semisolpr04

# semisolpr04 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 4 Solutions Manual Problem Solutions 37 Chapter 4 Problem Solutions 4.1 nN N E kT iC V g 2 = F H G I K J exp (a) Silicon TK ° () kT eV ncm i 3 bg 200 400 600 0 01727 . 0 03453 00518 7.68 10 4 x 2.38 10 12 x 9.74 10 14 x (b) Germanium (c) GaAs ° i 3 i 3 2.16 10 10 x 860 10 14 . x 382 10 16 . x 138 328 10 9 . x 572 10 12 . x 4.2 N E kT V g 2 = F H G I K J exp 10 2.8 10 104 10 300 112 12 2 19 19 3 bgb g b g = F H I K F H I K xx T kT .e x p . Then exp . 2.912 10 300 14 3 kT x T F H I K F H I K = By trial and error = 381 4.3 Computer Plot 4.4 NT E kT O V O g 2 3 =⋅ F H G I K J exp So nT T T E kT kT i i g 2 2 2 1 2 1 3 21 11 af a f =− F H G I K J F H G I K J L N M O Q P exp At T K kT eV 2 300 0 0259 =⇒ = At T K kT eV 1 200 0 01727 = Then 583 10 182 10 300 200 1 00259 1 0 01727 7 2 2 3 . . exp .. x x E g F H G I K J F H I K F H I K L N M O Q P or 1026 10 3 375 19.29 11 e x p xE g = which yields Ee V g = 125 . For = 300 , 300 7 2 3 x p . . xN N CO VO b g = F H I K or NN x CO VO = 115 10 29 . 4.5 (a) gf E E EE kT CF C F ∝− −− L N M O Q P exp a f − − L N M O Q P L N M O Q P kT kT C CC F exp exp a f a f Let EE x C −≡ Then x x kT F H I K exp Now, to find the maximum value dgf dx x x kT a f F H I K 1 2 12 / exp −⋅ = F H I K 1 0 kT x x kT / exp This yields 1 22 x x kT x kT / / = Then the maximum value occurs at kT C =+ 2 (b) E E kT VF V F 1 −∝ L N M O Q P a f exp L N M O Q P L N M O Q P kT kT V FV V exp exp a f a f Let x V

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 4 Solutions Manual Problem Solutions 38 Then gf x x kT VF 1 −∝ F H I K af exp To find the maximum value dg f dx d dx x x kT 1 0 = F H I K L N M O Q P exp Same as part (a). Maximum occurs at x kT EE V ==− 2 or kT V =− 2 4.6 nE kT kT C C C C 1 2 1 1 2 2 = −− L N M O Q P L N M O Q P exp exp where k T C 1 4 =+ and kT C 2 2 Then kT kT kT 1 2 12 4 2 = L N M O Q P exp F H I K L N M O Q P () 22 4 1 2 3 5 exp exp . or 1 2 00854 a f a f = . 4.7 Computer Plot 4.8 nA nB E kT E kT kT i i gA gB gA gB 2 2 F H G I K J F H G I K J L N M O Q P = = exp exp exp bg or kT i i gA gB L N M O Q P = exp 2 = = + L N M O Q P L N M O Q P exp . . exp . . 11 2 2 0 0259 020 2 0 0259 or i i = 47.5 4.9 Computer Plot 4.10 k T m m Fi midgap p n −= F H G I K J 3 4 ln * * Silicon: mm pO nO ** ., . == 056 108 e V Fi midgap 00128 . Germanium: . 037 055 e V Fi midgap 00077 . Gallium Arsenide: m m n O . 0 48 0 067 e V Fi midgap + 0038 . 4.11 (a) k T m m Fi midgap p n F H G I K J 3 4 ln * * =⇒ F H I K 3 4 00259 14 062 .l n . . e V F midgap i + 00158 . (b) Fi midgap F H I K 3 4 025 110 n . . e V Fi midgap 00288 . 4.12 k T N N Fi midgap V C F H G I K J 1 2 ln F H G I K J 1 2 104 10 2.8 10 0495 19 19 kT x x kT ln . .
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 4 Solutions Manual Problem Solutions 39 TK ° () kT eV EE e V Fi midgap 200 400 600 0 01727 . 0 03453 00518 00085 0017 00256 4.13 Computer Plot 4.14 Let gE K C == constant Then, ng E f E d E OC E C F = z () () = + F H I K z K kT dE F E C 1 1e x p −− L N M O Q P z K kT dE F E C exp af Let η = kT F so that dE kT d =⋅ We can write E E FC F C −= a f a f so that exp exp exp = ⋅− L N M O Q P L N M O Q P kT kT F

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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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semisolpr04 - Semiconductor Physics and Devices Basic...

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