ecee302hw3

ecee302hw3 - 4.2 1 n" = 1V C .NV exp k1" (1012 )2...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

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

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4.2 1 n" = 1V C .NV exp k1" (1012 )2 = (2.831019)(1.04.\‘1019)(fi) exp(_‘l:-;2) Then exp(£) = (2.9123-10“)(i) ‘7 300 By trial and error w _ex [—(1—1.2)]_ex[ +0.20 ] _ p 2(0.0259) _ p 2(0.0259) m(A) n,(B) = 47.5 4n ( ) 3 A h I”; a E.—E. =—”T1 . F1 mzdgap 4 . I” 3 ' 1.4 ' :—(0.0259)h1 — => 4 0.62 EB — 5mm = +0.0158 eV 4.18 ., nf (1.53-10”) (a) P0 = — = —4 => 12 5x10 0 p0 = 45.11015 CHI-3 . p0 > "0 => PST” (13) ER — EF = len[&) I). I 4.5x10” 2 —lo L5x10 or EFT. — E}r = 0.3266 eV 4.19 — E — E , p0 : NV exp|:#] [CT 19 —0.22 = 104.110 exp 0.0259 so p0 = 2.132(1015 ("m-3 Assuming EC — E}r = 1.12 —0.22 = 0.90 eV Then 01' [J —0.90 0.0259 no = NC expl: = 2.811018 exp( no = 2.271104 012'3 -(EC - Ex) J ...
View Full Document

This document was uploaded on 02/09/2010.

Page1 / 6

ecee302hw3 - 4.2 1 n" = 1V C .NV exp k1" (1012 )2...

This preview shows document pages 1 - 6. Sign up to view the full document.

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