semisolpr14

# Semisolpr14 - Semiconductor Physics and Devices Basic Principles 3rd edition Solutions Manual Chapter 14 Problem Solutions Chapter 14 Problem

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 14 Solutions Manual Problem Solutions 215 Chapter 14 Problem Solutions 14.1 (a) λµ = 124 . E m Then Ge: Ee V g =⇒ 066 . = 188 . m Si: V g 112 . = 111 . m GaAs: V g 142 = 0873 . m (b) E = . λ For 570 nm V = 2.18 For 700 nm V = 177 . 14.2 (a) GaAs he V ν = 2 ⇒= 062 . m so α 15 10 41 . xc m Then Ix I xx x O () =− exp exp . . 035 10 44 bg b g or I O = 059 . so the percent absorbed is (1-0.59), or 41% (b) Silicon Again V = 2 . m So 0 31 m Then I x O exp exp . 4 10 0 35 10 34 b g or I O = 087 . so the percent absorbed is (1-0.87), or 13% 14.3 ′ = g h For V m νλ µ = = 13 095 . . . For silicon, 310 21 m , Then for W cm () = 10 22 / we obtain g x x 310 10 16 10 19 b g .. ′ = −− gx c m s 144 10 19 3 1 . The excess concentration is δτ ng x = 10 19 6 . b g δ nx c m = 13 3 . 14.4 n-type GaAs, τ = 10 7 s (a) We want δδ np c m g g == = = 10 10 15 3 7 or gc m s 10 10 10 15 7 22 3 1 We have V m = = 19 065 . . . so that 13 10 . m Then = g h gh = 10 22 19 4 b gb g . x x or IW c m I O 00 2 3 4 2 ./ (b) I O ==− 020 4 .e x p . We obtain xm = .

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 14 Solutions Manual Problem Solutions 216 14.5 GaAs (a) For he V m νλ µ =⇒ = 165 075 .. So α 07 10 41 . xc m For 75% obsorbed, Ix I x O () ==− 025 .e x p Then xx x = F H I K F H I K ln ln . 1 1 1 4 or xm = 198 . (b) For 75% transmitted, I O 4 x p. bg we obtain = 041 . 14.6 GaAs For c m == 11 0 4 , we have 50% absorbed or 50% transmitted, then I x O 050 x p We can write =⋅ = F H I K F H I K F H I K 05 1 10 2 4 x ln . ln or = 069 10 . m This value corresponds to λµ ., . mE e V 14.7 The ambipolar transport equation for minority carrier holes in steady state is D dp dx G p p n L n p 2 2 0 δδ τ af +− = or dx p L G D nn p L p 2 22 −= where LD pp p 2 = The photon flux in the semiconductor is ΦΦ O ( ) =− exp and the generation rate is Gx x LO αα exp so we have dx p x p O p 2 Φ exp The general solution is of the form δ pA x L B x L n = + + F H G I K J F H G I K J exp exp ατ Φ Op p L x 1 exp At xp n →∞ = , 0 So that B = 0 , then x LL x n p p = F H G I K J exp exp Φ 1 At x = 0 , we have D dx sp p n n a f = 00 so we can write L n x p = 0 1 Φ and dx A n p p x = + 0 2 1 Φ Then we have −+ AD L D L sA s L p p Op p p p 2 Solving for A , we find A L sD sDL p p = + + L N M O Q P Φ 1 b g The solution can now be written as p L x L n p p p = + + R S T F H G I K J Φ 1 exp −− exp x r
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 14 Solutions Manual Problem Solutions 217 14.8 We have D dn dx G n n p L p n 2 2 0 δδ τ bg +− = or dx n L G D pp n L n 2 22 −= where LD nn n 2 = The general solution can be written in the form δτ nA x L B x L G p Ln =+ + F H G I K J F H G I K J cosh sinh For s =∞ at x = 0 means that δ n p 00 () = , Then 0 ⇒= AG A G ττ At xW = , == D dx sn n p op Now nW G W L B W L G pL n () F H G I K J F H G I K J =− + + cosh sinh and dx G L W L B L W L p n n = + F H G I K J F H G I K J sinh cosh so we can write GD L W L BD L W L Ln n n sinh cosh F H G I K J F H G I K J F H G I K J sG W L oL n n cosh ++ F H G I K J B W L G n sinh

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## This note was uploaded on 11/06/2008 for the course ECE 103 taught by Professor Wang during the Fall '08 term at UCSD.

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Semisolpr14 - Semiconductor Physics and Devices Basic Principles 3rd edition Solutions Manual Chapter 14 Problem Solutions Chapter 14 Problem

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