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EE_331F_2008_-_HW2S

# EE_331F_2008_-_HW2S - EE 331 Devices and Circuits I Problem...

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EE 331 Devices and Circuits I Autumn 2008 Problem Set #2 Solution 2.18 Since Ge is from column IV, acceptors come from column III and donors come from column V. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi 2.23 N A > N D : N A N D = 10 15 10 14 = 9 x 10 14 /cm 3 If we assume N A N D >> 2 n i = 10 14 / cm 3 : p = N A N D = 9 x 10 14 /cm 3 | n = n i 2 p = 2510 26 9 x 10 14 = 2 . 78 x 10 12 /cm 3 If we use Eq. 2.12 : p = 9 x 10 14 ± 9 x 10 14 ( ) 2 + 4 5 x 10 13 ( ) 2 2 = 9 . 03 x 10 14 and n = 2 . 77 x 10 12 /cm 3 . The answers are essentially the same. 2.29 (a) Arsenic is a donor, and boron is an acceptor. N D = 2 x 10 18 /cm 3 , and N A = 8 x 10 18 /cm 3 . Since N A > N D , the material is p-type. 3 3 18 6 20 2 3 18 i 3 18 3 10 i 7 16 10 6 10 and 10 6 So 2n >> / 10 6 and / 10 n re, temperatu room At (b) /cm . /cm x /cm p n n /cm x p cm x N N cm i D A = = = = = = 2.33 Indium is from column 3 and is an acceptor. N A = 7 x 10 19 /cm 3 . Assume N D = 0, since it is not specified.

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