# Solving the Poisson Equation for pn-Junctions

We have the general equation for the space charge r(x), and the Poisson equation:
r(x)  =  e · {nh(x) – ne(x) + N+D(x) – NA(x)}

e e0  ·   d2V(x)
dx2
=  r(x)
V(x) is the built-in potential resulting from the flow of majority carriers to the other side.
We consider a solution for the following conventions and approximations:
The zero point of the electrostatic potential is identical to the valence band edge in the p-side of the junction shown in the illustration.
All dopants are ionized, i.e. NA = NA = nh, and ND = N+D = ne. This is always valid as long as the Fermi level is not very close to a band edge.
For the carrier density we have the general expression
ne,h  =  Neffe,h  · exp – DE
kT
and DE was EDEF for electrons and EEA for holes.
If ED, A is a function of x because the bands are bent (while EF stays constant), we may write the energy difference as DE = DE0 + e · V(x) with DE = DE0 referring to the situation without band bending.
The carrier concentration than becomes
n   =   Neff · exp –  DE + eV(x)
kT
=  Neff · exp – DE
kT
· exp – eV(x)
kT

=   NA,D · exp – e · V(x)
kT

because the first term gives the concentration for V(x) = 0 and that is the dopant concentration in our approximation.
We thus have for the carrier concentrations in equilibrium anywhere in the junction:
nh(x)  =  NA · exp  –   e · V(x)
kT

ne(x)  =  ND · exp –  e · {V(x) - V(n}
kT
As soon as V(x) deviates noticeably from its constant value of  0 or V(n) - in other words: inside the space charge region - the carrier concentrations decrease exponentially from their values NA or ND far outside of the SCR. We therefore approximate their concentration by
nh for for for for
With dA, dD = boundaries of the space charge region with x = 0 at the geometrical junction
The space charge then is only given by the concentration of the dopants. That's where we could have started right away, just plugging in the usual assumptions. We have
r for for for
The Poisson equation then becomes
d2V dx2 for for for for
In addition we have the boundary conditions:
V ö ÷ ÷ ÷ ø ö ÷ ÷ ÷ ø
The solutions are easily obtained, they are

VA(x)  =  e
2ee0
· NA · (dA + x)2     for     d <  x  <  0

VD(x)    =  V(n)  –   e
2ee0
· NA · (dD  –  x)2   for         0  <  x  <  dD
V(n)       =    e
2ee0
(NA · dA2  +  ND · dD2)

The last equation comes from the condition of continuity at x = 0, i.e. VD(x = 0)  =  VA(x = 0.
The two limits of the space charge region, dA and dD, as well as the field strength E = – dV/dx in the SCR thus could be calculated if we would know V(n).
V(n), of course, is the difference of the potential across the SCR and thus identical to 1/e times the difference of the Fermi energies before contact in thermal equilibrium, we have
V(n)  =  –   E nF  –  E pF
e
If we superimpose an external voltage U, V(n) becomes (watch out for the correct sign!) .
V(n)  =  E nF  –  E pF
e
± eU
The following illustration shows the whole situation in one drawing.
We now can express the width dSCR of the space charge region as
dSCR  =  dA  +  dB   1  2ee0 =   e · [DEF + e ·Uex] · æ ç è æ ç è ö ÷ ø ö ÷ ø
DEF refers to the the difference of the Fermi energies before the contact and Uex is the external voltage.

3.4.1 Junction Diodes

2.2.4 Simple Junctions and Devices

2.3.5. Junction Reconsidered

Basic Equations

Potential Discontinuities and Dipole Layers

Solution to 2.3.5-1

© H. Föll (Semiconductor - Script)