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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 p x pO δ τ 5 10 10 13 6 or ′ = R x cm s 5 10 19 3 1 6.2 (a) R n nO O nO = τ and n n p cm O i O = = = 2 10 2 16 4 3 10 10 10 b g Then R x nO = 10 2 10 4 7 R x cm s nO = 5 10 10 3 1 (b) R n x n nO = = δ τ 10 2 10 12 7 or R x cm s n = 5 10 18 3 1 so R R R x x n n nO = = 5 10 5 10 18 10 R x cm s n 5 10 18 3 1 6.3 (a) Recombination rates are equal n p O nO O pO τ τ = n N cm O d = = 10 16 3 p n n x x cm O i O = = = 2 10 2 16 4 3 15 10 10 2.25 10 . b g So 10 2.25 10 20 10 16 4 6 τ nO x x = or τ nO x s = + 8 89 10 6 . (b) Generation Rate = Recombination Rate So G x x = 2.25 10 20 10 4 6 G x cm s = 1125 10 9 3 1 . (c) R G x cm s = = 1125 10 9 3 1 . 6.4 (a) E h hc x x x = = = ν λ 6 625 10 3 10 6300 10 34 8 10 . b gb g or E x J = 315 10 19 . This is the energy of 1 photon. Now 1 1 W J s = / 317 10 18 . x photons/s Volume = 1 01 01 3 ( )( ) = + . . cm Then g x = 317 10 01 18 . . g x e h pairs cm s = 317 10 19 3 . / (b) δ δ τ n p g x x = = = 317 10 10 10 19 6 . b gb g or δ δ n p x cm = = 317 10 14 3 . 6.5 We have = −∇ • + + p t F g p p p p τ and J e p eD p p p p = µ Ε The hole particle current density is F J e p D p p p p p + = + = ( ) µ Ε Now ∇ • = ∇ • ∇ • ∇ + ( ) F p D p p p p µ Ε We can write ∇ • = • ∇ + ∇ • ( ) p p p Ε Ε Ε and ∇ • ∇ = ∇ p p 2 so
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 6 Solutions Manual Problem Solutions 66 ∇ • = • ∇ + ∇ • + ( ) F p p D p p p p µ Ε Ε 2 Then = − • ∇ + ∇ • + ( ) p t p p D p p p µ Ε Ε 2 + g p p p τ We can then write D p p p p p • ∇ + ∇ • ( ) 2 µ Ε Ε + = g p p t p p τ 6.6 From Equation [6.18] = −∇ • + + p t F g p p p p τ For steady-state, = p t 0 Then 0 = −∇ • + + F g R p p p and for a one-dimensional case, dF dx g R x p p p + = = 10 2 10 20 19 dF dx x cm s p + = 8 10 19 3 1 6.7 From Equation [6.18], 0 0 2 10 19 = − + + dF dx x p or dF dx x cm s p + = − 2 10 19 3 1 6.8 We have the continuity equations (1) D p p p p p • ∇ + ∇ • ( ) ( ) 2 δ µ δ Ε Ε + g p p t p p = ( ) τ δ and (2) D n n n n n + • ∇ + ∇ • ( ) ( ) 2 δ µ δ Ε Ε + = ( ) g n n t n n τ δ By charge neutrality δ δ δ δ δ n p n n p = ⇒ ∇ = ∇ ( ) ( ) and = ∇ ( ) ( ) 2 2 δ δ n p and = ( ) ( ) δ δ n t p t Also g g g p n R n p p n = = , τ τ Then we can write (1) D n n p p p • ∇ + ∇ • ( ) ( ) 2 δ µ δ Ε Ε + = ( ) g R n t δ and (2) D n n n n n + • ∇ + ∇ • ( ) ( ) 2 δ µ δ Ε Ε + = ( ) g R n t δ Multiply Equation (1) by µ n n and Equation (2) by µ p p , and then add the two equations. We find µ µ δ n p p n nD pD n + ( ) b g 2 + • ∇ ( ) ( ) µ µ δ n p p n n Ε + + = + ( ) ( ) µ µ µ µ δ n p n p n p g R n p n t b g b g Divide by µ µ n p n p + b g , then µ µ µ µ δ n p p n n p nD pD n p n + + F H G I K J ( ) 2 + + • ∇ ( ) L N M O Q P ( ) µ µ µ µ δ n p n p p n n p n Ε + = ( ) ( ) g R n t δ
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