semisolpr06

# semisolpr06 - Semiconductor Physics and Devices: Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 6 Solutions Manual Problem Solutions 65 Chapter 6 Problem Solutions 6.1 n-type semiconductor, low-injection so that == R px pO δ τ 510 10 13 6 or ′ = −− Rxc m s 19 3 1 6.2 (a) R n nO O nO = and n n p cm O i O = 2 10 2 16 43 10 10 10 b g Then R x nO =⇒ 10 21 0 4 7 Rx c m s nO = 10 3 1 (b) R n x n nO 10 0 12 7 or m s n = 18 3 1 so RRR x x nnn O =− = 18 10 m s n 18 3 1 6.3 (a) Recombination rates are equal np O nO O pO ττ = nN c m Od 10 16 3 p n n x xc m O i O = 2 10 2 16 15 10 10 2.25 10 . b g So 10 2.25 10 20 10 16 4 6 nO x x = or nO xs = + 889 10 6 . (b) Generation Rate = Recombination Rate So G x x 2.25 10 20 10 4 6 Gx c m s = 1125 10 93 1 . (c) RG x c m s 1 . 6.4 (a) Eh hc xx x === ν λ 6625 10 3 10 6300 10 34 8 10 . b gb g or Ex J = 315 10 19 . This is the energy of 1 photon. Now 11 WJ s / 317 10 18 . x photons/s Volume = 1 01 01 3 () ( ) = + .. cm Then g x 18 . . gx e h p a i r s c m s 19 3 ./ (b) δδ g x x = 10 10 19 6 . bg b g or x c m 14 3 . 6.5 We have =−∇• + + p t Fg p pp p and Je pe D p p µ Ε The hole particle current density is F J e pD p p p + = + Ε Now ∇• = ∇•∇ + Fp D p p Ε We can write = •∇ + ∇• p ΕΕ Ε and ∇•∇ =∇ 2 so

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 6 Solutions Manual Problem Solutions 66 ∇• = •∇ + ∇• + () Fp p D p pp p µ ΕΕ 2 Then =− + p t D p 2 +− g p p p τ We can then write Dp p p ∇− ∇+∇ 2 +− = g t p p 6.6 From Equation [6.18] =−∇• + + p t Fg p p For steady-state, = p t 0 Then 0 + + FgR ppp and for a one-dimensional case, dF dx gR x p + =−= 10 2 10 20 19 dF dx xc m s p + −− = 810 19 3 1 6.7 From Equation [6.18], 00 2 1 0 19 + − + dF dx x p or dF dx m s p + 21 0 19 3 1 6.8 We have the continuity equations (1) p p + 2 δµ δ + g t p p −= and (2) Dn n n nn ∇+• + 2 g t n n By charge neutrality δδδ n p nnp =≡⇒ = () () and ∇= 22 δδ np and = n t p t Also ggg pn R =≡ , ττ Then we can write (1) n p + 2 n t and (2) n n + 2 n t Multiply Equation (1) by n n and Equation (2) by p p , and then add the two equations. We find µµ nD pD n +∇ bg 2 ()( ) n Ε ++ = + g R n t bgbg Divide by + , then nD pD n + + F H G I K J 2 + + •∇ L N M O Q P n Ε n t Define ′ = + + D nD pD = + + DD n p Dn D p and ′ = + Then we have ∇+ ′ •∇ + = ) n g R n t 2 Ε Q.E.D.
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 6 Solutions Manual Problem Solutions 67 6.9 For Ge: TK = 300 , nx c m i = 2.4 10 13 3 n NN n dd i =+ + F H I K 22 2 2 + 10 10 2.4 10 13 13 2 13 2 bg b g x or c m = 36 10 13 3 . Also p n n x x xc m i == = 2 13 2 13 13 3 2.4 10 16 10 b g . We have µµ np 3900 1900 , DD 101 49.2 Now ′ = + + () D DD n p Dn D p = + + ( ) 10 1 49 .

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## semisolpr06 - Semiconductor Physics and Devices: Basic...

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