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3 / 2
i
1.12 Applying Equation (1.8.11) to this problem yields
n
2
/ p = n = 2 p → 2 p
2
= n
2
→ p =
1
n
= 7.07 10
12
cm
−3
and n = 1.4110
13
cm
−3
.
i i
2
i
1.13 (a) B is a group III element. When added to Si (which belongs to Group IV), it acts as an
acceptor producing a large number of holes. Hence, this becomes a P-type Si film.
(b) At T = 300 K, since the intrinsic carrier density is negligible compared to the dopant
concentration, p=Na=410
16
cm
-3
, and n = (ni=10
10
cm
-3
)
2
/p = 2500 cm
-3
.
At T = 600 K,
n
i
= N
c (600 K )N
v
(600 K )e
−(E
g )/(2kT )
= 1.16 10
15
cm
−3
where
2 m
dn
kT
3 / 2
19 T
3 / 2
−3
19 −3
N
c
(T = 600 K ) = 2
h
2
= 2.810
300K
cm = 7.92 10 cm
and
2 m kT
3 / 2
T
N (T = 600 K ) = 2
dp
= 1.0410
19
cm
−3
= 2.94 10
19
cm
−3
.
v
h
2
300K
The intrinsic carrier concentration is no more negligible compared to the dopant
concentration. Thus, we have
p = N
a
+ n
i
= (4 10
16
+1.16 10
15
)cm
−3
= 4.12 10
16
cm
−3
, and
n = n
2
/ p = (1.16 10
15
cm
−3
)
2
/ 4.1210
16
cm
−3
= 3.27 10
13
cm
−3
.
The electron concentration has increased by many orders of magnitude.
(c) At high temperatures, there is enough thermal energy to free more electrons from
silicon-silicon bonds, and consequently, the number of intrinsic carriers increases.
(d) Using Equation 1.8.8, we calculate the position of the Fermi level with respect to Ev.
E
f
− E
v
= kT lnN
v (T )/ p(T ) = 0.34 eV , T = 600 K .
At 600 K, the Fermi level is located 0.34 eV above the valence band.
Incomplete Ionization of Dopants and Freeze-out
1.14 From Equation (1.9.1), we know that n + Na
-
= p + Nd
+
. Since Nd
+
is much larger than Na
-
,
all the samples are n-type, and n Nd
+
- Na
-
= 310
15
/cm
3
. This value is assumed to be
constant. Using the Equations (1.8.10) and (1.9.3b),