Transcription
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem SolutionsChapter 11.1ni BT 3 / 2 e(a) Silicon Eg / 2 kT 1.1 exp 2 ( 86 10 6 ) ( 250 ) 2.067 1019 exp [ 25.58]ni 1.61 108 cm 3(i)ni ( 5.23 1015 ) ( 250 )(ii)ni ( 5.23 1015 ) ( 350 )(b)GaAs(i)ni ( 2.10 1014 ) ( 250 )3/ 2 1.1 exp 2 ( 86 10 6 ) ( 350 ) 3.425 1019 exp [ 18.27 ]ni 3.97 1011 cm 33/ 23/ 2 1.4 exp 6 2 ( 86 10 ) ( 250 ) ( 8.301 1017 ) exp [ 32.56]ni 6.02 103 cm 3(ii)ni ( 2.10 1014 ) ( 350 )3/ 2 1.4 exp 2 ( 86 10 6 ) ( 350 ) (1.375 1018 ) exp [ 23.26]ni 1.09 108 cm 31.2a. Eg ni BT 3 / 2 exp 2kT 1.11012 5.23 1015 T 3 / 2 exp 6 2(86 10 )(T ) 6.40 103 1.91 10 4 T 3 / 2 exp T By trial and error, T 368 Kb.ni 109 cm 3 1.1 109 5.23 1015 T 3 / 2 exp 2 ( 86 10 6 ) (T ) 6.40 103 1.91 10 7 T 3 / 2 exp T By trial and error, T 268 K
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.3Silicon(a)ni ( 5.23 1015 ) (100 )3/ 2 1.1 exp 2 ( 86 10 6 ) (100 ) ( 5.23 1018 ) exp [ 63.95]ni 8.79 10 10 cm 3(b)ni ( 5.23 1015 ) ( 300 )3/ 2 1.1 exp 6 2 ( 86 10 ) ( 300 ) ( 2.718 1019 ) exp [ 21.32]ni 1.5 1010 cm 3(c)ni ( 5.23 1015 ) ( 500 )3/ 2 1.1 exp 2 ( 86 10 6 ) ( 500 ) ( 5.847 1019 ) exp [ 12.79]ni 1.63 1014 cm 3Germanium.(a)ni (1.66 1015 ) (100 )3/ 2 0.66 (1.66 1018 ) exp [ 38.37 ]exp 6 2 ( 86 10 ) (100 ) ni 35.9 cm 3(b)ni (1.66 1015 ) ( 300 )3/ 2 0.66 ( 8.626 1018 ) exp [ 12.79]exp 6 2 ( 86 10 ) ( 300 ) ni 2.40 1013 cm 3(c)ni (1.66 1015 ) ( 500 )3/ 2 0.66 (1.856 1019 ) exp [ 7.674]exp 6 2 ( 86 10 ) ( 500 ) ni 8.62 1015 cm 31.4(a) n-type; no 1015()()n22.4 1013cm ; po i no1015 322 5.76 1011 cm 3ni21.5 1010 2.25 10 5 cm 3no1015(b) n-type; no 1015 cm 3 ; po
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.5(a) p-type; p o 1016 cm 3 ; no (ni21.8 10 6 po1016)(2 3.24 10 4 cm 3)2ni22.4 1013 5.76 1010 cm 3po1016(b) p-type; p o 1016 cm 3 ; no 1.6(a)(b)n-typeno N d 5 1016 cm 310ni2 (1.5 10 )po 4.5 103 cm 3no5 10162(c)no N d 5 1016 cm 3From Problem 1.1(a)(ii) ni 3.97 1011 cm 3( 3.97 10 ) 11 2 3.15 106 cm 35 1016po1.7(a) p-type; p o 5 1016 cm 3 ; no ()()ni21.5 1010 po5 10162 4.5 10 3 cm 32ni21.8 10 6 6.48 10 5 cm 3po5 1016(b) p-type; p o 5 1016 cm 3 ; no 1.8(a) Add boron atoms(b) N a po 2 1017 cm 3()2ni21.5 1010 1.125 10 3 cm 3po2 1017(c) no 1.9(a)no 5 1015 cm 310n 2 (1.5 10 )po i po 4.5 104 cm 3no5 10152(b)n o p o n-type(c) no N d 5 1015 cm 3
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.10a.b.Add DonorsN d 7 1015 cm 3Want po 106 cm 3 ni2 / N dSo ni2 (106 )( 7 1015 ) 7 10 21 Eg B 2T 3 exp kT 2 1.1 7 1021 ( 5.23 1015 ) T 3 exp 6 ( 86 10 ) (T ) By trial and error, T 324 K1.11(a) I Aσ Ε 10 5 (1.5)(10) I 0.15 mA( )()Iρ 1.2 10 3 (0.4) 2.4 V/cm Aρ2 10 4(b) I AΕ Ε ()1.12J 120 1 6.67 (Ω cm)Ε 18σ(6.67)σ eμ n N d N d 3.33 1016 cm 3eμ n1.6 10 19 (1250)J σΕ σ ()1.13111 Nd 7.69 1015 cm 3eμ n N deμ n ρ 1.6 10 19 (1250)(0.65)Ε(b) J Ε ρ J (0.65)(160 ) 104 V/cmρ(a) ρ ()1.14σ1.5 9.375 1015 cm 3eμ n1.6 10 19 (1000)σ0.8 1.25 1016 cm 3(b) N a 19eμ p1.6 10 (400)(a) σ eμ n N d N d (())1.15(a) For n-type, σ eμ n N d (1.6 10 19 ) ( 8500 ) N dFor 1015 N d 1019 cm 3 1.36 σ 1.36 104 ( Ω cm ) 1(b) J σ E σ ( 0.1) 0.136 J 1.36 103 A / cm2
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.16Dn (0.026)(1250) 32.5 cm 2 /s; D p (0.026 )(450 ) 11.7 cm 2 /sJ n eDn 1016 1012dn 1.6 10 19 (32.5) dx 0 0.001(J p eD p) 52 A/cm 2 1012 1016dp 1.6 10 19 (11.7 ) dx 0 0.001() 18.72 A/cm 2 Total diffusion current densityJ 52 18.72 70.7 A/cm 21.17J p eD pdpdx 1 x eD p (10 15 ) exp Lp Lp (1.6 10 ) (15) (10 ) exp x 19Jp 15 Lp 10 10 4J p 2.4 e x / LpJ p 2.4 A/cm2(a)x 0(b)x 10 μ mJ p 2.4 e 1 0.883 A/cm 2x 30 μ mJ p 2.4 e 3 0.119 A/cm 2(c)1.18a.N a 1017 cm 3 po 1017 cm 3n 2 (1.8 10no i 1017pob.)6 2 no 3.24 10 5 cm 3n no δ n 3.24 10 5 1015 n 1015 cm 3p po δ p 1017 1015 p 1.01 1017 cm 3 N N1.19 Vbi VT ln a 2 d ni(a) (i) ()( 5 1015 5 1015Vbi (0.026 ) ln 2 1.5 1010(()) 0.661 V )( )() (10 )(10 ) (0.026 ) ln 0.937 V (1.5 10 ) (ii) 5 1017 1015 Vbi (0.026 ) ln 0.739 V10 2 1.5 10(iii)Vbi181810 2
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions()( 5 1015 5 1015Vbi (0.026 ) ln 2 1.8 10 6(b) (i)(()) 1.13 V )( ) 1.21 V() (10 )(10 ) (0.026 ) ln 1.41 V (1.8 10 ) (ii) 5 1017 1015Vbi (0.026 ) ln 2 1.8 10 6(iii)Vbi18186 21.20 N NVbi VT ln a 2 d nior (n ) exp V(1.5 10) exp 0.712 1.76 1016 cm 3bi V 1016Nd 0.026 T Na 1.212i2 N a (1016 ) N N Vbi VT ln a 2 d ( 0.026 ) ln 10 2 (1.5 10 ) ni For N a 1015 cm 3 , Vbi 0.637 VFor N a 1018 cm 3 , Vbi 0.817 V1.22 T kT (0.026) 300 000.09545.911,180.3
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions 1.4 ni ( 2.1 1014 )(T 3 / 2 ) exp 2 ( 86 10 6 ) (T ) N N Vbi VT ln a 2 d ni Tni2001.2562506.02 1033001.80 1063501.09 1084002.44 1094502.80 10105002.00 1011Vbi1.4051.3891.3701.3491.3271.3021.2771.23 V C j C jo 1 R Vbi 1/ 2 (1.5 10 16 )( 4 10 15 ) 0.684 VVbi ( 0.026 ) ln (1.5 10 10 ) 2 1 C j ( 0.4 ) 1 0.684 1/ 2(a)3 C j ( 0.4 ) 1 0.684 1/ 2(b) 0.255 pF 0.172 pF 1/ 25 0.139 pFC j ( 0.4 ) 1 0.684 (c)1.24(a) V C j C jo 1 R Vbi 1 / 25 For VR 5 V, C j (0.02) 1 0. 8 1 / 2 1. 5 For VR 1.5 V, C j (0.02) 1 0. 8 0.00743 pF 1 / 2 0.0118 pF
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions0.00743 0.0118 0.00962 pF2vC ( t ) vC ( final ) ( vC ( initial ) vC ( final ) ) e t / τC j (avg ) whereτ RC RC j (avg ) (47 103 )(0.00962 10 12 )orτ 4.52 10 10 sThen vC ( t ) 1.5 0 ( 5 0 ) e ti / τ5 r /τ 5 e 1 t1 τ ln 1.5 1.5 10t1 5.44 10 s(b)For VR 0 V, Cj Cjo 0.02 pF 1/ 2 3.5 For VR 3.5 V, C j ( 0.02 ) 1 0.00863 pF 0.8 0.02 0.00863C j (avg ) 0.0143 pF2τ RC j ( avg ) 6.72 10 10 svC ( t ) vC ( final ) ( vC ( initial ) vC ( final ) ) e t / τ(3.5 5 (0 5)e t2 /τ 5 1 e t2 /τso that t2 8.09 10 10)s1.25 VC j C jo 1 R Vbi 1 / 2()( )) 5 1015 1017 ; Vbi (0.026 ) ln 0.739 V10 2 1.5 10(For V R 1 V,Cj 0.6011 0.739 0.391 pFFor VR 3 V,Cj 0.6031 0.739 0.267 pFFor V R 5 V,0.60Cj (a)fo (b) f o 12π LC(51 0.739 ( 0.215 pF2π 1.5 102π 1.5 101 3)(0.391 10 ) 121 3)(0.267 10 ) 12 f o 6.57 MHz f o 7.95 MHz
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutionsfo 1 f o 8.86 MHz2π 1.5 10 0.215 10 12(c)( 3)()1.26a. V V I I S exp D 1 0.90 exp D 1 VT VT V exp D 1 0.90 0.10 VT VD VT ln ( 0.10 ) VD 0.0599 Vb.IFIR VF exp I VT S IS VR exp VT 0.2 1 exp 10.026 exp 0.2 1 1 0.026 2190 1IF 2190IR V1.27 I D I S exp D VT(a) (i)(ii)(iii)(iv) 1 ()()()()() 0.3 I D 10 11 exp 1.03 μ A 0.026 0.5 I D 10 11 exp 2.25 mA 0.026 0.7 I D 10 11 exp 4.93 A 0.026 0.02 12AI D 10 11 exp 1 5.37 10 0.026 (v) 0.20 11I D 10 11 exp 1 10 A0.026 (vi)I D 10 11 A(b) (i)(ii)(iii)()()()() 0.3 I D 10 13 exp 0.0103 μ A 0.026 0.5 I D 10 13 exp 22.5 μ A 0.026 0.7 I D 10 13 exp 49.3 mA 0.026
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions()(iv) 0.02 14I D 10 13 exp A 1 5.37 100.026 (v)I D 10 13 AI D 10 13 A(vi) I1.28 V D VT ln D IS 10 10 6(a) (i) V D (0.026 ) ln 11 10 0.359 V 100 10 6V D (0.026 ) ln 11 10 0.419 V 10 3 V D (0.026) ln 11 0.479 V 10 V (ii) 5 10 12 10 11 exp D 1 V D 0.018 V 0.026 10 10 6(b) (i) V D (0.026 ) ln 13 10 0.479 V 100 10 6V D (0.026 ) ln 13 10 10 3V D (0.026) ln 13 10 0.539 V 0.599 V V (ii) 10 14 10 13 exp D 1 V D 0.00274 V 0.026 1.29(a)(b)VD 0.7 10 3 I S exp 0.026 I S 2.03 10 15 AI D ( A ) ( n 1)I D ( A )( n 2 )0.19.50 101.39 10 140.24.45 10 129.50 10 140.32.08 10 106.50 10 13 90.49.75 104.45 10 120.54.56 10 73.04 10 11 50.62.14 102.08 10 100.710 31.42 10 9 14
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.30(a)I S 10 12 AVD(v)0.100.200.300.400.500.600.70(b)ID(A)4.68 10 112.19 10 91.03 10 74.80 10 62.25 10 41.05 10 24.93 10 1log10ID 10.3 8.66 6.99 5.32 3.65 1.98 0.307I S 10 14 AVD(v)ID(A)log10ID 13 12.30.104.68 10 11 10.660.202.19 10 9 8.990.301.03 10 8 7.320.404.80 10 6 5.650.502.25 10 4 3.980.601.05 10 3 2.310.704.93 101.31a. V VD1 ID2 10 exp D 2 I D1 VT ΔVD VT ln (10) ΔVD 59.9 mV 60 mVb.ΔVD VT ln (100 ) ΔVD 119.7 mV 120 mV1.32 2 (a) (i) V D (0.026) ln 0.539 V 9 2 10 20 (ii) V D (0.026) ln 0.599 V 9 2 10 0.4 (b) (i) I D 2 10 9 exp 9.60 mA 0.026 0.65 (ii) I D 2 10 9 exp 144 A 0.026 ()()
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.33 I 2 10 3 0.6347 VVD Vt ln D (0.026) ln 14 5 10 IS 2 10 3 0.5150 VVD (0.026) ln 12 5 10 0.5150 VD 0.6347 V1.34 0.30 8(a) 1.5 10 3 I S exp I S 1.46 10 A 0.026 0.35 (b) (i) I D 1.462 10 8 exp I D 10.3 mA 0.026 0.25 (ii) I D 1.462 10 8 exp I D 0.219 mA 0.026 ()()1.35()()()() 0.8 (a) I D 10 22 exp 2.31 nA 0.026 1.0 I D 10 22 exp 5.05 μ A 0.026 1.2 I D 10 22 exp 11.1 mA 0.026 0.02 23I D 10 22 exp 1 5.37 10 A0.026 For V D 0.20 V, I D 10 22 AFor V D 2 V, I D 10 22 A(b) 0.8 I D 5 10 24 exp 115 pA 0.026 1.0 I D 5 10 24 exp 0.253 μ A 0.026 1.2 I D 5 10 24 exp 0.554 mA 0.026 ()()()() 0.02 24AI D 5 10 24 exp 1 2.68 100.026 For V D 0.20 V, I D 5 10 24 AFor V D 2 V, I D 5 10 24 A
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.36IS doubles for every 5C increase in temperature.I S 10 12 A at T 300KFor I S 0.5 10 12 A T 295 KFor I S 50 10 12 A, (2) n 50 n 5.64Where n equals number of 5C increases.Then ΔT ( 5.64 )( 5 ) 28.2 KSo 295 T 328.2 K1.37I S (T ) 2ΔT / 5 , ΔT 155 CI S ( 55)I S (100) 2155 / 5 2.147 109I S ( 55)VT @100 C 373 K VT 0.03220VT @ 55 C 216 K VT 0.01865I D (100) (2.147 109 ) I D ( 55) 0.6 exp 0.0322 0.6 exp 0.01865 ( 2.147 10 )(1.237 10 )( 9.374 10 )9 813I D (100) 2.83 103I D ( 55)1.38(a) V PS I D R V D( )2.8 I D 10 6 VD ;() V I D 5 10 11 exp D 0.026 By trial and error,V D 0.282 V, I D 2.52 μ A(b)I D 5 10 11 A, VD 2.8 V
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.39 I 10 I D ( 2 10 4 ) VD and VD ( 0.026 ) ln D 12 10 Trial and error.VD(v)ID(A)VD(v) 40.500.51944.75 10 40.5170.51944.7415 10 40.51940.51944.740 10VD 0.5194 VI D 0.4740 mA1.40I s 5 10 13 A R2VTH R1 R2 30 (1.2) (1.2) 0.45 V80 I 0.45 I D RTH VD , VD VT ln D IS By trial and error:I D 2.56 μ A, VD 0.402 V
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.41(a) I D1 I D 2 1 mA(i) 10 3V D1 V D 2 (0.026 ) ln 13 10 0.599 V (ii) 10 3V D1 (0.026 ) ln 14 5 10 0.617 V 10 3V D 2 (0.026 ) ln 13 5 10 0.557 V (b) V D1 V D 2V D1 V D 2(ii)Ii 0.5 mA2 0.5 10 3 (0.026 ) ln 13 10I D1 I D 2 (i) 0.581 V II D15 10 14 S1 0.10I D 2 I S 2 5 10 13So I D1 0.10 I D 2I D1 I D 2 1.1I D 2 1 mASo I D 2 0.909 mA, I D1 0.0909 mANow 0.0909 10 3 0.554 VV D1 (0.026 ) ln 14 5 10 0.909 10 3 0.554 VV D 2 (0.026 ) ln 13 5 10 1.42() 0.635 (a) I D 3 6 10 14 exp 2.426 mA 0.026 0.635 0.635 mAIR 1I D1 I D 2 2.426 0.635 3.061 mA 3.061 10 3V D1 V D 2 (0.026 ) ln 14 6 10V I 2(0.641) 0.635 1.917 V 0.641 V (b) I D 3 2.426 mA0.635 1.27 mAIR 0.5I D1 I D 2 2.426 1.27 3.696 mA 3.696 10 3V D1 V D 2 (0.026 ) ln 14 6 10V I 2(0.6459 ) 0.635 1.927 V 0.6459 V
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.43(a) Assume diode is conducting.Then, VD Vγ 0.7 V0. 7 23.3 μ A301.2 0.7 50 μ AI R1 10Then I D I R1 I R 2 50 23.3So that I R 2 Or I D 26.7 μ A(b) Let R1 50 k Ω Diode is cutoff.30 (1.2) 0.45 V30 50Since VD Vγ , I D 0VD 1.44At node VA:5 VAV ID A(1)22At node V B V A Vγ(2)5 (VA Vr ) ID (VA Vr )225 (VA Vr ) 5 VA VA VA Vr So32 2 2Multiply by 6:10 2 (VA Vr ) 15 6VA 3 (VA Vr )25 2Vr 3Vr 11VA(a)Vr 0.6 V11VA 25 5 ( 0.6 ) 28 VA 2.545 V
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions5 VA VA 2.5 VA I D Neg. I D 022Both (a), (b) I D 0From (1) I D 2 5 2 V VD 0.50 V5VA 2.5, VB 1.45(a) VO I i (1) ; I D 0 ; for 0 I i 0.7 mAVO 0.7 V; I D (I i 0.7 ) mA; for I i 0.7 mA(b) VO I i (1) ; I D 0 ; for 0 I i 1.7 mAVO 1.7 V; I D (I i 1.7 ) mA; for I i 1.7 mA(c) VO 0.7 V; I D1 I i ; I D 2 0 ; for 0 I i 2 mA1.46Minimum diode current for VPS (min)I D (min) 2 mA, VD 0.7 VI2 0.75 0.7 4.3, I1 R2R1R1We have I1 I 2 I D4.3 0.7 2R1R2Maximum diode current for VPS (max)P I DVD 10 I D ( 0.7 ) I D 14.3 mAso (1)I1 I 2 I Dor(2)9.3 0.7 14.3R1R2Using Eq. (1),9.3 4.3 2 14.3 R1R1R1 0.41 kΩThen R2 82.5Ω 82.5Ω1.475 0.7 0.215 mA, VO 0.7 V205 0.6 0.220 mA, VO 0.6 V(ii) I 205 0.7 ( 5) 0.2325 mA, VO (0.2325)(20 ) 5 0.35 V(b) (i) I 405 0.6 ( 5)I 0.235 mA, VO (0.235)(20 ) 5 0.30 V(iii)40(a) (i) I
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions2 0.7 ( 8) 0.372 mA, VO 2 (0.372 )(5) 0.14 V252 0.6 ( 8) 0.376 mA, VO 2 (0.376 )(5) 0.12 V(ii) I 25(d) (i) I 0 , VO 5 V(ii) I 0 , VO 5 V(c) (i) I 1.48(a) I 5 VO V , I 5 10 14 exp D 20 0.026 By trial and error, V D VO 0.5775 V, I 0.221 mA()10 V D, VO 5 I (20 ) V D40I 0.2355 mA, V D 0.579 V, VO 0.289(b) I 10 V D, VO 2 I (5)25I 0.3763 mA, V D 0.5913 V, VO 0.1185(c) I (d) I 5 10 14 A, VO 5 V1.49(a)Diode forward biased VD 0.7 V5 (0.4)(4.7) 0.7 V V 2.42 V(b) P I VD (0.4)(0.7) P 0.28 mω1.50(a)0.65 0.65 mA I D11 2(0.65) 1.30 mAI R 2 I D1 ID2ID2 (b)VI 2Vr V0 5 3(0.65) 1.30 R1 2.35 KR1R10.65 0.65 mA18 3(0.65)ID2 I D 2 3.025 mA2I D1 I D 2 I R 2 3.025 0.65I D1 2.375 mAIR2 1.51
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem SolutionsV(0.026)τd T 0.026 kΩ 26Ωa.1I DQid 0.05 I DQ 50 μ A peak-to-peakvd idτ d (26)(50) μ A vd 1.30 mV peak-to-peakb.(0.026) 260Ω0. 1 5 μ A peak-to-peakFor I DQ 0.1 mA τ d id 0.05 I DQvd idτ d (260)(5) μ V vd 1.30 mV peak-to-peak1.52(a) rd VT0.026 1 kΩI DQ 0.0260.026 100 Ω0.260.026 10 Ω(c) rd 2.6(b) rd 1.53a.diode resistance rd VT /I VT /I rd vd vS V rd RS T RS I VT vd vs vo VT IRS b. vS RS 260Ω v0 VTv0.026 0 0.0909 vS VT IRS 0.026 (1)(0.26)vSvv0.026I 0.1 mA, 0 0 0.50vs 0.026 ( 0.1)( 0.26 )vSI 1 mA,I 0.01 mA.v0v0.026 0 0.909vS 0.026 (0.01)(0.26)vS
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem Solutions1.54pn junction diode 0.72 10 3 0.548 VV D (0.026 ) ln 13 5 10 Schottky diode 0.72 10 3 0.249 VV D (0.026 ) ln 8 5 10 1.55 V Schottky: I I S exp a VT I 0.5 10 3 Va VT ln (0.026) ln 7 5 10 IS 0.1796 VThenVa of pn junction 0.1796 0.30 0.4796I0.5 10 3 V 0.4796 exp a exp 0.026 VT I S 4.87 10 12 AIS 1.56(a)
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem SolutionsI1 I 2 0.5 10 3 V5 10 8 exp D VT VD 12 3 10 exp 0.5 10V T V5.0001 10 8 exp D VT 3 0.5 10 0.5 10 3 VD 0.2395VD (0.026) ln 8 5.0001 10 Schottky diode, I 2 0.49999 mApn junction, I1 0.00001 mA(b) V V I 10 12 exp D1 5 10 8 exp D 2 VT VT VD1 VD 2 0.9 V 0.9 VD1 10 12 exp D1 5 10 8 exp VT VT 0.9 VD1 5 10 8 exp exp V T VT 2V 5 10 8 0. 9 exp D1 exp 12 V10 0.026 T 5 10 8 0.9 1.18132VD1 VT ln 12 10 VD1 0.5907 pn junctionVD 2 0.3093 Schottky diode 0.5907 I 10 12 exp I 7.35 mA 0.026 1.57VZ VZ 0 5.6 V at I Z 0.1 mArZ 10Ω
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem SolutionsI Z rZ ( 0.1)(10 ) 1 mVVZ0 5.599a.RL IZ 10 5.5994.401 8.63 mAR rZ0.50 0.01VZ VZ 0 I Z rZ 5.599 ( 0.00863)(10 )VZ V0 5.685 Vb.11 5.599 10.59 mA0.51VZ V0 5.599 ( 0.01059 )(10 ) 5.7049 VVPS 11 V I Z 9 5.599 6.669 mA0.51VZ V0 5.599 ( 0.006669 )(10 ) 5.66569 VVPS 9 V I Z ΔV0 5.7049 5.66569 ΔV0 0.0392 Vc.I IZ ILVV V0V VZ 0, IZ 0I L 0 , I PSRLRrZ10 V0 V0 5.599 V0 0.500.010210 5.59911 1 V0 0.50 0.0100.500.0102 20.0 559.9 V0 (102.5)V0 5.658 V1.5810 6.8 6.4 mA0.5P I Z VZ (6.4)(6.8) 43.5 mW(b) I Z (0.1)(6.4) 0.64 mAI L 6.4 0.64 5.76 mA(a) I Z VZV6.8 RL Z 1.18 k ΩRLI Z 5.76IL 1.59I Z rZ ( 0.1)( 20 ) 2 mVVZ 0 6.8 0.002 6.798 Va.RL IZ 10 6.798 I Z 6.158 mA0.5 0.02
Microelectronics: Circuit Analysis and Design, 4th editionChapter 1By D. A. NeamenProblem SolutionsV0 VZ VZ 0 I Z rZ 6.798 ( 0.006158)( 20 )V0 6.921 Vb.I IZ IL10 V0 V0 6.798 V0 0.500.020110 6.79811 1 V0 0.30 0.020 0.50 0.020 1 359.9 V0 (53)V0 6.791 VΔV0 6.791 6.921ΔV0 0.13 V1.60For VD 0, I SC 0.1 A 0.2 1 For ID 0 VD VT ln 14 5 10 VD VDC 0.754 V1.61 V D 0, I D 0.2 AV D 0.60 V, I D 0.1995 AV D 0.65 V, I D 0.1964 AV D 0.70 V, I D 0.1754 AV D 0.72 V, I D 0.1468 AV D 0.74 V, I D 0.0853 AV D 0.7545 V, I D 01.62 V (a) 0.16 0.20 5 10 14 exp D 1 V D 0.7126 V 0.026 (b) P (0.16 )(0.7126 ) 0.114 W()
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem SolutionsChapter 22.1 1000 (a) For υ I 0.6 V, υ O (υ I 0.6 ) 1020 For υ I 0.6 V, υ O 0 1000 (b) (ii) υ O 0 [10 sin (ω t )1 0.6] 1020 0. 6Then sin (ω t )1 0.06 (ω t )1 3.44 0.01911π rad10Also (ω t )2 180 3.44 176.56 0.9809π radNowυ O (avg ) T11υ O (t )dt T 02π 12π2π [10 sin x 0.6]dx00.9809π 0.6 x 10 cos x 0.01911π0.01911π 0.9809π1[( 10 )( 0.9982 0.9982 ) 0.6(0.9809π 0.01911π )]2πυ O (avg ) 2.89 V 1000 π (iii) υ O ( peak ) 10 sin 0.6 9.2157 V; i d (max ) 9.2157 mA10202 (iv) PIV 10 V2.2v0 vI vD i vvD VT ln D and iD 0IR S v v0 vI VT ln 0 IS R
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions2.3 1 (a) υ S 120 2 16.97 V (peak) 10 υ O ( peak ) 16.27 V16.27 8.14 mA2(c) υ O 16.97 sin ω t 0.7(b) i D ( peak ) sin (ω t )1 0. 7 0.04125 (ω t )1 2.364 16.97(ω t )2 180 2.364 177.64 177.64 2.364 % 100% 48.7%360 (d)υ O (avg ) 12π0.9869π [16.97 sin x 0.7]dx0.01313π0.9869π0.9869π 1 0. 7 x( 16.97 ) cos x 2π 0.01313π0.01313π 1[( 16.97 )( 0.99915 0.99915) 0.7(0.9738π )] 2πυ O (avg ) 5.06 Vυ O (avg )5.06 2.53 mA22(e) i D (avg ) 2.4(a) υ R (t ) 15 sin ω t 0.7 9 15 sin ω t 9.7(ω t )1 sin 1 9.7 40.29 0.2238π(ω t )2rad 15 180 40.29 139.71 0.7762π radυ R (avg ) 12π0.7762π [15 sin x 9.7]dx0.2238π0.7762π0.7762π 1 1( 15) cos x 9.7 x 2π [( 15)( 0.7628 0.7628 ) 9.7(0.5523π )]2π 0.2238π0.2238π υ R (avg ) 0.9628 V i D (avg ) 0.8 0.9628 R 1.20 ΩR(b) 139.71 40.29 % 100% 27.6%360
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions2.5(a) i ( peak ) υ R ( peak ) R R(b) υ R (t ) 15 sin ω t 9.715 9.7 4.417 Ω1. 2(ω t )1 0.2238π ; (ω t )2 0.7762πυ R (avg ) 1π0.7762π [15 sin x 9.7]dx0.2238πOr from Problem 2.4, υ R (avg ) 2(0.9628) 1.9256 Vυ (avg ) 1.9256 0.436 Ai D (avg ) RR4.417(c) 139.71 40.29 % 100% 27.6%360 2.6(a) υ S ( peak ) 12 0.7 12.7 VN 1 120 2 13.4N212.712 60 Ω0 .2VM12 6667 μ FC 2 fRV r 2(60 )(60 )(0.25)(b) R (c) PIV 2υ S (max ) Vγ 2(12.7 ) 0.7 24.7 V2.7v0 vS 2Vγ vS ( max ) v0 ( max ) 2Vγa.b.v ( max ) 25 V vS ( max ) 25 2 ( 0.7 ) 26.4 VFor 0N1 160N 1 6.06N 2 26.4N2v ( max ) 100 V vS ( max ) 101.4 VFor 0N1N160 1 1.58N 2 101.4N2PIV 2vS ( max ) Vγ 2 ( 26.4 ) 0.7From part (a)PIV 2 (101.4 ) 0.7or PIV 52.1 V or, from part (b)or PIV 202.1 V2.8(a)vs (max) 12 2(0.7) 13.4 V13.4vs ( rms ) vs (rms) 9.48 V2
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions(b)Vr VMVM C 2 f RC2 f Vr RC 12 C 2222 μ F2 ( 60 )( 0.3)(150 )(c) 2VM 1 π Vr 2 (12 ) 12 1 π 150 0.3 id , peak 2.33 Aid , peak VMR2.9(a)vS ( max ) 12 0.7 12.7 VvS ( rms ) Vr (b)vS ( max )2 vS ( rms ) 8.98 VVMV12 C M fRCfRVr ( 60 )(150 )( 0.3)orC 4444 μ F VM 12 12 1 4π 1 4π iD , max 4.58 A2Vr 150 2 ( 0.3) (c) For the half-wave rectifieroriD , max VMR2.10(a) υ O ( peak ) 10 0.7 9.3 VVM9.3 620 μ FfRV r (60 )(500 )(0.5)(c) PIV 10 9.3 19.3 V(b) C 2.11(a) 10.3 υ O 12.3 V(b) Vr VM12.3 0.586 VfRC (60 )(1000 ) 350 10 6()10.3 0.490 V(60)(1000) 350 10 6So 0.490 V r 0.586 VVr ()VM12.3 513 μ FfRV r (60)(1000 )(0.4)(c) C
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions2.12( )(a) υ S ( peak ) 8.5 2 12.02 VVOmax 12.02 0.7 11.32 VVM11.32 0.03773 F2 f RV r 2(60 )(10 )(0.25)(c) PIV 2υ S ( peak ) Vγ 2(12.02 ) 0.7 23.34 V(b) C 2.13(a)vs ( peak ) 15 2 ( 0.7 ) 16.4 Vvs ( rms ) C 16.42 11.6 VVM15 2857 μ F2 f RVr 2 ( 60 )(125 )( 0.35 )(b)2.142.15(a) υ S 12.8 VN 1 120 2 13.3N212.812 24 Ω0 .5V r 3% V r (0.03)(12 ) 0.36 V(b) R C VM12 0.0116 F2 fRV r 2(60 )(24 )(0.36 ) 1 π 2V M Vr i D ( peak ) 13.3 A(c) i D ( peak ) VMR 12 1 π 2(12 ) 240.36
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions(d) i D (avg ) 1π2V r V M VMR π 1 2 2V MVr 1 π 2(0.36 ) 12 π 1 12 24 22(12 ) 0.36 i D (avg ) 0.539 A(e) PIV 12.8 12 24.8 V2.16(a) υ S 9 2(0.8) 10.6 VN 1 120 2 16N210.69 90 Ω0. 1VM9 4167 μ FC 2(60)(90)(0.2)2 fRV r(b) R (c) i D ( peak ) VMR(d) i D (avg ) 1π 1 π 2V M Vr 2V r V M VMR 9 1 π 2(9 ) 3.08 A 90 0.2 π 1 2 2V MVr 1 π i D (avg ) 0.1067 A(e) PIV υ S (max ) Vγ 10.6 0.8 9.8 V2(0.2 ) 9 π 1 9 90 22(9 ) 0.2 2.17For vi 0Vγ 0Voltage across RL R1 vi
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions RL 1 v0 vi viR R2 L1 Voltage Divider2.18Forvi 0, (Vγ 0 )a. R2 RL v0 vi R2 RL R1 R2 RL 2.2 6.8 1.66 kΩ 1.66 v0 vi 0.43 vi 1.66 2.2 v0 ( rms ) v0 ( max ) v0 ( rms ) 3.04 V2b.2.193 .9 0.975 mA420 3.9II 1.342 mA12I Z I I I L 1.342 0.975 0.367 mA(a) I L PZ I Z V Z (0.367 )(3.9) 1.43 mW3.9 0.39 mA10I Z 1.342 0.39 0.952 mA(b) I L PZ (0.952 )(3.9 ) 3.71 mW
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions2.20(a)40 12 0.233 A120P ( 0.233)(12 ) 2.8 WIZ (b)SoIR 0.233 A, IL (0.9)(0.233) 0.21 A120.21 RL 57.1ΩRLP ( 0.1)( 0.233)(12 ) P 0.28 W(c)2.21(a) PZ I Z V Z4 I z (15.4 ) I Z (max ) 259.74 mASo 15 I z 259.74 mA60 15.4 297.33 mA0.15So I L (max ) 297.33 15 282.33 mA(b) I I I L (min ) 297.33 259.74 37.59 mA15.4 54.55 Ω0.2823315.4R L (max ) 410 Ω0.03759So 54.55 R L 410 ΩThen R L (min ) 2.22a.20 10 I I 45.0 mA22210IL I L 26.3 mA380I Z I I I L I Z 18.7 mAII 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 mA1010RL RL 2 kΩ(c)I 57.1 mA I L 26.3 mAFor Ri 175Ω II Z 30.8 mAI Z ( max ) 40 mA I L ( min ) 57.1 40 17.1 mARL 10 RL 585Ω17.1
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions2.23a.From Eq. (2.30)500 [ 20 10] 50 [15 10]I Z ( max ) 15 ( 0.9 )(10 ) ( 0.1)( 20 )5000 2504I Z ( max ) 1.1875 AI Z ( min ) 0.11875 A From Eq. (2.28(b))b.Ri 20 10 Ri 8.08Ω1187.5 50PZ (1.1875 )(10 ) PZ 11.9 WPL I L ( max )V0 ( 0.5 )(10 ) PL 5 W2.24(a) I L 010 5.6 83.0 mA50 3V Z 5.6 (0.083)(3) 5.85 V V LIZ PZ I Z V Z (0.083)(5.85) 0.486 W(b)10 V L V L 5.6 V L 5032000.20 1.867 V L (0.02 0.3333 0.005)So V L 5.769 V5.769 28.84 mA0.210 5.769II 84.62 mA0.050And I Z I I I L 55.8 mAThen I L PZ (0.0558)(5.769 ) 0.322 W(c) I L 012 5.6 120.8 mA50 3V Z V L 5.6 (0.1208)(3) 5.962 VIZ PZ (0.1208)(5.962 ) 0.72 W(d)12 V L V L 5.6 V L 5032000.24 1.867 V L (0.02 0.333 0.005)So V L 5.88 V5.8812 5.88 29.4 mA; I I 122.4 mA0.200.05I Z 122.4 29.4 93 mAThen I L PZ (0.093)(5.88) 0.547 W
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem Solutions2.25(a) Set I Z 10 mA; I L V L 7.5 7.5 mA1RLI I 10 7.5 17.5 mAI I 17.5 12 7.5 Ri 257 ΩRi(b) 7.5 V ZO (0.01)(12 ) V ZO 7.38 VFor V I (1.1)(12 ) 13.2 V13.2 V L V L 7.38 V L 2571210000.05136 0.615 V L (0.00389 0.0833 0.001) V L 7.556 VFor V I (0.9 )(12 ) 10.8 V10.8 V L V L 7.38 V L 2571210000.04202 0.615 V L (0.08819 ) V L 7.450 V 7.556 7.450 Then, Source Reg 100% 4.42% 13.2 10.8 (c) For R L 1 k Ω , V L 7.50 V12 7.38 17.17 mA257 12V L 7.38 (0.01717 )(12 ) 7.586 VFor R L , I Z 7.586 7.50 Then , Load Reg 100% 1.15%7.50 2.26% Reg SoVL ( nom ) 100%VL ( nom ) I Z ( max ) rz (VL ( nom ) I Z ( min ) rz )VL ( nom ) I Z ( max ) I Z ( min ) ( 3 ) 0.056I Z ( max ) I Z ( min ) 0.1 ANowI L ( max ) Ri Now280 orVL ( max ) VL ( min )66 0.012 A, I L ( min ) 0.006 A5001000VPS ( min ) VZI Z ( min ) I L ( max )15 6 I Z ( min ) 0.020 AI Z ( min ) 0.012
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem SolutionsThenI Z ( max ) 0.1 0.02 0.12 AVPS ( max ) 6Ri andVPS ( max ) VZI Z ( max ) I L ( min ) VPS ( max ) 41.3 V0.12 0.006or280 2.27Using Figure 2.19VPS 20 25% 15 VPS 25 Va.ForVPS ( min ) :I I I Z ( min ) I L ( max ) 5 20 25 mARi b.ForVPS ( min ) VZIIVPS ( max ) 15 10 Ri 200Ω25 I I ( max ) 25 10 I I ( max ) 75 mARiForI L ( min ) 0 I Z ( max ) 75 mAVZ 0 VZ I Z rZ 10 ( 0.025 )( 5 ) 9.875 VV0 ( max ) 9.875 ( 0.075 )( 5 ) 10.25V0 ( min ) 9.875 ( 0.005 )( 5 ) 9.90ΔV0 0.35 V% Reg ΔV0 100% % Reg 3.5%V0 ( nom )c.2.28From Equation (2.28(a))VPS ( min ) VZ24 16Ri I Z ( min ) I L ( max ) 40 400Also Vr orRi 18.2ΩVMVM C 2 fRC2 fRVrR Ri rz 18.2 2 20.2ΩThenC 24 C 9901 μ F2 ( 60 )(1)( 20.2 )2.29VZ VZ 0 I Z rZ VZ ( nom ) 8 V8 VZ 0 ( 0.1)( 0.5 ) VZ 0 7.95 VIi VS ( max ) VZ ( nom )Ri 12 8 1.333 A3For I L 0.2 A I Z 1.133 AFor I L 1 A I Z 0.333 A
Microelectronics: Circuit Analysis and Design, 4th editionChapter 2By D. A. NeamenProblem SolutionsVL ( max ) VZ 0 I Z ( max ) rZ 7.95 (1.133)( 0.5
Microelectronics: Circuit Analysis and Design, 4. th. edition Chapter 1 By D. A. Neamen Problem Solutions
