ee606_s04_SDF_5.14

# ee606_s04_SDF_5.14 - Derivations of SDF Problem 5.14 ECE606...

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Derivations of SDF Problem 5.14 ECE606, Spring 2004 (Please report any typo to [email protected]) Assume that on both sides of the junction there are N A , N D doping. ρ = q ( p - n + N D - N A ) p = qn i [ p n i - n n i + N D n i - N A n i ] (1) For n side, N D > N A , we have N D = n i e [ E F - E i (+ )] /kT = n i e - U F N (2) N A = n i e - [ E F - E i (+ )] /kT = n i e U F N (3) n = N D e q ( V - V bi ) /kT = n i e - U F N - U BI + U (4) p = n 2 i n = n i e U F N + U BI - U (5) where U = V kT/q and U BI = V BI kT/q . Substituting (2) to (5) into (1). we have ρ = qn i [ e U F N - U + U BI - e U - U BI - U F N bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright sinh( U F N + U BI - U ) + e - U F N - e U F N bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright - sinh( U F N ) ] (6) For Electric field E , we have d 2 V dx 2 = - ρ 1 K S ǫ 0 d 2 U dx 2 ( kT q ) = - qn i 1 K S ǫ 0 ( ρ qn i ) d 2 U dx 2 = - 1 K S ǫ 0 kT 2 q 2 n i ( ρ 2 )( 1 qn i ) = - 1 ( L D ) 2 ( ρ 2 )( 1 qn i ) 1

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Multiplying both sides by 2 dU dx , we have, 2( dU dx )( d 2 U dx 2 ) = d dx ( dU dx ) 2 = - 2 ( L D ) 2 dU dx ( ρ 2 )( 1 qn i ) integraldisplay x d dx ( dU dx ) 2 = - 1 ( L D ) 2 integraldisplay x ρ dU dx 1 qn i dx = - 1 ( L D ) 2 integraldisplay
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