semisolpr10

# semisolpr10 - Semiconductor Physics and Devices: Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 10 Solutions Manual Problem Solutions 139 Chapter 10 Problem Solutions 10.1 Sketch 10.2 Sketch 10.3 (a) I eD A n x S nB EB O B = = −− () 16 10 20 10 10 10 19 4 4 4 . x b g b gb g or Ix A S = 32 10 14 . (b) (i) ix C =⇒ F H I K 05 00259 14 .e x p . . iA C = 7.75 µ (ii) C F H I K 06 14 x p . . im A C = 0368 . (iii) C F H I K 07 14 x p . . A C = 17.5 10.4 (a) β α = = 1 09920 1 0 9920 . . = 124 (b) From 10.3b (i) For C = 7.75 ; i i B C == 7.75 124 B = 00625 . , ii EC = + ⋅= F H G I K J F H I K 11 2 5 124 7.75 E = 7.81 (ii) For A C = ., B = 2.97 , A E = 0371 . (iii) For A C = 17.5 , A B = 0141 . , A E = 17.64 10.5 (a) i i C B 510 6 = 85 = + 1 85 86 = 09884 . iii ECB =+= +⇒ 510 6 E = 516 (b) 2.65 005 . = 53 53 54 = 09815 . i E =+⇒ 2.65 0 05 . A E = 2.70 10.6 (c) For A B = CB ( ) 100 0 05 A C = 5 We have vVi R CE CC C =−= 10 5 1 or vV CE = 5 10.7 (b) VI R V V CC C CB BE =+ + so 10 2 0 0 6 + I C or Im A C = 4.7 10.8 (a) n n N x xc m pO i B = 2 10 2 16 43 15 10 10 2.25 10 . b g At x = 0 , nn V V pp O BE t 0 F H G I K J = exp or we can write VV n n BE t p pO = F H G I K J ln 0

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 10 Solutions Manual Problem Solutions 140 We want nc m P 0 10% 10 10 16 15 3 () = , So V x BE = F H G I K J 00259 10 2.25 10 15 4 .l n or VV BE = 0635 . (b) At ′ = x 0, pp V V nn O BE t 0 F H G I K J = exp where p n N x xc m nO i E == = 2 10 2 17 33 15 10 10 2.25 10 . bg Then px n 02 . 2 5 1 0 3 F H I K =⇒ exp . . pc m n 01 0 14 3 () = (c) From the B-C space charge region, x e N NN N p bi R C BC B 1 1 12 21 = ∈+ + F H G I K J F H G I K J L N M O Q P af / We find V x V bi 1 16 15 10 2 10 10 L N M M O Q P P n . . b gb g Then x x x p 1 14 19 2117 88510 0635 3 16 10 = + ( ) L N M .. . . × + F H G I K J F H I K O Q P 10 10 1 10 10 15 16 15 16 / or xm p 1 0207 = . µ We find V x V bi 2 17 16 10 2 10 10 0754 L N M M O Q P P n . . b gb g Then x x x p 2 14 19 2 117 8 85 10 0 754 0 635 = ( ) L N M . . . × + F H G I K J F H I K O Q P 10 10 1 10 10 17 16 17 16 / or p 2 0118 = . Now xx BB Op p =− −=− 110 0 207 . or B = 0775 . 10.9 (a) p n N x x EO i E 2 10 2 17 510 . b g c m EO = 4.5 10 23 n n N x BO i B 2 10 2 16 10 . nx c m BO = 2.25 10 43 p n N x CO i C 2 10 2 15 10 . b g c m CO = 2.25 10 53 (b) V V O BE t 0 F H G I K J = exp = F H I K 2.25 10 0625 4 x exp . . or c m B 0 6 80 10 14 3 . Also V V EE O BE t 0 F H G I K J = exp = F H I K 4.5 10 2 x exp . . or c m E 0 136 10 13 3 . 10.10 (a) n n N x EO i E 2 10 2 18 10 . c m EO = 2.25 10 p n N x x BO i B 2 10 2 16 . b g
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 10 Solutions Manual Problem Solutions 141 px c m BO = 4.5 10 33 n n N x CO i C == 2 10 2 15 15 10 10 . b g nx c m CO = 2.25 10 53 (b) pp V V BB O EB t 0 () F H G I K J = exp = F H I K 4.5 10 0650 00259 3 x b g exp . . or c m B 0 357 10 14 3 () = . Also nn V V EE O EB t 0 F H G I K J = exp = F H I K 2.25 10 2 x bg exp . . or c m E 0 178 10 13 3 . 10.11 We have dn dx n x L V V O B B BE t δ a f =− F H G I K J F H G I K J L N M O Q P R S T sinh exp 1 × −− F H G I K J F H G I K J F H G I K J U V W 11 L xx LL x L B B B cosh cosh At x = 0 , dx n L x L V V O B B B BE t a f 01 F H G I K J F H G I K J L N M O Q P R S T = sinh exp ×+ F H G I K J U V W cosh x L B B At B = , dx x n L x L B B BO B B B af = F H G I K J sinh ×− + F H G I K J L N M O Q P F H G I K J R S T U V W exp cosh V V x L BE t B B Taking the ratio, dx x dx B B B a f 0 = −+ F H G I K J L N M O Q P F H G I K J F H G I K J

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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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semisolpr10 - Semiconductor Physics and Devices: Basic...

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