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Unformatted text preview: tor Devices Sang-Hyun Oh 17 UNIVERSITY OF MINNESOTA Potential via Poisson dV ρ(x) =− 2 dx K 0
2 Is V(xn) positive or negative? −qNa , −xp < x < 0 ρ(x) = qNd , 0 < x < xn
EE3161 Semiconductor Devices Sang-Hyun Oh V (−xp ) = 0, V (xn ) = Vbi 18 UNIVERSITY OF MINNESOTA Barrier Height
This only depends on the impurity concentration on either side of the junction and not on how it changes from one side to the other. On either side of the junctions:
nn = ni e
(EF −Ei,n )/kT pp = ni e(Ei,p −EF )/kT ND NA = n2 eqVbi /kT i nn pp = n2 e(Ei,p −Ei,n )/kT i kT Vbi = ln q
EE3161 Semiconductor Devices Sang-Hyun Oh ND NA n2 i 19 UNIVERSITY OF MINNESOTA Built-in Potential Vbi: Solve Poisson
dV (x) E (x) = − dx
Definition Vbi V (xn ) V (−xp ) dV = − JN xn −xp E (x)dx = V (xn ) − V (−xp ) ≡ Vbi Under equlibrium: dn = qµn nE (x) + qDN =0 dx
Eliminate D and µ DN dn/dx kT dn/dx E (x) = − =− µn n q n
EE3161 Semiconductor Devices Sang-Hyun Oh 20 UNIVERSITY OF MINNESOTA Built-in Potential Vbi: Solve Poisson
kT Vbi = − E (x)dx = q −xp
express using E-field & integral xn n(xn ) n(−xp ) dn kT = ln n q n(xn ) n(−xp ) express using n(xn) and n(-xp) For the case of step junction with ND and NA for n- and p-side doping: n(xn...
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This note was uploaded on 02/24/2010 for the course EE 3161 taught by Professor Prof.sang-hyunoh during the Spring '10 term at University of Minnesota Crookston.
- Spring '10