semisolpr09

semisolpr09 - Semiconductor Physics and Devices: Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 9 Solutions Manual Problem Solutions 123 Chapter 9 Problem Solutions 9.1 (a) We have ee V N N nt C d φ = F H G I K J ln == () F H G I K J 00259 2.8 10 10 0206 19 16 .l n . x eV (c) φφ χ BO m =− = 4.28 4.01 or BO V = 027 . and V bi BO n = 027 0206 .. or VV bi = 0064 . Also x V eN d bi d = L N M O Q P 2 12 / = ( ) L N M O Q P 2117 88510 16 10 10 14 19 16 . . / x x bg b gb g or xx c m d = 9.1 10 6 Then Ε max = eN x dd = −− 10 9.1 10 117 885 10 19 16 6 14 . x b g b g or Ε max ./ = 141 10 4 xV c m (d) Using the figure, Bn V = 055 . So V bi Bn n = 055 0206 or bi = 0344 We then find c m d = 2.11 10 5 and Ε max = 326 10 4 c m 9.2 (a) BO m 51 4.01 or BO V = 109 . (b) C d V N N = F H G I K J ln F H G I K J 2.8 10 10 0265 19 15 n . x V Then V bi BO n 109 0265 or bi = 0825 . (c) Wx V eN d bi d L N M O Q P 2 / = ( ) L N M O Q P 10 14 19 15 . . / x x b g or c m = 103 10 4 . (d) Ε max = eN x = 10 117 8 85 10 19 15 4 14 x b g b g or Ε max = 159 10 4 c m 9.3 (a) Gold on n-type GaAs = 4.07 V and m V = BO m 51 4.07 and BO V = 103 . (b) C d V N N = F H G I K J ln

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 9 Solutions Manual Problem Solutions 124 = () F H G I K J 00259 4.7 10 510 17 16 .l n x x or φ n V = 00580 . (c) V bi BO n =− φφ 103 0058 .. or VV bi = 0972 . (d) x eN d bi R d = ∈+ L N M O Q P 2 12 af / = + ( ) L N M O Q P 2131 88510 0972 5 16 10 5 10 14 19 16 . . / x xx bg b g or xm d = 0416 . µ (e) Ε max = eN x dd = −− 5 10 0 416 10 131 885 10 19 16 4 14 x x b g b g or Ε max / = 2.87 10 5 xV c m 9.4 Bn V = 086 .a n d n V = 0058 . (Problem 9.3) Then V bi Bn n = 086 0058 or bi = 0802 . and x eN d bi R d = L N M O Q P 2 / = + ( ) L N M O Q P 0802 5 14 19 16 . . / x b gb g or] d = 0410 . Also Ε max = eN x = 5 10 0 410 10 19 16 4 14 x x b g b g or Ε max / = 2.83 10 5 c m 9.5 Gold, n-type silicon junction. From the figure, Bn V = 081 . For Nx c m d = 15 3 , we have nt C d V N N = F H G I K J ln = F H G I K J 2.8 10 19 15 n x x = n V = 0224 Then bi = 0 81 0 224 0 586 . (a) Now ′ = + L N M O Q P C eN d bi R 2 / = 117 8 85 10 5 10 20586 4 19 14 15 . . / x L N M M O Q P P + b g b g or ′ = Cx F c m 9.50 10 92 / For Ax c m = 42 , CC A = So Cp F = 4.75 (b) For c m d = 16 3 , we find n x x V == F H G I K J 2.8 10 0164 19 16 n Then bi = 0 81 0164 0 646 . Now ′ = + L N M O Q P C x 117 8 85 10 5 10 20646 4 19 14 16 . . / b g b g or ′ = F c m 2.99 10 82 and A = so F = 15
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 9 Solutions Manual Problem Solutions 125 9.6 (a) From the figure, VV bi = 090 . (b) We find 1 310 0 20 9 103 10 2 15 15 = −− = F H I K () C V x x R . . and 2 15 . x eN d = Then we can write N xx x d = 2 16 10 131 8 85 10 19 14 15 .. bg b g or Nx c m d = 105 10 16 3 . (c) φ nt C d V N N = F H G I K J ln = F H G I K J 00259 4.7 10 17 16 .l n . x x or n V = 00985 . (d) φφ Bn bi n V =+= + 0 90 0 0985 or Bn V = 09985 . 9.7 From the figure, Bn V = 055 (a) C d V N N = F H G I K J ln == F H G I K J 2.8 10 10 0206 19 16 n . x V Then V bi Bn n =− = 055 0206 or bi = 0344 . We find x V eN d bi d = L N M O Q P 2 12 / = ( ) L N M O Q P 2117 88510 10 14 19 16 . . / x x b g or xm d = 0211 . µ Also Ε max = eN x dd = 10 0 211 10 117 885 10 19 16 4 14 x b g b g b g or Ε max ./ = 326 10 4 xV c m (b) Ε π = e 4 = L N M O Q P 41 1 78 8 5 1 0 19 4 14 / x or = 20 0 . mV Also x e m = 16 Ε = L N M O Q P 16 3 26 10 19 14 4 .

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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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semisolpr09 - Semiconductor Physics and Devices: Basic...

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