# Solution to Exercise 2.3.5-1

| We want to show that the following two equations are equivalent for equilibrium:
nep(U ) | ÷ ÷ | SCR edge | = | nen (U) | ÷ ÷ | SCR edge | · exp – | e(Vn + U) kT
n e p(U = 0) | = | ni2 nhp (U = 0)
--- | --- | ---
|  | The first equation then simplifies to
nep(U) | ÷ ÷ | SCR edge | = | nen(U) | ÷ ÷ | SCR edge | · exp – | eVn kT | = | nen(U ) | ÷ ÷ | SCR edge | · exp – | D EF kT
| 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 – | EF – EV p kT
--- | --- | ---
e·Vn | = | Difference of band edges | = | EVp – EVn = 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 D EF thus refers to the difference in Fermi energies before the contact !
| E Vp thus can be expressed as EV p = EVn + D EF .
|  | This brings you already to the n-side. However, you want to find nen in the equation, and for that you need a factor ECn – EF.
|  | So, express EV n in terms of E Cn via EV n = ECn – Eg with Eg = band gap. This yields
nhp(U = 0) | = Neff p · exp – | EF – ECn + Eg – D EF kT
--- | --- | ---
| You now have terms that occur in the definition of the electron density in n-Si [namely, EF – EC n = – (EC n – EF)] 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 Cn – EF)/(k T)]; thus, you have
nhp(U = 0) | = | ni2 nen | · exp | D EF 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 | ÷ ÷ ÷ | SCR edge | = | n i2 nhp | · exp | D EF kT | · exp – | D EF kT
Þ | nep | = | ni2 nhp
--- | --- | --- | ---
| That is exactly the second equation – Q.E.D.

With frame

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)