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Chapter 18
Electronic Materials
1
Objectives of Chapter 18
 To study electronic materials – insulators,
dielectrics, conductors, semiconductors,
and superconductors.
 To study conductivity in electronic
materials.
2
Chapter Outline
 18.1 Ohm’s Law and Electrical
Conductivity
 18.2 Band Structures of Solids
 18.3 Conductivity of Metals and Alloys
 18.4 Superconductivity
 18.5 Conductivity in Other Materials
 18.6 Semiconductors
 18.7 Applications of Semiconductors
 18.8 Insulators and Dielectric Properties
 18.9 Polarization in Dielectrics
 18.10 Electrostriction, Piezoelectricity,
Pyroelectricity, and Ferroelectricity
3
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Figure 18.1 Classification of technologically useful electronic
materials.
4
Section 18.1
Ohm’s Law and Electrical Conductivity
 Current density - The current flowing through per unit
cross-sectional area.
 Electric field - The voltage gradient or volts per unit
length.
 Drift velocity - The average rate at which electrons or
other charge carriers move through a material under the
influence of an electric or magnetic field.
 Mobility - The ease with which a charge carrier moves
through a material.
 Dielectric constant - The ratio of the permittivity of a
material to the permittivity of a vacuum, thus describing
the relative ability of a material to polarize and store a
charge; the same as relative permittivity.
5
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Figure 18.2 (a) Charge carriers,
such as electrons, are deflected by
atoms or defects and take an
irregular path through a
conductor. The average rate at
which the carriers move is the drift
velocity v. (b) Valence electrons in
the metallic bond move easily. (c)
Covalent bonds must be broken in
semiconductors and insulators for
an electron to be able to move. (d)
Entire ions must diffuse to carry
charge in many ionically bonded
materials.
6
7
TABLE 18-1 Electrical conductivity of selected materials at TF300 K* (Continued)
8
9
Example 18.1
Design of a Transmission Line
Design an electrical transmission line 1500 m long that will
carry a current of 50 A with no more than 5  105 W loss in
power. The electrical conductivity of several materials is
included in Table 18-1.
Example 18.1 SOLUTION
Electrical power is given by the product of the voltage and
current or:
From Equation 18-2:
10
Example 18.1 SOLUTION (Continued)
Let’s consider three metals—aluminum, copper, and silver—
that have excellent electrical conductivity. The table below
includes appropriate data and some characteristics of the
transmission line for each metal.
Any of the three metals will work, but cost is a factor as
well. Aluminum will likely be the most economical choice,
even though the wire has the largest diameter. However ,
other factors, such as whether the wire can support itself
between transmission poles, also contribute to the final
choice.
11
Example 18.2
A 42-volt Battery System for Cars
Some manufacturers of automotives are considering using a
42-volt battery system instead of the standard 12/14-volt
battery system. Why do you think this change is being
considered?
Example 18.2 SOLUTION
There is a need for increased overall power as newer cars
come equipped with many electronic features such as power
windows, power doors, power liftgates, and advanced
lighting, braking, and steering systems.
Increasing the voltage means less current is needed
(V = IR), so the total weight of the copper wires can be
reduced. Less weight in both the wiring and the motors
contributes to fuel efficiency.
12
Example 18.3
Drift Velocity of Electrons in Copper
Assuming that all of the valence electrons contribute to
current flow, (a) calculate the mobility of an electron in
copper and (b) calculate the average drift velocity for
electrons in a 100-cm copper wire when 10 V are applied.
Example 18.3 SOLUTION
1. The valence of copper is one: therefore, the number of
valence electrons equals the number of copper atoms in the
material.The lattice parameter of copper is 3.6151  10-8
cm and, since copper is FCC, there are 4 atoms/unit cell.
From Table 18-1
13
Example 18.3 SOLUTION (Continued)
2.The electric field is:
14
Section 18.2
Band Structures of Solids
 Valence band - The energy levels filled by electrons in
their lowest energy states.
 Conduction band - The unfilled energy levels into which
electrons can be excited to provide conductivity.
 Holes - Unfilled energy levels in the valence
band.Because electrons move to fill these holes, the
holes move and produce a current.
 Hybridization - When valence and conduction bands are
separated by an energy gap, leading to the
semiconductive behavior of silicon and germanium.
 Energy gap (Bandgap) - The energy between the top of
the valence band and the bottom of the conduction band
that a charge carrier must obtain before it can transfer a
charge.
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Figure 18.3 The energy levels broaden into bands as the
number of electrons grouped together increases.
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Figure 18.4 The
simplified band
structure for sodium.
The energy levels
broaden into bands.
The 3s band, which is
only half filled with
electrons, is
responsible for
conduction in sodium.
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Figure 18.5 (a) At absolute zero, all of the electrons in the
outer energy level have the lowest possible energy. (b)
When the temperature is increased, some electrons are
excited into unfilled levels. Note that the Fermi energy is
unchanged.
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Figure 18.6 The
band structure of
carbon in the
diamond form.
The 2s and 2p
levels combine to
form two hybrid
bands separated
by an energy gap,
Eg.
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Figure 18.7 Schematic of band structures for (a) metals, (b)
semiconductors, and (c) dielectrics or insulators. (Temperature
is 0 K.)
20
Section 18.3
Conductivity of Metals and Alloys
 Mean free path - The average distance that electrons can
move without being scattered by other atoms.
 Temperature Effect - When the temperature of a metal
increases, thermal energy causes the atoms to vibrate
 Effect of Atomic Level Defects - Imperfections in crystal
structures scatter electrons, reducing the mobility and
conductivity of the metal
 Matthiessen’s rule - The resistivity of a metallic material
is given by the addition of a base resistivity that
accounts for the effect of temperature (ρT), and a
temperature independent term that reflects the effect of
atomic level defects, including impurities forming solid
solutions (ρd).
 Effect of Processing and Strengthening
21
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Figure 18.8 Movement of an electron through (a) a perfect
crystal, (b) a crystal heated to a high temperature, and (c) a
crystal containing atomic level defects. Scattering of the
electrons reduces the mobility and conductivity.
22
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Figure 18.9 The effect of temperature on the electrical
resistivity of a metal with a perfect crystal structure. The
slope of the curve is the temperature resistivity coefficient.
23
24
Example 18.4
Resistivity of Pure Copper
Calculate the electrical conductivity of pure copper at (a)
400oC and (b) 100oC.
25
26
Example 18.4 SOLUTION
Since the conductivity of pure copper is 5.98  105 Ω-1 .
cm-1 the resistivity of copper at room temperature is 1.67
 10-6 ohm . cm. The temperature resistivity coefficient is
0.0068 ohm . cm/oC.
27
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Figure 18.10 The
electrical resistivity of a
metal is composed of a
constant defect
contribution ρd and a
variable temperature
contribution ρT.
28
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Figure 18.11 (a) the effect of solid-solution strengthening and
cold working on the electrical conductivity of copper, and (b)
the effect of addition of selected elements on the electrical
conductivity of copper.
29
30
Section 18.6
Semiconductors





Intrinsic semiconductor - A semiconductor in which
properties are controlled by the element or compound that
makes the semiconductor and not by dopants or impurities.
Extrinsic semiconductor - A semiconductor prepared by
adding dopants, which determine the number and type of
charge carriers.
Doping - Deliberate addition of controlled amounts of other
elements to increase the number of charge carriers in a
semiconductor.
Thermistor - A semiconductor device that is particularly
sensitive to changes in temperature, permitting it to serve
as an accurate measure of temperature.
Radiative recombination - Recombination of holes and
electrons that leads to emission of light; this occurs in
direct bandgap materials.
31
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Figure 18.16 When a voltage is applied to a semiconductor,
the electrons move through the conduction band, which the
electron holes move through the valence band in the
opposite direction.
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Figure 18.17 the distribution of electrons and holes in the
valence and conduction bands (a) at absolute zero and (b) at
an elevated temperature.
34
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Figure 18.18 The
electrical conductivity
versus temperature for
intrinsic semiconductors
compared with metals.
Note the break in y-axis
scale.
35
Example 18.7
Carrier Concentrations in Intrinsic Ge
For germanium at 25oC, estimate (a) the number of charge
carriers, (b) the fraction of the total electrons in the valence
band that are excited into the conduction band, and (c) the
constant n0.
36
37
Example 18.7 SOLUTION
1. From Equation 18-12:
38
2. The lattice parameter of diamond cubic germanium is
5.6575  10-8 cm. The total number of electrons in the
valence band of germanium is:
3. From Equation 18-14(a):
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Figure 18.19 When a dopant atom with a valence greater
than four is added to silicon, an extra electron is introduced
and a donor energy state is created. Now electrons are more
easily excited into the conduction band.
40
41
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Figure 18.20 The effect of temperature on the carrier concentration of an n-type
semiconductor. At low temperatures, the donor or acceptor atoms are not ionized.
As temperature increases, the ionization process is complete and the carrier
concentration increases to a level that is dictated by the level of doping. The
conductivity then essentially remains unchanged, until the temperature becomes
too high and the thermally generated carriers begin to dominate. The effect of
dopants is lost at very high temperatures and the semi-conductor essentially
shows “intrinsic” behavior.
42
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Figure 18.21 When a dopant atom with a valence of less than
four is substituted into the silicon structure, a hole is created
in the structure and an acceptor energy level is created just
above the valence band. Little energy is required to excite the
holes into motion.
43
Example 18.8
Design of a Semiconductor
Design a p-type semiconductor based on silicon, which provides
a constant conductivity of 100 ohm-1. cm-1 over a range of
temperatures. Comment on the level of purity needed.
Example 18.8 SOLUTION
In order to obtain the desired conductivity, we must dope the
silicon with atoms having a valence of +3, adding enough
dopant to provide the required number of charge carriers. If we
assume that the number of intrinsic carriers is small, then
The number of charge carriers required is:
44
Example 18.8 SOLUTION (Continued)
Let’s compare the concentration of donor atoms in Si with
the concentration of Si atoms. Assume that the lattice
constant of Si remains unchanged as a result, of doping:
Possible dopants include boron, aluminum, gallium, and
indium. High-purity chemicals and clean room conditions are
essential for processing since we need 26 dopant atoms
(apples) in a million silicon atoms (oranges).
45
Example 18.9
Creating the Color of a Light-Emitting
Diode Display
A light-emitting diode display made using a GaAs-GaP solid
solution of composition 0.4 GaP-0.6 GaAs has a direct bandgap
of 1.9 eV. What will be the color this LED display?
Example 18.9 SOLUTION
The wavelength of the light emitted is related to the bandgap
by:
46
Example 18.9 SOLUTION (Continued)
You can show that if the bandgap is in eV and λ is to be in
μm, then:
Therefore, λ = 0.652 μm or 652 nm. This is the
wavelength of red light, and, therefore, the LED display
would be red in color.
47
Section 18.7
Applications of Semiconductors





Diodes, transistors, lasers, and LEDs are made using
semiconductors. Silicon is the workhorse of very large scale
integrated (VLSI) circuits.
Forward bias - Connecting a p-n junction device so that the
p-side is connected to positive. Enhanced diffusion occurs
as the energy barrier is lowered, permitting a considerable
amount of current can flow under forward bias.
Reverse bias - Connecting a junction device so that the pside is connected to a negative terminal; very little current
flows through a p-n junction under reverse bias.
Avalanche breakdown - The reverse-bias voltage that
causes a large current flow in a lightly doped p-n junction.
Transistor - A semiconductor device that can be used to
amplify electrical signals.
48
Figure 18.22 Behavior of a p-n
junction device: (a) When no
bias is applied electron and hole
currents due to drift and
diffusion cancel out and there is
no net current and a built-in
potential V0 exists. (b) Under a
forward bias causes the
potential barrier to be reduced
to Vf - V0 and the depletion
region becomes smaller causing
a current to flow. (c) Under
reverse bias, the potential
barrier increases to Vr + V0 and
very little current flows. The
internal electric field that
develops is shown as E.
(Source: From Solid State
Electronic Devices, Third
Edition, by. B.G. Streetman,
Fig. 5-10. Copyright © 2000
Prentice Hall. Reprinted by
permission Pearson Education,
Inc., Upper Saddle River, NJ.)
49
Figure 18.23 (a) The current-voltage characteristic for a p-n
junction. Note the different scales in the first and third quadrants.
(b) If an alternating signal is applied, rectification occurs and only
half of the input signal passes the rectifier. (Source: From
Electronic Devices, Sixth Edition, by T.L. Floyd, Fig. 1-30. Copyright
© 2002 Prentice Hall. Reprinted by permission Pearson Education,
Inc., Upper Saddle River, NJ.)
50
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Figure 18.24 (a) A circuit for an n-p-n bipolar junction transistor. The
input creates a forward and reverse bias that causes electrons to move
from the emitter, through the base, and into the collector, creating an
amplified output. (b) Sketch of the cross-section of the transistor.
51
Figure 18.25 Operation of a n-p-n
transistor. (a) Schematic of a n-p-n
transistor. The dashes indicate
electrons, the + sign indicates
holes. Electrons in source and gate
regions are not shown. (b) The
gate has a positive voltage.
Electrons in the p-type substrate
move to the region between the
source and drain. (c) When positive
voltage is applied to a drain,
electrons from the source move to
the drain via the channel region
between the source and drain
(created by a positive gate
voltage). The transistor is now
‘‘on.’’ (d) When a gate voltage is
removed. The electron channel
between the source is broken and
the transistor is turned ‘‘off.’’
(Source: www.intel.com.)
52
Figure 18.25 (c) When
positive voltage is applied to
a drain, electrons from the
source move to the drain
via the channel region
between the source and
drain (created by a positive
gate voltage). The transistor
is now ‘‘on.’’ (d) When a
gate voltage is removed.
The electron channel
between the source is
broken and the transistor is
turned ‘‘off.’’ (Source:
www.intel.com.)
53
Figure 18.26 (a) Czochralski growth technique for
growing single crystals of silicon. (Source: From
Microchip Fabrication, Third Edition, by P. VanZant, Fig.
3-7. Copyright © 1997 The McGraw-Hill Companies.
Reprinted with permission The McGraw-Hill Companies.)
(b) Overall steps encountered in the processing of
semiconductors. (Source: From Fundamentals of Modern
Manufacturing, by M.P. Groover, p. 849, Fig. 34-3.
Copyright © 1996 Prentice Hall.)
54
Figure 18.26 (c) Production
of an FET semiconductor
device: (i) A p-type silicon
substrate is oxidized. (ii)
Photolithography, ultraviolet
radiation passing through a
photomask, exposes a
portion of the photoresist
layer. (iii) The exposed
photoresist is dissolved. (iv)
The exposed silica is
removed by etching. (v) An
n-type dopant is introduced
to produce the source and
drain. (vi) The silicon is
again oxidized. (vii)
Photolithography is repeated
to introduce other
components, including
electrical leads, for the
device.
55
Section 18.4
Superconductivity
 Superconductivity - Flow of current through a material
that has no resistance to that flow.
 Applications of Superconductors - Electronic circuits have
also been built using superconductors and powerful
superconducting electromagnets are used in magnetic
resonance imaging (MRI). Also, very low electrical-loss
components, known as filters, based on ceramic
superconductors have been developed for wireless
communications.
56
Superconductivity
Theoretical predictions of
resistivity of metals at
low T
Kamerlingh Onnes,
1911
57
perfect
conductivity
Superconductivity
Kamerlingh Onnes,
1911
58
R(T) at low temperatures
Experiment
Theoretical predictions
1) perfect conductivity
59
Superconductivity Transition Temperatures and
Critical Fields
Li Be
... 0.026
Transition temperature in Kelvin
Na
...
Mg
...
K
...
Ca
...
Sc
...
Ti
0.39
Rb
...
Sr
...
Y*
...
Zr
Nb
0.546 9.50
V
5.38
Cr*
...
Mn
...
Fe
...
Ra
...
C
...
N
...
O
...
Al
1.140
Si*
7
P*
5
S* Cl Ar
... ... ...
Ni Cu Zn
Ga
... ... 0.875 1.091
F Ne
... ...
Ge* As* Se* Br Kr
5
0.5 7 ... ...
Mo Tc Ru
Rh Pd Ag Cd
In
Sn Sb* Te* I Xe
0.90 7.77 0.51 0.0003 ... ... 0.56 3.4035 3.722 3.5 4 ... ...
Cs* Ba* La(fcc) Hf
Ta
W Re Os
1.5
5
6.00 0.12 4.483 0.012 1.4 0.655
Fr
...
Co
...
B
...
Ir
0.14
Pt Au Hg
... ... 4.153
Tl
2.39
Pb Bi* Po At Rn
7.193 8 ... ... ...
Ac
...
...
Ce*
2
Pr
...
Nd
...
Pm
...
Sm
...
Eu
...
Gd Tb
... ...
Dy
...
Ho
...
Er
...
Tm Yb Lu
...
... ... 0.1
...
Th
1.368
Pa
1.4
U*
2
Np
...
Pu
...
Am
...
Cm Bk
... ...
Cf
...
Es
...
Fm
...
Md No Lr
...
... ... ...
60
Isotope effect
Isotope
196
Hg
Hg
199
Hg
200
Hg
201
Hg
202
Hg
204
Hg
198
Natural abundance
(atom %)
0.15
9.97
16.87
23.10
13.18
29.86
6.87
Tc  M
M
61
-
Isotope effect
Isotope
196
Hg
Hg
199
Hg
200
Hg
201
Hg
202
Hg
204
Hg
198
Natural abundance
(atom %)
0.15
9.97
16.87
23.10
13.18
29.86
6.87
Tc  M
M
2) Lattice is important
62
-
Meissner effect
Perfect diamagnetism
Meissner and Ochsenfeld,
1933
perfect

conductor
superconductor
 = 0 => E =0
B=0
=>
dB/dt = 0 zero flux
constant
Meissner effect
flux
3) Perfect diamagnetism
63
Magnetization of superconductors
normal
metal
B paramagnetic
B=H
B=H
B>H
diamagnetic
B<H
H
64
Magnetization of superconductors
superconductor (type I)
B=H
B
B=0
B=0
Meissner
Hc
B=H
H
normal
65
Magnetization of superconductors
superconductor (type I)
B=H
B
B=0
B=0
Meissner
4M
Meissner
4M = - H
Hc
B=H
H
normal
Hc
4M = B - H
H
4M
normal
M=0
66
Magnetization of superconductors
Type I
Hc
H
4M
4M
Meissner
normal
B=0
B=H
67
Magnetization of superconductors
Type I
Type II
Hc
Hc1
H
4M
Hc2
H
4M
Meissner
normal
Meissner
mixed
normal
B=0
B=H
B=0
B<H
B=H
4) Magnetic field suppresses superconductivity
68
Magnetization of superconductors
Type I
Type II
Hc
Hc1
H
4M
Hc2
H
4M
Meissner
normal
Meissner
mixed
normal
B=0
B=H
B=0
B<H
B=H
69
Figure 18.13 (a) The effect of a magnetic field on the temperature
below which superconductivity occurs for a Type I superconductor.
(b) The critical surface of Nb3Sn, a Type II superconductor. The
superconducting envelope or surface showing the combined effects of
temperature, magnetic field, and current density on a Nb3Sn
superconductor. Conditions within the envelope produce
superconductivity.
70
Figure 18.13 (c) The effect of a magnetic field on the temperature
below which superconductivity occurs for a Type II superconductor. (d)
Critical-current density as a function of magnetic-flux density (in
Tesla) created by the applied magnetic fields (for different
superconductors).
71
Phase diagram
Type I
Type II
Hc2(T)
H
Hc
Normal
Hc(T)
H
metal
Normal
mixed state
metal
Hc1(T)
Meissner state
Meissner state
T
Tc
T
72
Tc
Tunneling experiments
Nobel 1973
norma
S
l
C
thin
insulato
r
Al
Al - Al2O3
Al - Al2O3 - Pb
73
Tunneling experiments
A
Pb
l
norma
S
l
C
thin
insulato
r
I
dI/dV
Pb
normal
D
Pb
superconducting
V
V
5) Energy gap in density of states
74
Superconducting energy gap
2D
3.5kTc (meV)
6) Tc and energy gap are related
75
Little-Parks
experiment
+
I
H
H
V
-
R
F0 = h/2e = 2.07x10-7 G
cm2
time
76
+
I
H
V
-
F0 = h/e
A
u
h/2
e
7) Quantization in units of h/2e
77
Critical temperature of
superconductors
Tc < 30 K
conventional superconductors
78
Critical temperature of
superconductors
High-temperature
superconductors (HTS)
Tc > 140 K
8) Superconductivity mechanism in HTS is different from
LTS
79
High-temperature superconductors
Nobel Prize in Physics 1987
Bi2Sr2CaCu2O8
Cu
O
Sr
J. Georg BednorzK. Alexander Müller
Bi
(La1.85Ba.15)CuO4
YBa2Cu3O7
Ca
Bi2Sr2CaCu2O8
Bi2Sr2Ca2Cu3O10
CuO2 double layer
Tl2Ba2Ca2Cu3O10
Hg0.8Tl0.2Ba2Ca2Cu3O8.33
80
Superconductivity phenomena
• Perfect conductivity
• Perfect diamagnetism

B=0
• Magnetic field suppresses superconductivity
Hc(T), Hc1(T), Hc2(T)
• Magnetic flux is quantized in units of h/2e
• Dynamics of the lattice is important Tc  M-
• Energy gap 2D
• Tc and energy gap are related
• Superconductivity mechanism in HTS is different from
LTS
81
82
Example 18.5
Design of a Superconductor System
Design the limiting magnetic field that will permit niobium to
serve as a superconductor at liquid helium temperatures.
Assume H0 = 1970 oersteds.
Example 18.5 SOLUTION
From Table 18-5, Tc = 9.25 K. We are given the value of
and H0 = 1970 oersted. Since the operating temperature
will be 4 K, we need to determine the maximum permissible
magnetic field, using Equation 18-9:
Therefore, the magnetic field (H) must remain below 1602
oersted in order for the niobium to remain superconductive.
83
84
Section 18.5
Conductivity in Other Materials
 Conduction in Ionic Materials - Conduction in ionic
materials often occurs by movement of entire ions, since
the energy gap is too large for electrons to enter the
conduction band. Therefore, most ionic materials behave
as insulators.
 Conduction in Polymers - Because their valence electrons
are involved in covalent bonding, polymers have a band
structure with a large energy gap, leading to lowelectrical conductivity.
85
Example 18.6
Ionic Conduction in MgO
Suppose that the electrical conductivity of MgO is determined
primarily by the diffusion of the Mg2+ ions. Estimate the
mobility of the Mg2+ ions and calculate the electrical
conductivity of MgO at 1800oC.
Example 18.6 SOLUTION
From Figure (5.18), the diffusion coefficient of Mg2+ ions in MgO
at 1800oC is 10-10 cm2/s. For MgO, Z = 2/ion, q = 1.6  10-19 C,
kB = 1.38  10-23 J/K, and T = 2073 K:
86
Example 18.6 SOLUTION (Continued)
Since the charge in coulomb is equivalent to Ampere .
second, and Joules is equivalent to Amp . sec . volts:
MgO has the NaCl structure, with four magnesium ions
per unit cell. The lattice parameter is 3.96  10-8 cm, so
the number of Mg2+ ions per cubic centimeter is:
87
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Figure 18.15 Effect of
carbon fibers on the
electrical resistivity of
nylon.
88
Section 18.8
Insulators and Dielectric Properties
 Materials used to insulate an electric field from its
surroundings are required in a large number of electrical
and electronic applications.
 Electrical insulators obviously must have a very low
conductivity, or high resistivity, to prevent the flow of
current.
 Porcelain, alumina, cordierite, mica, and some glasses
and plastics are used as insulators.
89
Section 18.9
Polarization in Dielectrics
 Capacitor - A microelectronic device, constructed from
alternating layers of a dielectric and a conductor, that is
capable of storing a charge.These can be single layer or
multi-layer devices.
 Permittivity - The ability of a material to polarize and
store a charge within it.
 Linear dielectrics - Materials in which the dielectric
polarization is linearly related to the electric field; the
dielectric constant is not dependent on the electric field.
 Dielectric strength - The maximum electric field that can
be maintained between two conductor plates without
causing a breakdown.
90
Figure 18.27 Polarization
mechanisms in materials: (a)
electronic, (b) atomic or
ionic, (c) high-frequency
dipolar or orientation
(present in ferroelectrics),
(d) low-frequency dipolar
(present in linear dielectrics
and glasses), (e) interfacialspace charge at electrodes,
and (f ) interfacial-space
charge at heterogeneities
such as grain boundaries.
(Source: From Principles of
Electronic Ceramics, L.L.
Hench and J.K. West, p. 188,
Fig. 5-2. Copyright © 1990
Wiley Interscience. Reprinted
by permission. This material
is used by permission of John
Wiley & Sons, Inc.)
91
Example 18.10
Electronic Polarization in Copper
Suppose that the average displacement of the electrons relative
to the nucleus in a copper atom is 1  10-8 Å when an electric
field is imposed on a copperplate. Calculate the electronic
polarization.
Example 18.10 SOLUTION
The atomic number of copper is 29, so there are 29 electrons in
each copper atom. The lattice parameter of copper is 3.6151 Å .
Thus
92
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Figure 18.28 (a) The oxygen ions are at face centers, Ba+2 ions
are at cube corners and Ti+4 is at cube center in cubic BaTi03.
(b) In tetragonal BaTi03 ,the Ti+4 is off-center and the unit cell
has a net polarization.
93
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Figure 18.29 A charge can be stored at the conductor plates in
a vacuum (a). However, when a dielectric is placed between
the plates (b), the dielectric polarizes and additional charge is
stored.
94
Figure 18.30 Frequency dependence of polarization mechanisms.
On top is the change in the dielectric constant with increasing
frequency, and the bottom curve represents the dielectric loss.
(Source: From Electroceramics: Material, Properties, Applications,
by A.J. Moulson and J.M. Herbert, p. 68, Fig. 2-38. Copyright ©
1990 Chapman and Hall. Reprinted with kind permission of Kluwer
Academic Publishers and the author.)
95
96
Section 18.10
Electrostriction, Piezoelectricity,
Pyroelectricity, and Ferroelectricity
 When any material undergoes polarization, its ions and
electronic clouds are displaced, causing the development
of a mechanical strain in the material. This effect is seen
in all materials subjected to an electric field and is
known as the electrostriction.
 Piezoelectrics - Materials that develop voltage upon the
application of a stress and develop strain when an
electric field is applied.
 Pyroelectric - The ability of a material to spontaneously
polarize and produce a voltage due to changes in
temperature.
 Ferroelectric - A material that shows spontaneous and
reversible dielectric polarization.
97
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Figure 18.31 The (a) direct and (b) converse piezoelectric
effect. In the direct piezoelectric effect, applied stress
causes a voltage to appear. In the converse effect (b), an
applied voltage leads to development of strain.
98
99
Example 18.11
Design of a Spark Igniter
A PZT spark igniter is made using a disk that has a 5-mm
diameter and 20-mm height.Calculate the voltage generated if
V m
the g coefficient for PZT used is 35  10
. Assume that
N
3
a compressive force of 10,000 N is applied on the circular face.
Example 18.11 SOLUTION
The g coefficient can be defined as:
100
Example 18.11 SOLUTION (Continued)
In this equation, X is the applied stress, E is the electric
field generated.
This field appears across a height (H) of 20 mm. The voltage
(V) generated will be:
101
Figure 18.32 Examples of ceramic capacitors. (a) Single-layer
ceramic capacitor (disk capacitors). (b) Multilayer ceramic
capacitor (stacked ceramic layers). (Source: From Principles of
Electrical Engineering Materials and Devices, by S.O. Kasap, p. 559,
Fig. 7-29. Copyright © 1997 Irwin. Reprinted by permission of The
McGraw-Hill Companies.)
102
Figure 18.33 (a) Different polymorphs of BaTiO3 and accompanying changes in lattice
constants and dielectric constants. (Source: From Principles of Electronic Ceramics, L.L.
Hench and J.K. West, p. 247, Fig. 6-5. Copyright © 1990 Wiley Interscience. This material is
used by permission of John Wiley & Sons, Inc.). The ferroelectric hysteresis loops for (b)
single crystal. (Source: From Electroceramics: Material, Properties, Applications, by A.J.
Moulson and J.M. Herbert, p. 76, Fig. 2-46. Copyright © 1990 Chapman and Hall. Reprinted
with kind permission of Kluwer Academic Publishers and the author.)
103
Figure 18.33 (b) single crystal. (Source: From Electroceramics: Material, Properties,
Applications, by A.J. Moulson and J.M. Herbert, p. 76, Fig. 2-46. © 1990 Chapman and Hall.
Reprinted with kind permission of Kluwer Academic Publishers and the author.) (c)
Polycrystalline BaTiO3 showing the influence of the electric field on polarization. (Source: From
Principles of Electrical Engineering Materials and Devices, by S.O. Kasap, p. 565, Fig. 7-35.
Copyright © 1997 Irwin. Reprinted by permission of The McGraw-Hill Companies.)
104
Figure 18.34 (a) The effect of temperature and grain size on the dielectric constant of
barium titanate. Above the Curie temperature, the spontaneous polarization is lost due to a
change in crystal structure and barium titanate is in the paraelectric state. The grain size
dependence shows that similar to yield-strength dielectric constant is a microstructure
sensitive property. (Source: From Electroceramics: Material, Properties, Applications, by
A.J. Moulson and J.M. Herbert, p. 78, Fig. 2-48. Copyright © 1990 Chapman and Hall.
Reprinted with kind permission of Kluwer Academic Publishers and the author.) (b)
Ferroelectric domains can be seen in the microstructure of polycrystalline BaTiO 3. (Courtesy
of Dr. Rodney Roseman, University of Cincinnati.)
105
Example 18.12
Polarization in Barium Titanate
Calculate the maximum polarization per cubic centimeter and
the maximum charge that can be stored per square centimeter
for barium titanate.
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Figure 18.28 (a) The oxygen ions are at face centers, Ba+2 ions are at
cube corners and Ti+4 is at cube center in cubic BaTi03. (b) In tetragonal
BaTi03 ,the Ti+4 is off-center and the unit cell has a net polarization.
106
Example 18.12 SOLUTION
In BaTiO3, the separations are the distances that the Ti4+
and O2- ions are displaced from the normal lattice points
(Figure 18.28). The charge on each ion is the product of q
and the number of excess or missing electrons. Thus, the
dipole moments are:
107
Example 18.12 SOLUTION (Continued)
Each oxygen ion is shared with another unit cell, so the
total dipole moment in the unit cell is:
The polarization per cubic centimeter is:
The total charge on a BaTiO3 crystal 1 cm  1 cm is:
108
Example 18.13
Design of a Multi-layer Capacitor
A multi-layer capacitor is to be designed using a BaTiO3-based
formulation containing SrTiO3. The dielectric constant of the
material is 3000. (a) Calculate the capacitance of a multi-layer
capacitor consisting of 100 layers connected in parallel using Ni
electrodes. The sides of the capacitor are 10  5 mm and the
thickness of each layer is 10 µm. (b) What is the role of SrTiO3?
(c) What processing technique will be used to make these?
Example 18.13 SOLUTION
(a) The capacitance of a parallel plate capacitor is given by:
109
Example 18.13 SOLUTION (Continued)
The permittivity of free space is ε0, 8.85  10-12 F/m.The
relative permittivity of BaTiO3 formulation εr or dielectric
constant (k) is 3000.
Capacitance per layer will be:
Note the conversion to SI units.
We have 100 layers connected in parallel. Capacitances
add up in this arrangement. All layers have same
geometric dimensions in this case.
110
Example 18.13 SOLUTION (Continued)
(b) The role of SrTiO3 addition to the formulation is to shift
the Curie temperature below room temperature. In this
way, the capacitor dielectric is in the high dielectric
constant, but paraelectric state. Thus, if it is subjected
to stress due to vibrations, it will not generate any
spurious voltages due to the piezoelectric effect present
in the tetragonal ferroelectric state.
(c) Tape casting will be the best way to make such thininterconnected layers.
111
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Figure 18.11 (Repeated
for Problems 18.15 and
18.17) (b) the effect of
selected elements on
the electrical
conductivity of copper.
112
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Figure 18.35
Ferroelectric
hysteresis loops
(for Problems
18.64 and 18.67).
113