# hw8_solutions - ECE 440 Homework VIII Due: Monday, October...

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ECE 440 Homework VIII Fall 2008 Due: Monday, October 27, 2008 1. An abrupt Si p-n junction has the following properties at 300 K: p-side n-side A = 10 -3 cm 2 N a = 3x10 17 /cm 3 N d = 6x10 15 /cm 3 τ n = 0.1 μs τ p = 10 μs μ p = 180 cm 2 /V-s μ n = 1160 cm 2 /V-s μ n = 540 cm 2 /V-s μ p = 420 cm 2 /V-s (a) Draw the band diagram qualitatively under forward and reverse bias showing the quasi- Fermi levels. (b) Calculate the reverse saturation current due to holes, due to electrons and the total reverse saturation current. The equation for reverse saturation current is equation (5-37b). (1) + = p n n n p p n L D p L D qA I 0 For the reverse saturation current due to holes we need the following quantities (2) s cm u q kT D p p / 87 . 10 420 0259 . 2 = = = (3) cm D L p p p 0104 . = = τ

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(4) 3 4 2 10 75 . 3 - × = = cm n n p n i n (5) A p L D qA I n p p sat p 15 4 3 19 , 10 6.2712 10 75 . 3 0104 . 87 . 10 10 10 6 . 1 - - - × = = = The same for the reverse saturation current due to electrons. (6) s cm u q kT D n n / 99 . 13 540 0259 . 2 = = = (7) cm D L n n n 3 10 2 . 1 - = = τ (8) 3 2 750 - = = cm p n n p i p (9) A n L D qA I p n n sat n 15 3 3 19 , 10 4 . 1 750 10 2 . 1 99 . 13 10 10 6 . 1 - - - - × = = = The total reverse saturation current can be found by evaluating (1) or by summing (5) and (9) as in (10). (10)
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## This note was uploaded on 10/14/2009 for the course ECE 440 taught by Professor Lie during the Spring '09 term at University of Illinois at Urbana–Champaign.

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hw8_solutions - ECE 440 Homework VIII Due: Monday, October...

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