
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 – 
DE
kT 




and DE 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 =
DE^{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_{ }
ee_{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_{ }
2ee_{0} 
· N_{A} · (d_{D} –
x)^{2} 

for
0
< x < d_{D} 
V(n) 
= 
e
2ee_{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. 


