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EE_331F_2009_-_HW2S

# EE_331F_2009_-_HW2S - specified N A N D material is p type...

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EE 331 Devices and Circuits I Fall 2009 Problem #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

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Unformatted text preview: specified. N A N D : material is p - type | N A N D 7 x 10 19 /cm 3 2 n i 2 x 10 10 /cm 3 p 7 x 10 19 /cm 3 | n= n i 2 p 10 20 7 x 10 19 1 . 43 /cm 3 N D N A 7 x 10 19 / cm 3 | Using Fig. 2.13, n 120 cm 2 V s and p 60 cm 2 V s 1 q p p 1 1.602 x 10 19 C 60 cm 2 V s 7 x 10 19 cm 3 1.49 m cm 2.37 Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsic because it contains impurities. Addition of the impurities has increased the resistivity....
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