Page 143 - Electrical Properties of Materials

Intrinsic semiconductors 125

we get
This formula is identical to
2 2 2 2 eqn (6.2) obtained from the
E = k + k + k z . (8.6)
y
x
2m ∗ free-electron model—well, nearly
identical.
The mass in the denominator is not the real mass of an electron but the
effective mass. But that is the only difference between eqn (8.6) and the
free-electron model. Thus, we are going to claim that electrons in the conduc-
tion band have a different mass but apart from that behave in the same way as
free electrons. Hence the formula derived for the density of states [eqn (6.10)]
is also valid, and we can use the same method to determine the Fermi level.
So we shall have the total number of electrons by integrating ... Wait, we for-
got about holes. How do we include them? Well, if holes are the same sort of
things as electrons apart from having a positive charge, then everything we said
about electrons in the conduction band should be true for holes in the valence
band. The only difference is that the density of states must increase downwards
for holes.
Choosing now the zero of energy at the top of the valence band, we may
write the density of states in the form
∗ 3/2
Z(E)= C e (E – E g ) 1/2 , C e =4π(2m ) /h 3 (8.7)
e
for electrons, and
∗ 3/2
Z(E)= C h (–E) 1/2 , C h =4π(2m ) /h 3 (8.8)
h
for holes, both of them per unit volume. This is shown in Fig. 8.1, where E
is plotted against Z(E). You realize of course that the density of states has E
meaning only in the allowed energy band and must be identically zero in the
gap between the two bands.
Let us return now to the total number of electrons. To obtain that we must
take the density of states, multiply by the probability of occupation (getting
thereby the total number of occupied states), and integrate from the bottom to E g
the top of the conduction band. So, formally, we have to solve the following Range of
integral: forbidden
energies
top of conduction band
0
N e = (density of states)(Fermi function) dE. (8.9) Z(E)
bottom of conduction band
There are several difficulties with this integral:
Fig. 8.1
1. Our solution for the density of states is valid only at the bottom of the band, Density of states plotted as a function
2. The Fermi function of energy for the bottom of the

–1 conduction (electrons) and top of the
E – E F
F(E)= 1+exp (8.10) valence (holes) bands. See eqns (8.7)
k B T and (8.8).
is not particularly suitable for analytical integration.
3. We would need one more parameter in order to include the width of the
conduction band.
We are saved from all these difficulties by the fact that the Fermi level lies
in the forbidden band, and in practically all cases of interest its distance from