21.2 Metal-Semiconductor Contacts
531
The integral starts at the lowest possible energy, the top of the Schottky barrier (no tunneling allowed
in this model!). The electron density dn in a small energy interval dE is
dn = D(E) f (E) dE .
(21.37)
For a bulk semiconductor density of states (6.71) and the Boltzmann distribution (F B kT )
dn =
1
2π 2
2m
∗
2
3/2
E − E C exp
−
E − E F
kT
dE .
(21.38)
For a given energy E, the carrier velocity v is determined by
E = E C +
m
∗ v
2
2
.
(21.39)
Thus we obtain
E − E C = v
m ∗ /2
(21.40)
and
dE = m
∗ v dv .
(21.41)
Also, with (21.39) and (21.1),
E − E F = (E − E C ) + (E C − E F ) =
m
∗ v
2
2
− e V n .
(21.42)
Therefore we write (21.38) as
dn = 2
m
∗
h
3
exp (βV n ) exp
−
m
∗ v
2
2 kT
4π v
2 dv .
(21.43)
Then the one-dimensional integral over 4π v
2 dv is converted into a three-fold integral over dv x dv y dv z .
Integration over all velocities in y and z directions yields a factor 2π kT /m
∗ . The integration over v x
runs from the minimum velocity v min,x necessary to pass the barrier,
∞
v min,x
exp
−
m
∗ v
2
x
2 kT
v x dv x =
kT
m ∗ exp
−
m
∗ v
2
min,x
2 kT
,
(21.44)
with the minimum velocity determined by
m
∗ v
2
min,x
2
= e (V bi − V ) .
(21.45)
Thus the current density is given as, using (21.5),
j s→m =
4π e m
∗ k
2
h 3
T
2 exp (−β(V bi − V n )) exp (βV )
= A
∗ T
2 exp
−
F Bn
kT
exp (βV ) ,
(21.46)