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We have the general equation for the
space
charge r(x), and
the Poisson equation: |
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| r(x) |
= |
e · {nh(x) –
ne(x) + N+D(x)
– N–A(x)} |
|
| e e0 |
· |
d2V(x)
dx2 |
= |
– r(x) |
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| |
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V(x) is the built-in potential resulting
from the flow of majority carriers to the other side. |
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We consider a solution for the following
conventions and approximations: |
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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. |
|
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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. |
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For the carrier density we have the general
expression |
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| ne,h |
= |
Neffe,h · exp – |
DE
kT |
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and DE was
ED – EF for electrons and
E – EA for holes. |
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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. |
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The carrier concentration than becomes |
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| n |
= |
Neff · exp – |
DE + eV(x)
kT |
= |
Neff · exp – |
DE
kT |
· exp – |
eV(x)
kT |
| |
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|
= |
NA,D · exp – |
e · V(x)
kT |
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because the first term gives the concentration for
V(x) = 0 and that is the dopant concentration in our
approximation. |
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We thus have for the carrier concentrations in equilibrium
anywhere in the junction: |
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|
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| nh(x) |
= |
NA · exp – |
e · V(x)
kT |
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| |
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| ne(x) |
= |
ND · exp – |
e · {V(x) - V(n}
kT |
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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 |
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| nh |
= |
NA |
|
for |
|
x < – dA |
| |
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| nh |
= |
0 |
|
for |
|
x > –
dA |
| |
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| ne |
= |
ND |
|
for |
|
x > d |
| |
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| ne |
= |
0 |
|
for |
|
x < d |
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With dA, dD
= boundaries of the space charge region with x = 0 at the
geometrical junction |
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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 |
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| r |
= |
NA |
|
for |
|
– dA < x
< 0 |
| |
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| r |
= |
ND |
|
for |
|
0 < x <
dN |
| |
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| r |
= |
0 |
|
for |
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everywhere else |
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The Poisson equation then becomes |
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d2V
dx2 |
= |
0 |
|
|
for |
|
– ¥ <
x < –dA |
| |
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|
d2V
dx2 |
= |
+ |
e
ee0 |
NA |
|
for |
|
– dA < x
< 0 |
| |
|
|
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|
d2V
dx2 |
= |
– |
e
ee0 |
ND |
|
for |
|
0 < x <
dD |
| |
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|
d2V
dx2 |
= |
0 |
|
|
for |
|
dD < x <
¥ |
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In addition we have the boundary conditions: |
|
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| V |
= |
0 |
ö
÷
÷
÷
ø |
for |
|
x = – dA |
| |
|
|
|
|
dV
dx |
= |
0 |
|
| |
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| V |
= |
V(N) |
ö
÷
÷
÷
ø |
for |
|
x = dD |
| |
|
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|
|
dV
dx |
= |
0 |
|
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| dA · NA |
= |
dD · ND |
|
Charge neutrality |
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The solutions are easily obtained, they are |
|
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|
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| VA(x) |
= |
e
2ee0 |
· NA · (dA
+ x)2 |
|
|
for – dA < x
< 0 |
| VD(x) |
= |
V(n) – |
e
2ee0 |
· NA · (dD –
x)2 |
|
for
0
< x < dD |
| V(n) |
= |
e
2ee0 |
(NA · dA2 +
ND · dD2) |
|
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|
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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). |
|
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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 |
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If we superimpose an external voltage U,
V(n) becomes (watch out for the correct sign!) . |
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| V(n) |
= |
E nF – E
pF
e |
± eU |
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The following illustration shows
the whole situation in one drawing. |
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© H. Föll (Semiconductor - Script)
2.2.4 Simple Junctions and Devices 
To index