semisolpr05

# semisolpr05 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 5 Solutions Manual Problem Solutions 53 Chapter 5 Problem Solutions 5.1 (a) nc m O = 10 16 3 and p n n x O i O == 2 6 2 16 18 10 10 . b g px c m O = −− 324 10 43 . (b) Jen nO = µ Ε For GaAs doped at Nc m d = 10 16 3 , n cm V s ≈− 7500 2 / Then Jx = () ( ) 16 10 7500 10 10 19 16 . bg b g or JA c m = 120 2 / (b) (i) pc m O = 10 16 3 , nx c m O = . (ii) For GaAs doped at m a = 10 16 3 , p cm V s 310 2 / Jep pO = Ε =⇒ ( ) 310 10 10 19 16 . x b g b g c m = 4.96 2 / 5.2 (a) VI R R =⇒= 10 01 . R = 100 (b) R L A L RA σ 10 100 10 3 3 =− 001 1 . cm (c) σµ eN nd or 1350 19 .. = xN d or Nx c m d = 4.63 10 13 3 (d) ≈⇒ ep 480 19 = xp O or c m N N N Oa d a = 130 10 10 14 3 15 or c m a = 113 10 15 3 . Note: For the doping concentrations obtained, the assumed mobility values are valid. 5.3 (a) R L A L A ρ and For c m d = 510 16 3 , n cm V s 1100 2 Then R xx = ( ) 1100 5 10 100 10 19 16 4 2 . . b g b g or Rx = 1136 10 4 . Then I V 5 4 . Im A = 044 . (b) In this case = 3 . Then I V 5 3 . A = 4.4 (c) Ε= V L For (a), = 5 010 50 . / Vc m And v dn ( ) Ε 1100 50 or vx c m s d = 55 10 4 ./ For (b), = = V L m 5 500 . And v d ( ) 1100 500 c m s d = 5

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 5 Solutions Manual Problem Solutions 54 5.4 (a) GaAs: R L A V I k L A == = ρ σ 10 20 05 . Now σµ eN pa For Nc m a = 10 17 3 , µ p cm V s ≈− 210 2 / Then () ( ) 16 10 210 10 3 36 19 17 1 .. xc m bg b g So LR A x 500 336 85 10 8 . or Lm = 14.3 (b) Silicon For m a = 10 17 3 , p cm V s 310 2 / Then ( ) 310 10 4.96 19 17 1 . m b g So A x ( ) 500 4.96 85 10 8 or = 211 . 5.5 (a) Ε= = = V L Vc m 3 1 3/ v v dn n d =⇒ = = µµ Ε Ε 10 3 4 or n cm V s =− 3333 2 / (b) v ( ) Ε 800 3 or vx c m s d = 2.4 10 3 / 5.6 (a) Silicon: For 1 kV cm / , c m s d = 12 10 6 ./ Then t d t d 10 4 6 . tx s t = 833 10 11 . For GaAs, c m s d = 7.5 10 6 / Then t d t d 10 7.5 10 4 6 s t = 133 10 11 . (b) Silicon: For 50 kV cm / , c m s d = 9.5 10 6 Then t d t d 10 9.5 10 4 6 s t = 105 10 11 . GaAs, c m s d = 71 0 6 Then t d t d 10 0 4 6 s t = 143 10 11 . 5.7 For an intrinsic semiconductor, ii n p en =+ (a) For NN c m da 10 14 3 , np cm V s cm V s = 1350 480 22 /, / Then i xx 15 10 1350 480 19 10 b g or i m 4.39 10 6 1 (b) For c m 10 18 3 , cm V s cm V s 300 130 / Then i 15 10 300 130 19 10 b g or i m 103 10 6 1 . 5.8 (a) GaAs ≈⇒ = ep x p pO 51 6 1 0 19 . b g From Figure 5.3, and using trial and error, we find p x cm cm V s Op ≈≈ 13 10 240 17 3 2 ., / Then
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 5 Solutions Manual Problem Solutions 55 n n p x x O i O == 2 6 2 17 18 10 13 10 . . b g or nx c m O = −− 2.49 10 53 (b) Silicon: σ ρ µ =≈ 1 en nO or n ex O n () ( ) 11 81 61 0 1 3 5 0 19 ρµ . bg or c m O = 579 10 14 3 . and p n n x x O i O 2 10 2 14 15 10 . . b g px c m O = 389 10 . Note: For the doping concentrations obtained in part (b), the assumed mobility values are valid. 5.9 σµ ii n p en =+ Then 10 16 10 1000 600 9 . xn i or nK x c m i 300 391 10 93 = .

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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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semisolpr05 - Semiconductor Physics and Devices Basic...

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