Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Second Edition ( 2001 McGraw-Hill)
Chapter 7
7.1 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 g/mol 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 εr depend on the frequency?
c. If the electronic polarizability of an isolated atom is given by
αe ≈ 4πεoro3
where ro is the radius of the atom, then calculate the electronic polarizability of an isolated Se atom,
which has r0 = 0.12 nm, and compare your result with that for an atom in a-Se. Why is there a
difference?
Solution
a
The polarization, P, is given by:
P = (ε o [ε r − 1])E
where E is the electric field.
The local field Eloc is given by:
Eloc = E +
P
3ε o
(ε [ε − 1])E = 1 + (ε − 1)  E =  3 + ε − 1 E
o
r
Substitute for P:
Eloc = E +
∴
ε + 2
Eloc =  r
E
 3 
3ε o
r


3



r
3

b
The relative magnitude of the local field refers to the local field compared to the applied field, i.e.:
Eloc / E. Therefore, with εr = 6.7:
Eloc  ε r + 2   6.7 + 2 
=
=
= 2.9
 3   3 
E
If D is the density then the concentration of Se atoms N is
7.1
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
3
3
23
−1
DN A ( 4.3 × 10 kg/m )(6.022 × 10 mol )
=
N=
Mat
(78.96 × 10 −3 kg/mol)
= 3.279 × 1028 m-3
The Clausius-Mossotti equation relates the relative permittivity to the electronic polarizability, αe:
N
εr − 1
=
αe
ε r + 2 3ε o
∴
−12
F/m )(6.7 − 1)
3ε o (ε r − 1) 3(8.854 × 10
αe =
=
28
N (ε r + 2 )
(3.279 × 10 m −3 )(6.7 + 2)
∴
α e = 5.31 × 10-40 F m2
This would be a type of electronic polarization, as Se is a covalent solid. εr is flat up to optical
frequencies.
c
The Se atom has a radius of about ro = 0.12 nm. Substituting into the given equation:
αe′ ≈ 4πεoro3 = 4π(8.854 × 10-9 F/m)(0.12 × 10-9 m)3
∴
α 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. In the solid, valence electrons are involved in bonding and these electrons contribute
to electronic bond polarization (the field can displace these electrons).
7.2 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 7Q2-1.
a. Explain why εr 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?
c. Plot the polarizability from part b against the bandgap energy, Eg. Is there a relationship?
d. Show that the refractive index n is √εr. When does this relationship hold and when does it fail?
e. Would your conclusions apply to ionic crystals such as NaCl?
Table 7Q2-1 Properties of diamond, Si, and Ge
Mat
Density
εr
(g cm-3 )
Diamond
5.8
12
3.52
Si
11.9
28.09
2.33
Ge
16
72.61
5.32
αe
7.2
Y
(GPa)
827
190
75.8
Eg
(eV)
5.5
1.12
0.67
n
2.42
3.45
4.09
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
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
For diamond, atomic concentration N is:
3
3
23
−1
DN A (3.52 × 10 kg/m )(6.022 × 10 mol )
=
N=
= 1.766 × 1029 m-3
−3
Mat
×
kg/mol
12
10
(
)
The polarizability can then be found from the Clausius-Mossotti equation:
εr − 1
N
αe
=
ε r + 2 3ε o
∴
−12
F/m )(5.8 − 1)
3ε o (ε r − 1) 3(8.854 × 10
αe =
=
29
N (ε r + 2 )
(1.766 × 10 m −3 )(5.8 + 2)
∴
α e = 9.256 × 10-41 F m2
The polarizability for Si and Ge can be found similarly, and are summarized in Table 7Q2-2:
Table 7Q2-2 Polarizability values for diamond, Si and Ge
α e (F m2 )
Diamond
1.766 × 1029 m-3
9.256 × 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
Polarizability per atom (F m2)
N (m-3)
6.00
10-40
5.00
10-40
4.00
10-40
3.00
10-40
2.00
10-40
1.00
10-40
0
0
200 400 600 800
Young's modulus (GPa)
1000
= 5.311 10-40 - (5.327 10-43)Y
Correlation coefficient = 0.9969
Figure 7Q2-1 Plot of polarizability per atom versus Young’s modulus.
7.3
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
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.
Polarizability per atom (F m2)
c
6.00
10-40
5.00
10-40
4.00
10-40
3.00
10-40
2.00
10-40
1.00
10-40
0
0
1
2
3
4
5
Bandgap, Eg (eV)
6
= 5.325 10-40 - (8.0435 10-41)Eg
Correlation coefficient = 0.9873
Figure 7Q2-2 Plot of polarizability versus bandgap energy.
There indeed seems to be a linear relationship between polarizability and bandgap energy.
d
To facilitate this proof, we can plot a graph of refractive index, n, versus relative permittivity, εr.
5
Refractive index, n
Log-log plot: Power law
2
4
10
20
Relative permittivity, r
Figure 7Q2-3 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 = Aεrx
(Correlation coefficient is 0.9987) where x = 0.513 ≈ 1/2 and A = exp(–0.02070) ≈ 1. Thus n = √(εr).
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 – 1015 Hz (ƒoptical). n and polarizability (or εr) 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 rapidly respond to the fast oscillations of the field.
7.4
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
The relationship between n and εr will not hold if we take εr at a low frequency (<< ƒoptical) where other
slow polarization contributions (such as ionic polarization, dipolar polarization, interfacial polarization)
also contribute to εr.
n = ε r would apply to ionic crystals if εr is taken at the corresponding optical frequency rather
than at frequencies below ƒoptical. Tabulated data for ionic crystals typically quote εr that includes ionic
polarization and hence this data does NOT conform to n = ε r .
e
7.3 Dipolar liquids
Given the static dielectric constant of water as 80, its high 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 (in the textbook) 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 6.1 × 10-30 C m. What is
your conclusion? What is εr calculated from the Clasius-Mossotti equation taking the true po (6.1 × 10-30
C m ) of a water molecule? (Note: Static dielectric constant is due to both orientational and electronic
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=
3
23
−3
−1
d N A (10 kg m )(6.022 × 10 mol )
=
= 3.35 × 1028 m- 3
−3
−1
Mmol
×
kg
mol
18
10
(
)
Using the high frequency dielectric constant εr_HV which is due only to electronic polarization and the
Clausius-Mossotti equation (Equation 7.15 in the textbook), we can calculate the electronic polarizability
αe of water molecules
αe =
(
) = 3(8.85 × 10
3ε 0 ε r HF − 1
(
)
−12
F m −1 )( 4 − 1)
(4 + 2)(3.35 × 10 28 m −3 )
ε r HF + 2 N
= 3.964 × 10-40 F m2
Now using this result and the static dielectric constant εr Stat which is due both to electronic and dipolar
polarization, we can solve the Clausius-Mossotti equation for the dipolar polarizability αd of water
αd =
=
[3ε (ε
0
r Stat
)
(
r Sat
+2 N
− 1 − N α e ε r Sat + 2
(ε
)
)] =
3(8.85 × 10 −12 F m −1 )(80 − 1) − (3.35 × 10 28 m −3 )(3.96 × 10 −40 F m 2 )(80 + 2)
(80 + 2)(3.35 × 10 28 m −3 )
= 3.674 × 10-40 F m2
The permanent dipolar moment per water p0 molecule can be calculated from Equation 7.19 (in the
textbook)
7.5
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
p0 = 3kT α d = 3 (1.38 × 10 −23 J K −1 )(300 K )(3.964 × 10 −40 F m 2 ) = 2.137 × 10-30 C m
This result is around 3 times smaller than the real permanent dipolar moment of the water molecule.
The same calculations performed on the base of the simple relationship in Equation 7.14 (in the textbook)
will result in
αe =
αd =
(
(
) = (8.85 × 10
ε 0 ε r HF − 1
F m −1 )( 4 − 1)
(3.35 × 1028 m −3 )
N
) − α = (8.85 × 10
ε 0 ε r Stat − 1
N
−12
−12
F m −1 )( 4 − 1)
(3.35 × 10
e
28
m
−3
)
= 7.928 × 10-40 F m2
− (7.298 × 10 −40 F m 2 ) = 2.009 × 10-38 F m2
and
p0 = 3kT α d = 3 (1.38 × 10 −23 J K −1 )(300 K )(2.009 × 10 −38 F m 2 ) = 1.58 × 10-29 C m
which is about 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, if
one uses the actual po = 6.1 × 10-30 C m in the Clasius-Mossotti equation, then εr turns out to be negative,
which is nonsense.
7.4 Dipole moment in a nonuniform electric field
Figure 7Q4-1 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 7Q4-1. Show that the net force acting on this dipole is given by
F=p
dE
dx
Net force on a dipole
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 of the textbook), 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 7Q4-1? (Remember that a
dielectric medium placed in a field develops polarization P directed along the field.)
7.6
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
E
p
F
Figure 7Q4-1
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.
Solution
E
P
F
E
E′ = E + δx(dE/dx)
E
F-
−Q
+Q
P
Figure 7Q4-2
When a dielectric is placed in a nonuniform
field it develop a polarization P along the
direction of the field. We can represent this
polarization by two oppositle charges +Q and
−Q separaed by δx.
F+
δx
x
x′ = x + δx
x
We can represent the dipole p as two opposite charges +Q and -Q separated by a distance δx as in
Figure 7Q4-2. Thus p = Qδx.
Let -Q be at x, then +Q is at x′ = x + δx. The field is nonuniform. If the field is E at x and then it
is E′ at x′, that is
E′ = E + δx(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 + δx(dE/dx)].
7.7
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
The net force along the +x direction is
F = F+ - F- = Q[E + δx(dE/dx)] - QE
or
F = Qδx(dE/dx) = p(dE/dx).
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 as shown in Figure 7Q4-2. We can of course represent the polarized dielectric by the surface charges
+Q and -Q as in Figure 7Q4-2. 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.
7.5 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 Cl- ions are 3.35 × 10-40 F m2
and 4.5 × 10-40 F m2 respectively, and the mean ionic polarizability per ion pair is 5.8 × 10-40 F m 2.
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=
1
n
28
=
m-3
3 = 1.258 × 10
3
−9
a
(0.430 × 10 m)
At low frequencies both ionic and electronic polarizability contribute to the relative permittivity.
Thus, from Equation 7.20 (in the textbook), (where αi is the mean ionic polarizability per ion pair, αeCs is
the electronic polarizability of Cs+ and αeBr is the electronic polarizability of Br-):
ε r ( low ) − 1
1
=
( Nα i + Nα eCs + Nα eBr )
ε r ( low ) + 2 3ε o
Remember that (Nαi + NαeCs + NαeBr) should be written as (Niαi + NCsαeCs + NBrαeCl), but since
there is a one-to-one ratio between the number of molecules and ions in CsCl, we can take all the N’s to be
the same.
∴
ε r ( low ) =
1
( Nα i + Nα eCs + Nα eBr ) ε r (low ) + 2 + 1
3ε o
(
7.8
)
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
∴
ε r ( low ) −
1
2
Nα i + Nα eCs + Nα eBr )ε r ( low ) =
(
( Nα i + Nα eCs + Nα eBr ) + 1
3ε o
3ε o
Isolate and simplify:
ε r ( low ) =
2 N (α i + α eCs + α eBr ) + 3ε o
3ε o − N (α i + α eCs + α eBr )
ε r ( low ) =
2(1.258 × 10 28 m −3 )(5.8 × 10 −40 F m 2 + 3.35 × 10 −40 F m 2 + 4.50 × 10 −40 F m 2 ) + 3(8.854 × 10 −12 F/m )
3(8.854 × 10 −12 F/m ) − (1.258 × 10 28 m −3 )(5.8 × 10 −40 F m 2 + 3.35 × 10 −40 F m 2 + 4.50 × 10 −40 F m 2 )
ε r(low) = 6.48
∴
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, εr(op), is:
ε r ( op ) =
2(1.258 × 10 28 m −3 )(3.35 × 10 −40 F m 2 + 4.50 × 10 −40 F m 2 ) + 3(8.854 × 10 −12 F/m )
3(8.854 × 10 −12 F/m ) − (1.258 × 10 28 m −3 )(3.35 × 10 −40 F m 2 + 4.50 × 10 −40 F m 2 )
ε r(op) = 2.77
∴
7.6 Electronic polarizability and KCl
KCl has the NaCl crystal structure with a lattice parameter of 0.629 nm. Calculate the relative
permittivity of a KCl crystal at optical frequencies given that the electronic polarizability of K+ is 1.264 ×
10-40 F m2 and that of Cl– is 3.408 × 10-40 F m 2. How does this compare with the measured value of
2.19?
Solution
The CsCl structure has a lattice parameter given by a = 0.629 nm, and there are 4 KCl ion pairs per
unit cell (see Table 1.3, pg. 48 in the textbook). The number of ion pairs, or individual ions, per unit
volume (N) is therefore:
N=
4
4
28
m-3
=
3 = 1.607 × 10
3
−
9
a
(0.629 × 10 m)
The electronic polarizability of the K+ ion is given as αeK = 1.264 × 10-40 F m2, and polarizability
of the Cl- ion is given as αeCl = 3.408 × 10-40 F m2. From Equation 7.20 (in the textbook), the relative
permittivity at optical frequencies, εr(op), can be found (see solution for question 7.5 for derivation):
ε r ( op ) =
∴
∴
ε r ( op ) =
2 N (α eK + α eCl ) + 3ε o
3ε o − N (α eK + α eCl )
2(1.607 × 10 28 m −3 )(1.264 × 10 −40 F m 2 + 3.408 × 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 )(1.264 × 10 −40 F m 2 + 3.408 × 10 −40 F m 2 )
ε r(op) = 2.18
This value is very close to the experimental value of 2.19.
7.9
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
7.7 Equivalent circuit of a polyester capacitor
Consider a 1 nF polyester capacitor that has a polymer (PET) film thickness of 1 µm. 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 εr′ is constant as
shown in Figure 7Q7-1. The relative permittivities at 50 ˚C and 120 ˚C can be found approximately from
Figure 7Q7-1. Upon inspection, at 50 ˚C, εr′ = 2.6 and at 120 ˚C εr′ = 2.8.
0.1
PET at f = 1kHz
2.8
0.01
tan
r' 2.7
0.001
2.6
r'
DEA
2.5
0
50
Loss tangent, tan
2.9
100
150
200
0.0001
250
Temperature (°C)
Figure 7Q7-1 Real part of the dielectric constant εr′ and loss tangent, tan δ, at 1 kHz versus
temperature for PET.
SOURCE: Data obtained by Kasap and Maeda (1955) using a dielectric analyzer (DEA).
From the equation for capacitance, the area of the dielectric can be found (where d is the thickness
of the film):
∴
C=
ε r′ε o A
d
A=
(1 × 10 −6 m)(1 × 10 −9 F) = 4.344 × 10-5 m2
dC
=
ε r′ε o (2.6)(8.854 × 10 −12 F/m )
Conductance = Gp = 1/Rp
C
Figure 7Q7-2 Equivalent circuit of a capacitor at one given frequency.
7.10
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Let tanδ equal the loss tangent or loss factor. From Figure 7Q7-1, at 50 ˚C, tanδ = 0.001 and at
120 ˚C tanδ = 0.01.
We need the equivalent parallel conductance, Gp, (or resistance Rp) at 50 ˚C and 120 ˚C. For
temperature T = 50 ˚C:
Gp = ωC tan δ = 2π (1000 Hz)(1 × 10 −9 F )(0.001) = 6.28 × 10-9 Ω - 1
R p = 1 / G p = 1.59 × 108 Ω or 159 MΩ
At 120 ˚C, the capacitance is different (C′) due to the change in εr′. The loss is higher which
means a lower equivalent parallel resistance Rp. Assuming no change in area A:
C′ =
(2.8)(8.854 × 10 −12 F/m)(4.344 × 10 −5 m 2 )
(1 × 10
−6
m)
= 1.077 × 10-9 F
A value 7.7% higher than at the lower temperature. The new conductance and resistance are:
Gp = ωC tan δ = 2π (1000 Hz)(1.077 × 10 −9 F )(0.01) = 6.77 × 10-8 Ω - 1
R p = 1 / G p = 1.48 × 107 Ω or 14.8 MΩ
7.8 Dielectric loss per unit capacitance
Consider the three dielectric materials listed in Table 7Q8-1 with the real and imaginary dielectric
constants, εr' and ε 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 7Q8-1 Dielectric properties of three insulators at 1 kHz
T =50 °C
′
Material
εr
ε r″
Polycarbonate
2.47
0.003
PET
2.58
0.003
PEEK
2.24
0.003
T = 120 °C
′
εr
2.535
2.75
2.25
ε r″
0.003
0.027
0.003
Solution
Since we are merely comparing values, assume voltage V = 1 V for calculation purposes. From
example 7.5 (in the textbook), the power dissipated per unit capacitance (Wcap) is given by:
Wcap = V 2ω
ε r′′
ε r′
where ω is the angular frequency (2πf) and εr′ and εr″ represent the real and imaginary components of the
relative permittivity εr, respectively. As a sample calculation, the power dissipated in polycarbonate is:
Wcap = (1 V) [2π (1000 Hz)]
2
(0.003)
= 7.63 W/F
(2.47)
Therefore, 7.63 W per F are dissipated at 50 ˚C at 1 V. The values for the other materials at both
50 ˚C and 120 ˚C are listed below in Table 7Q8-2:
7.11
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Table 7Q8-2 Power dissipated at different temperatures for the given materials.
50 ˚C
120 ˚C
Power Dissipated
Power Dissipated
(W per F)
(W per 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.
*7.9 TCC of a polyester capacitor
Consider the parallel plate capacitor equation
C=
ε oε r xy
z
where εr is the relative permittivity (or εr'), 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 ε 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 dε r
=
+λ
C dT ε r dT
Temperature coefficient of capacitance
where λ is the linear expansion coefficient defined by
λ=
1 dL
L dT
where L stands for any length of the material (x, y or z). Assume that the dielectric is isotropic, λ is the
same in all directions. Using εr' versus T behavior in Figure 7Q9-1 and taking λ = 50 × 10-6 K-1 as a
typical value for polymers, predict the TCC at room temperature and at 10 kHz.
Solution
The real part of the relative permittivity, εr′, is usually simply written as εr. We are given xy = area
(A) and z = thickness. From the definition of the linear expansion coefficient λ:
dx
= λx
dT
dy
= λy
dT
dz
= λz
dT
Now differentiate C with respect to T (remember that εr depends on temperature):
7.12
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
C=
Chapter 7
ε oε r xy
z
 ε xy dε r + ε ε y dx + ε ε x dy  z − ε ε xy dz
o r
o r
o r
dC  o dT
dT
dT 
dT
=
2
z
dT
∴
Substitute for the derivatives of x, y and z as according to the definition of λ and simplify:
dC
=
dT
dε r
+ λε oε r xy
dT
z
ε o xy
The temperature coefficient of capacitance is defined as:
TCC =
1 dC
C dT
ε o xy
substitute for dC/dT: TCC =
dε r
+ λε oε r xy
dT
Cz
dε r
+ λε oε r xy
dT
 ε oε r xy  z
 z 
ε o xy
substitute for C:
TCC =
simplify:
TCC =
1 dε r
+λ
ε r dT
2.60
2.5925
2.59
r'
2.58
2.57
2.56
PET, f = 10 kHz
20
30
40
60
50
Temperature (°C)
70
80
90
Figure 7Q9-1 Temperature dependence of εr′ at 10 kHz.
From the slope of the straight line in Figure 7Q9-1, we can estimate the value of dεr/dt:
dε r 2.5925 − 2.57
= 0.000375 ! C −1 or K-1
=
dT 80 ! C − 20 ! C
Using the given value of λ = 50 × 10-6 K-1 and εr = 2.57 (εr at room temperature from inspecting
Figure 7Q9-1):
TCC =
∴
1 dε r
1
0.000375 K −1 ) + 50 × 10 −6 K −1
+λ =
(
ε r dT
(2.57)
TCC = 0.000196 K-1 or ˚C- 1
This is 196 ppm per ˚C. PET capacitors are quoted to have typically 200 ppm/˚C.
7.13
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
7.10 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, V br , between two electrodes depends on the distance, d,
between the electrodes as well as the gas pressure, P, as shown in Figure 7Q10-1. V br 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 with
air at 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 greater? What is
your conclusion?
c. At what pressure is the breakdown voltage 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?
Solution
a
At pressure P = 1 atm = 1.013 × 105 Pa and air gap d = 5 mm, P × d = (1.013 × 105 Pa)(0.005 m)
= 506.5 Pa m. From Figure 7Q10-1, the corresponding values of breakdown voltage for air (Vair) and for
SF6 (VSF6) are:
V air = 21000 V or 21.0 kV
V SF6 = 50000 V or 50.0 kV
12
105
5
105
5
104
2.1
104
Breakdown voltage (V)
105
104
SF6
Air
103
5
102
10-1
1
101
102
5
5
103
5
Pressure Spacing (Pa m)
Figure 7Q10-1 Breakdown voltage versus (pressure × electrode spacing) (Paschen curves)
b
At P = 10 atm = 1.013 × 106 Pa and d = 5 mm, P × d = (1.013 × 106 Pa)(0.005 m) = 5065 Pa m.
Using linear extrapolation on Figure 7Q10-1:
V air = 500000 V or 500 kV
V SF6 = 1200000 V or 1200 kV
7.14
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Pressure increases by 10 times but the breakdown voltages increase by a factor of about 25 times this is a good improvement.
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 = 250 V, and the
minimum for SF6 is about VSF6 = 420 V. The corresponding values for P × d are (P × d)air = 0.62 Pa m
and (P × d)SF6 = 0.2 Pa m. From these we can determine the values of pressure needed for minimum
breakdown voltage:
Pair d = 0.62 Pa m
PSF6 d = 0.2 Pa m
0.62 Pa m
0.005 m
∴
Pair =
∴
Pair = 124 Pa or 0.00122 atm
PSF6 =
0.2 Pa m
0.005 m
PSF6 = 40.0 Pa or 0.000395 atm
A low pressure is 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:
Pdair = 0.62 Pa m
0.62 Pa m
1.013 × 10 5 Pa
∴
dair =
∴
d air = 6.12 × 10-6 m
This value corresponds to a breakdown voltage of 250 V. Therefore a gap of about 6 µm will only
need 250 V for breakdown.
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.
*7.11 Capacitor design
Consider a nonpolarized 100 nF capacitor design at 60 Hz operation. Note that there are three
candidate dielectrics, as listed in Table 7Q11-1.
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?
7.15
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Table 7Q11-1 Comparison of dielectric properties at 60 Hz (typical values)
Polymer Film
Ceramic
PET
TiO2
Name
Polyester
Polycrystalline
titania
3.2
90
εr ′
tan δ
5 × 10-3
4 × 10-4
-1
150
50
Ebr (kV cm )
Typical minimum
1-2 µm
10 µm
thickness
Chapter 7
High-K Ceramic
(BaTiO 3 based)
X7R
1800
5 × 10-2
100
10 µm
Solution
Note: All sample calculations are for Polymer film (PET). All methods of calculation for the
other materials are identical, and the obtained values are summarized in Table 7Q11-2.
a
To find the volume needed for C = 100 nF given that the dielectric has the minimum practical
thickness, d (Table 7Q11-1), find the capacitance per unit volume (Cvol):
−12
F/m )(3.2)
ε oε r (8.854 × 10
Cvol = 2 =
= 28.33 F/m 3
2
−
6
d
(1 × 10 m)
The volume V can now be found as follows:
V = C / C vol = (100 × 10-9 F) / (28.33 F/m3) = 3.53 × 10-9 m3
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:
1
V
Ebr = max
2
d
∴
d=2
500 V
Vmax


−5
= 2
 = 6.667 × 10 m
7


1.50 × 10 V/m
Ebr
Now the capacitance per unit volume can be found:
−12
Cvol =
∴
F/m )(3.2)
ε oε r (8.854 × 10
=
= 0.006374 F/m 3
2
2
−5
d
(6.667 × 10 m)
V = C / C vol = (100 × 10-9 F) / (0.006374 F/m3) = 1.57 × 10-5 m3
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
ωε oε r′ tan δ
η2
where η is the safety factor (assumed to be equal to 2) and ω = 2πf is the angular frequency. Evaluating:
7.16
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
2
(1.5 × 10 V/m) (2π (60 Hz))(8.854 × 10
W =
7
vol
∴
(2 )
Chapter 7
2
−12
F/m )(3.2)(5 × 10 −3 )
Wvol = 3004 W/m3
The power dissipated (W) is therefore:
W = WvolV = (3004 W/m3)(1.57 × 10-5 m3)
∴
W = 0.0472 W
Table 7Q11-2 Summarized values for volume and power of given capacitors.
Polymer Film
Ceramic
High-K Ceramic
PET
TiO2
(BaTiO3 based)
a
Low voltage volume (m3)
3.53 × 10- 9
1.25 × 10- 8
6.27 × 10- 1 0
b
High voltage volume (m3)
1.57 × 10- 5
5.02 × 10- 6
6.27 × 10- 8
c
Power dissipated (W)
0.0472
0.00377
0.471
Upon inspection we see that for part a and part b, high-K ceramic has the smallest volumes, and
for part c, ceramic has the lowest power dissipation.
*7.12 Dielectric breakdown in a coaxial cable
Consider a coaxial underwater high-voltage cable as in Figure 7Q12-1a. 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 =
ρLI 2
πa 2
Rate of heat generation
[1]
where ρ 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 – T o , between the inner and outer conductors;
on the sample geometry (a, b and L); and on the thermal conductivity κ of the dielectric. From
elementary thermal conduction theory, this is given by
Q′ = (Ti − To )
2πκL
b
ln 
 a
Rate of heat conduction
[2]
The inner core temperature, Ti, rises until, in the steady state, the rate of joule heat generation by
the electric current in Equation 1 is just removed by the rate of thermal conduction through the dielectric
insulation, given by Equation 2.
a. Show that the inner conductor temperature is
7.17
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
ρI 2
b
Ti = To + 2 2 ln 

a
2π a κ
Chapter 7
[3]
Steady state inner conductor temperature
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 7.10 in the textbook)
Emax =
V
b
a ln 
 a
[4]
Maximum field in a coaxial cable
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 7Q12-1b. 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 3. 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 nΩ m. The
insulation is 3 mm thick and is a polyethylene-based polymer whose long-term dc dielectric strength
is shown in Figure 7Q12-1b. 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?
[
]
Dielectric
V
a
b
Ti
Heat
Q'
To
Dielectric Strength (MV/m)
c. Rederive Ti in Equation 3 by considering that r depends on the temperature as ρ = ρ0 1 + α 0 (T − T0 )
(Chapter 2 of the textbook). Recalculate the maximum current in b given that
αo = 3.9 × 10-3 °C-1 at 25 °C.
60
50
40
30
20
10
0
-50
0
50
100
Temperature (°C)
150
(b)
(a)
Figure 7Q12-1
(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.
Solution
a
The resistance R of the inner conductor is:
R=
ρL ρL
=
A πa 2
where ρ is the resistivity of the conductor, L is the length and a is the radius.
7.18
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Joule heating power generated by the current I through the conductor (P) is given by:
P = I2R = I 2
ρL
πa 2
Eqn. [1]
The rate of heat conduction (Q′) from inner conductor to outer conductor is:
Q′ = (Ti − To )
2πκL
b
ln 
 a
Eqn. [2]
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
ρL
2πκL
= (Ti − To )
2
b
πa
ln 
 a
ρL
∴
(Ti − To ) = I 2 2κπ 2 a 2 L ln a 
∴
Ti = To +
b
ρI 2
b
ln 
2 2
2κπ a  a 
Note: A major assumption is the inner conductor resistivity ρ and thermal conductivity κ are
constant (do not change with temperature).
b
We are given the cable’s characteristics: resistivity ρ = 27 × 10-9 Ω, thermal conductivity κ = 0.3
-1
W K 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 4:
40 × 10 3 V
V
= 1.702 × 107 V/m or 17.02 MV/m
=
Emax =
m
0
008
b
.

a ln  (0.005 m ) ln
 a
 0.005 m 
60
50
40
Dielectric
Strength (MV/m)
30
20
90°C
10
0
-50
0
50
100
Temperature (°C)
150
Figure 7Q12-2 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 7Q12-2 above).
Now, from Equation 3:
7.19
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Ti = To +
Chapter 7
ρI 2
b
ln 
2 2
2κπ a  a 
2(0.3 W K −1 m −1 )π 2 (0.005 m ) (363 K − 298 K )
m
(27 × 10 −9 Ω m) ln 00..008
005 m 
2
2κπ 2 a 2 (Ti − To )
=
b

ρ ln
 a
∴
I=
∴
I = 871 A
c
As in a for 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 but now ρ is temperature dependent. Therefore:
I2
ρ0 [1 + α 0 (Ti − To )]L
πa
2
= (Ti − To )
2πκL
b
ln 
 a
(5)
Solving for Ti and simplifying we receive
Ti = To +
I2
2π 2 a 2κ
− I 2α 0
b
ρ0 ln 
 a
Now we have to calculate at what current Ti will reach 90 °C and hence thermal breakdown will occur.
Solving for I Equation (5) we have
I=
2π 2 a 2κ (Ti − To )
b
ρ0 [1 + α 0 (Ti − To )] ln 
 a
So that
2
2π 2 (5 × 10 −3 m ) (0.3 W K −1 m −1 )((90 − 25) K )
= 777.8 A
I=
 8
−9
−3
Ω
m
.
K
ln
27
10
1
3
9
10
90
25
×
+
×
−
(
)
)  5
(
) (
)(
[
]
7.13 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 (ρ ) 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 ρ = 2.65 g cm-3. For PZT, Y = 70 GPa and ρ = 7.7 g cm-3. Assume that the velocity
of mechanical oscillations in the crystal is v = Y ρ and the wavelength λ = v/f s. Consider only the
fundamental mode (n = 1).
7.20
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Solution
We are given the first resonance frequency, fs = 1 MHz. There is a second resonance frequency,
fa, related to the first by (see example 7.13, pg. 565 in the textbook)):
fa =
fs
1 − K2
where K is the electromechanical conversion factor. From Table 7.7 (pg. 561 in the textbook), Kquartz =
0.1 and KPZT = 0.72. The second resonance frequencies can then be found:
Quartz:
fa =
fs
1− K
(1 × 10 Hz) = 1.005 × 10 Hz
6
=
2
6
1 − (0.1)
2
The bandwidth Bquartz is defined as the difference between these two frequencies:
∴
B quartz = fa - fs = 1.005 × 106 Hz - 1 × 106 Hz = 5000 Hz
PZT:
fa =
∴
fs
1 − K2
(1 × 10 Hz) = 1.441 × 10 Hz
=
6
6
1 − (0.72)
2
B PZT = fa - fs = 1.441 × 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 (ρquartz = 2650 kg/m3, ρPZT = 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 Equation 7.50 (in the textbook).
Quartz:
Yquartz
(80 × 10 Pa) = 5494 m/s
=
(2650 kg/m )
9
vquartz =
ρquartz
∴
λquartz =
vquartz 5494 m/s
=
= 0.005494 m
fs
1 × 10 6 Hz
∴
1
1
Lquartz = n λquartz  = 1 (0.005494 m ) = 0.00275 m or 2.75 mm

2
 2
3
PZT:
(70 × 10 Pa) = 3015 m/s
(7700 kg/m )
9
vPZT =
YPZT
=
ρPZT
∴
λ PZT =
vPZT
3015 m/s
=
= 0.003015 m
fs
1 × 10 6 Hz
∴
1
1
LPZT = n λ PZT  = 1 (0.003015 m ) = 0.00151 m or 1.51 mm

2
 2
3
7.21
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
7.14 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 εoεr is the permittivity of the crystal, show that
d
ε oε r
g=
A BaTiO3 sample, along a certain direction (called 3), has d = 190 pC N-1, and its εr ≈ 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
F
F
Piezoelectric
A
L Piezoelectric V
Piezoelectric
F
F
(a)
(b)
Figure 7Q14-1 The piezoelectric spark generator.
The electric field E in the piezoelectric sample due to induced voltage V over the sample (see Figure
7Q14-1a) is
E=
V
L
The induced voltage is proportional to the induced charge Q by
Q=
V
,
C
where C is the capacitance of the sample.
The induced charge in his turn is related to the polarization P by
Q = AP
And finally, the capacitance of parallel plate capacitor is given by
C=
ε 0ε r A
L
7.22
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
Combining these expressions we receive
E=
P
V
Q
AP
=
=
=
A
L CL ε ε
L ε0 εr
0 r
L
(1)
The applied stress T and the induced polarization P are related through the piezoelectric voltage
coefficient d
P=dT
Thus substituting in (1) we can obtain the relation between the electric field E and T
E=
d
T
ε 0ε r
A direct comparison between this expression and the piezoelectric voltage coefficient definition shows that
g=
d
ε 0ε r
The piezoelectric voltage coefficient for the BaTiO3 sample described in the problem is
190 × 10 −12 C N −1 )
(
d
=
g=
= 0.0113 C -1 m2
ε oε r (8.8534 × 10 −12 F m −1 )(1900)
The received value is in pretty good agreement with the experimental one.
7.15 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 7Q15-1a, 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 the figure.
Show that the changes in the thickness and length are
δD = d33 V
Piezoelectric
effects
L
δL =   d31V
 D
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?
b. Consider two oppositely poled and joined ceramic plates, A and B, forming a bimorph, as shown in
Figure 7Q15-1b. 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
h=
3  L2
V
d31
2  D
Piezoelectric bending
What is the deflection of the cantilever for an applied voltage of 100 V? What is your conclusion?
7.23
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
W
Chapter 7
V
D Piezoelectric
V
L
3
D
2
L
2
A
B
1
h
0
(a)
(b)
Figure 7Q15-1
(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 =
δD/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 via Equation 7.47 (in the textbook), such that:
S 3 = d 33E 3
∴
V
δD
= d33
D
D
∴
δD = d33V
Along direction 1, the strain is the change in length divided by the length, S1 = δL/L. By the
piezoelectric effect (Equation 7.47 in the textbook):
S 1 = d 31E 3
∴
V
δL
= d31
D
L
∴
L
δL =   d31V
 D
Now, using the given values, we can find the change in thickness and length for 100 V:
δD = d33V = (500 × 10 −12 m/V)(100 V) = 5.00 × 10-8 m
0.02 m 
L
δL =   d31V = 
−250 × 10 −12 m/V)(100 V) = -2.00 × 10-6 m
 D
 0.00025 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
h=
∴
2
0.02 m 
3
3
L
d31   V = ( −250 × 10 −12 m/V) 
(100 V)
 0.00025 m 
 D
2
2
h = 0.000240 m or 0.240 mm
The cantilever configuration provides a greater displacement than the above cases.
7.24
Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001)
Chapter 7
7.16 Piezoelectricity
The wavelength, λ, of mechanical oscillations in a piezoelectric slab satisfies
1
n λ  = L
2 
where n is an integer, L is the length of the slab along which mechanical oscillations are set up, and the
wavelength λ is determined by the frequency f and velocity υ of the waves. The ultrasonic wave
velocity υ depends on Young's modulus Υ as
1/ 2
Y 
υ= 
ρ
where ρ is the density. For quartz, Y = 80 GPa and ρ = 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 ρ = 2650 kg/m3. We can then
calculate the ultrasonic wave velocity, υ:
υ=
Y
80 × 10 9 Pa
= 5494 m/s
=
2650 kg/m 3
ρ
First, consider f = 1 kHz:
λ=
υ 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 λ  = (1) × 5.494 m = 2.75 m
2 
2

A huge crystal, very impractical.
Now, consider f′ = 1 MHz:
λ′ =
∴
5494 m/s
υ
=
= 0.005494 m
f ′ 1 × 10 6 Hz
1
1
L′ = n λ  = (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.
7.25