Microelectronics - Circuit Analysis and Design 4th Sol.pdf-资料库

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

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.3 Silicon (a) n i = = = n i (b) n i = = = n i (c) n i = = = n i × 5.23 10 15 ( ( 5.23 10 18 8.79 10 10 − × × )( 100 ) exp cm 3 − [ 3/ 2 ) exp 63.95 − ⎡ ⎢ ⎢ ⎣ ] 1.1 − ( )( 100 2 86 10 6 − × ) ⎤ ⎥ ⎥ ⎦ × 5.23 10 15 ( ( 2.718 10 19 1.5 10 cm )( ) × 10 × 300 ) exp 3 − 3/ 2 exp ⎡ 1.1 − ⎢ ( )( 2 86 10 6 − ⎢ ⎣ ] 21.32 − × [ × 5.23 10 15 ( )( ( ) 5.847 10 19 × 1.63 10 cm 14 × 500 ) [ exp 3 − 3/ 2 exp ⎡ 1.1 − ⎢ ( )( 2 86 10 6 − ⎢ ⎣ ] 12.79 − × ⎤ ⎥ ⎥ ⎦ 300 ) ⎤ ⎥ ⎥ ⎦ 500 ) Germanium. ( 1.66 10 15 (a) n i = × )( 100 ) 3/ 2 exp n i = 35.9 cm 3 − (b) n i = ( 1.66 10 15 × )( 300 ) 3/ 2 exp n i = 2.40 10 cm × 13 3 − (c) n i = ( 1.66 10 15 × )( 500 ) 3/ 2 exp ⎡ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ ⎣ 0.66 − )( 100 6 − × ( 2 86 10 ) ⎤ ⎥ ⎥ ⎦ = ( 1.66 10 18 × ) exp 38.37 − [ ] 0.66 − )( 6 − × ( 2 86 10 300 ) 0.66 − )( 6 − × ( 2 86 10 500 ) ⎤ ⎥ ⎥ ⎦ ⎤ ⎥ ⎥ ⎦ = ( 8.626 10 18 × ) exp [ − 12.79 ] = ( 1.856 10 19 × ) exp [ − 7.674 ] n i = 8.62 10 cm × 15 3 − ______________________________________________________________________________________ 1.4 n 2 i n o n 2 i n o = = 2 ) 13 10 15 ( 4.2 ( 5.1 × 10 × 10 10 15 = 76.5 × 10 11 cm 3− 10 2 ) = 25.2 × 10 5 cm 3− (a) n-type; 1510=on 3− cm ; p o = (b) n-type; 1510=on 3− cm ; p o = ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5 (a) p-type; 1610=op 3− cm ; n = o n 2 i p o (b) p-type; 1610=op 3− cm ; n = o = = 26 ) ( 8.1 10 16 × 10 = 24.3 × 10 − 4 cm 3− n 2 i p o ( 4.2 10 16 × 10 13 2 ) = 76.5 × 10 10 cm 3− ______________________________________________________________________________________ 1.6 (a) n-type (b) N = − 3 n o p o = d n 2 i n o N 16 5 10 cm = × ( ) 2 1.5 10 10 × 5 10 16 × 16 − = 3 d = n o = × (c) From Problem 1.1(a)(ii) )2 5 10 cm in ( op = 3.97 10 11 × 5 10 16 × = 4.5 10 cm × 3 − 3 3.97 10 cm × 11 − 3 = = 3.15 10 cm × 6 3 − ______________________________________________________________________________________ 1.7 (a) p-type; 5×=op 1610 3− cm ; n (b) p-type; 5×=op 1610 3− cm ; n = = o o 2 ) 10 ( 5.1 10 × 16 5 10 × ( 8.1 10 × 16 10 5 × 26 ) = = n 2 i p o n 2 i p o = 5.4 × 10 3 cm 3− = 48.6 × 10 − 5 cm 3− ______________________________________________________________________________________ 1.8 a N (a) Add boron atoms 1710 (b) 10 × 17 10 × 2×= ( 5.1 2 p = o n 2 i p (c) = = n o o 3− cm ) 10 2 = 125.1 × 10 3 cm 3− ______________________________________________________________________________________ 1.9 (a) = × 3 − n o p o = = cm 5 10 15 ( 1.5 10 10 × 5 10 15 × n-type 5 10 15 n 2 i n o p N 2 ) ⇒ = p o 4.5 10 4 × cm− 3 3 d o n n o cm− = × ⇒> o ≅ (b) (c) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.10 a. Add Donors b. Want p )( So 6 exp n N / 2 = i 21 7 10 = × 7 10 cm 6 ( 10 B T 2 = × dN in 2 = = 15 3 − o d 3 3 − 10 cm = ) 7 10 15 × Eg −⎛ ⎜ kT ⎝ ⎞ ⎟ ⎠ ) 7 10 × 21 = ( 5.23 10 15 × 2 3 T exp ⎛ ⎜ ⎜ ⎝ 1.1 − 86 10 6 − × ( )( T ) ⎞ ⎟ ⎟ ⎠ ° T ≈ 324 K By trial and error, ______________________________________________________________________________________ 1.11 (a) mA =Ε = A I σ A Ε ρ = I ( )( )( ) I 10 5 15.0 105.1 − =⇒ )( ( ) I 3 10 2.1 4.0 − × ρ =Ε⇒ ) ( A 2 10 × = − 4 (b) = 4.2 V/cm ______________________________________________________________________________________ 1.12 J =⇒Ε= σσ 67.6 = = J Ε N d 120 18 σ e μ n − ) 1 ( cm −Ω ( ) 67.6 )( 19 10 1250 − × = ( 6.1 μσ n Ne ≅ d =⇒ = 33.3 × 16 10 cm 3− ) ______________________________________________________________________________________ 1.13 (a) =⇒ cm 3− 69.7 15 10 ≅ ρ N × = 1 Ne μ n Ε ρ d d = 1 e ρμ )( 160 n ( 65.0 ( 6.1 10 × ) 104 = 19 − 1 )( 1250 V/cm )( 65.0 ) (b) J = =Ε⇒ J ρ = ______________________________________________________________________________________ 1.14 (a) =⇒ cm 3− 375.9 Ne 15 10 N × = = ≅ μσ n d d 5.1 19 − (b) N a = σ e μ p = ( 6.1 × σ e μ n 8.0 )( 19 10 400 − ( 6.1 × 10 = 25.1 × ) )( 1000 10 ) cm 3− 16 ______________________________________________________________________________________ 1.15 (a) For n-type, σ μ n For dN 10 10 15 19 ≤ ) ( 0.1 (b) σ σ = ⇒ ______________________________________________________________________________________ ( 1.6 10 1.36 ⇒ J ≤ ≤ ≤ 1.36 10 3 d cm 0.136 ( × A cm / 2 d 1.36 10 )( σ ≤ × e N 8500 ≤ = Ω − ) 1 cm N E 19 − ≅ = × J ) − − 3 4

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.16 cm /s 5.32 7.11 = = 2 )( 450 ( .0 ) ) =nD J n = eD × 10 19 − 52 A/cm 2 )( 026 1250 dn ( 6.1 dx = n ⎛ ) ⎜⎜ ⎝ ( cm 2 /s; =pD 026.0 16 12 10 10 ⎞ )( − 5.32 −=⎟⎟ 001.00 − ⎠ 12 16 10 10 ⎛ ) ⎜⎜ 001.00 ⎝ 19 − − p p J × − 10 eD −= −= dp dx ( 6.1 )( 7.11 Total diffusion current density ______________________________________________________________________________________ 1.17 ⎞ −=⎟⎟ ⎠ A/cm 2 A/cm 2 72.18 −=J 72.18 7.70 −= 52 − ⎛ ⎜ ⎜ ⎝ ) 15 exp p 15 1 ⎞ ⎟ ⎟ ⎠ )( ⎛ − ) ⎜ ⎜ L ⎝ )( 15 10 10 10 4 − x L / 19 − × p − L x p exp ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ ⎞− x ⎟ ⎟ L ⎠ p J p = − eD p = − eD p dp dx ( 10 ( 1.6 10 × J p = − e J 2.4 = p x = 0 x x 2 2.4 2.4 pJ = 10 m μ= 30 m μ= 2.4 A/cm = = (a) (b) (c) ______________________________________________________________________________________ 1.18 a. 0.883 A/cm 0.119 A/cm e− 1 e− 3 pJ pJ = = N = 17 − 2 2 3 a 17 3 − = 10 cm p ⇒ = o )26 10 cm ( 1.8 10 × n 2 i p 10 17 o n 3.24 10 × δ + = p 10 10 15 17 + ⇒ = + δ = ⇒ = n o 5 − 3.24 10 cm × 5 − 3 − n 10 cm 10 15 + ⇒ = p 1.01 10 cm × 15 17 − 3 3 − n o = b. n p = = n o p o ______________________________________________________________________________________ 1.19 V bi = V T ln ⎛ ⎜⎜ ⎝ d NN 2 i a n ⎞ ⎟⎟ ⎠ (a) (i) =biV ( .0 026 ) ln (ii) =biV ( .0 026 ) (iii) =biV ( .0 026 ) ln ln 10 ) 2 15 ) ( 5 17 15 × × 10 10 × ( 5.1 ( 10 5 × ( 5.1 × ( )( 10 10 ( 10 5.1 10 )( 5 10 )( 10 ) 10 10 ) ⎤ ⎥ ⎥ ⎦ ) × 18 18 2 2 ⎡ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ ⎣ 15 ) ⎤ ⎥ ⎥ ⎦ = .0 661 V = .0 739 V ⎤ ⎥ ⎥ ⎦ = .0 937 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) (i) =biV ( 026.0 ) ln = 13.1 V 15 ) ⎤ ⎥ ⎥ ⎦ 17 15 15 ( 5 10 × ( 8.1 ( 5 10 × ( 8.1 × ( )( 10 10 ( 8.1 10 × )( 10 5 × ) 26 10 × )( 10 ) 26 10 ) ⎤ ⎥ ⎥ ⎦ ) ⎤ ⎥ ⎥ ⎦ 26 ) = 18 18 ⎡ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ ⎣ = 21.1 V 41.1 V (ii) =biV ( 026.0 ) ln (iii) =biV ( 026.0 ) ln ______________________________________________________________________________________ 1.20 ⎛ ⎜⎜ ⎝ d NN 2 i a n ⎞ ⎟⎟ ⎠ V bi = V T ln or N = a ) ( 2 n i N d ⎛ exp ⎜⎜ ⎝ V bi V T ⎞ =⎟⎟ ⎠ 2 ) ( 10 5.1 × 16 10 ⎛ exp ⎜ ⎝ .0 .0 712 026 ⎞ =⎟ ⎠ 76.1 × 10 16 cm 3− ______________________________________________________________________________________ 1.21 N N a n 2 i ln ⎛ ⎜ ⎝ cm 10 15 cm 10 18 3 − 3 − , , d = ⎞ ⎟ ⎠ V bi V bi V bi = V T N N a a = = For For ( 0.026 ln ) = = 0.637 0.817 ⎡ ⎢ ⎢ ⎣ V V ( )16 N 10 (1.5 10 ) 1 0 2 × a ⎤ ⎥ ⎥ ⎦ ______________________________________________________________________________________ 1.22 (0.026) kT = ⎞ ⎟ ⎠ T ⎛ ⎜ 300 ⎝ 200 250 300 350 400 450 500 kT 0.01733 0.02167 0.026 0.03033 0.03467 0.0390 0.04333 (T)3/2 2828.4 3952.8 5196.2 6547.9 8000.0 9545.9 11,180.3

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

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ . 0 00743 0 0118 pF = = ) . + 2 + ( final ) v C ( initial . 0 00962 ) v C − ( ) final ) e τ− t / RC RC avg ( ) = (47 10 )(0 00962 10 ) 1 2 − × × 3 . j C avg ( ( ) t j ( = = v C v C where = τ or . 4 52 10 = τ ( ) Cv . t Then 1 5 5 re 1 / 1.5 t = 1 5.44 10 × = = × τ + 10 − s 0 = + ⇒ = t 1 τ it / τ− ( ) 5 0 − 5 ⎛ ⎜ 1.5 ⎝ e ⎞ ⎟ ⎠ ln s 10 − (b) For VR = 0 V, Cj = Cjo = 0.02 pF 1/ 2 − ⎞ ⎟ ⎠ For VR = 3.5 V, 3.5 0.8 ⎛ 0.02 1 ⎜ ⎝ 0.02 0.00863 jC = + + ( ) = 0.0143 pF = p 0.00863 F ) ( = ( v C jC avg ) jRC avg = = τ ( ( ) final v t C e 3.5 5 (0 5) − = + − t 8.09 10 × = = 10 − 2 s 6.72 10 × ( ) initial v + C ( e 5 1 t / − τ − t ) ( so that ______________________________________________________________________________________ 1.25 ( = s v C final e τ− /t − ) 10 − ) ) / τ 2 2 2 C j = C jo For 1=RV + 1 ⎛ ⎜⎜ ⎝ V, 2/1 − R V V bi ⎞ ⎟⎟ ⎠ ; =biV ( .0 026 ) ln ⎡ ⎢ ⎢ ⎣ ( 10 5 × ( 5.1 × 15 )( 10 ) 10 10 17 2 ) = .0 739 V ⎤ ⎥ ⎥ ⎦ 57.6 MHz =jC For 3=RV V, =jC 1 1 60.0 1 + 739.0 60.0 3 739 .0 + For 5=RV V, =jC 1 60.0 5 739 .0 + = .0 391 pF = 267.0 pF = 215.0 pF (a) f o = 1 2 π LC = (b) f o = 2 π ( 5.1 × 1 )( 3 .0 × 10 − 391 × 10 ( 5.1 267 × 10 12 − ) =⇒ f o 2 π 1 )( 3 .0 − 10 12 − ) 95.7 =⇒ f o MHz

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Microelectronics - Circuit Analysis and Design 4th Sol.pdf

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