Page 149 - Electrical Properties of Materials

Extrinsic semiconductors 131

recollect that the N D donor electrons live at the level E D (at 0 K) all ready and
willing to become conduction electrons if somehow they can acquire (E g –E D )
+
joules of energy. N is hence a measure of how many of them have gone.
D
Therefore, If we multiply N D by the probabil-
+
N = N D {1 – F(E D )}. (8.28) ity of an electron not being at E D
+
D
we should get N .
D
For acceptor atoms the argument is very similar. The probability of an
electron occupying a state at the energy level E A is F(E A ). Therefore the
number of ionized acceptor levels is
–
N = N A F(E A ). (8.29)
A
We are now ready to calculate the position of the Fermi level in any
semiconductor whose basic properties are known; that is, if we know N A and
N D , the energy gap, E g , and the effective masses of electrons and holes. Sub-
+
–
stituting for N e , N h , N , N from eqns (8.17), (8.20), (8.28), and (8.29) into
A D
eqn (8.27) we get an equation that can be solved for E F . It is a rather cumber-
some equation but can always be solved with the aid of a computer. Fortunately,
we seldom need to use all the terms, since, as mentioned above, the dominant
impurity usually swamps the others. For example, in an n-type semiconductor,
+
–
usually N e N h and N N and eqn (8.27) reduces to
D A
~
+
N e = N . (8.30)
D
This, of course, implies that all conduction electrons come from the donor
levels rather than from host lattice bonds. Substituting eqns (8.17) and (8.28)
into eqn (8.30) we get ∗ ∗ Equation (8.31) is not, however, valid
at the limit of no impurity, because holes
–1
E g – E F E F – E D cannot then be neglected; nor is it valid
N c exp – ~ . (8.31) when N D is very large, because some
= N D 1+exp
k B T k B T
of the approximations [e.g. eqn (8.11)]
For a particular semiconductor eqn (8.31) is easily solvable, and we may are then incorrect, and in any case many
impurity atoms getting close enough to
plot E F as a function of N D or of temperature. Let us first derive a formula for each other will create their own impur-
the simple case when (E F –E D )/k B T is a large negative number. Equation (8.31) ity band.
then reduces to
E F ~
(constant) exp = N D . (8.32) E F increases with the logarithm
k B T of N D .
So we have already learned that the position of the Fermi level moves upwards,
and varies rather slowly with impurity concentration.
Let us consider now a slightly more complicated situation where the above
approximation does not apply. Take silicon at room temperature and arsenic as
the dopant with the data
22 –3
E g =1.15 eV, E g – E D = 0.049 eV, N D =10 m . (8.33)
We may get the solution easily by introducing the notation
E F
x =exp , (8.34)
k B T
reducing thereby eqn (8.31) to the form

N D
Ax = , (8.35)
1+ Bx