Download The goals of this chapter are to understand

A photograph of a silicon singlecrystal wafer with fabricated
computer processors. The wafer
is a thin slice from a silicon single
crystal, and the wafer is processed
to contain many computer processors. Each square is approximately
a half a centimeter across and is an
individual processor. The processors on this wafer use copper as the
conductor. The silicon single-crystal
wafer is approximately 30 cm (12 in.)
in diameter.
Courtesy of Intel Corporation
The goals of this chapter are to understand
●●
The relationship of electrical resistivity and conductivity to electrical resistance
●●
The variation of electrical conductivity across material classes
●●
The electrical conductivity and resistivity of metals
●●
The relationship among electrical resistivity, defects, and temperature in metals
●●
The distribution of free-electron energies in metals
●●
Thermionic emission of electrons from metals
●●
Superconductivity in metals and compounds
●●
The electrical conductivity of semiconductors
●●
The contact potential of junctions
●●
The diode equation for pn-junctions
●●
The function and applications of dielectric, ferroelectric, and piezoelectric materials
●●
The electrical conductivity of ionic materials
●●
The procedures for producing electronic integrated circuits
Chapter
16
Electrical Properties of Materials
16.1 Electrical Conductivity
and Resistivity of Materials
The resistance (R) of a specimen is determined by measuring the current (I ) in amperes (A)
with an applied voltage (V ) in volts (V) and using Ohm’s law, as shown in Equation 16.1.
V 5 IR
16.1
The resistance of a specimen is not a material property, since it depends upon both the
material and the dimensions of the specimen. The electrical resistivity () is a material
property independent of specimen dimensions, as given by Equation 16.2 for a specimen
with a uniform cross-sectional area A and length L.
5R
A
L
16.2
W-4
CHAPTER 16
The electrical conductivity () is also a material property, and the electrical conductivity is inversely
related to the electrical resistivity (), as shown in Equation 16.3.
5
1
16.3
Table 16.1 presents the electrical conductivity at room temperature of some materials with metallic,
covalent, and ionic bonding. FCC metals in Group IB, such as copper and silver, have the highest
electrical conductivity or lowest resistivity of the metals. These metals are good conductors of electricity,
and we will discuss their conductivity in Section 16.2. Pure silicon (Si) and germanium (Ge) have an
electrical conductivity that is approximately eight to eleven orders of magnitude less than those of silver
and copper, and these materials (Si and Ge) are called electrical semiconductors. We will discuss the
Table 16.1 The Electrical Conductivity of a Variety of Materials
Conductivity
Conductivity
(ohm21 ? cm21)
Material
Material
(ohm21 ? cm21)
SuperconductorsSemiconductors
Hg, Nb3Sn
Infinite below Group 4B elements
YBa2Cu3O72x a critical Si
4 3 1026
MgB2 temperature Ge
0.02
Metals Compound semiconductors
Alkali metals GaAs
Na
2.13 3 105 AlAs
K
1.64 3 10 SiC
2.5 3 1029
0.1
10210
5
Alkali earth metals
Insulators, Linear, and
Nonlinear Dielectrics
Mg
2.25 3 105
Ca
3.16 3 105 Polyethylene
Polymers
10215
Group 3B metals Polytetrafluoroethylene
10218
Al
3.77 3 10 Polystyrene
10
Ga
0.66 3 105 Epoxy
10212 to 10217
5
217
to 10219
Transition metals Ceramics
Fe
1.00 3 105 Alumina (Al2O3)10214
Ni
1.46 3 105 Silicate glasses
Group 1B metals Boron nitride (BN)
10217
10213
Cu
5.98 3 105 Barium titanate (BaTiO3)10214
Ag
6.80 3 105 C (diamond)
Au
4.26 3 105
, 10218
Unless specified otherwise, assumes high-purity material.
*
Based on data from Askeland, D.R., Fulay, P.P., and Wright, W.J., The Science and Engineering of Materials, 6th ed., Cengage Learning,
Stamford, CT. (2011), p. 723.
Electrical Properties of Materials
electrical conductivity of semiconductors in Section 16.5. Diamond and PE have an electrical conductivity
nearly twenty-four orders of magnitude less than those of silver and copper, and these materials are
electrical insulators.
Also listed in Table 16.1 are superconductors. Superconductors have an infinite electrical conductivity
or zero electrical resistivity below a critical temperature, as we will discuss in Section 16.4. Currently the
critical temperatures for all superconductors are well below room temperature. In this chapter we discuss
why materials have such different electrical conductivities.
Figure 16.1 is a photograph of an individual computer processor. Most processors are fabricated
on a thin single-crystal wafer of silicon, as shown in the photograph at the beginning of this chapter.
A computer processor is produced with a combination of metal (M) conductors, oxide (O) insulators,
and semiconductors (S). The computer processor is an integrated circuit (IC). In an IC, all of the
devices needed for operation of the processor are built into the wafer. The most common fabrication
architecture to produce ICs is called CMOS, an abbreviation for complementary metal-oxide
semiconductor. Complementary refers to the way in which the semiconductors are designed into the
circuit. The metal aluminum was used as the conductor in older processors; however, some newer
processors use copper, as shown in the photograph at the beginning of this chapter. The insulating
oxide is usually silica (SiO2), although other oxides are also utilized. The insulating oxide separates the
metal conductor from the semiconductor in areas where contact is not desired, and oxides are used for
capacitors and for memory.
Figure 16.1 A photograph of an Intel® Pentium® 4 computer processor. (Courtesy of Intel Corporation )
W-5
W-6
CHAPTER 16
16.2 Electrical Conductivity
of Metals and Metal Alloys
Most materials with a high electrical conductivity and a low resistivity are metals. Table 16.2 presents the
room-temperature electrical resistivity of some metals. In solid and liquid metals, the valence electrons
are free of the ion cores. The free electrons can move anywhere in the metal. When a voltage is applied to
a metal specimen, the free electrons are accelerated toward the positive terminal. The core electrons on
the atoms do not contribute to the current.
Since the free electrons can be found anywhere in the metal, each free electron has a probability of
interacting with every other free electron. The Pauli exclusion principle states that no two interacting
electrons with the same quantum numbers can occupy the same space. Therefore, in metals the Pauli
exclusion principle applies to all of the free electrons, and no two free electrons can have the same
quantum numbers in the entire metal specimen. The free electrons propagate through the metal
with momentum components in the three directions x, y, and z. Each free electron has a unique set
of quantum numbers (nx, ny, and nz) and a spin quantum number of 112 or 212 . If there are n free
electrons, then there are ne/2 different sets of quantum numbers (nx, ny, and nz). In a cubic meter of
metal, the number of free electrons (n) is typically 1028, and there would be 0.5 3 1028 different sets
of quantum numbers (nx, ny, and nz) for 1028 free electrons. The set of quantum numbers (nx, ny, and
nz) for a free electron is called a quantum state. The lowest set of quantum numbers is 1, 1, 1, and
Table 16.2 Room-Temperature Electrical Resistivity for Some Metals
Room-Temperature Resistivity
(ohm ? cm)
Metal
Temperature Resistivity Coefficient
(R) [ohm/(ohm ? °C)]
Be4.0 3 10260.0250
Mg4.45 3 10260.0037
Ca3.91 3 10260.0042
Al2.65 3 10260.0043
Cr12.90 3 1026 (0°C)
0.0030
Fe9.71 3 10 0.0065
26
Co6.24 3 10260.0053
Ni6.84 3 10260.0069
Cu1.67 3 10260.0043
Ag1.59 3 10260.0041
Au2.35 3 10260.0035
Pd10.8 3 10260.0037
W5.3 3 1026(27°C)0.0045
Pt9.85 3 10260.0039
Based on data from HANDBOOK OF ELECTROMAGNETIC MATERIALS: MONOLITHIC AND COMPOSITE VERSIONS AND THEIR
APPLICATIONS by P. S. Neelkanta. TAYLOR & FRANCIS GROUP LLC—BOOKS (1995).
W-7
Electrical Properties of Materials
Table 16.3 The Crystal Structure, Lattice Parameter (a), Free-Electron Density (n), and Fermi Energy
(EF) for Several Monovalent Metals
Metal Crystal Structure
a nm
1028 electrons/m3
EF eV
Cs
BCC 0.6130.868
1.53
K
BCC 0.5341.32
2.14
Na
BCC 0.4292.55
3.16
Li
BCC 0.3514.66
4.72
Ag
FCC 0.4095.85
5.51
Cu
FCC 0.3628.43
7.04
the quantum numbers increase until all electrons are assigned unique sets of quantum numbers. The
kinetic energy of each free electron is proportional to the sum of the squares of the quantum numbers
(n2x 1 n2y 1 n2z ). Figure 16.2a shows that the energy of a valence electron of a single isolated atom is
a single energy. However, the free-electron energy levels in a metal are spread into a band of energy
called the conduction band, as shown in Figure 16.2b. The band indicates that individual energy levels
are so close that they form a nearly continuous band of energies. The maximum kinetic energy of
free electrons at 0 K is called the Fermi energy (EF), as shown in Figure 16.2b. The Fermi energy of a
metal is typically in the range of 1.5 to 7 eV, as shown in Table 16.3. The Fermi energy is measured
with experimental techniques, such as low-energy X-ray spectroscopy, which is discussed in Section
15.3.1. Experimental measurements have confirmed the results of the quantum mechanical theory of
electrons. Below the energy of the conduction band in Figure 16.2b are the core electrons; however,
these electron energy levels are not normally indicated because they do not contribute to electrical
conductivity.
Ohm’s law can be written in a form that contains the electrical conductivity, as shown in Equation 16.4.
16.4
J 5 sEd
J is the current density in amperes per square meter (A/m2), and E is the magnitude of the applied
electric field. The magnitude of the electric field (E) in this book is printed as a capital in Roman type to
differentiate it from internal energy (E) that is written in italics. The vectors, such as the electric field (E),
Energy
–EF
(a)
(b)
Figure 16.2 (a) The discrete energy level of a single 3s electron on an isolated sodium atom. (b) The conduction
band of energy levels resulting from the free 3s electrons in sodium metal and the Fermi energy (EF).
W-8
CHAPTER 16
are printed in bold Roman type. Equation 16.5 gives the magnitude of the electric field (E ) for a metal
specimen of length L in the x direction with a uniform cross section and an applied voltage V.
E5
V dV
5
L
dx
16.5
The electric field is a vector that points from the positive terminal to the negative terminal of the applied
voltage. The current density in Equation 16.4 can also be written as the product of the free-electron
density (n), the charge on an electron (e), and the drift velocity of the free electrons (vd), as shown by
Equation 16.6.
J
electrons
m
5 n1
v
5 E
1coulombs
2 e 1coulombs
s?m 2
m
electron 2 1 s 2
2
d
3
16.6
The drift velocity of the free electrons is the electron velocity resulting from the applied electric field, and
the drift velocity for all free electrons is the same for a specified electric field applied to a specific metal.
Equation 16.6 is solved for the electrical conductivity in Equation 16.7:
5 ne
vd
E
5 nen
16.7
where n is the electron mobility. Table 16.3 gives values for the free-electron density (n) for some
metals, and the free-electron density (n) can be calculated as demonstrated in Example Problem 16.1.
Equation 16.4 is then rewritten in Equation 16.8 with the electrical conductivity from Equation 16.7 and
the magnitude of the electric field as the voltage gradient (dV/dx) from Equation 16.5.
J 5 nen
dV
dx
16.8
Equation 16.8 is similar to the steady-state diffusion Equation 4.37. The sign difference between
Equations 16.8 and 4.37 occurs because electrons go in the negative direction if the electric field is in the
positive direction.
Example Problem 16.1
Determine the free-electron density in sodium metal. Sodium is BCC and in Group I of the periodic
table, with a lattice parameter of 0.428 nm.
Solution
The number of conduction electrons per unit volume (n) for sodium is calculated by knowing that each atom
in the solid contributes one electron to the free-electron density. Sodium is in Group I of the periodic table;
therefore, there is one electron per atom contributed to the free-electron density. Sodium is BCC, there are two
atoms per unit cell, and the lattice parameter (a) is 0.428 nm. Since there is one free electron contributed by
each sodium atom, the number of atoms per unit volume is equal to the number of electrons per unit volume.
n5
2 atoms 1 electron
2 electrons
electrons
3
5
5 2.55 3 1028
a3
atom
m3
s0.428 3 1029 md3
In Equation 16.7 the mobility of the electrons is the primary variable for the electrical conductivity
of a metal. The charge on an electron (e) is a constant, and for a given metal, such as pure copper, the
free-electron density is a constant. The electron mobility depends upon a number of factors, including the
type of metal, the various possible defects in the metal, and the temperature. For example, silver and gold
W-9
Electrical Properties of Materials
are both in Group IB of the periodic table, and both have one valence electron per atom. Also, both have
FCC lattices with lattice parameters given in Appendix B of 0.40786 nm for gold and 0.40862 nm for silver.
The free-electron density for silver is slightly less than that for gold. However, the electrical conductivity
of the two elements is significantly different. Silver has an electrical conductivity calculated from the
inverse of the resistivity in Table 16.2 of 6.3 3 107 (V ? m)21, and gold has an electrical conductivity of
4.3 3 107 (V ? m)21. The only term in Equation 16.7 that can account for the difference in electrical
conductivity is the mobility of the free electrons. The electrons in silver are more mobile than the electrons
in gold. The mobility of free electrons and the electrical conductivity of a given metal, such as pure silver,
can be changed by altering the temperature, adding impurities, deforming the metal, or heat treating the
metal.
Equation 16.7 shows that the electron mobility is the drift velocity of the electrons divided by the
magnitude of the electric field. Since the electrical conductivity of a particular piece of metal, such as
annealed pure silver at room temperature, is a constant, and if the electric field is constant, the drift
velocity of the free electrons must also be a constant. However, there is a constant force on each free
electron equal to Ee, resulting in a constant acceleration. What keeps the electron drift velocity from
continually increasing as long as the force due to the electric field is applied?
Electron scattering provides a limit to the drift velocity. The electrical conductivity of a metal is
decreased by the addition of defects, as demonstrated in Figure 16.3 for copper additions to nickel.
Defects in a crystal scatter free conduction electrons, and conduction electrons are only accelerated for
the time () between scattering events. The shorter the time between scattering events (), the lower the
drift velocity. Conduction electrons are scattered by all of the possible structural defects in a crystal
and by lattice vibrations. According to Mathiessen’s rule in Equation 16.9, the resistivity is the sum of
resistivity components, where 0 is the resistivity due to all of the point (zero-dimension) defects, 1 is the
resistivity due to all the one-dimensional defects, 2 is the resistivity due to all of the two-dimensional
defects, 3 is the resistivity due to all three-dimensional defects, and th is the resistivity contribution of
the thermal vibrations of the atoms.
5 th 1 0 1 1 1 2 1 3 5
1
16.9
600
Resistivity (nano-ohm-m)
500
400
300
200
100
0
0
20
40
60
Atomic percent copper
80
100
Figure 16.3 The electrical resistivity in 1029 V ? m of copper-nickel alloys as a function of atomic-percent
copper. (Based on Ho, C.Y. et al. J. Phys. Chem. Ref. Data 12(2) (1983), p. 183.)
W-10
CHAPTER 16
The point-defect contribution to resistivity is the easiest relation to demonstrate. The electrical
resistivity of alloys is more than that of pure metals, as demonstrated by the increase in electrical resistivity
for copper-nickel alloys in Figure 16.3. Impurity atoms have an atomic size and electronegativity that
differ from those of the host atom, and impurity atoms scatter the free conduction electrons. Using
substitutional B-type atoms as an example, the distance between the substitutional atoms (lB) decreases
as the chemical composition of substitutional B-type atoms in atomic fraction (CB) increases, according
to the relationship in Equation 16.10:
lB 5
dA
CB
16.10
where dA is the shortest distance between the host A-type atoms. We can assume that each substitutional
B-type atom is a scattering center. The resistivity contribution for substitutional solid-solution B-type
atoms ( 0B ) is inversely proportional to the scattering distance. A short scattering distance results in a
high resistivity, and a large scattering distance results in a low resistivity, as shown in Equation 16.11:
0B 5
A1
lB
5 A2CB
16.11
where A1 and A2 are constants for a given metal. Equation 16.11 demonstrates Nordheim’s rule, that the
electrical resistivity of a metal alloy linearly increases with increases in the atomic fraction of impurity
atoms for dilute solutions. Equations 16.10 and 16.11 are applied to other point defects such as vacancies
or self-interstitials. Equations similar to Equation 16.10 relating defect densities to scattering distance
are also developed for all other defects in materials. In general, defects decrease the distance between
scattering events for conduction electrons and increase the resistivity. For example, annealed oxygen-free
high-conductivity (OFHC) copper has a low defect density, and it has a resistivity at room temperature
of 0.167 3 1029 V ? m. Drawing copper into a wire increases the number of defects, including interstitial
copper atoms, vacancies, dislocations, and grain boundaries. The terms 0, 1, 2, and 3 in Equation 16.9
are increased, and the electrical resistivity at room temperature for drawn OFHC copper wire is
0.33 3 1027 V ? m. The electrical resistivity of drawn OFHC copper wire is more than two orders of
magnitude higher than that of annealed OFHC copper.
Equation 16.10 applies to dilute solid solutions of B-type atoms into an A-type metal host. As can
be seen in Figure 16.3, when the concentration of nickel atoms in copper becomes large, the electrical
resistivity reaches a maximum and then starts to decrease. For large concentrations of B-type atoms in
an A-type host, Equation 16.10 is replaced by Equation 16.12:
0B 5 A3CBs1 2 CBd
16.12
where A3 is a constant for a given metal host and impurity-atom addition.
Equations 16.10 and 16.11 apply when the impurity atom is in the form of a solid solution. Some
specimens have two phases: and . Two phases can result from an alloy of metals with limited solid
solubility in each other, or from a composite material. The electrical resistivity of a material with a
mixture of and phases ( ) depends upon the volume fractions of each phase present (v ,v ) and the
electrical resistivity of each phase ( , ), as shown in Equation 16.13.
5 v 1 v 16.13
Equation 16.13 is the rule of mixtures. The rule of mixtures also applies to other physical properties of
two-phase materials, including thermal conductivity and some mechanical properties. However, the rule
of mixtures does not apply if the second phase is oriented in layers that are perpendicular to the direction
of the current flow, heat flow, or applied force. Chapter 12 covers the rule of mixtures and other rules that
govern the properties of two-phase materials.
W-11
Electrical Properties of Materials
As the temperature increases, the electrical resistivity of a metal increases, as shown in Figure 16.4.
The increase in electrical resistivity in excess of the value at 0 K is due to the thermal component of
electrical resistivity (th). As the temperature approaches 0 K for a metal that is not a superconductor,
there is a residual electrical resistivity (s) that is due to zero-, one-, two-, and three-dimensional structural
defects, as shown in Figure 16.4. We can assume that the thermal component of electrical resistivity (th)
in Equation 16.9 at 0 K is equal to 0. At any temperature (T ) above 0 K, the total resistivity is the sum of
the structural component of the resistivity at 0 K (s) plus the thermal component of resistivity (th), as
shown in Figure 16.4 and in Equation 16.14.
5 s 1 th 5
1
16.14
The thermal component of resistivity increases linearly with increasing temperature, because the
amplitude of the atom vibrations increases, and the conduction electrons are scattered more frequently
by the vibrating atoms. The distance between scattering events due to thermal vibrations (lT) is inversely
related to temperature (T ), as shown in Equation 16.15:
lT 5
b
T
16.15
where b is a constant for a given material that has units of m∙kelvin. Since electrical resistivity is
inversely proportional to the distance between scattering events, as expressed in Equation 16.11, the
thermal component of the electrical resistivity is directly proportional to the temperature, as expressed
in Equation 16.16:
16.16
5 BT
th
where B is a constant for a given material. Equation 16.16 predicts that the thermal component of
electrical resistivity increases linearly with temperature (T ), and experimentally we can observe this in
Figure 16.4. Inserting Equation 16.16 into Equation 16.14 results in the electrical resistivity as a function
of temperature, as expressed in Equation 16.17.
5 s 1 BT 5
1
16.17
160
Resistivity (nano-ohm-m)
140
120
100
80
60
40
20
0
0
200
400
600
800
Temperature (K)
1000
1200
Figure 16.4 The electrical resistivity of copper with 5 weight percent nickel (5.39 atom percent) as a function of
temperature. (Based on Ho, C.Y. et al. J. Phys. Chem. Ref. Data 12(2) (1983), p. 183.)
W-12
CHAPTER 16
Equation 16.17 predicts that at temperatures approaching 0 K, the electrical resistivity is equal to
the structural component of the electrical resistivity (s), as we can observe in Figure 16.4. Then, as
temperature is increased, the electrical resistivity increases linearly. Because the electrical resistivity is
not perfectly linear over large temperature ranges, and because the residual electrical resistivity at 0 K
is small in comparison to the room-temperature electrical resistivity, handbooks of electrical properties
of materials list the temperature coefficient of electrical resistivity (R) at reference temperatures. The
relationship between the temperature coefficient of electrical resistivity and the electrical resistivity
at a temperature different than the reference temperature is given by Equation 16.18, where DT is the
temperature difference (T 2 TR) between the actual temperature and the reference temperature, and R is
the electrical resistivity at the reference temperature.
5 Rs1 1 R DT d
16.18
Table 16.2 presents values of R for some metals at room temperature.
16.3 The Effect of Temperature
on Electron Energies in Metals
The source of electrons in free space necessary for the operation of many devices, such as X-ray tubes,
LEED systems, and electron microscopes, results from the temperature dependence of the free-electron
energies in metals. The schematic of the free electrons in the conduction band of a metal, shown in
Figure 16.2, is for 0 K. In a metal at 0 K, the electron energy levels are filled with a probability of 1 for
energies up to the Fermi energy (EF); and for energies above the Fermi energy the probability of
occupation is 0. The probability P(E) that a free electron has the energy E at temperature T in a metal is
given by the Fermi-Dirac statistical distribution, shown in Equation 16.19.
1
PsE d 5
1 1 exp
E 2 EF
16.19
kT
The Fermi-Dirac statistical distribution at 0 K is a step function that is equal to 1 for free-electron
energies less than the Fermi energy, and equal to 0 for free-electron energies greater than the
Fermi energy, as shown in Figure 16.5 and in Equations 16.20 and 16.21.
T 5 0 K, E , EF ,
PsE d 5
1
1
5 51
1 1 exp 2 ` 1
16.20
T 5 0 K, EF . EF ,
PsE d 5
1
1
5 50
1 1 exp ` `
16.21
For any finite temperature, the value of the Fermi-Dirac statistical distribution is 0.5 for free-electron
energies equal to the Fermi energy, as shown in Equation 16.22 and in Figure 16.5.
T . 0 K, E 5 EF ,
PsE d 5
1
1
1
5
5
1 1 exp 0 1 1 1 2
16.22
W-13
Electrical Properties of Materials
T=0K
Fermi-Dirac probability
P (E)
1.0
T = T1
T = T2> T1
0
Energy (E )
F
EF
Figure 16.5 A schematic of Fermi-Dirac probability distributions P(E ) plotted as a function of energy at 0 K
(dashed line), at a higher temperature (T1), and at an even higher temperature (T2). On the energy scale the
Fermi energy (EF) and the work function (F) are indicated.
The effect of increasing the temperature on the Fermi-Dirac statistical distribution is shown in
Figure 16.5. Some of the electrons that were at energies less than the Fermi energy are excited by
thermal vibrations to energies greater than the Fermi energy, as shown for temperature T1 and the
higher-temperature T2. Since P(E ) is the probability that free-electron energies are filled, then 1 2 P(E )
is the probability that an electron energy level E is unfilled. In Figure 16.5, the distance from the plot of
P(E ) to 1 is equal to 1 2 P(E ). P(E ) and 1 2 P(E ) are symmetric about P(E ) equal to 12 when plotted
as a function of energy.
16.3.1 Thermionic Emission of Electrons
from Metals
Free electrons outside a metal are developed by heating a filament with a current to a high temperature;
electrons are then emitted by thermionic emission. One material for filaments is tungsten, which has a
melting temperature of 3410°C (3683 K).
To understand thermionic emission of electrons by metals requires a further development to the model
of a metal. The band of conduction electron energy levels, shown in Figure 16.2b, does not include the
possibility of electron emission from the metal. Electron emission is included in Figure 16.6 by assuming that
electrons are in a finite potential well. The top of the potential well is the energy of free space, and the bottom
of the potential well is a free electron with the minimum kinetic energy inside the metal. If the electron has
the energy of free space, it is free of the metal. If the electron has less than the energy of free space, it is bound
inside the metal. The energy required to excite a Fermi-energy electron to the energy of free space is the work
function (F). Table 16.4 presents values of the work function for several metals. To remove any other energy
free electron inside the metal to the energy of free space requires the work function plus the energy of the
free electron relative to the Fermi energy. Free electrons with energies greater than the Fermi energy require
less energy than the work function for removal to the energy of free space, and electrons with energies less
than the Fermi energy require more energy than the work function for removal to the energy of free space.
The potential well in Figure 16.6 does not include temperature; this is a representation of the energies
at 0 K. At higher temperatures, it is possible to have electrons with energies greater than the Fermi energy.
W-14
CHAPTER 16
Free space
Energy
F = Work function
EF
E=0
Figure 16.6 The potential well model of a metal showing the conduction band (shaded), the Fermi energy (EF),
the work function (F), and the energy of free space.
Table 16.4 Values of the Work Function and the Parameter ARD
Metal
F (eV)
ARD (A/m2 ? K2)
W4.5 75 3 104
Ta4.2
55 3 104
Ni4.6
30 3 104
Cs1.8 160 3 104
Pt5.3 32 3 104
Cr4.6
48 3 104
Based on data from Herring, C. and Nichols, M.H., Revs. Mod. Phys. 21, (1949), p. 185.
Example Problem 16.2
Calculate the probability that an electron in tungsten at 3000 K has sufficient energy to be excited to
the energy of free space.
Solution
From Table 16.4, the work function of tungsten is 4.50 eV. The energy of free space is equal to EF 1 F. Putting
this energy and temperature into Equation 16.19.
PsEF 1 Fd 5
1 1 exp
1
5
EF 1 F 2 E F
PsEF 1 Fd 5
kT
1
F
1 1 exp
kT
1
4.5 eV
1 1 exp
0.259 eV
5
1
5
1 1 exp
4.5 eV
eV
8.62 3 1025
s3000 Kd
K
1
5 2.8 3 1028
1 1 exp 17.4
W-15
Electrical Properties of Materials
The free electrons that have sufficient energy for thermionic emission are those that have an energy greater
than Fermi energy plus the work function (EF 1 F), as shown in Figure 16.5.
The value of 1 in the denominator of the Fermi-Dirac statistical distribution is insignificant in
comparison to the exponential term in Example Problem 16.2, and little accuracy is lost by approximating
P(EF 1 F), as in Equation 16.23.
PsEF 1 Fd < exp
2F
kT
16.23
The thermionic current density (J ) in A/m2 through the surface of a metal at a temperature T is given by
the Richardson-Dushman equation, shown in Equation 16.24:
J < ARDT 2 exp
2F
kT
16.24
where ARD is a material constant whose value is given for some metals in Table 16.4.
16.4 Superconductors
Log of resistivity
In a superconductor, the electrical resistivity is 0 and the electrical conductivity is infinite below a
temperature called the critical temperature (Tc ). Superconductivity was first observed in mercury by Heike
Kamerlingh Onnes in 1908. As shown in Figure 16.7, the electrical resistivity of a metallic superconductor
as a function of temperature behaves like a normal metal for temperatures above the critical temperature
(Tc ). However, for temperatures below Tc, the electrical resistivity of the superconductor is 0. The
resistivity of a normal metal conductor, such as copper, does not go to zero as shown in Figures 16.4
and 16.7. Since its discovery in mercury, superconductivity has been found in many other materials, and
Table 16.5 lists the critical temperatures (Tc ) of some of these materials.
Superconductor
Metal conductor
Tc
0
Temperature (K)
Figure 16.7 A schematic of the electrical resistivity as a function of temperature for a normal metal conductor,
such as copper, and for a superconducting metal. Below Tc, the superconductor has 0 electrical resistivity, and
above Tc the superconducting metal behaves as a normal metal. The normal conductor, such as copper, has a
residual electrical resistivity at 0 K.
W-16
CHAPTER 16
Table 16.5 Critical Temperatures for Selected Superconducting Materials
Material
Critical Temperature (K)
Elements
V5.4
Nb9.3
Ta4.5
Al1.2
Sn3.7
Pb7.2
Ti0.4
W 0.02
Compounds
AlV311.8
AlNb.318
MgB239
GaV316.8
Nb3Sn18
Oxide compounds
YBa2Cu3O72
90
Ba2 CaCu2O8Tl2120
Bi2CaCu2O8Sr2110
Based on data from Haynes, W.M., ed., Handbook of Chemistry and Physics, 92nd ed., CRC Press, Boca Raton, FL. (2011-2012),
p. 12–60.
The superconducting elements and compounds and their values of Tc in Table 16.5 demonstrate
some of the interesting aspects of superconductivity. Although over 30 elements have been found to
be superconductors, the metals with the highest conductivities, such as copper, silver, and gold, are not
superconductors. Also, the only metal alloys that are superconductors are compounds. Solid solutions
are never superconductors. Once the model of superconductivity in metals is presented, we can explain
why solid solutions with randomly placed substitutional or interstitial atoms are not superconductors.
The materials with the highest critical temperatures are not metals; they are compound ceramic oxides.
Superconductivity has also been found in some polymers at temperatures up to approximately 12 K.
A theory for superconductivity in metals was developed by Bardeen, Cooper, and Schrieffer (BCS) in
1957. In the BCS theory, pairs of negative free electrons with spin up and spin down are coupled together
through interaction with the crystal positive ion cores. Figure 16.8 is a schematic of the interaction.
As an electron, such as the spin-up electron in Figure 16.8, moves through a solid that has strong
electron-ion coupling, there is an attraction between the positive ion cores and the negative electron.
As a result of the electron passing through the solid, the positive ions are displaced in the direction of
the electron. The electron has a much higher velocity than the positive ion cores do, and as a result the
ion cores are displaced in the direction of the spin-up electron after the electron has already passed.
The displaced positive ion cores attract the spin-down electron. The spin-up electron coupled with spindown electron through the displacement of the positive ion cores is called a Cooper pair of electrons. At
temperatures below the critical temperature, all of the free electrons in the crystal are coupled together
through the lattice of positive ion cores into Cooper pairs of electrons and the metal is a superconductor.
W-17
Electrical Properties of Materials
s = 1/2
+
+
ve
–
+
+
+
+
+
+
+
+
ve
–
+
+
s = –1/2
Figure 16.8 A schematic of the coupling of a Cooper pair of electrons (2) of opposite spin (s) with velocity ve
by interaction with the crystal lattice of positive ion cores (1) in a superconductor.
At temperatures below the critical temperature, the binding energy of the Cooper pair of electrons
through the displaced ion cores is greater than the thermal energy of the vibrating ions, which is equal
to approximately 3kT/2, and the Cooper pairs of electrons are not scattered by the ions vibrating in the
crystal. The stronger the coupling between the Cooper pair of electrons, the higher is the critical
temperature. The material becomes a normal conductor upon heating to temperatures above the critical
temperature when thermal vibrations of the atoms have sufficient energy to break the coupling between the
Cooper pair of electrons. The coupled pairs of electrons are like the coupling of automobile or bike
racers, where the second racer gets in the draft of the first racer, and this creates coupled pair. All of the
racers become coupled into one group where all of the racers travel at the same speed. When the electrons
of spin 112 and 212 form a Cooper pair, they no longer behave as individual particles with respective spins
of 112 and 212. The Pauli exclusion principle applies to individual electrons with spins of 112 and 212; it
does not apply to the Cooper pair. Therefore, in a superconductor, Cooper pairs of electrons can have the
same set of quantum numbers and move with the same velocity.
Now with the BCS theory we can explain some of the observations about superconducting materials.
Metals with a high electrical conductivity, such as copper, silver, and gold, are not superconductors,
because the coupling between the electrons and positive-ion cores is very weak. Superconductivity
requires a strong coupling between electrons and positive-ion cores. The superconductors are materials
such as compounds that have strong ion-electron coupling. Why are alloys with random substitutional
or interstitial atoms not superconductors? It is because the substitutional or interstitial atoms scatter
the Cooper pairs of atoms at any temperature, destroying the coupling of pairs of electrons. When an
ordered compound is in the superconducting state, the distance between the Cooper pair of electrons
adjusts to the regular spacing of the compound atoms and the Cooper pair is not scattered.
The higher the critical temperature, the easier it is to make superconducting products. Room
temperature is a desirable critical temperature; however, this has not been achieved to date.
Superconductivity has been achieved in some of the oxides at temperatures above the liquid nitrogen
equilibrium vaporization temperature of 77 K. The oxide superconductors are commercially viable in
some applications because liquid nitrogen is of reasonable cost. High-temperature superconductors have
not achieved widespread use, because it is difficult to make a durable wire of ceramic material that
can carry a high current density. Ceramic polycrystalline superconductor wires are brittle and subject
to cracking; therefore, they are encased in a metal such as silver. If the superconducting wire breaks, the
silver casing can carry current. Also, polycrystalline superconductor wires carry a lower current density
than single-crystal superconductors, because it is difficult for the electrons to cross the nonsuperconducting
grain boundaries. Because of the defects at the grain boundary of a superconductor, grain boundaries
are not normally superconducting. In spite of these problems, the city of Detroit, Michigan, in the
United States has installed 400 feet of high-temperature superconducting cable made of YBa2Cu3O72
that can deliver 100 million watts of power. Superconducting cable has the potential of delivering more
watts per unit area of cable without any loss of power over the length of the cable. In YBa2Cu3O72
the indicates that there must be some vacant oxygen sites for there to be superconductivity.
W-18
CHAPTER 16
Single-crystal thin films of superconductors can carry very high current densities of approximately
107 A/cm2. It is possible that an early production application of high-temperature superconductors will be
for single-crystal, thin-film interconnection applications, such as in supercomputers, that require a high
current density and can be cooled to liquid nitrogen temperatures.
The production of magnetic fields with superconductors provides the interesting phenomena of
magnetic levitation (maglev). In maglev a magnetic field is used to lift an object. We can observe maglev
when two similar poles of two magnets are brought together; they repel each other. Maglev trains are
lifted and propelled by magnetic fields. Maglev trains use superconducting coils to produce the magnetic
fields. Chapter 17 covers superconductors in relation to magnetic fields and maglev trains, and the
introductory photograph to Chapter 17 is of a maglev train.
16.5 Electrical Conductivity of
Semiconductors and Insulators
16.5.1 Room-Temperature Electrical
Conductivity of Intrinsic Semiconductors
The electrical conductivity of an intrinsic semiconductor is the same as the electrical conductivity of a pure
semiconductor. At high temperatures, semiconductors with impurity atoms have the same electrical conductivity
as a pure semiconductor, and these semiconductors are then considered to be intrinsic semiconductors, as we
will discuss in Section 16.7.3. As shown in Table 16.1, semiconductors have an electrical conductivity that is
typically from 1026 to 10215 times the electrical conductivity of metals such as copper. Table 16.6 lists some of
the different semiconductors and their intrinsic electrical conductivities at room temperature.
Table 16.6 The Energy Gap, Electrical Conductivity, Electron and Hole Mobility, and Density of Carriers
at Room Temperature for Some Typical Intrinsic Semiconductors
Energy Electrical Electron Hole Intrinsic
Gap
Conductivity
Mobility
Mobility Carrier
Material
(eV) (V ? m)21(m2/V ? s)
(m2/V ? s)
Density (m23)
Elements
Si
1.114 3 1024 Ge 0.672.2
0.14
0.05
1.0 3 1016
0.38 0.18 2.4 3 1019
III-V Compounds
GaAs1.42 1026
0.85 0.452 3 1012
InSb 0.172 3 10 7.7 0.07—
4
II-VI Compounds
CdTe 1.44 —
0.12 0.005—
ZnTe2.26 —
0.03 0.01 —
The Data is from a Variety of Sources
Electrical Properties of Materials
The electrical conductivity of intrinsic semiconductors is related to breaking the covalent bonds,
because if the covalent bonding is perfect, the covalently bonded electrons do not migrate and the
electrical conductivity is 0. Figure 16.9a shows the covalent bonding in Group IV elements, such as
silicon. Two electrons are localized between each pair of silicon atoms in a covalent electron pair bond,
as we discussed in Chapter 2. Each silicon atom is surrounded by four other silicon atoms, and between
each of these four atoms and the central silicon atom is a pair of covalently bonded electrons, resulting
in a total of eight electrons surrounding each silicon atom. Figure 16.9 is a planar view of the bonding;
the atoms are actually in a tetrahedral arrangement. The electrical conductivity of the semiconductor in
Figure 16.9a is 0. However, as shown in Figure 16.9b, if an electron is excited out of the covalent bond,
a free electron similar to the free electrons in a metal is created. All of the discussion of the electrical
properties of free electrons in metals applies to a free electron in a semiconductor. The free electron
can be conducted because it is free of the covalent bond, but it is bound inside the semiconductor.
To maintain charge neutrality, a positive charge called a hole remains on the atom where the negative
electron is excited from the covalent bond. The excitation of the electron out of the covalent bond can
be due to thermal vibrations, irradiation with photons, energetic electrons, or other energetic particles.
+4
+4
+4
+4
+4
+4
+4
+4
+4
(a) T = 0 K
+4
+4
+4
+4 +
+4
+4
+4
+4
–
+4
(b) T > 0 K
E
+4
+4
+4
+
–
+4
+4
+4
+4
+4
+4
(c) T > 0 K
Figure 16.9 (a) A two-dimensional schematic of the covalently bonded electrons (a pair of dashes) in silicon.
(b) The creation of a free conduction electron (red circle with 2 sign) by thermal excitation and the resulting hole
(yellow circle with 1 sign). (c) The application of an electric field (E) directed to the right moves the free electron
to the left, and the hole migrates to the right by diffusion through covalent bonds.
W-19
W-20
CHAPTER 16
This chapter considers excitation only by thermal vibrations, and Chapter 18 discusses how photons can
excite covalently bonded electrons.
An electric field (E) applied to the silicon produces a force on both the free conduction electron and the hole.
The conduction electron is accelerated in a direction opposite to the electric field, and the hole is accelerated
in the direction of the electric field. The conduction mechanism for the free electron in a semiconductor is
the same as for a free electron in a metal. The velocity of the electron is increased by the applied electric field,
creating a drift velocity of the free electrons. The mechanisms for electrical resistivity for the free conduction
electrons in a semiconductor are the same as for the conduction electrons in metals; it is the scattering of
conduction electrons by defects and lattice vibrations. The conduction mechanism for the hole is similar to
vacancy diffusion. The hole moves when an electron from a neighboring covalent bond jumps or diffuses
to the hole. This neutralizes the hole, and the jumping electron leaves behind another hole. In intrinsic
semiconductors the electrical conductivity is much lower than is observed in a metal. For example, at room
temperature the electrical conductivity of OFHC copper is 6.0 3 109 (V ? m)21 and the electrical conductivity
of pure silicon is 4.0 3 1024 (V ? m)21. This is a difference of 13 orders of magnitude in electrical conductivity.
Elemental semiconductors come from Group IV of the periodic table. Silicon and germanium
are intrinsic elemental semiconductors, as shown in Table 16.6. In addition, there are compound
semiconductors, such as the III-V (GaAs) and II-VI (CdTe) compounds. The Roman numbers III-V and
II-VI are the group numbers of the component atoms in the compound semiconductor, and the sum of
the group numbers is 8 or the average group number is 4. Eight is the same number of valence electrons
in the stable inert gas atoms. Compound semiconductors are utilized because of their unique properties.
For example, GaAs has a very high electron mobility, as shown in Table 16.6.
Figure 16.10 presents the band structure model of the conduction process in a pure elemental
semiconductor. In semiconductors and insulators, the covalently bonded electrons form the valence band
(VB). At 0 K the VB is filled with the covalently bonded electrons, as indicated by the blue shading in
Figure 16.10a, and the maximum energy electron in the VB is Ev. In the energy band model, there is an
energy gap (Eg) between Ev and the lowest possible energy conduction electron at energy Ec. The energy
gap is the minimum energy required to excite a valence electron into the conduction band. Table 16.6
presents the energy gap for some semiconductors. The conduction electrons are in the conduction band
of energies. The conduction band (CB) contains the possible energy levels for conduction electrons;
however, the energy levels are not necessarily occupied with conduction electrons. For example, at 0 K
there are no conduction electrons in the CB, as indicated by the white background in Figure 16.10a. The
intrinsic semiconductor is an insulator at 0 K, because there is no charge that can be conducted. As the
E
Eg
Energy
–
–
Ec
+
(a)
(b)
Conduction
Band (CB)
+
Ev
Valence
Band (VB)
(c)
Figure 16.10 (a) Electron band structure of an intrinsic semiconductor at 0 K. The VB is full as indicated by the
full blue shading. (b) Electron band structure at a temperature above 0 kelvin showing an electron (red circle with
a 2 sign) excited into the CB from the VB, leaving behind a hole (yellow circle with a 1 sign). (c) The conduction
electron and the hole both are both accelerated by an applied electric field (E), but in opposite directions. Electrons
in the CB are at a minimum energy Ec; electrons in the VB have a maximum energy of Ev, and the energy gap is
equal to Eg.
W-21
Electrical Properties of Materials
temperature increases, there is an increasing probability that electrons are thermally excited from the
covalent bonds in the VB into the CB, as shown in Figure 16.10b by the red circle with a minus 2sign,
and a hole is created in the valence band as indicated by the yellow circle with the 1 sign. The free
conduction electrons have a probability of being found anywhere in the semiconductor, and all of the free
electrons have a probability of interacting with each other. Therefore, the conduction electrons obey the
Pauli exclusion principle and Fermi-Dirac statistics. In semiconductors, the CB is similar to the band of
energies for metals shown in Figure 16.2.
The negative electron in the CB and the positive hole in the VB are both accelerated by an applied
electric field (E). Equation 16.7 calculates the electrical conductivity due to electrons for an intrinsic
semiconductor; however, a term for the hole conductivity must be added, as shown in Equation 16.25:
5 nen 1 pep
16.25
where p is the hole density and p is the hole mobility. For intrinsic semiconductors, the number of
conduction electrons and holes is equal because the electrons and holes are created in the same event.
Table 16.6 presents the density of intrinsic electrons and holes (n 5 p) at room temperature for some
semiconductors. In summing the electron and hole conductivities in Equation 16.25, the value of e is
the magnitude of the charge on an electron, and the current resulting from an electron going in the
opposite direction to the electric field adds positively to the current resulting from a hole going in the
direction of the electric field. The electron and hole mobility (n and p) are material properties that are
dependent upon the type of semiconductor, temperature, and defect density. Table 16.6 presents typical
room-temperature electron and hole mobility. If a semiconductor has a high number of defects, such
as impurity oxygen atoms or dislocations as a result of poor crystal growth procedures, the mobility of
the electrons and holes in the semiconductor is reduced. For conduction electrons in a semiconductor,
the distance between scattering events is reduced by defects, as we discussed for metals in Section 16.2,
and holes are trapped at negatively charged defects. Different semiconductors have different intrinsic
mobility. For example, GaAs is a compound semiconductor with an electron mobility of 0.85 m2/V ? s in
comparison to the electron mobility of 0.14 m2/V ? s for silicon. Electrons are six times as mobile in GaAs
as in silicon. This makes GaAs a desirable semiconductor for high-speed electronic applications, such as
fast computers; however, GaAs is more expensive than silicon because it is more difficult to produce.
Example Problem 16.3
With the values of mobility and carrier density given in Table 16.6, calculate the electrical conductivity
of pure germanium at room temperature.
Solution
The electrical conductivity is calculated with Equation 16.25 for elecrons (n) and holes (p).
5 nen 1 pep
5 2.4 3 1019
1
n
C
1.60 3 10219
3
m
n
210.38V ? s2 1 12.4 3 10 m 211.60 3 10 Cp210.18 Vm? s2
m2
19
p
3
5 1.5(V ? m)21 1 0.7(V ? m)21 5 2.2(V ? m)21
This is in agreement with the measured value of 2.2(V ? m)21 in Table 16.6.
2
219
W-22
CHAPTER 16
16.5.2 The Effect of Temperature upon the
Electrical Conductivity of Intrinsic Semiconductors
As is indicated in Figure 16.10, at 0 K there are no conduction electrons or holes in a covalently bonded
material. However, as temperature is increased there is an increasing probability, given by the FermiDirac statistical distribution, that an electron is excited from the VB into the CB, creating a negative
conduction electron and a positive hole. The Fermi-Dirac statistical distribution given in Equation 16.18
shows that this probability increases exponentially with temperature; therefore, the density of both holes
and electrons increases exponentially in an intrinsic semiconductor, and the electrical conductivity given
in Equation 16.25 increases exponentially. The temperature dependence of the electrical conductivity of
an intrinsic semiconductor is the opposite to that of a metal whose electrical conductivity decreases with
increases in temperature. In a metal, there is no change in the conduction electron density with temperature,
and the mobility of the electrons in a metal decreases with increases in temperature. The mobility of electrons
in a semiconductor also decrease with increases in temperature for the same reason that it does in a metal.
Intrinsic semiconductors are not utilized in electronic devices. Electronic devices utilize semiconductors
that are modified with impurity atoms, as we will discuss in Sections 16.5.4 and 16.5.5.
16.5.3 Electrical Conductivity of Insulators
An insulator has a high electrical resistivity and a low electrical conductivity because of the small density
of conduction electrons and holes. The density of conduction electrons and holes is small at room
temperature if the energy gap is large, because then there is a low probability for electrons in the VB to be
thermally excited into the CB. For a pure insulator with a large energy gap (Eg ), the number of conduction
electrons and holes is smaller than for an insulator with a small energy gap. This is demonstrated in
Table 16.6, where the density of intrinsic carriers at room temperature in germanium with a band gap
of 0.67 eV is 2.4 3 1019 m23; and for GaAs with a band gap of 1.42 eV the intrinsic carrier density is
2.0 3 1012 m23. The term carrier refers to the particles that transport charge. In an intrinsic semiconductor,
the carriers are both electrons and holes of equal number. The difference in carrier density between Ge
and GaAs is then reflected in the conductivities that are 2.2 (V ? m)21 and 1026(V ? m)21, respectively. The
electrical conductivity of an insulator also obeys Equation 16.25. Insulators have a large energy gap. For
example, diamond has the same crystal structure and bonding type as silicon, but diamond has a very
large energy gap of 5.5 eV in comparison to the energy gap of 1.1 eV for Si and other semiconductors
shown in Table 16.6. This results in pure diamond having one of the lowest conductivities of all materials
at room temperature. Diamond is an excellent insulator. The electrical conductivity of insulators increases
exponentially as temperature increases, as predicted by the Fermi-Dirac statistical distribution.
16.5.4 The Electrical Conductivity of n-Type
Extrinsic Semiconductors
Integrated circuits are not made from intrinsic semiconductors; they are made from extrinsic
semiconductors. An extrinsic semiconductor is one whose electrical properties are dependent upon the
addition of impurity atoms (doping) to a pure host crystal. An extrinsic n-type semiconductor is one
that has more negative charge carriers (free electrons) than holes. An n-type semiconductor is made, for
Electrical Properties of Materials
example, with silicon from Group IV by adding impurity atoms from a higher-number group, such as
phosphorus or arsenic from Group V. In the bond model of an n-type semiconductor, shown in Figure 16.11a,
a Group V substitutional atom contributes four electrons to the covalent pair bonds of the host silicon
crystal, but the Group V atom has an additional fifth valence electron. This fifth electron is not in a
covalent bond, because all of the eight bonding orbitals in the host silicon crystal are satisfied by four of
the five valence electrons of the Group V atom. In a host silicon crystal with dopant atoms of phosphorus,
the fifth valence electron is bonded to the phosphorus atom by a weak electrostatic bond of 0.045 eV that
maintains electrical neutrality, as shown in Table 16.7. If the fifth electron is removed from the Group V
atom by thermal energy, it is a free conduction electron, as shown in Figure 16.11b, and the Group V
atom becomes a positively charged ion. The positive charge on the Group V ion is not conducted
unless the ion diffuses. If an electric field (E) is applied to the n-type semiconductor, as shown in
+4
+4
+4
+4
+4
+5
+4
+4
+4
– Bound electron
(a) T = 0 K
+4
+4
+4
+4
+4
+5
+4
+4
+4
– Free electron
(b) T > 0 K
E
+4
+4
+4
+4
+4
+5
+4
+4
+4
–
(c) T > 0 K
Figure 16.11 The electron bond model of an n-type semiconductor. (a) When a Group V atom, such as
phosphorous, with five valence electrons substitutes for a silicon (Si) atom, only four of the group V atom’s valence
electrons are in covalent bonds (pairs of dashes) in the silicon lattice. The fifth valence electron (white circle with
a 2 sign) on the Group V atom has a weak electrostatic bond. (b) The fifth valence electron on the Group V atom is
excited by thermal vibrations to become a free electron (red circle with a 2 sign) in the CB. (c) An applied electric
field (E) accelerates the conduction electron in a direction opposite to the direction of the applied electric field.
W-23
W-24
CHAPTER 16
Table 16.7 Donor and Acceptor Impurity Levels in Silicon and Germanium
Silicon
Germanium
Dopant
Ec 2 Ed
Ea
Ec 2 Ed
P0.045
0.0120
As0.049
0.0127
Sb0.039
0.0096
Ea
B 0.0450.0104
Al0.0570.0102
Ga0.065 0.0108
In 0.1600.0112
Based on data from Askeland, D.R., Fulay, P.P., and Wright, W.J., The Science and Engineering of Materials, 6th ed., Cengage Learning,
Stamford, CT. (2011), p. 738.
Figure 16.11c, the free electron is accelerated by the electric field in a direction opposite to the direction
of the electric field.
Figure 16.12 shows the band model of an n-type semiconductor. The energies associated with the VB
and the CB are the same as in the intrinsic semiconductor. The energy level of the fifth valence electron
bound to the Group V atom is the donor level (Ed). The electron bound to the Group V atom has a
probability of being excited (donated) into the CB by thermal vibrations to provide conduction electrons,
as shown in Figure 16.12b. The conduction electrons are accelerated by an applied electric field, as shown
in Figures 16.11c and 16.12c. Some values of the donor level relative to the CB (Ec 2 Ed) are given in
Table 16.7 for some different donor atoms in silicon and germanium. It is possible for an electron in the CB
to move back to a donor atom or to recombine. When recombination happens, the positive and negative
charges of the ionized donor atom and the conduction electron are eliminated. Recombination is an
ongoing process, since this lowers the system energy by releasing an amount of energy equal to Ec 2 Ed.
However, a donor atom is again ionized, and the electron is excited into the CB by thermal vibrations to
produce an equilibrium number of conduction electrons.
Energy
E
–
–
–
–
–
–
–
+
–
CB
–
– –
+
–
Ec
Ed
Ev
VB
(a)
(b)
(c)
Figure 16.12 (a) The occupation of electron energy levels for an n-type semiconductor at 0 K. The VB is filled
up to the energy Ev, as indicted by the blue shading. The donor level at energy Ed is full of electrons, as indicated by
the symbols ⊝. The CB beginning at energy Ec is empty. (b) At a temperature above 0 K, donor level electrons are
excited into the CB, leaving behind a positive charge (∙) on the donor atom. (c) An electric field (E) accelerates the
free electron in the CB (red circle with a 2 sign).
W-25
Electrical Properties of Materials
Mobility (m2/ V-s)
100
10–1
n-GaAs
10–2
n-Si
p-GaAs
p-Si
–3
10
10–4 20
10
1022
1024
Impurity density (/m3)
1026
Figure 16.13 Mobility of electrons (n) and holes (p) in silicon and GaAs, as a function of impurity atom density
at 295 K. (Based on Kwok, H.L., Electronic Materials, PWS Publishing (1997), p. 111.)
The n-type semiconductors are normally designed to operate in the extrinsic temperature range, where
the density of conduction electrons is constant. In the extrinsic temperature range, all of the donor
atoms are ionized, because the energy difference between the donor level and the CB (Ec 2 Ed) is very
small, such as 0.01 eV. Also, there is an insignificant density of electrons excited from the VB into
the CB across the energy gap, because the energy gap is large (approximately 1 eV) in comparison to
0.01 eV. For example, in germanium doped with 6 3 1021 arsenic atoms per cubic meter, the extrinsic range
of operation extends from approximately 50 to 300 K. At temperatures higher than 300 K, a significant
number of electrons are excited from the valence band into the conduction band. In the extrinsic range
of operation of an n-type semiconductor made from an element in Group IV with donor atoms from
Group V, the concentration of electrons in the CB (n) is equal to the concentration of donor atoms (Nd),
as shown in Equation 16.26.
n 5 Nd
16.26
In Equation 16.26, we assume that each donor atom is singly ionized and that there are no holes in the
VB ( p 5 0). The mobility of electrons at the temperature and the donor concentration of the
semiconductor are needed for the electrical conductivity calculation in Equation 16.25. Figure 16.13 shows
that the mobility of electrons in n-type semiconductors at room temperature decreases as the impurity
concentration increases. This is because the ionized donor atoms scatter the conduction electrons, thereby
decreasing the electron mobility. The electrical conductivity of an n-type semiconductor is calculated
with Equation 16.25 at any temperature within the extrinsic range of operation where n 5 Nd and p 5 0,
if the electron mobility is known for that temperature and donor concentration (Nd).
Compound semiconductors, such as GaAs, are also made n-type. Gallium is in Group III and arsenic
is in Group V. If there are equal numbers of gallium and arsenic atoms, then each atom shares eight
electrons, and the semiconductor is intrinsic. GaAs is made n-type by replacing As from Group V with
an element such as sulfur from Group VI.
Example Problem 16.4
An n-type semiconductor is produced by doping silicon with 10 parts per billion phosphorous, which
is 10 atoms of phosphorous for 109 atoms of silicon. Calculate the electrical conductivity for this
semiconductor at room temperature using the data in Figure 16.13. Assume that the semiconductor
is operating in the extrinsic range.
W-26
CHAPTER 16
Solution
Phosphorous is in Group V of the periodic table; thus each donor atom contributes one electron to the CB of
silicon. Assume that the phosphorous atoms are all ionized at room temperature, which allows us to calculate
the electron density from the impurity composition.
Nd 5
1
2
1
2
2.33 3 103 kg of Si
10 P atoms
1 n 6.02 3 1023 Si atoms
1 mole of Si
23
9
m3
10 Si atoms P atom 1 mole of Si atoms 28.09 3 10 kg of Si
n 5 Nd 5 4.99 3 1020
n
m3
Taking the value of the electron mobility from Figure 16.13 to be 0.1 m2/V ? s results in
5 nen 5 4.99 3 1020
1
C
n
1.60 3 10 219
m3
n
21
0.1
2
m2
5 8 sV ? md 21
V?s
16.5.5 The Electrical Conductivity of p-Type
Extrinsic Semiconductors
Semiconductor devices are made by combining n-type and p-type extrinsic semiconductors. A p-type
extrinsic semiconductor is one that has more positive charge carriers (holes) than negative charge
carriers (electrons). An example of a p-type extrinsic semiconductor is Group IV silicon doped with an
element of a group number less than Group IV, such as the Group III elements boron, aluminum, or
gallium. A p-type semiconductor operating in the extrinsic temperature range has a constant density of
holes in the VB equal to the density of dopant atoms. As schematically shown in Figure 16.14a, when
a Group III element substitutes for a silicon atom, the Group III atom provides only three covalentbonding electrons (dashes) to share with the four silicon atoms that surround the Group III atom
in the silicon crystal. This leaves one vacant electron orbital (missing dash) in the covalent bonds
surrounding the Group III atom; however, the Group III atom in Figure 16.14a is electrically neutral. At
temperatures above 0 K, it is possible for a covalently bonded electron from a neighboring silicon atom
to diffuse into the previously vacant electron orbital at the Group III atom, as shown in Figure 16.14b.
This results in the Group III atom having an extra covalently bonded electron charging it negative.
A positively charged hole (yellow circle with a 1 sign) is associated with the silicon atom that is the
source of the extra electron on the Group III atom. The covalent molecular orbitals characteristic of
the silicon crystal exist at every atom site in the silicon crystal, even if an atom site is occupied by a
Group III atom. The extra electron associated with the negatively charged Group III ion is a covalently
bonded electron and not a mobile electron. In Figure 16.14c, an electric field (E) is applied to the p-type
semiconductor, and the hole migrates in the direction of the electric field by diffusion through covalent
bonds.
The energy of the fourth covalently bonded electron on the Group III atom is the acceptor level (Ea),
as shown in the band model of Figure 16.15. Figure 16.15a shows that at 0 K, all of the electrons in the
p-type semiconductors are in the VB. In Figure 16.15b, at temperatures above 0 K it is possible for an
electron to be excited into the acceptor energy level indicated by the symbol ⊝, and a hole is created in
the VB, as indicated by the yellow circle with the 1 sign. The acceptor level (Ea) is a little higher in energy
than the VB of the silicon atoms, because the fourth electron on the Group III ion is covalently bonded,
Electrical Properties of Materials
+4
+4
+4
+4
+4
+3
+4
+4
+4
(a) T = 0 K
+4
+4
+4
+4
+4
+3
+4
+4
+4
+
(b) T > 0 K
E
+4
+4
+4
+4
+4
+3
+4
+4
+4
+
(c) T > 0 K
Figure 16.14 The electron bonding model of a p-type semiconductor. (a) A Group III atom with three valence
electrons substitutes for a silicon (Si) atom, and there is one missing covalently bonded electron in the silicon
crystal lattice, indicated by the missing dash. The Group III atom is neutral in charge. (b) A covalently bonded
electron from a neighboring silicon atom diffuses to the Group III atom, creating a positive charge (hole) indicated
by the yellow circle with a 1 sign where the electron is missing from the silicon atom. There is a negative charge
at the Group III atom (acceptor atom), where there are now eight covalently shared electrons. (c) The hole with
a positive charge is accelerated by the applied electric field (E) in the direction of E, but the covalently bonded
negative charge on the Group III atom is not mobile.
but this fourth electron charges the Group III atom negatively rather than being neutral. Table 16.7
presents some acceptor levels in silicon and germanium semiconductors. It is possible for the electron
on the acceptor atom to move into a hole in the VB; when this recombination happens, the positive
and negative charges of the hole and acceptor electron cancel, and the two charges are eliminated.
Recombination in p-type semiconductors is an ongoing process, since this lowers the system energy by
releasing an amount of energy equal to the energy of the acceptor level above the top edge of the VB.
However, other holes in the VB are created by thermal vibrations exciting VB electrons to the acceptor
level to produce an equilibrium number of holes. An applied electric field (E) produces a force (Ee) on
the hole, and the positively charged hole moves in the direction of the electric field by diffusion, as shown
in Figure 16.14c and 16.15c.
W-27
W-28
CHAPTER 16
E
Eg
CB
Energy
Ec
(a)
–
+
–
(b)
(c)
Ea
+
Ev
VB
Figure 16.15 (a) The occupation electron energy levels in a p-type semiconductor at 0 K. (b) At a temperature
above absolute zero an electron is excited from the VB to the acceptor level (Ea) indicated by ⊝ associated with a
group three atom leaving a hole in the VB (yellow circle with a 1 sign). (c) An applied electric field (E) accelerates
the hole in the VB in the direction of the electric filed, but there is no conductivity of the covalently bonded
electron (⊝) in the acceptor level.
Equation 16.25 is used to calculate the electrical conductivity of a p-type semiconductor, and
simplifications are possible in the extrinsic range of operation. The energy difference between the
acceptor level and the VB (Ea 2 Ev) is very small, such as 0.01 eV, and the energy gap is large in
comparison (approximately 1 eV). Normally in semiconductor analysis the energy Ev is set equal to 0. In
the extrinsic temperature range of operation of a p-type semiconductor, all of the acceptor energy states
are filled, and there is an insignificant density of electrons excited from the VB into the CB across the
energy gap. For example, in silicon doped with 6.5 3 1023 boron atoms per cubic meter, the extrinsic range
of operation extends from approximately 200 to 700 K. In the extrinsic range of operation of a p-type
semiconductor made from a doped Group IV element, the concentration of holes in the VB (p) is equal
to the concentration of acceptor atoms from Group III (Na), as shown in Equation 16.27.
p 5 Na
16.27
In Equation 16.27, we assume that each acceptor atom is singly charged for Group III substitutional atoms
in silicon. The mobility of holes at the temperature and impurity concentration of the semiconductor
is needed for the electrical conductivity calculation in Equation 16.25. At any temperature within the
extrinsic range of operation, p 5 Na and n 5 0. The hole mobility must be known for that temperature
and acceptor concentration (Na). Figure 16.13 shows that the mobility of holes in p-type semiconductors
at room temperature decreases as the impurity concentration increases.
A p-type semiconductor can be produced in compound semiconductors such as GaAs by replacing
Ga from Group III by an element such as Be, Mg, or Zn from Group II. The Group II element provides
the acceptor atoms.
Example Problem 16.5
A semiconductor is produced by doping Group IV silicon with 1023 atoms per cubic meter of
Group III gallium. Assume that at room temperature this semiconductor is in the extrinsic range of
operation.
(a) What type of semiconductor is this?
(b) Calculate the room-temperature electrical conductivity of this semiconductor.
(c) State any assumptions you made in calculating the electrical conductivity in part (b).
Electrical Properties of Materials
Solution
a) Since the dopant is Group III gallium, the semiconductor is p-type.
b) In the extrinsic range of operation, the number of holes is equal to the number of acceptor atoms, resulting
in p 5 1023 holes/m3 and n 5 0. From Figure 16.13, the mobility of holes in p-type silicon is approximately
1.5 3 1022 m2/V ? s. From Equation 16.25, is the electrical conductivity of a p-type semiconductor is
5 pep 5 1 3 1023
1
2
p
C
m2
5 2.4 3 102 sV ? md 21
1.60 3 10 219
1.5 3 10 22
3
m
p
V?s
c) Assume n 5 0 and p 5 1023 atoms/m3.
16.6 Polymer Insulators,
Semiconductors, and Conductors
Most polymers such as polyethylene (PE) are insulators; PE has one of the highest values of electrical
resistivity, as shown in Table 16.1. The model of insulators, such as diamond, that we discussed in
Section 16.5.3 also applies to polymers. The covalently bonded electrons are in the VB, and electrons
must be excited out of covalent bonds to be conducted. PE has a large energy gap of 9 eV. Therefore, at
room temperature there is an insignificant density of conduction electrons and holes in pure PE.
PE and other polymers with high electrical resistivity are combined with conductive powders, such as
graphite or silver, to produce composite materials with increased electrical conductivity that are utilized
in resistors. The conductive powder is a second phase; it does not chemically combine with the polymer.
The polymer binds the conductive powder particles together into a composite solid. A current can pass
from conductive particle to conductive particle if the particles touch, in which case the electrical resistivity
is given by the rule of mixtures in Equation 16.13. However, if the particles do not touch, conduction
electrons must penetrate (tunnel) through the polymer in the gap between the conductive particles to
provide electrical conductivity. The gaps between particles result in a much higher electrical resistivity
than when the conductive particles touch. When the particles do not touch, the electrical resistivity of
the polymer-conductor composite material increases exponentially with increase in distance between the
conductive particles.
A conductive composite polymer with metal particles has recently been developed that has potential
biomedical applications as a skin material. This composite polymer is electrically conductive, it heals if
torn or cut, and it is sensitive to pressure. The polymer is produced by combining an olgiomer with urea
[CO(NH2)2]. An oligomer has only a few units, in contrast to a polymer, which has many units. Oligos
in Greek means “a few.” Urea links the oligomer into a network polymer in a manner similar to how
formaldehyde reacts with phenol molecules to form phenolformaldehyde (PF), as shown in Figure 7.28.
If the bond formed by the urea between the olgiomers is broken by cutting or tearing, the bond can
reform. However, broken PF bonds do not re-form. Micron-sized particles of nickel (Ni) are added to
the network polymer to provide electrical conductivity. There is pressure sensitivity because compression
forces the μNi particles closer together. The shorter distance between μNi particles increases the electrical
conductivity. An electrically conductive and pressure sensitive skin material can detect damage, and
self-healing can repair the damage.
Recently polymers such as polyacetylene have been found to be semiconductors and conductors.
The electrical conductivity of undoped polyacetylene is approximately 1026 (V ? m)21, and for doped
W-29
W-30
CHAPTER 16
polyacetylene it is approximately 104 (V ? m)21, depending upon the type and amount of dopant. As
shown in Table 16.6, the electrical conductivity of intrinsic silicon is 4 3 1024 (V ? m)21. Polyacetylene
has an energy gap of 1.8 eV in comparison to 1.1 eV for silicon. Figure 16.16a shows the molecular
structure of polyacetylene with perfect bonding. PE has two hydrogen atoms per carbon atom. However,
in polyacetylene each carbon atom has only one attached hydrogen atom. Because there is only one
hydrogen atom per carbon atom, every other carbon atom along the polyacetylene chain forms a double
bond with an adjacent carbon atom, giving the carbon atoms a total of eight shared electrons. Four
electrons are shared between two carbon atoms in a double bond. If the bonds on polyacetylene are
perfect, as shown in Figure 16.16a, polyacetylene is an insulator.
If there are defects in the bonding, polyacetylene conducts electricity. In polyacetylene, defects occur
in the bonding, as shown schematically in Figure 16.16b, where the alternation of single and double
bonds is incorrect, and two single bonds repeat to form a soliton. The carbon atom at the soliton is
electrically neutral if there is also an electron in a non–covalently bonded state indicated by the dot ( ) in
Figure 16.16b. The non–covalently bonded electron can be moved to the right along the polyacetylene
chain if one electron from the double bond to the right of the soliton moves to the left one spacing along
the chain; this moves the soliton along with the non–covalently bonded electron one spacing to the right.
The activation energy for this electron motion is estimated to be only 0.01 eV, and the current density is
calculated with an Arrhenius-type rate equation.
?
–
C
C
C
C
H
N
C
H
C
H
C
C
C
C
H
H
H
C
C
H
C
H
H
H
H
H
C
C
N
C
C
C
H
H
H
H
C
H
C
H
H
C
C
C
H
C
C
H
H
(d)
H
C
C
C
C
C
H
H
H
H
C
C
C
C
H
C
C
H
(c)
H
+
H
H
C
C
H
H
H
C
C
C
H
(b)
C
C
C
C
H
H
H
C
C
C
H
H
H
H
H
H
(a)
H
Figure 16.16 (a) The atomic structure of perfect polyacetylene. (b) The structure of neutral polyacetylene with
a defect in the bonding (soliton); and a single electron in a nonbonding state, indicated by ( ), resulting in an
electrically neutral chain. (c) A positive charge on polyacetylene (1) located at a soliton is created by the removal
of the nonbonded electron. (d) Negatively charged polyacetylene has two nonbonding electrons, indicted by ( ),
and a negative charge, indicated by (2). (Based on http://en.wikipedia.org/wiki/File:Trans-(CH)n.png)
?
??
Electrical Properties of Materials
Dopants of iodine, chlorine, or fluorine oxidize polyacetylene and produce a p-type semiconductor
with a positive charge by removing the non–covalently bonded electron from the soliton, as shown in
Figure 16.16c. The dopant does not substitute for an atom along the chain, as in the case of doped silicon
semiconductors. Rather, the dopant forms a second phase that is mixed with the polyacetylene. The non–
covalently bonded electron is transferred from the polyacetylene to the second phase. The positive charge
on the polyacetylene moves in the same way as described for the neutral soliton above.
Dopants such as sodium and lithium from Group IA mixed with polyacetylene produce an n-type
semiconductor by contributing electrons to the polyacetylene, as shown schematically in Figure 16.16d
by the two dots ( ), which represent two electrons. The second non–covalently bonded electron at the
soliton results in a negative charge, as shown in Figure 16.16d. The negative charge on the polyacetylene
moves in the same way as is described for the neutral soliton above.
The dopant is mixed with the polymer by several different processes. A solid polymer can be exposed
to a gas of the dopant, and the gas diffuses into the polymer. Liquid dopant is also mixed with liquid
polymer. If the dopant and the polymer are not soluble in each other, a two-phase mixture of polymer
and dopant results upon solidification.
Some possible advantages of polymer conductors and semiconductors relative to single-crystal
silicon are sustainable supplies, reduced costs, mechanical flexibility, and transparency. There have been
predictions that polymers with conductivities greater than that of copper can be developed. At very high
levels of doping, greater than 6 atom percent, polyacetylene has an electrical conductivity comparable
to that of some metals. There are conducting polymers other than polyacetylene; however, the simple
bonding in polyacetylene allows for demonstration of the conduction mechanism in polymers. Electronic
devices have been made from polymer semiconductors, and polymer semiconductors have already found
applications such as cell phone display screens.
??
16.7 Junctions and Electronic Devices
16.7.1 Introduction
Crystalline semiconductor devices such as diodes, light-emitting diodes (LEDs), diode lasers, and
photovoltaic solar cells are made from pn-junctions. The transistors that amplify a signal in an electronic
circuit are npn-junctions. Diodes control the direction of the current flow in an electrical circuit. Diodes
are also made from metal-polymer-metal (MPM) junctions; several uses of MPM-junctions are as LEDs
and solar cells.
16.7.2 Metal-Metal Junctions
Metal-metal junctions provide a foundation to understand pn-junctions and MPM-junctions. A metalmetal junction is formed when two different metals are in electrical contact. Two metals with different
work functions (FA and FB) are shown separated in Figure 16.17a. The work functions of some metals
are presented in Table 16.4.
If metals A and B are brought into electrical contact, as in Figure 16.17b, electrons from metal B flow
into metal A, as indicated by the arrow in Figure 16.17a, because there are lower unoccupied electron
energy levels in metal A. A positive charge results on metal B, and a negative charge results on metal A.
W-31
W-32
CHAPTER 16
Free space
(a)
FB
FA
EFB
Electrons
EFA
Electrons
FA – FB = eVc
Free space
(b)
EF
–+
–+
–+
–+
Metal A
Metal B
Figure 16.17 (a) The band structure of two separated metals A and B with work functions FA and FB.
(b) The band structure of metals A and B in electrical contact forming a junction and producing the contact
potential (VC). The horizontal dashed line is the energy of free space after contact.
The buildup of charge on metals A and B produces a contact voltage (VC ) between metals A and B.
Metals A and B are in equilibrium with each other when the Fermi energies of metals A and B are equal,
as shown in Figure 16.17b. To remove an electron from the Fermi energy of metal B to the energy of
free space requires an additional amount of energy equal to 0.5eVC relative to what is required before
contact. In metal A, to remove an electron from the Fermi energy to the free-space energy level requires
an amount of energy equal to 0.5 eVC less than is required before contact. Equation 16.28 is deduced
from Figure 16.17b.
FA 2 FB 5 eVC
16.28
Equation 16.29 then gives the contact voltage developed between metals A and B.
VC 5
FA 2 FB
e
16.29
Because the work function is the energy of free space minus the Fermi energy, Equation 16.30 is deduced
from Figure 16.17a:
FA 2 FB 5 EFB 2 EFA
16.30
where EFA and EFB are the Fermi energies of metals A and B. The contact voltage of the junction in terms
of the Fermi energy of metals A and B is then given in Equation 16.31.
VC 5
EFB 2 EFA
e
16.31
The contact voltage between the two metals A and B is not a voltage in the sense of a battery. If a
voltmeter is connected to a junction of two metals, the voltage reading is 0, because the electrons at the
Fermi energy on either side of the junction are at the same energy. The contact voltage is an internal
voltage associated with the junction interface.
W-33
Electrical Properties of Materials
Anyone who has metal fillings in their teeth has probably experienced the contact voltage when a fork has
touched one of the fillings and they received a tingle in the tooth. This is the result of the electron transfer
between the silver fork and the metal filling. The metal filling is a mercury-based amalgam of metals.
Example Problem 16.6
Mag wheels for automobiles are an alloy of magnesium and aluminum. Magnesium has a work
function of 3.68 eV. Mag wheels are bolted to the axle of a car that is made of an iron alloy that has
a work function of 4.50 eV.
(a) What contact voltage is developed where the iron alloy contacts the magnesium in the wheels?
(b) Which metal is negative and which is positive?
Solution
The contact voltage is calculated with Equation 16.29 because the work functions are known.
VC 5
FA 2 FB
e
5
4.50 eV 2 3.68 eV
5 0.82 V
1e
The value of e in this case is 1, since the energy unit of the work function is electron volts. If the work function
is in joules, then the value of e is 1.6 3 10219 C. (b) Because the magnesium alloy has the smaller work function,
electrons flow from this alloy to the iron alloy at contact. The magnesium alloy becomes positive and the iron
alloy takes on a negative charge. The positive charge on the magnesium alloy can contribute to corrosion by
Mg21 ions dissolving in an electrolytic solution, such as salt water.
16.7.3 pn-Junctions
A pn-junction is produced by joining p-type and n-type semiconductors in electrical contact. Figure 16.18a
shows the energy levels for separated p-type and n-type semiconductors, and Figure 16.18b shows the
energy levels for a pn-junction.
p-Type
Fp
Ea
– – – –
+ + + +
n-Type
Fn
– – – –
+ + + +
Ec
EF
Eg
EF
Ev
Free space
Ecp
Eo
p-Type
M
Eg
Donors in SCL
Ed
EF
Evp
n-Type
– – – – E
cn
EF
Acceptors in SCL
++
–––
–
+ + + +
eVc
++
Evn
(a)
Bulk
SCL
Bulk
(b)
Figure 16.18 (a) Separated p-type and n-type semiconductors at a temperature in the extrinsic range of
operation. (b) The pn-junction after contact, showing that the Fermi energy is continuous across the junction. The
original location of the junction is called the metallurgical junction, and it is the vertical dashed line (M).
W-34
CHAPTER 16
The values of the Fermi energy of the p-type and n-type semiconductors are necessary to calculate the
contact potential of a pn-junction. The value of the Fermi energy in p-type and n-type semiconductors
at 0 K is deduced by recalling that the Fermi energy has a probability of occupation of 12. In an n-type
semiconductor at 0 K, the electrons are all on the donor atoms and none of the electrons are in the
CB. Therefore, the donor energy level has a probability of 1 of being occupied, and the CB has a
probability of 0 of being occupied. The Fermi energy (EF) is halfway between the donor energy (Ed) and
the conduction-band energy (Ec), and EF is equal to (Ed 1 Ec )/2. In a p-type semiconductor at 0 K, the
valence electrons are all in the VB in covalent bonds, and no electrons are excited to acceptor atoms.
Therefore, the valence-band energy (Ev) has a probability of 1 of being occupied, and the acceptor energy
(Ea) has a probability of 0 of being occupied. EF is halfway between the valence-band energy (Ev) and the
acceptor energy (Ea), and EF is equal to (Ev 1 Ea)/2.
If the p-type and n-type semiconductors are raised to a sufficiently high temperature, eventually there
are many more electrons in the CB that are excited from the VB than there are acceptor atoms in the
p-type semiconductor or donor atoms in the n-type semiconductor. At these high temperatures the p-type
and n-type semiconductors have essentially the same number of electrons in the CB as a pure (intrinsic)
semiconductor, and the p-type and n-type semiconductors both behave as an intrinsic semiconductor. In
an intrinsic semiconductor the Fermi energy is always in the middle of the energy gap and equal to Eg/2.
This is because at 0 K the VB is full of electrons and the CB has zero electrons. At 0 K in an intrinsic
semiconductor, the probability of occupation of the energy level Ev is 1; the probability of occupation
of the energy level Ec is 0, and the probability of 12 is halfway between Ev and Ec, or in the middle of the
energy gap at Eg/2. As an intrinsic semiconductor is heated, any electron that is in the CB comes from the
VB, any increase in probability at Ec is balanced by a decrease in probability at Ev, and the probability of
1
2 occupation remains at Eg/2.
In an n-type semiconductor the Fermi energy is at (Ed 1 Ec)/2 at 0 K and decreases toward Eg/2 as
temperature is increased, because at sufficiently high temperature the n-type semiconductor behaves as
an intrinsic semiconductor. In a p-type semiconductor, the Fermi energy is at (Ev 1 Ea)/2 at 0 K and
increases toward Eg/2 as temperature is increased, because at sufficiently high temperature the p-type
semiconductor behaves as an intrinsic semiconductor. We assume in Figure 16.18 that the temperature is
in the extrinsic range of operation for the p-type and n-type semiconductors; therefore, the Fermi energy
of the p-type semiconductor is more than Ea/2 but less than Eg/2, and the Fermi energy of the n-type
semiconductor is less than (Ed 1 Ec)/2 but more than Eg/2.
After contact the Fermi energy, as shown in Figure 16.18b, is constant across the junction, as it is for
metals A and B in Figure 16.17b. The energy bands in the n-type and p-type semiconductors shift with
the Fermi energy. We can calculate the contact voltage for a pn-junction by using a modification of
Equation 16.31, as presented in Equation 16.32.
VC 5
Fp 2 Fn
e
5
EFn 2 EFp
16.32
e
The contact voltage results from electrons in donor levels and in the CB of the n-type semiconductor
in the junction region moving across the junction to lower-energy unoccupied states in the VB
of the p-type semiconductor in the junction region. This electron movement in the junction
region results in a positive charge (∙) in the n-type semiconductor and a negative charge in the p-type
semiconductor (⊝), as shown in Figure 16.18b. The separation of charge in the junction region is
called the space charge layer (SCL). The SCL creates an internal electric field (E0 ) pointing from
the positively charged n-type semiconductor to the negatively charged p-type semiconductor. The
internal electric field is what bends the electron energy bands, resulting in a constant Fermi energy
across the junction. Because the charges that produce the electric field are spread over a distance
across the junction, the bending of the electron energy bands occurs over a distance at the junction,
W-35
Electrical Properties of Materials
as shown in Figure 16.18b, resulting in a width to the pn-junction. Away from the SCL, the p-type
semiconductor has a hole density in the VB and the n-type semiconductor has an electron density in
the CB characteristic of bulk semiconductors.
16.7.4 The Diode Equation for a pn-Junction
A pn-junction in an electrical circuit is a diode. It allows current to flow in one direction in response to
an applied voltage, but not in the other direction. Why the pn- junction behaves as a diode is explained
by observing the changes that occur in the semiconductor electron energy band structure when a forward
and a reverse bias (voltage) is applied to the pn-junction, as shown in Figures 16.19a, 16.19b, and 16.19c.
In the upper part of Figure 16.19a, the band structure of a pn-junction is shown with no applied voltage,
and the lower part of Figure 16.19a shows schematically the location of conduction electrons (red circles
with a 2 sign) and holes (yellow circles with a 1 sign) in the pn-junction.
When a forward-bias voltage (Va) is applied to the pn-junction, the p-type end of the junction is
connected to the positive terminal of a battery, and the n-type end is connected to the negative terminal,
as shown in Figure 16.19b. The forward bias forces electrons in the n-type semiconductor and holes in
the p-type semiconductor toward the junction interface, as shown in the lower part of Figure 16.19b. The
forward bias creates a positive current flow of holes from the p-type to the n-type semiconductor, and
of electrons from the n-type to the p-type semiconductor. As shown in the upper part of Figure 16.19b,
p
n
eVc
Ecp
–
Ecp
e(Vc-Va)
–
– – – – Ecn
E1
E1
– – – –E
cn
EF
Evp + + + + +
EF
eVa
EFp
Evp + + + + + +
+
EFn
e(Vc-Va)
Ecp
E1
EFp
Evp + + + + +
– – – – Ecn
EFn
Evn
+
Evn
Evn
p
p
+ + +
+ + +
+
+ +
n
–
–
–
–
–
– –
+
+
+
n
+
+
+
+
+
–
–
–
–
– –
–
+
n
– – –
–
– –
–
–
I
Va
(a)
p
+ +
+ +
+
+ + +
(b)
Va
+
(c)
Figure 16.19 (a) The band structure of a pn-junction with no applied voltage, showing the contact potential
(Vc ) and the location of conduction electrons (red circles with 2 sign) and holes (yellow circles with 1 sign) in
the band structure. The schematic below shows the physical location of conduction electrons holes. (b) The band
structure of a pn-junction with a forward bias (Va), showing the displacement of the energy bands and the electron
and hole motion at the junction. In the schematic below, the electrons and holes are shown driven toward the
junction by the forward bias creating a positive current across the junction from left to right. (c) The band structure
of a pn-junction with a reverse bias (Va) showing the displacement of the energy bands. In the schematic below, the
electrons and holes are driven away from the junction, resulting in no current crossing the junction. (Adapted from
Kasap, S. O., Principles of Electronic Materials and Devices, 3rd ed., McGraw-Hill, Boston MA. (2006), p. 496.)
W-36
CHAPTER 16
the applied forward bias (Va) lowers the energy of the electrons in the p-type semiconductor, and it raises
the energy of the electrons in the n-type semiconductor. This reduces the height of the energy barrier for
electrons to migrate from the n-type to the p-type semiconductor at the pn-junction by an energy equal to
Vae. The reduction in barrier height results in more electrons in the n-type semiconductor with sufficient
energy to overcome the energy barrier that can flow to the p-type semiconductor. The electrons flowing
from the n-type to p-type semiconductor and holes flowing from the p-type to n-type creates a positive
current from left to right across the pn-junction.
If a reverse bias is applied, as in Figure 16.19c, the p-type semiconductor is charged negative and
the n-type semiconductor is charged positive. The reverse bias results in both the holes and electrons
being forced away from the junction, as shown in the bottom part of Figure 16.19c. The applied voltage
does not drive current through the junction. The electron current across the junction with a reverse bias
depends upon the rate at which electrons are thermally excited from the Fermi energy to the CB of the
p-type semiconductor (E1). The value of E1 is not affected by the application of the forward or reverse
bias, because all of the electron energy levels in the p-type semiconductor shift together as a result of an
applied voltage. If electrons are excited into the CB of the p-type semiconductor, they will flow down in
energy to the n-type semiconductor, with or without an applied voltage. It is the rate of thermal excitation
of electrons in the VB or the acceptor level by energy E1 into the CB of the p-type semiconductor that
controls the reverse electron current flow, and the magnitude of the applied reverse voltage has no effect.
A pn-junction allows current to flow with forward bias that is an exponential function of the applied
voltage; however, with a reverse bias the current does not increase with voltage, as shown in Figure 16.20.
The pn-junction functions as a diode; it allows current to flow in one direction in an electronic circuit but
not in the reverse direction. The current (I ) through a pn-junction is expressed in Equation 16.33a, where
I0 is the current with a large reverse bias. The current density through a pn-junction, in Equation 16.33b,
has the same form as Equation 16.33a.
Vae
16.33a
Vae
16.33b
3 1 kT 2 2 14
I 5 I0 exp
3 1 kT 2 2 14
J 5 J0 exp
An npn-junction is a transistor that amplifies an electrical current. In solid-state electronics the
transistor has replaced the vacuum-tube amplifier. The operation of transistors is discussed in books that
focus on the electronic properties of materials, such as those by Kasap and Kwok.
I
I0
–1.0
–0.5
12
10
8
6
4
2
–2
0.5
Voltage
1.0
Figure 16.20 A schematic of the current (I ) divided by the current with a large reverse bias I0, as a function of
applied voltage for a diode (pn-junction).
W-37
Electrical Properties of Materials
Example Problem 16.7
A pn-junction diode at 300 K produces a current density of 0.50 3 104 A/m2 with a forward bias of
0.100 V. Calculate the current density for a reverse bias of 0.100 V.
Solution
We use Equation 16.33b, the diode equation, to solve this. Everything in Equation 16.33b is known except J0.
Inserting the known values into Equation 16.33b for the forward bias allows solution of J0.
Vae
3 1 kT 2 214 5 0.5 3 10 mA 5 J 3exp 18.62 3 10 0.1eVeV? K
J 5 J0 exp
0.5 3 104
4
2
3 1
0
25
s300 Kd
21
2 214
2 4
A
0.1 eV
5 J0 exp
21 5 J0 [exp s3.86d 21g 5 J0 f47.5 2 1g 5 J0 f46.5g
m2
0.026 eV
Solving for J0,
J0 5
0.5 3 104 A/m2
A
5 1.08 3 102 2
47.5
m
The reverse current should be approximately 1.08 3 102 A/m2, and independent of applied voltage; however, we
can check this with the diode equation by inserting the reverse voltage of 20.1 eV.
3 1
2 4
J 5 1.08 3 102
A
20.1 eV
A
exp
21 5 1.08 3 102 2 fexp s23.86d 21g
m2
0.026 eV
m
J 5 1.08 3 102
A
A
A
f0.02 2 1g 5 1.08 3 102 2 f20.98g 5 21.05 3 102 2
2
m
m
m
The current with an applied voltage of 20.1 V is indeed approximately equal to 2J0.
Example Problem 16.8
A pn-junction diode at 300 K produces a current density of 0.50 3 104A/m2 with a forward bias of
0.100 V. Calculate the current density for a forward bias of 0.200 V.
Solution
We use Equation 16.33b, the diode equation, to solve this.
We have two unknowns, J2 and J0. We could solve for J0 as we did in Example Problem 16.7; however, because J0
occurs in the equations for both J1 and J2, if we take the ratio of J1 /J2, then J0 is eliminated, and we solve for J2
in terms of J1.
First determine the equations for J1 and J2.
Vae
3 1 kT 2 214 5 0.5 3 10 mA 5 J 3exp 18.62 3 10 0.1eVeV? K
J1 5 J0 exp
4
2
3 18.62 3 10
J2 5 J0 exp
0
25
2 4
s300 Kd
21
3 1
2 214 5 J f46.5g
0
2 4
0.2 eV
0.2 eV
21 5 J0 exp
21
0.026 eV
eV ? K21s300 Kd
25
J2 5 J0 f2191 2 1g 5 J0 f2190g
W-38
CHAPTER 16
Now take the ratio J1/J2.
J1
J2
5
J0 f46.5g
J0 f2190g
5
0.5 3 104A/m2
J2
Now cancel J0 and solve for J2.
J2 5 0.5 3 104
1 2
A 2190
A
5 23.6 3 104 2
m2 46.5
m
The exponential dependence of the diode current density on the applied voltage is demonstrated by this
calculation.
16.7.5 Metal-Polymer-Metal Junctions
and Electronic Devices
Electronic devices such as LEDs, solar cells and batteries, and visual displays have been produced
from MPM-junctions. Chapter 18 covers the photonic uses of MPM-junctions. Several advantages of
MPM-junctions over single-crystal semiconductor pn-junctions are that MPM-junctions are flexible, and
they have the potential of being relatively inexpensive. Figure 16.21 shows a schematic of the energy
levels of a MPM-junction.
Figure 16.21a shows the metal-1 (M1), polymer, and metal-2 (M2) before contact. The work
function of M1 (F1) is larger than that of M2 (F2), and the lowest energy in the polymer CB (Ec )
is above the Fermi energy of M1 and M2. After the junction is made and equilibrium established,
as shown in Figure 16.21b, the Fermi energies of the two metals are equal. The equality of
Fermi energies creates a contact voltage (Vc ) that results from a positive charge on M2 and a negative
charge on M1, due to electron flow that goes from M2 to M1. The polymer is a thin layer, and sufficient
charge is able to cross the polymer from M2 to M1 to establish equilibrium. As in the semiconductor
pn-junction, an internal electric field is created across the polymer due to the positive charge on M2
and the negative charge on M1. The electric field bends the energy bands of the polymer, as shown in
Figure 16.21b. The negative charge on M1 results in a lower energy for electrons in M1 to be excited to
free space, and the positive charge on M2 results in a higher energy for electrons in M2 to be excited
to free space. After equilibrium is established, electrons in M2 must overcome the energy barrier eVc
to flow to M1.
We can observe the diode action of this device by applying a forward bias voltage (Va ) to the
MPM-junction such that M2 is charged negative and M1 is charged positive, as schematically shown in
Figure 16.21c. The applied voltage reduces the energy barrier for electron flow from M2 to M1 provided
by the polymer to e(Vc 2Va ). The Fermi energy of M2 indicated by the top of the purple band is
the probability of occupation of 12 , and there are electrons with greater energy than the Fermi energy
that enter the polymer conduction band. The positive current flow is in the opposite direction to the
electron flow.
With a reverse bias, M1 is charged negative and M2 is charged positive. The reverse bias has no
effect on the reverse current flowing from M1 to M2, because the energy level of the CB of the polymer
(Ec) increases, along with the Fermi energy of the electrons in M1 as the reverse bias voltage (Va) is
increased. Therefore, the probability of an electron flowing from M1 to M2 is not increased by the
reverse bias.
Electrical Properties of Materials
Free space
Ec
F2
F1
EF2
Ev
EF1
Polymer
Metal 2
Metal 1
(a) Before Contact
Energy
Ec
eVc
–––
+++
Polymer
Metal 1
Metal 2
(b) After Contact
– –E – –
c
Polymer
Metal 1
+
Va
–
Electrons
Metal 2
I
(c) Applied Bias = Va
Figure 16.21 (a) Energy-level diagrams of separated metal-1 (M1) and metal-2 (M2) and a polymer, showing
the metal Fermi energies EF1 and EF2, the work functions F1 and F2, and the polymer valence-band and
conduction-band energies Ev and Ec. (b) The energy bands after contact and the contact potential Vc. (c) The
energy bands after the application of a forward bias Va and the flow of electrons from M2 to M1 and the positive
current I in the reverse direction.
16.8 Dielectric Materials
What happens in a material when an electric field or voltage is applied and there is no conduction of
electrons or holes? The center of the positive charge is displaced from the center of the negative charge.
These materials are called dielectrics, with dia meaning “through” or “across.” A dielectric material
allows penetration of an electric field through the material. Electric fields cannot penetrate metals. Supersecret electronic systems are enclosed in metal-lined rooms to prevent electric fields from entering or
escaping. Examples of dielectric materials are diamond, sapphire, glass, quartz, salt, water, and polymers
such as PE. Dielectric materials are utilized as insulators, capacitors, dynamic random-access memory
(DRAM), permanent memory, electro-optical light switches, micro electromechanical actuators, sound
and ultrasound transducers and detectors, thermistors, and sensors.
W-39
W-40
CHAPTER 16
16.8.1 Capacitance
Dielectrics are characterized by capacitance (C ), which is measured in farads (F), and 1 F is equal to
1 C/V. For the parallel-plate capacitor shown in Figure 16.22a, which has a vacuum between the plates,
an applied voltage (V ), and a total charge (Q0) on the plates measured in coulombs (C ), the capacitance
(C ) is given by Equation 16.34.
Q0
C5
16.34
V
If the vacuum capacitor has plates of area (A) with a distance between the plates of (l ), as shown in
Figure 16.22a, the capacitance is also given by Equation 16.35:
Q0
C5
V
5
0A
16.35
l
where 0 is the permittivity of a vacuum (8.85 3 10212 F/m). From the equality on the right side of
Equation 16.35, the charge density (D0) on the plates is calculated from Equation 16.36.
D0 5
Q0
5
A
0V
16.36
l
The term V/l is the magnitude of the electric field (E ) in volts/m across the capacitor. Inserting the
magnitude of the electric field into Equation 16.36 results in Equation 16.37.
16.37
D0 5 0E
In Figure16.23a, the electric field vector (E) is shown penetrating across the capacitor, because vacuum
is a dielectric.
+
D0 = 0E
+
l
+
+
+
–
–
Vacuum
–
–
–
–
–
–
–
E
–
–
+
V
+
–
–
–
–
–
–
(a)
D = 0E + P
+ + + + + + + + + +
– – – – – – – – – –
Dielectric
P
E
+ + + + + + + + + +
– – – – – – – – – –
+
+ –
+ –
+ –
+
+ –
+ –
–
+ –
+ –
+ –
–
V
(b)
Figure 16.22 (a) A parallel-plate capacitor with a vacuum, showing the surface charge density D0 and the
electric field vector E. (b) A parallel-plate capacitor with a dielectric material present, showing the polarization
vector P, the surface charge density D, and the electric field vector E.
W-41
Electrical Properties of Materials
Example Problem 16.9
A vacuum parallel-plate capacitor with an area of 1 cm2 and a gap of 1 mm has 1 V applied. Calculate
(a) the capacitance and (b) the surface charge density.
Solution
a) The capacitance is calculated from Equation 16.35.
C5
0 A
l
5
8.85 3 10212 F/ms1 3 1022 md2
5 8.85 3 10213 F
1.00 3 1023m
b) The total charge is calculated from the capacitance in Equation 16.35, and the surface charge density is
calculated from the surface charge, with Equation 16.36.
Q0 5 C V 5 8.85 3 10213 Fs1 Vd 5 8.85 3 10213 C
D0 5
Q0
A
5
8.85 3 10 213 C
C
5 8.85 3 10 29 2
m
1 3 10 24 m2
If a material replaces the vacuum in a capacitor with an applied voltage (V ), in analogy with
Equation 16.35, the capacitance is given by Equation 16.38:
C5
Q A
5
V
l
16.38
where Q is the charge on the plates with a dielectric material and is the permittivity of the material. It
follows from Equation 16.36 that the surface charge density on the plates (D) with a dielectric material is
given by Equation 16.39.
D5
Q V
5
5 E
A
l
16.39
The ratio of the permittivity of the dielectric material to the permittivity of a vacuum is the relative
permittivity (r ), as shown in Equation 16.40.
r 5
0
16.40
The relative permittivity is also called the dielectric constant. Table 16.8 presents the values of the relative
permittivity or dielectric constant of some dielectric materials at 60 Hz and at 106 Hz, along with values of
the dielectric strength and the electrical resistivity. As shown later in this section, the dielectric constant is
a function of frequency. Since the dielectric constant of these materials is greater than 1, the capacitance
with these dielectric materials for a constant voltage is greater than with a vacuum. Some materials,
such as barium titanate (BaTiO3), have dielectric constants approaching 104. The dielectric strength,
listed in Table 16.8, is the maximum static electric field that the material can withstand without dielectric
breakdown. The dielectric strength limits how much electrical energy can be stored in a capacitor.
Why does the capacitance increase when a dielectric material is inserted between the plates of
a capacitor? To answer this question, we must look at what happens to the dielectric material in the
capacitor when subjected to an electric field. As indicated in Figure 16.22b, the dielectric material develops
a surface negative charge on the side adjacent to the capacitor plate with a positive charge, and a surface
positive charge develops on the side of the material adjacent to the negative plate of the capacitor. When
W-42
CHAPTER 16
Table 16.8 The Dielectric Constant, Dielectric Strength, and Electrical Resistivity of Some
Dielectric Materials
Material
Dielectric Constant
(at 60 Hz)
(at 106 Hz) Dielectric Strength
(106 V ? m)
Resistivity
(ohm ? cm)
Polyethylene2.3 2.3
20
. 1016
Teflon
2.12.1
20
1018
Polystyrene 2.52.5
20
1018
PVC
3.53.2
40
1012
Nylon
4.03.6
20
1015
Rubber
4.03.2
24
Phenolic
7.04.9
12
1012
Epoxy
4.03.6
18
1015
Paraffin wax
2.3
10
1013–1019
Fused silica
3.8
3.8
10
1011–1012
Soda-lime glass
7.0
7.0
10
1015
Al2O3
9.0
6.5 6
10 –1013
14–110 8
1013–1018
TiO2
Mica7.0
40
BaTiO32000–5000
12
Water78.3
11
1013
10 –1015
8
1014
Based on data from Askeland, D.R., Fulay, P.P., and Wright, W.J., The Science and Engineering of Materials, 6th ed. Cengage Learning,
Stamford, CT. (2011), p. 754.
there is a separation of charge in the material, it is polarized. The magnitude of the polarization (P) of
the material is equal to the surface-charge density on the dielectric material, in coulombs per square
meter. The surface-charge density on the plates with a material inserted (D) is equal to the surface-charge
density with a vacuum (D0) plus the magnitude of the polarization (P), as shown in Equation 16.41.
D 5 D0 1 P 5 E
16.41
The polarization vector points from the negative charge in the dielectric material to the positive charge.
The charge density that is established on the plates of the capacitor with a dielectric material inserted
compensates for the opposite-charge density that develops on the dielectric material as a result of
polarization. The capacitor response to the polarization of the dielectric material results in the increased
capacitance.
Example Problem 16.10
The vacuum space in the capacitor in Example Problem 16.9 is filled with fused silica (SiO2). Calculate
the following for this capacitor: (a) the capacitance, (b) the surface-charge density on the capacitor
plates, and (c) the magnitude of the polarization of the fused silica.
W-43
Electrical Properties of Materials
Solution
a) The capacitance is calculated from Equations 16.38 and 16.40.
The value of the dielectric constant for fused silica is 3.8. Since the applied voltage is constant, the lowest
frequency value of the dielectric constant is used.
C5
A r0 A 3.8s8.85 3 10212F/mds1 3 1024 m2d
5
5
5 33.6 3 10213 F
l
l
1.00 3 1023 m
b) The surface charge is then calculated from the capacitance and the applied voltage using Equation 16.38.
Q 5 CV 5 (33.6 3 10213 F)(1.00 V) 5 33.6 3 10213 C
The surface-charge density is then calculated from Equation 16.39
Q 33.6 3 10213 C
C
5
5 33.6 3 1029 2
A
m
1 3 1024 m2
D5
c) The magnitude of the polarization is calculated from Equation 16.41.
P 5 D 2 D0 5 (33.6 2 8.85) 3 1029 C/m2 5 24.8 3 1029 C/m2
Polarization in a material results when it is possible to displace positive and negative charges relative
to each other. Polarization occurs when there is no flow of current. The types of polarization are shown
in Figure 16.23; they are (a) electronic, (b) ionic, and (c) orientational. Electronic polarization occurs
when the atomic centers of charge for the electrons and the nucleus are displaced from each other, as
E
E =0
+
–
–
+
(a)
–
–
–
–
+
+
–
–
–
–
(b)
–
+
–
–
+
+
+
–
–
+
–
+
(c)
Figure 16.23 Schematics of (a) electronic, (b) ionic, and (c) permanent polarization. The left side of the figure
shows the charge configurations without an electric field, and the right side shows the polarization in an electric
field (E).
W-44
CHAPTER 16
shown in Figure 16.23a. Electronic polarization is the only polarization type in a covalently bonded
element, such as silicon. In ionic polarization, an electric field displaces anions and cations relative to
their normal lattice positions, as shown in Figure 16.23b. Ionic polarization is present in ionic materials,
such as NaCl and MgO. H2O is a molecule with a permanent orientational dipole moment, as shown
in Figure 16.24. In a water molecule, the two hydrogen atoms attached to the oxygen are separated by
105°. The end of the hydrogen atoms away from the oxygen atom is a positively charged nucleus, because
the hydrogen electron is localized between the hydrogen and the oxygen in a covalent bond. The end
of the molecule toward the oxygen is negatively charged, because oxygen is more electronegative (3.5)
than hydrogen (2.1), and the electrons are more attracted to the oxygen. The positive and negative
charges on H2O are permanently displaced from each other. In H2O, if there is no applied electric
field, the orientation of the H2O molecules is random, and there is no net polarization, as shown in
Figure 16.23c on the left. However, in an applied electric field the H2O molecules are oriented as shown
in Figures 16.23c and 16.24.
Polarization results when a charge qi is displaced by a vector distance di from a position of zero dipole
moment to create the vector dipole moment pi, given by Equation 16.42.
pi 5 qi di
16.42
Both qi and di can be positive or negative, depending upon the charge and the direction of displacement.
Assume that the positive direction is the direction of movement of the positive charge. The polarization
vector (P) is the sum of the dipole moments per unit volume, as shown in Equation 16.43.
P5
pi
oV
16.43
i
The polarization vector (P) shown in Figure 16.22b is pointing in the direction of the displacement of
the positive charge. The units of polarization, as shown in Equation 16.43, are dipole moment per unit
volume or the equivalent coulombs per square meter.
In Table 16.8 the dielectric constant is given for an electric field oscillating at 60 Hz and 106 Hz.
The dielectric constant is a function of the frequency of the electric field, as shown in Figure 16.25,
because the displacement of atoms and molecules requires a certain response time. If the time for the
electric field cycle is less than the response time for a particular mode of polarization, then that mode of
polarization does not occur. As shown in Figure 16.25, at low frequency all of the modes of polarization
are present: electronic, ionic, and orientational. At frequencies above 108 Hz, there is insufficient time
for the molecules that produce orientational polarization to respond to the alternating electric field, and
only ionic and electronic polarization are present. An applied electric field of frequency above 1014 Hz
E
+
P
H
Oxygen
+
–
H
Figure 16.24 The permanent dipole of an H2O molecule oriented in an electric field (E) producing
polarization (P). (Based on Askeland, D.R., Fulay, P.P., and Wright, W.J., The Science and Engineering of Materials, 6th ed.,
Cengage Learning, Stamford, CT. (2011), p. 39.)
W-45
Dielectric constant
Electrical Properties of Materials
Orientationpermanent
Ionic
Electronic
104
108
1012
Frequency (Hz)
1016
Figure 16.25 A schematic of the dielectric constant (r) as a function of frequency for a dielectric material with
orientational, ionic, and electronic polarization.
is higher than the lattice vibration frequencies (~1013 Hz), and the ions in a crystal cannot be displaced
by applied electric fields of frequency higher than approximately 1013 Hz. The ionic component of the
dielectric constant disappears at frequencies above approximately 1013 Hz. Electrons have a much shorter
response time than atoms do, and the electronic component of dielectric constant continues to be present
to frequencies of 1017 Hz, but at frequencies above 1017 Hz even electronic polarization disappears.
In a normal dielectric material, the polarization is small, positive, and reversible. Polarization
is reversible if the polarization returns to 0 when the applied electric field returns to 0, as shown in
Figure 16.26. The magnitude of polarization (P) relative to the magnitude of the electric field (E ) is
proportional to the dimensionless dielectric susceptibility (d), given by Equation 16.44. The dimensionless
dielectric susceptibility is also the magnitude of the polarization of the dielectric material divided by the
surface-charge density of a vacuum capacitor, as shown by Equation 16.44.
d 5
P
P
5
0 E D0
16.44
The dielectric constant (r ) and the dielectric susceptibility are related through Equation 16.45.
r 5
D0 1 P
DE
5
5
5 1 1 d
0 D0E
D0
16.45
In Figure 16.26, the slope of the P versus E plot is equal to d 0, or 0(r 2 1), as shown by
Equations 16.44 and 16.45.
P
E
Figure 16.26 A schematic of the polarization (P) as a function of electric field (E) for a normal dielectric material.
W-46
CHAPTER 16
The material for DRAM in computers is the dielectric silica (SiO2). This is the memory that is active
only when the computer power is on. Computer memory is binary with values of 1 and 0 that represent
information. A value of 1 indicates a region of material with one direction of polarization, and a 0
represents a material polarized in the reverse direction. When the computer power is turned off, there
is no electric field; the polarization of all the material returns to 0, and all of the information is lost.
Dielectric materials that can permanently store information when the power is off are covered next.
16.8.2 Ferroelectric Materials
In ferroelectric materials, after a large electric field is applied and then reduced to 0, the polarization
does not return to 0, as it did in a normal dielectric material. The term ferroelectric comes from the
shape of the plot of the polarization as a function of an applied electric field, as schematically shown
in Figure 16.27. The shape is similar to the shape to the applied magnetic field-magnetization plot for
ferromagnetic materials, such as iron, shown in Figure 17.9. The shape of the P-vs.-E plot in Figure 16.27
results from ferroelectric materials having a permanent ionic dipole.
Ferroelectric materials, such as barium titanate (BaTiO3), shown in Figure 16.28, and lead zirconate
(PbZrO3) have ionic polarization. At temperatures above 130°C, BaTiO3 is a cubic crystal with no
permanent dipoles. At 130°C, the lattice parameter is 0.401 nm. However, below 130°C, the crystal is
tetragonal with the lattice parameters shown in Figure 16.28a. Figure 16.28b shows the ion displacements
relative to the barium ions at the corner positions. The ion displacements result in permanent polarization.
The Curie temperature is the temperature where the material changes from ferroelectric to paraelectric
upon heating. For barium titanate, the Curie temperature is approximately 130°C. In paraelectric
materials, the polarization returns to 0 when the electric field is reduced to 0; however, the relationship
between P and E can be nonlinear.
P
Ps
Pr
–Ec
0
Ec
E
Pr
Ps
Figure 16.27 The polarization (P) as a function of an electric field (E) for a ferroelectric material.
W-47
Electrical Properties of Materials
0.009 nm
0.403 nm
0.006 nm
0.398 nm
Ti4+
Ba2+
0.006 nm
0.398 nm
O2–
(a)
(b)
Figure 16.28 The barium titanate unit cell at room temperature. (a) An isometric view. (b) A front view showing
the displacements of the Ti41 and O22 ions relative to the barium atoms. (Based on Callister W.D., Materials Science and
Engineering, An Introduction 6th ed. John Wiley &Sons NY (2003), p. 645.)
The properties of a ferroelectric material are demonstrated by applying a large electric field in one
direction and then reversing the direction, as demonstrated in Figure 16.27. The origin in Figure 16.27
is a nonpolarized material (P 5 0) in a zero electric field (E 5 0). An increase in the electric field causes
an increase in polarization. Once all of the possible electrical dipoles are oriented in the direction of
the applied electric field, the polarization is saturated at Ps. In a ferroelectric material, when the applied
electric field is then reduced to 0, there remains a residual polarization (Pr) in the direction of the former
electric field. The polarization does not return to 0, because the internal electric field created by the dipoles
is sufficient to maintain the dipoles oriented in the direction of the original electric field. Equation 16.41
shows that if the applied electric field is 0, the sum of the polarization and vacuum surface charge density
is 0, but the polarization is not 0. The electric field in the reverse direction that reduces the polarization
to 0 is the coercive electric field (2Ec). When P 5 0 in a ferroelectric material, it does not mean that
there are no longer dipole moments in the material. In a ferroelectric material there are domains that are
regions with the polarization in the same direction. When P 5 0, there are as many domains oriented in
the original direction of the electric field as in the reversed field direction (2Ec ), as shown schematically
in Figure 16.29a. When the total polarization of the ferroelectric material changes, it is because the
polarization direction within individual domains changes. Figure 16.29b shows ferroelectric domains in a
crystal of barium titanate. The polarization of a ferroelectric material is saturated in the reverse direction
(P 5 2Ps) by application of a sufficient electric field in the reverse direction. If the applied electric field
is then reduced to 0, the polarization is the residual polarization 2Pr. The net material polarization is
forced to 0 by applying the coercive field (Ec) in the forward direction. At the coercive field, there are
as many domains with polarization in the forward direction as there are domains with polarization in
the reverse direction. The polarization is saturated in the forward direction by application of sufficient
electric field in the forward direction to orient all of the domains in the direction of E. The nonlinear
cycle of P as a function of E from 1Ps to 2Ps to 1Ps shown in Figure 16.27 is a hysteresis loop. Hystersis
loops are encountered in other cyclic plots such as stress and strain, and magnetic field and intensity
of magnetization as demonstrated in Chapter 17. If the electric field is cycled from saturation in the
forward direction to saturation in the reverse direction, the hysteresis loop is repeated. One way to return
the material to the origin (P 5 0, E 5 0) is to heat the material above the Curie temperature to disorder
W-48
CHAPTER 16
+ – + –
– + – +
(a)
Figure 16.29 (a) A schematic of ferroelectric domains, showing the polarization direction. (b) An image of
ferroelectric domains in barium titanate produced with a variant of atomic force microscopy (AFM) that applies
a voltage to the tip. Piezoelectric deformation in the ferroelectric domains resulting from the applied voltage is
detected by AFM. The width of the portion of the specimen in the image is 47 m. http://en.m.wikimedia.org/
wiki/File:Bulk_BTO_PFM_scan.png. (Tertib64)
the dipole moments, and then to cool the material to below the Curie temperature in the absence of an
electric field. With this procedure the domains are in random orientations, and the polarization is 0.
The polarized-unpolarized transformation in dielectric materials can be thermodynamically analyzed.
The internal energy change (dE) when the polarization changes by dP in an electric field (E ) is given
by dE 5 2EdP, and the entropy (S) is expressed in terms of the degree of polarization. The change in
the Gibbs free energy is then calculated from the internal energy change, the entropy change, and the
temperature.
Ferroelectric crystals are utilized in applications such as microdimensioned capacitors and permanent
memory for computers. Microcapacitors made from high dielectric constant materials, such as barium
titanate, have a high capacitance. In the permanent memory of a computer, polarization in the positive
direction corresponds to a 1 and polarization in the reverse direction corresponds to a 0. Writing is
accomplished with an electric field. If the power goes off and the electric field (E) is reduced to 0, the data
remains intact because of the residual polarization, and the data is preserved.
Example Problem 16.11
(a) Calculate the dipole moment in a unit cell of barium titanate with the room-temperature atom
positions shown in Figure 16.28b. (b) Calculate the polarization of barium titanate at saturation.
Solution
a) To solve this problem, we apply Equation 16.42 and sum the dipole moments over the unit cell. The charges
and displacements are as follows:
1. Two oxygen atoms of charge 22 move down 0.006 nm. The oxygen atoms in the face-centered positions are
shared between two unit cells. The positive direction is up.
Electrical Properties of Materials
q1 5 22(2e electrons) 5 24e electrons
d1 5 20.006 nm 5 26 3 10212 m
p1 5 q1d1 5 24e electrons(1.602 3 10219 C/electron)(26 3 10212 m)
p1 5 38.45 3 10231 C ? m
2. One oxygen ion of charge 22 moves down 0.009 nm.
q2 5 22e electrons
d2 5 20.009 nm 5 29 3 10212 m
p2 5 q2d2 5 22e electrons (1.602 3 10219 C/electron)(29 3 10212 m)
p2 5 28.84 3 10231 C ? m
3. One Ti atom of charge 5 14 moves up 10.006 nm.
q3 5 4e electrons
d3 5 0.006 nm 5 6 3 10212 m
p3 5 q3d3 5 4e electrons (1.602 3 10219 C/electron)(6 3 10212 m)
p3 5 38.45 3 10231 C ? m
The dipole moment per volume of a unit cell is the sum of calculations 1, 2, and 3.
p 5 (p1 1 p2 1 p3) 5 (38.45 1 28.84 1 38.45) 3 10231 C ? m 5 105.7 3 10231 C ? m
b) The dipole moment at saturation polarization corresponds to when all of the dipole moments are oriented
in the direction of the applied electric field. The polarization P is the sum of the dipole moments per unit
volume (V ).
P5
pi
o V 5 s0.403 3 10
i
1.057 3 10229C ? m
1.057 3 10229C ? m
5
29
29
0.638 3 10228 m3
mds0.398 3 10 mds0.398 3 10 md
29
P 5 0.165
C?m
m3
16.8.3 Piezoelectric Materials
Piezoelectric materials produce a voltage when deformed, and an applied voltage deforms the material.
The barium titanate crystal, shown in Figure 16.28 is a piezoelectric crystal. Other piezoelectric materials
include lead zirconate, quartz, and polymers such as polyvinylidene fluoride. The word piezo comes from
the Greek word piezein, meaning “to press or squeeze.” Piezoelectric crystals are utilized to convert the
bumps in the grooves of a record into a voltage that is amplified to produce music. Piezoelectric crystals
are also utilized to produce sound and ultrasound waves from a cyclic applied voltage, and they are
detectors of sound. Piezoelectric crystals are also utilized in micro-electromechanical systems (MEMS)
W-49
W-50
CHAPTER 16
to produce displacement, and they are utilized as the displacement control in scanning tunneling and
atomic-force microscopes, as discussed in Chapter 15.
Figure 16.28 shows the crystal structure of the piezoelectric crystal barium titanate at room temperature.
With the displacements shown in Figure 16.28b, the top of the crystal is charged positive, and bottom is
charged negative because the Ti14 has moved up and all of the O22 ions have moved down. If a voltage
is applied to the crystal such that the top of the crystal is charged positive and the bottom charged
negative, this causes the dipole displacement vectors d1, d2, and d3 in Example Problem 16.11 to decrease.
This decreases the 0.403-nm lattice parameter. Conversely, if the top of the crystal in Figure 16.28b is
charged negative and the bottom positive, this causes the displacement vectors (d1, d2, and d3) in Example
Problem 16.11 to increase. The increase in the displacement vectors causes the vertical lattice parameter
(c) to increase to more than 0.403 nm. The crystal is stretched or compressed by applying a voltage to the
crystal. The piezoelectric-strain coefficient (kp ) is the longitudinal strain () per unit of applied electric
field (E ), as shown in Equation 16.46.
kp 5
E
16.46
Table 16.9 presents values of kp for some materials. The piezoelectric-strain coefficient multiplied by the
magnitude of the electric field in V/m yields the dimensionless strain. Piezoelectric-strain coefficients as
high as 6.8 3 1029 m/V have been obtained for lead-lanthanum-zirconium-titanate. As long as the electric
field is less than the dielectric breakdown strength of the piezoelectric crystal, the displacement produced
by the voltage is very accurate and the response times are short.
To see how the voltage is generated by straining the crystal, we can look at the equations that relate
dipole moments to polarization and capacitance. When the crystal is mechanically strained in tension,
this increases the dipole displacements (d1, d2, and d3) and increases the dipole moments (p1, p2, and p3)
and the total polarization by DP according to Equations 16.42 and 16.43. The increase in the magnitude
of polarization results in an increase in the total surface charge density (D) given by Equation 16.41. A
larger surface-charge density produces a larger electric field (E) according to Equation 16.41, and a larger
voltage. Compressing the crystal causes a decrease in the dipole displacements, the polarization, surfacecharge density, and voltage. The piezoelectric materials allow for a direct conversion of electrical voltages
to mechanical displacements, or from mechanical displacements to electrical voltage.
There are also polymer ferroelectrics and piezoelectrics, such as polyvinylidene fluoride (PVDF).
Figure 16.30 shows the molecular structure of the trans form of PVDF. The structure of PVDF is like
that of PE, except that every other carbon atom is bonded to fluorine atoms rather than hydrogen.
Because of the strong affinity of fluorine for electrons, the fluorine atoms on the molecule are negatively
charged and hydrogen atoms are positively charged, creating a permanent dipole. The piezioelectric
strain coefficient for PVDF is typically 23 3 10212 m/V in comparison with 149 3 10212 m/V for BaTiO3,
as shown in Table 16.9.
Table 16.9 The Piezoelectric Strain Coefficient (kp) for Some Materials
Material
kp (m/V)
BaTiO3149 3 10212
PbZrO6 (PZT-4)
285 3 10212
PbNb2O6 85 3 10212
Quartz
2.3 3 10212
Based on data from the CRC Handbook of Tables for Applied Engineering Science, CRC Press, Boca Raton, Fl. (1973), p. 983.
W-51
Electrical Properties of Materials
Figure 16.30 The molecular structure of the piezoelectric polymer trans polyvinylidene fluoride. The red atoms
are carbon; the blue atoms are hydrogen, and the gold atoms are fluorine. (Based on http://en.wikipedia.org/wiki
/Ferroelectric_polymers)
Example Problem 16.12
A piezoelectric actuator for a micro electromechanical system of length 1 3 1024 m is produced from
barium titanate, which has a dielectric breakdown strength of 6 3 106 V/m. The piezoelectric strain
coefficient of the barium titanate is 1.49 3 10210 m/V. Assume that the maximum electric field applied
to the piezoelectric actuator is 1 3 106 V/m. What is the maximum displacement obtained from the
actuator?
Solution
Since the maximum electric field is 1 3 106 V/m, then the strain () is calculated from the coefficient of piezoelectric
strain.
kp 5
1
E
Dl 5 l0 kp E 5 1 3 1024 m 1.49 3 10210
m
V
2 11 3 10 m2 5 1.49 3 10
6
V
28
m
16.9 Electrical Conductivity
in Ionic Materials
Ionic conducting materials are utilized for devices such as thermistors and chemical sensors. Thermistors
use the electrical resistivity of a material to determine its temperature. Both thermistors and chemical
sensors utilize the electrical conductivity of ionic materials to measure properties of a system. Figure 16.31
shows the electrical conductivity of various mole percentages of Fe3O4 in solid solutions with MgCr2O4,
as a function of 1000/T. As the temperature increases, the electrical resistivity decreases. Most thermistors
are made from Mn, Ni, Fe, and Co oxides.
In ionic materials the electrical conductivity results primarily from the diffusion of ions. The ionic
contribution to the electrical conductivity (ion) is expressed in Equation 16.47, which is similar to
Equation 16.25.
ion 5
o Z en i
i
i
i
16.47
W-52
CHAPTER 16
–8
90%
100%
–7
80%
66.5%
49.4%
55%
–6
log10 conductivity ()
–5
39.4%
–4
25.5%
–3
–2
8.6%
5.4%
–1
1%
0
1
2
3
0
1
2
3
4
5
6
7
8
9
10
11
1000
T
Figure 16.31 The log of the electrical conductivity of various mole percentages of Fe3O4 in solid solution with
MgCr2O4, as a function of 1000/T. (Based on Verwey, E.J. et al., J. Chem. Phys., 15 (1947), p. 181.)
The ionic electrical conductivity is a sum of the electrical conductivity over all of the ions per unit
volume (ni ) with electron charge Zi e that have a mobility μi . The mobility of the ions is given by the
Nernst-Einstein Equation 16.48:
i 5
ZieDi
16.48
kT
where Di is the diffusion coefficient (diffusivity) of the i-type ions. Ionic electrical conductivity is directly
proportional to the diffusivity of the ions, and diffusivity of the ions is exponentially dependent upon
temperature. For example, in the temperature range from 1100°C to 1600°C, the diffusion coefficient for
oxygen ion diffusion in Ca0.15Zr0.85O2 is given by Equation 16.49.
1
DsO21 d 5 1 3 109 exp 2
2
0.84 eV
m2/s
kT
16.49
The electrical conductivity from ions, such as oxygen, is high in partially stabilized zirconia-calcia, because
of a high concentration of oxygen vacancies. The high oxygen-vacancy concentration results because
calcia (CaO) contributes only one oxygen atom for every calcium atom in a substitutional solid solution
in the zirconia structure (ZrO2) that requires two oxygen atoms for every zirconium atom. Materials such
as zirconia-calcia are used in devices such as oxygen sensors.
The oxygen sensor in an exhaust system of a gasoline powered engine operates on the principle that
the concentration of oxygen in a zirconia electrolyte is proportional to the 12 power of the partial pressure
of the oxygen in the atmosphere in contact with the zirconia electrolyte. Example Problem 4.12 discusses
this principle in the purification of hydrogen by diffusion through palladium. For oxygen sensors in
automobile exhaust systems called Nernst cells, the exhaust is on one side of the sensor and air is on the
Electrical Properties of Materials
Electrodes
(gas permeable)
Zirconia membrane
Exhaust Gas
Nernst cell
Heater
(option)
Reference air
0.2–0.8 V
Heater
(option)
Figure 16.32 A schematic of an oxygen sensor showing exhaust gas on one side of the electrolyte made of
partially stabilized zirconia, and air on the other side. Electrodes, normally porous platinum, plated on the
electrolyte are electrically connected to a voltage sensor. (Based on http://en.wikipedia.org/wiki/Oxygen_sensor)
other side, as shown in Figure 16.32. The different concentrations of oxygen on either side of the zirconia
electrolyte sensor create an electrical potential across the zirconia, because the Gibbs free energy of the
zirconia depends upon the oxygen concentration. The difference in the Gibbs free energy is converted
into a voltage, as given in Equation 10.12. The voltage from the sensor controls the air/fuel mixture
going into the engine for optimal combustion and minimal exhaust pollution. An output voltage of
0.2 V indicates a lean fuel to air mixture, and the concentration of oxygen is sufficient to produce CO2
rather than CO. An output voltage of 0.8 V indicates a high fuel-to-air mixture, and there is insufficient
oxygen to fully burn the fuel and CO is produced. An ideal voltage output is approximately 0.45 V. A
feedback system adjusts the fuel input to obtain the ideal fuel-to-air mixture.
16.10 Fabrication of Crystalline
Semiconductor Devices
Silicon is the most common elemental semiconductor in electronic devices. To have reproducible
electronic properties, the semiconductor material must be of very high purity. Silicon for semiconductor
applications is purified by zone refining until the impurity level is a few impurity atoms per billion (109)
silicon atoms. In zone refining, a zone of liquid is moved from one end of the crystal to the other, as
shown schematically in Figure 16.33. The silicon is purified, because the solubility of the impurities in the
liquid is greater than in the solid. Figure 16.34 shows a hypothetical phase diagram of some impurity in
silicon. The original impurity concentration in the silicon is C0. When the solid silicon is heated in a zone
to form a liquid, the liquid impurity concentration remains at C0. If the liquid is then cooled to form a
W-53
W-54
CHAPTER 16
Connection
to positive
high voltage
Specimen
Focusing
shield
Motion of
molten zone
Filament
Molten zone
Flux of
electrons
Figure 16.33 A schematic of an electron-beam zone-refining furnace. (Based on Guy, A.G., Introduction to Materials
Science, McGraw-Hill, New York (1972), p. 322.)
solid at Ts, the impurity concentration in the solid is Cs, and since Cs is much smaller than C0, the material
is purified. In semiconductor production, many liquid-solid zones simultaneously pass over a silicon rod
by having a series of heating elements.
In solid-state electronic systems, the semiconductor is normally a silicon single crystal, called an
ingot. The ingot is doped to produce a p-type semiconductor. The doped silicon ingots are grown
a
Liquid (L)
+
L
Ts
Temperature
a+L
Solid a
Cs
C0
Impurity concentration (Ci )
Figure 16.34 Part of the phase diagram near the melting temperature of silicon, with an impurity showing the
original composition (C0) and the composition of the solid (Cs ) that forms from the liquid of composition C0.
W-55
Electrical Properties of Materials
Rotary
chuck
Seed
crystal
Growing
crystal
Silicon
ingot
Molten
silicon
Crucible
Silicon
IC fabrication
wafer on wafer surface
Chip (die)
Packaged
chip
Heating
coils
(a)
SiO2
Exposed SiO2 is etched away
p-type silicon substrate
p-type silicon substrate
(i)
(iv)
Ultraviolet radiation
Photomask
Photoresist
SiO2
n-type dopant
n-type
n-type
p-type silicon substrate
(v)
SiO2
p-type silicon substrate
n-type
n-type
p-type silicon substrate
(ii)
(vi)
Exposed photoresist
is dissolved
SiO2
p-type silicon substrate
(iii)
(b)
Figure 16.35 (a) A schematic of the overall process of producing an integrated circuit (IC). Crystal growth by
the Czochralski technique produces a silicon ingot. The ingot is cut into wafers; followed by fabricating the metal,
oxide, and semiconductor materials into an IC (chip or die). Finally, the chip is packaged. (b) A schematic of
the steps in producing n-type and p-type regions in a semiconductor by a photolithographic process. ((a) Based on
Microchip Fabrication, Third Edition, by P. VanZant, The McGraw-Hill Companies. (1997), Fig. 3.7. (b) Based on Fundamentals of
Modern Manufacturing, by M.P. Groover, John Wiley, p. 849, Fig. 34-3.)
using the Czochralski technique, which we discussed in Section 13.2.7, and is shown schematically in
Figure 16.35a. At present, single crystals of silicon with a diameter up to 0.3m can be produced. The
p-type crystal is sliced into wafers with a diamond saw, and then the surface is polished for processing.
The insulating and n-type regions are produced by the following photolithographic process, shown
in Figure 16.35b. The p-type silicon is oxidized to produce an insulating layer of SiO2, as shown in
W-56
CHAPTER 16
Figure 16.35b(i). The oxide is then coated with a photoresist polymer that resists acid etching under
normal conditions; however, the photoresist can be etched by acid after exposure to UV light. In Figure
16.35b(ii), a mask allows UV light to impinge upon areas of the photoresist that are to be etched. In
Figure 16.35b(iii), the exposed photoresist polymer material is etched, and in Figure 16.35b(iv) the
insulating layer of SiO2 is etched to expose the p-type silicon. The unexposed photoresist is removed with
a solvent. In Figure 16.35b(v), donor atoms are added to the exposed p-type silicon to make it n-type. If
sufficient donor atoms are added to the p-type semiconductor, all of the acceptor levels and holes in the
VB are filled by electrons from the donor level, and there are still sufficient electrons to be excited into the
CB, producing an n-type semiconductor.
The n-type regions of the semiconductor can be produced by diffusion of donor atoms, as shown
in Figure 16.36a. However, now the preferred process is to use particle accelerators to implant
donor ions, as shown in Figure 16.36b. The location of the donor atoms is controlled with masks
that allow atoms to be diffused or implanted only in the desired locations. In ion implantation, the
penetration of the implanted ions is ballistic. Just as with shooting bullets into a soft wooden
target, there are no ions (bullets) that are stopped at the surface. Figure 16.35b(vi) shows the
npn-junction of a transistor. Conductors, such as aluminum and more recently copper, are deposited
through masks from a vapor in a vacuum system. Other dielectric oxides, such as iron oxide or zirconiumcalcia oxide (Zr-CaO2), are produced by vapor-depositing the metal and then oxidizing the metal. Through
the formation of the silicon semiconductor crystal (S) with dopant atoms, the insulator oxide (O), and
conductive metal (M), an integrated circuit (IC) is produced of MOS architecture. An IC integrates
all of the electronic devices into the silicon chip or die. The die is then packaged by embedding it into
a protective material, such as epoxy, as shown in Figure 16.35a. The packaged die is then electrically
connected to other devices in a product by metal conductors.
Gas of
dopant atoms
High velocity
dopant ions
Mask
Distance
O
O
Semiconductor
t
O
Dopant
concentration
Dopant
concentration
t
Distance
(a)
t
Distance
(b)
t
Figure 16.36 (a) A schematic of diffusion of dopant atoms into a silicon crystal through a mask. The lower
figure on the left is a schematic of the distribution of dopant atoms resulting from diffusion. (b) Ion implantation
of dopant atoms through a mask. The lower figure on the right is a schematic of the distribution of dopant atoms
resulting from ion implantation.
Electrical Properties of Materials
Heater
GaAs crystal
atomic or
molecular beams
Shutters
Furnaces
Ga
As
Figure 16.37 A schematic of molecular-beam epitaxial growth of a GaAs crystal.
Compound semiconductor crystals, such as GaAs, are produced by both the Bridgeman-Stockbarger
and Czochralski techniques, and by molecular-beam epitaxy (MBE), shown schematically in Figure 16.37.
In the MBE growth of a GaAs crystal, the Ga and the As are each vaporized from different furnaces.
The vaporized atoms are deposited on an existing GaAs crystal surface where they attach in the proper
locations to continue the epitaxial growth of the crystal. Shutters allow the furnaces to be closed. Other
furnaces can be added to deposit doping atoms to produce p-type and n-type semiconductors. Epitaxial
growth is discussed in Section 13.2.7.
16.11 Present and Future
Electronic Systems
Moore’s law states that the density of transistors in an integrated circuit will double every two years. This
is due to the reduction in the size of components in the MOS architecture. In 1971, the average size of a
state-of-the-art transistor was 10 m; in 1985 it was 1 m, in 2002 it was 90 nm, in 2007 it was 45 nm, and
22-nm devices were introduced in 2011. In the 45-nm transistor circuits, some of the insulators are as thin
as 1.2 nm, and in 22-nm transistors some of the insulators are so thin that they are porous. In the 45-nm
transistors, 30 million transistors fit on the head of a pin. Because of this small size, a 45-nm transistor
can switch on and off 300 billion times a second. With silicon-based integrated circuits, it is difficult to
produce circuits of further reduced size.
At some small size the silicon-based circuits may not perform well, and a change of technology may
be necessary. The integrated circuits of the future might be produced from carbon nanotubes or from
graphene. Conductor and semiconductor nanotubes are produced by changing the diameter of the
nanotube and by changing the orientation of the graphite structure relative to the axis of the nanotube.
Carbon nanotubes are also doped to produce p-type and n-type semiconductors. Electronic components,
such as diodes, transistors, and interconnections, have been produced from carbon nanotubes. Transistors
have also been made from graphene.
Polymer semiconductors will replace crystalline silicon semiconductors in applications that do not
require high-speed switching, because of their low cost. Organic light-emitting diodes (OLEDs) are
already utilized in the display screen in cell phones, smart phones, and digital cameras. OLEDs are
discussed in Chapter 18.
W-57
W-58
CHAPTER 16
Summary
●●
●●
●●
●●
●●
●●
●●
●●
●●
Electrical resistivity and electrical conductivity are material properties independent of the
specimen geometry, and they are inversely related to each other.
Metals, such as copper and silver, have the highest electrical conductivity of all materials at room
temperature. Semiconductors, such as pure silicon and germanium, have electrical conductivities
that are approximately eight to eleven orders of magnitude less than that of silver and copper at
room temperature. Insulators, such as diamond and polyethylene, have an electrical conductivity
nearly twenty-four orders of magnitude less than that of silver and copper at room temperature.
Superconductors have an infinite electrical conductivity, and zero electrical resistivity, below a
critical temperature that is well below room temperature.
In metals with an applied voltage, the movement of free electrons produces the current.
Because each free electron has a probability of being found anywhere in a metal, each free
electron has a probability of interacting with every other free electron. The Pauli exclusion
principle applies to interacting electrons and therefore to all of the free electrons. No two free
electrons can have the same set of quantum numbers in the entire metal specimen. Each free
electron has a unique set of quantum numbers (nx, ny, and nz) and a spin quantum number of
112 or –12. The kinetic energy of each free electron is proportional to the sum of the squares of
the quantum numbers. The free-electron energy levels in a metal are spread into a band of energy
called the conduction band. The maximum-kinetic-energy free electron in the conduction band
at 0 K is called the Fermi energy.
In a metal specimen subject to an electric field, the current density is equal to the product of
the free-electron density, the charge on an electron, and the drift velocity of the electrons. The
drift velocity is assumed constant for all free electrons in a specimen with a specified electric
field. Electron scattering provides a limit to the drift velocity. Conduction electrons are scattered
by the structural defects in a crystal and by lattice vibrations. For dilute concentrations of
point defects, the scattering distance is inversely related to the concentration of point defects.
Nordheim’s rule states that for a dilute solution, the electrical resistivity of a metal alloy linearly
increases with increases in the atomic fraction of impurity atoms.
The electrical resistivity of a composite material follows the rule of mixtures, unless the composite
has layers perpendicular to the current flow.
In a metal that is not a superconductor, as the temperature approaches 0 K, there is a residual
electrical resistivity due to structural defects, and the thermal component of electrical
resistivity approaches 0. The thermal component of electrical resistivity increases linearly
with temperature, because the amplitude of the atom vibrations increases. The electrons are
scattered more frequently by the vibrating atoms, and the distance between scattering events
decreases.
The probability that a free electron has the energy E at temperature T is given by the Fermi-Dirac
statistical distribution. Thermionic emission is when a material is heated to a high temperature
and electrons are emitted. The energy required to excite a Fermi energy electron to the energy of
free space is called the work function.
In a superconductor below the critical temperature, the electrical resistivity is 0 and the electrical
conductivity is infinite. In the BCS theory of superconductivity of metals, at temperatures below
the critical temperature electrons are coupled into pairs through interactions with the crystal
lattice.
Electrical Properties of Materials
●●
●●
●●
●●
●●
●●
●●
●●
The electrical conductivity of an intrinsic semiconductor is the same as the electrical conductivity
of a pure semiconductor, and the electrical conductivity is due to free electrons in the conduction
band and holes in the valence band.
An extrinsic semiconductor is one whose electrical conductivity depends upon the concentration of
foreign atoms. For example an extrinsic n-type semiconductor is made, with silicon from Group IV
by adding impurity atoms from a group of a higher number, such as phosphorus (P) from Group V.
In the extrinsic range of operation of an n-type semiconductor, such as Si doped with P, the
concentration of electrons in the conduction band is constant and equal to the concentration of P.
An example of a p-type extrinsic semiconductor is Group IV silicon doped with an element of a
group number less than Group IV, such as the Group III elements boron, aluminum, or gallium. In
the extrinsic range of operation of a p-type semiconductor made from Si and Group III atoms, the
concentration of holes in the valence band (p) is equal to the concentration of the Group III atoms.
Most polymers such as polyethylene are insulators. Insulating polymers are combined with
conductive powders, such as graphite or silver, to produce composite materials with increased
electrical conductivities that are utilized in resistors.
Polymers such as polyacetylene can be semiconductors and conductors. Polyacetylene conducts
electricity if there are defects in the bonding called solitons. Doping polyacetylene with iodine,
chlorine, or fluorine oxidizes polyacetylene and produces a p-type semiconductor by removing
electrons from the solitons. Dopants such as sodium and lithium mixed with polyacetylene
produce an n-type semiconductor by adding electrons to the solitons. At very high levels of
doping, polyacetylene behaves like a metal. Some possible advantages of polymer conductors
and semiconductors relative to metals and single-crystal silicon are sustainable supplies, reduced
costs, mechanical flexibility, and transparency.
When two metals with different work functions are brought into electrical contact, a few electrons
flow from the metal with the smaller work function to the metal with the larger work function,
producing a constant Fermi energy in the two metals. The electron flow produces a contact
voltage at the interface of the two metals, equal to the difference in the Fermi energy of the two
metals divided by the charge of an electron.
A pn-junction is produced by joining p-type and n-type material. In an electronic circuit, a
pn-junction functions as a diode that allows current to flow in one direction but not in the reverse
direction. An npn-junction is a transistor that acts as an amplifier in an electronic circuit. A
metal-polymer-metal junction is also a diode in an electronic circuit.
Dielectric materials are characterized by their capacitance. The capacitance of a parallel-plate
capacitor is the total electrical charge on the plates per applied volt. For a parallel-plate capacitor,
the capacitance is also the permittivity of the dielectric material between the plates times the
area of the plates, divided by the spacing between the plates. The ratio of the permittivity of a
dielectric material to the permittivity of a vacuum is called the relative permittivity. The relative
permittivity is also called the dielectric constant. The dielectric strength is the maximum static
electric field that the material can withstand without dielectric breakdown.
The capacitance of a capacitor with a dielectric material is increased relative to that of a vacuum
capacitor, due to polarization of the dielectric material. Polarization in a material results
when positive and negative charges are displaced relative to each other, but there is no flow of
current. The types of polarization are electronic, ionic, and orientational. In dielectric materials
the polarization is small, positive, and reversible; and when the applied electric field is 0, the
polarization is 0. In ferroelectric materials after a large electric field is applied and then reduced
to 0, a residual polarization is present because of ion displacements.
W-59
W-60
CHAPTER 16
●●
●●
●●
Piezoelectric materials produce a voltage when deformed, and an applied voltage deforms a
piezoelectric. Piezoelectric crystals are utilized to produce music from records, produce and detect
sound and ultrasound waves, and produce controlled displacement in micro electromechanical
systems, scanning tunneling microscopes, and atomic force microscopes.
In ionic materials electrical conductivity results primarily from the diffusion of ions. Ionic
conducting materials are utilized for devices such as thermistors and chemical sensors.
Silicon for electronic device applications is purified by zone-refining until the impurity level is a
few impurity atoms per billion (109) silicon atoms. Silicon single crystals for integrated circuits are
grown using the Czochralski technique. The silicon single crystals are cut into wafers, polished,
and processed into integrated circuits by photolithography. Metal conductors are deposited by
vapor deposition. Metal oxides are produced by vapor-depositing the metal, and then oxidizing
the metal. Compound semiconductor crystals, such as GaAs, are produced by crystal growth
techniques and by molecular-beam epitaxy (MBE).
Supplemental Reading: Subjects and Authors
Full references are listed at the end of the book.
General:
Askeland, Fulay, and Wright
Electronic properties of materials:
Gersten and Smith; Kasap; Kwok
Electronic properties of metals:Wilkes
Electronic properties of semiconductors:
Mayer and Lau; Kwok: Sze; Sze and Kwok
Electronic properties of superconductors:
Fossheim and Subdφ, Bardeen, Cooper, and Schrieffer
Electronic properties of ceramics:
Chiang, Birnie, and Kingery
Nano-electronic properties:
Dresselhaus, Dresselhaus, and Avouris; Turton
Electronic properties of polymers:
Hadziioannou and van Hutten; Heeger, Sariciftci, and Namdas
Integrated Circuit Technology:
Beadle, Tsai, and Plummer
Bioelectric composites:Benjamin
Homework
Concept Questions
1. Electrical resistivity and electrical ________________ are both material properties independent of the
specimen geometry, and they are inversely related to each other.
2. Materials with a room-temperature electrical conductivity eight to eleven orders of magnitude less than
metals are called __________________.
3. The maximum-kinetic-energy electron in the conduction band of a metal at 0 K is the ____________
energy.
Electrical Properties of Materials
4. According to Mathiessen’s rule, the electrical resistivity of a metal is the _________ of electrical resistivity
of zero-, one-, two-, and three-dimensional defects and thermal vibrations.
5. For dilute concentrations of point defects, the scattering distance for free electrons is _______________
related to the concentration of point defects.
6. Nordheim’s rule states that the electrical resistivity of a metal alloy increases ___________ with increases
in the atomic fraction of impurity atoms for a dilute solution.
7. _________________ emission occurs when a material is heated to a high temperature and electrons are
emitted.
8. The energy required to excite a(n) ___________ energy electron to the energy of free space is called the
work function.
9. In a superconductor, the electrical conductivity is infinite below the ______________ temperature.
10. In the BCS theory of superconductivity of metals, at temperatures below the critical temperature electrons
are coupled into ______________ pairs through interactions with the crystal lattice.
11. The electrical conductivity of a(n) _______________ semiconductor is the same as the electrical
conductivity of a pure semiconductor.
12. The electrical conductivity of an intrinsic semiconductor is due to the creation of free electrons in
the conduction band and __________ in the valence band.
13. A(n) _________________ semiconductor is one whose electrical conductivity depends upon the
concentration of foreign atoms added to the semiconductor.
14. The thermionic emission current density from a hot filament is in the form of a rate equation, with the
energy term equal to the ____________ __________.
15. Polyacetylene conducts electricity if it has bonding defects called ____________.
16. A mechanical advantage of polymer semiconductors relative to single-crystal silicon is that polymers are
_____________.
17. A metal with a work function of 5 eV and a metal with a work function of 3 eV are brought together in
electrical contact. They produce a contact voltage equal to ___________ volts.
18. A pn-junction in an electronic circuit functions as a(n) ____________.
19. Silicon for electronic device applications is purified by ____________ refining.
20. Silicon single crystals are grown using the ______________ technique.
21. Silicon single-crystal polished wafers are processed into integrated circuits with insulators, p-type and
n-type materials, and conductors by _______________________.
22. For a parallel-plate capacitor with a vacuum between the plates, the capacitance (C ) is the total charge
on the plates divided by the applied _______________.
23. The polarization in a vacuum capacitor is equal to __________.
24. The relative permittivity is also called the _______________ constant.
25. The dielectric _____________ is the maximum static electric field that the material can withstand without
dielectric breakdown.
26. ______________________ in a material results when it is possible to displace positive and negative charges
relative to each other, but there is no flow of current.
W-61
W-62
CHAPTER 16
27. In dielectric materials the polarization is small, _______________, and reversible.
28. The dimensionless dielectric ___________________ is the magnitude of the polarization of the dielectric
material divided by the surface-charge density of a vacuum capacitor.
29. In ____________________ materials, after an electric field is applied to saturation and then reduced to 0,
there is a residual polarization.
30. In a ferroelectric material there are ____________ that are regions with the polarization in the same
direction.
31. In a ferroelectric material, the cycle from saturation in the positive direction, to saturation in the negative
direction, to saturation in the positive direction again is a(n) _______________ loop.
32. In a ferroelectric material, the electric field in the reverse direction that reduces the polarization to 0 after
saturation in the forward direction is called the ______________ electric field.
33. Below the Curie temperature barium titanate is ferroelectric, and above the Curie temperature barium
titanate is _________________.
34. __________________ are used to determine temperature based upon the electrical resistivity of a material.
Engineer in Training–Style Questions
1.
In a metal that is not a superconductor, as the temperature approaches 0 K the electrical resistivity is due
to free-electron scattering by:
(a) Structural defects
(b) Thermal vibrations
(c) Core electrons
(d) Nuclei
2.
In a composite material, the electrical resistivity for current parallel to the axis of the 2nd phase follows
the rule of
(a) Mathiessen
(b) Nordheim
(c) Mixtures
(d) Fermi-Dirac
3.
In a metal specimen subjected to a constant electric field, the drift velocity of the free electrons is:
(a) The same for all free electrons
(b) A linearly increasing function of time for all free electrons
(c) Dependent upon the electron quantum numbers
(d) Dependent upon the electron energy
4.
_________________ have the highest electrical conductivity at room temperature.
(a) Metals
(b) Semiconductors
(c) Insulators
(d) Superconductors
5.
If a metal contains 1028 free electrons, the number of different quantum states (nx, ny, and nz) is equal to:
(a) 1028
(b) 2 3 1028
(c) 0.5 3 1028
(d) 1
Electrical Properties of Materials
6.
The electrical conductivity of a metal is not proportional to which of the following?
(a) Free-electron density
(b) Electron mobility
(c) Electron charge
(d) Atomic number
7.
In a metal that is not a superconductor, as the temperature increases which of the following is true?
(a) Free electrons are scattered more frequently by structural defects.
(b) Free electrons are scattered more frequently thermal vibrations.
(c) Free-electron drift velocity is increased because of scattering by vibrating atoms.
(d) Free-electron drift velocity is a constant.
8.
At any finite temperature, the probability that the Fermi energy is occupied with free electrons is equal to:
(a) 1
(b) 0
(c) 12
(d) A number less than 1 that is dependent upon the temperature.
9.
An extrinsic n-type semiconductor can be made from Group IV germanium by adding impurity atoms of:
(a) Group IV silicon
(b) Group III aluminum
(c) Gorup II magnesium
(d) Group V arsenic
10. In the extrinsic range of operation of a p-type semiconductor made from silicon with acceptor atoms of
gallium, the concentration of holes in the valence band (p) is equal to the concentration of:
(a) Gallium atoms
(b) Silicon atoms
(c) Donor atoms
(d) Electrons in the conduction band
11. A p-type semiconductor is produced in polyacetylene by doping with:
(a) Boron
(b) Aluminum
(c) Fluorine
(d) Lithium
12. An n-type semiconductor is produced in polyacetylene by doping with:
(a) Arsenic
(b) Phosphorus
(c) Fluorine
(d) Lithium
13. Which of the following materials is a dielectric?
(a) Copper
(b) n-type silicon
(c) Diamond
(d) Lithium-doped polyacetylene
14. Which of the following is not a type of polarization?
(a) Electron current flow
(b) Electronic
(c) Ionic
(d) Orientational
W-63
W-64
CHAPTER 16
15. If the applied electric field has a frequency of 1016 Hz, the type of polarization present in a dielectric is:
(a) Electron current flow
(b) Electronic
(c) Ionic
(d) Orientational
16. What type of material produces a voltage when deformed, and an applied voltage deforms the material?
(a) Dielectric
(b) Paraelectric
(c) Ferroelectric
(d) Piezoelectric
17. In a partially stabilized zirconia-calcia chemical sensor, the electrical conductivity is due primarily to:
(a) Electron flow in the conduction band
(b) Hole diffusion in the valence band
(c) Oxygen ion diffusion
(d) Zirconium ion diffusion
18. In an n-type semiconductor, the conduction-band energy is 1.1 eV and the donor energy level is 1.0 eV. At
temperatures approaching 0 K the Fermi energy is equal to:
(a) 1.1 eV
(b) 1.0 eV
(c) 1.05 eV
(d) 0.55 eV
19. In a semiconductor doped with donor atoms, the conduction-band energy is 1.1 eV and the donor energy
level is 1.0 eV. At a high temperature there are approximately as many electrons in the conduction band as
there are holes in the valence band. For this semiconductor, the Fermi energy is approximately equal to:
(a) 1.1 eV
(b) 1.0 eV
(c) 1.05 eV
(d) 0.55 eV
Problems
Problem 16.1 Copper is in Group IB of the periodic table, is FCC, and its room-temperature lattice parameter
is 0.362 nm.
(a) Calculate the free-electron density in Cu.
(b) The electrical conductivity of Cu at room temperature is 6.0 3 107 (V ? m)21. Calculate the
mobility of free electrons in Cu.
Problem 16.2 The electrical resistivity of pure copper metal at 0°C is 1.6 3 1028 V ? m. An alloy of 1.12 atom
percent nickel, which is substitutional in copper, produced with this copper has an electrical
resistivity at 0°C of 3 3 1028 V ? m. What is the electrical resistivity of an alloy produced from
this copper with 3.32 atom percent nickel in copper at 0°C?
Electrical Properties of Materials
Problem 16.3 The electrical conductivity of copper and nickel material at room temperature are 6.0 3 107
(V ? m)21 and 2.0 3 107 (V ? m)21, respectively. Copper and nickel form a solid solution at low
concentrations at room temperature. If an alloy of nickel plus 5 atom percent copper produced from
these materials has an electrical resistivity of 11 3 1028 V ? m, what would be the electrical resistivity
of an alloy of nickel plus 10 atom percent copper? You may assume that these are dilute solutions.
Problem 16.4 The room-temperature electrical conductivity of pure silver is 6.8 3 107 (V ? m)21, and for pure
copper the electrical conductivity is 6.0 3 107 (V ? m)21. An alloy is made of 40 volume percent
copper and 60 volume percent silver. Copper and silver form a eutectic phase diagram, with
no solubility of copper in silver or of silver in copper at room temperature. Predict the roomtemperature electrical resistivity of an alloy made from these materials.
Problem 16.5 The electrical resistivity of copper at 123 K is 0.5 3 1028 V ? m, and at 173 K it is 0.88 3 1028 V ? m.
Evaluate the B term in Equation 16.17.
Problem 16.6 The electrical resistivity of tantalum (Ta) at temperatures approaching 0 K is 1 3 1028 V ? m,
and at 150 K it is 7 3 1028 V ? m. Analytically predict the electrical resistivity of Ta at 295 K.
Problem 16.7 (a) For copper, determine the energies at which the probability of electron occupation is
0.9 and 0.1 at 300 K. (b) Are these energies symmetric about the Fermi energy of
7.04 eV?
Problem 16.8 A filament of tungsten wire is 0.001 m in diameter, and the filament is heated to 2500 K. The
work function of tungsten is 4.5 eV.
(a) Calculate the probability of electrons having sufficient energy to be emitted from the
filament.
(b) Calculate the emission current density from the tungsten wire.
(c) What length of wire can produce a thermionic electron current of 1 A?
Problem 16.9 With the values of mobility and carrier density given in Table 16.6, calculate the intrinsic
electrical conductivity of GaAs at 296 K, and compare the result with the listed value.
Problem 16.10 A semiconductor is produced by doping Group IV silicon with 1023 atoms per cubic meter of
Group III gallium. Assume that at room temperature this semiconductor is in the extrinsic
range of operation.
(a) What type of semiconductor is this?
(b) Calculate the room-temperature electrical conductivity of this semiconductor.
(c) Clearly state any assumptions you made in calculating the electrical conductivity in part b.
Problem 16.11 A semiconductor with an electrical conductivity of 300 (V ? m)21 at room temperature is
produced by doping Group IV silicon with Group V antimony. Assume that the mobility
of electrons in silicon is equal to 0.10 m2/V ? s and that the mobility of holes is equal to
0.001 m2/V ? s. You can assume that this semiconductor is in the extrinsic range of operation
at this temperature.
(a) What type of semiconductor is this?
(b) What doping concentration of antimony in atoms/m3 is necessary for this electrical
conductivity?
(c) State clearly any assumptions you make in this calculation.
W-65
W-66
CHAPTER 16
Problem 16.12 It is desired to produce a p-type semiconductor out of Group IV silicon that will have an
electrical conductivity of 100 (V ? m)21 at room temperature. Assume that the mobility of
electrons in silicon is equal to 0.10 m2/V ? s, and the mobility of holes is equal to 0.03 m2/V ? s.
You may assume that this semiconductor is in the extrinsic range of operation at this
temperature.
(a) What is the required density of holes in the valence band?
(b) What concentration of substitutional acceptor atoms will produce the desired electrical
conductivity, assuming that an element from Group III is substituted for the silicon
atoms?
Problem 16.13An n-type semiconductor is produced by doping Group IV silicon with 1 3 1022 atoms/m3
of Group V antimony. Assume that the semiconductor is at room temperature, and that this
semiconductor is in the extrinsic range of operation.
(a) What is the density of electrons in the conduction band?
(b) Calculate the room-temperature electrical conductivity for this semiconductor.
Problem 16.14 (a) For the metals with work functions listed in Table 16.4, what combination of metals
produces the largest contact voltage, and what is its magnitude? (b) In this couple, which
metal is charged positive and which is charged negative?
Problem 16.15A pn-junction is produced by doping silicon that has an energy gap of 1.11 eV with arsenic
(Group V) to produce the n-type material, and with boron (Group III) to produce the p-type
material.
(a) At temperatures approaching absolute zero, what is the value of the Fermi energy in the
n-type and p-type material?
(b) At temperatures approaching absolute zero, what would be the value of the contact
potential at a pn-junction of these two materials?
(c) At temperatures in the intrinsic range of operation for both of these materials, what is the
value of the contact potential?
Problem 16.16A pn-junction diode at 300 K produces a current density 1.00 3 104 A/m2, with a forward bias
of 0.100 V. Calculate the current density for a forward bias of 0.300 V.
Problem 16.17 A parallel-plate vacuum capacitor has an area of 1 3 1024 m2, a spacing between the plates of
1 3 1023 m, and a voltage of 1 V.
(a) Calculate the capacitance of this capacitor.
(b) A microcapacitor is replacing the capacitor in part (a). It must have the same capacitance;
however, the new area is 1 3 10210 m2, and the spacing between the plates is now 1 3 1026 m,
with 1 V applied.
(b1) What is the dielectric constant and dielectric strength required for the material in
the microcapacitor?
(b2) Are there any materials available that can meet these requirements, and, if so, what
are they?
(b3) What is the polarization (P) in this microcapacitor, and how does this polarization
compare to the surface-charge density if this were a vacuum capacitor?
Electrical Properties of Materials
Problem 16.18 The dynamic random access memory (DRAM) in computers is produced by oxidizing a silicon
wafer to produce a capacitor with SiO2 as the dielectric, and the memory corresponds to the
charge stored on the capacitor. Assume that the thickness of the dielectric is 1 3 1026 m, the
plates of the capacitor are square with an edge length of 10 3 1026 m, the dielectric constant or
relative permeability of SiO2 is 4, and the applied voltage is 10 V.
(a) What is the capacitance?
(b) What total charge is stored in this capacitor?
(c) What is the charge density on the capacitor plates?
(d) What is the polarization of the SiO2 in the capacitor?
Problem 16.19 A new polymer is developed, and to characterize the electrical properties of the polymer, the
capacitance is determined to be 20.36 3 10213 F in a parallel-plate capacitor that has square
plates with a side length of 1022 m, and the thickness of the polymer sheet and the spacing
between the plates is 1023 m.
(a) What is the dielectric constant or relative permittivity of this new material?
(b) If the capacitance measurement is made with an applied voltage of 10 V, what is the
magnitude of the polarization in the new polymer?
Problem 16.20A piezoelectric actuator is being produced for a scanning tunneling microscope. The
piezoelectric actuator is to be 1 3 1023 m in length and produced from barium titanate, which
has a dielectric breakdown strength of 6 3 106 V/m. The piezoelectric strain coefficient of
the barium titanate is 1.9 3 10210 m/V. Assume that the maximum electric field that can be
applied to the piezoelectric material is 1 3 106 V/m. (a) What is the required voltage to produce
an atomic-scale displacement of 0.1 nm? (b) What is the displacement produced by 1000 V?
(c) With 1000 V applied, is the piezoelectric safe from dielectric breakdown?
Problem 16.21 Cubic zirconia stabilized with CaO is used as an oxygen sensor material because of the high
electrical conductivity resulting from the fast diffusion of oxygen ions.
(a) Calculate the mobility of O21 ions in zirconia at 1373 K if the diffusivity of the oxygen
ions is given by
1
DsO 21 d 5 1 3 109 exp 2
2
0.84 eV
m2/s
kT
(b) What is the ratio of the electrical conductivity of the zirconia at 1373 K to that of an
n-type silicon semiconductor at room temperature with the same density of Group V
charge carriers as oxygen ions?
W-67