ee606_s04_SDF_5.14

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

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Unformatted text preview: 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 NA , ND doping. ρ = q (p − n + ND − NA ) p n ND NA ⇒ p = qni [ − + − ] ni ni ni ni For n side, ND > NA , we have ND = ni e[EF −Ei (+∞)]/kT = ni e−UF N NA = n i e −[EF −Ei (+∞)]/kT q (V −Vbi )/kT (1) (2) (3) (4) (5) = ni e UF N n = ND e = ni e 2 ni = ni eUF N +UBI −U p= n where U = V kT /q −UF N −UBI +U and UBI = VBI kT /q . Substituting (2) to (5) into (1). we have (6) ρ = qni [eUF N −U +UBI − eU −UBI −UF N + e−UF N − eUF N ] sinh(UF N +UBI −U ) − sinh(UF N ) For Electric field E , we have d2 V 1 = −ρ 2 dx KS ǫ0 2 U kT d 1 ρ ( ) = −qni ( ) 2 dx q KS ǫ0 qni d2 U 1 1 ρ1 ρ1 ⇒ = − K ǫ0 kT ( )( )=− ( )( ) 2 2 2 qn S dx 2 qni (LD ) i 2q 2 n i 1 Multiplying both sides by 2 dU , we have, dx dU d2 U )( ) dx dx2 x d dU ( )2 ⇒ ∞ dx dx dU 2 dU 2 ( ) |x − ( ) |∞ dx dx E2 −0 (kT )2 /q 2 2( With equation(6), − 1 qni x d dU 2 2 dU ρ 1 ( ) =− ( )( ) dx dx (LD )2 dx 2 qni x dU 1 1 1 =− dx = − ρ 2 (LD ) ∞ dx qni (LD )2 xρ 1 =− dU 2 (LD ) ∞ qni xρ 1 =− dU (LD )2 ∞ qni = x ∞ ρ dU qni (7) ρdU ∞ =− . . . 1 U (x) U (x) qni [{eUF N −U +UBI − eU −UBI −UF N }U (∞) + {e−UF N − eUF N }U (∞) ] qni = eUF N [e−(U −U BI ) + (U − UBI ) − 1] + e−UF N [e(U −UBI ) − (U − UBI ) − 1] = F 2 (U − UBI , UF N ) Substitute the above into (7), we have E 2 (x) = ( or for dV dx kT 2 1 ) F 2 (U − UBI , UF N ) q (LD )2 kT 1 E (x) = −( ) F (U − UBI , UF N ) q LD = −E . 2 ...
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This note was uploaded on 12/12/2010 for the course ECE 606 taught by Professor Staff during the Spring '08 term at Purdue.

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

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