Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Solution Manual for Principles of Electronic Materials
and Devices 4th Edition Kasap 0078028183
9780078028182
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Solutions Manual to
Principles of Electronic Materials and Devices
Fourth Edition
© 2018 McGraw-Hill
Safa Kasap
University of Saskatchewan
Canada
CHAPTER 7
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http://electronicmaterials.usask.ca
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
This electronic circuit board in a Tektronix oscilloscope clearly shows how prevalent and important
capacitors are in electronics engineering. There are several different types of capacitors such as
ceramic, polyester and electrolytic. (Photo by SK)
This circuit is a single channel 68 W audio amplifier designed to be mounted on to a heatsink through
the hole in the TO-220 package. (Photo by SK)
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Fourth Edition (© 2017 McGraw-Hill)
Chapter 7
7.1 Atomic polarizability and atomic radius Table 7.10 provides the radius and the polarizability of
atoms in Period 2 from Li (Z = 2) to Ne (Z = 10) and also for the inert gas atoms from He to Xe.
a.
Plot e versus ro3 and find the slope.
b.
Plot e versus ro on a log-log plot and find n in  e  ron
c.
Plot e and fo = o/2 versus Z on a log-log plot and find n in  e  Z n
d.
What are your conclusions for the above?
Table 7.10 Atomic radii and polarizability in Period 2 and for inert gases
Period 2
Li
Be
B
C
N
O
F
Ne
ro (pm)
167
112
87
67
56
48
42
38
e (×10− F m2)
27.1
6.23
3.37
1.86
1.22
0.892
0.621
0.434
Inert gases
He
Ne
Ar
Kr
Xe
Rn
ro (pm)
31
38
71
88
108
134
e (×10− F m2)
0.23
0.434
1.82
2.78
4.45
5.90
Note: Data for e from the CRC Handbook of Chemistry and Physics, 95th Edition, Ed. W.M. Haynes (Boca Raton). Rn is
radioactive
Solution (a) and (b)
Table 7Q01-1 Excel work sheet for calculations
Table 7Q01-1
o
8.85E-12
Element
Z
ro (pm)
x1E-40 F m2
x1E-40 F m2
x1E-40 F m2
Classical e
e (Experiment)
ro3 (pm3)
e(Experiment)
Z
fo (Hz)
2.979E+04
0.23
2
7.877E+15
He
2
3.10E+01
0.033
0.23
Ne
10 3.80E+01
0.061
0.434
5.487E+04
0.434
10 1.282E+16
Ar
18 7.10E+01
0.398
1.82
3.579E+05
1.82
18 8.401E+15
Kr
36 8.80E+01
0.758
2.78
6.815E+05
2.78
36 9.613E+15
Xe
54 1.08E+02
1.401
4.45
1.260E+06
4.45
54 9.306E+15
2.406E+06
5.9
86 1.020E+16
Rn
86 1.34E+02
2.676
5.9
Li
3
1.67E+02
5.180
27.1
4.657E+06
27.1
3
8.888E+14
Be
4
1.12E+02
1.563
6.23
1.405E+06
6.23
4
2.141E+15
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
B
5
8.70E+01
Chapter 7
0.732
3.37
6.585E+05
3.37
5
3.254E+15
3.008E+05
1.86
6
4.798E+15
C
6
6.70E+01
0.335
1.86
N
7
5.60E+01
0.195
1.22
1.756E+05
1.22
7
6.399E+15
O
8
4.80E+01
0.123
0.8921
1.106E+05
0.892
8
8.000E+15
F
9
4.20E+01
0.082
0.6207
7.409E+04
0.621
9
1.017E+16
Ne
10 3.80E+01
0.061
0.434
5.487E+04
0.434
10 1.282E+16
Period 2 from Li to Ne
Table 7Q01-1 shows an Excel worksheet with all the calculations we need for this question. The
classical polarizability e in column 4 is the simple classical polarizability prediction, calculated from ro
by using
 e  4o ro3
(1)
The 5th column is the reported experimental values (or calculated if no experimental values were
available) from the CRC Handbook. The last column fo is the resonant frequency in Equation 7.7
e =
Ze 2
meo2
1/ 2
 Ze 2 



(2)

m
 e e
Figure 7Q01-1 shows the experimental polarizability ae vs. ro3 on linear axes for Period 2, from Li to
Ne. The points for small ro for C to Ne are cluttered near the origin. Figure 7Q01-2 shows the same data
plotted on log-log axes to declutter the region with C to Ne points. The best straight line from Figure
7Q01-1, forced through the origin, has a slope given by
Slope = 5.69910−10 F m−1.
The simplest classical electronic polarizability is given by
 e  4o ro3
(1)
1
fo =
2
so that the expected slope for the classical theory should be
Slope = 4 o = 1.1110−10 F m−1
Clearly, the theory is out by a factor of ~5.
Figure 7Q01-3 is a plot of e vs ro on log-log axes and indicates that, experimentally,
 e  ro2.64
in which the index, 2.64, is not far out from 3, giving some support to the simple classical theory.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Figure 7Q01-1 The plot e versus ro and the best straight line forced through the origin. The points for small ro
3
are cluttered.
Figure 7Q01-2 The plot e versus ro on log-log axes, and the best straight line through the origin with
3
parameters give in the green box. The points for small ro are now decluttered.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Figure 7Q01-3 The plot e versus ro on log-log axes, and the best power law dependence.
Inert Gases from He to Rn
Figure 7Q01-4 shows the experimental polarizability e vs. ro3 on linear axes for the inert gases. (There
is no need to plot the data on log-log axes inasmuch as there is no severe cluttering of points.) The best
straight line from Figure 7Q01-4, forced through the origin, has a slope given by
Slope = 2.80710−10 F m−1.
The expected slope for the classical theory should be
Slope = 4 o = 1.1110−10 F m−1
Clearly, the theory is out by a factor of ~2.53.
Figure 7Q01-5 is a plot of e vs ro on log-log axes and indicates that, experimentally,
 e  ro2.23
The experiments do not follow the expected behavior in Equation (1), that is, e is not proportional to
ro3 .
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Chapter 7
Figure 7Q01-4 The plot e versus ro and the best straight line.
3
Inert Gases
Electronic polarizability x 1E-40 ( F
m2)
100
y = 1.204E-04x2.234E+00
R² = 9.800E-01
10
1
0.1
0.01
1.0E+01
Classical theory
y = 1.112E-06x3.000E+00
R² = 1.000E+00
1.0E+02
ro (pm)
1.0E+03
Figure 7Q01-5 The plot e versus ro on log-log axes, and the best straight power law dependence.
Solution (c) and (d)
Figures 7Q01-6 shows the plot e versus Z on log-log axes, and the best straight power law dependence.
Figure 7Q01-7 shows the plot fo versus Z on log-log axes, and the best straight power law dependence. It
is clear that the inert gases behave very differently than the elements in a given Period (row) in the
Periodic Table.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Figure 7Q01-6 The plot e versus Z on log-log axes, and the best straight power law dependence.
Figure 7Q01-7 The plot fo versus Z on log-log axes, and the best straight power law dependence.
Conclusions
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Consider the inert gases. These have filled shells and subshells. According to Figure 7Q01-6, e  Z0.93
or roughly e  Z. The resonant frequency fo remains relatively insensitive to Z. The classical
interpretation for the inert gases would lead to the following conclusion:
fo  constant in
1/ 2
  

f o = 2 
 Zme 
means that roughly   Z. As the nuclear charge increases with Z, so does the restoring force constant
() but as   Z rather than the intuitive   Z2. From Equation 7.7 in the textbook,
e =
Ze 2
[7.7]
me o2
since o is relatively constant, we expect e  Z.
Consider the row of elements in Period 2. According to Figure 7Q01-7, fo increases strongly with Z, that
is fo  Z2.1. Thus,  increases strongly with Z and it becomes more difficult for the applied field to
displace the electron clouds around the nucleus. Consequently, as Z increases, the restoring force 
becomes stronger, which means a smaller polarizability with Z as shown in Figure 7Q01-6
Comment: The two very different observations can be explained by considering the ionization energy of
each element, which is the energy we need to remove an outer electron from the atom. Put differently, it
represents a binding energy for an outer electron. Table 7Q01-2 lists these ionization energies, EI for the
two cases. Figure 7Q01-8 is a log-log plot of EI vs. Z for the two classes of elements. For the inert gases
EI actually decreases slightly with Z but for Period II, there is a large increase in EI as we move along Z
from Li to Ne. Consequently  increases strongly with Z, and so does fo. A strongly increasing  with Z
means a lower polarizability, as Z increases from Li to Ne.
Table 7Q01-2 Ionization energy vs. Z for Period II and inert gases (Data from the CRC Handbook of
Chemistry and Physics, 95th Edition, Ed. W.M. Haynes (Boca Raton))
Element
Z
Ionization
energy (eV)
Li
Be
B
C
N
O
F
Ne
3
4
5
6
7
8
9
10
5.39
9.32
8.29
11.26
14.53
13.62
17.4
21.56
Inert gas
Z
Ionization
energy eV
He
Ne
Ar
Kr
Xe
Rn
2
10
18
36
54
86
24.58
21.56
15.76
14
12.13
10.74
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Figure 7Q01-8 Ionization energy EI vs. Z for Period 2 (Li to Ne) and the inert gases.
7.2 SI, cgs, Debye and atomic units in electrostatics
a
The definitions of polarizability within the SI and cgs (cm-gram-second) unit systems are
p =  SI E
p = 4o cgs E
and
The cgs units are widely used for polarizability. Convert a polarizability of 1 F m2 to cgs units. The
polarizability SI of an Ar atom is 1.82×10− F m2. What is cgs for Ar in cm3 and Å3?
b.
Atomic polarizability vol is a dimensionless quantity in the cgs system obtained by dividing
cgs by an atomic volume, taken to be ao3 where ao is the Bohr radius in cm. What is vol atomic units
for Ar?
c.
The electric dipole unit in SI is simply C m (p = Qa). The atomic dipole moment is defined as
patomic = eao = 8.478×10− C m, where ao is the Bohr radius. One Debye (D) is a dipole unit within the
cgs system and corresponds to 3.3356×10− C m. Put differently, amounts of charge ±3.3356×10− C
(approximately ±0.21e) separated by 1 Å. Consider a molecule in a CsF vapor. Cs+ and F− in the CsF
molecule are separated by a bond length a that is 0.255 nm. Assume that Cs+ and F− are fully ionized in
forming the molecule. What is the permanent dipole moment po in Debye units? If the experimental
value is 7.88 D, what is actual charge on Cs+ and F−?
Solution
a. Consider
p =  SI E
Thus,
or
and
p = 4o cgsE
SI = 4o cgs
 cgs =
1
4 o
 SI =
1
(1 F m 2 ) = 8.99 109 m 3
4 (8.854 10 −12 F m −1 )
= 8.99  109 10 6 cm3 = 8.99 1015 cm3
where the last step was to convert m3 to cm3 since cgs involves the cm not m.
Given SI = 1.82×10− F m2 for Ar, we have
 cgs = (8.99 1015 cm3 )(1.82 10−40 ) = 1.64 10−24 cm3 = 1.64 Å3
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
b. The polarizability of Ar in atomic units is
 vol
1.64 10 −24 cm3
1.64 10 −24 cm3
= 5.8610−8
=
=
3
−7
3
ro
(0.529 10 cm)
c. Consider the CsF molecule in the gas phase; the molecules are are isolated from each other. Assume
simply that it consists of Cs+ and F− ions, each with a magnitude of change e, separated by an
equilibrium separation a of ions in the molecule.
psimple = Qa = (e)a = (1.602210−19 C)(0.25510−9 m) = 4.08610−29 C m in SI units.
Since one Debye is
1 D = 3.3356×10− C m
we have
psimple = 4.08610−29 C m / 3.3356×10− C m = 12.25 Debye or 12.25 D
Given the experimental value is 7.88 D, we can find the effective charge involved in the dipole, that is
pexperimental = Qeffectivea
and compare this with
psimple = (e)a
so that
Qeffective = e(pexperimental/psimple) =e(7.88 D / 12.25 D) = 0.64e
--------------Note: The ratio Qeffective/e represents the ionic character of the bond in CsF. Thus, the Cs+−F− bond is
64% ionic.
Static polarizability of inert gases are listed at
https://en.wikipedia.org/wiki/Noble_gas_(data_page) (29 October 2016). The experimental value of
7.88D for CsF from
http://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map
%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.4%3A_Elec
tronegativity_and_Dipole_Moment (29 October 2017)
7.3 Relative permittivity and polarizability
a. Show that the local field is given by
ε + 2
Eloc = E  r

 3 
Local field
b. Amorphous selenium (a-Se) is a high resistivity semiconductor that has a density of approximately
4.3 g cm−3 and an atomic number and mass of 34 and 78.96, respectively. Its relative permittivity at 1
kHz has been measured to be 6.7. Calculate the relative magnitude of the local field in a-Se. Calculate
the polarizability per Se atom in the structure. What type of polarization is this? How will r depend on
the frequency?
c. Calculate the electronic polarizability of an isolated Se atom, which has an atomic radius ro = 0.12
nm, and compare your result with that for an atom in a-Se. Why is there a difference? (See Example 7.1)
Solution
a. The polarization, P, is given by:
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P = (εo (εr − 1) )E
Chapter 7
(1)
where E is the electric field. The local field Eloc depends on the polarization P and is given by
E loc = E +
P
3ε o
(2)
Substitute for P from Equation (1) into (2) to find Eloc
ε + 2
Eloc =  r
E
 3 
(3)
The relative magnitude of the local field, Eloc / E is
Eloc  εr + 2   6.7 + 2 
=
=
 = 2.9
E
 3   3 
If d is the density then the atomic concentration N of Se atoms is
N=
(
)(
)
dN A
4.3 103 kg/m 3 6.022 10 23 mol −1
= 3.279  1028 m−3
=
M at
78.96 10 −3 kg/mol
(
)
The Clausius-Mossotti equation relates the relative permittivity to the electronic polarizability e,
εr − 1 N
=
αe
εr + 2 3εo
(
(
)
3εo (εr − 1) 3 8.854 10 −12 F/m (6.7 − 1)
=
N (εr + 2)
3.279 10 28 m −3 (6.7 + 2)

αe =

e = 5.31  10−40 F m2
)
This would be a type of electronic polarization and r would be flat up to optical frequencies.
c. The Se atom has a radius of about ro = 0.12 nm. Substituting into the given equation:
e  4oro3 = 4(8.854  10−9 F/m)(0.12  10−9 m)3

e  1.92  10−40 F m2
Comparing this value and our previous value:
α e 5.30  10 −40 F m 2
=
= 2.76
α e 1.92  10 − 40 F m 2
The observed polarizability per Se atom in the solid is 2.8 times greater than the polarizability of the
isolated Se atom. The simple classical theory is known to fail by a factor of 3−5 as shown in Question
7.1
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Chapter 7
Note: The atomic radius of Se is from WebElements
(https://www.webelements.com/selenium/atom_sizes.html, accessed 15 April 2017). Empirical atomic
radius 115 pm was rounded to 120 pm, same as the covalent radius.
7.4 Dielectric properties of diamond Consider the diamond crystal, which has a density of 3.52 g
cm−, a lattice parameter of 0.35670 nm and a low-frequency dielectric constant of 5.7. Calculate the
electronic polarizability per atom and also calculate the relative magnitude of the local field (see
Question 7.3). The polarizability of an isolated C atom is 1.86×10− F m2. Why is there a difference?
Solution
a. Given P as
P = (εo (εr − 1) )E
(1)
and the local filed dependence on P,
E loc = E +
P
3ε o
(2)
we can find the local field as
ε + 2
Eloc =  r
E
 3 
(3)
The relative magnitude of the local field, Eloc / E is
Eloc  ε r + 2   5.7 + 2 
=
=
 = 2.57
E
 3   3 
If d is the density, then the atomic concentration N of C atoms is
N=
(
)(
)
dN A 3.52 103 kg/m 3 6.022 10 23 mol −1
= 1.765  1029 m−3
=
M at
12.0110 −3 kg/mol
(
)
The Clausius-Mossotti equation relates the relative permittivity to the electronic polarizability, e,
εr − 1
N
=
αe
ε r + 2 3ε o
(
(
)

3εo (εr − 1) 3 8.854 10 −12 F/m (5.7 − 1)
αe =
=
N (εr + 2)
1.765 10 29 m −3 (5.7 + 2)

e = 0.92  10−40 F m2
)
This is electronic polarizability
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Chapter 7
The isolated C atom has a polarizability of 1.86×10− F m2. Clearly there is a factor of 2 difference. In
the crystal, each C atom uses 4 of its 6 electrons in Si-Si bonds. The 1s2 shell is the same as He which has
e = 0.23×10− F m2. The 4 electrons per atom in the bonds are in the valence band and belong to the
whole crystal (Ch. 4 and 5). Thus, the main contribution to the polarizability comes from these electrons
in the valence band, that is, electrons in the bonds, so this type of polarization is "electronic bond
polarization"
Note: The isolated C atom, according to the CRC Handbook of Chemistry and Physics, 95th Edition,
Ed. W.M. Haynes (Boca Raton), p10-188, has a polarizability of
1.67 ×10−24 cm3 which is (1.67 ×10−) / (8.898×10−) or 1.88×10− F m2.
7.5 Electronic polarization and SF6 Because of its high dielectric strength, SF6 (sulfur hexafluoride)
gas is widely used as an insulator and a dielectric in HV applications such as HV transformers, switches,
circuit breakers, transmission lines, and even HV capacitors. The SF6 gas at 1 atm and at room temperature
has a dielectric constant of 1.0015. The number of SF6 molecules per unit volume N can be found by the
gas law, P = (N/NA)RT . Calculate the electronic polarizability αe of the SF6 molecule. (Note: The SF6
molecule has no net dipole. Assume that the overall polarizability of SF6 is due to electronic polarization.)
Solution
Given the Gas Law
P=
NRT
NA
where the gas constant, R = 8.314 J mol −1 K −1 and N A = 6.023  10 23 mol −1 ,
At T = 300K and P = 1 atm =
1
N m − 2 so that
−6
9.869  10
N = 2.446 10 25 m −3 of SF6 molecules.
We know ε r = 1.0015 , εo = 8.854 10−12 CV −1m −1 and N = 2.446 10 25 m −3 so we can use the simple
relation in Equation 7.14,
εr = 1 +
Nαe
εo

1.0015 = 1 + (2.4461025 m−)e / (8.85410− F m−)

e = 5.42910− F m− or 5.4310− F m−
We do not actually need the Clasius-Mossotti equation because the concentration of SF6 molecules is low
compared to typical concentrations in solids. Consider, the Clausius-Mossotti equation
(ε r − 1) Nα e
=
(ε r + 2) 3ε o
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
αe =
Chapter 7
3εo (εr − 1)
= 5.426  10 − 40 F m 2 or 5.4310− F m−
N (εr + 2)
7.6 Electronic polarization in liquid xenon Liquid xenon has been used in radiation detectors. The
density of the liquid is 3.0 g cm−3. What is the relative permittivity of liquid xenon given its electronic
polarizability in Table 7.10? (The experimental εr is 1.96.)
Solution
-3
Given d = 3.0 g/cm we can calculate the number of Xe atoms per unit volume,
N=
N A d (6.023 10 23 )(3)
=
= 1.3762 10 22 cm−3
M at
131.3
With N = 1.3762  10 28 m −3 and the given values from Table 7.1, αe = 4.4510−40 F m 2 and
εo = 8.854 10−12 CV −1m −1 , so that
εr = 1 +
Nαe
= 1.692 or 1.70
εo
But, if we use the Clausius-Mossotti Equation,
(ε r − 1) Nα e
=
(ε r + 2) 3ε o

2 Nαe
3ε0
εr =
= 1.898 or 1.89
Nαe
1−
3ε0

εr (Xe) = 1.89
1+
which is closer to the experimental value.
Comment: Liquid Xe properties can be found at http://www.pd.infn.it/~conti/LXe.html, (29 October
2016). Dielectric properties are at http://www.pd.infn.it/~conti/images/LXe/xe_properties.txt. Liquid
xenon has been proposed as a detector for dark matter (see
http://www.nikhef.nl/pub/services/biblio/theses_pdf/thesis_R_Schon.pdf, 29 October 2016)
7.7 Relative permittivity, bond strength, bandgap and refractive index Diamond, silicon, and
germanium are covalent solids with the same crystal structure. Their relative permittivities are shown in
Table 7.11.
a. Explain why r increases from diamond to germanium.
b. Calculate the polarizability per atom in each crystal and then plot polarizability against the elastic
modulus Y (Young's modulus). Should there be a correlation?
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c. Plot the polarizability from part (b) against the bandgap energy, Eg. Is there a relationship?
d. Show that the refractive index n is  r . When does this relationship hold and when does it fail?
e. Would your conclusions apply to ionic crystals such as NaCl?
Table 7.11 Properties of diamond, Si, and Ge
Solution
a. In diamond, Si, and Ge, the polarization mechanism is electronic (bond). There are two factors that
increase the polarization. First is the number of electrons available for displacement and the ease with
which the field can displace the electrons. The number of electrons in the core shells increases from
diamond to Ge. Secondly, and most importantly, the bond strength per atom decreases from diamond to
Ge, making it easier for valence electrons in the bonds to be displaced.
b. Consider diamond. The atomic concentration N can be found from the density d and the atomic mass
Mat,
N=
(
)(
)
dN A 3.52 103 kg/m 3 6.022 10 23 mol −1
= 1.765  1029 m−3
=
M at
12.0110 −3 kg/mol
(
)
The polarizability can then be found from the Clausius-Mossotti equation:
εr − 1
N
=
αe
ε r + 2 3ε o
(
(
)
3εo (εr − 1) 3 8.854 10 −12 F/m (5.7 − 1)
=
N (εr + 2)
1.765 10 29 m −3 (5.7 + 2)

αe =

e = 9.19  10−41 F m2
)
The polarizability for Si and Ge can be found similarly, and are summarized in Table 7Q07-1:
Table 7Q07-1 Polarizabilities of Diamond, Si and Ge
N (m−3)
e (F m2)
Diamond
1.765  1029 m−3
0.919  10−41 F m2
Si
4.995  1028 m−3
4.170  10−40 F m2
Ge
4.412  1028 m−3
5.017  10−40 F m2
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Figure 7Q4-1: Plot of polarizability per atom versus Young’s modulus.
Figure 7Q07-1 shows a plot of e vs Young's modulus for the three crystals. As the polarization
mechanism in these crystals is due to electronic bond polarization, the displacement of electrons in the
covalent bonds depends on the flexibility or elasticity of these bonds and hence also depends on the elastic
modulus. (Remember that all three have the same crystal structure and the same type of bonding. Also,
consider Equation 7.5.)
Figure 7Q07-2: Plot of polarizability versus bandgap energy.
c. The plot of polarizability against the bandgap energy is given in Figure 7Q07-2. There indeed seems to
be a linear relationship between polarizability and bandgap energy for these three crystals.
d. To facilitate this proof (experimental verification), we can plot a graph of refractive index n versus
relative permittivity r as in Figure 7Q07-3 on log-log axes.
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Figure 7Q7-3: Logarithmic plot of refractive index versus relative permittivity.
The log-log plot exhibits a straight line through the three points. The best fit line is n = Arx (Correlation
coefficient is 0.9987) where x = 0.503 ≈ 1/2 and A = exp(0.02070) ≈ 1. Thus n is ε r .
The refractive index n is an optical property that represents the speed of a light wave, or an
electromagnetic wave, through the material (v = c/n). The light wave is a high frequency
electromagnetic wave where the frequency is of the order of 1014 to 1015 Hz (foptical). n and polarizability
(or r) will be related if the polarization can follow the field oscillations at this frequency (ƒoptical). This
will be the case in electronic polarization because electrons are light and can rapidly respond to the fast
oscillations of the field. The relationship between n and r will not hold if we take r at a low frequency
(<< ƒoptical) where other slow polarization contributions (such as ionic polarization, dipolar polarization,
interfacial polarization) also contribute to r but not to n. In the present case, all three exhibit the same
polarization mechanism.
e. n =  r would apply to ionic crystals if r is taken at the corresponding optical frequency rather than
at frequencies below ƒoptical. Tabulated data for most ionic crystals typically quote r that includes ionic
polarization and hence this data does NOT conform to n =  r since n is at optical frequencies.
7.8 Dipolar liquids
Given the static dielectric constant of water as 80, its optical frequency dielectric constant (due to
electronic polarization) as 4, its density as 1 g cm−3 calculate the permanent dipole moment po per water
molecule assuming that it is the orientational and electronic polarization of individual molecules that
gives rise to the dielectric constant. Use both the simple relationship in Equation 7.14 where the local
field is the same as the macroscopic field and also the Clausius-Mossotti equation and compare your
results with the permanent dipole moment of the water molecule which is 6.210−30 C m. What is your
conclusion? What is r calculated from the Clasius-Mossotti equation taking the true po (6.210−30 C m)
of a water molecule? (Note: Static dielectric constant is due to both orientational and electronic
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polarization. The Clausius-Mossotti equation does not apply to dipolar materials because the local field
is not described by the Lorentz field.)
Solution
We first need the number H2O molecules per unit volume. The molecular mass of water Mmol is 18  10−3
kg mol−1, its density is d = 103 kg m−3. The number of H2O molecules per unit volume is
N=
(
)(
)
d N A 103 kg m −3 6.022 10 23 mol −1
=
= 3.346  10 28 m −3
−3
−1
M mol
18 10 kg mol
(
)
I. Assuming the Clausius-Mossotti Equation
Using the high frequency dielectric constant rHF which is due only to electronic polarization and the
Clausius-Mossotti equation, we can calculate the electronic polarizability e of water molecules, that is,
using
(εr − 1) Nαe
=
(εr + 2) 3εo
at optical frequencies, so that
αe =
3εo (εrop − 1)
(ε
r op
+ 2)N
=
(
)
3 8.854 10 −12 F m −1 (4 − 1)
= 3.97  10 −40 Fm 2
28
−3
(4 + 2) 3.346 10 m
(
)
Using this result and the static dielectric constant rstat which is due both to electronic and dipolar
polarization, we can solve the Clausius-Mossotti equation for the dipolar polarizability d of water. At
low frequencies, we need to add the electronic and dipolar polarizabilities, and assuming that we can still
use the Clasius-Mossotti equation, we have
(εrstat − 1)
N
=
(αe + α d )
(εrstat + 2) 3εo
from which,
αd =
=
3ε (ε
o
rstat

− 1) − N αe (εrstat + 2)
(ε
rstat
+ 2)N
(
)
(
)(
)
3 8.854 10 −12 F m −1 (80 − 1) − 3.346 10 28 m −3 3.97 10 −40 F m 2 (80 + 2)
(80 + 2) 3.346 1028 m −3
(
)
= 3.68 10−40 F m2
The permanent dipolar moment per water po molecule can be calculated from Equation 7.20
2
1 p
d =   o
 3  kT
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
(
)
Chapter 7
(
)
po = 3kT αd = 3 1.3807 10 − 23 J K −1 (300 K ) 3.68 10 − 40 F m 2 = 2.14  10−30 C m
This result is around 3 times smaller than the real permanent dipolar moment of the water molecule.
Clearly, the application of the Clausius-Mossotti equation to water has failed.
Suppose we know use the true po and hence the corresponding d in the Clausius-Mossotti equation. First
find d we expect using the true po
2
(6.2 10 −30 C m)
 1  po  1 
= 3.09  10−39 F m2
d =  
= 
− 23
−1
 3  kT  3  (1.38110 J K )(300 K )
Applying the Clausius-Mossotti equation,

(εrstat − 1)
N
(3.346 10 28 m −3 )
=
( αe + α d ) =
3.97 10 −40 C m + 3.09 10 −39 C m
−12
−1
(εrstat + 2) 3εo
3(8.854 10 F m )

and solving, this gives

rstat = −2.9 (Nonsense)
II. Simple r− equation (Equation 7.14)
The same calculations performed using the simple relationship in Equation 7.14, that is,
εr = 1 +
Nαd Nαe
+
εo
εo
Apply the above at high frequencies (optical) where there is only e
αe =
εo (εrop − 1)
N
=
(8.854 10 F m )(4 −1)
(3.346 10 m )
−12
−1
28
−3
= 7.94  10−40 F m2
Now apply it at low frequencies (static) where we have bot e and d
εo (εrstat − 1)

αd =

ad = 2.01  10−38 F m2
N
− αe
(8.854 10 F m )(80 −1) − (7.94 10
=
(3.346 10 m )
−12
−1
28
− 40
−3
F m2
)
and
(
)
(
)
po = 3kT d = 3 1.3807 10 − 23 J K −1 (300 K ) 2.0110 −38 F m 2 = 1.58  10−29 C m
which is roughly 3 times bigger than the actual value.
Both are unsatisfactory calculations. The reasons for the differences are two fold. First is that the
individual H2O molecules are not totally free to rotate. In the liquid, H2O molecules cluster together
through hydrogen bonding so that the rotation of individual molecules is then limited by this bonding.
Secondly, the local field can neither be totally neglected nor taken as the Lorentz field. A better theory
for dipolar liquids is based on the Onsager theory which is beyond the scope of this book. Interestingly,
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if one uses the actual po = 6.210−30 C m in the Clasius-Mossotti equation, then r turns out to be negative,
which is nonsense.
Notes: po for H2O at Wikipedia is given as 1.85D (Debye), and from 7Q02, this is approximately 6.210−30
C m. (https://en.wikipedia.org/wiki/Properties_of_water)(29 October 2016)
7.9 Dielectric constant of water vapor or steam The isolated water molecule has a permanent dipole
po of 6.2 × 10−30 C m. The electronic polarizability αe of the water molecule under dc conditions is about
4 × 10−40 C m. What is the dielectric constant of steam at a pressure of 10 atm (10.1 × 105 Pa) and at a
temperature of 400 oC? [Note: The number of water molecules per unit volume N can be found from the
simple gas law, P = (N/NA)RT. The Clausius–Mossotti equation does not apply to orientational
polarization. Since N is small, use Equation 7.14.]
Solution
Given the gas law,
P=
NRT
NA
where the gas constant, R = 8.314 J.mol −1 .K −1 and N A = 6.022  10 23 mol −1 ,
At T = 673 K and P = 10 atm = 1.0110 6 Pa we will have,
N=
PN A
= 1.076 10 26 m −3 of H 2 O molecules.
RT
Since po = 6.210−30 Cm and k = 1.3807  10 −23 J K −1 , the dipolar polarizability will be,
po2
αd =
= 1.379 10 −39 F m 2
3kT
Therefore, given αe = 4 10−40 F m 2 , then we can use the simple permittivity-polarizability relationship in
Equation 7.14
εr = 1 +

N(αe + αd )
= 1.0216
ε0
r(steam) = 1.022 at 10 atm and T = 673 K (400 C).
7.10 Dipole moment in a nonuniform electric field Figure 7.62 shows an electric dipole moment p in
a nonuniform electric field. Suppose the gradient of the field is dE/dx at the dipole p, and the dipole is
oriented to be along the direction of increasing E as in Figure 7.62. Show that the net force acting on this
dipole is given by
F=p
dE
dx
Net force on a dipole
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Which direction is the force? What happens to this net force when the dipole moment is facing the
direction of increasing field? Given that a dipole normally also experiences a torque as described in
Section 7.3.2, explain qualitatively what happens to a randomly placed dipole in a nonuniform electric
field. Explain the experimental observation of bending a flow of water by a nonuniform field from a
charged comb as shown in the photograph in Figure 7.62? (Remember that a dielectric medium placed
in a field develops polarization P directed along the field.
Figure 7.62 Left: A dipole moment in a
nonuniform field experiences a net force F that
depends on the dipole moment p and the field
gradient dE∕dx. Right: When a charged comb (by
combing hair) is brought close to a water jet, the
field from the comb polarizes the liquid by
orientational polarization. The induced polarization
vector P and hence the liquid is attracted to the
comb where the field is higher. (Photo by S. Kasap)
Solution
Figure 7Q10-1: When a dielectric is placed in a nonuniform field it develops a polarization P along the direction
of the field. We can represent this polarization by two opposite charges +Q and −Q separated by δx.
We can represent the dipole p as two opposite charges +Q and −Q separated by a distance x as in Figure
7Q10-1. Thus p = Qx. Let −Q be at x, then +Q is at x = x + x. The field is nonuniform. If the field is E
at x and then it is E at x, that is
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E = E + x(dE/dx)
The force F− acting on −Q is along the −x direction and its magnitude is given by
F− = QE
and that on +Q is along +x and is given by
F+ = QE = Q[E + x(dE/dx)].
The net force F along the +x direction is
F = F+ − F− = Q[E + x(dE/dx)] − QE
F = Qx(dE/dx) = p(dE/dx).
or
The force is along the +x direction. The force changes direction if p is pointing in the opposite direction.
When a dipole moment is placed in a random direction in an electric field, it will always rotate to align
with the electric field. Thus, it will rotate until p is along E. Since the field is nonuniform, it will then
experience a net force towards higher electric fields. Thus, p rotates and moves towards higher fields.
When a dielectric is placed in a nonuniform electric field, it develops a polarization P along the field just
– it becomes polarized. This polarization P is p in Figure 7Q10-1. We can of course represent the polarized
dielectric by the surface charges +Q and −Q just as we used +Q and −Q for p in Figure 7Q10-1. Following
the above lines of argument, since induced P is along the field direction, the dielectric experiences a net
force F towards higher fields along +x direction,
F = P(dE/dx)
The water jet near a charged comb is attracted to the comb because water is a dielectric, develops a
polarization P along the field and becomes attracted towards the higher field region which is near the
comb.
7.11 Ionic and electronic polarization Consider a CsBr crystal that has the CsCl unit cell crystal
structure (one Cs+−Br− pair per unit cell) with a lattice parameter (a) of 0.430 nm. The electronic
polarizability of Cs+ and Br− ions are 2.710−40 F m2 and 5.310−40 F m2, respectively, and the mean ionic
polarizability per ion pair is 5.810−40 F m2. What is the low-frequency dielectric constant and that at
optical frequencies?
Solution
The CsBr structure has a lattice parameter given by a = 0.430 nm, and there is one CsBr ion pair per unit
cell. If n is the number of ion pairs in the unit cell, the number of ion pairs, or individual ions, per unit
volume (N) is
N=
n
1
=
3
a
0.430  10 −9 m
(
)
3
=1.258  1028 m−3
At low frequencies (subscript "LF") both ionic and electronic polarizability contribute to the relative
permittivity. We can use Equation 7.21 by taking i as the mean ionic polarizability per ion pair, eCs as
the electronic polarizability of Cs+ and eBr as the electronic polarizability of Br−, that is,
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Chapter 7
 rLF − 1
1
(N i + N eCs + N eBr )
=
 rLF + 2 3 o
Remember that (Ni + NeCs + NeBr) should be written as (Nii + NCseCs + NBreCl), but since there is a
one-to-one ratio between the number of molecules and ions in CsBr, we can take all the Ns to be the same.

εrLF =
1
(Nαi + NαeCs + NαeBr )(εrLF + 2) + 1
3εo

εrLF =
2 N (αi + αeCs + αeBr ) + 3εo
3εo − N (αi + αeCs + αeBr )
substituting
εrLF =

(
(
)(
) (
)
)
2 1.258 10 28 m −3 5.8 10 −40 F m 2 + 2.7 10 −40 F m 2 + 5.3 10 −40 F m 2 + 3 8.854 10 −12 F m −1
3 8.854 10 −12 F m −1 − 1.258 10 28 m −3 5.8 10 −40 F m 2 + 2.7 10 −40 F m 2 + 5.3 10 −40 F m 2
) (
)(
rLF = 6.67
At optical frequencies there is no contribution from ionic polarization. We only consider electronic
polarization of individual ions and therefore the relative permittivity at optical frequencies, rop, is
 rop =

(
(
)(
) (
)
)
2 1.258 10 28 m −3 2.7 10 −40 F m 2 + 5.3 10 −40 F m 2 + 3 8.854 10 −12 F m −1
3 8.854 10 −12 F m −1 − 1.258 10 28 m −3 2.7 10 −40 F m 2 + 5.3 10 −40 F m 2
) (
)(
rop = 2.83
Comment: The electronic polarizability is from the CRC Handbook of Chemistry and Physics, 94th Edition, 2013,
p12-14, converted to SI units. The refractive index of commercial CsBr crystals at a wavelength 1 m is 1.678,
which implies rop = 2.81 (See http://refractiveindex.info/?shelf=main&book=CsBr&page=Li, 19 February 2017).
The dielectric constant at 2 MHz (includes the ionic contribution) is quoted as 6.51 (See
http://www.korth.de/index.php/162/items/12.html; 19 February 2017). This is 2.5% smaller than the calculated
value above. Clearly i needs to be slightly smaller.
7.12 Ionic polarizability in KCl KCl has the same crystal structure as NaCl. KCl’s lattice parameter is
0.629 nm. The electronic polarizability of K+ is 0.92  10−40 F m2 and that of Cl− is 4.0  10−40 F m2. The
dielectric constant at 1 MHz is given as 4.80. Find the ionic polarizability i and dielectric constant at
optical frequencies rop.
Solution
The KCl structure has a lattice parameter given by a = 0.629 nm, and there are 4 KCl ion pairs per unit
cell . The number of ion pairs, or individual ions, per unit volume (N) is therefore:
N=
4
4
=
3
a
0.629  10 −9 m
(
)
3
= 1.607  1028 m−3
The electronic polarizability of the K+ ion is given as eK = 0.92  10−40 F m2, and polarizability of the
Cl− ion is given as eCl = 4.0  10−40 F m2. The relative permittivity at optical frequencies, rop, is given
by
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Chapter 7
 rop − 1
1
(N eK + N eCl )
=
 rop + 2 3 o
so that
 rop =
2 N ( eK +  eCl ) + 3 o
3 o − N ( eK +  eCl )

 rop =
2 1.607 10 28 m −3 0.92 10 −40 F m 2 + 4.0 10 −40 F m 2 + 3 8.854 10 −12 F/m
3 8.854 10 −12 F/m − 1.607 10 28 m −3 0.92 10 −40 F m 2 + 4.0 10 −40 F m 2

rop = 2.27
(
(
)(
) (
) (
)(
)
)
At low frequencies (e.g., 1 MHz), both ionic and electronic polarization contribute
 rLF − 1
1
(N i + N eK + N eCl )
=
 rLF + 2 3 o

(
)(
4.80 − 1 1.607 10 28 m −3  i + 0.92 10 −40 F m 2 + 4.0 10 −40 F m 2
=
4.80 + 2
3 8.854 10 −12 F m −1
(
)
)
solving this we find
i = 4.3  10−40 F m2
Comment: The electronic polarizability is from the CRC Handbook of Chemistry and Physics, 94th Edition, 2013,
p12-14, converted to SI units. The refractive index of commercial KCl crystals at a wavelength 1 m is 1.480,
and inc which implies rop = 2.19, close to the calculated value (by approximately 3.5%). The dielectric constant
at 1 MHz (includes the ionic contribution to the overall polarization) is quoted as 4.80 in K. Kamiyoshi and Y.
Nigara, Phys. Stat. Solidi A, 3, 735, 1970 (DOI: 10.1002/pssa.19700030320). (n = 1.480 at 1 m is from
https://refractiveindex.info/?shelf=main&book=KCl&page=Li; 19 February 2017)
7.13 Debye relaxation We will test the Debye equations for approximately calculating the real and
imaginary parts of the dielectric constant of water just above the freezing point at 0.2 °C. Assume the
following values in the Debye equations for water: εrdc = 87.46 (dc), ε r = 4.87 (at f = 300 GHz well
beyond the relaxation peak), and τ = 1/ωo = (2π9.18 GHz)−1 = 0.017 ns. Calculate the real and
imaginary,  r and  r , parts of εrdc for water at frequencies in Table 7.12, and plot both the
experimental values and your calculations on a linear–log plot (frequency on the log axis). What is your
conclusion? (Note: It is possible to obtain a better agreement by using two relaxation times or using
more sophisticated models.)
Table 7.12 Dielectric properties of water at 0.2 C
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Chapter 7
SOURCE: Data extracted from Buchner, R., et al., Chemical Physics Letters, 306, 57, 1999
Solution
Using Equations 7.34a and b and substituting all relevant parameters into these we can carry out the
calculations as follows
At f = 0.3 GHz
εr = εr +
εrdc − εr
87.46 − 4.87
= 4.87 +
= 87.38

1 + ( )
1 + (2π  0.3 109  0.017 10 −9 ) 2
and
 r =
(εrdc − εr )( ) (87.46 − 4.87)(2π  0.3  0.017)
=
= 2.64
1 + ( ) 2
1 + (2π  0.3  0.017) 2
Repeating the above calculations at different frequencies will yield the calculated values for the dielectric
constant as shown in the following table,
Table7Q13-1: Calculated and experimental values of dielectric constant r in terms of real and imaginary parts.
Freq. (GHz)
0.3
0.5
1
1.5
3
5
9.18
10
20
40
70
100
300
εr '
87.46
87.25
86.61
85.34
76.20
68.19
46.13
42.35
19.69
10.16
7.20
6.14
4.87
εr "
2.60
4.50
8.85
13.18
24.28
34.53
40.55
40.24
30.23
17.68
11.15
8.31
3.68
εr '
87.38
87.23
86.53
85.39
79.77
69.13
46.98
43.45
19.71
9.16
6.32
5.59
4.95
εr "
2.64
4.40
8.72
12.90
24.00
34.32
41.29
41.21
31.71
18.33
10.85
7.66
2.57
Expt.
Calc.
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Chapter 7
Figure 7Q13-1: Linear log plot of the real and imaginary parts of the relative permittivity (dielectric constant)
versus frequency. The solid lines are the calculated real and imaginary parts of the relative permittivity, and are
listed in table 7Q13-1. The circles (points) are the experimental values.
From Figure 7Q13-1, we can see that the dielectric relaxation of water can be approximated by Debye
equations.
7.14 Debye and non-Debye relaxation and Cole–Cole plots Consider the Debye equation
 r =  r +
( rdc −  r )
1 + j
Debye relaxation
and also generalized dielectric relaxation equation, which “stretches” (broadens) the Debye function,
 r =  r +
( rdc −  r )
[1 + ( j ) ]
Generalized dielectric relaxation
Take τ = 1, εrdc = 5, εr∞ = 2, and α = 0.8, and β = 1. Plot the real and imaginary parts of εr vs.
frequency (on a log scale) for both functions above from ω = 0, 0.1/τ, 1/3τ, 1/τ, 3/τ and 10τ. For the
same ω values, plot  r versus  r (Cole-Cole plot) for both functions using a graph in which the x and y
axes have the same divisions. What is your conclusion?
Solution
Using the two equations given and keeping in mind that  r =  r − j r , we will be able to generate the
following plots by substituting all the relevant parameters into the equations and solving the equations
numerically, as in the following sample calculation
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Chapter 7
For Debye relaxation
At ω = 0.1/τ, from Equation 7.34a and b,
 r =  r +

r −r
5−2
= 2+
= 4.9703
2
1 + ( )
1 + (0.1) 2

dc
and
 r =
( rdc −  r )( )
1 + ( )
2
=
(5 − 2)(0.1)
= 0.2970
1 + (0.1) 2
For the generalized dielectric relaxation function with  = 0.80, which represents a non-Debye relaxation,
we have
At ω = 0.1/τ,
r = r +
dc
r −r
5−2
3
= 2+
= 2+
 
0.8
[1 + ( j ) ]
[1 + ( j 0.1) ]
[1 + ( j 0.1) 0.8 ]
dc

Solving the above equation numerically on a good calculator and using a mathematics software (see Note
at the end) will yield,
 r = 4.8021 − j 0.4026
Similarly, the values for the dielectric constant at various frequencies can be calculated as in Table 7Q141 and data plotted as in Figure 7Q14-1. Clearly, the non-Debye relaxation has a broader loss (in r)
peak and a stretched r.
The cole-cole plot is generated by simply plotting εr" vs . εr, that is, the imaginary part vs. the real part
as in Figure 7Q14-2. The x-axis and the y-axis units have the same length so that the Debye relaxation is
a perfect semicircle in this plot; this is shown by the red circle passing through the red points. The nonDebye relaxation ( = 0.80) on this plot is not a semicircle, but half of an ellipse; it is shown as the blue
half ellipse passing through the blue square points. The conclusion is that non-Debye relaxations have "a
stretched circle" shape; the circle becomes and ellipse.
Table 7Q14-1 Calculation of real and imaginary parts of a complex dielectric constant for Debye and non-Debye
relaxation.
ω (rad/s)
Debye
Non-Debye
0.0000
0.1000
0.3333
1.0000
3.0000
10.0000
εr'
5.0000
4.9703
4.7000
3.5000
2.3000
2.0297
εr"
0.0000
0.2970
0.9000
1.5000
0.9000
0.2970
εr'
5.0000
4.8021
4.3687
3.5000
2.6313
2.1979
εr"
0.0000
0.4026
0.8290
1.0898
0.8290
0.4026
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Chapter 7
Figure 7Q14-1 Top: Linear-log plot of dielectric constant with its real and imaginary parts versus frequency.
Bottom: Log-log plot of the dielectric constant with its real and imaginary parts versus frequency. Red circle (⚫),
real part for Debye relaxation. Blue squares (◼), real part for non-Debye relaxation. Red diamond (◆), imaginary
part for Debye relaxation. Blue triangle (), imaginary part for non-Debye relaxation.
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Chapter 7
Figure 7Q14-2: Cole-Cole plot for the same frequencies. Red circle (⚫), Debye relaxation. Blue squares (◼),
non-Debye relaxation. The green square shows that y-axis and x-axis divisions are the same length. The red circle
is the Debye relaxation. The blue circle is the non-Debye relaxation. The green box is a nearly perfect square that
shows the axes have the same lengths for the divisions.
Note: We can easily do such complex number calculations online. For example on
http://www.wolframalpha.com
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Chapter 7
7.15 Equivalent circuit of a polyester capacitor Consider a 1 nF polyester capacitor that has a polymer
(PET) film thickness of 1 m. Calculate the equivalent circuit of this capacitor at 50 C and at 120 C for
operation at 1 kHz. What is your conclusion?
Solution
The capacitance is given as 1 nF at room temperature and also at 50 C where r is constant as shown in
Figure 7.39. The relative permittivities at 50 C and 120 C can be found approximately from Figure 7.39
as shown in Figure 7Q15-1. Upon inspection, at 50 C, r = 2.6 and at 120 C r = 2.8.
Figure 7Q15-1 Figure 7.39 showing the real part of the dielectric constant r and loss tangent, tan δ, at 1 KHz
versus temperature for PET. The values at the two temperatures of interest 50 and 120 C are shown.
From the equation for capacitance, the area of the dielectric can be readily found, If d is the thickness of
the film,

C=
ε r ε o A
d
A=
dC
1  10 −6 m 1  10 −9 F
= 4.344  10−5 m2
=
−12
ε r ε o (2.6) 8.854  10 F/m
(
(
)(
)
)
Let tan be the loss tangent or loss factor. From Figure 7Q15-1, at 50 C, tan  0.001 and at 120 C
tan  0.01.
We need the equivalent parallel conductance, Gp, (or resistance Rp) at 50 C and 120 C. At T = 50 C,
(
)
G p = ωC tan  = 2π(1000 Hz ) 110−9 F (0.001) = 6.28  10−9 −1

Rp = 1 / Gp = 1.59  108  or 159 M
At 120 C, the capacitance is different (C) due to the change in r. The loss is higher which means a
lower equivalent parallel resistance Rp. Assuming no change in the area A,
C =
(2.8)(8.854  10 −12 F/m )(4.344  10 −5 m 2 ) = 1.077  10−9 F or 1.077 nF
(1 10
−6
m
)
A value that is 7.7% higher than at the lower temperature. The new conductance and resistance are,
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(
Chapter 7
)
G p = ωC tan  = 2π(1000 Hz ) 1.077 10−9 F (0.01) = 6.77  10−8 −1
Rp = 1 / Gp = 1.48  107  or 14.8 M
Lower Rp means higher losses because this is a parallel equivalent resistance across the capacitor as shown
in Figure 7Q15-2
Figure 7Q15-2 The equivalent circuit of a capacitor with a parallel resistance Rp across an ideal capacitor C.
A variety of polyester capacitors (Photo by SK)
7.16 Student microwaves mashed potatoes A microwave oven uses electromagnetic waves at 2.45 GHz
to heat food by dielectric loss, that is, making use of r of the food material, which normally has
substantial water content. A student microwaves 60 cm3 of mashed potatoes for 40 seconds, then takes
them out and measures their temperature to be about 71 C. The room temperature is 23 C. The specific
heat capacity (cs) and density of mashed potatoes are approximately 3.8 J g−1 K−1 and 1.0 g cm−3. At 2.45
GHz, mashed potatoes have r  15. Assume that heat generated in mashed potatoes by the absorption of
microwaves increases the temperature, and ignore any heat conducted away. Calculate the rms electric
field Erms generated by the microwaves in the mash potatoes. (Note: You can use Erms instead of E in
Equation 7.32.)
Solution
First recall that from Equations 7.32, the power dissipated per unit volume is
2
Wvol = Erms
 o r
(1)
We need to know how much power per unit volume Wvol, we need to provide. The total heat Q that we
need to generate in the mash potatoes is given by
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Q = mcs T
Chapter 7
(2)
where cs is the specific heat capacity and m is the mass of the mash potatoes, that is m = V. This heat is
generated in the volume V in t seconds (duration of microwaving). The heat generated per second per
unit volume is
Q mcs T ( V )cs T
T
=
=
= cs
Vt
Vt
Vt
t
(3)
If we ignore thermal conduction losses from mash potatoes to the plate and air, this heating power per
unit volume must be Wvol.
Thus,
Wvol = csT/t = [(1 g cm−3)(3.8 J g−1 K−1)](77 – 23 C)] / (40 s) = 5.13 W cm−3
This much energy is generated as heat per unit time and per unit volume.
Now, this energy comes from Equation (1) (or Equation 7.32 in the textbook), and is due to the dielectric
loss represented by r. Thus,
2
2
Wvol = 5.13106 W m−3 = Erms
 o r = (2  2.45109 Hz)Erms
(8.85410−12 F m−1 )(15)
which we can solve for Erms,
Erms = 1.58 kV/m or 1.58 V/mm
We neglected thermal conduction losses from the mash potatoes to the plate.
Comments: A good reference on the science of microwaving food may be found in M. Vollmer, "Physics of the
microwave oven", Physics Education, 39, 74−81, 2004. It has the dielectric properties of mash potatoes and other
food items. Remember that the solution is only an estimate. The field inside the oven is not uniform because there
are standing waves within the oven. Values for cs and density of mash potatoes are from J. Wang, R. G. Olsen, J.
Tang and Z. Tang, "Influence of mashed potato dielectric properties and circulating water electric conductivity on
radio frequency heating at 27 MHz", Journal of Microwave Power & Electromagnetic Energy (online), 42 (No. 2),
pp31−46, 2008. Further r decreases with increasing temperature so that it would actually take longer than 40 s to
heat the mash potatoes to 77 C. Figure 5 in Physics Education, 39, 74−81, 2004, clearly shows the decrease in r
with temperature. The typical rms fields that heat the food in microwave ovens are usually in the range 1 − 2 kV
m−1.
7.17 Dielectric loss per unit capacitance Consider the three dielectric materials listed in Table 7.13
with the real and imaginary dielectric constants, r' and r''. At a given voltage, which dielectric will have
the lowest power dissipation per unit capacitance at 1 kHz and at an operating temperature of 50 C? Is
this also true at 120 C?
Table 7.13 Dielectric properties of three insulators at 1 kHz
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Chapter 7
SOURCE: Data taken using a DEA by Kasap and Nomura (1995)
Solution
Since we are merely comparing values, assume that voltage V = 1 V for calculation purposes. The power
dissipated per unit capacitance (Wcap), following Examples 7.5 and 7.6, is given by,
Wcap = V 2 ω
εr
εr
where  is the angular frequency (2f) and r and r represent the real and imaginary components of the
relative permittivity r, respectively. (See p686 where this equation is derived.) As a sample calculation,
the power dissipated in polycarbonate is
Wcap = (1 V ) 2π(1000 Hz )
2
(0.003)
(2.47)
= 7.63 W/F
Therefore, 7.63 W/F or, expressed more conveniently, 7.63 nW of power dissipated per nF at 50 C at 1
V at 1 kHz.
Carrying out similar calculations, we can list the values for all the materials at 50 C and 120 C as in
Table 7Q17-1:
Table 7Q14-2: Power dissipated at different temperatures for the given materials.
50 C
120 C
Power Dissipated
Power Dissipated
(W / F)
(W / F)
Polycarbonate
7.63
7.44
PET
7.31
61.7
PEEK
8.41
8.38
Material
At 50 C, all three are comparable in magnitude, but PET has the lowest power dissipation.
At 120 C, polycarbonate has the lowest dissipation, while PET is almost ten times worse. The PET
capacitor is particularly temperature sensitive compared to others. (See also Question 7.15.)
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Chapter 7
7.18 Parallel and series equivalent circuits Figure 7.63 shows simplified parallel and series
equivalent circuits for a capacitor. The elements Rp and Cp in the parallel circuit and the elements Rs and
Cs in the series circuit are related. We can write down the impedance ZAB between the terminals A and B
for both the circuits, and then equate ZAB(Parallel) = ZAB(Series). Show that
Rs =
Rp
1 + (R p C p )
2


1
Cs = C p 1 +
2
 (R p C p ) 
and
Equivalent series resistance and capacitance
and similarly by considering the admittance (1/impedance),


1
R p = Rs 1 +
2
 (Rs Cs ) 
and
Cp =
Cs
1 + (Rs Cs ) 2
Equivalent series resistance and capacitance
A 10 nF capacitor operating at 1 MHz has a parallel equivalent resistance of 100 kΩ. What are Cs and
Rs?
Figure 7.63 An equivalent parallel Rp and Cp circuit is equivalent to a series Rs and Cs circuit. The elements Rp
and Cp in the parallel circuit are related to the elements Rs and Cs in the series circuit.
Solution
The impedance of the parallel and series equivalent circuits must be the same at a given frequency.
Thus, we need ZAB−p = ZAB−s.
Consider the parallel equivalent circuit,
YAB−p = Gp + jCp
We can write the impedance across AB for the parallel equivalent circuit as
Z AB − p =
1
YAB − p
=
G p − jC p
1
= 2
G p + jC p G p + (ωC p ) 2
(1)
This has to be the same as the equivalent series circuit, that is ZAB−p = ZAB−s.
Consider,
Z AB − s = Rs +
1 + jRs Cs
j − Rs Cs Rs Cs − j
1
=
=
=
j C s
j C s
− C s
C s
(2)
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Chapter 7
so that equating Equations (1) and (2)
G p − jC p
Z AB − p =
G + (ωC p )
2
p
= Z AB − s =
2
Rs Cs − j
Cs
(3)
We now need to equate the real and imaginary parts. The reals parts give
Gp
G + (ωC p )
2
p

Rs =

Rs =
2
=
Rs Cs
Cs
Gp
G + (ωC p )
2
p
2
=
1/ Rp
1 / R + (ωC p ) 2
2
p
Rp
(4)
1 + (ωR p C p ) 2
Using the imaginary parts,
C p
1
=
2
G + (C p )
Cs
2
p
G p2 + (C p ) 2
Cs =

1 + (R p C p ) 2 
Cs = C p 

2 2 2
  R p C p 
 2C p
=
R p−2 + (C p ) 2

 2C p
=
1 + (R p C p ) 2
 2 R p2C p
(5)
Similarly, we can use the principle that at a given frequency, the admittance YAB−p of the parallel
equivalent circuit must be the same as that, YAB−s, of the series equivalent circuit.
Given,
Z AB − s = Rs +
1
j C s
we have

YAB − s =
1
Z AB − s
1
C s
1
1
=
=
=
1
1
1
Rs +
Rs − j
Rs2 +
j C s
C s
(Cs ) 2
Rs + j
Since we need YAB−p = YAB−s, we equate the two admittances
YAB − p =
1
+ jC p = YAB − s
Rp
1
C s
=
1
Rs2 +
(Cs ) 2
Rs + j
We need to equate the real and imaginary parts. Equating the real parts,
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Chapter 7




Rs
1
1
1


=
=
1
Rp R 2 + 1
Rs 1 +

s
 (Rs Cs ) 2 
(Cs ) 2



1
R p = Rs 1 +
2
 (Rs Cs ) 
(6)
Equating the imaginary parts we have
C p =

1
C s
Rs2 +
1
(C s ) 2


1
C p = Cs 
2
1 + (Rs Cs ) 
(7)
Clearly we can calculate the equivalent circuit parameters for one circuit from the other.
Given Rp = 100 k and Cp = 10.00 nF and f = 106 Hz and hence  = 2f = 6.28106 rad/s, we can
substitute these values into Equations (4) and (5) to find
Rs = 0.0025  and Cs = 10.00 nF
The equivalent series capacitance is the same as the parallel equivalent circuit capacitance and Rs =
0.0025 
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Chapter 7
7.19 Tantalum capacitors Electrolytic capacitors tend to be modeled by a series Rs + 1/(jωCs)
equivalent circuit. A nominal 22 μF Ta capacitor (22 μF at low frequencies) has the following
properties at 10 kHz, εr  20 (at this frequency), tanδ  0.05, dielectric thickness d = 0.16 μm, effective
area A = 150 cm2. Calculate Cp, Rp, Cs and Rs. (See Question 5.18 for Rs, Cs)
Solution
Use Equation 7.29 to find the capacitance at 10 kHz. Given r = 20 at f = 10 kHz,
C=
A o r (150 10 −4 m 2 )(8.854 10 −12 F m −1 )(20)
= 16.60 F
=
d
(0.16 10 −6 m)
Notice that this is smaller than the nominal value 22 F. This is now Cp. Since we know tan = 0.005,
r = rtan = (20)(0.005) = 1.00. From Equation 7.30 we have,
Gp =

A o r
d
(2 10 103 Hz)(150 10 −4 m 2 )(8.854 10 −12 F m −1 )(1.00)
=
(0.16 10 −6 m)
Gp = 0.05216 −1
which corresponds to
Rp = 1/Gp = 1/0.06216 = 19.17 
Using the parallel to series conversion equation in Question 7.18
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Rs =
Rp
1 + (ωR p C p )
2
=
Chapter 7
(19.17 )
1 + [(2 10 10 Hz)(19.17 )(16.6 10 −6 F)]2
3

Rs = 0.0478  or 48 m
and
1 + (R p C p ) 2 
 1 + [(2 10 103 Hz)(19.17 )(16.6 10 −6 F)]2 


Cs = C p 
=
(
16
.
60
μF
)

2 2 2
3
−6
2 
 [(2 10 10 Hz)(19.17 )(16.6 10 F)] 
  R p C p 

Cs = 16.64 F
Comment: The problem is somewhat oversimplified, because it gives the impression that the loss is totally due to
dipole relaxations (incorporated into r) in the dielectric; that is, the losses during polarization and depolarization.
There are also conductive losses (Joule heating) arising from a leakage current, represented by Ileakage = IL. The
leakage current involves the drift of ions, depends on the voltage and also on time – it decreases with time under a
dc bias. Typically, IL in data sheets refers to leakage currents after 5 minutes from the application of bias. We can
represent conductive (Joule) losses with a finite conductivity . Then, the imaginary part r'', in addition to dipole
polarization losses, will also depend on . More advanced textbooks incorporate both polarization and conduction
losses into r by
r = losses from polarization + (  o)
The problem, of course, is that  is field dependent, time dependent and frequency dependent. The quoted and
measured value tan  0.05 includes all losses that appear in the measurement of tan (i.e. polarization and
conduction losses).The measurement of dielectric properties at f cannot differentiate between the two losses. If we
double the r component, we would find Rp = 9.57  halved, and Rs = 0.095 , doubled.
A selection of tantalum capacitors (Photo by SK)
7.20 Tantalum versus niobium oxide capacitors Niobium oxide (Nb2O5) is a competing dielectric to
Ta2O5 (the dielectric in the tantalum capacitor). The dielectric constants are 41 for Nb2O5 and 27 for
Ta2O5. For operation at the same voltage, the Ta2O5 thickness is 0.17 µm, and that of Nb2O5 is 0.25 µm.
Explain why the niobium oxide capacitor is superior (or inferior) to the Ta capacitor. Use a quantitative
argument, such as the capacitance per unit volume. What other factors would you consider if you were
choosing between the two?
Solution
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Assuming the dielectrics are lossless, we can write the capacitance as
C=
A o r
d
Capacitance per unit volume is then
Cvol =
A o r
 
1

= o 2r
d
A d
d
Substituting all relevant parameters into the above equation, for Ta2O5, we have,
Cvol (Ta 2O5 ) =
or
 o r
d2
=
(8.854 10 −12 F m −1 )(27)
= 8.27103 F m−3
−6
2
(0.17 10 m)
Cvol(Ta2O5) = 8.27 F mm−3
and for Nb2O5
Cvol ( Nb 2O5 ) =
or
 o r
d2
=
(8.854 10 −12 F m −1 )(41)
= 5.81103 F m−3
−6
2
(0.25 10 m)
Cvol(Nb2O5) = 5.81 F mm−3
Based on the simple capacitance per unit volume calculations alone, it appears that the Nb2O5 capacitor
is inferior to the Ta2O5 capacitor. However we should also consider their respective cost and dielectric
losses at the operating frequency, dielectric breakdown and long-term stability as well. If the thickness
can be made the same, then it seems that the Nb-oxide capacitor is superior to the Ta-oxide capacitor in
terms of r. In addition, Nb-oxide capacitors with NbO-Nb2O5 exhibit superior stability and reliability
over a long time.
Comment: A good discussion is available in R. Hahn, M. Randall and J. Paulsen (KEMET Electronics Corporation),
"The Battle of Maximum Volumetric Efficiency – Part 1: When Technologies Compete, Customers Win",
Proceedings CARTS Europe 2007 Symposium, October-November 2007, Barcelona, Spain, pp63-73 (© 2007
Electronics Components, Assemblies, and Materials Association, Arlington, VA, USA). See also R.P. Deshpande,
Capacitors: Technology and Trends. Tata-McGraw-Hill Education, 2013, Chapter 10; some chapters available on
books.google.
*7.21 TCC of a polyester capacitor Consider the parallel plate capacitor equation
C=
 o  r xy
z
where r is the relative permittivity (or r'), x and y are the side lengths of the dielectric so that xy is the
area A, and z is the thickness of the dielectric. The quantities r, x, y and z change with temperature. By
differentiating this equation with respect to temperature, show that the temperature coefficient of
capacitance (TCC) is
TCC =
1 dC 1 d r
=
+
C dT  r dT
Temperature coefficient of capacitance
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Chapter 7
where  is the linear expansion coefficient defined by
=
1 dL
L dT
where L stands for any length of the material (x, y or z). Assume that the dielectric is isotropic,  is the
same in all directions. Using r' versus T behavior in Figure 7.64 and taking  = 50  10−6 K−1 as a typical
value for polymers, predict the TCC at room temperature and at 10 kHz.
Figure 7.64 Temperature dependence of εr at 10 kHz. (Data taken by Kasap and Maeda (1995))
Solution
The real part of the relative permittivity, r, is usually simply written as r. We are given xy = area (A)
and z = thickness. From the definition of the linear expansion coefficient :
dx
= x
dT
dy
= y
dT
dz
= z
dT
Now differentiate C with respect to T (remember that r depends on temperature):
C=

 o  r xy
z
d
dx
dy 
dz

+  o r x
  o xy r +  o  r y
 z −  o  r xy
dT
dT
dT 
dT
dC 
=
2
dT
z
Substitute for the derivatives of x, y and , according to the definition of  and simplify
dC
=
dT

dC
=
dT
 o xyz
 o xy
d r
+  o r xyz +  o r xyz −  o r xyz
dT
z2
d r
+  o  r xy
dT
z
The temperature coefficient of capacitance is defined as:
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TCC =
Chapter 7
1 dC
C dT
substitute for dC/dT:
TCC =
 o xy
d r
+  o  r xy
dT
Cz
substitute for C:
TCC =
Simplify,
TCC =
d r
+  o  r xy
dT
  o  r xy 

z
 z 
 o xy
1 d r
+
 r dT
Figure 7Q21-1: Temperature dependence of r at 10 kHz from Figure 7.64.
From the slope of the straight line in Figure 7Q21-1, we can estimate the value of dr/dt:
d r 2.5925 − 2.57
=
= 0.000375 ˚C −1 or K−1
dT
80 ˚C − 20 ˚C
Using the given value of  = 50  10−6 K−1 and r = 2.57 (r at room temperature from inspecting Figure
7.64):
TCC =

1 d r
1
+ =
0.000375 K −1 + 50  10 −6 K −1
(2.57 )
 r dT
(
)
TCC = 0.000196 K−1 or C−1
This is 196 ppm per C. PET capacitors are quoted to have typically 200 ppm/C. (Very close.)
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Chapter 7
7.22 Breakdown voltage of SF6 and N2 gaseous insulation Experiments have been carried on
breakdown between two spherical electrodes (5 cm in diameter) separated by 1 mm in two gases as
insulation: N2 and SF6. Table 7.14 summarizes the measurements of Vbr at different pressures P. Plot Vbr
versus Pd on a log-log plot and find x in Vbr  (Pd)x.
Table 7.14 Breakdown voltage between electrodes separated by 1 mm in N2 and SF6
N2
P (MPa)
0.74
1.48
2.14
2.83
3.48
4.31
Vbr (kV)
21.2
41.0
57.9
73.0
87.8
105.8
SF6
P (MPa)
0.76
1.47
2.17
2.77
3.41
4.49
Vbr (kV)
55.2
110.0
156.2
191.9
225.2
273.9
Note: Data extracted from Figure 5, D. Kosch, SF6 properties, and use in MV and HV switchgear, E/CT 188, Schneider
Electric (February 2003).
Solution
Table 7Q22-1 shows the Excel data sheets used in calculating the Pd product for each gas from Table
7.14. The electrode separation d is fixed; d = 1 mm. We can plot the results, Vbr vs. Pd on a log-log plot
as shown in Figure 7Q22-1. On a log-log scale, a straight line represents a power law dependence, that is
Vbr = A(Pd)x, where A and x are constants. We can use Excel's trend line feature and select a power law fit
to generate the best power law fits shown in Figure 7Q22-1. Indeed, the results in this voltage range
indicate Vbr = A(Pd)x type of behavior. The results of bet fits to two decimal places are
SF6,
x = 0.91,
Vbr  (Pd)0.91
N2,
x = 0.91,
Vbr  (Pd)0.91
Table 7Q22-1 Pressure, breakdown voltage Vbr and the Pd product from data given in Table 7.14. Blue area is for
SF6 and the pale red area is for N2. Note that d = electrode separation, given as 1 mm.
d=
1.00E-03
P (MPa)
7.61E-01
1.47E+00
2.17E+00
2.77E+00
3.41E+00
4.49E+00
SF6
Vbr (kV)
5.52E+01
1.10E+02
1.56E+02
1.92E+02
2.25E+02
2.74E+02
Pd (Pa m)
7.61E+02
1.47E+03
2.17E+03
2.77E+03
3.41E+03
4.49E+03
Vbr( kV)
55.2
110.0
156.2
191.9
225.2
273.9
P (MPa)
0.74
1.48
2.14
2.83
3.48
4.31
N2
Vbr
2.12E+01
4.10E+01
5.79E+01
7.30E+01
8.78E+01
1.06E+02
Pd (Pa m)
7.39E+02
1.48E+03
2.14E+03
2.83E+03
3.48E+03
4.31E+03
Vbr (V)
21.2
41.0
57.9
73.0
87.8
105.8
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Chapter 7
Figure 7Q22-1 Log-log plot of breakdown voltage Vbr vs. Pd.
7.23 Dielectric breakdown of gases and Paschen curves Dielectric breakdown in gases typically
involves the avalanche ionization of the gas molecules by energetic electrons accelerated by the applied
field. The mean free path between collisions must be sufficiently long to allow the electrons to gain
sufficient energy from the field to impact-ionize the gas molecules. The breakdown voltage, Vbr, between
two electrodes depends on the distance d between the electrodes as well as the gas pressure P, as shown
in Figure 7.65. Vbr versus Pd plots are called Paschen curves. We consider gaseous insulation, air and
SF6, in an HV switch.
a. What is the breakdown voltage between two electrodes of a switch separated by a 5 mm gap at 0.1
atm when the gaseous insulation is air and when it is SF6?
b. What are the breakdown voltages in the two cases when the pressure is 10 times larger? What is
your conclusion?
c. At what pressure is the breakdown voltage a minimum?
d. What air gap spacing d at 1 atm gives the minimum breakdown voltage?
e. What would be the reasons for preferring gaseous insulation over liquid or solid insulation?
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Chapter 7
Figure 7.65 Breakdown voltage versus (pressure × electrode spacing) (Paschen curves)
Solution
a. At pressure P = 0.1 atm = 1.013  104 Pa and air gap d = 5 mm, P  d = (1.013  104 Pa)(0.005 m) =
50.65 Pa m  50 Pa m. From Figure 7Q23-1, the corresponding values of breakdown voltage for air (Vair)
and for SF6 (VSF6) are:
Vair  2,900 V or 2.9 kV
VSF6  5,300 V or 5.3 kV
Figure 7Q23-1 Breakdown voltage versus (pressure  electrode spacing) (Paschen curves)
b. At P = 1 atm = 1.013  105 Pa and d = 5 mm, P  d = (1.013  105 Pa)(0.005 m) = 506.5 Pa m  500
Pa m. From Figure 7Q23-1,
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Chapter 7
Vair  18,000 V or 18 kV
VSF6  45,000 V or 45 kV
When the pressure increases by 10 times, the breakdown voltages for SF6 increases by a factor 8.5. Clearly
Vbr depends strongly on the Pd product for SF6 in this range. For air, there is also a strong increase, by a
factor of 6.2.
Conclusion: The breakdown voltage increases with pressure. (It is better to use gaseous insulation at a
higher pressure.)
c. With a gap length of 5 mm, we need to know the pressure at which the breakdown voltage is a minimum.
From the graph, the minimum breakdown voltage of air is about Vair  320 V, and the minimum for SF6
is about VSF6  500 V. The corresponding values for P  d are (P  d)air  0.70 Pa m and (P  d)SF6 = 0.25
Pa m. From these we can determine the values of pressure needed for minimum breakdown voltage:
Pair d = 0.65 Pa m
0.65 Pa m
0.005 m

Pair =

Pair = 130 Pa or 0.0013 atm
PSF6 d  0.23 Pa m
PSF6 =
0.23 Pa m
0.005 m
PSF6 = 46 Pa or 0.00046 atm
Low pressures are needed for minimum breakdown which explains why discharge tubes operate at a low
pressure.
d. At a set pressure P = 1 atm = 1.013  105 Pa, the air gap spacing d for minimum breakdown voltage
can be found in a similar manner to the one above, using the same values for P  d:
Pd air = 0.65 Pa m
0.65 Pa m
1.013 105 Pa

d air =

dair = 6.4  10−6 m or 6.4 m
This value corresponds to a breakdown voltage of 320 V. Therefore a gap of about 6.4 m will only need
320 V for breakdown. The corresponding field very roughly is ~320 V / 6.4 m = 50 MV m−1.
e. HV and high current switches or relays that have moving parts cannot be practically insulated using
solid dielectrics. Liquid dielectrics are not as efficient as gaseous dielectrics because some undergo
chemical changes under partial discharges. Further, they have a higher viscosity than gases that may affect
the efficiency of the moving parts. Gas naturally permeates all the necessary space or locations where
insulation is critical.
Note: There are obviously reading errors from the graph, especially on a log axis, but the values are roughly correct.
The Paschen curves lose their accuracy when the electrode spacing is a few microns. Further, the curves are different
for different electrode shapes and under dc or ac conditions.
Comments: The data for Figure 7.65 were extracted from two sources: (a) SF6: N.H. Malik and A.H. Qureshi,
"Breakdown Mechanisms in Sulphur-Hexafluoride", IEEE Transactions on Electrical Insulation, Vol. EI-13, pp.
135-145 (1978) (b) Air: Vasily Y. Ushakov, Insulation of High-Voltage Equipment, Springer (Heidelberg, 2004),
p29, Fig 3.1b
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Chapter 7
*7.24 Capacitor design Consider a nonpolarized 100 nF capacitor design at 60 Hz operation. Note that
there are three candidate dielectrics, as listed in Table 7.15.
a. Calculate the volume of the 100 nF capacitor for each dielectric, given that they are to be used under
low voltages and each dielectric has its minimum fabrication thickness. Which one has the smallest
volume?
b. How is the volume affected if the capacitor is to be used at a 500 V application and the maximum
field in the dielectric must be a factor of 2 less than the dielectric strength? Which one has the
smallest volume?
c. At a 500 V application, what is the power dissipated in each capacitor at 60 Hz operation? Which
one has the lowest dissipation?
Table 7.15 Comparison of dielectric properties at 60 Hz (typical values)
Solution
Note: All sample calculations are for the polymer film (PET). All methods of calculation for the other
materials are identical, and the obtained values are summarized in Table 7Q23-1.
a. To find the volume needed for C = 100 nF, given that the dielectric has the minimum practical thickness,
d (Table 7.15), find the capacitance per unit volume (Cvol),
Cvol =
 o r
d
2
=
(8.854 10
(110
−12
−6
)
F/m (3.2)
m
)
2
= 28.33 F/m 3 = 28.33 F cm−3 or 28.3 nF mm−.
The volume V can now be found as follows:
V = C / Cvol = (100  10−9 F) / (28.33 F/m3) = 3.53  10−9 m3 = 3.53 mm3
This is the volume at low voltage operation based on the minimum practical thickness.
b. Suppose that d is the minimum thickness (in m) which gives a maximum field of half of Ebr at 500 V.
Then,
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Chapter 7
V
1
E br = max
2
d

d =2
Vmax
500 V


−5
= 2
 = 6.667 10 m or 66.7 m
7
E br
 1.50 10 V/m 
Now the capacitance per unit volume can be found,
Cvol =

 o r
d
2
(8.854 10 F/m )(3.2) = 0.006374 F/m
=
(6.667 10 m)
−12
−5
3
2
V = C / Cvol = (100  10−9 F) / (0.006374 F/m3) = 1.57  10−5 m3 or 15.7 cm3
This is the dielectric volume at 500 V.
c. The power dissipation in the capacitor at 500 V (60 Hz operation) can be found by first obtaining the
power lost per unit volume Wvol. It is given by:
Wvol =
Ebr
2
2
 o r tan 
where  is the safety factor (assumed to be equal to 2) and  = 2f is the angular frequency. Thus,
substituting the values, we have
Wvol

(1.5  10
=
7
V/m
(2)2
) (2 (60 Hz))(8.854  10
2
−12
)
(
F/m (3.2) 5  10 −3
)
Wvol = 3004 W/m3
The total power dissipated (W) is therefore:
W = WvolV = (3004 W/m3)(1.57  10−5 m3)

W = 0.0472 W or 47.2 mW
The results are summarized in Table 7Q24-1. Upon inspection we see that for part a and part b, high-K
ceramic has the smallest volume, and for part c, regular ceramic has the lowest power dissipation.
Table 7Q24-1: Summarized values for volume and power of given capacitors.
Polymer Film
Ceramic
High-K Ceramic
PET
TiO2
(BaTiO3 based)
a
Low voltage volume (mm3)
3.53
12.5
0.627
b
High voltage volume (cm3)
15.7
5.02
0.0627
c
Power dissipated (W)
0.0472
0.00377
0.471
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Chapter 7
*7.25 Dielectric breakdown in a coaxial cable Consider a coaxial underwater high-voltage cable as in
Figure 7.66a. The current flowing through the inner conductor generates heat, which has to flow through
the dielectric insulation to the outer conductor where it will be carried away by conduction and
convection. We will assume that steady state has been reached and the inner conductor is carrying a dc
current I. Heat generated per unit second, Q = dQ /dt, by joule heating of the inner conductor is
Q  = RI 2 =
LI 2
a 2
Rate of heat generation
[7.98]
where  is the resistivity, a the radius of the conductor, and L the cable length.
This heat flows radially out from the inner conductor through the dielectric insulator to the outer
conductor, then to the ambient. This heat flow is by thermal conduction through the dielectric. The rate
of heat flow Q depends on the temperature difference Ti−To, between the inner and outer conductors;
Figure 7.66 (a) The Joule heat generated in the core conductor flows outward radially through the dielectric
material. (b) Typical temperature dependence of the dielectric strength of a polyethylene-based polymeric
insulation
on the sample geometry (a, b and L); and on the thermal conductivity  of the dielectric. From
elementary thermal conduction theory, this is given by
2L
Q  = (Ti − To )
Rate of heat conduction
[7.99]
b
ln  
a
The inner core temperature, Ti, rises until, in the steady state, the rate of joule heat generation by the
electric current in Equation 7.99 is just removed by the rate of thermal conduction through the dielectric
insulation, given by Equation 7.99.
a. Show that the inner conductor temperature is
I 2
b
Ti = To + 2 2 ln  
Steady state inner conductor temperature
[7.100]
2 a   a 
b. The breakdown occurs at the maximum field point, which is at r = a, just outside the inner
conductor, and is given by (see Example 7.12).
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E max =
V
b
a ln  
a
Maximum field in a coaxial cable
Chapter 7
[7.101]
The dielectric breakdown occurs when Emax reaches the dielectric strength Ebr. However the
dielectric strength Ebr for many polymeric insulation materials depends on the temperature, and
generally it decreases with temperature, as shown for a typical example in Figure 7.66b. If the load
current, I, increases, then more heat, Q, is generated per second and this leads to a higher inner core
temperature, Ti, by virtue of Equation 7.100. The increase in Ti with I eventually lowers Ebr so much
that it becomes equal to Emax and the insulation breaks down (thermal breakdown). Suppose that a
certain coaxial cable has an aluminum inner conductor of diameter 10 mm and resistivity 27 n m.
The insulation is 3 mm thick and is a polyethylene-based polymer whose long-term dc dielectric
strength is shown in Figure 7.64b. Suppose that the cable is carrying a voltage of 40 kV and the
outer shield temperature is the ambient temperature, 25 C. Given that the thermal conductivity of
the polymer is about 0.3 W K−1 m−1, at what dc current will the cable fail?
c. Rederive Ti in Equation 7.99 by considering that ρ depends on the temperature as
 =  o 1 +  o (T − To ) (Chapter 2). Recalculate the maximum current in b given that
o = 3.9  10−3 C−1 at 25 C.
Solution
a. The resistance R of the inner conductor is,
R=
L
A
=
L
a 2
where  is the resistivity of the conductor, L is the length and a is the radius.
Joule heating power generated by the current I through the conductor (P) is given by:
P = I2R = I 2
L
a 2
The rate of heat conduction (Q) from inner conductor to outer conductor is:
Q = (Ti − To )
2L
b
ln  
a
Rate of heat conduction
[7.99]
In the steady state, the rate of Joule heating of the inner conductor, P, is equal to the rate of heat flow,
Q, through the insulator. Therefore:
I2
L
2L
= (Ti − To )
2
a
b
ln  
a

(Ti − To ) = I 2
L
b
ln  
2 2
2 a L  a 
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
I 2
b
Ti = To +
ln  
2 2
2 a  a 
Steady state inner conductor temperature
Chapter 7
[7.100]
Note: A major assumption is that the inner conductor resistivity  and thermal conductivity  of the
insulation (dielectric) are constant, that is, do not change with temperature
b. We are given the cable’s characteristics: resistivity  = 27  10−9 , thermal conductivity  = 0.3 W
K−1 m−1 (polyethylene material at 25 C), inner conductor radius a = 0.005 m, and total radius b = 0.005
m + 0.003 m = 0.008 m. The outer temperature is given as To = 25 C + 273 = 298 K, and the applied
voltage is 40 kV. The maximum field Emax depends on the voltage V by Equation 7.100,
E max =
V
40 103 V
=
= 1.702  107 V/m or 17.02 MV/m
b
 0.008m 
a ln   (0.005m )ln 

a
 0.005m 
Figure 7Q25-1 Temperature dependence of the dielectric strength of a polyethylene-based polymeric insulation.
The breakdown field Ebr is equal to this maximum field Emax when the inner temperature Ti = 90 C =
363 K (see Figure 7Q25-1 above).
Now, from Equation 7.99,
Ti = To +
I 2
b
ln   = 90 C
2 2
2 a  a 
(
)
2 2 a 2 (Ti − To )
2 0.3 W K −1 m −1  2 (0.005 m ) (363 K − 298 K )
=
b
 0.008 m 
 ln  
27 10 −9  m ln 

a
 0.005 m 

I=

I = 871 A
(
2
)
(1)
Note: We do not need to convert C to K in Equation (1) since a temperature difference in C is the
same as K; it was done to keep track of units.
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Chapter 7
c. The resistivity depends on the temperature as  =  o 1 +  o (T − To ) . As in a under steady state
operation, the rate of Joule heating P (for power) of the inner conductor is equal to the rate of heat flow
Q through the insulator but now  is temperature dependent. Therefore,
I2
 o 1 +  o (Ti − To )L
2L
= (Ti − To )
2
a
b
(2)
ln  
a
Solving for Ti and simplifying we find
Ti = To +
I2
2 2 a 2
− I 2 o
b
 o ln  
a
Now we have to calculate at what current Ti will reach 90 C and hence thermal breakdown will occur.
Solving for I from Equation (2) we have
I=
2 2 a 2 (Ti − To )
b
 o 1 +  o (Ti − To )ln  
a
So that substituting all the values,
I=
(
)(
) (
)
2 2 5 10 −3 m 0.3 W K −1 m −1 ((90 − 25) K )
= 778 A
8
−9
−3
27 10  m 1 + 3.9 10 ((90 − 25) K ) ln  
5
(
2
)

Note: The operation of a HV cable would be such that there is a safety factor involved, for example, the
maximum field can be specified to be very roughly (1/2)Ebr. Put differently, at operating Ti, Ebr  2Emax
or 34 MV m−, which gives Ti = 65 C. This means the maximum current above from Equation (1)
would be smaller. The recalculations are left as an exercise.
7.26 Piezoelectricity Consider a quartz crystal and a PZT ceramic filter both designed for operation at fs
= 1 MHz. What is the bandwidth of each? Given Young's modulus (Y), density ( ) of each, and that the
filter is a disk with electrodes and is oscillating radially, what is the diameter of the disk for each material?
For quartz, Y = 80 GPa and  = 2.65 g cm−3. For PZT, Y = 70 GPa and  = 7.7 g cm−3. Assume that the
velocity of mechanical oscillations in the crystal is v = Y  and the wavelength  = v/fs. Consider
only the fundamental mode (n = 1).
Solution
We are given the first resonance frequency, fs = 1 MHz. There is a second resonance (antiresonant)
frequency fa related to the first by (see Example 7.15),
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fa =
Chapter 7
fs
1− k 2
where k is the electromechanical conversion factor. From Table 7.8, kquartz = 0.1 and kPZT = 0.72. The
second resonance frequency can then be easily calculated
Quartz
fa =
fs
1− k
2
(110
=
6
Hz
1 − (0.1)
2
) = 1.0050  106 Hz
The bandwidth Bquartz is defined as the difference between these two frequencies:

Bquartz = fa − fs = (1.0050  106 Hz) − (1  106 Hz) = 5000 Hz
PZT
fa =

fs
1− k
2
=
(110
6
Hz
)
= 1.4410  106 Hz
1 − (0.72)
2
BPZT = fa − fs = (1.4410  106 Hz) − (1  106 Hz) = 4.41  105 Hz
A much wider bandwidth than the quartz crystal.
Next, we need to find the diameter of the disks of each material. We are given Young’s modulus (Yquartz
= 80  109 Pa, YPZT = 70  109 Pa) and the density (quartz = 2650 kg/m3, PZT = 7700 kg/m3). From these
we can determine the velocity and wavelength of the mechanical oscillations in the crystal, and then the
diameter (L), from the mechanical resonance condition n(/2) = L,
Quartz
vquartz =
Yquartz
quartz
vquartz
=
(8010 Pa )
(2650 kg/m )
9
3
= 5494 m/s
5494 m/s
= 0.005494 m (5.494 mm)
1106 Hz

quartz =

1
 1

L quartz = n quartz  = 1 (0.005494 m ) = 0.00275 m or 2.75 mm
2
 2

fs
=
PZT
v PZT =

PZT =
YPZT
 PZT
=
(70 10 Pa )
(7700 kg/m )
9
3
= 3015 m/s
v PZT 3015 m/s
=
= 0.003015 m (3.015 mm)
fs
110 6 Hz
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
Chapter 7
1
 1

L PZT = n  PZT  = 1 (0.003015 m ) = 0.00151 m or 1.51 mm
2
 2

7.27 Piezoelectric voltage coefficient The application of a stress T to a piezoelectric crystal leads to a
polarization P and hence to an electric field E in the crystal such that
E = gT
Piezoelectric voltage coefficient
where g is the piezoelectric voltage coefficient. If or is the permittivity of the crystal, show that
g=
d
 o r
A BaTiO3 sample, along a certain direction (called 3), has d = 190 pC N−1, and its r  1900 along this
direction. What do you expect for its g coefficient for this direction and how does this compare with the
measured value of approximately 0.013 C−1 m2?
Solution
Figure 7.42 The piezoelectric spark generator.
The electric field E n the piezoelectric sample due to induced voltage V across the sample is
E=
V
L
The induced voltage and the induced charge Q are related through the capacitance C of the sample by
Q=
V
,
C
The induced charge on the dielectric surfaces is related to the polarization P by
Q = AP
The capacitance of parallel plate capacitor is given by
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Chapter 7
 o r A
C=
L
Combining these expressions above,
V
Q
AP
P
=
=
=
L C L   A L  o r
o r
L
E=
(1)
The applied stress T and the induced polarization P are related through the piezoelectric coefficient d.
P = dT
(2)
Thus substituting in equation (1) into (2), we can obtain the relation between the electric field E and T
E=
d
 o r
T
A direct comparison between this expression and the piezoelectric voltage coefficient definition shows
that
g=
d
 o r
The piezoelectric voltage coefficient for the BaTiO3 sample described in the problem is
g=
d
 o r
=
(190 10
(8.8534 10
−12
−12
)
C N −1
= 0.0113 C−1 m2
−1
F m (1900 )
)
The received value is in good agreement with the experimental one.
7.28 Piezoelectricity and the piezoelectric bender
a.
Consider using a piezoelectric material in an application as a mechanical positioner where the
displacements are expected to be small (as in a scanning tunneling microscope). For the piezoelectric
plate shown in Figure 7.67a, we will take L = 20 mm, W = 10 mm, and D (thickness) = 0.25 mm.
Under an applied voltage of V, the plate changes length, width, and thickness according to the
piezoelectric coefficients dij, relating the applied field along i to the resulting strain along j.
Suppose we define direction 3 along the thickness D and direction 1 along the length L, as shown in
Figure 7.67a. Show that the changes in the thickness and length are
D = d33 V
Piezoelectric effects
L
d 31V
D
L = 
Given d33  500  10−12 m V−1 and d31  −250  10−12 m V−1, calculate the changes in the length and
thickness for an applied voltage of 100 V. What is your conclusion?
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Chapter 7
b. Consider two oppositely poled and joined ceramic plates, A and B, forming a bimorph, as shown in
Figure 7.67b. This piezoelectric bimorph is mounted as a cantilever; one end is fixed and the other
end is free to move. Oppositely poled means that the electric field elongates A and contracts B, and
the two relative motions bend the plate. The displacement h of the tip of the cantilever is given by
2
3 L
h = d 31   V
2 D
Piezoelectric bending
What is the deflection of the cantilever for an applied voltage of 100 V? What is your conclusion?
Figure 7.67 (a) A mechanical positioner using a piezoelectric plate under an applied voltage of V. (b) A
cantilever-type piezoelectric bender. An applied voltage bends the cantilever.
Solution
a. The strain along direction 3 is given by the change in thickness divided by the thickness, S3 = D/D,
and the electric field along 3 is given by E3 = V/D where V is the voltage. The piezoelectric effect relates
the strain S and the field E through
S3 = d33E3


D
D
= d 33
V
D
D = d 33V
Along direction 1, the strain is the change in length divided by the length, S1 = L/L. The piezoelectric
effect related this to the field,
S1 = d31E3


L
L
= d 31
V
D
L
d 31V
D
L = 
Now, using the given values, we can find the change in thickness and length for 100 V:
D = d 33V = (500  10−12 m/V)(100 V) = 5.00  10−8 m or 50.0 nm
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(
Chapter 7
)
L
 0.02 m 
−12
d 31V = 
 − 250  10 m/V (100 V )
D
 0.00025 m 
L = 
= −2.00  10−6 m or −2.00 m
There is a much greater change in the length than in the thickness of the plate.
b. The cantilever tip is displaced by h, whose magnitude is:
2
(
)
2
3
3
L
 0.02 m 
h = d 31   V = − 250 10 −12 m/V 
 (100 V )
2
2
D
 0.00025 m 

h = 0.000240 m or 0.240 mm or 240 m
The cantilever configuration provides a greater displacement than the above cases.
A piezoelectric bender (actuator). An applied voltage of 90 V can produce a
deflection of 315 m if the deflecting end is free. The bender piezoelectric
length is 2.86 cm
(Courtesy of Piezo Systems Inc., USA)
7.29 Piezoelectricity The wavelength  of mechanical oscillations in a piezoelectric slab satisfies
1 
n   = L
2 
where n is an integer, L is the length of the slab along which mechanical oscillations are set up, and the
wavelength  is determined by the frequency f and velocity v of the waves. The ultrasonic wave velocity
v depends on Young's modulus  as
Y 
v = 
 
1/ 2
where  is the density. For quartz, Y = 80 GPa and  = 2.65 g cm−3. Considering the fundamental mode
(n = 1), what are practical dimensions for crystal oscillators operating at 1 kHz and 1 MHz?
Solution
We are given the characteristics of quartz, Y = 80  109 Pa and  = 2650 kg/m3. We can then calculate the
ultrasonic wave velocity v,
Y 
v = 
 
1/ 2
=
80 109 Pa
= 5494 m/s
2650 kg/m 3
First, consider f = 1 kHz:
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=
Chapter 7
v 5494 m/s
=
= 5.494 m
f 1000 Hz
The length L of the quartz crystal at 1 kHz at the fundamental mode (n = 1) is therefore:
1 
1

L = n   = (1)  5.494 m  = 2.75 m
2 
2

A huge crystal, very impractical.
Now, consider f = 1 MHz,
 =

v 5494 m/s
=
= 0.005494 m or 5.494 mm
f  110 6 Hz
1 
1

L  = n   = (1)  0.005494 m  = 0.00275 m or 2.75 mm
2 
2

This is a much more practical value.
Note: Using n > 1 increases the size of the quartz crystal.
7.30 Pyroelectric detectors Consider two different radiation detectors using PZT and PVDF as
pyroelectric materials whose properties are summarized in Table 7.16. The receiving area is 4 mm2. The
thicknesses of the PZT ceramic and the PVDF polymer film are 0.1 mm and 0.005 mm, respectively. In
both cases the incident radiation is chopped periodically to allow the radiation to pass for a duration of
0.05 s.
a. Calculate the magnitude of the output voltage for each detector if both receive a radiation of intensity
10 W cm−2. What is the corresponding current in the circuit? In practice, what would limit the
magnitude of the output voltage?
b. What is the minimum detectable radiation intensity if the minimum detectable signal voltage is 10
nV?
Table 7.16 Properties of PZT and PVDF
Solution
a. The change in voltage of a pyroelectric detector is given by (see Example 7.17)
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
V =
Chapter 7
pIt
c r  o
where p is the pyroelectric coefficient,  is the density, c is the specific heat capacity, I is the incident
radiation intensity (all assumed to be absorbed), and t is the duration of radiation. The output voltages
for the two materials can then be calculated:
VPZT =

(380  10 C m
(7700 kg/m )(300 J K
−6
3
)(
)
)
K −1 0.1 W/m 2 (0.05 s )
−1
kg −1 (290 ) 8.854  10 −12 F/m
−2
(
)
VPZT = 0.000320 V or 0.320 mV
Similarly,
VPVDF =

(27  10 C m K )(0.1 W/m )(0.05 s)
(1760 kg/m )(1300 J K kg )(12)(8.854  10
−6
3
−2
−1
−1
2
−1
−12
F/m
)
VPVDF = 0.000555 V or 0.555 mV
The corresponding currents can be found in two ways:
(1) Neglecting thermal conduction, the current is equal to the change in polarization P over time, or:
i=
dP
dT
pdH
=p
=
dt
dt mcs dt
where p is the pyroelectric coefficient, m is the mass, cs is the specific heat capacity and dH is the change
in heat (enthalpy). The change in heat can be expressed as dH = IAdt, where I is the incident radiation
intensity, and the mass can be found from the density,  = m/V, i.e. m = Ad. Substituting into the previous
equation, we obtain:
i=

pI
dc s
i PZT =
(380  10
−6
(7700 kg m )(
−3
)(
)(
)
C m −2 K −1 0.1 W m −2
= 1.65  10−7 A or 0.165 A
0.1  10 −3 m 300 J K −1 kg −1
)
similarly, we can repeat the above calculation for PVDF
iPVDF = 2.36  10−7 A or 0.236 A
Heat losses, e.g. thermal conduction, would prevent the absorbed heat from raising the temperature
indefinitely. The calculation assumes that the rise in the temperature is small so that the heat losses,
proportional to the temperature difference, can be neglected.
b. Rearrange the equation for change in voltage and isolate the radiation intensity I:
I=
Vc r  o
tp
The minimum signal is given as V = 10  10−9 V. Substituting to find the detectable light intensity:
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I PZT

(10  10
=
−9
)(
)(
(
)
(
)
V 7700 kg/m 3 300 J K −1 kg −1 (290 ) 8.854  10 −12 F/m
(0.05 s ) 380  10 −6 C m −2 K −1
IPZT = 3.12  10−6 W/m2
Chapter 7
)
(Small – a distinct advantage of pyroelectric detectors.)
For the PVDF,
I PVDF =

(10  10
−9
)(
)(
(
) (
)
V 1760 kg/m 3 1300 J K −1 kg −1 (12) 8.854  10 −12 F/m
(0.05 s ) 27  10 −6 C m −2 K −1
IPVDF = 1.80  10−6 W/m2
)
(About the same order of magnitude as PZT)
Note: Notice that we did not need the pyroelectric material thickness. This is because we neglected the
thermal conduction loss. If we were to include heat losses via thermal conduction then we would indeed
need the thickness - this problem is treated in advanced texts.
7.31 LiTaO3 pyroelectric detector LiTaO3 (lithium tantalate) detectors are available commercially.
LiTaO3 has the following properties: pyroelectric coefficient p ≈ 200 × 10−6 C m−2 K−1, density ρ = 7.5 g
cm−3, specific heat capacity cs = 0.43 J K−1 g−1. A particular detector has a cylindrical crystal with a
diameter of 10 mm and thickness of 0.2 mm. Suppose we chop the input radiation and allow the radiation
to fall on the detector for short periods of time. Each input radiation pulse has a duration of Δt = 10 ms.
(The time between the radiation pulses is long, so consider only the response of the detector to a single
pulse of radiation.) Suppose that all the incident radiation is absorbed. If the input radiation has an
intensity of 10 µW cm−2, calculate the pyroelectric current, and the maximum possible output voltage that
can be generated assuming that the input impedance of the amplifier is sufficiently large to be negligible.
What is the current responsivity of this detector? What are the major assumptions in your calculation of
the voltage signal?
Solution
Assuming all the radiation energy H is absorbed and "spent" in raising the temperature by T , we will
have
H = ( AL )cs T
H
( AL )cs
Thus,
T =
and
T
1
H
=
t ( AL )cs t
If the radiation is constant during t , then substituting all of the given parameters into Jp
Jp = p
dT
T
p
H
p H / A
p
=p
=
=
=
I
dt
t ( AL )cs t
( L )cs t
( L ) c s
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
Chapter 7
200  10 −6
Jp =
(10  10 −6  10 4 ) = 3.1×10−9 A m−2
−6
3
3
(2  10 )(7.5  10 )(0.43  10 )
and the pyroelectric current is
Ip = JpA = 2.4×10−13 A or 0.24 pA

RI =
Jp
I
=
p
= 3.1×10−9 A/W or 3.1 nA/W
( L ) c s
From Table 7.9, we can see that r= 47. If C is the capacitance of the pyroelectric crystal (area A and
thickness L), then the pyroelectric current flowing into the capacitance C of the pyroelectric crystal
generates a voltage whose rate of change is given by
C
V
= Ip
t
A r o
pI
)V = I p = AJ p = A(
)
L
cs r o

(

V = (
p
c s  r'  o
)It = 0.00015 V or 0.15 mV
There are various assumptions, most important of which are the following. All incident radiation is
absorbed (emissivity of the surface = 1); there are no heat losses so that the absorbed heat simply raises
the temperature of the crystal; and the pyroelectric current develops a voltage across the pyroelectric
crystal's capacitance without being shunted by leakage, that is Rp (Rp is assumed to be very large); or the
input of the amplifier across the crystal does not load the crystal.
Pyroelectric detectors (Model QS-THZ), which can be used to detect
radiation over the wavelength range 0.1–1000 μm.
Courtesy of Gentec Electro Optics, Inc.
*7.32 Pyroelectric detectors Consider a typical pyroelectric radiation detector circuit as shown in Figure
7.68. The FET circuit acts as a voltage follower (source follower). The resistance R1 represents the input
resistance of the FET in parallel with a bias resistance that is usually inserted between the gate and source.
C1 is the overall input capacitance of the FET including any stray capacitance but excluding the
capacitance of the pyroelectric detector. Suppose that the incident radiation intensity is constant and equal
to I. Emissivity  of a surface characterizes what fraction of the incident radiation that is absorbed. I is
the energy absorbed per unit area per unit time. Some of the absorbed energy will increase
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Chapter 7
Figure 7.68 A pyroelectric detector with an FET voltage follower
the temperature of the detector and some of it will be lost to surroundings by thermal conduction and
convection. Let the detector receiving area be A, thickness be L, density be , and specific heat capacity
(heat capacity per unit mass) be c. The heat losses will be proportional to the temperature difference
between the detector temperature, T, and the ambient temperature, To, as well as the surface area A (much
greater than L). Energy balance requires that
Rate of increase in the internal energy (heat content) of the detector
= Rate of energy absorption − Rate of heat losses
that is,
( AL )c dT
dt
= AI − KA(T − To )
where K is a constant of proportionality that represents the heat losses and hence depends on the thermal
conductivity . If the heat loss involves pure thermal conduction from the detector surface to the detector
base (detector mount), then K = /L. In practice, this is generally not the case and K = /L is an
oversimplification.
a. Show that the temperature of the detector rises exponentially as
T = To +
I 
 t
1 − exp −
K
  th



Detector temperature
where th is a thermal time constant defined by th = Lc/K. Further show that for very small K, the
above equation simplifies to
T = To +
I
t
Lc
b. Show that temperature change dT in time dt leads to a pyroelectric current, ip, given by
i p = Ap
 t
dT ApI
=
exp −
dt
Lc
  th



Pyroelectric current
where p is the pyroelectric coefficient. What is the initial current?
c. The voltage across the FET and hence the output voltage v(t) is given by
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  t
v(t ) = Vo exp −
   th

 t 
 − exp − 

  el 
Chapter 7
Pyroelectric detector output voltage
where V0 is a constant and el is the electrical time constant given by R1Ct , where Ct , total
capacitance, is (C1 + Cdet), where Cdet is the capacitance of the detector. Consider a particular PZT
pyroelectric detector with an area 1 mm2, and thickness 0.05 mm. Suppose that this PZT has r = 250,
 = 7.7 g cm−3, c = 0.3 J K−1 g−1, and  = 1.5 W K−1 m−1. The detector is connected to an FET circuit
that has R1 = 10 M and C1 = 3 pF. Taking the thermal conduction loss constant K as /L, and  = 1,
calculate th and el. Sketch schematically the output voltage. What is your conclusion?
Solution
a. Energy balance requires the following:
ALc
dT
= AI − KA(T − To )
dt
(1)
Consider the given expression for the temperature with thermal time constant th = Lc/K:
T = To +
I 
 t
1 − exp −
K 
  th



(2)
We can show that Equation (2) is a solution by substituting Equation (2) into (1) and verifying that it is
indeed a solution.
Substitute for T in Equation (1),
ALc

Lc

Lc
 t
d 
I 
To +
1 − exp −

dt 
K 
  th
I 
1
 −
K   th

 t
 − exp −

  th

 
 t
I 
 = AI − KA To +
1 − exp −


K 
 
  th

 

 t
 = I − I 1 − exp −


  th
 


 − To 



 

 
 t 
I  1   t 
  exp −  = I exp − 
K   th     th 
  th 
substitute for th,
Lc
simplify,
1   t

 exp −
K  Lc K     th
I 

 t
 = I exp −

  th



I = I
As the right hand side equals the left hand side, Equation (1) is satisfied and Equation (2) is a solution.
For small K, we can expand the exponential of Equation (2) (exp(−t /th) = exp(−Kt / LC)), i.e. assuming
x is small and neglecting second order and higher order terms (x2 and higher), exp(−x) = 1 − x.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
T = To +
Chapter 7
I 
 Kt 
I  
Kt 
 = To +

1 − exp −
1 − 1 −
K 
K   Lc 
 Lc 

T = To +
I  Kt 


K  Lc 

T = To +
I
t
L c
(3)
Therefore, for small loss (K small) or for sufficiently short periods of time (t small), the temperature rise
is linear with time.
b. The pyroelectric current ip is given by
i p = Ap
dT
dt
(4)
This is so because in time dt, the temperature changes by dT. The change in the polarization is p = dP/dT.
This changes the charge on the pyroelectric crystal by an amount dQ = AdP = A(pdT). The current ip is
dQ/dt, and hence the above equation.
Substitute Equation (2) into (4),
i p = Ap

i p = Ap

ip =
 t
d 
I 
To +
1 − exp −

dt 
K 
  th
I 
1
 −
K   th

 t
− exp −

  th
 t
ApI
exp −
Lc
  th
 

 

 t
I 
1 
 = Ap
 −
− exp −
K  Lc K 

  th






(5)
To find the initial current, substitute t = 0,

i p (0) =
ApI
ApI
exp(0 ) =
Lc
Lc
i p (0) =
ApI
Lc
(6)
c. To find the electrical time constant el, we need the capacitance of the detector Cdet, which can be found
from the relative permittivity r = 250, area A = 1 mm2, thickness L = 0.05 mm as follows
Cdet =
 o r A (8.854 10 −12 F/m )(250 )(110 −6 m 2 )
=
= 4.427  10−11 F or 44.27 pF
-3
L
0.05 10 m
The FET circuit capacitance C1 is given as 3 pF. Thus, the total capacitance Ct can be found,
Ct = C1 + Cdet = 3  10−12 F + 4.427  10−11 F = 4.727  10−11 F = 47.27 pF
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Chapter 7
and the electrical time constant with the FET circuit resistance R1 = 10 M is
el = R1Ct = (10  106 )(4.727  10−11 F) = 0.000473 s or 0.473 ms
To find the thermal time constant th, we need the constant of proportionality K (where  is the thermal
conductivity and L is the thickness):
K =  / L = (1.5 W K−1 m−1) / (0.05  10−3 m) = 30000 W K−1

 th =
(
)(
)(
)
Lc 0.05 10 −3 m 7700 kg/m 3 300 J K −1 kg −1
=
= 0.00385 s or 3.85 ms
K
30000 W K −1
For the sketch, we can set Vo to 1 for simplicity. Note: that the ntensity is not required as the output is a
schematic sketch. The magnitude of the output voltage is:
  t
v(t ) = Vo exp −
   th

 t 
 − exp 

  el 
Consider the case in which the radiation intensity is switched on suddenly at t = 0, i.e. it is a step excitation.
The output peaks and drops as =shown−in Figure
This is because R1 leaks the induced charge on
− 7Q32-1.
−

th

el
the pyroelectric sample to ground at a rate determined by the electrical time constant el. If we were to set
el =  (R1 = ) then there would be no decay in the output voltage v(t), at least in theory. However, in
practice there is always some leakage or finite R1.
0.8
0.6
0.4
voltage
0.2
0.002
t
0.004
0.006
0.008
Figure 7Q32-1 Schematic plot of output voltage vs time for a suddenly turned radiation (step excitation). The
time axis is in seconds.
7.33 Spark generator design Design a PLZT piezoelectric spark generator using two back-to-back PLZT
crystals that provide a 60 J spark in an air gap of 0.5 mm from a force of 50 N. At 1 atmosphere in an
air gap of 0.5 mm, the breakdown voltage is about 3000 V. The design will need to specify the dimensions
of the crystal and the dielectric constant. Assume that the piezoelectric voltage coefficient is 0.023 V m
N−1.
Solution
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Figure 7Q33-1 Back-to-back piezoelectric crystals for a spark generator
Figure 7Q33-1 shows a spark generator consisting of two back-to-back piezoelectric ceramics. Given, F
= applied force = 50 N to be applied through a mechanical lever arrangement.
Two ceramics are used to obtain a greater energy in the spark. Energy stored in each ceramic just before
breakdown has to be 30 J to provide a total spark energy of 2  30 J
The stress T induces an electric field E and hence a voltage V given by,
E = gT
(1)
where g is the piezoelectric voltage coefficient. Thus, at breakdown, when V = Vbr,
Vbr
F
=g
L
A

L Vbr
=
A gF

L
(300 V)
=
= 2608 m−1.
−1
A (0.203 V m N )(50 N)
When V reaches Vbr and the air breaks down, the energy stored on the piezoelectric sample capacitance is
very roughly the energy E in the spark,
 A
1
E = CV 2  o r Vbr2
2
2L
(2)
2E  L 
2(30 10 −6 J)
r 
(2.6 103 m −1 ) = 1,958
=
2 
−12
−1
2
 oVbr  A  (8.85 10 F m )(3000 V)
This is the minimum dielectric constant needed. Recall the relationship between the units: [J] = [F][V]2.
Taking a crystal length L of 5 mm gives, A = 0.0055/2.6  103 m2 or 1.92 mm2.
This means a sample diameter of 1.56 mm.
Note: g determines the force needed to spark the gap and hence determines the crystal dimensions. r
determines the amount of energy in the gap. This is the reason for providing g and finding L/A and r.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Alternatively one can provide r and find g. The problem can also be approached in terms of d and r but
it is more usual to use g when a voltage needs to be generated; of course g = d/(or).
7.34 Ionic polarization resonance in CsCl Consider a CsCl crystal which has the following properties.
The optical dielectric constant is 2.62, the dc dielectric constant is 7.20, and the lattice parameter a is
0.412 nm. There is only one ion pair (Cs+–Cl−) in the cubic-type unit cell. Calculate (estimate) the ionic
resonance absorption frequency and compare the value with the experimentally observed resonance at
3.10 × 1012 Hz. What effective value of Q would bring the calculated value to within 10 percent of the
experimental value?
Solution
Q here is the magnitude of the ionic charge, that of Cs+or Cl− as in Section 7.12. There is only one ion
pair (Cs+–Cl−) in the cubic-type unit cell. Given a, the concentration of ion pairs is
Ni =
1
=1.42×1028 m−3
3
a
Ionic dc polarizability i(0) or simply i is given by
 rdc − 1  rop − 1 N i i
−
=
 rdc + 2  rop + 2 3 o
(1)
Substituting for εrdc, εrop and Ni we find, αi = 6.005×10−40 F m2
The reduced mass is
M +M −
10 −3
 132.91 35.45 
Mr =
=
= 4.6472 10 −26 kg

23
M + + M −  132.91 + 35.45  6.022 10
Apply Equation 7.91
 Q2 
I = 

 M r i 
or
1/ 2
(
)
2


1.6 10 −19
=

− 26
6.005 10 −40 
 4.647 10
(
)(
)
1/ 2
= 3.03  1013 rad s−1.
fI =  I /2π = 4.82  1012 Hz
This is 1.56 times greater than the experimental value, 3.10 × 1012 Hz.
If we change the calculated frequency fI to be within 10% of the experimental value, that is
f I = 1.1 f experiment = 3.41  1012 Hz ,
then
Qeffective = [(2 f I ) 2  i M r ]1 / 2 = 0.7074Q = 1.132  10−19 C
or
Qeffective = 0.71e− (71% of elementary charge)
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
Note: Equation (1) comes out from the following. At optical frequencies, only the electronic polarization
contributes,
 rop − 1 N i e
=
 rop + 2 3 o
(2)
At low frequencies (dc) both ionic and electronic polarizations contribute,
 rdc − 1 N i
=
( e +  i )
 rdc + 2 3 o
(3)
Substituting Equation (2) into (3) leads to (1)
7.35 Bruggeman mixture rule The Bruggeman mixture rule gives the overall effective relative
permittivity reff of a dielectric with dispersed spherical particles (r1) in a host medium (r2) as
v1
 r1 −  reff
 −  reff
+ (1 − v1 ) r 2
=0
 r1 + 2 reff
 r 2 + 2 reff
Bruggeman mixture rule
[7.102]
where v1 is the volume fraction of spherical particles (1) dispersed in medium (2) as in Figure 7.60.
Suppose that the continuous phase has r2 = 3.9 (SiO2). Using Bruggeman, Maxwell-Garnett and
Lichtenecker formulas, estimate the porosity that would result in reff = 3.1 (20 percent lower than r2)
Solution
Consider the Bruggeman equation,
v1
 r1 −  reff
 −  reff
+ (1 − v1 ) r 2
=0
 r1 + 2 reff
 r 2 + 2 reff
where v1 is the volume fraction of air pores in this example. Given, reff = 3.1, r2 = 3.9 (host) and r1 =
1 (air pores), we have
v1
1 − 3.1
3.9 − 3.1
+ (1 − v1 )
=0
1 + 2(3.1)
3.9 + 2(3.1)
Solving the above, we have
v1 = 0.21 or 21% (Bruggeman rule)
Consider the Maxwell-Garnett mixture rule,
v1
 − r2
 r1 −  r 2
= reff
 r1 + 2 r 2  reff + 2 r 2
Substitute reff = 3.1, r2 = 3.9 (host) and r1 = 1 (air pores),
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
v1
Chapter 7
1 − 3.9
3.1 − 3.9
=
1 + 2(3.9) 3.1 + 2(3.9)
Solving the above, we have
v1 = 0.22 or 22% (Maxwell-Garnett rule)
Consider the Lichtenecker logarithmic rule,
ln( reff ) = v1 ln( r1 ) + (1 − v1 ) ln( r 2 )
Substitute reff = 3.1, r2 = 3.9 (host) and r1 = 1 (air pores),
ln( 3.1) = v1 ln(1) + (1 − v1 ) ln( 3.9)
Solving the above, we have
v1 = 0.17 or 17% (Lichtenecker rule)
The range of values are 17 – 22 % which indicates the volume of porosity needed to bring 3.9 down to
3.1
7.36 Low- porous dielectrics for microelectronics Interconnect technologies need lower εr interlayer
dielectrics (ILDs) to minimize the interconnect capacitances. These materials are called low-κ dielectrics.
Consider fluorinated silicon dioxide, also known as fluorosilicate glass (FSG), which has an εr of 3.2.
Using Equations 7.97 and 7.102, calculate the expected effective dielectric constant if the ILD is 30
percent porous? What should be the starting εr2 if we need an effective εreff less than 2 and the porosity
cannot exceed 30 percent?
Solution
Part I: Given r1 = 1 (air pores), r2 = 3.2 (host dielectric), and v1 = 0.3 (i.e. 30 % volume)
Lichtenecker equation
ln( reff ) = v1 ln( r1 ) + (1 − v1 ) ln( r 2 )
(1)
substituting r1 = 1, r2 = 3.2, and v1 = 0.3, we have
ln( reff ) = v1 ln(1) + (1 − v1 ) ln( 3.2)

reff = 2.257
Maxwell-Garnett equation
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
v1
 − r2
 r1 −  r 2
= reff
 r1 + 2 r 2  reff + 2 r 2
Chapter 7
(2)
Rearranging Equation (2), we have

 reff =

 r1 −  r 2
) + 1
 r1 + 2 r 2


 r1 −  r 2
1 − v1 (
)
 r1 + 2 r 2
 r 2 2v1 (
with r1 = 1, r2 = 3.2, and v1 = 0.3 gives
 reff

1 − 3.2


3.2  2  0.3  (
) + 1
1 + 2  3.2


=
1 − 3.2
1 − 0.3  (
)
1 + 2  3.2
reff = 2.414
Bruggeman equation
v1
 r1 −  reff
 −  reff
+ (1 − v1 ) r 2
=0
 r1 + 2 reff
 r 2 + 2 reff
(3)
with r1 = 1, r2 = 3.2, and v1 = 0.3 gives
(0.3)
1 −  reff
3.2 −  reff
+ (1 − 0.3)
=0
1 + 2 reff
3.2 + 2 reff

−2(reff)2 + 3.42(reff) + 3.2 = 0

reff = 2.382
The Bruggeman and Maxwell-Garnett equations give effective dielectric constant values that are quite
close, within 1.3%. The Lichtenecker equation predicts a significantly lower reff; by roughly 5.2%
Part II: Find εr2 so that the effective εreff = 2 and the porosity v1 = 30 percent. r1 = 1 as before (air pores)
Lichtenecker equation
ln( reff ) = v1 ln( r1 ) + (1 − v1 ) ln( r 2 )
(1)
substituting r1 = 1, reff = 2, and v1 = 0.3, we have
ln( 2) = v1 ln(1) + (1 − v1 ) ln( r 2 )
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
r2 = 2.692

Maxwell-Garnett equation
v1
 − r2
 r1 −  r 2
= reff
 r1 + 2 r 2  reff + 2 r 2
(2)
gives
(0.3)
1− r2
2 − r2
=
1 + 2 r 2 2 + 2 r 2
−1.4(reff)2 + 3.0(reff) + 1.4 = 0

solving we find
r2 = 2.537
Bruggeman equation
v1
 r1 −  reff
 −  reff
+ (1 − v1 ) r 2
=0
 r1 + 2 reff
 r 2 + 2 reff
(3)
gives
(0.3)
1− 2
 − (2)
+ (1 − 0.3) r 2
=0
1 + 2(2)
 r 2 + 2(2)
which is a linear equation in r2 so that
r2 = 2.563

The worst case deviation is between the Lichtenecker and Bruggeman equation, which is 5.0%. Again,
the Bruggeman and the Maxwell-Garnett results are very close, within 1%
Notes: There has always been interest in the dielectric properties of mixtures. Many dielectrics are invariably
mixtures of one kind or another. Some recommended references on the subject are:
1. E. Tuncer, "Dielectric Mixtures – Importance and Theoretical Approaches", IEEE Electrical Insulation
Magazine (IEEE), 29 (6) 49 – 58 (November/December, 2013)
2. V. Markel, "A tutorial on Maxwell Garnett approximation. I. Introduction", Journal of the Optical
Society of America A, 33 (7) 1244-1256 (2016) https://doi.org/10.1364/JOSAA.33.001244
3. K.K. Kärkkäinen,et al "Effective Permittivity of Mixtures: Numerical Validation by the FDTD Method",
IEEE Transactions on Geoscience and Remote Sensing, 38 (3), 13030 – 1308 (May 2000)
4.
APPENDICES
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
A. Dipole moments (permanent) of some molecules in the gas phase
(Data from
http://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Ch
ang)/12%3A_The_Chemical_Bond/12.4%3A_Electronegativity_and_Dipole_Moment) (29 October 2016)
Molecule
Equilibrium
separation (nm)
Experimental dipole
moment
Bond iconicity
(%)
(D)
LiF
LiCl
LiBr
LiI
NaCl
KF
KCl
KBr
KI
CsF
CsCl
0.152
0.202
0.217
0.238
0.236
0.217
0.267
0.282
0.305
0.255
0.291
6.33
7.13
7.27
7.43
9.00
8.60
10.27
10.41
10.80
7.88
10.42
86.7
73.5
69.8
65.5
79.4
82.5
80.1
76.9
73.7
64.4
74.6
B. A few useful references and reviews
Atomic polarizability
"Theory and applications of atomic and ionic polarizabilities", J. Milroy, M.S. Safronova and C.W.
Clark, J. Phys. B: At. Mol. Opt. Phys., 43 (2010) 202001 (38pp) doi:10.1088/0953-4075/43/20/202001
[Includes tables of polarizabilities.]
"Table of experimental and calculated static dipole polarizabilities for the electronic ground states of the
neutral elements (in atomic units) Last Update: February 11, 2014", Peter Schwerdtfeger,
http://ctcp.massey.ac.nz/index.php?menu=dipole&page=dipole (15 April 2017)
Capacitors
"Introduction to Electrochemical Capacitor Technology", John R Miller, IEEE Insulation Magazine
(DEIS), 26 (July/August) pp40-47, 2010, DOI: 10.1109/MEI.2010.5511188
"A Brief Introduction to Ceramic Capacitors", M-J Pan and C.A. Randall, IEEE Insulation Magazine
(DEIS), 26 (May/June) pp44-50, 2010, DOI: 10.1109/MEI.2010.5482787
"Overview of Laminar Dielectric Capacitors", S.A. Boggs, J. Ho and T.R. Jow, IEEE Insulation
Magazine (DEIS), 29 (March/April) pp8-15, 2013, DOI: 10.1109/MEI.2013.6457595
"The modern era of aluminum electrolytic capacitors", J. Both, IEEE Insulation Magazine (DEIS), 31
(July/August) pp24-34, 2015, DOI: 10.1109/MEI.2015.7126071
Breakdown
" 50+ Years of Intrinsic Breakdown", Y. Sun, C. Bealing and S. Boggs, IEEE Insulation Magazine
(DEIS), 26 (March/April) pp7-13, 2010, DOI: 10.1109/MEI.2010.5482550
Dielectric mixture rules
"Dielectric Mixtures—Importance and Theoretical Approaches", E. Tuncer,
IEEE Insulation Magazine (DEIS), 29 (November/December) pp49-50, 2013, DOI:
10.1109/MEI.2013.6648753
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
Solutions to Principles of Electronic Materials and Devices: 4th Edition (16 April 2017)
Chapter 7
"Evaluation of Mixing Rules for Dielectric Constants of Composite Dielectrics by MC-FEM
Calculation on 3D Cubic Lattice", Y. Wu, X. Zhao, F. Li and Z. Fan, Journal of Electroceramics, 11,
227–239, 2003
"Effective Permittivity of Mixtures: Numerical Validation by the FDTD Method", K. K. Kärkkäinen, A.
H. Sihvola, and K. I. Nikoskinen, IEEE Transactions on Geoscience and Remote Sensing, 38, No. 3
(May), pp1303-1308, 2000
"A tutorial on Maxwell Garnett approximation. I. Introduction", V. Markel, Journal of the Optical
Society of America A, 33, Issue 7, pp1244-1256, 2016 (https://doi.org/10.1364/JOSAA.33.001244 or
https://hal.archives-ouvertes.fr/hal-01282105; 15 April 2017)
Capacitors with energy in the range 100 to 300 mJ. The energy stored is (½)CV2. Because of the V2 dependence, a
small-value capacitor (such as the ceramic capacitors at far left) that has a high breakdown voltage can store as
much energy as a high-value capacitor with a low breakdown voltage (such as the electrolytic capacitors at far
right). (Photo by SK)
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.