1. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 1 1.1 3/ 2 Eg / 2kT in BT e = (a) Silicon (i) ( )( ) ( )( ) = 5.23 10 15 250 3/ 2 exp 1.1 [ ] 6 19 8 3 2 86 10 250 2.067 10 exp 25.58 1.61 10 cm n i n i = = = 5.23 10 350 exp 1.1 (ii) ( 15 )( ) 3/ 2 ( )( ) [ ] 6 19 11 3 2 86 10 350 3.425 10 exp 18.27 3.97 10 cm n i n i = = (b) GaAs (i) ( )( ) ( )( ) = 2.10 10 14 250 3/ 2 exp 1.4 ( ) [ ] 6 17 3 3 2 86 10 250 8.301 10 exp 32.56 6.02 10 cm n i n i = = = 2.10 10 350 exp 1.4 (ii) ( 14 )( ) 3/ 2 ( )( ) ( ) [ ] 6 18 8 3 2 86 10 350 1.375 10 exp 23.26 1.09 10 cm n i n i = = ______________________________________________________________________________________ 1.2 a. n BT 3/ 2 exp Eg = 2 i kT 10 12 5.23 10 15 3 / 2 exp 1.1 = 2(86 10 6 )( ) T T 3 1.91 10 4 T 3/ 2 exp 6.40 10 = T By trial and error, T 368 K b. 109 cm 3 i n = = 10 5.23 10 exp 1.1 ( )( ) 9 15 3/ 2 2 86 10 6 T T 3 1.91 10 7 T3/ 2 exp 6.40 10 = T By trial and error, T 268 K ______________________________________________________________________________________

2. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.3 Silicon (a) = ( 5.23 10 15 )( 100 ) 3/ 2 exp 1.1 ( )( ) ( ) [ ] 6 18 10 3 2 86 10 100 5.23 10 exp 63.95 8.79 10 cm n i n i = = = 5.23 10 300 exp 1.1 (b) ( 15 )( ) 3/ 2 ( )( ) ( ) [ ] 6 19 10 3 2 86 10 300 2.718 10 exp 21.32 1.5 10 cm n i n i = = = 5.23 10 500 exp 1.1 (c) ( 15 )( ) 3/ 2 ( )( ) ( ) [ ] 6 19 14 3 2 86 10 500 5.847 10 exp 12.79 1.63 10 cm n i n i = = Germanium. ( 15 )( ) 3/ 2 (a) = 1.66 10 100 exp 0.66 ( = ( 1.66 10 18 ) exp [ 38.37 ] 6 )( ) 3 2 86 10 100 35.9 cm n i n i = = = 1.66 10 300 exp 0.66 8.626 10 exp 12.79 (b) ( 15 )( ) 3/ 2 ( 18 ( ) [ ] 6 )( ) 13 3 2 86 10 300 2.40 10 cm n i n i = = = 1.66 10 500 exp 0.66 1.856 10 exp 7.674 (c) ( 15 )( ) 3/ 2 ( 19 ( ) [ ] 6 )( ) 15 3 2 86 10 500 8.62 10 cm n i n i = ______________________________________________________________________________________ 1.4 2 ( 13 ) 2 (a) n-type; no = 1015 cm 3 ; p 2.4 10 = 11 cm3 15 5.76 10 10 i n = = o o n 2 ( 10 ) 2 (b) n-type; no = 1015 cm 3 ; 1.5 10 = 5 p cm3 15 2.25 10 10 i n = = o o n ______________________________________________________________________________________

3. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5 2 ( 6 ) 2 (a) p-type; po = 1016 cm 3 ; n 1.8 10 = 4 cm3 16 3.24 10 10 i n = = o o p 2 ( 13 ) 2 (b) p-type; po = 1016 cm 3 ; 2.4 10 = 10 n cm3 16 5.76 10 10 i n = = o o p ______________________________________________________________________________________ 1.6 (a) n-type (b) 16 3 5 10 cm 1.5 10 ( ) n N 10 2 2 3 3 16 4.5 10 cm 5 10 o d i o o p n n = = = = = (c) 5 1016 cm 3 o d n = N = From Problem 1.1(a)(ii) 3.97 1011 cm 3 i n = ( )11 2 5 10 o p 6 3 3.97 10 16 3.15 10 cm = = ______________________________________________________________________________________ 1.7 2 ( 10 ) 2 (a) p-type; po = 51016 cm 3 ; n 1.5 10 = 3 cm3 16 4.5 10 5 10 i n = = o o p 2 ( 6 ) 2 (b) p-type; po = 51016 cm 3 ; 5 1.8 10 = n cm3 16 6.48 10 5 10 i n = = o o p ______________________________________________________________________________________ 1.8 (a) Add boron atoms (b) Na = po = 21017 cm 3 2 ( 10 ) 2 (c) n 1.5 10 = 3 cm3 17 1.125 10 2 10 i n = = o o p ______________________________________________________________________________________ 1.9 (a) 15 3 n cm ( ) 10 2 2 4 3 p n p 15 5 10 1.5 10 4.5 10 5 10 o i o o n o cm = = = = (b) > n-type no po (c) 5 1015 3 o d n N = cm ______________________________________________________________________________________

4. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.10 a. Add Donors 7 1015 cm 3 d N = b. Want p = = n N So 2 ( 6 )( 15 ) 106 cm 3 2 / o i d 10 7 10 7 10 = = = 2 3 exp i n B T Eg kT 21 = 7 10 5.23 10 exp 1.1 ( ) ( )( ) 21 15 2 3 86 10 6 T T By trial and error, T 324 K ______________________________________________________________________________________ 1.11 (a) I = A = (105 )(1.5)(10) I = 0.15 mA ( )( ) (b) 1.2 10 0.4 ( ) 2.4 2 10 4 3 = I A I = = = A V/cm ______________________________________________________________________________________ 1.12 J = = J cm = 120 = 6.67( ) 1 18 ( ) 6.67 = e N N cm3 ______________________________________________________________________________________ 1.13 (a) ( )( )( ) = = ( )( ) 16 19 3.33 10 1.6 10 1250 n n d d e 15 1 1 1 = 19 7.69 10 N = = 1.6 10 1250 0.65 n d e N n d e cm3 (b) J = = J = (0.65)(160) = 104 V/cm ______________________________________________________________________________________ 1.14 (a) e N N = = 1.5 ( = 9.375 10 15 cm3 1.6 10 19 )( 1000 ) n n d d e 0.8 = (b) ( )( ) 16 19 1.25 10 = = 1.6 10 400 p a e N cm3 ______________________________________________________________________________________ 1.15 (a) For n-type, (1.6 10 19 )(8500) n d d e N = N For 1015 1019 3 1.36 1.36 104 ( ) 1 dN cm cm (b) J = E = (0.1)0.136 J 1.36103 A/ cm2 ______________________________________________________________________________________

5. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.16 = (0.026)(1250) = 32.5 cm2 /s; Dn = (0.026)(450) = 11.7 Dp cm /s 2 16 12 ( )( ) 52 J eD dn 1.6 10 32.5 10 10 cm2 n n A/ 19 0 0.001 = = = dx 12 16 ( )( ) 18.72 J dp p eD p 1.6 10 19 11.7 10 10 = A/cm 2 0 0.001 Total diffusion current density = = dx J = 52 18.72 = 70.7 A/cm2 ______________________________________________________________________________________ 1.17 J eD dp 10 1 exp ( 15 ) p p p ( 1.6 10 19 )( 15 )( 10 15 ) 4 / exp 10 10 2.4 p p p p p x L p dx eD x L L J x L J e = = = = (a) x = 0 2.4 A/cm2 p J = (b) x =10 m 2.4 1 0.883 A/cm2 pJ = e = (c) x = 30 m 2.4 3 0.119 A/cm2 pJ = e = ______________________________________________________________________________________ 1.18 a. 1017 cm 3 1017 cm 3 a o N = p = ( )6 2 2 5 3 1.8 10 17 3.24 10 cm 10 i o o o n n n p = = = b. 5 15 15 3 n n n n p p p p 3.24 10 10 10 cm 10 10 1.01 10 cm 17 15 17 3 o o = + = + = = + = + = ______________________________________________________________________________________ N N 1.19 a d V V = 2 ln i bi T n ( ) ( 15 )( 15 ) (a) (i) 0.026 ln 5 10 5 1010 2 = Vbi V ( ) = 0.661 1.5 10 ( ) ( 17 )( 15 ) (ii) 0.026 ln 5 10 1010 2 ( ) = 0.739 1.5 10 = Vbi V ( ) ( 18 )( 18 ) (iii) 0.026 ln 10 10 10 2 = Vbi V ( ) = 0.937 1.5 10

6. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( ) ( 15 )( 15 ) (b) (i) 0.026 ln 5 10 5 10 6 2 = Vbi V ( ) = 1.13 1.8 10 ( ) ( 17 )( 15 ) (ii) 0.026 ln 5 10 106 2 ( ) = 1.21 1.8 10 = Vbi V ( ) ( 18 )( 18 ) (iii) 0.026 ln 10 10 6 2 = Vbi V ( ) = 1.41 1.8 10 ______________________________________________________________________________________ 1.20 N N a d V V = 2 ln i bi T n or ( 2 ) ( ) 2 16 1.5 exp 10 = N cm3 16 1.76 10 exp 0.712 0.026 10 = = bi n a V T i d V N ______________________________________________________________________________________ 1.21 ( ) ( 16 ) 10 N N N a d a = ln 2 = 0.026 ln (1.5 10 1 ) bi T i V V n 0 2 V V For 1015 3 , 0.637 a bi N = cm V = For 1018 3 , 0.817 a bi N = cm V = ______________________________________________________________________________________ 1.22 kT = (0.026) T 300 kT (T)3/2 200 0.01733 2828.4 250 0.02167 3952.8 300 0.026 5196.2 350 0.03033 6547.9 400 0.03467 8000.0 450 0.0390 9545.9 500 0.04333 11,180.3

7. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ = 2.1 10 exp 1.4 ( 14 )( 3/ 2 ) ( )( ) 6 2 86 10 in T T N N 2 ln a d = bi T i V V n T ni Vbi 200 1.256 1.405 250 6.02 103 1.389 300 1.80 106 1.370 350 1.09 108 1.349 400 2.44 109 1.327 450 2.80 1010 1.302 500 2.00 1011 1.277 ______________________________________________________________________________________ 1.23 1/ 2 C C V 1 R = + j jo V bi ( )( ) ( ) 16 15 1.5 10 4 10 = = 0.026 ln 0.684 V ( ) 10 2 1.5 10 bi V 1/ 2 0.4 1 1 0.255 pF = + = (a) ( ) 0.684 j C 1/ 2 0.4 1 3 0.172 pF = + = (b) ( ) 0.684 j C 1/ 2 0.4 1 5 0.139 pF = + = (c) ( ) 0.684 j C ______________________________________________________________________________________ 1.24 V 1 / 2 (a) C C 1 = + R j jo V bi For VR = 5 V, 1 2 (0 02) 1 5 0 00743 = + = 0 8 / j C . . p . F For VR = 1.5 V, 1 2 (0 02) 1 1 5 0 0118 = + = C . . . p 0 8 / j . F

8. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ C ( avg ) = 0 . 00743 + 0 . 0118 = 0 . 00962 pF 2 j ( ) ( ) ( ( ) ( )) t / C C C C v t = v final + v initial v final e where ( ) (47 103 )(0 00962 10 1 ) j = RC = RC avg = . 2 or = 4.521010 s Then ( ) 1 5 0 (5 0) ti / C v t = . = + e 5 ln 5 1.5 1.5 = e + r / 1 t = 1 10 1 t = 5.4410 s (b) For VR = 0 V, Cj = Cjo = 0.02 pF For VR = 3.5 V, ( ) 1/ 2 0.02 1 3.5 0.00863 = + = 0.8 jC p F + C ( avg ) 0.02 0.00863 0.0143 pF j 2 = = ( ) 6.72 10 10 j = RC avg = s ( ) ( ) ( ( ) ( )) t / C C C C v t = v final + v initial v final e 3.5 = 5 + (0 5)et2 / = 5(1 et2 / ) so that 10 2 t = 8.0910 s ______________________________________________________________________________________ 1.25 V 1/ 2 ( ) ( )( ) C C 1 ; = + R j jo V bi 15 17 0.026 ln 5 10 1010 2 ( ) = 0.739 1.5 10 = Vbi V For = 1 V, VR Cj = 0.391 pF 0.60 = + 1 1 0.739 For = 3 V, VR Cj = 0.267 pF 0.60 = + 1 3 0.739 For = 5 V, VR C = pF j 0.215 0.60 = + 1 5 0.739 1 = = o o f (a) 6.57 2 2 ( 1.5 10 )( 0.391 10 ) 1 3 12 = LC f MHz 1 3 12 (b) fo = ( fo = 7.95MHz 2 1.5 10 )( 0.267 10 )

9. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 3 12 = o o f f (c) = 8.86 2 ( 1.5 10 )( 0.215 10 ) MHz ______________________________________________________________________________________ 1.26 V a. I I V exp D 1 0.90 exp D 1 = = S V V T T V = = V exp D 1 0.90 0.10 T ln (0.10) 0.0599 V D T D V = V V = b. exp 1 exp 0.2 1 0.026 exp 0.2 1 exp 1 0.026 2190 1 2190 F F S T R S R T F R V I I V I I V V I I = = = = ______________________________________________________________________________________ 1.27 D = exp 1 D S V T I I V (a) (i) ( ) 1.03 3 . 0 exp 10 11 = I D A 0.026 (ii) ( ) 2.25 5 . 0 exp 10 11 = I D mA 0.026 (iii) ( ) 4.93 7 . 0 exp 10 11 = I D A 0.026 (iv) ( 11 ) 1 5.37 10 12 10 exp 0.02 = = I D A 0.026 (v) ( 11 ) 1 10 11 10 exp 0.20 = I D A 0.026 (vi) = (1011 ) A ID (b) (i) ( ) 0.0103 3 . 0 exp 10 13 = I D A 0.026 (ii) ( ) 22.5 5 . 0 exp 10 13 = I D A 0.026 (iii) ( ) 49.3 7 . 0 exp 10 13 = I D mA 0.026

10. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (iv) ( 13 ) 1 5.37 10 14 10 exp 0.02 = = I D A 0.026 (v) 1013 A I D (vi) 1013 A I D ______________________________________________________________________________________ I 1.28 V = V ln D S D T I 0.026 ln 10 1011 (a) (i) ( ) 0.359 10 6 = = VD V 0.026 ln 100 10 11 ( ) 0.419 10 6 = = VD V 0.026 ln 10 11 ( ) 0.479 VD V 10 3 = = = exp 10 10 5 11 12 = (ii) 1 0.018 0.026 D D V V V 0.026 ln 10 1013 (b) (i) ( ) 0.479 10 6 = = VD V 0.026 ln 100 10 13 ( ) 0.539 10 6 = = VD V 0.026 ln 10 13 ( ) 0.599 VD V 10 3 = = = exp 10 10 13 14 = (ii) 1 0.00274 0.026 D D V V V ______________________________________________________________________________________ 1.29 (a) 10 3 exp 0.7 0.026 S = I 2.03 10 15 A S I = (b) D V ( )( 1) DI A n = ( )( 2) DI A n= 0.1 9.501014 1.391014 0.2 4.4510 12 9.501014 0.3 2.0810 10 6.501013 0.4 9.75109 4.4510 12 0.5 4.56107 3.041011 0.6 2.14105 2.0810 10 0.7 10 3 1.42109 ______________________________________________________________________________________

11. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.30 (a) 10 12 S I = A VD(v) ID(A) log10ID 0.10 4.681011 10.3 0.20 2.19109 8.66 0.30 1.03107 6.99 0.40 4.80106 5.32 0.50 2.25104 3.65 0.60 1.05102 1.98 0.70 4.93101 0.307 (b) 10 14 S I = A VD(v) ID(A) log10ID 0.10 4.681013 12.3 0.20 2.191011 10.66 0.30 1.03109 8.99 0.40 4.80108 7.32 0.50 2.25106 5.65 0.60 1.05104 3.98 0.70 4.93103 2.31 ______________________________________________________________________________________ 1.31 a. I V V I V V V V 10 exp ln (10) 59.9 mV 60 mV = = D D D D T 2 2 1 1 = = D T D b. ln (100) 119.7 mV 120 mV D T D V = V V = ______________________________________________________________________________________ 1.32 (a) (i) ( ) 0.539 0.026 ln 2 = 9 = VD V 2 10 0.026 ln 20 = 9 (ii) ( ) 0.599 = VD V 2 10 (b) (i) ( ) 9.60 2 10 9 exp 0 . 4 0.026 = I D mA (ii) ( ) 144 65 . 0 exp 10 2 9 = I D A 0.026 ______________________________________________________________________________________

12. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.33 V V I . . 3 14 ln (0 026) ln 2 10 0 6347 V = = = 3 12 5 10 D D t S I (0 026) ln 2 10 0 5150 V V = . = . D 5 10 . V . 0 5150 0 6347 V D ______________________________________________________________________________________ 1.34 (a) I S IS A 1 . 5 10 = exp 0 . 30 = 3 1.46 10 8 0.026 (b) (i) ( ) 10.3 35 . 0 exp 10 462 . 1 8 = = I D I D mA 0.026 (ii) ( ) 0.219 25 . 0 exp 10 462 . 1 8 = = I D I D mA 0.026 ______________________________________________________________________________________ 1.35 (a) ( ) 2.31 8 . 0 exp 10 22 = I D nA 0.026 ( ) 5.05 0 . 1 exp 10 22 = I D A 0.026 ( ) 11.1 2 . 1 exp 10 22 = I D mA 0.026 ( ) 22 1 5.37 10 23 10 exp 0.02 = = I D A 0.026 For = 0.20 V, A VD = 1022 I D For = 2 V, A VD = 1022 I D (b) ( ) 115 8 . 0 exp 10 5 24 = I D pA 0.026 ( ) 0.253 0 . 1 exp 10 5 24 = I D A 0.026 ( ) 0.554 2 . 1 exp 10 5 24 = I D mA 0.026 ( ) 24 1 2.68 10 24 5 10 exp 0.02 = 0.026 = I D A For = 0.20 V, A VD = 51024 I D For = 2 V, A VD = 51024 I D ______________________________________________________________________________________

13. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.36 IS doubles for every 5C increase in temperature. 10 12 S I = A at T = 300K For 0.5 10 12 T 295 K SI = A = For 50 10 12 , (2)n 50 5.64 SI = A = n = Where n equals number of 5C increases. Then T = (5.64)(5) = 28.2 K So 295 T 328.2 K ______________________________________________________________________________________ 1.37 I T / 5 ( ) S T S 2 , 155 C ( 55) T I = = 155/ 5 9 (100) 2 2.1471 ( 55) I I S S = = 0 @100 C 373 K 0.03220 T T V V = @ 55 C 216 K 0.01865 T T V V = ( )( ( ) ) 9 9 8 13 3 exp 0.6 (100) (2.147 10 ) 0.0322 ( 55) exp 0.6 0.01865 2.147 10 1.237 10 9.374 10 (100) 2.83 10 ( 55) D D D D I I I I = = = ______________________________________________________________________________________ 1.38 (a) VPS = I DR +VD 2 . 8 = ID ( 10 6 ) +VD ; I = ( 5 10 11 ) V exp D 0.026 D By trial and error, = 0.282 V, VD = 2.52 I D A (b) 51011 A, V I D = 2.8 VD ______________________________________________________________________________________

14. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.39 V I 10 (2 104 ) D D = I +V and ( ) 12 0.026 ln 10 D D = Trial and error. VD(v) ID(A) VD(v) 0.50 4.75104 0.5194 0.517 4.7415104 0.5194 0.5194 4.740104 0.5194 0.5194 V 0.4740 mA V I D D = = ______________________________________________________________________________________ 1.40 5 10 13 A s I = = = = + V R 2 1 2 (1.2) 30 (1.2) 0.45 V TH R R 80 0.45 , ln D = + = D TH D D T S I R V V V I I By trial and error: 2.56 A, 0.402 V D D I = V = ______________________________________________________________________________________

15. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.41 (a) 1 2 1 mA = = I D I D 0.026 ln 10 13 (i) ( ) 0.599 VD VD V 10 3 1 2 = = = 3 0.026 ln 10 14 (ii) ( ) 0.617 1 = VD V 5 10 = 3 0.026 ln 10 13 ( ) 0.557 2 = VD V 5 10 = (b) VD1 = VD2 1 2 2 = = i = (i) 0.5 D D I I I mA 0.026 ln 0.5 10 13 ( ) 0.581 10 3 1 2 = = = VD VD V 5 10 14 S I I (ii) D 1 = 0.10 = = 5 10 13 1 2 2 S D I I So I D1 = 0.10I D2 1 2 1.1 2 1 mA + = = I D I D I D So 2 0.909 mA, mA = I D 1 0.0909 = I D Now 0.026 ln 0.0909 10 14 ( ) 0.554 5 10 3 1 = = VD V 0.026 ln 0.909 10 13 ( ) 0.554 5 10 3 2 = = VD V ______________________________________________________________________________________ 1.42 (a) = ( 6 10 14 ) exp 0.635 2.426 I D mA 3 0.026 I = 0.635 = R 0.635 mA 1 1 2 2.426 0.635 3.061 mA = = + = I D I D 0.026 ln 3.061 1014 ( ) 0.641 6 10 3 1 2 = = = VD VD V = 2(0.641)+ 0.635 = 1.917 V VI (b) 3 2.426 mA = I D I = 0.635 = R 1.27 mA 0.5 1 2 2.426 1.27 3.696 mA = = + = I D I D 0.026 ln 3.696 10 14 ( ) 0.6459 6 10 3 1 2 = = = VD VD V = 2(0.6459)+ 0.635 = 1.927 V VI

16. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 1.43 (a) Assume diode is conducting. Then, V = V = 0.7 V D So that I = 0 . 7 23 . 3 A 2 R 30 I . . A 1 1 2 0 7 50 = R 10 Then 1 2 50 23 3 D R R I = I I = . Or 26 7 DI = . A (b) Let Diode is cutoff. 1 R = 50 k 30 (1 2) 0 45 V = . = D 30 + 50 . V Since , 0 D D V V I < = ______________________________________________________________________________________ 1.44 At node VA: (1) 5 V I V = + A A D 2 2 At node VB VA V = (2) 5 ( V V ) ( V V ) A r A r + = D I 2 2 So 5 ( ) 5 V V A r V V V V + A A = A 3 2 2 2 r ) Multiply by 6: 10 2( ) 15 6 3( A r A A r V V + V = V V A 25 2 r 3 r 11 + V + V = V (a) r 0.6 V V = 11 25 5(0.6) 28 2.545 V A A V = + = V =

17. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ From (1) 5 = = Neg. ID = 0 I 2.5 V I D V V A A D A 2 2 Both (a), (b) 0 D I = VA = 2.5, 2 5 2 V 0.50 V 5 B D V = = V = ______________________________________________________________________________________ 1.45 (a) VO = Ii (1); = 0 ; for mA I D 0 0.7 Ii = 0.7 V; mA; for mA O V (= 0.7 I D Ii ) 0.7 Ii (b) VO = Ii (1); = 0 ; for I D 0 1.7 Ii mA 7 (V = 1.7 V; I = ) O D Ii 1. mA; for Ii 1.7 mA (c) = 0.7 V; ; ; for VO I D = Ii 1 2 0 = I D 0 2 Ii mA ______________________________________________________________________________________ 1.46 Minimum diode current for VPS (min) (min) 2 , 0.7 D D I = mA V = V I = 0.7 , I = 5 0.7 = 4 3 2 1 R R 2 1 . R 1 We have 1 2 D I = I + I so (1) 4.3 = 0.7 + 2 R R 1 2 Maximum diode current for VPS (max) 10 (0 7) 14 3 D D D D P = I V = I . I = . mA 1 2 D I = I + I or (2) 9.3 = 0.7 + 14.3 R R 1 2 9.3 4.3 2 14.3 R 0.41 k R R Using Eq. (1), 1 = + = 1 1 Then R = 82.5 82.5 2 ______________________________________________________________________________________ 1.47 5 (a) (i) 0.7 = 0.215 I = mA, = 0.7 VO V 20 5 0.6 = I = mA, = 0.6 VO V (ii) 0.220 20 (b) (i) 5 0.7 ( 5 ) = 0.2325 I = mA, = (0.2325)(20) 5 = 0.35 VO V 40 5 (iii) 0.6 ( 5 ) = 0.235 I = mA, = (0.235)(20) 5 = 0.30 VO V 40

18. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) (i) 2 0.7 ( 8 ) = 0.372 I = mA, = 2 (0.372)(5) = 0.14 VO V 25 2 (ii) 0.6 ( 8 ) = 0.376 I = mA, = 2 (0.376)(5) = 0.12 VO V 25 (d) (i) I = 0 , = 5 V VO (ii) I = 0 , = 5 V VO ______________________________________________________________________________________ 1.48 5 (a) VO 20 I = , I = ( 5 10 14 ) V exp D 0.026 By trial and error, = VD = 0.5775 VO V, I = 0.221 mA (b) I 10 VD = , VO = 5 I ( 20 ) VD 40 I = 0.2355 mA, VD = 0.579 V, VO = 0.289 (c) I 10 VD = , VO = 2 I (5) 25 I = 0.3763 mA, VD = 0.5913 V, VO = 0.1185 (d) I = 51014 A, 5 V VO ______________________________________________________________________________________ 1.49 (a) Diode forward biased VD = 0.7 V 5 = (0.4)(4.7) + 0.7 +V V = 2.42 V (b) D (0.4)(0.7) 0.28 m P = I V = P = ______________________________________________________________________________________ 1.50 (a) 2 1 1 I = I = = = I R D D I = = 2 I V V V R 0.65 0.65 mA 1 2(0.65) 1.30 mA 2 5 3(0.65) 1.30 2.35 K = 0 = = = 2 1 1 1 D I r D R R (b) 2 0.65 0.65 mA 1 8 3(0.65) 3.025 mA = = 2 2 2 1 2 2 1 3.025 0.65 2.375 mA R D D D D R D I I I I I I I = = = = = ______________________________________________________________________________________ 1.51

19. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ a. V . . I (0 026) 0 026 k 26 1 = = = = = = = = = i . I v i v . 0 05 50 A peak-to-peak (26)(50) A 1 30 mV peak-to-peak T d DQ d DQ d d d d I . . b. For 0 1 mA (0 026) 260 = = = DQ d 0 . 1 0 05 5 A peak-to-peak d DQ i = . I = (260)(5) V 1 30 mV peak-to-peak d d d d v = i = v = . ______________________________________________________________________________________ 1.52 (a) 1 r V = = 0.026 = k 0.026 T DQ d I 0.026 (b) = 100 0.26 rd 0.026 (c) = 10 2.6 rd ______________________________________________________________________________________ 1.53 a. diode resistance d T r =V /I v r v V I d T r R V R d S d S T S T d s o T S I v V v v V IR / = = + + = = + S v b. 260 S R = I v V . v . 1 mA, 0 026 0 0909 = 0 = T = 0 = v V + IR 0 . 026 + (1)(0 . 26) v S T S S = = = I . v . v . 0 1 mA, 0 0 026 0 0 50 ( )( ) v . . . v 0 026 + 0 1 0 26 s S I . v . v . 0 01 mA. 0 026 0 909 = 0 = 0 = v . . . v 0 026 + (0 01)(0 26) S S ______________________________________________________________________________________

20. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.54 pn junction diode 0.026 ln 0.72 1013 ( ) 0.548 5 10 3 = = VD V Schottky diode 0.026 ln 0.72 108 ( ) 0.249 5 10 3 = = VD V ______________________________________________________________________________________ 1.55 V Schottky: exp a S T I I V 3 7 V V I ln (0.026) ln 0.5 10 = = 5 10 0.1796 a T S I V = Then a V = + of pn junction 0.1796 0.30 0.4796 = 0 5 10 3 exp 0 4796 exp 0 026 I I . S V . V . a T = = 4 87 10 12 A SI = . ______________________________________________________________________________________ 1.56 (a)

21. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3 I + I = 0.510 1 2 5 10 8 exp D 10 12 exp D 0 5 10 V V . V V + = T T 3 . V . 5 0001 10 8 exp D 0 5 10 3 = V T 3 8 V . . V . (0 026) ln 0 5 10 0 2395 = D 5 . 0001 10 D = Schottky diode, 2 I = 0.49999 mA pn junction, 1 I = 0.00001 mA (b) I V V 10 12 exp D1 5 10 8 exp D2 = = V V T T 1 2 0.9 D D V +V = V = V V V 10 exp 5 10 exp 0.9 12 1 8 1 D D T T 5 10 exp 0.9 exp V 8 D 1 V V T T = V 8 = . V . 2 exp 5 10 exp 0 9 1 12 10 0 026 D T 8 2 ln 5 10 0 9 1 1813 V = V + . = . D 1 T 10 12 1 0 5907 pn junction DV = . 2 0 3093 Schottky diode DV = . I . I 10 12 exp 0 5907 7 35 mA = = . 0 026 . ______________________________________________________________________________________ 1.57 0 5 6 V at 0 1 mA Z Z Z V = V = . I = . 10 Z r =

22. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (0 1)(10) 1 mV Z Z I r = . = VZ0 = 5.599 a. L R I . . . 10 5 599 4 401 8 63 mA Z R r 0 . 50 0 . 01 Z = = = + + ( )( ) 0 5.599 0.00863 10 Z Z ZZ V =V + I r = + 0 5 685 V Z V =V = . V I . . b. 11 V 11 5 599 10 59 mA PS Z 0 . 51 = = = ( )( ) 0 5.599 0.01059 10 5.7049 V Z V =V = + = V I . . 9 V 9 5 599 6 669 mA PS Z 0 . 51 = = = ( )( ) 0 5.599 0.006669 10 5.66569 V Z V =V = + = 0 0 V = 5.7049 5.66569V = 0.0392 V c. I = IZ + IL I V I V V I V V 0 , PS 0 , 0 Z 0 = = = L Z R R r L Z V V . V . . 10 5 599 0 0 0 50 0 010 2 = + 0 . V 10 5 599 1 1 1 0 50 0 010 0 50 0 010 2 + = + + 0 . . . . 20.0 + 559.9 = V0 (102.5) 0 V = 5.658 V ______________________________________________________________________________________ 1.58 10 (a) 6.8 = 6.4 IZ = mA 0.5 P = I ZVZ = (6.4)(6.8) = 43.5 mW (b) = (0.1)(6.4) = 0.64 mA I Z = 6.4 0.64 = 5.76 I L mA I V = = = 6.8 = 1.18 k 5.76 Z Z L Z L I L R V R ______________________________________________________________________________________ 1.59 (0.1)(20) 2 mV Z Z I r = = 0 6 8 0 002 6 798 V ZV = . . = . a. L R = I . I . 10 6 798 6 158 mA = = Z 0 . 5 + 0 . 02 Z

23. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( )( V0 =VZ =VZ 0 + IZ rZ = 6.798+ 0.006158 20) 0 V = 6.921 V b. Z L I = I + I V V . V . . 10 6 798 0 0 0 50 0 020 1 = + 0 . V 10 6 798 1 1 1 0 30 0 020 0 50 0 020 1 + = + + 0 . . . . 359.9 = V0 (53) V0 = 6.791 V 0 V = 6.791 6.921 0 V = 0.13 V ______________________________________________________________________________________ 1.60 For VD = 0, I = 0.1 A SC For ID = 0 ln 0.2 1 14 V = V + D T 5 10 V =V = 0.754 V D DC ______________________________________________________________________________________ 1.61 = 0, = 0.2 A VD I D = 0.60 V, A VD = 0.1995 I D = 0.65 V, A VD = 0.1964 I D = 0.70 V, A VD = 0.1754 I D = 0.72 V, A VD = 0.1468 I D = 0.74 V, A VD = 0.0853 I D = 0.7545 V, VD = 0 I D ______________________________________________________________________________________ 1.62 (a) = ( ) 1 0.7126 0 . 16 0 . 20 5 10 14 exp = 0.026 D VD V V (b) P = (0.16)(0.7126) = 0.114 W ______________________________________________________________________________________

24. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 2 2.1 (a) For > 0.6 I 1000 = O I V, ( 0.6) 1020 For < 0.6 I V, = 0 O 0 1000 1 = = t O (b) (ii) [10 sin( ) 0.6] 1020 sin 0.6 1 t = = ( ) 3.44 0.01911 1 t = rad Then ( ) 0.06 10 Also ( ) 180 3.44 176.56 0.9809 2 t = = rad Now 1 1 ( ) (t)dt [ x ]dx T avg T O = O = 2 0 0 10sin 0.6 2 1 x x 10 cos 0.6 = 0.9809 0.01911 0.9809 0.01911 2 1 ) = [( )( ) ( ] 10 0.9982 0.9982 0.6 0.9809 0.01911 2 O (avg) = 2.89 V 1000 = = (iii) ( ) 0.6 9.2157 2 10 sin 1020 O peak V; (max) = 9.2157 id mA (iv) PIV =10 V ______________________________________________________________________________________ 2.2 v v v 0 v V i i v 0 I R v v V v 0 0 ln and ln I D D D T D S I T I R S = = = = ______________________________________________________________________________________

25. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.3 (a) 16.97 1 2 120 = = S V (peak) 10 O (peak ) = 16.27 V i peak = 16.27 = D mA (b) ( ) 8.14 2 (c) O = 16.97 sin t 0.7 sin 0.7 1 1 t t ( ) = = 0.04125( ) = 2.364 16.97 ( ) = 180 2.364 = 177.64 2 t 100% 48.7% % = 177 . 64 2 . 364 = 360 (d) 1 (avg) = ]O [ 16.97 sin x 0.7 dx 0.9869 0.01313 2 1 ( ) x x [( )( ) ( ] 16.97 cos 0.7 = 0.9869 0.01313 0.9869 0.01313 2 = 1 ) 16.97 0.99915 0.99915 0.7 0.9738 2 O (avg) = 5.06 V (e) ( ) ( ) 2.53 avg = = 5.06 = 2 2 i avg O D mA ______________________________________________________________________________________ 2.4 (a) R (t) = 15sin t 0.7 9 = 15sin t 9.7 sin 1 9.7 ( t ) = = 40.29 0.2238 rad 1 15 ( t ) = 180 40.29 = 139.71 0.7762 rad 2 1 (avg) = R [ 15sin x 9.7 ]dx 0.7762 0.2238 2 1 0.7762 15 cos 9.7 1 ( ) [( )( ) ( ] = x x ) 15 0.7628 0.7628 9.7 0.5523 2 2 0.2238 0.7762 0.2238 = R (avg) = 0.9628 V ( ) = 0.8 = 0.9628 R = 1.20 R iD avg (b) % 139.71 40.29 = 100% 27.6% 360 = ______________________________________________________________________________________

26. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.5 (a) ( ) ( ) = R 15 9.7 peak = = 4.417 1.2 R i peak R (b) R (t) = 15sin t 9.7 ( ) 0.2238 1 t = ; ( ) 0.7762 2 t = 1 15sin 9.7 R (avg) = [ x ]dx 0.7762 0.2238 Or from Problem 2.4, R (avg) = 2(0.9628) = 1.9256 V ( ) ( ) 0.436 avg = = 1.9256 = 4.417 R i avg R D A (c) % 139.71 40.29 = 100% 27.6% = 360 ______________________________________________________________________________________ 2.6 (a) S (peak ) =12 + 0.7 = 12.7 V 13.4 120 2 1 = = N 12.7 2 N R 12 (b) = = 60 0.2 12 ( )( )( ) 6667 2 60 60 0.25 V = = 2 r M fRV C F (c) = 2 (max) = 2(12.7) 0.7 = 24.7 PIV S V V ______________________________________________________________________________________ 2.7 ( ) ( ) 0 0 2 max max S S v v V v v V2 = = + a. For ( ) ( ) ( ) 0 max 25 V max 25 2 0.7 = 26.4 V S v = v = + N N N N 160 6.06 26.4 = = 1 1 2 2 b. For ( ) ( ) 0 max 100 V max 101.4 V S v = v = N N N N 160 1.58 101.4 = = 1 1 2 2 From part (a) 2 (max) 2(26.4) 0. S PIV v V = = 7 or PIV = 52.1 V or, from part (b) PIV = 2(101.4) 0.7 or PIV = 202.1 V ______________________________________________________________________________________ 2.8 (a) v v v (max) 12 2(0.7) 13.4 V rms 13.4 (rms) 9.48 V ( ) = + = = = 2 s s s

27. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) V = V C = V M M f R f V R 2 2 12 2222 F ( )( )( ) 2 60 0.3 150 r C r C = C = (c) i V V = M + M ( ) , peak 1 2 12 2 12 1 150 0.3 = + , peak 2.33 A d r d R V i = ______________________________________________________________________________________ 2.9 (a) ( ) ( ) ( ) ( ) v max 12 0.7 12.7 V = + = v max = = V v v rms rms 8.98 2 S S S S 12 = = = ( )( )( ) (b) 60 150 0.3 M M r r V V C V fRC fRV or C = 4444 F (c) For the half-wave rectifier 1 4 12 1 4 12 = M + M = + i = 4.58 A or D, max , max 2 150 20.3 ( ) D r i V V R V ______________________________________________________________________________________ 2.10 (a) (peak ) = 10 O 0.7 = 9.3 V V (b) = = 9.3 ( )( )( ) 620 60 500 0.5 r M fRV C F (c) PIV = 10 + 9.3 = 19.3 V ______________________________________________________________________________________ 2.11 (a) 10.3 O 12.3 V (b) V = VM = 12.3 = r 0.586V fRC ( 60 )( 1000 )( 350 10 6 ) 10.3 = Vr V ( )( = 0.490 60 1000 )( 350 10 6 ) So 0.490 Vr 0.586 V V (c) = = 12.3 ( )( )( ) 513 60 1000 0.4 r M fRV C F ______________________________________________________________________________________

28. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.12 (a) S (peak ) = 8.5( 2 )= 12.02 V 12.02 0.7 11.32 max = = VO V 11.32 V (b) = = ( )( )( ) = 0.03773 2 60 10 0.25 2 r M f RV C F (c) PIV= 2 ( ) = 2(12.02) 0.7 = 23.34 S peak V V ______________________________________________________________________________________ 2.13 (a) ( ) ( ) ( ) peak 15 2 0.7 16.4 V rms 16.4 11.6 V 2 s s v v = + = = = 15 2857 F = M = = (b) 2 2 ( 60 )( 125 )( 0.35 ) r C V f RV ______________________________________________________________________________________ 2.14 ______________________________________________________________________________________ 2.15 (a) = 12.8 S V 13.3 120 2 1 = = N 12.8 2 N R 12 = 3% = (0.03)(12) = 0.36V Vr Vr (b) = = 24 0.5 12 ( )( )( ) 0.0116 2 60 24 0.36 V = = = 2 r M fRV C F (c) ( ) ( ) + = V M M i peak iD (peak ) = 13.3A 1 V = + 1 2 12 0.36 2 12 24 r D R V

29. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d) ( ) ( ) ( ) + V V 1 2 2 1 2 0.36 = M M r i avg iD (avg) = 0.539 A = + D V 2 12 0.36 2 1 12 24 12 2 1 M r V R V (e) PIV = 12.8 +12 = 24.8V ______________________________________________________________________________________ 2.16 (a) = 9 + 2(0.8) = 10.6 S V 16 120 2 1 = = N 10.6 2 N R 9 (b) = = 90 0.1 = 9 F = ( )( )( ) 4167 r V M fRV C 2 2 60 90 0.2 (c) ( ) ( ) 3.08 + = 1 2 9 V 1 = i peak A 0.2 2 9 M M V = + 90 r D R V (d) ( ) ( ) ( ) + V V 1 2 2 1 2 0.2 = M M r i avg iD (avg) = 0.1067 A = + D V 2 9 0.2 2 1 9 90 9 2 1 M r V R V (e) PIV = (max) = 10.6 0.8 = 9.8 S V V ______________________________________________________________________________________ 2.17 For 0 i v > V 0 = Voltage across L 1 i R + R = v

30. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Voltage Divider v R v = = 0 R + R 1 1 2 L i i L v ______________________________________________________________________________________ 2.18 For 0, ( 0) iv V > = a. = + = = = = + v R R 2 v 0 L L 2 1 2 0 || || i || 2.2 || 6.8 1.66 k 1.66 0.43 1.66 2.2 L i i R R R R R v v v b. ( ) ( max ) 0 ( ) 0 0 rms rms 3.04 2 v v = v = V ______________________________________________________________________________________ 2.19 (a) 0.975 = 3.9 = I L mA 4 I = I 1.342 mA 20 3.9 = 12 = = 1.342 0.975 = 0.367mA I Z I I I L = = (0.367)(3.9) = 1.43mW PZ I ZVZ = 3.9 = I L mA (b) 0.39 10 = 1.342 0.39 = 0.952mA I Z = (0.952)(3.9) = 3.71mW PZ ______________________________________________________________________________________

31. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.20 (a) 40 12 0.233 A 120 0.233 12 2.8 W ( )( ) Z I P = = = = (b) IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A So 0.21 12 57.1 L = = L R R (c) P = (0.1)(0.233)(12) P = 0.28 W ______________________________________________________________________________________ 2.21 (a) PZ = I ZVZ 4 = I (15.4) I (max) = z Z 259.74 mA So 15 I z 259.74 mA 60 (b) 15.4 = 297.33 I = mA I 0.15 So I (max) = L 297.33 15 = 282.33mA (min) = 297.33 259.74 = 37.59mA I L min 15.4 RL ( ) = = 410 Then ( ) = = 54.55 0.28233 max 15.4 RL 0.03759 So 54.55 RL 410 ______________________________________________________________________________________ 2.22 a. 20 10 45.0 mA 222 10 26.3 mA 380 = = 18.7 mA I I I I I I I I I I = = = = L L Z I L Z b. max 400 mW max 400 40 mA ( ) ( ) ( ) ( ) ( ) 10 P I I I I I Z Z min max 45 40 min 5 mA 10 2 k L I Z L L L R R = = = = = = = = (c) For 175 i R = 57.1 mA 26.3 mA 30.8 mA I L Z I = I = I = (max) 40 mA (min) 57.1 40 17.1 mA I I = = = = = Z L 10 585 17.1 R R L L ______________________________________________________________________________________

32. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.23 a. From Eq. (2.30) ( ) [ ] [ ] ( )( ) ( )( ) ( ) ( ) 500 20 10 50 15 10 max 15 0.9 10 0.1 20 5000 250 4 max 1.1875 A min 0.11875 A Z Z Z I I I = = = = From Eq. (2.28(b)) 20 10 8.08 i 1187.5 50 i R R = = + b. ( )( ) ( ) ( )( ) 0 P P P I V P 1.1875 10 11.9 W max 0.5 10 5 W = = = = = Z Z L L L ______________________________________________________________________________________ 2.24 (a) = 0 I L I = Z 83.0 mA 10 5.6 50 + 3 = 5.6 + (0.083)(3) = 5.85V VZ = VL = = (0.083)(5.85) = 0.486W PZ I ZVZ (b) 10 VL VL 5.6 + VL 3 200 50 = 0.20 +1.867 = (0.02 + 0.3333 + 0.005) VL So = 5.769V VL Then 28.84 = 5.769 = I L mA 0.2 I = I 84.62 mA 10 5.769 = 0.050 And = = 55.8 I Z I I I L mA = (0.0558)(5.769) = 0.322W PZ (c) = 0 I L I = Z 120.8 mA VZ = VL = 5.6 + (0.1208)(3) = 5.962 V 12 5.6 50 + 3 = (0.1208)(5.962) = 0.72W PZ (d) 12 VL VL 5.6 + VL 3 200 50 = 0.24 +1.867 = (0.02 + 0.333+ 0.005) VL So = 5.88V VL Then 29.4 = 5.88 = I L mA; 122.4 0.20 12 5.88 = = I I mA 0.05 = 122.4 29.4 = 93 mA I Z = (0.093)(5.88) = 0.547W PZ ______________________________________________________________________________________

33. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.25 V (a) Set I = Z 10 mA; I = = 7.5 = 7.5 mA 1 L L L R = 10 + 7.5 = 17.5 mA I I = 17.5 = 12 7.5 i 257 I R = i R I (b) 7.5 = + (0.01)(12) = 7.38 V VZO VZO For = (1.1)(12) = 13.2V VI 13.2 VL VL 7.38 + VL 12 1000 257 = 0.05136 + 0.615 = (0.00389 + 0.0833 + 0.001) VL = 7.556 VL V For = (0.9)(12) = 10.8 V VI 10.8 VL VL 7.38 + VL 12 1000 257 = 0.04202 + 0.615 = (0.08819) = 7.450 VL VL V Then, Source Reg 100% 4.42% 7 . 556 7 . 450 = 13.2 10.8 = (c) For = 1 k , V RL = 7.50 VL 12 7.38 For = , RL 17.17 257 + 12 = I Z mA = 7.38 + (0.01717)(12) = 7.586 VL V Then , Load Reg 100% 1.15% 50 . 7 586 . 7 = = 7.50 ______________________________________________________________________________________ 2.26 ( ) ( ) V V max min L L % Reg = 100% ( ) V nom ( ) ( ) ( ( ) ( ) ) V nom + I max r V nom + I min r = = = ( ) ( max ) ( min ) ( 3 ) 0.05 6 L L Z z L Z L Z Z V nom I I z A So (max) (min) 0.1 Z Z I I = Now (max) 6 0.012 , (min) 6 0.006 L 500 L 1000 I = = A I = = A Now ( ) ( ) ( ) V min V PS Z min max i Z L R I I = + 280 15 6 ( ) ( min ) 0.020 Z or min 0.012 Z I A I = = +

34. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then (max) 0.1 0.02 0.12 and ZI = + = A ( ) ( ) ( ) V max V PS Z max min i Z L R I I = + or (max) 6 ( ) PS 280 max 41.3 0.12 0.006 PS V V V = = + ______________________________________________________________________________________ 2.27 Using Figure 2.19 a. 20 25% 15 25 V PS PS V = V For (min) : PS V ( ) ( ) ( ) I I I min max 5 20 25 mA min 15 10 200 = + = + = R R 25 I Z L V V PS Z i i I I = = = I (max) 25 = 10 (max) I (max) = 75 mA I I b. For V PS R i For (min) 0 (max) 75 mA L Z I = I = ( )( ) Z Z ZZ V V Ir V V V 0 0 0 0 10 0.025 5 9.875 V = = = ( ) ( )( ) ( ) ( )( ) max = 9.875 + 0.075 5 = 10.25 min = 9.875 + 0.005 5 = 9.90 0.35 V = V % Reg 0 100% % Reg 3.5% c. V ( nom ) 0 = = ______________________________________________________________________________________ 2.28 From Equation (2.28(a)) ( ) ( ) ( ) V V min 24 16 PS Z min max 40 400 i Z L R I I = = + + or 18.2 i R = Also V V C V = M = M 2 fRC 2 fRV + = + = 18.2 2 20.2 r r R R r i z Then 24 9901 C = C = F ( )( )( ) 2 60 1 20.2 ______________________________________________________________________________________ 2.29 ( ) nom 8 V V V I r V = + = = + = 0 0 0 ( )( ) ( ) ( ) V V V V 8 0.10.5 7.95 V max nom 12 8 1.333 A = = = 3 Z Z ZZ Z Z Z S Z i i I R For 0.2 A 1.133 A L Z I = I = For 1 A 0.333 A L Z I = I =

35. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( ) ( ) V V I r = + = + = = + = + = max max L Z Z Z ( )( ) 0 7.95 1.133 0.5 8.5165 ( ) ( ) V min V I min r L Z Z Z ( )( ) 0 7.95 0.333 0.5 8.1165 0.4 V = % Reg = = 0.4 % Reg = 5.0% ( ) 0 nom 8 = = 2 2 = + = + = 3 0.5 3.5 L L M M r r i z V V V V V C V fRC fRV R R r 12 0.0357 C = C = F ( )( )( ) Then 2 60 3.5 0.8 ______________________________________________________________________________________ 2.30 For 6.3 = I 3 V, O I For > 3 I V, 3 I = I 1.5 and O I I (0.5) = ( ) 0.667 1.0 = I 1.5 3 5 . 0 + = I O I For < 6.3 I V, + 6.3 2.5 I = I and O I I (0.5) = 6.3 + ( ) 0.8 1.26 5 . 0 = = I 2.5 I O I I 0______________________________________________________________________________________ 2.31 (a) For 10 v 0, both diodes are conducting v = 0 I O For v 3, i = 0, v = 0 Zener not in breakdown, so 1 O v v > i = mA v v v = ( ) = For 3 3 1 20 3 10 1 1.5 20 2 I I I o I At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA

36. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) For 0, both diodes forward biased I v < I i v 1 0 . 10 = At 1 10 , 1 Iv = V i = mA For v i v 1 3 3, I . 20 I > = At 1 10 , 0.35 Iv = V i = mA ______________________________________________________________________________________ 2.32 (a) V = 1 15 = 5 V for v 5.7, v = v 1 3 I 0 I For 5.7 Iv > V ( ) v V V V v V v v v v 0.7 15 , 0.7 1 1 1 1 2 1 15 ( 0.7 ) 0.7 1 2 1 15.7 0.7 1 1 1 2.5 1 2 1 1 2 1 ( ) v v v 8.55 2.5 1 3.42 ( ) 0 1 0 0 0 0 0 v v v v 0 0 v v v v 0 0 2.5 5.7 5.7 15 9.42 I I I I I I I + + = = + + = + + = + + = + = = + = = = =

37. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) 0 for D i = 0 5 I v Then for 5.7 Iv > V .7 I + 3.42 I = I O = 2.5 1 1 D v v i v v or i v 0.6 3.42 1 I D = For vI = 15, iD = 5.58 mA ______________________________________________________________________________________ 2.33 (a) (i) = 1.8 V VB For 1.1 I V, O I = For 1.1 I V, = 1.1 O V (ii) = 1.8V VB For 2.5 I V, O I = For 2.5 I V, = 2.5 O V (b) (i) = 1.8 V VB For 2.5 I V, = 2.5 O V For 2.5 I V, O I = (ii) = 1.8V VB For 1.1 I V, = 1.1 O V For 1.1 I V, O I = ______________________________________________________________________________________ 2.34 For 30 V, i 30 10.7 0.175 A 100 10 I v = = = + 0 v = i(10) +10.7 = 12.5 V

38. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. ______________________________________________________________________________________ 2.35 (a) (i) = 5 V VB For 5.7 I V, = 5.7 O I For 5.7 I V, = 0 O (ii) = 5 V VB For 4.3 I V, = + 4.3 O I For 4.3 I V, = 0 O (b) (i) = 5 V VB For 4.3 I V, = 0 O For 4.3 I V, = 4.3 O I (ii) = 5 V VB For 5.7 I V, = 0 O For 5.7 I V, = + 5.7 O I ______________________________________________________________________________________ 2.36 a. 0 V = 0.6 V = b. 0 V =

39. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.6 V = ______________________________________________________________________________________ 2.37 ______________________________________________________________________________________ 2.38 One possible example is shown. 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. ______________________________________________________________________________________ 2.39 (a) Square wave between +40V and 0. (b) Square wave between +35V and 5 V. (c) Square wave between +5 V and 35V. ______________________________________________________________________________________ 2.40 a. For 0 2.7 V x V V = = b. For 0.7V 2.0 V x V = V = ______________________________________________________________________________________

40. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.41 Circuit similar to Figure 2.31(a) with = 10 VB V. ______________________________________________________________________________________ 2.42 In steady-state, O = (10 sin t + 5) V ______________________________________________________________________________________ 2.43 (i) = 5 V, In steady-state, VB O = (10sin t 5) V (ii) = 5 V, In steady-state, VB O = (10 sin t 15) V ______________________________________________________________________________________ 2.44 a. 10 0.6 0.94 mA 0 9.5 0.5 I I V I V = = D D 1 1 + ( ) 9.5 8.93 V = = D 0 1 0 D2 I = b. 5 0.6 0.44 mA 0 9.5 0.5 I I V I V = = D D 1 1 + ( ) 9.5 4.18 V = = D 0 1 0 D2 I = c. Same as (a) d. ( I )( ) ( ) ( ) 0 0 I I 10 0.5 0.6 9.5 0.964 mA = + + = 2 = 9.5 = 9.16 V = = = = V I V I I I I I 1 2 1 2 0.482 mA 2 D D D D ______________________________________________________________________________________ 2.45 a. 1 2 0 0 1 D D I = I = I = V = 0 b. ( ) ( ) D D I I I I I 10 9.5 0.6 0.5 0.94 mA 0 = + + = = = = = ( ) 2 1 V I V 10 9.5 1.07 V 0 0 c. ( ) ( ) D D I I I I I 10 9.5 0.6 0.5 5 0.44 mA 0 = + + + = = = = ( ) 2 1 V I V 10 9.5 5.82 V 0 0 =

41. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ d. I I I ( ) ( ) 10 = 9.5 + 0.6 + 0.5 = 0.964 mA 2 I I I I I V I V = = = = = = D D D D 1 2 1 2 ( ) 0 0 0.482 mA 2 10 9.5 0.842 V ______________________________________________________________________________________ 2.46 a. V V D D D V I I 0 , , , on 4.4 V = = = 1 2 1 2 3 0 10 4.4 0.589 mA 9.5 = = 4.4 0.6 7.6 mA 0.5 I I I I I I I I I = = = = = + = = D D D D 1 2 1 2 ( ) 2 7.6 0.589 14.6 mA D D D D 3 1 2 3 b. V V D D D 5 V and on, off = = = + + + = 1 2 1 2 3 I I I ( ) ( ) 10 9.5 0.6 0.5 5 0.451 mA 2 I I I I I = = = = D D D D 1 2 1 2 3 0.226 mA 2 0 D I = ( ) ( )( ) V I V 10 9.5 10 0.451 9.5 5.72 V = = = 0 0 c. V1 = 5 V, V2 = 0 D1 off, D2, D3 on 0 V = 4.4 V 10 4.4 0.589 mA 9.5 4.4 0.6 7.6 mA 0.5 I I = = I I = = D D 2 2 D 1 I I I I I 3 2 3 0 7.6 0.589 7.01 mA D D D = = = = d. V1 = 5 V, V2 = 2 V D1 off, D2, D3 on 0 V = 4.4 V 10 4.4 0.589 mA 9.5 4.4 0.6 2 3.6 mA I I = = I I = = D D 2 2 D 1 0.5 I I I I I 3 2 3 0 3.6 0.589 3.01 mA D D D = = = = ______________________________________________________________________________________ 2.47 (a) V = 4.4 V, I = 0.2 = 10 0.6 4.4 R = 25 k 1 D 1 1 R 1 2 0.2 0.3 0.5mA = + = I R V = 4.4 ( 0.6 ) 2 0.6V, 0.5 2 10 I = = R R k 2 = 2 R

42. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3 0.5 0.5 1.0mA = + = I R ( ) 4.4 1.0 0.6 5 R = k 3 = (b) Assume all diodes conducting 10 0.6 4.4 1 4.4 V, V = 0.5 I = = mA D 1 10 4.4 ( 0.6 ) V = 2 0.6V, 1.25 = I R mA 2 = 4 Then 2 1.25 0.5 0.75 mA = = I D ( ) 2 2.2 0.6 5 I = = mA R 3 Then 3 2 1.25 0.75mA = = I D (c) Diode D2 cutoff 2 0 = I D 10 0.6 ( 0.6 ) V = 10 2 0.6V, 1.11 I D mA 1 = = 9 + 1 2 = R R 1 10 0.6 (1.11)(3) 6.07 V V = = ( ) 1.76 2.5 0.6 5 I = = mA R 3 Then 3 1.76 1.11 0.65mA = = I D (d) Diode D3 cutoff 3 0 = I D 10 0.6 4.4 1 4.4 V, V = 0.833 I = = mA D 1 6 ( ) 1.044 4.4 5 9.4 I R mA = R R 2 = = 9 + 2 3 2 (1.044)(6) 5 1.27 V V = = Then 2 1.044 0.833 0.211mA = = I D ______________________________________________________________________________________ 2.48 (a) 1 2 2.5mA = = I D I D 0.7 0.7 ( 2 ) 2.5 = 2 0.8 I = = D R k R (b) I D1 = 0.2I D2 , 1 2 5 + = I D I D 0.2 2 2 5 2 4.167 + = = I D I D I D mA 2 4.167 2 0.48 = = R = I D k R (c) I D1 = 5I D2 , 1 2 5 + = I D I D 5 2 2 5 2 0.833 + = = I D I D I D mA 2 0.833 2 2.4 = = R = I D k R ______________________________________________________________________________________

43. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.49 (a) D1 and D2 on ( ) 0.7 5 1.1 5 0.7 + VA VA 2 5 = + = VA + VA 2.30 V 5 2.15 3.909 1 1 = 2 1.1 5 0.7 2.3 = I D mA Then 1.0 1 = 2 ( ) 6.0 2.3 0.7 5 I = = mA D 2 1.1 (b) D1 cutoff, 1 0 = I D 2 5mA, = I D = 0.7 + (5)(2.5) 5 = 8.2 VA V 5 0.7 0 (c) = 0 , VA 2.15 = I D mA 1 = 2 Then 2 5 2.15 7.15mA = + = I D 0 0.7 ( 5 ) 7.15 2 0.60 I = = D R k 2 = 2 R ______________________________________________________________________________________ 2.50 (a) (i) = 5 I V, D1 and D2 on ( ) 0.6 5 0.6 0.5 + O O O O 5 0.5 5 5 = + + 0.88 +1.0 +1.2 = (0.20 + 0.20 + 2.0 + 2.0) = 0.7 O O V (ii) = 5 I V = V O I 0.455 5 . 0 = 0.5 5 + (b) (i) = 5 I V, = 4.4 O V (ii) = 5 I V, = 0.6 O V ______________________________________________________________________________________ 2.51

44. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For 0. when D I v > 1 and D4 turn off 10 0.7 0.465 mA 20 10 k 4.65 V I v I = = = = ( ) 0 0 = for 4.65 4.65 I I v v v ______________________________________________________________________________________ 2.52 (a) All diodes on ( ) ( ) VA VA VA VA 15 0.7 10 24 0.7 5 14 2 0.7 6.15 + + = 2.439 + 0.35 0.307 0.3875 = VA (0.1626 + 0.50 + 0.0714 + 0.0417) VA = 2.70 V Then 2.70 0.7 1.0 = I D mA 1 = 2 ( ) 0.50 2.70 0.7 5 = I D mA 2 = 14 ( ) 0.50 2.70 0.7 10 = I D mA 3 = 24 (b) D1 cutoff, 1 0 = I D ( ) ( ) VA VA VA 15 0.7 10 5.2 0.7 5 3.3 6.15 + = 2.439 1.303 1.788 = (0.1626 + 0.303 + 0.1923) = 0.991 VA VA V Then 0.991 0.7 ( 5 ) 1.0 I = = mA D 2 3.3 ( ) 1.60 0.991 0.7 10 I = = mA D 3 5.2 (c) D1 and D2 cutoff, I = = D 1 I D 2 0 ( ) 3.25 15 0.7 10 24.3 I D mA 3 = 6.15 1.32 1 4 + = + = R R = 15 (3.25)(6.15) = 5 V VA ______________________________________________________________________________________

45. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.53 a. 1 2 R = 5 k, R = 10 k D1 and D2 on 0 V = 0 ( ) 1 1 10 0.7 0 10 1.86 1.0 5 10 0.86 mA D D I I = = = b. D R R D D I I V IR V 10 k , 5 k , off, on 0 10 0.7 10 = = = 1 2 1 2 ( ) = = = = 0 2 0 1.287 15 10 3.57 V 1 ______________________________________________________________________________________ 2.54 If both diodes on (a) V V I A O ( ) ( ) 1 2 1 1 2 1 1 0.7 V, 1.4 V 10 0.7 1.07 mA 10 1.4 15 2.72 mA 5 2.72 1.07 1.65 mA R R R D R D D I I I I I I = = = = = = + = = = (b) D1 off, D2 on ( ) ( )( ) I = I = = R 1 R 2 + V I R V V V D o I = = = = + = = O R 2 2 O A 1 D 1 10 0.7 15 1.62 mA 5 10 15 1.62 10 15 1.2 V 1.2 0.7 1.9 ff , 0 ______________________________________________________________________________________ 2.55 (a) D1 on, D2 off 1 10 0.7 0.93 mA 10 15 V I V D O = = =

46. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) D1 on, D2 off 1 10 0.7 1.86 mA 5 15 V I V D O = = = ______________________________________________________________________________________ 2.56 ( V ) 0 V V 0 0 15 + 0.7 + 0.7 10 20 20 15 0.7 0.7 1 1 1 4.0 10 10 20 0 10 20 20 0 20 0 6.975 V 0 0.349 mA 20 D D V V V I V I = + = + + = = = = ______________________________________________________________________________________ 2.57 (a) Diode is cutoff, = 0 , I D = 0 VD = = 3 V VA VB (b) Diode is conducting, = 0.7 V VD VB VB VB VB 5 0.7 2 10 0.7 10 10 10 + = + 0.50 + 0.07 + 0.27 = (0.10 + 0.10 + 0.10 + 0.10) = 2.1 VB VB V and = 1.4 VA V D V V = + 10 10 B B I 5 5 2.1 2.1 = So 0.08 = I D mA 10 10 (c) Diode is cutoff, = 0 I D = 1 = VA V, (4) 2.0 (5) 2.5 2 = 1 = VB V 2 = 2 2.5 = 0.5 V VD (d) Diode is conducting, = 0.7 V VD VB VB VB VB 8 0.7 2 10 0.7 10 10 10 + = + 0.80 + 0.07 + 0.27 = VB (0.40) VB = 2.85 V and VA = 2.15 V 8 Then = 1 [8 2(2.85)] = 0.23 10 = B B 10 10 D V V I mA ______________________________________________________________________________________

47. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.58 0, D I v = 1 off, D2 on 10 2.5 0.5mA 15 10 0.5 5 7.5V for 0 7.5V o o I v v ( )( ) = = = I v I For > v 7.5 V , Both D 1 and D2 on I o o o v v v v 2.5 10 = + 15 10 5 or (5.5) 33.75 I o v = v When vo = 10 V, D2 turns off (10)(5.5) 33.75 21.25 V I v = = For 21.25 V, 10 V I o v > v = ______________________________________________________________________________________ 2.59 (a) For = 0.5 I V, 1 2 3 0 = = = I D I D I D , = 0.5 O V (b) For = 1.5 I V, D1 on; 2 3 0 = = I D I D I = D 0.0667 mA 1.5 0.7 1 = 4 + 8 = 0.7 + (0.0667)(8) = 1.23 O V (c) For = 3 I V, D1 and D2 conducting, 3 0 = I D 3 6 1.7 O O O 8 0.7 4 + = 0.75 + 0.0875 + 0.2833 = (0.25 + 0.125 + 0.1667) = O O 2.069 V 2.069 Then 0.7 0.171 I = = D mA 1 8 0.0615 2.069 1.7 2 = 6 = I D mA (d) For = 5 I V, all diodes conducting 5 4 2.7 O O O O 6 1.7 8 0.7 4 + + = 1.25 + 0.0875 + 0.2833 + 0.675 = (0.25 + 0.125 + 0.1667 + 0.25) O So = O 2.90 V 2.90 Then 0.7 0.275 I = = D mA 1 8 I = D 0.20 mA 2.90 1.7 2 = 6 I = D 0.05 mA 2.90 2.7 3 = 4 I ______________________________________________________________________________________ 2.60 (a) I = < D 2 0 for 4.5 V 2 100 mA for = I D > 4.5 I V (b) 2 0 for = I D < 9 I V 2 100 mA for = I D > 9 I V ______________________________________________________________________________________

48. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.61 a. V01 =V02 = 0 b. 01 02 V = 4.4 V, V = 3.8 V c. 01 02 V = 4.4 V, V = 3.8 V Logic 1 level degrades as it goes through additional logic gates. ______________________________________________________________________________________ 2.62 a. 01 02 V =V = 5 V b. 01 02 V = 0.6 V, V =1.2 V c. 01 02 V = 0.6 V, V =1.2 V Logic 0 signal degrades as it goes through additional logic gates. ______________________________________________________________________________________ 2.63 ( ) ( 1 2 3 V AND V OR V AND V ) 4 ______________________________________________________________________________________ 2.64 10 1.5 0.2 12 mA 0.012 10 10 8.3 691.7 0.012 681.7 I R R R = = = + + = = = ______________________________________________________________________________________ 2.65 I V V V = I = = = ( ) 10 1.7 8 0.75 10 1.7 8 0.75 2.3 V I I ______________________________________________________________________________________ 2.66 R + rf = 0.220 k rf = 20 R = 200 5 1.7 = 15 ______________________________________________________________________________________

49. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.67 V I V V 1 V, 0.8 mA 1 ( 0.8 )( 2 ) 2.6 V R PS PS = = = + = ______________________________________________________________________________________ 2.68 Ph I e A = = = 3 ( )( 19 )( ) 0.6 10 1 1.6 10 10 2 2 3.75 10 cm 17 A A ______________________________________________________________________________________

50. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 3 3.1 mA/V 2 n 0.75 10 0.8 120 k W = 2 2 = L K n (a) (i) = 0 I D (ii) = (0.75)[2(1 0.4)(0.1) (0.1)2 ]82.5 I D A (iii) = (0.75)[2(2 0.4)(0.1) (0.1)2 ]= 0.2325 I D mA (iv) = (0.75)[2(3 0.4)(0.1) (0.1)2 ]= 0.3825 I D mA (b) (i) = 0 I D (ii) = (0.75)(1 0.4)2 = 0.27mA I D (iii) = (0.75)(2 0.4)2 =1.92mA I D (iv) = (0.75)(3 0.4)2 = 5.07mA I D ______________________________________________________________________________________ 3.2 I D = Kn [2(VGS VTN )VDS VD2S ] 0.5 = Kn [2(0.6)VDS VD2S ] 1.0 = Kn [2(1.0)VDS VD2S ] Take ratio V V = VDS 0.5VD2S =1.2VDS VD2S DS DS V V 2 2 1.2 2 0.5 DS DS or 1 0.5VDS = 1.2 VDS which yields = 0.4 V VDS Then 0.5 = [(1.2)(0.4) (0.4)2 ] Kn = 1.56 Kn mA/V 2 ______________________________________________________________________________________ 3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now ( ) ( ) ( ) ( ) 2 K K K K K K K K K K K K 0.03 = 2 1.5 = 0.25 = 0.12 n n n 0.15 = 3 1.5 2 = 2.25 = 0.0666 n n n 0.39 = 4 1.5 2 = 6.25 = 0.0624 n n n 0.77 = 5 1.5 2 = 12.25 = 0.0629 n n n From last three, (Avg) 0.0640 mA/V2 n K = (c) 2 2 i i i i V V (sat) = 0.0640(3.5 1.5) (sat) = 0.256 mA for = 3.5 V (sat) = 0.0640(4.5 1.5) (sat) = 0.576 mA for = 4.5 V D D D D GS GS ______________________________________________________________________________________

51. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.4 0 V = = = = a. ( ) ( ) V sat V V 0 2.5 2.5 V GS DS GS TN i. V I I 0.5 V Biased in nonsaturation 1.1 2 0 ( 2.5) 0.5 0.5 2.48 mA = = ( ) ( )( ) ( )2 = DS D D V I I 2.5 V Biased in saturation 1.1 0 2.5 6.88 mA DS = = = ii. ( )( ( ))2 D D iii. VDS = 5 V Same as (ii) 6.88 mA D I = b. VGS = 2 V (sat) 2 ( 2.5) 4.5 V DS V = = V I I 0.5 V Nonsaturation 1.1 2(2 ( 2.5))(0.5) (0.5) 4.68 mA DS = = = i. ( ) 2 D D V I I 2.5 V Nonsaturation 1.1 2(2 ( 2.5))(2.5) (2.5) 17.9 mA DS = = = ii. ( ) 2 D D V I I 5 V Saturation 1.1 2 2.5 22.3 mA DS = = = iii. ( )( ( ))2 D D ______________________________________________________________________________________ 3.5 (a) ( ) = = 2.2 0.4 = 1.8 VDS sat VGS VTN V 2.2 = VDS > VDS (sat) = 1.8Saturation (b) ( ) = = 1 0.4 = 0.6 V VDS sat VGS VTN = 0.6 (1) = 0.4 V VDS < VDS (sat) = 0.6 V Nonsaturation (c) VGS = 11 = 0 Cutoff ______________________________________________________________________________________ 3.6 (a) VSG = 2.2 2.2 = 0Cutoff (b) = 2V, V VSG = 2 (1) = 3 VSD ( ) = + = 2 + ( 0.4) = 1.6 VSD sat VSG VTP V So VSD = 3 > VSD (sat) = 1.6Saturation (c) = 2V, V VSG = 2 1 = 1 VSD ( ) = + = 2 + ( 0.4) = 1.6 VSD sat VSG VTP V So VSD = 1 < VSD (sat) = 1.6Nonsaturation ______________________________________________________________________________________ 3.7 ( )2 k W n D V V I 2 L GS TN = W = 5.79 0.5 0.12 [0 ( 1.2)]2 = 2 L W L ______________________________________________________________________________________

52. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.8 n ox (600)(3.9)(8.85 10 14 ) 2.071 10 10 = = = =t n n ox ox ox ox k C t t (a) 500 A 41.4 A/V2 n k = (b) 250 82.8 A/V2 n k = (c) 100 207 A/V2 n k = (d) 50 414 A/V2 n k = (e) 25 828 A/V2 n k = ______________________________________________________________________________________ 3.9 ( 20 10 4 )( 650 )( 3.9 )( 8.85 10 14 ) (a) ( )( ) 1.40 n ox W 2 4 8 2 0.8 10 200 10 = = = ox n Lt K mA/V 2 (b) I = Kn ( VGS VTN )2 = (1.40)(2 0.4 ) 2 D Or I = D 3.58 mA (c) VDS ( sat ) = VGS VTN = 2 0.4 = 1.6 V ______________________________________________________________________________________ 3.10 k (a) n W ( )2 D V V I 2 L GS TN = W W 0.6 0.12 2 = (1.4 0.8) 27.8 2 = L L Or W = (27.8)(0.8) = 22.2 m (b) (27.8)[2(1.4 0.8)(0.4) (0.4) ] 0.534 12 . 0 2 = = I D mA 2 (c) ( ) = = 1.4 0.8 = 0.6 VDS sat VGS VTN V ______________________________________________________________________________________ 3.11 ( 600 )( 3.9 )( 8.85 10 ) 8 14 200 10 = = = n ox ox k C n n ox t k 2 mA/V 1035 . 0 = n ( )2 k W n D V V I 2 L GS TN = W W 1 . 2 0 . 1035 2 = (3 0.6) 4.026 2 = L L Then W = (4.026)(0.8) = 3.22 m ______________________________________________________________________________________

53. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.12 I D =WCox (VGS VTN ) sat ( 3.9 )( 8.85 10 ) 7 C F/cm 8 14 1.726 10 200 10 = = = ox ox ox t = (3.22104 )(1.726107 )(3 0.6)(2107 ) I D = 2.67mA I D ______________________________________________________________________________________ 3.13 ( )2 k p W I + D V V 2 L SG TP = W + 0.225 0.05 VTP (2 )2 2 L = W + 0.65 0.05 VTP (3 )2 2 L = 3 + 0.65 = = TP V V Then 1.70 0.571 = TP 2 0.225 + TP V V W W 0 . 225 = 0 . 05 2 = And (2 0.571) 4.41 2 L L ______________________________________________________________________________________ 3.14 V V V V V V sat V V 5 V, 0 5 V 0.5 50.5 4. = = = = = + = =5 V S G SG ( ) TP SD SG TP a. V V I I 0 5 V Biased in saturation 2 5 0.5 40.5 mA = = = = D SD ( )2 D D b. V V I I 2 V 3 V Nonsaturation 2 2 5 0.5 3 3 36 mA = = = = D SD ( )( ) ( )2 D D c. V V I I 4 V 1 V Nonsaturation 2 2 5 0.5 1 1 16 mA = = = = D SD ( )( ) ( )2 D D d. 5 V 0 0 D SD D V = V = I = ______________________________________________________________________________________ 3.15 (a) Enhancement-mode (b) From Graph VTP = + 0.5 V ( ) ( ) ( ) ( ) 2 = = = p p p = 2 = = 2 = = 2 = 2 0.45 2 0.5 2.25 0.20 1.25 3 0.5 6.25 0.20 2.45 4 0.5 12.25 0.20 4.10 5 0.5 20.25 0.202 Avg 0.20 mA/V p p p p p p p k K K k K k K k K K =

54. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) 2 2 (sat) 0.20 (3.5 0.5) 1.8 mA (sat) 0.20 (4.5 0.5) 3.2 mA i i D D = = = = ______________________________________________________________________________________ 3.16 VSD (sat ) =VSG +VTP (a) ( ) 1 2 ( ) 1 SD SD V sat = + V sat = V (b) ( ) 0 2 ( ) 2 SD SD V sat = + V sat = V (c) ( ) 1 2 ( ) 3 SD SD V sat = + V sat = V k W k W I V V V sat ( ) ( ) 2 2 p p = + = D SG TP SD L L 2 2 (a) 0.040 (6)(1)2 0.12 2 D D I = I = mA (b) 0.040 (6)(2)2 0.48 2 D D I = I = mA (c) 0.040 (6)(3)2 1.08 2 D D I = I = mA ______________________________________________________________________________________ 3.17 K 0.375 mA/V 2 (a) Nonsaturation 12 k W = p 0.8 50 = 2 2 = p p K L = (0.375)[2(2 0.5)(0.2) (0.2)2 ]= 0.21 I D mA (b) Nonsaturation = (0.375)[2(2 0.5)(0.8) (0.8)2 ]= 0.66 I D mA (c) Nonsaturation = (0.375)[2(2 0.5)(1.2) (1.2)2 ]= 0.81 I D mA (d) Saturation = (0.375)(2 0.5)2 = 0.844 I D mA (e) Saturation = (0.375)(2 0.5)2 = 0.844 I D mA ______________________________________________________________________________________ 3.18 ( )( )( 14 ) 11 = = = = . . . k C p ox 250 3 9 8 85 10 8 629 10 t t 0 0 p p ox x x 0x t (a) 500 17 3 A/V2 ox p t = k = . (b) 250 34 5 A/V2 pk = . (c) 100 86 3 A/V2 pk = . (d) 50 173 A/V2 p k = (e) 25 345 A/V2 p k =

55. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 3.19 ( )( 14 ) = = = t ( ) ( )( ) ( ) ( )( ) 8 2 8 0 8 2 8 2 3.9 8.85 10 6.90 10 F/cm 500 10 675 6.90 10 46.6 A/V 375 6.90 10 25.9 A/V ox C ox x k C k C n n ox p p ox = = = = PMOS: = + k W I V V ( ) W W L L = ( ) = = = L W K K K = ( ) = 2 2 2 2 L 0.8 0.0259 5 0.6 3.19 2 4 m 12.8 m 0.0259 3.19 41.3 A/V 2 p D SG TP p p p p = p p n Want Kn = Kp k n W k = p W = 41.3 L L = = = = 2 N 2 p 46.6 W 41.3 W 1.77 2 L L N N L W 4 m 7.09 m N ______________________________________________________________________________________ 3.20 ( )( ) V I r r 2 V, 0.2 2 1.2 0.128 mA 1 1 781 k = = = = = = GS D ( 0.01 )( 0.128 ) ( )( ) I V I r r 4 V, 0.2 4 1.2 1.57 mA = = = = = 1 63.7 k ( )( ) V V = = = ( ) 2 0 0 2 0 0 0.01 1.57 1 1 100 V 0.01 D GS D A A ______________________________________________________________________________________ 3.21 0.12 = ( ) (4)(2 0.5) 0.54 I mA 2 k W = GS TN 2 2 2 = n D V V L 1 1 1 ( )( ) 0.00926 r = = = 3 3 = 200 10 0.54 10 o I r I DQ o DQ V 1 Then = 1 = 108 A V V ______________________________________________________________________________________

56. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.22 = + + V V V V V 2 2 TN TNO f SB f ( ) ( ) = 2 = 0.8 TN 2 + 2 0.35 f SB ( ) V V 2.5 0.837 2 0.35 10.4 SB SB + = + = V ______________________________________________________________________________________ 3.23 2 2 TN TNo f SB f ( ) ( ) 0.75 0.6 2 0.37 3 2 0.37 0.75 0.6 [ 1.934 0.860 ] 1.39 V TN (sat) 2.5 1.39 1.11 V DS V V r V V V = + + = + + = + = = = (a) Sat Region (15) 0.08 (2.5 1.39)2 2 0.739 mA D D I I = = (b) Non-Sat (15) 0.08 2(2.5 1.39)(0.25) (0.25)2 2 0.296 mA D D I I = = ______________________________________________________________________________________ 3.24 = 6106 V/cm ox (a) (i) = = (6106 )(120108 )= 7.2V VG ox tox = 7.2 = VG V (ii) 2.4 3 (b) (i) = (6106 )(200108 )=12 V VG = 12 = VG V (ii) 4 3 ______________________________________________________________________________________ 3.25 Want ( )( ) ( )ox tox tox 3 24 = = 6106 t = 1.2 10 5 cm = 1200 Angstroms 0 x ______________________________________________________________________________________

57. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.26 = = = + + V R V 18 2 ( 10 ) 3.6 V G R R DD 18 32 1 2 Assume transistor biased in saturation region I V V V K V ( ) = S = G GS = D n R R V V ( )( )( ) 2 2 3.6 = 0.5 2 0.8 2 V V 1.6 0.64 S S GS GS = + GS GS GS TN V V V 0.6 2.96 0 0.6 0.6 4 2.96 = ( ) ( ) V V GS GS I V V I D D ( ) GS GS G GS V V I R R ( )( ) ( ) 2 2 2.046 V 2 3.6 2.046 0.777 mA 2 R 10 0.777 4 2 5.34 V sat S DS DD D D S DS DS DS V V V + = = = = = + = + = > = ______________________________________________________________________________________ 3.27 (a) 4 V (sat) 4 0 8 3 2 V GS DS V = V = . = . ( )2 If Sat 0.25 4 0.8 2.56 1.44 I V D DS = = = Non-Sat ( ) 4 2 4 0.25 1 2 4 0.8 4 2.6 0.25 0.25 2.6 4 0 D D DS n D GS T DS DS DS ( )( ) ( ) ( ) 2 2 2 2 2.6 6.76 4 1.88 V 2 0.25 4 1.88 2.12 mA 1 DS DS DS DS DS DS DS DS D I R V KR V V V V V V V V V V V V V I = + = + = + = + = = = = =

58. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Non-Sat region ( ) = + = + = + = I R V KR V V V V V 5 2 5 0.25 3 2 5 0.8 5 7.3 0.75 D D DS n D GS T DS DS DS ( )( ) ( ) 2 2 2 DS DS DS DS DS V V V V V 0.75 2 7.3 5 0 DS DS 7.3 53.29 15 ( ) 2 0.75 0.741 V 5 0.741 1.42 mA 3 DS DS D V V V V I + = = = = = ______________________________________________________________________________________ 3.28 12 . 0 8 . 0 2 = = (80)( VGS 0.4) VGS 0.808 V 2 1 V = GS Rin VDD R 1 0.80825 1 (200)(1.8) 1 445 = R = R 1 k 1 2 200 = = R R Rin k 2 363 k R = ______________________________________________________________________________________ 3.29 (a) 3.5 (sat) 3.5 0.8 2.7 V SG DD SD V =V = V = = ( )( )2 If biased in Sat region, 0.2 3.5 0.8 D I = = 1.46 mA 3.5 (1.46)(1.2) 1.75 V SD V = = Biased in Non-Sat Region.

59. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( ) SD D D SD p D SG TP SD SD ( )( ) ( ) SD SD SD SD SD SD SD SD ( ) SD SD ( ) ( )( ) ( ) 2 2 2 2 2 3.5 2 3.5 0.2 1.2 2 3.5 0.8 3.5 1.296 0.24 0.24 2.296 3.5 0 2.296 5.272 3.36 use sign 1.90 V 2 0.24 0.2 2 3.5 0.8 1.9 1.9 D V I R V KR V V V V V V V V V V V V V V I = + = + + = + = + + = + = = = = 0.2 [ 10.26 3.61 ] 3.5 1.90 1.33 mA 1.2 1.33 mA D D I I = = = (b) 5 V (sat) 5 0.8 4.2 V SG DD SD V =V = V = = If Sat Region ( )( )2 0.2 5 0.8 3.53 mA, 0 D S I = = V D< Non-Sat Region. ( ) = + + = + = + V KR V V V V V V V V V V V V 5 2 5 0.2 4 2 5 0.8 5 6.72 0.8 0.8 7.72 5 0 SD p D SG TP SD SD ( )( ) ( ) SD SD SD SD SD SD + = SD SD V V ( ) 2 2 2 2 7.72 59.598 16 use sign 0.698 V SD SD 2 0.8 5 0.698 1.08 mA I I D D 4 = = = = ______________________________________________________________________________________ 3.30 22 = (6) 3 1.40 = G V V 22 8 + ( ) = K pRS VSG +VTP +VSG +VG 3 2 3 = (0.5)(0.5)( 2 1.6 + 0.64)+ +1.40 VSG VSG VSG 0.25 2 + 0.6 1.44 = 0 =1.483V VSG VSG VSG = (0.5)(1.483 0.8)2 = 0.2332mA I D = 6 (0.2332)(0.5 + 5) = 4.72V VSD ______________________________________________________________________________________

60. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ S V 3.31 VG = 0, VSG = Assume saturation region ( ) I K V V = = + = = + = = = = = = = D p SG TP ( )( V ) V V S S V I R ( )( ) V V V V 2 ( ) 2 0.4 0.4 0.2 0.8 S 0.4 0.8 2.21 V 0.2 5 0.4 5 5 3 V 2.21 3 5.21 V D D D SD S D SD (sat) SD SD V > V ______________________________________________________________________________________ 3.32 I mA 0.18 3.3 1.6 0.8 = 5 = V V V = DD DSQ RS D D R W W 0 . 18 0 . 12 2 = (0.8 0.4) 18.75 2 = L L RS = 0.8 = 4.44 k 0.18 R 0.8 0.8 1.6 2 (3.3) 1 2 + = + = = R R VG Now I = (0.05)(0.18) R 9 A So 3.3 367 1 2 R + R = = k 0.009 = R Then (3.3) 178 2 = 1.6 2 367 R k and 1 189 R = k ______________________________________________________________________________________ 3.33 I = 0.2 = D 0.2 mA 1 = 1.8 (0.2)(4 +1) = 0.8 VDS V Now VDS (sat) = 0.8 0.4 = 0.4 V ( ) VDS sat =VGS VTN 0.4 =VGS 0.4VGS = 0.8 V ( )2 k W n I D V V 2 GS TN L = W W 0 . 2 0 . 12 2 = (0.8 0.4) 20.8 2 = L L Now = + = 0.8 + (0.2)(1) = 1.0 VG VGS I DRS V 1 1 1 (200)(1.8) 1 360 VG = = = R = in DD k R R V R 1 1 1 2 200 = = R R Rin k 2 450 k R =

61. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 3.34 (a) (50)( 0.4) 0.742 12 . 0 35 . 0 2 = = GS GS V V V 2 (b) = 1.8 (0.35)(2) = 1.1V VDS ( ) = = 0.742 0.4 = 0.342 VDS sat VGS VTN V VDS > VDS (sat)Saturation _____________________________________________________________________________________ 3.35 6 = (10) 5 2 = VG V 6 14 + ( ) 5 k W 5 2 = + = + S GS TN 2 n G GS D S GS R V V L V V I R V (25)(0.5)( 0.8 0.16) 5 2 = VGS + 0 . 12 VGS 2 VGS + 2 Or 0.75 2 + 0.4 2.88 = 0 =1.71V VGS VGS VGS 12 . 0 2 = I = (25)(D 1.71 0.4) 2.58 mA 2 = 10 (2.58)(1.2 + 0.5) = 5.62V VDS ______________________________________________________________________________________ 3.36 L Let 20 = W for example, then 05 . 0 20 . 0 2 = = (20)( VSG 0.6) VSG 1.232 V 2 Then = 1.232 = (0.2) = 6.16 VRS RS RS k RD = 8.84 k 6 1.232 3 = 0.2 = 31.232 1.232 = 0.536 V VG 6 (0.1)(0.2) 0.02 1 2 I = = mA, R + = R 1 R 2 300 k = R + R = = R = ( ) (6) 3 R 0.536 6 3 2 2 300 R R 1 2 + VG Or 2 176.8 k and k R = 1 123.2 R = ______________________________________________________________________________________ 3.37 (a) (i) ( ) 2 50 500 1.2 1.516 5 1.516 6.516 I V V V V = = = = = = Q GS GS ( ) DS DS V V (ii) ( )( ) ( ) 2 1 0.5 1.2 2.61 5 2.61 7.61