semisolpr07

# semisolpr07 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 7 Solutions Manual Problem Solutions 83 Chapter 7 Problem Solutions 7.1 VV NN n bi t ad i = F H G I K J ln 2 where t = 00259 . and nx c m i = 15 10 10 3 . We find (a) For Nc m d = 10 15 3 bi () iN c m a = 10 15 3 ii N cm a = 10 16 3 iii N cm a = 10 17 3 iv N cm a = 10 18 3 0575 . V 0635 0695 . 0754 . (b) For m d = 10 18 3 bi c m a = 10 15 3 ii N cm a = 10 16 3 iii N cm a = 10 17 3 iv N cm a = 10 18 3 . V 0814 0874 . 0933 . 7.2 Si: c m i = 10 3 Ge: c m i = 2.4 10 13 3 GaAs: c m i = 18 10 63 n bi t i = F H G I K J ln 2 and t = (a) m m da == −− 10 10 14 3 17 3 , Then Si V V bi :. = , Ge V V bi = 0253 , GaAs V V bi = 110 (b) N x cm N x cm 510 16 3 16 3 Then Si V V bi = 0778 , Ge V V bi = 0396 , GaAs V V bi = 125 (c) m m 10 10 17 3 17 3 Then Si V V bi = , Ge V V bi = 0432 , GaAs V V bi = 128 7.3 Computer Plot 7.4 Computer Plot 7.5 (a) n-side: EE k T N n FF i d i −= F H G I K J ln = F H G I K J 15 10 .l n . x x or e V i 03294 . p-side: EEk T N n Fi F a i F H G I K J ln = F H G I K J 10 17 10 n . x or e V Fi F 04070 . (b) V bi =+ 0 3294 0 4070 .. or bi = 07364 . (c) n bi t i = F H G I K J ln 2 = L N M M O Q P P 10 5 10 17 15 10 2 n . b gb g bg x x or bi = 07363 .

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 7 Solutions Manual Problem Solutions 84 (d) x V e N NN N n bi a da d = + F H G I K J F H G I K J L N M O Q P 21 12 / = () ( ) L N M 2117 88510 0736 16 10 14 19 .. . . x x bg × + F H G I K J F H I K O Q P 10 510 1 10 5 10 17 15 17 15 xx / or xm n = 0426 . µ Now x x x p = ( ) L N M 14 19 . . × + F H G I K J F H I K O Q P 10 1 10 5 10 15 17 17 15 x x / or p = 00213 . We have Ε max = eN x dn = −− 5 10 0 426 10 117 8 85 10 19 15 4 14 x x b g b g or Ε max ./ = 329 10 4 xV c m 7.6 (a) n-side EE x x FF i −= F H G I K J 00259 0 15 10 16 10 .l n . e V i 03653 . p-side x x Fi F F H G I K J 0 16 10 n . e V Fi F . (b) V bi =+⇒ 0 3653 0 3653 .. VV bi = 07306 . (c) n bi t ad i = F H G I K J ln 2 = L N M M O Q P P 0 0 16 16 10 2 n . x b gb g or bi = 07305 . (d) x x x n = ( ) L N M 2 117 8 85 10 0 7305 14 19 . . × + F H G I K J F H I K O Q P 0 0 1 0 0 16 16 16 16 x x / or n = 0154 . By symmetry p = . Now Ε max = eN x = 2 10 0154 10 117 8 85 10 19 16 4 14 x x b g b g or Ε max / = 4.76 10 4 c m 7.7 (b) nN kT OC CF = L N M O Q P exp a f = F H I K 2.8 10 021 19 x exp . . or xc m Od == 843 10 15 3 . (n-region) pN kT OV FV = L N M O Q P exp a f = F H I K 104 10 018 19 .e x p . . x or m Oa 9.97 10 15 3 (p-region) (c) n bi t i = F H G I K J ln 2
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 7 Solutions Manual Problem Solutions 85 = () L N M M O Q P P 00259 9.97 10 8 43 10 15 10 15 15 10 2 .l n . . xx x b gb g bg or VV bi = 0690 . 7.8 (a) GaAs: bi = 120 ., nx c m i = 18 10 63 . xWx x pn p == + 02 .. or x x p n = 025 Also Nx x x N N dn ap p n d a =⇒ = = Now NN n bi t ad i = F H G I K J ln 2 or 0 0259 2 2 l n . = F H G I K J N n a i Then 2 2 . exp . . N n a i = F H I K or Nn ai = L N M O Q P 2 2 0 0259 exp . . or c m a = 4.14 10 16 3 (b) da c m d = 104 10 16 3 . (c) x V e N N n bi a d = + F H G I K J F H G I K J L N M O Q P 21 12 / = 2131 88510 16 10 14 19 . . L N M x x × + F H I K F H I K O Q P 4 1 1 4.14 10 16 16 . / or xm n = 0366 .

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## This note was uploaded on 05/07/2008 for the course ECE E-302 taught by Professor Nabet during the Spring '08 term at Drexel.

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semisolpr07 - Semiconductor Physics and Devices Basic...

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