Page 148 - Electrical Properties of Materials

130 Semiconductors

In the present case, the third interpretation is the most convenient one—to
start with. A group III atom has three valence electrons. So when it replaces
a silicon atom at a certain atomic site, it will try to contribute to bonding as
much as it can. However, it possesses only three electrons for four bonds. Any
electron wandering around would thus be welcomed to help out. More aggress-
ive impurity nuclei might even consider stealing an electron from the next site.
The essential point is that a low energy state is available for electrons—not
as low an energy as at the host atom but low enough to come into consideration
when an electron has acquired some extra energy and feels an urge to jump
somewhere. Therefore (changing now to band theory parlance), the energy
levels due to group III impurities must be just above the valence band. Since
these atoms accept electrons so willingly, they are called acceptors and the
corresponding energy levels are referred to as acceptor levels.
A real material will usually have both electron donors and acceptors present
(not necessarily group V and group III elements; these were chosen for simpli-
city of discussion and because they are most often used in practice). However,
usually one type of impurity exceeds the other, and we can talk of impurity
semiconductors as n- (negative carrier) or p- (positive carrier) types accord-
ing to whether the dominant charge carriers are electrons or holes. If we had
some silicon with 10 20 atoms per cubic metre of trivalent indium, it would be
a p-type semiconductor. If we were somehow to mix in 10 21 atoms per cubic
metre of pentavalent phosphorus, the spare phosphorus electrons would not
only get to the conduction band but would also populate the acceptor levels,
thus obliterating the p characteristics of the silicon and turning it into an n-type
semiconductor.
Let us calculate now the number of electrons and holes for an extrinsic semi-
conductor. We have in fact already derived formulae for them, as witnessed by
eqns (8.17) and (8.20); but they were for intrinsic semiconductors. Did we
make any specific use of the fact that we were considering intrinsic semicon-
ductors? Perhaps not. We said that only electrons at the bottom of the band mat-
ter, and we also said that the bottom of the band is many times k B T away from
the Fermi level in energy but this could all be equally valid for extrinsic semi-
conductors. It turns out that these approximations are valid, apart from certain
exceptional cases (we shall meet one exception when we consider devices).
But how can we determine the Fermi level? It is certainly more difficult
for an extrinsic semiconductor. We have to consider now all the donors and
acceptors. The condition is that the crystal must be electrically neutral, that is,
the net charge density must be zero. Let us consider what sort of charges we
may meet in an extrinsic semiconductor. There are our old friends, electrons
and holes; then there are the impurity atoms that donated an electron to the
conduction band, and are left with a positive charge; and finally there are the
N – is the number of ionized ac- acceptor atoms that accepted an electron from the valence band and thus have
A
ceptor atoms, which accepted an a negative charge. Hence the formula for overall charge neutrality is
electron from the valence band, – +
+
and N is the number of ionized N e + N N h + N . (8.27)
A
D
D
donor atoms, which donated an We have written eqn (8.27) with the sign used by chemists to show that
electron to the conduction band. it is a dynamic equilibrium, rather than a once and for all equation.
+
–
How can we find the number of ionized impurities, N and N ,fromthe
A D
actual number of impurity atoms N A and N D ? Looking at Fig. 8.3(b) you may