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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