# Solution to Exercise 2.3.5-1

We want to show that the following two equations are equivalent for equilibrium:
ne p(U ) SCR edge ÷ ÷ ÷ ÷

n e p(U = 0)  =  ni2
nhp (U = 0)
The first equation then simplifies to
nep(U) SCR edge  = ÷ ÷ ÷ ÷ ÷ ÷

Start with the equation for the majority carrier concentration nhp(U = 0) in general and the definitions of the energies:
nhp(U = 0)  =  Neff p · exp – EFEV p
kT

Vn  =  Difference of
band edges
=  EVp   –  EV n  =  ECp   –  ECn   =  D E F
Consult the solution to the Poisson equation if you are unsure (the relevant diagram is reprinted below) and recall that in the band diagram, the energy scale refers to electrons, which carry a negative electric charge – so that the electrostatic potential contributes with a negative sign.
Also note that EF , of course, is constant in equilibrium, and DEF thus refers to the difference in Fermi energies before the contact !
E Vp thus can be expressed as EV p = EVn + DEF .
This brings you already to the n-side. However, you want to find nen in the equation, and for that you need a factor ECnEF.
So, express EV n in terms of E Cn via EV n = ECnEg with Eg = band gap. This yields
nhp(U = 0)  =  Neff p · exp – EFE Cn + EgD EF
kT
You now have terms that occur in the definition of the electron density in n-Si [namely, EFEC n = – (EC nEF)] and for the intrinsic carrier density (namely, Eg).
So, multiply with Neff n / Neff n, remember that ni2 = Neffp · Neffn · exp – Eg/(kT), and 1/nen = 1/N effn · exp[(E CnEF)/(k T)]; thus, you have
nhp(U = 0)  =  ni2
ne n
· exp  DEF
kT
This gives for nen:
nen(U = 0)  =  ni2
nhp
· exp D EF
kT
We now can substitute nen in our first equation and obtain
 ne p ÷ ÷ ÷

Þ     nep  =  ni2
nh p
That is exactly the second equationQ.E.D.

Exercise 2.1-1 Free Electron Gas with Constant Boundary Conditions

Solving the Poisson Equation for p-n-Junctions

Exercise 2.3.5-1

© H. Föll (Semiconductors - Script)