 |
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) – N–A(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 = N– A
= 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 | = |
N effe,h · exp – |
D E
kT |
|
|
|
|
and D E was ED – EF
for electrons and E – EA 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 = D
E0 referring to the situation without band bending. |
|
|
The carrier concentration than becomes |
| |
| n | = |
Neff · exp – |
DE + eV( x)
kT
| = | N eff · 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 | = |
NA | | for |
|
x < – dA |
| | | |
| |
| |
| n h | = | 0 |
| for | |
x > – dA |
| | | |
| |
| |
| ne | = |
ND | | for
| |
x > d | | |
| | |
| |
| | ne |
= | 0 | |
for | |
x < d |
|
|
|
|
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 | = |
N A | |
for | |
– dA < x < 0 |
| | | |
| | |
| | r |
= | ND |
| for | |
0 < x < dN |
| | | |
| | |
| | r |
= | 0 | |
for | |
everywhere else |
|
|
|
The Poisson equation then becomes |
|
|
d2V
dx2 |
= | 0 |
| | for |
|
– ¥ < x < –d
A | | | |
| |
| |
| |
d2V
dx 2 |
= | + |
e
ee0 |
NA | | for |
|
– d A < x < 0 |
| | |
| |
| | |
|
d2V
dx2 |
= | – |
e
e e0 |
ND | | for |
|
0 < x < dD |
| | |
| |
| |
| |
d2V
dx2 |
= | 0 |
| | for |
|
dD < x < ¥
|
|
|
|
|
In addition we have the boundary conditions: |
|
|
| V | = | 0
|
ö
÷
÷
÷
ø |
for | |
x = – dA |
| | | |
| |
dV
dx | = |
0 | |
| | | |
| |
| |
| V | = |
V( N) |
ö
÷
÷
÷
ø |
for | |
x = dD |
| | | |
| |
dV
dx | = |
0 | |
| | |
| |
| |
| d A · NA |
= | dD · N D |
|
Charge neutrality |
|
|
|
The solutions are easily obtained, they are |
|
|
| VA(x) | = |
e
2ee0
|
· NA · (dA + x)2
| | |
for – dA
< x < 0 |
| VD(x) | = |
V(n) – | e
2 ee0 |
· NA · (dD – x )2 |
|
for 0
< x < dD |
| V(n) |
= |
e
2e e0 |
(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 |
| |
|
|
|
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. |
| |
|
© H. Föll (Semiconductors - Script)
2.2.4 Simple Junctions and Devices 
With frame