1. Ramey Soft Innovation with Excellence. Date:- ______________ Theres more to a product than just code. We understand there is more that goes into a project than coding. Your software design, strategy, tools, and processes all have an impact on how successful your custom software project or product will be. Our software consulting services are broken into three offers to assist with those key areas of success. Exceed client expectations by going beyond software to provide solutions that transform data into knowledge, enabling them to solve problems and better serve our clients. Also counseling in graduate and post graduate projects. All kinds of software source codes available. Our expertises are: Web Services Data Base Design Web Development Managements Systems (Hospitals, Warehouses, Commercial businesses, Super Stores and so on) Desktop Application Development Android Application Development Java, Asp.net, C/C++, PHP, C#, SQL, MySQL, Graphics, Logo Design Skype: rameysoft +92-311-1379796 +92-334-7595788 Contact Details : Skype: rameysoft Email: rameysoft@gmail.com Contact No. +92-311-1379796 +92-334-7595788

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/ 2 3/ 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 2 EX1.3 For n-type, drift current density J en nE or 200 = (1.6 1019 ) ( 7000 ) (1016 ) E which yields E = 17.9 V / cm EX1.4 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 = 19 Jp 16 2 103 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 and 1/ 2

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 pF 1/ 2 = C jo ( 7.61) 1/ 2 EX1.7 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 V EX1.9 (a) ID = (b) ID = Then R = (c) VPS V R VPS V R 5 0.7 I D = 1.08 mA 4 VPS V R= ID = 8 0.7 = 6.79 k 1.075

4. ID(mA) Diode curve 1.25 1.08 Load lines (b) (a) 0 0.7 2 4 VD(v) 6 8 EX1.10 PSpice analysis EX1.11 Quiescent diode current I DQ = VPS V = 10 0.7 = 0.465 mA 20 R 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 mA Also I = 10 5.6 = 1.79 R = 2.46 k R Test Your Understanding Exercises TYU1.1 (a) T = 400K Eg Si: ni = BT 3 / 2 exp 2kT ni = ( 5.23 1015 ) ( 400 ) or ni = 4.76 1012 cm 3 3/ 2 1.1 exp 6 2 ( 86 10 ) ( 400 )

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

38. 0 20 2.13 For vi > 0, (V = 0 ) i R1 0 R2 RL a. R2 || RL v0 = vi R2 || RL + R1 R2 || RL = 2.2 || 6.8 = 1.66 k 0 1.66 v0 = vi = 0.43 vi 1.66 + 2.2 4.3 v0 ( rms ) = b. v0 ( max ) 2 v0 ( rms ) = 3.04 V 2.14 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 mW 2.15 (a) 40 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 W IZ = (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) 2.16 Ri VI II IZ VZ V0 RL IL VI = 6.3 V, Ri = 12, VZ = 4.8

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 192 b. PZ = I Z VZ = (100 )( 4.8 ) PZ = 480 mW PL = I LV0 = (120 )( 4.8 ) PL = 576 mW 2.17 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 10 10 RL RL = 2 k (c) For Ri = 175 I I = 57.1 mA I L = 26.3 mA I Z = 30.8 mA I Z ( max ) = 40 mA I L ( min ) = 57.1 40 = 17.1 mA RL = 10 RL = 585 17.1 2.18 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 + 50 b. PZ = (1.1875 )(10 ) PZ = 11.9 W PL = I L ( max ) V0 = ( 0.5 )(10 ) PL = 5 W 2.19 (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 V

40. % Reg = b. 2.20 % Reg = = 12.375 10.2375 100% % Reg = 21.4% 10 VL ( 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 A 6 6 = 0.012 A, I L ( min ) = = 0.006 A 500 1000 VPS ( min ) VZ Now I L ( max ) = Now Ri = or 280 = I Z ( min ) + I L ( max ) 15 6 I Z ( min ) = 0.020 A I Z ( min ) + 0.012 Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri = or 280 = VPS ( max ) 6 0.12 + 0.006 VPS ( max ) VZ I Z ( max ) + I L ( min ) VPS ( max ) = 41.3 V 2.21 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 mA Ri = b. VPS ( min ) VZ II = 15 10 Ri = 200 25 For VPS ( max ) I I ( max ) = 25 10 I I ( max ) = 75 mA Ri For I L ( min ) = 0 I Z ( max ) = 75 mA VZ 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 ) = 9.90 V0 = 0.35 V c. % 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 fRVr R Ri + rz = 18.2 + 2 = 20.2 Then

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 3 For 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 ) = 8.5165 VL ( min ) = VZ 0 + I Z ( min ) rZ = 7.95 + ( 0.333)( 0.5 ) = 8.1165 VL = 0.4 V VL 0.4 = % Reg = 5.0% % Reg = V0 ( nom ) 8 Vr = VM VM C = 2 fRC 2 fRVr R = Ri + rz = 3 + 0.5 = 3.5 Then 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 3.5 10 (b) 3.0 For vI < 0, both diodes forward biased 0 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 = 10 I(V)

42. i1(mA) 0.35 10 10 I(V) 3 1 2.25 (a) 1K I 0 1K 2K 15 V1 1 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) 9.42 5.7 0 5.7 15 I (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 1

43. iD(mA) 5.58 5.7 15 I (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) 9.3 3.33 0 (b) 4.03 10 I (V) For vI 4.03 V , iD = 0 10 vo vo ( 10 ) = 10 20 3 Which yields iD = vI 0.605 20 For vI = 10, iD = 0.895 mA For vI > 4.03, iD + iD(mA) 0.895 0 2.27 30 4.03 O 12.5 10.7 10.7 30 I 30 30 10.7 = 0.175 A 100 + 10 v0 = i(10) + 10.7 = 12.5 V For vI = 30 V, i = 10 I(V)

44. b. O 12.5 10.7 0 30 2.28 5 I O R 6.8 K V = 0.6 V vI = 15sin t O 4.4 19.4 2.29 a. V = 0 0 3 V V = 0.6 0 2.4 b. V = 0 20 5 V = 0.6 19.4 5 2.30 O

45. 10 6.7 0 4.7 10 2.31 One possible example is shown. Ri Ii DZ Vign L D VZ 14 V RADIO VRADIO L 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 5 2.33 C I O Vx a. For V = 0 Vx = 2.7 V b. For V = 0.7 V Vx = 2.0 V 2.34 C I 2.35 10 V O

46. 20 O 10 VB 0 0 I 10 20 O 13 10 3 0 VB 3 V I VB 7 20 O 10 7 VB 3 V 0 3 I VB 13 2.36 For Figure P2.32(a) 10 I 0 10 O 20 2.37 a. 10 0.6 I D1 = 0.94 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) V0 = 8.93 V I D1 = ID2 = 0 b. 5 0.6 I D1 = 0.44 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) V0 = 4.18 V I D1 = c. d. 10 = ID2 = 0 Same as (a) (I ) 2 ( 0.5 ) + 0.6 + I ( 9.5 ) I = 0.964 mA V0 = I ( 9.5 ) V0 = 9.16 V I D1 = I D 2 = 2.38 a. b. I I D1 = I D 2 = 0.482 mA 2 I = I D1 = I D 2 = 0 V0 = 10

47. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) I = I D 2 = 0.94 mA I D1 = 0 V0 = 10 I ( 9.5 ) V0 = 1.07 V c. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 I = I D 2 = 0.44 mA I D1 = 0 V0 = 10 I ( 9.5 ) V0 = 5.82 V d. 10 = I ( 9.5 ) + 0.6 + I ( 0.5) I = 0.964 mA 2 I I D1 = I D 2 = 0.482 mA 2 V0 = 10 I ( 9.5 ) V0 = 0.842 V I 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 mA I= I D1 I D3 b. 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 V V1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 V c. 10 4.4 9.5 4.4 0.6 = 0.5 I= I D2 I = 0.589 mA I D 2 = 7.6 mA I D1 = 0 I D 3 = I D 2 I = 7.6 0.589 I D 3 = 7.01 mA V1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 V d. 10 4.4 9.5 4.4 0.6 2 = 0.5 I= I D2 I = 0.589 mA I D 2 = 3.6 mA I D1 = 0 I D 3 = I D 2 I = 3.6 0.589 I D 3 = 3.01 mA 2.40 (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+6

48. V1 = 10 0.6 (1.25 )( 2 ) V1 = 6.9 V I R3 = I D3 0.6 ( 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 R1 or I D1 = 0.833 mA I R2 = 4.4 ( 5 ) R2 + R3 = 9.4 = 0.94 mA 10 I 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 V All diodes are on V1 = 4.4V , V2 = 0.6 V (c) I D1 = 0.5 mA = 10 0.6 4.4 R1 = 10 k R1 I R 2 = 0.5 + 0.5 = 1 mA = I R 3 = 1.5 mA = 4.4 ( 0.6 ) 0.6 ( 5 ) R3 R2 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 = 0.06 vI 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 on 2.42 10 V 10 K D2 D1 I 0 D3 D4 10 K 10 V 10 K

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= 0 4.65 10 4.65 4.65 10 I 4.65 v0 = vI for 4.65 vI 4.65 2.43 a. R1 D2 10 V V0 D1 ID1 R1 = 5 k, R2 = 10 k D1 and D2 on V0 = 0 R2 10 V 10 0.7 0 ( 10 ) = 1.86 1.0 5 10 = 0.86 mA I D1 = I D1 b. 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 V 2.44 If both diodes on (a) VA = 0.7 V, VO = 1.4 V I R1 = IR2 I R1 + I D1 10 ( 0.7 ) = 1.07 mA 10 1.4 ( 15 ) = = 2.72 mA 5 = I R 2 I D1 = 2.72 1.07 I 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 = 0 2.45

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 V 2.46 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 20 2.47 10 K V1 Va VD 10 K Vb V2 ID 10 K 10 K a. V1 = 15 V, V2 = 10 V Diode off Va = 7.5 V, Vb = 5 V VD = 2.5 V ID = 0 b. V1 = 10 V, V2 = 15 V Diode on V2 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 V 2.48 vI = 0, D1 off, D2 on 10 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 V For vI > 21.25 V, vo = 10 V

51. 2.49 a. V01 = V02 = 0 b. V01 = 4.4 V, V02 = 3.8 V c. V01 = 4.4 V, V02 = 3.8 V Logic 1 level degrades as it goes through additional logic gates. 2.50 a. V01 = V02 = 5 V b. V01 = 0.6 V, V02 = 1.2 V c. V01 = 0.6 V, V02 = 1.2 V Logic 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 K VR = 1 V, I = 0.8 mA VPS = 1 + ( 0.8 )( 2 ) VPS = 2.6 V 2.55 I Ph = eA 0.6 103 = (1) (1.6 1019 )(1017 ) A A = 3.75 102 cm 2

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 2 2 (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 V VSD ( sat ) = VSG + VTP = 3 2 = 1 V (a) (b) (c) VSD = 0.5 V Nonsaturation VSD = 2 V Saturation VSD = 5 V Saturation EX3.3 R2 160 VG = (VDD ) = (10 ) = 3.636 V = VGS 160 + 280 R1 + R2 I D = 0.25 ( 3.636 2 ) = 0.669 mA 2 VDS = 10 ( 0.669 )(10 ) = 3.31 V P = I DVDS = ( 0.669 )( 3.31) = 2.21 mW EX3.4 I DQ = K P (VSG + VTP ) 2 1.2 = 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 R2 682 R1 = 200 R1 = 283 K 682 + R1 RD = 10 4 =5K 1.2 EX3.5 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 VGS

53. 2 ( 5 1) VGS = ( 0.5 )(1) (VGS 2VGS + 1) 2 2 0.5VGS 3.5 = 0 VGS = 7 VGS = 2.646 V I D = ( 0.5 )( 2.646 1) I D = 1.354 mA VDS = 10 (1.354 )( 3) = 5.937 V 2 4 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 = 0 VGS = 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 ) 2 2 0.475VGS + 0.0975VGS 3.5713 = 0 VGS = 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) 2 2 0.525 VGS + 0.0025VGS 3.5262 = 0 VGS = 0.0025 0.00000625 + 7.40502 2 ( 0.525) = 2.5893 V I D = ( 0.525)( 2.5893 0.95) = 1.411 2 VDS = 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 = 0 0.0025 0.00000625 + 6.60499 2(0.475) = 2.7027 VGS = VGS I D = (0.475)(2.7027 1.05) 2 = 1.2973 mA VDS = 10 I D (3) = 6.108 V 1.297 I DQ 1.411 mA 5.768 VDS 6.108 V EX3.6

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 D ID = 4.286 VSG 1.2 I D = K p (VSG + VTP ) 2 ( 2 4.286 VSG = (1.2 )( 0.25 ) VSG 2VSG ( 1) + ( 1) 2 4.286 VSG = ( 0.3) V 0.6VSG + 0.3 ) 2 SG 2 0.3VSG + 0.4VSG 3.986 = 0 VSG = ( 0.4 ) 0.4 + 4 ( 0.3)( 3.986 ) 2 2 ( 0.3) Must use + sign VSG = 3.04 V I D = ( 0.25 )( 3.04 1) I D = 1.04 mA 2 VSD = 10 I D ( RS + RD ) = 10 (1.04 )(1.2 + 4 ) VSD = 4.59 V VSD > VSD ( sat ) , Yes EX3.7 VSD = 10 I DQ ( RS + RP ) VSD = 10 K P (VSG + VTP ) ( RS + RP ) 2 Set VSD = VSG + VTP VSG + VTP = 10 ( 0.25 )(VSG + VTP ) ( 5.2 ) 2 1.3 (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 V EX3.8 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 = 0

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 V EX3.9 10 VSG 2 ID = and I D = K p (VSG + VTP ) RS 0.12 = ( 0.050 )(VSG 0.8 ) VSG = 2.35 V 2 10 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 ) 8 KP 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 = 0 VSG = 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 = 0 VSG = 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 = 0

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 = 0 VSG = 4.0873 16.706 + 95.242 2 ( 3.028 ) VSG = 2.422 V I D = 0.119 mA VSD = 8.11 V Summary 0.119 I D 0.121 mA 7.89 VSD 8.11 V EX3.10 V VGS , I D = DD RS I D = K n (VGS VTN ) 2 2 2 10 VGS = (10 )( 0.2 ) (VGS 2VGSVTN + VTN ) 2 10 VGS = 2VGS 8VGS + 8 2 2VGS 7VGS 2 = 0 VGS = 7 (7) + 4 ( 2) 2 2 ( 2) 2 Use + 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 ] 2 2 5 2 ( 4 1) VD VD = ( 5 VD 1) = ( 4 VO ) = 16 8VO + VO2 2 2 2 6VD 38VO + 16 = 0 VD = 38 1444 384 2 ( 6) VD = 0.454 V (b) VI = 2 V Driver: Sat K nD [VI VTND ] = K nL [VDD VO VTNL ] 2 2 5 [ 2 1] = [5 VO 1] 2 2 5 = 4 VO VO = 1.76 V EX3.12 If the transistor is biased in the saturation region

57. I D = K n (VGS VTN ) = K n ( VTN ) 2 2 I D = ( 0.25 )( 2.5 ) I D = 1.56 mA 2 VDS = 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 ) 2 K 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 ) 2 2 K nL = 50 A / V 2 and K nD = 103 A / V 2 EX3.14 For M N I DN = I DP K n (VGSN VTN ) = K p (Vscop + VTP ) 2 2 VGSN = 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 V VDD VO = VSD ( sat ) = VSGP + VTP = ( 5 3.25 ) 1 = 0.75 V So Vot = 5 0.75 Vot = 4.25 V EX3.15 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 mW EX3.16 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

58. 1.5V02 13V0 + 5 = 0 V0 = 13 (13) 4 (1.5 )( 5) V0 = 0.40 V 2 (1.5 ) 2 5 0.40 I R = I D1 = 0.153 mA 30 V1 = V2 = 5 V I 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) 2 5 0.205 I R = 0.160 mA 30 = I D 2 = 0.080 mA IR = I D1 EX3.17 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 2 0.4 = 0.6 (VGS 1 1) VGS1 = 1.82 V 2 EX3.18 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 V EX3.19 (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 V 2 0.1 = K n 2 ( 2 1) 2 (c) 2 K n 2 = 0.1 mA / V 2 0.2 = K n 3 ( 2 1) K n 3 = 0.2 mA / V 2 2 0.2 = K n 4 ( 3 1) K n 4 = 0.05 mA / V 2 2 EX3.20 2

59. VS 2 = 5 5 = 0 RS 2 = I D 2 = K n 2 (VGS 2 VTN 2 ) 5 = 16.7 K 0.3 2 0.3 = 0.2 (VGS 2 1.2 ) VGS 2 = 2.425 V VG 2 = VGS 2 + VS = 2.425 V 2 5 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 K I D1 = K n1 (VGS 1 VTN 1 ) 2 0.1 = 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 + R2 EX3.21 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 = 0.25 EX3.22 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 2 2 EX3.23 Assume the transistor is biased in the saturation region.

60. V I D = I DSS 1 GS VP 2 2 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 ) = 2.33 Yes, the transistor is biased in the saturation region. EX3.24 I D = 2.5 mA V I D = I DSS 1 GS VP 2 2 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 ) 2 R1 + R2 = 2 R1 + R2 = 200 k VG = 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 k EX3.25 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 2 VGS = 4 16 4 ( 0.375 )( 6 ) 2 ( 0.375 ) VGS = 8.86 or VGS = 1.806 V impossible ID = VGS = 1.806 mA RS

61. VD = I D RD 5 = (1.81)( 0.4 ) 5 = 4.278 VSD = 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 + R2 I DQ = 5 mA, VS = I DQ RS = ( 5 )(1.2 ) = 6 V VSDQ = 12 V VD = VS VSDQ = 6 12 = 18 V RD = 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 V 2 VG = 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 V

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 Saturation I 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 2 VTN = 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 Nonsaturation 2 I D = ( 0.274 ) 2 ( 2 + 1.2 )(1) (1) I D = 1.48 mA (iii) VDS = 5 V Saturation I D = ( 0.274 )( 2 + 1.2 ) I D = 2.81 mA 2 TYU3.3 (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 ) 2 I 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 mA TYU3.5 2

63. (a) = 0, VDS ( sat ) = 2.5 0.8 = 1.7 V For 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 2 I 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 , 1 2 2 ro = K n (VGS VTN ) = ( 0.02 )( 0.1)( 2.5 0.8 ) 1 or 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 ) 2 0.4 = 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 ) , Yes TYU3.8

64. VS ( 5 ) ID = and VS = VGS RS So RS = 5 VGS 0.1 I D = K n (VGS VTN ) 2 0.1 = ( 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 2 VD = 2.18 5 VD 5 2.18 RD = = RD = 28.2 k ID 0.1 VDS > VDS ( sat ) , Yes TYU3.9 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) 2 2 K n = 0.389 mA / V = W n Cox 2 L W ( 389 )( 2 ) W = = 19.4 L L ( 40 ) TYU3.10 (a) The transition point is ( VDD VTNL + VTND 1 + K nD / K nL VIt = = ) 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 2 TYU3.11 2 ( VDD VTNL + VTND 1 + K nD /K nL VIt = ) 1 + K nD /K nL 2.5 = ( 5 1 + 1 1 + K nD /K nL ) 1 + K nD /K nL 2.5 + 2.5 K nD /K nL = 5 + K nD /K nL K nD /K nL = b. For VI = 5, driver in nonsaturated region. 5 2.5 = 1.67 K nD /K nL = 2.78 1.5

65. I DD = I DL K nD 2 (VI VTND ) VO VO2 = K nL (VGSL VTNL ) 2 K nD 2 2 (VI VTND ) VO VO2 = [VDD VO VTNL ] K nL 2.78 2 ( 5 1) V0 V02 = [5 V0 1] 2 22.24V0 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 ) 2 TYU3.12 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 2 W 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 ) 2 2 VDSL ( 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 V TYU3.14 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 mA RD = VDD Vo 10 0.35 = RD = 30.3 k ID 0.319 TYU3.15 (a) Transistor biased in the nonsaturation region

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 V Then R = 5 1.5 0.374 R = 261 12 TYU3.16 a. ID = 5 VO = K n 2 (V2 VTN ) VO VO2 RD 5 ( 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 ) 2 TYU3.17 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 V VDS = VD VS = 4.5 0.050 VDS = 4.45 V TYU3.18 2 I D = K 2 (VGS VTN ) VDS VDS 2 = 100 2 ( 0.7 0.2 )( 0.1) ( 0.1) ID = 9 A RD = 2.5 0.1 RD = 267 k 0.009

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 V 2 I D = 0.333 2 ( 2 0.8 )( 0.1) ( 0.1) = 0.0767 mA (d) VGS = 3 V 2 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 mA 2 2 2 3.3 (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 n K n = 0.0666 0.39 = K n ( 4 1.5 ) = 6.25 K n K n = 0.0624 0.77 = K n ( 5 1.5 ) = 12.25 K n K n = 0.0629 2 2 2 From 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 = 0 VDS ( sat ) = VGS VTN = 0 ( 2.5 ) = 2.5 V i. VDS = 0.5 V Biased in nonsaturation ii. 2 I D = (1.1) 2 ( 0 (2.5) )( 0.5 ) ( 0.5 ) I D = 2.48 mA VDS = 2.5 V Biased in saturation I D = (1.1) ( 0 ( 2.5 ) ) I D = 6.88 mA 2 iii. VDS = 5 V Same as (ii) I D = 6.88 mA b. 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

68. VDS = 2.5 V Nonsaturation ii. I D = (1.1) 2(2 (2.5))(2.5) (2.5) 2 I D = 17.9 mA VDS = 5 V Saturation iii. I D = (1.1) ( 2 ( 2.5 ) ) I D = 22.3 mA 2 3.5 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) 25 2.071 1010 tox kn = 41.4 A/V 2 (b) = kn = 828 A/V 2 3.7 a. Cox = Kn = ox ( 3.9 ) ( 8.85 10 = t0 x 450 108 14 ) ox t0 x = 7.67 108 F/cm 2 n Cox W 2 L 1 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 2 3.8 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 2 2

69. Kn = n Cox W L 1 W = ( 600 ) ( 8.63 108 ) 2 2.5 2 K n = (1.036 105 ) W I D = K n (VGS VTN ) 2 1.2 10 3 = (1.036 10 5 ) W ( 5 1) W = 7.24 m 2 3.10 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 + VTP W 2 0.040 W Then 0.225 = = 25.1 ( 3 2.33) L 2 L 3.11 VS = 5 V, VG = 0 VSG = 5 V VTP = 0.5 V VSD ( sat ) = VSG + VTP = 5 0.5 = 4.5 V a. VD = 0 VSD = 5 V Biased in saturation I D = 2 ( 5 0.5 ) I D = 40.5 mA 2 b. VD = 2 V VSD = 3 V Nonsaturation c. 2 I D = 2 2 ( 5 0.5 )( 3) ( 3) I D = 36 mA VD = 4 V VSD = 1 V Nonsaturation d. 2 I D = 2 2 ( 5 0.5 )(1) (1) I D = 16 mA VD = 5 V VSD = 0 I D = 0 3.12 (a) (b) Enhancement-mode From Graph VTP = + 0.5 V 0.45 = k p ( 2 0.5 ) = 2.25 K p K p = 0.20 1.25 = k p ( 3 0.5 ) = 6.25 K p 0.20 2.45 = k p ( 4 0.5 ) = 12.25 K p 0.20 2 2 2 4.10 = k p ( 5 0.5 ) = 20.25 K p 2 (c) 0.202 Avg K p = 0.20 mA/V 2 iD (sat) = 0.20 (3.5 0.5) 2 = 1.8 mA iD (sat) = 0.20 (4.5 0.5) 2 = 3.2 mA

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 V ID = (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 mA 3.15 k p = p Cox = p ox t0 x = ( 250 )( 3.9 ) (8.85 1014 ) t0 x (a) tox = 500 k = 17.3 A/V 2 p (b) 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) = 25 k = 345 A/V 2 p 3.16 14 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 2 k = ( p Cox ) = ( 375 ) ( 6.90 108 ) 25.9 A/V 2 p PMOS: 8.629 1011 t0 x

71. ID = k W 2 p (VSG + VTP ) 2 L p 2 0.0259 W W 0.8 = ( 5 0.6 ) = 3.19 2 L p L p L = 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 m 3.17 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 ) 2 VGS = 4 V, I D = ( 0.2 )( 4 1.2 ) = 1.57 mA 1 r0 = r = 63.7 k ( 0.01)(1.57 ) 0 2 VA = 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 0.00646 3.19 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 V 3.20 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 V VDS (sat) = 2.5 1.39 = 1.11 V

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 3 3.22 Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 106 ) t0 x t0 x = 1.2 105 cm = 1200 Angstroms 3.23 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 + 0.64 2 2 VGS 0.6VGS 2.96 = 0 VGS = 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 2 VDS = VDD I D ( RD + RS ) = 10 ( 0.777 )( 4 + 2 ) VDS = 5.34 V VDS > VDS ( sat ) 3.24

73. ID(mA) 4 (a) Q-pt Q-pt 1.67 (b) 4 5 V (V) DS VGS = 4 V VDS (sat) = 4 0.8 = 3.2 V (a) If Sat I D = 0.25 ( 4 0.8 ) = 2.56 2 VDS = 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 = 0 2.6 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 = 0 VDS = 7.3 53.29 15 2 ( 0.75 ) VDS = 0.741 V 5 0.741 ID = = 1.42 mA 3 3.25 ID(mA) 2.92 (a) Q-pt 1.25 (b) 3.5 5 V (V) SD

74. VSG = VDD = 3.5 (a) VSD ( sat ) = 3.5 0.8 = 2.7 V If biased in Sat region, I D = ( 0.2 )( 3.5 0.8 ) = 1.46 mA VSD = 3.5 (1.46 )(1.2 ) = 1.75 V 2 Biased 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 = 0 VSD = +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 mA VSG = 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 2 Non-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 = 0 VSD = ID = 7.72 59.598 16 use sign VSD = 0.698 V 2 ( 0.8 ) 5 0.698 I D = 1.08 mA 4 3.26 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 ) 2 2 5.33 VSG = 0.5 (VSG 4VSG + 4 ) 2 0.5VSG VSG 3.33 = 0 VSG = 1 (1) 2 + 4 ( 0.5 )( 3.33) 2 ( 0.5 ) VSG = 3.77 V

75. VSD 10 ( 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 V ID = 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 ) 2 0.4 + 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 V Then 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 + R2 Then 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) 2 3 VGS = 0.469 (VGS 2VGS + 1) 2

76. 2 0.469 VGS + 0.0625 VGS 2.53 = 0 VGS = 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 2 VDS = 10 (1.49 )(1.7 ) VDS = 7.47 V 3.30 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 ) RS W 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 250 3.31 (a) (i) I Q = 50 = 500 (VGS 1.2 ) VGS = 1.516 V 2 VDS = 5 ( 1.516 ) = VDS = 6.516 V (ii) I Q = 1 = ( 0.5 )(VGS 1.2 ) VGS = 2.61 V 2 VDS = 5 ( 2.61) VDS = 7.61 V (b) (i) Same as (a) VGS = VDS = 1.516 V (ii) VGS = VDS = 2.61 V 3.32 I D = K n (VGS VTN ) 2 0.25 = ( 0.2 )(VGS 0.6 ) 2 0.25 + 0.6 VGS = 1.72 V VS = 1.72 V 0.2 VD = 9 ( 0.25 )( 24 ) VD = 3 V VGS = 3.33 (a)

77. ID(mA) 1.0 0.808 Q-pt 0.5 3.81 RD = 10 V (V) DS 5 1 RD = 8 K 0.5 I DQ = 0.5 = 0.25 (VGS 1.4 ) VGS = 2.81 V 2 RS = 2.81 ( 5 ) RS = 4.38 K 0.5 Let RD = 8.2 K, RS = 4.3 K (b) Now VGS ( 5 ) 5 VGS = I D = 0.25 (VGS 1.4 ) 4.3 2 = 1.075 (VGS 2.8 VGS + 1.96 ) 2 2 1.075 VGS 2.01 VGS 2.89 = 0 VGS = 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 2 VDS = 10 ( 0.504 )( 8.2 + 4.3) VDS = 3.70 V (c) If RS = 4.3 + 10% = 4.73 K 2 5 VGS = 1.18 (VGS 2.8VGS + 1.96 ) 2 1.18 VGS 2.31 VGS 2.68 = 0 VGS = 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 2 If Rs = 4.3 10% = 3.87 K 2 5 VGS = ( 0.9675 ) (VGS 2.8VGS + 1.96 ) 2 0.9675VGS 1.71VGS 3.10 = 0 VGS = 1.71 2.924 + 12.0 = 2.88 V 2 ( 0.9675 ) I D = ( 0.25 )( 2.88 1.4 ) = 0.548 mA 2 3.34 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 m

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 + R2 I DQ = K p (VSG + VTP ) 2 1.25 = 0.5 (VSG + 1.5 ) 2 1.25 1.5 = VSG 0.5 VSG = 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-pt RD = VD ( 5 ) I DQ 5 V (V) SD = 52 RD = 12 K 0.25 2 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 p W ID = L 5 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 = 0

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 V ID = VSD k = 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 = 0 VSG = 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 V ID = VSD 3.37 ID = VSD ( 10 ) RD 5= 6 + 10 RD = 0.8 k RD I D = K P (VSG + VTP ) 5 = 3 (VSG 1.75 ) 2 VSG = 2 5 + 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 + R2 3.38 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) 2 For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D1 2 30 ( 1.8 ) = 120 2 ( 3 0.8 ) vO vO 2 2 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 2

80. 2 We find 4vO 33.6vO + 3.24 = 0 vO = 0.0976 V 3.39 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 1 3.40 a. M1 and M2 in saturation K 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 2 2 I D = (15 )( 40 )( 2.5 0.8 ) I D = 1.73 mA 2 b. W W > VGS1 < VGS 2 L 1 L 2 40 (VGS1 0.8 ) = (15 )(VGS 2 0.8 ) 2 2 VGS 2 = 5 VGS 1 1.633 (VGS 1 0.8 ) = ( 5 VGS1 0.8 ) 2.633VGS 1 = 5.506 VGS 1 = 2.09 V VGS 2 = 2.91 V, V0 = VGS1 = 2.91 V I D = (15 )(15 )( 2.91 0.8 ) I D = 1.0 mA 2 3.41 (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 3 V2 = 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

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 ) 2 2 Take 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 + 2.64 From (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 V 3.42 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 D 3.43 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 D 3.44 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 L 3.45 2

82. 5 = I D RD + V + VDS 5 = (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 V 3.48 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 = L VD1 = 0.75 + 2 = 1.25 V RD = 2.5 1.25 RD = 12.5 K 0.1

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 V RD = 2.7 ( 5 ) 0.25 RD = 9.2 K 3.50 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 = L 3.51 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 mA b. V 1 I D = I DSS 1 GS = 6 1 I D = 2.67 mA VP 3 2 2 c. d. 2 2 I D = 6 1 I D = 0.667 mA 3 ID = 0

84. 3.53 V I D = I DSS 1 GS VP 1 2.8 = I DSS 1 VP 2 2 3 0.30 = I DSS 1 VP 1 1 2.8 VP = 0.30 3 1 VP 2 2 = 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 2 1 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 VP So VDD 2.5 V V I D = 2 = I DSS 1 GS VP 2 2 V 2 = 6 1 GS VGS = 1.06 V VS = 1.06 V 2.5 3.55 I D = K n (VGS VTN ) 2 18.5 = K n ( 0.35 VTN ) 86.2 = K n ( 0.5 VTN ) 2 2 Then ( 0.35 VTN ) 18.5 = 0.2146 = VTN = 0.221 V 2 86.2 ( 0.50 VTN ) 2 18.5 = K n ( 0.35 0.221) K n = 1.11 mA / V 2 2 3.56 I D = K (VGS VTN ) 2 250 = K ( 0.75 0.24 ) K = 0.961 mA / V 2 2

85. 3.57 2 V V V I D = I DSS 1 GS = S = GS VP RS RS 2 V 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 V VGS 1.17 = I D = 5.85 mA RS 0.2 VD = 20 ( 5.85 )( 2 ) = 8.3 V VDS = VD VS = 8.3 1.17 VDS = 7.13 V 3.58 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 2 2 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 + R2 3.59 V I D = I DSS 1 GS VP 2 2 V 5 = 8 1 GS VGS = 0.838 V 4 VSD = VDD I D ( RS + RD ) = 20 ( 5 )( 0.5 + 2 ) VSD = 7.5 V

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 + R2 3.60 V I D = I DSS 1 GS VP 2 2 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 k VS = 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 2 VGS = 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) 3.62

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 2 3.63 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 ID 3.64 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 2 2 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 k 3.65

88. VDS = VDD I D ( RS + RD ) 2 = 3 ( 0.040 )(10 + RD ) RD = 15 k I D = K (VGS VTN ) 2 40 = 250 (VGS 0.20 ) VGS = 0.60 V 2 VG = 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 k 3.66 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 2 2

89. Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS VTN ) and I D = K n (VGS VTN ) 2 0.75 = 0.5 (VGS 0.8 ) VGS = 2.025 V g m = 2 ( 0.5 )( 2.025 0.8 ) g m = 1.22 mA / V 2 ro = 1 I DQ 1 (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/V Av = ( 0.8944 )(10 ) = 8.94 EX4.3 (a) R2 320 VGS = VDD = ( 5 ) = 1.905 V R1 + R2 520 + 320 I DQ = 0.20 (1.905 0.8 ) = 0.244 mA 2 g m = 2 K n I DQ = 2 ( 0.2 )( 0.244 ) = 0.442 mA/V ro = (b) Av = g m RD = ( 0.422 )(10 ) = 4.22 (c) (d) Ri = R1 R2 = 520 320 = 198 K RO = RD = 10 K EX4.4 At transition point, I D = 1 mA I D = K n (VGSt VTN ) = K n (VDS ( sat ) ) 2 2 1 = 0.2 (VDS ( sat ) ) VDS ( sat ) = 2.236 V 2 5 2.236 + 2.236 = 3.62 V 2 5 3.62 RD = = 2.76 K 0.5 Want 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 R1

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 = 1.75 EX4.5 (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 ) 2 2 = 1.385 V I 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 RS g 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 = 3.05 EX4.6 VSDQ = 3 V and I DQ = 0.5 mA RD = I DQ = K P (VSG VTP ) 53 RD = 4 k 0.5 2 0.5 = 1(VSG 1) VSG = 1.71 V VGG = 5 1.71 VGG = 3.29 V 2 Av = g m RD g m = 2 K P I DQ = 2 (1)( 0.5 ) g m = 1.414 mA/V Av = (1.414 )( 4 ) Av = 5.66 Av = v0 vsd 0.46sin t = = = 5.66 vi = 0.0813sin t vi vi vi EX4.7 a. VSG = 9 I DQ RS , I DQ = K P (VSG VTP VSG = 9 ( 2 )(1.2 )(VSG 2 ) 2 2 = 9 2.4 (VSG 4VSG + 4 ) 2 2.4VSG 8.6VSG + 0.6 = 0 ) 2

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 = 2 8.6 b. V0 VSG Vi gmVSG RD RS CS ( 2 )( 4.57 ) = 6.046 mA/V g m = 2 K P I DQ = 2 V0 = g mVSG RD Av = g m RD = ( 6.046 )(1) Av = 6.05 EX4.8 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 ) 2 2 R2 VGS = 2.025 V = VG VS = VG 5 VG = 7.025 V = R1 + R2 So R2 = 281 k, R1 = 119 k Neglecting RSi , Av = g m ( RS 1 + g m ( RS 1 r0 ) r0 ) 1 r0 = 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 ) 2 3 = 2 (VSG 2 ) VSG = 3.22 V 5 VSG 5 3.22 I DQ = 3= RS = 0.593 k RS RS 2 1 1 r0 = 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 = 2 For r0 RS = 16.7 0.593 = 0.573 k R2 10 VDD = 400

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 ) x x = 0.402 = 0.573 RL 0.573RL = 0.402 ( 0.573 0.402 ) RL = ( 0.402 )( 0.573) RL = 1.35 k RL + 0.573 EX4.10 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 = ) 2 5 VS RS Then ( 0.4 )( 5 )(VS 0.581 0.8 ) = 5 VS 2 2 (VS 1.381) = 5 VS 2 2 (VS2 2.762VS + 1.907 ) = 5 VS 2VS2 4.52VS 1.19 = 0 VS = 4.52 ( 4.52 ) + 4 ( 2 )(1.19 ) 2 ( 2) 2 VS = 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 / V g 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 + 0.5 Neglecting RSi , Av = 0.817 R0 = RS 1 1 =5 = 5 1.12 R0 = 0.915 k 0.894 gm EX4.11 gmVsg V0 Vi RS Vsg RD RL

93. V0 = g mVsg ( RD RL ) and Vsg = Vi Av = g m ( RD RL ) I DQ = 5 VSG = K p (VSG VTP RS 5 VSG = (1)( 4 )(VSG 0.8 ) ) 2 2 2 5 VSG = 4 (VSG 1.6VSG + 0.64 ) 2 4VSG 5.4VSG 2.44 = 0 VSG = 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 = RS 1 1 =4 = 4 0.552 Rin = 0.485 k gm 1.81 EX4.12 Kn2 = n Cox W Av = 2 = ( 0.015 )( 2 ) = 0.030 mA / V 2 L 2 K n1 K = 6 n1 = 36 Kn2 Kn2 K 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 V

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]) 2 2 2 10 (VGS1 0.8 ) = (1.0209 ) = 1.482 (VGS1 0.8 ) 2 = 0.145 VGS1 = 1.18 V I DQ = K n1 (VGS1 0.8 ) (1 + ( 0.01)( 2.094 ) ) 2 b. 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 1 EX4.15 (a) RO = So g m1 = 1 1 ro 2 ro1 g m1 g m1 1 1 = = 0.5 mA/V R0 2 g m1 = 2 K n I D ( 0.2 ) I D 0.5 = 2 (b) Av = g m1 ( ro1 ro 2 ) 1 + g m1 ( ro1 ro 2 ) ro1 = ro 2 = Av = I D = 0.3125 mA 1 = 320 K ( 0.01)( 0.3125 ) 0.5 ( 320 320 ) 1 + ( 0.5 )( 320 320 ) Av = 0.988 EX4.16

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 D 2.4 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 K EX4.17 R0 = RS 2 1 gm2 g m 2 = 0.632 mA/V, RS 2 = 8 k R0 = 8 1 = 8 1.58 R0 = 1.32 k 0.632 EX4.18 a. I DQ1 = K n1 (VGS 1 VTN 1 ) 2 1 = 1.2 (VGS 1 2 ) VGS 1 = VGS 2 = 2.91 V 2 RS = 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 k Then 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:

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 = 6.57 EX4.19 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 2 2 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 = 2.87 gm = EX4.20 a. 2 2 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 = 1.75 Also I DQ = VGS ( 10 ) RS 2= 1.75 + 10 RS = 5.88 k RS b. Vi gmVgs Vgs V0 RS

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 ( RS 1 + g m ( RS ( 2.29 )( RS 1 + ( 2.29 )( RS RL ) ro RL ro ) RL ro ) RL ro ) = ( 0.80 )( 0.9234 ) = 0.7387 = 0.7387 RS RL ro = 1.235 k RS ro = 5.261 k 5.261RL = 1.235 k RL = 1.61 k 5.261 + RL TYU4.1 g m = 2 K n (VGS VTN ) and I D = K n (VGS VTN ) VGS VTN = 2 and g m = 2 K n I DQ Kn I DQ Kn = 2 K n I DQ ( 3.4 ) g2 = 1.445 mA / V Kn = m = 4 I DQ 4 ( 2) 2 K n = n Cox W 2 L W 1.445 = ( 0.018 ) L W = 80.3 L 1 2 ro = K n (VGS VTN ) = I DQ 1 1 ro = ( 0.015 )( 2 ) ro = 33.3 k TYU4.2 a. I DQ = K n (VGS VTN ) 2 0.4 = 0.5 (VGS 2 ) VGS = 2.894 V 2 VDSQ = VDD I DQ RD = 10 ( 0.4 )(10 ) VDSQ = 6 V b. g m = 2 K n (VGS VTN ) = 2 ( 0.5 )( 2.894 2 ) g m = 0.894 mA/V 1 r0 = I DQ , = 0 r0 = v0 Av = = g m RD = ( 0.894 )(10 ) Av = 8.94 vi c. vi = 0.4sin t vds = ( 8.94 )( 0.4 ) sin t vds = 3.58sin t

98. At VDS 1 = 6 3.58 = 2.42 VGS1 = 2.89 + 0.4 = 3.29 VGS1 VTN = 3.29 2 = 1.29 = VDS ( sat ) So VDS 1 > VGS 1 VTN Biased in saturation region TYU4.3 I DQ = K n (VGS VTN ) = ( 0.25 )( 2 0.8 ) I DQ = 0.36 mA 2 a. 2 VDSQ = VDD I DQ RD = 5 ( 0.36 )( 5 ) VDSQ = 3.2 V b. 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 vi TYU4.4 vi = vgs = 0.1sin t id = g m vgs = ( 0.6 )( 0.1) sin t id = 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 = vDS TYU4.5 a. VSDQ = VDD I DQ RD 7 = 12 I DQ ( 6 ) I DQ = 0.833 mA I DQ = K P (VSG VTP ) 2 0.833 = 2 (VSG 1) VSG = 1.65 V 2 g 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 gmVsg VSG RD TYU4.6 I DQ = K n (VGS VTN ) VGS VTN = 2 g m = 2 K n (VGS VTN ) = 2 K n So g m = 2 K n I DQ TYU4.7 = 2 2 f + vSB I DQ Kn I DQ Kn

99. = (a) 0.40 2 2 ( 0.35 ) + 1 = (b) = 0.153 0.40 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 = 2 For (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 RD VGS = g m = 2 ( 0.5 )( 2.5 1.5 ) = 1 mA/V For Av = 4.0 RD = 4 k VD = 5 ( 0.5 )( 4 ) = 3 V VDSQ = 3 ( 2.5 ) = 5.5 V TYU4.9 a. With RG VGS = VDS transistor biased in sat. region I D = K n (VGS VTN ) = K n (VDS VTN ) 2 2 VDS = VDD I D RD = VDD K n RD (VDS VTN ) VDS = 15 ( 0.15 )(10 )(VDS 1.8 ) 2 2 2 = 15 1.5 (VDS 3.6VDS + 3.24 ) 2 1.5VDS 4.4VDS 10.14 = 0 VDS = 4.4 ( 4.4 ) 2 + ( 4 )(1.5 )(10.14 ) 2 (1.5 ) VDSQ = 4.45 V I DQ = ( 0.15 )( 4.45 1.8 ) I DQ = 1.05 mA 2 b. 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 ) = 2 VSG VSG = 4 RS 2 3.2 ( 4 4VSG + VSG ) = VSG 2 3.2VSG 13.8VSG + 12.8 = 0

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 mA VSG = VSG 2 13.8 b. VDSQ = VDD I DQ ( RD + RS ) 6 = 10 ( 0.338 )( RD + 4 ) RD = 10 ( 0.338 )( 4 ) 6 0.338 RD = 7.83 k c. V0 Vgs Vi gmVgs RD RS Vi = Vgs + g mVgs RS Vgs = Vi 1 + g m RS V0 = 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 RS TYU4.11 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 k b. V0 Vi VSG gmVsg r0 RD V0 = 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 = 3.71

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 Ix I x + g mVgs = I x = g mVx + Vx Vx and Vgs = Vx r0 Vx 1 1 R0 = r0 = 50 R0 0.25 k 4 r0 gm With RS = 4 k Av = (b) r0 gmVgs r0 || Rs = 50 || 4 = 3.7 k Av = g m ( r0 RS ) 1 + g m ( r0 RS ) ( 4 )( 3.7 ) 1 + ( 4 )( 3.7 ) Av = 0.937 TYU4.13 (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 2 W 1.25 = ( 0.020 ) L L W = 62.5 L I DQ = K n (VGS VTN ) 0.8 = 1.25 (VGS 2 ) VGS = 2.8 V 2 2 b. 1 1 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 2 TYU4.14

102. Rin = 1 = 0.35 k g m = 2.86 mA/V gm V0 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 ) 2 0.5 = 4.09 (VGS 1) VGS = 1.35 V VS = 1.35 V, VD = 5 ( 0.5 )( 6 ) = 2 V 2 VDS = VD VS = 2 ( 1.35 ) = 3.35 V We 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 Kn2 W L C W = n ox 2 L Av = 2 = ( 0.020 )( 80 ) = 1.6 mA / V 1 2 = ( 0.020 )(1) = 0.020 mA / V 2 K n1 1.6 = Av = 8.94 Kn2 0.020 The transition point is determined from vGSt VTND = VDD VTNL K n1 ( vGSt VTND ) Kn2 vGSt 0.8 = ( 5 0.8 ) ( 8.94 )( vGSt 0.8 ) vGSt = ( 5 0.8) + ( 8.94 )( 0.8) + 0.8 1 + 8.94 vGSt = 1.22 V For Q-point in middle of saturation region VGS = 1.22 0.8 + 0.8 VGS = 1.01 V 2 TYU4.16 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 ) 2 2 = 1(VGS 2 2 ) VSG 2 = 3.41 V VG 2 = 3.41 V 2 10 3.41 RD1 = 3.29 k 2 = 10 V VS 1 = 3.41 10 = 6.59 V Then RD1 = For VDSQ1 Then RS 1 = 6.59 ( 10 ) 2 I D1 = K n1 (VGS 1 VTN 1 ) RS1 = 1.71 k 2 2 = 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 R1

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/V b. 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 = 8.03 1 1 RS 2 = 5 = 0.3536 5 R0 = 0.330 k gm2 2.828 R0 = 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 = 4.01 TYU4.18 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 2 2 2 20 VS1 17.2 VS 1 V = 6 1 + = 6 S 1 7.6 and I DQ1 = 2 2 2 4 2 20 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) 2 VS 1 = 14.2 V VGS 1 = 17.2 14.2 = 3 V > VP

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 V on I DQ1 = VSDQ1 VD1 = 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 ) 2 2 2 V 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 Then VS 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 V For 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 V I DQ 2 = VDS 2 b. Vg2 gm2Vgs2 Vgs2 Vi RD1 V0 Vsg1 gm1Vsg1 RS2 Vg 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 ) = RL

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 2 4.2 p Cox W gm = 2 K p I D = 2 ID 2 L (a) 2 W 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 2 4.3 I D = K n (VGS VTN ) (1 + VDS ) 2 I 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 0.4 4.4 r0 = ID = 1 ID 1 r0 = 1 ( 0.012 )(100 ) I D = 0.833 mA 4.5 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 1

106. ro = VDS 2 = ro = 100 K 0.02 I D 4.6 (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 ro I D 1.0 2% = 50 ID I D = (ii) VDS 1 = = 0.01 10 A ro 100 I D 10 = 2% ID 500 4.7 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 0.5 For 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 = 1 c. = 1 r = 133 k ( 0.015)( 0.50 ) o Av = g m ( r0 RD ) = ( 0.835 )(133 8 ) Av = 6.30 I DQ 4.10 2 2 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 =

107. 2 K nVgs Ratio of signal at 2 to that at : 2 cos 2 t 2 K n (VGSQ VTN ) Vgs sin t The coefficient of this expression is then: Vgs 4 (VGSQ VTN ) 4.11 0.01 = Vgs 4 (VGSQ VTN ) So Vgs = ( 0.01)( 4 )( 3 1) Vgs = 0.08 V 4.12 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 R2 Vgs = 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 RS 2 4.8 VGS = (1)( 2 ) (VGS 4VGS + 4 ) 2 2VGS 7VGS + 3.2 = 0 VGS = 7 (7) 2 4 ( 2 )( 3.2 ) 2 ( 2) = 2.96 V I D = (1)( 2.96 2 ) I D = 0.920 mA 2 VDS = VDD I D ( RD + RS ) = 12 ( 0.92 )( 3 + 2 ) VDS = 7.4 V (b) Vo = g mVg ( RD RL ) 1 + g m RS where Vg = Then R1 R2 R1 R2 + RSi Vi = 300 200 300 200 + 2 Vi = 120 Vi = ( 0.9836 ) Vi 120 + 2

108. Av = g m ( 0.9836 )( RD || RL ) 1 + g m RS g m = 2 (1)( 2.96 2 ) = 1.92 mA / V So Av = (1.92 )( 0.9836 )( 3 || 3) 1 + (1.92 )( 2 ) Av = 0.585 c. AC load line 1 1 Slope 3.5 K 33 2 0.92 7.4 12 1 i D = vds 3.5 k i D = 0.92 mA vds = ( 0.92 )( 3.5 ) = 3.22 6.44 V peak-to-peak 4.15 a. I D Q = 3 mA VS = I DQ RS = ( 3)( 0.5 ) = 1.5 V I DQ = K n (VGS VTN ) 2 3 = ( 2 )(VGS 2 ) VGS = 3.225 V 2 VG = 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 RS g m = 2 ( 2 )( 3.225 2 ) = 4.90 mA / V Av = ( 4.90 )( 2 || 5 ) 1 + ( 4.90 )( 0.5 ) Av = 2.03 4.16 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 = 2.83

109. AC load line 1 1 Slope 33 1.5 K 0.92 7.4 12 1 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 K 4.18 (a) Av = g m RD 1 + g m RS (1) 8 = g m RD 1 + g m (1) (2) 16 = g m RD Then 8 = 16 1 + g m (1) g m = 1 mA/V RD = 16 K (b) Av = 10 = (1)(16 ) 1 + (1) RS RS = 0.6 K 4.19 a. AC load line 1 Slope 5K IDQ VDS(sat) VDSQ

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 Q I DQ = b. ( 0.5 )( 0.4 ) = 0.8944 mA / V g 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 = 4.38 4.20 (a) I DQ = K n (VGSQ VTN ) 2 0.1 = 0.85 (VGSQ 0.8 ) 2 VGSQ = 1.143 V RS = 1.143 ( 5 ) RS = 38.6 K 0.1 V = I ( RD RL ) ro ro = 1 = 0.1( RD RL ) ro RD RL ro = 10 K = 1 = 500 K ( 0.02 )( 0.1) 40 500 RD 40 500 + RD 37.04 RD = 10 37.04 + RD RD = 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 = 5.83 4.21 a. 2 I D = K n (VGS VTN ) 2 = 4 (VGS ( 1) ) 2 VGS = 0.293 V VS = 0.293 V = I D RS = (2) RS RS = 0.146 k

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 RS g 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 = 2.98 c. AC load line 1 1.852 0.146 1 1.107 K Slope 2 mA 10 6 v0 = ( iD )(1.85 || 2 ) = ( 2 )(1.85 || 2 ) = 1.92 V vi = 1.92 = 0.645 Vi = 0.645 V 2.98 4.22 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 ) = 2 VGS 2.5 RS VGS = I DQ RS = 2 + RS RS 2 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 + RS 2 = 2.5 2 4 ( 4 6 RS + 2.25 RS ) = 2.5 ( 2 + RS ) 2 9 RS 26.5RS + 11 = 0 RS = 26.5 ( 26.5) 4 ( 9 )(11) 2 (9) 2 RS = 0.5 k or 2.44 k But RS = 2.44 k VGS = 1.37 Cut off. RS = 0.5 k, I DQ = 1.0 mA

112. b. Av = g m ( RD || RL ) 1 + g m RS ( 4 )(1) = 4 mA / V g m = 2 K n I DQ = 2 Av = ( 4 )( 2 || 2 ) 1 + ( 4 )( 0.5 ) Av = 1.33 4.23 a. 5 = I DQ RS + VSDQ + I DQ RD 5 5 = I DQ RS + 6 + I DQ (10 ) 5 1. I DQ = 4 RS + 10 VS = VSDQ + I DQ RD 5 = VSGQ 2. 1 + I DQ (10 ) = VSGQ 3. I DQ = K p (VSGQ 2 ) 2 Choose RS = 10 k I DQ = 4 = 0.20 mA 20 VSGQ = 1 + (0.2)(10) = 3 V 0.20 = K P (3 2) 2 K P = 0.20 mA / V 2 b. 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 = 2.0 g m = 2 K P I DQ = 2 c. 4 = 0.133 mA 30 = 1 + (0.133)(10) = 2.33 V Choose RS = 20 k I DQ = VSGQ 0.133 = 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 = 4.03 A larger gain can be achieved. 4.24 (a) I DQ = K p (VSGQ + VTP ) 2 0.25 = 0.8 (VSGQ 0.5 ) 2 VSGQ = 1.059 V 3 1.059 RS = RS = 7.76 K 0.25 VD = VS VSDQ = 1.059 1.5 = 0.441 V RD = (b) 0.441 ( 3) 0.25 RD = 10.2 K

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 = 2 Av = 1.50 (c) VO = I ( RD || RL ) = 0.25 (10.2 || 2 ) = 0.418 So VO = 0.836 peak-to-peak 4.25 I DQ = K n (VGSQ VTN ) 2 g 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 2 VDSQ = 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 = 0 RD = 0.47 + ( 0.47 ) + 4 ( 2.2 )( 3.93) RD = 1.23 k, 2 ( 2.2 ) = 2.8 + ( 6 )( 0.1) = 3.4 V 2 VG = VGS + VS 1 1 VG = Rin VDD = (100 )(18 ) = 3.4 R1 = 529 k R1 R1 529 R2 = 100 R2 = 123 k 529 + R2 4.26 a. VSD(sat) VSDQ V1