Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.3 Silicon (a) ( )( ) ( )( ) ( ) 3/ 215 6 18 10 3 1.15.23 10 100 exp 2 86 10 100 5.23 10 exp 63.95 8.79 10 cm i i n n − − − ⎡ ⎤−⎢ ⎥= × ×⎢ ⎥⎣ ⎦ = × − = × (b) ( )( ) ( )( ) ( ) 3 / 215 6 19 10 3 1.15.23 10 300 exp 2 86 10 300 2.718 10 exp 21.32 1.5 10 cm i i n n − − ⎡ ⎤−⎢ ⎥= × ×⎢ ⎥⎣ ⎦ = × − = × (c) ( )( ) ( )( ) ( ) 3 / 215 6 19 14 3 1.15.23 10 500 exp 2 86 10 500 5.847 10 exp 12.79 1.63 10 cm i i n n − − ⎡ ⎤−⎢ ⎥= × ×⎢ ⎥⎣ ⎦ = × − = × Germanium. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.10 a. Add Donors 15 37 10 cm dN −= × 6 3 210 cm / o i d n N−= = b.

• Microelectronics Circuit Design 4th Edition Solution Manual
• Microelectronic Circuit Design 4th Solution

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 0 00743 0 0118( ) 0 00962 2 j.C avg.

PF+= = ( ) ( ) ( ) ( )( ) t /C C C Cv t v final v initial v final e τ−= + − where 3 1( ) (47 10 )(0 00962 10 ) jRC RC avg.τ −= = = × × 2 or 104 52 10. Sτ −= × Then ( ) ( )1 5 0 5 0 it /Cv t. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions (c) ( )( ) 86.8.12 1 123 =⇒ ×× = −− oo ff π MHz 1.26 a. Exp 1 0.90 exp 1 D DS T T V VI I V V ⎡ ⎤⎛ ⎞ ⎛ ⎞ = − − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎣ ⎦ − exp 1 0.90 0.10 D T V V ⎛ ⎞ = − =⎜ ⎟ ⎝ ⎠ ( )ln 0.10 0.0599 V D T DV V V= ⇒ = − b.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.30 (a) 1210 SI A −= VD(v) ID(A) log10 ID 0.10 114 68 10. −× 10 3.− 0.20 92 19 10. −× 8 66.− 0.30 71 03 10.

−× 6 99.− 0.40 64 80 10. −× 5 32.− 0.50 42 25 10. −× 3 65.− 0.60 21 05 10. −× 1 98.− 0.70 14 93 10. −× 0 307.− (b) 1410 SI A −= VD(v) ID(A) log10 ID 0.10 134 68 10. −× 12 3.− 0.20 112 19 10.

−× 10 66.− 0.30 91 03 10. −× 8 99.− 0.40 84 80 10. −× 7 32.− 0.50 62 25 10. −× 5 65.− 0.60 41 05 10. −× 3 98.− 0.70 34 93 10. −× 2 31.− 1.31 a. 2 2 1 1 10 exp ln (10) 59.9 mV 60 mV D D D D T D T D I V V I V V V V ⎛ ⎞− = = ⎜ ⎟ ⎝ ⎠ Δ = ⇒ Δ = ≈ b.

( )ln 100 119.7 mV 120 mV D T DV V VΔ = ⇒ Δ = ≈ 1.32 (a) (i) ( ) 539.0 102 2ln026.0 9 =⎟⎠ ⎞ ⎜ ⎝ ⎛ × = − DV V (ii) ( ) 599.0 102 20ln026.0 9 =⎟⎠ ⎞ ⎜ ⎝ ⎛ × = − DV V (b) (i) ( ) 60.9 026.0 4.0exp102 9 ⇒⎟ ⎠ ⎞ ⎜ ⎝ ⎛×= − DI mA (ii) ( ) 144 026.0 65.0exp102 9 ⇒⎟ ⎠ ⎞ ⎜ ⎝ ⎛×= − DI A. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.33 3 14 3 12 2 10ln (0 026) ln 0 6347 V 5 10 2 10(0 026) ln 0 5150 V 5 10 0 5150 0 6347 V D D t S D D IV V.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.36 IS doubles for every 5C increase in temperature. 1210 SI A −= at T = 300K For 120.5 10 T 295 K SI A −= × ⇒ = For 1250 10, (2) 50 5.64 nSI A n −= × = ⇒ = Where n equals number of 5C increases. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.39 ( )410 2 10 D DI V= × + and ( ) 120.026 ln 10 D D I V − ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ Trial and error. VD(v) ID(A) VD(v) 0.50 44.75 10−× 0.5194 0.5 10−× 0.5194 0.5194 44.740 10−× 0.5194 0.5194 V 0.4740 mA D D V I = = 1.40 135 10 A sI −= × 2 1 2 30(1.2) (1.2) 0.45 V 80 TH R V R R ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠ 0.45, ln DD TH D D T S I I R V V V I ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠ By trial and error: 2.56 A, 0.402 V D DI Vμ= =.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 1.43 (a) Assume diode is conducting. Then, 0.7 DV V Vγ= = So that 2 0 7 23 3 30 R.I.

Aμ= ⇒ 1 1 2 0 7 50 10 R.I Aμ−= ⇒ Then 1 2 50 23 3 D R RI I I.= − = − Or 26 7 DI. Aμ= (b) Let Diode is cutoff. 1 50 R k= Ω 30 (1 2) 0 45 30 50 D V.= ⋅ = +. V Since, 0 D DV V Iγ. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions From (1) 5 2.5 2 2 A A D A V V DI V I − = − = − ⇒ Neg. 0 DI⇒ = Both (a), (b) 0 DI = VA = 2.5, 2 5 2 V 0.50 V 5 B D V V= ⋅ = ⇒ = 1.45 (a);; for mA ( )1 iO IV = 0= DI 7.00 ≤≤ iI V; mA; for mA 7.0= OV ( 7.0−= iD II ) 7.0≥ iI (b);; for ( )1 iO IV = 0= DI 7.10 ≤≤ iI mA V; mA; for mA 7.1= OV ( 7.1−= iD II ) 7.1≥ iI (c) V;;; for 7.0= OV iD II =1 02 = DI 20 ≤≤ iI mA 1.46 Minimum diode current for VPS (min) (min) 2, 0.7 D DI mA V V= = 2 1 2 1 0 7 5 0 7 4 3,.I I R R − = = = 1.

R We have 1 2 DI I I= + so (1) 1 2 4 3 0 7 2. R R = + Maximum diode current for VPS (max) ( )10 0 7 14 3 D D D DP I V I. MA= = ⇒ = 1 2 DI I I= + or (2) 1 2 9 3 0 7 14 3.

R R = + Using Eq. (1), 1 1 1 9 3 4 3 2 14 3 0 41 Ω. K R R = − + ⇒ = Then 2 82 5 82 5 R.= Ω Ω 1.47 (a) (i) 215.0 20 7.05 = − = I mA, 7.0= OV V (ii) 220.0 20 6.05 = − = I mA, 6.0= OV V (b) (i) ( ) 2325.0 40 57.05 = −−− = I mA, ( )( ) 5.0 −=−= OV V (iii) ( ) 235.0 40 56.05 = −−− = I mA, ( )( ).0 −=−= OV V. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions a. (0 026) 0 026 k 26 1 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 V.

Τ μ τ μ = = = Ω = Ω = = = = ⇒ = b. For (0 026)0 1 mA 260 0 1 DQ d.I. Τ= ⇒ = = Ω 0 05 5 A peak-to-peak d DQi. I μ= = (260)(5) V 1 30 mV peak-to-peak d d d dv i v.τ μ= = ⇒ = 1.52 (a) 1 026.0 026.0 DQ T d I Vr k Ω (b) Ω⇒= 100 26.0 026.0 dr (c) Ω⇒= 10 6.2 026.0 dr 1.53 a. Diode resistance d Tr V I= / d T d S Td S S T d s o T S r V Iv v Vr R R I Vv v v V IR ⎛ ⎞ ⎜ ⎟⎛ ⎞ / = = ⎜ ⎟⎜ ⎟+⎝ ⎠ ⎜ ⎟+⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = =⎜ ⎟+⎝ ⎠ Sv b. 260 SR = Ω ( ) ( ) 0 0 0 0 0 0 0 0261 mA, 0 0909 0 026 (1)(0 26) 0 0260 1 mA, 0 50 0 026 0 1 0 26 0 0260 01 mA. 0 909 0 026 (0 01)(0 26) T S T S S s S S S v vV.I.

V ⎛ ⎞ = = = ⇒ =⎜ ⎟+ +⎝ ⎠ = = ⇒ = + = = ⇒ = +. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions 31 2 0 5 10 I I.

−+ = × 8 125 10 exp 10 exp 0 5 10 D D T T V V. V V − −⎛ ⎞ ⎛ ⎞× + =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3−× 8 35 0001 10 exp 0 5 10 D T V. V − −⎛ ⎞× =⎜ ⎟ ⎝ ⎠ × 3 8 0 5 10(0 026) ln 0 2395 5 0001 10 D D.V. − − ⎛ ⎞× = ⇒⎜ ⎟×⎝ ⎠ = Schottky diode, 2 0 49999 mA I.= pn junction, 1 0 00001 mA I.= (b) 12 81 210 exp 5 10 exp D D T T V V I V V − −⎛ ⎞ ⎛= = ×⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎞ ⎟ ⎠ 1 2 0.9 D DV V+ = 12 81 1 8 1 0.910 exp 5 10 exp 0.95 10 exp exp D D T T D T T V V V V V V V − − − ⎛ ⎞ ⎛ ⎞− = ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ − = × ⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎞ ⎟ ⎠ 8 1 12 2 5 10 0 9exp exp 0 02610 D T V.

− − ⎛ ⎞ ⎛ ⎞× ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ 8 1 12 5 102 ln 0 9 1 1813 10 D T V V. − − ⎛ ⎞× = + =⎜ ⎟ ⎝ ⎠ 1 0 5907 pn junction DV.= 2 0 3093 Schottky diode DV.= 12 0 590710 exp 7 35 mA 0 026.I I.

− ⎛ ⎞= ⇒ =⎜ ⎟ ⎝ ⎠. 1.57 0 5 6 V at 0 1 mA Z Z ZV V. I.= = = 10 Zr = Ω. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions ( )( )0 1 10 1 mV Z ZI r.= = VZ0 = 5.599 a. LR →∞⇒ 10 5 599 4 401 8 63 mA 0 50 0 01 Z Z.I. − = = = + + ( )( )0 5.599 0.00863 10 Z Z Z ZV V I r= + = + 0 5 685 V ZV V.= = b.

Microelectronics Circuit Design 4th Edition Solution Manual

11 5 59911 V 10 59 mA 0 51 PS Z.V I. − = ⇒ = = ( )( )0 5.599 0.01059 10 5.7049 V ZV V= = + = 9 5 5999 V 6 669 mA 0 51 PS Z.V I.

Microelectronic Circuit Design 4th Solution

− = ⇒ = = ( )( )0 5.599 0.006669 10 5.66569 V ZV V= = + = 0 05 7049 5 66569 0 0392 V V. V.Δ = − ⇒ Δ = c. I = IZ + IL 0 0 0, 0 PS ZL Z L Z V V V V VI I I R R r − − = = = 0 010 5 599 0 50 0 010 2 V V. − − = + 0 0 10 5 599 1 1 1 0 50 0 010 0 50 0 010 2. ⎡ ⎤+ = + +⎢ ⎥⎣ ⎦ 20.0 + 559.9 = V0 (102.5) 0 5.658 V V = 1.58 (a) 4.6 5.0 8.610 = − = ZI mA mW ( )( ) 5.438.64.6 ZZVIP (b) mA ( )( ) 64.04.61.0 ZI 76.564.04.6 =−= LI mA 18.1 76.5 8.6 ⇒= Z Z L L Z L I VR R VI kΩ 1.59 ( )( )0.1 20 2 mV Z ZI r = = 0 6 8 0 002 6 798 V ZV.= − = a.

LR = ∞ 10 6 798 6 158 mA 0 5 0 02 Z Z.I I. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. Neamen Problem Solutions ( )( )0 0 6.798 0.006158 20 Z Z Z ZV V V I r= = + = + 0 6 921 V V.= b. Z LI I I= + 0 010 6 798 0 50 0 020 1 V V.

− − = + 0 0 10 6 798 1 1 1 0 30 0 020 0 50 0 020 1. Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D.

Neamen Problem Solutions Chapter 2 2.1 (a) For 6.0 Iυ V, ( )6. −⎟ ⎠ ⎞ ⎜ ⎝ ⎛= IO υυ For 6.0.