Page 142 - Electrical Properties of Materials

124 Semiconductors

To be specific, let us think about silicon, although most of our remarks
will be qualitatively true of germanium and other semiconductors. Silicon has
the diamond crystalline structure; the four covalent bonds are symmetrically
arranged. All four valence electrons of each atom participate in the covalent
bonds, as we discussed before. But now, having learned band theory, we may
express the same fact in a different way. We may say that all the electrons are
in the valence band at 0 K. There is an energy gap of 1.1 eV above this before
the conduction band starts. Thus, to get an electron in a state in which it can
take up kinetic energy from an electric field and can contribute to an electric
current, we first have to give it a package of at least 1.1 eV of energy. This can
come from thermal excitation, or by photon excitation quite independently of
temperature.
Let us try to work out now the number of electrons likely to be free to
take part in conduction at a temperature T. How can we do this? We have
already solved this problem for the one-dimensional case: eqn (7.51) gives us
the effective number of electrons in a partly filled band; so all we need to do
is to include the Fermi function to take account of finite temperature and to
generalize the whole thing to three dimensions. It can be done, but it is a bit
too complicated. We shall do something else, which is less justifiable on strictly
theoretical grounds, but is physically much more attractive. It is really cheating
because we use only those concepts of band theory that suit us, and instead of
solving the problem honestly, we shall appeal to approximations and analogies.
It is a compromise solution that will lead us to easily manageable formulae.
First of all we shall say that the only electrons and holes that matter are
those near the bottom of the conduction band and the top of the valence band,
respectively. Thus, we may assume that

k x a, k y b, k z c 1, (8.1)

and we may expand the cosine term in eqn (7.34) to get the energy in the form,

1 2 2
1 2 2
1 2 2
E = E 1 –2A x 1– k a –2A y 1– k b –2A z 1– k c . (8.2)
2 x
2 y
2 z
Using our definition of effective mass, we can easily show from the above
equation that
2 2 2
∗
∗
∗
m = , m = , m = . (8.3)
z
x
y
2A z a 2 2A y b 2 2A z c 2
2
2
2
Substituting the values of A x a , A y b , and A z c from eqn (8.3) back into
eqn (8.2) and condensing the constant terms into a single symbol, E 0 ,wemay
now express the energy as
2
2 k 2 x k y k 2 z
E = E 0 + + + . (8.4)
2 m ∗ m ∗ m ∗
x y z
Taking further E 0 = 0, and assuming that everything is symmetric, that is
∗
∗
∗
∗
m = m = m = m , (8.5)
y
x
z