
We have the general equation for the space charge
r(x ), and the Poisson equation: 
 
r(x)  = 
e · {n^{h}(x) – n^{e}(x ) + N^{+}
_{D}(x) – N^{–}_{A}(x )} 

e e_{0}  · 
d^{2}V(x) dx^{2}  = 
– r (x) 




V(x) is the builtin 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 pside of the junction shown in the illustration. 


All dopants are ionized, i.e. N_{A} = N^{–} _{ A}
= n^{h}, and N_{D} = N^{+}_{ D}
= n^{e}. 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 
 
n^{e,h}  = 
N_{ eff}^{e,h} · exp – 
D E kT 




and D E was E_{D} – E_{F}
for electrons and E – E_{A} for holes. 


If E_{D, A} is a function of x because the bands are bent (while E_{F}
stays constant), we may write the energy difference as DE = DE^{0}
+ e · V(x) with DE = D
E^{0} referring to the situation without band bending. 


The carrier concentration than becomes 
 
n_{ }  =_{ } 
N_{eff} · exp – 
DE + eV( x) kT
 =  N _{eff} · exp – 
DE kT 
· exp – 
eV(x) kT 
  
  
  
 =_{ } 
N_{A,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: 
 
n^{h} (x)  = 
N_{A} · exp – 
e · V(x) kT 
  
 
n^{e}(x)  = 
N_{D} · 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 N_{A} or N_{D} far outside of the SCR. We therefore
approximate their concentration by 
 
n^{h}  = 
N_{A}   for 

x < – d_{A} 
  
 
 
n^{ h}  =  0 
 for  
x > – d_{A} 
  
 
 
n^{e}  = 
N_{D}   for
 
x > d  
  
 
 n^{e} 
=  0  
for  
x < d 




With d_{A}, d_{D} = 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  = 
N _{A}  
for  
– d_{A} < x < 0 
  
  
 r 
=  N_{D} 
 for  
0 < x < d_{N} 
  
  
 r 
=  0  
for  
everywhere else 



The Poisson equation then becomes 


d^{2}V dx^{2} 
=  0 
  for 

– ¥ < x < –d_{
A}   
 
 
 
d^{2}V dx^{ 2} 
=  + 
e_{ }
ee_{0} 
N_{A}   for 

– d_{ A} < x < 0 
 
 
  

d^{2}V dx^{2} 
=  – 
e_{ }
e e_{0} 
N_{D}   for 

0 < x < d_{D} 
 
 
 
 
d^{2}V dx^{2} 
=  0 
  for 

d_{D} < x < ¥





In addition we have the boundary conditions: 


V  =  0

ö ÷ ÷ ÷ ø 
for  
x = – d_{A} 
  
 
dV dx  = 
0  
  
 
 
V  = 
V( N) 
ö ÷ ÷ ÷ ø 
for  
x = d_{D} 
  
 
dV dx  = 
0  
  
 
 
d_{ A} · N_{A} 
=  d_{D} · N_{ D} 

Charge neutrality 



The solutions are easily obtained, they are 


V_{A}(x)  = 
e_{ } 2ee_{0}

· N_{A} · (d_{A} + x)^{2}
  
for – d_{A }
< x < 0 
V_{D}(x)  = 
V(n) –  e_{ }
2 ee_{0} 
· N_{A} · (d_{D} – x )^{2} 

for 0
< x < d_{D} 
V(n) 
= 
e 2e e_{0} 
(N_{A} · d_{A}^{2} + N_{D} ·
d_{D}^{2}) 




The last equation comes from the condition of continuity at x = 0, i.e. V_{D}(x
= 0) = V_{A}(x = 0. 

The two limits of the space charge region, d_{A} and d_{D}
, 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 ^{n}_{F} – E
^{p}_{ F} e_{ } 




If we superimpose an external voltage U, V(n) becomes (watch out for
the correct sign!) . 
 
V(n )  = 
E ^{n}_{F} – E
^{p}_{F} e_{ } 
± eU_{ } 



The following illustration shows the whole situation in one drawing. 
 
