semisolex05

# semisolex05 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 5 Solutions Manual Exercise Solutions 51 Chapter 5 Exercise Solutions E5.1 nx c m o =−= 10 10 9 10 15 14 14 3 so p n n x x xc m o i o == = 2 10 2 14 53 15 10 91 0 2.5 10 . b g Now Je n p e n drf n o p o n o =+ µµ µ bg ΕΕ = () ( ) 16 10 1350 9 10 35 19 14 . xx b g or JA c m drf = 680 2 ./ E5.2 p drf p o Ε Then 120 480 20 19 = ( ) . xp o so px c m N oa 7.81 10 16 3 E5.3 Use Figure 5.2 (a) (i) n cm V s ≅− 500 2 / , (ii) 1500 2 cm V s / (b) (i) p cm V s 380 2 / ,(ii) 200 2 cm V s E5.4 Use Figure 5.3 [Units of cm V s 2 / ] (a) For Nc m I = 10 15 3 ; n 1350 , p 480: (b) Nx c m I = 17 3 .; n 700 , p 300: (c) c m I = 11 10 17 3 n 800 , p 310: (d) c m I = 21 0 17 3 ; n 4500 , p 220 E5.5 (a) For c m I = 71 0 16 3 ; n cm V s 1000 2 /, p cm V s 350 2 / (b) σµ eNN nd a a f =⇒ 1000 3 10 19 16 . b g σ =− 4.8 1 cm ρ == ⇒ 11 4.8 0208 . cm E5.6 eN 1 so 1 01 10 19 . . xN Then = 625 10 19 . Using Figure 5.4a, c m d 0 16 3 Then n cm V s ≈− 695 2 / E5.7 (a) R V I k === 5 2 2.5 (b) Rx x 2.5 10 12 10 10 3 3 6 . b g 2.08 cm (c) From Figure 5.4a, c m a 0 15 3 E5.8 D dn dx eD x L diff n n n F H G I K J F H G I K J 10 10 15 4 exp Dc m s n = 25 2 Lc m m n 10 1 4 Then J x Ac m diff F H I K 40 1 2 exp / (a) x = 0 ; c m diff 40 2 / (b) xm = 1 ; c m diff 14.7 2 / (c)

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semisolex05 - Semiconductor Physics and Devices Basic...

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