 (VGS 4 VTN ) I1 = R1 10 VGS 4 = ( 0.1)( 80 )(VGS 4 0.8 )22 10 VGS 4 = 8 (VGS 4 1.6VGS 4 + 0.64 ) 2 8VGS 4 11.8VGS 4 4.88 = 0VGS 4 =11.8 (11.8) + 4 ( 8)( 4.88) 2 (8) 2= 1.81 V10 1.81 = 0.102 mA 80 0.102 = ID2 = = 0.0512 mA 2I1 = I Q = I D1= K n1 (VGS 1 VTN )20.0512 = 0.050 (VGS 1 0.8 ) VGS 1 = 1.81 V 2v01 = v02 = 5 ( 0.0512 )( 40 ) = 2.95 V Max vcm : VDS 1 ( sat ) = VGS1 VTN= 1.81 0.8 = 1.01 V vcm ( max ) = v01 VDS 1 ( sat ) + VGS 1 = 2.95 1.01 + 1.81 vcm ( max )= 3.75 VMin vcm : VDS 4 ( sat ) = VGS 4 VTN = 1.81 0.8 = 1.01 Vvcm ( min )= VGS 1 + VDS 4 ( sat ) 5 = 1.81 + 1.01 5vcm ( min )= 2.18 V2.18 vcm 3.75 VEX11.11(1)( 2 ) g f ( max ) = 1 mA / V 2 ( 2) = g f RD = (1)( 5 ) Ad = 5g f ( max ) = AdK n IQ=EX11.12 K Kn iD1 1 = + vd 1 n 2 IQ IQ 2 2IQ 2 vd Using the parameters in Example 11.11, K n = 0.5 mA / V 2 , I Q = 1 mA, then 0.5 2 iD1 1 0.5 = 0.90 = + vd 1 2 (1) vd 2 2 (1) IQ By trial and error, vd = 0.894 V EX11.13 a.www.elsolucionario.net 425. gm =IQ 2VT=0.5 = 9.615 mA/V 2 ( 0.026 )r02 =VA 2 125 = = 500 k I C 2 0.25r04 =VA 4 85 = = 340 k I C 4 0.25Ad = g m ( r02 r04 ) = ( 9.615 ) ( 500 340 ) Ad = 1946b.c.(Ad = g m r02 r04 RL)Ad = ( 9.615 ) 500 340 100 Ad = 644 VT (150 )( 0.026 ) r = = = 15.6 k I CQ 0.25Rid = 2r Rid = 31.2 kd.R0 = r02 r04 = 500 340 R0 = 202 kEX11.14 80 80 2 2 I REF = (10 )(VGS 4 0.5 ) = ( 0.33)(VGS 5 0.5 ) 2 2 VGS 4 + 2VGS 5 = 6 1 VGS 5 = ( 6 VGS 4 ) 2 10 1 (VGS 4 0.5) = ( 6 VGS 4 ) 0.5 0.33 2 6.0VGS 4 = 5.252 VGS 4 = 0.8754 V 80 2 I REF = (10 )( 0.8754 0.5 ) = 53.37 A 2 Also I Q = I REF = 53.37 A 80 g m = 2kn I Q = 2 (10 )( 53.37 ) 0.2066 mA/V 2 1 1 = = 1873.7 K ro1 = ro8 = ID 0.05337 ( 0.02 ) 2 Ad = g m ( ro1 ro8 ) = ( 0.2066 )(1873.7 1873.7 ) Ad = 194EX11.15 Ad = g m ( ro 2 RO ) 400 = g m ( 500 101000 ) g m = 0.8 mA / V g m = 2 K n I DQ 0.08 W W W 0.8 = 2 ( 0.1) = = 40 2 L 1 L 1 L 2EX11.16www.elsolucionario.net 426. I e 6 = (1 + ) I b 6 = I b 7 I c 7 = I b 7 = (1 + ) I b 6 Ic7 = (1 + ) = (100 )(101) = 1.01 104 Ib6EX11.17 r + 300 ROA = A 1+ I CA = 1+ = 1+ I EA 1 (1) 9.803 A 1 + (100 )( 0.026 ) r A = = 265.2 K 0.009803 265.2 + 300 ROA = = 5.596 K 101 r + ROA RO = B 10 1+ r B =(100 )( 0.026 )= 2.626 K 0.99 2.626 + 5.596 10 RO = 101 RO = 81.4 104 = 80.7 EX11.18 10 0.7 ( 10 ) I1 = = 0.6 R1 = 32.2 K R1 I I Q R2 = VT ln 1 I Q .( 0.2 ) R2 = ( 0.026 ) ln 0.6 R2 = 143 0.2 I R 6 = I1 R3 = 0 10 = I C1 RC + VCE1 0.7 10.7 = ( 0.1) RC + 4 RC = 67 K vo 2 = 0.7 + 4 = 3.3 V vE 4 = 3.3 1.4 = 1.9 V I R4 =vE 4 1.9 R4 = R4 = 3.17 K 0.6 R4vC 3 = vO 2 1.4 + vCE 4 = 3.3 1.4 + 3 = 4.9 V I R 5 = I R 4 = 0.6 =10 4.9 R5 = 8.5 K R5vE 5 = 4.9 0.7 = 4.2 V R6 = R7 =0 ( 10 ) 54.2 0.7 = 5.83 K 0.6 R7 = 2 Kwww.elsolucionario.net 427. EX11.19 Ri 2 = r 3 + (1 + ) r 4(100 )( 0.026 )r 4 =0.6 2VTr 3 I R4= 4.333 K(100 ) ( 0.026 ) 2=0.6= 433.3 KRi 2 = 433.3 + (101)( 4.333) Ri 2 = 871 K Ri 3 = r 5 + (1 + R6 + r 6 + (1 + ) R7 (100 )( 0.026 )r 5 =0.6 (100 )( 0.026 )r 6 4.333 K= 0.52 K 5 Ri 3 = 4.333 + (101) 5.83 + 0.52 + (101)( 2 ) Ri 3 = 21.0 M gm 0.1 ( RC Ri 2 ) g m = 0.026 = 3.846 mA/V 2 3.846 Ad 1 = ( 67 871) = 119.64 2 I 0.6 A2 = R 4 ( R5 Ri 3 ) = (8.5 21000 ) = 98.037 2 ( 0.026 ) 2VT Ad 1 =A = Ad 1 A2 = (119.64 )( 98.037 ) = 11, 729EX11.20 fz =1 1 = 2 RO CO 2 (10 106 )( 0.2 1012 )f z = 79.6 kHz fp =1 2 Req COReq = 51.98 from Example 11.20 1 fp = 2 ( 51.98 ) ( 0.2 1012 ) f p = 15.3 GHzTYU11.1 Vd = V1 V2 = 2 + 0.005sin t ( 0.5 0.005sin t ) Vd = 1.5 + 0.010sin t ( V ) V1 + V2 2 2 + 0.005sin t + 0.5 0.005sin t = Vcm = 1.25 V 2Vcm =TYU11.2 For v1 = v2 = +4 V Minimum vC1 = vC 2 = 4 V IQ I C1 = I C 2 = = 1 mA 2 10 4 RC = RC = 6 k 1www.elsolucionario.net 428. TYU11.3 From Equation (11.41) g R CMRR = m O RC RC For CMRR dB = 75 dB CMRR = 5623.4 Then 5623.4 =( 3.86 )(100 ) RC (10 )Or RC = 0.686 K TYU11.4 From Equation (11.49) 1 + 2 RO g m CMRR = g 2 m gm For CMRR dB = 90 dB CMRR = 31622.8 1 + 2 (100 )( 3.86 ) Then 31622.8 = ( 3.86 ) 2g m g m 0.0472 = = 0.0122 1.22% Or g m = 0.0472 mA/V or 3.86 gmTYU11.5 For v1 = v2 = 5 V min vC1 = vC 2 = 5 V So I C1 RC = 10 5 = 0.25 RC RC = 20 k Ad =I Q RC=4VT( 0.5 )( 20 ) Ad 4 ( 0.026 )= 96.2 Let I Q = 0.5 mAC M RRdB = 95 db C M RR = 5.62 104 Acm =Acm96.2 Acm = 1.71 103 5.62 104 I Q RC 2VT = = 1.71 103 (1 + ) I Q R0 1 + VT ( 0.5 )( 20 ) 2 ( 0.026 ) = 1.71 103 ( 201)( 0.5 ) R0 1 + ( 0.026 )( 200 ) 1 + 19.33R0 = 1.125 105 R0= 5.82 103 k = 5.82 MWe have R0 = r04 1 + g m 4 ( R2 r 4 ) r04 =VA 125 = = 250 k I Q 0.5gm4 = r 2 =IQ VT VT IQ=0.5 = 19.23 mA/V 0.026=( 200 )( 0.026 ) 0.5= 10.4 kwww.elsolucionario.net 429. 5.820 = 250 1 + g m 4 ( R2 r 4 ) 19.23 ( R2 r 2 ) = 22.28(R2r 2 ) = 1.159 kR2 (10.4 )R2 + 10.4= 1.159R2 (10.4 1.16 ) = (1.16 )(10.4 ) R2 = 1.30 k Let I1 = 1 mA I I Q R2 I1 R3 = VT ln 1 I Q ( 0.5)(1.304 ) (1) R3 = ( 0.026 ) ln 1 R3 = 0.634 k 0.5 If VBE ( Q3 ) 0.7 V10 0.7 ( 10 )R1 + R3 =1= 19.3 R1 18.7 kTYU11.6 a. v0 = Ad vd + Acm vcm vd = v1 v2 = 0.505sin t 0.495sin t = 0.01sin t v +v 0.505sin t + 0.495sin t vcm = 1 2 = 2 2 = 0.50sin t v0 = ( 60 )( 0.01sin t ) + ( 0.5 )( 0.5sin t ) v0 = 0.85sin t ( V )b. vd = v1 v2 = 0.5 + 0.005sin t ( 0.5 0.005sin t ) = 0.01sin t v +v vcm = 1 2 2 0.5 + 0.005sin t + 0.5 0.005sin t = 2 = 0.5 v0 = ( 60 )( 0.01sin t ) + ( 0.5 )( 0.5 ) v0 = 0.25 + 0.6sin t ( V ) TYU11.7a. b.I B1 = I B 2 = r = VT I CQIQ / 2=1 I B1 = I B 2 = 6.62 A 151(1 + ) (150 )( 0.026 )=1= 3.9 kRid = 2r = 2 ( 3.9 ) = 7.8 k Ib =c.Vd 10sin t ( mV ) = I b = 1.28sin t ( A ) Rid 7.8 kRicm 2 (1 + ) R0 = 2 (151)( 50 ) 15.1 M Ib =Vcm 3sin t = I b = 0.199sin t ( A ) Ricm 15.1 Mwww.elsolucionario.net 430. TYU11.8 Ad = g f RD 8 = g f ( 4 ) g f ( max ) = 2 mA / V g f ( max ) =( 2)2Kn IQ 2 4 = K n K n = 2 mA / V 2 2TYU11.9 From Example 11-10, I Q = 0.587 mA Kn2 IQ( 0.1)( 0.587 ) (16) Ad = 2.74 2 1 1 = R0 = 85.2 k For M 4 , R0 = 4 I Q ( 0.02)( 0.587 ) Ad =2 RD =g m = 2 K n (VGS 2 VTN ) = 2 ( 0.1)( 2.71 1) = 0.342 mA/V Acm = ( 0.342)(16) g m RD = Acm = 0.0923 1 + 2 g m R0 1 + 2 ( 0.342)(85.2) 2.74 C M RRdB = 20 log10 C M RRdB = 29.4 dB 0.0923 TYU11.10 1 CMRR = 1 + 2 2 K n I Q Ro 2 CMRRdB = 60 dB C M RR = 1000 1 1000 = 1 + 2 2 ( 0.1)( 0.2 ) R0 2 2000 = 1 + 0.4 R0 R0 5 M TYU11.11 Ro = ro 4 + ro 2 (1 + g m 4 ro 4 )Assume I REF = I O = 100 A and = 0.01 V 1 ro 2 = ro 4 =1 1 = 1 M I D ( 0.01)( 0.1)Let K n ( all devices ) = 0.1 mA / V 2( 0.1)( 0.1) = 0.2 mA / V Ro = 1000 + 1000 (1 + ( 0.2 )(1000 ) ) 202 M Then g m 4 = 2 K n I D = 2Now VGS1 = VGS 2 =ID 0.05 + VTN = + 1 = 1.707 V 0.1 KnVDS1 ( sat ) = VGS 1 VTN = 1.707 1 = 0.707 V So vo1 ( min ) = +4 VGS 1 + VDS 1 ( sat ) = 4 1.707 + 0.707vo1 ( min ) = 3 V = 10 I D RD = 10 ( 0.05 ) RD RD = 140 kFor a one-sided output, the differential gain is:www.elsolucionario.net 431. Ad =1 g m1 RD where g m1 = 2 K n I D 2=2( 0.1)( 0.05) = 0.1414 mA / V1 ( 0.1414 )(140 ) Ad = 9.90 2 The common-mode gain is: 2 K n I Q RD 2 ( 0.1)( 0.1) (140 ) Acm = = Acm = 0.0003465 1 + 2 2 K n I Q Ro 1 + 2 2 ( 0.1)( 0.1) ( 202000 ) Ad =Ad CMRRdB = 89.1 dB AcmThen CMRRdB = 20 log10 TYU11.12 I B5 =a.IQ (1 + )=0.5 15.3 nA (180 )(181)So I 0 = 15.3 nA b. For a balanced condition VEC 4 = VEC 3 = VEB 3 + VEB 5 VEC 4 = 1.4 V VCE 2 = VC 2 VE 2 = (10 1.4 ) ( 0.7 ) VCE 2 = 9.3 VTYU11.13 Ad = 2 g f ( r02 r04 ) gf =IQ 4VT=0.2 = 1.923 mA/V 4 ( 0.026 )r02 =VA 2 120 = = 1200 k I C 2 0.1r04 =VA 4 80 = = 800 k I C 4 0.1Ad = 2 (1.923) (1200 800 ) Ad = 1846TYU11.14 P = ( I Q + I REF ) ( 5 ( 5 ) )10 = ( 0.1 + I REF )(10 ) I REF = 0.9 mA R1 =5 0.7 ( 5 ) I REF=9.3 R1 = 10.3 k 0.9I I Q RE = VT ln REF I Q 0.026 0.9 ln RE = RE = 0.571 k 0.1 0.1 ro 2 =VA 2 120 = 2.4 M I C 2 0.05ro 4 =VA 4 80 = 1.6 M I C 4 0.05gm =0.05 = 1.923 mA / V 0.026()()Ad = g m ro 2 ro 4 RL = (1.923) 2400 1600 90 Ad = 158www.elsolucionario.net 432. TYU11.15 a. R0 = r02 r04 120 r02 = = 1.2 M 0.1 80 r04 = = 0.8 M 0.1 R0 = 1.2 0.8 R0 = 0.48 Mb. Ad ( open circuit ) = 2 g f ( r02 r04 )(Ad ( with load ) = 2 g f r02 r04 RL For Ad ( with load ) =)1 Ad ( open circuit ) RL = ( r02 r04 ) RL = 0.48 M 2TYU11.16 2Kn 1 Ad = 2 I Q ( 2 + 4 ) =22 ( 0.1) 0.11( 0.01 + 0.015) Ad = 113TYU11.17 For the MOSFET, I D = 25 g m1 = 2 K n I D = 275 = 24.26 A 101( 20 )( 24.26 ) = 44.05 A/VFor the Bipolar, I E = 100 25 = 75 A 100 IC = ( 75 ) = 74.26 A 101 (100 )( 0.026 ) r 2 = = 35.0 K 0.07426 0.07426 gm2 = = 2.856 mA/V 0.026 g (1 + g m 2 r ) ( 44.05 ) 1 + ( 2.856 )( 35 ) C g m = m1 = 1 + g m1r 1 + ( 0.04405 )( 35 ) C g m = 1.75 mA/VTYU11.18 From Figure 11.43 80 r04 = = 160 k 0.5 R0 r04 = (150 )(160 ) k R0 = 24 MFrom Figure 11.44 1 1 r06 = = r06 = 160 k I D ( 0.0125 )( 0.5 ) 0.5 = 0.5 (VGS 1) VGS = 2 V 2g m 6 = 2 K n (VGS VTN ) = 2 ( 0.5 )( 2 1) = 1 mA / V r04 = 160 kR0 = ( g m 6 )( r06 )( r04 ) = (1)(160 )(150 )(160 ) R0 = 3.840 Mwww.elsolucionario.net 433. TYU11.19 From Equation (11.126) 2 (1 + ) VT 2 (121)(120 )( 0.026 ) Ri = 1.51 M Ri = = IQ 0.5 r 11 = VT=IQ(120 )( 0.026 ) 0.5= 6.24 k RE = r 11 R3 = 6.24 0.1 = 0.0984 k g m11 = r011 =IQ VT=0.5 = 19.23 mA/V 0.026VA 120 = = 240 k I Q 0.5 Then RC11 = r011 (1 + g m11 RE )= 240 1 + (19.23)( 0.0984 ) = 694 kr 8 = VT IC 8=(120 )( 0.026 ) 2= 1.56 kRb8 = r 8 + (1 + ) R4 = 1.56 + (121)( 5 ) = 607 k Then RL 7 = RC11 Rb8 = 694 607 = 324 k 0.5 IQ Then Av = ( 324 ) Av = 3115 RL 7 = 2 ( 0.026 ) 2VT r +Z R0 = R4 8 1+ Z = RC11 RC 7 V 120 RC 7 = A = = 240 k I Q 0.5 Z = 694 240 = 178 k 1.56 + 178 R0 = 5 = 5 1.48 R0 = 1.14 k 121 TYU11.20 IQ Av = 2VT RL 7 0.5 103 = 2 ( 0.026 ) RL 7 RL 7 = 104 k www.elsolucionario.net 434. Chapter 11 Problem Solutions 11.1 (a) 0.7 ( 3) RE= 0.1 RE = 23 K3 1.5 = 0.05 RC = 30 K RC(b)vCE 2 = 6 iC 2 ( RC + 2 RE ) = 6 iC 2 ( 76 )(c)vcm ( max ) vCB 2 = 0 vCE 2 = 0.7 VSo 0.7 = 6 iC 2 ( 76 ) iC 2 = 69.74 A( v ( max ) 0.7 ) ( 3) = 2( 0.06974 ) vCM ( max ) = 0.908 V 23 ( min ) VS = 3 V vCM ( min ) = 2.3 VCMvCM 11.2Ad = 180, C M RRdB = 85 dB Ad 180 = Acm = 0.01012 Acm Acm Assume the common-mode gain is negative. v0 = Ad vd + Acm vcm C M RR = 17, 783 180vd 0.01012vcm v0 = 180 ( 2sin t ) mV ( 0.01012 )( 2sin t ) V v0 = 0.36sin t 0.02024sin tIdeal Output:v0 = 0.360sin t ( V )Actual Output:v0 = 0.340sin t ( V )11.3 a.www.elsolucionario.net 435. I1 =10 2 ( 0.7 )IC 2 = I1 = 1.01 mA8.5 I12 1+ (1 + )=1.01 I C 2 1.01 mA 2 1+ (100 )(101) 100 1.01 IC 4 = I C 4 0.50 mA 101 2 VCE 2 = ( 0 0.7 ) ( 5 ) VCE 2 = 4.3 V VCE 4 = 5 ( 0.5 )( 2 ) ( 0.7 ) VCE 4 = 4.7 V b. For VCE 4 = 2.5 V VC 4 = 0.7 + 2.5 = 1.8 V 5 1.8 IC 4 = I C 4 = 1.6 mA 2 1+ 101 IC 2 + ( 2IC 4 ) = ( 2 )(1.6 ) I C 2 = 3.23 mA 100 I1 I C 2 = 3.23 mA R1 =10 2 ( 0.7 ) 3.23 R1 = 2.66 k11.4 a.Neglecting base currents 30 0.7 R1 = 73.25 k I1 = I 3 = 400 A R1 = 0.4 VCE1 = 10 V VC1 = 9.3 V 15 9.3 RC = 28.5 k RC = 0.2 b. (100 )( 0.026 ) r = = 13 k 0.2 50 r0 ( Q3 ) = = 125 k 0.4 We have Ad =(100 )( 28.5) Ad 2 ( r + RB ) 2 (13 + 10 ) RC 62 RC 1 Acm = 2r0 (1 + ) r + RB 1+ r + RB (100 )( 28.5 ) 1 = Acm = 0.113 13 + 10 2 (125 )(101) 1+ 13 + 10 62 C M RRdB = 20 log10 C M RRdB = 54.8 dB 0.113 c.www.elsolucionario.net 436. Rid = 2 ( r + RB ) = 2 (13 + 10 ) Rid = 46 k 1 r + RB + 2 (1 + ) r0 2 1 = 13 + 10 + 2 (101)(125 ) Ricm = 12.6 M 2Ricm =11.5 vCM ( max ) VCB = 0 so that vCM ( max ) = 5 (a)vCM ( max ) = 3 VIQ 2( RC ) = 5 (b) Vd I CQ Vd 0.25 0.018 = = = 0.08654 mA 2 VT 2 0.026 2 = I RC = ( 0.08654 ) ( 8 ) = 0.692 VI = g m VC 2 (c) 0.25 0.010 I = = 0.04808 mA 0.026 2 VC 2 = ( 0.04808 )( 8 ) = 0.385 V11.6 P = ( I1 + I C 4 ) (V + V )I1 I C 4 so 1.2 = 2 I1 ( 6 ) I1 = I C 4 = 0.1 mA R1 =3 0.7 ( 3) 0.1 R1 = 53 k For vCM = +1V VC1 = VC 2 = 1 V RC =3 1 RC = 40 k 0.05One-sided output 1 0.05 Ad = g m RC where g m = = 1.923 mA / V 2 0.026 Then 1 Ad = (1.923)( 40 ) Ad = 38.5 2 11.7 a. IE ( 2 ) + I E (85) 5 2 5 0.7 IE = I E = 0.050 mA 85 + 1 I E 100 0.050 I C1 = I C 2 = = 1 + 2 101 2 0 = 0.7 +Or I C1 = I C 2 = 0.0248 mA VCE1 = VCE 2 = 5 I C1 (100 ) ( 0.7 ) So VCE1 = VCE 2 = 3.22 Vb.vcm ( max ) for VCB = 0 and VC = 5 I C1 (100 ) = 2.52 VSo vcm ( max ) = 2.52 Vvcm ( min ) for Q1 and Q2 at the edge of cutoff vcm ( min ) = 4.3 V(c) Differential-mode half circuitswww.elsolucionario.net( 0.5) 2(8) 437. V vd = V + + g mV .RE 2 r (1 + ) = V 1 + RE r Then V = ( vd / 2 ) (1 + ) RE 1 + r vo = g mV RC Ad = r = VT=I CQ RC 1 2 r + (1 + ) RE(100 )( 0.026 ) = 105 k RE = 2 k 0.0248Then Ad =11.8 a. RE =1 (100 )(100 ) Ad = 16.3 2 105 + (101)( 2 )For v1 = v2 = 0 and neglecting base currents 0.7 ( 10 ) 0.15 RE = 62 kb. Ad = r = Ad =v02 RC = vd 2 ( r + RB ) VT I CQ=(100 )( 0.026 ) 0.075(100 )( 50 ) Ad 2 ( 34.7 + 0.5 )= 34.7 k= 71.0 RC 1 Acm = r + RB 2 RE (1 + ) 1 + r + RB 1 = Acm = 0.398 34.7 + 0.5 2 ( 62 )(101) 1 + 34.7 + 0.5 71.0 C M RRdB = 45.0 dB C M RRdB = 20 log10 0.398 c. Rid = 2 ( r + RB )(100 )( 50 ) Rid = 2 ( 34.7 + 0.5 ) Rid = 70.4 kCommon-mode input resistance 1 Ricm = r + RB + 2 (1 + ) RE 2 1 = 34.7 + 0.5 + 2 (101)( 62 ) Ricm = 6.28 M 2 11.9www.elsolucionario.net 438. (a) v1 = v2 = 1 V VE = 1.6 IE =9 1.6 18.97 A 390IE = 9.49 A I C1 = I C 2 = 9.39 A 2 vC1 = vC 2 = ( 9.39 )( 0.51) 9 = 4.21 V (b) 9.39 gm = 0.361 mA/V 0.026 V I = g m d = ( 0.361 103 ) ( 0.005 ) = 1.805 A 2 vC = (1.805 106 )( 510 103 ) = 0.921 V vC 2 = 4.21 + 0.921 3.29 V vC1 = 4.21 0.921 5.13 V11.10 (a) v1 = v2 = 0 I E1 = I E 2 6 A = 60 I C1 = I C 2 = 5.90 A vC1 = vC 2 = ( 5.90 )( 0.360 ) 3 = 0.875 V VEC1 = VEC 2 = +0.6 ( 0.875 ) = 1.475 V (b) (i) 5.90 gm = 0.227 mA/V 0.026 Ad = g m RC = ( 0.227 )( 360 ) = 81.7 Acm = 0 (ii) ( 60 )( 0.026 ) g R Ad = m C = 40.8 r = 2 0.0059 = 264 K Acm = ( 0.227 )( 360 ) = 0.0442 2 ( 61)( 4000 ) 1+ 26411.11www.elsolucionario.net 439. For v1 = v2 = 0.20 V I C1 = I C 2 = 0.1 mA vC1 = vC 2 = ( 0.1)( 30 ) 10 = 7 V 0.1 gm = = 3.846 mA/V 0.026 v I = g m d = ( 3.846 )( 0.008 ) 30.77 A 2 vC = I RC = ( 30.77 106 )( 30 103 ) = 0.923 V v2 I C 2 vC 2 vC1= 7 + 0.923 = 6.077 VvC 2 = 7 0.923 = 7.923 V11.12 RC = 50 K For v1 = v2 = 0 IE =0.7 ( 10 )75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026 (120 )( 0.026 ) r = = 50.7 K 0.0615 Differential Input v V v1 = d v2 = d 2 2 Half-circuit. V R I = + g m d vC1 = I RC + 2 2 vo = vC1 vC 2R vC 2 = +I RC 2 R R = I RC + I RC 2 2 = 2IRC V = 2 g m d RC 2 Ad = g m RC = ( 2.365 )( 50 ) = 118.25 Common-mode input.www.elsolucionario.net 440. V vcm = V + + g mV ( 2 RE ) r vcm V = 1 + 1 + ( 2 RE ) r I = g mV =g m vcm vcm = r + (1 + )( 2 RE ) 1+ 1+ ( 2 RE ) r R RC + vcm 2 vC1 = IR1 = r + (1 + )( 2 RE ) vC 2R RC vcm 2 = IR2 = r + (1 + )( 2 RE )vo = vC1 vC 2R R RC + vcm vcm + RC 2 2 =[ ][ ] R 2 vcm 2 = r + (1 + )( 2 RE )Acm = (120 )( 0.5 ) R = r + (1 + )( 2 RE ) 50.7 + (121)( 2 )( 75 )= 0.0032966 118.25 C M RR = = 35,870.5 0.0032966 C M R R = 91.1 dB dB11.13 v1 = v2 = 0 IE =0.7 ( 10 )75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026 g m = 0.01 gm g m1 = 2.377 mA/V g m 2 = 2.353 mA/V r =(120 )( 0.026 ) 0.0615= 50.7 Kwww.elsolucionario.net 441. Vd 2 V = g m1 d Rc 2 Vd = + gm2 Rc 2I = g m vC1 vC 2vo = vC1 vC 2 = g m1Vd V RC g m 2 d RC 2 2Vd RC ( g m1 + g m 2 ) 2 R 50 Ad = C ( g m1 + g m 2 ) = ( 2.377 + 2.353) Ad = 118.25 2 2 Common-Mode g m1 RC vcm g m 2 RC vcm vC1 = vC 2 = 1+ 1+ 1+ 1+ ( 2 RE ) ( 2 RE ) r r = ( g m1 g m 2 ) RC ( 2.377 2.353) ( 50 ) vo = Acm = = vcm 1+ 121 1+ 1+ ( 2 )( 75 ) ( 2 RE ) 50.7 r 1.2 = Acm = 0.003343 358.99 C M R R = 91 dB dB11.14 (a) v1 = v2 = 0 vE = +0.7 V 5 0.7 IE = = 4.3 mA 1 I C1 = I C 2 = 2.132 mA vC1 = vC 2 = ( 2.132 )(1) 5 = 2.87 V (b) v1 = 0.5, v2 = 0 Q2 on Q1 off 120 I C1 = 0, I C 2 = 4.3 mA = 4.264 mA 121 vC1 = 5 V vC 2 = ( 4.264 ) (1) 5 vC 2 = 0.736 V2.132 = 82.0 mA/V 0.026 (82.0 ) v V I = g m d vC = I RC = g m d RC = Vd (1) = 41.0Vd 2 2 2 Vd = 0.015 vc = 0.615 V(c)vE 0.7 Vgm =vC 2 vC1 vC1 = 2.87 + 0.615 = 2.255 V vC 2 = 2.87 0.615 = 3.485 V 11.15www.elsolucionario.net 442. (a) gm =IC 1 = = 38.46 mA/V VT 0.026Ad =vo 1 = = 100 vd 0.01Ad = g m RC 100 = 38.46 RC Rc = 2.6 K(b) With v1 = v2 = 0 vC1 = vC 2 = 10 (1)( 2.6 ) = 7.4 V vcm ( max ) = 7.4 V11.16 a. i.( v01 v02 ) = 0ii. I C1 = I C 2 = 1 mA v01 v02 = V + I C1 RC1 V + I C 2 RC 2 = I C ( RC 2 RC1 ) = (1)( 7.9 8 ) v01 v02 = 0.1 Vb. v I 0 = ( I S 1 + I S 2 ) exp BE VT v 2 103 So exp BE = 13 13 VT 10 + 1.1 10 = 9.524 109 v I C1 = I S 1 exp BE VT 13 9 = (10 )( 9.524 10 ) I C1 = 0.952 mA I C 2 = (1.1 1013 )( 9.524 109 ) I C 2 = 1.048 mAi. v01 v02 = I C 2 RC 2 I C1 RC1 v01 v02 = (1.048 0.952 )( 8 ) v01 v02 = 0.768 V ii. v01 v02 = (1.048 )( 7.9 ) ( 0.952 )( 8 ) v01 v02 = 8.279 7.616 v01 v02 = 0.663 V 11.17 From Equation (11.12(b)) IQ iC 2 = 1 + evd / VT 1 0.90 = 1 + evd / VT 1 So evd / VT = 1 = 0.111 0.90 vd = VT ln ( 0.111) = ( 0.026 ) ln ( 0.111) vd = 0.0571 V 11.18 From Example 11.2, we havewww.elsolucionario.net 443. vd ( max ) 1 vd ( max ) / 0.026 4 ( 0.026 ) 1 + e = 0.02 v ( max ) 0.5 + d 4 ( 0.026 )0.5 + v ( max ) 1 0.98 0.5 + d = vd ( max ) / 0.026 4 ( 0.026 ) 1 + e 0.490 + 9.423vd ( max ) =By trial and error vd ( max ) = 23.7 mV1 1+ e vd ( max ) / 0.02611.19 a. For I1 = 1 mA, VBE3 = 0.7 V 20 0.7 R1 = R1 = 19.3 k 1 I 0.026 1 V R2 = T ln 1 = ln R2 = 0.599 k I 0.1 IQ 0.1 Q b. (180 )( 0.026 ) r 4 = = 46.8 k 0.1 0.1 gm = = 3.846 mA/V 0.026 100 r04 = 1 M 0.1 From Chapter 10 R0 = r04 1 + g m ( RE r 4 ) RE r 4 = 0.599 46.8 = 0.591 R0 = (1) 1 + ( 3.846 )( 0.591) = 3.27 M r01 =100 2 M 0.05 r Ricm (1 + ) R0 (1 + ) 01 2 = (181)( 3.27 ) (181)(1) = 592 181 Ricm = 139 M(c)From Eq. (11.32(b)) g m RC Acm = 2 (1 + ) Ro 1+ r + RB 0.05 = 1.923 mA / V 0.026 (180 )( 0.026 ) r = = 93.6 k 0.05 RB = 0gm =Then Acm = (1.923)( 50 ) Acm = 0.00760 2 (181)( 3270 ) 1+ 93.6www.elsolucionario.net 444. 11.19 For vCM = 3.5 V and a maximum peak-to-peak swing in the output voltage of 2 V, we need the quiescent collector voltage to be VC = 3.5 + 1 = 4.5 V Assume the bias is 10 V , and I Q = 0.5 mA.Then I C = 0.25 mA 10 4.5 RC = 22 k 0.25 (100 )( 0.026 ) = 10.4 k In this case, r = 0.25 Then (100 )( 22 ) Ad = = 101 So gain specification is met. 2 (10.4 + 0.5 ) Now RC =For CMRRdB = 80 dB 1 (1 + ) I Q Ro 1 (101)( 0.5 ) Ro Ro = 1.03 M 1 + = 1 + 2 VT 2 ( 0.026 )(100 ) Need to use a Modified Widlar current source. Ro = ro 1 + g m ( RE1 r ) CMRR = 104 =If VA = 100V , then ro = r =(100 )( 0.026 )100 = 200 k 0.5= 5.2 k 0.5 0.5 gm = = 19.23 mA / V 0.026 Then 1030 = 200 1 + (19.23)( RE1 r ) RE1 r = 0.216 k = RE1 5.2 RE1 = 225 Also let RE 2 = 225 and I REF 0.5 mA11.20 (a)RE =0.7 ( 10 ) 0.25 RE = 37.2 k (b)www.elsolucionario.net 445. 1+ V 1 V V Ve + g mV 1 + 2 + g mV 2 = e or (1) (V 1 + V 2 ) = r r RE r RE r V 1 V1 Ve = V 1 = (V1 Ve ) r RB + r r + RB V 2 = V2 Ve Then 1+ r r V (V1 Ve ) + (V2 Ve ) = e r + RB RE From this, we find(1) V1 + Ve =r + RB V2 r r + RB r + RB +1+ r RE (1 + )Now Vo = g mV 2 RC = g m RC (V2 Ve ) We have r =(120 )( 0.026 ) 0.125 25 k ,gm =0.125 = 4.81 mA / V 0.026(i) Set V1 =Vd V and V2 = d 2 2Then 25 + 0.5 Vd 1 25 ( 0.02 ) Ve = = 2 2.026 25 + 0.5 25 + 0.5 +1+ 25 ( 37.2 )(121) Vd 2So Ve = 0.00494Vd Now V V Vo = ( 4.81)( 50 ) d ( 0.00494 )Vd Ad = o = 119 Vd 2 (ii) Set V1 = V2 = Vcm Then 25 + 0.5 Vcm 1 + V ( 2.02 ) 25 Ve = = cm 2.02567 25 + 0.5 25 + 0.5 +1+ 25 ( 37.2 )(121) Ve = Vcm ( 0.9972 ) Then Vo = ( 4.81)( 50 ) Vcm Vcm ( 0.9972 ) or Acm =Vo = 0.673 Vcm11.21 From Equation (11.18)www.elsolucionario.net 446. v0 = vC 2 vC1 = g m RC vd gm =I CQ VTFor I Q = 2 mA, I CQ = 1 mA 1 = 38.46 mA/V 0.026 Now 2 = ( 38.46 ) RC ( 0.015 )Then g m =So RC = 3.47 k Now VC = V + I C RC= 10 (1)( 3.47 ) = 6.53 VFor VCB = 0 vcm ( max ) = 6.53 V11.22 The small-signal equivalent circuit isA KVL equation: v1 = V 1 V 2 + v2 v1 v2 = V 1 V 2 A KCL equation V 1 V + g mV 1 + 2 + g mV 2 = 0 r r 1 + g m = 0 V 1 = V 2 r (V 1 + V 2 ) Then v1 v2 = 2V 1 V 1 =1 1 ( v1 v2 ) and V 2 = ( v1 v2 ) 2 2At the v01 node: v01 v01 v02 + + g mV 1 = 0 RC RL 1 1 1 v01 + v02 RL RC RL At the v02 node: 1 = g m ( v2 v1 ) 2(1)v02 v02 v01 + + g mV 2 = 0 RC RL 1 1 1 1 v02 + v01 = g m ( v1 v2 ) RC RL RL 2 From (1):(2)www.elsolucionario.net 447. R 1 v02 = v01 1 + L g m RL ( v2 v1 ) RC 2 Substituting into (2) R 1 1 1 1 1 1 + + v01 1 + L g m RL ( v2 v1/ ) v01 RC RC RL 2 RC RL RL 1 = g m ( v1 v2 ) 2 RL 1 RL 1 1 + 2 + + 1 v01 = g m ( v1 v2 ) 1 RC RC RC 2 RC v01 RL 1 RL 2+ = gm ( v1 v2 ) RC RC 2 RC For v1 v2 = vd 1 g m RL v01 Av1 = = 2 vd RL 2+ RC 1 g m RL v02 From symmetry: Av 2 = = 2 vd RL 2+ RC Then Av =v02 v01 g m RL = vd RL 2+ RC 11.23 The small-signal equivalent circuit isKVL equation: v1 = V 1 V 2 + v2 or v1 v2 = V 1 V 2 KCL equation:www.elsolucionario.net 448. V 1 V + g mV 1 + g mV 2 + 2 = 0 r r 1 + g m = 0 V 1 = V 2 r (V 1 + V 2 ) Then v1 v2 = 2V 2 or V 2 = Nowv0 = g mV 2 ( RC 1 = g m ( RC 2For v1 v2 vd Ad =RL )1 ( v1 v2 ) 2RL )( v1 v2 ) v0 1 = g m ( RC vd 2RL )11.23 a. 10 7 RD = 6 k 0.5 I Q = I D1 + I D 2 I Q = 1 mA RD =b. 10 = I D ( 6 ) + VDS VGS and VGS =ID + VTN KnFor I D = 0.5 mA, VGS =0.5 + 2 = 3.12 V 0.4and VDS = 10.12Load line is actually nonlinear. c. Maximum common-mode voltage when M 1 and M 2 reach the transition point, or VDS ( sat ) = VGS VTN = 3.12 = 2 = 1.12VThen vcm = v02 vDS ( sat ) + VGS = 7 1.12 + 3.12 Or vcm ( max ) = 9 V Minimum common-mode voltage, voltage across I Q becomes zero. So vcm ( min ) = 10 + 3.12 vcm ( min ) = 6.88 V11.24www.elsolucionario.net 449. We have VC 2 = g mV 2 RC = g m (Vb 2 Ve ) RC and VC1 = g mV 1 RC = g m (Vb1 Ve ) RC Then V0 = VC 2 VC1= g m (Vb 2 Ve ) RC g m (Vb1 Ve ) RC = g m RC (Vb1 Vb 2 )Differential gain Ad =V0 = g m RC Vb1 Vb 2Common-mode gain Acm = 0 11.25 (a) vcm = 3 V VC1 = VC 2 = 3 V 10 3 Then RC = RC = 70 k 0.1 (b) CMRRdB = 75 dB CMRR = 5623 Now CMRR = 5623 =1 (1 + ) I Q Ro 1 + VT 2 1 (151)( 0.2 ) Ro 1 + Ro = 1.45 M 2 (150 )( 0.026 ) Use a Widlar current source. Ro = ro [1 + g m RE ] Let VA of current source transistor be 100 V. 100 0.2 = 500 k , g m = = 7.69 mA / V Then ro = 0.2 0.026 (150 )( 0.026 ) r = = 19.5 k 0.2 So 1450 = 500 1 + ( 7.69 ) RE RE = 0.247 k Now RE = RE r 0.247 = RE 19.5 RE = 250 I Then I Q RE = VT ln REF I Q I REF I REF = 1.37 mA ( 0.2 ) ( 0.2 )( 0.250 ) = ( 0.026 ) ln Then R1 =10 0.7 ( 10 )11.26 At terminal A. RTHA = RA R =1.37 R1 = 14.1 k R (1 + ) RR (1 + ) + R=R (1 + ) 2+R = 5 k 2Variation in RTH is not significant RA + R (1 + )( 5 ) 5 (1 + ) VTHA = = V = R (1 + ) + R 2+ RA + R www.elsolucionario.net 450. At terminal B. R = 5 k 2 R + VTHB = V = 2.5 V R+R From Eq. (11.27) RC (V2 V1 ) VO = where V2 = VTHB and V1 = VTHA 2 ( r + RB ) RTHB = R R =RB = 5 k , r = So VO =(120 )( 0.026 )0.25 (120 )( 3)(V2 V1 ) 2 (12.5 + 5 )= 12.5 k = 10.3 (V2 V1 )We can find V2 V1 = VTHB VTHA 5 (1 + ) VTHB VTHA = 2.5 2+ 2.5 ( 2 + ) 5 (1 + ) 2.5 5 = = 2+ 2+ 2.5 = 1.25 2 Then VO = (10.3)( 1.25 ) = 12.9So for 0.01 0.01 We have 0.129 VO 2 0.129 V 11.27 a. Rid = 2r r =(180 )( 0.026 )0.2 So Rid = 46.8 k= 23.4 kAssuming r , thenb.Ricm (1 + ) R0 Ricm = (181)(1) = 181 Ricm = 181 M11.28 (a) 10 0.7 ( 10 ) I1 = = 0.5 R1 = 38.6 K R1 R2 =0.026 0.5 ln R2 = 236 0.14 0.14 (b)www.elsolucionario.net 451. Ricm (1 + ) Ro 0.14 = 5.385 mA/V 0.026 (180 )( 0.026 ) r 4 = = 33.4 K 0.14 RE = 33.4 0.236 = 0.234 K Ro = ro 4 (1 + g m 4 RE ) g m 4 =100 = 714 K 0.14 Ro = 714 1 + ( 5.385 )( 0.234 ) ro 4 1614 KRicm = (181)(1614 ) 292 M(c) Acm = g m1 RC 2 (1 + ) Ro 1+ r 1g m1 =r 1 =0.07 = 2.692 mA/V 0.026(180 )( 0.026 ) 0.07 ( 2.692 )( 40 ) 2 (181)(1614 ) 1+ 66.86 Acm = 0.0123= 66.86 KAcm =11.29 Ad 1 = g m1 ( R1 r 3 ) g m1 = r 3 = Ad 2 =I Q1 / 2 VT VT IQ2 / 2= 19.23I Q1 =2 (100 )( 0.026 ) IQ 2=5.2 IQ 2IQ 2 / 2 g m 3 R2 , g m3 = = 19.23I Q 2 2 VT(19.23) I Q 2 R2 I Q 2 R2 = 3.12 V 2 Maximum vo 2 vo1 = 18 mV for linearity Then 30 =vo3 ( max ) = ( 18 )( 30 ) mV 0.54 V so I Q 2 R2 = 3.12 V is OK.From Ad 1 :www.elsolucionario.net 452. 5.2 R1 I Q2 20 = 19.23I Q1 ( R1 r 3 ) = 19.23I Q1 R + 5.2 1 IQ 2 19.23I Q1 R1 ( 5.2 ) 20 = I Q 2 R1 + 5.2 I Q1Let2 R1 = 5V I Q1 R1 = 10 VThen 20 =19.23 (10 )( 5.2 ) I Q 2 R1 + 5.2Now I Q1 R1 = 10 R1 = I Q 2 R1 = 44.8 V10 I Q1 10 IQ2 = 4.48 So I Q 2 = 44.8 I I Q1 Q1 Let I Q1 = 100 A, I Q 2 = 448 A Then I Q 2 R2 = 3.12 R2 = 6.96 k I Q1 R1 = 10 R1 = 100 k 11.30 a. 20 VGS 3 2 I1 = = 0.25 (VGS 3 2 ) 50 2 20 VGS 3 = 12.5 (VGS 3 4VGS 3 + 4 ) 2 12.5VGS 3 49VGS 3 + 30 = 0VGS 3 =49 ( 49 )2 4 (12.5 )( 30 )2 (12.5 ) VGS 3 = 3.16 V20 3.16 I1 = I Q = 0.337 mA 50 IQ = I D1 = 0.168 mA 2I1 = I D10.168 = 0.25 (VGS 1 2 ) VGS1 = 2.82 V VDS 4 = 2.82 ( 10 ) VDS 4 = 7.18 V 2VD1 = 10 ( 0.168 )( 24 ) = 5.97 VVDS1 = 5.97 ( 2.82 ) VDS 1 = 8.79 V(b)(c)www.elsolucionario.net 453. Max vCM VDS 1 = VDS 2 = VDS ( sat ) = VGS1 VTN 2.82 2 = 0.82 V Now VD1 = 10 ( 0.168 )( 24 ) = 5.97 VVS ( max ) = 5.97 VDS1 ( sat ) = 5.97 0.82 VS ( max ) = 5.15 VvCM ( max ) = VS ( max ) + VGS1 = 5.15 + 2.82 vCM ( max ) = 7.97 VvCM ( min ) = V + VDS 4 ( sat ) + VGS 1VDS 4 ( sat ) = VGS 4 VTN = 3.16 2 = 1.16 V Then vCM ( min ) = 10 + 1.16 + 2.82 vCM ( min ) = 6.02 V11.31 a. I D1 = I D 2 = 120 A = 100 ( VGS1 1.2 ) VGS 1 = VGS 2 = 2.30 V 2For v1 = v2 = 5.4 V and VDS1 = VDS 2 = 12 V 5.4 2.30 + 12 = 4.3 V = VD 10 4.3 RD = 47.5 k 0.12 I Q = I D1 + I D 2 I Q = I1 = 240 A RD =I1 = 240 = 200 (VGS 3 1.2 ) VGS 3 = 2.30 V 2R1 =20 2.3 R1 = 73.75 k 0.24b. r04 =1 IQI Q 1( 0.01)( 0.24 )= 416.7 k1 5.4 VDS = I Q 13 A r04 416.711.32 (a) I Q = 160 A k W 2 I D = n (VGS VTN ) 2L 80 2 80 = ( 4 )(VOS 0.5 ) 2 80 = 160 (Vo5 0.5 )280 + 0.5 = 1.207 V 160 52 = 37.5 K VDS = 2 ( 1.207 ) = 3.21 V RD = 0.08 (c) VDS ( sat ) = VGS VTN = 1.207 0.5 = 0.707 V VGS =Then VS = VO 2 VDS ( sat ) = 2 0.707 = +1.29 V And v1 = v2 = vcm = VGS + VS = 1.207 + 1.29 vcm = 2.50 V(b)www.elsolucionario.net 454. 11.33 vD = 5 ( 0.2 )( 8 ) = 3.4 V VGS =ID + VTN Kn0.2 + 0.8 = 1.694 V 0.25 VDS ( sat ) = VGS VTN = 1.694 0.8 = 0.894 V VS = VD VDS ( sat ) = 3.4 0.894 = 2.506 =vCM = VS + VGS = 2.506 + 1.694 vCM = 4.2 V(b) vD = I D RDI D = g m Vd 2gm = 2 Kn I D =2( 0.25)( 0.2 ) = 0.4472 mA/VI D = ( 0.4472 )( 0.05 ) 22.36 AvD = ( 22.36 106 )( 8 103 ) = 0.179 V vD 2 = 3.4 + vD vD 2 = 3.4 + 0.179 vD 2 = 3.58 V (c) vd = 50 mVI D = ( 0.4472 )( 0.025 ) 11.18 AvD = (11.18 106 )( 8 103 ) = 0.0894 V vD 2 = 3.4 0.0894 vD 2 = 3.31 V11.34 a. I D1 = I D 2 = 0.5 mA v01 v02 = V + I D1 RD1 V + I D 2 RD 2 v01 v02 = I D 2 RD 2 I D1 RD1 = I D ( RD 2 RD1 )i.RD1 RD 2 = 6 k, v01 v02 = 0ii.RD1 = 6 k, RD 2 = 5.9 kv01 v02 = ( 0.5 )( 5.9 6 ) v01 v02 = 0.05 Vb.www.elsolucionario.net 455. K n1 = 0.4 mA / V 2 , K n 2 = 0.44 mA / V 2 VGS1 = VGS 2 I Q = ( K n1 + K n 2 )(VGS VTN )21 = ( 0.4 + 0.44 )(VGS VTN ) (VGS VTN ) = 1.19 22I D1 = ( 0.4 )(1.19 ) = 0.476 mA I D 2 = ( 0.44 )(1.19 ) = 0.524 mA i. RD1 = RD 2 = 6 kv01 v02 = ( 0.524 0.476 )( 6 ) v01 v02 = 0.288 Vii. RD1 = 6 k,RD 2 = 5.9 kv01 v02 = ( 0.524 )( 5.9 ) ( 0.476 )( 6 )= 3.0916 2.856 v01 v02 = 0.236 V11.35 (a)From Equation (11.69) K Kn iD 2 1 = vd 1 n 2IQ IQ 2 2IQ 0.90 = 0.50 2 vd 0.1 2 0.1 vd 1 vd 2 ( 0.25 ) 2 ( 0.25 ) 2 +0.40 = ( 0.4472 ) vd 1 ( 0.2 ) vd 2 0.8945 = vd 1 ( 0.2 ) vdSquare both sides 2 2 0.80 = vd (1 [ 0.2] vd )( 0.2 ) ( vd2 ) 2 vd =22 vd + 0.80 = 01 1 4 ( 0.2 )( 0.80 ) 2 ( 0.2 )= 4V 2 or 1V 2Then vd = 2 V or 1 V But vdmax=IQ kn=0.25 = 1.58 0.1So vd = 1V, vd = 1Vb.From part (a), vd ,max = 1.58 V11.36www.elsolucionario.net 456. i d D1 I Q= dvd K Kn 1 n 2I 2IQ Q) vdvd =0Kn 2IQ= So linear 2 vd + ( Kn iD1 1 = + vd IQ 2 2 IQ1 K 2 Kn Kn 1 + vd ( max ) + vd ( max ) 1 n vd ( max ) 2 2IQ 2 IQ 2 2I n = 0.02 Then Kn 1 + v 2 2 I Q d ( max ) 1 1 K Kn Kn 0.98 + vd ( max ) = + vd ( max ) 1 n 2I 2IQ 2IQ 2 2 Q 0.49 + 0.98 2 vd ( max ) 1 0.15 2 0.15 0.15 vd ( max ) = + vd ( max ) 1 vd ( max ) 2 ( 0.2 ) 2 ( 0.2 ) 2 ( 0.2 ) 2 2 0.49 + 0.600 vd ( max ) = 0.50 + 0.6124 vd ( max ) 1 ( 0.6124 ) vd ( max ) 2 0.600 vd ( max ) = 0.010 + 0.6124 vd ( max ) 1 ( 0.6124 ) vd ( max )By trial and error vd ( max ) 0.429 V 11.37 (b) gm = 2 K p I D = 2( 0.05 )( 0.008696 )= 0.0417 mA/V Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 vD = ( 0.002085 )( 510 ) = 1.063 I = g mvD 2 vD 2 = 1.063 4.565 = 3.502 V vD1 = 1.063 4.565 = 5.628 V 9 = I S RS + VSG + 1 I S = 2I D 8 = 2 K P RS (VSG + VTP ) + VSG 28 = ( 2 )( 0.05 )( 390 )(VSG 0.8 ) + VSG 22 8 = 39 (VSG 1.6VSG + 0.64 ) + VSG 2 39VSG 61.4VSG + 16.96 = 0VSG =61.4 3769.96 4 ( 39 )(16.96 ) 2 ( 39 )= 1.217 V VS = 2.217 I S = 0.01739 mAI D1 = I D 2 8.696 AvD1 = vD 2 = ( 8.696 )( 0.510 ) 9 = 4.565 V(b)www.elsolucionario.net 457. g m = 2 K P I DQ = 2( 0.05 )( 0.008696 ) = 0.0417 mA/VVd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 vD = ( 0.002085 )( 510 ) = 1.063 V vD = I D RDI D = g m v1 , I D1 , vD1 vD1 = 4.565 1.063 = 5.628 V vD 2 = 4.565 + 1.063 = 3.502 V 11.38 (a) v1 = v2 = 0 I D = K n (VSG + VTP ) ID = 6 A26 + 0.4 = VSG 30 VSG = 0.847 V VS = +0.847 V vD = I D RD 3= ( 6 )( 0.36 ) 3 = 0.84 VVSD = VS vD = 0.847 ( 0.84 ) vSD = 1.69 V (b) (i) Ad = g m RD g m = 2 K n I D=2( 30 )( 6 ) = 26.83 A/VAd = ( 26.83)( 0.36 ) Ad = 9.66 Acm = 0 (ii) ( 26.83)( 0.36 ) g R Ad = m D = Ad = 4.83 2 2 ( 26.83)( 0.36 ) g m RD Acm = = = 0.0448 1 + 2 g m RO 1 + 2 ( 26.83)( 4 )11.39www.elsolucionario.net 458. For v1 = v2 = 0.30 V I D1 = I D 2 = 0.1 mA VSG =ID VTP KP0.1 +1 = 2 V 0.1 = vD 2 = ( 0.1)( 30 ) 10 = 7 V =vD1gm = 2 K p I D = 2( 0.1)( 0.1) = 0.2 mA/VV I D = g m d = ( 0.2 )( 0.1) = 0.02 mA 2 vD = ( I D ) RD = ( 0.02 )( 30 ) = 0.6 V vD 2 vD 2 = 7 + 0.6 vD 2 = 6.4 V vD1 = 7 0.6 vD1 = 7.6 V11.40 For v1 = v2 = 0 0 = VGS + 2 I D RS 10 10 = VGS + 2 K n RS (VGS VTN )2= VGS + 2 ( 0.15 )( 75 )(VGS 1)22 22.5VGS 44VGS + 12.5 = 0So VGS = 1.61 V and I D = ( 0.15 )(1.61 1) 55.9 A 2gm = 2 Kn I D = 2( 0.15 )( 0.0559 )g m = 0.1831 mA/V Use Half-circuits Differential gain R V vD1 = g m d RD + 2 2 R V vo 2 = g m d RD 2 2 vo = vD1 vD 2 = g mVd RD v Ad = o = g m RD Vd Now Common-Mode Gainwww.elsolucionario.net 459. Vi = Vgs + g mVgs ( 2 RS ) = Vcm Vcm Vgs = 1 + g m ( 2 RS ) vD1vD 2R g m RD + D Vcm 2 = 1 + g m ( 2 RS ) R gm RD D Vcm 2 = 1 + g m ( 2 RS )vO = vD1 vD 2 So vo = Acm = g m ( RD ) Vcm 1 + g m ( 2 RD ) g m ( RD ) vo = Vcm 1 + g m ( 2 RS )Then Ad = ( 0.1831)( 50 ) = 9.16 Acm = ( 0.1831)( 0.5 )1 + ( 0.1831)( 2 )( 75 )= 0.003216C M R R = 69.1 dB bB11.41 a. Ad = g m ( r02 r04 ) r02 =VA 2 150 = = 375 k I C 2 0.4r04 =VA 4 100 = = 250 k I C 4 0.4gm =IC 2 0.4 = = 15.38 mA/V VT 0.026Ad = (15.38 ) ( 375 250 ) Ad = 2307b. RL = r02 r04 = 375 250 RL = 150 k11.41 From 11.40 I D1 = I D 2 = 55.9 A g m = 0.183 mA/Vwww.elsolucionario.net 460. +V vD 2 = + g m 2 d RD 2 V V vO = vD1 vD 2 = g m1 d RD g m 2 d RD 2 2 V V g g vO = d RD ( g m 2 + g m1 ) = d RD g m m + g m m 2 2 2 2 Ad = g m RD = ( 0.183) ( 50 ) = 9.15 Ad : vD1 = g m1Vd RD 2g g g m + M RD vcm g m M RD vCM 2 2 = + 1 + g m ( 2 RS ) 1 + g m ( 2 RS )ACM : vO = vD1 vD 2 g m RD vO = vcm 1 + g m ( 2 RS )Acm = Acm =g m = ( 0.01) ( 0.183) = 0.00183 ( 0.00183) ( 50 )1 + ( 0.183)( 2 ) ( 75 )= 0.003216C M R R = 69.1 dB dB11.42 (a) v1 = v2 = 0 5 = 2 I D RS + VSG 5 = 2 K p RS (VSG + VTP ) + VSG 22 5 = 2 ( 0.5 )( 2 ) (VSG 1.6VSG + 0.64 ) + VSG 2 5 = 2VSG 2.2VSG + 1.28 2 2VSG 2.2VSG 3.72 = 0VSG =2.2 4.84 + 4 ( 2 )( 3.72 ) 2 ( 2)VSG = 2.02 V vS = 2.02 V,5 2.02 = 1.49 mA 2 = I D 2 = 0.745 mAIS = I D1vD1 = vD 2 = ( 0.745 (1) 5 ) vD1 = vD 2 = 4.26 V(b) 5 = I S RS + VSG 2 5 = ( I D1 + I D 2 ) RS + VSG 2 2 2 5 = K p (VSG1 + VTP ) + K p (VSG 2 + VTP ) RS + VSG 2 VSG1 = VSG 2 15 = ( 0.5 )( 2 ) (VSG 2 1.8 ) + (VSG 2 0.8 ) + VSG 2 2 2 5 = VSG 2 3.6VSG 2 + 3.24 + VSG 2 1.6VSG 2 + 0.64 + VSG 2 222 5 = 2VSG 2 4.2VSG 2 + 3.88 2 2VSG 2 4.2VSG 2 1.12 = 0VSG 2 =4.2 17.64 + 4 ( 2 ) (1.12 ) 2 ( 2)www.elsolucionario.net 461. VSG 2 = 2.339 V VSG1 = 1.339 V vS = 2.339 V I D1 I D1 vD1 vD1= 0.5 (1.339 0.8 ) = 0.1453 mA = ( 0.1453)(1) 5 = 4.855 V2I D2 I D2 vD 2 vD 2= 0.5 ( 2.339 0.8 ) = 1.184 mA = (1.184 ) (1) 5 = 3.816 V2(c) I = g mVd 2gm = 2 K p I DvS 2.02 V = 2( 0.5 )( 0.745 )g m = 1.22 mA/V I = (1.22 )( 0.1) = 0.122 mA vD = ( I ) RD = ( 0.122 )(1) = 0.122 V vD 2 vD1 vD1 = 4.26 + 0.122 vD 2 = 4.26 0.122 vD1 = 4.138 V vD 2 = 4.382 V11.43a.gf =IQ 4VT I Q = g f ( 4VT ) = ( 8 )( 4 )( 0.026 ) I Q = 0.832 mANeglecting base currents. 30 0.7 R1 = R1 = 35.2 k 0.832 V 100 r04 = r02 = A = = 240 k b. I CQ 0.416www.elsolucionario.net 462. gm =I CQ VT=0.416 = 16 mA / V 0.026Ad = g m ( r02 r04 ) = 16 ( 240 240 ) Ad = 1920 Rid = 2r , r =(180 )( 0.026 ) 0.416= 11.25 k Rid = 22.5 k R0 = r02 r04 R0 = 120 kc. Max. common-mode voltage when VCB = 0 for Q1 and Q2 . Therefore vcm ( max ) = V + VEB ( Q3 ) = 15 0.7 vcm ( max ) = 14.3 V Min. common-mode voltage when VCB = 0 for Q5 . Therefore vcm ( min ) = 0.7 + 0.7 + ( 15 ) = 13.6 V So 13.6 vcm 14.3 V 1 (1 + )( 2 R0 ) 2 V 100 R0 = A = = 120 k I Q 0.832 Ricm Ricm = (181)(120 ) Ricm = 21.7 M11.43 (a) gm = 2 Kn I D =2( 0.4 )(1)g m = 1.265 mA/V v 1 Ad = o = = 10 vd 0.1 Ad = g m RD 10 = (1.265 ) RD RD = 7.91 K(b) Quiescent v1 = v2 = 0 vD1 = vD 2 = 10 (1)( 7.91) = 2.09 V VGS =ID 1 + VTN = + 0.8 = 2.38 V Kn 0.4VDS ( sat ) = 2.38 0.8 = 1.58 So vcm = vD VDS ( sat ) + VGS = 2.09 1.58 + 2.38 vcm = 2.89 V11.44www.elsolucionario.net 463. g m RD 2 For vCM = 2.5 V IQ = 0.25 mA I D1 = I D 2 = 2 Ad =Let VD1 = VD 2 = 3 V , then RD = Then 100 =g m ( 28 )2 k W And g m = 2 n 2L10 3 RD = 28 k 0.25 g m = 7.14 mA / V ID 0.080 W 7.14 = 2 ( 0.25 ) 2 L W W = = 1274 (Extremely large transistors to meet the gain requirement.) L 1 L 2 Need ACM = 0.10 From Eq. (11.64(b)) g m RD ACM = 1 + 2 g m Ro( 7.14 )( 28) Ro = 140 k 1 + 2 ( 7.14 ) Ro For the basic 2-transistor current source 1 1 Ro = ro = = = 200 k I Q ( 0.01)( 0.5 ) So 0.10 =This current source is adequate to meet common-mode gain requirement. 11.45 Not in detail, Approximation looks good. a. V ( 5 ) 2 I S = GS 1 and I S = 2 I D = 2 K n (VGS 1 VTN ) RS 5 VGS 1 2 = 2 ( 0.050 )(VGS 1 1) 20 2 5 VGS 1 = 2 (VGS1 2VGS1 + 1) 2 2VGS1 3VGS 1 3 = 0VGS1 =3( 3)2+ 4 ( 2 )( 3)2 ( 2) VGS1 = 2.186 V5 2.186 I S = 0.141 mA 20 I I D1 = I D 2 = S I D1 = I D 2 = 0.0704 mA 2 v02 = 5 ( 0.0704 )( 25 ) v02 = 3.24 V IS =b. g m = 2 K n (VGS VTN ) = 2 ( 0.05 )( 2.186 1) g m = 0.119 mA/V 1 1 r0 = = = 710 k I DQ ( 0.02 )( 0.0704 )www.elsolucionario.net 464. Vgs1 = v1 VS , Vgs 2 = v2 VS v01 v V + g mVgs1 + 01 S = 0 RD r0 1 V 1 v01 + + g m ( v1 VS ) S = 0 r0 RD r0 v02 v VS + g mVgs 2 + 02 =0 RD r0(1) 1 V 1 v02 + + g m ( v2 VS ) S = 0 RD r0 r0 v V v VS V + g mVgs 2 = S g mVgs1 + 01 S + 02 r0 r0 RS(2)g m ( v1 VS ) +v01 v02 2VS V + + g m ( v2 VS ) = S r0 r0 r0 RSg m ( v1 + v2 ) +v01 v02 + = VS r0 r0 2 1 2 g m + + r0 RS (3)From (1) 1 VS g m + g m v1 r0 v01 = 1 1 + RD r0 Then 1 VS g m + g m v1 r0 v 2 1 g m ( v1 + v2 ) + + 02 = VS 2 g m + + (3) r0 r0 RS 1 1 r0 + RD r0 www.elsolucionario.net 465. 1 1 1 1 1 2 1 1 1 g m ( v1 + v2 ) r0 + + VS g m + g m v1 + v02 + = VS 2 g m + + r0 + r0 r0 RS RD r0 RD r0 RD r0 1 r r0 1 2 1 1 g m ( v1 + v2 ) 1 + 0 g m v1 + v02 + = VS 2 g m + + 1 + gm + RD RD r0 r0 RS RD r0 1 r r r r 1 2 1 2 1 g m v1 0 + v2 + v2 0 + v02 + = VS 2 g m + + + 2gm 0 + + 0 gm RD r0 RS RD RD RS RD r0 RD RD r0 1 r r r0 2 1 1 1 g m v1 0 + v2 + v2 0 + v02 + = VS 2 g m + + (1 + g m r0 ) (4) 1 + + RD r0 RS RD RD RD RD r0 1 1 1 Then substituting into (2), v02 + + g m v2 = VS g m + r0 RD r0 710 1 710 1 + v2 + v2 (4) Substitute numbers: ( 0.119 ) v1 + v02 25 + 710 25 25 1 1 710 2 = VS 0.119 + + 1 + + 1 + ( 0.119 )( 710 ) 710 20 25 25 ( 0.119 ) [ 28.4v1 + 29.4v2 ] + ( 0.0414 ) v02 = VS {0.1204 + 1.470 + 6.8392} = VS ( 8.4296 ) or VS = 0.4010v1 + 0.4150v2 + 0.00491v02 1 1 1 (2) Then v02 + + ( 0.119 ) v2 = VS 0.119 + 710 25 710 v02 ( 0.0414 ) + v2 ( 0.119 ) = ( 0.1204 ) [ 0.401v1 + 0.4150v2 + 0.00491v02 ] v02 ( 0.0408 ) = ( 0.04828 ) v1 ( 0.0690 ) v2 v02 = (1.183) v1 (1.691) v2vd 2 vd v2 = vcm 2 v v So v02 = (1.183) vcm + d (1.691) vcm d 2 2 Or v02 = 1.437vd 0.508vcm Ad = 1.437, Acm = 0.508 Nowv1 = vcm + 1.437 C M R RdB = 20 log10 C M R RdB = 9.03 dB 0.508 11.46KVL:www.elsolucionario.net 466. v1 = Vgs1 Vgs 2 + v2 So v1 v2 = Vgs1 Vgs 2KCL: g mVgs1 + g mVgs 2 = 0 Vgs1 = Vgs 2 1 1 So Vgs1 = ( v1 v2 ) , Vgs 2 = ( v1 v2 ) 2 2 Now v02 v02 v01 + = g mVgs 2 RD RL 1 1 v01 = v02 + RD RL RL v01 v01 v02 + = g mVgs1 RD RL 1 1 v02 = v01 + RD RL RL R From (1): v01 = v02 1 + L + g m RLVgs 2 RD (1)(2)Substitute into (2): 1 v02 R 1 1 1 g mVgs1 = v02 1 + L + + + g m RL Vgs 2 RD RD RL RD RL RL 1 R 1 R 1 g m ( v1 v2 ) + g m 1 + L ( v1 v2 ) = v02 + L + 2 RD 2 RD RD RD 1 g m RL v02 v02 RL 1 RL gm = 2 ( v1 v2 ) = 2 + Ad 2 = 2 RD RD RD v1 v2 RL 2+ RD 1 g m RL v01 From symmetry Ad 1 = = 2 v1 v2 RL 2+ RD Then Av =v02 v01 g m RL = v1 v2 RL 2+ RD 11.47www.elsolucionario.net 467. v1 v2 = Vgs1 Vgs 2 and g mVgs1 + g mVgs 2 = 0 Vgs1 = Vgs 2 Then v1 v2 = 2Vgs 2 Or Vgs 2 = 1 ( v1 v2 ) 2v0 = g mVgs 2 ( RD RL ) = Or Ad =gm ( RD RL ) ( v1 v2 ) 2gm ( RD RL ) 211.48 From Equation (11.64(a)), Ad = We need Ad = Then 10 =Kn IQ 2 RD2 = 10 0.2K n ( 0.5 ) RD or K n RD = 20 2 If we set RD = 20 k , then K n = 1 mA / V 2For this case VD = 10 ( 0.25 )( 20 ) = 5 V 0.25 + 1 = 1.5 V 1 VDS ( sat ) = VGS VTN = 1.5 1 = 0.5 V VGS =Then vcm ( max ) = VD VDS ( sat ) + VGS = 5 0.5 + 1.5 Or vcm ( max ) = 6 V11.49 Vd 1 = g mVgs1 RD = g m RD (V1 Vs ) Vd 2 = g mVgs 2 RD = g m RD (V2 Vs ) Now Vo = Vd 2 Vd 1 = g m RD (V2 Vs ) ( g m RD (V1 Vs ) ) Vo = g m RD (V1 V2 ) Define V1 V2 Vd V Then Ad = o = g m RD and Acm = 0 Vd11.49 Ad = g m ( r02 r04 ) g m = 2 kn I DQ=2( 0.12 )( 0.075 )= 0.1897 mA/V 1 1 r02 = = = 889 k n I DQ ( 0.015 )( 0.075 ) r04 =1 p I DQ=1 = 667 k 0.02 )( 0.075 ) (Ad = ( 0.1897 ) ( 889 667 ) Ad = 72.311.50 (a)www.elsolucionario.net 468. K W K n1 = K n 2 = n 2 L VGS1 = VGS 2 = 0.080 2 = (10 ) = 0.40 mA / V 2 ID 0.1 + VTN = + 1 = 1.5 V Kn 0.4VDS1 ( sat ) = 1.5 1 = 0.5 VFor vCM = +3 V VD1 = VD 2 = vCM VGS 1 + VDS 1 ( sat )= 3 1.5 + 0.5 VD1 = VD 2 = 2 V RD =10 2 RD = 80 k 0.1(b) 1 g m RD and g m = 2 ( 0.4 )( 0.1) = 0.4 mA / V 2 1 Then Ad = ( 0.4 )( 80 ) = 16 2 16 C M R RdB = 45 C M R R = 177.8 = Acm Ad =So Acm = 0.090 Acm =g m RD 1 + 2 g m Ro0.090 =( 0.4 )(80 ) 1 + 2 ( 0.4 ) Ro Ro = 443 k If we assume = 0.01 V 1 for the current source transistor, then 1 1 ro = = = 500 k I Q ( 0.01)( 0.2 ) So the CMRR specification can be met by a 2-transistor current source. W W Let = = 1 L 3 L 4 IQ 0.2 0.080 2 Then K n 3 = K n 4 = + VTN = + 1 = 3.24 V (1) = 0.040 mA / V and VGS 3 = 2 0.04 K n3 For vCM = 3 V , VD 3 = 3 VGS1 = 3 1.5 = 4.5 V VDS 3 ( min ) = 4.5 ( 10 ) = 5.5 V > VDS 3 ( sat )So design is OK. W On reference side: For 1, VGS ( max ) = 3.24 V L 20 VGS 3 = 20 3.24 = 16.76 VThen16.67 = 5.17 We need six transistors in series. 3.24www.elsolucionario.net 469. 20 3.24 = 2.793 V 6 2 K W = n (VGS VTN ) 2 L VGS = I REF2 0.080 W W 0.2 = ( 2.793 1) = 1.56 for each of the 6 transistors. 2 L L11.51Ad =1 g m RD 2gm = 2 Kn I D = 2( 0.25 )( 0.25) = 0.50 mA / V1 ( 0.50 )( 3) = 0.75 2 From Problem 11.26 Ad =www.elsolucionario.net 470. V1 = VA =5 (1 + ) 2+, V2 = VB = 2.5 V and V1 V2 = 1.25Then Vo 2 = Ad (V1 V2 ) = ( 0.75 )(1.25 ) = 0.9375So for 0.01 0.01 9.375 Vo 2 9.375 mV11.52 From previous results v v Ad 1 = o 2 o1 = g m1 R1 = 2 K n1 I Q1 R1 = 20 v1 v2 and Ad 2 = SetI Q1 R1 2vo3 1 1 2 K n3 I Q 2 R2 = 30 = g m 3 R2 = 2 vo 2 vo1 2 I Q 2 R2= 5 V and2= 2.5 VLet I Q1 = I Q 2 = 0.1 mA Then R1 = 100 k , R2 = 50 k 2 0.06 W 20 W W Then 2 ( 0.1) = = = 6.67 2 L 1 100 L 1 L 2 2 2 ( 30 ) 0.060 W W W and 2 = = 240 ( 0.1) = 2 L 3 L 3 L 4 50 11.53 a.iD1 v = I DSS 1 GS 1 VP v iD 2 = I DSS 1 GS 2 VP iD1 iD 222 v v = I DSS 1 GS 1 I DSS 1 GS 2 VP VP I DSS=( vGS 2 vGS1 )VPI DSS=VP vd =I DSS( VP ) vdiD1 + iD 2 = I Q iD 2 = I Q iD1(iD1 I Q iD1)2=I DSS( VP )22 vdiD1 2 iD1 ( I Q iD1 ) + ( I Q iD1 ) = Then iD1 ( I Q iD1 ) =I DSS( VP )22 vd I 1 2 I Q DSS 2 vd 2 ( VP ) Square both sideswww.elsolucionario.net 471. 2 I 1 2 i iD1 I Q + I Q DSS 2 vd = 0 4 ( VP ) 2 D1 I 1 2 I Q I 4 I Q DSS 2 vd 4 ( VP ) 22 QiD1 =22 2 2 1 2 2 2 I Q I DSS vd I DSS vd iD1 = IQ IQ + 2 2 2 2 ( VP ) ( VP ) Use + signIQiD11 2 I Q I DSS 2 = + vd 2 2 ( VP )2iD1 =IQIQ 2+ I 2 DSS 2 vd ( V ) P 22222222 I DSS I DSS 1 IQ vd I 2 ( VP ) IQ Q vd V P 2 I DSS I DSS vd I IQ Q vd V P 2 I DSS I DSS vd I IQ Q vd V POr iD1 1 1 = + I Q 2 2VP We had iD 2 = I Q iD1 Then iD 2 1 1 = I Q 2 2VP b. If iD1 = I Q , then 1 1 1= + 2 2VP 2 I DSS I DSS vd I IQ Q2 I DSS I DSS VP = vd I IQ Q Square both sides2 vd V P2 vd V P22www.elsolucionario.net 472. VP2 I DSS I Q2 vd = 2I I = v DSS DSS I IQ Q 2 d22 2 vd V P 2 1 2 2 2 I DSS 2 vd + VP ( vd ) V IQ P 2I 2 I DSS DSS I IQ Q2 I 4 DSS I Q 2I 2 DSS IQ 22=0 2 1 2 (VP ) V P2 1 VP 22 IQ 2 vd = (VP ) I DSS 1/ 2 IQ Or vd = VP I DSS c. For vd small, IQ 1 IQ 2 I DSS + vd iD1 IQ 2 2 ( VP ) gf =diD1 d vdvd 0=2 I DSS 1 IQ IQ 2 ( VP ) 1 I Q I DSS Or g f ( max ) = 2 VP 11.53 Ad = g m ( ro 2 Ro ) Want Ad = 400 From Example 11.15, ro 2 = 1 M Assuming that g m = 0.283 mA / V for the PMOS from Example 11.15, then Ro = 285 M . k W So 400 = g m (1000 285000 ) g m = 0.4014 mA / V = 2 n I DQ 2 L 1 0.080 W W W 0.04028 = ( 0.1) = = 10.1 2 L 1 L 1 L 211.54 a. I Q = I D1 + I D 2 I Q = 1 mA v0 = 7 = 10 ( 0.5 ) RD RD = 6 kb. 1 I Q I DSS g f ( max ) = 2 VP 1 (1)( 2 ) g f ( max ) = g f ( max ) = 0.25 mA/V 2 4 c. g R Ad = m D = g f ( max ) RD 2 Ad = ( 0.25 )( 6 ) Ad = 1.5www.elsolucionario.net 473. 11.55 a. IS =VGS ( 5 ) RS V = ( 2 ) I DSS 1 GS VP V 5 VGS = ( 2 )( 0.8 )( 20 ) 1 GS ( 2 ) 1 2 5 VGS = ( 2 )16 1 + VGS + VGS 4 2 8VGS + 33VGS + 27 = 0 VGS =2233 1089 4 ( 8 )( 27 ) 2 (8)= 1.125 V IS =5 ( 1.125 )20 = 0.306 mAI D1 = I D 2 = 0.153 mA vo 2 = 1.17 V (b)11.56 Equivalent circuit and analysis is identical to that in problem 11.36. 1 g m RL Ad 2 = 2 RL 2+ RD 1 g m RL Ad 1 = 2 RL 2+ RD Av =v02 v01 g m RL = vd RL 2+ RD 11.57 (a) Ad = g m ( ro 2 ro 4 ) 0.1 = 3.846 mA/V 0.026 120 ro 2 = = 1200 K 0.1 80 ro 4 = = 800 K 0.1 Ad = ( 3.846 ) (1200 800 ) gm =Ad = 1846(b)www.elsolucionario.net 474. For Ad = 923 = ( 3.846 ) (1200 800 RL ) 240 = 480 RL =480 RL RL = 480 K 480 + RL11.58 (a) 2 I Q = 250 A I REF = I Q 1 + 2 = 250 1 + = 252.8 A 180 5 ( 0.7 ) ( 5 ) R1 = R1 = 36.8 K 0.2528 (b) 0.125 = 4.808 mA/V Ad = g m ( ro 2 ro 4 ) gm = 0.026 150 = 1200 K ro 2 = 0.125 100 = 800 K Ad = ( 4.808 ) (1200 800 ) ro 4 = 0.125 Ad = 2308(c) Rid = 2r =2 (180 )( 0.026 ) 0.125 Rid = 74.9 KRo = ro 2 ro 4 = 1200 800 = 480 K = Ro(d) vcm ( max ) = 5 0.7 = 4.3 V vcm ( min ) = 0.7 + 0.7 5 = 3.6 V 11.59 a. IQ 1 I 0 = I B3 + I B 4 2 2 I Q 0.2 I0 = = I0 = 2 A 100 b. V 100 = 1000 k r02 = r04 = A = I CQ 0.1 gm =I CQ VT=0.1 = 3.846 mA/V 0.026Ad = g m ( r02 r04 ) = ( 3.846 ) (1000 1000 ) Ad = 1923c. Ad = g m r02 r04 RL()Ad = ( 3.846 ) (1000 1000 250 ) Ad = 64111.60 a.www.elsolucionario.net 475. Ad = g m ( r02 r04 RL ) gm =I CQ VT=IQ 2VTV 125 r02 = A 2 = I CQ I CQ r04 =VA 4 80 = I CQ I CQIf I Q = 2 mA, then g m = 38.46 mA/V r02 = 125 k, r04 = 80 k So Ad = 38.46 125 80 200 Or Ad = 1508 For each gain of 1000. lower the current level For I Q = 0.60 mA, I CQ = 0.30 mA 0.3 gm = = 11.54 mA/V 0.026 125 r02 = = 417 k 0.3 80 r04 = = 267 k 0.3 Ad = 11.54 417 267 200 = 1036 So I Q = 0.60 mA is adequateb. For V + = 10 V, VBE = VEB = 0.6 V For VCB = 0, vcm ( max ) = V + 2VEB = 10 2 ( 0.6 ) Or vcm ( max ) = 8.8 V 11.61 a.From symmetry.VGS 3 = VGS 4 = VDS 3 = VDS 4 =0.1 +1 0.1Or VDS 3 = VDS 4 = 2 V 0.1 +1 = 2 V 0.1 = VSD 2 = VSG1 (VDS 3 10 ) = 2 ( 2 10 )VSG1 = VSG 2 = VSD1Or VSD1 = VSD 2 = 10 Vb. r0 n = r0 p =1n I DQ 1= =1 1 M 0.01)( 0.1) ( 1 0.667 M( 0.015 )( 0.1) g m = 2 K p (VSG + VTP ) = 2 ( 0.1)( 2 1) = 0.2 mA / V Ad = g m ( ron rop ) = ( 0.2 ) (1000 667 ) Ad P I DQ= 80(c)www.elsolucionario.net 476. IQI D 2 = I D1 =2= 0.1 mA1 1 = = 1000 k n I D 4 ( 0.01)( 0.1)ro 4 =1ro 2 =P I D 2=1( 0.015)( 0.1)= 667 k Ro = ro 2 ro 4 = 667 1000 = 400 k 11.62 Ad = g m ( ro 4 ro 2 ) 0.08 gm = 2 ( 2.5 )( 0.05 ) 2 = 0.1414 mA/V 1 ro 4 = = 1000 K ( 0.02 )( 0.05 ) ro 2 =1 = 1333 K ( 0.015)( 0.05 )Ad = ( 0.1414 ) (1000 1333) Ad = 80.811.63 R04 = r04 1 + g m 4 ( R r 4 ) 80 = 800 K 0.1 0.1 gm4 = = 3.846 0.026 (100 )( 0.026 ) r 4 = 0.1 = 26 K r04 =R r 4 = 1 26 = 0.963 K Assume = 100 r 3 =(100 )( 0.026 ) 0.1= 26 k0.1 = 3.846 mA/V 0.026 R04 = 800 1 + ( 3.846 )( 0.963) 3.763 M g m3 = R0 = 3.763M Then Av = g m ( r02 R0 ) 120 = 1200 k 0.1 0.1 gm = = 3.846 mA/V 0.026 Av = ( 3.846 ) 1200 3763 Av = 3499 r02 =b. Forwww.elsolucionario.net 477. R = 0, r04 = Av = g m ( r0280 = 800 k 0.1 r04 )= ( 3.846 ) 1200 800 Av = 1846 (c)For part (a), Ro = ( 3.763 1.2 ) = 0.910 M For part (b), Ro = (1.2 0.8 ) = 0.48 M 11.64 I B5 =IE5 I +I I +I = B3 B4 = C 3 C 4 1+ 1+ (1 + )Now I C 3 + I C 4 I Q IQ So I B 5 (1 + ) I B6 =I Q1 IE6 = 1 + (1 + )For balance, we want I B 6 = I B 5 So that I Q1 = I Q11.65 Resistance looking into drain of M4.Vsg 4 I X R1 I X g m 4Vsg 4 =VX Vsg 4 r04 R V I X 1 + g m 4 R1 + 1 = X r04 r04 R Or R0 = r04 1 + g m 4 R1 + 1 r04 a.www.elsolucionario.net 478. Ad = g m 2 ( ro 2 Ro ) g m 2 = 2 K n I DQ = 2( 0.080 )( 0.1)= 0.179 mA / V 1 1 ro 2 = = = 667 k n I DQ ( 0.015 )( 0.1) g m 4 = 2 K P I DQ = 2ro 4 =1 p I DQ( 0.080 )( 0.1)= 0.179 mA / V 1 = = 500 k 0.02 )( 0.1) (1 R0 = 500 1 + ( 0.179 )(1) + = 590.5 k 500 Ad = ( 0.179 ) 667 590.5 Ad = 56.06 b. When R1 = 0, R0 = r04 = 500 k Ad = ( 0.179 ) 667 500 Ad = 51.15 (c)For part (a), Ro = ro 2 Ro = 667 590.5 Ro = 313 k For part (b), Ro = ro 2 ro 4 = 667 500 Ro = 286 k 11.66 Let = 100, VA = 100 Vwww.elsolucionario.net 479. ro 2 =VA 100 = = 1000 k I CQ 0.1 Ro 4 = ro 4 [1 + g m RE ] where RE = r RE Now r =(100 )( 0.026 )= 26 k 0.1 0.1 gm = = 3.846 mA / V 0.026 RE = 26 1 = 0.963 k Then Ro 4 = 1000 1 + ( 3.846 )( 0.963) = 4704 k Ad = g m ( ro 2 Ro 4 ) = 3.846 (1000 4704 ) Ad = 317211.67 (a) For Q2, Q4Vx V 4 V + g m 2V 2 + g m 4V 4 + x ro 2 ro 4(1)Ix =(2)g m 2V 2 +(3)V 4 = V 2From (2)Vx V 4 V = 4 ro 2 r 4 r 2 1 Vx 1 = V 4 + + gm2 ro 2 r 4 r 2 ro 2 www.elsolucionario.net 480. Now 120 = ( 0.5 ) = 0.496 mA 121 120 IQ 1 I C 2 = 0.0041 mA = = ( 0.5 ) (121)2 2 1+ 1+ IC 4 = 1+ IC 2 IQ 2So r 2 =(120 )( 0.026 )= 761 k 0.0041 0.0041 gm2 = = 0.158 mA/V 0.026 100 ro 2 = 24.4 M 0.0041 (120 ) ( 0.026 ) r 4 = = 6.29 k 0.496 0.496 gm4 = = 19.08 mA / V 0.026 100 ro 4 = = 202 k 0.496 Now Vx Vx 1 1 = V 4 + + 0.158 which yields V 4 = ro 2 ( 0.318) ro 2 6.29 761 24400 From (1), V V 1 I x = x + x + V 4 g m 4 g m 2 ro 2 ro 4 ro 2 1 19.08 0.158 Ix 1 V 1 24400 which yields Ro 2 = x = 135 k = + + Vx 24400 202 Ix ( 0.318)( 24400 ) 80 Now ro 6 = = 160 k 0.5 Then Ro = Ro 2 ro 6 = 135 160 Ro = 73.2 k (b) c c Ad = g m Ro where g m =i vd / 2www.elsolucionario.net 481. vd 2i = g m1V 1 + g m 3V 3 and V 1 + V 3 = V Also 1 + g m1V 1 r 3 = V 3 r 1 1+ So V 1 r 3 = V 3 r 1 121 Or V 1 ( 6.29 ) = V 3 V 1 761 v v Then 2V 1 = d V 1 = d 2 4v v So i = ( g m1 + g m 3 ) V 1 = ( 0.158 + 19.08 ) d = 9.62 d 4 2 i c = 9.62 Ad = ( 9.62 )( 73.2 ) Ad = 704 So g m = vd / 2 Now Rid = 2 Ri where Ri = r 1 + (1 + ) r 3 Ri = 761 + (121)( 6.29 ) = 1522 k Then Rid = 3.044 M 11.69 (a) Ad = 100 = g m ( ro 2 ro 4 ) Let I Q = 0.5 mA 1 1 ro 2 = = = 200 k n I D ( 0.02 )( 0.25 ) ro 4 =1 1 = = 160 k P I D ( 0.025 )( 0.25 )Then 100 = g m ( 200 160 ) g m = 1.125 mA / V K W gm = 2 n 2 L ID 0.080 W W 1.125 = 2 ( 0.25 ) = 31.6 2 L L n W W Now somewhat arbitrary. Let = 31.6 L P L P11.70www.elsolucionario.net 482. Ad = g m ( ro 2 ro 4 )P = ( I Q + I REF ) (V + V ) Let I Q = I REF Then 0.5 = 2 I Q ( 3 ( 3) ) I Q = I REF = 0.0417 mA ro 2 =1 1 = = 3205 k n I D ( 0.015 )( 0.0208 )ro 4 =1 1 = = 2404 k P I D ( 0.02 )( 0.0208 )Then Ad = 80 = g m ( 3205 2404 ) g m = 0.0582 mA/V k W gm = 2 n I D 2 L n 0.080 W W 0.0582 = 2 ( 0.0208 ) = 1.02 2 L n L n11.71 Ad = g m ( ro 2 Ro ) g m ro 2 1 ro 2 = n I D =1 = 666.7 K ( 0.015)( 0.1)Ad = 400 = g m ( 666.7 ) g m = 0.60 mA/V k W = 2 n 2 L ID 0.08 W 0.60 = 2 ( 0.1) 2 L W 0.090 = 0.004 L W W = = 22.5 L 1 L 211.72www.elsolucionario.net 483. Ad = g m ( Ro 4 Ro 6 ) where Ro 4 = ro 4 + ro 2 [1 + g m 4 ro 4 ]Ro 6 = ro 6 + ro8 [1 + g m 6 ro 6 ] We have 1 = 1667 k ro 2 = ro 4 = ( 0.015 )( 0.040 ) ro 6 = ro8 =1 = 1250 k 0.02 )( 0.040 ) ( 0.060 gm4 = 2 (15 )( 0.040 ) = 0.268 mA/V 2 0.025 gm6 = 2 (10 )( 0.040 ) = 0.141 mA/V 2 Then Ro 4 = 1667 + 1667 1 + ( 0.268 )(1667 ) 748 M Ro 6 = 1250 + 1250 1 + ( 0.141)(1250 ) 222.8 M (a) Ro = Ro 4 Ro 6 = 748 222.8 Ro = 172 M (b) Ad = g m 4 ( Ro 4 Ro 6 ) = ( 0.268 )(172000 ) Ad = 46096 11.73 Ad = g m ( ro 2 ro 4 ) ro 2 = ro 4 = =1 ID 1( 0.02 )( 0.1)gm = 2 Kn I D = 2= 500 K( 0.5)( 0.1)= 0.4472 mA/V Ad = ( 0.4472 ) ( 500 500 ) Ad = 112 Ro = ro 2 ro 4 = 500 500 Ro = 250 K11.74 (a) I DP = K p (VSG + VTP )20.4 + 1 = VSG 3 = 1.894 V 0.5 I DN = K n (VGS VTN )20.4 + 1 = VGS 1 = 1.894 V 0.5 VDS1 ( sat ) = VGS1 VTN = 1.894 1 = 0.894 V V + = VSG 3 + VDS1 ( sat ) VGS 1 + vCM V + = 1.894 + 0.894 1.894 + 4 V + = 4.89 V = V (b)www.elsolucionario.net 484. Ad = g m ( ro 2 ro 4 ) ro 2 = ro 4 =1 IDgm = 2 Kn I D1 166.7 K( 0.015 )( 0.4 ) = 2 ( 0.5 )( 0.4 ) = 0.8944 mA/VAd = ( 0.8944 ) (166.7 166.7 ) Ad = 74.511.75 (a) For vcm = +2V V + = 2.7 V If I Q is a 2-transistor current source, V = vcm 0.7 0.7 V = 3.4 V V + = V = 3.4 V (b) 100 = 1000 K Ad = g m ( ro 2 ro 4 ) ro 2 = 0.1 60 = 600 K ro 4 = 0.1 0.1 = 3.846 mA/V gm = 0.026 Ad = ( 3.846 ) (1000 600 ) Ad = 144211.76 (a) (b)V + = V = 3.4 V75 = 1250 K 0.06 40 = 666.7 K ro 4 = 0.06 0.06 = 2.308 mA/V gm = 0.026 Ad = ( 2.308 ) (1250 666.7 ) ro 2 =Ad = 100411.77 g m1 = 2 K n I Bias1 = 2 gm2 = r 2 =I CQ VT VT I CQ= =( 0.2 )( 0.25 ) = 0.447 mA/V0.75 = 28.85 mA/V 0.026(120 )( 0.026 ) 0.75= 4.16 kwww.elsolucionario.net 485. i0 = g m1Vgs1 + g m 2V 2 V 2 = g m1Vgs1r 2 and vi = Vgs1 + V 2 i0 = Vgs1 ( g m1 + g m 2 g m1r 2 )vi = Vgs1 + g m1Vgs1r 2 and Vgs1 = i0 = vi C gm =g m1 (1 + )vi 1 + g m1r 21 + g m1r 2( 0.447 )(121) i0 g m1 (1 + ) = = vi 1 + g m1r 2 1 + ( 0.447 )( 4.16 )C g m = 18.9 mA/V11.78 r0 ( M 2 ) = r0 ( Q2 ) =1n I DQ=1( 0.01)( 0.2 )= 500 kVA 80 = = 400 k I CQ 0.2g m ( M 2 ) = 2 K n I DQ = 2( 0.2 )( 0.2 )= 0.4 mA/V Ad = g m ( M 2 ) r0 ( M 2 ) r0 ( Q2 ) = 0.4 500 400 Ad = 88.9 If the IQ current source is ideal, Acm = 0 and C M RRdB = 11.79 a.b.Assume RL is capacitively coupled. Thenwww.elsolucionario.net 486. I CQ + I DQ = I Q I DQ =VBE 0.7 = = 0.0875 mA R1 8I CQ = 0.9 0.0875 = 0.8125 mA g m1 = 2 K P I DQ = 2 (1)( 0.0875 ) g m1 = 0.592 mA/V gm2 = r 2 =I CQ VT VT I CQ=0.8125 g m 2 = 31.25 mA/V 0.026=(100 )( 0.026 ) 0.8125 r 2 = 3.2 kc. V0 = ( g m1Vsg g m 2V 2 ) RL Vi + Vsg = V0 Vsg = V0 Vi V 2 = ( g m1Vsg ) ( R1 r 2 )V0 = g m1Vsg + g m 2 g m1Vsg ( R1 r 2 ) RL V0 = (V0 Vi ) g m1 + g m 2 g m1 ( R1 r 2 ) RL g m1 + g m 2 g m1 ( R1 r 2 ) RL V0 = Vi 1 + g m1 + g m 2 g m1 ( R1 r 2 ) RL We find Av =g m1 + g m 2 g m1 ( R1 r 2 ) = 0.592 + ( 31.25 )( 0.592 ) ( 8 3.2 ) = 42.88 Then Av =( 42.88 )( RL ) 1 + ( 42.88 )( RL )11.80 a. I DQ I CQAssume RL is capacitively coupled. 0.7 = = 0.0875 mA 8 = 1.2 0.0875 = 1.11 mAg m1 = 2 K p I DQ = 2 (1)( 0.0875 ) g m1 = 0.592 mA/V gm2 = r 2 =I CQ VT VT I CQ= =1.11 g m 2 = 42.7 mA/V 0.026(100 )( 0.026 ) 1.11 r 2 = 2.34 kb.www.elsolucionario.net 487. Vsg = VX I X = g m 2V 2 + g m1Vsg(gVm1 sg)(R1r 2 ) = V 2I X = VX g m1 + g m 2 g m1 ( R1 r 2 ) VX 1 R0 = = IX g m1 + g m 2 g m1 ( R1 r 2 ) =10.592 + ( 0.592 )( 42.7 ) ( 8 2.34 ) R0 = 21.6 11.81 (a)(1)g m 2V +(2)g m 2V +Then V =Vo ( V ) ro 2 Vo ( V ) ro 2=0 = g m1Vi + 1 1 V V + or 0 = g m1Vi V + ro1 r ro1 r g m1Vi 1 1 + ro1 r From (1)www.elsolucionario.net 488. Vo 1 =0 g m 2 + V + ro 2 ro 2 1 gm2 + ro 2 1 Vo = ro 2 g m 2 + V = ro 2 g m1Vi ro 2 1 1 + ro1 r 1 g m1ro 2 g m 2 + ro 2 V Av = o = Vi 1 1 + ro1 r Now( 0.25)( 0.025 ) = 0.158 mA / Vg m1 = 2 K n I Q = 2 gm2 = ro1 =VT 1 IQ= =0.025 = 0.9615 mA / V 0.026 1( 0.02 )( 0.025)= 2000 k VA 50 = = 2000 k I Q 0.025ro 2 = r =IQ VT IQ=(100 )( 0.026 ) 0.025= 104 k Then 1 ( 0.158 )( 2000 ) 0.9615 + 2000 Av = 30039 Av = 1 1 + 2000 104 To find Ro; set Vi = 0 g m1Vi = 0www.elsolucionario.net 489. I x = g m 2V +Vx ( V )V = I x ( ro1 r )ro 2Then V 1 I x = g m 2 + ( I x ) ( ro1 r ) + x ro 2 ro 2 Combining terms, Ro = Vx 1 = ro 2 1 + ( ro1 r ) g m 2 + Ix ro 2 1 = 2000 1 + ( 2000 104 ) 0.9615 + Ro = 192.2 M 2000 (b)Vo ( Vgs 3 )(1)g m 3Vgs 3 +(2)g m 3Vgs 3 +(3)Vgs 3 ( V 2 ) ( V 2 ) V 2 + g m 2V 2 + = g m1Vi + r 2 ro 2 ro1From (2), V 2 =ro3 Vo ( Vgs 3 ) ro3=0 = g m 2V 2 +Vgs 3 ( V 2 ) ro 2 1 Vgs 3 or 0 = V 2 g m 2 + ro 2 ro 2 Vgs 3 1 ro 2 g m 2 + ro 2 Then (3)Vgs 3 1 1 1 + gm2 + + = g m1Vi + V 2 ro 2 ro1 ro 2 r 2orwww.elsolucionario.net 490. Vgs 3 1 1 1 + = g m1Vi + + gm2 + ro 2 ro1 ro 2 1 r ro 2 g m 2 + 2 ro 2 Vgs 3 Vgs 3 1 1 1 104 + 0.9615 + 2000 + 2000 = 0.9615Vi + 2000 1 2000 0.9615 + 2000 Then Vgs 3 = 1.83 105 Vi Vgs 3 V 1 1 5 From (1), g m 3 + Vgs 3 = o or Vo = 2000 0.158 + (1.83 10 ) Vi ro 3 ro3 2000 V Av = o = 5.80 107 ViTo find RoVx ( Vgs 3 )(1)I x = g m 3Vgs 3 +(2)g m 3Vgs 3 +(3)V 2 = I x ( ro1 r 2 )ro3Vx ( Vgs 3 ) ro 3= g m 2V 2 +Vgs 3 ( V 2 ) ro 2 1 V From (1) I x = Vgs 3 g m 3 + + x ro 3 ro3 1 Vx I x = Vgs 3 0.158 + + 2000 2000 V Ix x 2000 So Vgs 3 = 0.1585www.elsolucionario.net 491. From (2), 1 1 V 1 + + x = V 2 g m 2 + Vgs 3 g m 3 + ro 3 ro 2 ro 3 ro 2 1 1 Vx 1 + Vgs 3 0.158 + + 2000 = V 2 0.9615 + 2000 2000 2000 Vx I Vx / 2000 Then x ( 0.159 ) + 2000 = I x ( 2000 104 ) ( 0.962 ) 0.1585 V We find Ro = x = 6.09 1010 Ix11.82 Assume emitter of Q1 is capacitively coupled to signal ground. 80 I CQ = 0.2 = 0.1975 mA 81 0.2 I DQ = = 0.00247 mA 81 (80 )( 0.026 ) r = = 10.5 k 0.1975 0.1975 g m ( Q1 ) = = 7.60 mA / V 0.026 gm ( M1 ) = 2 K n I D = 2g m ( M 1 ) = 0.0445 mA / V( 0.2 )( 0.00247 )Vi = Vgs + V and V = g m ( M 1 ) Vgs r or Vgs =V g m ( M 1 ) r 1 Then Vi = V 1 + g (M )r m 1 Vi or V = 1 1 + g (M ) r m 1 g m ( Q1 ) RC V Vo = g m ( Q1 ) V RC Av = o = Vi 1 1 + g (M )r m 1 ( 7.60 )( 20 ) Then Av = Av = 48.4 1 1 + ( 0.0445 )(10.5 ) www.elsolucionario.net 492. 11.83 Using the results from Chapter 4 for the emitter-follower: r 9 + r07 R011 r 8 + 1+ R0 = R4 1+ VT (100 )( 0.026 ) r 8 = = = 2.6 k 1 IC8 IC 8IC 9 r 9 1 = 0.01 mA 100(100 )( 0.026 )= 260 k 0.01 V 100 r07 = A = = 500 k I Q 0.2r011 =VA 100 = = 500 k I Q 0.2 R011 = r011 [1 + g m RE ] , g m = r 11 =(100 ) ( 0.026 )0.2 = 7.69 0.026= 13 k 0.2 RE = 0.2 13 = 0.197 k R011 = 500 1 + ( 7.69 )( 0.197 ) = 1257 k Then 260 + 500 1257 2.6 + 101 R0 = 5 101 = 5 0.0863 R0 = 0.0848 K 84.8 11.84 Ri = r 1 + (1 + ) r 2 r 2 =(100 )( 0.026 ) 0.5= 5.2 k(100 )( 0.026 ) (100 ) ( 0.026 ) = = 520 k 0.5 ( 0.5 /100 ) Ri = 520 + (101)( 5.2 ) Ri 1.05 M (100 )( 0.026 ) r 3 + 50 2r 1 =R0 = 5 R0 = 5101, r 3 =1= 2.6 k2.6 + 50 = 5 0.521 R0 = 0.472 k 101www.elsolucionario.net 493. V V0 = 3 + g m 3V 3 ( 5 ) r 3 1+ V0 = V 3 ( 5) r 3 (1)(V V ) V 3 = g m 2V 2 + 0 3 50 r 3 1 1 V g m 2V 2 = V 3 + 0 r 3 50 50 V V 2 = 1 + g m1V 1 r 2 r 1 1+ = V 1 r 1 r 2 and Vin = V 1 + V 2 gm2 =(2)(3)(4)0.5 = 19.23 mA/V 0.026Then 101 V0 = V 3 ( 5 ) V 3 = V0 ( 0.005149 ) 2.6 And 1 V 1 + 0 19.23V 2 = V0 ( 0.005149 ) 2.6 50 50 = V0 ( 0.02208 )(1)(2)Or V 2 = V0 ( 0.001148 )And V 1 = Vin V 2 = Vin + V0 ( 0.001148 )(4)So 101 V0 ( 0.001148 ) = Vin + V0 ( 0.001148 ) 520 ( 5.2 ) V0 ( 0.001148 ) V0 ( 0.001159 ) = Vin (1.01) Av =(3) V0 = 438 Vin11.85www.elsolucionario.net 494. I2 =5 = 1 mA 51 + 0.8 = 2.21 V 0.5 2.21 ( 5 ) I1 = = 0.206 mA 35VGS 2 =V0 = ( g m 2Vgs 2 ) ( R2 r02 )Vgs 2 = ( g m1Vsg1 ) ( r01 R1 ) V0 and Vsg1 = Vin So Vgs 2 = ( g m1Vin ) ( r01 R1 ) V0 Then V0 = g m 2 ( R2 r02 ) ( g m1Vin ) ( r01 R1 ) V0 Av =V0 g m 2 ( R2 r02 ) g m1 ( r01 R1 ) = Vin 1 + g m 2 ( R2 r02 )gm2 = 2 Kn2 I D 2 = 2( 0.5 )(1) = 1.414 mA / Vg m1 = 2 K p1 I D1 = 2( 0.2 )( 0.206 ) = 0.406 mA / Vr01 =11 I D1r02 1( 0.01)( 0.206 )= 485 k1 1 = = 100 k 2 I D 2 ( 0.01)(1)R2 r02 = 5 100 = 4.76 k R1 r01 = 35 485 = 32.6 k Then Av = (1.414 )( 4.76 )( 0.406 )( 32.6 ) 1 + (1.414 )( 4.76 )So Av = 11.5Output ResistanceFrom the results for a source follower in Chapter 6. 1 1 R0 = R2 r02 = 5 100 gm2 1.414 = 0.707 4.76 So R0 = 0.616 k11.86 a.www.elsolucionario.net 495. R2 =5 R2 = 10 k 0.5 I D2 0.5 VTP 2 = + 1 = 2.41 V K p2 0.25VSG 2 = R1 =5 ( 2.41) 0.1 R1 = 74.1 kb.V0 = ( g m 2Vsg 2 ) ( r02 R2 )Vsg 2 = V0 ( g m1Vgs1 ) ( r01 R1 ) and Vgs1 = Vin Av =V0 ( g m 2 ) ( r02 R2 ) ( g m1 ) ( r01 R1 ) = Vin 1 + ( g m 2 ) ( r02 R2 )g m1 = 2 K n1 I D1 = 2 gm2 = 2 K p 2 I D 2 = 2 r01 = r02 =11 I D1=( 0.1)( 0.1) = 0.2 mA / V ( 0.25)( 0.5) = 0.707 mA / V1( 0.01)( 0.1)= 1000 k1 1 = = 200 k 2 I D 2 ( 0.01)( 0.5 )r02 R2 = 200 10 = 9.52 k r01 R1 = 1000 74.1 = 69.0 k Then Av = ( 0.707 )( 9.52 )( 0.2 )( 69 ) 1 + ( 0.707 )( 9.52 )So Av = 12.0 R0 =1 1 10 200 R2 r02 = 0.707 gm2= 1.414 9.52 Or R0 = 1.23 k11.87 a. I C 2 = 0.25 mA 52 R= R = 12 k 0.25 v VBE ( on ) 2 0.7 I C 3 = 02 RE1 = RE1 = 2.6 k RE1 0.5 RC =5 v03 5 3 = RC = 4 k IC 3 0.5 v03 VBE ( on ) ( 5 ) IC 4 = RE 2 RE 2 =3 0.7 + 5 RE 2 = 2.43 k 3www.elsolucionario.net 496. b.Input resistance to base of Q3, Ri 3 = r 3 + (1 + ) RE1 r 3 =(100 )( 0.026 )= 5.2 k 0.5 Ri 3 = 5.2 + (101)( 2.6 ) = 267.8 k v 1 Ad 1 = 02 = g m 2 ( R Ri 3 ) vd 2 0.25 = 9.62 mA/V 0.026 1 Ad 1 = ( 9.62 ) (12 267.8 ) Ad 1 = 55.2 2 ( RC Ri 4 ) v Now 03 = v02 r 3 + (1 + ) RE1 gm2 =where Ri 4 = r 4 + (1 + ) RE 2 and(1 + ) RE 2 v0 = v03 r 4 + (1 + ) RE 2 (100 )( 0.026 )r 4 0.867 k 3 (101)( 2.43) v0 = = 0.9965 v03 0.867 + (101)( 2.43)Ri 4 = 0.867 + (101)( 2.43) = 246.3 k r 3 = 5.2 k Sov03 (100 ) ( 4 246.3) = = 1.47 5.2 + (101)( 2.6 ) v02 v0 = ( 55.2 )( 0.9965 )( 1.47 ) Ad = 80.9 vd Using Equation (11.32b) g m 2 ( R Ri 3 )So Ad =c. Acm1 =1+ r 2 =2 (1 + ) R0 r 2(100 )( 0.026 )= 10.4 k 0.25 ( 9.62 ) (12 267.8 ) = 0.0569 = Acm1 Acm1 = 2 (101)(100 ) 1+ 10.4 v0 v03 Then Acm = Acm1 v03 v02 = ( 0.9965 )( 1.47 )( 0.0569 ) Acm = 0.08335 80.9 C M RRdB = 20 log10 C M RRdB = 59.7 dB 0.08335 11.88 a. RC1 =10 v01 10 2 = RC1 = 80 k I C1 0.1RC 2 =10 v04 10 6 = RC 2 = 20 k IC 4 0.2www.elsolucionario.net 497. b. Ad 1 =v01 v02 = g m1 ( RC1 r 3 ) vd0.1 = 3.846 mA/V 0.026 (180 )( 0.026 ) r 3 = = 23.4 k 0.2 Ad 1 = ( 3.846 ) ( 80 23.4 ) Ad 1 = 69.6g m1 =Ad 2 =v04 1 = g m 4 RC 2 v01 v02 20.2 = 7.692 mA/V 0.026 1 Ad 2 = ( 7.692 )( 20 ) = 76.9 2 Then Ad = ( 76.9 )( 69.6 ) Ad = 5352 gm4 =11.89 a.Neglect the effect of r0 in determining the differential-mode gain. v02 1 Ad 1 = = g m 2 ( RC Ri 3 ) where Ri 3 = r 3 + (1 + ) RE vd 2 A2 =I1 = RC 2 r 3 + (1 + ) RE12 0.7 ( 12 ) R1=23.3 = 1.94 mA I C 5 121 (1.94 ) gm2 = 2 = 37.3 mA/V 0.026 ( 200 )( 0.026 ) r 3 = IC 3 1 (1.94 )(8) = 4.24 V 2 4.24 0.7 = 1.07 mA IC 3 = 3.3 ( 200 )( 0.026 ) = 4.86 k r 3 = 1.07 Ri 3 = 4.86 + ( 201)( 3.3) = 668 kv02 = 12 Ad 1 =1 ( 37.3) 8 668 = 147.4 2Then Ad = Ad 1 A2 = (147.4 )( 1.197 ) Ad = 176 R0 = r05 = Acm1 = g m 2 ( RC Ri 3 ) 1+r 2 =VA 80 = = 41.2 k I C 5 1.942 (1 + ) R0 r 2( 200 )( 0.026 ) 1 (1.94 ) 2= 5.36 kwww.elsolucionario.net 498. Acm1 = ( 37.3) ( 8 668 )2 ( 201)( 41.2 ) 1+ 5.36 A2 = 1.197= 0.09539Acm = ( 0.09539 )( 1.197 ) Acm = 0.114b. vd = v1 v2 = 2.015sin t 1.985sin t vd = 0.03sin t ( V ) v +v vcm = 1 2 = 2.0sin t 2 v03 = Ad vd + Acm vcm = ( 176 )( 0.03) + ( 0.114 )( 2 ) Or v03 = 5.052sin t Ideal, Acm = 0 So v03 = Ad vd = ( 176 )( 0.03) v03 = 5.28sin tc. Rid = 2r 2 = 2 ( 5.36 ) Rid = 10.72 k 2 Ricm 2 (1 + ) R0 (1 + ) r0 VA 80 = = 82.5 k 1 IC 2 (1.94 ) 2 2 Ricm = 2 ( 201)( 41.2 ) ( 201)( 82.5 ) r0 16.6 M 16.6 M So Ricm = 4.15 M11.90 a. 24 VGS 4 2 = kn (VGS 4 VTh ) I1 = R1 24 VGS 4 = ( 55 )( 0.2 )(VGS 4 2 )22 24 VGS 4 = 11 (VGS 4 4VGS 4 + 4 ) 2 11VGS 4 43VGS 4 + 20 = 0VGS 4 =43 ( 43)2 4 (11)( 20 )2 (11)= 3.37 V24 3.37 = 0.375 mA = I Q 55 0.375 v02 = 12 ( 40 ) = 4.5 V 2 v02 VGS 3 2 = I D 3 = kn (VGS 3 VTh ) R5 I1 =www.elsolucionario.net 499. 2 4.5 VGS 3 = ( 0.2 ) ( 6 ) (VGS 3 4VGS 3 + 4 ) 2 1.2VGS 3 3.8VGS 3 + 0.3 = 0VGS 3 = I D3 =3.8 ( 3.8)2 4 (1.2 ) ( 0.3)2 (1.2 )= 3.09 V4.5 3.09 = 0.235 mA 6gm2 = 2 Kn I D2 = 2( 0.2 ) 0.375 2 = 0.387 mA/V 1 1 g m 2 RD = ( 0.387 )( 40 ) Ad 1 = 7.74 2 2 g m 3 RD 2 A2 = 1 + g m 3 R5Ad 1 =g m3 = 2 K n I D3 = 2( 0.2 )( 0.235 )= 0.434 mA/V A2 = ( 0.434 ) ( 4 )1 + ( 0.434 )( 6 )= 0.482So Ad = Ad 1 A2 = ( 7.74 ) ( 0.482 ) Ad = 3.73 R0 = r05 = Acm1 =1 1 = = 133 k I Q ( 0.02 )( 0.375 ) ( 0.387 ) ( 40 ) g m 2 RD = 1 + 2 g m 2 R0 1 + 2 ( 0.387 ) (133)= 0.149Acm = ( 0.149 )( 0.482 ) Acm = 0.0718b. vd = v1 v2 = 0.3sin t v +v vcm = 1 2 = 2sin t 2 v03 = Ad vd + Acm vcm = ( 3.73)( 0.3) + ( 0.0718 )( 2 ) v03 = 0.975sin t ( V ) Ideal, Acm = 0 v03 = Ad vd = ( 3.73)( 0.3)Or v03 = 1.12sin t ( V ) 11.91 The low-frequency, one-sided differential gain iswww.elsolucionario.net 500. r v02 1 = g m RC vd 2 r + RB 1 RC = 2 r + RBAv 2 =r =(100 )( 0.026 ) 0.5= 5.2 k1 (100 )(10 ) Av 2 = 87.7 Av 2 = 2 5.2 + 0.5 CM = C (1 + g m RC ) 0.5 = 19.23 mA/V 0.026 CM = 2 1 + (19.23)(10 ) CM = 387 pF gm =1 2 r RB ( C + CM ) 1 = So f H = 883 kHz 3 2 5.2 0.5 10 ( 8 + 387 ) 1012fH =11.92 From Equation (11.117), f Z =a.1 1 = 2 R0 C0 2 ( 5 106 )( 0.8 1012 )Or f Z = 39.8 kHz b. From Problem 11.69, f H = 883 kHz. From Equation (11.116(b)), the low-frequency common- mode gain is g m RC Acm = RB 2 (1 + ) R0 1 + + r r r = 5.2 k, g m = 19.23 mA/V So Acm = (19.23)(10 ) 0.5 2 (101) ( 5 106 ) 1 + + 5.2 103 5.2 4 = 9.9 10 87.7 C M RRdB = 20 log10 = 98.9 dB 4 9.9 10 www.elsolucionario.net 501. 11.93 a.From Equation (7.72), fT =gm2 ( C + C )1 = 38.46 mA/V 0.026 38.46 103 Then 800 106 = 2 ( C + C ) gm =Or C + C = 7.65 1012 F = 7.65 pF And C = 6.65 pF CM = C (1 + g m RC ) = 1 1 + ( 38.46 )(10 ) = 386 pF 1 fH = 2 r RB ( C + CM ) r =(120 )( 0.026 ) 1= 3.12 k1 2 3.12 1 10 ( 6.65 + 386 ) 1012 Or f H = 535 kHz fH =b.3From Equation (11.140), f Z =1 1 = 2 R0 C0 2 (10 106 )(1012 )Or f Z = 15.9 kHz 11.94 The differential-mode half circuit is:v g m d RC 2 or Av = v02 = r 1+ 1+ RE r r =(100 )( 0.026 ) 0.51 RC 2 + (1 + ) RE= 5.2 k1 (100 )(10 ) 500 2 = Av = 5.2 + (101) RE 5.2 + (101) REwww.elsolucionario.net 502. a.For RE = 0.1 k : Av = 32.7b.For RE = 0.25 k : Av = 16.4www.elsolucionario.net 503. Chapter 12 Exercise Solutions EX12.1 A 1 + A A A A = 1 1 + A = Af Af Af =a.=1 1 1 1 = = 0.05 0.0001 = 0.0499 Af A 20 104Afb.=20(1/ ) (1/ 0.0499 )=20 = 0.998 20.040EX12.2 A 1 + A 1 1 1 1 = 6 = 0.01 106 = Af A 100 10Af = = 0.009999 dA Af dA = Af (1 + A) A A A dAf 100 dAf = 6 ( 20 ) % = 0.002% Af Af 10 dAf=1EX12.3 Bandwidth = H (1 + A0 ) A 105 = H 0 = ( 2 )(10 ) 100 At = ( 2 ) (104 ) rad / sec f = 10 kHz EX12.4 (a) vOA = A1 A2 vi + A2 vn = (100 )(10 ) vi + (10 ) vn So 1000vi S = = 100 i No 10vn Ni (b) A1 A2 A2 vOC = vi + vn 1 + 1 2 1 + 1 2=105 10 v + vn 5 i 1 + ( 0.001)10 1 + ( 0.001) (105 )So 103 vi S = = 104 i N o 0.1vn Niwww.elsolucionario.net 504. EX12.5 a. V = VS V fb = 100 99 = 1 m V 5 Av = 5000 V/V 0.001 V fb 0.099 V fb = V0 = = = 0.0198 V/V V0 5 V0 = AvV Av =Avf =Av 1 + Av=5000 Avf = 50 V/V 1 + ( 0.0198 )( 5000 )Rif = Ri (1 + Av ) = ( 5 ) 1 + ( 0.0198 )( 5000 ) Rif = 500 k b.R0 f =R0 4 = R0 f 40 1 + Av 1 + ( 0.0198 )( 5000 )EX12.6 a. I = I S I fb = 100 99 = 1 A Ai Aif =I0 5 = Ai = 5000 A/A I 0.001 I fb I0=0.099 = 0.0198 A/A 5Ai 5000 Aif = 50 A/A = 1 + Ai 1 + ( 5000 )( 0.0198 ) Rif =b.Ri 5 = Rif 50 1 + Ai 1 + ( 0.0198 )( 5000 )R0 f = (1 + Ai ) R0 = 1 + ( 0.0198 )( 5000 ) ( 4 ) R0 f = 400 k EX12.7 Avf =Avf =Av 104 = Avf = 3.9984 Av 104 1+ 1+ 1 + ( R2 / R1 ) 1 + ( 30 /10 ) 105 = 3.99984 105 1+ 1 + ( 30 /10 )3.99984 3.9984 100% 0.0360% 3.9984EX12.8 Use a non inverting op-amp. R R 1 + 2 = 15 2 = 14 R1 R1Let=R2 = 140 K R1 = 10 K 1 = 0.066667 R2 1+ R1www.elsolucionario.net 505. Input resistance. Rif = 5 ( 0.06667 ) ( 5 103 ) 1.67 M Rof =50( 0.066667 ) ( 5 103 ) 0.15EX12.9 RE ii R + rk E 1+ h FE 80 )( 0.026 ) ( rk = = 4.16 k 0.5 Then rk 4.16 = = 0.0514 k 1 + hFE 81 Then we want io RE 80 = 0.95 = ii 81 RE + 0.0514 h io = FE 1 + hFEor RE = 0.9619 RE + 0.0514 which yields RE ( min ) = 1.30 k and 81 V + = I E RE + 0.7 = ( 0.5 )(1.3) + 0.7 V + ( min ) = 1.36 V 80 EX12.10 Use the configuration shown in figure 12.20. RS = 500 , RL = 200 Let 1 +RF = 15 R1For example, let R1 = 2 K RF = 28 KEX12.11 a. R2 20 VG = (10 ) 5 = (10 ) 5 20 + 30 R1 + R2 VG = 1 V VS = 1 VGS ID =VS ( 5 ) RS= K n (VGS VTN )2( 1) VGS + 5 = (1.5)( 0.4 )(VGS 2 )2www.elsolucionario.net 506. 4 VGS = 0.6 (V 2GS 4VGS + 4 )0.6V 2GS 1.4VGS 1.6 = 0 VGS =1.4 (1.4 )2+ 4 ( 0.6 )(1.6 )2 ( 0.6 )VGS = 3.174 Vg m = 2 K n (VGS VTN ) = 2 (1.5 )( 3.17 2 ) g m = 3.52 mA / V RD 2 I0 = ( g mVgs ) = ( 3.52 ) Vgs 2+2 RD + RL I 0 = 1.76Vgs Vi = Vgs + g mVgs RS Vgs =Vgs =Vi 1 + g m RSVi = ( 0.4153) Vi 1 + ( 3.52 )( 0.4 )I 0 = (1.76 )( 0.4153)Vi Agf =I0 = 0.731 mA / V Vib. For K n = 1 mA / V 2 From dc analysis: 4 VGS = (1)( 0.4 )(VGS 2 )24 VGS = 0.4 (V 2GS 4VGS + 4 ) 0.4V 2GS 0.6VGS 2.4 = 0VGS =0.6 ( 0.6 ) + 4 ( 0.4 )( 2.4 ) 2 ( 0.4 ) 2VGS = 3.31V g m = 2 (1)( 3.31 2 ) = 2.623 mA / V 2 I0 = ( 2.623) Vgs = 1.311Vgs 2+2 Vi Vgs = = Vi ( 0.488 ) 1 + ( 2.623)( 0.4 )I 0 = (1.31)( 0.488 )Viwww.elsolucionario.net 507. Agf =I0 = 0.6398 mA / V Vi% change =0.731 0.6398 12.5% 0.731EX12.12 Use the circuit with the configuration shown in Figure 12.27. The LED replaces RL . 1 RE = 100 REAgf = 10 mS = 10 103 =EX12.13 dc analysis: 10 V0 V0 = ID + 4.7 47 + 20 I D = K n (VGS VTN )(1)2(2) 20 VGS = (3) V0 = 0.2985V0 20 + 47 2.13 V0 ( 0.213) = I D + ( 0.0149 ) V0 (1) I D = 2.13 V0 ( 0.2279 )From (2): 2.13 V0 ( 0.2279 ) = 1 ( 0.2985V0 ) 1.5 22.13 V0 ( 0.2279 ) = 0.0891V0 0.8955V0 + 2.25 20.0891V0 0.6676V0 + 0.12 = 0 2V0 =0.6676 ( 0.6676 ) 4 ( 0.0891)( 0.12 ) 2 ( 0.0891) 2V0 = 7.31 V 10 7.31 7.31 = 0.572 0.109 4.7 67 I D = 0.463 mA ID =VGS =a.0.463 + 1.5 = 2.18 1 g m = 2 K n (VGS VTN ) = 2 (1)( 2.18 1.5 ) g m = 1.36 mA/Vwww.elsolucionario.net 508. V0 Vgs V0 + g mVgs + = 0 (1) RD RFVgs Vi RS+Vgs V0 RF=0(2) 1 1 V0 Vi + + Vgs = RS RF RF RS V 1 V 1 Vgs + = 0 + i 20 47 47 20 Vgs ( 0.0713) = V0 ( 0.0213) + Vi ( 0.050 ) Vgs = V0 ( 0.299 ) + Vi ( 0.701)From (1): V0 V Vgs + (1.36 ) Vgs + 0 =0 4.7 47 47 V0 ( 0.213) + (1.36 ) Vgs + V0 ( 0.213) Vgs ( 0.0213) = 0 V0 ( 0.234 ) + Vgs (1.34 ) = 0 V0 ( 0.234 ) + (1.34 ) (V0 )( 0.299 ) + (Vi )( 0.701) = 0 V0 ( 0.635 ) + Vi ( 0.939 ) = 0 Avf =V0 = 1.48 Vib. For K n = 1.5 mA / V 2 From dc analysis: 2.13 V0 ( 0.2279 ) = 1.5 ( 0.2985V0 ) 1.5 2= 1.5 0.0891V0 0.8955V0 + 2.25 2= 0.1337V0 1.343V0 + 3.375 20.1337V0 1.115V0 + 1.245 = 0 2V0 = V0 =1.115 (1.115) 4 ( 0.1337 )(1.245 ) 2 ( 0.1337 ) 21.115 0.7597 V0 = 7.01 V 2 ( 0.1337 )10 7.01 7.01 = 0.636 0.105 4.7 67 I D = 0.531 mA ID =0.531 + 1.5 = 2.09 1.5 g m = 2 K n (VGS VTN ) = 2 (1.5 )( 2.09 1.5 )VGS 1.77 mA/VFrom ac analysis: V0 V Vgs + (1.77 )Vgs + 0 =0 4.7 47 47www.elsolucionario.net 509. V0 ( 0.213) + (1.77 ) Vgs + V0 ( 0.0213) Vgs ( 0.0213) = 0 V0 ( 0.234 ) + Vgs (1.75 ) = 0 V0 ( 0.234 ) + (1.75 ) V0 ( 0.299 ) + Vi ( 0.701) = 0 V0 ( 0.757 ) + Vi (1.23) = 0 Avf =% change =V0 = 1.62 Vi1.62 1.48 9.46% 1.48EX12.14 a. Input resistance.V0 Vgs V0 + g mVgs + = 0 (1) RD RF Ii =Vgs V0 RF(2)So V0 = Vgs I i RF 1 1 1 V0 + + Vgs g m = 0 (1) RF RD RF 1 1 1 Vgs I i ( 47 ) 4.7 + 47 + Vgs 1.36 47 = 0 Vgs I i ( 47 ) ( 0.234 ) + Vgs (1.34 ) = 0 Vgs Vgs (1.57 ) = I i (11.0 ) Rif = = 7.0 k IiOutput Resistance.www.elsolucionario.net 510. IX =VX VX + g mVgs + RD RS + RF RS 20 Vgs = VX = VX = 0.2985VX 20 + 47 RS + RF V VX I X = X + (1.36)(0.2985)VX + 4.7 20 + 47 I X = VX [ 0.213 + 0.406 + 0.0149] R0 f =b.VX = 1.58 k IXFrom part (a)1 Vgs I i ( 47 ) ( 0.234 ) + Vgs 1.77 = 0 47 Vgs Vgs (1.98 ) = I i (11) Rif = = 5.56 k IiIX =VX VX + (1.77 )( 0.2985 ) VX + 4.7 20 + 47I X = VX [ 0.213 + 0.528 + 0.0149] R0 f =VX = 1.32 k IXEX12.15 Use the circuit with the configuration shown in Figure 12.41 Let RF = 10 K EX12.16 5.5 VTH = (10 ) = 0.9735 V 5.5 + 51 RTH = 5.5 51 = 4.965 kI BQ =0.973 0.7 = 0.00217 mA 4.96 + (121)(1)I CQ = 0.2605 mA r = 11.98 k, g m = 10.02 mA / V Req = RS R1 R2 r= (10 ) 51 5.5 12 = 2.598 kFrom Equation (12.99(b)): Req T = ( g m RC ) R +R +R F eq C 2.598 = (10 )(10 ) T = 2.75 10 + 82 + 2.598 EX12.17 Computer Analysis EX12.18www.elsolucionario.net 511. V = Vt , V0 = AvV = AvVt R1 Ri R1 Ri Vr = V0 = ( AvVt ) R1 Ri + R2 R1 Ri + R2 T = R1 Ri Vr = Av Vt R1 Ri + R2 or T=Av R 1+ 2 R1 RiEX12.19 f Phase = 180 = 3 tan 1 5 10 f or tan 1 5 = 60 f = 1.732 105 Hz 10 (100 ) T ( f ) = 1 = 3 2 f 1+ 5 10 (100 ) = 0.08 = 3 1 + (1.732 )2 EX12.20 Afo =1000 = 7.092 1 + ( 0.140 )(1000 )www.elsolucionario.net 512. ( 0.140 )(1000 )T =1= f f 1 + 3 . 1 + 4 10 5 10 2 f 1+ 6 10 2By trial and error, at f = 7.65 104 Hz( 0.140 )(1000 ) 2 2 1 + ( 76.5 ) 1 + (1.53) 1 + ( 0.0765 ) ( 0.140 )(1000 ) = = 1.00 76.5 )(1.828 )(1.0 ) (T =2Then = tan 1 (76.5) + tan 1 (1.53) + tan 1 (0.0765) = [89.25 + 56.83 + 4.37 ] = 150.45Phase Marge = 180 150.45 = 29.5 EX12.21 The new loop gain function is 105 1 T ( f ) = f f 1 + j f 1 + j f 1 + j 1 + j 107 5 108 f PD 5 105 f f f f 1 Phase = tan 1 + tan 1 7 + tan 1 + tan 5 8 5 10 10 5 10 f PD For a phase margin 45 Phase = 135, the poles are far apart so this will occur at approximately f = 5 105 Hz. Then 105T ( f ) = 1 =2 5 105 1+ 1+1 1 1 f PD 1051=(1.414 ) 5 105 1+ f PD 22 5 105 9 1+ = 5 10 f PD f PD 5 105 5 109 f PD = 7.07 HzEX12.22 Af ( 0 ) =A0 1 + A0=2 1051 + ( 0.05 ) ( 2 105 ) Af (0) 20f C = f PD (1 + A0 ) = 100 1 + ( 0.05 ) ( 2 105 ) fC 1 MHz EX12.23 Phase Margin = 45 Phase = 135 This will occur at approximately f = 107 Hzwww.elsolucionario.net 513. 105T ( f ) = 1 =2 107 1+ 2 1 f PD 2 107 9 1+ = 5 10 f PD 107 f PD = 141 Hz f PD 5 109TYU12.1 A Af = 1 + A A f + A f A = AAf = A (1 Af ) A=Af 1 Af =80 1 ( 80 )( 0.0120 )A = 2000TYU12.2 dAf dA Af dA 1 = = . Af (1 + A) A A A dA dAf = A Af A A f 5 105 dA = 5% = ( 0.001) A 100 TYU12.3 Af f H = A0 f1 Af =6 A0 f1 (10 ) ( 8 ) Af ( 0 ) = 32 = fH 250 103TYU12.4 V = VS V fb = 100 99 = 1 mV Ag Agf =I 0 5 mA = Ag = 5 A/V V 1 mV V fb=I099 mV = 19.8 V/A 5 mAAg 1 + Ag=5 Agf = 0.05 A/V = 50 mA/V 1 + (19.8 )( 5 )TYU12.5www.elsolucionario.net 514. I = I S I fb = 100 99 = 1 A Az Azf =V0 5V = Az = 5 106 V/A I 1 A I fb=V099 A = 1.98 105 A/V 5VAz 1 + AzTYU12.6 r =a. gm =I CQ=VT=5 1061 + (1.98 105 )( 5 106 ) Azf = 5 104 V/A = 50 V/mAhFEVT (100 )( 0.026 ) = = 5.2 k I CQ 0.50.5 = 19.23 mA / V 0.0261 1 + 19.23 ( 2 ) + g m RE r 5.2 = Avf = 1 1 + 19.23 ( 2 ) 1 + + g m RE 1 + 5.2 r (19.42 )( 2 ) Avf = 0.97490 1 + (19.42 )( 2 ) = r + (1 + hFE ) RE = 5.2 + (101)( 2 ) Rif =Rifr 5.2 =2 R0 f = 0.0502 k 50.2 1 + hFE 101R0 f = REb.= 207.2 khFE = 150 r = 7.8 k, g m = 19.23mA/V 1 + 19.23 ( 2 ) (19.36 )( 2 ) 7.8 Avf = = 1 + (19.36 )( 2 ) 1 1+ + 19.23 ( 2 ) 7.8 Avf = 0.97482 0.0082% change in Avf Rif = 7.8 + (101)( 2 ) = 209.8 k 1.25% change in Rif R0 f = REr 7.8 =2 = 2 0.0517 1 + hFE 151R0 f = 50.36 0.319% change in R0 fTYU12.7www.elsolucionario.net 515. V0 = ( g mVgs ) RS Vi = Vgs + g m RSVgs Vi Vgs = 1 + g m RS g m = 2 K n I DQ = 2( 0.2 )( 0.25 ) = 0.447 mA / VAvf =V0 g m RS = Vi 1 + g m RSAvf =( 0.447 )( 5 ) Avf 1 + ( 0.447 )( 5 )= 0.691Rif = I X = g mVgs =VX RSVgs = VX 1 I X = VX g m + RS 1 1 R0 f = RS = 5 gm 0.447 R0 f = 1.55 kTYU12.8 Computer Analysis TYU12.9 Computer Analysis TYU12.10a.Agf =b.Agf =hFE Ag1 + ( hFE Ag ) RE=( 200 ) (103 ) Agf 1 + ( 200 ) (103 )(103 )( 200 ) (104 ) Agf 1 + ( 200 ) (104 )(103 )= 103 A/V=1 mA/V 1 mA/Vwww.elsolucionario.net 516. Percent change is negligible. ( 4.5 107 % )( 200 ) (104 ) ( 200 ) (103 ) 1 + ( 200 ) (104 )(103 ) 1 + ( 200 ) (103 )(103 ) 103= =( 200 ) (104 ) 1 + ( 200 ) (104 )(103 ) (10 )( 2 10 )( 2 10 ) 398(200)(103 )[1 + (200)(104 )(103 )] (103 )(2 109 )(2 108 ) 200 104 200 103(103 )( 2 109 )( 2 108 )TYU12.11 T = Ai =T ( f1 )Ai 0 = 4.5 109(10 ) ( 0.01) 5= f f 1 + j 1 + j 10 f1 3 10 =1= 2 f 1+ E 10 f 1 + E = 106 10 2f E = 10 106 1 f E 104 Hz 104 f Phase = = tan 1 E = tan 1 10 10 = tan 1 (103 ) 90 Phase Margin = 180 90 Phase Margin = 90 TYU12.12 0 f f 1+ j 1 + j f1 f2 f f Phase = tan 1 + tan 1 f1 f2 T = Ai =Phase Margin = 60 Phase = 120 f f 120 = tan 1 4 + tan 1 5 10 10 At f = 7.66 104 Hz, Phase = tan 1 ( 7.66 ) + tan 1 ( 0.766 ) = [82.56 + 37.45] = 120www.elsolucionario.net 517. (10 ) 5T ( f ) = 1 =1+ ( 7.66 ) 1 + ( 0.766 ) 2(10 ) 251=( 7.725 )(1.26 ) = 9.73 105TYU12.13 Phase Margin = 60 Phase = 120 f Phase = 120 = 3 tan 1 5 10 f tan 1 5 = 40 f = 0.839 105 Hz 10 (100 ) T ( f ) = 1 = 3 2 f 1+ 5 10 (100 ) = = 0.0222 3 1 + ( 0.839 )2 www.elsolucionario.net 518. Chapter 12 Problem Solutions 12.1 (a) Af =A 1 + A120 =5 105 1 + (5 105 ) 120 + (120)(5 105 ) = 5 105 = 0.008331 (b) 5 103 120 = 1 + (5 103 ) 120 + (120)(5 103 ) = 5 103 = 0.008133 12.2Af =(a)A 1 + A = 0.15T = A T = (i) A = 80 dB A = 104 T = 1.5 103 (ii) (iii) T = 15 1 (i) Af = = 6.667(ii)Af = 6.662(iii)Af = 6.25(b) (i) (ii) (iii) (i)T = T = 2.5 103 T = 25 1 Af = = 4.00(ii)Af = 3.9984(iii)Af = 3.84612.3 (a) A 1 = 125 1 + A = 0.0080 (b) Af =www.elsolucionario.net 519. Af = (125)(0.9975) = 124.6875 124.6875 =A 1 + (0.008) A124.6875 [1 + (0.008) A] = A 124.6875 = A [1 0.9975] A = 49,87512.4 (a)100 =104 1 + (104 ) = 9.9 103 (b) Af dA = A A AfdAfdAf 100 = 4 (0.10) = 0.001 = 0.10% Af 10 12.5 Af dA = A A AfdAf0.001 = Af = 500 500 =Af 5 104(0.10)5 104 1 + (5 104 ) = 1.98 103 12.6 (a) For Fig. P12.6(a) vo1 vo 200 10 = = vi 1 + 200 1 vo1 1 + 10 1 200 10 Af = = 40 1 + 200 1 1 + 10 1 50 = (1 + 200 1 )(1 + 10 1 ) = 1 + 210 1 + 2000 12 2000 12 + 210 1 49 = 0 210 44100 + 4(2000)(49) 2(2000) 1 = 0.1126 For Fig P12.6(b) (200)(10) 40 = 1 + (200)(10) 21 = 2 = 0.0245 Fig. P12.6(a) (b)www.elsolucionario.net 520. A1 = 180 180 10 Af = 1 + (10)(0.1126) = (8.4634)(4.704) 1 + (180)(0.1126) Af = 39.81 dAf Af=39.81 40 0.475% 40Fig. P12.6(b) (180)(10) = 39.91 Af = 1 + (180)(10)(0.0245) dAf 39.91 40 = 0.225% 40 Af (c)Fig. P12.6(b) is a better feedback circuit.12.7 (a)VO = (10)(15)(20)V = 3000VV = VO + VS So VO = 3000( VO + VS ) We find V 3000 Avf = O = VS 1 + 3000 For Avf = 120 = (b) Then Avf =3000 = 0.008 1 + 3000 Now VO = (9)(13.5)(18)V = 2187V 2187 2187 = = 118.24 1 + 2187 1 + 2187(0.008)% change =120 118.24 100 1.47% change 12012.8 (105 )(4) = (50) f B f B = 8 kHz 12.9 (a)(50) f3 dB = (105 )(4) f 3 dB = 8 kHz(b)(10) f3 dB = (105 )(4) f3 dB = 40 kHz12.10 (50)(20 103 ) = 5A0 so A0 = 2 105 12.11 Low freq. Af =A0 1 + A0 5000 = 0.0098 1 + (5000) Freq. response 100 =www.elsolucionario.net 521. Af =A 1 + A5000 f f 1 + j f 1 + j f 1 2 = (5000)(0.0098) 1+ f f 1 + j f 1 + j f 1 2 5000 f f 1 + j f 1 + j f + 49 1 2 5000 = f f jf jf 1+ j + j + f1 f 2 f1 f 2 + 49 5000 f f jf jf 50 + j + j + f1 f 2 f1 f 2 Also Af =Af 0 f f 1 + j f 1 + j f A B =100 f f f f 1+ j j +j + j fA fB f A fB So 100 100 = 1 jf jf f f f f f f +j + j j 1 + j +j + 1+ j fA f B f A f B 50 f1 50 f 2 50 f1 f 2 Then 1 1 1 1 + = + f A f B 50 f1 50 f 2and1 1 = f A f B 50 f1 f 2f1 = 10 and f 2 = 2000 1 1 1 1 + = + = 0.002 + 0.000010 = 0.002010 f A f B 50(10) 50(2000) and f 1 1 1 = = B f A f B (50)(10)(2000) f A 106ThenfB 1 + = 0.002010 6 fB 10106 f B2 + 1 = 2.01 + 103 f B 106 f B2 2.01 103 f B + 1 = 0 fB =2.01 103 4.0401 10 6 4(106 )(1) 2(106 )fB =2.01 103 2.0025 104 2(106 )+ signf B = 1.105 103 Hz+ signf A = 9.05 102 Hzwww.elsolucionario.net 522. 12.12 (a)Fig. P12.6(a) 200 f 1+ j f1 10 Af = 1 + (10)(0.1126) 200 1 + (0.1126) 1+ j f f1 200 = (4.704) f 1 + j + 22.52 f1 =940.73 23.52 + jf f1=940.73 1 f 23.52 1 + j (23.52) f140 f 3dB = (23.52)(100) 2.352 kHz jf 1+ (23.52) f1 Fig P12.6(b) (200)(10) f 1+ j f1 2000 = Af = f (0.0245)(200)(10) 1+ 1 + j + 49 f f1 1+ j f1 (b)2000 1 f 3dB = (50)(100) 5 KHz 50 1 + j f (50) f1 Overall feedback wider bandwidth.12.13 v0 = A1 A2 vi + A1vn v0 = (100)vi + (1)vn = (100)(10) + (1)(1) S0 1000 = = 1000 N0 112.14 (a)www.elsolucionario.net 523. (b)Circuit (b) less distortion 12.15 (a) Low input R Shunt input Low output R Shunt output Or a Shunt-Shunt circuit (b) High input R Series input High output R Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a)Ri (max) = Ri (1 + T ) = 10(1 + 104 ) Ri (max) 105 k www.elsolucionario.net 524. Ri 10 = 103 k 1 + T 1 + 104 Or Ri (min) = 1 Ri (min) =Ro (max) = Ro (1 + T ) = 1(1 + 104 ) Ro (max) 104 k (b)Ro 1 = 104 k 1 + T 1 + 104 Or Ro (min) = 0.1 Ro (min) =12.17 io Series output = current signal and Shunt input = current vi signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.Overall Transconductance Amplifier, Ag =12.18 R1 Ri V = Vx R1 Ri + R2 12.19 V = Vi V fb = 50 48 = 2mV Av Vo 5 = = 2.5 103 V/V V 0.002 V fb Vo=0.048 = 0.0096 V/V 5V 5 Arf = o = = 100 V/V Vi 0.0512.20www.elsolucionario.net 525. R R Avf 1 + 2 = 20 2 = 19 R1 R1 vd = iS RivS vd (vs vd ) v0 + R1 R2(1)v0 A0 L vd v0 (vs vd ) + =0 R0 R2(2)is = 1 (v v ) 1 A v v0 + = 0 L d + S d R0 R2 R0 R2 A0 L vd (vS vd ) + R0 R2 v0 = 1 1 + R0 R2 From (1): 1 A0 L vd (vS vd ) + R2 vS vd vS vd R2 R0 + iS = R1 R2 1 1 + R0 R2 A0 L 1 1 1 1 R0 R2 R2 1 1 vd + + iS = vS + R R1 R2 R1 R2 1 + R2 1+ 2 R0 R0 vd = iS Ri 1 1 1 R A0 L 1 1 R2 1 Ri + 1 + 2 + + 1 + R1 R2 R0 R0 R2 R1 R2 R0 R2 iS 1 + = vS R2 R 1+ 1+ 2 R0 R0 R 1 R 1 1 R2 1 1 A0 L 1 iS 1 + 2 + Ri + 2 + + = vS + + R0 R1 R1 R0 R0 R0 R1 R1 R0 R0 R R R R iS R0 + R2 + Ri 0 + 1 + 2 + A0 L = vS 0 + 1 + 2 (1) R1 R1 R1 R1 Let R2 = 190 k , R1 = 10 k 0.1 0.1 iS 0.1 + 190 + 100 + 20 + 105 = vS + 20 10 10 7 iS (1.000219 10 ) = vS (20.01) vS 5 105 k Rif 500 M iS Output Resistance Rif =www.elsolucionario.net 526. IX =VX A0 L vd VX + R0 R2 + R1 Rivd = R1 Ri VX R1 Ri + R2A0 L R1 Ri IX 1 1 1 = = + + VX R0 f R0 R0 ( R1 Ri + R2 ) R2 + R1 Ri R1 Ri = 10 100 = 9.091 1 105 9.09 1 = + + R0 f 0.1 0.1 9.09 + 190 190 + 9.09 = 10 + 4.566 104 + 0.00502 R0 f = 2.19 105 k R0 f = 0.0219 12.21 a.vS vd v0 (vS vd ) v = and vd = 0 R1 R2 Awww.elsolucionario.net 527. 1 vS vS v 1 + = 0 + vd + R1 R2 R2 R1 R2 v0 v0 1 1 + + R2 A R1 R2 1 1 v 1 R vS + = 0 1 + 1 + 2 R1 R1 R2 R2 A = R2 1 + R1 v0 = vS 1 R 1 + 1 + 2 A R1 which can be written as v A Avf = 0 = vS R 1 + A / 1 + 2 R1 1 b. = R 1+ 2 R1 20 =c.105 1 + (105 ) 105 1 So = 20 5 = 0.04999 10 R2 1 R 1 Then = 1 = 1 2 = 19.004 R1 0.04999 R1 A 9 104 9 104 Af = = 19.99956 1 + (9 104 )(0.04999)d.Af Af=Af 4.444 104 = 2.222 10 3 % = 0.005% 20 Af12.22 Aif = Rif =Ai 1000 = = 90.9 A/A 1 + i Ai 1 + (1000)(0.01) Ri 1 + i Ai=1 Rif = 90.9 1 + (0.01)(1000)Rof = Ro (1 + i Ai ) = 10 [1 + (0.01)(1000) ] Rof = 110 k12.23www.elsolucionario.net 528. I = I i I fb = 50 47.5 = 2.5 A Ai =i =Io 5 = = 2000 A/A I 0.0025 I fb Io=0.0475 = 0.0095 A/A 5I 5 Aif = o = = 100 A/A I i 0.0512.24 a.Assume that V1 is at virtual ground. V0 = I fb RF Now I fb = I 0 +I fb RF V0 = I0 R3 R3I fb = I S I and I 0 = Ai I =I0 Aiso I0 Ai From above R I fb 1 + F = I 0 R3 I fb = I S I 0 RF I S 1 + = I0 Ai R3 R 1 R I S 1 + F = I 0 1 + 1 + F R3 R3 Ai orwww.elsolucionario.net 529. Aif = RF 1 + R3 = 1 RF 1 + 1 + R3 Ai Ai = = Aif Ai 1+ RF 1 + R3 I0 ISb.i =c.25 =1 RF 1 + R3 105 1 + (105 ) i105 1 so i = 25 5 i = 0.03999 10 RF R 1 1 = 1 = 1 F = 24.0 so R3 i R3 0.03999Ai = 105 (0.15)(105 ) = 8.5 104d.so Aif = soAifAif8.5 104 = 24.9989 1 + (8.5 104 )(0.03999) =1.10 103 = 4.41 10 5 4.41 10 3 % 2512.24 b. V0 = ( g m1 Vgs + g m 2 V ) RL (1) Vi = Vgs + V0(2) 1 + hFE V1 V V + g m 2 V + 1 = 0 V =0 + r RE 2 r RE 2 V V V = g m1 Vgs + 1 = 0 r RD1 or V V V1 = RD1 + g m1 Vgs (4) r RD1 Then 1 + hFE V r(3) RD1 + RE 2 1 + hFE V r Let V V g m1 Vgs = 0 (3) + r RD1 RD1 RD1 1 + = g m1 + Vgs RE 2 r RE 2 RE 2 www.elsolucionario.net 530. V R 1 = g m1 D1 Vgs Req RE 2 R so V = g m1 Req D1 Vgs RE 2 Then R V0 = g m1 Vgs + g m1 g m 2 Req D1 Vgs RL RE 2 so R V0 = g m1 RL 1 + g m 2 Req D1 (Vi V0 ) RE 2 so R g m1 RL 1 + g m 2 Req D1 V RE 2 Av = 0 = Vi R 1 + g m1 RL 1 + g m 2 Req D1 RE 2 c. Set Vi = 0I X + g m1 Vgs + g m 2V =(1)VX RLVgs = VXFrom part (b), we have R R V = g m1 Req D1 Vgs = g m1 Req D1 VX RE 2 RE 2 Then R IX 1 1 = = + g m1 g m 2 Req D1 VX R0 RL RE 2 or R0 = RL1 g m11 R g m1 g m 2 Req D1 RE 2 12.25I S = I + I fb , V1 = I Ri I fb = I 0 +V0 and V0 = V1 I fb RF R3I 0 = Ai I I =I0 Aiwww.elsolucionario.net 531. Now I fb = Ai I +1 (V1 I fb RF ) R3 R V I fb 1 + F = Ai I + 1 R3 R3 I fb = I S I e R V ( I S I e ) 1 + F = Ai I e + 1 R3 R3 R V R I S 1 + F = I e 1 + F + Ai + 1 R3 R3 R3 V I = 1 Ri 1 1 RF + Ai + = V1 1 + R3 Ri R3 RF 1 + R3 V1 Rif = = I S 1 RF 1 1 + + Ai + Ri R3 R3 R I S 1 + F R3The 1/ R3 term in the denominator will be negligible. Using the results of Problem 12.15: 25 Rif = 1 5 (25) + 10 2 Rif 5 104 k Rif = 0.5 Output Resistance (Let Z L = 0)IX =VX VX + Ai I + R3 RF + RiI =VX RF + Riso A + 1 RF IX 1 1 , = = + i = 24 VX R0 f R3 RF + Ri R3 Let RF = 240 k, R3 = 10 k 1 1 105 + 1 = + Rof 10 240 + 2www.elsolucionario.net 532. so R0 f RF + Ri 240 + 2 R0 f 2.42 103 k or R0 f 2.42 = 5 10 + 1 Ai + 112.26 Agf =Ag 5 = = 0.2 A/V 1 + z Ag 1 + (4.8)(5)Rif = Ri [1 + z Ag ] = (10) [1 + (4.8)(5) ] = 250 kRof = Ro [1 + z Ag ] = (10) [1 + (4.8)(5) ] = 250 k12.27 V = Vi V fb = 40 38 = 2.0 mV I o 8 mA = = 4 A/V V 2 mVAg =V fbz =I0Agf 38 mV = 4.75 V/A 8 mAI0 8 mA = = 0.2 A/V Vi 40 mV12.28IE =V V (1 + hFE ) I0 = S hFE REAlso I 0 = hFE ( AgV ) so V =I0 hFE AgThen V I0 1 + hFE I0 = S hFE RE hFE Ag RE 1 + hFE VS 1 + I0 = hFE hFE Ag RE RE Ag (1 + hFE ) RE + 1 VS I0 = hFE Ag RE RE I0 1 = VS RE hFE Ag RE hFE Ag I0 1 + Ag (1 + hFE ) RE VS 1 + (hFE Ag ) RE www.elsolucionario.net 533. b.Bz = REc.10 =5 105 1 + (5 105 ) z5 105 1 z = 10 5 z = RE = 0.099998 k 5 10 d. If Ag 5.5 105 thenAgf = AgfAgf5.5 105 = 10.0000182 1 + (5.5 105 )(0.099998) =1.82 105 1.82 104 % 1012.29I E = (1 + hFE ) AgV , I E =Now (1 + hFE ) Ag I S Ri =V I S and V = I S Ri , V = VS V = VS I S Ri RE1 (VS I S Ri ) I S RE V Ri + 1 I S = S (1 + hFE ) Ag Ri + RE RE VS R = RE (1 + hFE ) Ag Ri + i + 1 IS RE From Problem 12.16: (1 + hFE ) Ag hFE Ag = 5 105 mSRif =RE 0.1 k 20 +1 so Rif = (0.1) (5 105 )(20) + 0.1 or Rif = 106 kwww.elsolucionario.net 534. V = AgV r I X = g mV +VX (V ) R0(1)V = ( I X + AgV )( RE Ri )(2)or V 1 + Ag ( R Ri ) = I X ( RE Ri ) Now: V V I X = g m Ag r V + X + (1) R0 R0 1 I X ( RE Ri ) VX I X = g m Ag r + + R0 1 + Ag ( RE Ri ) R0 VX R0 f = IX 1 ( RE Ri ) = R0 1 + g m Ag r + R0 1 + Ag ( RE Ri ) g m r Ag = hFE Ag = 5 105 mSLet hFE = 100 so Ag = 5 103 mS RE Ri = 0.1 20 0.1 k Then 1 0.1 R0 f = 50 1 + 5 105 + 3 50 1 + (5 10 )(0.1) or R0 f = 5.04 M12.30 Azf =Az 1 + g Az=5 = 0.2 V/A 1 + (4.8)(5)Rif =Ri 1 = Rif = 40 1 + g Az 1 + (4.8)(5)Rof =Ro 1 = Rof = 40 1 + g Az 1 + (4.8)(5)12.31www.elsolucionario.net 535. I = I i I fb = 40 38 = 2 A Az =Vo 8 V = =4 I 2 Ag = Azf =I fb Vo=38 A = 4.75 8 VVo 8 V = = 0.2 I i 40 A12.32 a.Assuming V1 is at virtual ground (V0 ) = I fb RF and (V0 ) = Az I I =V0 AzI fb = I S I V So V0 = ( I S I ) RF = I S RF 0 RF Az R V0 1 + F = I S RF Az V RF AR so Azf = 0 = = z F I S RF Az + RF 1 + A z Az Az = or Azf = 1 1 + Az g 1 + Az RF 1 RFb.g =c.5 104 =5 106 1 + (5 106 ) g5 106 1 4 g = 5 10 6 g = 1.98 105 5 10 1 RF = RF = 50.5 kgd.Az = (0.9)(5 106 ) = 4.5 106www.elsolucionario.net 536. Azf = Azf Azf4.5 106 = 4.994 104 1 + (4.5 106 )(1.98 105 ) =55.4939 = 1.11 103 0.111% 5 10412.33V1 = I Ri , V0 = Az I V0 = Ax I I fb = I S I and V0 = V1 I fb RF Az I = V1 ( I S I ) RF V V Az 1 = V1 I S RF + 1 RF Ri Ri A R I S RF = V1 1 + z + F Ri Ri V RF Rif = 1 = I S Az RF 1 + R + R i i Or, using the results from problem 12.18. 50.5 103 Rif = 5 106 50.5 103 1 + 10 103 + 10 103 =50.5 103 Rif = 99.79 [1 + 500 + 5.05]12.34 Assume I CQ = 0.2 mA Then r =(100)(0.026) = 13 k 0.2www.elsolucionario.net 537. (1) 1 1 Vi VA VA VA Vo V V 1 = + i + o = VA + + Ri R1 R2 Ri R2 Ri R1 R2 Now Vi Vo 1 1 1 + = VA + + Vi + Vo = VA (12) 10 10 10 1 10 1 or VA = (Vi + Vo ) 12 AvV Vo Vo VA (2) (1 + hFE ) = R2 Ro + r where V = Vi VA Then Av (Vi VA ) Vo Vo VA (1 + hFE ) = R0 + r R2 we find AvVi (1 + hFE ) Vo (1 + hFE ) Vo AvVA (1 + hFE ) VA = Ro + r Ro + r R2 Ro + r R2 Then (5 103 )(101)Vi Vo (101) Vo (5 103 )(101) 1 = VA 14 14 10 14 10 Rearranging terms, we find V Avf = o =10.97 Vi Vi Vi Vi = = Ri I i Vi VA Vi VA Ri 1 1 VA = (Vi + Vo ) = (Vi + 10.97Vi ) = 0.9975Vi 12 12 Then 1 Rif = (10 k ) Rif = 4 M 1 0.9975 To find the output resistance: Rif =www.elsolucionario.net 538. Ix +( AvV )(1 + hFE ) Vx = Ro + r R2 + R1 Ri R1 Ri V = Vx R1 Ri + R2 Now R1 Ri = 1 10 = 0.909 Then V = 0.0833Vx Now 3 1 (5 10 )(0.0833) + 1 I x = Vx (101) + 1 + 13 10 + 0.909 = Vx {3.012 103 + 0.0917}Or Vx = Rof = 3.32 104 k Rof = 0.332 Ix 12.35 a. Neglecting base currents I C 2 = 0.5 mA, VC 2 = 12 (0.5)(22.6) = 0.7 V I C1 = 0.5 mA v0 = 0Then I C 3 = 2 mA b.r 1 = r 2 =hFE VT (100)(0.026) = = 5.2 k I C1 0.50.5 = 19.23 mA / V 0.026 (100)(0.026) r 3 = = 1.3 k 2 2 g m3 = = 76.92 mA / V 0.026 g m1 = g m 2 =www.elsolucionario.net 539. V 1 V + g m1V 1 + g m1V 2 + 2 = 0 r 1 r 1 1 (V 1 + V 2 ) + g m1 = 0 V 1 = V 2 r 1 V 1 ( RS + r 1 ) V 2 + Vb 2 Vi = r 1(1) R or Vi = V 1 1 + S V 2 + Vb 2 r 1 But V 2 = V 1 so R Vi = V 1 2 + S + Vb 2 (2) r 1 V02 V V + g m1V 2 + 02 0 = 0 RC r 3(3)V 3 V V V + g m 3V 3 = 0 + 0 b 2 r 3 RL R2 V 3 = V02 V0 so 1 + hFE (V02 V0 ) r 3 1 1 V + b2 = V0 RL R2 R2 (4)Vb 2 V0 Vb 2 V 2 + + = 0 (5) R2 R1 r 1 Substitute numbers into (2), (3), (4) and (5): 1 Vi = V 2 2 + + Vb 2 5.2 Vi = V 2 (2.192) + Vb 2 (2) 1 1 1 + V02 + (19.23)V 2 V0 = 0 22.6 1.3 1.3 V02 (0.8135) + (19.23)V 2 (0.7692)V0 = 0 (3) 101 101 1 1 1 V02 + + Vb 2 = V0 1.3 1.3 4 50 50 V02 (77.69) = V0 (77.96) Vb 2 (0.02) (4) 1 1 1 1 Vb 2 + V0 + V 2 =0 50 10 50 5.2 Vb 2 (0.120) V0 (0.020) + V 2 (0.1923) = 0(5)www.elsolucionario.net 540. From (2): Vb 2 = Vi + V 2 (2.192). Substitute in (4) and (5) to obtain: V02 (77.69) = V0 (77.96) [Vi + V 2 (2.192)](0.02) (4) [Vi + V 2 (2.192)](0.120) V0 (0.020) + V 2 (0.1923) = 0 (5) So we now have the following three equations: V02 (0.8135) + (19.23)V 2 (0.7692)V0 = 0 (3) V02 (77.69) = V0 (77.96) Vi (0.02) V 2 (0.04384) (4) (0.120)Vi + V 2 (0.4553) V0 (0.020) = 0 (5) From (3): V02 = V0 (0.9455) V 2 (23.64). Substitute for V02 in (4) to obtain: (77.69)[Vo (0.9455) V 2 (23.64)] = V0 (77.96) Vi (0.02) V 2 (0.04384) or 0 = V0 (4.504) Vi (0.02) + V 2 (1836.5) Next, solve (5) for V 2 : (0.120)Vi + V 2 (0.4553) V0 (0.020) = 0 V 2 = V0 (0.04393) Vi (0.2636) Finally, 0 = V0 (4.504) Vi (0.02) + (1836.5)[V0 (0.04393) Vi (0.2636)] 0 = V0 (85.18) Vi (484.12) So V 484.12 Avf = 5.68 Avf = 0 = Vi 85.1812.36 a.RTH = R1 R2 = 400 75 = 63.2 k R2 75 VTH = VCC = (10) = 1.579 V R1 + R2 75 + 400 1.579 0.7 I BQ1 = = 0.007106 mA 63.2 + (121)(0.5) I CQ1 = 0.853 mAVC1 = 10 (0.853)(8.8) = 2.49 V 2.49 0.7 = 0.497 mA 3.6 = 10 (0.497)(13) = 3.54 VIC 2 VC 2IC 3 3.54 0.7 = 2.03 mA 1.4Thenwww.elsolucionario.net 541. (120)(0.026) = 3.66 k 0.853 0.853 g m1 = = 32.81 mA / V 0.026 (120)(0.026) r 2 = = 6.28 k 0.497 0.497 gm2 = = 19.12 mA / V 0.026 (120)(0.026) r 3 = = 1.54 k 2.03 2.03 g m3 = = 78.08 mA / V 0.026 b. r 1 =Vi = V 1 + V 1 V 1 = Vi V 1(1)V 1 V V V + g m1V 1 = 1 + 1 0 r 1 RE1 RF(2)V 2 = ( g m1V 1 )( RC1 r 2 )(3)g m 2V 2 +V 3 + V0 V 3 + =0 RC 2 r 3(4)V 3 V V V + g m 3V 3 = 0 + 0 1 (5) r 3 RE 3 RF Substitute numbers in (2), (3), (4) and (5): 1 V 1 1 V 1 + 32.81 = (Vi V 1 ) + 0 3.66 0.5 10 10 (2) or V 1 (35.18) = Vi (2.10) V0 (0.10) V 2 = (32.81)V 1 (88 6.28) or V 2 = V 1 (120.2) (3) (19.12)V 2 +V 3 V0 V 3 + + =0 13 13 1.54or V 2 (19.12) + V 3 (0.7263) + V0 (0.07692) = 0 (4) 1 V V 1 1 V 3 + 78.08 = V0 + i 1 10 1.54 1.4 10 or V 3 (78.73) = V0 (0.8143) Vi (0.10) + V 1 (0.10) (5) Now substituting V 2 = V 1 (120.2) in (4): (19.12)[V 1 (120.2)] + V 3 (0.7263) + V0 (0.07692) = 0www.elsolucionario.net 542. or V 1 (2298.2) + V 3 (0.7263) + V0 (0.07692) = 0 Then V 3 = V 1 (3164.3) V0 (0.1059) Substituting V 3 = V 1 (3164.3) V0 (0.1059) in (5): (78.73)[V 1 (3164.3) V0 (0.1059)] = V0 (0.8143) Vi (0.10) + V 1 (0.10) or V 1 (2.49 105 ) V0 (9.152) = Vi (0.10) Then V 1 = V0 (3.674 105 ) Vi (4.014 107 ) Now substituting V 1 = V0 (3.674 105 ) Vi (4.014 107 ) in (2):(35.18)[V0 (3.674) 105 ) Vi (4.014 10 7 )] = Vi (2.10) V0 (0.10) or V0 (0.1013) = Vi (2.10) SoV0 = 20.7 Vi Rif =c. I RB1 = I b1 =Vi and I i = I RB1 + I b1 IiVi RB1V 1 r 1Now V 1 = (20.7Vi )(3.674 105 ) Vi (4.014 107 ) V 1 = Vi (7.60 104 ) Then Vi Rif = Vi V (7.60 104 ) + i 63.2 3.66 1 = 0.01582 + 2.077 104or Rif = 62.4 k d.To determine R0 f :Equation (1) is modified to V 1 + Ve1 = 0 (Vi = 0) Equation (5) is modified to: V 3 (78.73) + I X = V0 (0.8143) + V 1 (0.10) (5) Now V 1 (35.18) = V0 (0.10) (2) V 2 = V 1 (120.2) (3) V 2 (19.12) + V 3 (0.7263) + V0 (0.07692) = 0 (4) Now V 1 = V0 (0.002843) sowww.elsolucionario.net 543. V 2 = (V0 )(0.002843)(120.2) V 2 = V0 (0.3417) Then V0 (0.3417)(19.12) + V 3 + (0.7263) + V0 (0.07692) = 0 or V 3 = V0 (9.101) (4) So then V0 (9.101)(78.73) + I X = V0 (0.8143) + (0.10)(V0 )(0.002843) or I X = V0 (717.3) (5) or V R0 f = 0 = 0.00139 k R0 f = 1.39 IX12.37 (a)(1)Vi VA V V V + g m1V 1 = A + A O r 1 RE RF(2)VB V + g m1V 1 + B = 0 RC1 r 2(3)VC V Vo + g m 2V 2 + C =0 RC 2 r 3(4)g m 3V 3 +VC VO VO VA = r 3 RFV 1 = Vi VA V 2 = VB V 3 = VC VOwww.elsolucionario.net 544. 1 V V V + g m1 = A + A O r 1 RE RF (1)(Vi VA ) (2) 1 1 VB + + g m1 (Vi VA ) = 0 RC1 r 2 (3) 1 VO 1 VC + =0 + g m 2VB r 3 RC 2 r 3 (4)(VC VO ) g m3 +(1) (2) (3) (4)(Vi VA )(555.5) = VA (20) + (VA VO )(0.8333) VB (5.109) + 550(Vi VA ) = 0 VC (3.257) + 178VB VO (1.718) = 0 (VC VO )(173.7) = (VO VA )(0.8333)(1) (2) (3) (4)Vi (555.5) + VO (0.8333) = VA (576.3) VB (5.109) + 550Vi VA (550) = 0 VC (3.257) + 178VB VO (1.718) = 0 VC (173.7) + VA (0.8333) = VO (174.5)1 VO VA = r 3 RF (100)(0.026) 14.3 r 1 = = 0.182 K g m1 = = 550 mA/V 14.3 0.026 (100)(0.026) 4.62 r 2 = = 0.563 K g m 2 = = 178 mA/V 4.62 0.026 (100)(0.026) 4.47 r 3 = = 0.582 K g m 3 = = 172 mA/V 4.47 0.026 V V 1 V + 550 = A + A O (1) (Vi VA ) 1.2 0.182 0.05 1 1 (2) VB + + (550)(Vi VA ) = 0 0.3 0.563 VO 1 1 VC + =0 (3) + 178 VB 0.65 0.582 0.582 1 VO VA (4) (VC VO ) 172 + = 0.582 1.2 VB = VA (107.7) Vi (107.7) From (2) From (4) VC = VO (1.0046) VA (0.004797) Substitute into (3) (3.257) [VO (1.0046) VA (0.004797) ] +(178)[VA (107.7) Vi (107.7)] Vo (1.718) = 0 VO (3.272) VA (0.01562) + VA (19170.6) Vi (19170.6) VO (1.718) = 0 VA (19170.6) = Vi (19170.6) VO (1.554) VA = Vi (1.00) VO (0.00008106) Substitute into (1)www.elsolucionario.net 545. Vi (555.5) + Vo (0.8333) = (576.3) [Vi (1.00) Vo (0.00008106) ] = Vi (576.3) Vo (0.0467) Vo (0.880) = Vi (20.8) Vo = Avf = 23.6 Vi Ideal R + RE 1.2 + 0.05 = = 25.0 Avf = F 0.05 RE Rif =(b)Vi V V VA and I i = 1 = i r 1 r 1 IiWe have VA = Vi (1.00) Vo (0.00008106) = Vi (1.00) (23.6)Vi (0.00008106) VA = Vi (0.99809) Vi (1 0.99809) = Vi (0.01051) 0.182 Vi Rif = Rif = 95.1 K Vi (0.01051)Then I i =To find Rof , set Vi = 0 I X + g m 3V 3 +V 3 Vx VA = r 3 RFV 3 = VC VX I X + (VC VX )( g m 3 +V VA 1 )= X r 3 RFFor Vi = 0, we have VC = VX (1.0046) VA (0.004797) VA (576.3) = VX (0.8333) VA = VX (0.001446) VC = VX (1.0046) VX (0.001446)(0.004797) VC = VX (1.0046) 1 VX (1 0.004797) I X + VX (1.0046 1.0) 172 + = 0.582 1.2 I X + VX (0.7991) = VX (0.8293) I X = VX (0.03024) Rof =VX = 33.1 K IX12.38 RD1 = RE1 = 8 K VGG = VGS + ( I D + I C ) RL ID =10 VD 8IC =10 (VD + 0.7) 8www.elsolucionario.net 546. VGG =4.5 = 36 =ID + VTN + ( I D + I C ) RL kn(10 VD ) 10 VD 9.3 VD +1+ (1.8) + (1.8) 8(0.4) 8 8 8 3.210 VD + 8 + 18 1.8VD + 16.74 1.8VD36 = 4.472 10 VD + 42.74 3.6VD 3.6VD 6.74 = 4.472 10 VD 2 12.96VD 48.528VD + 45.4276 = 19.999(10 VD ) 2 12.96VD 28.528VD 154.6 = 028.528 813.847 + 4(12.96)(154.6) 2(12.96) VD = 4.726 V VD =10 4.726 = 0.6593 mA 8 9.3 4.726 IC = 8 = 0.5718 mA 0.6593 VGS = + 1 = 2.284 V 0.4 VO = 4.5 2.284 = 2.216 V ID =VDS = VD VO = 4.726 2.216 = 2.51V(V ( sat ) = 1.28 V ) DSAlso VO = ( I D + I C ) RL = (0.6593 + 0.5718)(1.8) = 2.216 V OK VECQ 2 = (4.726 + 0.7) 2.216 = 3.21 V (b)www.elsolucionario.net 547. (1)Vi = Vgs + Vo(2)V VA + g m1Vgs = RD1 r 2(3)VB V + + g m 2V = 0 RE 2 r 2 V = VB VAVgs = Vi Vo(2)VA V VA + g m1 (Vi Vo ) = B RD1 r 2(3)VB VB VA + + g m 2 (VB VA ) = 0 RE 2 r 2(4)g m1Vgs + g m 2V =VO RLg m1 (Vi Vo ) + g m 2 (VB VA ) =VO RLg m1 = 2 K n I D1 = 2 (0.4)(0.6593) = 1.027 mA/V gm2 = r 2 =I C 2 0.5718 = = 21.99 mA/V VT 0.026(100)(0.026) = 4.547 K 0.5718www.elsolucionario.net 548. (2) (3)VA V VA + (1.027)(Vi Vo ) = B 8 4.547 VB VB VA + + (21.99)(VB VA ) = 0 8 4.547 0.125VA + 1.027Vi 1.027Vo = 0.2199VB 0.2199VA(2) (3) 0.125VB + 0.2199VB 0.2199VA + 21.99VB 21.99VA = 0 (2) 0.3449VA 0.2199VB = 1.027Vi + 1.027Vo (3) 22.21VA = 22.335VB VA = 1.0056VB Then (0.3449)(1.0056)VB 0.2199VB = 1.027Vi + 1.027Vo 0.1269VB = 1.027Vi + 1.027Vo VB = 8.093Vi + 8.093Vo VA = 8.138Vi + 8.138Vo Then (4)(1.027)(Vi Vo ) + (21.99) [ 8.093Vi + 8.093Vo + 8.138Vi 8.138Vo ] =Vo 1.81.027Vi 1.027Vo 177.965Vi + 177.965Vo + 178.955Vi 178.955Vo = 0.5556Vo 2.017Vi 2.5726Vo = 0 Vo = 0.784 = AV ViSet Vi = 0 I X + g m 2V + g m1Vgs =VX RLI X + g m 2 (VB VA ) + g m1 (VX ) =VX RLVB = 8.093VX VA = 8.138VX I X + (21.99) [8.093VX 8.138VX ] + 1.027( VX ) = VX (0.5556) I X = 2.572VX VX = Rof = 0.389 K IX12.39 a.g m = 2 K n I DQ = 2 (0.5)(0.5) = 1 mA / Vwww.elsolucionario.net 549. V0 = ( g mVgs ) RSVi = Vgs + V0 so Vgs = Vi V0 Then V0 = g m RS (Vi V0 ) Av =g m RS 1(2) = Av = 0.667 1 + g m RS 1 + (1)(2)To determine R0 f I X + g mVgs =VX and Vgs = VX RSIX 1 1 = = gm + VX R0 f RS so R0 f =1 1 RS = 2 R0 f = 0.667 k 1 gmFor K n = 0.8 mA / V 2b.g m = 2 (0.8)(0.5) = 1.265 mA / V Av =(1.265)(2) = 0.7167 1 + (1.265)(2)Af=Af0.7167 0.667 7.45% increase 0.6671 2 = 0.7905 2 1.265 = 0.5666 kR0 f = R0 fR0 f R0 f=0.5666 0.667 15.05% decrease 0.66712.40 dc analysis: RTH 1 = 150 47 = 35.8 k , 47 VTH 1 = (25) = 5.96 V 47 + 150 RTH 2 = 33 47 = 19.4 k , 33 VTH 2 = (25) = 10.3 V 33 + 47 5.96 0.7 I B1 = = 0.0187 mA 35.8 + (51)(4.8) I C1 = (50)(0.0187) = 0.935 mA 10.3 0.7 = 0.03705 mA 19.4 + (51)(4.7) = (50)(0.03705) = 1.85 mAI B2 = IC 2(50)(0.026) = 1.39 k; 0.935 (50)(0.026) = 0.703 k r 2 = 1.85 r 1 =www.elsolucionario.net 550. 0.935 = 35.96 mA / V 0.026 1.85 = = 71.15 mA / V 0.026g m1 = gm2VS = V 1 + Ve(1)V 1 V V V0 + g m1V 1 = e + e r 1 R1 RF(2)g m1V 1 +V 2 V 2 V 2 + + =0 RC1 RB 2 r 2(3)V0 V0 Ve (4) + =0 RC 2 RF Substitute numerical values in (2), (3) and (4): Ve = VS V 1 (1) g m 2V 2 +V 1 1 1 1 + (35.96)V 1 = (VS V 1 ) + V0 1.39 0.1 4.7 4.7 or V 1 (46.89) = VS (10.213) V0 (0.2128) (2) 1 1 1 + (35.96)V 1 + V 2 + =0 10 19.4 0.703 or (35.96)V 1 + V 2 (1.574) = 0(3)1 1 1 + (71.15)V 2 + V0 (VS V 1 ) =0 4.7 4.7 4.7 or (71.15)V 2 + V0 (0.4255) VS (0.2128) + V 1 (0.2128) = 0 (4) From (3): V 2 = V 1 (22.85) Then substitute in (4): (71.15)V 1 (22.85) + V0 (0.4255) VS (0.2128) + V 1 (0.2128) = 0 or V 1 (1625.6) + V0 (0.4255) VS (0.2128) = 0 From (2): V 1 = VS (0.2178) V0 (0.004538) Then (1625.6)[VS (0.2178) V0 (0.004538)] + V0 (0.4255) VS (0.2128) = 0 or VS (354.3) + V0 (7.802) = 0 Finally V 0 = 45.4 VSwww.elsolucionario.net 551. 12.41 For example, use a z-stage amplifier. Each stage shown in Fig. 12.29. 12.42For M 3 : K n 3 = kn 2W W Let = 25 L 3 L 3 0.080 2 Then K n 3 = (25) = 1 mA / V 2 Want vo = 0 for vi = 0, so thatI D 3 = I Q 2 = 0.4 = 1 (VGS 3 VTN ) 2 0.4 + 2 = 2.63 V 1 For VGS 3 = 2.63V VG 3 = 12 I D 2 RD Or 2.63 = 12 (0.1) RD RD = 93.7 k Then VGS 3 =g m 3 = 2 K n 3 I D 3 = 2 (1)(0.4) = 1.26 mA / V R1 10 VA = (Vo ) = (Vo ) = 0.0769Vo R1 + R2 120 + 10 (Small amount of feedback) (1) Vi = Vgs1 Vgs 2 + VAwww.elsolucionario.net 552. g mVgs1 + g mVgs 2 = 0 Vgs1 = Vgs 2(2) ThenVi = 2Vgs 2 + VA Vgs 2 =1 (VA Vi ) 2Vgs 2 = 0.0384 Vo 0.5Vi(3)VB = g mVgs 2 RD = g m RD [0.03846Vo 0.5Vi ](4)Vgs 3 = VB Vo and Vo = g m 3Vgs 3 [ RL ( R1 + R2 )]So Vo = g m 3 [ RL ( R1 + R2 ) ] (VB Vo ) Then Vo = g m 3 RL ( R1 + R2 ) g m RD ( 0.03846Vo 0.5Vi ) Vo Or Vo 1 + g m 3 [ RL ( R1 + R2 )][ g m RD (0.03846) + 1] = g m 3 [ RL ( R1 + R2 ) ][ 0.5 g m RD ] Vi Now RL ( R1 + R2 ) = 4 130 = 3.88 So Vo 1 + (1.26)(3.88) [ g m (93.7)(0.03846) + 1] = (1.26)(3.88)(0.5) g m (93.7)Vi Rearranging terms, we find Vo 229 g m = = 10 g m = 1.11 mA / V Vi 5.89 + 17.6 g m We have k W gm = 2 n 2 L 0.080 W I D 1.11 = 2 2 L W W (0.1) = = 77 L 1 L 212.43 Assuming an ideal op-amp, then from Equation (12.58) I0 R 20 = 1+ 1 = = 100 IS R2 0.2 ThenR1 = 99 R2For example, set R2 = 5 k and R1 = 495 k 12.44 h 100 I C1 = FE I E1 = (0.2) = 0.198 mA 101 1 + hFE VC1 = 10 (0.198)(40) = 2.08 V(a)2.08 0.7 = 1.38 mA 1 100 = (1.38) = 1.37 mA 101 IE2 = IC 2For Q1 : (100)(0.026) = 13.1 k 0.198 0.198 = = 7.62 mA / V 0.026r 1 = g m1www.elsolucionario.net 553. For Q2 : (100)(0.026) = 1.90 k 1.37 1.37 = = 52.7 mA / V 0.026r 2 = gm2(b)IS =V 1 V 1 V 1 Ve + + RS r 1 RFg m1V 1 +V 2 + Ve V 2 + =0 RC1 r 2(1) (2)V 2 V V V (3) + g m 2V 2 = e + e 1 r 2 RE RF Substitute numerical values in (1), (2), and (3): 1 1 1 1 + Ve I S = V 1 + 10 13.1 10 10 I S = V 1 (0.2763) Ve (0.10) (1) 1 1 1 (7.62)V 1 + V 2 + + Ve = 0 40 1.90 40 (7.62)V 1 + V 2 (0.5513) + Ve (0.025) = 0(2) 1 1 1 1 + 52.7 = Ve + V 1 V 2 1.90 1 10 10 (3) V 2 (53.23) = Ve (1.10) V 1 (0.10) From (3), Ve = V 2 (48.39) + V 1 (0.0909) Substituting into (1), I S = V 1 (0.2763) (0.10) [V 2 (48.39) + V 1 (0.0909)]or I S = V 1 (0.2672) V 2 (4.839) (1) and substituting into (2), (7.62)V 1 + V 2 (0.5513) + ( 0.025 ) V 2 ( 48.39 ) + V 1 ( 0.0909 ) = 0 or (7.622)V 1 + V 2 (1.761) = 0 V 1 = V 2 (0.2310) Then substituting (2) into (1), we obtain I S = (0.2672)(V 2 )(0.2310) V 2 (4.839) or I S = V 2 (4.901) Now(2)www.elsolucionario.net 554. RC 2 I O = g m 2V 2 RC 2 + RL 2 = (52.7) V 2 = (42.16)V 2 2 + 0.5 Then IS I O = (42.16) 4.901 or I Aif = O = 8.60 Is(c)Ri =V 1 and Ri = RS Rif ISWe had V 1 = V 2 (0.2310) and I S = V 2 (4.901) so V 1 IS = (4.901) = V 1 (21.22) 0.2310 Then V 1 = 0.04713 Ri = 1 = IS 21.22 Finally 10 Rif 0.04713 = Rif = 47.4 10 + Rif 12.45 (a) V 1 V V + 1 e2 RF RS RB1 r 1(1)Ii =(2)g m1V 1 +(3)V 2 V V V + g m 2V 2 = e 2 + e 2 1 r 2 RE 2 RF(4) RC 2 I o = ( g m 2V 2 ) RC 2 + RL VC1 V + 2 = 0 RC1 RB 2 r 2Now (1)Ii =V 1 V e2 RS RB1 r 1 RF RF RF V I R So Ve 2 = RS RB1 r 1 RF 1 i F Now, from (2) V + Ve 2 V 2 g m1V 1 + 2 + =0 Rc1 RB 2 r 2(2) V 2 Ve 2 1 + =0 g m1 + V 1 + r 2 RC1 RB 2 r 2 RC1 RB 2 www.elsolucionario.net 555. Also (3) 1 V 1 1 1 = Ve 2 + gm2 + V 2 + r 2 RF RE 2 RF And I R + RL V 2 = O C 2 g m 2 RC 2 Substitute (1) into (2) and (3)(4) RF V 1 I i RF = 0 RS RB1 r 1 RF V 1 1 RF 1 1 = + V 1 I i RF (3) gm2 + V 2 + r 2 RF RE 2 RF RS RB1 r 1 RF Solve for V 1 from (2) and substitute into (3). Also use Equation (4).(2) V 2 1 1 + g m1 + V 1 + r 2 RC1 RB 2 r 2 RC1 RB 2 (b)RB1 = R1 R2 = 20 80 = 16 K 20 VTH 1 = (10) = 2 V 100 2 0.7 I BQ1 = = 0.0111 mA 16 + (101)(1) I CQ1 = 1.11 mA RTH 2 = 15 85 = 12.75 K 15 VTH 2 = (10) = 1.5 V 100 1.5 0.7 I BQ 2 = = 0.01265 mA 12.75 + (101)(0.5) I CQ 2 = 1.265 mA 1.11 = 42.69 mA/V 0.026 1.265 gm2 = = 48.65 mA/V 0.026 (100)(0.026) r 1 = = 2.34 K 1.11 (100)(0.026) r 2 = = 2.06 K 1.265 Now RC1 RB 2 = 2 12.75 = 1.729 Kg m1 =RS RB1 r 1 RF RB1 r 1 RF = 16 2.34 10 = 1.695 K RC1 RB 2 r 2 = 1.729 2.06 = 0.940 KNow (2)(3)V 2 1 1 10 + V 1 I i (10) 42.69 + V 1 + 2.06 0.940 1.729 1.695 46.587V 1 + 1.064V 2 5.784 I i = 0 V 1 1 1 10 101 = + V 1 I i (10) V 2 + 2.06 10 0.5 10 1.695 49.03V 2 = 12.29V 1 21I iwww.elsolucionario.net 556. From (2) V 1 = (0.1242) I i (0.02284)V 2 Then 49.03V 2 = 12.29 [ (0.1242) I i (0.02284)V 2 ] 21I i (3) 49.31V 2 = 19.47 I iFrom (4) I 4 + 4 V 2 = o = (0.0411) I o 48.65 4 Then (49.31) [ (0.0411) I o ] = 19.47 I i Io = Ai = 9.61 Ii12.46 a.RTH = 13.5 38.3 = 9.98 k 13.5 VTH = (10) = 2.606 V 13.5 + 38.3 (120)(2.606 0.7) I C1 = = 1.75 mA 9.98 + (121)(1)VC1 = 10 (1.75 )( 3) = 4.75 V 4.75 0.7 = 0.50 mA 8.1 (120)(0.026) r 1 = = 1.78 k 1.75 1.75 g m1 = = 67.31 mA / V 0.026 (120)(0.026) r 2 = = 6.24 k 0.50 0.50 gm2 = = 19.23 mA / V 0.026 b. IC 2 VS V 1 V 1 V V = + 1 e2 RS RB r 1 RF g m1V 1 +V 2 + Ve 2 V 2 + =0 RC1 r 2V 2 V V V + g m 2V 2 = 2 + 2 1 r 2 RE 2 RF and(1) (2) (3)www.elsolucionario.net 557. V0 = ( g m 2V 2 ) RC 2 (4) Substitute numerical values in (1), (2), and (3) VS V 1 1 1 Ve 2 = V 1 + + 0.6 0.6 9.98 1.78 1.2 1.2 VS (1.67) = V 1 (4.011) Ve 2 (0.8333)(1)1 Ve 2 1 (67.31)V 1 + V 2 + =0 + 3 6.24 3 or V 1 (67.31) + V 2 (0.4936) + Ve 2 (0.3333) = 0(2)V V 1 V V 1 + 19.23 = e 2 + e 2 2 6.24 8.1 1.2 1.2 or V 2 (19.39) = Ve 2 (0.9568) V 1 (0.8333) (3) From (1) Ve 2 = V 1 (4.813) VS (2.00) Then V 1 (67.31) + V 2 (0.4936) + (0.3333)[V 1 (4.813) VS (2.00)] = 0 or V 1 (68.91) + V 2 (0.4936) VS (0.6666) = 0 (2) and V 2 (19.39) = (0.9568) [V 1 (4.813) VS (2.00) ] V 1 (0.8333)or V 2 (19.39) = V 1 (3.772) VS (1.914) (3) We find V 1 = VS (0.009673) V 2 (0.007163) Then V 2 (19.39) = (3.772)[VS (0.009673) V 2 (0.007163)] VS (1.914) V 2 (19.42) = VS ( 1.878) or V 2 = VS (0.09670) so that V0 = (19.23)(4)(VS )(0.09670) Then V0 = 7.44 VS 12.47 Using the circuit from Problem 12.32, we have Rif =V 1 . ISVS V 1 RS From Problem 12.32 V 1 = VS (0.009673) V 2 (0.007163)Where I S VS (0.009673) (0.007163)(VS )(0.09670) = VS (0.01037)So Rif =VS (0.01037) (0.6) = 0.00629 k VS VS (0.01037)www.elsolucionario.net 558. or Rif = 6.29 12.48 RTH = 1.4 17.9 = 1.298 k 1.4 VTH = (10) = 0.7254 V 1.4 + 17.9 0.7254 0.7 = 0.0196 mA I B1 = 1.298 I C1 = (50)(0.0196) = 0.98 mA Neglecting dc base currents, VB 2 = 10 (0.98)(7) = 3.14 V3.14 0.7 = 3.25 mA 0.25 + 0.5 50 I C 2 = (3.25) = 3.19 mA 51 (50)(0.026) r 1 = = 1.33 k 0.98 0.98 g m1 = = 37.7 mA / V 0.026 (50)(0.026) r 2 = = 0.408 k 3.19 3.19 gm2 = = 123 mA / V 0.026 IE2 =IS =V 1 V V + 1 1 R1 R2 r 1 RFg m1V 1 +V 2 V 2 + Ve 2 + =0 r 2 RC1V 2 V V + g m 2V 2 = e 2 1 r 2 RE1(1) (2) (3)Ve 2 V 1 V V V (4) = 1 + 1 1 RE1 RE 2 RF Enter numerical values in (1), (2), (3) and (4): V 1 V V IS = + 1 1 17.9 1.4 1.33 5 orwww.elsolucionario.net 559. I S = V 1 (1.722) V1 (0.20) (1) (37.7)V 1 +V 2 V + Ve 2 + 2 =0 0.408 7or V 1 (37.7) + V 2 (2.594) + Ve 2 (0.1429) = 0 V 2 V V + (123)V 2 = e 2 1 0.408 0.25 or V 2 (125.5) = Ve 2 (4) V1 (4)(2)(3)Ve 2 V1 V V V = 1 + 1 1 0.25 0.50 5 or Ve 2 (4) = V1 (6.20) V 1 (0.20) (4) From (4): Ve 2 = V1 (1.55) V 1 (0.05) Then substituting in (3): V 2 (125.5) = (4)[V1 (1.55) V 1 (0.05)] V1 (4) or V 2 (125.5) = V1 (2.20) V 1 (0.20) (3) and substituting in (2): V 1 ( 37.7 ) + V 2 ( 2.594 ) + ( 0.1429 ) V1 (1.55 ) V 1 ( 0.05 ) = 0 or V 1 (37.69) + V 2 (2.594) + V1 (0.2215) = 0 Now V1 = V 1 (170.16) V 2 (11.71) Then substituting in (1): I S = V 1 (1.722) (0.20)[ V 1 (170.16) V 2 (11.71)] or I S = V 1 (35.75) + V 2 (2.342) and substituting in (3): V 2 (125.5) = (2.20)[ V 1 (170.16) V 2 (11.71)] V 1 (0.20) or V 2 (151.3) = V 1 (374.55) so that V 1 = V 2 (0.4040)Then I S = (35.75)[V 2 (0.4040)] + V 2 (2.342) I S = V 2 (12.10) RC 2 I 0 = ( g m 2V 2 ) RC 2 + RL 2.2 = (123) V 2 = (64.43)V 2 2.2 + 2 or V 2 = (0.01552) I 0 Then I0 I 1 = 0 = 5.33 I S (0.01552)(12.10) IS12.49www.elsolucionario.net 560. For example, use the circuit shown in Figure P12.30 12.50 r 1 = 6.24 k, r 2 = 3.12 k, r 3 = 1.56 k g m1 = 19.23mA / V , g m 2 = 38.46 mA / V, g m 3 = 76.92 mA / VVS = V 1 + Ve1(1)V 1 V V V + g m1V 1 = e1 + e1 e 3 r 1 RE1 RF V 2 = g m1V 1 ( RC1 r 2 ) g m 2V 2 +(2)(3)V 3 + Ve 3 V 3 + =0 RC 2 r 3(4)V 3 V V V (5) + g m 3V 3 = e 3 + e 3 e1 r 3 RE 2 RF Enter numerical values in (2)-(5): V 1 1 1 1 + (19.23)V 1 = Ve1 + Ve 3 6.24 0.1 0.8 0.8 or V 1 (19.39) = Ve1 (11.25) Ve 3 (1.25) (2) V 2 = (19.23)V 1 (5 3.12) = (36.94)V 1 (3) 1 1 1 (38.46)V 2 + V 3 + + Ve 3 = 0 2 1.56 2 or V 2 (38.46) + V 3 (1.141) + Ve 3 (0.5) = 0(4)1 1 1 1 V 3 + 76.92 = Ve 3 + Ve 3 1.56 0.1 0.8 0.8 or V 3 (77.56) = Ve 3 (11.25) Ve1 (1.25) (5) From (1) V 1 = VS Ve1 Then (VS Ve1 )(19.39) = Ve1 (11.25) Ve3 (1.25) or VS (19.39) = Ve1 (30.64) Ve 3 (1.25) (2) V 2 = VS (36.94) + Ve1 (36.94)(3)(38.46)[VS (36.94) + Ve1 (36.94)] + V 3 (1.141) + Ve 3 (0.5) = 0(4)From (5): Ve 3 = V 3 (6.894) + Ve1 (0.1111) Then VS (19.39) = Ve1 (30.64) (1.25)[V 3 (6.894) + Ve1 (0.1111)]www.elsolucionario.net 561. or VS (19.39) = Ve1 (30.50) V 3 (8.6175) (2) and VS (1420.7) + Ve1 (1420.7) + V 3 (1.141) + (0.5)[V 3 (6.894) + Ve1 (0.1111)] = 0 or VS (1420.7) + Ve1 (1420.76) + V 3 (4.588) = 0 (4) From (2): Ve1 = VS (0.6357) + V 3 (0.2825) Then substituting in (4): VS (1420.7) + (1420.76)[VS (0.6357) + V 3 (0.2825)] + V 3 (4.588) = 0 VS (517.5) + V 3 (405.95) = 0 Now I 0 = g m 3V 3 = 76.92V 3 or V 3 = I 0 (0.0130) Then VS (517.5) + I 0 (0.0130)(405.95) = 0 or I0 = 98.06 mA / V VS12.52 (100)(0.026) = 5.2 k 0.5 0.5 g m1 = g m 2 = = 19.23 mA / V 0.026 (100)(0.026) r 3 = = 1.3 k 2 2 g m3 = = 76.92 mA / V 0.026 r 1 = r 2 =V 1 V + g m1V 1 + g m 2V 2 + 2 = 0 r 1 r 2(1)Since r 1 = r 2 and g m1 = g m 2 , then V 1 = V 2 VS = V 1 V 2 + Ve 3 = 2V 2 + Ve 3 (2) g m 2V 2 +V 3 V 3 + Ve 3 + =0 r 3 RC 2V 3 V V + g m 3V 3 = e3 + 2 r 3 RF r 2(3)(4)www.elsolucionario.net 562. RC 3 I0 = ( g m 3V 3 ) RC 3 + RL From (2): Ve 3 = VS + 2V 2(19.23) V 2 +(5)V 3 V 3 1 + + (VS + 2V 2 ) = 0 1.3 18.6 18.6or (19.23)V 2 + (0.8230)V 3 + (0.05376)VS = 0 (3) V 1 1 + 76.92 = (VS + 2V 2 ) + 2 V 3 1.3 10 5.2 or (77.69)V 3 = (0.3923)V 2 + (0.1)VS (4) 2 I0 = (5) (76.92)V 3 = (51.28)V 3 2 +1 From (3): V 2 = (0.04255)V 3 (0.002780)VS Then (77.69)V 3 = (0.3923)[ (0.04255)V 3 (0.002780)VS ] + (0.1)VS(77.71)V 3 = (0.0989)VS or V 3 = (0.001273)VS so that I 0 = (51.28)(0.001273)VS or I0 = (0.0653) mA/V VS12.52 A=5 103 2f f 1 + j 4 1 + j 5 10 10 f f Phase = = tan 1 4 2 tan 1 5 10 10 By trial and error, when f = 1.095 105 Hz , 180For T = 1 at f = 1.095 105 Hz, 1= ( 5 103 ) f 1+ 4 10 2 f 1 + 5 10 2 1= ( 5 103 )(10.996 )( 2.199 ) = 4.84 10312.53 Use the basic circuit shown in Figure 12.27.www.elsolucionario.net 563. For the ideal case I0 1 = Vi RE we want I0 = 103 A/V = 1 mA/V Vi Set RE = 1 k Since the op-amp has a finite gain, finite input resistance, and finite output resistance, the closed-loop gain is slightly less than the ideal. RE will need to be slightly decreased to increase the gain. 12.54 dc analysis I E RE + VEB (on) + I B RB + VCC = 0 5 0.7 = 0.0343 100 + (51)(0.5) I C = (50)(0.0343) = 1.71 mA IB =(50)(0.026) = 0.760 k 1.71 1.71 gm = = 65.77 mA/V 0.026 a.Then r =To determine Rif : IS +V V (V ) + 0 =0 RB r RF(1)www.elsolucionario.net 564. V0 V0 (V ) (2) + RC RF Now from (2): V 1 1 (65.77)V = V0 + 10 1 10 (65.67)V = V0 (1.10) or V0 = (59.7)V and from (1): V V V + + 0 =0 IS + 100 0.760 10 10 g mV =I S + V (0.8543) + (0.1)(59.7)V = 0 I S = V (6.824) Now (V ) Rif = 147 Rif = ISTo determine R0 f :IX =VX VX + g mV RC RF + RB r(3) ( RB r ) V = ( R r ) + R (VX ) (4) F B Now (100 0.760) V = (100 0.760) + 10 (VX ) = (0.07014)VX so 1 1 I X = VX + + (65.77)(0.07014) 1 10.754 VX R0 f = 175 R0 f = IX b. From part (a), we find I V = S 6.824 then IS V0 = (59.7) 6.824 orwww.elsolucionario.net 565. V0 = 8.75 k VSc.If capacitance is finite, a phase shift will be introduced.12.55 dc analysis: VGS = VDS ID =VDD VGS = K n (VGS VTN ) 2 RD10 VGS = (0.20)(8)(VGS 2) 2 2 10 VGS = 1.6(VGS 4VGS + 4) 2 1.6VGS 5.4VGS 3.6 = 05.4 (5.4) 2 + 4(1.6)(3.6) = 3.95 V 2(1.6) 10 3.95 ID = = 0.756 mA 8 g m = 2 K n I D = 2 (0.2)(0.756) g m = 0.778 mA/VVGS =a. Vgs VS RS+Vgs V0 RF=0 1 1 VS V0 Vgs + + = RS RF RS RF V0 Vgs V0 + g mVgs + =0 RD RF(1) 1 1 1 V0 + g m (2) = Vgs RD RF RF So from (1): 1 VS V0 1 Vgs + + = 10 100 10 100 or Vgs (0.11) = VS (0.10) + V0 (0.010) Vgs = VS (0.909) + V0 (0.0909)Then from (2):www.elsolucionario.net 566. 1 1 1 V0 + 0.778 = Vgs 8 100 100 V0 (0.135) = Vgs (0.768) = (0.768)[VS (0.909) + V0 (0.0909)] V0 (0.2048) = VS (0.6981) so V Av = 0 = 3.41 VSb. We have Vgs = VS (0.909) + V0 (0.0909) = VS (0.909) + (0.0909) (3.41VS ) = 0.599VS Now V V0 (3.41VS ) RS Azf = 0 = = I S VS Vgs VS 0.599VS RS or (3.41)(10) Azf = Azf = 85.0 V/ma 0.401 Vgs Vgs 0.401VS 0.599VS c. Rif = = = (10) Rif = 14.9 k 0.401VS IS RS d.IX =VX VX + g mVgs + RD RS + RF RS 10 Vgs = VX = VX RS + RF 10 + 100 = (0.0909)VX 1 1 I X = VX + (0.778)(0.0909) + 10 + 100 8 IX 1 = = 0.2048 R0 f = 4.88 k VX R0 f12.56 As g m ,V0 RF 100 = = = 10 VS RS 10www.elsolucionario.net 567. To be within 10% of ideal, V0 = 10(0.9) = 9 VS From Problem 12.41, we had Vgs = VS (0.909) + V0 (0.0909) = VS (0.909) + ( 9VS )(0.0909) = 0.0909VS Also from Problem 12.41, we had V0 (0.135) = Vgs (0.010 g m )or (9VS )(0.135) = (0.0909)VS (0.010 g m ) 1.215 = 0.000909 0.0909 g m or g m = 13.36 mA/V 12.57 dc analysis RTH = 24 150 = 20.7 k 24 VTH = (12) = 1.655 V 24 + 150 1.655 0.7 I BQ = = 0.00556 mA 20.7 + (151)(1)so I CQ = 0.834 mA (150)(0.026) = 4.68 k 0.834 0.834 gm = = 32.08 mA / V 0.026r =VS V V V V0 = + RS RB r RF(1)V0 V0 V + = 0 (2) RC RF From (1): 1 VS 1 1 V0 = V + + 5 5 20.7 4.68 RF RF g mV +orwww.elsolucionario.net 568. 1 V0 VS (0.20) = V 0.4620 + RF RF From (2): 1 1 1 32.08 V + V0 + =0 RF 6 RF so 1 V0 0.1667 + RF (2) V = 1 32.08 RF Then 1 V0 0.1667 + RF 1 VS (0.20) = 0.4620 + RF 1 32.08 RF V0 RF Neglect the RF in the denominator term. Now V0 V = 5 VS = 0 = V0 (0.20) 5 VS V ( 0.1667 RF + 1) V0 (0.20)(0.20) RF = (0.4620 RF + 1) 0 V0 32.08RF 2 1.283RF = (0.4620 RF + 1)(0.1667 RF + 1) 32.08 RF 2 1.206 RF 32.71RF 1 = 032.71 (32.71) 2 + 4(1.206)(1) 2(1.206) so that RF = 27.2 k RF =12.58 dc analysis RTH = 4 15 = 3.16 k = RB 4 VTH = 12 = 2.526 V 4 + 15 2.526 0.7 I BQ = = 0.00251 3.16 + (181)(4) I CQ = 0.452 mA (180)(0.026) = 10.4 k 0.452 0.452 gm = = 17.4 mA/V 0.026r =www.elsolucionario.net 569. Vi V 1 V 1 V V = + 1 0 RS RB r RF(1)g mV 1 +V 2 =0 RC RB r(2)g mV 2 +V 3 =0 RC RB r(3)g mV 3 +V0 V0 V0 V 1 + + =0 RC RL RF(4)Now RC RB r = 8 3.16 10.4 = 1.86 k RB r = 3.16 10.4 = 2.42 k Now substituting in (2): V (17.4)V 1 + 2 = 0 or V 2 = (32.36)V 1 1.86 and substituting in (3): V (17.4)V 2 + 3 = 0 1.86 V (17.4)[(32.36)V 1 ] + 3 = 0 1.86 or V 3 = (1047.3)V 1 Substitute numerical values in (1): 1 Vi 1 1 V0 = V 1 + + 10 10 2.42 RF RF or 1 V0 Vi (0.10) = V 1 0.513 + RF RF Substitute numerical values in (4): 1 1 1 V 1 =0 (17.4)(1047.3)V 1 + V0 + + 8 4 RF RF 1 1 V 1 1.822 104 + V0 0.375 + =0 RF RF 1 V0 0.375 + RF V 1 = 1 1.822 104 RF so thatwww.elsolucionario.net 570. 1 V0 0.375 + RF V0 1 Vi (0.10) = 0.513 + RF 1.822 104 1 RF RF V V We have 0 = 80 or Vi = 0 80 Vi 1 0.375 + RF 1 (0.10) 1 = 0.513 + 80 RF 1.822 104 1 RF RF Neglect that 1/ RF term in the denominator. (0.513RF + 1)(0.375 RF + 1) (0.00125 RF ) = 1 1.822 104 RF 2 22.775 RF = (0.513RF + 1)(0.375 RF + 1) + 1.822 10 4 RF We find 2 22.58 RF 1.822 104 RF 1 = 0RF =1.822 104 (1.822 10 4 ) 2 + 4(22.58)(1) 2(22.58)or RF = 0.807 M 12.59a.VS (Vd ) Vd V1 = RS RFor 1 1 VS V1 + + =0 Vd + RS RF RS RF V0 V1 V1 V1 (Vd ) = + R1 R2 RF or 1 V0 1 1 Vd = V1 + + + R1 R1 R2 RF RF(1)(2)www.elsolucionario.net 571. V0 A0 L Substitute numerical values in (1) and (2): V0 1 1 VS V1 + + + = 0 104 5 10 5 10 or V0 (0.3 104 ) + VS (0.20) + V1 (0.10) = 0and V0 = A0 LVd or Vd =(1)V0 1 1 1 V 1 = V1 + + + 04 50 50 10 10 10 10 or V0 (0.02 105 ) = V1 (0.22) (2) 0.02 105 Then V1 = V0 0.22 and 0.02 105 V0 (0.3 104 ) + VS (0.20) + (0.10) V0 = 0 0.22 V0 0.3 104 0.4545 105 + 0.00909 + VS (0.20) = 0 Then b.V0 V 0.20 = 0 = 21.94 3 VS 9.115 10 VS Rif =Now Vd =Vd V R = d S VS (Vd ) VS + Vd RSV0 21.94VS = A0 L 104(21.94 104 )(5) 1 21.94 104 or Rif = 1.099 102 k Rif = 10.99 Then Rif =c. Because of the A0LVd source, R0 f = 0 12.60 For example, use the circuit shown in Figure 12.41 12.61 Break the loopwww.elsolucionario.net 572. It = INow Ai I t +V0 V0 + =0 R1 RF + RS Ri RS V0 Ir = RS + Ri RF + RS Ri R + Ri or V0 = I r S ( RF + RS Ri ) RS Then 1 RS + Ri 1 Ai I t + + Ir ( RF + RS Ri ) = 0 R R +R R F S i RS 1 Ai I T = r T = It 1 RS + Ri 1 + ( RF + RS Ri ) R1 RF + RS Ri RS 12.62V 1 V V V + g m1V 1 = 1 + 1 0 r 1 RE1 RF g m1V 1 +Vr RC1 r 2(1)= 0 Vr = ( g m1V 1 )( RC1 r 2 )(2)V 2 = Vt so that g m 2Vt +V 3 + V0 V 3 + = 0 (3) RC 2 r 3V 3 V V V + g m 3V 3 = 0 + 0 1 (4) r 3 RE 3 RF From (4): 1 1 V 1 V0 + + g m3 + 1 = V 3 RE 3 RF r 3 RFBut V 1 = V 1 1 V + g m3 1 V 3 r 3 RF so V0 = 1 1 + RE 3 RF Thenwww.elsolucionario.net 573. 1 1 1 + V 1 + g m1 = RE1 RF r 1 1 V V 3 + g m3 + 1 r 3 RF 1 1 RF + RE 3 RF (1)and 1 V + g m3 1 V 3 1 1 r 3 RF = 0 + g m 2Vt + V 3 + 1 1 RC 2 r 3 + RC 2 RE 3 RF (3)From (3), solve for V 3 and substitute into (1). Then from (1), solve for V 1 and substitute into (2). Then T =Vr . Vt12.63Vr Vr Vr Ve + + =0 RS r 1 RF g m1Vt +(1)V 2 + Ve V 2 + = 0 (2) RC1 r 2V 2 V V Vr + g m 2V 2 = e + e (3) r 2 RE RF Using the parameters from Problem 12.29, we obtain 1 1 V 1 Vr + + e = 0 10 15.8 10 10 or Vr (0.2633) = Ve (0.10) (1) 1 Ve 1 =0 (7.62)Vt + V 2 + + 40 2.28 40 or Vt (7.62) + V 2 (0.4636) + Ve (0.025) = 0(2) 1 1 1 V V 2 + 52.7 = Ve + r 2.28 1 10 10 or V 2 (53.14) = Ve (1.10) Vr (0.10) Then V 2 = Ve (0.0207) Vr (0.001882) (3) Substituting in (2): Vt (7.62) + (0.4636)[Ve (0.0207) Vr (0.001882)] + Ve (0.025) = 0www.elsolucionario.net 574. or Vt (7.62) + Ve (0.03460) Vr (0.0008725) = 0 From (1) Ve = Vr (2.633) Then Vt (7.62) + Vr (2.633)(0.03460) Vr (0.0008725) = 0 Vt (7.62) + Vr (0.09023) = 0 Vr = 84.45 Vt Now V T = r T = 84.45 Vtor12.64V = Vtg mV =V0 V0 + RC RF + RB r(1)and RB r (2) Vr = V R r +R 0 F B Now 1 1 (65.77)V = V0 + 1 10 + 100 0.760 or (65.77)V = V0 (1.0930) and 0.754 Vr = V0 = (0.07011)V0 10 + 0.754 so V0 = (14.26)Vr Then (65.77)(Vt ) = (14.26)Vr (1.0930) Vr = 4.22 so that T = 4.22 Vt12.64 Want f1 = 12 MHz for a phase margin of 45 TdB ( 0 ) = 80 dB T ( 0 ) = 104Thenwww.elsolucionario.net 575. T ( 0)T(f)= f 1 + j f PD Set f = f1 and T f 1 + j 6 12 10 =1So 104T =1=2 12 106 1+ 2 f PD which yields 12 106 104 = f PD = 1.70 kHz f PD 212.65 a. f 1 f = tan 1 2 tan 4 5 102 10 or f f 180 = tan 1 180 2 2 tan 1 180 f180 1.05 104 Hz 4 5 10 10 (105 ) T ( f180 ) = 1 = b. 2 2 1.05 104 1.05 104 1+ 1 + 2 4 5 10 10 5 (10 ) or 1= (21.02)(2.105) = 4.42 104 12.65 A0 = 80 dB A0 = 104 A0 Af ( 0 ) = 1 + A0 or 5 =1041 + (104 ) 0.2Then T ( 0 ) = A0 = 0.2 104 Inserting a dominate pole f 1 f 1 f = tan 1 tan 6 tan 7 10 f PD 10 If we want a phase margin of 45, then f f 135 90 tan 1 6 tan 1 7 10 10 By trial and error, f 0.845 MHz Then 0.2 104 1 T =1= 2 2 2 0.845 106 0.845 0.845 1+ 1+ 1+ 10 f PD 1 www.elsolucionario.net 576. 0.845 106 0.2 104 f PD (1.309 )(1.0036 )so f PD = 555 Hz 12.66 (103 )T = Av =(a)f 1 + j 4 10 f f = 2 tan 1 4 tan 1 5 10 10 Set = 1802f 1 + j 5 10 By trial and error, f = 4.58 10 4 Hz Set T = 1 at f = 4.58 10 4 Hz(b) (103 )1=2 4 2 4 2 1 + 4.58 10 1 + 4.58 10 4 5 10 10 (103 ) = 0.02417 1= (21.976)(1.10) 103 = 39.7 1 + (10 )(0.02417)(c)Avfo =(d)Wait T < 1 at f = 4.58 104 Hz, so system is stable for smaller values of .312.67 f f f tan 1 tan 1 5 4 4 10 5 10 10 4 At f = 8.1 10 Hz, = 180.28 = tan 1 Determine T ( f ) at this frequency. T = (103 ) =a.1 8.1 104 1+ 4 10 21 8.1 104 1+ 4 5 10 21 8.1 104 1+ 5 10 2 (103 ) (8.161)(1.904)(1.287) For = 0.005T ( f ) = 0.250 < 1 Stableb.For = 0.05T ( f ) = 2.50 > 1 Unstable12.68 (b)Phase margin = 80 = 100 f f tan 1 3 4 10 5 10 3 By trial and error, f = 1.16 10 Hz = 100 = 2 tan 1 www.elsolucionario.net 577. Then (5 103 )T =1=2 3 2 3 2 1 + 1.16 10 1 + 1.16 10 3 4 10 5 10 (5 103 ) = = 4.7 104 (2.35)(1.00)12.69For = 0.005,c.T ( f ) = 1(0 dB) at f 2.10 104 HzThen 4 4 2.10 104 1 2.10 10 1 2.10 10 tan tan 3 4 5 10 10 10 = tan 1 or= 87.27 64.54 11.86 = 163.7 System is stable. Phase margin = 16.3 For = 0.05, T ( f ) = 1 (0 dB) at f 6.44 104 Hzwww.elsolucionario.net 578. Then 6.44 104 6.44 104 6.44 104 tan 1 tan 1 3 4 5 10 10 10 = tan 1 or= 89.11 81.17 32.78 = 203.1 System is unstable. 12.70 (105 )T = A =f f f 1 + j 5 1 + j 1 + j 4 5 10 10 5 105 Phase Margin = 60 = 120 So f f f 120 = tan 1 tan 1 5 tan 1 4 5 5 10 10 5 10 By trial and error, at f = 105 Hz , 120 Then (105 ) T = 1= 2 2 2 105 105 105 1+ 5 1+ 1+ 4 5 5 10 10 5 10 1= (105 )( 2.236 )(1.414 )(1.02 ) = 3.22 10512.71 (a)100 =105 = 9.99 103 1 + (105 ) T = 1 = Av =1=(9.99 103 )(105 ) f 1+ 3 10 999 22 f 1+ 5 10 f f 1+ 3 1+ 5 10 10 5 f = 3.08 10 Hz22Phase f f tan 1 5 3 10 10 5 3.08 10 3.08 105 = tan 1 tan 1 103 105 = 89.81 72.01 = 161.8 Stable (b) Phase Margin = 180 161.8 = 18.2 = tan 112.72www.elsolucionario.net 579. = tan 1(a)For = 180f f f tan 1 6 tan 1 4 5 10 10 5 107By trial and error, f180 = 7.25 106 Hz T =(b)=(0.10)(105 ) 7.25 106 1+ 4 5 10 2 7.25 106 1+ 6 10 2 7.25 106 1+ 7 5 10 2104 (145)(7.319)(1.01)T = 9.33 System is unstable T =1=104 f 1+ 4 5 10 2 f 1+ 6 10 2 f 1+ 7 5 10 2f = 2.14 107 Hz 7 7 2.14 107 1 2.14 10 1 2.14 10 tan tan 4 6 7 5 10 10 5 10 = tan 1 = 89.87 87.32 23.17 = 200.4For f = 7.25 106 Hz(c)(0.0010)(105 )T =2 7.25 106 7.25 106 1+ 1+ 4 6 5 10 10 100 = (145)(7.319)(1.01)2 7.25 106 1+ 7 5 10 2T = 0.0933 System is stable. T =1=100 f 1+ 4 5 10 2 f 1+ 6 10 2 f 1+ 7 5 10 2f = 2.13 106 Hz 2.13 106 2.13 106 2.13 106 tan 1 tan 1 5 104 106 5 107 = 88.66 64.85 2.44 = 155.9 = tan 112.73 (a)f180 f 2 tan 1 180 5 103 105 = 1.05 105 Hz = 180 = tan 1 f180www.elsolucionario.net 580. (0.0045)(2 103 )T =(b)2 2 1.05 105 1.05 105 1+ 1 + 3 5 5 10 10 9 = (21.02)(2.1025)0 T = f180 = 0.204System is stable 9T =1= f 1+ 3 5 10 2 f 2 1 + 5 10 f = 3.88 104 Hz 3.88 104 3.88 10 4 2 tan 1 3 5 10 105 = 82.66 42.41 = 125.1 = tan 1(c)T (0.15)(2 103 ) 1.05 105 1+ 3 5 10 2 1.05 105 2 1 + 5 10 300 (21.02)(2.1025)T = 6.79 System is unstable T = 1=300 f 1+ 3 5 10 2 f 2 1 + 5 10 f = 2.33 105 Hz2.33 105 2.33 105 2 tan 1 5 103 105 = 88.77 133.54 = tan 1 = 222.3 12.74 Phase Margin = 45 = 135 = 135 f f f f = tan 1 3 tan 1 4 tan 1 5 tan 1 6 10 10 10 10 4 At f = 10 Hz, = 135.6www.elsolucionario.net 581. T =11= (103 ) 21 104 104 1+ 3 1+ 4 10 10 1 1 2 4 2 10 104 1+ 5 1+ 6 10 10 1=2 (103 ) (10.05)(1.414)(1.005)(1.00)or = 0.01428 12.75 1 1 f f 1 + j 3 1 + j 300 10 f PD 1 1 f f 1 + j 1 + j 6 2 10 25 106 Phase Margin = 45 = 135 at f = 300 kHz T = 5000 3 300 103 1 300 10 135 = tan 1 00 tan 3 300 10 f PD = 90 45 Now T = 1@ f = 300 kHzT =15000 2 300 103 1+ 2 1 1 f PD 2 300 103 5000 1+ = f PD 2 f PD 12.76 (a)2300 103 2 f PD = 84.8 Hz 5000At f = 106 Hz,106 106 2 tan 1 6 104 10 = 89.4 90 180 = tan 1www.elsolucionario.net 582. T 103 106 1+ 4 10 2 106 2 1 + 6 10 103 =5 (100)(2)T > 1 at f = f180 = 106 Hz, System is unstable.103T=(b) f f f 1 + j 1 + j 4 1 + j 6 f PD 10 10 Phase margin = 45 = 135 = 135 = tan 1f f PD tan 12f f 2 tan 1 6 4 10 104f 10 Hz T =1=1103 104 1+ f PD 2 104 1+ 4 10 2 104 2 1 + 6 10 103 104 (1.414)(1.0) f PD (104 )(1.414) f PD = 103 f PD = 14.14 Hz12.77 (a) f180 f f tan 1 180 4 tan 1 180 4 5 10 5 10 10 = 180 = tan 1 f180 8.06 104 Hz(b)500T =2 8.06 10 8.06 104 1+ 1+ 4 4 10 5 10 500 = (8.122)(1.897)(1.284) 42 8.06 104 1+ 5 10 2T = 25.3 500 f f f f 1 + j 1 + j 4 1 + j 1 + j 5 10 5 104 10 f PD Phase Margin = 60 = 120(c)T=120 = tan 1f f f f tan 1 4 tan 1 tan 1 5 10 5 104 10 f PDAssume tan 1f f PD 90www.elsolucionario.net 583. Then f 4.2 103 Hz 500T =1=2 4.2 103 4.2 103 1+ 1+ 4 10 f PD 5001=2 4.2 103 1+ 4 5 10 2 4.2 103 1+ 5 10 2 4.2 103 1+ (1.085)(1.004)(1.0) f PD 4.2 103 500 (1.0846)(1.0035)(1.0) f PD f PD = 9.14 Hz12.78 50 =(a) T=104 = 0.0199 1 + (104 ) (0.0199)(104 ) f f 1 + j 1 + j 5 10 f PD Phase margin = 45 = 135 135 = tan 1f f PD tan 1f 1055f = 10 Hz T =1=(0.0199)(104 ) 105 1+ f PD 2 105 1+ 5 10 2105 (0.0199)(104 ) = f PD 1.414 f PD = 711 Hz 20 =(b)T=104 = 0.0499 1 + (104 ) (0.0499)(104 ) f f 1 + j 1 + j 5 711 10 T =1=(0.0499)(104 ) f 1+ 711 2 f 1+ 5 10 2f = 1.76 105 Hz 5 1.76 105 1 1.76 10 tan 5 711 10 = tan 1 = 89.77 60.40 = 150.2 Phase Margin = 180 150.2 = 29.8www.elsolucionario.net2 584. 12.79 (a) 20 =AO = 100 dB AO = 105105 = 0.04999 1 + (105 ) (0.04999)(105 ) f f f 1 + j 1 + j 6 1 + j 7 f PD 10 10 Phase Margin = 45 = 135 T=135 = tan 1f f f tan 1 6 tan 1 7 10 10 f PDf 106 Hz T =1=1=(0.04999)(105 ) 106 1+ f PD 2 106 1+ 6 10 2 106 1+ 7 10 2(0.04999)(105 ) 2 106 1+ (1.414)(1.005) f PD 106 (0.04999)(105 ) = f PD (1.414)(1.005) f PD = 2.84 Hz(b) T =1=5=105 = 0.19999 1 + (105 ) (0.19999)(105 ) f 1+ 284 2 f 1+ 6 10 2 f 1+ 7 10 2f = 2.25 106 Hz 6 6 2.25 106 1 2.25 10 1 2.25 10 tan tan 6 7 284 10 10 = tan 1 = 89.99 66.04 12.68 = 168.7 Phase Margin = 180 168.7 = 11.3 12.80 a. T( f ) =T (0) = 100 dB T (0) = 105 105 f f f 1+ j 1 + j 1 + j 6 10 5 10 10 106 T =1= = 105 1 f 1+ 10 21 f 1+ 6 5 10 21f 1+ 6 10 10 2www.elsolucionario.net 585. By trial and error f = 0.976 MHz 0.976 106 1 0.976 1 0.976 tan tan 10 5 10 = tan 1 = 90 11.05 5.574 = 106.6 Phase Margin = 180 106.6 = 73.4b.f P1 1 10 75 so = CF f P1 20or f P1 = 2.67 Hz Now T =1= = 105 1 f 1+ 2.67 21 f 1+ 6 5 10 21 2f 1+ 6 10 10 By trial and error f 2.66 105 Hz then 2.66 105 1 0.266 1 0.266 = tan 1 tan tan 5 10 2.67 = 90 3.045 1.524 = 94.57 Phase Margin = 180 94.57 = 85.412.81 1 where = ( Ro1 Ri 2 ) Ci 2 = (500 1000) 103 2 1012 = 6.67 107 s(a)f 3 dB =Then 1 f 3 dB = 239 kHz 2 (6.67 107 ) (b) For 1 1 f PD = 10 Hz, = = = 0.0159 s 2 f PD 2 (10)f 3 dB =Then = ( Ro1 Ri 2 ) ( Ci + CM )0.0159 = ( 500 1000 ) 103 ( Ci + CM )or ( Ci + CM ) = 4.77 108 = 2 x1012 + CM CM = 0.0477 F 12.82 Assuming a phase margin of 45.www.elsolucionario.net 586. f f 135 90 tan 1 tan 1 6 6 2 10 25 10 By trial and error, f 1.74 MHz Then T =1 1= 5000 2 1.74 106 1+ f PD 1 1 2 2 1.74 1.74 1+ 1+ 2 25 or 1.74 106 5000 f PD (1.325)(1.0024) so f PD = 462 Hz 12.83 20 =105 = 0.04999 1 + (105 ) Phase Margin = 60 = 120 f120 = tan 1Assume tan 1f PD tan 1f f tan 1 6 10 5 107f = 90 f PDf 30 f = 5.77 105 Hz 106 (0.04999)(105 ) T = 1= 2 2 2 5.77 105 5.77 105 5.77 105 1+ 1+ 1+ 6 7 f PD 10 5 10 tan 11=5.77 105 f PD4.999 103 2 5.77 105 1+ (1.155)(1.0) f PD 4.999 103 (1.155)(1.0)f PD = 133.3 Hzwww.elsolucionario.net 587. Chapter 13 Exercise Solutions EX13.1 I C1 = I C 2 9.5 A 9.5 A I B1 = I B 2 = = 0.0475 A I B1 = I B 2 = 47.5 nA 200 EX13.2 5 0.6 0.6 (5) = 0.22 mA I REF = 40 I C17 = I C13 B = 0.75I REF = (0.75)(0.22) = 0.165 mA 0.165 (0.165)(0.1) + 0.6 + I C16 = 200 50 = 0.000825 + 0.01233 I C16 = 13.2 A EX13.3V 0.18 103 = 1014 exp D VT 0.18 103 VD = VT ln 14 10 0.18 103 = (0.026) ln 14 10 VD = 0.6140 VBB = 2VDD 1.228 V V /2 I C14 = I C 20 = I S exp BB VT 0.6140 = 3 1014 exp 0.026 I C14 = I C 20 = 0.541 mAEX13.4 100 ro 6 = 10.5 M 0.0095 Then, using results from Example 13.4www.elsolucionario.net 588. Ract1 = ro 6 [1 + g m 6 ( R2 r 6 ) ] = 10.5 [1 + (0.365)(1 547) ] = 14.3 M V 100 ro 4 = A = 10.5 M I C 4 0.0095 9.5 Ad = (10.5 14.3 4.07 ) = 889 0.026 EX13.5 100 = 556 K 0.18 Ri 3 = r 22 + (1 + P )[ R19 R20 ] = 7.22 + (51)(556 111) 4.73 M 100 Ract 2 = = 185 K 0.54 100 Ro17 = = 185 K 0.54 (200)(201)(50)(185 4730 185) Av 2 = 4070[50 + [9.63 + (201)(0.1)]] 182358786.9 = 32450.1 Av 2 = 562 R19 = ro13 A =EX13.6 100 100 ro17 = = 185 K ro13 B = = 185 K 0.54 0.54 RC17 = 185 [1 + (20.8)(0.1 9.63) ] RC17 = 566 K 7.22 + 566 185 = 2.88 K Re 22 = 51 100 = 556 K RC19 = 0.18 0.65 + 2.88 556 Re 20 = = 0.0689 51 = 68.9 RO = 22 + 68.9 = 90.9 EX13.7 Ci = C1 (1+ A2 ) = 30(1 + 562) = 16890 pF Ri 2 = 4.07 M Ro1 = Ract1 ro 4 = 14.3 10.5 = 6.05 M Then Req = Ro1 Ri 2 = 6.05 4.07 = 2.43 MThen f PD =1 1 = 6 2 Req Ci 2 (2.43 10 )(16890 1012 )= 3.88 Hzwww.elsolucionario.net 589. EX13.8 For M 5 , M 6 40 K P = (12.5) = 250 A / V 2 2 5 + 5 VSG 5 0.25(VSG 5 0.5) 2 = 225 2 56.25(VSG 5 VSG 5 + 0.25) = 10 VSG 5 2 56.25VSG 5 55.25VSG 5 + 4.0625 = 055.25 3052.5625 914.0625 2(56.25) = 0.902 VVSG 5 = VSG 510 0.902 40.4 A 225 Current in M 1 M 4 = 20.2 A I set = I Q =EX13.9 40 K P1 = K P 2 = (12.5) = 250 A/V 2 2 From Exercise Ex 13.8, I D1 = I D 2 = 20.2 A ro 4 = ro 2 =1 1 = = 2.48 M I D (0.02)(0.0202)Ad = 2 K P1 I Q (ro 2 ro 4 ) = 2(0.25)(0.0404)(2480 2480) Ad = 176 80 K n 7 = (12.5) = 500 A / V 2 2 g m 7 = 2 K n 7 I Q = 2 (0.50)(0.0404) = 0.284 mA/V ro 7 = ro8 =1 ID7=1 1.24 M (0.02)(0.0404)Av 2 = g m 7 (ro 7 ro8 ) = (0.284)(1240 1240) = 176 AV = Ad AV 2 = (176)(176) = 30,976EX13.10 (a) k W I Q1 g m1 = 2 n 2 L 1 2 0.08 0.25 =2 (20) 2 2 = 0.6325 mA/V1 = 800 K (0.01)(0.25/ 2) 1 ro 4 = = 533.3 K (0.015)(0.25/ 2) ro 2 =Ad 1 = g m1 (ro 2 ro 4 ) = (0.6325)(800 533.3) Ad 1 = 202.4www.elsolucionario.net 590. k W g m5 = 2 n IQ 2 2 L 5 0.04 =2 (80)(0.25) 2 = 1.265 mA/V1 = 266.7 K (0.015)(0.25) 1 ro9 = = 400 K (0.01)(0.25) ro 5 =A2 = g m 5 (ro 5 ro 9 ) = (1, 265)(266.7 400) A2 = 202.4Overall gain (assuming A3 = 1) A = Ad 1 A2 = (202.4)(202.4) = 40,966 EX13.11 I D1 = I D 2 = 25 A g m1 = g m8 = 2k W p 2 L 40 I DQ = 2 (25)(25) g m1 = g m8 = 224 A / V 2 k W gm6 = 2 n 2 L 80 I DQ = 2 (25)(25) g m 6 = 316 A / V 2 1 1 = = 2 M ro1 = ro 6 = ro8 = ro10 = I D (0.02)(25)ro 4 =1 ID4=1 =1 M (0.02)(50)Ro8 = g m8 (ro8 ro10 ) = (224)(2)(2) = 896 M Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 ) = 316(2) (1 2 ) = 421 M Then Ad = g m1 ( Ro 6 Ro8 ) = 224 ( 421 896 ) Ad = 64,158 EX13.12 V + V = VEB1 + VEB 6 + VBE 7 + I1 R1 = 0.6 + 0.6 + 0.6 + (0.236)(8) = 3.69 V So V + = V = 1.845 V EX13.13 Ad = 2 K P I Q 5 ( Ri 2 ) = 2(1)(0.2)(26) Ad = 16.4EX13.14www.elsolucionario.net 591. r 13 = VT I C13=(200)(0.026) 0.20= 26 k Ri 2 = r 13 + (1 + ) RE13 = 26 + 201(1) = 227 k r010 = r012 = g m12 = r 12 =1 I D10=1 = 500 k (0.02)(0.1)VA 50 = = 500 k I C12 0.1 I C12 0.1 = = 3.85 mA / V VT 0.026 VT I C12=(200)(0.026) = 52 k 0.1Ract1 = r012 [1 + g m12 (r 12 R5 )] = 500[1 + (3.85)(52 0.5)] = 1453 k Ad = 2 K n I Q 5 ( ro10 Ract1 R12 ) = 2(0.6)(0.2) (500 1453 227) = (0.490)(141) Ad = 69.1EX13.15 From Example 13.15, f PD = 265 Hz Av = Adi A2 = (16.4)(1923) = 31,537 fT = f PD Av = (265)(31,537) = 8.36 MHzTYU13.1 Computer Analysis TYU13.2 Computer Analysis TYU13.3 Vi N (max) = V + VEB (on) = 15 0.6 = 14.4 V Vi N (min) 4VBE (on) + V + = 4(0.6) 15 = 12.6 V 12.6 Vi N (cm) 14.4 V TYU13.4www.elsolucionario.net 592. V0 (max) V + 2VBE (on) = 15 2(0.6)a.V0 (max) = 13.8 V V0 (min)= 3VBE (on) + V = 3(0.6) 15V0 (min) 13.2 V 13.2 V0 13.8 V V0 (max) = 5 1.2 = 3.8 Vb.V0 (min) 3VBE + V = 3(0.6) 5 = 3.2 V 3.2 V0 3.8 VTYU13.5 15 2(0.6) (15) = 0.72 mA I REF 40 I 0.72 103 VBE = VT ln REF = (0.026) ln 14 10 IS = 0.650 VSo I REF =30 2(0.65) I REF = 0.718 mA 40 VBE11 = 0.650 VI I C10 R4 = VT ln REF I C10 0.718 I C10 (5) = (0.026) ln I C10 By trial and error: I C10 = 18.9 AVBE10 = VBE11 I C10 R4 = 0.650 (0.0189)(5) VBE10 0.556 V I C10 18.9 = = 9.45 A 2 2 I 9.45 106 = VT ln C 6 = (0.026)ln VBE 6 = 0.537 V 14 10 IS IC 6 VBE 6TYU13.6 10 0.6 0.6 (10) I REF = I REF = 0.47 mA 40 I I C10 R4 = VT ln REF I C10 0.47 I C10 (5) = (0.026) ln I C10 By trial and error: I C10 17.2 A IC 6 I C13 BI C10 I C 6 = 8.6 A 2 = (0.75) I REF I C13 B = 0.353 mAI C13 A = (0.25) I REF I C13 A = 0.118 mAwww.elsolucionario.net 593. TYU13.7R0 = R6 + RE14 RE14 =r 14 + R0 d R013 A1 + n The diode resistance can be found as V I D = I S exp D VT 1 1 I D = = IS rd VD VT VD exp VT ID = VTor rd =VT V 0.026 = T = 144 I D I C13 A 0.18RE 22 =r 22 + R017 R013 B 1+ PR013 B = r013 B = 92.6 k R017 = r017 1 + g m17 ( R8 r 17 ) = 283 k From previous calculations RE 22 = 1.51 k R0 d = 2rd + RE 22 = 2(0.144) + 1.51 = 1.80 k R013 A = r013 A = 278 k V (200)(0.026) r 14 = n T = = 1.04 k I C14 5 1.04 + 1.8 278 14.1 201 R0 = R6 + RE14 = 27 + 14.1 R0 41 RE14 =www.elsolucionario.net 594. TYU13.8 For M 6 we have VSG 5 = VSG 6 = 1.06 V So VSD 6 (sat) = 1.06 0.5 = 0.56 V For M 1 and M 2 ID =IQ= K P (VSG1 + VTP )22 0.0397 = 0.125(VSG1 0.5) 2 VSG1 = 0.898 V 2 So maximum input voltage = V + VSD 6 (sat) VSG1 = 5 0.56 0.898 Vi N (max) = 3.54 VFor M 3, K p = (6.25)(20) = 125 A / V 2 I D3 =IQ=39.7 A 22 39.7 = 125(VGS 3 VTN ) 2 2 VTN = 0.5V VGS 3 = 0.898 V VSD1 (sat) = 0.898 0.5 = 0.398 V Vi N (min) = V + VGS 3 + VSD1 (sat) VSG1 = 5 + 0.898 + 0.398 0.898 Vi N (min) = 4.60 V4.60 Vi N (cm) 3.54 V TYU13.9 V0 (max) = V + VSD 8 (sat) VSG 8 = VSG 5 = 1.06 V VSD 8 (sat) = 1.06 0.5 = 0.56 V V0 (max) = 5 0.56 = 4.44 V V0 (min) = V + VDS 7 (sat) VGS 7 = 1.06 VDS 7 (sat) = 1.06 0.5 = 0.56 V0 (min) = 5 + 0.56 = 4.44 4.44 V0 4.44 VTYU13.10 (a) For M 5, K p 5 = 125 A / V 2www.elsolucionario.net 595. V + V VSG 5 RsetK p 5 (VSG 5 + VTP ) 2 =5 + 5 VSG 5 100 2 12.5(VSG 5 VSG 5 + 0.25) = 10 VSG 5 0.125(VSG 5 0.5) 2 =2 12.5VSG 5 11.5VSG 5 6.875 = 011.5 (11.5) 2 + 4(12.5)(6.875) 2(12.5) = 1.33 VVSG 5 = VSG 5 Then10 1.33 86.7 A 100 IQ = ID4 = = 43.35 A 2I REF = I Q = I D 8 = I D 7 = I D1 = I D 2 = I D 3K p1 = K p 2 = 125 A / V 2(b)ro 2 = ro 4 =I ID=1 = 1153 k (0.02)(0.04335)Input stage gain Ad = 2 K p1 I Q (ro 2 ro 4 ) = 2(0.125)(0.0867) (1153 1153) Ad = 84.9Transconductance of M 7 g m 7 = 2 K n 7 I D 7 = 2 (0.250)(0.0867) = 0.294 mA / V ro 7 = ro8 =1 I D7=1 = 577 k (0.02)(0.0867)Second stage gain Av 2 = g m 7 (ro 7 ro8 ) = (0.294)(577 577) Av 2 = 84.8 Overall gain = Ad Av 2 = (84.9)(84.8) A = 7, 200TYU13.11 (a) Ad = Bg m1 ( ro 6 ro8 ) k W g m1 = 2 n 2 L g m1 = 400 A / Vro 6 = ro8 =1P I D 6= 80 I D1 = 2 (20)(50) 21 = 0.333 M (0.02)(150)1 1 = = 0.333 M n I D 8 (0.02)(150)Ad = 3(400) ( 0.333 0.333) Ad = 200(b)f PD =1 2 Ro (CL + CP )where Ro = ro 6 ro8 = 0.333 0.333 M www.elsolucionario.net 596. f PD =1 f PD = 477 kHz 2 ( 0.333 0.333) 106 2 1012f PD Ad = (477 103 )(200) 95.5 MHzTYU13.12 (a) From Exercise TYU 13.11, g m1 = 400 A / V ro 6 = ro8 = ro10 = ro12 = 0.333 M k g m10 = 2 P 2W 40 L I D10 = 2 2 (20)(150) g m10 = 490 A / V k W 80 g m12 = 2 n I D12 = 2 (20)(150) g m12 = 693 A / V 2 2 L Ro10 = g m10 (ro10 ro 6 ) = (490)(0.333)(0.333) = 54.4 M Ro12 = g m12 (ro12 ro8 ) = (693)(0.333)(0.333) = 77.0 M Ad = Bg m1 ( Ro10 Ro12 ) = 3(400)(54.4 77.0) Ad = 38, 254 Ro = Ro10 Ro12 = 54.4 77.0 = 31.9 M (b)1 = 2.50 kHz 2 (31.9 106 )(2 1012 ) f PD Ad = (2.5 103 )(38, 254) 95.6 MHz f PD =TYU13.13 (a) Ad = g m1 ( Ro 6 Ro8 )From Example 13.10, g m1 = 316 A / V , Ro8 = 316 M Now Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) ro1 = 1 M , ro 4 = 0.5 M I 50 gm6 = C 6 = 1.923 mA / V VT 0.026 ro 6 =VA6 80 = = 1.6 M I C 6 50Then Ro 6 = (1.923)(1600)(0.5 1) = 1026 M Ad = (316)(1026 316) Ad = 76,343 (b)1 f PD = 329 Hz 2 (316 1026) 106 2 1012 f PD Ad = (329)(76,343) 25.1 MHz f PD =TYU13.14 For Q7 and R1 VSG = VBE 7 + I1 R1 = 0.6 + I1 (5) For M 8 : I 2 = K p (VSG + VTP ) 2 I 2 = 0.3(VSG 1.4) 2 By trial and error:www.elsolucionario.net 597. VSG = 2.54 V I1 = I 2 = 0.388 mATYU13.15 For J 6 biased in the saturation region I C 3 = I DSS = 300 A Q1 , Q2 , Q3 are matched I C1 = I C 2 = I C 3 = 300 Awww.elsolucionario.net 598. Chapter 13 Problem Solutions 13.1Computer Simulation13.2Computer Simulation13.3(Ad = g m1 ro 2 ro 4 Ri 6(a))g m1 =I C1 20 = 0.769 mA / V VT 0.026ro 2 =VA 2 80 = = 4 M I C 2 20ro 4 =VA 4 80 = = 4 M I C 2 20Ri 6 = r 6 + (1 + n ) R1 r 7 (120)(0.026) r 7 = = 15.6 k 0.2 V (on) 0.6 = = 0.030 mA I C 6 BE 20 R1 (120)(0.026) = 104 k 0.030r 6 =Then Ri 6 = 104 + (121) 20 15.6 1.16 M Then()Ad = 769 4 4 1.16 Ad = 565Now R1 Vo = I c 7 ro 7 = ( n I b 7 )ro 7 = n ro 7 R1 + r 7 Ic6 R1 Vo1 = n (1 + n )ro 7 I b 6 and I b 6 = R1 + r 7 Ri 6 Then n (1 + n )ro 7 R1 V Av 2 = o = Vo1 Ri 6 R1 + r 7 ro 7 =VA 80 = = 400 k I C 7 0.2So Av 2 =(120)(121)(400) 20 Av 2 = 2813 1160 20 + 15.6 Overall gain = Ad Av 2 = (565)(2813) A = 1.59 106(b)Rid = 2r 1 and r 1 =(80)(0.026) = 104 k 0.020Rid = 208 k (c)f PD =1 and CM = (10)(1 + 2813) = 28,140 pF 2 Req CMReq = ro 2 ro 4 Ri 6 = 4 4 1.16 = 0.734 M f PD =1 = 7.71 Hz 2 (0.734 10 )(28,140 1012 ) 6www.elsolucionario.net 599. Gain-Bandwidth Product = (7.71)(1.59 106 ) 12.3 MHz13.4 a. b.Q3 acts as the protection device. Same as part (a).13.5 If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5 So breakdown voltage 56.4 V. 13.6 (a)I REF =15 0.6 0.6 (15) = 0.50 R5 = 57.6 k R5I I C10 R4 = VT ln REF I C10 0.026 0.50 R4 = ln R4 = 2.44 k 0.030 0.030 5 0.6 0.6 (5) I REF = 0.153 mA 57.6 0.153 I C10 (2.44) = (0.026) ln I C10 (b)I REF =By trial and error, I C10 21.1 A13.7 (a)I REF 0.50 mAI 0.50 103 VBE = VT ln REF = (0.026) ln VBE11 = 0.641V = VEB12 14 10 IS Then 15 0.641 0.641 (15) R5 = 57.4 k R5 = 0.50 0.026 0.50 ln R4 = R4 = 2.44 k 0.030 0.030 0.030 103 VBE10 = 0.026 ln VBE10 = 0.567 V 14 10 (b)From Problem 13.6, I REF 0.15 mA 0.15 103 VBE11 = VEB12 = 0.026 ln = 0.609 V 14 10 5 0.609 0.609 (5) I REF = 0.153 mA Then I REF = 57.4 Then I C10 21.1 A from Problem 13.613.8 5 0.6 0.6 (5) I REF = 0.22 mA 40 I I C10 R4 = VT ln REF I C10 a.I REF = 0.22 I C10 (5) = (0.026) ln I C10 www.elsolucionario.net 600. By trial and error; I C10 14.2 A I C10 I C 6 = 7.10 A 2 = 0.75 I REF I C17 = 0.165 mAIC 6 I C17I C13 A = 0.25I REF I C13 A = 0.055 mA(b) Using Example 13.4 r 17 = 31.5 k RE = 50 [31.5 + (201)(0.1)] = 50 51.6 = 25.4 k r 16 = nVT I C16and0.165 (0.165)(0.1) + 0.6 + = 0.0132 mA 200 50 r 16 = 394 k Then Ri 2 = 394 + (201)(25.4) 5.5 M r 6 = 732 k 0.00710 gm6 = = 0.273 mA / V 0.026 50 r06 = = 7.04 M 0.0071 Then Ract1 = 7.04[1 + (0.273)(1 732)] = 8.96 M 50 r04 = = 7.04 M 0.0071 Then 7.1 Ad = (7.04 8.96 5.5) 0.026 or Ad = 627 Gain of differential amp stage I C16 =Using Example 13.5, and neglecting the input resistance to the output stage: V 50 Ract 2 = A = = 303 k I C13 B 0.165 Av 2 =(200)(201)(50)(303) (303)(5500)[50 + 31.5 + (201)(0.1)] or Av 2 = 545 Gain of second stage13.9 I C10 = 19 A From Equation (13.6) 2 + 2 P + 2 (10) 2 + 2(10) + 2 = 2I I C10 = 2 I P 2 2 (10) + 3(10) + 2 P + 3 P + 2 122 = 2I 132 So 132 2 I = (19) = 20.56 A 122 I C 2 = I = 10.28 Awww.elsolucionario.net 601. IC 9 =I B92I=20.56 I C 9 = 17.13 A 2 1+ 10 2 1 + P I 17.13 = C9 = I B 9 = 1.713 A P 10IB4 =10.28 I = I B 4 = 0.9345 A (1 + P ) 11 IC 4 = I P 1+ P 10 = (10.28) I C 4 = 9.345 A 11 13.10 VB 5 V = VBE (on) + I C 5 (1) = 0.6 + (0.0095)(1) = 0.6095 0.6095 IC 7 = I C 7 = 12.2 A 50 I C 8 = I C 9 = 19 A I REF = 0.72 mA I E13 = I REF = 0.72 mA I C14 = 138 A Power = (V + V ) [ I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 ] = 30[0.0122 + 0.019 + 0.019 + 0.72 + 0.72 + 0.138] Power = 48.8 mW Current supplied by V + and V = I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 = 1.63 mA13.11 (a)vcm (min) = 15 + 0.6 + 0.6 + 0.6 + 0.6 = 12.6 Vvcm (max) = +15 .6 = 14.4 V So 12.6 vcm 14.4 V(b)vcm (min) = 5 + 4(0.6) = 2.6 Vvcm (max) = 5 0.6 = 4.4 V So 2.6 vcm 4.4 V 13.12 If v0 = V = 15 V , the base voltage of Q14 is pulled low, and Q18 and Q19 are effectively cut off. As a first approximation 0.6 I C14 = = 22.2 mA 0.027 22.2 I B14 = = 0.111 mA 200 Then I C15 = I C13 A I B14 = 0.18 0.111 = 0.069 mA Now I VBE15 = VT ln C15 15 0.069 103 = (0.026) ln 14 10 = 0.589 Vwww.elsolucionario.net 602. As a second approximation 0.589 I C14 = I C14 = 21.8 mA 0.027 21.8 I B14 = = 0.109 mA 200 and I C15 = 0.18 0.109 I C15 = 0.071 mA 13.13 a. Neglecting base currents: I D = I BIAS Then I VBB = 2VD = 2VT ln D IS 0.25 103 = 2(0.026) ln 14 2 10 or VBB = 1.2089 V V / 2 I CN = I CP = I S exp BB VT 1.2089 = 5 1014 exp 2(0.026) So I CN = I CP = 0.625 mAFor vI = 5 V, v0 5 Vb.5 = 1.25 mA 4 As a first approximation I CN iL = 1.25 mA iL 1.25 103 VBEN = (0.026) ln = 0.6225 V 14 5 10 Neglecting base currents, VBB = 1.2089 VThen VEBP = 1.2089 0.6225 = 0.5864 V 0.5864 I CP = 5 1014 exp I CP = 0.312 mA 0.026 As a second approximation, I CN = iL + I CP = 1.25 + 0.31 I CN 1.56 mA13.14 R1 + R2 =VBB 1.157 = = 64.28 k (0.1) I BIAS 0.018I VBE = VT ln C IS (0.9) I BIAS = (0.026) ln IS 0.162 103 = (0.026) ln 14 10 www.elsolucionario.net 603. VBE = 0.6112 V R2 VBE = VBB R1 + R2 R 0.6112 = 2 (1.157) 64.28 So R2 = 33.96 kThen R1 = 30.32 k 13.15(Ad = g m ro 4 ro 6 Ri 2(a))From example 13.4 9.5 gm = = 365 A / V , ro 4 = 5.26 M 0.026 Now ro 6 = ro 4 = 5.26 M Assuming R8 = 0, we find Ri 2 = r 16 + (1 + n ) RE = 329 + (201) ( 50 9.63) 1.95 M Then()Ad = (365) 5.26 5.26 1.95 Ad = 409(b) Av 2 =From Equation (13.20),( n (1 + n ) R9 Ract 2 Ri 3 R017{})Ri 2 R9 + r 17 + (1 + n ) Rg For Rg = 0, Ri 2 = 1.95 M Using the results of Example 13.5 Av 2 =(200(201)(50) 92.6 4050 92.6 (1950){50 + 9.63})Av2= 79213.16 Let I C10 = 40 A, then I C1 = I C 2 = 20 A. Using Example 13.5, Ri 2 = 4.07 M (200)(0.026) = 260 k 0.020 0.020 gm6 = = 0.769 mA/V 0.026 50 r06 = 2.5 M 0.02 Then Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 M 50 2.5 M r06 = 0.02 Then r 6 =www.elsolucionario.net 604. I CQ Ad = (r04 Ract1 Ri 2 ) VT 20 = (2.5 4.42 4.07) 0.026 So Ad = 88213.17 From Problem 13.8 I1 = I 2 = 7.10 A, I C17 = 0.165 mA, I C13 A = 0.055 mA I C16 I B17 +I E17 R8 + VBE17 R9=0.165 (0.165)(0.1) + 0.6 + 200 50= 0.000825 + 0.01233I C16 = 0.0132 mA (200)(0.026) = 31.5 K 0.165 1 RE = R9 [ r 17 + (1 + ) R8 ] = 50 [31.5 + (201)(0.1)]r 17 50 51.6 = 25.4 K r 16 =(200)(0.026) = 394 K 0.0132Then 1 Ri 2 = r 16 + (1 + ) RE = 394 + (201)(25.4) 5.50 M Now (200)(0.026) r 6 = = 732 K 0.0071 0.0071 gm6 = = 0.273 mA/V 0.026 50 ro 6 = 7.04 M 0.0071 Ract1 = ro 6 [1 + g m 6 ( R r 6 )] = 7.04[1 + (0.273)(1 732)] = 8.96 M 50 ro 4 = 7.04 M 0.0071 Then Ad = g m1 (ro 4 Ract1 Ri 2 ) 7.10 = (7.04 8.96 5.5) 0.026 Ad = 62750 50 303 K Ro17 = = 303 K 0.165 0.165 From Eq. (13.20), assuming Ri 3 Now Ract 2 =Av 2 = (1 + ) R9 ( Ract 2 R017 ) Ri 2 { R9 + [r 17 + (1 + ) R8 ]} (200)(201)(50)(303 303)(5500)[50 + 31.5 + (201)(0.1)] Av 2 = 545=3.045 108 5.588 105Overall gain Av = (627)(545) = 341, 715www.elsolucionario.net 605. 13.18 Using results from 13.17 100 Ri 2 = 5.50 M, Ract1 [1 + (0.273)(1 732)] 17.93 M 0.0071 100 14.08 M ro 4 = 0.0071 7.10 Ad = (14.08 17.93 5.50) 0.026 Ad = 885 Now 100 100 Ract 2 = = 606 K Ro17 = = 606 K 0.165 0.165 (200)(201)(50)(606 606) 6.09 108 Av 2 = = (5500)[50 + 31.5 + (201)(0.1)] 5.588 105 Av 2 = 1090 Overall gain Av = (885)(1090) = 964, 650 13.19 Now Re14 =r 14 + R01 and R0 = R6 + Re14 1+ PAssume series resistance of Q18 and Q19 is small. Then R01 = r013 A Re 22where Re 22 =r 22 + R017 r013 B 1+ Pand R017 = r017 [1 + g m17 ( R8 r 17 )] Using results from Example 13.6, r 17 = 9.63 k r 22 = 7.22 k g m17 = 20.8 mA/V r017 = 92.6 k Then R017 = 92.6[1 + (20.8)(0.1 9.63)] = 283 k 50 r013 B = = 92.6 k 0.54 Then 7.22 + 283 92.6 = 1.51 k Re 22 = 51 R01 = r013 A Re 22 = 278 1.51 = 1.50 k (50)(0.026) = 0.65 k r 14 = 2 Then 0.65 + 1.50 Re14 = = 0.0422 k 51 or Re14 = 42.2 Then R0 = 42.2 + 27 R0 = 69.2 13.20www.elsolucionario.net 606. r Rid = 2 r 1 + (1 + n ) 3 1+ P n = 200, P = 10 (a) I C1 = 9.5 A (200)(0.026) = 547 K 0.0095 (10)(0.026) r 3 = = 27.4 K 0.0095 Then (201)(27.4) Rid = 2 547 + 11 Rid 2.095 M r 1 =(b) I C1 = 7.10 A (200)(0.026) r 1 = = 732 K 0.0071 (10)(0.026) r 3 = = 36.6 K 0.0071 (201)(36.6) Rid = 2 732 + 11 Rid 2.80 M 13.21 We can write A0 f f 1 + j 1 + j f PD f1 181, 260 = f f 1 + j 1 + j 10.7 f1 Phase: f 1 f = tan 1 tan 10.7 f1 A( f ) =For a Phase margin = 70, = 110 So f 1 f 110 = tan 1 tan 10.7 f1 Assuming f10.7, we have f f tan 1 = 20 = 0.364 f1 f1 At this frequency, A( f ) = 1, sowww.elsolucionario.net 607. 1181, 260 2 f 2 1+ 1 + (0.364) 10.7 170,327 2 f 1+ 10.7 f = 170,327 f = 1.82 MHz or 10.7 Then, second pole at f f1 = f1 = 5 MHz 0.36413.22 a.Original g m1 and g m 2W C K p1 = K p 2 = P ox = (12.5)(10) L 2 = 125 A / V 2 So IQ g m1 = g m 2 = 2 K p1 2= 2 (0.125)(10) = 0.09975 mA/VW If is increased to 50, then L K p1 = K p 2 = (50)(10) = 500 A / V 2So g m1 = g m 2 = 2 (0.5)(0.0199) = 0.1995 mA/V b. Gain of first stage Ad = g m1 (r02 r04 ) = (0.1995)(5025 5025)or Ad = 501 Voltage gain of second stage remains the same, or Av 2 = 251 Then Av = Ad Av 2 = (501)(251) or Ad = 125, 751 13.24 a.K p = (10)(20) = 200 A / V 2 = 0.2 mA / V 2I REF = I SET10 VSG (10) 200 = k P (VSG 1.5) 2 =2 20 VSG = (0.2)(200)(VSG 3VSG + 2.25) 2 40VSG 119VSG + 70 = 0VSG =119 (119) 2 4(40)(70) VSG = 2.17 V 2(40)Thenwww.elsolucionario.net 608. 20 2.17 I REF = 89.2 A 200 M 5 , M 6 , M 8 matched transistors so that I REF =I Q = I D 7 = I REF = 89.2 Ab. Small-signal voltage gain of input stage: Ad = 2 K p1 I Q ( ro 2 ro 4 ) 1 = 1.12 M 89.2 (0.02) 2 1 1 r04 = = = 2.24 M n I D 89.2 (0.01) 2 Then Ad = 2(200)(89.2) (1.12 2.24) r02 =1 = P I Dor Ad = 141 Small-signal voltage gain of second stage: Av 2 = g m 7 (r07 r08 ) K n 7 = (20)(20) = 400 A / V 2 So g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.0892) = 0.378 mA/V 1 1 r08 = = = 561 k P I D 7 (0.02)(0.0892) r07 =1n I D 7=1 = 1121 k (0.01)(0.0892)So Av 2 = (0.378)(1121 561) Av 2 = 141 Then overall voltage gain Av = Ad Av 2 = (141)(141) Av = 19,881 13.25 Small-signal voltage gain of input stage: Ad = 2 K p1 I Q ( ro 2 ro 4 ) K p1 = (10)(10) = 100 A / V 2 1 1 = = 1000 k IQ 0.2 P (0.01) 2 2 1 1 r04 = = = 2000 k IQ 0.2 (0.005) n 2 2 r02 =Then Ad = 2(0.1)(0.2) (1000 2000) or Ad = 133 Small-signal voltage gain of second stage: Av 2 = g m 7 ( r07 r08 ) K n 7 = (20)(20) = 400 A / V 2 Sowww.elsolucionario.net 609. g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.2) = 0.566 mA/V 1 1 r08 = = = 500 k P I D 7 (0.01)(0.2) r07 =1n I D 7=1 = 1000 k (0.005)(0.2)So Av 2 = (0.566)(1000 500) Av 2 = 189 Then overall voltage gain is Av = Ad Av 2 = (133)(189) Av = 25,137 13.26 f PD =1 2 Req Ciwhere Req = r04 r02 and Ci = C1 (1 + Av 2 ) We can find that Av 2 = 251 and r04 = r02 = 5.025 M Now Req = 5.025 5.025 = 2.51 M and Ci = 12(1 + 251) = 3024 pF So 1 f PD = 2 (2.51 106 )(3024 1012 ) or f PD = 21.0 Hz 13.27 f PD =1 2 Req Ciwhere Req = r04 r02 From Problem 13.22, r02 = 1.12 M, r04 = 2.24 M and Av 2 = 141 So 1 8= 2 (1.12 2.24) 106 Ci or Ci = 2.66 108 = C1 (1 + Av 2 ) = C1 (142) or C1 = 188 pF 13.28 R0 = r07 r08 We can find that r07 = r08 = 2.52 M Then R0 = 2.52 2.52 or R0 = 1.26 Mwww.elsolucionario.net 610. 13.29 a.V0 = ( g m1Vgs1 )(r01 r02 ) VI = Vgs1 + V0Then V0 = g m1 (r01 r02 )(VI V0 ) or Av =g m1 (r01 r02 ) 1 + g m1 (r01 r02 )b.I X + g m1Vgs1 =VX VX and Vgs1 = VX + r02 r01R0 =1 r r g m1 01 0213.30 (a)(b)2 80 I Q 2 = (20) [1.1737 0.7 ] 2 I Q 2 = 180 A 2 80 I D 6 = (25) (VGS 6 0.7 ) = 25 VGS 6 = 0.8581 V 2 2 40 I D 7 = (50) (VSG 7 0.7 ) = 25 VSG 7 = 0.8581 V 2 Set VSG 8 P = VGS 8 N = 0.8581 V 40 W W 180 = (0.8581 0.7) 2 = 360 2 L 8 P L 8 P 80 W W 180 = (0.8581 0.7) 2 = 180 2 L 8 N L 8 N13.31www.elsolucionario.net 611. VGS11 2 80 200 = (20) (VGS 11 0.7 ) 2 VGS 11 = 1.20 VLet M 12 = 2 transistors in series. Than 5 1.20 = 1.90 V 2 2 80 W W W 200 = (1.90 0.7 ) = = 3.47 L 12 A L 12 B 2 L 12 VGS12 =13.32 (a) 2 80 I Q 2 = 250 A = (5) (VGS 8 0.7 ) 2 VGS 8 = 1.818 V 1.818 VGS 6 = VSG 7 = = 0.909 V 2 80 I D 6 = I D 7 = (25)(0.909 0.7) 2 = 43.7 A 2 (b) 80 250 g m1 = 2 (15) 0.5477 mA/V 2 2 1 ro 2 = = 800 K ( 0.01)( 0.125)r04 =1= 533.3K( 0.015)( 0.125) Ad 1 = g m1 ( ro 2 ro 4 ) = ( 0.5477 ) ( 800533.3)Ad 1 = 175 Second stage:www.elsolucionario.net 612. A2 = g m 5 (ro 5 ro 9 ) 40 g m 5 = 2 (80)(250) 1.265 mA/V 2 1 r05 = = 266.7 K (0.015)(0.25) 1 r09 = = 400 K (0.01)(0.25) A2 = (1.265)(266.7 400) A2 = 202 Assume the gain of the output stage 1, then Av = Ad 1 A2 = (175)(202) Av = 35,35013.33 (a)Ad = g m1 ( Ro 6 Ro8 )g m1 = 2 K n I DQ = 2 (0.5)(0.025) 224 A / V g m1 = g m8 g m 6 = 2 (0.5)(0.025) 224 A / V ro1 = ro 6 = ro8 = ro10 = ro 4 =1 1 = = 2.67 M I DQ (0.015)(25)1 1 = 1.33 M I D 4 ( 0.015 )( 50 )Now Ro8 = g m8 (ro8 ro10 ) = (224)(2.67)(2.67) = 1597 M Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) = (224)(2.67)(2.67 1.33) Ro 6 = 531 M Then Ad = (224)(531 1597) Ad = 89, 264 (b)Ro = Ro 6 Ro8 = 531 1597 Ro = 398 M (c)f PD =1 1 = f PD = 80 Hz 2 Ro CL 2 ( 398 106 )( 5 1012 )GBW = (89, 264)(80) GBW = 7.14 MHz13.34 (a) ro1 = ro8 = ro10 = ro 6 = ro 4 =1n I D=1 1 = = 2 M p I D (0.02)(25)1 = 2.67 M (0.015)(25)1 1 = = 1.33 M n I D 4 (0.015)(50) 35 W W g m1 = 2 (25) = 41.8 = g m8 2 L 1 L 1 80 W W g m 6 = 2 (25) = 63.2 2 L 6 L 6 Ro = Ro 6 Ro8 = [ g m 6 (ro 6 )(ro 4 ro1 )] [ g m8 (ro8 ro10 )]www.elsolucionario.net 613. W W Define X 1 = and X 6 = L 1 L 6 Then Ro = 63.2 X 6 ( 2.67 ) (1.33 2 ) 41.8 X 1 ( 2 )( 2 ) 22,539 X 1 X 6 = 134.8 X 6 167.2 X 1 = 134.8 X 6 + 167.2 X 1 Ad = g m1 Ro 22,539 X 1 X 6 = (41.8 X 1 ) 134.8 X 6 + 167.2 X 1 = 10, 0001 W W Now X 6 = = = 0.674 X 1 L 6 2.2 L 1 We then find W W X 12 = = 4.06 = L 1 L pand W = 1.85 L n 13.35 Let V + = 5V , V = 5V P = IT (10) = 3 IT = 0.3 mA I REF = 0.1 mA = 100 A 1 ro1 = ro8 = ro10 = = 1 M (0.02)(50) 1 ro 6 = = 1.33 M (0.015)(50) 1 ro 4 = = 0.667 M (0.015)(100) 35 W g m1 = 2 (50) = 59.2 X 1 = g m8 2 L 1 W where X 1 = L 1 Assume all width-to-length ratios are the same. 80 W g m 6 = 2 (50) = 89.4 X 1 2 L Now Ro = Ro 6 Ro8 = g m 6 ( ro 6 ) ( ro 4 ro1 ) g m8 ( ro8 ro10 ) = 89.4 X 1 (1.33) ( 0.667 1) 59.2 X 1 (1)(1) ( 47.6 X 1 )( 59.2 X 1 ) = [ 47.6 X 1 ] [59.2 X 1 ] = 47.6 X 1 + 59.2 X 1So Ro = 26.4 X 1 Now Ad = g m1 Ro = ( 59.2 X 1 )( 26.4 X 1 ) = 25, 000 So that X 12 =W = 16 for all transistors L13.36www.elsolucionario.net 614. (a)Ad = Bg m1 (ro 6 ro8 ) ro 6 = ro8 =1 1 = = 0.741 M I DQ (0.015)(90) k W g m1 = 2 n I D1 = 2 (500)(30) = 245 A / V 2 L Ad = (3)(245)(0.741 0.741) Ad = 272(b)Ro = ro 6 ro8 = 0.741 0.741 Ro = 371 k (c)f PD =1 1 = f PD = 85.8 kHz 2 Ro C 2 (371 103 )(5 1012 )GBW = (272)(85.8 103 ) GBW = 23.3 MHz13.37 (a)1 = 0.5 M (0.02)(2.5)(40) 1 ro8 = = 0.667 M (0.015)(2.5)(40) Ad = Bg m1 ( ro 6 ro8 )ro 6 =400 = (2.5) g m1 ( 0.5 0.667 ) g m1 = 560 A / V 80 W W g m1 = 560 = 2 (40) = 49 2 L L Assume all (W/L) ratios are the same except for W W M 5 and M 6 . = = 122.5 L 5 L 6(b) Assume the bias voltages are V + = 5V , V = 5V .W W Assume = = 49 L A L B 80 I Q = (49)(VGSA 0.5) 2 = 80 VGSA = 0.702 V 2 Then 80 W I REF = 80 = (VGSC 0.5) 2 2 L C For four transistorswww.elsolucionario.net 615. 10 0.702 = 2.325 V 4 80 W W 80 = (2.325 0.5) 2 = 0.60 2 L C L CVGSC =(c)f 3 dB =1 2 Ro CRo = 0.5 0.667 = 0.286 M 1 = 185 kHz 2 (286 103 )(3 1012 ) GBW = (400)(185 103 ) 74 MHz f 3 dB =13.38 (a) From previous results, we can write Ro10 = g m10 (ro10 ro 6 ) Ro12 = g m12 (ro12 ro8 ) Ad = Bg m1 ( Ro10 Ro12 ) Now ro10 = ro 6 =1 1 = = 0.5 M (0.02)(2.5)(40) P B ( I Q / 2 )ro12 = ro8 =1 1 = = 0.667 M (0.015)(2.5)(40) n B ( I Q / 2 )Assume all transistors have the same width-to-length ratios except for M 5 and M 6 . W Let L Then 2 = X k W 35 p g m10 = 2 ( I DQ10 ) = 2 X 2 (2.5)(40) 2 2 L 10 = 83.67 X k W 80 g m12 = 2 n ( I DQ12 ) = 2 X 2 (2.5)(40) 2 2 L 12 = 126.5 X 80 g m1 = 2 X 2 (40) = 80 X 2 Then Ro10 = (83.67 X )(0.5)(0.5) = 20.9 X M Ro12 = (126.5 X )(0.667)(0.667) = 56.3 X M We want 20, 000 = (2.5)(80 X )[20.9 X 56.3 X ] (20.9 X )(56.3 X ) 2 = 200 X = 3048 X 20.9 X + 56.3 X Then W X 2 = 6.56 = L Then W W = = (2.5)(6.56) = 16.4 L 6 L 5(b)Assume bias voltages are V + = 5V , V = 5Vwww.elsolucionario.net 616. W W Assume = = 6.56 L A L B 80 I Q = 80 = (6.56)(VGSA 0.5) 2 VGSA = 1.052 V 2 Need 5 transistors in series 10 1.052 VGSC = = 1.79 V 5 Then 80 W W I REF = 80 = (1.79 0.5) 2 = 1.20 2 L C L C(c)f 3 dB =1 where Ro = Ro10 Ro12 2 Ro CNow Ro10 = 20.9 6.56 = 53.5 M Ro12 = 56.3 6.56 = 144 M Then Ro = 53.5 144 = 39 M 1 = 1.36 kHz 2 (39 106 )(3 1012 ) GBW = (20, 000)(1.36 x103 ) GBW = 27.2 MHz f 3 dB =13.39 Ad = g m ( M 2 ) ro 2 ( M 2 ) ro 2 (Q2 ) 40 g m ( M 2 ) = 2 (25)(100) = 447 A / V 2 1 1 ro 2 ( M 2 ) = = = 500 k I DQ (0.02)(0.1) ro 2 (Q2 ) =VA 120 = = 1200 k I CQ 0.1Then Ad = 447(0.5 1.2) Ad = 158 13.40www.elsolucionario.net 617. Ad = g m ( M 2 ) ro 2 ( M 2 ) ro 2 (Q2 ) 80 g m ( M 2 ) = 2 (25)(100) = 632 A / V 2 1 1 ro 2 ( M 2 ) = = = 667 k I DQ (0.015)(0.1) VA 80 = = 800 k I CQ 0.1ro 2 (Q2 ) =Ad = (632) ( 0.667 0.80 ) Ad = 23013.41 (a)I REF = 200 A Ad = g m1 ( Ro 6 Ro8 )K n = K p = 0.5 mA / V 2n = p = 0.015 V 1where Ro8 = g m8 (ro8 ro10 ) Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 ) Now g m8 = 2 K P I D 8 = 2 (0.5)(0.1) = 0.447 mA/V 1 1 ro8 = = = 667 k P I D 8 (0.015)(0.1) ro10 =1 = 667 k P I D 8gm6 =IC 6 0.1 = = 3.846 mA/V 0.026 VTro 6 =VA 80 = = 800 k I C 6 0.1ro 4 =1 1 = = 333 k n I D 4 (0.015)(0.2)ro1 =1 p I D1=1 = 667 k (0.015)(0.1)g m1 = 2 K P I D1 = 2 (0.5)(0.1) = 0.447 mA/V So Ro8 = (0.447)(667)(667) 198.9 M Ro 6 = (3.846)(800)(333 667) 683.4 M Then Ad = 447(198.9 683.4) Ad = 68,865 13.42 Assume biased at V + = 10V , V = 10V . P = 3I REF (20) = 10 I REF = 167 A Ad = g m1 ( Ro 6 Ro8 ) = 25, 000 kn = 80 A / V 2 , k = 35 A / V 2 pn = 0.015V 1 , p = 0.02 V 1 W W Assume = 2.2 L p L nwww.elsolucionario.net 618. Ro8 = g m8 ( ro8 ro10 ) Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) ro8 = ro10 =1P I D 8 1P I D 8=1 = 0.60 M (0.02)(83.3)= 0.60 M k W 35 p g m8 = 2 I D 8 = 2 (2.2) X 2 (83.3) 2 2 L 8 = 113.3 X W where X 2 = L n VA 80 ro 6 = = = 0.960 M I C 6 83.3ro 4 = ro1 = gm6 =1n I D 4=1 = 0.40 M (0.015)(167)1 1 = = 0.60 M p I D1 (0.02)(83.3) IC 6 83.3 = = 3204 A / V 0.026 VT kp W 35 g m1 = 2 I D1 = 2 (2.2) X 2 (83.3) 2 2 L 1 = 113.3 X Now Ro 6 = (3204)(0.960) 0.40 0.60 = 738 M Ro8 = (113.3 X )(0.60)(0.60) = 40.8 X M Then Ad = 25, 000 = (113.3 X ) 738 40.8 X 30,110 X = (113.3 X ) 738 + 40.8 X which yields X = 2.48 or W X 2 = 6.16 = L nand W + (2.2)(6.16) = 12.3 L P 13.43 For vcm (max), assume VCB (Q5 ) = 0. Then VS = 15 0.6 0.6 = 13.8 V 0.236 I D 9 = I D10 = = 0.118 mA 2 Using parameters given in Example 13.11 I 0.118 VSG = D 9 VTP = + 1.4 = 2.17 V 0.20 KP Then vcm (max) = 13.8 2.17 vcm (max) = 11.6 Vwww.elsolucionario.net 619. For vcm (min) , assume VSD ( M 9 ) = VSD ( sat ) = VSG + VTP = 2.17 1.4 = 0.77 V Now VD10 = I D10 (0.5) + 0.6 + I D10 (0.5) 15 = 0.118 + 0.6 15 VD10 = 14.28 V Then vcm (min) = 14.28 + VSD (sat) VSG = 14.28 + 0.77 2.17 = 15.68 V Then, common-mode voltage range 15.68 vcm 11.6 Or, assuming the input is limited to 15 V, then 15 vcm 11.6 V 13.44 For I1 = I 2 = 300 A, VSG = VBE + (0.3)(8) = 0.6 + 2.4 = 3.0 V Then I 2 = K P (VSG + VTP ) 2 0.3 = K P (3 1.4)2 K P = 0.117 mA / V 2 13.45 For VCB = 0 for both Q6 and Q7 , then VS = 0.6 + 0.6 + VSG + (VS ) So 2VS = 1.2 + VSG Now I1 0.6 + I 2 R1 = VSG = + VTP and I1 = I 2 KP Also I1 = I 2 = K P (VSG + VTP ) 2 so 0.6 + (0.25)(8)(VSG 1.4) 2 = VSG 2 0.6 + 2(VSG 2.8VSG + 1.96) = VSG 2 2VSG 6.6VSG + 4.52 = 0 6.6 (6.6) 2 4(2)(4.52) = 2.33 V 2(2) Then 2VS = 1.2 + 2.33 = 3.53 and VS = 1.765 V VSG =13.46 I C 5 = I C 4 = 300 A Using the parameters from Examples 13.12 and 13.13, we have V (200)(0.026) Ri 2 = r 13 = n T = = 17.3 k 0.3 I C13 Ad = 2 K n I Q 5 ( Ri 2 ) = 2(0.6)(0.3) (17.3)or Ad = 10.38 Nowwww.elsolucionario.net 620. I C13 0.3 = = 11.5 mA/V 0.026 VTg m13 = r013 =VA 50 = = 167 k I C13 0.3Then Av 2 = g m13 r013 = (11.5)(167) or Av 2 = 1917 Overall gain: Av = (10.38)(1917) = 19,895 Assuming the resistances looking into Q4 and into the output stage are very large, we have13.47 Av 2 = R013 r 13 + (1 + ) RE13where R013 = r013 1 + g m13 ( RE13 r 13 ) I C13 = 300 A, r013 =50 = 167 k 0.30.3 = 11.5 mA / V 0.026 (200)(0.026) = = 17.3 k 0.3g m13 = r 13So R013 = (167) 1 + (11.5) (1 17.3) 1.98 M Then Av 2 =(200)(1980) = 1814 17.3 + (201)(1)Now Ci = C1 (1 + Av 2 ) = 12 [1 + 1814] Ci = 21, 780 pF 1 f PD = 2 Req Ci Req = Ri 2 r012 r010Neglecting R3 , r010 =1 1 = = 333 k I D10 (0.02)(0.15)Neglecting R5 , 50 r012 = = 333 k 0.15 Ri 2 = r 13 + (1 + ) RE13= 17.3 + (201)(1) = 218 kThen f PD =1 2 218 333 333 103 ( 21, 780 ) 1012 or f PD = 77.4 Hz Unity-Gain Bandwidth Gain of first stage:www.elsolucionario.net 621. Ad = 2 K n I Qs ( R12 ro12 ro10 ) = 2(0.6)(0.3) (218 333 333) = (0.6)(218 333 333)or Ad = 56.6 Overall gain: Av = (56.6)(1814) = 102, 672 Then unity-gain bandwidth = (77.4)(102, 672) 7.95 MHz 13.48 Since VGS = 0 in J 6 , I REF = I DSS I DSS = 0.8 mA 13.49 a.Ri 2 = r 5 + (1 + ) [ r 6 + (1 + ) RE ](100)(0.026) = 13 k 0.2 I 200 A C6 = = 2 A 100 r 6 = IC 5So r 5 =(100)(0.026) = 1300 k 0.002Then Ri 2 = 1300 + (101) [13 + (101)(0.3) ] or Ri 2 = 5.67 M Av = g m 2 ( r02 r04 Ri 2 )b. gm2 =2 I D I DSS VP=2 (0.1)(0.2) 3= 0.0943 mA / V r02 =1 1 = = 500 k I D (0.02)(0.1)r04 =VA 5.0 = = 500 k I C 4 0.1Then Av = (0.0943)[500 500 5670] or Av = 22.6 13.50 a.Need VSD (QE ) VSD ( sat ) = VP For minimum bias 3 VSet VP = 3 V and VZK = 3 V I REF 2 =VZK VD1 R33 0.6 R3 = 24 k 0.1 Set bias in QE = I REF 2 + I Z 2 = 0.1 + 0.1 = 0.2 mA Therefore,so that R3 =www.elsolucionario.net 622. I DSS = 0.2 mAb.Neglecting base currents 12 0.6 I 01 = I REF 1 = 0.5 mA = R4 so that R4 = 22.8 k13.51 a. gm2We have 2 = I D I DSS VP =2 (0.5)(1) 4= 0.354 mA/V r02 = r04 =1 ID=1 = 100 k (0.02)(0.5)VA 100 = = 200 k I D 0.50.5 = 19.23 mA/V 0.026 (200)(0.026) = 10.4 k r 4 = 0.5 So R04 = r04 1 + g m 4 ( r 4 R2 ) gm4 200 1 + (19.23) (10.4 0.5 ) = 2035 kAd = g m 2 ( r02 R04 RL )For RL Ad = 0.354 (100 2035 ) = 33.7 b.With these parameter values, gain can never reach 500. Similarly for this part, gain can never reach 700.www.elsolucionario.net 623. Chapter 14 Exercise Solutions EX14.1 a.ACL =b.dACL ACL ACL =50 ACL = 49.949 1 1 + 5 104 ( 51) dA 51 = 10 CL = 0.0102% ACL 5 10450 ACL = 49.943 51 1 + 4.5 104 EX14.2 a. For R0 = 0 1 1 1 = + (1 + 104 ) = 0.1 + 103 Ri f 10 10 Ri f = 103 k = 1 b.For R0 = 10 k1 1 1 1 + 104 + 1 104 = + 0.1 + Ri f 10 10 1 + 1 + 1 3 (10 ) Rif = 3 103 k Ri f = 3 EX14.3 40 40 (1 + 104 ) + 99 1 + 1 Ri f = 99 1+ 1 5 4 10 + 4.059 103 100 Ri f = 4.04 103 k Ri f = 4.04 MEX14.4 R 1 + 2 = 100 R1 a.1 1 105 = = 10 R0 f = 0.1 R0 f 100 100 b.1 1 105 2 = = 10 R0 f 10 100 R0 f = 102 k R0 f = 10 EX14.5 From Equation (14.43)www.elsolucionario.net 624. ACL ( f ) ACL 0 1+ j f f PD ( A0 /ACL 0 ) 25f 1+ j ( 50 ) (104 / 25)=25 1+ j f 2 104f = 2 kHza.v0 = 25 v0 ( peak ) = 1.25 mV vI f = 20 kHzb.v0 1 = 25 v0 ( peak ) = 0.884 mV vI 2 c. f = 100 kHz v0 25 25 = = = 4.90 2 vI 5.099 1 + (100 / 20 ) v0 = 0.245 mVEX14.6 Full-scale response = 1 5 = 5 V t=5 t = 2.5 s 2EX14.7 a.FPBW =SR 0.63 106 = 2 V0 ( max ) 2 (1)FPBW = 1.0 105 FPBW = 100 kHzb.FPBW =0.63 106 = 1.0 104 2 (10 ) FPBW = 10 kHzEX14.8 I 1.85 1014 V0 S = VT ln S 2 = (0.026) ln 14 2 10 I S1 V0 S = 2.03 mVEX14.9 We need iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V By Equation (14.60(a)) v 10 iC1 = I S 1 exp BE1 1 + VT 50 v 0.6 = I S 3 exp EB 3 1 + 50 VT By Equation (14.60(b))www.elsolucionario.net 625. v 10 iC 2 = I S 2 exp BE 2 1 + VT 50 v 0.6 = I S 4 exp EB 4 1 + 50 VT I S1 = I S 2, take the ratio: v v I exp BE1 BE 2 = S 3 VT IS 4 vBE1 vBE 2 = V0 SI = VT ln S 3 IS 4 = 0.026 ln (1.05 ) V0 S = 1.27 m VEX14.10 VOS = 0.020 =I Q K n 1 2 2Kn Kn 1 150 K n 2 2 ( 50 ) 50 K n = 1.63 A / V 2 K n 1.63 = 3.26% 50 KnEX14.11 R5 + Want V = 5 m V R5 + R4 R5R5 =R4 soR5 V + = 0.005 R4( 0.005)(100 )= 0.05 k 10 R5 = 50 EX14.12 R1 = 25 1 = 0.9615 k R2 = 75 1 = 0.9868 k For I Q = 100 A iC1 = iC 2 = 50 A From Equation (14.75) 50 106 ( 0.026 ) ln 14 + ( 0.050 )( 0.9615 ) 10 i = ( 0.026 ) ln C 2 + ( 0.050 )( 0.9868 ) IS 4 0.58065 + 0.048075 i = ( 0.026 ) ln C 2 + 0.04934 IS 4 www.elsolucionario.net 626. i ln C 2 = 22.284 IS 4 50 106 = 4.7625 109 IS 4 I S 4 1.05 1014 AEX14.13 From Equation (14.79) R v0 = I B1 R2 I B 2 R3 1 + 2 R1 For v0 = 0 100 0 = (1.1 106 ) (100 k ) (1.0 106 ) R3 1 + 10 R3 (11) = (1.1)(100 k ) R3 = 10 kTYU14.1 v1CM ( max ) = V + VSD1 ( sat ) VSG1 v1CM ( min ) = V + VDS 4 ( sat ) + VSD1 ( sat ) VSG1 We have: I REF = 100 A, kn = 80 A / V 2 , k = 40 A / V 2 , p W L For = 25 M1 :2 40 I D = 50 = ( 25 )(VSG1 + VTP ) 2 So 50 = 500 (VSG1 0.5 ) VSG1 = 0.816 V 2VSD1 ( sat ) = 0.816 0.5 = 0.316 VThen vCM ( max ) = V + 0.316 0.816 = V + 1.13 V For M 4 : 2 80 I D = 100 = ( 25 )(VGS 4 VTN ) 2So 100 = 1000 (VGS 4 0.5 ) VGS 4 = 0.816 V 2VDS 4 ( sat ) = 0.816 0.5 = 0.316 VVCM ( min ) = V + 0.316 + 0.316 0.816 = V 0.184 So V 0.184 vCM V + 1.13 VTYU14.2 vo ( max ) = V + VSD 8 ( sat ) VSG10 vo ( min ) = V + VDS 4 ( sat ) + VDS 6 ( sat ) Nowwww.elsolucionario.net 627. 50VSG 8 = VSG10 =+ 0.5 = 0.816 V( 40/ 2 )( 25) VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V So vo ( max ) = V + 0.316 0.816 = V + 1.13 Also VGS 6 =50 + 0.5 = 0.724 V (80/ 2 )( 25)VGS 4 =100 + 0.5 = 0.816 V (80/ 2 )( 25)VDS 6 ( sat ) = 0.724 0.5 = 0.224 V VDS 4 ( sat ) = 0.816 0.5 = 0.316 V So vo ( min ) = V + 0.316 + 0.224 = V + 0.54Then V + 0.54 vo V + 1.13 V TYU14.3 500 = 25 20 Within 0.1% 25 + ( 0.001)( 25 ) ACL = 24.975 ACL ( ideal ) = 24.975 =25 26 1 + A 0L 25 26 = 1 = 0.0010 A0 L 24.975 A0 L = 25.974TYU14.4ACL ( )a.ACL =b.dACL 100 = 10 5 = 0.01% 10 ACL A () 1 + CL A0 L R 495 ACL ( ) = 1 + 2 = 1 + = 100 R1 5 100 ACL = ACL = 99.90 100 1+ 5 10 ACL ( ) = 100ACL = 99.90 ( 0.0001)( 99.90 ) ACL = 99.89TYU14.5www.elsolucionario.net 628. ACL ( ) ACL= 1ACL ( )ACL = 1 ACL ( )1 A () 1 + CL A0 LACL ( ) ACL ( ) 1 A0 L A0 L So 0.001 = = A () A () 1 + CL 1 + CL A0 L A0 L 1+0.001 = 0.999 ACL ( ) =ACL ( ) A0 L0.001 0.001 A0 L = (104 ) ACL ( ) = 10.010 0.999 0.999or ACL = (1 0.001)(10.010 ) ACL = 10.0 TYU14.6 ii Rif = I1 Ri iI 0.1 a. = = 1 105 i1 104 iI 10 = = 1 103 i1 104b.TYU14.7 Voltage follower R2 = 0, R1 = Rif = Ri (1 + A0 L ) = 10 (1 + 5 105 ) 5 106 k Rif = 5000 MTYU14.8 f3dB =(105 ) (10 ) 20 kHz fT = 50 ACL 0f max = f 3dB = V0 ( max ) =SR 2 V0 ( max )SR 0.8 106 = 2 f3dB 2 ( 20 103 ) V0 ( max ) = 6.37 VTYU14.9 v0 = I B1 R3 = (106 )( 200 103 ) a. v0 = 0.20 Vb.R4 = R1 R2 R3 = 100 50 200 R4 = 28.6 kwww.elsolucionario.net 629. Chapter 14 Problem Solutions 14.1 Ad =vo = 80 vivo (max) = 4.5 vi (max) = 56.25 mVSo vi (max)rms =56.25 2= 39.77 mV14.2(a) 4.5 = 0.028125 mA 160 4.5 iL = = 4.5 mA 1 Output Circuit = 4.528 mA v 4.5 vi = o = vi = 0.05625 V A 80 (b) v 4.5 io 15 mA = o = RL RL i2 = RL (min) = 300 14.3 (1) (2)v2 = 12.5 mV(3) (4) (5)AOL = 2 104 v1 = 8 V AOL = 1000vo = 2 V14.4 From Eq. (14.4)www.elsolucionario.net 630. ACL =15 = R2 / R1 1 R2 1+ 1 + AoL R1 R2 / R1 1 R2 1+ 1 + R1 2 103 R 15 2 1 R1 = R2 15 1 + + R1 2 103 2 103 15.0075 =R2 (0.9925) R1R2 = 15.12 R114.5vI v1 v1 v0 v1 = + and v0 = A0 L v1 R1 R2 Ri so that v1 = v0 A0 L 1 vI v0 1 1 + = v1 + + R1 R2 R1 R2 Ri So 1 vI 1 1 1 1 = v0 + + + R1 R2 A0 L R1 R2 Ri Then v0 (1/ R1 ) = = ACL vI 1 1 1 1 1 + + + R2 A0 L R1 R2 Ri From Equation (14.20) for RL = and R0 = 01 1 1 (1 + A0 L ) = + Rif Ri R2 1For Ri = 1 ka.(1/ 20) 1 1 1 1 1 100 + 103 20 + 100 + 1 0.05 = [0.01 + 1.06 103 ]ACL =orwww.elsolucionario.net 631. ACL = 4.52 1 1 1 + 103 = + Rif = 90.8 Rif 1 100For Ri = 10 kb. ACL (1/ 20) 1 1 1 1 1 100 + 103 20 + 100 + 10 0.05 [0.01 + 1.6 104 ]or ACL = 4.92 1 1 1 + 103 = + Rif = 98.9 Rif 10 100For Ri = 100 kc. ACL (1/ 20) 1 1 1 1 1 100 + 103 20 + 100 + 100 0.05 [0.01 + 7 105 ]or ACL = 4.965 1 1 1 + 103 = + Rif = 99.8 Rif 100 10014.6 R2 1 + R1 v ACL = o = vi 1 R2 1 + 1 + R1 AOL For the ideal: R2 0.10 = 50 1 + = R1 0.002 vo (actual ) = (0.10)(1 0.001) = 0.0999 Sowww.elsolucionario.net 632. 0.0999 50 = = 49.95 1 0.002 1 + (50) AOL which yields AOL = 100014.7 From Equation (14.18) A 1 OL v Ro R2 Avf 1 = o1 = v1 1 1 1 + + RL Ro R2 Or 5 103 1 1 100 (4.99999 103 ) vo1 = v1 = v1 1.11 1 1 1 + + 10 1 100 vo1 = 4.504495 103 v1 Now i1 vi v1 = K v1 R1v1 Then vi v1 = KR1v1 which yields vi v1 = KR1 + 1 Now, from Equation (14.20) 1 1 + 5 103 + 1 1 10 K= + 10 100 1 + 1 + 1 10 100 5.0011 103 = (0.1) + (0.01) = 45.15495 1.11 Then vi vi v1 = = ( 45.15495)(10 ) + 1 452.5495We find vi vo1 = 4.504495 103 452.5495 Or v Avf 1 = o1 = 9.9536 viFor the second stage, RL = www.elsolucionario.net 633. 5 103 1 1 100 vo 2 = v1 = 4.950485 103 v1 1 1 + 1 100 1 1 1 + 5 103 K + = 49.61485 10 100 1 + 1 100 vo1 vo1 vo1 v1 = = = KR1 + 1 (49.61485)(10) + 1 497.1485 Then vo 2 4.950485 103 = = 9.95776 497.1485 vo1 So v Avf = o 2 = (9.9536)(9.95776) Avf = 99.12 vi14.8 a.v1 vI v v v + 1 + 1 0 =0 R3 + Ri R1 R2(1) 1 vI 1 1 v v1 + + = 0 + R3 + Ri R1 R2 R2 R3 + Ri v0 v0 A0 L vd v0 v1 + + =0 RL R0 R2 or 1 A v 1 1 v v0 + + = 1 + 0L d RL R0 R2 R2 R0 v v vd = I 1 Ri R3 + Ri So substituting numbers: vI 1 v 1 1 v1 + + = 0 + 10 + 20 10 40 40 10 + 20 or v1[0.15833] = v0 [0.025] + vI [0.03333] 1 v (104 )vd 1 1 v0 + + = 1 + 0.5 1 0.5 40 40(2)(3)(1)(2)www.elsolucionario.net 634. or v0 [3.025] = v1 [ 0.025] + ( 2 104 ) vd v v vd = I 1 20 = 0.6667 ( vI v1 ) 10 + 20 So v0 [3.025] = v1 [ 0.025] + ( 2 104 ) ( 0.6667 )( vI v1 )(3) (2)or v0 [3.025] = 1.333 10 4 vI 1.333 104 v1 From (1): v1 = v0 ( 0.1579 ) + vI ( 0.2105 ) Then v0 [3.025] = 1.333 104 vI 1.333 104 v0 ( 0.1579 ) + vI ( 0.2105 ) v0 2.1078 103 = vI 1.0524 104 or v ACL = 0 = 4.993 vI To find Rif : Use Equation (14.27) 0.5 0.5 iI 1 + + 1 40 3 1 1 0.5 0.5 0.5 (10 ) vd = v1 + 1 + + 1 40 (40) 2 40 10 40 iI (1.5125) = v1{(0.125)(1.5125) 0.0003125} 25vd or iI (1.5125) = vI {0.18875} 25vd Now vd = iI Ri = iI (20) and v1 = vI iI (20) So iI (1.5125) = [vI iI (20)] [0.18875] 25iI (20)iI [505.3] = vI (0.18875) or vI = 2677 k iI Now Rif = 10 + 2677 Rif = 2.687 M To determine R0 f : Using Equation (14.36) 1 1 A0 L = R2 R0 f R0 1 + R R 1 i or R0 f = 3.5 1 103 = 0.5 1 + 40 10 20 Then R0 f = 1 k 3.5 R0 f = 3.49 b. Using Equation (14.16) dACL dA 5 = (10) 3 CL = (0.05)% ACL ACL 10 14.9www.elsolucionario.net 635. v0 A0 L vd v0 vI + = 0 and vd = vI v0 R0 Ri So v0 A0 L v v (vI v0 ) + 0 I = 0 R0 R0 Ri Ri1 A 1 A 1 v0 + 0 L + = vI + 0 L R0 R0 Ri Ri R0 1 (104 ) 1 1 (104 ) + + + v0 = vI 0.2 0.2 100 100 0.2 v0 [5.000501 104 ] = vI [5.000001 10 4 ]So ACL =Set vI = 0b. i0 =v0 = 0.9999 vIv0 A0 L vd v0 + and vd = v0 R0 Ri1 A 1 i0 = v0 + 0 L + R0 R0 Ri Then 1 1 A0 L 1 = + + R0 f R0 R0 Rior 1 1 (104 ) 1 = + + R0 f 0.2 0.2 100 which yields R0 f 0.02 14.10www.elsolucionario.net 636. vI 1 v1 vI 2 v1 v1 v0 + = 20 10 40 vI 1 vI 2 v0 1 1 1 + + = v1 + + 20 10 40 20 10 40 v and v0 = A0 L v1 so that v1 = 0 A0L Then 1 1 7 vI 1 (0.05) + vI 2 (0.10) = v0 + 3 40 2 10 40 = v0 [2.50875 102 ] v0 = 1.993vI 1 3.986vI 2 v0 2 1.993 v = 0 = 0.35% 2 v0 v014.11 40 4 vB = v2 = v2 = 0.8v2 40 + 10 5 v1 v A v A v0 = 10 40 v1 v0 1 1 + = vA + 10 40 10 40 v1 (0.1) + v0 (0.025) = vA (0.125) v0 = A0 L vd = A0 L (vB v A ) or v0 = A0 L [0.8v2 v A ] v0 0.8v2 = v A A0 L v A = 0.8v2 (1)(2) (3)v0 A0 LThenwww.elsolucionario.net 637. v v1 (0.1) + v0 (0.025) = (0.125) 0.8v2 0 A0 L 0.125 v1 (0.1) v2 (0.1) = v0 0.025 + 103 2 = v0 [2.5125 10 ] Ad = v0 = 3.9801 v2 v1Ad 0.0199 = 0.4975% Ad 414.12 a.Considering the second op-amp and Equation (14.20), we have 1 1 1 1 + 100 101 = + = 0.10 + (0.1)(11) Rif 2 10 0.1 1 + 1 0.1 So Rif 2 = 0.0109 k The effective load on the first op-amp is then RL1 = 0.1 + Rif 2 = 0.1109 kAgain using Equation (14.20), we have 1 1 + 100 + 1 1 1 0.1109 = 0.10 + 110.017 = + 11.017 Rif 10 1 1 + 1 + 1 0.1109 1 so that Rif = 99.1 b.To determine R0 f :For the first op-amp, we can write, using Equation (14.36) 1 100 1 1 A0 L = = R2 1 40 R0 f 1 R0 1+ 1+ R1 Ri 1 10 which yields R0 f 1 = 0.021 k For the second op-amp, then A0 L 1 1 = R2 R0 f R0 1 + ( R + R ) R 1 0f1 i 1 100 = 0.10 1 1 + (0.121) 10 or R0 f = 18.4 c.To find the gain, consider the second op-amp.www.elsolucionario.net 638. v01 (vd 2 ) vd 2 vd 2 v02 + = 0.1 Ri 0.1(1)v01 v02 1 1 1 + vd 2 + + = 0.1 0.1 0.1 10 0.1 or v01 (10) + vd 2 (20.1) = v02 (10) v02 A0 L vd 2 v02 (vd 2 ) + =0 R0 0.1(2)v02 100 1 v02 vd 2 =0 + 1 1 0.1 0.1 v02 (11) vd 2 (90) = 0 or vd 2 = v02 (0.1222) Then Equation (1) becomes v01 (10) + v02 (0.1222)(20.1) = v02 (10) or v01 = v02 (1.246) Now consider the first op-amp.vI (vd 1 ) vd 1 vd 1 v01 + = 1 Ri 1(1)1 1 1 vI (1) + vd 1 + + = v01 (1) 1 10 1 or vI (1) + vd 1 (2.1) = v01 (1) v01 v A0 L vd 1 v01 (vd 1 ) + 01 + =0 0.1109 R0 1(2)1 1 1 100 1 v01 + + vd 1 =0 0.1109 1 1 1 1 v01 (11.017) vd 1 (99) = 0 or vd 1 = v01 (0.1113) Then Equation (1) becomeswww.elsolucionario.net 639. vI (1) + v01 (0.1113)(2.1) = v01or vI = v01 (1.234) We had v01 = v02 (1.246) So vI = v02 (1.246)(1.234) orv02 = 0.650 vIv02 =1 vI So ratio of actual to ideal = 0.650.d.Ideal14.13 (a)For the op-amp. A0 L f 3dB = 106106 = 50 Hz 2 104 For the closed-loop amplifier. 106 f 3dB = = 40 kHz 25 (b) Open-loop amplifier. 2 104 2 104 A = A= 2 f f 1+ j 1+ f3dB f3dB f 3dB =f = 0.25 f 3dB A = f = 5 f 3 dB A =2 104 1 + (0.25)2= 1.94 1042 104= 3.92 103 1 + (5) 2 Closed-loop amplifier 25 f = 0.25 f 3dB A = = 24.25 1 + (0.25) 2 f = 5 f 3 dB A =25 1 + (5) 2= 4.9014.14 The open loop gain can be written as A0 A0 L ( f ) = f f 1 + j 1 + j f PD 5 106 where A0 = 2 105. The closed-loop response is A0 L ACL = 1 + A0 L At low frequency, 2 105 100 = 1 + (2 105 ) So that = 9.995 103. Assuming the second pole is the same for both the open-loop and closed-loop, then f f 1 = tan 1 tan 6 f PD 5 10 www.elsolucionario.net 640. For a phase margin of 80, = 100. So f 100 = 90 tan 1 6 5 10 or f = 8.816 105 Hz Then A0 L = 1 2 105= 8.816 105 1+ f PD 2 8.816 105 1+ 6 5 10 2or 8.816 105 1.9696 105 f PD or f PD = 4.48 Hz 14.15 (a) 1st stage (10) f3 dB = 1 MHz f 3 dB = 100 kHz 2nd stage (50) f3 dB = 1 MHz f 3 dB = 20 kHz Bandwidth of overall system 20 kHz (b) If each stage has the same gain, so 2 K = 500 K = 22.36 Then bandwidth of each stage (22.36) f 3 dB = 1 MHz f 3 dB = 44.7 kHz 14.16A=Ao 1+ jA=f f3 dB Ao f 1+ f3 dB 2 200 =5 104 104 1+ f3 dB 2 f3 dB = 40 HzThen fT = (5 104 )(40) fT = 2 MHz 14.17www.elsolucionario.net 641. (5 104 ) f PD = 106 f PD = 20 Hz (25) f 3 dB 106 f3 dB = 40 kHz Av =Avo 1+ jf Av =f3 dB25 f 1+ 3 40 10 2At f = 0.5 f 3 dB = 20 kHz Av =25 1 + (0.5) 2= 22.36At f = 2 f 3 dB = 80 kHz Av =25 1 + (2) 214.18 (20 103 ) Avf= 11.18MAX= 106 AvfMAX= 5014.19 From Equation (14.55), SR 10 106 F P BW = = 2 VP 0 2 (10) or F P BW = f max = 159 kHz 14.20 a. VP 0Using Equation (14.55), 8 106 = 2 (250 103 )or VP 0 = 5.09 V b.1 1 = = 4 106 s f 250 103 One-fourth period = 1 sPeriod T =Slope =VP 0 = SR = 8 V/ s 1 s VP 0 = 8 V14.21 For input (a), maximum output is 5 V.www.elsolucionario.net 642. S R = 1 V/s soFor input (b), maximum output is 2 V.For input (c), maximum output is 0.5 V so the output is14.22 For input (a), max v01 = 3 V.Then v02max= 3(3) = 9 VFor input (b), max v01 = 1.5 V.www.elsolucionario.net 643. Then v02max= 3 (1.5 ) = 4.5 V14.23 f MAX = 20 kHz, SR = 0.8 V / s V po =SR 0.8 106 = V po = 6.37 V 2 f MAX 2 (20 103 )14.24 V V I1 = I S 1 exp BE1 , I 2 = I S 2 exp BE 2 VT VT Want I1 = I 2 , so V 5 1014 (1 + x) exp BE1 I1 VT =1= I2 V 5 1014 (1 x) exp BE 2 VT = V V (1 + x) exp BE1 BE 2 VT (1 x) Or V VBE1 VOS 1+ x = exp BE 2 = exp 1 x VT VT 0.0025 = exp = 1.10 0.026 Now 1 + x = (1 x )(1.10) x = 0.0476 4.76% 14.25 From Equation (14.62), vCE 2 vCE1 1+ V I 1+ V AN AN = S3 1 + vEB I S 4 1 + vEC 4 V VAP AP For vCE 2 = 0.6 V, then vEC 4 = 5 V. We have vCE1 = 5 V sowww.elsolucionario.net 644. 5 0.6 1 + 80 I S 3 1 + 80 = 1 + 0.6 I S 4 1 + 5 80 80 or I S 3 (1.0625) 2 = = 1.112 I S 4 (1.0075) 2 So I S 3 = (1014 ) (1.112 )or I S 3 = 1.112 1014 A 14.26By superposition: R vo (vi ) = 2 vi = 50vi R1 R vo (vos ) = 1 + 2 vos = 51vos R1 So vo = vo ( vi ) + vo ( vos ) = 50vi + 51vosFor vi = 20 mV and vos + 2.5 mV vo = 50(0.02) + 51(0.0025) = 0.8725 V For vi = 20 mV and vos = 2.5 mV vo = 50(0.02) + 51(0.0025) = 1.1275 V So 1.1275 vo 0.8725 V 14.27 vo = 50vi = 50 2.5 mV + sin t ( mV ) vo = [ 0.125 0.25sin t ] ( v )14.28www.elsolucionario.net 645. 0.5 103 = 5 108 A 104 Also i dV 1 I I = C o Vo = Idt = t dt C0 C Then 5 108 5= t t = 103 s 10 106 I=14.29 a. 100 v01 = 10 1 + or v01 = 110 mV 10 Then 50 v02 = v01 (5) + 10 1 + = (110)(5) + (10)(6) 10 or v02 = 610 mV14.30 v0 due to vI 1 v0 = (0.5) 1 + = 0.9545 V 1.1 Wiper arm at V + = 10 V, (using superposition) R1 R5 0.0909 v1 = (10) = (10) R1 R5 + R4 0.0909 + 10 = 0.090 1 Then v01 = (0.090) = 0.090 1 Wiper arm in center, v1 = 0 and v02 = 0Wiper arm at V = 10 V, v1 = 0.090 So v03 = 0.090 Finally, total output v0 : (from superposition) Wiper arm at V + , v0 = 0.8645 V Wiper arm in center, v0 = 0.9545 V Wiper arm at V ,www.elsolucionario.net 646. v0 = 1.0445 V14.31 a. R1 = R2 = 0.5 25 = 0.490 k or R1 = R2 = 490 b.From Equation (14.75), 125 106 (0.026) ln + (0.125) R1 14 2 10 125 106 = (0.026) ln + (0.125) R2 14 2.2 10 0.586452 + (0.125) R1 = 0.583974 + (0.125) R2 0.002478 = (0.125)( R2 R1) So R2 R1 = 0.0198 k 19.8 Then R2 (1 x) Rx R Rx 1 = 0.0198 R2 + (1 x) Rx R1 + xRx(0.5)(1 x)(50) (0.5)(50) x = 0.0198 (0.5) + (1 x)(50) (0.5) + x(50) 25(1 x) 25 x = 0.0198 50.5 50 x 0.5 + 50 x (0.5 + 50 x)(25 25 x) (25 x)(50.5 50 x) = 0.0198 (50.5 50 x )(0.5 + 50 x)25 {0.5 0.5 x + 50 x 50 x 2 50.5 x + 50 x 2 } = 0.0198 {25.25 + 2525 x 25 x 2500 x 2 } 25 {0.5 x} = 0.0198 {25.25 + 2500 x 2500 x 2 } 0.5 x = 0.019998 + 1.98 x 1.98 x 2 1.98 x 2 2.98 x + 0.48 = 0 x=2.98 (2.98) 2 4(1.98)(0.48) 2(1.98)So x = 0.183 and 1 x = 0.817 14.32 R1 = R1 15 = 0.5 15 = 0.4839 k R2 = R2 35 = 0.5 35 = 0.4930 k From Equation (14.75), i i (0.026) ln C1 + iC1 R1 = (0.026) ln C 2 + iC 2 R2 IS3 IS 4 i (0.026) ln C1 = iC 2 R2 iC1 R1 iC 2 i i R (0.026) ln C1 = iC 2 R2 1 C1 1 iC 2 iC 2 R2 i i (0.026) ln C1 = iC 2 (0.4930) 1 (0.9815) C1 iC 2 iC 2 By trial and error:www.elsolucionario.net 647. iC1 = 252 A and iC 2 = 248 Aor iC1 = 1.0155 iC 2 14.33 From Eq. (14.79), we have R vo = I B1 R2 I B 2 R3 1 + 2 R1 I B1 = 1 A I B 2 = 2 A Setting vo = 0, we have 200 0 = (106 )( 200 103 ) ( 2 106 ) R3 1 + 20 200 103 R3 = R = 9.09 K ( 2 106 ) (11) 314.341+R2 = 80 R1 R1 = 6.329 KV f = vI = 5sin t ( mV ) I1 + I B = I 2(a)VI = 0 VX = 0 I 2 = I B VO = (106 )(500 103 ) vo = 0.50 V(b)v VX VX + IB = o R1 R2 1 1 v 1 1 VX + + I B = 0 vo = R2 + vI + I B R2 R1 R1 R2 R1 R2 vo = 80 5sin t ( mV ) + (106 )( 500 103 ) vo = [ 0.5 + 0.4sin t ] ( v )14.35 a.www.elsolucionario.net 648. For I B 2 = 1 A, then v0 = (106 )(104 ) orv0 = 0.010 Vb.If a 10 k resistor is included in the feedback loopNow v0 = I B 2 (10) + I B1 (10) = 0 Circuit is compensated if I B1 = I B 2 .14.36 From Equation (14.83), we have v0 = R2 I 0S where R2 = 40 k and I 0 S = 3 A. Then v0 = ( 40 103 )( 3 106 ) or v0 = 0.12 V 14.37 a. Assume all bias currents are in the same direction and into each op-amp. v01 = I B1 (100 k ) = (106 )(105 ) v01 = 0.1 V Then v02 = v01 ( 5 ) + I B1 ( 50 k )= ( 0.1)( 5 ) + (106 )( 5 104 ) = 0.5 + 0.05or v02 = 0.45 V b. Connect R3 = 10 100 = 9.09 k resistor to noninverting terminal of first op-amp, and R3 = 10 50 = 8.33 k resistor to noninverting terminal of second op-amp.www.elsolucionario.net 649. 14.38 a. For a constant current through a capacitor. 1 t v0 = I dt C 0 0.1 106 or v0 = t v0 = (0.1)t 106 v0 = 1 V b. At t = 10 s, c. Then 100 1012 v0 = t v0 = (104 )t 6 10 At t = 10 s, v0 = 1 mV 14.39 a. Assume all bias currents are into the op-amp. v01 = I B1 ( 50 k ) = (10 106 )( 50 103 ) or v01 = v02 = 0.5 V v03 = ( 1)( v01 ) + (10 106 )( 20 103 )or v03 = 0.3 V b.RA = 10 50 RA = 8.33 kRB = 20 20 RB = 10 k c. Assume the worst case offset current, that is, I 0 S = I B1 I B 2 or I 0 S = I B 2 I B1 . From Equation (14.83), v01 = R2 I 0 S = ( 50 103 )( 2 106 )or v01 = v02 = 0.1 V v03 = ( 1) v01 I 0 S R2= ( 1)( 0.1) ( 2 106 )( 20 103 )or v03 = 0.14 V 14.40 a. Using Equation (14.79), Circuit (a), 50 v0 = ( 0.8 106 )( 50 103 ) ( 0.8 106 )( 25 103 ) 1 + 50 or v0 = 0Circuit (b), 50 v0 = ( 0.8 106 )( 50 103 ) ( 0.8 10 6 )(103 ) 1 + 50 2 = 4 10 1.6 or v0 = 1.56 Vb. Assume I B1 = 0.7 A and I B 2 = 0.9 A, then using Equation (14.79): Circuit (a),www.elsolucionario.net 650. 50 v0 = ( 0.7 106 )( 50 103 ) ( 0.9 106 )( 25 103 ) 1 + 50 = 0.035 0.045orv0 = 0.010 VCircuit (b), 50 v0 = ( 0.7 106 )( 50 103 ) ( 0.9 106 )(106 ) 1 + 50 = 0.035 1.8orv0 = 1.765 V14.41 a.If R = 0, 100 v0,max = 1 + V0 S + I B (100 k) 10 = (11)(10 103 ) + (2 106 )(100 103 ) v0,max = 0.110 + 0.20 v0,max = 0.310 Vb.R = 10.1 100 = 9.17 k = R14.42 a. Ri (15) = 0.010 V Ri + R2 15 = 0.0006667 15 + R2 15(1 0.0006667) = 0.0006667 R2 Then R2 = 22.48 Mb.R1 = Ri RF = 15 10 R1 = 6 k14.43 a. Assume the offset voltage polarities are such as to produce the worst case values, but the bias currents are in the same direction. Use superposition:www.elsolucionario.net 651. Offset voltages 100 v01 = 1 + (10) = 110 mV = v01 10 50 v02 = (5)(110) + 1 + (10) 10 v02 = 610 mVBias Currents: v01 = I B (100 k) = (2 106 )(100 103 ) = 0.2 V Then v02 = (5)(0.2) + (2 106 )(50 103 ) = 0.9 V Worst case: v01 is positive and v02 is negative, then v01 = 0.31 V and v02 = 1.51 V b.Compensation network:If we want RB + + V = 20 mV and V = 10 V RB + RC 8.33 (10) = 0.020 8.33 + RC or RC 4.15 M14.44 Assume bias currents are in same direction, but assume polarity of offset voltages are such as to produce the worst case output. a. Let I B1 = 5.5 A, I B 2 = 4.5 A Bias Current Effects: v01 = I B1 (50 k) = 0.275 V v02 = 0.275 V v03 = I B1 (20 k) v01 v03 = 0.165 V Offset Voltage Effects: 50 v01 = (5) 1 + = 30 mV v02 = 30 mV 10 20 v03 = v01 5 1 + v03 = 40 mV 20 Total Effect: v01 = 0.305 V and v02 = 0.305 V v03 = 0.205 V 14.45 For circuit (a), effect of bias current: v0 = (50 103 )(100 109 ) 5 mV Effect of offset voltagewww.elsolucionario.net 652. 50 v0 = (2) 1 + = 4 mV 50 So net output voltage is v0 = 9 mVFor circuit (b), effect of bias current: Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79), 50 v0 = (450 109 )(50 103 ) (550 109 )(106 ) 1 + 50 2 = 2.25 10 1.1 or v0 = 1.0775 V If the offset voltage is negative, then v0 = (2)(2) = 4 mV So the net output voltage is v0 = 1.0815 V14.46 a. At T = 25C, V0 S = 2 mV so the output voltage for each circuit is v0 = 4 mV b. For T = 50C, the offset voltage for is V0 S = 2 mV + (0.0067)(25) = 2.1675 mV so the output voltage for each circuit is v0 = 4.335 mV 14.47 a.At T = 25C,V0 S = 1 mV, then 50 v01 = (1) 1 + v01 = 6 mV 10 and 60 60 v02 = v01 1 + + (1) 1 + 20 20 = 6(4) + (1)(4) v02 = 28 mVb. At T = 50C, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then v01 = (1.0825)(6) v01 = 6.495 mV and v02 = (6.495)(4) + (1.0825)(4) or v02 = 30.31 mV 14.48 25C; I B = 500 nA, I 0 S = 200 nA 50C, I B = 500 nA + (8 nA / C)(25C) = 700 nA I 0 S = 200 nA + (2 nA / C)(25C) = 250 nA a. Circuit (a): For I B , bias current cancellation, v0 = 0 Circuit (b): For I B , Equation (14.79), 50 v0 = (500 109 )(50 103 ) (500 10 9 )(106 ) 1 + 50 = 0.025 1.00 v0 = 0.975 Vb. Due to offset bias currents. Circuit (a):www.elsolucionario.net 653. v0 = (200 109 )(50 103 ) v0 = 0.010 VCircuit (b): Let I B 2 = 600 nA I B1 = 400 nA Then 50 v0 = (400 109 )(50 103 ) (600 109 )(106 ) 1 + 50 = 0.020 1.20 v0 = 1.18 Vc. Circuit (a): Due to I B , v = 0 Circuit (b): Due to I B , 50 v0 = (700 109 )(50 103 ) (700 109 )(106 ) 1 + 50 = 0.035 1.40 v0 = 1.365 VCircuit (a): Due to I 0 S , v0 = (250 109 )(50 103 ) v0 = 0.0125 VCircuit (b): Due to I 0 S , Let I B 2 = 825 nA I B1 = 575 nA Then 50 v0 = (575 109 )(50 103 ) (825 109 )(106 ) 1 + 50 = 0.02875 1.65 v0 = 1.62 V14.49 25C; I B = 2 A, I 0 S = 0.2 A 50C, I B = 2 A + (0.020 A / C)(25C ) = 2.5 A I 0 S = 0.2 A + (0.005 A / C)(25C) = 0.325 A a. Due to I B : (Assume bias currents into op-amp). v01 = I B (50 k) = (2 106 )(50 103 ) v01 = 0.10 V 60 60 v02 = v01 1 + + I B (60 k) I B (50 k) 1 + 20 20 3 6 6 = (0.1)(4) + (2 10 )(60 10 ) (2 10 )(60 103 )4orv02 = 0.12 Vb. Due to I 0 S : 1st op-amp. Let I B1 = 2.1 A 2nd op-amp. Let I B1 = 2.1 A I B 2 = 1.9 A v01 = I B1 (50 k) = (2.1 106 )(50 103 ) v01 = 0.105 V 60 60 v02 = v01 1 + + I B1 (60 k) I B 2 (50 k) 1 + 20 20 3 6 = (0.105)(4) + (2.1 10 )(60 10 ) (1.9 10 6 )(50 103 )(4) or v02 = 0.166 Vc.Due to I B :www.elsolucionario.net 654. v01 = (2.5 10 6 )(50 103 ) v01 = 0.125 V 60 60 v01 = v02 1 + + I B (60 k) I B (50 k) 1 + 20 20 6 3 = (0.125)(4) + (2.5 10 )(60 10 ) (2.5 106 )(50 103 (4) or v02 = 0.15 VDue to I 0 S : Let I B1 = 2.625 A I B 2 = 2.3375 A v01 = I B1 (50 k) = (2.6625 106 )(50 103 ) v01 = 1.133 V 60 60 v02 = v01 1 + + I B1 (60 k) I B 2 (50 k) 1 + 20 20 = (0.133)(4) + (2.6625 106 )(60 103 ) (2.3375 106 )(50 103 )(4) or v02 = 0.224 V14.50 R4 R2 vB = vI 1 and v0 (vI 2 ) = vB 1 + R1 R3 + R4 or R4 R2 v0 (vI 2 ) = 1 + vI 2 R1 R3 + R4 or vI 1 . v0 (vI 1 ) = R2 vI R1Then R4 R2 R2 v0 = 1 + vI 2 vI 1 R1 R1 R3 + R4 V V we can write vI 2 = vcm + d and vI 1 = Vcm d Then 2 2 R4 R2 Vd R2 Vd v0 = 1 + Vcm + Vcm R1 2 R1 2 R3 + R4 Common-mode gain R4 R2 R2 v Acm = 0 = 1 + Vcm R3 + R4 R1 R1www.elsolucionario.net 655. Differential mode gain v 1 R4 R2 R2 Ad = 0 = 1 + + Vd 2 R3 + R4 R1 R1 Then A CM RR = d Acm 1 R4 R2 R2 1 + + R1 R1 2 R3 + R4 = R4 R2 R2 1 + R1 R1 R3 + R4 1 R4 2 R3 R2 R2 1 1 + + R1 R1 R4 1+ R3 CM RR = R2 R2 R4 1 1 + R3 R4 R1 R1 1 + R3 Minimum CMRR Maximum denominator R R maximum 4 and minimum 2 . Then R3 R1 R4 (1.02)(50) = = 5.204 R3 (0.98)(10) R2 (0.98)(50) = = 4.804 R1 (1.02)(10) Then 1 5.204 (5.804) + (4.804) 2 6.204 CMRR = 5.204 6.204 (5.804) (4.804) 1 (9.6725) = 2 (0.06447) CMRR = 75.0 CMRRdB = 20 log10 (75.0) CMRRdB = 37.5 dB14.51 Use the results of Problem 14.50: R 1 + x 50 Let 4 = (1 + 2 x)(5) R3 1 x 10 R 1 x 50 Let 2 = (1 2 x)(5) R1 1 + x 10 Thenwww.elsolucionario.net 656. CMRR = 1 (1 + 2 x ) 5 ( 6 10 x ) + (1 2 x )( 5 ) 2 6 + 10 x (1 + 2 x ) 5 ( 6 10 x ) (1 2 x )( 5 ) 6 + 10 x 1 2 30 + 10 x 100 x + 30 10 x 100 x 2 = 2 2 2 30 + 10 x 100 x ( 30 10 x 100 x ) a. 20 x =b. 20 x =1 [60 200 x 2 ] 30 100 x 2 2 = = 20 x 20 x For CMRRdB = 90 dB CMRR = 31, 623 x will be small, neglect the x 2 term. Then 30 x = 0.0000474 = 0.00474% 31, 623 For CMRRdB = 60 dB CMRR = 1000. Then 30 x = 0.0015 = 0.15% 1000www.elsolucionario.net 657. Chapter 15 Exercise Solutions EX15.1 For the circuit shown in Figure 15.7 1 f 3dB = 2 RC or 1 1 RC = = = 3.979 106 2 f3dB 2 ( 40 103 ) For R = 75 K Then C = 5.311011 = 53.1 pF We have C3 = 1.414C = 75.1 pF C4 = 0.707C = 37.5 pFEX15.2 1 fC = CReq or C=1 1 = f c Req (105 )( 20 106 )C = 0.5 pFEX15.3 Low-frequency gain: T = f 3dB =C1 30 = = 6 C2 53 12 fC C2 (100 10 )( 5 10 ) = f 3dB = 6.63 kHz 2 CF 2 (12 1012 )EX15.4 1f0 =2 3RC 1 1 RC = = = 6.13 106 3 2 f 0 3 2 (15 10 ) 3 Let C = 0.001 F = 1 nFThen R = 6.13 k so R2 = 8R = 49 k EX15.5 f0 = C=1 1 C = 2 RC 2 f 0 R 12 ( 800 ) (104 ) C 0.02 FR2 = 2 R1 = 2 (10 ) R2 = 20 kEX15.6www.elsolucionario.net 658. R1 VTH = VH R1 + R2 R1 2= (12) R1 + 20 2 ( R1 + 20 ) = 12 R140 = 10 R1 R1 = 4 kEX15.7 R1 VTH VTL = (VH VL ) R1 + R2 R1 0.10 = (10 [ 10]) R1 + R2 1+R2 R 20 = = 200 2 = 199 R1 0.10 R1 R2 VS = VREF R1 + R2 R 1 VREF = 1 + 1 VS = 1 + (1) VREF = 1.005 V R2 199 I=VH VBE ( on ) VR + 0.1 10 0.7 0.7 R + 0.1 = = 43 k 0.2 R = 42.9 kEX15.8 At t = 0 , let v0 = 5 so v X = 2.5. For t > 0 t v X = 10 + ( 2.5 10 ) exp rX When v X = 5.0, output switches t 5.0 = 10 12.5 exp 1 rX t 10 5 5.0 = exp 1 = rX 12.5 12.5 t 12.5 12.5 exp + 1 = t1 = rX ln t1 = rX ( 0.916 ) 5.0 rX 5.0 During the next part of the cycle t v X = 5 + ( 5 [ 5]) exp rX When v X = 2.5, output switcheswww.elsolucionario.net 659. t 2.5 = 5 + 10 exp 2 rX t 5 2.5 2.5 exp 2 = = 10 10 rX t 10 10 exp + 2 = t2 = rX ln t2 = rX (1.39 ) rX 2.5 2.5 1 Period = t1 + t2 = T = ( 0.916 ) + (1.39 ) rX = 2.31rX Frequency = 2.31rX rX = ( 50 103 )( 0.01 10 6 ) = 5 10 4 s f = 866 Hz Duty cycle =( 0.916 ) t1 100% = 100% Duty cycle = 39.7% t1 + t2 ( 0.916 ) + (1.39 )EX15.9 a. rX = RX C X R1 10 vY = v0 = (12 ) = 1.2 V 10 + 90 R1 + R2 R1 = 0.10 = R1 + R2 0.7 1 + 12 1 + V /VP T = rX ln = rX ln 1 1 (0.10) T = 50 106 = rX ln [1.18] = (0.162) rX50 106 RX = 3.09 k (0.1 106 )(0.162) b. Recovery time t v X = VP + (1.2 VP ) exp rX RX =When v X = V , t = t2www.elsolucionario.net 660. t 0.7 = 12 + ( 1.2 12 ) exp 2 rX t 12 0.7 = 0.856 exp 2 = 13.2 rX 1 t2 = rX ln = ( 0.155 ) rX 0.856 rX = ( 3.09 103 )( 0.1 106 ) = 3.09 104 t2 = 48.0 sEX15.10 T = 1.1 RC T = 75 106 Let C = 10 nF Then 75 106 R= = 6.82 K (1.1) (10 109 )EX15.11 1 1 = f = 802 Hz 0.693 ( RA + 2 RB ) C ( 0.693) 20 + 2 ( 80 ) 103 ( 0.01 106 ) R + RB 20 + 80 Duty cycle = A 100% = 100% Duty cycle = 55.6% 20 + 2 ( 80 ) RA + 2 RB f =EX15.12 P=a.1 VP2 2 RLVP = 2 RL P = 2 ( 8 )(1) VP = 4 V IP =VP 4 = I P = 0.5 A RL 8VCE = 12 4 = 8 Vb.I C 0.5 ASo P = I C VCE = ( 0.5 )( 8 ) P = 4 W EX15.13 VP = 2 RL PL = 2 ( 8 )(10 ) = 12.65 V V PS = VS P RL VS =PS R2 (10 ) ( 8 ) = VP 12.65VS = 19.9 VEX15.14Line regulation =dV0 dV dV = 0 Z+ + dV dVZ dVNowwww.elsolucionario.net 661. dV0 10 = 1 + = 2 dVZ 10 dVZ rZ 10 = = 0.00227 = + dV rZ + R1 10 + 4400 So Line regulation = ( 2 )( 0.00227 ) = 0.004540.454% EX15.15V1 V0 V1 1 1 V = V1 + = 0 10 10 10 10 10 V 2 V V1 = 0 V0 = 2V1 V1 = 0 10 10 2 V0 V1 V0 V0 A0 L (VZ V1 ) + + =0 RL R0 10 V0 V0 V0 A0 LVZ V1 A0 LV1 + + = R0 R0 10 RL R0 10 =V0 A V 0L 0 2(10) 2 R0V0 V 1000 ( 6.3) V0 (1000 ) V0 + I0 + 0 = 10 0.5 0.5 20 2 ( 0.5 ) V0 [0.10 + 2.0 0.05 + 1000] + I 0 = 12, 600 V0 (1002.05) + I 0 = 12, 600 For I 0 = 1 mA V0 = 12.5732For I 0 = 100 mA V0 = 12.4744 Load reg =V0 ( NL ) V0 ( FL ) V0 ( NL ) 100%12.5732 12.4744 100% 12.5732 Load reg = 0.786%=EX15.16 a.www.elsolucionario.net 662. IC 3 =VZ 3VBE ( on )IC 3 =R1 + R2 + R3 5.6 3 ( 0.6 )3.9 + 3.4 + 0.576 I I C 4 R4 = VT ln C 3 IC 4 =3.8 I C 3 = 0.482 mA 7.88 0.482 I C 4 (0.1) = (0.026) ln IC 4 By trial and error I C 4 = 0.213 mA VB 7 = 2(0.6) + (0.482)(3.9) VB 7 = 3.08 Vb. R13 V0 = VB 8 = VB 7 R13 + R12 2.23 (5) = 3.08 2.23 + R12 ( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08) R1211.15 = 6.868 = 3.08R12 R12 = 1.39 kTYU15.1 1 2 RC 1 1 RC = = = 1.59 105 2 f3dB 2 (104 ) f 3dB =Let C = 0.01 F R = 1.59 k Then C1 = 0.03546 F C2 = 0.01392 F C3 = 0.002024 F 1T =6=1 f 20 1+ 1+ 10 f3d T = 0.124 or T = 18.1 dB6TYU15.2 f 3dB = RC =1 1 RC = 2 RC 2 f3dB 12 ( 50 103 )= 3.18 106Let C = 0.001 F = 1 nF R = 3.18 k Thenwww.elsolucionario.net 663. R1 = 2.94 k R2 = 3.44 k R3 = 1.22 k R4 = 8.31 k T = 0.01 =1 f 1 + 3 dB f 882 f 1 4 1 + 3 dB = = 10 f 0.01 2 f3 dB f 3dB f 15.8 kHz 10 f = 10 f TYU15.3 1-pole2-pole3-pole4-poleT =T =T =T =1 12 1+ 10 12 12 1+ 10 14 12 1+ 10 61 12 1+ 10 8 3.87 dB 4.88 dB 6.0 dB 7.24 dBTYU15.4 1 Req = fC Cor f C C =1 1 = = 2 107 Req 5 106If C = 10 pF fC = 20 kHz TYU15.5 1 1 f0 = = f 0 65 kHz 4 2 6 RC 2 6 (10 )(100 1012 ) R2 = 29 R = 29 (104 ) R2 = 290 kTYU15.6www.elsolucionario.net 664. f0 =1 CC 2 L 1 2 C1 + C2 =1 (109 ) 2 2 (10 ) 9 2 10 6 f 0 = 7.12 MHzC2 = gm R C1 C2 1 1 = g m = 0.25 mA / V C1 R 4 103 We have k W g m = 2 (VGS VTh ) 2 L k 20 A / V 2 , VGS VTh 1 V gm =SoW 0.25 103 = = 12.5 L ( 20 106 ) (1)and a value of W / L = 12.5 is certainly reasonable. TYU15.7 R VTH = 1 VL R2 R R 0.10 = 1 ( 10 ) 1 = 0.010 R2 R2 Let R1 = 0.10 k then R2 = 10 kTYU15.8 a. R2 10 VS = VREF = ( 2) 1 + 10 R1 + R2 VS = 1.82 V R1 1 VTH = VS + VH = 1.82 + (10 ) R1 + R2 1 + 10 VTH = 2.73 V VTL = R1 1 VS + VL = 1.82 + ( 10 ) R1 + R2 1 + 10 VTL = 0.91 Vb.www.elsolucionario.net 665. TYU15.9 R VS = 1 + 1 VREF R2 R R VTH = VS 1 VL and VTL = VS 1 VH R2 R2 R Hysteresis Width = VTH VTL = 1 (VH VL ) R2 R R 2.5 = 1 ( 5 [ 5]) = 10 1 R2 R2 R So 1 = 0.25 R2Then R VS = 1 = 1 + 1 VREF = (1 + 0.25)VREF VREF = 0.8 V R2 Then VTH = 1 ( 0.25 )( 5 ) VTH = 0.25 V VTL = 1 ( 0.25 )( 5 ) VTL = 2.25 VTYU15.10 R1 1 10 vX = v0 = v0 = v0 3 10 + 20 R1 + R2 10 t = 0, vX = 3 t 10 v X = 10 + 10 exp 3 rX Output switches when v X =10 3www.elsolucionario.net 666. 10 = 10 13.33 exp 3 t1 rX t 10 3.33 6.67 = exp 1 = 13.33 13.33 rX t 13.33 2 exp + 1 = rX 6.67 t1 = rX ln (2) = (0.693)rX T = 2(0.693)rX f =1 2(0.693)rXrX = RX C X = (104 )( 0.110 6 ) = 110 3 f = 722 Hz Duty cycle = 50%TYU15.11 R1 20 = 0.333 = = R1 + R2 20 + 40 rX = RX C X = (104 )( 0.01 106 ) = 1 104 1 + V / VP T = rX ln 1 0.7 1+ 8 4 T = 48.9 s = (1 10 ) ln 1 0.333 Recovery timewww.elsolucionario.net 667. R1 20 vY = v0 = (8) = 2.667 V R1 + R2 20 + 40 t 0.7 = 8 + ( 2.667 8 ) exp 2 rX t 8 0.7 exp 2 = = 0.6844 rX 10.66 1 t2 = rX ln t2 = 37.9 s 0.685 TYU15.12 f =1( 0.693)( RA + RB ) CRA + RB =1 ( 0.693) fCLet C = 0.01 F, RA + RB =f = 1kHz 1( 0.693) (103 )( 0.01106 )Duty cycle = 55 1.443 105RA + RB 100% RA + 2 RB(1.443 10 ) (100 ) (1.443 10 ) + R (1.443 10 ) (100 55) R = 555 =5B5RB55B= 118 k so RA = 26.2 kTYU15.13 v01 R2 30 a. = 1 + = 1 + = 2.5 vI R1 20 v02 R 50 = 4 = = 2.5 20 vI R3(b)P=1 VL2 1 [12 (12)]2 = = 240 mW 2 RL 2 1.2Or P = 0.24 W c.12 = V pi = 4.8 V 2.5www.elsolucionario.net 668. Chapter 15 Problem Solutions 15.1 (a)For example:vo R2 R = 1 + = 10 2 = 9 vi R1 R1 Corner Frequency: 1 f = = 5 103 RC = 3.18 105 2 RC (b) For Example:Low-Frequency:1 R2 j C vo R 1 = = 2 vi R1 R1 1 + j R2 C So, set R2 = 15 For example, R1 = 10 k , R2 = 150 k R1 R2 C =1 1 = = 1.06 105 2 f3 dB 2 (15 103 )Then C = 70.7 pF 15.2 (a)Av =1 f 1+ f3 dB 2=1 1 + (2) 2= 0.447 Av = 7 dBwww.elsolucionario.net 669. (b)(c)Av =Av =1 f 1+ f3 dB 12 f 1+ f3 dB 21 1 + (2) 4 1 1 + (2)6= 0.2425 Av = 12.3 dB= 0.1240 Av = 18.1 dB15.3 (a) Figure 15.6 1 f = 2 RC 1 1 RC = = = 7.958 106 2 f 2 (20 103 ) Let R1 = R2 = R = 10 K C = 795.8 pF So C4 = 0.707C = 562.6 pF C3 = 1.414C = 1.125 nF (b) 1 T = = 0.777 (i) 4 18 1+ 20 1 T = = 0.707 (ii) 4 20 1+ 20 1 T = = 0.637 (iii) 4 22 1+ 20 15.4 Use Figure 15.10(b) 1 f3 dB = 2 RC or 1 RC = = 3.18 106 2 (50 103 ) For example, let C = 100 pF Then R = 31.8 k And R1 = 8.97 k R2 = 22.8 k R3 = 157 k From Equation (15.26) 1 T = 6 f 1 + 3dB f www.elsolucionario.net 670. We find f kHz 30 35 40 45 T 0.211 0.324 0.456 0.58915.5 From Equation (15.7). Y1Y2 T (s) = Y1Y2 + Y4 (Y1 + Y2 + Y3 ) For a high-pass filter, let Y1 = Y2 = sC , Y3 =1 1 , and Y4 = R3 R4Then s 2C 2 1 1 s 2 C 2 + sC + sC + R4 R3 1 = 1 1 1+ 2+ sR4 C sR3C T (s) =Define r3 = R3C and r4 = R4 C 1 1 1 1+ 2+ sr4 sr3 Set s = j T (s) =1 1 1 1+ 2+ j r4 j r3 1 = j j 1 2 r4 r3 T ( j ) 1 1 2j 1 2 r3 r4 r4 1/ 22 1 4 ( j ) = 1 2 T + 2 2 r3 r4 r4 For a maximally flat filter, we want dT =0 d Taking the derivative, we find d T ( j ) d2 1 1 4 = 1 2 + 2 2 2 r3 r4 r4 3 / 2 1 2 4(2) 2 1 2 3 + 3 2 r3 r4 r3 r4 r4 www.elsolucionario.net 671. or d T ( j ) d=0 4 1 8 = 3 1 2 3 2 r3 r4 r3 r4 r4 =4 1 3 r3 r4 Then 1 1 2 1 2 2 r3 r4 r3 r4 r4 1 2 1 2 2 r3 r4 r4 =0 1 2 = 2r3 = r4 r3 r4 Then the transfer function can be written as:So that2 1 4 T ( j ) = 1 2 2 + 2 2 (2r3 ) (4r3 ) 1/ 2 1 1 1 = 1 2 2 + + r3 4( 2 r32 ) 2 2 r32 1/ 21/ 2 1 = 1 + 2 2 2 4( r3 ) 3 dB frequency 1 2 2 r32 = 1 or = = 2r3 Define 1 = RC So that R R3 = 21 2 R3CWe had 2r3 = r4 or 2( R3C ) = R4 C R4 = 2 R3 So that R4 = 2 R 15.6 From Equation (15.25) 1 T = 25 dB T = 0.0562 2N f 1+ f 3 dB f f 3 dB So=20 =2 10www.elsolucionario.net 672. 0.0562 =1 1 + (2) 2 N1 + (2) 2 N = 316.6 (2)2 N = 315.6 2 N ln (2) = ln (315.6) N = 4.15 N = 5 A 5-pole filter15.7 T =1 f 1+ f 3dB 2NAt f = 12 kHz, T = 0.9 10.9 = f 1+ f 3dB 12 f3dB Also2N=2N= 14 f3dB 12 f3dB 2N1 14 1+ f 3dB 2N1 1 = 9999 (0.01) 22N2N 12 1+ f 3dB 1 1 = 0.2346 (0.9) 20.01 = 14 f3dB 1=2N 14 = 12 2N=9999 = 4.262 104 0.2346(1.16667) 2 N = 4.262 104 N = 35 Then0.9 = 12 f3dB 1 12 1+ f 3dB 2N2N= 0.2346 12 2N = (0.2346)0.014286 = (0.2346) f3dB = 0.9795 So f 3dB = 12.25 kHz 115.8www.elsolucionario.net 673. T =1 2N f 1+ f 3dB (a) For N = 3 1 T = = 0.2841 1 + (1.5)6For N = 5 1 T = = 0.1306 1 + (1.5)10(b)For N = 7 1 T = = 0.05843 1 + (1.5)14(c)15.9 ConsiderFor low-frequency:vo R2 + R3 = vi R1 + R2 + R3For high-frequency:vo R2 = vi R1 + R2So we need R2 R2 + R3 = 25 R1 + R2 + R3 R1 + R2 Let R1 + R2 = 50 k and R2 = 1.5 k R1 = 48.5 k Thenwww.elsolucionario.net 674. 1.5 + R3 1.5 = 25 R3 = 144 k 50 + R3 50 Connect the output of this circuit to a non-inverting op-amp circuit.At low-frequency: vo1 =R2 + R3 1.5 + 144 vi = vi = 0.75vi R1 + R2 + R3 48.5 + 1.5 + 144Need to have vo = 25. R R R vo = 25 = 1 + 5 vo1 = 1 + 5 (0.75)vi 5 = 32.3 R4 R4 R4 To check at high-frequency. R2 1.5 vo1 = vi = vi = 0.03vi R1 + R2 1.5 + 48.5 vo = (1 + 32.3)vo1 = (33.3)(0.03)vi = (1.0)vi which meets the design specification Consider the frequency response. 1 R2 + R3 vo1 sC = 1 vi R1 + R2 + R3 sC Now R3 1 = R3 sC 1 + sR3C Then, we find vo1 R3 + R2 (1 + sR3C ) = vi R3 + ( R1 + R2 )(1 + sR3C ) which can be rearranged as ( R2 + R3 ) (1 + s ( R2 R3 )C ) vo1 = vi ( R1 + R2 + R3 ) 1 + s ( R3 ( R1 + R2 ) ) C()So fL 1 1 1 = = 3 2 ( R2 R3 ) C 2 (1.5 144 ) 10 C ( 9.33 103 ) CfH 1 1 = 2 ( R3 ( R1 + R2 ) ) C 2 (144 50 ) 103 C=1( 2.33 10 ) C 5Setwww.elsolucionario.net 675. fL + fH 1 1 1 = + 3 5 2 2 ( 9.33 10 ) C ( 2.33 10 ) C Which yields C = 2.23 nF 25 kHz =15.10Av =1 f 1+ f3 dB 100 dB 1052NSo 105 =1 770 1+ 12 2Nor 2 1 1 + (64.2) 2 N = 5 = 1010 10 or (64.2) 2 N 1010 NowN 1 2 3Left Side 4.112 103 1.7 107 7 1010So, we need a 3rd order filter. 15.11www.elsolucionario.net 676. Low-pass: 50 dB 3.16 103 Then 1 1 3.16 103 = = 4 4 f 60 1+ 1+ fL fL We find f L = 3.37 Hz High Pass: 1 1 3.16 103 = = 4 4 fH f 1+ H 1+ 60 f We find f H = 1067 Hz Bandwidth: BW = f H f L = 1067 3.37 BW 1064 Hz 15.12 a. v vI = 02 R4 R3v0 1 R1 sC (1)v0 v = 01 (2) R2 1 sC v01 v (3) = 02 v01 = v02 R5 R5 Then 1 v0 v = + 02 or v02 = v0 (2) R2 1 sR2 C sC Andwww.elsolucionario.net 677. 1 v0 sR2 C 1 R1 sC v vI = 0 R4 R3 1 1 = v0 + R3 ( sR2 C ) R1 (1/ sC ) R1 + (1/ sC ) (1) 1 + sR1C 1 = v0 + R1 R3 ( sR2 C ) R + (1 + sR1C )( sR2 R3C ) = v0 1 ( sC ) R1 R2 R3 Then v0 ( sC )( R1 R2 R3 ) 1 = vI R4 R1 + sR2 R3C + s 2 R1 R2 R3C 2 or Av ( s ) =1 R4v0 = 1 1 vI + sC + R1 sCR2 R3 Av ( j ) =b.1 R41 1 + j C + R1 j CR2 R3or Av ( j ) =1 R4 1 1 + j C CR2 R3 R1 R1 1 = R4 R1 1 + j R1C CR2 R3 R1 1 Av ( j ) = 2 1/ 2 R4 R1 1 + R1C CR2 R3 Av R1 when R1C =0 CR2 R3 maxThen Avmax=R1 85 = Av 3 R4max= 28.3Now R1C 1 1 1 = 0 or = 2 C R2 R3 C R2 R3 2Thenwww.elsolucionario.net 678. f =1 2 C R2 R3=1 2 (0.1 106 ) (300) 2So f = 5.305 kHz To find the two 3 dB frequencies, R1 R1C = 1 CR2 R3 2 R1 R2 R3 C 2 R1 = R2 R3 C 2 (85 103 )(300) 2 (0.1 106 ) 2 85 103 = (300) 2 (0.1 106 ) 2 (7.65 105 ) 85 103 = (9 103 ) 2 (7.65 105 ) (9 103 ) 85 103 = 0 (9 103 ) 2 + 4(7.65 105 )(85 103 ) (9 103 ) 2(7.65 105 ) 2(7.65 105 ) We find f = 5.315 kHz and f = 5.296 kHz=15.13 a. vI v A v = A (1) R 1 sC vI vB vB v0 = (2) R R and v A = vB So vI 1 1 + sRC = v A + sC = v A (1) R R R or vI vA = 1 + sRC Then 2vI vI + v0 = 2vB = 2vA = (2) 1 + sRC 2 1 sRC v0 = vI 1 = vI 1 + sRC 1 + sRC Now v0 1 j RC = A( j ) = 1 + j RC vI A=1 + 2 R2C 2 1 + 2 R2C 2 A =1Phase: = 2 tan 1 ( RC ) b. f 0RC = (104 )(15.9 109 ) = 1.59 104 0www.elsolucionario.net 679. 11.4 53.1 90 157 16910 2 5 103 1/ 2 RC = 103 Hz 5 103 10 415.14 a. Vi Vi V0 + =0 R1 R2 (1/ sC ) Vi Vi V0 + =0 R1 R2 1 + sR C 2 R2 1 (Vi ) + Vi = V0 R1 1 + sR2 C V0 R2 + R1 (1 + sR2 C ) ( R2 + R1 ) [1 + s ( R1 R2 )C ] = = Vi R1 (1 + sR2 C ) R1 (1 + sR2 C ) V0 R2 1 + s ( R1 R2 )C = 1 + Vi R1 (1 + sR2 C ) f 3dB1 =1 2 R2 C f 3dB2 =1 2 ( R1 R2 )Cb. Vi V V0 + i =0 R1 (1/ sC ) R2 Vi V V + i = 0 R1 R2 R2 1 + sR1C R Vi 2 (1 + sR1C ) + 1 = V0 R1 Vi [ R2 + R1 + sR1 R2 C ] = V0 R1 V0 R2 + R1 V R 1 = [1 + s ( R1 R2 )C ] 0 = 1 + 2 [1 + s ( R1 R2 )C ] f3dB = 2 ( R1 R2 )C Vi R1 Vi R1 15.15 a.www.elsolucionario.net 680. V0 Vi = R1 + (1/ sC1 ) R2 (1/ sC2 ) sC1 1 + sR2 C2 Vi = V0 1 + sR1C1 sC2 V0 sR2 C1 sR2 C1 = = Vi (1 + sR1C1 )(1 + sR2 C2 ) 1 + sR1C1 + sR2 C2 + s 2 R1 R2 C1C2 V0 R sC1 = 2 Vi R1 1 R2 C2 2 + sC1 1 + + s R2 C1C2 R1 C1 R1 or V0 R2 1 = T (s) = Vi R1 1 R2 C2 + 1 + + sR2 C2 R1 C1 sR1C1 b. T ( j ) = R2 1 2 2 1/ 2 R1 1 R2 C2 1 + . + R2 C2 R1 C1 R1C1 1 when R2 C2 = 0, we want R1C1 R 1 T ( j ) = 50 = 2 R1 R2 C2 1 + R1 C1 At the 3 dB frequencies, we want R2 C2 1 R2 C2 = 1 + R1C1 R1 C1 For f = 5 kHz, use + sign and for f = 200 Hz, use sign.1 = 2 (200) = 1257 2 = 2 (5 103 ) = 3.142 104 Define r2 = R2 C2 and r1 = R1C1 Then 50 =R2 R11 r 1+ 2 r1 r2 1 2 r2 = + 1 + 2 r1 r1 r2 1 1 r2 = 1 + 1 r1 r1 From (2) 2 2 r1r2 1 r1 + r2 = 2 r1 r1(1)(2) (3)www.elsolucionario.net 681. or 2 r1 r2 12= r1 + r2r1 ( 2 r2 1) =12+ r2So 1+r2 r1 = 2 2 r2 1Substituting into (3), we find r ( r 1) 1 1r2 = 1 + 2 2 2 1 1 + r2 + r2 2 1 2 2 r2 1 1 1 1 1r2 + r2 ( 2 r2 1) = + r2 + r2 ( 2 r2 1) 2 1 2 1 2 1 1 r2 + 1r22 r2 + = r2 2 r22 + r22112 1 1 (1 + 2 )r22 + 1 2 r2 + + =0 1 2 2 1 (3.2677 104 )r22 24.96r2 + 8.273 104 = 0 (24.96) 2 4(3.2677 10 4 )(8.273 104 ) 24.96 2(3.2677 10 4 ) 2(3.2677 104 ) Since 2 is large, r2 should be small so use minus sign: r2 =r2 = 3.47 105 Then 3.18 105 + 3.47 105 r1 = r1 = 7.32 104 9.09 102 Now R 1 50 = 2 R1 3.47 105 1+ 7.32 104 Then R2 = 52.37 or R2 = 524 k R1 Also r1 = R1C1 so that C1 = 0.0732 F r2 = R2 C2 so that C2 = 66.3 pF15.16 Gain = 10 dB Gain = 3.162 For example, we may havewww.elsolucionario.net 682. WantR4 = 2.162 R3For example, letR3 = 50 k, R4 = 108 kf1 =1 = 200 2 R1C1So R1C1 =1 = 0.796 103 2 (200)For example, let R1 = 200 k A large input resistance C1 = 0.00398 F 1 1 f2 = = 50 103 R2 C2 = = 3.18 106 2 R2 C2 2 (50 103 ) For example, let R2 = 10 k and C2 = 318 pF15.17 f C = 100 kHz 1 Req = fC C a.For C = 1 pF, Req = 10 Mb.For C = 10 pF, Req = 1 Mc.For C = 30 pF, Req = 333 k15.18 a. From Equation (15.28), V1 V2 Q= TC Req and f C = 100 kHz so that TC =1 10 s 100 103Now Req =1 1 = 1 M 3 f C C (100 10 )(10 1012 )Sowww.elsolucionario.net 683. Q=(2 1)(10 106 ) = 10 1012 C 106or Q = 10 pC Q 10 1012 or I eq = 1 A = TC 10 106b.I eq =c.Q = CV so find the time that V0 reaches 99% of its full value.V0 = V1 (1 e t / r ) where r = RC Then 0.99 = 1 e t / r or e t / r = 0.01 or t = r ln (100) r = RC = (103 )(10 1012 ) = 108 s Then t = 4.61108 s15.19 Low frequency gain = 10 f 3dB = 10 103 Hz =C1 = 10 C2fC C2 2 CFSet f C = 10 f3dB = 100 kHz Then C2 2 (10 103 ) = = 0.628 CF 100 103 The largest capacitor is C1 , so let C1 = 30 pF Then C2 = 3 pF and CF = 4.78 pF 15.20 a. Req =Time constant = Req CF = r where 1 1 = = 2 106 12 3 f C C1 (100 10 )(5 10 )Then r = (2 106 )(30 1012 ) or r = 60 s b.v0 = 1 vI dt ror v0 =(1)TC 1 ,TC = r fCwww.elsolucionario.net 684. So v0 =1 (60 10 )(100 103 ) 6or v0 = 0.167 V c. Now v0 = 13 = N (0.167) or N = 78 clock pulses 15.21 1 1 = 2 3RC 2 3(104 )(0.10 1012 ) f O = 91.9 MHz fO =R2 = 8R = 80 k15.22 a. sRCV R v0 = v0 R + (1/ sCV ) 1 + sRCV R sRC v = v v2 = 1 1 1 + sRC 1 R+ sC R sRC v3 = v = v 1 2 1 + sRC 2 R+ sC R2 v0 = v3 R Then 2 R sRC sRCV v0 = 2 v0 R 1 + sRC 1 + sRCV v1 =Set s = j j RCV 2 R 2 C 2 1 + 2 j RC 2 R 2 C 2 1 + j RCV The real part of the denominator must be zero. 1 2 R 2 C 2 2 2 R 2 CCV = 0 so 1 0 = R C (C + 2CV ) 1= R2 Rb. f 0,max =1 2 (10 ) (10 411)(1011 + 2[1011 ])f 0,max = 919 kHz f 0,min =1 2 (10 ) (10 411)(1011 + 2[50 1012 ])f 0,min = 480 kHzwww.elsolucionario.net 685. 15.23 1fO = RC =2 6 RC 1 2 6 fORC = 2.32 10=1 2 6(28 103 )62.32 106 R = 18.56 k 125 1012 R2 = 29 R R2 = 538.4 k R=15.24 v0 v1 v1 v1 v2 (1) = + 1 1 R sC sC v or (v0 v1 ) sC = 1 + (v1 v2 ) sC R v1 v2 v2 v2 (2) = + 1 1 R +R sC sC v v ( sC ) or (v1 v2 ) sC = 2 + 2 R 1 + sRC v0 v2 (3) = 1 R2 +R sC v v sC or 2 = 0 1 + sRC R2 so v0 v2 = (1 + sRC ) sR2 C From (2) 1 sC v1 ( sC ) = v2 sC + + R 1 + sRC or v (1 + sRC ) 1 1 1 + + v1 = 0 sR2 C sRC 1 + sRC From (1) 1 v0 ( sC ) = v1 sC + + sC v2 ( sC ) R Then v0 1 v0 (1 + sRC ) 1 + sRC 1 v0 = 2 + + + sR C (1 + sRC ) sRC sR2 C 1 + sRC sRC 2 2 1 + 2sRC 1 + sRC (1 + sRC ) + sRC 1 + sRC 1 = sRC sR2 C ( sRC )(1 + sRC ) sR2 C1 =(1 + 2 sRC )(1 + 2sRC + s 2 R 2 C 2 + sRC ) (1 + sRC )( sRC ) 2 ( sRC ) 2 ( sR2 C ) ( sRC ) 2 ( sR2 C )www.elsolucionario.net 686. Set s =(1 + 2 j RC ) (1 + 3 j RC + 2 R 2 C 2 ) (1 + j RC ) ( 2 R 2C 2 ) j 1 = ( 2 R 2C 2 ) ( j R2C ) ( 2 R 2C 2 ) ( j R2C )The real part of the numerator must be zero. 1 2 R 2 C 2 6 2 R 2 C 2 + 2 R 2 C 2 = 0 6 2 R 2 C 2 = 1 so that 1 0 = 6 RC Condition for oscillation: 2 j RC + 3 j RC 2 j 3 R 3 C 3 + j 3 R 3 C 3 1 = ( 3 R 2 C 2 )( j R2 C ) 1=5 2 R2 C 2 ( RC )( R2 C )But = 0 =1 6RCThen 1 2 2 1 5 (6 R C ) 6 6 1= = ( RC )( R2 C ) RR2 C 2 2 2 6R C 29 (6 R ) R 6 1= or 2 = 29 R2 R 515.25 Let RF 1 = RF 2 = RF 3 RF 1 1 + 5R vo C R 1 Vo 2 = 1 + F vo1 R 1 + 5 RC vo3 + vo 2 v v + o3 + o3 = 0 R 1/ sC R 2 v vo3 + sC = o 2 R R R Vo1 = 1 + F R 1 vo3 = vo 2 2 + 5RC R vo = F vo 3 R R 1 RF 1 RF 1 vo = F vo 1 + 1 + R 2 + 5 RC R 1 + 5 RC R 1 + 5 RC 1= RF R RF 1 1 1 1 + R 2 + 5R 1 + 5R 1 + 5R C C C 2www.elsolucionario.net 687. Let S = j 1= RF R 1 1 1 RF 1 + R 2 + j R 1 + j R 1 + j R C C C =RF R 1 1 RF 1 + R 2 + j R 2 2 2 C 1 + 2 j RC R C =RF R 1 RF 1 + R 2 2 2 2 2 2 3 3 3 2 + 4 j RC 2 R C + j RC 2 R C j R C 222 RF RF 1 1 + R 2 4 2 R 2 C 2 + 5 j RC j 3 R 3C 3 R Imaginary Term must be zero 3 5 j 0 RC j 0 R 3 C 3 = 0 21= 2 5 j 0 R 2 C 2 = 00 =5 RCThen R 1= F R1= RF R 2 1 RF 1+ R 4R2 C 2 5 2 R2C 2 2 RF 1 1 RF 1 + R 2 20 = 18 R RF 1 + R 22R R R 18 = F 1 + F F = 2 R R R15.26 a. v0 v01 v v v = 01 + 01 02 R R 1 sC v01 v02 v v v = 02 + 02 03 R R 1 sC v02 v03 v v = 03 + 03 R 1 R sC R v0 = F v03 R We can write the equations as v0 v01 = v01 ( sRC ) + v01 v02 v01 v02 = v02 ( sRC ) + v02 v03 v02 v03 = v03 ( sRC ) + v03 and R v0 = F v03 R(1)(2)(3)(4) (1) (2) (3)(4)www.elsolucionario.net 688. Combining terms, we find v0 = v01 (2 + sRC ) v02 (1) v01 = v02 (2 + sRC ) v03 (2) v02 = v03 (2 + sRC ) (3) and R v0 = F v03 (4) R Combining Equations (3) and (2) v01 = v03 (2 + sRC ) 2 v03 = v03 (2 + sRC ) 2 1 (2)Then Equation (1) is v0 = v03 (2 + sRC ) 2 1 (2 + sRC ) v03 (2 + sRC ) Using Equation (4), we find R F v03 = V03 (2 + sRC ) 2 1 (2 + sRC ) (2 + sRC ) R To find the frequency of oscillation, set s = j and set the imaginary part of the right side of the equation to zero. We will have R F = (2 + j RC )[4 + 4 j RC 2 R 2 C 2 1 1] R Then j RC (2 2 R 2 C 2 ) + 8 j RC = 0 or j RC 2 2 R 2 C 2 + 8 = 0 {}Then the frequency of oscillation is 1 10 f0 = 2 RC The condition to sustain oscillations is determined from R F = 2[2 2 R 2 C 2 ] 4 2 R 2 C 2 R or R F = 4 6 2 R 2 C 2 R 10 Setting 2 = 2 2 , we have RC RF = 4 6(10) R or RF = 56 R b. For R = 5 k and f 0 = 5 kHz, we find 10 C = 0.02 F 2 (5 103 )(5 103 ) and RF = 56(5) RF = 280 kC=15.27 a.We can writewww.elsolucionario.net 689. R1 vA = R1 + R2 Zp v0 and vB = Z +Z s pwhere Z p = RB and Z s = RA + v0 RB 1 = sCB 1 + sRB CB 1 + sRA C A 1 = sC A sC ASetting v A = vB , we have RB R1 1 + sRB CB = RB 1 + sRAC A R1 + R2 + sC A 1 + sRB CB R1 sRB C A = (1) R1 + R2 sRB C A + (1 + sRAC A )(1 + sRB CB ) To find the frequency of oscillation, set s = j and set the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is j RB C A + (1 + j RAC A )(1 + j RB CB ) or j RB C A + 1 + j RAC A + j RB CB 2 RA RB C ACB (2) Then from (2), we must have 2 1 0 RA RB C ACB = 0 or f0 =1 2 RA RB C ACBb.To find the condition for sustained oscillation, combine Equations (1) and (2). Then R1 j RB C A = R1 + R2 j RB C A + j RAC A + j RB CB )or R2 R C = 1+ A + B R1 RB C A Then R2 RA CB = + R1 RB C A 1+15.28 a. We can write R1 vA = v0 R1 + R2 and R sL vB = v0 R sL + R + sL Setting v A = vB , we havewww.elsolucionario.net 690. sRL R1 R + sL = v0 sRL R1 + R2 + R + sL R + sL R1 sRL = (1) R1 + R2 sRL + ( R + sL) 2 To find the frequency of oscillation, set s = j and se the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is: j RL + ( R + j L) 2 or j RL + R 2 + 2 j RL 2 L2 (2) Then 2 R 2 0 L2 = 0 or R 1 f0 = L 2 b. To find the condition for sustained oscillations, combine Equations (1) and (2). R1 j RL 1 = = R1 + R2 j RL + 2 j RL 3 Then R 1+ 2 = 3 R1 so that R2 =2 R115.29 1 2 RC 1 1 RC = = 2 fO 2 (28 103 ) fO =RC = 5.684 106 For example, Let R = 20 K Then C = 284.2 pFAlsoR2 =2 R115.30 From Equation (15.59) 1 f0 = CC 2 L 1 2 C1 + C2 and from Equation (15.61) C2 = gm R C1www.elsolucionario.net 691. Now, g m = 2 kn I DQ = 2 (0.5)(1) = 1.414 mA / V We have C1 = 0.01 F, R = 4 k, f 0 = 400 kHz So C2 = g m RC1 = (1.414)(4)(0.01) or C2 = 0.0566 F and 400 103 =1 (0.01)(0.0566) 2 L 106 0.01 + 0.0566 2 1 L(8.5 109 ) = = 1.58 1013 3 2 (400 10 ) Then L = 18.6 H15.31 V = V0 V0 1 sC2 +V0 V0 V1 + = g mV = g mV0 RL 1 sC1 1 V0 sC2 + sC1 + + g m = V1 ( sC1 ) RL V1 V0 V1 + + g mV = 0 sL 1 sC1 (1) (2) 1 V1 + sC1 = V0 ( sC1 + g m ) sL V0 ( sC1 + g m ) V1 = 1 + sC1 sL Then V ( sC1 )( sC1 + g m ) 1 V0 s (C1 + C2 ) + + gm = 0 RL 1 + sC1 sL 1 1 + g m + sC1 = sC1 ( sC1 + g m ) s (C1 + C2 ) + RL sL g C1 + C2 sC 1 + s 2 C1 (C1 + C2 ) + + 1 + sg m C1 + m = s 2 C12 + sg m C1 L sRL L RL sL C1 + C2 sC g 1 + s 2 C1C2 + + 1 + m =0 L sRL L RL sL Set s = j g C1 + C2 j C1 1 2 C1C2 + + + m =0 L j RL L RL j Lwww.elsolucionario.net 692. Then2 =C1 + C2 C1 + C2 0 = C1C2 L C1C2 Land gm C1 1 + = L RL L RL Then gm (C + C2 )C1 1 + = 1 L RL L C1C2 LRLgm +1 C1 + C2 = RL C2 RLg m RL + 1 =C1 C + 1 or 1 = g m RL C2 C215.32 a. V0 V0 V0 + + g mV + =0 (1) 1 sL1 R + sL2 sC sL2 V = (2) V0 1 + sL 2 sC Then g (s2 L C ) 1 sC 1 V0 + + + m 2 2 =0 2 sL1 R 1 + s L2 C 1 + s L2 C R (1 + s 2 L2 C ) + ( sL1 )(1 + s 2 L2 C ) s 2 RL1C + g m ( sRL1 )( s 2 L2 C ) + =0 ( sRL1 )(1 + s 2 L2 C ) ( sRL1 )(1 + s 2 L2 C ) Set s = j . Both real and imaginary parts of the numerator must be zero.R(1 2 L2 C ) + j L1 (1 2 L2 C ) 2 RL1C + ( j g m RL1 )( 2 L2 C ) = 0 Real part: R (1 2 L2 C ) 2 RL1C = 0 R = 2 RC ( L1 + L2 ) or 1 0 = C ( L1 + L2 ) b. Imaginary part: j L1 1 2 L2C j gm RL1 2 L2C = 0()((L1 = L1 L2C + gm RL1 L2C 22))www.elsolucionario.net 693. Now 2 =1 ( L1 + L2 )1=1 [ L2 C + g m RL2 C ] C ( L1 + L2 )1=L2 L (1 + g m R ) 1 = (1 + g m R ) L1 + L2 L2or L1 = gm R L215.33 0 = 2 (800 103 ) =1 C ( L1 + L2 )or C ( L1 + L2 ) = 3.96 1014 AlsoL1 = gm R L2For example, if R = 1 k, thenL1 = (20)(1) = 20 L2So L1 = 20 L2 Then C (21L2 ) = 3.96 1014 or CL2 = 1.89 1015 C = 0.01 F If thenL2 = 0.189 HandL1 = 3.78 H15.34 v0 v1 v1 v1 vB = + R 1 R sC and vB v v + B 1 =0 R 1 sC or(1)(2)1 v vB sC + = 1 v1 = vB (1 + sRC ) R R From (1) 2 v v0 ( sC ) = v1 sC + B R R or v0 ( sRC ) = vB (1 + sRC )(2 + sRC ) vB = vB [ (1 + sRC )(2 + sRC ) 1]Nowwww.elsolucionario.net 694. R R2 sRC sRC T ( s ) = 1 + 2 = 1 + R 2 2 2 R1 (1 + sRC )(2 + sRC ) 1 2 + 3sRC + s R C 1 1 or R sRC T ( s ) = 1 + 2 2 2 2 R1 s R C + 3sRC + 1 R j RC T ( j ) = 1 + 2 R1 1 2 R 2 C 2 + 3 j RC Frequency of oscillation: 1 f0 = 2 RC Condition for oscillation: R j RC 1 = 1 + 2 R1 3 j RC or R2 =2 R115.35vb vo vb vb va + + =0 1 1 R sC sC vb vo (1) + 2vb sC va sC = 0 R Va Vb Va (2) + =0 1 R sC 1 1 + sRC Va sC + = vb sC vb = va R sRC From (1) 1 v vb + 2 sC = o + va sC R R Substitute (2) into (1)www.elsolucionario.net 695. 1 + sRC 1 + 2 sRC vo va = + va sC R sRC R (1 + sRC )(1 + 2 sRC ) v va sC = o ( sRC ) R R (1 + sRC )(1 + 2 sRC ) va sRC = vo sRC 2 2 2 vo (1 + sRC )(1 + 2 sRC ) s R C = va sRC va sRC = vo 1 + 3sRC + 2( sRC ) 2 s 2 R 2 C 2 va sRC = vo 1 + 3sRC + ( sRC ) 2 R sRC T ( s ) = 1 + 2 R1 1 + 3sRC + ( sRC ) 2 R j RC T ( j ) = 1 + 2 R1 1 2 R 2 C 2 + 3 j RC 2 So 1 0 R 2 C 2 = 0So f O =1 2 RC R 1 Also 1 = 1 + 2 R1 3 R So 2 = 2 R115.36 v0 v1 v1 v1 vB = + sL R R sL vB = v1 R + sL or(1) (2) R + sL v1 = vB sL Then v0 1 2 v = v1 + B sL sL R R or v0 R + sL 1 2 vB = + vB sL sL sL R R R + sL R + 2 sL 1 = vB sL sRL R Then(1)www.elsolucionario.net 696. vB =v0 1 sL ( R + sL)( R + 2 sL) ( sL) 2 ( sL)( sRL) Now R sRL T ( s ) = 1 + 2 2 R1 R + 3sRL + 2 s 2 L2 s 2 L2 or R sRL T ( s ) = 1 + 2 2 2 R1 s L + 3sRL + R 2 And R j RL T ( j ) = 1 + 2 2 R1 R 2 L2 + 3 j RL R Frequency of oscillation: f 0 = 2 L Condition for oscillation: R 1 1 = 1 + 2 R1 3 or R2 =2 R115.37 From Equation (15.65(b)), the crossover voltage is R vI = 2 VREF R1 Let R2 = RVAR + RF where RVAR is the potentiometer and RF is the fixed resistor. Let VREF = 5 V, RF = 10 k, and RVAR = 40 k Then we have R 10 vI = F VREF = (5) = 1 V R1 50 and 50 vI = (5) = 5 V 50 15.38 R1 VTH VTL = (VH VL ) R1 + R2 R1 R1 0.2 = (13 (13) ) = 26 R1 + R2 R1 + R2 R1 So = 0.007692 R1 + R2 13 = 0.25 R1 + R2 = 52 R1 + R2 Then I=www.elsolucionario.net 697. R1 = (0.007692)(52) = 0.4 k = R1So R2 = 51.6 k 15.39 a. R1 10 VTH = VH = (10) R1 + R2 10 + 40 so VTH = 2 V R1 10 VTL = VL = (10) 10 + 40 R1 + R2 so VTL = 2 Vb.vI = 5sin t15.40 a. Upper crossover voltage when v0 = +VP , Now R1 vB = (+VP ) R1 + R2 and RA RB vA = VREF + VTH RA + RB RA + RB v A = vB so that R1 RA VP = R1 + R2 RA + RB or RB VREF + RA + RB VTH R + RB R1 RA VTH = A VP VREF R1 + R2 RB RB Lower crossover voltage when v0 = VP So R + RB R1 RA VTL = A VP VREF R1 + R2 RB RB b. 10 + 20 5 10 VTH = (10) (2) 5 + 20 20 20 or VTH = 2 Vand 10 + 20 5 VTL = (10) 1 VTL = 4 V 5 + 20 20 15.41 a.www.elsolucionario.net 698. vB VREF vB v0 vB = + R1 R3 R2 1 v 1 1 V + = REF + 0 vB + R3 R2 R1 R2 R3 VTH = vB when v0 = +VP and VTL = vB when v0 = VP So VREF VP + R3 R2 VTH = 1 1 1 + + R1 R2 R3 and VTL =VREF VP R3 R2 1 1 1 + + R1 R2 R3 b. VREF 1 1 1 + R3 + R1 R2 R3 10 5 = 1 1 1 + 10 + R1 R2 10 VS =1 1 1 1 + = = 0.10 R1 R2 5 10 VT = VTH VTL =0.2 =2VP R2 1 1 1 + + R1 R2 R3 2(12) R2 (0.10 + 0.10)So R2 = 600 k Then 1 1 + = 0.10 R1 R2 1 1 + = 0.10 R1 = 10.17 k R1 600 c. VTH = 5 + 0.1 = 4.9VTL = 5 0.1 = 5.1 15.42 a.If the saturated output voltage is VP < 6.2 V, then the circuit behaves as a comparatorwhere v0 < 6.2 V.www.elsolucionario.net 699. If the saturated output voltage is VP > 6.2 V, the output will flip to either +VP or VP and the input has no control. b. Same as part (a) except the curve at vI 0 will have a finite slope. c. Circuit works as a comparator as long as v01 < 8.7 V and v02 > 3.7 V. Otherwise the input has no control. 15.43 a.Switching point is when v0 = 0. Then R2 v+ = vI VS = VREF R1 + R2 VTH occurs when vo = VH , then by superposition R1 R2 v+ = VTH = VH + VREF R1 + R2 R1 + R2 or R1 VTH = VS + VH R1 + R2 VTL occurs when v0 = VL , then by superposition R1 R2 v+ = VTL = VL + VREF R1 + R2 R1 + R2 or R1 VTL = VS + VL R1 + R2 b. NowFor VTH = 2 V and VTL = 1 V, then VS = 1.5 V 10 2 = 1.5 + (10) 10 + R2 0.5 10 = R2 = 190 k 10 10 + R2 190 Now VS = 1.5 = VREF 10 + 190 so VREF = 1.579 V15.44 a. NowSwitching point when v0 = 0. R2 v+ = VREF = vI where vI = VS . R1 + R2 Then R + R2 R1 VS = 1 VREF = 1 + VREF R2 R2 Now upper crossover voltage for v1 occurs when v0 = VL and v+ = VREF . Thenwww.elsolucionario.net 700. VTH VREF VREF VL = R1 R2 or VTH = R1 R VL + VREF 1 + 1 R2 R2 R1 VL R2 Lower crossover voltage for vI occurs when v0 = VH and vI = VREF . Then VH VREF VREF VTL = R2 R1 or VTH = VS or VTL = R1 R VH + VREF 1 + 1 R2 R2 or VTL = VS R1 VH R2For VTH = 1 and VTL = 2, VS = 1.5 V. Then VTL = VS b.R1 R VH 2 = 1.5 1 (12) R2 20so that R1 = 0.833 k Now R VS = 1 + 1 VREF R2 0.833 1.5 = 1 + VREF 20 which gives VREF = 1.44 V15.45 (a) vo (max) = 4.7 + 0.7 = 5.4 V vo (min) = 5.4 V R1 VTH TTL = ( 5.4 (5.4) ) R1 + R2 2 0.8 = (10.8) 2 + R2 2 + R2 = 27 R2 = 25 K(b) R=Neglecting current in R1 and R2 , 13.0 5.4 R = 15.2 K 0.515.46 a. v0 = VREF + 2V 5 = VREF + 2(0.7) or VREF = 3.6 V b.www.elsolucionario.net 701. R1 VTH = (VREF + 2V ) R1 + R2 R1 0.5 = (5) R1 + R2 R2 R = 10 2 = 9 R1 R1 For example, let R2 = 90 k and R1 = 10 k c. For vI = 10 V, and v0 is in its low state. D1 is on and D2 is off. or 1 +v1 (v1 + 0.7) VREF v1 v1 v0 + = 100 1 1 For v1 = 0.7, then 10 0 3.6 (0.7) 0.7 v0 + = 100 1 1 or v0 = 5.1 V15.47 For v0 = High = (VREF + 2V ). Then switching point is when. R1 vI = vB = v0 R1 + R2 R1 or VTH = (VREF + 2V ) R1 + R2 Lower switching point is when R1 v1 = vB = v0 and v0 = (VREF + 2V ) R1 + R2 so R1 VTL = (VREF + 2V ) R1 + R2 15.48 By symmetry, inverting terminal switches about zero. Now, for v0 low, upper diode is on. VREF v1 = v1 v0 v0 = 2v1 VREF where v1 = V so v0 = (VREF + 2V ) Similarly, in the high state v0 = (VREF + 2V ) Switching occurs when non-inverting terminal is zero. So for v0 low. VTH 0 0 (VREF + 2V ) = R1 R2 or VTH =R1 (VREF + 2V R2)www.elsolucionario.net 702. By symmetry R VTL = 1 (VREF + 2v ) R2 15.49 f =1 2.2 RX C X1 1 = 2.2 f (2.2)(12 103 ) RX C X = 3, 788 105RX C X =For example, Let RX = 56 K C X = 680 pF Within1 of 1% of design specification. 215.50 (a) R1 20 vx1 = vo = vo 20 + 5 R1 + R2 So vH = 9.6 Also vx2 = 12 + (9.6 12)e t / x = 12 21.6e t / x Set vx1 = vx29.6 = 12 21.6e t / x T = x ln (9) 2 1 1 f = = T 2 x ln (9)e+ t / x = 9 t1 =1 2(22 10 )(0.2 106 ) ln (9) f = 51.7 Hz Duty cycle = 50% f =315.51 (a) R1 20 + vx1 = VH = (15) = 12 V 20 + 5 R1 + R2 20 vx1 = (10) = 8 V 20 + 5 + vx 2 = 15 + (8 15)e t / x = 15 23e t / x Then 12 = 15 23e t1 / xe t1 / x = 7.667 t1 x ln (7.667) t2 t1 x vx 2 = 10 + (12 + 10)e Thenwww.elsolucionario.net 703. 8 = 10 + 22e t2 t1 x ( t2 t1 )ex= 11 t2 t1 = x ln (11)Period = t2 = T = x [ ln (7.667) + ln (11) ] = x (4.435) x = (22 103 )(0.2 106 ) = 4.4 103T = 1.95 102 1 f = = 51.2 Hz T t1 = x ln(7.667) = (4.4 10 3 ) ln (7.667)t1 = 8.962 103 t1 8.962 103 = T 1.951 102 Duty Cycle = 45.9%Duty cycle =15.52 t1 = 1.1RX C X = (1.1)(10 4 )(0.1 106 ) t1 = 1.1 ms 0 < t < t1 , vY = 10(1 e t / rY ) rY = RY CYNow= (2 103 )(0.02 106 ) = 4 105 st1 = 2.75 rY CY completely charges during each cycle.15.53 a. Switching voltage R1 + R3 10 + 10 vX = VP = (10) 10 + 10 + 10 R1 + R3 + R2 So v X = 6.667 V Using Equation (15.83(b)) 2 2 v X = VP + VP VP e t1 / rX = VP 3 3 5 2 Then 1 e t1 / rX = 3 3 1 5 t1 / rX or t1 = rX ln (5) = e 3 3 T 1 1 t1 = = t1 = 0.001 s = 2 2 f 2(500) 103 = rX ln (5) rX = 6.21 104 = RX (0.01 106 ) So RX = 62.1 k b.Switching voltage R1 vX = (VP ) R1 + R3 + R2 10 1 = (VP ) = (VP ) 3 10 + 10 + 10 www.elsolucionario.net 704. Using Equation (15.83(b)) 1 1 v X = VP + VP VP e t1 / rX = VP 3 3 4 t1 / rX 1 = Then 1 e 3 3 2 4 t1 / rX = e 3 3 t1 = rX ln (2) = (6.21 104 ) ln (2) = 4.30 104 s T = 2t1 = 8.6 104 s 1 f = f = 1.16 kHz T 15.54 From Equation (15.92) V 1+ V T = rX ln P 1 R1 10 where = = = 0.2857 R1 + R2 10 + 25 so 0.7 1+ 5 100 = rX ln 1 0.2857 so X = 213.9 s = RX C X For example, RX = 10 k, C X = 0.0214 F R1 10 vY = VP = (5) = 1.43 V 10 + 25 R1 + R2 and v X = 0.7 V To trigger the circuit, vY must be brought to a voltage less than v X . Therefore minimum triggering pulse is 0.73 V. Using Equation (15.82) for T < t < T v X = VP + (0.2857VP VP )e t / rXRecovery period is when v X = V = 0.7 V. 0.7 = 5 + (6.43)e t / rX 6.43e t / rX = 4.3 or t = rX ln (1.495) rX = 213.9 s so t = T T = 86.1 s15.55 (a)www.elsolucionario.net 705. 1 + (Vr / VP ) T =x 1 x = Rx Cx = 1 2 T 0.69 x = (0.69)(47 103 )(0.2 106 ) T = 6.49 ms Recovery Time 0.4 x (b) = 3.76 ms15.56 a. From Equation (15.95) T = 1.1 RC For T = 60 s = 1.1 RC then RC = 54.55 s For example, let C = 50 F and R = 1.09 M b.Recovery time: capacitor is discharged by current through the discharge transistor. 5 0.7 If V + = 5 V, then I B = 0.043 mA 100 If = 100, I C = 4.3 mA VC =1 Ic IC dt = C t C2 + V = 3.33 V 3 V C (3.33)(50 106 ) So that t = C = IC 4.3 103Capacitor has charged toSo recovery time t 38.7 ms 15.57 T = 1.1 RC 5 106 = 1.1 RC so RC = 4.545 106 s For example, let C = 100 pF and R = 45.5 k From Problem (15.53), recovery time V C (3.33)(100 1012 ) t C = IC 4.3 103 or t = 77.4 ns 15.58 From Equation (15.102), 1 f = (0.693)(20 + 2(20)) 103 (0.1 106 ) or f = 240.5 Hz Duty cycle =20 + 20 100% = 66.7% 20 + 2(20)www.elsolucionario.net 706. 15.59 f =1 (0.693)( RA + 2 RB )CRA = R1 = 10 k, RB = R2 + xR3 So 10 k RB 110 k 1 = 627 kHz (0.693)(10 + 2(110)) 103 (0.01 106 ) 1 = = 4.81 kHz (0.693)(10 + 2(10)) 103 (0.01 106 )f min = f maxSo 627 Hz f 4.81 kHz Duty cycle =RA + RB 100% RA + 2 RBNow 10 + 10 100% = 66.7% 10 + 2(10) and 10 + 110 100% = 52.2% 10 + 2(110) So 52.2 Duty cycle 66.7% 15.60 1 k RA 51 k 1 k RB 51 k 1 = 1.40 Hz (0.693)(1 + 2(51)) 103 (0.01 106 ) 1 f max = = 2.72 kHz (0.693)(51 + 2(1)) 103 (0.01 106 ) or 1.40 kHz f 2.72 kHz f min =Duty cycle =RA + RB 100% RA + 2 RB1 + 51 100% = 50.5% 1 + 2(51) or 51 + 1 100% = 98.1% 51 + 2(1) or 50.5% Duty cycle 98.1%15.61 a. I E3 = IE 4 =V + 3VEB R1 A + R1BAssume VEB = 0.7www.elsolucionario.net 707. I E3 = I E 4 =22 3(0.7) = 0.398 mA 25 + 25Now 20 I C 3 = I C 4 = I C 5 = I C 6 = (0.398) 21 I C 3 = I C 4 = I C 5 = I C 6 = 0.379 mA 0.398 20 I C1 = I C 2 = 0.018 mA 21 21 I D = 0.398 mA, current in D1 and D2I C1 = I C 2 =b.I 0.398 103 VBB = 2VD = 2VT ln D = 2(0.026) ln 13 10 IS or VBB = 1.149 V = VBE 7 + VEB 8 Now IC 7 IC 4 + IC 9 + I E8 I C 4 = 0.379 mA 20 I B9 = IC 8 = I E 8 21 So I I E 8 = 1.05 I B 9 = 1.05 C 9 100 100 21 IC 7 = IC 4 + I E 8 + I E 8 = I C 4 + (96.24) I C 8 1.05 20 So I C 7 = 0.379 mA + 101I C 8 and I I VBE 7 = VT ln C 7 ; VEB 8 = VT ln C 8 IS IS Then I I 1.149 = 0.026 ln C 7 + ln C 8 I S IS I (0.379 103 ) + 101I C 8 44.19 = ln C 8 (1013 ) 2 2 (1013 ) 2 exp (44.19) = 101I C 8 + 3.79 10 4 I C 8 2 7 1.554 10 = 101I C 8 + 3.79 104 I C 8IC 8 =(3.79 104 ) 2 + 4(101)(1.554 107 ) 3.79 104 2(101) 2(101) I C 8 = 37.4 AI C 7 = 0.379 + 101(0.0374) I C 7 = 4.16 mA 21 I C 9 = 4.16 0.379 0.0374 20 I C 9 = 3.74 mAc.P = (0.398 + 0.398 + 4.16)(22) P = 109 mWwww.elsolucionario.net 708. 15.62 a. b.From Figure 15.47, 3.7 W to the load V + 19 Vc.P=1 VP2 2 RLor VP = 2 RL P = 2(10)(3.7) VP = 8.6 V15.63 1 VP2 P= 2 RL so VP = 2 RP = 2(10)(20) 20 V peak-to-peak output voltage Maximum output voltage of each op-amp = 10 V. Current is (20 /10) = 2 A. Bias op-amps at 12 V. For A1 ,v01 R2 R = 1 + = 15 2 = 14 vI R1 R1For A2 ,v02 R = 4 = 15 vI R3For example, let R1 = R3 = 10 k, and R2 = 140 k and R4 = 150 k. 15.64 a.v01 = iR2 + vI where i =vI R1Then R v01 = vI 1 + 2 R1 Now R v02 = iR3 = vI 3 R1 So R R vL = v01 v02 = vI 1 + 2 vI 3 R1 R1 Av =b.vL R R = 1+ 2 + 3 vI R1 R1Want Av = 10 R2 R3 + =9 R1 R1 R R Also want 1 + 2 = 3 R1 R1 R2 R2 R + 1 + = 9 so 2 = 4 R1 R1 R1 For R1 = 50 k, R2 = 200 kThenandwww.elsolucionario.net 709. R3 = 5 so R3 = 250 k R1 P=c.1 VP2 2 RLor VP = 2 RL P = 2(20)(10) = 20 VSo peak values of output voltages are v01 = v02 = 10 V Peak load current =20 =1 A 2015.65 (a) R vo1 = 1 + 2 vI R1 R2 1 + vI R1 vL = vo1 vo 2 vo 2 = Av =R4 R3vL R2 R4 = 1 + 1 + vI R1 R3 For v01 12 V, vo 2 = 12 V when R3 = R4(b) So vL = 24sin + (V ) R R 10 = 1 + 2 1 + 4 R1 R3 Let R3 = R4(c)ThenR2 =4 R115.66 (a) From Problem 15.65 vO R2 R4 = 1 + 1 + vI R1 R3 For vo1 = vo 2 R3 = R4 Then R vO = 2 1 + 2 vI R1 R R 15 = 2 1 + 2 2 = 6.5 R1 R1 (b)Peak output voltageVP = 2 RL PL = 2(16)(20) = 25.3 VThen Vo1 = Vo 2 = 12.65 V iL (peak) =25.3 = 1.58 A 16www.elsolucionario.net 710. 15.67 Line regulation =V0 V +Now I =V + and VZ = rZ I and V0 = 10VZ R1So V0 = 10 rZ V + R1So Line regulation =V0 10(15) = 1.61% V + 930015.68 R0 f = V0 I 0So R0 f =(10 103 ) 1or R0 f = 10 m 15.69 For V0 = 8 V V + (min) = V0 + I 0 (max) R11 + VBE11 + VBE10 + VEB 5This assumes VBC 5 = 0. Then V + (min) = 8 + (0.1)(1.9) + 0.6 + 0.6 + 0.6 V + (min) = 9.99 V 15.70 a. IC 3 = IC 5 =VZ 3VBE (npn) R1 + R2 + R36.3 3(0.6) = 0.571 mA 0.576 + 3.4 + 3.9 1 0.6 IC 8 = = 0.106 mA 2 2.84 Neglecting current in Q9 , total collector current and emitter current in Q5 is 0.571 + 0.106 = 0.677 Now I Z 2 R4 + VEB 4 = VEB 5 IC 3 = IC 5 =I VEB 4 = VT ln Z 2 I5 I VEB 5 = VT ln C 5 2I S www.elsolucionario.net 711. I Then I Z 2 R4 = VT ln C 5 2I Z 2 0.677 0.026 R4 = ln 0.25 2(0.25) or R4 = 31.5 b.From Example 15.16, VB 7 = 3.43 V . Then R13 V0 = VB 8 = VB 7 R12 + R13 or 2.23 (12) = 3.43 2.23 + R12 3.43(2.23 + R12 ) = (2.23)(12) which yields R12 = 5.57 k15.71 Line regulation =V0 V +Now VB 7 = I C 3 R1 R13 and (V0 ) = VB 7 = I C 3 R1 R12 + R13 VZ I Z rZ and I C 3 = = R1 + R2 + R3 R1 + R2 + R3and I Z =V V + where r0 = A IZ r0Then 0.015 (0.4288)( V0 ) = I C 3 (3.9) = (3.9) I Z 7.876 50 = 87.6 k r0 = 0.571 Then V + (0.4288)(V0 ) = (0.00743) 87.6 So V0 = 0.0198% V +15.72 a.www.elsolucionario.net 712. IZ =25 5 = 10 R1 + rZ20 = 2 k = R1 + 0.01 R1 = 1.99 k 10 b. In the ideal case; R3 + R4 V0 = VZ R2 + R3 + R4 So R1 + rZ = 2 +1 V0 = 5 V0 = 6.67 V 2 +1+1 and R4 V0 = VZ R2 + R3 + R4 1 V0 = 5 V0 = 20 V 2 +1+1 So 6.67 V0 20 Vc.3 3 V0 so vd = 5 V0 4 4 and V0 = A0 L vd VBEV1 =I and VBE = VT ln 0 IS Now 3 V0 = A0 L 5 V0 VBE 4 3 V0 1 + A0 L = 5 A0 L VBE 4 5 A0 L VBE V0 = 3 1 + A0 L 4www.elsolucionario.net 713. Load regulation =V0 (NL) V0 (FL) V0 (NL)5 A0 L VBE (NL) 5 A0 L VBE (FL) (1 + 3 A0 L ) (1 + 3 A0 L ) 4 4 = 5 A0 L VBE (NL) (1 + 3 A0 L ) 4 I (FL) VT ln 0 VBE (FL) VBE (NL) I 0 (NL) = = 5 A0 L VBE (NL) 5 A0 L VBE (NL) V0 6.67 = = 1.67 mA 4 k 4 k 1 (0.026) ln 3 1.67 10 3.33 104 % Load regulation = 4 5(10 ) 0.7 I 0 (FL) = 1 A, I 0 (NL) =15.73 V 5.6 IE = Z = = 1.12 mA 5 R2 I0 = 1+ 100 IE = (1.12) I 0 = 1.109 mA Load current 101 For VBC = 0 V0= 20 VZ 0.6 = 20 5.6 0.6or V0 = 13.8 V Then V 13.8 RL = 12.4 k RL = 0 = I 0 1.109 So 0 RL 12.4 kwww.elsolucionario.net 714. Chapter 16 Exercise Solutions EX16.1 COX = =( 3.9 ) (8.85 1014 ) 200 108= 1.726 107 F / cm 22 (1.6 1019 ) (11.7 ) ( 8.85 10 14 )(1015 ) 1.726 1071= 0.1055 V 2 VTN = r 0.576 + VSB 0.576 = ( 0.1055 ) 0.576 + 5 + 0.576 VTN = 0.169 VEX16.2 (a) vo = VDD I D RD k W vo = 3 n 2 L vo = 0.1 2 2 ( 3 0.5 ) vo vo RD 2 0.06 0.1 = 3 ( 5 ) ( 5 )( 0.1) ( 0.1) RD 2 0.1 = 3 0.0735RDRD = 39.5 K (b) 2 0.06 ( 5 )( 39.5 )(VIt 0.5 ) + (VIt 0.5 ) 3 = 0 2 5.925 (VIt 0.5 ) + (VIt 0.5 ) 3 = 0 2(VIt 0.5 ) = VOt=1 1 + 4 ( 5.925 )( 3) 2 ( 5.925 )VOt = 0.632 V VIt = 1.132 VEX16.3 (a) (i) vo = VDD VTNL = 3 0.4 vo = 2.6 V (ii) 2 W W 2 2 ( vI 0.4 ) vo vo = [VDD vo 0.4] L D L L 2 16 2 ( 2.6 0.4 ) vo vo = 2 [3 vo 0.4] 22 2 35.2vo 8vo = 6.76 5.2vo + vo 2 9vo 40.4vo + 6.76 = 0vo =40.4 1632.16 4 ( 9 )( 6.76 ) 2 (9)vo = 0.174 Vwww.elsolucionario.net 715. (b) 2 60 iD = (2) [3 0.174 0.4] 2 iD = 353.1 AP = iD VDD = 1.06 mWEX16.4 (a) 2 W W 2 2 ( vI 0.4 ) vo vo = ( ( 0.8 ) ) L D L L 2 6 2 ( 3 0.4 ) vo vo = 2 ( 0.64 ) 2 6vo 31.2vo + 1.28 = 0vo =31.2 973.44 4 ( 6 )(1.28 ) 2(6)vo = 41.4 mV (b) 2 2 W W ( vIt 0.4 ) = ( (0.8) ) L D L L 6 ( vIt 0.4 ) = 2 ( 0.64 ) 2 vIt = 0.862 V Driver vOt = 0.862 0.4 = 0.462 V vIt = 0.862 V Load vOt = VDD + VTNL = 3 0.8 = 2.2 V (c) 2 60 iD = (2) ( ( 0.8 ) ) = 38.4 A 2 P = iD VDD = 115.2 WEX16.5 We have{} 0.73 } VOH = VDD VTNLO + r 2 fP + VSB 2 fP {VOH = 5 0.8 + 0.35 0.73 + VOH VOH 4.499 = 0.35 0.73 + VOH Squaring both sides 2 VOH 8.998VOH + 20.241 = 0.1225(0.73 + VOH ) 2 VOH 9.1205VOH + 20.15 = 0 9.1205 83.1835 4(20.15) 2 = 3.76 VVOH = VOHEX16.6 a. i. A = logic 1 = 10 V, B = logic 0 A driver in nonsaturation. B driver offwww.elsolucionario.net 716. 2 kn W k W 2 2 L ( VTNL ) = 2 L 2 ( vI vTND ) VOL VOL D L 2 ( 3) = (10 ) 2 (10 1.5 ) V0 L V02L 29 = 5 (17V0 L V02L )5V02L 85V0 L + 9 = 0(85 )85 2 4 ( 5 )( 9 ) V0 L = 0.107 V 2(5) ii. A = B = logic 1 2 kn W k W 2 2 L ( VTNL ) = 2 2 L 2 ( vI vTND )VOL VOL D L V0 L =2 ( 3) = ( 2 )(10 ) 2 (10 1.5 ) V0 L V02L 29 = 10 (17V0 L V02L )10V02L 170V0 L + 9 = 0 V0 L =170 (170 ) 4 (10 )( 9 ) V0 L = 0.0531 V 2 (10 ) 2b.Both cases. 35 2 iD = ( 2 )( 3) = 315 A P = iD VDD P = 3.15 mW 2EX16.7 800 = 160 A 5 35 W 2 W W iD = 160 = (1.4 ) = 34.3 = 4.66 L L 2 L L L L P = iD VDD iD =iD = 160 =35 W 2 W 2 ( 5 0.8 )( 0.12 ) ( 0.12 ) = 9.20 L D 2 L D EX16.8 (a)VOPt VONt(b)VDD 2.1 = = 1.05 V 2 2 = VIt VTD = 1.05 (0.4) = 1.45 V = VIt VTN = 1.05 0.4 = 0.65 VVIt =VIt = VOPt2.1 + (0.4) + 0.5(0.4)1 + 0.5 = 1.16 + 0.4 = 1.56 V= 1.16 VVONt = 1.16 0.4 = 0.76 V(c)VIt = VOPt VONt2.1 + (0.4) + 2(0.4)= 0.938 V 1+ 2 = 0.938 + 0.4 = 1.338 V = 0.538 VEX16.9www.elsolucionario.net 717. 2 P = f CL VDD( 0.10 10 ) = f ( 0.5 10 ) ( 3) 6122f = 2.22 104 Hz f = 22.2 kHzEX16.10 a. K n / K p = 200 / 80 = 2.5 VIt =10 2 + 2.5(2) 1 + 2.5 VIt = 4.32 VV0 Pt = 6.32 V V0 Nt = 2.32 Vb. VIL = 2 + 10 2 2 2.5 2 1 VIL = 3.39 V 2.5 1 2.5 + 3 1 {(1 + 2.5)(3.39) + 10 (2.5)(2) + 2} 2 = 9.43 VV0 HU = V0 HUVIH = 2 + 10 2 2 2(2.5) 1 VIH = 4.86 V 2.5 1 3(2.5) + 1 (4.86)(1 + 2.5) 10 (2.5)(2) + 2 2(2.5) = 0.802 VV0 LU =V0 LU c. NM L = VIL V0 LU = 3.39 0.802 NM L = 2.59 V NM H = V0 HU VIH = 9.43 4.86 NM H = 4.57 VEX16.11 3 PMOS in series and 3 NMOS in parallel. Worst Case: Only one NMOS is ON in Pull-down mode same as the CMOS inverter Wn = W . All 3 PMOS are on during pull-up mode W p = 3(2W ) = 6W . EX16.12 NMOS: Worst Case, M NA , M NB on, Wn = 2(W ) or M NC , M ND or M NC , M NE on Wn = 2(W ). PMOS: M PA and M PC on or M PA and M PB on WP = 2(2W ) = 4W If M PD and M PE on, need WP = 2(4W ) = 8W EX16.13 a. vI = = 5 V v0 = 4 V b.vI = 3 V, = 5 V v0 = 3 Vc.vI = 4.2 V, = 5 V v0 = 4 Vd.vI = 5 V, = 3 V v0 = 2 VEX16.14 (a) vI = 8V , = 10V vGSD = 8 V M D in nonsaturationwww.elsolucionario.net 718. 2 K D 2(vGSD VTND )vO vO K L [VDD vO VTNL ]2KD K 2 2 2 ( 8 2 )( 0.5 ) ( 0.5 ) = [10 0.5 2] D = 9.78 KL KL (b) vI = = 8V vGSD = 6 V KD K 2 2 2 6 2 )( 0.5 ) ( 0.5 ) = [10 0.5 2] D = 15 ( KL KLEX16.15 16 K 16384 cells Total Power = 125 mW = (2.5) IT IT = 50 mA50 mA I = 3.05 A 16384 V 2.5 R = 0.82 M or R = DD = 3.05 IThen, for each cell, I =Now, I VDD RTYU16.1 P = iD VDD iD =750 = 150 A 535 W 2 (5 0.2 0.8) 2 L L W W 150 = 280 = 0.536 L L L L 150 = k W 2 iD = n 2(vI VTND )vO vO 2 L D 35 W 150 = 2(4.2 0.8)(0.2) (0.2) 2 2 L D W W 150 = 23.1 = 6.49 L D L DTYU16.2 P = iD VDD I D =350 = 70 A 52 k W iD = n ( VTNL ) 2 L L 35 W 2 W 70 = ( 2 ) = 1 2 L L L L35 W 2 2 ( 5 0.8 )( 0.05 ) ( 0.05 ) 2 L D W W 70 = 7.31 = 9.58 L D L DiD =TYU16.3www.elsolucionario.net 719. 800 = 160 A 5 35 W 2 W iD = 160 = (1.4 ) = 4.66 2 L L L L P = iD VDD iD =iD = 160 A =35 1 W 2 W 2 ( 5 0.8 )( 0.12 ) ( 0.12 ) = 27.6 L D 2 3 L D TYU16.4 a. From the load transistor: kn W 35 2 2 I DL = (VGSL VTNL ) = ( 0.5 )( 5 0.15 0.7 ) 2 2 L L or I DL = 150.7 A Maximum v0 occurs when either A or B is high and C is high. For the two NMOS is series, the effective k N is cut in half, so I DL = 1 k n W 2 2 (VGSD VTND ) VDS VDS 2 2 L D or 1 35 W 2 2 ( 5 0.7 )( 0.15 ) ( 0.15 ) 2 2 L D which yields W = 13.6 L D 150.7 =b.P = iD VDD = (150.7 )( 5 ) P = 753 WTYU16.5 a. v0 (max) occurs when A = B = 1 and C = D = 0 or A = B = 0 and C = D = 1 1 W W 2 2 (VTNL ) = 2(vI VTND )vO vO 2 L D L L 1 W (0.5)(1.2)2 = 2(5 0.7)(0.15) (0.15) 2 2 L D W W 0.72 = (0.634) = 1.14 L D L Db. 2 k W 35 iD = n (VTNL ) 2 = (0.5) [ (1.2) ] 2 2 L L iD = 12.6 AP = iD VDD = (12.6)(5) P = 63 WTYU16.6 a. K n = K p = 50 A / V 2 VIt = 2.5 V iD (max) = K n (VIt VTN ) 2 = 50(2.5 0.8) 2 iD (max) = 145 Awww.elsolucionario.net 720. b. K n = K p = 200 A / V 2 VIt = 2.5 V iD (max) = (200)(2.5 0.8) 2 iD (max) = 578 ATYU16.7 a. 5 2 + (1)(0.8) VIt = 1+1 VIt = 1.9 V V0 Pt = 3.9 V V0 Nt = 1.1 V b. 3 VIL = 0.8 + [5 2 0.8] VIL = 1.63 V 8 1 V0 HU = {2(1.63) + 5 0.8 + 2} 2 V0 HU = 4.73 V 5 VIH = 0.8 + (5 2 0.8) VIH = 2.18 V 8 1 V0 LU = {2(2.18) 5 0.8 + 2} 2 V0 LU = 0.275 V c. NM L = VIL V0 LU = 1.63 0.275 NM L = 1.35 V NM H = V0 HU VIH = 4.73 2.18 NM H = 2.55 VTYU16.8www.elsolucionario.net 721. TYU16.9 NMOS 2 transistors in series Wn = 2 (W ) = 2W PMOS 2 transistors in series W p = 2 ( 2W ) = 4WTYU16.10 The NMOS part of the circuit is:www.elsolucionario.net 722. TYU16.11 The NMOS part of the circuit is:TYU16.12 Exclusive-ORA 0 1 0 1B 0 0 1 1f 0 1 1 0www.elsolucionario.net 723. TYU16.13NMOS conducting for 0 vI 4.2 V NMOS Conducting: 0 t 8.4 s NMOS Cutoff: 8.4 t 10 s PMOS cutoff for 0 vI 1.2 V PMOS Cutoff: 0 t 2.4 s PMOS Conducting: 2.4 t 10 sTYU16.14 (a) 1 K 32 32 array Each row and column requires a 5-bit word 6 transistors per row and column, 32 6 + 32 6 = 384 transistors plus buffer transistors. (b) 4 K 64 64 array Each row and column requires a 6-bit word 7 transistors per row and column 64 7 + 64 7 = 896 transistors plus buffer transistors. (c) 16 K 128 128 array Each row and column requires a 7-bit word 8 transistors per row and column 128 8 + 128 8 = 2048 transistors plus buffer transistors. TYU16.15 From Equation (16.84) 2 (W / L )nA 2 (VDDVTN ) 3VTN 2(2.5)(0.4) 3(0.4) 2 = = = 0.526 2 2 (W / L )n1 (VDD 2VTN ) ( 2.5 2(0.4) ) From Equation (16.86)www.elsolucionario.net 724. (W / L ) p (W / L )nB=2 2(2.5)(0.4) 3(0.4) 2 kn 2 (VDDVTN ) 3VTN = (2.5) = 0.862 2 k (2.5 0.4) 2 (VDD + VTP ) pW W So of transmission gate device must be < 0.526 times the of the NMOS transistors in the L L W W inverter cell. The of the PMOS transistors must be < 0.862 times the of the transmission gate L L W W devices. Then the of the PMOS devices must be < 0.453 times of NMOS devices in cell. L LTYU16.16 Initial voltage across the storage capacitor = VDD VTN = 3 0.5 = 2.5 V . Now dV I I = C or V = t + K dt C 2.5 = 1.25 V , and C = 0.05 pF . Then where K = 2.5 V , t = 1.5 ms, V = 2 I (1.5 103 ) 1.25 = 2.5 (0.05 1012 ) I = 4.17 1011 A I = 41.7 pAwww.elsolucionario.net 725. Chapter 16 Problem Solutions 16.1 (a) VTN = Cax =2e s N a Cax 2 fp + VSB 2 fp ax (3.9)(8.85 10 14 ) = = 7.67 108 450 108 tax2e s N a = 2 (1.6 10 19 ) (11.7 ) ( 8.85 1014 )( 8 1015 ) Then 5.15 108 VTN = 2(0.343) + VSB 2(0.343) 7.67 108 For VSB = 1 V :1/ 2= 5.15 10 8VTN = 0.671 1.686 0.686 VTN = 0.316 V For VSB = 1 V : VTN = 0.671 2.686 0.686 VTN = 0.544 V (b) For VGS = 2.5 V, VDS = 5 V, transistor biased in the saturation region. ID For VSB ID For VSB= K n (VGS VTN ) 2 = 0, = 0.2(2.5 0.8) 2 = 0.578 mA = 1,I D = 0.2 ( 2.5 [ 0.8 + 0.316]) = 0.383 mA 2For VSB = 2, I D = 0.2 ( 2.5 [ 0.8 + 0.544]) = 0.267 mA 216.2 (a) ID =VDD vO 2 = K n 2(VGS VTN )vO vO RD5 (0.1) 2 = K n 2 ( 5 0.8 )( 0.1) ( 0.1) 3 40 10 5 8 10 W or K n = 1.476 104 A / V 2 = 2 L W So that = 3.69 L b. From Equation (16.10).K n RD [VIt VTN ] + [VIt VTN ] VDD = 0 2(0.1476)(40) [VIt 0.8] + [VIt 0.8] 5 = 0 21 (1) 2 + 4(0.1476)(40)(5) 2(0.1476)(40) or [VIt 0.8] = 0.839 or [VIt 0.8] =www.elsolucionario.net 726. So that VIt = 1.64 V P = I D (max) VDD5 (0.1) = 0.1225 mA 40 or P = 0.6125 mWand I D (max) =16.3 a. From Equation (16.10), the transistor point is found from K n RD (VIt VTN ) 2 + (VIt VTN ) VDD = 0 K n = 50 A / V 2 , RD = 20 k , VTN = 0.8 V (0.05)(20)(VIt VTN ) 2 + (VIt VTN ) 5 = 0 1 1 + 4(0.05)(20)(5) = 1.79 V So VIt = 2.59 V 2(0.05)(20) V0t = 1.79 V Output voltage for vI = 5 V is determined from Equation (16.12): VIt VTN =2 v0 = 5 (0.05)(20) 2(5 0.8)v0 v0 2 v0 9.4v0 + 5 = 0So v0 =b.9.4 (9.4) 2 4(1)(5) = 0.566 V 2(1) For RD = 200 k,1 1 + 4(0.05)(200)(5) = 0.659 V So VIt = 1.46 V 2(0.05)(200) V0t = 0.659 V(VIt VTN ) =2 v0 = 5 (0.05)(200) 2 ( 5 0.8 ) v0 v0 2 or 10v0 85v0 + 5 = 0v0 =16.4 (a)85 (85 ) 4 (10 )( 5 ) = 0.0592 V 2 (10 ) 2P = IVwww.elsolucionario.net 727. 0.25 = I (33) I = 75.76 A 3.3 0.15 R= R = 41.6 K 0.07576 2 k W I = n (VGS VTN ) 2 L 2 80 W W 75.76 = ( 3.3 0.8 ) = 0.303 2 L L (b) VDS (sat) = VGS VTN V VDS (sat) 2 I D = K n (VGS VTN ) = DD R 3.3 (VGS 0.8 ) 0.08 2 ( 0.303) (VGS 1.6VGS + 0.64 ) = 41.6 2 2 0.504 (VGS 1.6VGS + 0.64 ) = 4.1 VGS 2 0.504VGS + 0.1936VGS 3.777 = 00.1936 0.03748 + 4(0.504)(3.777) 2(0.504) = 2.55 VVGS = VGSFor 0.8 VGS 2.55 V Transistor biased in saturation region 16.5 (a) P = I VDD 0.25 = I (3.3) I = 75.76 A For Sat Load 2 80 W W I = 75.76 = ( 3.3 0.15 0.8 ) = 0.343 2 L L L L 2 2 80 80 W ( 0.343)( 3.3 0.15 0.8 ) = I = 75.76 = 2 ( 2.5 0.8 )( 0.15 ) ( 0.15 ) 2 2 L D W = 3.89 L DEq 16.21 3.89 3.3 0.8 + 0.8 1 + 0.343 5.994 = VIt = 4.3677 3.89 1+ 0.343 0.8 VGS 1.372 V16.6 (a) From Equation (16.23) KD K 2 2 2 3 0.5 )( 0.25 ) ( 0.25 ) = ( 3 0.25 0.5 ) D = 4.26 ( KL KLwww.elsolucionario.net 728. KD K 2 2 2 2.5 0.5 )( 0.25 ) ( 0.25 ) = ( 3 0.25 0.5 ) D = 5.4 ( KL KL(b)iD = K L (VGSL VTNL ) = K L (VDD vO VTNL ) 2 2(c) 0.080 2 = (1)(3 0.25 0.5) iD = 0.203 mA 2 P = iD VDD = (0.203)(3) P = 0.608 mWfor both parts (a) and (b). 16.7 P = 0.4 mW = iD VDD = iD (3) iD = 0.1333 mA iD = K L (VDD vO VTNL ) 2 2 0.080 W W 0.1333 = ( 3 0.1 0.5 ) = ( 0.2304 ) 2 L L L L W So = 0.579 L LKD 2 2 2 2.5 0.5 )( 0.1) ( 0.1) = ( 3 0.1 0.5 ) ( KL KD W = 14.8 so that = 8.55 KL L DVIt =(3 0.5 + 0.5 1 + 14.8)1 + 14.8 or VIt = 1.02 V , VOt = 0.52 V16.8 We have KD 2 2 2 ( vI VTND ) vO vO = (VDD vO VTNL ) KL(W / L )D 2 2 2 V V VTN )( 0.08VDD ) ( 0.08VDD ) = (VDD 0.08VDD VTN ) ( DD TN (W / L )L (W / L )D 2 2 2 2 (VDD 2 ( 0.2 )VDD ) ( 0.08VDD ) 0.0064VDD = ( 0.92 0.2 )VDD = 0.5184VDD (W / L )L (W / L )D (W / L )D = 5.4 [0.096] = 0.5184 (W / L )L (W / L )L 16.9 VOH = VB VTN = Logic 1 So (a) VB = 4 V VOH (b) VB = 5 V VOH (c) VB = 6 V VOH (d) VB = 7 V VOH For vI = VOH= 3V = 4V = 5V = 5 V ,since VDS = 0www.elsolucionario.net 729. 2 K D 2 ( vI VT ) vO vO = K L [VB vO VT ] 2Then (a)2 (1) 2 ( 3 1)VOL VOL = ( 0.4 ) [ 4 VOL 1] (b) (1) 2 ( 4 1)VOL V(c)2 (1) 2 ( 5 1)VOL VOL = ( 0.4 ) [6 VOL 1] 22 OL VOL = 0.657 V = ( 0.4 ) [5 VOL 1] VOL = 0.791 V 22 VOL = 0.935 V(d) Load in non-sat region iDD = iOL 2 2 (1) 2 ( 5 1) VOL VOL = ( 0.4 ) 2 ( 7 VOL 1)( 5 VOL ) ( 5 VOL ) 2 2 8VOL VOL = ( 0.4 ) 2 ( 6 VOL )( 5 VOL ) ( 25 10VOL + VOL ) 2 2 = ( 0.4 ) 2 ( 30 11VOL + VOL ) 25 + 10VOL VOL 2 2 = ( 0.4 ) 60 22VOL + 2VOL 25 + 10VOL VOL 2 2 8VOL VOL = 14 4.8VOL + 0.4VOL 2 1.4VOL 12.8VOL + 14 = 0VOL =12.8 163.84 4 (1.4 )(14 ) 2 (1.4 )VOL = 1.27V For load VDS ( sat ) = 7 1.27 1 = 4.73V VDS = 5 1.27 = 3.73 non-sat16.10 a.For load VOt = VDD + VTNL = 5 2 = 3 VKD (VIt VTND ) = VTNL KL 500 (VIt 0.8 ) = ( 2 ) 100 VIt = 1.69 V Load VOt = 3 V Driver: VOt = VIt VTND = 1.69 0.8 = 0.89 V VIt = 1.69 V Driver V0t = 0.89 V From Equation (16.29(b)):b. 2 500 2 2(5 0.8)v0 v0 = ( 2 ) 100 2 5v0 42v0 + 4 = 0 v0 =c.42 ( 42 ) 4 ( 5)( 4 ) v0 = 0.0963 V 2 (5) 22iD = K L ( VTNL ) = 100 ( 2 ) iD = 400 A 2www.elsolucionario.net 730. 16.11 2 2 500 2 ( 3 0.5 )( 0.1) ( 0.1) = ( VTNL ) 50 So( VTNL )2= 4.9 VTNL = 2.21 V16.12 (a) P = iD VDD 150 = iD ( 3) iD = 50 A iD = K L (VTNL ) 2 2 80 W W 50 = ( 1) = 1.25 2 L L L L 2 KD 2 2 ( 3 0.5 )( 0.1) ( 0.1) = ( 1) KL K D (W / L ) D W = = 2.04 = 2.55 K L (W / L ) L L D For the Load: VOt = VDD + VTNL = 3 1 VOt = 2 V2.04 (VIt 0.5 ) = ( 1) VIt = 1.20 V For the Driver: VOt = VIt VTND = 1.20 0.5 VOt = 0.70 V VIt = 1.20 V(b)NM L = VIL VOLU NM H = VOHU VIHVIL = 0.5 + ( 1) = 0.902 V ( 2.04 )(1 + 2.04 )2 ( 1) = 1.31 V VIH = 0.5 + 3 ( 2.04 ) Then VOHU = ( 3 1) + ( 2.04 )( 0.902 0.5 ) = 2.82 VVOLU =(1.31 0.5)= 0.405 V 2 NM L = 0.902 0.405 NM L = 0.497 V NM H = 2.82 1.31 NM H = 1.51 V16.13 a. From Equation (16.29(b)): 2 W W 2 2 ( 2.5 0.5 )( 0.05 ) ( 0.05 ) = [ ( 1)] L L L D W =1 L Lwww.elsolucionario.net 731. W = 5.06 L DThen2 80 iD = (1) ( 1) 2 or iD = 40 Ab.P = iD VDD = ( 40 )( 2.5 ) P = 100 W16.14 a.vI = 0.5 V iD = 0 P = 0 vI = 5 V, From Equation (16.12),i. ii.2 v0 = 5 ( 0.1)( 20 ) 2 ( 5 1.5 ) v0 v0 2 2v0 15v0 + 5 = 0v0 =15 (15 ) 4 ( 2 )( 5 ) v0 = 0.35 V 2 ( 2) 25 0.35 = 0.2325 mA 20 P = iD VDD = ( 0.2325 )( 5 ) P = 1.16 mWiD =b. ii.vI = 0.25 V iD = 0 P = 0 i. vI = 4.3 V, From Equation (16.23),2 100 2 ( 4.3 0.7 ) v0 v0 = 10 [5 v0 0.7 ] 22 2 10 7.2v0 v0 = 18.49 8.6v0 + v0 Then 2 11v0 80.6v0 + 18.49 = 0 v0 =80.6 (80.6 ) 4 (11)(18.49 ) v0 = 0.237 V 2 (11) 2Then iD = 10 [5 0.237 0.7 ] = 165 A 2P = iD VDD = (165 )( 5 ) P = 825 Wc.i. ii.vI = 0.03 V iD = 0 P = 0 vI = 5 V 2iD = K L ( VTNL ) = (10 ) ( 2 ) = 40 A 2P = iD VDD = ( 40 )( 5 ) P = 200 W16.15 vo1 = 3.8V Load & Driver in Sat, region, ML1, MD1www.elsolucionario.net 732. iDL = iDD 2 2 W W ( vGSL VTNL ) = ( vGSD VTND ) L L L D(1) ( 5 vo1 0.8 ) = (10 )( vI 0.8 ) 20.16 = 10 ( vI 0.8 )22vI = 0.9265VNow MD2: Non Sat and ML2: Sat iDL = iDD 2 W W 2 ( vGSL VTNL ) = 2 ( vo1 VTND ) vo 2 vo 2 L L L D(1)( 5 vo 2 0.8 ) ( 4.2 vo 2 )222 = (10 ) 2 ( 3.8 0.8 ) vo 2 vo 2 2 = 10 6vo 2 vo 2 2 2 17.64 8.4vo 2 + vo 2 = 60vo 2 10vo 2 2 11vo 2 68.4vo 2 + 17.64 = 0vo 2 =68.4 4678.56 4 (11)(17.64 ) 2 (11)vo 2 = 0.270 V16.16 a. VIHFrom Equation (16.41), 2 ( 2 ) V = 1.95 V = v = 0.8 + IH 01 3( 4)M D 2 in non-saturation and M L 2 in saturation. 2 K D 2 ( v01 VTND ) v02 v02 = K L ( VTNL ) 2 4 2 (1.95 0.8 ) v02 v02 = (1) ( 2 ) 222 4v02 9.2v02 + 4 = 0v02 =9.2 ( 9.2 ) 4 ( 4 )( 4 ) v02 = 0.582 V 2 ( 4) 2Both M D1 and M L1 in saturation region. From Equation (16.28(b)). 4 ( vI 0.8 ) = ( 2 )or vI = 1.8 V b.VIL = 0.8 +( +2 ) = 1.25 V = v01 4 (1 + 4 )M D 2 in saturation, M L 2 in non-saturationwww.elsolucionario.net 733. 2 2 K D [ vO1 VTND ] = K L 2 ( VTNL )( 5 vO 2 ) ( 5 vO 2 ) 4 (1.25 0.8 ) = 2 ( 2 )( 5 v02 ) ( 5 v02 ) 2( 5 v02 ) 5 v02 =22 4 ( 5 v02 ) + 0.81 = 0( 4)42 4 (1)( 0.81)2 (1)= 0.214 Vso v02 = 4.786 V To find vI : 4 ( v01 0.8 ) = (1) ( ( 2 ) ) 22v01 0.8 = 1 v01 1.8 = V VIH = 1.95 V, VIL = 1.25 V c.16.17 a. i. Neglecting the body effect, v0 = VDD VTN Assume VDD = 5 V, then v0 = 4.2 V ii.Taking the body effect into account: From Problem 16.1. VTN = VTN 0 + 0.671 0.686 + VSB 0.686 and VSB = v0 Then v0 = 5 0.8 + 0.671 0.686 + v0 0.686 ()v0 = 4.756 0.671 0.686 + v0 0.671 0.686 + v0 = 4.756 v0 2 0.450 ( 0.686 + v0 ) = 22.62 9.51v0 + v0 2 v0 9.96v0 + 22.3 = 0v0 =b.9.96 ( 9.96 )2 4 ( 22.3) v0 = 3.40 V 2 PSpice results similar to Figure 16.13(a).16.18 Results similar to Figure 16.13(b). 16.19 a. M X on, M Y cutoff. From Equation (16.29(b)): 2 KD 2 2 5 0.8 )( 0.2 ) ( 0.2 ) = ( 2 ) ( KL or b.KD = 2.44 KLFor v X = vY = .5 Vwww.elsolucionario.net 734. 2 2 ( 2.44 ) 2 ( 5 0.8 ) v0 v0 = ( 2 ) 22 4.88v0 41.0v0 + 4 = 0v0 =41 ( 41) 4 ( 4.88 )( 4 ) 2 ( 4.88 ) 2or v0 = 0.0987 V c. 2 80 iD = (1) ( 2 ) = 160 A 2 P = (160 )( 5 ) P = 800 Wfor both parts (a) and (b). 16.20 (a)Maximum value of vO in low state- when only one input is high, then,2 KD 2 2 ( 3 0.5 )( 0.1) ( 0.1) = ( 1) KL KD = 2.04 KL(b) P = iD VDD 0.1 = iD (3) iD = 33.3 A 2 k W iD = n ( VTNL ) 2 L L 2 80 W W 33.3 = ( 1) = 0.8325 2 L L L LW Then = 1.70 L D(c)22 3 ( 2.04 ) 2 ( 3 0.5 ) vO vO = ( 1) vO = 0.0329 V 16.21 (a)One driver in non-sat, 2 2 I D = K L ( VTNL ) = K D 2 (VGS VTND ) VDSD VDSD 2 K 2 ( 1) = D 2 ( 3.3 0.5 )( 0.1) ( 0.1) KLKD = 1.82 KL(b)www.elsolucionario.net 735. P = VDD 0.1 = ( 3.3) I = 30.3 A 2 80 W W 30.3 = I = ( 1) = 0.7575 2 L L L LW = 1.38 L D(c) 0.1 = 0.05 V 2 0.1 = 0.0333 V Three inputs High, vo 3 0.1 = 0.025 V Four inputs High, vo 4Two inputs High, vo (i) (ii)(iii)16.22 a. P = iD VDD 250 = iD ( 5 ) iD = 50 k' W 2 iD = n [ VTNL1 ] 2 L ML1 2 60 W 50 = ( 2 ) 2 L ML1 W So that = 0.417 L ML1KD 2 2 2 ( vI VTND ) vO vO = [ VTNL ] KL 2 KD 2 2 5 0.8 )( 0.15 ) ( 0.15 ) = ( 2 ) ( KLorKD W = 3.23 = 1.35 KL L MD1b. For v X = vY = 0 v01 = 5 and v03 = 4.2 Then 2 2 2 K D 2 2 ( vO1 VTND ) vO 2 vO 2 + K D 3 2 ( vO 3 VTND ) vO 2 vO 2 = K L 2 [ VTNL 2 ] K D 2 8, K D 3 8, K L 2 1 2 2 8 2 ( 5 0.8 ) v02 v02 + 8 2 ( 4.2 0.8 ) v02 v02 = (1) ( 2 ) 22 2 67.2v02 8v02 + 54.4v02 8v02 = 4 Then 2 16v0 121.6v0 + 4 = 0v02 =121.6 (121.6 ) 4 (16 )( 4 ) 2 (16 ) 2So v02 = 0.0330 V 16.23www.elsolucionario.net 736. a.We can write2 2 K x 2 ( v X VTN ) vDSX vDSX = K y 2 ( vY vDSX VTN ) vDSY vDSY = K L [VDD vO VTN ] 2where v0 = vDSX + vDSY We have v X = vY = 9.2 V , VDD = 10 V , VTN = 0.8 V 2 2 As a good first approximation, neglect the vDSX and vDSY terms. Let v0 2vDSX . Then from the first and third terms in the above equation.9 2 ( 9.2 0.8 ) vDSX (1)(10 2vDSX 0.8 ) (151.2 ) vDSX2 84.64 36.8vDSXSo that vDSX = 0.450 V From the first and second terms of the above equation. 9 2 ( 9.2 0.8 ) vDSX 9 2 ( 9.2 vDSX 0.8 ) vDSY or (16.8)( 0.45) = 2 ( 9.2 0.45 0.8) vDSY which yields vDSY = 0.475 V Then v0 = vDSX + vDSY = 0.450 + 0.475 or v0 = 0.925 V We have vGSX = 9.2 V and vGSY = 9.2 vDSX = 9.2 0.45 or vGSY = 8.75 V b. Since v0 is close to ground potential, the body effect will have minimal effect on the results. From a PSpice analysis: For part (a): vDSX = 0.462 V, vDSY = 0.491 V, v0 = 0.9536 V, vGSX = 9.2 V, and vGSY = 8.738 V For part (b): vDSX = 0.441 V, vDSY = 0.475 V, v0 = 0.9154 V, vGSX = 9.2 V, and vGSY = 8.759 V 16.24 a. We can write 2 2 2 K x 2 ( v X VTNX ) vDSX vDSX = K y 2 ( vY vDSX VTNY ) vDSY vDSY = K L [ VTNL ] 2 From the first and third terms, (neglect vDSX ),4 2 ( 5 0.8 ) vDSX = (1) ( 1.5 ) 2or vDSX = 0.067 V 2 From the second and third terms, (neglect vDSY ),4 2 ( 5 0.067 0.8 ) vDSY = (1) ( 1.5 ) 2or vDSY = 0.068 V Now vGSX = 5, vGSY = 5 0.067 vGSY = 4.933 V and v0 = vDSX + vDSY v0 = 0.135 V Since v0 is close to ground potential, the body-effect has little effect on the results. 16.25www.elsolucionario.net 737. (a)0.2 = 0.05 V 4 2 VDSD We have VDS of each driver K L [ VTNL ] = K D 2 (VGSD VTN ) VDSD 22 K 2 ( 1) = D 2 ( 3.3 0.4 )( 0.05 ) ( 0.05 ) KL KD = 3.478 KL(b) P = I VDD 0.15 = I ( 3.3) I = 45.45 A 2 80 W W 45.45 = ( 1) = 1.14 2 L L L LW = 3.95 L D16.26 Complement of (B AND C) OR A ( B C ) + A 16.27 Considering a truth table, we find A B Y 0 0 0 0 1 1 1 0 1 1 1 0 which shows that the circuit performs the exclusive-OR function. 16.28 ( A + B )(C + D) 16.29 (a)Carry-out = A ( B + C ) + B Cwww.elsolucionario.net 738. (b)For vO1 = Low = 0.2 V2 KD 2 2 ( 5 0.8 )( 0.2 ) ( 0.2 ) = ( 1.5 ) KL W W For = 1, then = 1.37 L L L D W So, for M 6 : = 1.37 L 6To achieve the required composite conduction parameter, W For M 1 M 5 : = 2.74 L 1 5 16.30AE: VDS 0.075 2 2 W (1) ( 1) 2 ( 3.3 0.4 )( 0.075 ) ( 0.075 ) L D W W W W W = = 2.33 = = = 4.66 L A L E L B L C L D16.31 Not givenwww.elsolucionario.net 739. 16.32 a.From Equation (16.43), 5 0.8 + 0.8 vI = VIt = = VIt = 2.5 V 1+1 p channel, V0 Pt = 2.5 ( 0.8 ) V0 Pt = 3.3 V n channel, V0 Nt = 2.5 0.8 V0 Nt = 1.7 VcFor vI = 2 V, NMOS in saturation and PMOS in nonsaturation. From Equation (16.49),2 = 2 ( 5 2 0.8 )( 5 v0 ) ( 5 v0 ) 1.44 = 4.4(5 v0 ) (5 v0 ) 2( 2 0.8)2So ( 5 v0 ) 4.4 ( 5 v0 ) + 1.44 = 0 2( 5 v0 ) =4.4 ( 4.4 )2 4 (1)(1.44 )2 or 5 v0 = 0.356 v0 = 4.64 VBy symmetry, for vI = 3 V, v0 = 0.356 V 16.33 (a) 80 K n = ( 2 ) = 80 A / V 2 2 40 K p = ( 4 ) = 80 A / V 2 2 VDD + VTP +(i)VIt =Kn VTN Kp Kn Kp1+=3.3 0.4 + (1)(0.4) 1+1VIt = 1.65 VPMOS: VOt = VIt VTP = 1.65 ( 0.4 ) VOt = 2.05 V NMOS: VOt = VIt VTN = 1.65 ( 0.4 ) VOt = 1.25 V (iii) For vO = 0.4 V : NMOS: Non-sat: PMOS:Sat 2 K n 2 (VGSN VTN )VDS VDS = K p [VSGP + VTP ] 22 ( vI 0.4 )( 0.4 ) ( 0.4 ) = ( 3.3 vI 0.4 ) vI = 1.89 V 22For vO = 2.9 V , By symmetry vI = 1.65 (1.89 1.65 ) vI = 1.41 V(b) 80 K n = ( 2 ) = 80 A/V 2 2 40 K p = ( 2 ) = 40 A/V 2 2 www.elsolucionario.net 740. 80 ( 0.4 ) 40 VIt = 1.44 V 80 1+ 403.3 0.4 +VIt =(i)PMOS: VOt = 1.44 ( 0.4 ) VOt = 1.84 V NMOS: VOt = 1.44 0.4 VOt = 1.04 V For vO = 0.4 V(iii) = 40 3.3 vI 0.4]2 vI = 1.62 V ( )[ For vO = 2.9 V : NMOS: Sat, PMOS: Non-sat (80 ) 2 ( vI 0.4 )( 0.4 ) ( 0.4 )(80 ) [vI 0.4]216.34 (a) (i)2 = ( 40 ) 2 ( 3.3 vI 0.4 )( 0.4 ) ( 0.4 ) vI = 1.18 V From Eq. (16.43), switching voltage VDD + VTP +vI t =(ii)2Kn VTN KpKn 1+ Kp=2 ( 4) ( 0.4 ) 3.2266 12 = vIt = 1.776 V 1.8165 2 ( 4) 1+ 123.3 + ( 0.4 ) +v0 = 3.1 V , PMOS, non-sat; NMOS, sat kp W 2 kn W 2 2 (VSG + VTP ) VSD VSD = (VGS VTN ) 2 L p 2 L n 2 2 40 80 (12 ) 2 ( 3.3 vI 0.4 )( 3.3 3.1) ( 3.3 3.1) = ( 4 ) [ vI 0.4] 2 2 12 [1.16 0.4vI 0.04] = 8 vI2 0.8vI + 0.16 8vI2 1.6vI 12.16 = 0 vI =(iii)1.6 2.56 + 4 ( 8 ) (12.16 ) 2 (8) vI = 1.337 Vv0 = 0.2 V PMOS: sat, NMOS, non-sat.2 2 40 80 (12 ) [3.3 vI 0.4] = ( 4 ) 2 ( vI 0.4 )( 0.2 ) ( 0.2 ) 2 212 8.41 5.8vI + vI2 = 8 [ 0.4vI 0.2] 12vI2 72.8vI + 102.52 = 0 vI =72.8 5299.84 4 (12 )(102.52 ) 2 (12 )vI = 2.222 V (b)www.elsolucionario.net 741. vIt =(i)2 ( 6) ( 0.4 ) 3.5928 4 = 2.732 2 (6) 1+ 43.3 + ( 0.4 ) +vIt = 1.315 V(ii) From (a), (ii) 4 [1.16 0.4vI 0.04] = 12 vI2 0.8vI + 0.16 12vI2 8vI 2.56 = 0 vI =8 64 + 4 (12 )( 2.56 ) 2 (12 ) vI = 0.903 V(iii) From (a), (iii) 8.41 5.8vI + vI2 = 12 [ 0.4vI 0.2] 4 4vI2 28vI + 36.04 = 0 vI =28 784 4 ( 4 )( 36.04 ) 2 ( 4)16.35 a. vI = 1.70 VFor vO1 = 0.6 < VTN vO 2 = 5 VN1 in nonsaturation and P in saturation. From Equation (16.45), 1 2 ( vI 0.8 )( 0.6 ) ( 0.6 )2 = [5 vI 0.8]2 1.2vI 1.32 = 17.64 8.4vI + vI2 or vI2 9.6vI + 18.96 = 0vI =9.6 ( 9.6 )2 4 (1)(18.96 ) 2or vI = 2.78 V V0 Nt v02 V0 Pt b. From symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 0.8 = 1.7 V So 1.7 v02 3.3 V16.36 a. V0 Nt v01 V0 Pt By symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 0.8 = 1.7 V So 1.7 v01 3.3 V b.For vO 2 = 0.6 < VTN vO 3 = 5 VN 2 in nonsaturation and P2 in saturation. From Equation (16.57),www.elsolucionario.net 742. 2 ( vI 2 0.8 )( 0.6 ) ( 0.6 )2 = [5 vI 2 0.8]2 1.2vI 2 1.32 = 17.64 8.4vI 2 + vI22 or vI22 9.6vI 2 + 18.96 = 0 So vI 2 = v01 = 2.78 VFor v01 = 2.78, both N1 and P in saturation. Then 1 vI = 2.5 V 16.37 a. iPeak = K n ( vI VTN ) iPeak = 0.1 ( 2.5 0.8 ) = 0.538 ( mA )1/ 2b.iPeak = 0.1 (1.65 0.8 ) = 0.269 ( mA )1/ 216.38 (a) 50 K n = ( 2 ) = 50 A / V 2 2 25 K p = ( 4 ) = 50 A / V 2 2 I D , peak = K n ( v1 VTN ) = 50 ( 2.5 0.8 ) 22or I D , peak = 144.5 A (b)K n = 50 A / V 2 , K p = 25 A/V 2From Equation (16.55), 50 5 0.8 + (0.8) 25 vIt = = 2.21 V 50 1+ 25 Then I D , peak = K n (VIt VTN ) 2 = 50 ( 2.21 0.8 )2or I D , peak = 99.4 A 16.39www.elsolucionario.net 743. (a)Switching Voltage, Eq. (16.43)2 ( 4) ( 0.4 ) 8 vIt = = 1.65 V = vIt 2 ( 4) 1+ 8 80 2 iD, peak = ( 4 )(1.65 0.4 ) iD , peak = 250 A 2 (b) 3.3 0.4 +2 ( 4) ( 0.4 ) 4 = 1.436 V = vIt vIt = 2 ( 4) 1+ 4 2 80 iD , peak = ( 4 )(1.436 0.4 ) iD , peak = 172 A 2 (c) 3.3 0.4 +2 ( 4) ( 0.4 ) 12 vIt = vIt = 1.776 V 2 ( 4) 1+ 12 2 80 iD , peak = ( 4 )(1.776 0.4 ) iD , peak = 303 A 2 3.3 0.4 +16.40 2 a. P = fCLVDD For VDD = 5 V, P = (10 106 )(0.2 1012 )(5) 2 or P = 50 W For VDD = 15 V, P = (10 106 )(0.2 1012 )(15) 2 or P = 450 W b. For VDD = 5 V, P = (10 106 )(0.2 1012 )(5) 2 or P = 50 W 16.41 (a)2 P = fCLVDD = (150 106 )( 0.4 1012 ) ( 5 ) = 1.5 10 3 W / inverter 2Total power: PT = ( 2 106 )(1.5 103 ) PT = 3000 W !!!! (b)For f = 300 MHz2 1.5 10 = ( 300 106 )( 0.4 1012 ) VDD VDD = 3.54 V316.42 3 = 3 107 W 7 10(a)P=(b)2 P = fCLVDD CL =P 2 fVDDwww.elsolucionario.net 744. (i)CL =(ii)CL =(iii)CL =3 107( 5 10 ) ( 5) 6 CL = 0.0024 pF23 1072 CL = 0.00551pF2 CL = 0.0267 pF( 5 10 ) ( 3.3) 63 107( 5 10 ) (1.5) 616.43 10 = 2 106 W 5 106 P CL = 2 fVDD P=(a) (b)2 106 CL = 0.01 pF(i)CL =(ii)CL =(iii)CL =16.44 (a)For vI VDD , NMOS in nonsaturation(8 10 ) ( 5) 622 1062 CL = 0.023 pF2 CL = 0.111 pF(8 10 ) ( 3.3) 62 106(8 10 ) (1.5) 62 iD = K n 2 ( vI VTN ) vDS vDS and vDS 0 Sodi 1 = D K n 2 (VDD VTN ) rds dvDSOr rds =1 kn W 2 L 2 (VDD VTN ) nor 1rds =W kn L For vI 0, (VDD VTN ) n PMOS in nonsaturation2 iD = K p 2 (VDD vI + VTP ) vSD vSD and vSD 0 for vI 0. So k W di 1 p = D 2 (VDD + VTP ) rsd dvSD 2 L p or rsd =1 1 k pW Lp (VDD + VTP ) www.elsolucionario.net 745. W W (b) For = 2, = 4 L n L p rds = rsd =1 rds = 2.38 k 50 )( 2 )( 5 0.8 ) ( 1( 25 )( 4 )( 5 0.8 ) rsd = 2.38 k W For = 2,. L p rsd =1 rsd = 4.76 k 25 )( 2 )( 5 0.8 ) (Now, for NMOS: vds = id rds or id =vds 0.5 = id = 0.21 mA rds 2.38For PMOS: For rsd = 2.38 k , id =vsd 0.5 = id = 0.21 mA rsd 2.38For rsd = 4.76 k , id =vsd 0.5 = id = 0.105 mA rsd 4.7616.45 From Equation (16.63) 3 VIL = 1.5 + (10 1.5 1.5 ) VIL = 4.125 V 8 and Equation (16.62) 1 V0 HU = 2 ( 4.125 ) + 10 1.5 + 1.5 2 or V0 HU = 9.125 V From Equation (16.69) 5 VIH = 1.5 + (10 1.5 1.5 ) VIH = 5.875 V 8 and Equation (16.68) 1 V0 LU = 2 ( 5.875 ) 10 1.5 + 1.5 2 or V0 LU = 0.875 V Now NM L = VIL V0 LU = 4.125 0.875 NM L = 3.25 V NM H = V0 HU VTH = 9.125 5.875 NM H = 3.25 V16.46 From Equation (16.71) VIL = 1.5 + 100 50 1 = 1.5 + 7 2 ( 0.632 ) 1 2 100 + 3 50 (10 1.5 1.5 ) 100 1 50 www.elsolucionario.net 746. or VIL = 3.348 V From Equation (16.70) 1 100 100 V0 HU = 1 + ( 3.348 ) + 10 (1.5 ) + 1.5 2 50 50 or V0 HU = 9.272 V From Equation (16.77)VIH 100 2 (10 1.5 1.5 ) 50 = 1.5 + 1 = 1.5 + 7 [1.51 1] 100 100 1 3 +1 50 50 or VIH = 5.07 V From Equation (16.76) 100 100 ( 5.07 ) 1 + 10 (1.5 ) + 1.5 50 50 V0 LU = 100 2 50 or V0 LU = 0.9275 V Now NM L = VIL V0 LU = 3.348 0.9275 or NM L = 2.42 V NM H = V0 HU VIH = 9.272 5.07 or NM H = 4.20 V16.47 (a) Kn = KP 3 (VDD + VTP VTN ) 8 3 = 0.4 + ( 3.3 0.4 0.4 ) VIL = 1.3375 V 8 1 VOHu = {2 (1.3375 ) + 3.3 0.4 + 0.4} 2 VOHu = 2.9875 VVIL = VTN +VIH = 0.4 +5 ( 3.3 0.4 0.4 ) VIH = 1.9625 V 81 {2 (1.9625) 3.3 0.4 + 0.4} 2 = 0.3125 VVOLu = VOLuNM H = VOHu VIH = 2.9875 1.9625 NM H = 1.025 V NM L = VIL VOLu = 1.3375 0.3125 NM L = 1.025 V(b)www.elsolucionario.net 747. 2 ( 4) ( 3.3 0.4 0.4 ) 2.5 12 VIL = 0.4 + 2 1 = 0.4 + ( 0.147 ) 2 ( 4) ( 2 )( 4 ) ( 0.333) +3 1 12 12 VIL = 1.505 V VOHu = 2 ( 4) 1 ( 2 )( 4 ) 1 (1.505 ) + 3.3 ( 0.4 ) + 0.4 = {2.5083 + 3.3 0.2667 + 0.4} 1 + 2 12 2 12 VOHu = 2.9708 V 2 ( 4) 2 ( 3.3 0.4 0.4 ) 12 2.5 1 = 0.4 + VIH = 0.4 + [ 0.2302] VIH = 2.1282 V 2 ( 4) ( 0.333) ( 2 )( 4 ) + 1 1 3 12 12 2 ( 4) 2 ( 4) ( 2.1282 ) 1 + 3.3 ( 0.4 ) + 0.4 12 3.547 3.3 0.2667 + 0.4 12 VOLu = = VOLu = 0.2853 V 1.333 2 ( 4) 2 12 NM H = VOHu VIH = 2.9708 2.1282 NM H = 0.8426 V NM L = VIL VOLu = 1.505 0.2853 NM L = 1.22 V16.48 a. v A = vB = 5 V N1 and N 2 on, so vDS1 vDS 2 0 V P and P2 off 1 So we have a P3 N 3 CMOS inverter. By symmetry, vC = 2.5 V (Transition Point). b. For v A = vB = vC vI Want K n ,eff = K p ,eff kn W k 3W = P 2 3L n 2 L P With kn = 2k P , then 2 1 W 1 W = 3 2 3 L n 2 L P 9 W W Or = L n 2 L P c.We have k W 2k 9 W p Kn = n = 2 L n 2 2 L p k W p Kp = 2 L p Then from Equation (16.55)www.elsolucionario.net 748. Kn ( 0.8 ) Kp5 + ( 0.8 ) + VIt = 1+Kn KpNow Kn 9 = ( 2) = 9 Kp 2 Then VIt =5 + ( 0.8 ) + 3 ( 0.8 ) 1+ 3 VIt = 1.65 V16.49 By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V. State N1 N2 N3 N4 N5v01 2 3 40 0 5 0off off on onon off on onoff on off offon on off onLogic function ( v X OR vY ) ( v X AND vZ ) Exclusive OR of ( vX OR vY ) with ( vX AND vZ ) 16.50 W NMOS in Parallel = 2 L n W 4-PMOS in series = 4 ( 4 ) = 16 L p(b)CL doubles current must double to maintain switching speed.W =4 L n W = 32 L p16.51 W 4-NMOS in series = 4 ( 2 ) = 8 L n W 4-PMOS in parallel = 4 L p(b)W L W L = 16 n =8 p16.52www.elsolucionario.neton off on on 749. W NMOS in parallel = 2 L n W 3-PMOS in series = 3 ( 4 ) = 12 L P(a)(b)W L W L =4 n = 24 p16.53 W 3-NMOS in series = 3 ( 2 ) = 6 L n W 3-PMOS in parallel = 4 L p(a)(b)16.54 (a) (b)W L W L = 12 n =8 pY = A( B + C )( D + E )W For NMOS in pull down mode, 3 in series = 3 ( 2 ) = 6 L n For PMOS W =4 L P, A(c)W = 2 ( 4) = 8 L P , B ,C , D , Ewww.elsolucionario.net 750. 16.55 (a) (b)Y = A( BD + CE )(c)NMOS: 3 transistors in series for pull down mode. W For twice the speed: = 2 ( 3)( 2 ) = 12 L n W PMOS: = 2 ( 4 ) = 8 L P, AW = 2 ( 2 )( 4 ) = 16 L P , B ,C , D , E 16.56 (a) (b)Y = A + BC + DEW W NMOS: = 2 =4 L n, A L n , B ,C , D , E PMOS: 3 transistors in series for the pull-up mode(c)www.elsolucionario.net 751. W = 3 ( 4 ) = 12 L p16.57 (a) (b)Y = A + ( B + D)(C + E )(c)W NMOS: = 2 ( 2 ) = 4 L n, AW = ( 2 )( 4 ) = 8 L n , B ,C , D , EPMOS: W = ( 2 ) 3 ( 4 ) = 24 L p 16.58 (a) A classic design is shown:www.elsolucionario.net 752. A, B, C signals supplied through inverters. W W (b) For Inverters, = 1 and = 2 L p L n W W For PMOS in Logic function, let = 1 , then for NMOS in Logic function, = 2.25 L p L n16.59 (a)A classic design is shown:www.elsolucionario.net 753. (b)W W =2 = 1, L ND L NA, NB , NC W W = 8, =4 L PA, PB L PC , PD16.60( A OR B )AND C16.61 W 5-NMOS in series = 5 ( 2 ) = 10 L n W 5-PMOS in parallel = 4 L p16.62 By definition: NMOS off if gate voltage = 0 NMOS on if gate voltage = 5 V PMOS off if gate voltage = 5 V PMOS on if gate voltage = 0 State 1 2N1 off onP 1 on offNA off onNB off offNC off offv01 5 5www.elsolucionario.netN2 on onP2 off offv02 0 0 754. 3 4 5 6off on off onon off on offoff off off offoff off off onoff on off on5 5 5 0on on on offoff off off onLogic function is v02 = ( v A OR vB ) AND vC 16.63 State 1 2 3 4 5 6 Logic function: v03 = ( vX OR vZ ) AND vYv01v02v035 0 5 5 5 05 0 5 0 5 50 5 0 5 0 016.64www.elsolucionario.net0 0 0 5 755. 16.65www.elsolucionario.net 756. 16.66www.elsolucionario.net 757. 16.672 I = CdVC dtSo 1 ( 2I ) t C For VC = 0.5 V VC = www.elsolucionario.net 758. 0.5 = 16.68 (a) (i) (ii) (iii) (b) (i) (ii) (iii) 16.69 (a) (i) (ii) (iii) (b) (i) (ii) (iii)2 ( 2 x 1012 ) t 25 x 1015 t = 3.125 msvO = 0 vO = 4.2 V vO = 2.5 V vO = 0 vO = 3.2 V vO = 2.5 Vvo = 0 vo = 2.9 V vo = 2.4 V vo = 0 vo = 2.0 V vo = 2.0 V16.70 Neglect the body effect. v01 (logic 1) = 4.2 V , v02 (logic 1) = 5 V a. vI = 5 V vGS 1 = 4.2 V b. M 1 in nonsaturation and M 2 in saturation. From Equation (16.23) W L W L Or W L2 W 2 2 ( vGS 1 VTND ) vO1 vO1 = (VDD vO1 VTNL ) L L D 2 2 2 ( 4.2 0.8 )( 0.1) ( 0.1) = (1) [5 0.1 0.8] D W ( 0.67 ) = 16.81 = 25.1 D L DNow v01 = 4.2 V vGS 3 = 4.2 V M 3 in nonsaturation and M 4 in saturation. From Equation (16.29(b)). W L W L2 W 2 2 ( vGS 3 VTND ) vO 2 vO 2 = [ VTNL ] D L L 2 2 2 ( 4.2 0.8 )( 0.1) ( 0.1) = ( 2 ) ( 1.5 ) DW (0.67) = 2.25 L Dwww.elsolucionario.net 759. W Or = 3.36 L D16.71 A B Y 0 0 1 0 1 0 1 0 0 1 1 indeterminate 0.1 Without the top transistor, the circuit performs the exclusive-NOR function. 16.72 B A A 0 1 0 0 1 1 1 0 0 1 0 1 Y = A + AB = A + B Z = Y or Z = ABB 1 0 1 0Y 0 1 1 1Z 1 0 0 016.7316.74 For = 1, = 0, then Y = B. And for = 0, = 1, then Y = A . A multiplexer. 16.75 Y = AC + BC 16.76 Y = AB + AB = A B 16.77www.elsolucionario.net 760. A 0 1 0 1B 0 0 1 1Y 0 1 1 0Exclusive-OR function. 16.78 This circuit is referred to as a two-phase ratioed circuit. The same width-to-length ratios between the driver and load transistors must be maintained as discussed previously with the enhancement load inverter. When 1 is high, v01 becomes the complement of vI . When 2 goes high, then v0 becomes the complement of v01 or is the same as vI . The circuit is a shift register. 16.79 Let Q = 0 and Q = 1 ; as S increases, Q decreases. When Q reaches the transition point of the M 5 M 6 inverter, the flip-flop with change state. From Equation (16.28(b)), KL VIt = ( VTNL ) + VTND KD where K L = K 6 and K D = K 5 . Then 30 VIt = ( 2 ) + 1 VIt = Q = 2.095 V 100 This is the region where both M 1 and M 3 are biased in the saturation region. Then S=K3 ( VTNL ) + VTND = K130 ( 2 ) + 1 200 or S = 1.77 V This analysis neglects the effect of M 2 starting to turn on at the same time. 16.80 Let vY = R, v X = S , v02 = Q, and v01 = Q. Assume VThN = 0.5 V and VThP = 0.5 V. For S = 0, we have the following:www.elsolucionario.net 761. If we want the switching to occur for R = 2.5 V, then because of the nonsymmetry between the two circuits, we cannot have Q and Q both equal to 2.5 V. Set R = Q = 2.5 V and assume Q goes low. For the M 1 M 5 inverter, M 1 in nonsaturation and M 5 in saturation. Then 2 2 K n 2 ( 2.5 0.5 ) Q Q = K p [ 2.5 0.5] Or 2 Kp 4Q Q = 4 Kn For the other circuit, M 2 M 4 in saturation and M 6 in nonsaturation. Then()2 K n ( 2.5 0.5 ) + K n (Q 0.5) 2 = K p 2 5 Q 0.5 ( 2.5 ) ( 2.52 ) Combining these equations and neglecting the Q 3 term, we find KpQ = 1.4 V andkn= 0.916.81 3.3 + ( 0.4 ) + 0.5 vIt = = 1.7 V 1+1 vI = 1.5 V NMOS Sat; PMOS Non Sat 2 = 2 ( 3.3 vI 0.4 )( 3.3 vo1 ) ( 3.3 vo1 ) vo1 = 2.88 V vI = 1.6 V vo1 = 2.693 V vI = 1.7 V vo1 = variable (switching region)( vI 0.5)2vI = 1.8 VNMOS Non Sat; PMOS Sat( 3.3 VI 0.4 )22 = 2 ( vI 0.5 ) vo1 vo1 vo1 = 0.607 V Now vI = 1.5 V, vo1 = 2.88 V vo 0VvI = 1.6 V, vo1 = 2.693 V NMOS Non Sat; PMOS Sat 2 ( 3.3 vo1 0.4 ) = 2 ( vo1 0.5 ) vo vo2 vo = 0.00979 V vI = 1.7 V, v o1 = Switching Mode v0 = Switching Mode. vI = 1.8 V, vo1 = 0.607 V NMOS Sat; PMOS Non Sat( v01 0.5)22 = 2 ( 3.3 v01 0.4 )( 3.3 v0 ) ( 3.3 v0 ) v0 = 3.298 V 16.82 For R = = VDD and S = 0 Q = 0, Q = 1 For S = = VDD and R = 0 Q = 1, Q = 1 The signal is a clock signal. For = 0, The output signals will remain in their previous state. 16.83 a.Positive edge triggered flip-flop when CLK = 1, output of first inverter is D and then Q = D = D .www.elsolucionario.net 762. b.For example, put a CMOS transmission gate between the output and the gate of M 1 driven by aCLK pulse.16.84 For J = 1, K = 0, and CLK = 1; this makes Q = 1 and Q = 0 .For J = 0, K = 1, and CLK = 1 , and if Q = 1, then the circuit is driven so that Q = 0 and Q = 1. If initially, Q = 0, then the circuit is driven so that there is no change and Q = 0 and Q = 1. J = 1, K = 1, and CLK = 1, and if Q = 1, then the circuit is driven so that Q = 0. If initially, Q = 0 , then the circuit is driven so that Q = 1. So if J = K = 1, the output changes state. 16.85 For J = v X = 1, K = vY = 0, and CLK = vZ = 1, then v0 = 0. For J = v X = 0, K = vY = 1, and CLK = vZ = 1, then v0 = 1. Now consider J = K = CLK = 1. With v X = vZ = 1, the output is always v0 = 0, So the output does not change state when J = K = CLK = 1. This is not actually a J K flip-flop. 16.86 64 K 65,536 transistors arranged in a 256 256 array. (a) Each column and row decoder required 8 inputs. (b) (i) Address = 01011110 so input = a7 a6 a5 a4 a3 a2 a1a0 (ii) Address = 11101111 so input = a7 a6 a5 a4 a3 a2 a1a0 (c) (i) (ii)Address = 00100111 so input = a7 a6 a5 a4 a3 a2 a1a0 Address = 01111011 so input = a7 a6 a5 a4 a3 a2 a1a016.87 (a)1-Megabit memory = 1, 048,576 1024 1024 Nuclear & input row and column decodes lines necessary = 10(b) 250K 4 bits 262,144 4 bits 512 512 For 512 lines 9 row and column decoder lines necessary. 16.88 Put 128 words in a 8 16 array, which means 8 row (or column) address lines and 16 column (or row) address lines. 16.89 Assume the address line is initially uncharged, then dV 1 I I = C C or VC = Idt = t C C dt 12 VC C ( 2.7 ) ( 5.8 10 ) = I 250 106 t = 6.26 10 8 s 62.6 nsThen t =16.90www.elsolucionario.net 763. (a) W or L (b)5 0.1 35 W = 1 2 L2 2 ( 5 0.7 )( 0.1) ( 0.1) = 0.329 16 K 16,384 cells2 = 2 A 1 Power per cell = (2 A)(2 V ) = 4 W Total Power = PT = (4 W )(16,384) PT = 65.5 mW iD Standby current = (2 A)(16,384) IT = 32.8 mA 16.91 16 K 16,384 cells 200 12.2 W 16,384 V P 12.2 2.5 iD = = = 4.88 A DD = R = 0.512 M VDD R R 2.5 PT = 200 mW Power per cell =If we want vO = 0.1 V for a logic 0, then k W 2 iD = n 2 (VDD VTN ) vO vO 2 L 2 35 W 4.88 = 2 ( 2.5 0.7 )( 0.1) ( 0.1) 2 L W So = 0.797 L16.92 Q = 0, Q = 1 So D = Logic 1 = 5 V A very short time after the row has been addressed, D remains charged at VDD = 5 V . Then M p 3, M A, and M N 1 begin to conduct and D decreases. In steady-state, all three transistors are biased in the nonsaturation region. Then 2 2 2 K p 3 2 (VSG 3 + VTP 3 ) VSD 3 VSD 3 = K nA 2 (VGSA VTNA ) VDSA VDSA = K n1 2 (VGS 1 VTN 1 ) VDS 1 VDS 1 Or 2 2 K p 3 2 (VDD + VTP 3 )(VDD D ) (VDD D ) = K nA 2 (VDD Q VTNA )( D Q ) ( D Q ) = K n1 2 (VDD VTN 1 ) Q Q 2 (1) Equating the first and third terms: 2 20 2 40 (1) 2 ( 5 0.8 )( 5 D ) ( 5 D ) = ( 2 ) 2 ( 5 0.8 ) Q Q (2) 2 2 As a first approximation, neglect the ( 5 D ) and Q 2 terms. We find 2Q = 1.25 0.25 D (3) Then, equating the first and second terms of Equation (1): 2 2 20 40 (1) 2 ( 5 0.8 )( 5 D ) ( 5 D ) = (1) 2 ( 5 Q 0.8 )( D Q ) ( D Q ) 2 2 www.elsolucionario.net 764. Substituting Equation (3), we find as a first approximation: D = 2.14 V Substituting this value of D into equation (2), we find 2 8.4 ( 5 2.14 ) ( 5 2.14 ) = 4 8.4Q Q 2 We find Q = 0.50 V Using this value of Q, we can find a second approximation for D by equating the second and third terms of equation (1). We have 2 20 2 ( 4.2 Q )( D Q ) ( D Q ) = 40 2 ( 4.2Q ) Q 2 Using Q = 0.50 V , we find D = 1.79 V 16.93 Initially M N 1 and M A turn on. M N 1, Nonsat; M A , sat. K nA [VDD Q VTN ] = K n1 2 (VDD VTN 1 ) Q Q 2 22 40 40 2 (1) [5 Q 0.8] = ( 2 ) 2 ( 5 0.8 ) Q Q 2 2 which yields Q = 0.771 VInitially M P 2 and M B turn on Both biased in nonsaturation reagion 2 2 K P 2 2 (VDD + VTP 3 ) VDD Q VDD Q = K nB 2 (VDD VTNB ) Q Q 2 2 20 40 ( 4 ) 2 ( 5 0.8 ) 5 Q 5 Q = (1) 2 ( 5 0.8 ) Q Q 2 2 () (() ())which yields Q = 3.78 V Note: (W / L) ratios do not satisfy Equation (16.86) 16.94 For Logic 1, v1:( 5 )( 0.05 ) + ( 4 )(1) = (1 + 0.05 ) v1 v1 = 4.0476 V v2 : (5)(0.025) + (4)(1) = (1 + 1.025)v2 v2 = 4.0244 VFor Logic 0, v1: (0)(0.05) + (4)(1) = (1 + 0.05)v1 v1 = 3.8095 V v2 : (0)(0.025) + (4)(1) = (1 + 0.025)v2 v2 = 3.9024 V16.95 Not given 16.96 Not given 16.97 Not givenwww.elsolucionario.net 765. 16.98 Quantization error =(a)Or LSB 0.10 V For a 6-bit word , LSB = 1 LSB =(b)1 LSB 1% 0.05 V 25 = 0.078125 V 645 = 0.078125 V 643.5424 64 = 45.34 n = 45 5 Digital Output = 101101 45 5 = 3.515625 64 . 1 = 3.5424 3.515625 = 0.026775 < LSB. 2(c)16.99 Quantization error =(a)1 LSB = 0.10 V1 LSB 0.5% 0.05 V 210 = 0.078125 V 128 1 LSB = 0.078125 VFor a 7-beit word, LSB = (b)3.5424 128 = 45.34272 n = 45 10 Digital output = 0101101(c)Now45 10 = 3.515625 128 = 3.5424 3.515625 = 0.026775 V