Solution to Exercise 2.3.5-1

W want to show that the following two equations are equivalent for equilibrium
npe(U) ÷
÷
edge
SCR
 =  nne(U) ÷
÷
edge
SCR
 · exp –   e(Vn + U)
kT

npe(U = 0)  =  ni2
nph(U = 0)
The first equation than simplifies to.
npe(U) ÷
÷
edge
SCR
 =  nne(U) ÷
÷
edge
SCR
 · exp –   eVn
kT
 =   nne(U) ÷
÷
edge
SCR
 · exp –   DEF
kT
 
Start with the equation for the majority carrier concentration nph(U = 0) in general and the definitions of the energies:
nph(U = 0)  = N peff · exp – EFE pV
kT  

Vn  =  Difference of
band edges
 =  E pV  –  E nV  =  E pC  –  E nC  =  DEF
Consult the solution to the Poisson equation if you are unsure (the relevant diagram is reprinted below) and recall that multiplying the potential V by the elementary charge e reverses the signs formally. Also note that EF, of course, is constant in equilibrium, and DEF thus refers to the difference in Fermi energies before the contact!
Junction
EpV thus can be expressed as E pV = E nV + DEF.
This brings you already to the n-side. However, you want to find nne in the equation and for that you need a factor EnC – EF. So express EnV in terms of EnC via
EnV= EgEnC with Eg = band gap. This yields
n ph(U = 0)  =  N peff · exp – EFE nC + EgDEF
kT
You now have terms that occur in the definition of the electron concentration in n-Si (EFE nC = – (E nCEF) and for the intrinsic carrier concentration (Eg).
So multiply with N neff /N neff , remember that ni2 = N peff ·Nneff · exp –Eg/kT, and 1/nne = 1/Nneff · exp(EnCEF)/kT, and you have
n ph(U = 0)  =  ni2
nne
 · exp  DEF
kT
This gives for nne:
nne(U = 0)  =  ni2nph  · exp DEF
kT
We now can substitute nne in our first equation and obtain
npe ÷
÷
edge
SCR
 =  ni2
nph
· exp – DEF
kT
· exp  DEF
kT

npe  =  ni2
nph
That is exactly the second equation - Q.E.D

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go to Solving the Poisson Equation for p-n-Junctions

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© H. Föll (Semiconductor - Script)