EE 33807 HW8_Solutions

EE 33807 HW8_Solutions - 11/15/2007 EE 338 Fall 2007 HW#5...

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11/15/2007 EE 338 Fall 2007 HW#5 Solutions 6.1 E FS E i E V E C E C E V E i E FS (b) p-type, depletion (a) p-type, inversion E C E V E i E FS E C E V E i E FS (d) n-type, inversion (c) p-type, accumulation 1
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11/15/2007 6.2 (a) For T K = 300 Silicon: φ Fp t a i V N n =− F H G I K J ln () F H G I K J 0 0259 10 15 10 0347 16 10 .l n . . x V Now x eN dT Fp a = L N M O Q P 4 12 / = ( ) L N M O Q P 4 117 8,85 10 0 347 16 10 10 14 19 16 .. . / x x b g b gb g or xm dT = 030 . μ Also = Qe N SD a dT max x g = −− 10 030 10 19 16 4 xx bg b g b or = Qx C SD max / 4.8 10 82 c m GaAs : Fp x V F H G I K J 0 0259 10 18 10 0581 16 6 n . . and x x x dT = ( ) L N M O Q P 4131 88510 14 19 16 . . / b g or dT = 0410 . Then = SD max . / 656 10 Germanium C c m Fp x V F H G I K J 00259 10 2.4 10 0156 16 13 n . Then x x x dT = ( ) L N M O Q P 4 16 8 85 10 14 19 16 . / b g b gb g or dT = 0 235 . Then = SD max . / 376 10 C c m (b) For T K = 200 , VV t == F H I K 0 0259 200 300 0 01727 Silicon : nx c i = 7.68 10 43 m We obtain and Fp V 0442 . Q x C dT SD = = 0388 54 10 ., m a x . / c m GaAs : nc i = 138 3 . m We obtain and Fp V 0 631 . Q x C dT SD = = 0428 685 10 m a x . / c m Germanium : c i = 2.16 10 10 3 m Fp V 0225 . We obtain and Q x C dT SD = = 0 282 4.5 10 m a x / c m 2
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11/15/2007 6.4 p-type silicon, Nx c m a = 61 0 15 3 Then, from Figure 6.21: (a)
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EE 33807 HW8_Solutions - 11/15/2007 EE 338 Fall 2007 HW#5...

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