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semisolpr08

semisolpr08 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 8 Solutions Manual Problem Solutions 101 Chapter 8 Problem Solutions 8.1 In the forward bias II eV kT fS F H I K exp Then I I I I eV kT eV kT e kT VV f f S S 1 2 1 2 12 =⋅ = F H I K F H I K L N M O Q P exp exp exp af or kT e I I f f 1 2 −= F H I K F H G I K J ln (a) For I I f f 1 2 10 =⇒ V V mV mV 59.9 60 (b) For I I V V mV mV f f 1 2 100 119.3 120 = 8.2 eV kT S =− F H I K L N M O Q P exp 1 or we can write this as I I eV kT S += F H I K 1e x p so that V kT e I I S =+ F H I K F H G I K J ln In reverse bias, I is negative, so at I I S 090 . , we have V () ( ) 00259 1 090 .l n . or Vm V 59.6 8.3 Computer Plot 8.4 The cross-sectional area is A I J x xc m == = 10 10 20 510 3 42 We have JJ V V S D t ≈⇒ F H G I K J exp 20 065 = F H I K J S exp . . so that Jx A c m S = 2.52 10 10 2 / We can write Je n N D N D Si a n nO d p pO + L N M O Q P 2 11 ττ \ We want 1 010 N D N D N D a n nO a n nO d p pO ⋅+ = τ . or 5 5 0 7 77 Nx a ad −− = + = 7.07 10 7.07 10 4.47 10 3 33 x x N N x a d bg . which yields N N a d = 14.24 Now x x S 2.52 10 16 10 15 10 10 19 10 2 .. b gb g ×⋅ + L N M O Q P 1 14.24 25 0 dd We find c m d = 7.1 10 14 3 and c m a = 101 10 16 3 .
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 8 Solutions Manual Problem Solutions 102 8.5 (a) J JJ eD n L eD n L eD p L n np O n O n pn O p + = + = ⋅+ Dn N N D n N n nO i a n nO i a p pO i d τ ττ 2 22 = +⋅ F H G I K J 1 1 D D N N O O a d We have D D p n p n == µ 1 2.4 and nO pO = 1 01 . so J N N n a d + = F H G I K J 1 1 1 2.4 1 . or J N N n a d + = + () F H G I K J 1 12 . 0 4 (b) Using Einstein&s relation, we can write J e L n N e L n N e L n N n n n i a n n i a p p i d + = 2 = eN L L nd n p pa µµ We have σµ nn d = and pp a = Also L L D D n p O O = 2.4 4.90 . Then J n + = + σσ bg 4.90 8.6 For a silicon + junction, IA e n N D Si d p pO =⋅ 2 1 −− 10 16 10 15 10 1 10 12 10 41 9 1 0 2 16 7 b g b g .. xx or Ix A S = 394 10 15 . Then II V V x DS D t F H G I K J F H I K exp . exp . . 050 00259 15 or A D = 9.54 10 7 8.7 We want J n + = 095 . = + = + eD n L eD n L eD p L D LN D D O n O n O p n na n p pd = D L D L D L N N n n n n p p a d We obtain LD x n O 25 01 10 6 . Lm n = 158 . x p O 10 01 10 6 . p = 10 Then
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 8 Solutions Manual Problem Solutions 103 095 25 158 25 10 10 . . . = +⋅ F H G I K J N N a d which yields N N a d = 0083 . 8.8 (a) p-side: EEk T N n Fi F a i −= F H G I K J ln =⇒ () F H G I K J 00259 510 15 10 15 10 .l n . x x EE e V Fi F 0329 . Also n-side: k T N n FF i d i F H G I K J ln F H G I K J 10 17 10 n . x e V i 0407 . (b) We can find Dc m s n == ( ) 1250 0 0259 32.4 2 ./ m s p ( ) 420 0 0259 10 9 2 .. / Now Je n N D N D Si a n nO d p pO =+ L N M O Q P 2 11 ττ = 16 10 15 10 19 10 2 xx b gb g ×+ −− L N M O Q P 1 32.4 10 1 10 10 9 10 15 6 17 7 x . or Jx A c m S = 4.48 10 11 2 / Then IA J x SS 10 4.48 10 41 1 b gb g or Ix A S = 4.48 10 15 We find II V V S D t = F H G I K J exp = F H I K 4.48 10 05 15 x bg exp .
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