Elements of Modern Physics
278
The knowledge of ε f allows us to calculate the number of carrier electrons and
holes, and the conductivity of intrinsic semiconductors. Substituting for ε f in
Eqs. (8.36) and (8.40), gives
n e = n h =
3/ 2
3/ 4
3
2(2
) ( * *) exp [(
) / 2 ]
π
ε − ε
e
h
v
c
kT
m m
kT
h
(8.44)
For deducing conductivity, it is noted that the current density j is given by
j = (
)
+
e e
h h
e n v n v
(8.45)
where e
v and h
v are the magnitudes of the average drift velocities of the electrons
and holes, respectively. The conductivity of the semiconductor is therefore given
by
σ = e(n e µ e + n h µ h )
(8.46)
where the mobilities µ e and µ h are defined by
µ e = / ,
/
µ =
e
h
h
v E
v E
(8.47)
E being the magnitude of the electric field. Substituting the expressions for
the carrier densities,
σ = e(µ e + µ h )
3/ 2
3/ 4
3
2(2
) ( * *) exp [(
) / 2 ]
π
ε − ε
e
h
v
c
kT
m m
kT
h
(8.48)
which leads to
ln σ =
1 3 ln
2
2
ε − ε


−
+
+




c
v
T c
k
T
(8.49)
where c is a constant. Here, it has been assumed that the mobilities are
independent of T. Actually, they do vary as a function of temperature. However,
the main variation is due to the 1/T term and a plot of ln σ as a function of 1/T
gives an approximate straight line. The slope of the straight line gives an
estimation of the energy gap (ε c – ε v ) of the semiconductor.
It may be noted m e * ≈ 0.25 m, m h * ≈ 0.3 m for Si and m e * ≈ m h * ≈ 0.1 m for
Ge, m being the electron mass. These values imply that at a temperature of
300 K, the carrier concentrations are about 2.3 × 10
15
m
–3
for Si and about 10
18
m
–3
for Ge. The intrinsic conductivity at this temperature has the values of about
10
–4
(Ω m)
–1
for Si and about 0.1 (Ω m)
–1
for Ge.
ε
ε ε
ε ε f for Extrinsic Semiconductors
In an extrinsic semiconductor, the donor or acceptor levels play an important
role in the determination of the Fermi energy and the conductivity of the
semiconductor.