EE_331F_2007_-_HW2S

EE_331F_2007_-_HW2S - EE 331 Devices and Circuits I Problem...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 331 Devices and Circuits I Autumn 2007 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 + 45 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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, μ n = 120 cm 2 V s and p
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

EE_331F_2007_-_HW2S - EE 331 Devices and Circuits I Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online