semisolex03

semisolex03 - Semiconductor Physics and Devices: Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 3 Solutions Manual Exercise Solutions 21 Chapter 3 Exercise Solutions E3.1 −= + 11 0 sin cos α a a a By trial and error, ar a d = 5305 . Now 2 2 2 mE a h ⋅= . so E ma x xx 2 2 2 2 2 34 2 31 10 2 2 5 305 1054 10 29 . 1110 510 == () −− . .. h b g bg b g or Ex Je V 2 19 686 10 4.29 Also 2 1 2 mE a h π so E ma x 1 22 2 2 34 2 31 10 2 2 1054 10 . ()() h . b g b g or J e V 1 19 2.41 10 150 Then EE E =−= 21 4.29 or Ee V = 2.79 E3.2 gE m h EE c n c =− 42 32 3 * / b g Then g m h EE d E T n C E Ek T c c + z 3 12 * / / b g af F H I K + 2 3 3 m h n c c c E T * / / b g or g m h kT T n = F H I K 2 3 3 * / / which yields g x x T = F H I K 1 0 89 . 1 1 1 0 6625 10 2 3 31 34 3 . . / b g × 00259 16 10 19 / x b g which yields gx m x c m T 2.12 10 2.12 10 25 3 19 3 E3.3 We have g m h EEd E T p v T E v v z 3 * / / b g which yields g m h T p v v v T E = F H I K 2 3 3 * / / b g or g m h kT T p = F H I K 2 3 0 3 * / / = F H I K 2 3 3 m h kT p * / / Then g x x T = F H I K 0 5 69 . 1 1 1 0 2 3 31 34 3 . . / b g × 19 / x b g or m x c m T 7.92 10 7.92 10 24 3 18 3 E3.4 (a) f kT kT F Fc F = + = + F H I K F H I K 1 1 1 1 exp exp or f F = + F H I K 1 1 030 00259 exp . . fx F = 9.32 10 6
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 3 Solutions Manual Exercise Solutions 22 (b) f F = + + F H I K 1 1 0 30 0 0259 00259 exp .. . fx F = 343 10 6 . E3.5 (a) 11 1 1 −= + F H I K f EE kT F F exp = + = + F H I K F H I K F H I K exp exp exp kT kT kT F FF 1 1 1 Then 1 1 1 035 + F H I K f F exp . . so 1 135 10 6 F . (b) 1 1 1 0 35 0 0259 + + F H I K f F exp . or 1 4.98 10 7 F E3.6 kT == () F H I K 400 300
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This note was uploaded on 11/06/2008 for the course ECE 103 taught by Professor Wang during the Fall '08 term at UCSD.

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semisolex03 - Semiconductor Physics and Devices: Basic...

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