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


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




The first equation than simplifies to. 


n^{p}_{e}(U) 
÷
÷ 
edge
SCR 
= 
n^{n}_{e}(U) 
÷
÷ 
edge
SCR 
· exp – 
eV^{n}
kT 
= 
n^{n}_{e}(U) 
÷
÷ 
edge
SCR 
· exp – 
DE_{F}
kT 




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


n^{p}_{h}(U = 0) 
= N ^{p}_{eff} · exp – 
E_{F} – E ^{p}_{V}
kT 
e·V^{n} 
= 
Difference of
band edges 
= 
E ^{p}_{V} – E
^{n}_{V} = E ^{p}_{C}
– E ^{n}_{C} = DE_{F} 




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
E_{F}, of course, is constant in equilibrium, and
DE_{F} thus refers to the
difference in Fermi energies before the
contact! 




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


This brings you already to the nside.
However, you want to find n^{n}_{e} in the
equation and for that you need a factor E^{n}_{C} –
E_{F}. So express E^{n}_{V} in
terms of E^{n}_{C} via 


E^{n}_{V}=
E_{g} – E^{n}_{C} with
E_{g} = band gap. This yields 


n ^{p}_{h}(U = 0) 
= N ^{p}_{eff} · exp – 
E_{F} – E ^{n}_{C} +
E_{g} – DE_{F}
kT 



You now have terms that occur in the
definition of the electron concentration in nSi
(E_{F} – E ^{n}_{C} = –
(E ^{n}_{C} – E_{F}) and for the
intrinsic carrier concentration (E_{g}). 


So multiply with N
^{n}_{eff} /N ^{n}_{eff} , remember
that n_{i}^{2} = N ^{p}_{eff}
·N^{n}_{eff} · exp
–E_{g}/kT, and 1/n^{n}_{e} =
1/N^{n}_{eff} ·
exp(E^{n}_{C} –
E_{F})/kT, and you have 


n ^{p}_{h}(U = 0) 
= 
n_{i}^{2}
n^{n}_{e} 
· exp 
DE_{F}
kT 



This gives for
n^{n}_{e}: 


n^{n}_{e}(U = 0) 
= 
n_{i}^{2}n^{p}_{h} 
· exp 
DE_{F}
kT 




We now can substitute
n^{n}_{e} in our first equation and obtain 


n^{p}_{e} 
÷
÷ 
edge
SCR 
= 
n_{i}^{2}
n^{p}_{h} 
· exp – 
DE_{F}
kT 
· exp 
DE_{F}
kT 
n^{p}_{e} 
= 
n_{i}^{2}
n^{p}_{h} 



That is exactly the
second equation 
Q.E.D 