
What we would need first, are defined boundary conditions so we can tackle the
differential equation. There are no obvious candidates, so we have to think a
little harder now. 


Accumulation means that we have some surface charge r
that we put on the surface of the semiconductor (with our
fictive thin insulating
layer in between). 


We thus need to refomulate the differential equation so that
surface charge can be included. the (not
overly obvious) way to do this is to introduce the electrical field strength E(x) as a new variable besides
DE_{C}. 


For that we use the relation 


d^{2}DE_{C}
dx^{2} 
= 
d
dx 
æ
ç
è 
dDE_{C}
dx 
ö
÷
ø 
= 
d
dx 
[e · E(x)] 
= e · 
dE(x)
dDE_{C} 
· 
dDE_{C}
dx 
= e^{2} · E(x) · 
dE(x)
dDE_{C} 




We also made use of the equaltity dDE_{C}/dx = e ·
E(x) with E(x) = field strength. 

Inserting and separating the variables (and
omitting the "(x)" for clarity) gives 


e^{2} · E · dE 
= 
kT
L^{2}_{D} 
· exp – 
DE_{C}
kT 
· dDE_{C} 




Tricky, but worth it. Now we can integrate both sides. The
integrations run from far inside the bulk , i.e. from E = 0, to some value of E, and that means from dDE_{C} = 0 to some corresponding value
dDE_{C}. 

Omitting the integration (which is trivial), we
obtain 


E^{2}
2 
= 
æ
ç
è 
kT
e · L_{D} 
ö
÷
ø 
2 
· 
æ
ç
è 
exp – 
DE_{C}
kT 
– 1 
ö
÷
ø 




i.e. an equation relating the amount of band bending at some
position x to the electrical field strenght at this point, which
is 


E(x) 
= ± 
kT
e · L_{D} 
æ
ç
è 
2exp – 
DE_{C}(x)
kT 
– 1 
ö
÷
ø 
1/2 




For ntype semiconductors, which we are considering,
DE_{C} is negative and large
(i.e. DE_{C} >>
kT)  and we may neglect the – 1, obtaining 


E(x) 
» ± 
kT
e · L_{D} 
· 
æ
ç
è 
2exp – 
DE_{C}(x)
kT 
ö
÷
ø 
1/2 



While this is fine, we still don't have the
solution we want. We must now remember that there is a simple relation tying surface charge to volume
charge. 


This is Gauss law,
stating that the flux of the electrical field through a surface S is the
integral over the components of E
perpendicular to the surface. 


The charge is usually expressed in terms of charge density
r(x,y,z). Gauss law then states: 


ó
õ

ó
õ

E ·
n · da 
= 
1_{ }
ee 
· 
ó
õ

ó
õ

ó
õ

r(x,y,z) · dV 
S 
V 




With n = normal vector of the surface S,
da = surface increment, dV = volume increment. For
more
details use the link. 

For our case it means that we could replace the
total charge r contained in a slice between x =
¥ (where there is no charge and the field
strength is E = E_{ bulk} = 0) and x,
by a surface (or better areal) charge s_{
area}(x) at x given by 


s_{ area}(x) 
= 
ee_{0}
· (E(x) –
E_{ bulk}) 
= 
ee_{0} ·
E(x) 
s_{ area}(x) 
= ± 
ee · kT
e · L_{D} 
· 
æ
ç
è 
2exp – 
DE_{C}(x)
kT 
ö
÷
ø 
1/2 



The total
amount of bandbending induced by a real external
surface charge s_{ex} is
simply DE_{C}(x = 0)
which we call DE_{C}^{0}: 


DE_{C}^{0} 
= ± 2kT · ln 
s_{ex} · e ·
L_{D}
2^{1/2} · ee_{0} · kT 




So we have all we need. The +/ sign came from the two
solutions of the square root; we have to pick the correct one depending on the
situation (holes or electrons considered). 


