Page 191 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers

Statistics
doping with total donor concentration N
D
N
n +
tion N . In this case (5.44) turns into n = , and zero acceptor concentra-
.
A e h D +
In order to calculate N we must know the fraction of electron occupy-
D +
ing the donor levels, which is given by
N D
n = ----------------------------------------- (5.45)
D E – µ
D
1 + exp  ---------------- 
 k T 
B
(5.45) is calculated by applying integration in energy space of a Fermi-
Dirac distribution (5.36) inside the band-gap. Usually there are no energy
levels present. Due to doping we obtain at E a density of states given
D
(
by DE() = δ E – E ) , where the delta function accounts for the fact
D
that only the energy level E in the band-gap can be occupied. This
D
gives N + = N – n .
D D D
We are left with the task to calculate n . We do not know the chemical
h
potential, but we know from (5.39a) and (5.39b) that, wherever it comes
to lie
 – E G  2
n n = D D exp  ---------- = n i (5.46)
k T 
C
V
e h
B
2
holds. Thus we use n = n ⁄ n and obtain a second order equation for
h i e
n reading
e
2
2
n – n N D + – n = 0 (5.47)
e
i
e
Assuming that at the given temperature all donor atoms are ionized
N = N we obtain
D D +
N  4n 
2
i
D

n = ------- 1 + 1 + -------- (5.48)
e
2  N 
2
D
The value in (5.48) must equal that in (5.39a). This gives us the chemical
potential
188 Semiconductors for Micro and Nanosystem Technology