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Final Exam Practice Problem Solutions

Final Exam Practice Problem Solutions - Chapter 8 Problems...

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1 Chapter 8 Problems 8.15 (a) For 0 2 10 6   t x s n g t nO nO 1 exp a f 10 10 1 20 6 b gb g exp t nO or n t nO 10 1 14 exp a f where nO s 10 6 At t x s 2 10 6 n s 2 10 1 2 1 14 a f b g a f exp or n s x cm 2 0865 10 14 3 a f . For t x s 2 10 6 n x t x nO   L N M O Q P 0865 10 2 10 14 6 . exp b g (b) (i) At t 0 , n 0 (ii) At t x s 2 10 6 , n x cm 0 865 10 14 3 . (iii) At t   , n 0 _______________________________________ 8.18 p-type so electrons are the minority carriers D n n g n n t n n nO        2 For steady state,   n t 0 and for x 0 , g 0 0 , , so we have D d n dx n n nO 2 2 0  or d n dx n L n 2 2 2 0   where L D n n nO 2 The solution is of the form n A x L B x L n n exp exp a f a f The excess concentration n must remain finite, so that B 0 . At x n cm  0 0 10 15 3 , , so the solution is n x L n 10 15 exp a f We have that n cm V s 1050 2 / , then D kT e cm s n n F H I K  1050 0 0259 27.2 2 . / Then L D x n n nO 27.2 8 10 7 b g L m n 46 6 . (a) Electron diffusion current density at x 0 J eD d n dx n n x   0 eD d dx x L n n x 10 15 0 exp a f eD L x x n n 10 1 6 10 27.2 10 46 6 10 15 19 15 4 b g b g b g . . or J A cm n  0 934 2 . / Since p n and since excess holes diffuse at the same rate as excess electrons, then J x A cm p  0 0 934 2 . / (b) At x L n , J eD d n dx n n n x L     eD L n n 10 1 15 b g a f exp   16 10 27.2 10 46 6 10 1 19 15 4 .
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