CHAPTER 12
ELECTRICAL PROPERTIES
PROBLEM SOLUTIONS
12.11 For each of the following pairs of semiconductors, decide which will have the smaller band gap
energy, Eg, and then cite the reason for your choice. (a) C (diamond) and Ge, (b) AlP and InSb, (c) GaAs and ZnSe,
(d) ZnSe and CdTe, and (e) CdS and NaCl.
Solution
(a) Germanium will have a smaller band gap energy than C (diamond) since Ge is lower in row IVA of the
periodic table (Figure 2.6) than is C. In moving from top to bottom of the periodic table, Eg decreases.
(b) Indium antimonide will have a smaller band gap energy than aluminum phosphide. Both of these
semiconductors are III-V compounds, and the positions of both In and Sb are lower vertically in the periodic table
(Figure 2.6) than Al and P.
(c) Gallium arsenide will have a smaller band gap energy than zinc selenide. All four of these elements are
in the same row of the periodic table, but Zn and Se are more widely separated horizontally than Ga and As; as the
distance of separation increases, so does the band gap.
(d) Cadmium telluride will have a smaller band gap energy than zinc selenide. Both are II-VI compounds,
and Cd and Te are both lower vertically in the periodic table than Zn and Se.
(e) Cadmium sulfide will have a smaller band gap energy than sodium chloride since Na and Cl are much
more widely separated horizontally in the periodic table than are Cd and S.
Extrinsic Semiconduction
12.16 The following electrical characteristics have been determined for both intrinsic and p-type extrinsic
gallium antimonide (GaSb) at room temperature:
σ (Ω -m)–1
n (m–3)
p (m–3)
Intrinsic
8.9 × 104
8.7 × 1023
8.7 × 1023
Extrinsic (p-type)
2.3 × 105
7.6 × 1022
1.0 × 1025
Calculate electron and hole mobilities.
Solution
In order to solve for the electron and hole mobilities for GaSb, we must write conductivity expressions for
the two materials, of the form of Equation 12.13—i.e.,
 = n| e | e + p| e |  h
For the intrinsic material

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8.9  104 ( - m)-1 = (8.7  1023 m-3)(1.602  10-19 C)  e
+ (8.7  1023 m-3)(1.602  10-19 C)  h
which reduces to
0.639 = e +  h

Whereas, for the extrinsic GaSb
2.3  105 ( - m)-1 = (7.6  1022 m-3)(1.602  10-19 C)  e

+ (1.0  1025 m-3)(1.602  10-19 C)  h
which may be simplified
to

0.1436 = 7.6  10-3  e +  h

Thus, we have two independent expressions with two unknown mobilities. Upon solving these equations
simultaneously, we get e = 0.50 m2/V-s and h = 0.14 m2/V-s.

Factors That Affect Carrier Mobility
12.20 Calculate the room-temperature electrical conductivity of silicon that has been doped with 10 23 m–3
of arsenic atoms.
Solution
Inasmuch as As is a group VA element in the periodic table (Figure 2.6) it acts as a donor in silicon. Thus,
this material is n-type extrinsic, and it is necessary to use Equation 12.16), with n = 1023 m-3 since at room
temperature all of the As donor impurities are ionized. The electron mobility, from Figure 12.18 at an impurity
concentration of 1023 m-3, is 0.065 m2/V-s. Therefore, the conductivity is equal to
  n | e | e  (1023 m3 )(1.602  1019 C)(0.065 m2 / V s)  1040 (  m)1
Semiconducting Devices

12.23 Briefly describe electron and hole motions in a p–n junction for forward and reverse biases; then
explain how these lead to rectification.
The explanations called for are found in Section 12.15.
Capacitance
12.26 A parallel-plate capacitor using a dielectric material having an r of 2.2 has a plate spacing of 2 mm
(0.08 in.). If another material having a dielectric constant of 3.7 is used and the capacitance is to be unchanged,
what must be the new spacing between the plates?
Solution
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We want to compute the plate spacing of a parallel-plate capacitor as the dielectric constant is increased form 2.2 to
3.7, while maintaining the capacitance constant. Combining Equations 12.26 and 12.27 yields
A
A
C    r 0
l
l
Now, let us use the subscripts 1 and 2 to denote the initial and final states, respectively. Since C1 = C2, then
r10 A

And, solving for l2
l2 =
l1
r2 l1
r1
=
=
r2 0 A
l2
(3.7)(2 mm)
= 3.36 mm
2.2
Field Vectors and Polarization
Types of Polarization

12.28 For CaO, the ionic radii for Ca2+ and O2–ions are 0.100 and 0.140 nm, respectively. If an externally
applied electric field produces a 5% expansion of the lattice, compute the dipole moment for each Ca 2+–O2– pair.
Assume that this material is completely unpolarized in the absence of an electric field.
Solution
Shown below are the relative positions of Ca2+ and O2- ions, without and with an electric field present.
Now,
d = r 2+ + r 2- = 0.100 nm + 0.140 nm = 0.240 nm
Ca
O
and
d = 0.05 d = (0.05)(0.240 nm) = 0.0120 nm = 1.20  10 -11 m

From Equation 12.28, the dipole moment, p, is just
p = q d

= (1.602  10-19 C)(1.20  10-11 m)


= 1.92  10-30 C-m
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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