Final Exam
August 9, 2002
Assume Jn = 00 = q n un (-dV/dx) + q Dn (dn/dx)
Dn = Vt un
un n dV = un Vt dn
dV/Vt = dn/n
Integrate
[V(wn) - V(0)]/Vt = ln [n(wn)/n(0)]
Van = V(0) - V(wn) = Vt ln [ ND + Pn(0)/ND]
True, since the electron concentration at the contact, n(wn), is ND and,
at the edge of the depletion region, n(0) = ND + pn(0)n(0) = ND + pn(0) is a statment of charge neutrality. Negative charge, n(0), equals the positive charge, ND + pn(0).
If pn(0) equals ND, Van is significant since, Van = Vt ln(2) = 18mV. 18mV of the applied voltage appears across the neutral region rather than across the junction. The resulting current is 1/2 what it would be if this fraction of the applied voltage had appeared across the junction.