
We want to show that the following two equations are equivalent for equilibrium: 


n_{e}^{ p}(U ) 
÷ ÷ 
SCR edge  = 
n_{e}^{n} (U) 
÷ ÷ 
SCR edge 
· exp – 
e(V^{n} + U) kT 
n _{e} ^{p}(U = 0)  =
 n_{i}^{2}
n_{h}^{p} (U = 0) 




The first equation then simplifies to 
 
n_{e}^{p}(U) 
÷ ÷ 
SCR edge  = 
n_{e}^{n}(U) 
÷ ÷ 
SCR edge 
· exp –  eV^{n}
kT  = 
n_{e}^{n}(U ) 
÷ ÷ 
SCR edge 
· exp – 
DE_{F} kT 




Start with the equation for the majority carrier concentration n_{h}^{p}(U
= 0) in general and the definitions of the energies: 


n_{h}^{p}(U = 0) 
= N_{eff} ^{p} · exp – 
E_{F} – E_{V} ^{p}
kT 
e·V^{n}  = 
Difference of band edges  = 
E_{V}^{p} – E_{V}^{ n} = E_{C}^{p}
– E_{C}^{n} = 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 E_{F}
, of course, is constant in equilibrium, and DE_{F}
thus refers to the difference in Fermi energies before the contact ! 
 


E
_{V}^{p} thus can be expressed as E_{V}
^{p} = E_{V}^{n} + DE_{F} . 


This brings you already to the nside. However, you want to find n_{e}^{n}
in the equation, and for that you need a factor E_{C}^{n} – E_{F}.



So, express E_{V} ^{n} in terms of E
_{C}^{n} via E_{V}^{ n} = E_{C}^{n} – E_{g}
with E_{g} = band gap. This yields 


n_{h}^{p}(U = 0) 
= N_{eff} ^{p} · exp – 
E_{F} – E_{ C}^{n} + E_{g} – D
E_{F} kT 



You now have terms that occur in the definition of the electron density in nSi
[namely, E_{F} – E_{C}^{ n} = – (E_{C}
^{n} – E_{F})] and for the intrinsic carrier density (namely, E_{g}).



So, multiply with N_{eff} ^{n} / N_{eff}
^{n}, remember that n_{i}^{2} = N_{eff}^{p} · N_{eff}^{n}
· exp – E_{g}/(kT), and 1/n_{e}^{n} = 1/N_{ eff}^{n}
· exp[(E _{C}^{n} – E_{F})/(k T)]; thus, you have 
 
n_{h}^{p}(U = 0)  =
 n_{i}^{2} n_{e}^{ n}
 · exp 
DE_{F} kT 



This gives for n_{e}^{n}: 
 
n_{e}^{n}(U = 0)  =
 n_{i}^{2} n_{h}^{p}
 · exp 
D E_{F} kT 




We now can substitute n_{e}^{n} in our first
equation and obtain 
 
n_{e} ^{p} 
÷ ÷
÷  SCR edge 
=  n _{i}^{2}
n_{h}^{ p}  · exp 
DE_{F} kT 
· exp – 
DE_{F} kT 
Þ 
n_{e}^{p}  = 
n_{i}^{2} n_{h}^{ p} 



That is exactly the second equation – Q.E.D. 