Basic Electric Circuit Theory
A One-Semester Text
I. D. Mayergoyz
University of Maryland
Department of Electrical Engineering
College Park, Maryland
W. Lawson
University of Maryland
Department of Electrical Engineering
College Park, Maryland
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Library of Congress Cataloging-in-Publication Data
Mayergoyz, I.D.
Basic electric circuit theory : a one-semester text /1. Mayergoyz, W. Lawson.
p. cm.
Includes index.
ISBN-13: 978-0-12-480865-2 ISBN-10: 0-12-480865-4 (alk. paper)
1. Electric circuits. I. Lawson, W. (Wes) II. Title.
TK454.M395
1996
621.319'2—<Ic20
96-18904
CIP
ISBN-13: 978-0-12-480865-2
ISBN-10: 0-12-480865-4
PRINTED IN THE UNITED STATES OF AMERICA
06 07 SB 9 8 7
To our wives Deborah and Kathy
with gratitude for their patience and inspiration.
Preface
You have in your hands an undergraduate text on basic electric circuit theory. As
such, it contains no new material for distinction or long remembrance, but it does
reflect the current state of instruction in basic circuit theory in electrical engineering
(EE) departments in the United States. And this is a state of transition. This transition
is brought about by the necessity to introduce new topics into the undergraduate
electrical engineering curriculum in order to accommodate the important recent developments in electrical engineering. This can be accomplished only by restructuring
classical courses such as basic circuit theory. As a result, there is increasing pressure
to find new ways to teach basic circuit theory in a concise manner without compromising the quality and the scope of the exposition of this theory. This text represents
an attempt to explore such new approaches.
The most salient feature of this book is that it is designed as a one-semester
text on basic circuit theory, and it was used as such in our teaching of the topic at
the University of Maryland. Since this is a one-semester text, some traditional topics
which are usually presented in other books on electric circuit theory are not covered
in this text. These topics include Fourier series, Laplace and Fourier transforms, and
their applications to circuit analysis. There exists a tacit consensus that these topics
should belong to a course on linear systems and signals. And this is actually the case
at many EE departments, where only a one-semester course on basic circuit theory is
offered.
Another salient feature of this text is its structure. Here, we deviated substantially from the existing tradition, in which resistive circuits are introduced first and
numerous analysis techniques are presented first only for these circuits. In this text,
resistors, capacitors, and inductors along with independent sources are introduced
from the very beginning and ac steady-state analysis of electric circuits with these
basic elements is then developed.
It is known that the ac steady-state equations and the basic equations for resistive
electric circuits have identical mathematical structures. As a result, the analysis
techniques for ac steady-state and resistive circuits closely parallel one another and
are almost identical. The only difference is that in the case of ac steady-state one
deals with phasors and impedances, whereas in the case of resistive circuits one deals
with instantaneous currents (voltages) and resistances. For this reason, the analysis
of resistive circuits can be treated as a particular case of ac steady-state analysis. This
is the approach which is adopted in this text.
xi
Xll
Preface
There are several important reasons for this approach.
First, we believe that EE undergraduate students are well prepared for this style
of exposition of the material. They usually have (or should have) sufficient familiarity
with the basic circuit elements from a physics course on electricity and magnetism.
Second, this approach allows one to introduce the phasor technique and the
notion of impedance at the very beginning of the course and to use them frequently
and systematically throughout the course. As our teaching experience suggests, this
results in better comprehension and absorption of the phasor technique and the
impedance concept by the students. This is crucially important because the notions
of phasor and impedance are central and ubiquitous in modern electrical engineering.
When the traditional approach to the exposition of electric circuit theory is practiced
in the framework of a one-semester course, phasors and impedances are usually
introduced toward the end of the course. As a result, students do not have sufficient
experience with and exposure to these very important concepts.
Third, the approach adopted here allows one to present numerous analysis techniques (e.g., equivalent transformations of electric circuits, superposition principle,
Thevenin's and Norton's theorems, nodal and mesh analysis) in phasor form. In the
traditional approach, these techniques are first presented for resistive circuits and
subsequently modified for ac steady-state analysis. As a result, this style of exposition requires more time, which is very precious in the framework of a one-semester
course.
Finally, the introduction of phasors at the very beginning of the course allows
one to use them in the analysis of transients excited by ac sources. This makes the
presentation of transients more comprehensive and meaningful. Furthermore, the
machinery of phasors paves the road to the introduction of transfer functions, which
are then utilized in the analysis of transients, and the discussion of Bode plots and
filters.
Another salient feature of the structure of this text is the consolidation of the
material concerned with dependent sources and operational amplifiers. In many textbooks, this material is scattered over several chapters, which somewhat undermines
its integrity and importance. In this text, this material is consolidated in one chapter
where dependent sources are introduced as linear models for semiconductor devices
on the basis of small-signal analysis. Then, electric circuits with dependent sources
and operational amplifiers are systematically studied.
Finally, we have not completely avoided the temptation to introduce new topics
in our textbook. These topics include the use of symmetry in the analysis of electric
circuits, the Thevenin theorem for resistive electric circuits with single nonlinear
resistors, diode bridge rectifier circuits with RL and RC loads, the transfer function
approach to the analysis of transients in electric circuits, active RC filters, and the
synthesis of transfer functions by using RC operational amplifier circuits. These
topics are either not discussed or barely covered in the existing textbooks. We realize
that the choice of new topics is always debatable. However, we feel that these topics
are of significant educational importance, which prompted our decision to introduce
them in this text.
Preface
xin
Usually, the basic circuit theory course is the first electrical engineering course
taken by undergraduates. For this reason, we believe that it is incumbent upon this
course to give students a "taste" of electrical engineering, to kindle their curiosity
and enthusiasm about electrical engineering, and to prepare them psychologically
for future courses. Probably, the appropriate way to achieve this is to emphasize the
connections of electric circuit theory with various areas of electrical engineering.
This is exactly what we have tried to accomplish in this text. For instance, we have
stressed the connections of basic circuit theory with the area of linear systems and
signals when we covered such topics as unit impulse and step responses of linear
circuits, the convolution integral technique, the concept of transfer functions and
utilization of their poles and zeros in transient analysis, Bode plots, and synthesis of
transfer functions by RC circuits with operational amplifiers. We have emphasized
the connections of basic circuit theory with electronics when we covered dependent
sources as linear models for transistors. Finally, we have also stressed that circuit
theory has close ties with electromagnetic theory. In basic circuit theory, it is assumed
that the values of resistances, inductances, and capacitances are given. The calculation
of these quantities is the task of electromagnetic field theory. Furthermore, Kirchhoff ' s
laws, which are treated as basic axioms in circuit theory, can be derived (can be
proved) from Maxwell's equations of electromagnetic field theory. More important,
by using electromagnetic field theory, the approximate nature of Kirchhoff's laws can
be clearly elucidated and the limits of applicability of these laws (and circuit theory)
can be established.
There is ongoing discussion concerning the place and role of SPICE (or MicroSim® PSpice®) simulators in a basic circuit theory course. We believe that these
circuit simulators should play a complementary role in this course. It is important
to emphasize the usefulness of these computer aided design tools and that their
effectiveness increases in the hands of "educated consumers." It is equally important to
stress that these tools are not a substitute for sound knowledge of electric circuit theory
and to provide this knowledge is the ultimate goal of the basic circuit theory course.
In other words, we would like to warn against undue invasion of the basic circuit
theory course by SPICE and PSpice simulators, an invasion which may compromise
the very goals of this course. For this reason, PSpice examples are confined to the
final sections of some of the later chapters and a list of PSpice references is relegated
to Appendix C.
In undertaking this project, we wanted to produce a student-friendly textbook.
We have come to the conclusion that students' interests will be best served by a short
book which will closely parallel the presentation of material in class. We have not
avoided the discussion of complicated concepts; on the contrary, we have tried to
introduce them in a straightforward way and strived to achieve clarity and precision
in exposition. We believe that material which is carefully and rigorously presented
is better absorbed. From our teaching experience, we have found that there are some
topics which are more difficult for students to digest than others. We have observed
that the mathematical form of circuit theory is not the major obstacle. Students
usually encounter more difficulties in reading connectivity of the electric circuits
than in understanding the mathematics of circuit equations. For instance, we have
XIV
Preface
found that it is difficult for students to recognize even simple series and parallel
connections if they are masked (obscured) by the drawing of the electric circuit. For
this reason, we have made a special effort to explain carefully these "psychologically"
difficult topics. It is for the students to judge to what extent we have succeeded.
In writing this book, we have been assisted by several of our students and
colleagues. In particular, C. Buehler aided us in the development of the first version
of our lecture notes. D. Kerr and Chung Tse helped us in further modifications of our
manuscript. Our colleagues, Professors T. Antonsen, N. Goldsman, and C. Striffler
read our manuscript and provided us with their suggestions and constructive criticism.
We are specially thankful to Professor C. Striffler for using our manuscript in his basic
circuit theory class. Mrs. P. Keehn patiently and diligently typed several versions of
our manuscript. We are very grateful to our students and colleagues mentioned above
for their invaluable help in our work on this book.
Chapter 1
Basic Circuit Variables
and Elements
1.1
Introduction
Before we discuss the equations which describe the operation of electric circuits, we
must first review a few fundamental physical concepts that will be needed in our study
and then define the basic elements which are the building blocks of electric circuits.
We begin with basic circuit variables. The reader should have some familiarity with
these circuit variables from physics courses on electromagnetism. These variables are
the electric charge, electric and displacement currents, voltage and electric potential,
electric energy and power, and magnetic flux linkage. Throughout this text we will
always use the international system of units for these variables, which is denoted as SI
or MKS A for meter-kilogram-second-ampere. The values of the physical quantities
will range over many orders of magnitude, so we will make liberal use of the common
multiplying factors and abbreviations as given in Table 1.1.
Afterward, we will introduce the concept of a two-terminal element and define
the convention for reference directions. It is crucial that the reader adhere to this
convention for every circuit problem she or he encounters; failure to do so may result
Table 1.1: Common multiplying factors.
Factor
12
10
109
106
103
10- 3
Prefix name
Symbol
tera
T
G
M
k
m
giga
mega
kilo
milli
Factor
Prefix name
Symbol
10-6
micro
nano
pico
femto
atto
M
n
Kr9
10-12
10" 15
10-18
1
P
f
a
2
Chapter 1. Basic Circuit Variables and Elements
in many embarrassing sign errors. The basic two-terminal elements which will be
discussed below are the resistor, capacitor, inductor, ideal voltage source, and ideal
current source. We will explain the operation of these circuit elements and will derive
the terminal relationships between voltage and current for each element. We will
also derive expressions for the energy stored and power dissipated in terms of each
element's voltage, current, and physical characteristics. Understanding the terminal
relationships for these elements is essential in order to master the material in the
following chapters.
1.2
Circuit Variables
1.2.1
Electric Charge
Electric charge is one of the most fundamental quantities in physics. As such, it
cannot be defined—it can only be described. Electric charge is a property of particles
which manifests itself through what is termed the electromagnetic interaction. This
electromagnetic interaction occurs between charged particles at a distance, without
actual contact as in the case of mechanical interaction. As a result of the electromagnetic interaction, forces appear. These forces act on charged particles and have
two distinct components. The first of these components is due to the instantaneous
positions of the charged particles, while the second component is due to both the
instantaneous positions and the velocities of the charged particles.
In order to describe the first component of the forces, the notion of the electric
field is introduced. Each charged particle creates an electric field which is distributed
in space and interacts with other charged particles. This interaction is called the
Coulomb interaction and is characterized by Coulomb's law. To describe the second
component of the interactive forces, the notion of the magnetic field is introduced.
Magnetic fields are created by moving charged particles and exert forces on moving
charged particles. By using the notions of electric and magnetic fields, interaction at
a distance can be described.
The circuit notation for electric charge is q(t), implying that the charge may vary
with time. The MKS A unit for electric charge is the coulomb, denoted by C:
[q] = C
(1.1)
Electric charges can be either positive or negative. For example, electrons are negatively charged particles, while protons are positively charged particles. The sign of
the charge can be distinguished through the Coulomb interaction force, by which like
charges repel and opposite charges attract. As a unit of electric charge, the coulomb
is quite large in comparison to the elementary charges of an electron or a proton. One
coulomb in comparison with the charge of one proton is
1 C = 6.24 X 1 0 % ,
(1.2)
3
1.2. Circuit Variables
where qp is the positive charge of one proton. An electron has a charge equal in
magnitude to a proton but opposite in sign, so that it takes approximately 6.24 X 1018
electrons to make — 1 C.
The principle of conservation of electric charge is a fundamental property of
nature. It states that in a closed system the total charge does not change with time.
However, equal amounts of positive and negative charges can be simultaneously
created or annihilated without disrupting the total balance of electric charge. This
creation/annihilation phenomenon occurs only in very high energy systems; in our
study of electric circuits we can assume that the total number of both positive and
negative charges does not change with time.
1.2.2
Electric and Displacement Currents
By definition, electric charges in motion constitute an electric current. To characterize
electric current quantitatively, we assume that a net charge q(t) flows through an
arbitrary surface 5. Then the current through the (open) surface is defined to be the
instantaneous time rate of change of the net charge flow through the surface (see
Figure 1.1). This can be expressed mathematically as:
lit) = ——.
(1.3)
at
By integrating equation (1.3) we can find the total charge through S in terms of
electric current:
q(t) = q(t0)+
/ i(r)dr.
Figure 1.1: Surface S with current flow.
(1.4)
4
Chapter 1. Basic Circuit Variables and Elements
The MKS A unit of electric current is the ampere, which is defined as a coulomb
per second:
[i] = A = C/s.
(1.5)
An electric current always creates a magnetic field because it is (by definition) the
motion of electric charges.
EXAMPLE 1.1 Suppose that under steady-state conditions, 1012 electrons flow
through a surface S every microsecond. What current does this represent?
According to the previous definition, we have:
/ = dq/dt = Aq/At
= (electron charge) X (#electrons)/(time interval) (1.6)
/ = -1.602 X 10" 1 9 CX 10 12 /10~ 6 s = - 0 . 1 6 A.
(1.7)
In circuit theory, the electric current normally flows in metal wires, so the surfaces
we construct to apply the above definition are often ones that simply cut through the
wire. These cuts are called cross sections. For convenience, we often just drop the
picture of the surface S altogether since the current flowing through the wire is a
fairly straightforward concept.
In addition to the previously described electric current, there is another type
of current which is not associated with the flow of electric charges. This current is
called displacement current. Displacement currents occur due to the time variation
of electric fields. These currents also create magnetic fields just as the currents due
to the motion of electric charges do. As shown below, one important example of
displacement current is the current through a capacitor. Displacement currents are
also responsible for electromagnetic wave propagation through empty space.
Another very important concept is the principle of continuity of current, which
states that the total current (electric + displacement) through any closed surface at
any instant of time is always zero. If the currents entering the surface are taken with
positive signs, and the currents leaving the surface are taken with negative signs, then
the principle of continuity of current can be expressed mathematically as follows:
£ >
= 0.
(1.8)
k
From the principle of continuity of current, it is evident that displacement currents
are continuations (extensions) of currents due to the motion of electric charges. For
example, consider the case of a capacitor, shown in Figure 1.2 with the surface S
enclosing one of the two plates of the capacitor. The principle of continuity of current
asserts that the current entering S from the wire must equal the current leaving S on
the right. But the only current leaving S on the right is due to the time variation of
1.2. Circuit Variables
Figure 1.2: Capacitor with displacement current.
the capacitor's electric field, and therefore it is a displacement current. Thus, we can
see that the displacement current is actually a continuation (extension) of the current
through the wire, which is due to the motion of electric charges.
EXAMPLE 1.2 To illustrate the principle of continuity of current (1.8) consider the
following example. There are currents i\ and /4 flowing through surface S into the
volume in Figure 1.3 and i2 and z3 flowing out of the volume through separate wires.
If ¿i = 10 A, i2 = 6 A, and i3 = 7 A, find i4.
Formula (1.8) states that i\ — i2 — h + U = 0, so /4 = 6 + 7 — 10 = 3 A. Note
that if i\ had been 15 A, then 14 would have been —2 A. This means that, contrary to
the picture, current is actually leaving S through the fourth wire. We will talk more
about assumed (reference) directions for current later in this chapter.
■
Figure 1.3: Example of conservation of current.
6
1.2.3
Chapter 1. Basic Circuit Variables and Elements
Electric Energy
In order to move electric charges and produce current, work must be performed
against interaction forces, resulting in an expenditure of energy. This energy is called
electric energy. Of course, this is only one of the many forms of energy, which include
light energy, mechanical energy, heat energy, chemical energy, etc. However, electric
energy has some attractive features which distinguish it from other forms of energy.
Electric energy is relatively cheap and easy to produce, because it can be centrally
generated in large quantities at power plants. However, thisfirstproperty would hardly
be an advantage if electric energy were not so easily transmittable over large distances
through power transmission lines to almost anywhere it is needed. Electric energy
is also extremely versatile—it can be easily converted into other forms of energy,
such as light energy or mechanical energy, or even be used to encode and process
information.
The mathematical notation for energy (work) is w(t) and the MKSA unit is the
joule:
M = J.
1.2.4
(1.9)
Voltage
Voltage is normally discussed as existing between two different points (terminals).
Consider two points P and Q in Figure 1.4. By definition, voltage (denoted v(/)) is
the work done on a unit positive charge by moving it from point P to point Q.
Therefore,
wit) = qv(t)
(1.10)
is the work done on moving an arbitrary charge q. From equation (1.10) we find
v(0 = ^ .
q
The MKSA unit for voltage is the volt, defined as a joule per coulomb:
[ v ] = J / C = V.
Figure 1.4: Diagram of an arbitrary path between two points.
(1.11)
(1.12)
1.2. Circuit Variables
7
An important property of voltage is its path independence. For static and quasistationary fields, the voltage is independent of the path along which the charge is
moved. Thus, voltage will depend only on the position of the two endpoints of the
path and can be expressed as a difference of the potentials at each point,
v(f) = <¡>Q(t) - MO-
(1.13)
This aspect of voltage will be exploited later in the book when the method of nodal
potentials is used for the analysis of electric circuits.
1.2.5
Electric Power
Electric power is defined as the rate of energy expenditure. Mathematically it means
that the power p(t) is given by:
dw(t)
p{t) = -^-.
(1.14)
The MKS A unit of electric power is the watt, defined as a joule per second:
[p] = J/s = W.
(1.15)
The expression for power can be integrated with respect to time to find the energy:
w(t) = w(t0) + / p(r)dr.
Jto
(1.16)
A very important consideration to keep in mind is that there are no sources of
infinite power. Thus, power is always finite. The importance of this fact becomes more
evident later when the continuity of voltage across capacitors and the continuity of
current through inductors are derived.
Now, by using the expression relating energy and voltage, we can derive the
expression for electric power in terms of voltage and current. Consider a two-terminal1
electric device (shown in Figure 1.5) to which a voltage v(t) is applied and through
which a current i(t) flows. Note that the current is pictured in Figure 1.5 as flowing
into the positive terminal and also that an equal current must be flowing out of the
negative terminal. Then, in an infinitesimally small time period from t to t + dt,
an infinitesimally small charge dq passes through the device. From the previous
relationship between charge and work (equation (1.10)) we have
dw = v{t)dq,
1
(1.17)
A terminal is a location on a device (usually a metal contact or wire) where connections
are generally made to other devices.
Chapter 1. Basic Circuit Variables and Elements
0M
Electric
v(t)
Device
oFigure 1.5: Electric device connected to a power source.
where dw is the infinitesimally small amount of work done on moving dq. Dividing
both sides of equation (1.17) by dt, we have
dw
dq
(1.18)
Recall from equations ( 1.3) and (1.14) that the time derivative of charge is the current
/(f), and the time derivative of work is the power p(t), so
p{t) = v(t)i(t).
(1.19)
This expression for power is especially useful because voltage and current are the
two most often encountered circuit variables and are readily measurable.
1.2.6
Flux Linkages
The concept of flux linkages will be crucial when we analyze inductors as circuit
elements. To define flux linkages we first consider a simple one-turn coil of wire with
a current passing through it (see Figure 1.6). As stated previously, the current in the
wire creates a magnetic field, which can be represented by magnetic field lines. Recall
that the right-hand rule gives the direction of the field lines. These magnetic field
lines enclose the electric current and form the magnetic flux <ï> which links the coil.
This flux is defined mathematically to be the integral of the magnetic flux density B
over the surface 5 bounded by the coil: <E> = JSB • d~s. When the coil of wire has
several closely spaced turns, the flux linkage is defined as
^=NO,
where TV is the number of turns and O is the magnetic flux linking one turn.
(1.20)
9
1.2. Circuit Variables
Figure 1.6: A one-turn coil carrying a current i{t) and some resulting field lines.
Flux linkages are very important in the consideration of Faraday's law, which
states that the time variation of flux linkage induces voltage. Mathematically, this is
expressed as
d^
(1.21)
dt
We can obtain an expression for flux linkages through integration of equation (1.21):
v(f)
(1.22)
¥(f) = ^(fo) + / v{r)dr.
The MKSA unit for flux linkages is the weber:
m = Wb = V • s.
(1.23)
A collection of all the standard units which will be used in this book is given in
Table 1.2.
Table 1.2: Basic circuit quantities and associated units.
Variable
Symbol
Unit
Notation
Charge
Current
Energy
Voltage
Power
Flux linkage
Resistance
Capacitance
Inductance
q{t)
i(t)
w{t)
v(i)
Pit)
coulomb
ampere
joule
volt
watt
weber
ohms
farads
henries
C
A
J
V
W
Wb
O
F
H
■KO
R
C
L
10
Chapter 1. Basic Circuit Variables and Elements
1.3
Reference Directions
Most of the circuit elements that we will encounter in this text will have two terminals
and can be represented schematically as in Figure 1.7. Examples of these elements
include resistors, capacitors, inductors, and sources.
Each two-terminal element can be completely characterized by the voltage across
the terminals of the element and the current through the element. In order to write
meaningful equations relating the voltages and currents in electric circuits, it is
necessary to assign a direction to the current and a polarity to the voltage for each
two-terminal element. These assigned directions are called reference directions, and
they are used in determining the signs given to the circuit variables when setting up
Kirchhoff's equations (which are discussed in the following chapter).
It is important to stress that these reference directions are assigned entirely arbitrarily. The actual directions of the currents and the polarities of the voltages in
a circuit are not generally known a priori} They can be found only by analyzing
the circuit, which (as we will soon demonstrate) cannot proceed until the reference
directions for each element are assumed. Finally, we note that whereas reference
directions for circuit variables are fixed, the actual currents and voltages often reverse their directions and polarities as the state of the circuit evolves with time. For
example, in ac circuits (which are introduced in Chapter 3 and figure prominently in
electrical engineering), voltages and currents periodically alter their actual polarities
and directions.
In circuit notation, the reference direction for a current is indicated by an arrow
(Figure 1.7a), while the reference polarity for a voltage is shown by using plus and
¡(t)
v(t)
(a)
(b)
Figure 1.7: Two-terminal element with (a) reference direction for current and (b)
reference polarity for voltage.
2
As one gains familiarity with the analysis of electric circuits, inspection of a particular
circuit layout will often seem to suggest the appropriate reference directions. While the student
should be encouraged to make an educated guess as to the actual current directions and voltage
polarities and to assign the reference directions accordingly, remember that ultimately it is the
solution of Kirchhoff's equations that provides the correct answers.
11
1.4. The Resistor
minus signs placed next to element terminals (Figure 1.7b). Although they are chosen
arbitrarily, the reference directions and polarities are coordinated for passive circuit
elements such as resistors, inductors, and capacitors. This coordination means that the
reference direction of the current is always chosen to be from the positive reference
terminal to the negative reference terminal. By adopting this passive sign convention,
we know that a positive value for the power p{t) = v{t)i{t) always means that energy
is flowing into the element at that instant in time.
Reference directions are used in order to write Kirchhoff 's equations for currents
and voltages. First, the arbitrary reference directions and polarities are assigned to all
elements in the circuit. Then the circuit equations are set up by using these reference
directions. These equations are solved and the signs of the circuit variables are found.
If, after solving these equations, a current is found to be positive
/(f) > 0 ,
(1.24)
then the actual direction ofthat current at time t coincides with the reference direction.
If, on the other hand, the current is negative
i(0 < 0 ,
(1.25)
then the actual direction of the current at time t is opposite to the reference direction.
The same rule applies to the reference polarity of the voltage. If the voltage is
determined to be positive,
v(f)>0,
(1.26)
then the actual polarity coincides with the reference polarity. If the voltage is found
to be negative,
v(f)<0,
(1.27)
then the actual polarity is opposite to the reference polarity. Thus, the reference
directions allow one to write circuit equations, to solve them, and then to find the
actual directions and polarities for circuit variables.
1.4 The Resistor
The circuit notation for a resistor is shown in Figure 1.8. The resistor is a twoterminal element which impedes the flow of electric current through it by converting
some of the electrical energy to heat. The degree to which it impedes the current
flow is characterized by resistance. Every normal conductor possesses this property
to some extent. For an idealized resistor, the value of resistance is independent of
the applied voltage and current and is a function only of the conductor geometry and
physical properties of the materials from which it is made.
As stated previously, it is very important to know the terminal relationship
between current and voltage for each of the circuit elements. For a resistor, this
12
Chapter 1. Basic Circuit Variables and Elements
¡(t)
+
R
v(t)
Figure 1.8: Circuit notation for a resistor.
relationship is given by Ohm's law:
(1.28)
v(t) = Ri(t).
Thus, this law states that the voltage drop across the resistor is directly proportional
to the current through it. The MKSA unit for resistance is the ohm, denoted by Í 1
The ohm is by definition a volt per ampere:
[R] = v/A = a
(1.29)
In effect, resistance is a coefficient of proportionality between voltage and current, as
indicated in Figure 1.9. Solving equation (1.28) for current, we have the equivalent
relationship
v(t)
R'
(1.30)
m=
A resistor can also be characterized by a quantity called conductance G, defined
as the reciprocal of the resistance:
(1.31)
G =
R = slope
Figure 1.9: The voltage-current relationship for an ideal linear resistor.
13
2.4. The Resistor
In terms of conductance, equation (1.30) can be rewritten as
i(t) = Gv(t),
(1.32)
which is sometimes a more useful form of Ohm's law, depending on the particular
application. The unit for conductance is the mho (ohm spelled backward). The mho
is defined as an ampere per volt and is also sometimes referred to as a Siemens (S):
[G] - A/V = U.
(1.33)
We now recall that instantaneous power was derived earlier as pit) = v(t)i(t).
From this equation and equation (1.28) we see that power can be expressed as:
p(t) =Ri2it).
(1.34)
The same formula can be rewritten in the form:
pit) = Gv2(i),
(1.35)
or equivalently as
V
p(t) =
-^-.
(1.36)
As shown by these equations, the instantaneous power in a resistor is always
positive. This means that the resistor always consumes power—it can never give it
back to the source. Therefore, the resistor is an energy-dissipating element. Because
of this fact, resistors are often used to model irreversible losses of electric energy.
In circuit theory, it is customarily assumed that the values for resistances are
given. However, when solving real-world problems, these values are not always
known, and they must be computed (or measured) by the engineer in order to undertake a circuit analysis. The problem of the calculation of resistance belongs to
electromagnetic field theory. However, in the case of a conductor of uniform cross
section (shown in Figure 1.10), the following simple formula is valid:
R= \
aA
L
A'
Figure 1.10: A uniform, cross-sectional area conductor.
(1-37)
14
Chapter 1. Basic Circuit Variables and Elements
where L is the length of the conductor, A is the cross-sectional area, and cr is
a physical constant called conductivity which depends on the material composition
and temperature. From this equation we see that the resistance is directly proportional
to length and inversely proportional to cross-sectional area. Practically, this means
that longer wires have larger resistances while thicker wires have smaller resistances.
However, if the cross section is not uniform it is very difficult to calculate the
resistance, and we must invoke electromagnetic field theory to solve this problem.
Standard off-the-shelf resistor values range from the milliohm level up to the
multi-megohm level. Several different methods of fabrication are required to span
such a large range. In addition to the nominal resistance value, resistors are usually
categorized by type of construction, power handling capability, and the tolerance
of the resistance value. Fabrication materials for resistors include wire-wound metal
film, carbon film, and carbon composition. The latter type are by far the most common
in low-power packages. By varying the carbon concentration, a wide range of resistance values can be achieved in the same geometry. For example, standard 1/4 watt
resistors come in cylindrical casings 0.635 cm long and 0.229 cm in diameter, while
2 watt resistors are 1.746 cm long and 0.794 cm in diameter. Standard tolerances for
resistance values are ±10%, ±5%, and ± 1 % (tighter tolerance means higher cost).
The carbon composition resistors come in standard values from 1.0 up to 9.1 in each
decade from one ohm up to the multi-megohm level. The standard values as well as a
color code scheme for identifying the values can be found in Electronic Components
and Measurements by B. D. Wedlock and J. K. Roberge (Prentice-Hall, 1969). The
basic idea is that each standard resistor has a resistance about 10% (or less) larger
than the one below it. If one makes resistors with 10% tolerance, then in principle all
values of resistances will be fabricated.
EXAMPLE 1.3 What is the resistance of a cylindrical carbon composition resistor
that is L = 1 cm long, has a radius of r = 1 mm, and has a conductivity of
a = 2 X 104 U/m?
From equation (1.37) we have:
R = L/crA = L/airr2
= 0.01/(2 X 104 X rr X (10 -3 ) 2 ) - 0.16 Í1 (1.38)
EXAMPLE 1.4 As an application of a resistor, consider a heater which is connected
to some fixed voltage source (see Figure 1.11). The power of this heater (i.e., how
much heat is produced) is given by the expression
theater =
^
,
^heater
(1.39)
and it is obvious that a smaller resistance is desired for a greater heating effect.
Such heaters are common in many household items including toasters, incandescent
15
1.4. The Resistor
v(t)
R
heater
Figure 1.11: Simple diagram of a heater.
lightbulbs, electric ovens, electric clothes dryers, electric water heaters and furnaces,
and hair dryers.
■
EXAMPLE 1.5 Consider the transmission of electric power through power lines
(Figure 1.12). In this case, there is a fixed amount of power intended for transmission:
(1.40)
Arans = v(f )*(*)•
There is also some power that is lost due to the resistance i?Wire of the transmission
line. This power is given by the expression:
(1.41)
Plost ~~ ^nvire* ( 0 -
Therefore, to minimize the power loss, we have two alternatives:
1. Reduce i?Wire> which, according to (1.37), requires an increase in the crosssectional area A. This makes for a very expensive alternative.
2. Make the current /(f) as small as possible. Since ptmns is fixed, the voltage v(t)
must be high in order to transmit the same power with the desired small current.
The latter approach is the basic idea behind high-voltage power transmission lines.
O
+
'(t)
i(t)
P
-AAAA^
wire
Electric
Device
Figure 1.12: Simple diagram of power transmission.
16
1.5
Chapter 1. Basic Circuit Variables and Elements
The Inductor
An inductor is a two-terminal element which stores energy in magnetic fields and is
characterized by inductance. The circuit notation for an inductor is given in Figure
1.13. In practice, an inductor is a coil of wire consisting of many closely spaced
turns. In order to enhance the inductance of the coil, an iron or ferrite core is usually
inserted because of the high magnetic permeability of these substances. Inductance,
denoted by L, is by definition the ratio of total magnetic flux linking the inductor
turns to the current flowing through the inductor. This can be written mathematically
as follows:
i{t)
However, the inductance does not depend on either the flux linkage or the
current—it depends solely on the design of the inductor itself. A general formula for
inductance is too complicated for this text; however, it can be shown that inductance
is determined by the square of the number of turns, the geometric dimensions of
the inductor, and the magnetic permeability of the core. In effect, inductance is a
coefficient of proportionality between flux linkage and current:
*(f) = Li(t).
(1.43)
The MKS A unit of inductance is the henry,
Wb
V•s
[L] = H = — = — = f l - s .
(1.44)
To determine the relationship between current and voltage in the inductor, we
use Faraday's law:
dW
vit) = — .
dt
¡(t)
o—
+
v(t)
Figure 1.13: Circuit notation for an inductor.
(1.45)
17
1.5. The Inductor
By substituting (1.43) into (1.45) and using the product rule for differentiation, we
find:
dL
v(t)=—i(t)
at
di(t)
+ L-^.
at
(1.46)
In this text, inductance is assumed to be constant (time invariant). Consequently, the
last equation leads to the following terminal relationship:
v« = LdM.
(1 .47)
In the special case of dc (direct current), there is no time variation of i{t\ and
consequently, there is no voltage induced because the flux linkage is constant.
Knowing the voltage across the inductor, we can derive an expression for current
through it by integrating equation (1.47):
m = i(t0)+j
f v(r)dr.
(1.48)
To derive an expression for power delivered to the inductor, we use the expression
for instantaneous power in terms of voltage and current and equation (1.47):
diit)
p(t) = v(t)i(t) = Li(t)-^.
at
(1.49)
From this equation we can conclude that the current through an inductor must be
a continuous function of time. To draw this conclusion, we must recall an earlier
statement that there are no sources of infinite power. Because power can never be
infinite, di(t)/dt is never infinite as well, which means that i(i) must always be
differentiable. And from calculus we know that differentiability implies continuity.
Mathematically, this means that the value of the current immediately before an instant
of time i0 must equal the value of the current immediately after i0:
i(fo+) = ¿Co-).
(1.50)
This is the principle of continuity of electric current through an inductor. This
principle will be extensively used throughout the text.
By using the chain rule for differentiation in equation (1.49), power can also be
expressed as:
m
Ldtfjt))
~2-dT-
(1 51)
'
Thus, we can conclude that power is positive if the square of the current increases
with time and that it is negative if the square of the current decreases with time.
Mathematically, if
K
,
dt
>0,
(1.52)
18
Chapter 1. Basic Circuit Variables and Elements
then the absolute value of the current is monotonically increasing with time and the
power is positive. However, if
^ » < 0 ,
(1.53)
at
then the absolute value of the current is monotonically decreasing with time and the
power is negative.
From the above discussion we conclude that, unlike the resistor, the inductor
is an energy storage element and can give energy back as well as absorb and store
it. If the absolute value of the current is increasing with time, then power is being
consumed by the inductor and energy is being stored in the inductor's magnetic field.
If the absolute value of the current is decreasing with time, then power is given back
at the expense of the energy previously stored in the magnetic field. This suggests
a practical application of inductors for energy storage. However, since inductors
are made from wires, practical inductors often have a resistance associated with
them and this leads to some energy dissipation into heat. We will usually ignore
the problem by just assuming that the inductor wire has a —+ oo. This is actually
the case in superconducting materials, when the materials are cooled to the point
where they offer no resistance to current flow. The energy storage application is
especially attractive when superconducting wires are used, since there is no power
loss associated with the circulating current in the inductor.
Recalling that power is the time derivative of energy, and using this in equation
(1.51), we can determine the energy stored in the magnetic field of the inductor by
integration. Assuming that at zero time we have
w(0) - i(0) = 0,
(1.54)
we find through integration of (1.51) that the energy stored in the inductor is
Li2(t)
w(t) = -±±.
(1.55)
EXAMPLE 1.6 The current through a 10 mH inductor is given (in amperes) by
i(t) = 2t2e~^10 when t > 0. Find (a) the voltage across the inductor, (b) the power
absorbed by the inductor, and (c) the energy stored in the inductor.
(a)
V(t)
= L-=
Mte-t/l° - .002i V / 1 0 V
dt
= .04fó- r/10 (l - í / 2 0 ) V .
(b) p{t) = v(t)i(t) = .04te~ r/10 (l - t/20)2t2e-t/l°
W
/5
= .08fV (l -f/20)W.
(c)
w(t) = ^
= . 0 2 f V / 5 J.
■
19
1.6. The Capacitor
o—
i(t)
-
+
v(t)
Figure 1.14: Circuit notation for a capacitor.
1.6
The Capacitor
A capacitor is a two-terminal element which stores energy in the electric field and is
characterized by capacitance, denoted C. Its circuit notation is shown in Figure 1.14.
A capacitor is formed by two conductors separated by some distance. Sometimes
a dielectric material may be placed between the conductors to provide a means of
separation and to enhance the capacitance. Both conductors of the capacitor possess
charges of equal magnitude but opposite sign (see Figure 1.15):
(1.56)
q\ = -qi
These charges create an electric field between the two conductors. We wish to relate
the voltage across the capacitor to the charge of the capacitor. Recall that voltage
is usually specified as the potential difference between two points. However, since
each conductor under static (or quasistatic) conditions has the same potential at every
point, we do not need to specify two points to determine voltage—we can just speak
of the voltage between the two conductors. We now define capacitance as the ratio
between the charge and this voltage:
git)
(1.57)
v(t)'
Although the capacitance is the ratio of charge to voltage, it does not depend
on either of these two variables. Capacitance is determined only by the geometric
C =
Figure 1.15: Two conductors with the corresponding electric field.
20
Chapter 1. Basic Circuit Variables and Elements
dimensions of the conductors, the separation distance between the conductors, and
the permittivity of the dielectric material between the conductors. In this respect, the
capacitance serves as a constant of proportionality between charge and voltage, much
as inductance is between flux linkages and current:
q(t) = Cv(t).
(1.58)
Physically, capacitance determines the ability of the capacitor to accumulate charge.
Thus, the larger the capacitance, the greater the charge a capacitor can store for the
same applied voltage.
The MKS A unit for capacitance is the farad, defined as a coulomb per volt:
A farad is a huge unit of capacitance and in practice we usually use micro-, nano-,
and picofarads.
In order to obtain the terminal relationship between voltage and current in a
capacitor, we use the product rule to differentiate equation (1.58) with respect to
time:
da
i(t) =
dv(t)
c
ir ^r
dC
+
* *>•
(160>
-
Assuming capacitance is not a function of time (its derivative is equal to zero), we
arrive at the terminal relationship:
¡(0 = C*f.
(1.61)
From this equation we clearly see that ac voltage (alternating voltage) can cause a
current through a capacitor because this voltage is time varying. However, dc voltage
cannot cause a current because it is a constant function of time. This is consistent
with the notion that the current through a capacitor is a displacement current which
exists only when the electric field in the capacitor varies with time.
We can derive an expression for voltage by integrating equation (1.61):
v(f) = v ( i 0 ) + i f i(T)dT.
(1.62)
We also find the expression for instantaneous power delivered to a capacitor:
dv(t)
p(t) = v(t)i(t) = CV(í)-¿^(1.63)
dt
This last equation shows that the voltage across a capacitor must be a continuous
function of time because power can never be infinite. Indeed, the voltage must be
differentiable and differentiability implies continuity. From this we find that the
voltage immediately before any instant of time to must equal the voltage immediately
21
1.6. The Capacitor
after this instant of time:
v(r0+) = v(fo-).
(1.64)
This is the principle of continuity of voltage across a capacitor, a principle which will
be extensively used throughout the text.
By using the chain rule for differentiation in equation (1.63), we derive another
expression for power:
Cd{v\t))
p{t)=
2^r-
(L65)
From this equation we see that the power is positive when the square of the voltage
is monotonically increasing,
^
> 0,
(1.66)
at
and it is negative when the square of the voltage is monotonically decreasing,
d(y\t))
dt
<0.
(1.67)
This suggests that the capacitor is also an energy storage element; however, it
stores energy in the electric field. When the power is positive, energy is consumed
and stored in the electric field, and when the power is negative, the energy is given
back from the electric field. This also suggests that capacitors can be utilized for
energy storage, as is the case for many pulse power applications. It is important to
stress that energy storage in capacitors is not usually associated with energy loss as
in the case of inductors. This is because dielectrics are much closer to ideal than are
conducting wires.
To find the energy stored in the capacitor in terms of voltage, we integrate
equation (1.65) by using the (assumed) initial conditions
w(0) - v(0) = 0
(1.68)
w(t) = —^.
(1.69)
and derive that
The capacitance is usually a difficult quantity to calculate, and this is done by
using electromagnetic field theory. However, there are some widely used expressions
for a few standard configurations. Consider the capacitor consisting of two parallel
plates separated by a dielectric slab of permittivity e = e r e 0 , where e0 = 8.854 X
10" 12 F/m is the permittivity of vacuum and er is the relative dielectric constant. If
Chapter 1. Basic Circuit Variables and Elements
22
the plate area is A and the plate separation is d, the capacitance is given by
(1.70)
Standard off-the-shelf capacitance values range from the picofarad level up to
the thousands of microfarads level. The parallel plate configuration cannot be used
to span the entire range. There is a practical limit to how large the plate area A
can be and a physical limit to how large the permittivity e can be. Likewise, if you
make the separation distance d too small, the capacitor will not be able to sustain
high voltage. In addition to the nominal capacitance value, capacitors are usually
categorized by type of construction, voltage handling capability, and the tolerance of
the capacitance value. Fabrication methods for capacitors include parallel plates with
ceramic or mica dielectrics, electrolytic, and rolled paper/foil. Standard tolerances
for capacitance values are ± 10%, ±5%, and ± 1 % .
EXAMPLE 1.7 What is the capacitance of a system of two circular disks of radius
r = 1 cm separated by a 1 mm thick piece of alumina (er = 6)?
From equation (1.70) we have:
C - eA/d = 67rr2/d = 6X 8.854 X 10~12 X T T X (10" 2 ) 2 /10 - 3
= 16.68 pF.
(1.71)
■
EXAMPLE 1.8 The current through a 1 /xF capacitor is shown in Figure 1.16.
Assuming that the initial capacitor voltage is v(0) = 0, plot the instantaneous power
and electric energy stored in the capacitor.
The current indicated in the figure can be expressed as:
2
0 < t <2 ms,
2 < t < 3 ms,
0
3 < t < 4 ms,
i(t) (mA) - { - 2
t - 6 4 < t < 6 ms,
2
t > 6 ms.
(1.72)
We can obtain the capacitor voltage by integrating [see equation (1.62)]:
v(0 (V)
It
4
4 - lit - 3)
(i - 6 ) 7 2
2(f - 6)
0 < t < 2 ms,
2 < í < 3 ms,
3<i<4ms,
4 < t < 6 ms,
r > 6 ms.
The power is found by using (1.72) and (1.73) in equation (1.19):
(1.73)
23
1.6. The Capacitor
3
2
<1
c
1o
I"
1
2
;I
4
5 / 6
7
8
9
10
-2
-3 -
Time (ms)
Figure 1.16: The capacitor current.
0 < t < 2 ms,
2 < í < 3 ms,
3<i<4ms,
4 < f < 6 ms,
í >6ms
' At
0
pit) (mW) = I - 2 ( 1 0 - 2 0
(i - 6) 3 /2
At -24
(1.74)
and is plotted in Figure 1.17a. The total stored energy comes by integrating the power:
3
4
5
6
Time (ms)
Figure 1.17: Physical quantities for the capacitor: (a) power and (b) energy.
24
Chapter 1. Basic Circuit Variables and Elements
0 < / < 2 ms,
It2
2<i<3ms,
8
w(i)(/U) = < (10 - 2i) 2 /2 3 < i < 4 m s ,
4 < í < 6 ms,
(/ - 6) 4 /8
(4/ - 24) 2 /8 / > 6 ms.
(1.75)
This result for the energy is shown in Figure 1.17b.
1.7 Ideal Independent Voltage and Current Sources
By definition, an ideal voltage source is a two-terminal element with the property
that the voltage across the terminals is specified at every instant in time. This voltage
does not depend on the current through the source. That is, any current in any direction
could possibly flow through the source. This current will be determined solely by the
circuit elements connected to this source and it can be computed according to the
rules which we will learn in the next chapter. Depending on the actual direction of
the current through the source, the voltage source can either provide power or absorb
it. The circuit notation for an ideal voltage source is given in Figure 1.18a. If the
specified voltage is a constant, we often refer to the source as a battery and use the
symbol shown in Figure 1.18b.
An ideal current source is by definition a two-terminal element with the property
that the current flowing through the device is specified at every instant in time.
This current does not depend on the voltage across the source. The voltage will be
determined solely by the circuit elements which are connected to this source and
it can be computed according to the rules which we will learn in the next chapter.
Depending on the actual polarity of the voltage across the source, the current source
can either provide or absorb power. The circuit notation for an ideal current source is
given in Figure 1.19.
All sources fall into two main categories: nonperiodic and periodic. The nonperiodic sources will be treated in Chapter 7 when we study transient analysis. For
Ô v.»
T
(a)
(b)
Figure 1.18: Circuit notation for an ideal voltage source: (a) general symbol and (b)
symbol for a battery.
25
1.8. Summary
O) ..»
Figure 1.19: Circuit notation for an ideal current source.
example, such a source could be described by exponential or polynomial time dependences. Examples of four different periodic source functions are given in Figure 1.20.
The ac or sinusoidal source will be the key source that we analyze in detail starting
in Chapter 3. The dc source can be treated as a special case of the ac source when
the period approaches infinity (and the frequency approaches zero). Finally, the more
complicated periodic sources like the ramp (sawtooth) source and the pulse (square
wave) are often encountered in applications.
v(t)
v(t)
(a)
v(t)
(b)
t
(c)
(d)
Figure 1.20: (a) dc source, (b) ac source, (c) ramp source, and (d) pulse source.
1.8
Summary
In this chapter we have first discussed the key circuit variables from an electromagnetic theory point of view. The convention for assigning the assumed (reference)
26
Chapter 1. Basic Circuit Variables and Elements
directions for voltages and currents in passive two-terminal elements has also been
presented.
We have then introduced the three principal passive two-terminal elements (the
resistor, the inductor, and the capacitor) and discussed the connections between the
circuit variables for each. There is a linear dependence between two of the circuit
variables for each element, with the proportionality constant being determined by the
geometry and the material composition of the element. For the resistor, the ratio of
voltage to current is the resistance R. For the inductor, the inductance L is the ratio
of flux linkages to the current. For the capacitor, the ratio of the charge to the voltage
is the capacitance C.
Of primary importance are the terminal relationships which describe the connections between voltage and current for each element. If we examine the terminal
relationships for capacitors and inductors from a purely mathematical standpoint,
we can see that these equations reveal some duality. That is, the equation for the
inductor is mathematically identical to that of the capacitor, with the roles of voltage
and current being reversed (and interchanging C and L). This parallelism is useful to
note and may prove helpful in remembering these equations.
Furthermore, we have introduced the ideal independent voltage and current
sources. We have also briefly mentioned the types of source functions that we will
investigate in the upcoming chapters. Because a change in current through an ideal
voltage source doesn't affect the voltage, we often say that an ideal voltage source
has zero internal resistance. Likewise, we say that an ideal current source has infinite
internal resistance or zero internal conductance because a change in the applied
voltage does not affect the current.
Finally, we point out that all five elements that have been studied so far are ideal
elements. Our treatment in this chapter of resistors, inductors, capacitors, and sources
ignores some small parameters and other deviations from ideal behavior which occur
for real-world circuit elements. For example, some capacitors and resistors have
measurable inductances and most inductors have some resistance. Furthermore, many
of the devices exhibit nonlinear effects when some parameters become large. Nonideal
circuit elements and their models as well as nonlinear resistors will be treated in
Chapter 5.
1.9
Problems
1. Convert the following derived units to the four basic SI units (m, kg, s, A): (a) joules,
(b) coulombs, (c) volts, (d) webers.
2. Express the following quantities in terms of multiplying factors: (a) 2.54 X 101 x electrons,
(b) -5.63 X 1(T14 C, (c) 0.0035 Wb, (d) 0.00000915 A.
3. Find the number of electrons required to generate a total charge of: (a) -1.6 X 10~7 C,
(b) -1.17 X 10~17C, (c) 12/i,C,(d) -3.73 pC.
27
1.9. Problems
4.
If a —3 ja A current flows through a surface S, how many electrons pass through S in (a)
I s , (b)3jas, (c)53.4fs?
5.
If the current flowing through a surface S has the time variation shown in Figure PI-5,
plot the time dependence of net charge accumulating on the other side of S. Assume zero
initial charge.
3
4
5
Time (ms)
Figure Pl-5
6.
The total charge in a volume V as a function of time is shown in Figure PI-6. Plot the
time variation of electric current leaving V.
2
3
Time (ms)
Figure Pl-6
7.
The time dependences of the voltage across and the current through a two-terminal
electric device are shown in Figure PI-7. Plot the power flowing into the device and the
total stored energy as a function of time. Assume w(0) = 0.
Chapter 1. Basic Circuit Variables and Elements
(b)
c
CD
Ü
3
4
5
Time (s)
Figure Pl-7
The voltage developed across a 12-turn coil of wire is shown in Figure Pl-8. Plot the
time variation of the magnetic flux linking one turn of the device. Assume zero initial
flux.
4
6
Time (|is)
Figure Pl-8
29
1.9. Problems
-°
£
¿0
x
_3
LL -1 h
/
_3l
/
.
0
1
/
/
/
/
\
\
,
2
,
3
,
4
1
5
\
\
\
\
1
6
/
1
7
/
1
8
9
/
/
/
/
.
1
10
Time (JIS)
Figure Pl-9
9.
The total magnetic flux linking a circuit is shown in Figure Pl-9 as a function of time.
Plot the resulting voltage developed across the device.
10.
The current through a circuit is given by i(t) = (3 + t) A. Determine the flux (at time t)
that must be linked through the device if the power consumed is a constant 1 W. Assume
zero initial flux at f = 0.
11.
If the current through a closed surface is given by i(i) = 1 + 5t2 + cos(3i) A, what is
the expression for the charge accumulating in the volume? Assume q(0) = 0.
12.
If the voltage across a two-terminal device is v(t) = te~l V and the current through it is
i(t) = (1 — f)e~x A, plot the energy stored in the device as a function of time (for t > 0).
Assume w(0) = 0.
13.
Calculate the nominal resistance of the following automotive hardware (which run off a
12 V battery): (a) a 12 W dome light, (b) a 50 W horn, and (c) a 300 W headlight.
14.
To reach a new customer, the local electric company must transport five miles a current
of i(t) = 200cos(377i) amperes. What is the required diameter of a copper wire that
would keep the average power transmission losses below 2 kW? The conductivity of
copper is ( j = 5 . 8 X 107 (flm)" 1 .
15.
Assume you have a carbon composition resistor with a conductivity of or = 2.0 X
103(flm)~1. If you use this material to make 2 cm long resistors, what is the required
diameter to produce a resistance of: (a) 4 mil, (b) 1 Í1, (c) 1 kil, and (d) 1 Mil? Which
of these designs do you think are practical ones? How might you actually construct the
impractical designs, if any?
16.
Calculate the resistance value for each of the following cases of a circular-cross-section
wire:
Resistor
Conductance (ílm) !
Length (cm)
Radius (mm)
a
b
c
5
2 X 103
5.8 X 107
1
2
100
2
1
.003
30
Chapter 1. Basic Circuit Variables and Elements
17.
Find the average power absorbed by a 20 Q resistor when a current of i{t) = 3/4 sin(3f)
A is applied.
18.
A 10 ¿¿H inductor is driven by a current i(t) = 30cos(500i) mA. Calculate the power
supplied to the inductor.
19.
The voltage across a 2 mH inductor is shown in Figure PI-19. Plot the current flowing
through the inductor assuming an initial current of 10 mA.
4
5
Time (ms)
6
7
Figure Pl-19
20.
The current flowing through a 39 fiH inductor is shown in Figure PI-20. Plot the power
in the inductor as a function of time.
3
4
5
6
Time (ms)
Figure Pl-20
21.
The current through a 91 /xH inductor is given by i(t) = 7e~t/20 A. Find expressions
for the flux linked by the inductor and the voltage across the inductor. What is the total
charge that passes through the inductor from t = 0 to t = oo ? Assume q(0) = 0.
31
2.9. Problems
22.
23.
Calculate the capacitance for each of the following cases of a parallel plate capacitor:
Capacitor
Relative dielectric constant er
Plate gap (mm)
Radius (cm)
a
b
c
1
2.3
32.4
2
.1
.003
1
4
10
The voltage across a 10 /xF capacitor is shown in Figure PI-23. Plot the power supplied
to the capacitor as a function of time.
4
6
Time (JIS)
Figure Pl-23
24.
A 47 pF capacitor is driven by a voltage v(t) = 1 + 6e 2i V. Calculate the power supplied
to the capacitor.
25.
The current through a 20 /xF capacitor is given by i(i) = 2e - 3 r A for r > 0. Find
expressions for the voltage, accumulated charge, instantaneous power, and electric energy
stored in the capacitor. Assume zero initial charge.
26.
The voltage across a 5 nF capacitor is given by v(t) = 1 + / + 3t2 V for / > 0. Find
expressions for the current, accumulated charge, instantaneous power, and electric energy
stored in the capacitor.
27.
The voltage across a 5 mH inductor is given by v(t) = 3t2 + 5t4 V for t > 0. Find
expressions for the current, magnetic flux, instantaneous power, and electric energy
stored in the inductor. Let /(0) = 0.
28.
The current through a 33 /xH inductor is given by i(t) = 20i cos(30i) mA for t > 0.
Find expressions for the voltage, magnetic flux, instantaneous power, and electric energy
stored in the inductor.
32
Chapter 1. Basic Circuit Variables and Elements
29.
The current through a 1 kil resistor is given by i(t) — 2e~t sin(lOi) mA for t > 0. Find
expressions for the voltage across the resistor and the instantaneous power dissipated by
the resistor. What is the maximum power dissipated in the resistor?
30.
The voltage across a 1.4 fí resistor is given by v(t) = e _r/1 ° — e~t/xm V for t > 0. Find
expressions for the current through the resistor and the instantaneous power dissipated
by the resistor. Plot the power dissipated in the resistor.
Chapter 2
Kirchhoff's Laws
2.1
Introduction
What exactly is an electric circuit? So far circuit variables and circuit elements
have been discussed, but we still have not defined precisely what an electric circuit
is. Basically, an electric circuit is an interconnection of circuit elements. In this
definition two things are important: the circuit elements themselves and the way in
which they are connected. In circuit theory there exist two types of relationships
between currents and voltages that correspond to the two facets of electric circuits.
In the previous chapter, we derived the terminal relations which are determined by
the physical nature of the circuit elements. In this chapter we will present the rules
that govern the ramifications of the circuit interconnections. These relationships are
sometimes referred to as topological ones. Before presenting these rules we will first
have to characterize the topology (layout) of the circuit and introduce the terminology
related to various aspects of the circuit. The notion of the graph of the electric circuit
will be introduced as a way to describe the circuit interconnections. We will then
present the two fundamental laws of circuit analysis, known as Kirchhofes laws,
that will be used in conjunction with the terminal relations to determine the voltages
and currents everywhere in a circuit. Afterward, we will demonstrate a method for
assembling the proper number and types of equations that will guarantee unique
solutions for electric circuit problems. We will defer the discussion of the techniques
needed to solve these systems of equations to the following chapters. The chapter
will close with the discussion of resistive circuits which contain only sources and
resistors.
33
34
Chapter 2. Kirchhoff's Laws
¡s«
¡s«
(a)
(b)
Figure 2.1: Two example circuits with identical elements.
2.2
Circuit Topology
Consider the following two examples of electric circuits, shown in Figure 2.1. Although they are both made from the same set of elements they are different as circuits.
It is their interconnections that make the difference. As we will soon see, these differences will give rise to different voltages and currents in these two circuits. This
example emphasizes the importance of both aspects of the definition of an electric
circuit.
In order to characterize circuit interconnections, we must examine the definitions
and rules of circuit topology. There are several important terms that we will use in
our discussion of electric circuit topology. These terms are defined in the following
paragraphs.
A node of an electric circuit is a point where two or more elements are connected
together. For both circuits in Figure 2.1, there are four nodes. An example of a node is
the point where the inductor L¡ and resistors R2 and R^ come together. A node where
only two elements are connected together will be called a trivial node. It will be clear
from the subsequent discussion that trivial nodes can be easily excluded from the
circuit analysis.
A branch of an electric circuit is a two-terminal element. There are five branches
in each circuit of Figure 2.1, which represent the individual circuit elements.
A loop is defined as a set of branches that form a closed path with the property
that each node is encountered only once as the loop is traced. For planar circuits (i.e.,
circuits which can be drawn without any intersections) we can define meshes. They
form a subset of loops and have the property that they do not enclose any branches.
There are only three different loops and two different meshes in Figure 2.1. The
leftmost mesh of circuit (a) includes the inductor L¡, resistor R2, and the current
source. The other mesh traces through the inductor L4 and the two resistors. The last
35
2.2. Circuit Topology
Figure 2.2: Oriented graph of previous circuits.
loop traces the outermost elements of the circuit and is not a mesh because it encloses
the resistor 7?2.
To characterize the connectivity of a circuit, we introduce the notion of a circuit
graph. In a circuit graph we ignore the physical nature of the elements and represent
only their interconnections. To achieve this, each branch is shown as a line. If we
introduce directions which coincide with the reference directions of currents, we
arrive at what is called a directed or oriented graph. Figure 2.2 shows the oriented
graph that corresponds to the previous circuits of Figure 2.1.
A graph tree of an electric circuit is a subset of the branches of the graph with
the property that all nodes of the circuit are connected together, but there are no
closed loops formed by these branches. You can obtain a graph tree from the graph
of a circuit by removing branches until there are no closed loops. The graph trees are
not unique; two examples which correspond to the circuits of Figure 2.1 are given in
Figure 2.3.
A cut set is a subset of the branches of a circuit with the following two properties:
(1) if you remove all the branches of the cut set, the circuit will be divided into two
disconnected parts; (2) if you then place any one of the removed branches back into
the circuit, it is no longer disconnected. Cut sets are also not unique. For circuit (b)
in Figure 2.1, one cut set includes the two resistors and the inductor L4 while another
cut set contains the inductor L\9 the resistor R2, and the current source.
Figure 2.3: Two examples of graph trees.
36
Chapter 2. Kirchhoff's Laws
23
Kirchhoff 's Laws
2.3.1
Kirchhoff 's Current Law
Kirchhoff's current law is generally abbreviated as KCL. KCL states that the algebraic
sum of electric currents at any node of an electric circuit is equal to zero at every
instant of time. The term "algebraic sum" implies that some currents are taken with
positive signs while others are taken with negative signs. It is by convention that
we assign positive signs to the currents entering the node and negative signs to the
currents leaving the node. Expressed mathematically, KCL is written as
5>(0 = 0.
(2.1)
k
In circuit theory, KCL is considered to be an axiom or postulate. It is to be
accepted on faith and considered as a mathematical generalization of numerous
observations and experimental data. However, it should be noted that KCL can be
proved by using electromagnetic field theory, namely, the principle of continuity of
electric current which was described in the last chapter.
As an example, consider the node shown in Figure 2.4. The currents i\ and ¿3 are
entering the node while the remaining currents are leaving the node. An application
of equation (2.1) results in the expression: i\ — i2 + h ~ h ~~ h ~ 0.
With the aid of the principle of continuity of electric current, we can also see that
KCL applies to cut sets of a circuit. In this case, KCL would state that the algebraic
sum of currents for any cut set is equal to zero. However, in most cases discussed in
this text it will be sufficient to consider only KCL for nodes.
EXAMPLE 2.1 Consider the directed graph shown in Figure 2.5. KCL for node A
requires that i\ + i2 — U = 0, KCL for node B requires that —i\ — z'3 = 0, KCL
for node C yields z3 + i5 — i2 = 0, and KCL for node D yields i6 + /8 — i5 = 0.
Consider the surface Si which cuts the circuit into two distinct parts via the fourth,
sixth, and eighth branches of the circuit. These three branches constitute a cut set.
Figure 2.4: A node with currents entering and leaving.
37
23. Kirchhofes Laws
Figure 2.5: KCL example.
The corresponding KCL equation is U — k — h — 0. The minus signs in front of
the latter two terms arise because the currents from those branches are leaving the
volume inside S\. KCL yields ¿5 — ¿ 4 = 0 for the cut set generated by the surface
S2 indicated in the figure. There are two interesting points to note here. First, we
can arrive at the KCL equation for a cut set by summing the KCL equations for the
nodes inside the cut set. For the surface S2> we sum the KCL equations for nodes A,
B, and C and get i5 — /4 = 0 as claimed. Second, note that the negative sum of the
two cut set equations is equal to the KCL equation for node D (the only node which
is not contained within a cut set). This example clearly demonstrates that the cut set
equations follow from the node equations (and vice versa) and leads us to the notion
of independent equations, which will be discussed later in this chapter.
■
2.3.2
Kirchhoff 's Voltage Law
Kirchhoff 's voltage law is the second fundamental axiom of circuit theory and is
often abbreviated as KVL. KVL is applied to voltages in loops and states that the
algebraic sum of branch voltages around any loop of an electric circuit is equal to
zero at every instant of time. Mathematically, it can be written as follows:
£ > * ( ' ) = 0.
(2.2)
k
To actually write KVL equations, we shall need the following rule for voltage
polarities. We start by introducing reference voltage polarities for each branch. (Remember that these reference voltages are coordinated with the reference currents for
the passive elements.) Then, we trace each loop in an arbitrary direction (it is helpful
to trace every loop in the same direction—so let us always agree to go clockwise
in this text). If, while tracing through the loop, we enter the "plus" terminal of an
element and exit its "minus" terminal, then we take the voltage across this element
with a positive sign. If, on the other hand, we enter the negative terminal while tracing
the loop, we take the corresponding voltage with a negative sign.
To demonstrate the above rule, consider the example circuit shown in Figure 2.6
with the loops already labeled (keep in mind that these are not the only possible loops
38
Chapter 2. Kirchhoffs Laws
Figure 2.6: Example circuit with four loops shown as I, II, III, IV.
for this circuit). KVL for loop I in Figure 2.6 yields —vi + v3 + v2 = 0. Likewise,
for loop III we will get — V5 — v6 + v9 = 0 from KVL.
Since for passive elements, reference polarities for voltage and reference directions for current are coordinated, we can formulate the rule for determining the signs
of branch voltages in a KVL equation in terms of the reference current directions.
Namely, if the tracing direction coincides with the reference current direction, the
branch voltage is taken with a positive sign in the KVL equation. Otherwise, the
branch voltage requires a minus sign in the KVL equation. In loop I, the tracing
direction coincides with reference directions of ¿2 and ¿3 but is opposite to ¿i, so we
would get — vi + V3 + V2 = 0 as we must. Because the reference current directions
and voltage polarities are not necessarily coordinated for sources, one should always
use the previous rule for voltage polarities to evaluate the signs of the voltages across
the sources in the KVL equations.
EXAMPLE 2.2 Consider the circuit shown in Figure 2.7. Find all of the unknown
voltages and currents.
Solution: KVL for loops (I)-(IV) yield, respectively:
- 1 2 + V! +2.5 = 0,
-2.5 + v3 + v2 = 0,
4 - 5 - v3 = 0,
- v 4 + 2 - 4 - 0.
The first equation gives v\ = 9.5 V; the third yields v3 = —IV; and the fourth
requires v4 = - 2 V. Plugging v3 into the second equation results in v2 = 3.5 V.
2.4. Linearly Independent Kirchhoff Equations
39
Figure 2.7: Example circuit with unknown currents and voltages.
KCL for nodes B, C, and D, respectively, yield:
iA + 3.5 - 4.75 - 0,
ÍB ~ 3.5 — ic = 0,
2 + 2 +ic = 0.
The first equation requires iA = 1.25 A and the third implies that ic = — 4 A.
Inserting this value into the second equation completes the solution of this problem
by yielding i# = — 1 / 2 A. KCL at node A can be used to check for algebraic mistakes.
The resulting equation is
4.75 - 1.25 A - (-.5 A ) - 2 A - 2 A = 0,
which checks out correctly.
2.4
Linearly Independent Kirchhoff Equations
2.4.1
General Circuits
■
By writing equations according to Kirchhoff's laws, we end up with a system of linear
algebraic equations. The task is to guarantee that the resulting Kirchhoff equations
are linearly independent (i.e., solvable and without redundant information). We will
define linearly independent equations as those in which each subsequent equation
contains a variable (an unknown) which the previous equations in the system do
not contain. Therefore, each equation cannot be obtained from the previous ones. In
this sense, each equation contains essentially new information. The given definition
of linearly independent equations is a very special (particular) case of the general
definition of linearly independent equations used in linear algebra. However, the
40
Chapter 2. Kirchhoff's Laws
above definition will be quite sufficient in order to find a way to write linearly
independent Kirchhoff equations.
Consider how linearly independent KCL equations can be written. To be specific,
we will analyze the electric circuit shown in Figure 2.8. For nodes A, B, C, and D
we, respectively, obtain:
hit) + i2(t) - hit) = 0,
(2.3)
hit) - hit) - i5it) = 0,
(2.4)
kit) + hit) - hit) = 0,
(2.5)
hit) - hit) - hit) = 0.
(2.6)
The equations for nodes A, B, and C are linearly independent because each subsequent
equation has a variable that the previous ones do not contain. However, the equation
at node D is linearly dependent. Indeed, if we sum up the first three equations and
multiply the resulting sum by —1, we can produce the equation for node D. Thus,
this equation can be formed as a linear combination of the others, which means it is
not linearly independent. From this example we can see that, in general, if we have
n nodes we can write n — 1 linearly independent Kirchhoff current equations. This
implies a general approach to writing linearly independent KCL equations: write one
equation for each node except the last.
Consider the same electric circuit redrawn in Figure 2.9, and let us write the
Kirchhoff voltage equations for the loops I, II, III, and IV, respectively:
v i ( f ) - v 2 ( 0 = 0,
(2.7)
v2(f) + v3(i) + v4(i) + v6(f) = 0,
(2.8)
v5(0 - nit) = 0,
(2.9)
vi(í) + v3(í) + v5(í) + V6(í) = 0.
(2.10)
A ¡ 3 (t )
o >' I
MM
i 2 (t)f
B U(t)
o—>■
(t)
's «
D
Figure 2.8: Sample circuit for KCL equations.
41
2.4. Linearly Independent Kirchhoff Equations
Figure 2.9: Sample circuit for KVL equations.
We see that the voltage equation for loop IV is not linearly independent because it
can be obtained by summing up voltage equations for loops I, II, and III.
Since there are many possible loops for any given circuit, determining how many
and which loops should be traced in order to find linearly independent KVL equations
is not entirely obvious. One method which will always produce the correct number
of linearly independent KVL equations is based on the use of a graph tree. Figure
2.10 shows a graph tree of the circuit shown in Figure 2.9. All four nodes A, B, C, D
are connected together, and no loops are formed—exactly as the graph tree definition
states. Now, as the figure shows, if a circuit has n nodes, then its graph tree will have
n — 1 branches. This holds for all circuits and can be proved very simply. To make a
graph tree of an arbitrary circuit, we first connect two nodes—in order to do this only
one branch is necessary. There are now n — 2 nodes left to connect. To connect every
other node to the tree only one branch has to be used, resulting inn— \ branches for
the graph tree:
1 + ( / i - 2 ) = n-
1.
(2.11)
Graph trees can be used to determine linearly independent KVL equations in a
simple manner. Just add the missing branches and create loops (remember from the
definition of a graph tree that there were no loops before). By adding a new branch
we create a new loop and a new KVL equation, which will always contain a new
A
Q
Ô-
D
C
Figure 2.10: Graph tree of the previous circuit.
42
Chapter 2. Kirchhoff's Laws
A
-o
B
-Q
11
+5
I
I
L.
Figure 2.11: The same graph tree, with all linearly independent loops formed by
using dashed branches.
variable in comparison with previously written KVL equations. This new variable is
the voltage across the added branch. Consider Figure 2.11 as an example. By adding
branch 1, we create loop I and the corresponding KVL equation (2.7) contains v\(t).
By adding branch 3, we create loop II and the corresponding KVL equation (2.8)
contains v^(t), which is not present in the previous KVL equation. Next, by adding
branch 5, we create loop III, and the corresponding KVL equation (2.9) contains
v 5 (0, which is not present in the previous equations. If the total number of branches
is b, then the total number of linearly independent Kirchhoff voltage equations will
be equal to the total number of added branches, which is b — (n — 1).
Although by using graph trees we can produce linearly independent KVL equations for any circuit, this approach can be somewhat overcomplicated for most of the
simple circuits encountered in practice. A much easier way to write the linearly independent KVL equations for planar circuits (circuits which can be drawn out without
any lines crossing) is to write one KVL equation for each mesh. In the previously
illustrated example, the circuit was planar, and the graph tree approach produced the
loops which are meshes. In a way, the graph tree technique can be used to justify the
mesh technique for writing KVL equations; however, we shall not delve further into
this matter.
Thus, by using KVL and KCL, one arrives at b linearly independent equations:
{n — 1) linearly independent KCL equations and b — (n — 1) linearly independent
KVL equations. However, we have 2b unknowns, which are the branch currents and
branch voltages. Thus, b additional equations are needed to solve the problem. Fortunately, these equations can be written by using the terminal relationships between
voltages and currents which we derived in the previous chapter for the various circuit
elements. These relationships can be written for resistors, inductors, and capacitors,
respectively, as follows:
vAt) = Rk,iki{t)9
dikn(t)
dt
dvkm(t)
*V"(0 — Q dt
Vkn(t) = Lk
(2.12)
(2.13)
(2.14)
43
2.4. Linearly Independent Kirchhoff Equations
The variables on the right-hand side of the above equations can be used as state
variables in the sense that we can write the KVL and KCL equations in terms of
these variables. In order to achieve this, we substitute expressions (2.12) and (2.13)
in the appropriate KVL equation (2.2) and expression (2.14) in the appropriate KCL
equation (2.1). Once this is done, the result will be a system of linear differential
equations which is written in terms of currents through the resistors and inductors and
in terms of voltages across the capacitors. In this way, we reduce the total number of
unknowns to b, which is the total number of equations. These differential equations
should be supplemented by the initial conditions for ikn(t) and v¿///(í), respectively,
*'*"(0+) = 4"(0-),
(2.15)
v*'"(0+) = v*///(0_).
(2.16)
These initial conditions follow from the principles of continuity of electric currents
through inductors and voltages across capacitors.
EXAMPLE 2.3 We shall illustrate the above discussion with an example of an
electric circuit, shown in Figure 2.12. Note that the source voltage dependence vs(t)
is assumed to be specified. We want to write b linearly independent equations with
b unknowns. The first step is to specify the nodes and to assign reference directions.
Next, we shall write Kirchhoff equations. By using KCL, we see that the current
equations for nodes A, B, and C, respectively, are:
hit) + hit) ~ hit) = 0,
(2.17)
h{t) - i4(t) - hit) = 0,
(2.18)
hit) + hit) - i6it) = 0.
(2.19)
Notice that we do not write an equation for node D because this equation would not
be linearly independent. Notice also that there is a trivial node E that connects only
the capacitor and the negative terminal of the voltage source. Rather than add another
Figure 2.12: Example circuit, with nodes and reference directions shown.
44
Chapter 2. Kirchhoff's Laws
equation and state variable (which for this example would be the current is through
the source), we simply replace is by i\ wherever it occurs.
We next write the voltage equations by using KVL for the three meshes in the
circuit (since the circuit is planar, there is no need for a graph tree). The equations
for meshes I, II, and III are
v i ( 0 - v J ( i ) - v 2 ( i ) = 0,
(2.20)
v 2 (0 + v3(i) + v4(t) + v6(i) = 0,
(2.21)
-v 4 (i) + v5(i) = 0.
(2.22)
We chose the state variables to be v\{t\ Í2(t), /3(f), U(0> h(0> a n d v6(t). We can now
supplement the previous Kirchhoff equations by the terminal relationships for the
circuit elements:
dvi(t)
at
(2.23)
v2(f) = R2Í2(0,
(2.24)
(2.25)
L
V 3 (í) =
3—J—>
at
v4(r) = R4i4(t),
(2.26)
j
di5{t)
(t,
v 5 (i) =
L5—7—,
(2.27)
at
. _
i6(t)
=
r
dv6(t)
C6——.
(2.28)
substituting (2.23) and (2.28) into the KCL equations, we find:
dvi(f)
.
C\ —j— + i2(t) - iÁt)
z
= 0,
at
hit) - i4(t) - i5(t) --= 0,
(2.29)
■-= 0.
(2.31)
w) + hit) - C6
dt
(2.30)
By substituting (2.24)-(2.27) into the KVL equations, we obtain:
Viit) - R2i2it) = vsit),
Rihit) + L3 - ^
at
+ R4i4it) + v6it) = 0,
(2.32)
(2.33)
-R4i4(t) + L 5 ^ = 0 .
(2.34)
dt
The equations (2.29)-(2.34) form a system of six linear differential and algebraic
equations with six unknowns. These equations should be complemented with initial
conditions which come from the continuity conditions for capacitors and inductors:
2.4. Linearly Independent Kirchhoff Equations
45
v 1 (0 + ) = v1(0_),
(2.35)
i 3 (0 + ) - i 3 (0-),
(2.36)
i 5 (0 + ) = i 5 (0-),
(2.37)
v 6 (0 + ) = v6(0_).
(2.38)
The circuit variables in the right-hand side of equations (2.35)-(2.38) are equal to
zero if there was no electromagnetic process in the electric circuit prior to the time
t = 0. Thus, in this case we will have zero initial conditions. If there was some
electromagnetic process in the circuit before the time í = 0, then the values of the
above circuit variables should be determined from the analysis of this process.
■
Differential equations (2.29)-(2.34) together with initial conditions (2.35)-(2.38)
form the initial value problem. The essence of this problem is to find the solution to
the above differential equations which satisfies the prescribed initial conditions. If
this solution is somehow found, then by using expressions (2.23)-(2.28) we can also
compute the currents through the capacitors and voltages across the inductors and
resistors. In other words, we can find all voltages and currents in the electric circuit.
Initial value problems for ordinary differential equations are studied in mathematics and various techniques for their solutions have been developed. These techniques
will be extensively used later in the text and they constitute an important component
of the overall circuit analysis. However, it is not the goal of this discussion to present
here the complete analysis of electric circuits. It is rather to demonstrate that Kirchhoff equations (2.1) and (2.2) together with terminal relationships (2.12)-(2.14) and
initial conditions (2.15) and (2.16) allow one to reduce the analysis of electric circuits
to well-studied problems in mathematics, i.e., initial value problems.
EXAMPLE 2.4 The derivation of the differential equations for the circuit shown in
Figure 2.12 has been quite general in nature. In some cases this derivation can be
simplified by exploiting a particular structure of an electric circuit. To demonstrate
this, consider the electric circuit shown in Figure 2.13.
This circuit has four passive elements: two inductors and two resistors. For this
reason, we may expect to end up with four coupled linear differential and algebraic
equations with respect to the currents through the inductors and resistors. However,
the analysis of this circuit can be significantly simplified by observing that nodes B
and D are trivial (i.e., only two elements are connected by these nodes). By applying
KCL to these nodes, we obtain:
hit) = is(t),
(2.39)
¿3(0 = M(0.
(2.40)
Equation (2.39) tells us that the current i\{t) is essentially known because the source
current is(t) is given. Equation (2.40) further reduces the number of unknowns because
/3(f) can be replaced by ¿4(i). Thus, we essentially have only two unknown currents:
46
Chapter 2. Kirchhoff's Laws
Figure 2.13: The electric circuit considered in Example 2.4.
¿2(0 and /4O). To derive the coupled differential and algebraic equations for these
currents, we shall apply KCL to node A:
and KVL to mesh I:
¿i(i) - i2(t) - h{t) = 0,
(2.41)
v3(i) + v4(i) - v2(i) = 0.
(2.42)
Next, we shall use the terminal relationships:
v2(t) = R2i2(t),
(2.43)
v3(t) = R3i3(t),
(2.44)
di^it)
dt
(2.45)
v 4 (0 = L4
By substituting (2.43)-(2.45) into (2.42) and by using (2.39) and (2.40) in (2.41) and
(2.42) we end up with the following coupled equations for i2{t) and i4(t):
i2(t) + i4(t) = is{t),
(2.46)
R3U(t) + Ud-^ß- - R2i2(t) = 0.
dt
(2.47)
Equations (2.46) and (2.47) should be complemented with the initial condition
i 4 (0 + ) = ¿4(0-)
(2.48)
which follows from the principle of continuity of electric current through an inductor.
Equations (2.46) and (2.47) together with initial condition (2.48) constitute the
initial value problem. If this problem is solved, then by using (2.40) and (2.43)(2.45) we can determine i3(t) and all voltages. After all branch voltages and currents
in the circuit are determined, the voltage vx(t) across the current source can be found.
2.4. Linearly Independent Kirchhoff Equations
47
Indeed, by applying KVL to mesh II, we arrive at:
vx(t) + v2(t) - vx(t) = 0,
(2.49)
v,(f) = v,(í) + v2(í).
(2.50)
which can be rewritten as:
By using (2.39) and the terminal relationship for Li, we obtain:
Vl(f)
= L i ^ .
at
By substituting (2.43) and (2.51) into (2.50), we derive:
(2.51)
vx(t) = L , ^ + R2i2(t),
(2.52)
at
which is the expression for the voltage across the current source in terms of branch
current i2, which is presumed to be found. It is important to note that the reference
polarity for vx(t) can be chosen arbitrarily. The actual polarity ofvx(t) at any instant of
time / can always be found from the sign of vx(t) at this instant of time and its reference
polarity. The reference polarity of vx(t) should not necessarily be coordinated with
the reference direction of the current source. However, the coordination of current
reference directions and voltage reference polarities is very important for passive
elements. This is because their terminal relationships (2.12), (2.13), and (2.14) have
been derived for coordinated reference directions. For current and voltage sources,
on the other hand, terminal relationships are specified by the expressions for is(f) and
v5(i), respectively. Thus, these terminal relationships do not depend on coordination
of reference directions.
For this reason, one usually chooses a reference current for a voltage source to be
in the expected direction of the actual current, which is typically out of the positive
terminal. An analogous selection is made for a reference voltage polarity of a current
source.
■
2.4.2
Resistive Circuits
Next, we shall consider purely resistive circuits. These circuits contain only sources
and resistors. We start with KCL and KVL equations (2.1) and (2.2), respectively,
and rewrite them in a slightly different form by moving all known driving (voltage
and current) sources to the right-hand side of these equations:
5 > ( 0 = -J>*(i),
k
(2.53)
k
X>(') = -J2vsk(t\
(2.54)
48
Chapter 2. Kirchhoff'$ Laws
We can always write n—\ linearly independent KCL equation (2.53) and b — (n — 1)
linearly independent KVL equations (2.54). Thus, altogether we can write b linearly
independent equations. However, we have 2b unknowns, which are currents through
and voltages across the resistors. To circumvent this difficulty, we shall use the
terminal relationships
v*(f) = Rkik(t)
(2.55)
for the resistors and substitute them in the KVL equation (2.54). In this way, we
arrive at the b linearly independent algebraic equations
J^ik(t)= ~5>*(0,
k
(2.56)
k
Y^ik(t)Rk = - £ % s k ( i )
k
(2.57)
k
with respect to b unknown currents ik(t). Thus, we can see that the analysis of purely
resistive circuits is reduced to the solution of simultaneous algebraic equations rather
than to the solution of differential equations. This clearly suggests that calculus and
differential equations enter circuit analysis when energy storage elements are present.
We shall illustrate the application of the equations (2.56) and (2.57) to the analysis
of resistive circuits by the following two examples.
EXAMPLE 2.5 Consider the electric circuit shown in Figure 2.14. This circuit has
three resistors and we want to write three linearly independent equations for the
currents through these resistors. We first specify the nodes and introduce reference
directions. Next, we shall write KCL and KVL equations. This circuit has only two
nontrivial nodes, so only one KCL equation is linearly independent:
M(i)-12(f)-¿3(0 = 0.
(2.58)
This circuit is planar and has two meshes: I and II. KVL equations (2.57) for these
meshes are:
Rih(t) + R2i2(t) = vsl(t\
(2.59)
- R2i2(t) + R3i3(t) = v,2(i).
(2.60)
R
1
i,(t)
Ut)
rAAAA^-f^
R
3
Figure 2.14: Resistive circuit with three resistors.
49
2.4. Linearly Independent Kirchhoff Equations
Thus, we end up with three linear algebraic equations, which is what is expected
since there are only three resistors in the circuit. Systems of linear simultaneous
equations can be solved in a variety of ways. For small systems of equations, the
choice of solution methods is debatable. However, for large systems of equations the
Gaussian elimination technique is a very attractive method of choice. This method is
presented in Appendix B.
As far as equations (2.58)-(2.60) are concerned, the attractive method of solution
is the method of substitutions. Indeed, from (2.59) and (2.60) we can find i\{t) and
13(f) in terms of Í2(t):
hit) = -iiit)—
+ ——,
R
+, —s2(t)
—.
• ,,x
• ,f\ 2
l3Í0
= llit)—
^3
(2.61)
V
(2.62)
^3
By substituting (2.61) and (2.62) into (2.58), we end up with the following equation
for i2(t):
.
R2
.
R2
. ,.
v sl (f)
vs2(t)
- i2(t)— - i2it)— - i2(t) = — - — + — — ,
(2.63)
from which we find:
■vS2(t)Ri +
vsl(t)R3
(2.64)
7?l/?2 ~^~ ^1^3 ~^~ R2R3
After ¿2(0 is computed, expressions (2.61) and (2.62) can be used to calculate i\(t)
and h{t). This concludes the analysis of the above circuit.
■
hit)
EXAMPLE 2.6 Consider the resistive circuit shown in Figure 2.15. This circuit
has four nodes (A, B, C, and D) and three meshes (I, II, and III). For this reason,
one may expect to end up with the coupled linear algebraic equations for currents
i\(t), Í2(t), Í2>(i), Í4(t), ix(t) and voltage vx(t). However, the analysis of this circuit can
be simplified by observing that node C is a trivial one. By applying KCL to this node,
.(*)
vO Q
l
»m
(m) v«>(p
Figure 2.15: More complicated resistive circuit.
50
Chapter 2. Kirchhoff's Laws
we obtain:
i 4 (0 = /,(').
(2-65)
We can also observe that mesh I is a trivial mesh (which we define to be a mesh with
only two branches). By applying KVL to this mesh, we find:
vi(i) = v,(f),
(2.66)
= v,(i).
(2-67)
which can be rewritten as:
iiWi
From (2.67), we obtain:
hit) = ^ .
(2.68)
Equations (2.65) and (2.68) tell us that the currents hit) and ¿i(i) are essentially
known, which reduces the total number of unknown currents for which we have to
solve. By writing KCL equations for nodes A and B and KVL equation for mesh II,
we arrive at:
hit) ~ hit) - hit) = 0,
(2.69)
i2{t) - hit) + kit) = 0,
(2.70)
-ii(r)Äi + hit)R2 + i3(f)/?3 = 0.
(2.71)
By using (2.65) in (2.70), (2.67) in (2.71), and (2.68) in (2.69), we obtain the following
simultaneous linear algebraic equations for hit), hit), and ix(t)\
hit) - hit) = -£A
vÁt)
(2.72)
hit) - hit) = -hit),
(2.73)
hit)R2 + hit)R3 = Vj(i).
(2.74)
The above equations can be solved by the method of substitutions. Indeed, from
(2.73) we find:
¿3(0 = hit) + hit)-
(2.75)
By substituting (2.75) into (2.74), we arrive at the following equation for hit):
hit)R2 + h(t)R3 + hit)R3 = vsit),
(2.76)
which can be further transformed as
hit)[R2 + R3] = Vsit) - is(t)R3.
(2.77)
From the last equation we get:
.
12(0
=
vsjt) - isjt)R3
R2 + R3 •
(2 78)
-
51
2.5. Summary
By substituting (2.78) into (2.75), we find the expression for i3(t):
W> = * ' ) + ; f * * .
(2.79)
Finally, from (2.72) and (2.78), we derive:
. ,,.
v,(i)
v,(Q - ¿,(Q/?3
=
+
^
T~
jg 2 + ig 3 •
(2 80)
-
Now, we can also find the voltage vx(t) across the current source is(t). By applying
KVL to mesh III, we obtain:
vx(t) - i3(t)R3 - ÍA{Í)RA = O.
(2.81)
By using (2.65) and (2.79) in (2.81), we find:
Vx(0 = h(t)R4 +
P
This concludes the analysis of the above circuit.
2.5
, P
.
(2.82)
■
Summary
In this chapter we have stressed the importance of the interconnections of circuit
elements in determining the voltages and currents in electric circuits. Several key
topological terms have been introduced to characterize the connectivity of electric
circuits.
The two main relations for currents and voltages due to circuit topology have
been presented. They are Kirchhoff 's current and voltage laws. KCL states that the
algebraic sum of electric currents at any node of an electric circuit is equal to zero at
every instant of time. KVL tells us that the algebraic sum of branch voltages around
any loop of an electric circuit is equal to zero at every instant of time. By using KCL
we can write n — Í linearly independent equations of the form:
] T ik(t) = 0,
(2.83)
where n is the total number of nodes of the electric circuit.
Similarly, by using KVL we can write b — (n—\) linearly independent equations:
£ v * ( i ) = 0,
(2.84)
where b is the total number of branches of the electric circuit. In a planar circuit, the
number of independent KVL equations corresponds exactly to the number of meshes.
Thus, by using KVL and KCL, one arrives at b linearly independent equations.
However, these equations have 2b unknowns, which are the branch currents and
52
Chapter 2. Kirchhoff's Laws
branch voltages. To reduce the number of unknowns to ¿>, the terminal relationships
for resistors, inductors, and capacitors are used.
The chapter is closed by the discussion of resistive circuits which contain only
sources and resistors. It is shown that the analysis of resistive circuits is reduced to
the solution of simultaneous linear algebraic equations, and some examples of this
analysis are given.
It should be stressed that the Kirchhoff equations (2.83) and (2.84), together with
the terminal relationships (2.12)—(2.14) and the initial conditions (2.15) and (2.16),
form the mathematical foundation of electric circuit theory. In the following chapters
we will develop some powerful machinery that will help us to arrive at the solutions
to circuit problems. This machinery will always be based on the foundation laid out
in this chapter.
It is also clear from equations (2.29)-(2.38) that a circuit can be excited in
various ways: by the initial conditions in the circuit, by sources, or by both. If a
circuit is excited by initial conditions (such as some initial voltage across a capacitor
which is going to be discharged), the response will gradually decay with time and
the time evolution of circuit variables is called the transient response. If the circuit
is driven by an ac source, the process in the circuit will go on indefinitely and with
time the variation of circuit variables becomes repeatable and is called the steadystate response. In this text, the ac steady state is considered first in the next four
chapters and the analysis of transient responses is delayed until Chapter 7. It will be
shown in the next chapter that the basic circuit equations for the ac steady state are
mathematically similar to the basic equations (2.56) and (2.57) for resistive electric
circuits. This will allow us to treat the analysis of resistive circuits as a particular case
of the ac steady-state analysis.
2.6
Problems
1. Draw the directed graphs which correspond to the circuits shown in Figure P2-1.
fAAA/V^
(c)
Figure P2-1
2. Use your imagination and thefivebasic circuit elements to create circuits which have the
topology of the directed graphs shown in Figure P2-2.
53
2.6. Problems
3.
Clearly label the nodes in the circuit shown in Figure P2-3 and write KCL for each node.
4.
Clearly label the nodes in the circuit shown in Figure P2-4 and write KCL for each node.
5.
Choose three different cut sets of the elements in Figure P2-4. Write KCL for each cut
set.
6.
Choose three different cut sets of the elements in Figure P2-3. Write KCL for each cut
set.
*AAA/Vi
VWV
Figure P2-3
7.
Figure P2-4
Consider the circuit shown in Figure P2-7. The currents listed in each individual row of
the following table represent a different solution to the KCL equations. For each row of
currents, fill in the missing currents.
54
Chapter 2. Kirchhoff's Laws
h
il
h
¿4
h
h
h
h
i9
1
?
?
2
?
1
7
7
3
?
0
1
2
7
7
1
?
?
?
?
-1
1
-1
1
?
7
?
4
?
7
?
7
3
7
7
9
Figure P2-7
8. Determine the number of independent KCL equations for each circuit in Figure P2-1.
9. Write KVL for each loop indicated in the circuit shown in Figure P2-4.
10. Write KVL for each loop indicated in the circuit shown in Figure P2-3.
11.
Determine the number of independent KVL equations for each circuit shown in Figure
P2-1.
12. Consider the circuit shown in Figure P2-7. The voltages listed in each individual row of
the following table represent a different solution to the KVL equations. For each row of
voltages, fill in the missing voltages, (vi is the voltage across the component indicated
by z'i, etc.)
Vl
V2
^3
V4
V5
n
v7
V8
v9
1
1
?
-1
2
2
?
?
?
0
?
3
?
?
?
-2
4
2
?
1
-1
2
?
2
2
?
?
1
1
?
0
?
1
-1
?
?
55
2.6. Problems
13.
Draw three different graph trees for the circuit shown in Figure P2-3.
14. Draw three different graph trees for the circuit shown in Figure P2-4.
In Problems 15-20, derive the complete system of first-order differential equations which
describe the circuit shown in thefigureindicated. Clearly identify the number of branches, label
the nodes and meshes, and list the KCL equations, KVL equations, and terminal relationships
used in the derivation. Use v\ to denote the voltage drop across branch that has i\ flowing
through it, etc.
15.
Figure P2-15.
16.
Figure P2-16.
'1
R l
¡2
11,
L
"
ii
©v . R
Figure P2-15
17.
Figure P2-17.
Figure P2-17
Figure P2-16
18. Figure P2-18.
Figure P2-18
56
Chapter 2. Kirchhoff's Laws
19. Figure P2-19.
R.
R,
r>^AAA^f^AAA^rAAAA^
+ it
á)\
^0
Figure P2-19
20. Figure 2.1(b) (page 34).
In Problems 21-26, find the currents through any voltage sources and the voltages across any
current sources in the circuits shown in the figures indicated. Label nodes and meshes and
define reference directions when needed. Assume the passive sign convention for all elements.
21.
Figure P2-21.
0Vs
2R
Figure P2-21
22. Figure P2-22.
3Q
23.
3Q
Figure P2-22
Figure P2-23.
3a
3a
57
2.6. Problems
24.
Figure P2-24.
r V W W A A A A n
3 il
>
2Q
Ô3V
5V
1 Q
Figure P2-24
25.
Figure P2-25.
7 Q
>
5 Q
(P3VS2Q 3üS
2A
Figure P2-25
26.
Figure P2-26.
Ôv.
R,
R
LAA/VV-I—0
Figure P2-26
ô
Chapter 3
AC Steady State
3.1
Introduction
In this chapter, we shall begin our study of electric circuits driven by ac sources. Under
steady-state conditions, all voltages and currents in such circuits will be sinusoidal.
A special and very powerful analysis technique exists which exploits this fact. It is
known as the phasor technique and it uses complex numbers to represent sinusoidal
time-varying quantities. The main power of the phasor technique is that it reduces
the operations of calculus on sinusoidal quantities to simple algebraic operations on
complex numbers (phasors). As a result, the basic differential equations of electric
circuits discussed in the previous chapter are reduced to linear algebraic equations
with respect to phasors. This significantly simplifies the analysis of electric circuits
under ac steady-state conditions.
The phasor technique leads to the very important concept of impedance (and/or
admittance). The notions of impedance as well as phasors themselves are ubiquitous
in modern electrical engineering. For this reason, this text is structured in such a way
that the phasors and impedances are introduced very early in the book. It is believed
that this approach will allow students to become more familiar with these important
notions and firmly absorb the related material.
This chapter includes a brief discussion of the important subject of ac power. The
notions of root-mean-square (rms) values of voltage and current will be introduced
along with the definition of power factor. We will demonstrate how phasors can be
used to calculate average power and we will examine the conditions for maximum
power transfer to a load. This chapter will close with a generalization of the phasor
technique to complex frequency. This topic will be of special importance when we
study transients.
58
59
3.2. AC Quantities
3.2
AC Quantities
In this chapter we will consider the steady-state behavior of electric circuits driven
by ac (sinusoidal) voltage or current sources.
Any ac voltage can be expressed mathematically as
v(t) = Vm cos((x)t + 4>y).
(3.1)
In this equation Vm represents the peak or amplitude of the voltage, and co represents
the angular frequency. Angular frequency is related to frequency / by a factor of 277:
CO = 2 7 7 /
(3.2)
Frequency is defined as the reciprocal of the period:
(3.3)
The unit for frequency is s - 1 and is called Hz (hertz). For example, in the United
States, the frequency of the ac voltage in conventional power networks is 60 Hz,
which corresponds to an angular frequency of about 377 s _ 1 (or rad/s).
The initial phase of the voltage is 4>v. When time is equal to zero, the expression
for ac voltage becomes
V(0) = Vm COS 4>y.
(3.4)
This is the initial value of the voltage, which is determined by the amplitude and the
initial phase. The graphical representation of ac voltage is shown in Figure 3.1.
The expression for ac current is very similar and is given by:
i(t) = ImCOS(ù)t + (/)/),
(3.5)
where Im is the peak value of the current and <p¡ is its initial phase. From equations
(3.1) and (3.5) we see that if co is zero then the voltage and current expressions are
v(0)
v(t)
*
Figure 3.1: Graphical representation of ac voltage.
60
Chapter 3. AC Steady State
reduced to
v{t) = ymcosc^>v = constant
(3.6)
(3.7)
i(t) = Imcos4>i = constant.
These expressions yield constant values, and from this it is easy to deduce that dc
(nonvarying) voltages and currents are simply particular cases of ac (sinusoidal)
voltages and currents. As a result, dc analysis is a particular case of ac steady-state
analysis, and it will be treated as such in this text.
Most often the frequency has some fixed predetermined value. Since the frequency is fixed, peak values and the initial phases completely determine the ac
currents and voltages. For this reason, we would like to know the relationships between the peak values and initial phases of ac currents and voltages for every circuit
element that we have studied.
33
Amplitude and Phase Relationships
for Circuit Elements
We begin with a resistor. By using the expressions for ac voltage and ac current
and substituting them into Ohm's law, we can derive the relationships between peak
values and initial phases for a resistor:
v(t) = Vmcos(a)t + (f)Vl
(3.8)
i(t) = 7mcos(coi + <J>7),
(3.9)
fa).
(3.10)
Vm cos(cot + (f)V) = Ri(t) = RIm cos(cot +
From the last equation it follows that the peak and initial phase values of the current
and voltage should be related as follows:
Vm=RIm,
(3.11)
<t>v = */.
(3.12)
Therefore, it is evident that the voltage and current through a resistor have the same
initial phase angle; in other words, they are in phase. The graph of ac voltage and
current in Figure 3.2 illustrates this fact.
Figure 3.2: AC voltage and current in the case of a resistor.
3.3. Amplitude and Phase Relationships for Circuit Elements
61
Now, we consider the case of an inductor. Recall that the terminal relationship
between the voltage and current in the case of an inductor is:
di(t)
(3.13)
v(i) = L
dt
By using this equation, we shall find the desired relationships. First, we take the time
derivative of the current:
d
(I cos(a)t + (/)/)) = — ú)Imsm(ü)t + fa).
(3.14)
dt m
By substituting (3.1) and (3.14) into (3.13), we find:
Vm cos(ù)t + c/)y) = —LcoIm sin(coi + </>/).
(3.15)
By using the trigonometric identity
cos ( a + —- l = — sin ay
(3.16)
the last equation can be rewritten as follows:
Vm cos(coi + (f)V) = L(oIm cos l(ot + (f)i + —j.
(3.17)
From this equation we find that:
Vm = o)LImy
(3.18)
4>v = <Pi + 2".
(3.19)
Consequently, the ac voltage and current are not in phase, and the ac voltage across
the inductor leads the ac current through the inductor by ir/2. Another way to express
the peak value and phase relationships in an inductor is
l = YüL
(3.20)
(x)L
$1
=
77
</>V "
(3.21)
The last expression suggests that the current lags behind the voltage by TT/2. Figure
3.3 gives a graphical representation of ac voltage and current in the case of an inductor.
Figure 3.3: AC voltage and current in the case of an inductor.
62
Chapter 3. AC Steady State
Finally, we consider the case of a capacitor. Recall that the terminal relationship
between the current and voltage in the case of a capacitor is:
i(i) - C
dv(t)
dt
(3.22)
We first take the time derivative of the voltage:
d
(V cos(cof + </>y)) = — coVmsm((x)t + <f>v).
dt m
(3.23)
By substituting (3.23) and (3.5) into (3.22), we find:
Imcos(o)t + (/>/) = -Co)Vm sin(co/ + <j>v).
(3.24)
By making use of trigonometric identity (3.16), we transform (3.24) as follows:
Im cos(o)i + </>/) = CcoVm cos ( cot + 4>y + — ) .
(3.25)
From this expression we find that the peak values and phases are related by:
Im = (oCVm,
77
<f>! = (f)y + - ,
(3.26)
(3.27)
which can also be written as follows:
V
=
—
o)C
(3.28)
(3.29)
Thus, in the case of capacitors, we see that the ac voltage lags behind current by
77/2, or, in other words, the ac current leads the ac voltage by 77/2. Figure 3.4 gives
a graphical representation of ac voltage and current in the case of a capacitor.
EXAMPLE 3.1 Let's assume that we have a sinusoidal voltage source with a fixed
amplitude of Vm = 5 V and a fixed frequency of / = 1 MHz. What would the
Figure 3.4: AC voltage and current in the case of a capacitor.
63
3.4. Phasors
maximum current be through (a) a 5 Í1 resistor, (b) a 5 /xH inductor, and (c) a 5 /xF
capacitor?
Solution: (a) From (3.11) Im = Vm/R = 5 V/5 ft = 1 A, (b) from (3.20)
Im = Vm/(oL = 5 V/(2TT X 106 Hz X 5 X 1CT6 H) = 1/(2TT) A = 159 mA, (c)
from (3.26) lm = coCVm - 2TT X 106 Hz X 5 X 10~6 F X 5 V - 50TT A = 157 A U
Based on the previous discussion, we can conclude that ac voltages across
resistors, inductors, and capacitors result in ac currents, and vice versa. This fact
follows from the linearity of the terminal relationships for resistors, capacitors, and
inductors. Consequently, if all the voltage and current sources in a circuit are ac
sources with the same frequency, then voltages and currents everywhere in the circuit
are sinusoidal functions with that frequency. In the next few sections, we will present
a circuit analysis method based on this fact.
3.4
Phasors
In the previous chapter, we discussed KCL and KVL for electric circuits and observed
that they result in a system of b linear differential equations, where b is the number
of branches in the circuit. By using these differential equations, we can solve for the
currents and voltages in a circuit, and indeed, in some cases, it is the only method of
choice. However, when we deal with sinusoidal voltages and currents (this is the case
of ac steady state), there is a method which avoids solving a system of differential
equations, a task which can be very difficult and time consuming. This method is
called the phasor technique and its basic idea is to reduce differential equations for
circuit variables to algebraic equations for phasors.
A phasor is a complex number1 used for the representation of a sinusoidal
quantity. To start the discussion, we consider an arbitrary sinusoidal quantity:
g(t) = Gmcos(o)t + (/>).
(3.30)
We wish to represent this sinusoidal quantity by a complex number. To do this, we
make use of Euler's formula
eje = cos 0 + y sino
(3.31)
where j is \J — 1 (electrical engineers often use j instead of the usual symbol /
because the latter is used to represent currents). Notice that the right-hand side of
Euler's formula is a complex number whose real part is cos 9 and whose imaginary
part is sin 6. So it is clear that
Re (eje) = cos 0,
(3.32)
where Re means the real part of the complex number. From this equation, we can see
how to rewrite the sinusoidal quantity g(t) in complex form:
g{t) = Gm cos(œt + <£>) = Gm Re(^' (cor+ ^).
1
A brief review of complex numbers is included in Appendix A.
(3.33)
64
Chapter 3. AC Steady State
By using simple rules for exponents and complex numbers, we can rewrite the last
equation as:
g(t) = Gm Re ( ^ V n
(3.34)
g(t) = Re ( G , ^ V ' n
(3.35)
g(t) = Re (Ge^\
(3.36)
G = Gmej+.
(3.37)
where
Equation (3.37) is the expression for the phasor of g(t). (In this text, a phasor will
be denoted by a capital letter with a "hat.") Phasors can be completely characterized
by two quantities: the magnitude (absolute value) and the polar angle. According to
equation (3.37) the magnitude is Gm and the polar angle is 4>. The magnitude of the
phasor is equal to the peak value of the sinusoidal quantity, while the polar angle of
the phasor is equal to the initial phase of the sinusoidal quantity. It is important to
note that the phasor G does not depend on the frequency co of g(t), and, therefore, we
can conclude that phasors can be used only to describe sinusoidal quantities of the
same fixed frequency.
We can freely convert from phasor representation to sinusoidal representation
and vice versa using the following rules:
g(t) = Gm cos(cot + </>) — G = Gmej*,
G = Gmej(t> — g(t) = Re (Gejù)t) = Gm cos(atf + (¿>).
(3.38)
(3.39)
In ac steady-state analysis of electric circuits, the equations are written and
solved in terms of phasors, and then the final answers are converted back into the
time domain form by using (3.39).
Next, we consider some examples.
EXAMPLE 3.2 Given a sinusoidal voltage:
v(i) - 50 cos Í50t + f ) V,
(3.40)
find the expression for the phasor.
We know that the peak value of v(t) is equal to the magnitude of the phasor and
that the initial phase of v(i) is equal to the polar angle of the phasor. So, the expression
for the phasor is:
V = 5 0 ^ v / 6 = 50 icos ^ + j sin | ) - 25\/3 + 25; V,
(3.41)
where the phasor V has been written in algebraic form by using Euler's formula.
■
EXAMPLE 3 3 Given the phasor representation of the sinusoidal voltage:
V - 30 + 30; V,
find the expression for the corresponding ac voltage.
(3.42)
65
3.4. Phasors
We start by finding the magnitude (or absolute value) of the phasor from (3.42):
\V\ = \/30 2 + 302 = 30\/2 V.
(3.43)
We then find the polar angle:
tantf>v = | j = 1,
(3.44)
4>v = j .
(3.45)
Now we can write the phasor expression as
V = 30v 7 ^ 77 " 74 V.
(3.46)
Using equation (3.39) we can convert the phasor into a sinusoidal quantity:
v(f) = 3 0 ^ c o s (cot + j \ V.
(3.47)
We choose the angular frequency arbitrarily to be co because it will usually be specified
in the problem and we do not have to worry about it.
■
Now consider another form of a sinusoidal voltage:
v(t) = A cos cot - B sin cot
(3.48)
We want to find the phasor expression for v(t). This problem appears somewhat
different from the previous ones, and we do not yet have a formula which will
immediately give the answer. We start by observing that v(t) is written as a sum of
sine and cosine terms of the same frequency. It is known that such a sum can always
be reduced to the form:
v(i) = Vmcos(cot + 4>v)
(3.49)
in a simple manner. Equation (3.49) is in a form suitable to be converted into a phasor
using one of the previously derived formulas. To convert (3.48) into form (3.49), we
shall use the following transformation of (3.48):
v(f) = \l A2 + B2 [
cos cot -
.
\VA2+B2
Now if we introduce the notations:
.
VA2+B2
sin cot ) .
Vm - VA2 + B2,
cos^v =
VA 2 + B2
(3.50)
(3.51)
.
VA 2 + B2
sin 4>y =
J
(3.52)
=,
(3.53)
equation (3.50) will become:
v(i) = Vm(cos (f>y cos cot — sin <j>v sin cot).
(3.54)
66
Chapter 3. AC Steady State
By using the trigonometric identity
cos(a + jß) = cos a cos ß — sin a sin ß,
(3.55)
we can see that (3.50) is equivalent to
v(f) = Vm cos(cot + </>v).
(3.56)
Once the sinusoidal quantity is in this form, it is trivial to convert it into a phasor.
By repeating the same line of reasoning as before we can show that a sinusoidal
voltage
v(t) = A cos cot + B sin cot
(3.57)
v(f) = Vmcos((ot - <f>v)
(3.58)
can be reduced to
with the same expression as before for Vm and <f>y.
EXAMPLE 3,4 Consider the following problem. Express the voltage
v(t) = 40 cos cot - 40 sin out V
(3.59)
in phasor form.
Comparing equation (3.59) to equation (3.50) we identify A = 40 and B = 40.
Next, we must find the magnitude of the phasor from equation (3.51):
\V\ = v/402 + 402 = 4 0 v ^ V.
(3.60)
Finally, we need the polar angle, which we can determine in many ways, one of which
is to use equation (3.52)
COS(/)y =
40
1
■= — —T= ,
40 V 2
(3.61)
V2
4>v = \
(3.62)
Now we have all the information we need to express v{t) as either a "single" sinusoidal
quantity or a phasor:
v(i) = 40V2cos (iùt + - ) V,
(3.63)
V = 4 0 \ / 2 ^ V / 4 V.
3-5
(3.64)
Impedance and Admittance
In Section 3.3 we discussed the relationships between ac voltages and currents for the
basic two terminal elements. This time we shall express these relationships in phasor
form.
67
3.5. Impedance and Admittance
To find the relationship between the phasors of ac current and voltage in the case
of a resistor, we recall equation (3.10):
Vm cos(cot + <})y) = RIm cos((x)t + (/>/).
(3.65)
Converting this equation into phasor form yields:
VmeJ<h = RImej4>i,
(3.66)
V = RL
(3.67)
or written more succinctly:
We define the impedance of a circuit branch to be the ratio of the phasor voltage
across the branch to the current through the branch. We denote impedances with an
uppercase Z. The impedance for a resistor is simply equal to its resistance value:
(3.68)
Z = V/I = R.
Note that the units for impedance are ohms.
Now, consider the case of an inductor. We start with the sinusoidal relationship
(see (3.17)):
Vm COS(COÍ + 4>y) — (x)LIm COS (u)t+(f>i
+ —J.
(3.69)
Converting (3.69) into phasor form and manipulating algebraically yields:
VmeJ<h = (oLImej^I+7T/2)
where we made use of the fact that e^2
= a>LImej(f)1 ej7r/2 = jù)UmeJ4>i,
(3.70)
= j . The last expression is equivalent to:
V = jcoLI,
(3.71)
Z = V/Î = jù)L.
(3.72)
which results in an impedance of
In the last derivation, the power of phasors is made evident. Recall from earlier
sections that the voltage across an inductor is related to the current through an
inductor by the equation
But when using phasors we have
di(t)
v(i) = L - ^ .
at
(3.73)
V = JÍÚÜ.
(3.74)
Therefore the operation of differentiation is reduced to multiplication by a factor of
j(o:
j , - *»
(3 75)
-
68
Chapter 3. AC Steady State
Thus, the differential operations on sinusoidal functions are replaced by algebraic
operations on corresponding phasors, which are complex numbers.
To find the impedance of a capacitor, we begin with the sinusoidal relationship
(see (3.25)):
Vm COS ( COt + <\)y + — ) = —^- C0S(0)í + </)/).
V
2/
coC
Converting (3.76) into phasor form and manipulating algebraically yields
Vmej{<i>v+^
= — e**',
Vmej(t)ve^ = —e^\
coC
jVmeJ<t>v
= JmeJ(hf
cot
V = —tí,
coC
Z = V/Î = 1/jœC,
(3.76)
(3.77)
(3.78)
(3J9)
(3.80)
(3.81)
where again use was made of the identity ej77//2 = j .
Now, let us recall from earlier sections the expression relating the voltage across
the capacitor to the current through the capacitor:
■H-
v(0 = T; / Mat.
(3.82)
By comparing the previous equations we can see that the operation of integration of
sinusoidal functions is reduced to multiplication of the corresponding phasors by a
factor of —j/co (which is also equivalent to division by jœ):
+ —.
CO
(3.83)
Thus, the true power of the phasor technique is evident—it reduces the operations of
calculus on sinusoidal quantities to those of simple algebra on phasors.
EXAMPLE 3,5 Calculate the impedances of a 50 fi resistor, a 1 /xH inductor, and a
100 /xF capacitor at the following frequencies: f = 0 Hz, 1 kHz, 1 MHz, 1 GHz, oo.
Solution: The results are given in Table 3.1 as a function of angular frequency
and come from plugging the parameter values into equations (3.68), (3.72), and
(3.81). We note that for dc signals (co = 0), inductors have zero impedance and can
be replaced by short circuits (wires), while capacitors have infinite impedance and
can be replaced by open circuits. As the frequency increases, inductor impedances
increase while capacitor impedances decrease. At very high frequencies, inductors
can be modeled by open circuits while capacitors can be approximated by short
circuits.
69
3.5. Impedance and Admittance
Table 3.1: Sample impedances of passive components.
o) (rad/s)
R = 50 H
0
50 O
50 a
50 a
50 Û
2TT X 10
2TT X 10
2TT X 10
3
6
9
00
C = 100 ¿iF
L=\¡xñ
0
7-6.28 mil
76.28 Í1
76.28 kO
so a
00
-71.59 a
-7'1.59ma
-71.59 ^ a
0
00
The next discussion may seem trivial to some, but it is very important if we are to
use phasors to represent sinusoidal quantities in the circuit equations. The question:
Is the phasor of a sum of sinusoidal quantities equal to the sum of the phasors of the
same sinusoidal quantities? The answer is affirmative.
Assume g{t) is a sum of sinusoidal quantities:
git) = V ] Gmk cos(cot + <j)k),
(3.84)
k
then we want to prove that
(3.85)
G = Y,Gk,
where G is the phasor of g{t), while G¿ are the phasors of the corresponding sinusoidal
terms of the sum in (3.84).
Proof:
YGmkRe(eJ(«t+^)
g(t) =
k
= ^Re(GmfcÉ>V^)
k
= ^Re(GK>f)
Re
Ysóke
K
k
Re
J<ût
= Re [Öeja*l
By looking at the last two lines of this expression, one can see that the phasor of the
sum of sinusoidal quantities is indeed equal to the sum of the phasors of sinusoidal
quantities. Now that this little proof is secure, we can proceed to the systematic use
of phasors in ac steady-state analysis of electric circuits.
Chapter 3. AC Steady State
70
Kt)
+
R
L
AA/W
V(t )
v L (t)
+
vc(t)
V(t)
Figure 3.5: A general RLC branch.
To continue our study of ac circuits, we now consider a general branch. A
diagram of this branch is shown in Figure 3.5. It is simply a resistor, a capacitor, and
an inductor connected in series, which means that the same current flows through
each of these elements. This general branch is called an RLC branch, and it is the
basic branch of many ac circuits. We intend to show that each RLC branch can be
characterized by an impedance just as we did with individual elements.
We wish to determine an expression for impedance in terms of resistance, inductance, and capacitance. To start, we apply KVL to the loop shown in Figure
3.5:
(3.86)
v(i) = vR{t) + vL{t) + vc(t\
where v(t) has been moved to the left-hand side of the equation. By assuming that all
quantities in the branch are sinusoidal and by using (3.85), we now convert equation
(3.86) into phasor form:
(3.87)
V=V» + V,+ VC.
We next use the terminal relationships for resistors, inductors and capacitors in the
phasor form:
VR = RI,
(3.88)
VL = jcoLl
(3.89)
Vi
j
/.
coC
(3.90)
As a result, equation (3.87) becomes:
V = RI + jcoLÎ + — / - / R +
coC
j[ù)L-
Ù)C
(3.91)
Remember that the current / is the same for each element; that is why we factored it
out in the last expression. Now, we define the impedance Z as:
R + j (coL
1
coC
(3.92)
and substitute (3.92) back into (3.91), which leads to:
V = ÎZ.
(3.93)
71
3.5. Impedance and Admittance
Thus, we can see that every RLC branch can be characterized by impedance, defined
by equation (3.92), and the phasor of the branch voltage is related to the phasor of the
branch current by expression (3.93), which is mathematically similar to Ohm's law.
As we can see from (3.92), impedance is a complex number whose real part is
the resistance and whose imaginary part, called the reactance, is a combination of
two terms which are due to inductance and capacitance, respectively. The reactance
is due to the energy storage elements of the circuit and is denoted as X, leading to an
alternative expression for impedance:
Z = R + jX,
(3.94)
X= LL'-^Y
(3.95)
where
Since impedance is a complex number, we can represent it in the polar form,
Z = \Z\eJ*,
(3.96)
where \Z\ is the absolute value and 4> is the polar angle of the impedance. These
quantities can be found by converting from Cartesian to polar form (see Appendix
A). This leads to the following important formulas:
\Z\ = VR2 + X2 = \R2 + (<oL - ^- j ,
tmcf) = - .
R
(3.97)
(3.98)
Rewriting equation (3.93) in polar form yields
Vmej4* = Imej4>l\Z\ej*.
(3.99)
We can now relate the peak values and initial phases:
Vm=Im\Z\,
(3.100)
4>v = </>/ + &
4>v- (/>/ = </>.
(3.101)
(3.102)
From (3.102) we can see that the polar angle of the impedance is the phase difference
between the ac branch voltage and ac branch current.
EXAMPLE 3.6 Now, we illustrate the previous discussion by the following example.
Consider the circuit shown in Figure 3.5. Given v(t) = Vm cos(cot + cpv) and the values
ofR, L, and C, we wish to find the expression for the current i{t) by using the phasor
method.
We begin with converting all known quantities into phasor-impedance form.
First, we convert the given voltage v(t) into a phasor form:
72
Chapter 3. AC Steady State
v(r) — V = Vmej(pv.
(3.103)
Then, we introduce the impedance given by the expression:
Z=R
+ j(uL--^)=
\Z\e*.
(3.104)
We have represented the impedance in polar form to make the calculations easier. In
general, when multiplying or dividing phasors, the polar form is more desirable, but
when adding or subtracting phasors, the algebraic form is convenient. We can now
use equation (3.93) to relate the phasors of branch voltage and current:
(3.105)
V = IZ,
V pjw
V
V
; = - = ^r-r
= IZe***-».
(3.106)
v
;
Z
\Z\eJ*
\Z\
We now have the phasor of the branch current, and we want to find the expression
for the current in the time domain. Thus, we must convert the phasor of the current
into the sinusoidal quantity which it represents:
/ = Y^eMv-<t>) _
í(í) =
^
C0S(0Jt
+ 9v _ ^
(3-107)
Expression (3.107) is the solution to the problem. By using the phasor technique, we
were able to obtain the expression for the current through the branch without resorting
to any calculus. All ac steady-state problems will be solved in a similar fashion by
using the basic phasor techniques outlined in this and subsequent sections.
■
If any one of the elements is removed from the general RLC branch, the new
general impedance is found simply by deleting from (3.92) the term which corresponds to the deleted element. For example, the RC branch shown in Figure 3.6 has
the impedance
Z = R-j±~.
OJC
(3.108)
Similarly, an RL branch would have an impedance of
Z - R + jcoL
(3.109)
and an LC branch would have a purely imaginary impedance of
Z = jf(oL~-^Y
(3.110)
To calculate the impedance for a general branch, one must simply know the elements
of the branch and take them into account according to equations (3.92), (3.108)—
(3.110), or (3.68), (3.72), or (3.81) for single-element branches.
3.5. Impedance and Admittance
R
i(t)
v(t)
AAAAn
Ô
Figure 3.6: An RC branch driven by a voltage source.
EXAMPLE 3.7 An RC branch has a current i(t) - 50 sin(377i) mA flowing through
it when a voltage of v{t) = 1 2 0 y 2 cos(377f — 150°) V is applied across it (see Figure
3.6). Find the resistance and the capacitance of the branch elements.
Solution: The impedance of the RC branch is given by (3.108):
R - J -
= V/I
120y/2g-^ 5 0 0
.05e~J90°
a
or
R - -±- = 2,400V2e" 7 6 0 ° = 1,697 - ^2,939 a
(X)C
The real part of the previous equation yields the resistance:
R « 1.7 k i l
The imaginary part of the equation, coupled with the angular frequency o>
s:
377 r a d / s , yields:
C
1
377(2939)
0.9 juF.
In addition to impedance, there is another quantity which can be used to characterize any branch at ac steady state. This quantity is called admittance and is defined
as the reciprocal of impedance:
1
Y = -,
Z
(3.111)
where Y is used to denote admittance. Admittance is analogous to the conductance
when dealing with dc resistive circuits. However, the admittance is in general a
complex number made up of both real and imaginary parts:
Y = G + jB,
(3.112)
where G is the conductance and B is called the susceptance. Like impedance, the
admittance can also be used to relate the phasors of branch current and voltage:
/ = VY.
(3.113)
74
Chapter 3. AC Steady State
There is a simple relationship between conductance/susceptance and resistance/
reactance. Note that:
J
Y
G + jB
(G + jB) (G - jB)
G2 + B2
v
'
By equating real and imaginary parts we find that
and X=
-&T¥-
*=GÏTBÏ
(3115)
Thus, the conductance is equal to the reciprocal of the resistance only if the susceptance is zero. Admittances will be used more frequently later in the text, when the
method of node potentials is studied.
3.6
AC Steady-State Equations
We begin with the phasor versions of Kirchhoff 's laws. Recall from earlier sections
that the KCL and KVL equations are written in time domain form as
£ / * ( ' ) = 0,
(3.116)
k
$ > * ( f ) = 0.
(3.117)
k
We now rewrite them in a slightly different form by moving all known driving (voltage
and current) sources to the right-hand side of these equations:
Y, *k(t) = - $ ^ * ( i ) '
k
(3.118)
k
X>*(0 = " X>,*(0k
(3.119)
k
In the case of ac steady state, all variables in equations (3.118) and (3.119) are
sinusoidal. Consequently, by using the proven fact that the phasor of the sum of
sinusoidal quantities is equal to the sum of phasors of the same sinusoidal quantities,
the above equations can be converted into the phasor form:
Y,lk = -I>*'
(3 12
X^* = " X ^ -
(3 121)
k
k
k
k
- °)
-
For each branch, phasors of branch voltage and current are related to one another
through the impedance of the branch:
Vk=IkZk.
(3.122)
75
3.6. AC Steady-State Equations
By substituting (3.122) into (3.121), we arrive at the basic ac steady-state equations:
£4 = - £ i *
k
(3.123)
k
(3.124)
These are simultaneous linear algebraic equations with respect to phasors of branch
currents. The total number of these equations is equal to the total number of branches,
which is b. By solving these equations, we can find 4 and then, by using (3.122),
we can determine phasors of branch voltages Vk. As soon as we know the phasors
of branch voltages and currents, we can compute their instantaneous values using
the rule (3.39) of conversion of phasors into sinusoidal functions. Equations (3.123)
and (3.124) can be solved numerically by using the techniques developed in linear
algebra, for instance, the Gaussian elimination technique.
Now, we summarize how the basic equations (3.123) and (3.124) can be used in
circuit analysis. We assume that an electric circuit is given. This means resistances,
capacitances, and inductances as well as the driving sources in the electric circuit are
specified. Our goal is to solve for the branch voltages and currents.
We begin by introducing reference directions and reference polarities for all the
branch currents and voltages. We then convert the driving sources into phasors. Next,
we characterize each branch by its impedance. Finally, we write the basic equations
(3.123) and (3.124) for all independent nodes and loops (meshes), respectively. We
solve these equations and convert phasors into sinusoidal quantities. We illustrate this
summary by the following examples.
EXAMPLE 3.8 Consider the circuit shown in Figure 3.7. We are given:
v,i(0
= V^ICOSCCÜÍ +
(3.125)
<ki),
Vj2(f ) = Vmsl C0S(0tf + 4>s2),
(3.126)
and L\,Ri, C2, R3, L3, and we want to find all currents and voltages.
R1
R.
"3
AA/WrAAA/V
*-«©
4t)
Figure 3,7: Example circuit.
Ut)
CK «
Chapter 3. AC Steady State
Figure 3.8: The same circuit in phasor notation.
First, we begin by introducing reference directions, already shown in Figure 3.7.
Now, we convert the source voltages into phasors:
v,i(i)->fr,i = Vmsle*«,
(3.127)
Vs2{t)^Vs2 = Vms2e^\
(3.128)
Then we characterize each branch by its impedance. It helps to redraw the circuit in
phasor notation to easily discern all the branches. Figure 3.8 shows the same circuit,
this time drawn in phasor notation with boxes indicating the three different branch
impedances. The branch currents are also indicated in phasor form this time. We can
easily calculate the three branch impedances:
Zj =RX + j(oLh
(3.130)
J
Z2=
(3.129)
o)C2
Z3 = R3 + j<oL3.
(3.131)
Once the impedances are found, the circuit equations can be set up. This circuit has
only two nontrivial nodes, so only one KCL equation is linearly independent:
îi-î3-î2=0.
(3.132)
The circuit is also planar, resulting in two KVL equations—one for each mesh.
Mesh I:
hZx + I2Z2 = Vsl
(3.133)
- Î2Z2 + 73Z3 = Vs2
(3.134)
Mesh II:
That is all there is to it. We ended up with three equations, which is what was to be
expected since there are only three branches.
We can quickly arrive at the solution of this problem by using the method of
substitution. Indeed, from (3.133) and (3.134) we can find îi and 73 in terms of l2:
/, =
Zx
z3
Zi
z3
(3.135)
(3.136)
3.6. AC Steady-State Equations
77
By substituting (3.135) and (3.136) into (3.132), we end up with the following
equation for/27
— i~
from which we find:
Z2
Zi
- Z2
— i~
- _
— ¿2 — —
Z3
h =
Vsi
. VS2
-r
Zi
Z3
(3.137)
,
yc9z, + vi fl l^z3
(3.138)
Z,Zj +Z1Z1
1^3 +ZoZ
-2^ 3
After 72 is computed, expressions (3.135) and (3.136) can be used to calculate I\
and h■
EXAMPLE 3.9 Consider the circuit shown in Figure 3.9. It is assumed that sources
vs(t) and is(t) are given:
vs(t) = Vmscos((ot + <pvs),
(3.139)
hit) = lms cos((ot + <pIs),
(3.140)
as well as parameters R\, C\, R2, L2, R3, L4. We want to find all currents and voltages.
First, we begin by introducing reference directions, already shown in Figure 3.9.
Next, we convert the ac sources into phasors:
(3.141)
v,(f) - Vs = Vmsej^,
(3.142)
Then, we characterize each branch by its impedance:
J
Zi
=Ri
Z2 = /?2 +
y*)
(3.143)
(oC\
(3.144)
jwLi,
Z3 = R3,
(3.145)
Z4 = R4 + JC0L4.
(3.146)
L
P
R
B
4
¡4«
i 3(t)
a
Figure 3.9: Example circuit.
fl(bi¡ (t)
s
78
Chapter 3. AC Steady State
Figure 3.10: The previous circuit in phasor notation.
It is helpful to redraw the circuit in impedance-phasor notations to easily discern all
the branches. The redrawn circuit is shown in Figure 3.10.
The analysis of this circuit is facilitated by the following observations. It is clear
that node C is a trivial one. Consequently, by using KCL for this node, we obtain:
(3.147)
h = Is
Next, it is also clear that mesh I is a trivial one. By using KVL for this mesh, we
obtain:
/iZi = Vs,
(3.148)
7 _ ys
(3.149)
which leads to
Now, we shall apply KCL to nodes A and B and KVL to mesh II. As a result, we
arrive at the following equations:
(3.150)
= 0,
(3.151)
+ 72Z2 + 23Z3 = 0.
(3.152)
¡2-h+h
-îiZi
By using (3.147) in (3.151), (3.148) in (3.152), and (3.149) in (3.150), we obtain the
following simultaneous linear equations:
-Vs
(3.153)
h - h = -L
(3.154)
72Z2 + / 3 Z 3 = Vs.
(3.155)
I
To solve these equations, we first use (3.154) in order to express 73 in terms of 22:
h=h+l
(3.156)
79
3.6. AC Steady-State Equations
By substituting (3.156) into (3.155), we arrive at the following equation for l2:
I2Z2 + I2Z3 + ÎSZ3 = Vs,
(3.157)
which can be further transformed as follows:
72(Z2 + Zj) = Vs - ISZ3.
(3.158)
The last equation yields the following expression for I2:
*
Vs - ISZ3
z2 + z3-
(3.159)
By substituting the last expression into (3.156), we find I3:
h = ^ - ^ .
z 2 -t- z 3
Finally, by using (3.159) in (3.153), we derive:
1 = -1 + _i
Ll.
Zi
Z2+Z3
(3.160)
(3,161)
To find the voltage V* across the current source /?, we apply KVL to loop III:
Vx - I3Z3 - /4Z4 = 0.
(3.162)
By using (3.147) and (3.160) in (3.162) we obtain:
yx = ISZ4 +
Vs SZ3 + LZ
2Z3
I
*2 \
z 2 -t- z 3
This concludes the analysis of the above circuit.
(3.163)
■
In conclusion, we would like to stress the mathematical similarity between the
basic ac steady-state equations (3.123) and (3.124) and the equations (2.56) and (2.57)
for resistive circuits. It is clear that these equations have identical mathematical forms.
This explains why the analysis of electric circuits considered in Examples 3.8 and 3.9
closely parallels the analysis of resistive circuits considered in Examples 2.5 and 2.6.
The only difference is that in the case of ac steady state we deal with current phasors
and impedances, while in the case of resistive circuits we deal with instantaneous
currents and resistances. This similarity suggests that the analysis of resistive circuits
can be regarded as a particular case of ac steady-state analysis. Consequently, all the
analysis techniques which are developed for ac steady state can be directly used for
the analysis of resistive circuits. This is the approach which is adopted in this text.
It allows one to avoid duplication in the exposition of circuit theory and, at the same
time, to emphasize the generality of circuit analysis techniques.
80
3.7
Chapter 3. AC Steady State
AC Power
Consider a two-terminal electric device (Figure 3.11) whose terminal voltage v(r) and
current i(t) are periodic functions of time with period T. The instantaneous electric
power supplied to the device is given by:
pit) = v(t)i(t).
(3.164)
This power varies with time. For this reason, it is customary to characterize power
consumption by average power Pav which is defined as follows:
1 fT
r / P(t)dt.
1
(3.165)
Jo
First, let us consider a simple case in which the device can be electrically modeled
as a pure resistor. In this case, we have the following terminal relationship:
v(f) = Riit).
(3.166)
By substituting expression (3.166) into formula (3.164) and then into (3.165), we end
up with:
»r
P = *
i2(t)dt.
(3.167)
1
-/0
Next, we introduce the important notion of root-mean-square (rms) value of electric
current. This value is denoted as Irms and it is defined as follows:
(3.168)
By using (3.168) in (3.167), we obtain:
— L>T2
Pav = Rlrms-
(3-169)
Hence, the root-mean-square value of a periodic current is equal to the dc value of
the current which delivers the same average power to a resistor.
By rewriting expression (3.166) as
i(t) =
O
+ U
v(t)
-
O
v(t)
R
¡(t)
fe
>
Electric
Device
u
Figure 3.11: Electric device connected to a power network.
(3.170)
81
37. AC Power
and by substituting (3.170) into (3.164) and then into (3.165), we obtain:
Pav=
Sv2(t)dt)-
ïi\
(3 i7i)
This leads to the following definition of the rms values of a periodic voltage:
-
v2
(3.172)
Pav = ^ f •
(3.173)
Vrms= Jj¡f
«)dt.
By using (3.172) in (3.171), we arrive at:
Again, we see that the root-mean-square value of a periodic voltage is equal to the dc
value of the voltage which delivers the same average power to a resistor. Thus, the
idea behind the notion of root-mean-square value of a periodic time-varying quantity
is to replace this time-varying quantity by a dc quantity which will guarantee the
same average power.
For the sake of notational simplicity, in the subsequent discussion we shall omit
the subscript "rms" and we shall use capital letters without any subscripts for the
notations of root-mean-square values. In other words, we introduce the following
convention:
Irms = I
Vrms = V.
(3.174)
By using this convention, expressions (3.169) and (3.173) can be written as follows:
Pav = RI2 = —.
(3.175)
It is important to stress here that root-mean-square values (3.168) and (3.172) can be
introduced for any periodic current and voltage. However, in the case of a sinusoidal
quantity, a very simple relation can be established between its rms and peak values.
To see this, consider a sinusoidal current i(t):
i(t) = Imcos(a>t + cpi).
(3.176)
i2(t) = I2cos2(cot + <p¡) = ^ [ 1 + cos(2cof + 2(p/)].
(3.177)
Then:
By substituting (3.177) into (3.168), we find
i rT
1 + - / cos(2wi + 2<pj)dt
T Jo
(3.178)
It is clear that cos(2otf + 2<p/) is a periodic function with period T/2. Consequently,
cos(2o>í + 2<p¡)dt = 0.
/
Jo
(3.179)
82
Chapter 3. AC Steady State
By substituting (3.179) into (3.178), we derive:
/ = ~ .
(3.180)
By using the same line of reasoning, we can show that in the case of a sinusoidal
voltage we have:
V = -P=.
(3.181)
V2
Next, let us consider the general case of ac power supplied to an electric device
shown in Figure 3.11. In the general case, this device cannot be modeled as a pure
resistor. Hence, its terminal voltage and current are not in phase. Suppose that the
voltage and current are given by the expressions:
v(f) = Vmcos(cot + cpv),
(3.182)
/(f) = Imcos(o)t + <p7).
(3.183)
By substituting (3.182) and (3.183) into (3.164), we obtain:
Pit) = VmImcos((ot + çv)cos(o)t + <p/).
(3.184)
Now, we shall use the following trigonometric identity:
cos a cos ß = -[cos(a — ß) + cos(a + ]3)].
(3.185)
By assuming that a = cot + cpy and ß = oof + <p¡ and by using (3.185) in (3.184),
we transform the latter as follows:
pit) =
"L m [cos((pv - (pi) + cos(2cof + <pv + <p/)].
(3.186)
By substituting (3.186) into (3.165) and by taking into account that
we obtain:
Jo
cos(2cof + cpv + <p¡)dt = 0,
Pav = J]~
COS((pV - <pi).
(3.187)
(3.188)
Finally, by recalling (3.180) and (3.181) and by introducing the phase difference cp
between terminal voltage and current
<P - cpv - <Pi>
(3.189)
we arrive at the following very important formula:
Pav = VI COS (p.
(3.190)
This formula can be used to compute an average ac power supplied to (or consumed
by) an electric device when the rms values of terminal current and voltage as well
83
3.7. AC Power
as their phase difference are known. The same average power can also be computed
by using the current and voltage phasors. To derive the appropriate expression, we
introduce the phasor V of ac terminal voltage (3.182) and the complex conjugate /*
of the phasor of the ac terminal current (3.183):
V = Vmejipv,
(3.191)
(3.192)
Now, we define the complex power S as follows:
S = -VI*.
2
From (3.191), (3.192), and (3.193) we find:
S = \vmlme^~*¡\
(3.193)
(3.194)
By recalling (3.180), (3.181), and (3.189), we transform the last expression as follows:
S = VIej(p.
(3.195)
By comparing (3.190) with (3.195), we conclude that:
Re{S}.
(3.196)
Pav= Re (\vî*Y
(3.197)
Pav=
From (3.196) and (3.193), we obtain:
The last expression allows one to compute average power by using the phasors of the
terminal ac voltage and current.
At this point, we shall return to expression (3.190) and shall discuss its practical
implications. The factor coscp in (3.190) is called the power factor and it has a
significant economic impact as far as distribution and consumption of ac power are
concerned. To understand this impact, we remark that ac power is usually supplied
at specified constant rms voltage values. Actually, this is one of the main technical
challenges for the electric power utility industry: to supply ac power at more or less
constant rms voltage values in the face of permanently changing load (consumption).
Since V in (3.190) is constant, then the same average power Pav can be delivered
at: 1) larger rms current values / if the power factor cos <p is small and 2) smaller
rms current values / if the power factor cos cp is close to one. Clearly, the first option
is undesirable (and in many cases unacceptable) from the economic point of view.
This is because the larger rms values / of the terminal current will result in larger
ohmic losses [see expression (3.169)] in connecting distribution and transmission
lines. Consequently, power utility companies have to generate more power to supply
the same average power to a customer with a low power factor than would be required
if the customer's power factor were high. This explains why the same average power
84
Chapter 3. AC Steady State
delivered to a customer with a lower power factor should cost more. This is the reason
why the power companies insist that their customers should maintain sufficiently high
power factors, and usually adjust their rates to penalize consumers who do not meet
their requirements. It is also important to keep in mind that line losses represent
electric energy converted into heat and benefit no one. Actually, these losses have
negative environmental effects directly and indirectly because they require more
electric energy production.
The above discussion suggests that it is important to find an efficient way to
raise the power factor of a load. It turns out that the power factor can be corrected by
connecting a capacitor across the terminals of the device as shown in Figure 3.12a.
This technique achieves its goal when the device has overall inductive reactance,
that is, when the terminal current lags behind the terminal voltage. In this case, the
device can be electrically modeled as a series connection of an inductor and a resistor
(see Figure 3.12b). This situation is typical for electric motors, actuators, and various
devices which contain multiturn coils. We shall next show how the capacitance C can
be chosen to adjust the power factor to one.
For the terminal current / in Figure 3.12b we have:
I = l + h,
(3.198)
where Ic and I¿ are the phasors of electric currents through the capacitor and the
device, respectively. Since the same voltage V is applied across the capacitor and RL
branch, it is clear that the following expressions are valid for the phasors Ic and ld :
Ic = jœCV,
(3.199)
V
o)L
R
V.
R + jcoL
\R2 + co2L2 JJ-R2 + co2L2
By substituting (3.199) and (3.200) into (3.198), we derive:
R
( „
coL
V.
2
2 2 V + jlcoC
2
R + co L
R + œ2L2
(3.200)
(3.201)
A
+0
I
^
A
I
- o
Ï
R
Electric
Device
(b)
Figure 3.12: Correction of a power factor.
85
3.7. AC Power
If the capacitance C is chosen to be
C =
then from (3.201) we find:
R2 + co2L2'
(3.202)
R
(3.203)
■V.
R2 + co2L2
Since R/(R2 + o)2L2) is a real number, it is clear from expression (3.203) that the
terminal current and voltage are in phase. Consequently,
cp = 0, cos cp = 1.
(3.204)
/ =
It is also clear from (3.201) that the peak (and rms) value of the terminal current
achieves its minimum value when the capacitance is chosen according to the expression (3.202).
Finally, the following two remarks are in order. First, by connecting a capacitor
across the device terminals, we do not affect the overall average power consumed by
the device. This is because a capacitor is an energy storage element and consumes no
average power. This is also clear from the fact that for a capacitor cp = TT/2, COS cp = 0
and, according to (3.190), the average power is equal to zero. Second, by connecting
a capacitor across the device, we do not affect the device terminal voltage. This would
not be the case if the capacitor were connected in series with the device.
In some applications, it is of interest to find the maximum power that can be
delivered to a load by a power network. To find this power, we will model the power
network by a voltage source Vs and an impedance Zs. Justification behind this model
is the Thevenin theorem, which will be discussed in Chapter 5. By using the above
model, we can determine the maximum power that the power network can supply
and the manner in which to adjust the load to achieve maximum power transfer.
Consider the circuit shown in Figure 3.13. Here, the power network is represented by the voltage source Vs and the impedance Z5, while the load is modeled by
impedance ZL. All the nodes in the circuit are trivial ones, so with a quick application
of KVL around the one mesh we can find:
// =
Vr - V,
V,
(3.205)
ZL
(3.206)
+
VsÔ
A
V,
Figure 3.13: Representation of a power network by the voltage source and impedance.
86
Chapter 3. AC Steady State
From (3.205), we obtain the complex conjugate T[ of the load terminal current:
/* =
.
Yi
(3.207)
Now, by using formula (3.197), from (3.205) and (3.207) we derive:
P„v = ^ l 2 ï # ~ E .
(3.208)
2
\ZS + ZL\Á
The last expression can be transformed as follows:
1
/?
Pav =
(3 209)
2V2ms(Rs+R^
+ (Xs+XL)r
'
Next, we would like to find such values of RL and X¿ for which Pav achieves its
maximum. It is clear from (3.209) that for any fixed value of RL average power
achieves its maximum if
XL = -Xs.
(3.210)
By using (3.210) in (3.209), we find
1
/?
(3 211)
i {Rs + RLr
Now, we shall find the value of R¡_ for which Pav given by (3.211) achieves its
maximum. To this end, we differentiate (3.211) with respect to RL:
Pav =
y2ms
dPav = 1 2 (Rs + RL)2 - 2RL(RS + RL) = 1 2 Rs - RL
dRL
2 Vms
(Rs + RLf
2 Vms (Rs + RL? '
It is clear from (3.212) that for
RL = Rs,
^'Z1 }
(3.213)
the above derivative is equal to zero and changes its sign from plus to minus (as RL
is increased above Rs). This means that under condition (3.213) the average power in
(3.211) achieves its maximum value, which is equal to
max{/U = ^f.
(3.214)
By combining (3.210) and (3.214), we find that the above maximum power transfer
occurs when
ZL = z;.
(3.215)
Expression (3.215) is typical of matching conditions which are encountered in similar
forms in various areas of electrical engineering.
3.8 Complex Frequency
In our previous discussions of ac steady-state analysis of electric circuits, we have
dealt with sinusoidal voltages and currents and we have used the phasor technique
to calculate these voltages and currents. The main idea of the phasor technique is to
87
3.8. Complex Frequency
reduce the operations of calculus on sinusoidal quantities to algebraic operations on
their phasors. It turns out that this idea can be extended to a broader class of voltages
and currents.
Consider the following voltage:
v(t) = Vmeat cos(ú)t + <pv).
(3.216)
By using the Euler formula, we can transform this voltage as follows:
v(f) = Vmeat Re [^<<*+**>] = Re [Vmej(pve(a+ja))t)] .
(3.217)
Now, we introduce the voltage phasor V:
V - VmeJ<pv,
(3.218)
as well as the quantity s, which is called a complex frequency:
s = a + j(ú.
(3.219)
The real part of this frequency characterizes the exponential decay (or growth) of the
peak value of the voltage (3.216), while the imaginary part is the angular frequency.
By using (3.218) and (3.219), we represent (3.217) in the form:
v(f) = Re [Vest]
(3.220)
This formula extends the notion of phasors to voltages (3.216) which are characterized
by complex frequency s. In a particular case when a = 0, this formula is reduced to
the familiar formula for sinusoidal voltages:
v(f) = Vmcos(cot + cpv) = Re [Vej0)t] .
(3.221)
i(i) = Imeat cos(coi + <p/) - Re [lmej<Plest],
(3.222)
If we have a current:
then, by introducing phasor 7:
7 = ImeJip>,
(3.223)
we end up with a similar representation for /(f):
i(t) = Re [Iest] .
(3.224)
Next, we shall establish phasor terminal relationships for resistors, inductors, and
capacitors for the case of voltages and currents of the same complex frequency s.
In the case of a resistor, we have:
v(f) = Vmeat cos((ot + cpv) = Ri(t) = Rime*7* cos((ot + <p7).
(3.225)
By expressing (3.225) in the phasor form, we find:
Re [Ve5t] = Re [RIest],
(3.226)
V = RÎ.
(3.227)
which suggests that:
Chapter 3. AC Steady State
Next, we consider an inductor and begin with its terminal relationship:
di(t)
(3.228)
v(r) = Ldt
By substituting (3.222) into (3.228) and by performing differentiation, we derive:
vit) = aLImeat cos(cot + q>¡) - toLIme(7t sm(cot + <p¡),
(3.229)
v(i) = crLIme(r' cos(cot + (p¡) + coLImeat cos ( cot + cp¡ +
(3.230)
By using the Euler formula and the definition of current phasor, we transform (3.230)
as follows:
v(i) = Re [<jLlest] + Re \coLIeJ7T/2est
(3.231)
v(f) = Re [(a + jto)Lles'] = Re [sLIest]
(3.232)
By comparing (3.232) with (3.220), we derive:
(3.233)
V = sLÎ.
By using the same line of reasoning, we can establish the following relationship
between the voltage and current phasors in the case of a capacitor:
(3.234)
/ = sCV.
The last formula can also be written as:
(3.235)
In the particular case of a = 0 , the last three formulas are reduced to the familiar
expressions:
V = jtoLI,
I = jtoCV,
V=
1
jcoC
(3.236)
Now, we consider the circuit shown in Figure 3.14, which consists of an RLC branch
and the voltage source:
v(i) = Vme** cos(cot + <pv)+ V
i-AAA/V
R
Vme
ot
cos( (Ot +<pv)Q
(3.237)
+ V
+
Figure 3.14: An RLC branch driven by a complex frequency source.
89
3.8. Complex Frequency
We look for the current in the form:
i(t) = lmem cos(a>f + <p/).
(3.238)
We shall use the phasor technique and formulas (3.227), (3.233), and (3.235) in order
to find Im and <p¡ in (3.238). We begin by writing the KVL equation in phasor form:
V = VR + VL + Vc.
(3.239)
Voltage phasors in the right-hand side of (3.239) can be written in terms of the current
phasor as follows:
VR = RÍ,
VL = sLÎ,
Vc = —I.
sC
(3.240)
By substituting (3.240) into (3.239), we derive:
V = ( R + sL + — I /.
sC
(3.241)
We shall next introduce the impedance Z(s) as a function of complex frequency s:
Z(s) = R + sL+ —.
(3.242)
sC
In the particular case of a = 0, the last expression is reduced to the familiar formula:
Z(j(o) =R + ja>L + - Í - .
jùiC
(3.243)
By using (3.242), we can rewrite (3.241) as follows:
V = Z{s)I,
(3.244)
i = ^-v
(3.245)
which can now be solved for /:
Thus, in order tofindthe current phasor /, we first find the voltage phasor V = Vmej(pv
from (3.237), then we evaluate the impedance Z(s) for complex frequency s = a + yco,
and finally we use (3.245) to find 1. Knowing /, we can easily find Im and <p/. It is clear
that the described calculations are almost identical to the calculations which we have
performed for ac steady-state analysis. The only difference is that the impedance
Z should be evaluated at the complex frequency s = er + ja) rather than at jco.
This suggests that the analysis of electric circuits excited by voltage and/or current
sources of complex frequency can be performed by using exactly the same techniques
which we have developed for ac steady-state analysis.
In applications, it is quite rare that electric circuits are excited by sources of
complex frequency. However, voltages and currents of complex frequency regularly
appear in the case of transients in electric circuits. For this reason, the notion of
90
Chapter 3. AC Steady State
complex frequency as well as the notion of impedance Z(s) as a function of complex
frequency s will be useful in the analysis of transients in electric circuits. That is why
the notion of complex frequency is extensively used in Chapters 7, 8, and 9.
3-9
Summary
In this chapter, we introduced the concept of phasors, which were used to transform
the complete set of differential equations for ac source-driven circuits into a linear
system of algebraic equations.
The phasor representation of a current source, for example, is a complex number
whose magnitude is equal to the peak value of the current source and whose polar
angle is equal to the initial phase of the current source: Is = Imse^ls if and only if
is(t) = Ims cos(otf + 4>[s). A similar representation holds for voltage sources and for all
sinusoidal voltages and currents in electric circuits under ac steady-state conditions.
The phasor method tacitly assumes that any and all voltages and currents are
varying with the same angular frequency w. Problems for which this assumption is
not true must be deferred until the discussion of the superposition method in the next
chapter.
We also introduced the concept of impedance, which is a far-reaching generalization of resistance in the case of the steady-state problems. For any branch of an
electric circuit the impedance Z is defined as the ratio of the phasor of the branch
voltage V to the phasor of the branch current /:
\/Y = 7 7 ^ .
(3.246)
G + jB
It is clear that the resistance, R, and the reactance, X, are defined as the real and
imaginary parts of the impedance, while the admittance Y is the reciprocal of the
impedance. The conductance, G, and the susceptance, 2?, are the real and imaginary
parts of the admittance. Table 3.2 summarizes the expressions for these quantities for
various passive circuit elements.
Z = V/I = R + jX=
Table 3.2: Expressions for impedances and admittances.
Parameter
Impedance
Resistance
Reactance
Admittance
Conductance
Susceptance
Resistor
Inductor
Capacitor
R(ü)
L(H)
C(F)
1/jtoC
0
R
R
0
\IR
IIR
0
j(x)L
0
u)L
l/j(oL
0
-1/coL
-1/ÍOC
jo)C
0
o)C
91
3.10. Problems
With the aid of the impedance concept, we transformed KCL and KVL equations
into the standard ac steady-state equations:
¿4 = -£**'
k
X>Z, = -]TVV
k
(3.247)
k
(3.248)
k
These equations are mathematically similar to the basic equations (2.56) and
(2.57) for resistive electric circuits. As a result, the analysis techniques developed
for ac steady-state can be directly used for the analysis of resistive circuits. The only
difference is that in the case of ac steady-state we deal with phasors and impedances
(admittances) while in the case of resistive circuits we deal with instantaneous values
and resistances (conductances).
It is important to stress here that the linear equations produced by the KCL
and KVL equations are somewhat special. They are usually "sparse" because not
every equation contains every variable. When standard matrix notations are used, this
corresponds to the coefficient matrix containing many zeroes. Sparsity of Kirchhoff
equations is closely related to the "connectivity" of electric circuits, that is, to the
way circuit elements are interconnected. In circuit theory special analysis techniques
have been developed to take advantage of the sparsity of the basic equations and
reduce the number of needed computations. These techniques will be discussed in
the following three chapters.
The discussion of ac power was presented. The notion of root-mean-square (rms)
values was introduced and the expression for the average ac power was derived in
terms of rms values of terminal voltage and current. This expression contains the
power factor, which has a significant economic impact on distribution and consumption of ac power. The means of adjusting the power factor to unity were discussed
along with the means of achieving the maximum power transfer from a power network
to a load.
The chapter closed with a brief discussion of complex frequency. It was demonstrated that the phasor technique can be extended to the analysis of electric circuits
excited by sources with complex frequency. It was remarked that voltages and currents of complex frequency regularly appear in the analysis of transients in electric
circuits and it is there that the notion of complex frequency is most useful. That is
why this notion will be extensively used in Chapters 7, 8, and 9.
3.10
Problems
1. The voltage across a 43 mH inductor is given by v(t) = 50 cos(o)t + 6) V. Find the peak
current at a frequency / = (a) 1/50 Hz, (b) 1 Hz, (c) 30 Hz, and (d) 1 MHz.
2. The current through a 10 /xF capacitor is given by i{t) = 0.1 sin(2atf) A. Find the peak
voltage at a frequency / = (a) 1/20 Hz, (b) 2 Hz, (c) 1 kHz, and (d) 30 MHz.
3. Find the phasors which represent the following sinusoidal currents: /(/) =
(a) 6cos(otf + 1/2) A, (b) 4sin(otf + TT/3) mA, (c) 3cos(coi - 3TT/8) A, and
(d) 2cos(cuf) - 4sin(cüí) A.
92
Chapter 3. AC Steady State
4.
Find the phasors which represent the following sinusoidal voltages: v(t) = (a)
l/2cos(cot - IT) V, (b) 7cos(2otf + TT/2) mV, (c) 2sin(o>i ~ u/1) kV, and (d)
3 sin(ûtf) — cos(œt) V.
5.
Transform the following voltage phasors to the standard form v(t) = Vm cos((ot + <¿>v) V:
and (d) 5 + jl (all in volts).
V = (a) 20é^ / 6 , (b) 5 e " ^ / 2 , (c) 3(1/2 - j\ß/2),
6. Transform the following current phasors to the standard form i{t) = Im cos(œt + <\>j) A:
/ = (a) O.OléT2', (b) 3eJir/49 (c) 1/2 - j3, and (d) 2.7 (1/3 + jly/l/3)
(all in amps).
7.
Find the impedance of a 4.7 pF capacitor at the following frequencies: (a) 1 Hz, (b) 5
kHz, (c) 1 MHz, and (d) 2 GHz.
8. Find the admittance of a 1.1 mH inductor at the following frequencies: (a) 10 mHz, (b)
3.3 Hz, (c) 27 kHz, and (d) 5 MHz.
9. A voltage v(t) — 10cos(10i + TT/3) V is applied across a two-terminal device with an
impedance Z = 8 — j6 fl. Find the expression for the time variation of the ac current
flowing through the device.
10. A current i(t) = 0.02sin(1000i + TT/4) A flows through a two-terminal device with an
impedance Z = 50 — 7*40 ÎÎ. Find the expression for the time variation of the ac voltage
across the device.
11.
The impedance of a branch is given by the expression Z = 50 + j40 fl. Find the:
(a) reactance, (b) admittance, (c) conductance, and (d) susceptance of the branch,
12. The admittance of a branch is given by F = 12 — j5 mil. Find the: (a) susceptance,
(b) impedance, and (c) reactance of the element.
13.
A voltage v(i) = 120v//2cos(377/ + 120°) V is applied across an RL branch with
R = 10 fl and L = 10 mH. Find an expression for the time variation of the current
through the branch.
14. A current of i(t) = 3 cos(f) — 4 sin(r) A is made to flow through an RLC branch with
R = 2 fl, L = 2 H, and C = 1 F. Find an expression for the time variation of the voltage
across the branch.
15. A voltage v(t) = 100sin(50r) mV is applied across an RC branch with R = 5 kfl and
C — 10 jU,F. Find an expression for the time variation of the current through the branch.
16. Find the time dependence of the voltage source that must be applied to an RLC branch
(R = 10 fl, L = 1 H, and C = 3/2 F ) in order to achieve a current of i(t) = cos(lOf)
mA through the branch.
17.
A voltage v(i) = 120y2cos(2400i) V is applied across an LC branch with L = 1 mH
and C — \ ¿¿F. Find an expression for the time variation of the current through the
branch.
18. When a voltage of v{t) = 20cos(100i + 30°) V is applied across an RL branch, the
resulting current through the branch is i{t) == 5cos(100i - 45°) A. Find the values of
the resistance and inductance for this branch.
19.
Consider the circuit shown in Figure P3-19a. Figure P3-19b represents a concise version
of the circuit where some of the elements have been combined into single branches. By
93
3.20. Problems
comparing the two figures, write the impedance for each branch of the circuit in Figure
P3-19b. Let Z[ be the impedance for the branch with ix flowing through it, etc.
L
q x v »0
-CD
(a) original circuit
(b) impedance representation
Figure P3-19
20.
Consider the circuit shown in Figure P3-20a. Figure P3-20b represents a concise version
of the circuit where some of the elements have been combined into single branches. By
comparing the two figures, write the impedance for each branch of the circuit in Figure
P3-20b. Let Z\ be the impedance for the branch with ix flowing through it, etc.
(a) original circuit
(b) impedance representation
Figure P3-20
Problems 21-25. For the circuits shown in the figures indicated, label the nodes and meshes
and write the ac steady-state KCL and KVL equations. Write all equations in terms of the
phasor currents and the source phasors.
21.
Figure P3-21.
22.
Figure P3-22.
10 Q
10 Û
Figure P3-21
Figure P3-22
Chapter 3. AC Steady State
94
23.
Figure P3-23.
Figure P3-23
24.
Figure P3-24.
r^VWW^ \k
Jcos(2t)V
Figure P3-24
25.
Figure P3-25.
2 F
rj5cos(t)
2,
>2Q 1H
Figure P3-25
Problems 26-30. An alternate form for the basic ac steady-state equations arises from substituting into KCL the expression lk = YkVk for each branch:
]TW = -£/,,; X> = -E^
95
3.10. Problems
For the circuits shown in the figures indicated, label the nodes and meshes and write the ac
steady-state KCL and KVL equations. Write all equations in terms of the voltage phasors and
the source phasors.
26. Figure P3-25.
27.
Figure P3-27.
Figure P3-27
28. Figure P3-28.
2cos(t/2) A
2sin(t/2) A
Figure P3-28
29.
Figure P3-29.
Figure P3-29
30.
Figure P3-30.
Figure P3-30
96
Chapter 3. AC Steady State
31.
Find the rms value of the periodic ramp voltage source shown in Figure 1.20c. The
expression for the voltage when |i| > T/2, where T is the period, is given by v(f) =
2Vmt/T.
32.
The periodic pulse voltage source, shown in Figure 1.20d, is given in the interval 0 <
t < 7/by
w
10
T<t<T
(where T is the period). Find an expression for the rms voltage as a function of T/T and
show that it approaches a reasonable value as T/T —* 1.
33.
Consider the circuit shown in Figure P3-33. Find the rms current and power factor of the
load for the parameters listed in the following table.
R
Qv m cos(cot)
Figure P3-33
34.
Case
V,„ (V) ta (rad/s)
Ä(il)
L(H)
C(F)
a
b
c
d
169.7
169.7
100
100
20
Ik
5
100
0.01
0.3
0
2
0.001
377
377
2513
10
00
1 0 0 jLL
0.005
Consider the circuit shown in Figure P3-34. Find the rms voltage and power factor of
the load for the parameters listed in the following table.
ô'
cos( cot)
Figure P3-34
Case
/m(A)
io (rad/s)
R(il)
L(H)
C(F)
a
b
c
d
1
10
lm
25 m
377
377
IM
10
5
50
5k
50
0
0.1
0.2 m
5
470 ix
OO
100 p
0.002
97
3.20. Problems
35.
Consider the circuit shown in Figure P3-35. Find the impedance of the RLC branch and
the branch current for the cases indicated in the table below.
Qv m e ot cos((ot+4> v )
C
Figure P3-35
36.
Case
Vm(V)
a
b
c
10
50
100
<T(S-1)
-1
-50
-0.2 M
hi (rad/s)
1
Ik
IM
4>v (deg) ß(il)
30
45
75
10
50
300
L(H)
cm
1
0
0.01m
12m
47 ju
CO
Consider the circuit shown in Figure P3-35. Find the current through the RLC branch for
the cases indicated in the table below.
Case
V«(V)
vis'1)
to (rad/s)
<£v (deg)
J?(ÍX)
¿(H)
cm
a
b
c
170
5
12
-2
-2
377
-90
-45
30
50
10
300
0.01
0
0.1m
470 \x
0.01
10 \x,
-2 M
4TT
10 M
37.
A voltage source v(t) = 10e~r sin(7¿)V generates a current of /(/) = 0.2e~T cosÇ/t — 30°)
A when connected to an RC branch. Find the values of resistance and capacitance of the
branch.
38.
When a current source i{t) = e~20t cos(80i + 45°) A is connected to an RL branch, the
resulting voltage across the source is measured to be v(f) = SOy/ïe-™
sin(S0t) V.
Find the values of resistance and inductance of the branch.
Chapter 4
Equivalent Transformations
of Electric Circuits
4.1
Introduction
This chapter deals with equivalent transformations of electric circuits. Through the use
of equivalent transformations, complex circuits requiring numerous computations for
their analysis can be reduced to simpler equivalent circuits, sometimes dramatically
reducing the computational effort required to analyze them.
We first consider passive circuits and derive general formulas for transforming
an arbitrary number of series and parallel elements. We next introduce rules which
are known as the current divider and the voltage divider rules. We then consider the
concept of input impedance and give several examples of equivalent transformations
of electric circuits which utilize alternating applications of series and parallel transformations. We end the study of transformations of passive circuits with a discussion
of symmetry.
For active circuits, we introduce the superposition principle and demonstrate via
several examples how it can be used to simplify circuit analysis. We also detail the
solution technique for circuits with ac sources which operate at different frequencies.
The chapter is closed with an introduction to the MicroSim PSpice circuit simulation package. This introduction walks the reader through Microsoft® Windows™based PSpice simulations of three examples from this chapter.
4.2
Series and Parallel Connections
Consider the branch shown in Figure 4.1, which is connected to the rest of the electric
circuit. The impedances shown in this branch are said to be connected in series. This
means that they all have the same identical current flowing through them. This fact
98
99
4.2. Series and Parallel Connections
A
Rest
of
Electric
Circuit
r~\
I
Z1
Z2
Zn
U
A
V
Figure 4.1: Branch with impedances connected in series.
can be easily concluded from KCL and happens because each node that connects two
elements together is a trivial one.
We wish to simplify this branch by representing all the impedances with a single equivalent impedance, shown in Figure 4.2. But how can we guarantee that the
replacement by one equivalent impedance will not affect the rest of the circuit? The
answer is that the electric circuit in the box will not be affected by this equivalent
transformation if this transformation preserves the terminal relationship between the
phasors of branch voltage V and branch current /. Indeed, if the above terminal relationship is preserved, then the circuits shown in Figures 4.1 and 4.2 will be described
by the identical sets of linear algebraic equations. To prove this, we remark that the
electric circuits in the boxes are the same and, consequently, they are described by
identical sets of basic ac steady-state equations. These basic equations are complemented by the same terminal (V versus /) relationship for the branches outside the
box. This makes the overall sets of ac steady-state equations for the above circuits
identical Consequently, by solving these equations, we find the same branch currents
for both circuits. This proves that the rest of the circuit will not be affected by the
equivalent transformation.
In a way, we have already formulated and proved a very general principle of
equivalent transformation. According to this principle, we can divide an electric
circuit into two parts (I and II) which are connected through terminals A and B
Figure 4.2: The equivalent impedance.
100
Chapter 4. Equivalent Transformations of Electric Circuits
A f
1
+
II
A
V
B
Figure 4.3: An electric circuit divided into two parts.
(see Figure 4.3). Then, any transformation of part II, which preserves the terminal
V-I relationship, will not affect the currents and voltages in part I of the circuit.
In this sense, this transformation is an equivalent one. This principle of equivalent
transformations based on preserving voltage-current terminal relationships will be
frequently used throughout this text. We shall first apply this principle to the circuits
shown in Figures 4.1 and 4.2.
We write KVL for the loop in Figure 4.1 :
V = V{ + V2 + •
(4.1)
Now, by using the formula Vk = IZk, we find:
V = IZi+ IZ2 + •
+ IZn,
(4.2)
V = I(ZX + Z2 + •
+ Zn).
(4.3)
From Figure 4.2 we obtain:
(4.4)
V = IZ<eq>
t
Clearly, the V-I terminal relationships (4.3) and (4.4) will be the same if:
+ Zn = > Zk.
Zeq — Zi + Z2 +
(4.5)
k
Therefore, the equivalent impedance of a series connection of impedances is merely
the sum of those impedances. From equation (4.5) and the definition of impedance
we can see that the equivalent resistance of a branch composed of resistors connected
in series is:
Req — /
jRk
(4.6)
For a branch composed of inductors connected in series, we have:
j(oLeq = V " jo)Lh
(4.7)
k
which leads to:
(4.8)
-eq
k
101
4.2. Seríes and Parallel Connections
Similarly, for a branch composed of capacitors connected in series, we find:
J
coCeq
=Y.-^->
o)Ck
^
which results in:
- = £^eq
(4.10)
^k
r
or:
{spa
£*i/<V
(4.11)
In the particular case of a series connection of identical impedances Z, from (4.5) we
find:
^ea
(4.12)
TlZjy
where n is the total number of impedances.
Now, consider the circuit shown in Figure 4.4. The impedances in this circuit are
said to be connected in parallel, which means that the same voltage is applied across
each of them. This happens when corresponding terminals of the circuit elements are
connected together to a pair of nodes by wires.
Again, we wish to derive an equivalent transformation which allows one to
replace the circuit with just one equivalent impedance (Figure 4.2). We begin by
writing the KCL equation for the circuit shown in Figure 4.4:
/i + h +
+ /*
(4.13)
Note that there are only two nodes in this circuit, one on the top and one on the bottom
where all the branches are connected together. This is because the lines connecting
circuit elements have zero impedance and therefore can all be consolidated into one
node on the top and bottom, respectively.
Applying the relation Ik = V/Zk, we derive:
V
V
V
Z\
Z2
Zn
/ = — + — + ••• + —,
Figure 4.4: Circuit with impedances connected in parallel.
(4.14)
102
Chapter 4. Equivalent Transformations of Electric Circuits
?
+ + +
= K¿ ¿ - ¿''
(415)
Now we can see that the V-Î terminal relationships (4.16) and (4.4) will be the same,
if:
1
1
~eq
i
Zi
1i
Z2
1 ,,, 1 i
Zn
V i '
2-^k Zk
(4-17)
In a particular case of equal impedances connected in parallel, from (4.17) we derive:
Zeq = -,
(4.18)
n
where n is the total number of impedances.
From equation (4.17) and the definition of impedance we find that if the branches
in parallel are all composed of resistors, then:
1 _ 1
1
1
(4.19)
•+
Req
R\
R2
Rn
they contain only inductors, then:
1
Leq
1
L\
1
L2
1
"Tn
(4.20)
nd in the case of capacitors, we have:
Ceq = C\ + C2 + * •• + cn.
(4.21)
It is sometimes more convenient to describe the relationship for parallel connections in terms of admittance. Equation (4.17) can be rewritten as \/Zeq — Y^k 1/Zjb
and since we defined the admittance as the reciprocal of impedance, we get Yeq —
l/Zeq = ^2k Yk. In other words, the equivalent admittance of a parallel connection
is simply the sum of the individual admittances.
EXAMPLE 4.1 Consider a 10 ÍÍ, a 22 Í1, and a 47 Í1 resistor. What is the equivalent
resistance if they are all connected in series? What is the equivalent resistance if they
are connected in parallel?
For a series connection we have Req = 10 Í1 + 22 Í1 + 47 Í1 = 79 fi. For a
parallel connection Req = 1/(1/10 + 1/22 + 1/47) « 6 f t .
■
EXAMPLE 4.2 Consider a 10 ÍÍ resistor, a 470 jaF capacitor, and a 1.1 mH inductor.
What is the equivalent impedance if these elements are connected in series and driven
with a 60 Hz source? What is the equivalent impedance if these elements are connected
in parallel and driven by a 400 Hz source (a frequency often used in aviation)?
The impedance of the capacitor at 60 Hz is Zc = —j/coC = —J/(2TT X
60 X 470 X 10"6) = - j'5.644 Í1 and the impedance of the inductor is ZL =
103
4.3. Voltage and Current Divider Rules
jo)L = jlir X 60 X 1.1 X 10 3 = y'0.415 fl. The equivalent series impedance is
Zeq = 10 - y-5.644 + jO.415 = 10 - 7*5.229 Ü.
The admittance of the capacitor at 400 Hz is Yc = ja>C = 2TT X 400 X
470 X 10~6 = 7I.I8I Ö and the admittance of the inductor is YL = -j/coL =
-j/ilir
X 400 X 1.1 X 10~3) = -jO.362 U. The equivalent parallel admittance is
Yeq = 0.10 + 7 U 8 I - 7'0.362 = 0.10 + jf0.819 Ö. The equivalent impedance is
Zeq = \/Yeq = (0.10 - y0.819)/0.676 = 0.148 - 71.212
fí.
■
4.3 Voltage and Current Divider Rules
Sometimes in circuit analysis, only the voltage across one specific element in the
circuit is desired. To find the voltage across one impedance which is in series with
many other impedances, we can utilize the voltage divider rule. The voltage divider
rule is based on the fact that the total voltage is proportionately divided between
impedances in series. Consider the circuit in Figure 4.5. Assume we wish to find
V\. We know that the voltage across Z\ is simply the product of the current and the
impedance:
V« = /Z,
(4.22)
Vs = I(ZX + Z2 + • • • + Zn).
(4.23)
We also know that (see (4.3)):
We now look at the ratio of V\ to Vs:
V,
V,
/z,
(4.24)
I(Zl + Z2 + • • • + Zn)
which leads to:
Vx =VS
+
v
i -
(4.25).
Z\ +Zo + • • • + Z„
+ V
+ V
Figure 4.5: Circuit with impedances in series.
104
Chapter 4. Equivalent Transformations of Electric Circuits
Similarly, it can be derived that the voltage across the £th element is:
Vk = Vs-
—^
—.
(4.26)
The current divider rule is useful for finding the currents through impedances
that are connected in parallel. If the total current and all the impedances are known,
then the current divider rule can be used to determine the individual currents through
each impedance. It is derived in a fashion similar to the voltage divider rule. Consider
the circuit shown in Figure 4.6. Assume we want to find I\. We know that the current
through Z\ is the voltage across it divided by the impedance:
Í,=^ .
(4.27)
We also know that (see (4.15)):
i
i
h =V — + — +
z, z2
We now find the ratio of I\ to Is:
+—
7
h =
(¿ + Í + ---H)'
l
(4.28)
(4.29)
which leads to:
z¡
+ z4+
2
i •
+ z„
(4.30)
By using the relationship between impedance and admittance, from (4.30) we
find:
Yi + Y2 + ■ • • + Yn
(4.31)
Similarly, it can be derived that the current through the Ath element is:
h
=L
Yk
7i + Yi +
+ Yn
,cb
Figure 4.6: Circuit with impedances in parallel.
(4.32)
105
43. Voltage and Current Divider Rules
The current divider rule is quite often used in the case of two impedances
connected in parallel. In this case the rule is very easy to remember. Consider the
simple circuit in Figure 4.7. From (4.30) we can find the expression for I\ :
7
i - "-^T —
'
+ T—
z,
(4.33)
z?
which can be simplified by multiplying the numerator and denominator by Z{Z2
%z2
(4.34)
Zx + Z2
Therefore, whenever two impedances are connected in parallel, the current through
one is equal to the total current times the adjacent impedance divided by the sum of
the two impedances connected in parallel.
h -
EXAMPLE 4.3 Consider a voltage divider consisting of only two resistors: R\ =
10 O and R2 = 5 fl. Find the ratio of the voltage across R2 to the input voltage Vs.
If instead you configure the two resistors as a current divider, what is the ratio of the
current through R2 to the input current Isl
From equation (4.26), we find V2/Vs = Z2/(Z{ + Z2) = 5/(5 + 10) = 1/3.
On the other hand, the current divider rule for two elements (4.34) yields I2/Is =
ZX/{ZX + Z2) = 10/(5 + 10) - 2/3.
■
EXAMPLE 4.4 Design a capacitive voltage divider to measure a 50,000 V signal
with a 10 V probe. The largest capacitor you have available to use has a capacitance
of 1 ¡xF.
For this problem it is most convenient to write the voltage divider rule for two
elements in terms of admittances. By using (4.26), it is easy to show that V2/Vs =
Y\/(Yi + Y2) - jo)Cx/{j(oCx + jo)C2) = Ci/(C{ + C2). For the design problem, we
need to find values for Cx and C2. We need to have V2/Vs - 10/50,000 = 0.0002.
This will require that C2 be much larger than C\. Let C2 = 1 /xF. Then 0.0002 =
C\ /(C\ + 1 JUF), which can be solved to find C\ — 200 pF. Similar capacitive voltage
dividers are quite useful in high-voltage, pulsed power systems. Note that C\ must
physically be very large to sustain (without breakdown) over 50,000 volts. It is often
made by immersing two large electrodes in a dielectric oil. C2 can be a standard
off-the-shelf capacitor.
■
Figure 4.7: Simple circuit with two parallel impedances.
106
Chapter 4. Equivalent Transformations of Electric Circuits
IJ
O
Figure 4.8: Example circuit.
EXAMPLE 4,5 Consider a two-branch current divider consisting of a capacitor C
and a resistor R. Derive the expression for the magnitude of the ratio of the output
current through the resistor to the input current as a function of angular frequency co.
What is the divider ratio at the limit of very low and very high frequencies?
Equation (4.32) for this example becomes I2/Is = Y2/(Yl+Y2) =
(l/R)/(l/R+
jœC) = 1/(1 + jcoRC). The magnitudeof the ratio is just \í2/ís\ = \/\J\ +((oRC)2.
As a) —► 0, the divider ratio approaches 1 and the output signal I2 equals the input
signal. As co —► oo, the output signal goes to zero. This device is called a passive
low-pass filter and will be discussed in Chapter 9.
■
We now demonstrate how combinations of equivalent transformations can be
used to simplify problems in circuit analysis.
Consider the circuit shown in Figure 4.8. It is assumed that VS,Z\,Z2, and Z3 are
given and we want to find all the currents I\, 72, and / 3 . First, we find I\. To do this,
we notice that Z2 and Z3 are connected in parallel, and we use the equivalent transformation for parallel connections to combine them into one equivalent impedance
Z23 (see Figure 4.9):
^23
Z2Z3
Zo +Z*
(4.35)
To further simplify the circuit, we use the equivalent transformation for series connections to combine Z\ and Z23 into one equivalent impedance Zeq. Thus, the total
Figure 4.9: Z2 and Z3 combined into one impedance.
107
4.4. Input Impedance
Figure 4.10: All impedances combined into Zeq.
impedance of the circuit (Figure 4.10) is:
->eq
zx + z 23 = z, + ZJ+ZT,
(4.36)
We can now find I\ :
/. =
V,
(4.37)
->eq
Now, to find I2 and / 3 we use the current divider rule, because from Figure 4.8 it is
evident that Z2 and Z3 are connected in parallel and we already know I\ (the total
current). So, we have
h =
/1Z3
Z2 + Z3'
(4.38)
/iZ 2
(4.39)
Z2+Z3
The circuit is now completely analyzed, and no system of equations has been set up or
solved. This is the advantage of using equivalent transformations for circuit analysis.
These transformations explicitly exploit the connectivity of electric circuits.
h =
4.4
Input Impedance
The input impedance is defined as the equivalent impedance with respect to the input
terminals of the source. It can sometimes be found through the successive use of the
equivalent transformations of parallel and series connections, although the circuits
may originally be drawn in such a way as to obfuscate the connections between the
elements. Consider the following examples.
EXAMPLE 4.6 Consider the circuit shown in Figure 4.11. This circuit may seem
to be complex with impedance Z4 connected diagonally between the other four
impedances, but in actuality this circuit is no more difficult to analyze than any of
the others we have already considered. As a proof, we redraw the above circuit (see
108
Chapter 4. Equivalent Transformations of Electric Circuits
Figure 4.11: Example circuit.
Figure 4.12). It is the same circuit as before, just represented in a more familiar
manner which reveals the parallel and series connections. This circuit is easy to
transform by using the methods already discussed. Indeed, it is now easy to see that
impedances Z5 and Z6 are in series and that they together are in parallel with Z4.
The three together are in series with Z3, and that whole group is then in parallel with
Z2. Finally, it is obvious that the group of five impedances mentioned above is in
series with Zj and Z7. Therefore, the whole circuit can be reduced to one equivalent
impedance:
7
_(|S^+z 3)z2
= Z4(Z +Z )
5
6
Z4 ~t~25 ~^~Z^
+ Z^+Zi
+ Zx + Z7.
(4.40)
Although writing a formula like this one may seem complicated, one should try to
become fluent at reading circuits and recognizing the connections. Figuring out input
impedances should become natural with practice.
■
EXAMPLE 4.7 Consider the circuit shown in Figure 4.13. This is another circuit
which might be confusing atfirstglance. This circuit also features diagonally arranged
impedances, which give some appearance to this circuit of being more complex than
it really is. Again, we can redraw this circuit in a more "enlightening" manner which
Figure 4.12: The example circuit redrawn more clearly.
109
4.4. Input Impedance
Figure 4.13: Another example circuit.
reveals its hidden simplicity (Figure 4.14). By joining the nodes connected only by
wires with no impedances, we can now clearly see that Z2 and Z3 are in parallel, as
well as Z4 and Z5. Those two groups are in series with each other and together are in
parallel with Z6. Impedances Z\ and Z7 are then connected in series with the rest of
the circuit. Therefore, the whole circuit can be reduced to the equivalent impedance:
Z4Z5
7
+ z4+z5
+ Zi + Z 7
=
Z* + z z +
z4+z5
2 3
ZAZ$
(4.41)
EXAMPLE 4.8 The circuit for this example is known as the R—2R ladder circuit
and is depicted in Figure 4.15a for a three-stage configuration. What is the equivalent
input resistance of this system?
The two rightmost resistors have resistance R and are connected in series to
yield an equivalent resistance of 2R. This equivalent resistor is in parallel with a 27?
resistor as indicated in Figure 4.15b and the equivalent resistance of this combination
is 2R X 2R/(2R + 2R) = R. The cycle repeats itself as this equivalent R resistor is
in series with another R resistor and that combination is in parallel with a 2R resistor
(Figure 4.15c) to produce an R equivalent resistance which is in series with the final
j
y
Figure 4.14: The previous circuit redrawn.
Chapter 4. Equivalent Transformations of Electric Circuits
110
(a)
(b)
R
Z„
< 2R <. 2R
(c)
2R
(d)
Figure 4.15: AnR-2R ladder.
R resistor. Thus the equivalent input resistance is 2R as shown in Figure 4.15d. It is
left as an exercise to show that the equivalent resistance of an infinite R—2R ladder
is also equal to 2R.
■
4.5
Symmetry
Sometimes no matter how a circuit is redrawn, it is just not possible to simplify
it any further by using only the tools we have already developed. For some of
these circuits, symmetry can be successfully used for analysis. Consider the circuit
shown in Figure 4.16a. The input impedance for this circuit is not immediately
apparent. The impedance connected between the nodes A and B prevents the use
of the equivalent transformation of parallel connections, so something else must be
done. In this example, we are aided by the observation that all the impedances in the
vertical branches are of the same value, Z. For this reason, the two vertical paths that
the current could possibly take are identical (they are indistinguishable, which is a
reflection of their symmetry). As a result, the current will be evenly divided between
111
4.5. Symmetry
A
\
o—[
1Z
AM
\-¿B
o-HZZht
X
z
iM
xnhL
AJCD^B
i
T
7
I
(a)
(b)
Figure 4.16: A circuit solvable by symmetry: (a) original circuit and (b) circuit redrawn
to show rotational symmetry.
them. Another way to look at the symmetry of the circuit is to redraw it as shown in
Figure 4.16b. The circuit looks exactly the same if we rotate it around the indicated
axis through 180°, even though this interchanges the currents paths for I\ and ^
Thus, these paths are indistinguishable and the currents must be the same. Therefore,
the voltage drops across the two top impedances will be the same. This means that
the potentials of nodes A and B are equal. In other words, the potential difference
between these points is zero. If the potential difference between A and B is zero then
no current flows between these nodes, and the impedance Z3 between them can be
omitted. Thus, we can just redraw the circuit as shown in Figure 4.17 and analyze it
easily.
The key to the above argument is the observation that the currents flowing
through each vertical branch will be the same, since there is nothing to distinguish
n
ï ï
_r
Figure 4.17: Circuit with middle branches removed.
112
Chapter 4. Equivalent Transformations of Electric Circuits
>
<
Ï
-T
Figure 4.18: The circuit with equipotential points connected.
one branch from another. This is an example of a symmetry argument. Most symmetry
arguments use similar approaches to eliminate unnecessary branches or to connect
different nodes together into one. For example, the circuit in Figure 4.16 can also
be redrawn as shown in Figure 4.18. This is because the equipotential nodes can be
connected together into one node. Of course, analyzing the circuit in this manner will
yield the same input impedance. Indeed, from Figure 4.17 we find:
2Z
(4.42)
zin = — +zl+z2 = z + z{+ z2.
From Figure 4.18 we derive:
Z
Z
Z + Z\ + Zo.
Zin — — + -^ + Z\ + Z2
(4.43)
Thus, we arrive at identical expressions for Zin, which supports our previous assertion.
Sometimes a circuit is drawn in a manner which hides the symmetry it contains.
As an example, consider the circuits shown in Figure 4.19. The left circuit contains the
same type of symmetry present in the Figure 4.16, although it may not be immediately
visible. But when redrawn as the right circuit, we can clearly see it. By the same
argument as in the first example, we can conclude that there will be no current flow
through the horizontal branch, and, thus, we can remove it without altering the input
impedance. The problem is now reduced to a simple parallel transformation, yielding
A
O
-
z.
I
+-
1
Z
rh
Z
Figure 4.19: A circuit with hidden symmetry revealed.
X
113
4.5. Symmetry
the input impedance:
(4.44)
7—7
We can alter the previous circuit in a small way to demonstrate another aspect
of symmetry. Consider the same circuit as before, but this time with impedances
that are not all equal (refer to the circuit on the left in Figure 4.20). The symmetry
argument can still be applied to find the input impedance, although the impedances
are no longer all equal. First we observe that there is symmetry between the upper
and lower parts of the circuit, that is, the symmetry with respect to impedance Z2. In
other words, the upper and lower parts of the circuit are identical. Because ofthat, the
voltage drops over the upper and lower parts are the same. Thus, the potentials at the
nodes A and B are equal, meaning that no current will flow between them and that
we can remove the impedance in between. This results in the circuit shown on the
right in Figure 4.20. Now we can find the input impedance using series and parallel
transformations:
(2Z)(2Z0
2Z + 2ZX
2ZZ!
Z + Zx
(4.45)
EXAMPLE 4.9 This is a somewhat more challenging problem with symmetry. Figure 4.21 depicts a three-dimensional cube with each branch containing an impedance
Z. We wish to find the input impedance between points 1 and 7. This circuit cannot
be simplified by using standard transformations, so we shall try to use a symmetry
argument. We start by considering the current entering at node 1. By symmetry, we
can see that the three paths the current can take are equivalent, so the current will
be evenly divided. (This is because a rotation of the circuit through 120° or 240°
around an axis which goes through nodes 1 and 7 leaves the circuit unchanged.)
Consequently, the potentials at nodes 2, 4, and 5 will be equal, and we can connect
these three nodes together. A similar argument can be used for nodes 3, 6, and 8 by
considering the incoming current at node 7, which is also evenly divided. This means
that these three nodes can be connected together as well. This results in the circuit
Î
A
+-EZht B
z.
X—J
O-
X
Figure 4.20: Symmetrical circuit with different impedances.
114
Chapter 4. Equivalent Transformations of Electric Circuits
Figure 4.21: A cube of impedances.
shown in Figure 4.22. Now, we can see that the input impedance is:
Z
Z
Z
5Z
Zinm = - + - + - =
.
3
6
3
6
(4.46)
EXAMPLE 4.10 Consider the problem of measuring the resistance of resistor R connected to some (unknown) circuit by using an ohmmeter and only one measurement.
By using the conventional technique and connecting the ohmmeter to the terminals 1
and 2 of the resistor, we will produce a reading which corresponds to the equivalent
resistance of R in parallel with the input resistance of the circuit. Thus, the conventional technique does not work. It may first seem that the posed problem cannot be
solved without disconnecting the resistor. However, the solution of the problem can
be accomplished by using the measuring scheme shown in Figure 4.23. Here, the
ohmmeter has one of its terminals connected to nodes 1 and 2, while the other termi-
2,4,5
3,6,8
Figure 4.22: The simplified impedance cube.
115
4.6. The Superposition Principle
ho
Rest
of
Circuit
ho
Figure 4.23: Ohmmeter connected to a resistor.
nal is connected to the middle of the resistor, at point 3. With the ohmmeter "hooked"
in this manner, the resistor R is "electrically disconnected" from the rest of the circuit
and its resistance can be easily measured.1 To see this, consider the potentials of
nodes 1 and 2. Since they are both connected to the same terminal of the ohmmeter,
they are at the same potential. Thus, no current will flow between them through the
connected circuit in the box. The current will flow in symmetrical fashion from node
3 to nodes 1 and 2. Therefore, we can redraw the diagram as shown in Figure 4.24.
From Figure 4.24 it is clear that the ohmmeter will record a measurement of R/4
since the two halves of the resistor are in parallel. In this manner, we can measure
the resistance of the resistor R without having it physically disconnected from the
circuit.
JR
2
v
1 IAA/W
k/vv\
Figure 4.24: Equivalent circuit for the measuring technique.
■
4.6 The Superposition Principle
In the previous sections, we analyzed electric circuits containing only one source.
We used equivalent transformations to reduce these circuits to forms which yielded
1
(Of course, not all resistors are exposed in the middle.)
116
Chapter 4. Equivalent Transformations of Electric Circuits
easily obtainable solutions. This approach can be generalized to electric circuits with
several sources. This can be done by using the principle of superposition.
The superposition principle is a very important concept in the circuit theory. This
principle can be formulated as follows. Consider an electric circuit which contains
many sources. We can subdivide these sources into several (nonintersecting) groups
and consider different regimes of the original circuit when only sources of one group
are active while all other sources are set to zero. It is clear that the number of different
regimes is equal to the number of the groups into which all sources are divided. The
superposition principle states that the actual branch currents in the original circuit are
equal to the superposition (the algebraic sum) of the corresponding branch currents
for all different regimes of the circuit. This principle is based on the linearity of the
basic circuit equations and can be easily understood (or proved) from the following
mathematical fact. If the right-hand side of the basic equations (3.123) and (3.124) is
represented as an algebraic sum of several right-hand sides, then the solution of the
basic equations is equal to the algebraic sum of solutions of these equations obtained
for each of the right-hand sides. Since the right-hand side of the basic equations
(3.123) and (3.124) contains only sources, the subdivision of their right-hand side
into several right-hand sides is equivalent to the subdivision of all circuit sources into
several groups. This remark clearly suggests that the above-mentioned mathematical
fact is equivalent to the superposition principle.
There are several facts which should be clearly understood while using the
superposition principle. First, the superposition principle is valid for any subdivision
of sources into several groups. A choice of subdivision is usually suggested by the
structure (connectivity) of the electric circuit being analyzed. There is nothing which
prohibits having each group consist of only one source. In this case, we shall have as
many regimes as the total number of sources in the electric circuit. Second, when we
set all sources (except those which belong to an active group) to zero, it means that
we set the corresponding voltages and currents of those sources to zero. If a voltage
source is set to zero, it means that it is replaced by a short-circuit branch (see Figure
4.25), because the voltage across this branch is equal to zero. If a current source is set
to zero, it means that it is replaced by an open branch (see Figure 4.26), because the
current through an open branch is equal to zero. These replacements of sources by
short-circuited and open branches often lead to significant simplifications of electric
circuits, simplifications which allow one to exploit a particular connectivity of the
electric circuit being analyzed.
v.© —>
Figure 4.25: Voltage source set to zero.
117
4.6. The Superposition Principle
sk
©
Figure 4.26: Current source set to zero.
EXAMPLE 4.11 Consider the circuit shown in Figure 4.27. It is not much different
from the circuit shown in Figure 4.8, except that it is driven by two sources. We want
to find all the branch currents.
We begin, as before, by representing the circuit in phasor-impedance notation
(Figure 4.28) and by introducing reference directions (these directions are especially
important in the superposition principle). We calculate all the impedances and convert
the sources into phasor form:
Z, = Ri + joLi = 1 + 7(2)
J
"C2
Zo =
Z3=R3+
J
(2)(i)
=
= i + ya
(4.47)
-2jil,
(4.48)
1 + 3 y ft,
jcoL3 = 1 + i(2)
(4.49)
(4.50)
V!(f) = c o s 2 i ^ V , = IV,
v2(r) = \/2cos(2i - 45°) — V2 = y/2e~j*
=
1-;V.
(4.51)
Now we apply the superposition principle. We will consider two separate regimes,
each with one source. Because both sources in the above circuit are voltage sources, we
set them to zero by replacing the corresponding sources with short-circuit branches.
We can now consider two separate circuits, corresponding to the two separate regimes.
These circuits are denoted as a and b, respectively, in Figure 4.29. The first circuit
contains only the left voltage source, while the second circuit has only the right one.
1/2 H
©v,(0
1a
1a
/ V r W W ia(t)
i t (t)A A AMt)
+
cos( It) V
1/4 F
3/2 H
Q v 2 (/) = 4Ïcos( It - 45°) V
Figure 4.27: A circuit excited by two sources.
118
Chapter 4. Equivalent Transformations of Electric Circuits
-CD-
ex
à
Figure 4.28: The same circuit represented in phasor-impedance notation.
It is important to introduce reference directions for these circuits, but they do not
have to be the same as those introduced for the original circuit. For this problem, it is
convenient to reverse the direction of the current through Z2 in circuit b to conform
to the model of the current divider. Note that the branch currents shown in the new
circuits are not the same as the original branch currents. They represent two different,
separate components of actual branch currents which must be added together to find
the actual currents.
Now circuits a and b can be analyzed by using the techniques of equivalent
transformations. We omit the details since the procedure is identical to the one being
used for the analysis of the circuit shown in Figure 4.8. For circuit a, we first find 1°
by dividing the voltage of the source by the equivalent impedance with respect to the
terminals of the first source:
Vi
(4.52)
z2
Z,1 + zz33+Z2
Now ¡2 and 1% can be found by applying the current divider rule to the parallel
branches:
TCI
/f(Z 3)
(4.53)
Z, + Zo
Circuit a:
Circuit b:
CX
Figure 4.29: Separate circuits a and b.
119
4.6. The Superposition Principle
ja = Í L Í ^ _ .
(4.54)
The currents in circuit b can be found in a similar manner. Dividing the voltage by
the equivalent impedance of the circuit with respect to the terminals of the second
source, we find/^:
i" = ~ïzT-
(4-55)
Z, + Zx+Z2
Next, we use the current divider rule:
/f = ^LL y,
Z\ +z2
(4.56)
j* = È^L.
2
zx + z 2
(4.57)
I\ = /f + /?,
(4.58)
h = %- %
(4.59)
1
(4.60)
Now that the branch currents for each of the two separate regimes have been found,
we must add them with the proper signs to find the actual currents. When adding the
regime currents, the proper signs are determined by comparing the initial reference
directions with the reference directions for each regime. If the reference directions of
the currents in circuits a and b coincide with the reference directions in the original
circuit, then the currents are taken with a plus sign; otherwise they are taken with a
minus sign. By examining the reference directions in circuit a, we can see that all
of them coincide with the reference directions in the original circuit. However, for
circuit b, the reference direction for l\ is opposite to the reference direction of I2,
so it will be taken with a minus sign. Consequently, the expressions for the branch
currents are
h = H + Il
The numerical values obtained according to equations (4.52)-(4.60) are listed ( in
amperes) below:
11
"
6 '
l{
3
ja _ 1 + v
~
6 '
/» = !
H = -j/3,
n = -JA
h
12
A
3'
- d + j)
6
-1+3;
h= 6 '
h=
h = -4/76.
ick to the time domain we arrive at:
ix{t) ~--
6
cos(2i -- 135'')A,
120
Chapter 4. Equivalent Transformations of Electric Circuits
hit) = ^— cos(2i + 108.4°) A,
6
2
hit) = - c o s ( 2 f - 9 0 ° ) A .
This concludes the analysis of the circuit.
■
As is clear from the above example, the superposition principle presents a way to
analyze electric circuits with multiple current and voltage sources. This method will
also work for circuits with multiple sources of different frequencies. In this case, we
first use the superposition principle and then convert a circuit into phasor-impedance
form for each regime (frequency) separately. Next, we analyze the separate regimes
as before (using different frequencies to calculate the impedances) and convert the
phasors back into time domain form before summing for the actual currents. This
procedure will preserve the proper frequency dependence of the various sources and
their respective responses.
EXAMPLE 4.12 Consider the same circuit in Figure 4.27 but this time with the
source:
v2(f) = V ^ c o s ^ i - 45°) V.
(4.61)
Regime (a) of the problem remains exactly the same as before. We convert the currents
for this regime to the time domain separately to get:
Í1
ia\it) = ^ r - cos(2/ + 45°) A,
6
(4.62)
iiit) = ^ ~ cos(2/ + 71.6°) A,
6
(4.63)
if (i) = ~ cos(2i - 90°) A.
(4.64)
For regime (b), the phasor V2 is still equal to 1 — j . However, we get new values for
the impedances when we insert œ = 4 into equations (4.47)-(4.49):
Z{ = R{ + jo)Lx = 1 + 2 . /
Z2 = -j/uG2
= -j a
Z 3 = R3 + > C 3 = 1 + 6 . / a
ft,
(4.65)
(4.66)
(4.67)
Inserting these values into equations (4.55)-(4.57) results in:
/? = (3 - j)/\5 A,
(4.68)
l| = (-l+y)/3A,
(4.69)
/ | = -2(1 - 2 7 V 1 5 A.
(4.70)
When we convert these phasors to the time domain and add to the results of the first
regime, we arrive at the final answer:
121
4.6. The Superposition Principle
cos(2i + 45°) + ^ - cos(4/ - 18.4°) A,
15
(4.71)
10
\/2
h(t) = ^—- cos(2r + 71.6°) - ^— cos(4i + 135°) A,
6
3
(4.72)
hit) = i cos(2í - 90°) + = y ^ cos(4r + 116.6°) A.
(4.73)
h(t) = ^
6
EXAMPLE 4.13 We now consider a rectangular grid of impedances which extends
to infinity (practically this can be realized by using a large grid of impedances, while
only considering measurements away from the edges).
We want to find the input impedance between nodes A and B, shown in Figure
4.30. To solve this problem, we must make use of both symmetry and the superposition
principle. We begin by writing the expression for Zin in terms of the voltage applied
across A and B and the current flowing through the grid:
r(A,B) = Y^l
I
(4.74)
This is actually the definition of input impedance.
The approach we take is to find VAB in terms of the current I by using the
superposition principle. We consider two separate regimes, illustrated in Figure 4.31.
In the first regime the current / is introduced into the grid at the node A (and flows
out of the circuit infinitely far away), while in the second regime the current / is
0
0
Ft * !
-XQXQXQJ Figure 4.30: An infinite grid of impedances.
122
Chapter 4. Equivalent Transformations of Electric Circuits
Figure 4.31: Two regimes for infinite grid problem.
taken from the grid at node B. In the first regime, we conclude from symmetry that
the current will be evenly divided into four equal parts since no path from A can be
distinguished from another. Thus, the current flowing through the impedance between
points A and B is 7/4 and the voltage is then:
(1)
t>
=- Z
V
AB
(4.75)
4^ -
For the second regime, we can make the same argument to show that:
V
AB
(4.76)
4¿-
Thus, by using the superposition principle we can find that the total voltage is the
sum of these two voltages:
VAS = «SB + «2 = i z + i z = | Z
(4.77)
If we substitute (4.77) into (4.74), we will find the desired input impedance:
y(A,B) =
-7
IT.
/
=
7
_
(4.78)
2*
EXAMPLE 4.14 As a final example of the superposition principle we consider the
three-stage digital-to-analog (D/A) signal converter shown in Figure 4.32a. We want
to convert a binary number into an output voltage (or current). We will use ¿>0 to denote
the least significant digit, b\ to represent the intermediate digit, and b2 to indicate the
most significant digit. We can express the binary number as b2b\b§. Thus, the voltage
sources b0,bi, and b2 will either be set to zero volts or set to one volt to reflect their
digital nature. For three stages we can go from 0 (000) up to 7 (111). The output
voltage Vb is the voltage across the output 2R resistor as indicated in the figure. We
want to show that this voltage is proportional to the number represented by b2bib0.
There are three independent voltage sources and so we will consider three separate regimes. First, let us set b\ and b0 to zero and find the dependence of V0 on
b2. The resulting circuit is shown in Figure 4.32b. The part of the circuit enclosed
123
4.6. The Super-position Principle
2R
V. ¿R
b 2 Q (c)
A/WVi
R
R
2R.
+
V
2R<
2R'
>,Q 2 R < 2R'
(d)
(e)
Figure 4.32: A three-stage D/A converter.
in the box is just the R—2R ladder that was discussed in Section 4.4. Its equivalent
resistance of 2R is in parallel with the output resistance. The simplified equivalent
circuit, shown in Figure 4.32c, is just a voltage divider. Using equation (4.26), we
see that the sought relationship is given by VQ = b2 X R/(R + 2R) = b2/'i.
124
Chapter 4. Equivalent Transformations of Electric Circuits
The circuit that results when we set b2 and b0 to zero is shown in Figure 4.32d.
The equivalent resistance of the resistors in both boxes is 2R. We must be careful when
we make this simplification because we lose direct information about V0. However,
by the voltage divider rule, we know that VQ will be one-half the voltage across the
equivalent 2R resistance. The simplified equivalent circuit for this case looks just
like that in Figure 4.32c if we replace b2 by b\ and Vb by 2V0. Therefore, another
application of the voltage divider formula results in V0 = b\/6. The circuit that
results when we set b2 and b\ to zero is shown in Figure 4.32e. The part of the circuit
enclosed in the box is another R—2R ladder. Again we lose direct information about
VQ when simplifying, but careful analysis will yield the expression Vb — bo/12.
The final result comes from summing up the three individual answers, V0 =
(4b2+2b\+bo)/\2, and gives us the proper conversion formula (with a proportionality
constant).
■
4.7
An Introduction to Electric Circuit Simulation
with MicroSim PSpice
The use of computer simulation tools for the analysis of electric and electronic circuits
pervades the modern workplace. These codes are typically employed when the circuits
contain a sufficient number of elements to render impractical the computation of
circuit voltages and currents by hand. This is the case for many circuits of interest.
While most of these circuits can be broken down into subcircuits that can be analyzed
by hand, analysis of the performance of the complete circuit is often of paramount
importance and can be done only via computer. These simulation tools are also useful
for investigating circuits of any size which have nonlinear elements. Furthermore,
they can also be used to evaluate the dependence of output quantities on the parameter
tolerances or the nonideal properties of various circuit elements.
In this book we will make use of numerical simulations only to examine the
relatively small, linear, electric circuits that we can also analyze by hand. The purpose
of this is threefold. First, because of the prevalent use of simulators in industry, you
should have a working knowledge of at least one circuit simulation package and have
a general understanding of their capabilities. Second, once you have the ability to
simulate circuits, you can use the codes to check your hand calculations and perhaps
even to investigate deviations from ideal circuit performance. Finally, if you utilize a
code which contains a good graphics package (like the one described here), you can
readily obtain visual information about any circuit variable in either the time or the
frequency domain.
It cannot be overemphasized that a circuit simulator is not a substitute for the
knowledge of basic circuit theory principles. It will not design circuits from scratch
for you. It will not troubleshoot a circuit for you if it is not performing according
to expectations. Perhaps most important, the code will likely, without hesitation or
complaint, give you incorrect or unreasonable answers if you give it incorrect or
unreasonable input data. The only way you can trust the simulation results is to
4.7. An Introduction to Electric Circuit Simulation with MicroSim PSpice
125
have a solid understanding of Kirchhoff's laws, terminal relations, and the rest of the
electric circuit theory concepts.
In this section we will give a brief introduction, via several examples, to the
operation of the MicroSim PSpice circuit simulator and auxiliary routines. At the end
of Chapters 6 through 10 you will find examples and homework problems which are
based on this simulator. PSpice was chosen for two reasons. First, it is based on the
SPICE2 simulator developed at University of California Berkeley in the 1970s, which
is the basic engine (albeit often modified) for a number of simulators that are widely
used in industry. Second, there is available an evaluation copy of MicroSim's family
of products, which is comprised of a schematic generator, the PSpice simulator for
analog and digital components, and the PROBE graphics package. This software is
available free to professors and can be copied for students with the encouragement
of the manufacturer, MicroSim Corporation.2 This discussion is based on the IBMPC compatible version 6.2 for Microsoft Windows. Furthermore, there are often site
licenses for the regular SPICE simulator on the various workstations that are typically
available to students. If one is familiar with the operating system and can locate the
SPICE code, this introduction should be sufficient to enable one to simulate circuits
in that environment as well.
This introduction is by no means a substitute for an extensive operating manual
for PSpice. It contains only a bare-bones description designed to give the user the
ability to perform the basic operations of circuit simulation. There are a number of
excellent texts dedicated to PSpice simulation. Several of them are listed in Appendix
C. The Reference guide from MicroSim Corporation is also indicated there. However,
after mastering the basics, one can also learn more about the code's capabilities by
roaming through the various drop-down menus and experimenting with the various
options.
Before proceeding, the evaluation copy must be loaded onto an IBM-PC. Version
6.2 is usually supplied on five 3 1/2 inch 1.44 MB floppy disks or a CD-ROM. Installation follows the usual procedure for any Windows 3.1 software.3 Disk 1 is loaded
into the floppy drive and the setup.exe program on that disk is executed either by
double-clicking on the "setup.exe" listing in the File Manager or by using the "Run"
command on the "File" drop-down menu in the Program Manager. (Windows 95 users
can use the Windows Explorer or the control panel to load the program.) Follow the
directions that appear on the screen to load the software, generate a Program Group
entitled "MicroSim Eval 6.2," and make any necessary changes to your autoexec.bat
file. The files will nominally be saved in a directory called "MSIMPR62." The pro-
submit a request on educational letterhead to: Product Marketing Department, MicroSim
Corporation, 20 Fairbanks, Irvine, CA 92718. Their phone number is (714) 770-3022.
3
Users of older versions of Windows may have to also install "Win32s" in order to run
PSpice. This code is also supplied on the PSpice CD-ROM and can be transferred to two floppy
disks for distribution. Alternatively, one can use an older version of MicroSim's products.
Version 6.0, for example, has most of the Version 6.2 features and runs without "Win32s."
126
Chapter 4. Equivalent Transformations of Electric Circuits
gram group should contain at least five items: 1) MicroSim Schematics, 2) MicroSim
PSpice, 3) MicroSim Probe, 4) MicroSim Stimulus Editor, and 5) MicroSim Parts.
There will also be a "README" file which has some information about the latest
versions of the codes. We will always begin our examples by double-clicking on the
Schematics icon; all of the other programs can be invoked at the proper time from the
Schematics window. (Unless otherwise stated, "clicking" always refers to pressing
the left button on a two-button mouse.)
The first circuit we will analyze is the current divider of Example 4.5. The first
step will be to draw the schematic of the circuit. Then we must select the types of
analyses that we would like PSpice to perform. In this example we claim that this circuit functions as a low-pass (frequency) circuit, so we will plot the current through the
resistor as a function of frequency. Third, we must run the PSpice simulator. Finally,
we will use Probe to draw the resistor current over the specified frequency range.
We begin by opening the "MicroSim Eval" Program Group in Windows and
double-clicking on the Schematics icon. (Windows 95 (or later) users must use the
start button and move through the various menus to arrive at the "schematics" label.)
From the drop-down menus at the top of the window, click on "Draw" and then on
"Get New Part..." A window entitled "Add Part" will appear and in the dialog box
you should enter "C" and then press the "OK" button (by clicking on it). The box
will disappear and the cursor arrow will have a capacitor attached to it. Move the
capacitor to a desirable location and place it by clicking the left mouse button. You
can continue to add capacitors by repeating the last two operations. Click the right
mouse button to stop adding capacitors. If you have placed any parts unintentionally,
you can delete them by clicking on them (they change color to show that they have
been selected) and then hitting the delete key. Next click the "Edit" drop-down menu
and select "Rotate" to get the capacitor vertical on the page.
Default values for the capacitor's name and capacitance value are indicated in
the schematic. The capacitor's name is typically Cn, where n is an ascending integer.
The name can be changed by double-clicking on it so that the "Edit Reference
Designator" dialog box appears. Type in a new name and hit "OK." Double-click on
the capacitance (whose default is typically 1 nF) to invoke the "Set Attribute Value"
dialog box. For this example set the capacitance to 1 F by entering a " 1 " (farad is
assumed — if you type 1 F it will assume that you meant one femtofarad). The
multiplying factors that can be entered to scale the units are essentially the same
as those listed in Table 1.1. Two exceptions are that "u" is used instead of "/x" to
mean 10~6 and "MEG" must be used to represent 106 because "m" and "M" are both
interpreted by PSpice to mean 10~3.
Add a resistor to the circuit by repeating the steps above for the capacitor, except
that in the "Add Part" dialog box you should enter "R" for a resistor. Change the
resistance to " 1 " (ohms are assumed) and move the part parallel to the capacitor by
"dragging" it with the mouse (holding the left mouse button down while moving the
mouse). Connect the two parts in parallel using the "Draw" drop-down menu and
selecting "Wire." Click once after moving the pointer to the top of the capacitor and
once again at the top of the resistor. Note that the wires are only drawn orthogonally
and that they "snap" to the grid points. You can change this in the drop-down
4.7. An Introduction to Electric Circuit Simulation with MicroSim PSpice
"Options" menu by selecting "Display Options," but the settings are adequate for our
purposes. You can select "Repeat" from the "Draw" menu to place the wire between
the two lower ends of the elements.
It is important to note that PSpice will automatically number the nodes in a
circuit, but that the user must supply the ground node in order for the simulation to
run. You do this by selecting "AGND" in the "Add Part" dialog box. Connect this
ground to the base of the capacitor. The final element of this circuit is the current
source. The notation for this element in the "Add Part" dialog box is "ISRC." Rotate
it twice to get the direction arrow pointing up and then add two wires to place it in
parallel with the other elements. To set the parameters of the current source, doubleclick on the arrow to bring up the window which is titled "II Part name: ISRC."
There are many options that can be adjusted. Because we want to perform an ac
analysis, click on the line that says "AC= " There are two boxes at the top of the
window. The left one is entitled "Name" and "AC" should appear in that box. Click
somewhere in the box on the right, enter a " 1 " for one ampere, and press the "Save
Attr" button. Finally, press the "OK" button. The circuit has now been completely
drawn and should look like the circuit in Figure 4.33.4 Note that you can use the
"Text" option of the "Draw" drop-down menu if you want to add comments to your
circuit.
You are now ready to save this schematic on disk. This is accomplished by using
the "File" drop-down menu and clicking on the "Save" option. The usual file extension
for Schematics drawings is *.SCH; we will call our schematic examplel.sch. The
schematic can be printed (if a printer is connected to your computer) by selecting
"File" and "Print." After selecting any options you want in the window that appears,
press the "OK" button to initiate the hardcopy generation.
Once the file is saved, go to the "Analysis" drop-down menu and click on "Create
Netlist." This will convert the schematic to the tabular form that is required by PSpice.
Also in the "Analysis" menu you can click on "Examine Netlist" to view the data
in this form. This will launch the Windows Notepad editor to allow you to examine
Figure 4.33: The PSpice schematic for Example 4.5.
4
In this edition, it was not possible to incorporate the PSpice screen images directly into the
text. Instead, graphics packages were used to imitate the PSpice output. Consequently, there
will be slight differences occasionally between the figures and your computer screen images.
127
128
Chapter 4. Equivalent Transformations of Electric Circuits
Table 4.1: The circuit Netlist for Example 4.5.
*Schematics Netlist*
C_C1
R_R1
LI
0
0
0
N_0001
N_0001
$N_0001 AC
1
1
1
the Netlist, which has been saved on disk as examplel.net. The result is shown in
Table 4.1. For the capacitor and the resistor (the second and third rows in the Netlist,
respectfully), the first column is the name of the element, the next two entries are the
nodes to which the elements are connected, and the fourth and final entry is the value
(capacitance and resistance, respectfully). The first node entered is always considered
the "plus" node in terms of the assumed reference voltage polarity. Note that PSpice
uses the passive sign convention for all elements, including sources! The plus node in
this example is always the ground node and is indicated by a zero. The only other node
in this example is given the name $N_0001. The final row in the Netlist represents
the current source. The columns are similar to those of the passive elements except
that an "AC" precedes the current value. We could have also assigned currents for dc
analysis, transient analysis, etc. in the appropriate dialog box and they would have
been indicated on this line as well.
Before PSpice can be executed we must indicate the type of analyses to be run.
Go to the "Analysis" drop-down menu and select "Setup." In the "Analysis Setup"
dialog box that appears, normally only "dc Bias Point Detail" is selected, as indicated
by the "x" in the box to the left of that button. Calculation of the dc bias point is
always the first step of a PSpice run. Enable the ac analysis by clicking on the box
to the left of "AC Sweep" so that an "x" appears. Next, push the ac sweep button. A
window will appear entitled "AC Sweep and Noise Analysis" that will allow us to
modify the parameters of the frequency sweep. First, change the "AC Sweep Type" to
"Decade" by clicking in the appropriate circle. Next, in the "Sweep Parameter Box"
set the "Pts/Decade" to 21, the "Start Freq." to 0.01 (Hz), and the "End Freq." to 100.
Press the "OK" button at the bottom of that window. Finally, press the "Close" button
in the "Analysis Setup" window to return to the schematic. We are now ready to run
PSpice.
From the "Analysis" drop-down menu select "Simulate." First a small window
entitled "Schematics" will appear announcing that the Netlist is being generated
(unless this has already been done). Next, a "PSpice" window will appear that will
update you on the status of the simulation. First it will say that the program is reading
and checking the circuit. If any errors are encountered, an error message will be
printed in the window and the simulation will stop. The user must then determine
the error and correct it before attempting another simulation. If there are no errors,
PSpice will proceed with the bias point calculation and then move on to the AC
analysis.
4.7. An Introduction to Electric Circuit Simulation with MicroSim PSpice
After the successful completion of the simulation, the "PROBE" program will
be initiated automatically. There are two points to note. First, even though "PROBE"
usually comes up "full screen," the schematics generator, PSpice, and PROBE are
all running at the same time in different windows and one can move back and forth
between them as necessary. Second, PROBE need not be launched automatically after
PSpice is run; this option is selected in the "Analysis" drop-down menu under "Probe
Setup
" When the PROBE window appears, the frequency axis is drawn but no
curves are drawn. Notice that the independent axis has a log scale as requested in an
earlier dialog box. Select the "Trace" drop-down menu and click on "Add." In the
window that appears, double-click on "I(R1)." The dependence of the resistor current
on frequency is then displayed, along with a dependent axis scale and a legend. The
current is nearly equal to the source current (of 1 A) at low frequency and goes to
zero as the frequency is increased. This is the principal characteristic of the low-pass
filter. Note that we can delete any trace if necessary by clicking on the appropriate
name in the legend (its color will then change) and hitting the delete key. There are
many modifications that we could make to this plot, but we will just modify the labels
and leave other changes for later examples. Click on the "Edit" drop-down menu and
select "Modify Title
" In the resulting dialog box enter "Example 1" and press
"OK." From the "Plot" drop-down menu select "Y Axis Settings," type "Resistor
Current" in the "Axis Title" box, and then hit "OK " Selecting "File" and then "Print"
will bring up a window that can be used to generate a hard copy. This graph is shown
in Figure 4.34.
We conclude this rather long example with a brief discussion of the files that
are generated during a typical circuit simulation. The files examplel.sch and exam-
1.0A
R
e
s
i
s
t
o
r
C
0.5A
e
n
OA
10mHz
lOOmHz
KR1)
1.0Hz
10Hz
Frequency
Figure 4.34: The probe output for Example 4.5.
100Hz
129
130
Chapter 4. Equivalent Transformations of Electric Circuits
Table 4.2: The PSpice input file for Example 4.5.
* D:\ CIRCUITS\ EXAMPLE1.SCH
* Schematics Version 6.2 - April 1995
*SunJun25 15:31:14 1995
** Analysis setup **
.acDEC21 .01 100
.OP
* From [SCHEMATICS NETLIST] section of msim.ini:
.lib nom.lib
■INC "EXAMPLEl.net"
JNC "EXAMPLE Lais"
.probe
.END
plel.net have already been described. The input to PSpice is actually read from the file
examplel.cir, a listing of which is given in Table 4.2. The first line is automatically a
comment line and gives the name of the schematic file. Additional comment lines are
generated by placing an asterisk in the first position. Comments can also be placed at
the end of input lines by separating them from the data by a semicolon. The first two
noncomment lines tell PSpice what types of analyses to run. The first line initiates
the ac analysis and the trailing parameters indicate the method of determining the
frequency points. They come directly from our earlier dialog box entries. The next
line tells PSpice to find the dc operating point. The line which begins with "Jib"
indicates the name of a library that will be used to specify the parameters of various components. This is discussed in detail at the end of Chapter 7 after diodes are
introduced. The two lines beginning with ".inc" tell PSpice to read in the data from
the two files indicated. We have already seen the Netlist file. The other file contains
aliases, i.e., a list of alternate ways to identify nodes. We will examine this file in the
next example. The second to last line invokes the PROBE program at the end of the
simulation and the final line identifies the end of the PSpice data.
The final file that we will point out is the example.out file. This file contains
the output from the PSpice run and can be loaded into Notepad from the schematics
window via the drop-down "Analysis" window by clicking on "Examine Output "
It can also be examined from the PSpice window from the "File" drop-down menu
by clicking on "Examine Output." Scrolling through it, one can find a listing of the
input information as well as some details of the simulation. Error messages that are
generated if the input file is not correct will be given in this file. One curious point is
that next to total power dissipation the output shows 0.00 W. This is because PSpice
calculates only the power dissipated by voltage sources. Another thing that you may
notice is that nowhere are the frequency dependences of the circuit voltages and
currents to be found in this output file. These numbers are stored in another file that
is typically readable by PROBE, but they are not in a simple text format. If we want
to generate a table of the resistor current as a function of frequency in the output file,
4.7. An Introduction to Electric Circuit Simulation with MicroSim PSpice
for example, we must insert the line: ".PRINT AC I(R_R1)" into the example.cir file
somewhere before the ".END" statement and rerun PSpice. Notepad, or any other
editor that is at your disposal, can be readily used to accomplish the required change.
Next, we will use PSpice to analyze the circuit from Example 4.11. We will
perform a transient analysis on this circuit and compare the simulations with the
analytic results. The circuit redrawn by the schematic editor is shown in Figure 4.35.
The resistors, the capacitor, and ground are added and their values are modified as
described in the previous example. The inductors are added by selecting "L" in the
"Add Part" dialog box. Their values (in henries) are modified by double-clicking on
the default values in the schematic.
The name for the voltage sources in the "Add Part" dialog box is "VSIN." By
double-clicking on the left voltage source one brings up a window entitled "VI Part
Name: VSIN." Click on the "Voff = " line and then move to the upper right box and
type a "0" and push the "Save Attr" button. This sets the source's dc offset to zero.
Next, click on the "Vampl = " line, type a " 1 " in the box, and hit "Save Attr" to set
the sine wave amplitude to 1 V. Repeat the steps outlined above to set the frequency
to Freq = 0.31831 (2 rad/s) and the phase to 90°. Note that zero degrees gives a
sine function for the source, so 90 is required to achieve a cosine dependence. When
adding the right voltage source, be sure to rotate it twice to get the positive terminal
facing downward. For this element we need to enter Voff = 0, Vampl = 1.414213
( v 2), F re q = 0.31831, and Phase = 45°. The Netlist for this problem is shown in
Table 4.3. There is a total of six nodes (including ground) because PSpice must label
even the trivial ones.
To select the analysis type we click on "Setup" in the "Analysis" drop-down
menu. We enable the Transient analysis by clicking on the box to the left of the
"Transient... " button. Then, by depressing this button a window appears that allows
us to define the transient analysis. Click on "Skip initial transient solution." Because
the period is about 3.14 seconds, set the "Final Time" to 10 (seconds). That will
allow over three full periods to be simulated. For a total of about 200 printed points,
select the "Print Step" to be 50m. Then click the "OK" button in that window and the
"Close" button in the "Analysis Setup" window. Save this circuit as example2.sch.
We are now ready to run the second simulation.
L1
0.5
V1
R1
R2
L2
1
1
1.5
AAAArVVWV
C1
.25
V 2 v±.
<7,
Figure 4.35: The PSpice schematic for Example 4.11.
131
132
Chapter 4. Equivalent Transformations of Electric Circuits
Table 4.3: The circuit Netlist for Example 4.11.
R_R1
R_R2
L-Ll
L_L2
C_C1
V_V1
+ SIN
V_V2
+ SIN
N_0002
N.0001
N_0004
N_0003
0
N_0004
0
0
0
N_0001
N_0003
N_0002
N_0005
N_0001
0
1
N-0005
1.414213
1
1
0.5
1.5
0.25
0.31831
0
0
90
0.31831
0
0
45
From the "Analysis" drop-down menu select "Simulate." The usual sequence of
windows will come and go and, if there are no errors, the "PROBE" program will
be initiated automatically. The input file for PSpice is displayed in Table 4.4, and
the alias list is given in Table 4.5. The only new input in the example2.cir file is the
directive for the transient analysis, which includes the print step, the final time, and
an instruction to skip the initial conditions. The alias file just gives a list of alternate
expressions for the nodes in terms of the circuit elements. For example, the first line
after the .ALIASES line means that Rl : 1 should be interpreted as $N_0002 and Rl :2
is the same as $N_0001. In the next line and in the seventh line, R2:l and Cl:2 are
also made equivalent to $N_0001, respectively, since they are also connected to that
node.
When the PROBE window appears, the time axis is drawn but no curves are
present. Select the "Trace" drop-down menu and click on "Add." In the window that
appears, one can select time, five voltages that correspond to each of the nonzero
node potentials via aliases, a current through any of the passive or active elements,
or the ground potential. One can also select combinations of the variables. For
example, the voltage across the resistor Rl can be plotted as follows. Click on the
"Trace Command" line and then enter "V(R1:1)-V(R1:2)" and press "OK." We can
Table 4.4: The PSpice input file for Example 4.11.
* D:\ CIRCUITS\EXAMPLE2.SCH
* Schematics Version 6.2 - April 1995
* Mon Jun 26 02:29:11 1995
** Analysis setup **
.tran 50m 10 SKIPBP
.OP
* From [SCHEMATICS NETLIST] section ofmsim.ini:
.lib nom.lib
.INC "EXAMPLE2.net"
.INC "EXAMPLE2.als"
.probe
.END
4.7. An Introduction to Electric Circuit Simulation with MicroSim PSpice
Table 4.5: The PSpice alias file for Example 4.11.
* Schematics Aliases *
.ALIASES
R_R1
R.R2
L_L1
L_L2
C_C1
V_V1
V_V2
.ENDALIASES
R1(1=$N_0002 2=$N_0001 )
R2(1=$N_0001 2=$N_0003 )
Ll(l =$N_0004 2 = $N_0002 )
L2(1=$N_0003 2=$N_0005 )
C1(1=0 2=$N_0001)
Vl(+=$N_0004-=0)
V2(+=0-=$N_0005)
plot the analytic result for comparison as follows. Select the "Trace" drop-down
menu again and click on "Add." Click on the "Trace Command" line and then enter
"0.2357 * cos(2 * time - 2.35619)" and then press "OK." [The analytic result is
vÄ1(i) = (v^/6)cos(2f-135°).]
Probe can be used to make multiple plots. Click on "Add Plot" in the "Plot" dropdown menu. Repeat the above sequences, but enter "V(R2:1)-V(R2:2)" to plot the
voltage across R2. The analytic result for R2 is vRl{t) — (2/3) sin(2i). The resulting
plots are shown in Figure 4.36. Note that the curves do not agree well at early times
but are nearly indistinguishable at later times. The difference between the curves is
called the transient response and is the subject of Chapter 7.
Our final example will be that of the three-stage digital-to-analog converter that
was analyzed in Example 4.14. We will perform only a detailed bias point analysis,
but we will use the "Parametric" analysis to change the voltage output of some of
the sources. The circuit as it is redrawn in PSpice is shown in Figure 4.37. The
resistors and the ground are drawn and the resistances are modified as in the previous
examples. The name for the voltage sources in the "Add Part" dialog box is "VSRC."
By double-clicking on the left voltage source one brings up a window entitled "VI
Part Name: VSRC." Click on the "DC = " line then move to the upper right box and
type "{Vt}" and push the "Save Attr" button and then "OK." Vt will be the name of the
parameter that we will assign two values to: 0 V for "off" and 5 V for "on." Symbolic
values will not work properly without the braces { }. Repeat the above procedure for
the middle voltage source. Set the "DC = " voltage on the rightmost voltage source
to 5 V. Don't use braces around the 5, of course, since it is a numeric value.
From the "Analysis" drop-down menu select "Setup." Click the box to the left of
the "Parametric" button to place an "x" and activate this sweep. Press the "Parametric"
button and a window will appear that will allow us to define Vt. Set the "Swept Var.
Type" to "Global Parameter" and set the "Sweep Type" to "Value List" by clicking
on the circle to the left of each name (or on the name itself). Click on the "Name"
box in the upper right corner and enter "Vt." Click on the "Value" box in the lower
right corner and enter "0, 5." Finally, hit the "OK" and "Close" buttons and save the
133
134
Chapter 4. Equivalent Transformations of Electric Circuits
1.0 r
SEL»
Os
2s
D v(r2:1)-v(r2:2)
O
-500m L
Os
2s
D v(r1:1)-v(r1:2)
O
4s
6s
0.666*sin(2*t¡me)
4s
6s
8s
0.2357*cos(2*time-2.35619)
500m
Time
Figure 4.36: The PROBE output for Example 4.11.
R1%2
R8 S 1
Figure 4.37: The PSpice schematic for Example 4.14.
10s
4.7. An Introduction to Electric Circuit Simulation with MicroSim PSpice
Table 4.6: The analysis and Netlist sections of the output file for the PSpice Example
4.14.
** Analysis setup **
.STEP PARAM Vt LIST
+ 0,5
.OP
* Schematics Netlist *
R.R1
R_R2
R.R3
R_R4
R_R5
R_R6
R_R7
R_R8
V.V1
+{vt}
V.V2
+{Vt}
V_V3
0 $N_0001
$N_0002 $N_0001
$N_0001 $N_0003
$N_0004 $N_0003
$N_0003 $N_0005
$N_0006 $N_0005
$N_0005 $N_0007
0 $N_0007
$N_0002
2
2
1
2
1
2
1
1
0
DC
$N_0004
0
DC
$N_0006
0
DC
5
schematic as example3.net. (Note that you cannot run PSpice until the schematic has
been saved.) We are now ready to perform the simulation.
Select "Simulate" from the "Analysis" drop-down menu to execute PSpice. When
the analysis is complete, select "Examine Output" from the "File" drop-down menu
in the PSpice window. The example3.out file will be loaded into the Notepad program
for you to view. Selected lines from that file are listed in Table 4.6, and Table 4.7.
The ".STEP" line in the Analysis setup section in Table 4.6 indicates the values of
Vt at which the circuit variables will be computed. The lines for V_V1 and V_V2 in
the Schematics Netlist section show the parameterized dc voltage.
Table 4.7 contains the results of the two DC bias point analyses for the different
values of Vt. Note that the first case corresponds to the digital number 1 because V3
is the only non-zero voltage source. The output resistor is Rl and the expected output
voltage for this case according to Example 4.14 is one-twelfth of the "on" voltage.
We do in fact see 5/12 = 0.4167 V for $N_0001 in the node voltage table. Note that
the current flowing through V3 is given as —1.667 A. This source is supplying 8.33
W so clearly current is flowing out of the plus terminal of the source. The negative
sign is a direct consequence of the passive sign convention. Current is flowing into
the other two sources, but no power is consumed since their voltages are zero.
All the voltage sources are "on" in the second case. This corresponds to a digital
7 and the output voltage at node $N_0001 is seven times that of the first case. Note
that all voltage source currents are now negative as they are all supplying power to
the circuit.
135
136
Chapter 4. Equivalent Transformations of Electric Circuits
Table 4.7: The DC bias result sections of the output file for the PSpice Example 4.14.
**** 06/28/95 03:51:28 ******* Win32s Evaluation PSpice (April 1995) *********
* D:\CIRCUITS\EXAMPLE3.SCH
*** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C
*** CURRENT STEP PARAM VT = 0
***********************************************************************
NODE VOLTAGE NODE VOLTAGE
($N_0001) .4167 ($N_0002) 0.0000
($N_0003) .8333 ($N_0004) 0.0000
($N_0005) 1.6667 ($N_0006) 5.0000
($N_0007) .8333
VOLTAGE SOURCE CURRENTS
NAME CURRENT
V.V1 2.083E-01
V_V2 4.167E-01
V.V3-1.667E+00
TOTAL POWER DISSIPATION 8.33E+00 WATTS
*** 06/28/95 03:51:28 ******* Win32s Evaluation PSpice (April 1995) *********
D:\CIRCUITS\EXAMPLE3.SCH
*** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C
*** CURRENT STEP PARAM VT - 5
**********************************************************************
NODE VOLTAGE NODE VOLTAGE
($N_0001) 2.9167 ($N_0002) 5.0000
($N_0003) 3.3333 ($N_0004) 5.0000
($N_0005) 2.9167 ($N_0006) 5.0000
($N_0007) 1.4583
VOLTAGE SOURCE CURRENTS
NAME CURRENT
V_V1 -1.042E+00
V.V2 -8.333E-01
V.V3-1.042E+00
TOTAL POWER DISSIPATION 1.46E+01 WATTS
4.8
Summary
In this chapter we introduced a number of important concepts and rules that can be
used to simplify the analysis of electric circuits. Along with these concepts came many
definitions which we will continue to use throughout the text. The main concepts and
definitions are summarized below.
• Equivalent impedance is the impedance of a single element that can be used to
replace a collection of elements at a pair of nodes without affecting any of the
voltages and currents throughout the rest of the circuit.
137
4.9. Problems
• The principle of equivalent transformations is to preserve the terminal V-I
relationships under these transformations.
• Series connection—two (or more) elements are said to be connected in series
if they always have identical currents flowing through them.
• The equivalent impedance of a series connection is equal to the sum of the
individual impedances.
• Parallel connection—two (or more) elements are said to be connected in parallel
if they always have the same voltage across them.
• The equivalent admittance of a parallel connection is equal to the sum of the
individual admittances.
• The voltage divider formula gives the voltage across an individual element (say
the jth one) which is in series with n elements in terms of the applied voltage
and the impedances: V) = VsZj/Zeq.
• The current divider formula gives the current through an individual element
(say the jth one) which is in parallel with n elements in terms of the total
current and the admittances: îj = IsYj/Yeq.
• The input impedance of a circuit is the equivalent impedance with respect to
the input terminals.
• The symmetry of some electric circuits can be exploited to simplify their
analysis. The main idea in exploiting symmetry of electric circuits is to identify
equipotential nodes in symmetric electric circuits and connect them together
into one node.
• The superposition principle states that the voltages and currents in a complex
circuit with multiple sources can be found by summing the voltages and currents
found for several regimes of the original circuit when only some subsets of
original sources are active while other sources are set to zero. These subsets of
sources are formed as a result of subdivision of all original sources into several
nonintersecting groups.
• When voltage sources are set to zero, they are replaced by short-circuit
branches.
• When current sources are set to zero, they are replaced by open-circuit branches.
4.9
Problems
Problems 1-6. Find the equivalent impedances of the circuits shown in thefiguresindicated.
1. Figure P4-1.
3. Figure P4-3. Assume / = 2 kHz.
5. Figure P4-5.
2. Figure P4-2.
4. Figure P4-4. Assume / = 10 kHz.
6. Figure P4-6.
138
Chapter 4. Equivalent Transformations of Electric Circuits
2Q. \ 4 i i % 6 Q \ 8 Q
Figure P4-1
N10Q
Figure P4-2
<>AAAAr
'4.1 ml-P ; 11 |iF>-2.7 m H > 9 . 1 Q
7
"i n
25 Í2
1 JLLF
8 Q
2 |iF
-3I—
■2mH
oAAA/V
Figure P4-3
Figure P4-6
Figure P4-5
The divider ratios in Problems 7-10, denoted a : b, indicate the relative proportion of the
source quantity to the output quantity. For example, Va/Vs = b/a for a voltage divider.
7. Design a resistive voltage divider with the input resistance and divider ratio given in the
following table. (Find R\ and R2.)
Divider
Input resistance (Í1)
Divider ratio
a
50
2:1
b
1 k
10:1
2M
10,000:1
8. Design a resistive current divider with the input resistance and divider ratio given in the
following table. (Find R\ and R2.)
Divider
Input resistance (Í1)
Divider ratio
a
50
5:2
b
300
50:1
c
10k
300:1
139
4.9. Problems
9.
Design an inductive current divider with the input impedance magnitude and divider ratio
given in the following table. (Find L\ and L2.)
Divider
Input impedance (il )
Frequency (radVs)
Divider ratio
a
50
10k
3:1
b
2k
IM
20:1
180 k
22 M
100:1
10. Design a capacitive voltage divider with the input impedance magnitude and divider ratio
given in the following table. (Find C\ and Cz.)
11.
Divider
Input impedance (il)
Frequency (rad/s)
Divider ratio
a
50
7
7:3
b
20 k
Ik
40:1
47 k
IM
50,000:1
Find the ac current i{t) in the circuit shown in Figure P4-11.
12. Find the ac voltage v(t) in the circuit shown in Figure P4-12.
r ^ W W
8a
1mF
Figure P4-11
0.2 H
C)v s (t)=5cos(40t+ it/3) V ^ 2 . 5 m F
v(t)
Figure P4-12
Problems 13-18. Find the equivalent admittances of the circuits shown in the figures indicated.
13. Figure P4-13.
15. Figure P4-15.
17. Figure P4-17 (use symmetry arguments).
14. Figure P4-14.
16. Figure P4-16.
18. Figure P4-18 (use symmetry arguments).
140
Chapter 4. Equivalent Transformations of Electric Circuits
Figure P4-13
Figure P4-14
Figure P4-15
Figure P4-16
ww-
-AA/W
Figure P4-18
Figure P4-17
In Problems 19 and 20, assume that each branch has an impedence of Z.
19.
Find the input admittance of the symmetric cube across a side as shown in Figure P4-19.
20.
Find the input impedance of the symmetric cube across a face as shown in Figure P4-20.
CZr-
CD-ED
U^
0
CZ}
-CD
Figure P4-19
Figure P4-20
D
141
4.9. Problems
21.
Prove that the input impedance of an R-2R ladder with an infinite number of stages is
2R.
22.
Find the input impedance of an R-3R ladder with an infinite number of stages.
23.
Design a five-stage D/A converter that produces 1 A through the output 2R resistor when
the input voltages (either 0 V or 1 V) correspond to the number 30 (decimal). That is,
draw the circuit and indicate the values of all resistances.
In Problems 24-29, use the superposition principle to find the required circuit variable.
24.
Find the expression for the current indicated in the circuit shown in Figure P4-24.
Figure P4-24
25.
Find the expression for the time dependence of the voltage indicated in the circuit shown
in Figure P4-25.
+
v
Figure P4-25
26.
Find the phasor current / indicated in the circuit shown in Figure P4-26,
A/WV
i(t)
2 £1
Q v , , ( 0 = ^ % cos(/ + 7t/4) V \ 1 Ç1
SH
1F
1H
v v 2 (0 = cos(í + 7T /3) V
Figure P4-26
142
27.
Chapter 4. Equivalent Transformations of Electric Circuits
Find the voltage V0 indicated in the circuit shown in Figure P4-27.
Figure P4-27
28.
Find the time-dependent current i0 indicated in the circuit shown in Figure P4-28.
3Q
7Q
AAAA,
rAAA/V
© v(/) = 15cos(2í) vyiQ.
3Q
* J L -0
—W\A,
i(t) = 5cos(3i) A
Figure P4-28
29.
Find the voltage v indicated in the circuit shown in Figure P4-29.
rAA/VW—O—t-—0-
0
I
4£2 +<^
4V
40
v
-2V
< 8Q
6Q
Figure P4-29
<>
OV
10 Q
15 Q
Chapter 5
Thevenin's Theorem and
Related Topics
5.1
Introduction
In this chapter we first discuss the properties of "real-world" circuit components
and define nonideal circuit elements (active and passive). Next, we demonstrate the
equivalent nature of nonideal voltage and current sources.
We then present and prove two most central theorems in circuit theory, which are
attributed to Thevenin and Norton. The essence of these theorems is that any linear
circuit with numerous sources can be replaced at a given pair of nodes by a nonideal
independent source. After proving these theorems, we present several examples of
their applications.
The chapter is closed by a discussion of resistive circuits with single nonlinear
resistors and the Thevenin theorem for these circuits is then developed.
5.2
Nonideal Two-Terminal Circuit Elements
All the circuit elements so far studied in this text have been ideal elements. However,
real circuit elements are always nonideal, which means that they possess certain
imperfections which make their observed behavior deviate slightly from the mathematical definitions we assigned to ideal elements. It turns out that we can model
some of the deviations from ideal behavior by using impedances and resistors. In this
way, we arrive at nonideal elements. Although nonideal circuit elements are important in modeling real-life circuit elements, their usefulness is not limited only to this
purpose. It turns out that nonideal elements play an important role in the theory of
equivalent transformations of electric circuits.
We shall first describe the model of a nonideal voltage source. The circuit
notation for a nonideal voltage source is shown in Figure 5.1a. A nonideal voltage
143
144
Chapter 5. Thevenin's Theorem and Related Topics
*.ô
f.(b
(a)
J
(b)
(c)
(d)
Figure 5.1: Nonideal voltage source (a), nonideal current source (b), nonideal inductor
(c), and nonideal capacitor (d).
source consists of an ideal voltage source connected in series with a source impedance.
This impedance allows for a slight variation in voltage across the source output due
to a variation in the current through the source. When used as a model for a real-life
voltage source, this source impedance is usually very small.
Next, we consider a nonideal current source. The circuit notation for nonideal
current source is shown in Figure 5.1b. A nonideal current source consists of an
ideal current source connected in parallel with a source admittance. This admittance
allows for a slight variation in source output current due to a variation in the voltage
across the source. The small current which flows through the admittance's branch is
called the leakage current. This leakage current is a function of the voltage across
the source. In this way the dependence of the total (external) current on the voltage is
introduced. When used as a model for a real-life current source, the source admittance
is usually small.
Now, we discuss a nonideal inductor. The circuit notation for the low-frequency
model of a nonideal inductor is shown in Figure 5.1c. This model consists of an
ideal inductor connected in series with a resistor. This resistor represents the small
resistance of the wire from which the coil of the inductor is made. Since the energy
losses of the inductor are proportional to the square of the current, the resistor is
connected in series. A figure of merit of the quality of an inductor is given by the
ratio L/R (which is related to a decay time as we will see in Chapter 7).
Finally, we consider a nonideal capacitor. The circuit notation for the lowfrequency model of a nonideal capacitor is shown in Figure 5. Id. This model consists
of an ideal capacitor connected in parallel with a resistor. In real capacitors there are
dielectric losses which are taken into account by the resistor. Because these losses
are proportional to the square of the voltage, the resistor is connected in parallel. At
high frequencies, one must consider the inductance of the leads as well. A figure of
merit for the quality of a capacitor is given by the ratio C/G = RC. Usually, real
capacitors are closer to ideal than are real inductors, i.e., C/G » L/R. Because
145
5.3. Equivalent Transformations ofNonideal Voltage and Current Sources
of this, the shock hazard from capacitors should never be underestimated. For this
reason, high-voltage capacitors should always be stored with their terminals shorted.
In conclusion, we remark that resistors are also not ideal. Resistors that are constructed by winding a fine wire in a helical coil, for example, often have appreciable
inductances and must be modeled with the circuit in Figure 5.1c. They are usually
not suitable for high-frequency applications.
5.3
Equivalent Transformations of Nonideal
Voltage and Current Sources
Consider the equivalent transformation of a nonideal voltage source into a nonideal
current source, and vice versa. Such a transformation may first seem superfluous.
However, it has proved to be very useful in circuit analysis, especially in the node
and mesh analysis techniques, which will be discussed in the next chapter.
According to the previously discussed general principle of equivalent transformations, the transformation of a nonideal voltage source into a nonideal current
source will be equivalent if it preserves the terminal V versus / relationship. Being
guided by this principle, we first find the terminal relationship for the left circuit in
Figure 5.2. By using KVL, we derive:
- Vs + IZS + V = 0.
(5.1)
V = V, - /Z,.
(5.2)
Thus,
In order to find the terminal relationship for the right circuit in Figure 5.2, we use
KCL at node A:
Î = 1 - V.
(5.3)
I = h - VYS
(5.4)
Consequently,
Figure 5.2: Nonideal voltage source - nonideal current source transformation.
146
Chapter 5. Thevenin's Theorem and Related Topics
where the expression V = VYS was used in (5.4). Equation (5.4) can be solved for
voltage, which yields:
1
V
(5.5)
Comparing terminal relationships (5.2) and (5.5) we see that they will be identical
when:
S
(5.6)
y
1
S
'-I
(5.7)
Is^s»
Formulas (5.6) and (5.7) constitute the rules for the equivalent transformation of
nonideal sources. The polarities of the equivalent nonideal sources are also critical
and are indicated in Figure 5.2. For example, when replacing a nonideal voltage
source with a nonideal current source, the "tail" of the current arrow is connected to
the node that was originally connected to the negative terminal of the voltage source.
5.4 Thevenin's Theorem
Thevenin's theorem is the central result of basic circuit theory. For this reason, it
is essential to understand this theorem, know how to prove it and how to use it in
solving problems.
Consider a branch with impedance Z connected to an arbitrary active circuit
(network). Thevenin's theorem basically states that as far as the current through this
impedance is concerned, the active linear network can be replaced by an equivalent
nonideal voltage source. This theorem is illustrated in Figure 5.3. In the following
section, the proof of Thevenin's theorem is given.
A
A
0
U '.*
o '.
Active
Network
—* H
z —
0
!-0-
—o
Figure 5.3: Graphical illustration of Thevenin's theorem.
1
147
5.4. Theveniris Theorem
5.4.1
Proof of Theveniris Theorem
The proof consists of the following three steps:
Step I:
First, we introduce two voltage sources into the branch with impedance Z (see Figure
5.4). Both have the same peak value of voltage but opposite polarities. We can see
that doing this will not affect the current through Z at all, because the two voltage
sources will cancel out in any KVL equations due to their opposite polarities.
Step II:
Next, we shall use the superposition principle. To do this, we divide all the sources
in the network into two groups:
1. The first group consists of all the sources in the active network and the left voltage
source introduced into the branch with impedance Z.
2. The second group consists of only the right voltage source in the same branch.
Now, we can consider two separate regimes, shown in Figure 5.5. Thefirstregime
contains the sources of the first group, and the second regime contains the only source
of the second group. Note that, in the second regime, the active network has been
replaced by a passive network. This passive network is formed when the sources in
the active network are set to zero. This is accomplished by replacing voltage sources
with short-circuit branches and current sources with open branches.
By using the superposition principle we find that the total current through Z is
equal to the sum of the branch currents in the above two regimes:
Step III:
Consider the first regime. We choose Vs in a way that causes 7(1) to be zero. This is
equivalent to having the branch with Z open. Consequently, the voltage across the
A
o
U.
Active
Network
z
A
A
Vs
Vs
Figure 5.4: Introduce two voltage sources.
148
Chapter 5. Thevenirís Theorem and Related Topics
Regime 1
0—
A
Regime 2
fd)
Active
Network
0—
A
Passive
Network
z
-G—'
[-l-l
B
0—
B
0—
f(2)
y
HD—I
Figure 5.5: The two regimes.
terminals A and B will be equal to the open-circuit voltage Voc. This situation is
shown in Figure 5.6. To find such a Vs which guarantees that 7(1) will be equal to
zero, we write the KVL for the loop shown in Figure 5.6:
Vs - Voc = 0,
(5.9)
Vs = Voc.
(5.10)
Therefore, Vs must be equal to the open-circuit voltage in order to force 7(1) = 0.
According to equation (5.8), we find that the total current is determined by the second
regime:
/
r(2)
(5.11)
Next, consider the second regime. The passive network contains no sources and
can be replaced by the equivalent input impedance with respect to the terminals A
and B. This replacement is shown in the rightmost circuit in Figure 5.7. This series
combination of Zs and Vs constitutes a nonideal voltage source which (according to
Figure 5.6: Regime 1.
149
5.4. Thevenin's Theorem
A
>
Z£
e-^
Figure 5.7: Regime 2.
(5.11)) generates the same current through the load impedance Z that the original
network does. This concludes the proof of the theorem.
As a useful by-product, we have also found the following expressions for the
source voltage and source impedance:
Vs = Voc,
(5.12)
7=7
(5.13)
By using these expressions and the Thevenin equivalent circuit, we derive the
following formula:
/ =
V0
Vs
Z-j<i
I
Z-i
¿-¿in
' *-*<
(5.14)
which relates the branch current to the open-circuit voltage and the input impedance.
This equation is essential in the application of Thevenin's theorem to circuit analysis.
5,4.2
U s i n g Thevenin's Theorem in Analysis
Thevenin's theorem and equation (5.14) can be successfully used in circuit analysis to
solve for currents through specific branches. All we need to do is tofindtwo quantities,
the open-circuit voltage and the input impedance, and use them in equation (5.14) to
arrive at the solution. Finding these quantities is fairly straightforward, and the entire
method can be reduced to the following three steps.
Step I:
To calculate the current through any particular branch, the open-circuit voltage must
be found. Consequently, the first step is to remove the branch through which the
current is desired and determine the voltage with respect to the open terminals (see
Figure 5.8). It is this first step of removing the branch which usually simplifies the
circuit.
150
Chapter 5. Thevenin's Theorem and Related Topics
A
0
'.
0
+
Active
Network
z —
^W
Active
Network
A
0
U
Figure 5.8: Step I: Open branch, find open-circuit voltage.
Step II:
The second step is to find the input impedance with respect to the open terminals of
the removed branch. The rest of the circuit must be transformed to a passive network
by setting all sources to zero (see Figure 5.9). Recall that a voltage source that is
set to zero is replaced by a short-circuit branch, while a current source set to zero is
replaced by an open branch.
0
\
0
Active
Network
Passive
Network
n
vJ
1
z
0
Figure 5.9: Step II: Find input impedance of the passive network.
Step III:
Once the open-circuit voltage and the input impedance are known, finding the current
through the branch with equation (5.14) is simple. The justification behind using this
equation is Thevenin's theorem itself.
EXAMPLE 5.1 As an example of using Thevenin's theorem in the analysis of electric circuits, we consider a "bridge" circuit, pictured on the left in Figure 5.10. It is
assumed that Z\, Z2, Z3, Z4, Z5, Vs are given. We want to find 25.
We start the solution of this problem with step one, namely, we remove the
branch in question and find the open-circuit voltage. Once we remove the branch,
the circuit becomes the one shown on the right in Figure 5.10, which is now much
simplified. To find the open-circuit voltage, we write KVL for the loop shown:
Voc-hZ*
+ LZ} =0,
(5.15)
151
5.4. Thevenin's Theorem
Figure 5.10: A bridge circuit, with center branch removed.
We want to express the open-circuit voltage in terms of the given source voltage, so
we must find 1\ and 73 in terms of this source voltage. Notice that the circuit is now
quite simple; it contains two branches in parallel with the same voltage Vs across
them. (One branch consists of the series combination of Z\ and Z2 and the other is
the series combination of Z3 and Z4.) Hence, we can calculate the currents through
these branches by dividing Vs by the impedances of the branches:
/, =
h =
Vx
Z1+Z2'
(5.16)
Vx
(5.17)
z3+z4
Substituting (5.16) and (5.17) into (5.15), we find the open-circuit voltage to be:
Vnr =
Vx
Z3
ZJ'X
\ /JA
Zi
ZJ\ H~ ZJ^
(5.18)
Now, we move on to step two, finding the input impedance. The circuit must be
replaced by a passive network—in this case the voltage source must be replaced
by a short-circuit branch. The resulting passive network (see Figure 5.11) may still
seem complicated; however, redrawing the circuit reveals its simplicity. The essence
of redrawing is to merge into one node the two nodes which are connected by the
short-circuit branch. In Figure 5.11 the passive network is redrawn to emphasize its
simplicity. In this case, the input impedance is easy to calculate. It is merely two
parallel connections in series with one another:
7. =
¿-¡in
ZiZ2
Z3Z4
z, +z9 + z» + z 4
(5.19)
152
Chapter 5. Thevenin's Theorem and Related Topics
Figure 5.11: The passive network.
Now that the open-circuit voltage and the input impedance have been found, equation
(5.14) can be used to find the current 1$. This constitutes the final step:
U = Z3Z4
:r+\/'+Z2j.
z{z2 + Zs
Z3+Z4 +
(5.20)
Z}+Zo
EXAMPLE 5.2 As a second example of using Thevenin's theorem in the analysis
of electric circuits, consider the same bridge circuit as in the previous example. This
time we want to find the current through one of the side branches, say I\ (Figure
5.12).
Again, we follow the same three steps as before. First, we open the branch with
Z] and find the open-circuit voltage. Writing KVL for the circuit shown on the right
hand in Figure 5.12 yields
Vac = I5Z5 + 73Z3.
Figure 5.12: Circuit for Example 5.2.
(5.21)
153
5.4. Thevenin's Theorem
Now, the currents Î5 and I3 must be found in terms of the given source voltage. From
Figure 5.12 it is apparent that I3 is equal to the source voltage divided by the total
circuit impedance with respect to the terminals of the given voltage source:
Vs
(5.22)
(Z2+Z5)Zt
z3 + z +z +z
2
4
5
or
h =
VS(Z2 + Z4 + Z5)
Z3(Z2 + Z4 + Z5) + (Z2 + Z5)Z4
(5.23)
To facilitate the understanding of how the total impedance has been evaluated, we
remark that according to the right-hand circuit in Figure 5.12 impedances Z5 and
Z2 are connected in series and together are in parallel with Z4, and then they are all
together in series with Z3.
Once I3 is found, it can be used to calculate /5 by using the current divider rule:
= Vx
Z9 +ZA + Z*
' Z3(Z2 + Z4 + Z5) + (Z2 + Z5)Z4
(5.24)
Now, the open-circuit voltage can be found in terms of the known source voltage by
substituting (5.23) and (5.24) into (5.21), which yields:
Vnr =
Z3(Z2 + Z4 + Z5) + Z4Z5
'Z 3 (Z 2 +Z4+Z5) + (Z 2 +Z5)Z 4 '
(5.25)
V,
Now, according to step two, the input impedance must be calculated by setting the
source to zero and forming a passive network. In this case, redrawing the circuit
is also helpful in determining the proper input impedance (Figure 5.13). From the
redrawn circuit, it is clear that Z4 and Z3 are connected in parallel with each other
and that combination is in series with Z5. This group of three impedances is then
zin
Z
<X Z a
]z 5 y
^ Ny^
z
4
1
1
0—t
—1
Z
3
HUH
z4
Figure 5.13: The passive network.
Z
2
Z
5
,
1
H
r
154
Chapter 5. Thevenin's Theorem and Related Topics
connected in parallel with Z2. So, the resulting input impedance is:
(5.26)
Z3Z4
+ Zs+Zi
z +z
3
which after a simple transformation yields:
Zin =
(Z3Z4 + Z5Z3 + Z5Z4)Z2
^ ^
—
s_v_¿
Z3Z4 + Z5Z3 + Z3Z2 + Z4Z5 + Z 4 Z 2
.
(5.27)
In the third step, we use (5.25) and (5.27) in equation (5.14), and after simple
transformations we find the desired current I\ :
h—
Zin + Z\
j = ç
1
s
Z3(Z2 + Z4 + Z5) + Z4Z5
(z3z4 + z3z5 + z4z5)(z2 + zo + z{(z3z2 + z4z2y
We have seen from the preceding examples how Thevenin's theorem is useful
from an analytical standpoint. Now, we discuss how this theorem can be useful from
an experimental point of view. Recall that to find the current through a branch with
impedance Z by using equation (5.14), the open-circuit voltage must be determined.
This can be accomplished by simply measuring this voltage with a voltmeter. But how
could the input impedance be found experimentally? The answer is to short-circuit
the branch with impedance Z and measure the short-circuit current, Isc. According to
Thevenin's theorem and (5.14), the open-circuit voltage and the short-circuit current
are related by the following expression:
L = y^-Zin = ^ .
A/7
ISC
(5.29)
Thus, if the open-circuit voltage and short circuit current are both measured, the input
impedance can be found. This technique enables the experimental use of Thevenin's
theorem in some practical applications. Measuring the open-circuit voltage and the
short-circuit current is accomplished by using two "extreme" tests shown in Figures
5.14 and 5.15, respectively. By using the result of these two tests, the current / through
the branch with any impedance Z can be predicted:
I = -rJ^.
Y +z
(5.30)
It is important to note that the described experimental approach does not require any
a priori knowledge concerning the active network. For this reason, the open-circuit
and short-circuit tests are widely used for characterizations of many devices such as
transformers, induction motors, and synchronous generators.
155
5.5. Norton's Theorem
Figure 5.14: Open-circuit test.
5.5
Figure 5.15: Short-circuit test.
Norton's Theorem
Norton's theorem is very similar to Thevenin's theorem. Basically it states that as
far as the current through an impedance Z is concerned any active network can be
replaced by a non-ideal current source. It is graphically represented in Figure 5.16.
This theorem is trivial to prove by using Thevenin's theorem.
Proof: We know that according to Thevenin's theorem the active network in the
left circuit of Figure 5.16 can be equivalently replaced by a nonideal voltage source.
Since any nonideal voltage source can be equivalently transformed to a nonideal
current source (see Figure 5.17), Norton's theorem is proven.
We know the equivalence between nonideal voltage and current sources means
that
*-b
i.< =
Vx
(5.31)
(5.32)
1 sr •
Therefore, the current source in Norton's theorem is equal to the short-circuit current,
and the admittance is equal to the input admittance of the corresponding passive
-o- 1 -»-
üb
Figure 5.16: Norton's theorem.
156
Chapter 5. Thevenin's Theorem and Related Topics
-O-2-*-
>
e—
i
Figure 5.17: "Proof" of Norton's theorem.
network:
I, = ha
(5.33)
Ys = ^r = Yin.
(5.34)
¿in
Norton's theorem can be used in a manner similar to Thevenin's theorem.
EXAMPLE 5.3 We want to find the Norton equivalent nonideal current source for
the circuit shown in Figure 5.18.
First, we can find the equivalent impedance by setting the ideal sources to
zero. For this example, that entails shorting the voltage source Vs and opening the
current source Is. The resulting passive circuit is given in Figure 5.19. Because Zx is
disconnected on the left side, it cannot have any current flow through it and thus has
no effect on the equivalent impedance. Thus, the elements Z2 and Z3 are effectively
in series and that combination is in parallel with Z4. The resulting Norton equivalent
admittance is:
NO
1
Z2 + Z3
1
+ —
z2 + z3+ z 4
Z4
(5.35)
(Z2 + Z3)Z4
-o A
■(OA(T)
o
-oB
v,(0 V
Figure 5.18: Circuit for the first Norton example.
157
5.5. Norton's Theorem
-o A
Z
4
h
Y
in
-oB
Figure 5.19: The passive circuit used to find the input admittance.
Next, we will use the principle of superposition to find the Norton current source.
The two circuits that we need to analyze are given in Figures 5.20a and b. In both
cases, we short-circuit the output terminals. This results in zero voltage across Z4 and
subsequently zero current through that impedance. Therefore, it has no effect on the
circuit and we don't need to consider it further. We must find the short-circuit current
for each of the two problems and add the results to get the current of the equivalent
non-ideal current source. Problem (a) resembles a current divider and the current
we need is also the current through Z3. The appropriate formula for current dividers
yields: /£ = ISZ2/(Z2 + Z3). Problem (b) is best solved with KVL by taking a path
through the voltage source, Z2, Z3, and the short-circuit branch. The resulting current
is found to be: lhsc = VS/(Z2 + Z3). Therefore, the Norton equivalent current is:
wo
(Vs + / y Z 2 )/(Z 2 + Z3).
(b)
o
(5.36)
¿,(0 A
v,(0 V
Figure 5.20: The two problems that arise from the superposition principle.
158
Chapter 5. Thevenin's Theorem ana Related Topics
Notice that Z\ does not enter into the equivalent circuit in any way. Can you explain
why this is so?
■
EXAMPLE 5.4 Consider a nonideal voltage source with v^ = 120v2cos(377i +
120°) V and an internal resistance of 0.1 ii, connected in parallel with a 1 JLLF
capacitor and having a lead inductance of 0.1 mH. The resulting circuit is given in
Figure 5.21. Again, we want to find the Norton equivalent circuit.
This time, we shall find the Thevenin equivalent circuit and arrive at the final
result via the rules for the transformation of nonideal sources. When we convert
the circuit to phasor form, we find that Vs = I20y/lej27r/3 V, Rs = 0.1il,Z L =
7*0.0377 ii, and Zc = —7*2.653 kii. To find the Thevenin impedance, we short
the voltage source and notice that the equivalent impedance is equal to the series
combination of ZL with the parallel combination of Rs and Zc:
Zin = ZTH =ZL+
RsZc/(Rs + Zc) « 0.1 + 0.03777 A
(5.37)
To find the Thevenin voltage, we notice that for the open-circuit configuration, there
is no current through the inductor and therefore no voltage drop across it. That means
that the open-circuit voltage is equal to the voltage across the capacitor. We can find
this voltage by the voltage divider rule:
Vac = VJB = VC = VsZc/(Rs + Zc) « Vs(\ - 7/26, 525) - 1 2 0 A / 2 ^ / 3 V.
(5.38)
The Norton current just comes from the relation:
I NO — VTH/ZTH —
12y/le'2"'3
- 1.588^ L734 kA.
0.1 +0.03777
(5.39)
Notice that in this example the capacitance is sufficiently small that it doesn't significantly affect the equivalent circuit.
■
Figure 5.21: Circuit for the second Norton example.
159
5.6. Nonlinear Resistive Circuits
5.6
Nonlinear Resistive Circuits
In the previous sections, we have analyzed electric circuits containing only linear
circuit elements such as resistors, inductors, and capacitors. The term "linear" means
that the values of resistances, inductances, and capacitances do not depend on voltages across the corresponding circuit elements or currents through these elements.
For these reasons, the terminal relationships for such elements are linear and the
circuits with these elements are described by linear equations such as equations
(2.29)-(2.34). It has also been assumed in our previous discussions that the values
of resistances, inductances, and capacitances do not change with time. Such circuit
elements are called time-invariant and circuits with linear and time-invariant elements
are described by linear equations with constant coefficients. Equations (2.29)-(2.34)
can be again used as an example. Linear and time-invariant (LTI) circuits are important representatives of LTI systems and comprehensive mathematical study of these
systems belongs to a course on linear systems and signals. However, it is useful to
stress here the two most fundamental properties of such circuits and systems. The first
property is the invariance with respect to time shifting, while the second property
is the superposition principle. To express these properties mathematically, consider
an LTI electric circuit with two terminals shown in Figure 5.22. Suppose that terminal
voltage v(f ) results in terminal current response /(f). Then, the invariance with respect
to time shifting means that terminal voltage v(t — r) will result in terminal current
response i(t — r), where r is an arbitrary shift in time. This property is transparent
from the physical point of view and simply states that for linear time-invariant circuits there is nothing which singles out the origin of time. This origin can be chosen
arbitrarily without affecting the mathematical description of LTI electric circuits, and
the change of time origin is equivalent to time shifting.
To demonstrate the superposition principle, let us suppose that two terminal voltages V] (f ) and v 2 (0 result in two terminal current responses i{(t) and i2(t), respectively.
Then, according to the superposition principle, the terminal voltage
v(f) = c1vi(f) + c2v2(i)
(5.40)
i(t)
(t)
Linear
Electric
Circuit
Figure 5.22: Input voltage and terminal current response for an LTI electric circuit.
160
Chapter 5. Thevenin's Theorem and Related Topics
will result in the terminal current response
i(t) = cMt) + c2i2(t),
(5.41)
where c\ and c2 are arbitrary numbers.
It has been stressed before that the superposition principle is a mathematical
consequence of linearity of circuit equations.
The time shifting invariance and the superposition principle can be combined into
the following statement: if i\(t) and i2(t) are terminal current responses to terminal
voltages vi(i) and v2{t), respectively, then
i(t) = cih(t - TI) + c2i2{t - r2)
(5.42)
will be the current response to the terminal voltage
v(f) = civi(t - TO + c2v2(t - T2),
(5.43)
where r{ and T2 are arbitrary time shifts. The last statement can be considered as
another definition of LTI circuits, the definition which is extensively used in the linear
system theory.
In many applications, the values of resistances, inductances, and capacitances
may depend on voltages across or currents through the circuit elements. Such circuit
elements are nonlinear, and electric circuits with nonlinear elements are described by
nonlinear differential and algebraic equations for which the superposition principle
is not valid. For this reason, the theory of nonlinear electric circuits is much more
complex. In this section, we consider only some simple facts and results related
to nonlinear resistive circuits. Such circuits are described by nonlinear algebraic
equations.
First, we shall recall the definition of linear resistors. These resistors are characterized by the following terminal relationship:
v = Ri,
(5.44)
where v and / are the instantaneous voltage across and the current through a resistor,
and R is the resistance, which does not depend on the values of v and i.
Expression (5.44) can be graphically illustrated by a straight line shown in Figure
5.23. The slope of this straight line is equal to the value of the resistance:
R = - = tan a.
(5.45)
/
The circuit notation for a nonlinear resistor which will be used throughout this section
is shown in Figure 5.24. The nonlinear resistor cannot be characterized completely
by one value of the resistance. Usually, the nonlinear resistor is characterized by a
"v(0" curve like the one shown in Figure 5.25. It is clear from this figure that if we
define the resistance as the ratio
v(i)
R(i) = — ,
(5.46)
/
the value of this resistance will vary with the current through the resistor. This fact is
emphasized by the notation R(i). Indeed, the values of the resistance for two different
5.6. Nonlinear Resistive Circuits
Figure 5.23: v versus / representation of a linear resistor.
values ¿i and i2 of the current through the resistor will be equal to two different slopes:
R(i\) = — = tan a\,
h
v
2
R(h) = — = tan a2.
(5.47)
(5.48)
12
This explains why v(/) curves are used for characterization of nonlinear resistors.
There are many physical mechanisms which lead to a nonlinear dependence of
resistance on electric current through a resistor. One simple mechanism, for example,
is based on the fact that the resistance generally increases with temperature. As the
current through a resistor is increased, the amount of power dissipated (as heat)
increases, and its temperature goes up. This results in the increase of resistance.
Analysis of electric circuits with nonlinear resistors requires the solution of
nonlinear algebraic equations. For this reason, this analysis encounters some difficulties which can be somewhat circumvented by using graphical techniques. We shall
illustrate this by an example of a simple nonlinear circuit shown in Figure 5.26.
+
v(i)
Figure 5.24: Circuit notation for a nonlinear resistor.
162
Chapter 5. Theveniris Theorem and Related Topics
al a2
ix
Figure 5.25: v(/) curve for a nonlinear resistor.
By applying KVL to this circuit, we end up with the equation:
vs — Rsi + v\(i).
(5.49)
This is a nonlinear algebraic equation. The nonlinearity of this equation stems from
the last term vi(/), which is the "voltage versus current" description of the nonlinear
resistor. We shall solve this nonlinear equation graphically. To this end, we transform
it as follows:
(5.50)
vv — Rsi = Vj(/).
Next, we shall plot the left-hand side and right-hand side of this equation as shown
in Figure 5.27. The straight line which corresponds to the left-hand side of (5.50) is
usually called the "load line." The solution of equation (5.50) is the value / of the
current for which both sides in (5.50) are equal. It is clear from Figure 5.27 that
rVWV-^
VsÔ
v/i)
Figure 5.26: Electric circuit with a nonlinear resistor.
5.6. Nonlinear Resistive Circuits
Figure 5.27: Graphical technique for the analysis of a nonlinear electric circuit.
both sides in (5.50) will be equal to one another for the value of the current which
corresponds to the intersection of the plots of these sides. Thus, we can graphically
find the current / through the circuit and the voltage v across the nonlinear resistor
(see Figure 5.27).
The discussed problem is a very simple one. However, arbitrary resistive electric circuits with single nonlinear resistors can be reduced to the circuit shown in
Figure 5.26. This statement is the Thevenin theorem for resistive circuits with single
nonlinear resistors.
To prove this theorem, consider the circuit shown in Figure 5.28. Suppose / is
the current through the nonlinear resistor. The voltage v across the terminals "A-B"
will remain the same if we replace this resistor with a current source is = i (see
Figure 5.29). This is true because this replacement preserves the current through the
terminals A-B of the active linear resistive circuit. In other words, this linear resistive
Active
Linear
Resistive
Circuit
Figure 5.28: Resistive circuit with a single nonlinear resistor.
Chapter 5. Thevenin's Theorem and Related Topics
Active
Linear
Resistive
Circuit
CD'.-
Figure 5.29: Replacement of a nonlinear resistor with a current source.
circuit cannot distinguish between the nonlinear resistor and the current source if
each of them results in the same terminal current. Consequently, in both cases the
voltage across the terminals A-B will be the same. From a purely mathematical point
of view, this is true because the linear resistive circuits in Figure 5.28 and Figure 5.29
are described by identical linear algebraic equations. Consequently, by solving these
equations, we find the same voltage across the terminals A-B.
Next, we shall apply the superposition principle to the circuit in Figure 5.29. To
do this, we divide all the sources in this circuit into two groups: 1) all the sources in
the active linear resistive circuit, 2) a single current source is = i. Now, we consider
two separate regimes shown in Figure 5.30. According to the superposition principle,
the voltage v across the terminals A-B is equal to the sum of voltages v(1) and v(2)
across the same terminals for the first and second regimes, respectively:
-
M) + V1,(2)
(5.51)
v=v
It is clear that v(1) is equal to the open-circuit voltage
v(1) - v
v
(5.52)
v
oc>
Regime 2
Regime 1
A ¡
A
Active
Linear
Resistive
Circuit
+
v<1>
-
_ O
Passive
Linear
Resistive
Circuit
0
B
Figure 5.30: Two regimes.
>
i
+
v<2> ( J
CD'.-
r^
U
B
165
5.6. Nonlinear Resistive Circuits
while v(2) can be expressed in terms of the equivalent input resistance Rin of the
passive linear resistive circuit and the current / as follows:
v(2) - -RinL
(5.53)
Please note that the minus sign in (5.53) has appeared because the reference directions
of the current / through R¡n and voltage v(2) across the same resistance are not
coordinated (are opposite).
By substituting (5.52) and (5.53) into (5.51), we obtain:
v = Voc - Rini-
(5.54)
If we consider the Thevenin equivalent circuit shown in Figure 5.26, we find that:
v = v, - Rsi.
(5.55)
Formula (5.54) is the expression for the voltage versus current at the terminals A-B
of the original circuit shown in Figure 5.28, while formula (5.55) is the expression for
the voltage versus current at the terminals A-B of the circuit shown in Figure 5.26.
These two expressions will be identical if:
v s = vOCy
(5.56)
Rs = Rin-
(5.57)
Under conditions (5.56) and (5.57), the voltage versus current relationship at the
terminals A-B in the circuits shown in Figures 5.28 and 5.26 will be the same. This
guarantees the same currents through the nonlinear resistor for both circuits. Thus,
the proof of the Thevenin theorem for resistive circuits with single nonlinear resistors
is concluded.
The given proof of the Thevenin theorem is more subtle than the one presented
in Section 5.4. The main distinction is that this proof uses the superposition principle
only for the active (linear) part of the circuit rather than for the entire circuit.
We have also established already familiar expressions (5.56) and (5.57) for the
Thevenin voltage vs and resistance Rs.
By using the Thevenin theorem, we can formulate the following two-step algorithm for the analysis of the generic circuit shown in Figure 5.28.
Step I:
Find the open circuit voltage across the terminals A-B for the active linear resistive circuit and the equivalent input resistance across the same terminals for the
corresponding passive linear resistive circuit.
Step II:
Replace the active linear resistive circuit by the Thevenin equivalent circuit (see
Figure 5.26) and use the graphical technique shown in Figure 5.27 for the analysis of
the transformed circuit.
166
Chapter 5. Thevenin's Theorem and Related Topics
rA/WV
v,«
Figure 531: Voltage regulator circuit.
We shall illustrate the above algorithm by the following two examples.
EXAMPLE 5.5 A Voltage Regulator Circuit. Consider the circuit shown in Figure
5.31, where a load resistor R¿ is connected in parallel with a nonlinear resistor
characterized by the vi(i) curve shown in Figure 5.32. This curve exhibits "voltage
saturation." In other words, it has an almost horizontal (flat) portion which starts from
small current values. We would like to find all currents and voltages in this circuit.
First, we redraw this circuit to represent it in the form similar to the generic
circuit in Figure 5.28. The redrawn circuit is shown in Figure 5.33. Next, we shall
find the Thevenin voltage v^ which according to (5.56) is equal to the open circuit
voltage, that is, the voltage across the load resistor RL when the nonlinear resistor is
removed. By using the voltage divider rule, it is easy to find that:
Vx
(5.58)
R + RL
\d)
Figure 5.32: Voltage-current relationship for a nonlinear resistor used in a voltage
regulator circuit.
167
5.6. Nonlinear Resistive Circuits
R
A
rAA/VV*
ov.O)
Figure 5.33: Redrawn voltage regulator circuit.
Now, we shall find the Thevenin resistance Rs, which is equal to the resistance
across the terminals A-B when the nonlinear resistor is removed and dc voltage source
Vs is short-circuited. It is easy to see that this resistance is equal to the equivalent
resistance ofR and RL connected in parallel:
Rs — Rjn —
RRL
R + RL
(5.59)
Thus, we have found the parameters of the equivalent circuit shown in Figure 5.26
and now we can use the graphical technique demonstrated in Figure 5.27. For our
case, this technique is illustrated in Figure 5.34. The current i through the nonlinear
resistor corresponds to the point of intersection of the straight load line and vj(0
Figure 5.34: Graphical analysis of a voltage regulator circuit.
168
Chapter 5. Thevenirís Theorem and Related Topics
curve. The voltage v¿ which corresponds to this point is the voltage across the load
resistance. Thus, by dividing this voltage by 7?¿, we find the current ii through the
load resistance. By summing up this current with /, we find the current through the
resistor R, and this concludes the analysis of the above circuit.
Now, we shall demonstrate a peculiar feature of this circuit. Suppose that the
source voltage Vs does not remain constant and changes somewhat with time. Then,
the load line will change its position but will remain parallel to the original load line
(see Figure 5.34). New load voltage v¿ will correspond to a new intersection point.
However, since the vi(/) curve exhibits voltage saturation, this new load voltage will
be practically the same as before. This explains why this circuit is called a voltage
regulator circuit. To build circuits like this, nonlinear resistors with the property of
voltage saturation are needed. It turns out that this property is exhibited by Zener
diodes, which can be used for the design of the voltage regulator circuits. Zener
diodes are usually studied in some detail in courses on electronic circuits.
The regulator circuit of the type shown in Figure 5.33 is often regarded as
inefficient. The reason is the following. If Vs tends to increase, this results in a
substantial increase in the current / through the nonlinear resistor (Zener diode).
Since the load voltage and, consequently, load current remain the same, this leads
to an increase in the current through the resistor R. This causes an increase in the
voltage drop across this resistor which offsets the increase in V5. Thus, we can see
that the voltage regulation is achieved at the expense of increases in currents through
the nonlinear resistor and resistor R and, consequently, at the expense of increase in
power dissipation in these resistors. For this reason, these voltage regulator circuits are
mostly used in low-power applications. For high-power applications, more efficient
voltage regulator circuits have been designed.
■
EXAMPLE 5.6 A Bridge Circuit. This circuit is shown in Figure 5.35. It would be
very difficult to analyze this circuit without the use of the Thevenin theorem. The
utilization of this theorem simplifies the analysis appreciably.
> p (i)
v
o
Figure 5.35: "Bridge" circuit with a nonlinear resistor.
5.6. Nonlinear Resistive Circuits
Figure 5.36: Electric circuit for the calculation of
First, we shall find the Thevenin voltage vs which is equal to the open-circuit
voltage, that is, the voltage across the terminals A-B when the nonlinear resistor is
removed (see Figure 5 3 6 ) . It is easy to see that
l\ = *3 =
12 — H —
vo
R\ +R3
vo
(5.60)
(5.61)
R2+R4
By using (5.60), (5.61), and KVL for the loop consisting of Ru the open terminals,
and R2, we derive:
Vs = voc = R2i2 - R\i\ = v0
R2
Ri + RA
R\
R] + 7?^
(5.62)
Now, we shall find the Thevenin resistance Rs which is equal to the input resistance
across the terminals A-B when the voltage source vo is short-circuited (see Figure
5.37). It is easy to see that this resistance is given by:
R* — Ri
R\R3
R\
+ ÍV3
R2R4
7?2 "I" R4
Figure 5.37: Electric circuit for the calculation of i?„
(5.63)
170
Chapter 5. Thevenin's Theorem and Related Topics
Finally, by using the graphical technique demonstrated in Figure 5.27, we can find
the current through and the voltage across the nonlinear resistor. Then it is easy to
find all other voltages and currents in the circuit.
■
5.7
Summary
In this chapter we introduced a number of important concepts and rules that can be
used to simplify the analysis of electric circuits. Along with these concepts came some
definitions which we will continue to use throughout the text. The main concepts and
definitions are summarized below.
• A nonideal voltage source is modeled by an ideal independent voltage source
in series with a (normally small) impedance.
• A nonideal current source is modeled by an ideal independent current source
in parallel with a (normally small) admittance.
• Nonideal voltage and current sources can be converted equivalently into one
another via the formulas: Zs = \/Ys and Vs = ISZS.
• Thevenin's theorem states that any active network can be replaced at a given
pair of nodes by a nonideal voltage source. The equivalent voltage source is
equal to the open-circuit voltage VOC9 while the source impedance is equal to
the input impedance Zin.
• Norton's theorem states that any active network can be replaced at a given
pair of nodes by a nonideal current source. The equivalent current source is
the short-circuit current Isc, while the source admittance is equal to the input
admittance Yi„.
• The input impedance is related to the open-circuit voltage and short-circuit
current by the equation: Zin = Voc/lsc.
• Thevenin's theorem can be extended to nonlinear circuits with single nonlinear
resistors.
5-8
Problems
1. A nonideal inductor is connected in series with a nonideal capacitor. Find the input
impedance if the circuit parameters are as indicated in the table.
171
5.8. Problems
Circuit
Frequency
Inductor
CU (rad/s) L(mH)
2.
Capacitor
Rdtail)
C(/iF)
Gc (mö)
a
377
.001
100.
15
.01
b
2k
2.2
10.
.001
1
c
3.1 M
470
1.
47
100
A nonideal inductor is connected in parallel with a nonideal capacitor. Find the input
admittance if the circuit parameters are as indicated in the table.
Circuit
Inductor
Frequency
Capacitor
a) (rad/s)
L(mH)
RL (mil)
C(/iF)
Gc (raü)
a
1.1 M
.003
1000.
.0047
.002
b
5k
91
10.
1.0
5
c
377
20
1000.
56
20
3.
A nonideal current source has a phasor current /, = 20y3e- /V/ ' 3 . Find the equivalent
nonideal voltage source if Ys = (a) 1 W, (b) 2.1 MU, (c) -j'0.07 MU, and (d) 15 + 3j
kö.
4.
A nonideal voltage source has a phasor voltage Vs = 12 y 2ei7r^. Find the equivalent
nonideal current source if Z$ — (a) 3 kfl, (b) 20 /xil, (c) jO.07 Cl, and (d) 33 — lOy mil.
In Problems 5-11, find the Thevenin equivalent circuits with respect to the terminals indicated
in the following figures.
5. Figure P5-5.
6. Figure P5-6.
7. Figure P5-7.
9. Figure P5-9.
10. Figure P5-10.
11. Figure P5-11.
1 Q
AAA/V
8. Figure P5-8.
1/V3 H
o A
(T)\sl (0 = V2 cos( 3i) A \ 1 Q
O
2 Q
2.5Q
0.5 F
oB
v l 2 (i) = 5sin(3f) V
Figure P5-5
Figure P5-6
172
Chapter 5. Thevenin's Theorem and Related Topics
5 ft
2 ft
1 ft
1 ft
©1+JA
3Ü
2ft
<
L-AAA/v-i
Figure P5-7
Figure P5-10
-oA
3ft
Ql8cos(iot)<6ft
G)sin(cot)
3Ü
AAAAr
-OB
2 ft
r^AA/V
0.5 F
—lh-
U>os(2t)
?1H
I
1—O-
Figure P5-11
12.
Q-
Figure P5-8
Figure P5-9
6ft
7V
-o A
2ft
-o B
Figure P5-19
Prove Norton's theorem from basic principles (without invoking Thevenin's theorem).
In Problems 13-19, find the Norton equivalent circuits with respect to the terminals indicated
in the following figures.
13. Figure P5-6.
14. Figure P5-7.
15. Figure P5-8.
17. Figure P5-10.
18. Figure P5-11.
19. Figure P5-19.
16. Figure P5-9.
20.
Find the value of resistance R in the circuit shown in Figure P5-20 that results in a
Thevenin equivalent voltage of 5 V at the terminals indicated. What is the corresponding
equivalent resistance?
21.
An RL load is connected to a voltage source as shown in the circuit in Figure P5-21. For
the parameters listed in the following table, find the capacitance which, when added in
parallel to the load, adjusts the power factor to unity. Find the average power delivered
by the source after the capacitor is connected.
173
5.8. Problems
2 Q
R Q
Qvmcos(cot)
Figure P5-20
22.
L
Figure P5-21
Case
Vm{\)
to (rad/s)
R(£l)
L(mH)
a
5
10
1
10
b
12
2513
10
1
c
170
377
100
10
d
294
377
Ik
500
Consider the circuit shown in Figure P5-22. Find the value of ZL which maximizes the
power transferred to this element. What is the average power dissipated in the load?
AA/W
0.1 Q
(T)l20A/2 cos( 377 0 V
Figure P5-22
1 mH
1mF Z L n
Figure P5-24
23.
The resistance of a nonlinear resistor can be described by R = 10 + .5/ Í1, where / is the
current through the resistor in amperes. Use the graphical technique to find the voltage
across the load when it is connected to a nonideal voltage source with Vs = 169.7 V and
Rs = 10 fî.
24.
A varistor is an element which is designed to protect sensitive elements from large
current spikes. It has high resistance at low current and low resistance at high current. If
a varistor's resistance is given by R = 1000e - ' /2 Í1, where / is the current in amperes,
find the current through the 50 il resistor in Figure P5-24 when (a) Is — 1 A, (b) Is = 10
A, and (c) Is = 17.5 A. Use the graphical technique to solve the problem.
Chapter 6
Nodal and Mesh Analysis
6.1
Introduction
It has been shown in earlier chapters that the basic method for ac steady-state analysis
involves setting up and solving a system of linear equations derived from Kirchhoff's
current and voltage laws. Here, we recall that each branch in a circuit can be characterized by either impedance or admittance (the latter is merely the reciprocal of the
former). Given these impedances or admittances, the KCL and KVL equations can
be set up for the nodes and loops of the circuit in the following manner.
First, we use KCL:
k
k
which results in n — 1 linearly independent equations, where n is the number of nodes.
Then, we use KVL:
YJA = -J2^>
k
k
(6 2)
-
which results in b — (n — 1) linearly independent equations, where b is the number of
branches. In total, b equations will be produced, which could be a very large system
of equations for complex circuits. Fortunately, these equations are generally sparse,
a property which gives rise to special techniques for circuit analysis. Two of these
techniques are nodal analysis and mesh current analysis. They are presented in this
chapter.
In the final section, the PSpice circuit simulation package is used to analyze
several of the examples from this chapter.
6.2
Nodal Analysis
The main idea of nodal analysis is to represent KCL equations in terms of node
potentials. To describe the method of nodal analysis (also called the method of
node potentials), we consider an example circuit shown in Figure 6.1. Note that the
174
175
6.2. Nodal Analysis
Figure 6.1: Sample circuit for method of node potentials.
branches are characterized by admittances—this is not accidental. When using nodal
analysis, it is important to convert impedances into admittances before beginning
circuit analysis. This circuit also contains only current sources that are in parallel
with other elements. This is to simplify our initial discussion, and later this restriction
on sources will be removed.
To start the analysis, we introduce three new variables, v\, i>2> and v3. These
variables are the node potentials; that is, they are the potentials of the corresponding
nodes. Now, we recall from Chapter 1 that voltages can be expressed as potential
differences. Thus, we can write the voltage across any branch in the circuit in terms
of these node potentials. For instance, the voltage across the branch Y3 (between
nodes 1 and 2) is just the difference in the potentials at nodes 1 and 2. The values
of the node potentials must be taken with respect to some reference. Because we are
only interested in the difference in potentials and not their actual values, we are free
to choose whatever reference we wish. To simplify things, we set the potential of
some chosen node equal to zero. This is called the reference node, and it is generally
chosen to be the node with the most branches connected to it. In the example circuit,
the node with the greatest connectivity is node 3, so its potential is set to zero:
h = o.
Now, we write the KCL equations for nodes 1 and 2:
(6.3)
(6.4)
(6.5)
=
h ~ hcurrent,
~ h voltage
/s3 ~~ ï?4Recalling the relationship between
and admittance,
h = YkVk,
(6.6)
we can write the branch currents in terms of node potentials and admittances:
h = Yiih - h) = Yih,
(6-7)
h = Yiih - v3) = Y2v2,
(6.8)
176
Chapter 6. Nodal and Mesh Analysis
h = Y3(v{ - v2),
(6.9)
(6.10)
I4 = Y4(v{-hl
Using the above relations, we can now express the KCL equations (6.4) and (6.5) in
terms of node potentials and admittances,
Yin + y3(vi - h) + r4(vi - h) = i i - U
(6.11)
Y2v2 - F3(vi - h) - YA{vx - v2) = U ~ Î4After combining similar terms in the above equations, we have:
(6.12)
(7i + F3 + 74)vi - (Y3 + YA)h = 75l - U
(6.13)
-(Y3 + r4)vi + to + ^3 + n)v2 = i 3 - U
(6.14)
The last two equations represent the essence of the method of node potentials. These
equations can be solved for v\ and O2, and, once these node potentials are known, all
the currents in the circuit can be found by using expressions (6.7)-(6.10).
We rewrite the obtained equations in matrix form in order to make the benefits
of nodal analysis even more apparent. The matrix form is as follows:
Yv = t
(6.15)
where Y is an admittance matrix, and v and Is are the vector of unknown node
potentials and the "source" vector, respectively:
Y =
" F11
.
F
21
Yn '
Y22
Vl
. ^2 . '
L=
(6.16)
(6.17)
(6.18)
By using the introduced matrix notations, equations (6.13) and (6.14) can be written
as follows:
] r v,
-Y4-Y3
Yi + Y3 + Y4
/si — /s2
(6.19)
F
+
FS
+
*4
h
-Y4-Y3
/s3 ~~ IsA
2
Thus, we can see that:
(6.20)
Yn = Yi + Y3 + Y4,
Yn = Y2l = -(Y3 + Y4),
(6.21)
Y22 ™ Y2 + Y3 + Y4,
(6.22)
/5(1) = i i - hi,
(6.23)
(6.24)
Writing the equations in matrix form is useful because it reveals the pattern that the
matrices and source vectors of nodal analysis follow. For the admittance matrix in the
example, we can see that Y¡ 1 is equal to the sum of all branch admittances connected
to node 1. Likewise, 722 is equal to the sum of all branch admittances connected to
177
6.2. Nodal Analysis
node 2. The other two admittances, YX2 and F 2 i> are identical and equal to the negative
of the sum of the branch admittances connected between nodes 1 and 2. In the case of
the source vector, 7V(1) is equal to the algebraic sum of all current sources connected
to node 1, while 1^ is equal to the algebraic sum of all current sources connected to
node 2. In this case, "algebraic sum" means that currents entering the node are taken
with positive signs while those leaving the node are taken with negative signs.
This pattern for forming the coefficient matrices and the source vectors of the
nodal equations can be generalized to an arbitrary circuit with any number of nodes.
To write the matrix equations for any circuit, three simple rules must be followed. It is
this fast and easy way to produce the matrix equations as well as the relatively small
number of these equations that makes the method of node potentials so attractive.
Consider the general case of a circuit with n + 1 nodes. This circuit can be
characterized by n + 1 node potentials, which give a total of n equations (one node
is taken to be the reference node; its potential is set to zero and does not appear in
the equations). Then, a general form of the matrix equations for a circuit with n + 1
nodes is
YU
Y2i
Y12
Y22
/(I)
Yin
Y2n
/(2)
(6.25)
■nl
The three rules for determining the elements of the admittance matrix and the source
vector are as follows.
For diagonal matrix elements we have:
r„ = 5 >
(6.26)
which means that a diagonal element Ya of the coefficient matrix is the sum of all
branch admittances connected to node /, and superscript "/" in (6.26) suggests that
the summation is performed over all these branches.
For off-diagonal matrix elements we have:
//
v
ß
(6.27)
k
which means that an element which is on an intersection of the /th row and y'th column
of the matrix is the negative of the sum of all branch admittances directly connected
between nodes / and 7, and superscript "//" in (6.27) indicates that the summation is
performed over all branches between those nodes.
For the source vector elements we have:
Hi)
sky
k
(6.28)
178
Chapter 6. Nodal and Mesh Analysis
which means that the /th element of the source vector is the algebraic sum of all
current sources connected to node /.
Only these three simple rules are required to form the matrix equations for
the method of node potentials. It is clear that even for very complex circuits, these
equations can easily be formed by inspecting an electric circuit. We shall illustrate
the previous discussion by the following example.
EXAMPLE 6.1 Consider the circuit shown in Figure 6.2. We would like to write the
node potential equations in order to find all branch currents.
This circuit is already given in phasor-admittance form, but, if this had not been
the case, the first step in the analysis would have been to convert it to this form.
Recall that for the method of node potentials, the branches must be characterized by
admittances. These can be found by taking the reciprocal of the impedances.
We now determine how many nodes the circuit has. In this case, there are four
nodes, which means there will be three equations or, equivalently, a 3 X 3 admittance
matrix.
Once the nodes are identified and labeled (done already in Figure 6.2), the
node potentials are introduced, and the reference node is chosen. For our example,
we choose node 4 to be the reference node since it has the greatest connectivity.
Therefore, we set v4 equal to zero and proceed to set up the matrix equations to solve
for the remaining three node potentials. To set up the matrix, we simply follow the
three rules outlined above. In this way, we end up with:
+ Y4 + Y6
-Y4
-Y6
-YA
-Y6
-Y5
Y2 + Y4 + Y5
Y
+
Y5 + Y3
-Y5
6
Vl
v2
.
V
=
¡si ~~ hi
0
L A3
3 .
(6.29)
These matrix equations can be solved by using a variety of methods. In most cases,
the recommended method of solving linear systems of equations is the Gaussian
elimination technique (see Appendix B). This method involves triangulation of the
coefficient matrix and back-substitution of the variables.
i
i
1
i1
'S2V
1
i>
A
1
Yf i
Ig
\ i1 »
\
'A
i
1
A
1
1
1
1
r
'
-5
'i
1
Y2[
«t
(>
'
1
l3
'
1
Y5
Y3[
(>
Figure 6.2: Example circuit for nodal analysis.
d
'S3
179
6.2. Nodal Analysis
Once the node potentials are found, the branch currents can be determined by
using the relationship between currents, admittances, and node potentials:
h = YiVh
(6.30)
h = Y2v2,
(6.31)
h = Y3v3,
(6.32)
h = Y4(v{ - v2),
(6.33)
k = Y5(v2 - V3),
(6.34)
h = Y6(v\ - v3).
(6.35)
EXAMPLE 6.2 Consider the circuit shown in Figure 6.3. We want to use nodal
analysis to find the current through the 1/2 O resistor.
Comparing Figure 6.3 with Figure 6.2, we see that the circuits are identical if
we let Y6 = 2 15, Y2 = 3 U, Yx = Y3 = Y4 = Y5 = 1 U, Isl = 2 A, Is2 = 7 A, and
7s3 = 4 A. By plugging these values into the matrix equation (6.29), we obtain:
4
-1
-2
-1
5
-1
-2
-1
4
Vl
V2
V3
=
;
0
4
(6.36)
To solve the above equations, we can use the Gaussian elimination technique
that is described in Appendix B. Following the procedure in the appendix, we can
arrive at the following upper-triangular matrix:
4
0
0
-1
-2
19/4 - 3 / 2
0
48/19
Vl
V2
V3
=
-5
-5/4
21/19
(6.37)
We then use back-substitution to find v3 = 0.4375 V, v2 = —0.125 V, and v¡
-1.0625 V.
A/VW
1/2 a
Figure 6.3: A numerical example of nodal analysis.
180
Chapter 6. Nodal and Mesh Analysis
Finally, by using equation (6.35) we find the desired current: i$ = 2(—1.0625 —
0.4375) = - 3 A.
This concludes the analysis of the problem.
■
In the above treatment of nodal analysis, only circuits containing current sources
connected in parallel with the admittances have been considered. We now consider the
three other possibilities: (1) branches with current sources in series with admittances,
(2) branches with voltage sources in parallel with admittances, and (3) branches with
voltage sources in series with admittances. The development of techniques to deal
with these cases will make the method of node potentials applicable to any general
circuit.
The case of branches with current sources in series with admittances is the
simplest of the three—in fact, it is trivial. This is because admittances in series with
current sources will have no effect on the branch currents and, consequently, on the
KCL equations. Thus, they will not affect the matrix equations of nodal analysis.
This means that, for the purpose of finding node potentials, we can treat the series
admittance as though it wasn't there (i.e., replace it with a short). However, this
does not mean that these admittances have no effect on the circuit at all. Actually,
these admittances affect how much power is absorbed or supplied by current sources.
Indeed, the voltage drops across the current sources must be known to calculate power.
The currents are already known (they are the source currents) and the voltages can be
found as follows. First, we find the potential differences between the nodes to which
the above branches are connected; then the voltage drops across the admittances in
these branches are determined. Subtracting these two quantities will yield the voltages
across the current sources, which can then be used to calculate power.
The second case of branches with only voltage sources is somewhat more complicated. This case is shown in Figure 6.4, which shows the same circuit as in Example
6.1 except that it has one added voltage source between nodes 2 and 3. For this case,
we recall that voltage sources define the potential difference between the nodes
U CD'
Figure 6.4: A voltage source connected in parallel.
181
6.2. Nodal Analysis
to which they are connected. Since the potentials at nodes 2 and 3 are v2 and v 3 ,
respectively, we can conclude that
(6.38)
h - v2 = vs,
(6.39)
h = h + vs.
Thus, because of the voltage source, v2 and v3 are no longer independent variables,
and we can redefine v3 in terms of v2 and Vs as shown in (6.39). This effectively
reduces the number of unknown variables in the equations by one. However, there
is a complication: the current Ix through voltage source Vs is not known and it
cannot be expressed through known quantities. To circumvent this difficulty, we shall
temporarily treat Ix as a known quantity and write the nodal equations, which are
similar to (6.29):
Yi + Y4 + Y6
-I4
Y2 + Y4 + Y5
-Y4
-Y6
-Y5
-Y6
-Y5
Y6 + Y5 + F 3
I si ~ hi
Vl
v2
.
V3
h
=
_ Is3 - h
_
(6.40)
We next eliminate Ix by summing up the second and third equations in (6.40). This
results in the following set of two equations with three unknowns:
Vl
i+Y4 + Y6
-Y4
-Y6
-Y4-Y6
Y2 + Y4 Y3 + Y6 _
v2
V3
/si — hi
Is
(6.41)
Finally, we reduce the number of unknowns by replacing v3 by v2 + Vs (see (6.39)).
After simple transformations, we end up with the following two equations with two
unknowns:
Y\ + Y4 + Y6
-Y4-Y6
-Y4-Y6
Y2 + Y3 + Y4 + Y6
vi
v2
hi ~ hi + Y^VS
I3 - (Y3 + Y6)VS
(6.42)
By solving these equations, we can find all nodal potentials and then compute all
currents.
The third case of voltage sources connected in series with admittances is somewhat easier than the previous one. As an example, consider the circuit shown on the
left in Figure 6.5. Note that the voltage sources in this circuit can actually be considered as nonideal voltage sources since they are connected in series with impedances.
For this reason, we simply use the equivalent transformation of nonideal voltage
sources into nonideal current sources. After performing the transformation, the circuit will look like the one shown on the right in Figure 6.5. This circuit has current
sources in parallel with admittances. The values for the new current sources are
VsiYl9
(6.43)
hi = Vs2Y2.
(6.44)
182
Chapter 6. Nodal and Mesh Analysis
Figure 6.5: Example of voltage sources in series with admittances.
The direction of the current sources can be determined from the polarity of the voltage
sources as described at the end of Section 5.3. In our case, this means that the new
current sources will be entering nodes 1 and 3.
Once the voltage sources have been transformed, we end up with precisely the
same case which has been considered in the original derivation of nodal equations.
Therefore, we already know how to write these nodal equations. They are as follows:
Yx+Y3 + Y6
~Y3
-Y6
-Y3
Y3 + Y4 + Y5
-Y5
-Y6
-Y5
Y2 + Y5 + Y6
rii i
" Vl
0
v2
V
.
. (6.45)
3
By substituting (6.43) and (6.44) into (6.45), we end up with:
F! + F3 + ^6
-Y3
-Y6
-Y3
Y3 + Y4 + Y5
~Y5
-Y6
-Y5
Y2 + Y5 + Y6
Vl
h
.
V
3
=
r Y,VS1 i
0
[ Y2Vs2 \
(6.46)
By comparing equations (6.46) with the original (not transformed) circuit shown in
Figure 6.5, we conclude that we can write these matrix equations just by inspection
of the original circuit. The first two rules concerning the formation of the admittance
matrix remain the same. However, the third rule, which is used for the formation of
the source vector, should be somewhat modified. According to the modified rule, the
elements of the source vector are given by the expression:
Sk
k
+ ¿JVPrf.
(6.47)
Here, the first term in the right-hand side of (6.47) has the same meaning as in (6.28)
and it accounts for all current sources connected to node /. The second term in (6.47)
is the algebraic sum of products Y^Vsk and summation is taken over all voltage sources
connected to node /. The word "algebraic" means that voltage sources are taken with
183
6.2. Nodal Analysis
positive signs if their "positive" terminals are connected to node /, and voltage sources
are taken with negative signs if their "negative" terminals are connected to node /.
By using rules (6.26), (6.27), and (6.47) we can immediately write the nodal
equations in matrix form just by inspecting an electric circuit. This is true for all
circuits except those which contain branches with voltage sources alone. These
"exceptional" circuits require special treatment which has already been discussed.
EXAMPLE 6.3 As an application of nodal analysis, we consider a three-phase circuit, pictured in Figure 6.6. This three-phase circuit consists of three voltage sources
and three impedances arranged into "star" connections. Each of the voltage sources
has the same peak value and frequency, but they are out of phase with each other
by 120°. There is a neutral wire connecting nodes 1 and 2 which keeps them at
almost the same potential if the impedance Zn of the neutral wire is very small.
This wire is present so that variations of any (load) impedance in the circuit will
not affect significantly voltages across other load impedances. In other words, the
neutral wire is used to maintain almost constant voltages across load impedances.
Such three-phase circuits are used by utility companies to supply electric power at
almost constant voltages in the face of permanently and drastically changing loads.
Ideally, the impedance of the neutral wire should be zero, but realistically there is
always some small impedance present. We would like to analyze this circuit by using
nodal techniques in order to understand the influence of Zn on the variation of load
voltages.
We begin by introducing node potentials v2 and vi for each of the two nodes in
the circuit. Because there are only two nodes, we will end up with one nodal equation.
We choose v2 to be the reference node, so
v2 = 0.
(6.48)
Now, we must convert the impedances into admittances, because nodal analysis is
carried out for circuits characterized by admittances. Therefore, we have:
Figure 6.6: A three-phase network.
184
Chapter 6. Nodal and Mesh Analysis
We now form the node equation. In our case, the admittance matrix consists of only
one element, which is equal to the sum of all admittances connected to node 1. We
also note that this circuit contains voltage sources in series with admittances, so we
can use the modified rule (6.47). This leads to:
(Y{ + Y2 + Y3 + Yn)vx = Y{Vsl + Y2Vs2 + Y3Vs3,
(6.50)
.
YXVS1 + Y2VS2 + Y3VS3
vi =
.
, , . n
(6.51)
which results in:
Equation (6.51) is the central formula of three-phase circuit theory. By using this
formula, we can make a few general observations.
First, when the load is balanced, that is, when all the admittances are equal, then
the neutral wire is not needed. To demonstrate this, we take advantage of the fact that
in the case of a balanced load:
Y{=Y2
(6.52)
= Y3 = Y.
Then equation (6.51) is reduced to
vi =
.
(6.53)
v
J
3Y + Yn
Although it is not immediately obvious, this equation means that when the load is
balanced, v\ will be equal to zero and, consequently, it will be equal to v2 for any
value of Yn. To prove this, recall that the three voltage sources have the same peak
values and are out of phase by 120°. Thus, we can write them in phasor form as
follows:
Vsl = Vmt
(6.54)
Vs2 - Vmej27r/\
(6.55)
Vs3 = Vme**/3.
(6.56)
Through a little algebraic manipulation, we can show that the sum of these three
expressions is zero, making vi in (6.53) also equal to zero. Indeed,
Vm + Vme^/3
/
+ Vme^lz
i
(
2/7T
Z.7T
4 / 7T
= Vm Í1 + cos — + j sin — + cos —+j
4/7T\
sin — J,
(6.57)
VS] + VS2 + VV3 = Vm 11 - .5 + ^-j
- .5 - ^-j)
= 0.
(6.58)
Therefore, the value of Yn does not matter when the load is balanced, and the neutral
wire can be omitted.
185
6.2. Nodal Analysis
Next, consider the case of an unbalanced load (i.e., when (6.52) is not valid). In
this case, if \Zn\ is very small, \Yn\ is very large, and, in the limit of \Zn\ approaching
zero, from (6.51) we find:
(6.59)
vi = 0.
Thus, in the case of an unbalanced load but an ideally conducting neutral wire we
still have
(6.60)
vi = v2,
which should be expected from the configuration of the three-phase circuit. In this
ideal case, the voltages across load impedances Z* are equal to V^, respectively, and
they do not depend on load impedances.
If Yn is finite, then from (6.51) we can find \>i and voltages across loads Zk:
(6.61)
Vt = V,v
Expressions (6.51) and (6.61) allow one to estimate variations of load voltages due
to an unbalanced load.
Why do we need three-phase power? One reason is that, in order to generate electricity or to convert it into mechanical power, synchronous generators and induction
motors are utilized. The principle of operation of these electric machines is based on
uniformly rotating magnetic fields. It is easy to create these rotating magnetic fields
by using three-phase circuits. Another reason is that the total power consumed by
the loads in a balanced three-phase system is constant in time. This can be seen by
summing the expressions for instantaneous power (3.184) for each of the three loads
and performing a few simple algebraic manipulations. This property is desirable for
power plants, which prefer to generate energy at a fairly constant rate.
■
EXAMPLE 6.4 As an example of nodal analysis of general circuits, let us find the
current through the 1 il resistor in the circuit shown in Figure 6.7a.
We see that the 4 V voltage source has a series resistor and consequently, it can
be converted into a nonideal current source. The 3 Í1 resistor that is connected in
series with the 3 A current source can be ignored for the purpose of calculating node
potentials. The 3 V voltage source cannot be converted, so we assume that some
unknown current Ix is flowing through it. We consider the node at the bottom of the
circuit to be the reference node. The modified circuit is shown in Figure 6.7b and the
corresponding matrix equation is given below:
5/6
-1/2
0
-1/2
5/6
-1/3
0
-1/3
4/3
Vl
V2
V3
=
'2-1
1
(6.62)
h
Because the 3 V voltage source is connected between node 1 and node 3, vj = 3 + v3.
Now, we can reduce the number of equations to two by adding the first and third rows
and by inserting the above expression for v\ into the unknown node potential vector
186
Chapter 6. Nodal and Mesh Analysis
2A
r^WW
O4V
2Q
WW4
3Q
T
vww
(a)
(b)
Figure 6.7: A general circuit for nodal analysis: (a) the original circuit and (b) an
equivalent circuit.
and shifting the constants to the source column:
5/6
-5/6
-5/6
13/6
v2
V3
By using Gaussian elimination, we can find the desired current to be ¿m = 1.5 A. ■
6.3
Mesh Current Analysis
Mesh current analysis is a technique which is dual in form to the method of node
potentials. The basic idea of the mesh current technique is to represent KVL equations
in terms of mesh currents.
To arrive at the method of mesh currents, we consider the circuit shown in Figure
6.8.
+ z
- A
*
.0
L-.
-
*
-
*
+z
- A
i
1 »g
K J Wv&
)
■
Figure 6.8: Example circuit for mesh current analysis.
6.3. Mesh Current Analysis
187
To simplif
y ou
r discussion
, we hav
e deliberatel
y constructe
d a circui
t tha
t con
tain
s onl
y voltag
e source
sn
i serie
s wit
h impedances
. Later
, we wil
l exten
d th
e mes
h
curren
t techniqu
eo
t includ
e othe
r sourc
e configuration
s a
s well
. We begi
n wit
h KCL.
Ther
e ar
e tw
o non-trivia
l nodes
,s
o ther
e wil
l be onl
y on
e KCL equation
:
(6.64
)
or, equivalently
,
(6.65
)
Now , we writ
e KVL fo
r th
e tw
o meshe
s ni th
e circuit
:
hZx + I3Z3 - Vsl = 0,
(6.66
)
-73Z 3 +I2Z2 + Vs2 = 0.
(6.67
)
If we substitut
e (6.65
) int
o th
e KVL equations,
we wil
l en
d up wit
h tw
o equation
sn
i
term
s of onl
y tw
o unknowns
:I\ an
d I2.
IXZX + (/
, - 72)Z3 = Vs],
(6.68
)
- (!/ -I2)Z3 + I2Z2 = -Vs2.
(6.69
)
Rearrangin
g th
e terms
, we obtain
:
/1 ( Z1 + Z 3 ) -2/Z 3 = Vs[,
(6.70
)
-hz3 + i2iz2 + z3) = -vs2.
(6.71
)
These tw
o equation
s ar
e th
e sam
e equation
s tha
t wil
l be produce
d belo
w by usin
g
mesh curren
t analysis
. However
, th
e mes
h curren
t analysi
s provide
s a much easie
r
way o
t do thi
s by jus
t lookin
g a
t th
e circuit
.
Mesh curren
t analysi
s use
s ne
w mathematica
l quantitie
s calle
d mesh currents.
These ar
e fictitious
, virtua
l circulatin
g current
s whic
h ar
e introduce
d o
t simplif
y
calculations
. The
y hav
e no physica
l meanin
g bu
t ar
e ver
y helpfu
ln
i writin
g equation
s
for circuit
s whic
h contai
n many meshes
. We introduc
e on
e mes
h curren
t fo
r eac
h
mesh of th
e circuit
, show
n n
i Figur
e 6.9
. Not
e tha
t referenc
e direction
s fo
r al
l mes
h
current
s ar
e coordinated
. We shal
l als
o choos
e a tracin
g directio
n fo
r eac
h loo
p
».6 (\
)*-0 (T
)9
Figur
e 6.9
: Circui
t wit
h mes
h currents
.
Chapter 6. Nodal and Mesh Analysis
188
(mesh
) coincidin
g wit
h th
e referenc
e directio
n of th
e mes
h curren
t throug
h thi
s loop
.
The following rules are introduced n
i orde
r o
t writ
e KVL equation
s n
i term
s of
mesh currents
. When mes
h curren
t Xk flowsthroug
h impedanc
e Zk, thi
s result
s n
i
the voltag
e dro
p TkZk. Thi
s voltag
e dro
p s
i take
n wit
h positiv
e sig
n fi th
e referenc
e
e mes
h curren
t coincide
s wit
h th
e tracin
g directio
n of a loop
, otherwis
e
directio
n of th
it s
i take
n wit
h negativ
e sign
. Fo
r instance
, fi we conside
r th
e first
loo
pn
i Figur
e 6.9
,
the
n th
e voltag
e dro
p n
i thi
s loo
p du
e o
t mes
h curren
t% s
i%{Z\ + Z3), whil
e th
e
voltag
e dro
p du
e o
t mes
h curren
t% s
i —Z2Z3. By usin
g th
e abov
e rules
, we obtai
n
the followin
g KVL equations
:
XKZ j + Z3) - X2Z3 = VSh
(6.72
)
-I{Z3
(6.73
)
+ X2(Z3 + Z2) = -Vs2.
It is clear that equations (6.72) and (6.73) are mathematically identical to equations
(6.70) and (6.71). Thus
, we ca
n conclud
e that
:
h = Z\,
(6.74
)
h = %i,
(6.75
)
73 = Jj -12.
(6.76
)
The abov
e thre
e equation
s relat
e th
e purel
y mathematica
l mes
h current
so
t th
e actua
l
branc
h current
sn
i th
e circuit
. Fro
m thi
s example
, we ca
n formulat
e tw
o genera
l rule
s
of relatin
g mes
h current
so
t actua
l branc
h currents
.
1. f
I a branc
h belong
so
t onl
y on
e mesh
, the
n th
e branc
h curren
ts
i equa
lo
t th
e mes
h
curren
t fi thei
r referenc
e direction
s coincide
, otherwis
e t
is
i th
e negativ
e of th
e
mesh current
.
2. f
I abranc
h s
i share
d betwee
n tw
o meshes
, the
n th
e branc
h curren
ts
i th
e algebrai
c
sum of th
e tw
o mes
h currents
. The wor
d "algebraic
" mean
s tha
t a mes
h curren
t
is take
n wit
h a positiv
e sig
n fi it
s referenc
e directio
n coincide
s wit
h th
e referenc
e
directio
n of th
e branc
h current
; otherwis
e th
e mes
h curren
ts
i take
n wit
h a minu
s
sign
.
Like th
e metho
d of nod
e potentials
, th
e mes
h curren
t analysi
s generate
s a syste
m
of linea
r equation
s whic
h ca
n be represente
d ni th
e matri
x form
:
Z X = Vs.
(6.77
)
For a two-mes
h circuit
, th
e abov
e for
m of th
e matri
x equation
s is
:
Z\[
Z 2[
Z\2
Z22
=
r t> (1) 1
v
.
s
V 2)
s
.
In ou
r example
, thes
e equation
s ca
n be specifie
d a
s follows
:
r z, + z 3 - z 3
I
- z3
z3 + z2
1\
V S1
£2
-VS2
(6.79
)
6.3. Mesh Current Analysis
189
Comparin
g th
e las
t matri
x equatio
n wit
h equatio
n (6.78
) we find
that
:
Zi i = Zj + Z3,
(6.80
)
z3 + z2,
(6.81
)
-22
Z12 — Z 2 i — —Z 3
(6.82
)
t>(1) - Vvl,
(6.83
)
Vj2) = -Vs2.
(6.84
)
As n
i noda
l analysis
, a specifi
c pattern emerge
s when th
e mes
h curren
t equation
s ar
e
writte
n n
i matri
x form
. Indeed
,Z\ \ s
i equa
lo
t th
e su
m of al
l impedance
sn
i th
e first
mesh, whil
e Z22 s
i equa
lo
t th
e su
m of al
l impedance
sn
i th
e secon
d mesh
. Impedance
s
Z12 an
d Z21 ar
e th
e sam
e an
d equa
lo
t th
e negativ
e of th
e su
m of impedance
s share
d
betwee
n th
e tw
o meshes
. Voltage
s Vs(1) an
d Vv(2) ar
e equa
lo
t th
e algebrai
c sums of
voltag
e source
s aroun
d meshe
s 1 an
d 2, respectively
.
Of course
, thes
e pattern
s ca
n be generalize
d fo
r an
y plana
r circui
t wit
h an
y
number of meshes
. As wit
h noda
l analysis
, ther
e ar
e thre
e simpl
e rule
so
t follow
.
For a genera
l circui
t wit
h m meshes
, th
e matri
x equatio
n ha
s th
e form
:
z„
z2]
Z12
Z22
^ra
2
z lw i r 1\% i
%.
'-"mm
j
Tm
j
_ -L,
■J-'YY
r vil)
t>(2)
(6.85
)
j
y(m)
The thre
e rule
s fo
r writin
g thes
e matri
x equation
s ar
e th
e followin
g
For diagona
l element
s of th
e impedanc
e matri
x we have
:
(6.86
)
Z/7 - / ^Zk,
which mean
s tha
t a diagona
l elemen
t Zz/ of th
e coefficien
t matri
x s
i th
e su
m of al
l
branc
h impedance
sn
i mes
h .
/
For off-diagona
l matri
x element
s we have
:
Z/7 -
- >
Zh
(6.87
)
k
which mean
s tha
t a
n elemen
t on th
e intersectio
n of th
e /t
h ro
w an
d y't
h colum
n of
the coefficien
t matri
x s
i th
e negativ
e of th
e su
m of al
l branc
h impedance
s common
to meshe
s /an
d j.
For th
e sourc
e vecto
r element
s we have
:
W = YA
(6.88
)
k
which mean
s tha
t th
e /t
h elemen
t of th
e sourc
e vecto
rs
i th
e algebrai
c su
m of al
l
voltag
e source
s n
i th
e /t
h mesh
. A voltag
e sourc
e s
i take
n wit
h a minu
s sig
n fi th
e
mesh curren
t referenc
e directio
n s
i pointin
g ou
t of ("exiting"
) th
e negativ
e termina
l
Chapter 6
. Nodal and Mesh Analysis
«.©
(T
)
•
,6
z.(
Z
5
Z
l
5
/
6
6
l6
Figur
e 6.10
: Exampl
e circui
t fo
r mes
h curren
t analysis
.
of th
e voltag
e source
, an
d ti s
i take
n wit
h a plu
s sig
n fi th
e referenc
e directio
n of th
e
mesh curren
ts
i exitin
g th
e positiv
e termina
l of th
e voltag
e source
.
The abov
e thre
e rule
s shoul
d be complemente
d by th
e previousl
y discusse
d
two rule
s whic
h relat
e virtua
l mes
h current
so
t th
e actua
l branc
h currents
. We shal
l
illustrat
e th
e previou
s discussio
n by th
e followin
g example
.
EXAMPL E 6.
5 Conside
r th
e circui
t show
n n
i Figur
e 6.10
. Our goa
ls
io
t writ
e th
e
matri
x mes
h curren
t equation
s an
d find
al
l branc
h currents
.
The circui
ts
i give
n n
i phasor-impedanc
e for
m a
s t
is
i appropriat
e fo
r mes
h
analysis
. t
Is
i clea
r fro
m th
e figure
tha
t th
e circui
t s
i plana
r an
d contain
s thre
e
meshes
. Therefore
, we wil
l nee
d o
t se
t up a 3 X 3 impedanc
e matrix
. The resultin
g
matri
x equation
s are
:
+ z2 + z 3
~z2
-z3
-z2
z2 + z4 + z5
-z4
-z 3
-z 4
Z3 + Z4 +Z^
1\
ll
Vsi
=
Y% J
vs2
[ -Vs3
(6.89
)
These equation
s ca
n no
w be solve
d by usin
g th
e Gaussia
n eliminatio
n techniqu
e o
t
findth
e mes
h currents
. Usin
g th
e mes
h current
s an
d th
e tw
o rule
s fo
r relatin
g mes
h
current
so
t actua
l branc
h currents
, we ca
n calculat
e th
e branc
h current
s ni th
e circuit
:
(6.90
)
(6.91
)
(6.92
)
U — X2 — J3
,
(6.93
)
6.3. Mesh Current Analysis
191
I5 = J
2,
(6.94
)
h = ±3
.
(6.95
)
This conclude
s th
e solutio
n of th
e problem
.
In th
e abov
e discussio
n of mes
h curren
t analysi
s we deal
t onl
y wit
h circuit
s
tha
t containe
d voltag
e source
s n
i serie
s wit
h impedances
. We no
w conside
r thre
e
more possibilitie
s whic
h may arise
: a curren
t sourc
en
i paralle
l wit
h a
n impedance
,a
curren
t sourc
e n
i serie
s wit
h a
n impedance
, an
d a voltag
e sourc
e n
i paralle
l wit
h a
n
impedance
.
The cas
e of a curren
t sourc
e ni paralle
l wit
h a
n impedanc
e s
i relativel
y eas
y
to remedy
. Thi
s situatio
n s
i illustrate
d on th
e lef
t n
i Figur
e 6.11
. Not
e tha
t thi
s
connectio
n ca
n be simpl
y construe
d a
s a nonidea
l curren
t source
, whic
h ca
n be
equivalentl
y transforme
d o
t a nonidea
l voltag
e source
. Now, th
e sourc
e s
in
i a for
m
which we alread
y kno
w ho
w o
t handl
e an
d we als
o en
d up wit
h a circui
t whic
h ha
s
fewer meshes
.
The secon
d cas
e s
i a bi
t mor
e complicate
d o
t handle
.f
I a branc
h contain
s a
curren
t source
, the
n we alread
y kno
w th
e valu
e of th
e branc
h current—i
ts
i th
e sourc
e
curren
t itself
. Therefore
, thi
s curren
t sourc
e eithe
r define
s on
e of th
e mes
h current
s
(i
f th
e branc
h s
i no
t share
d by tw
o meshes)
, or t
i define
s th
e differenc
e betwee
n tw
o
mesh current
s (i
f th
e branc
h s
i common o
t tw
o meshes)
.n
I an
y case
, th
e curren
t
sourc
e wil
l reduc
e th
e numbe
r of unknowns
. As a
n example
, conside
r th
e circui
t
shown n
i Figur
e 6.12
, whic
h s
i th
e sam
e a
s th
e circui
tn
i Figur
e 6.1
0 excep
t fo
r a
curren
t sourc
e connecte
d n
i serie
s wit
h impedanc
e Z4. We kno
w 14 s
i equa
lo
t th
e
curren
t source
, an
d t
is
i als
o equa
lo
t th
e differenc
e betwee
n X2 an
d X3. So we have
:
h = I = T2-
%,
(6.96
)
%=T2-1.
(6.97
)
Clearly
,X2 an
d X3 ar
e no longe
r independen
t s
o th
e actua
l numbe
r of unknown
s si
reduce
d by one
. But what seem
s o
t presen
t a proble
m s
i th
e unknow
n voltag
e Vx
acros
s th
e curren
t source
. However
, thi
s voltag
e ca
n be eliminate
d when composin
g
the mes
h curren
t equations
. Conside
r th
e mes
h equation
s se
t up wit
h thi
s voltag
e
—f
Z
—i-
bi
/ +\ A
A
0Vs='sZ
Figur
e 6.11
: Curren
t sourc
en
i parallel
wit
h impedance
.
Chapter 6. Nodal and Mesh Analysis
192
6
r^—i—r
3
1 '
1
Figur
e 6.12
: Curren
t sourc
e adde
d n
i series
.
include
d a
s a voltag
e source
.t
I the
n appear
s ni th
e sourc
e vector
:
V,i
Vs2 -Vx
1
.
(6.98
)
-vs3 + vx J
(The coefficien
t matri
x s
i no differen
t tha
n before
; tha
ts
i why t
is
i omitte
d here.
)
W e ca
n easil
y eliminat
e thi
s unknow
n voltag
e by summin
g up th
e botto
m tw
o equa
tions
. Thi
s result
sn
i tw
o equation
s wit
h no unknow
n voltag
e source
s present
. Thes
e
equation
s are
:
z{+z2+
z3
-z2
~(Z2 + Z3) z2+~z5
-z3
z3+z6
%
12
% -1
Vs2 - Vs3
(6.99
)
By usin
g simpl
e transformations
, equatio
n (6.99
) ca
n be reduce
d o
t th
e followin
g
form:
A+Z2+
Z3
X,
V5i ~ ISZ3
-z2 - z3
V,
1
v
-z2 - z3 z2+z3 + z5+ z6
2
s2
s3 + Uz3 + z
6)
(6.100
)
These equation
s ca
n no
w be solve
d by employin
g th
e conventiona
l linea
r algebr
a
techniques
.t
Is
i worthwhil
e o
t not
e tha
t th
e mes
h curren
t analysi
s of circuit
s wit
h
curren
t source
ss
i dua
lo
t th
e noda
l analysi
s of circuit
s wit
h voltag
e sources
.
The final
cas
e of voltag
e source
sn
i paralle
l wit
h impedance
ss
i th
e simples
t of
the three
. Thi
ss
i becaus
e th
e value
s fo
r th
e current
s throug
h thes
e impedance
s ar
e
immediat
e an
d thes
e impedance
s do no
t affec
t an
y of th
e othe
r mes
h currents
. Thi
s
means, fo
r th
e purpos
e of finding
mes
h currents
, tha
t we ca
n trea
t paralle
l impedance
s
6.3, Mesh Current Analysis
r W W f
cos(t) ( T )
in
>
193
It
1H
F
2 11Q
OrH )
AAA/V
1/2 a
Q
3j1Q>(4J1/2Q
>2sin(t)
Figure 6.13: A circui
t fo
r mes
h analysis
.
as thoug
h the
y weren'
t ther
e (i.e.
, replac
e the
m wit
h a
n ope
n branch)
. Still
, thes
e
impedance
s do affec
t th
e circuit—the
y modif
y ho
w much powe
r s
i absorbe
d or
supplie
d by th
e voltag
e sources
.
EXAMPLE 6.6 Conside
r th
e circui
t show
n n
i Figur
e 6.1
3 wit
h al
l sourc
e current
s
give
n n
i amps
. Fin
d al
l th
e phaso
r mes
h current
sn
i th
e circuit
.
W e star
t by convertin
g th
e nonidea
l curren
t sourc
eo
t th
e righ
t of th
e fourt
h mes
h
int
o a nonidea
l voltag
e sourc
e an
d the
n combin
e th
e tw
o 1/2(
1 resistors
, whic
h
become connecte
d n
i series
. We the
n conver
t th
e source
s o
t phaso
r for
m (co = 1)
,
and findth
e impedance
s of al
l branches
. As a result
, we arriv
e a
t th
e circui
t show
n
in Figur
e 6.14
. By usin
g th
e standar
d mes
h analysi
s procedure
, we obtai
n th
e matri
x
equation
:
2
-1
0
0
- 1 0 0
2-j
-1
0
-1
2 + j -1
0 - 12
" 1 "
%
%
1% J
r -vx i
0
0
(6.101)
-j
Becaus
e th
e curren
t sourc
es
i containe
d onl
y ni th
e first
mesh
,T{ s
i specifie
d an
d we
reduc
e thi
s proble
m o
t thre
e equation
s by simpl
y ignorin
g th
e equatio
n represente
d
by th
e first
ro
w an
d by shiftin
g th
e first
colum
n of th
e matri
x (multiplie
d by 1 A) o
t
-AAA/V
1 Q.
1 Q
V
6 1 (T)i a
A
1 Q
V
o
Figure 6.14: An equivalen
t circui
t fo
r th
e mes
h analysi
s problem
.
Chapter 6. Nodal and Mesh Analysis
194
the sourc
e column
:
2-j
1
0
-1
2+j 1
0 - 1 2
\%. ]
T,
=
[% J
1
0
. -j _
(6.102
)
t solv
e fo
r th
e mes
h currents
. The uppe
r triangula
r
W e us
e Gaussia
n eliminatio
n o
matri
x is
:
2-j
0
0
1
4(2 + j)/5
0
0
1
6
( +j)/4
?2
lz
%J
=
1
(2 +j)/5
(6.103
)
[ 1/4 -j
W e complet
e th
e proble
m wit
h back-substitutio
n o
t findJ
( — 25j)/319 X3 =
4 =2
=
(4 — 13j)/37
, andX
(1
9 + 3y)/37
. Thi
s conclude
s th
e analysi
s of th
e problem
.
2
EXAMPLE 6.7 Conside
r acircui
t simila
ro
t th
e on
e show
n n
i Figur
e 6.1
3 excep
t
tha
t th
e curren
t sourc
es
i common o
t meshe
s 1 an
d 2a
s show
n ni Figur
e 6.15
.n
I thi
s
cas
e th
e standar
d mes
h analysi
s procedur
e produces
:
2
-1
0
0
-1
2-j
-1
0
0
0
-1
0
-1
2 +7
-1
2
\% 1
T2
X3
L^4 J
r -vx l
+vx
(6.104)
0
-j
Now we must ad
d row
s 1 an
d 2o
t eliminat
e th
e unknow
n voltage
:
1 1 -7
0
1
0
0
rA/WV
1 a
1
0
2 + 7 -I
- 12
1H
x2
x3
(6.105
)
1A
t-A/Wv—f
1/2 a
1 Q > (^3J 1 ^ S (^4j i/2 H
Q
Figure 6.15: Anothe
r exampl
e circui
t fo
r mes
h analysis
.
>2sin(t)
6A. MicroSim PSpice Simulations
195
A o
t eliminat
e X\:
W e us
e th
e fac
t tha
t 2\ - 12 = 1
0
-1
(6.106
)
~J
The applicatio
n of Gaussia
n eliminatio
n o
t determin
e th
e mes
h current
ss
i lef
t a
s a
n
exercis
e fo
r th
e reader
.
6.4
MicroSi
m PSpic
e Simulation
s
In thi
s section
, we wil
l us
e th
e PSpic
e circui
t simulatio
n cod
e o
t explor
e th
e perfor
mance of thre
e of th
e exampl
e circuit
s fro
m thi
s chapter
.f
I yo
u di
d no
t wor
k throug
h
the example
s discusse
d a
t th
e en
d of Chapte
r 4 wit
h th
e PSpic
e code
, we strongl
y
urge yo
u o
t go bac
k an
d do tha
t now, s
o tha
t yo
u ca
n maximiz
e you
r understandin
g of
thi
s simulatio
n tool
.f
I necessary
, Appendi
x C list
s severa
l reference
s whic
h contai
n
considerabl
y mor
e informatio
n abou
t th
e code
.
The firs
t circui
t tha
t we conside
r s
i th
e three-phas
e networ
k of Exampl
e 6.3
.
The circuit
,a
s redraw
n by th
e PSpic
e schemati
c generator
,s
i show
n ni Figur
e 6.16
.
The thre
e voltag
e source
s represen
t househol
d power
, hav
e th
e par
t name "VSIN,
"
and ar
e give
n magnitude
s (VAMPL) of 169.
7 V, offse
t voltage
s (VOFF) of zero
, an
d
frequencie
s (FREQ) of 60 Hz. The phase
s (PHASE) of vl
, v2
, an
d v3 ar
e 0°
, 120°
,
and 240°
, respectively
. The neutra
l wir
es
i give
n aresistanc
e of 1
0 £1 an
d th
e nomina
l
resistanc
e of th
e thre
e load
ss
i 1 kfl
. The groun
d (AGND) s
i connecte
d o
t th
e secon
d
node a
sn
i th
e example
.
PARAMETERS:
1k
rv
Figur
e 6.16
: The schemati
c drawin
g o
f th
e three-phas
e network
.
196
Chapter 6. Nodal and Mesh Analysis
W e wil
l us
e PSpic
e o
t investigat
e th
e curren
t throug
h th
e neutra
l wir
e (R4
)
and th
e instantaneou
s powe
r dissipatio
n when th
e loa
d s
i no
t balanced
. A transien
t
analysi
ss
i selecte
d wit
h a final
tim
e of 50 ms 3
( cycles
) an
d a prin
t ste
p of 0.
2 ms. A
parametri
c analysi
s wil
l be use
d o
t var
y th
e resistanc
e of th
e thir
d leg
. The resisto
r
R3 si assigne
d a valu
e of "rv
" an
d v
r s
i define
d o
t be a "Globa
l Parameter
" n
i th
e
"Parametric...
" sectio
n of th
e "Analysi
s Setup...
" menu
. "Valu
e List
"s
i chose
n a
s
the "Swee
p Type
" an
d we giv
e v
r th
e value
s of 790
, 990
, 1190
, an
d 1390
. We must
als
o assig
n a valu
e o
t R3 fo
r th
e dc bia
s poin
t analysi
s an
d thi
s s
i don
e by typin
g
"PARAM " int
o th
e "Ad
d Part
" dialo
g box
. We plac
e th
e resultin
g ico
n anywher
e on
the schemati
cn
i th
e usua
l way an
d double-clic
k on th
e wor
d "Parameter
"o
t brin
g up
the "PM1 Par
t Name: PARAM " dialo
g box
. We se
t Namel equa
lo
tv
r an
d se
t Valu
e1
equa
lo
t k
I (remembe
ro
t hi
t th
e "Sav
e Attr
" butto
n fo
r eac
h change)
. Afte
r selectin
g
the "OK" button
, th
e initia
l valu
e of k
I fo
r v
r shoul
d be liste
d unde
r "Parameter
s "
Afte
r savin
g th
e schematic
, PSpic
e ca
n be ru
n by selectin
g "Simulate
" fro
m th
e
"Analysis
" drop-dow
n menu
.
When Prob
es
i launche
d a
t th
e en
d of th
e simulation
, adialo
g bo
x wil
l appea
r tha
t
allow
s th
e use
ro
t selec
t whic
h of th
e parametri
c result
so
t plot
. Al
l fou
r simulation
s
can be plotte
d by selectin
g "All
" an
d "OK.
" The curren
tn
i th
e neutra
l wir
e (I(R4)
) si
plotte
d n
i Figur
e 6.17a
. The dashe
d lin
e si th
e 79
0 ftcase
, th
e soli
d lin
e correspond
s
to th
e 99
0 ftcase
, th
e dot-dashe
d lin
e s
i th
e 119
0 ftcase
, an
d th
e dotte
d lin
e s
i th
e
1390 ftcase
. The curren
ts
i almos
t zer
o fo
r th
e 99
0 ftcas
e sinc
e th
e loa
d s
i nearl
y
balanced
. The pea
k valu
e of abou
t 1.6
6 mA (foun
d by expandin
g th
e y-axis
) agree
s
wit
h (6.51
) when th
e circui
t parameter
s ar
e inserte
d int
o th
e equation
. Not
e als
o tha
t
the phas
e of th
e curren
t flips
by 180
° when th
e resistanc
e of R3 switche
s fro
m bein
g
to
t bein
g greate
r tha
n 1 kft
.
les
s tha
n 1 kf
The ne
t powe
r dissipate
d by th
e thre
e voltag
e source
s si plotte
d n
i Figur
e 6.17b
.
Again
, th
e dashe
d lin
e s
i th
e 79
0 ftcase
, th
e soli
d lin
e correspond
s o
t th
e 99
0 ft
case
, th
e dot-dashe
d lin
e s
i th
e 119
0 ftcase
, an
d th
e dotte
d lin
es
i th
e 139
0 ftcase
.
The powe
r si foun
d by multiplyin
g th
e relevan
t voltage
s an
d current
s fo
r eac
h of th
e
power supplie
s an
d addin
g the
m together
. The fac
t tha
t th
e powe
rs
i negativ
e s
ia
consequenc
e of th
e passiv
e sig
n convention
. The ne
t powe
rs
i nearl
y a constan
t fo
r
the 99
0 ftcas
e an
d th
e variatio
n wit
h tim
e si du
e mainl
y o
t th
e powe
r supplie
d by
the thir
d voltag
e source
. Thi
s ca
n be see
n by finding
th
e analyti
c expressio
n fo
r th
e
tota
l power
:
2
2
2
Pit) = -Vo
[sin
(co0/#
i + sin
(co
? + 120°)/R2 + sin2 (cot + 240°)/R3)]
2
2
= -(V
2 + sin
(ot
f + 240°)(
1 -R1/R3)]
0 /#i)[3/
(6.107
)
where th
e secon
d equalit
y follow
s onl
y fi Rl = R2. Fo
r th
e perfectl
y matche
d cas
e
(Rl= R2 = R3 = 1 kft
) an
d th
e parameter
s of thi
s proble
m (V
7 V an
d
0 = 169.
co = 37
7 rad/s)
, expressio
n (6.107
) yield
s P(i) = -43.
2 W, a
s indicate
d n
i th
e
figure.
The secon
d circui
t come
s fro
m Exampl
e 6.
4 an
d th
e PSpic
e schemati
c s
i repro
duce
dn
i Figur
e 6.18
. The goa
l of th
e exampl
es
i simpl
y o
t find
th
e curren
t throug
h th
e
1 ftresistor
,s
o we nee
d onl
y perfor
m a dc bia
s poin
t solution
. The par
t name
s fo
r th
e
6.4. MicroSim PSpice Simulations
197
0.05
/ *
0.03
»/
V
-0.03
\
-0.05
0.00
0.0
1
\ /
/
V V ^
•
*
*
*
*
0.02
#
"
*
0.03
"
V
%
0.04
0
'
0.05
(b)
-10
-20
-30
<D
o
Q_
-4
0
-50 N/
-60
0.00
\ /
0.0
1
\
\ /
0.0
2
*t
'
0.0
3
Time (s)
w
H
w
0.0
4
*^
r
1
0.0
5
Figur
e 6,17
: The simulate
d output
: (a
) th
e neutra
l wir
e curren
t an
d (b
) th
e ne
t powe
r
supplie
d by thre
e voltag
e sources
.
voltag
e an
d curren
t source
s ar
e "VSRC" an
d "ISRC,
" respectively
. The 1Cl resisto
r
is R4 an
d th
e PSpic
e node
s $N_000
1 throug
h $N_000
3 ar
e chose
n o
t correspon
d o
t
nodes 1 throug
h 3n
i Figur
e 6.7
. Selecte
d line
s fro
m th
e outpu
t file
ar
e give
n n
i Tabl
e
6.1. The curren
t throug
h th
e 1f
l resisto
rs
i foun
d by takin
g th
e differenc
e ni nod
e
potential
s acros
s th
e resisto
r an
d dividin
g by th
e impedance
.n
I thi
s case
, th
e resisto
r
curren
tn
i ampere
ss
i equivalen
to
t th
e potentia
l of nod
e N$_0002
. The tabl
e reveal
s
thi
s potentia
lo
t be 1.
5 V, ni agreemen
t wit
h th
e analyti
c resul
t foun
d n
i th
e example
.
Chapter 6. Nodal and Mesh Analysis
198
R2
rA/WV
V1
2
R5>3
Q-
R4 > 1
Figure 6.18: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 6.4
.
Table 6.1: Selecte
d line
s fro
m th
e PSpic
e outpu
t file
.
** Analysi
s setu
p **
.OP
R.R1 $N_000
3 $N_000
2 3
R_R2$N_0001$N_000
4 2
R_R3 0 $N_000
5 3
R.R4 0 $N_000
3 1
R_R5 0$N_0001
3
V_V1 $N_000
4 $N_000
2 DC 4 AC 0
V_V 2 $N_000
1 $N_000
3 DC 3 AC 0
U l $N_000
5 $N_000
2 DC 3 AC 0
.prob
e
.END
*** SMALL SIGNAL BIAS SOLUTION
NOD E VOLTAG E
($N_0001)4.500
0
($N_0002
) 4.500
0
($N_0003
) 1.500
0
($N_0004
) 8.500
0
($N_0005
) -9.000
0
VOLTAG E SOURCE CURRENT S
NAM E CURRENT
NAM E CURRENT
V_V1 -2.000E+0
0
V.V2 5.000E-0
1
6.5. Summary
199
1
rAAAArf
R1
>
.5
1
5h
L1
C1
R3%1
AAAA^
R5
R4% 1
R6
I2
Figure 6.19: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 6.6
.
The remainin
g nod
e potential
s an
d th
e current
s throug
h th
e voltag
e source
s ar
e als
o
give
n n
i th
e table
. Not
e tha
t th
e curren
t throug
h V2 s
i positive
, indicatin
g tha
t powe
r
is bein
g absorbe
d by th
e source
.
The final
PSpic
e simulatio
n come
s fro
m th
e mes
h analysi
s proble
m n
i Exampl
e
6.6
. The PSpic
e schemati
c s
i redraw
n n
i Figur
e 6.19
. Not
e tha
t PSpic
e use
s a non
standar
d symbo
lo
t represen
t th
e a
c curren
t sourc
e (ISIN)
. The defaul
t name
s I an
d
12 indicat
e tha
t th
e part
s ar
e curren
t sources
. Any doub
t ca
n be remove
d by double
clickin
g on th
e par
to
t brin
g up th
e "ISIN
" dialo
g box
. We wil
l us
e PSpic
e o
t find
the curren
t throug
h th
e inducto
r an
d compar
e tha
t resul
to
t th
e analyti
c steady-stat
e
solutio
n foun
d n
i th
e example
:
iLAt) = 0.367
6 cos(
? -72.897°)
.
(6.108
)
A transien
t analysi
ss
i performe
d wit
h a final
tim
e of 20 san
d a prin
t ste
p of 10
0
ms. The analyti
c an
d simulate
d result
s ar
e plotte
d n
i Figur
e 6.2
0 on pag
e 200
. Ther
e
is a discrepanc
y n
i th
e curve
s a
t earl
y times
, bu
t th
e result
s becom
e nearl
y identica
l
afte
r th
e first
cycle
. The differenc
e betwee
n th
e tw
o curve
s involve
s th
e transien
t
solutio
n of th
e proble
m an
d s
i th
e subjec
t of investigatio
n n
i th
e nex
t chapter
.
6.5
Summar
y
The noda
l an
d mes
h curren
t technique
s allo
w on
e o
t writ
e by inspection matri
x
equation
s whic
h ca
n be use
do
t quickl
y analyz
e complicate
d a
c steady-stat
e problems
.
These technique
s hav
e bee
n derive
d fro
m th
e KCL an
d KVL equation
s by expressin
g
the
m respectivel
y n
i term
s of nod
e potential
s an
d circulatin
g mes
h currents
. Thes
e
technique
s hav
e bee
n presente
d fo
r variou
s sourc
e configurations
, suc
h a
s curren
t
or voltag
e source
s connecte
d n
i serie
s or paralle
l wit
h impedances
. However
, th
e
solutio
n metho
d varie
s onl
y slightl
y fo
r eac
h typ
e of sourc
e configuration
.
In th
e cas
e of noda
l analysis
, we hav
e first
considere
d th
e standar
d n + 1 i*od
e
circui
t wit
h admittance
s an
d curren
t sources
. The matri
x equatio
n fo
r thi
s circui
ts
i
Chapter 6. Nodal and Mesh Analysis
0.5
Figur
e 6.20
: The analyti
c (dashe
d line
) an
d th
e simulate
d (soli
d line
) result
s fo
r th
e
curren
t throug
h th
e inducto
rn
i th
e circui
tn
i Figur
e 6.19
.
eY s
i th
e n X n admittanc
e matrix
, vs
i th
e colum
n vecto
r of nod
e
Y • v = 75, wher
potentials
, an
d Is s
i th
e colum
n vecto
r of sources
. The diagona
l element
s Yu of th
e
admittanc
e matri
x ar
e equa
lo
t th
e su
m of al
l th
e admittance
s connecte
d o
t th
e ith
l th
e admittance
s
node. The off-diagona
l element
s Yjj = F/
; ar
e th
e negativ
e su
m of al
which ar
e connecte
d o
t bot
h th
e ith an
d jt
h nodes
. The sourc
e vecto
r componen
t 1^
is equa
lo
t th
e algebrai
c su
m of curren
t source
s connecte
d o
t th
e jth nod
e (th
e sourc
e
current
s flowing
int
o th
e nod
e ar
e take
n wit
h positiv
e signs)
.
In th
e cas
e of mes
h curren
t analysis
, we hav
e first
considere
d th
e standar
d plana
r
circui
t wit
h m meshe
s whic
h ha
s onl
y impedance
s an
d voltag
e sources
. The matri
x
equatio
n fo
r thi
s circui
ts
i Z •1 = Vs, wher
e Zs
i th
e m Xm impedanc
e matrix
,1 s
i
i th
e colum
n vecto
r of sources
.
the colum
n vecto
r of fictitious
mes
h currents,
an
d Vs s
The diagona
l element
s Z,-,
- of th
e impedanc
e matri
x ar
e equa
lo
t th
e su
m of al
l th
e
impedance
s n
i th
e ith mesh
. The off-diagona
l element
s Ztj = Zy{ ar
e equa
lo
t th
e
negativ
e su
m of th
e impedance
s whic
h ar
e common o
t bot
h th
e /t
h an
d jt
h meshes
.
The sourc
e vecto
r componen
t V^ s
i foun
d by takin
g th
e algebrai
c su
m of voltag
e
source
sn
i th
e jth mes
h (positiv
e mean
s th
e mes
h curren
ts
i flowin
g ou
t of th
e positiv
e
termina
l of th
e source)
.
The modification
s of th
e describe
d standar
d form
s of noda
l an
d mes
h curren
t
equation
s fo
r variou
s sourc
e configuration
s hav
e bee
n discusse
d n
i th
e chapte
r an
d
illustrate
d by specifi
c examples
.
6.6. Problems
6.6
1.
201
Problem
s
Fin
d al
l th
e nod
e potential
sn
i th
e circui
t show
n n
i Figur
e P6-1
.
■e-
O
4A
4cos(3t) A
vvwv
K
2F
N2
-o—
5 Q.
10 Q ( T ) 1 A 15 Q < 3 A ( T ) 1 0 Q<
^ 3 F V I / 3 Q(T)sin(3t)A <>1/5 Q(T)2cos(3t) A M / 3 H
Figure P6-1
Figure P6-2
2.
Fin
d th
e tim
e dependenc
e of th
e secon
d circui
t nod
e potentia
l show
n n
i Figur
e P6-2
.
3.
Se
t up
, bu
t do no
t solve
, th
e noda
l matri
x equatio
n fo
r th
e circui
t show
n n
i Figur
e P6-3
.
4.
Se
t up
, bu
t do no
t solve
, th
e noda
l matri
x equatio
n fo
r th
e circui
t show
n n
i Figur
e P6-4
.
e-
A / V W 1/3 Q.
AA/V\n
l3cos( cot) I
2A
R3
-e2A
wvwWwv
1/5 Q
2 Q
R,
C 2 (t)l 2 sin(cot)
C3
L 1 ^l 1 cos(tot+60)(T)
Figure P6-3
1£2N5A(T)
1/3 Q S
Figure P6-4
1/2QS1A(T)
202
Chapter 6. Nodal and Mesh Analysis
— e
K-
l 3 cos( oot+300)
C3
-tWWl
FL
C1
L, h1cos(cot-60°)
Figure P6-5
5.
CD
Figure P6-6
Fin
d th
e curren
t throug
h C\ ni th
e circui
t show
n ni Figur
e P6-5
.
6. For th
e circui
t show
n n
i Figur
e P6-6
, find
th
e valu
e of th
e resistanc
e R tha
t result
sn
i a
thir
d nod
e potentia
l of ~2 V.
7. Fin
d th
e phaso
r of th
e potentia
l of th
e thir
d nod
e fo
r th
e circui
t show
n ni Figur
e P6-7
.
8. Fin
d al
l th
e nod
e potential
s fo
r th
e circui
t show
n ni Figur
e P6-8
.
o
2cos(5t) V
0.4 F ;
N2
Vf
<>AAAAX>VVW
N3
1 Q.
0.2 H > cos( 5t) A
Figure P6-7
CD
0.4 Q > 0 . 4 O >
0.2 Q."
Figure P6-8
9.
Fo
r th
e circui
t show
n n
i Figur
e P6-9
, find
th
e valu
e of th
e resistanc
e R tha
t result
s ni a
thir
d nod
e potentia
l of 5 V.
10.
Fin
d th
e curren
t throug
h th
e 1
0 f
l resisto
r ni th
e circui
t show
n ni Figur
e P6-10
.
Figure P6-9
Figure P6-10
6.6. Problems
203
Figure P6-12
Figure P6-11
11.
Fin
d th
e powe
r dissipate
d by th
e 5 ftresiste
rn
i th
e circui
t show
n n
i Figur
e P6-11
.
12.
Fin
d th
e averag
e powe
r dissipate
d by th
e 1 H resiste
r ni th
e circui
t show
n n
i Figur
e
P6-12
.
13.
Se
t up
, bu
t do no
t solve
, th
e mes
h analysi
s equation
s fo
r th
e circui
t show
n ni Figur
e
P6-13
.
14.
Se
t up
, bu
t do no
t solve
, th
e mes
h analysi
s equation
s fo
r th
e circui
t show
n n
i Figur
e
P6-14
.
i—VWv—t-AAA/V-n
1 Q.
3Q
03cos(10t)V C ) s i n ( 1 0 t ) V
-le-
0.2 H
0.5 F
20
:0.1 F
Q2sin(10t)V
1a
Figure P6-13
-wwv-
Figure P6-14
15.
Fin
d th
e curren
ti{t) throug
h th
e 2 F capacito
r ni th
e circui
t show
n n
i Figur
e P6-1
5 wit
h
all sourc
e current
s give
n ni amps
.
16.
For th
e circui
t show
n ni Figur
e P6-16
, find
th
e valu
e of th
e resistanc
e R tha
t result
s ni
secon
d mes
h curren
t of —2 A.
3 Q.
Ft Q.
:iaQiov(i)<6a(2)C )20V
4a
A/VW
Figure P6-15
Figure P6-16
204
Chapter 6. Nodal and Mesh Analysis
M A A n
2 ft
1Q
_)3cos(10t)V > 1 H 0.1 F
7)
>
(^
Q
1_H
■K—
0.05 F
2sjn(10t)V
(3) 4Q
Figure P6-17
1Q
A/VW
Figure P6-18
17.
Fin
d th
e phaso
r curren
t of th
e thir
d mes
h fo
r th
e circui
t show
n n
i Figur
e P6-17
.
18.
Fin
d al
l th
e mes
h current
s fo
r th
e circui
t show
n ni Figur
e P6-18
.
19.
Fin
d th
e phaso
r curren
t throug
h th
e 2Cl resisto
r ni th
e circui
t show
n ni Figur
e P6-19
.
20.
Fo
r th
e circui
t show
n ni Figur
e P6-20
, fin
d th
e valu
e of th
e resistanc
e R tha
t result
s ni a
firs
t mes
h curren
t of 1 A.
6Q
1 Q
Figure P6-19
R Q
AAA/V
Figure P6-20
21.
Fin
d th
e curren
t flowin
g throug
h th
e neutra
l wir
e of th
e three-phas
e circui
t show
n ni
Figur
e P6-21
.
22.
Fin
d th
e tota
l instantaneou
s powe
r dissipate
d ni al
l thre
e resistor
s of th
e three-phas
e
circui
t indicate
d ni Figur
e P6-22
.
6.6. Problems
205
100eJ 120 V
100 Q
100e' 120 V
100 V
10 a
(T
)
-o
10 Q.
100e'120 V
Figure P6-21
9.9 Q
100e"i120 V
O-
ww
Figure P6-22
23.
Conside
r th
e circui
t show
n n
i Figur
e P6-23
. Al
l voltage
s ar
en
i volts
; al
l current
s ar
e ni
amps. Dra
w th
e phaso
r equivalen
t circui
ts
o tha
tt
is
i read
y fo
r th
e genera
l mes
h analysi
s
. Writ
e down bu
t do no
t solv
e
formalism
. Clearl
y indicat
e al
l th
e circulatin
g mes
h currents
the genera
l mes
h equation
sn
i matri
x form
.
24.
Conside
r th
e circui
t show
n n
i Figur
e P6-24
. Solv
e fo
r al
l th
e mes
h currents
.
(St)©
4£0.5 F
r-vwvi
1 £2
(p2cos(t)v(T)sin(t)V
0.25 F
3H
Figure P6-23
Figure P6-24
25.
Conside
r th
e circui
t show
n ni Figur
e P6-25
. Writ
e th
e matri
x equation
s fo
r th
e genera
l
node analysi
s problem
. Fin
d th
e valu
e of th
e resistanc
e R tha
t result
s ni a voltag
e of 3 V
acros
s th
e 2ft resistor
.
26.
Conside
r th
e circui
t show
n ni Figur
e P6-25
. Writ
e th
e matri
x for
m fo
r th
e genera
l mes
h
analysi
s problem
. Fin
d th
e valu
e of th
e resistanc
e R tha
t result
s ni a voltag
e of 3 V acros
s
the 2fl resistor
.
27.
Conside
r th
e circui
t show
n n
i Figur
e P6-27
. Al
l voltage
s ar
en
i volts
; al
l current
s ar
e ni
amps. Writ
e down bu
t do no
t solv
e th
e genera
l noda
l equation
s ni matri
x form
.
Chapter 6. Nodal and Mesh Analysis
206
rAAAA/
1
e-
O
0.2 H
w
7Q
- G-
2sin(4t)
6A
+-WVV -
AAA/V
3cos(4t)
0.4 H
2Q
A A / V V
1 Q
>1 Q
Q 2 A > 0 . 2 fi 3 V > 2 Q 10 V'
:0.6F 0.7 F
0.2 H >4cos(4t)(T)
1n
R (Q)
*AAAAA-
Figure P6-27
Figure P6-25
28.
Use genera
l mes
h analysi
so
t find
th
e curren
t throug
h th
e 3O, resisto
r ni th
e circui
t show
n
in Figur
e P6-28
.
29.
Use noda
l analysi
so
t find
th
e valu
e of th
e curren
t flowing
throug
h th
e 1/
2 fl resisto
rn
i
the circui
t show
n n
i Figur
e P6-29
.
O
AAA/V
0
1 £1
2A
1Q
3AN20
4fl'
3Q
AAA/V
2fl
O
1 A
4H
AAA/V^WW- 1
Figure P6-29
Figure P6-28
30.
Use noda
l analysi
so
t find
th
e time-dependen
t potential
s of th
e tw
o node
s (relativ
eo
t th
e
ground
)n
i th
e circui
t show
n n
i Figur
e P6-30
. Al
l voltage
s ar
en
i volts
; al
l current
s ar
en
i
amps.
1F
N1
1/2 F
(T)cos(2t)
O
sin(2t)
>1 Q
1&
fAAAAr-
_N2
1 Q.
1/2 H
1F
VJcos(2
t + 45°
)
0
Figure P6-30
6.6. Problems
207
Use th
e PSpic
e cod
eo
t solv
e Problem
s 31-38
.
31.
Fin
d al
l th
e nod
e potential
sn
i th
e circui
t show
n n
i Figur
e P6-1
.
32.
Fin
d th
e curren
t throug
h th
e 1
0 ftresisto
rn
i th
e circui
t show
n n
i Figur
e P6-10
.
33.
Fin
d th
e valu
e of th
e curren
t flowing
throug
h th
e 1/2(
1 resisto
rn
i th
e circui
t show
n n
i
Figur
e P6-29
.
34.
For th
e circui
t show
n n
i Figur
e P6-6
, find
th
e valu
e of th
e resistanc
e R tha
t result
sn
i a
thir
d nod
e potentia
l of —2 V.
35.
Fo
r th
e circui
t show
n n
i Figur
e P6-20
, find
th
e valu
e of th
e resistanc
e R tha
t result
sn
i a
first
mes
h curren
t of 1 A.
36.
Fo
r th
e circui
t show
n n
i Figur
e P6-25
, find
th
e valu
e of th
e resistanc
e R tha
t result
sn
i a
voltag
e of 3 V acros
s th
e 2 ftresistor
.
37.
Plo
t th
e curren
t throug
h Cl ni th
e circui
t show
n ni Figur
e P6-5
. Le
t co = 37
7 rad/s
,
Cx = C2 = 2C3 = 1
0 /xF
,RI = 1 HI, LI = 1 mH, and/
j - 2/
2 = 3/
3 = 6 A.
38.
Plo
t th
e curren
t flowing
throug
h th
e neutra
l wir
e 1
( ft)of th
e three-phas
e circui
t show
n
in Figur
e P6-21
.
Chapte
r 7
Transien
t Analysi
s
7.1
Introductio
n
Transients ar
e non-steady-stat
e tim
e variation
s of voltage
s an
d current
s whic
h occu
r
t of switchin
g circui
t element
s an
d (or
) changin
g interconnection
s of electri
c
as a resul
circuits
. The transient
s occu
r becaus
e of th
e presenc
e of energ
y storag
e element
s
(i.e.
, inductor
s an
d capacitors)
. We recal
l tha
t a curren
t throug
h a
n inducto
r an
d a
voltag
e acros
s a capacito
r ar
e bot
h continuou
s function
s of time
. Thi
s continuit
y
determine
s initia
l condition
s fo
r energ
y storag
e element
s a
t moment
s of switchings
.
When switching
s occur
, th
e energ
y storag
e element
s usuall
y hav
e initia
l condition
s
tha
t ar
e differen
t fro
m ne
w steady-stat
e conditions
. Due o
t th
e sam
e continuity
,t
i take
s
some tim
e fo
r th
e energ
y storag
e element
so
t reac
h th
e ne
w steady-stat
e conditions
.
This gradua
l (i
n time
) adjustmen
t of energ
y storag
e element
s o
t ne
w steady-stat
e
condition
s result
sn
i transient
sn
i electri
c circuits
.
In thi
s chapter
, we wil
l stud
y circuit
s wit
h on
e or tw
o energ
y storag
e elements
.
First
, we wil
l conside
r circuit
s wit
h on
e energ
y storag
e elemen
t whic
h ar
e called^zm
order circuits. Thi
ss
i becaus
e th
e applicatio
n of Kirchhoff'
s law
s lead
so
t a first-order
differentia
l equatio
n whic
h describe
s th
e tim
e evolutio
n of circui
t variables
. We first
investigat
e transient
s ni circuit
s whic
h ar
e excite
d by initia
l condition
s an
d the
n move
on o
t th
e discussio
n of circuit
s whic
h ar
e excite
d by sources
.t
I wil
l be show
n tha
t th
e
metho
d of phasor
s ca
n be use
d o
t simplif
y th
e analysi
s of transient
s tha
t ar
e excite
d
by a
c sources
. The analysi
s of first-order
circuit
s wil
l be conclude
d wit
h a discussio
n
of circuit
s tha
t ar
e excite
d by bot
h initia
l condition
s an
d sources
.
Second-order circuits contai
n tw
o energ
y storag
e element
s an
d ar
e describe
d
by second-orde
r differentia
l equations
. First
, we wil
l develo
p solutio
n technique
s fo
r
circuit
s tha
t ar
e excite
d by initia
l conditions
. Then
, we wil
l discus
s circuit
s excite
d by
sources
.t
I wil
l be show
n tha
t th
e natur
e of th
e transient
s depend
s strongl
y on th
e valu
e
of th
e resistanc
e n
i th
e circuit
. We wil
l conside
r separatel
y fou
r cases
: overdampe
d
circuits
, criticall
y dampe
d circuits
, underdampe
d circuits
, an
d undampe
d circuits
. We
shal
l als
o demonstrat
e ho
w th
e metho
d of phasor
s ca
n be applie
d o
t th
e analysi
s of
transient
sn
i second-orde
r circuit
s drive
n by a
c sources
.
208
7.2. First-Order Circuits
209
W e wil
l the
n presen
t tw
o powerfu
l technique
s tha
t ca
n be use
d o
t solv
e transien
t
problems
. First
, we wil
l defin
e th
e transfe
r functio
n of a circui
t an
d presen
t a
n
algebrai
c approac
h o
t th
e solutio
n of transien
t problem
s whic
h utilize
s th
e transfe
r
functio
n a
s a functio
n of th
e comple
x frequenc
y s. Next
, convolutio
n integral
s wil
l be
introduce
d alon
g wit
h th
e concep
t of uni
t ste
p respons
e an
d uni
t impuls
e response
.
It wil
l be demonstrate
d tha
t th
e convolutio
n integra
l provide
s a systemati
c approac
h
for th
e calculatio
n of circui
t response
so
t arbitrar
y sources
.
Finally
, we introduc
e diodes an
d simpl
e diod
e circuit
s an
d demonstrat
e ho
w
transien
t analysi
s ca
n be usefu
l fo
r th
e analysi
s of diod
e circuits
. The chapte
r s
i
close
d wit
h a collectio
n of PSpic
e examples
.
7.2
First-Orde
r Circuit
s
7.2,1
Circuits Excited b y Initial Conditions
The first
typ
e of circuit
s we conside
r ar
e thos
e whic
h hav
e onl
y on
e energ
y storag
e
elemen
t an
d whic
h ar
e drive
n onl
y by th
e initia
l condition
s fo
r thes
e energ
y storag
e
elements
.
Conside
r th
e circuit
s show
n n
i Figur
e 7.1
. The drawin
g on th
e lef
t show
s th
e
circui
t befor
e switching
. Here
, a capacito
rs
i connecte
d o
t a voltag
e source
. At tim
e
/ = 0 th
e switche
s ar
e thrown
, an
d th
e voltag
e sourc
e s
i disconnecte
d leavin
g th
e
charge
d capacito
r connecte
d n
i serie
s wit
h th
e resistor
. Thi
ss
i show
n by th
e drawin
g
on th
e right
. The capacito
r no
w proceed
s o
t discharg
e itsel
f throug
h th
e resistor
,
producin
g th
e transien
t process
.
Now , we shal
l analyz
e thi
s transien
t proces
s mathematically
. The analysi
s pro
cedur
e generall
y involve
s tw
o distinc
t steps
:
1. Determinatio
n of th
e initia
l condition
s fo
r th
e energ
y storag
e elements
. Thi
s ste
p
ofte
n require
s analysi
s of th
e circui
t befor
e switching
.
2. Analysi
s of th
e circui
t afte
r switchin
g by usin
g th
e previousl
y foun
d initia
l condi
tions
. Thi
s ste
p normall
y involve
s solvin
g a
n initia
l valu
e proble
m fo
r differentia
l
equations
.
I
v
s ( t )j(
°T<)"
+
^
t <0
t >0
Figur
e 7.1
: Exampl
e RC circui
t drive
n by initia
l conditions
.
Chapter 7
. Transient Analysis
210
W e begi
n wit
h ste
p one
: finding
th
e initia
l conditions
. We examin
e th
e circui
t
befor
e switchin
g (t < 0)
, show
n on th
e lef
t ni Figur
e 7.1
.n
I thi
s cas
e ther
e s
i on
e
capacito
r ni th
e circuit
,s
o we nee
d o
t find
th
e initia
l voltag
e acros
s th
e capacitor
. We
use KVL fo
r th
e singl
e loop
, whic
h result
s in
:
v c ( 0 - v0, (= 0,
vc(t) = vs(t),
f( < 0 ,
)
(7.1
)
(*=S0)
.
(7.2
)
Thus, we kno
w tha
t th
e voltag
e acros
s th
e capacito
ra
t time
s t < 0s
i th
e sam
e a
s th
e
sourc
e voltag
e a
t f < 0. Therefore
, we ca
n conclud
e that
:
) = v,(0) = Vbvc(0-
(7.3
)
Sinc
e a voltag
e acros
s a capacito
rs
i a continuou
s functio
n of time
, we have
:
vc(0
) = V0.
+ ) = vc(0_
(7.4
)
Equatio
n (7.4
)s
i th
e initia
l conditio
n fo
r th
e pose
d problem
. We shal
l no
w analyz
e
the circui
t afte
r switching
, tha
t is
, fo
r t > 0. Applyin
g KVL yields
:
v*(0 +vc(t) = 0,
(7.5
)
where v#(/
) an
d vc(t) ar
e th
e voltage
s acros
s th
e resisto
r an
d th
e capacitor
, respec
tively
. By substitutin
g n
i (7.5
) th
e termina
l relationshi
p betwee
n voltag
e an
d curren
t
for a resistor
, we find:
i{t)R + vc(t)
= 0.
(7.6
)
o we nee
d o
t eliminat
e i(t).
The las
t equatio
n ha
s tw
o unknowns
, i{t) an
d vc(t), s
Sinc
e th
e sam
e curren
ti(t) s
i flowing
throug
h bot
h th
e resisto
r an
d th
e capacitor
, we
can expres
st
in
i term
s of th
e capacito
r voltag
e by usin
g th
e termina
l relationshi
p fo
r
a capacitor
:
dvr(t)
Kt) - C - £A
at
(7.7
)
Substitutin
g (7.7
) int
o (7.6
) produce
s th
e followin
g differentia
l equation
:
dvr(t)
+ vdO = o.
(7.8
)
RC^r1
at
Now , by combinin
g (7.8
) wit
h (7.4)
, we en
d up wit
h th
e followin
g initia
l valu
e
problem
: find
th
e solutio
n of th
e differentia
l equatio
n
RC^l
at
+ Vc{t) = 0
(7.9
)
subjec
to
t th
e initia
l condition
:
vc(0
+) = V0.
(7.10
)
111
7.2. First-Order Circuits
The abov
e equatio
ns
i know
n a
s afirst-order,
linea
r homogeneou
s differentia
l equatio
n
wit
h constan
t coefficient
s an
d ca
n be solve
d by usin
g th
e followin
g method
. We loo
k
for th
e solutio
n n
i th
e form
;
vc(t) = Aest
(7.11
)
e (7.11
) an
d it
s
where A an
d s ar
e some unknow
n constants
. To find
s, we substitut
first
derivativ
e int
o (7.9)
, whic
h yield
s sequentially
:
sRCAest +Aest
= 0,
(7.12
)
Aest(sRC+
1
) = 0,
(7.13
)
sRC + 1 - 0,
(7.14
)
* + -L
(7.15
)
=0.
sC
Equatio
n (7.14
)s
i calle
d th
e characteristic equation of th
e differentia
l equatio
n an
d
can be solve
d fo
rs:
s
= -w
(716)
By th
e way
, ther
e s
i a quic
k way o
t findth
e characteristi
c equatio
n jus
t by lookin
g
at th
e differentia
l equation
. To do this
, we simpl
y replac
e th
e first
derivativ
e wit
h th
e
facto
r s an
d th
e functio
n itsel
f wit
h 1
.
If we writ
e th
e characteristi
c equatio
n a
s (7.15)
,t
i ca
n be interprete
d n
i term
s of
impedance
. Indeed
, settin
g s =jco yield
s
(7.17
)
R+-^-=Z(j<o),
jcoC
which s
i th
e familia
r impedanc
e of th
e serie
s RC circuit
. Thi
s justifie
s th
e notation
:
R+^
=Z(s\
(IAS)
sC
Thus, th
e solution
s of th
e characteristi
c equatio
n ar
e zero
s of th
e impedanc
e a
s a
functio
n of comple
x frequenc
y s:
Z(s) = 0.
(7.19
)
Recal
l tha
t th
e ter
m "comple
x frequency
" refer
so
t th
e fac
t tha
ts may assum
e arbitrar
y
(not onl
y imaginary
) values
.
Returnin
g o
t th
e origina
l proble
m an
d substitutin
g (7.16
) int
o (7.11)
, we finda
genera
l solutio
n o
t equatio
n (7.9)
:
vc(t)
=Ae't/RC.
(7.20
)
This solutio
n s
i calle
d "general
" becaus
et
i satisfie
s (7.9
) fo
rany constan
tA. To find
thi
s constant
, we emplo
y initia
l conditio
n (7.10)
, whic
h yields
:
Chapter 7. Transient Analysis
212
vc(0) =Ae~0/RC = A = V0,
(7.21
)
A = V0.
(7.22
)
Thus, th
e solutio
n o
t th
e initia
l valu
e proble
m (7.9)-(7.10
) is
:
vc(t) = Voe~t/RC.
(7.23
)
The las
t formul
a give
s th
e expressio
n fo
r th
e voltag
e acros
s th
e capacito
r fo
r al
lt ^ 0
.
It ca
n be use
d o
t fin
d th
e expression
s fo
r th
e voltag
e acros
s th
e resisto
r an
d th
e curren
t
throug
h th
e circuit
:
-V0e-'/RC,
vR(t) = -vc(t)=
(7.24
)
_Voe-t/*Cm
{125)
at
R
It s
i apparen
t fro
m (7.24
) an
d (7.25
) tha
t al
l circui
t variable
s deca
y exponentiall
y
wit
h time
. Thes
e exponentia
l decay
s ca
n be characterize
d by th
e sam
e time constant
r, whic
h s
i define
d by
:
m
= c*c«) =
T =RC.
(7.26
)
The tim
e constan
t alway
s ha
s th
e uni
to
f time
. Indeed
:
[r
] = ft^
=
.
s
(7.27
)
For exponentia
l decay
, Ts
i suc
h tha
t a
t tim
e t = r th
e quantit
y s
i decrease
d b
y a
facto
r o
f1/e. Indeed
,a
t tim
e t — r =RC equatio
n (7.23
) yields
:
V C (T) =V0e^RC
= V0e-RC/RC = *
°
e
(7.28
)
In term
s o
f th
e tim
e constant
, equatio
n (7.23
)s
i writte
n as
:
vc(0
=V0e-t/T.
(7.29
)
The tim
e constan
ts
i ameasur
e o
f exponentia
l decay
. Indeed
, sinc
er s
i th
e tim
e
neede
d fo
r aquantit
y o
t deca
y b
y afacto
r o
f 1/e, we conclud
e tha
t th
e smalle
r th
e
time constan
t th
e faste
r th
e decay
. The grap
h presente
d n
i Figur
e 7.
2 show
s th
e deca
y
ofvc(t) fo
r variou
s value
s o
f T.
Anothe
r way o
t interpre
t th
e relationshi
p betwee
n ran
d resistanc
eR s
i by usin
g
power considerations
. Fo
r ou
r circuit
, powe
r dissipatio
n s
i define
d by
:
Pit) = - ^ .
-
(7.30
)
It s
i clea
r fro
m th
e las
t equatio
n tha
t asmalle
r valu
e ofR wil
l lea
d o
t alarge
r powe
r
dissipation
, whic
h wil
l resul
tn
i faste
r decay
. Thi
s s
i consisten
t wit
h th
e expressio
n
of T =RC. A smalle
rR mean
s asmalle
r T, whic
h als
o correspond
s o
t faste
r decay
.
It s
i als
o clea
r fro
m (7.23
) tha
t wit
h tim
e (theoreticall
y speaking
, wit
h infinit
e
time
) th
e voltag
e acros
s th
e capacito
r wil
l adjus
to
t it
s ne
w zer
o steady-stat
e condition
.
7.2. First-Order Circuits
21
3
Time (t)
Figur
e 7.2
: Grap
h o
f exponentia
l decay
:T\ < 2r < T3.
EXAMPL E 7.
1 n
I Figur
e 7.1
, le
t C = 1 /JLF
, ?/ = 10 kft
, an
d V0 = 10 V. We want
to find
th
e tim
e constan
t fo
r th
e circui
t an
d th
e tim
et
i take
s th
e capacito
ro
t discharg
e
to 1 V. Also
, find
th
e instantaneou
s powe
r dissipate
d n
i th
e resisto
r when t = 1 ms.
The tim
e constan
t fo
r th
e circui
ts
i r = RC = 1 /x
F X 10 kf
l = .0
1 s= 1
0 ms.
If th
e capacito
r ha
s discharge
d o
t 1 V by th
e tim
e t0, the
n equatio
n (7.29
) yield
s
o/0i
, whic
h ca
n be solve
d fo
r t0: t0 = .0
1 X ln(10)
s = 23.0
3 ms.
vc(h) = 1 = 10(?~'
Note tha
t we use
d th
e fac
t tha
t ln(l/x
) = — ln(x
)o
t ge
t th
e final
answer
. Finally
,
fro
m equatio
n (7.30
) we find
tha
t th
e instantaneou
s powe
r dissipate
d ni th
e resisto
r
is:
V2
200
02
p(t) - (V0e~t/T)2/R
= -±e~2t/T = O.Ole"
' W - O.Ole"
' W =8.1
9 mW .
R
(7.31
)
EXAMPL E 7.
2 The previou
s exampl
e show
s ho
w o
t analyz
e first-order
circuit
s
excite
d by initia
l condition
s tha
t contai
n on
e capacito
r an
d on
e resistor
. Now, we
conside
r a slightl
y mor
e complicate
d problem
, show
n n
i Figur
e 7.3
; le
t us find
expression
s fo
r th
e curren
tn
i al
l th
e resistors
.
This proble
m s
i simila
ro
t th
e first
RC circui
t problem
; however
,t
i contain
s mor
e
tha
n on
e resistor
. We star
tn
i th
e usua
l manner
, by finding
th
e initia
l condition
. Sinc
e
thi
s circui
t befor
e switchin
g s
i identica
lo
t th
e first
on
e we considered
, th
e initia
l
conditio
n s
i als
o th
e same
:
vc(0+) = Vb-
(7.32)
Afte
r th
e circui
ts
i switched
,t
i look
s lik
e th
e on
e show
n on th
e righ
tn
i Figur
e
7.3
. Now, th
e circui
t appear
s o
t be somewha
t differen
t fro
m th
e first
RC circui
t we
Chapter 7
. Transient Analysis
214
"W^ -1—°^ X> A/\A/
V
v
WWA
s ( t )j(
t >0
t <0
Figur
e 7.3
: A mor
e complicate
d RC circuit
.
considered
. However
,n
i actualit
y t
is
i ver
y similar
. We simpl
y hav
eo
t us
e equivalen
t
transformation
s o
t combin
e al
l th
e resistor
s int
o on
e equivalen
t resisto
r Re. Fro
m
Figur
e 7.
3 t
is
i eviden
t tha
t th
e equivalen
t resistanc
e (show
n n
i Figur
e 7.4
) is
:
RP = R] +
R2R3
R2 + R3
(7.33)
i, W
t >0
Figur
e 7.4
: An equivalen
t circui
t when al
l resistor
s ar
e combine
d int
o one
.
Now , th
e circui
ts
i mathematicall
y identica
lo
t th
e first
RC circui
t an
d ha
s th
e sam
e
solutio
n excep
t tha
t th
eR s
i replace
d by Re:
vc(t) = V0e-t/R<c.
Next, al
l th
e current
sn
i th
e circui
t ca
n be found
. First
, we calculat
e i\(t)\
dv
, m _r c(0
_ _V0 t/ReC
(7.34
)
(7.35)
Usin
g i\(t) an
d th
e curren
t divide
r rule
, th
e current
s i2(t) an
d 3/(0 ar
e found
:
(7.36
)
Re /?
2 +R3
hit)
Vo R2
Re R2 + R3
-t/ReC
(7,37
)
EXAMPL E 7.
3 We no
w conside
r th
e analysi
s of a circui
t wit
h a
n inducto
r an
d a
resisto
r excite
d by initia
l conditions
. The tw
o configuration
s of th
e circuit
, befor
e
Notic
e tha
t afte
r switchin
g
switchin
g an
d afte
r switching
, ar
e illustrate
d ni Figur
e 7.5.
7.2. First-Order Circuits
215
iL(t
)
-
t<0
+
L
t>
0
Figur
e 7.5
: Exampl
e o
fa
n RL circui
t drive
n by initia
l conditions
.
ther
e ar
e no source
s n
i th
e circuit—i
ts
i drive
n solel
y by th
e energ
y store
d n
i th
e
inductor
.
W e begi
n th
e analysi
s by determinin
g th
e initia
l condition
s fo
r th
e energ
y storag
e
element
.n
I particular
, we want o
t find
th
e curren
t throug
h th
e inducto
ra
t tim
e t = 0.
From th
e circui
t on th
e lef
tn
i Figur
e 7.
5 t
is
i obviou
s tha
t th
e curren
t throug
h th
e
inducto
rs
i equa
lo
t th
e sourc
e curren
t sinc
e the
y ar
e connecte
d n
i series
. So we kno
w
tha
ta
t an
y tim
e t < 0,
(t < 0)
.
(7.38)
iL (0) = i,(0) = /<>
.
(7.39
)
k(t) = is(t\
Therefore
, we find
, we obtai
n th
e
Sinc
e th
e curren
t throug
h a
n inducto
rs
i a continuou
s functio
n of time
initia
l conditio
n
*L(0+) = I L ( 0)- = o/
(7.40)
Once th
e initia
l conditio
n s
i found
, we ca
n analyz
e th
e circui
t afte
r switching
, tha
t is
,
the circui
t show
n on th
e righ
tn
i Figur
e 7.5
. An applicatio
n of KVL yield
s
vR(t) +vL{t) = 0.
(7.41
)
W e woul
d lik
e o
t expres
s thi
s equatio
n n
i term
s of ii(f) sinc
e thi
s s
i th
e functio
n
for whic
h we hav
e th
e initia
l condition
. Thi
s s
i eas
y o
t do by usin
g th
e termina
l
relationship
s fo
r resistor
s an
d inductors
. Afte
r substitutio
n of thes
e relationships,
th
e
las
t equatio
n look
s like
:
dirii)
L - ^ + RiL(t) = 0.
at
(7.42)
This equation
, alon
g wit
h th
e initia
l conditio
n (7.40)
, make
s up th
e initia
l valu
e
proble
m o
t be solved
. The abov
e equatio
n s
i als
o a first-order,
linea
r homogeneou
s
differentia
l equatio
n wit
h constan
t coefficients
. Therefore
, we loo
k fo
r a solutio
n n
i
Chapter 7
. Transient Analysis
216
the form
:
iL(t) = Aest.
(7.43
)
By substitutin
g (7.43
) int
o (7.42)
, we find
th
e characteristi
c equation
:
Ls + R = 0.
(7.44
)
Notic
e tha
t th
e solutio
n o
t th
e characteristi
c equatio
n s
i onc
e agai
n a zer
o of th
e
impedanc
e a
s a functio
n of comple
x frequency
. Thi
ss
i clea
r fi we le
ts = jco; then
:
Z(jco) = R + ja)L.
(7.45
)
Ls + R = Z(s).
(7.46
)
This justifie
s th
e notation
:
Consequently
, th
e characteristi
c equatio
n (7.44
) ca
n be writte
n a
s follows
:
Z{s) = 0.
(7.47
)
The characteristi
c equatio
n s
i easil
y solve
d fo
rs\
s=-f-
(7.48
)
Now , th
e solutio
n of equatio
n (7.42
) is
:
iL(t) = Ae-$'
(7.49
)
and finding
th
e constan
tA si al
l tha
ts
i require
d o
t complet
e th
e solution
. Thi
s constan
t
is foun
d fro
m th
e initia
l condition
. Substitutin
g t = 0 int
o equatio
n (7.49
) an
d settin
g
it equa
lo
t th
e initia
l conditio
n (7.40
) give
s th
e valu
e fo
r A:
iL (0+) = I0 = A.
(7.50
)
iL(t) = V "f r.
(7.51
)
Now , th
e solutio
n s
i known
:
From equatio
n (7.51)
, th
e voltag
e acros
s th
e inducto
r an
d th
e resisto
r ca
n be foun
d
by usin
g th
e voltage-curren
t relationship
s fo
r thes
e elements
:
vR(t) = RiL(t) = IoRe-*f,
(7.52
)
vdt) = L
(7.53
)
^
= -IoRe-i<.
From th
e abov
e expressions
, we find
th
e tim
e constant
:
r - -.
(7.54
)
R
Althoug
h t
is
i define
d differentl
y tha
n n
i th
e previou
s problem
, th
e tim
e constan
t stil
l
has th
e uni
t of time
,a
st
i alway
s must
:
7.2. First-Order Circuits
217
[T] = - ^
= .
s
(7.55
)
In term
s of th
e tim
e constant
, equatio
n (7.51
) ca
n be writte
n as
:
iL(t) = he~t/T.
(7.56
)
In thi
s case
, th
e large
r th
e valu
e ofR, th
e smalle
r r an
d th
e faste
r th
e decay
. Once
again
, thi
s ca
n als
o be see
n by considerin
g th
e powe
r dissipation
. Fo
r thi
s circuit
, th
e
power dissipatio
n s
i give
n by th
e expression
:
p(t) = il(t)R,
(7.57
)
which show
s tha
t th
e large
r th
e resistance,
th
e greate
r th
e powe
r dissipatio
n an
d th
e
faste
r th
e decay
.
■
EXAMPLE 7.4 A nonidea
l inductor
, wit
h a
n inductanc
e of 10 mH an
d a serie
s
r t < 0 an
d the
n
resistanc
e of 10 11
, was connecte
d o
t a curren
t sourc
e of 1 A fo
short-circuite
d a
t t= 0. How lon
g wil
lt
i tak
e fo
r th
e curren
to
t deca
y o
t 1 mA? How
lon
g wil
lt
i tak
e befor
e t
i decay
s o
t 1 /xA
? How lon
g woul
d t
i tak
e o
t reac
h thos
e
current
s fi th
e serie
s resistanc
e was onl
y 10 mH?
A mode
l fo
r th
e shorte
d nonidea
l inducto
rs
i give
n by th
e rightmos
t circui
tn
i
Figur
e 7.5
. The instantaneou
s curren
tn
i thi
s inducto
r ca
n be foun
d fro
m equatio
n
(7.56)
. When we plu
gn
i th
e number
s fo
r th
e first
part
of ou
r problem
, we find
/L(*O
) =
Next
, we inver
t th
e equatio
n o
t ge
tt0 = .00
1 Xln(1000)
s - 6.9
1
0.00
1 - le~imt0.
6
ms. For iL(t0) = 1 JUA, we ge
tt0 = .00
1 X ln(10
) s = 13.8
2 ms. We observ
e tha
t
A a
st
i di
d o
t ge
to
t 1 mA. f
I th
e resistanc
es
i
it take
s onl
y twic
e a
s lon
g o
t ge
to
t 1 /x
onl
y 10 mfl
, th
e tim
e constan
ts
i 100
0 time
s large
r tha
n ni th
e first
case
,s
o th
e circui
t
would tak
e 100
0 time
s longe
ro
t decay
. Thus
, 1 mA woul
d be achieve
d n
i 6.9
1 san
d
1 JUA woul
d be lef
t afte
r 13.8
2 s
.
■
7.2.2
Circuits Excited b y Sources
Now , we conside
r electri
c circuit
s excite
d by sources
.n
I situation
s wher
e ther
e s
i
a dc or a
c source
, th
e circui
t respons
e eventuall
y reache
s th
e appropriat
e steady
stat
e condition
. Thi
ss
i becaus
e th
e transien
t par
t of th
e respons
e drive
n by a
n initia
l
conditio
n alway
s decay
s o
t zero
, whil
e th
e steady-stat
e (forced
) componen
t of th
e
respons
e whic
h s
i drive
n by a sourc
e remains
.
Consider
,a
s th
e first
example
, th
e RL circui
t picture
d n
i Figur
e 7.6
. The proble
m
is approache
d n
i th
e sam
e way a
s an
y proble
m involvin
g transients—th
e initia
l
condition
s must be foun
d first.
Examinin
g th
e circui
t befor
et
is
i switche
d (lef
t circui
t
in Figur
e 7.6
) suggest
s tha
t th
e initia
l curren
t throug
h th
e inducto
r si zero
. Thi
s s
i
becaus
e th
e circui
ts
i ope
n befor
e switchin
g an
d th
e curren
t throug
h th
e inducto
rs
ia
continuou
s functio
n of time
. Therefore
, th
e initia
l conditio
n is
:
I'L(O) = I'L(0
.
+) = 0
(7.58
)
Chapter 7. Transient Analysis
218
'.«
© O
va (t)
t<0
t>
0
Figur
e 7.6
: Exampl
e o
fa
n RL circui
t excite
d by a source
.
Now , we shal
l analyz
e th
e circui
t afte
r switchin
g (righ
t circui
tn
i Figur
e 7.6)
. Writin
g
KV L fo
r thi
s circui
t yields
:
vi(
0 +vR(t) - vs(t) = 0.
(7.59
)
W e want o
t expres
s equatio
n (7.59
) n
i term
s of th
e curren
t throug
h th
e inductor
,
becaus
e thi
ss
i th
e physica
l quantit
y fo
r whic
h we hav
e th
e initia
l condition
. To thi
s
end, we us
e th
e termina
l relationships
:
diL(t)
vL{t) = Ldt
(7.60
)
vR(t) =
(7.61
)
and substitut
e the
m int
o (7.59)
. Thi
s yields
:
dkit)
(7.62
)
L——
+ RiL(t) vs{t).
dt
Differentia
l equatio
n (7.62
) combine
d wit
h initia
l conditio
n (7.58
) constitute
s th
e
initia
l valu
e problem
.
The differentia
l equatio
n (7.62
)s
i a first-orde
r linea
r differentia
l equatio
n wit
h
constan
t coefficients
, bu
tt
is
i no longe
r homogeneou
s (i.e.
, th
e right-han
d sid
e of
the equatio
n s
i no
t zero)
.n
I fact
, thi
s equatio
n s
i simila
ro
t th
e equatio
n we derive
d
earlie
r fo
r th
e RL circui
t excite
d onl
y by initia
l conditions
, excep
t fo
r th
e appearanc
e
of th
e voltag
e sourc
e on th
e right-han
d side
.n
I general
, circuit
s excite
d by source
s
wil
l lea
d o
t nonhomogeneou
s differentia
l equation
s becaus
e thei
r source
s wil
l alway
s
appea
r on th
e right-han
d sid
e of th
e equations
.
It s
i know
n fro
m mathematic
s tha
t a complet
e solutio
n of a nonhomogeneou
s
linea
r differentia
l equatio
n ca
n be represente
d a
s asu
m of a genera
l solutio
n of th
e cor
respondin
g homogeneou
s equatio
n an
d a particula
r solutio
n of th
e nonhomogeneou
s
equation
:
kit)
= ih(t) + ip(t),
(7.63
)
where ih(t) stand
s fo
r th
e genera
l solutio
n of th
e homogeneou
s equatio
n
T dih{t)
dt
+ Rih{t) = 0,
(7.64
)
72. First-Order Circuits
219
s th
e particula
r solutio
n of th
e nonhomogeneou
s equatio
n
whil
e ip(t) represent
di„(t)
L^^+ RUt)
= vs(t).
(7.65
)
Equatio
n (7.64
)s
i mathematicall
y identica
lo
t (7.42)
. Consequently
, it
s solutio
n ca
n
be writte
n n
i th
e form
:
ih(t) = Ae~^.
(7.66
)
Pleas
e not
e tha
t th
e constan
t A need
s o
t be determined
. However
, we must dela
y
solvin
g fo
rA unti
l we hav
e foun
d th
e particula
r solution
,because the initial condition
must be applied to the complete solution (7.63).
The metho
d of finding
th
e particula
r solutio
n (7.65
) depend
s on th
e mathematica
l
for
m ofvs(t) on th
e right-han
d sid
e of th
e nonhomogeneou
s equation
. We begi
n wit
h
the simpl
e problem
, when ou
r circui
ts
i excite
d by a dc voltag
e source
:
vv(0 = V0 = constant
.
(7.67
)
In thi
s case
, equatio
n (7.65
) ha
s th
e form
:
dipit)
L - ^ + Rip(t) = Vb.
dt
(7.68
)
The particula
r solutio
n of thi
s equatio
n ca
n be foun
d by usin
g th
e metho
d of un
determine
d coefficients
. The basi
c ide
a of thi
s metho
d s
io
t loo
k fo
r th
e solutio
n n
i
the for
m whic
h mimic
s th
e for
m of th
e right-han
d side
.t
I mean
s tha
t ni th
e cas
e of
equatio
n (7.68)
, we hav
eo
t loo
k fo
r th
e solutio
n n
i th
e form
:
ip(t) = IQ = constant
.
(7.69
)
Now , we substitut
e (7.69
) int
o equatio
n (7.68)
, whic
h result
s in
:
Rio = V0,
7 7
(
- °)
iP(t) = o' = ^r
-
( 7 / 7)1
Combinin
g th
e homogeneou
s an
d particula
r solution
s give
s th
e complet
e solutio
n
to equatio
n (7.62)
:
kit) = AeS
+ f
K
(7.72
)
However
, th
e constan
tA s
i stil
l unknown
. To find
it
, we evok
e th
e initia
l conditio
n
(7.58)
. Accordin
g o
t th
e initia
l condition
:
IL ( 0 +) = 0 - A + ^,
A
= -~i-
(7.73
)
(7 74)
-
220
Chapter 7
. Transient Analysis
So, th
e final
solutio
n o
t th
e initia
l valu
e proble
m is
:
w
R
(7.75
)
R
It s
i clea
r fro
m equatio
n (7.75
) tha
t ther
e ar
e tw
o component
s of th
e solution
: componen
t Vo/R correspond
so
t th
e steady-stat
e solutio
n an
d doe
s no
t deca
y wit
h time
,
whil
e componen
t — (Vo/R)e~Rt^L correspond
so
t th
e transien
t response
, an
d t
i doe
s
deca
y wit
h time
. As tim
e goe
s o
t infinity
, th
e transien
t goe
s o
t zero
, an
d onl
y th
e
steady-stat
e respons
e remains
.
Thus, we ca
n dra
w th
e followin
g conclusio
n (whic
h is
, by th
e way
, ver
y genera
l
in nature)
: the homogeneous solution has the physical meaning of transient (free)
response, while the particular solution has the physical meaning of steady-state
(forced) response. The latte
r suggest
s tha
t we ca
n find
th
e particula
r solutio
n withou
t
invokin
g th
e metho
d of undetermine
d coefficient
s bu
t by usin
g technique
s fo
r steady
stat
e analysi
s of electri
c circuits
. Fo
r instance
, ni ou
r exampl
e circuit
, inducto
r L ha
s
zer
o voltag
e dro
p a
t dc stead
y state
; consequently
, th
e sourc
e voltag
e si applie
d acros
s
the resisto
rR an
d th
e dc steady-stat
e curren
ts
i equa
lo
t
In =
Vo
R'
(7.76
)
which coincide
s wit
h th
e particula
r solution
.
W e ca
n als
o find
th
e voltage
s acros
s th
e resisto
r an
d th
e inducto
r usin
g equatio
n
(7.75)
:
vR(t) = RiL(t) = V0 - V0e I1
(7.77
)
vL(t) = L—j— = V0e L
dt
(7.78
)
One way o
t ge
t a goo
d feelin
g fo
r ho
w th
e circui
t behave
s ove
r tim
es
io
t grap
h
it
s voltage
s an
d currents
. Figur
e 7.
7 show
s th
e graph
s of th
e current
, th
e voltag
e
acros
s th
e resistor
, an
d th
e voltag
e acros
s th
e inductor
. Thes
e graph
s sho
w tha
t th
e
curren
t throug
h an
d th
e voltag
e acros
s th
e resisto
r bot
h star
ta
t zer
o an
d the
n becom
e
large
r unti
l the
y reac
h thei
r steady-stat
e values
. On th
e othe
r hand
, th
e voltag
e acros
s
Time (t
)
Time (t
)
Figur
e 7.7
: Graph
s o
fiL(t), vR{t), an
d vL(t).
Time (t
)
72. First-Order Circuits
221
the inducto
r start
s a
t it
s maximu
m valu
e an
d the
n decay
s o
t zer
o a
t steady-stat
e
conditions
. Thi
s s
i a goo
d illustratio
n of th
e propertie
s of a
n inductor
. Initiall
y th
e
inducto
r behave
s a
s a
n open circuit an
d t
i ha
s al
l th
e sourc
e voltag
e applie
d acros
s
it
. Thi
ss
i becaus
e ther
es
i no curren
t flowing
throug
h th
e inducto
ra
s a resul
t of zer
o
initia
l condition
. When th
e dc steady-stat
e conditio
n s
i reached
, th
e inducto
r act
sa
sa
short circuit an
d it
s voltag
e dro
p s
i zer
o an
d th
e curren
t flow
s throug
h t
i unrestricted
.
W e shal
l se
e late
r tha
ta
n uncharge
d capacito
r behave
sn
i th
e opposit
e manner
.
A t tim
e / = 0 th
e capacito
r act
s a
s a shor
t circuit
, whil
e a
t tim
e / = °° t
i act
s a
s a
n
open circuit
.
EXAMPL E 7.
5 Conside
r th
eRL circui
t again
, bu
t thi
s tim
e wit
h a sinusoida
l voltag
e
source
. Thi
s sourc
e ha
s th
e form
:
vs(t) = Vm cos(co
r + <f)s).
(7.79
)
The differentia
l equatio
n s
i nearl
y th
e sam
e a
s before—onl
y th
e right-han
d sid
e of
the equatio
n s
i different
:
dij (t)
at
Clearl
y th
e "homogeneous
" solutio
n o
t thi
s equatio
n wil
l be identica
lo
t th
e previou
s
one (se
e (7.66))
,s
o we nee
d onl
y conside
r th
e particula
r solution
. To us
e th
e metho
d
of undetermine
d coefficient
s fo
r thi
s problem
, we hav
e o
t assum
e th
e for
m of th
e
particula
r solutio
n whic
h mimic
s th
e for
m of th
e right-han
d sid
e of th
e equatio
n
(7.80)
:
ip(i) = Imcos(a)t + (/>/)
.
(7.81
)
However
,t
is
i no
t necessar
y o
t us
e th
e metho
d of undetermine
d coefficient
s o
t solv
e
for th
e particula
r solutio
n when th
e sourc
es
i sinusoidal
. We recal
l tha
t th
e particula
r
solutio
n correspond
so
t th
e a
c steady-stat
e respons
e of th
e circuit
. For thi
s reason
,n
i
orde
ro
t find
th
e particula
r solutio
n we ca
n emplo
y th
e phaso
r technique
, whic
h ha
s
been studie
d n
i detai
ln
i previou
s chapters
.
A s alway
s wit
h th
e phaso
r technique
, th
e first
ste
p s
io
t conver
t th
e sourc
e int
o
phaso
r form
:
V5 = Vmej+*.
(7.82
)
Then, we represen
t th
e unknow
n curren
t (7.81
)n
i th
e phaso
r for
m a
s well
:
I = Imej(t)I.
(7.83
)
The tota
l impedanc
e of ou
r circui
ts
i
Z = R + jc*L.
(7.84
)
To find
th
e unknow
n current
, we us
e th
e relationshi
p betwee
n th
e phasor
s of voltag
e
and current
:
/ =\
(7.85
)
222
Chapter 7. Transient
Analysis
Thus, we ca
n findth
e pea
k valu
e Im an
d phas
e angl
e <f>j of th
e unknow
n curren
t as
follows
:
vv
. vvm
(7.86
)
2 m 2 2
+ 0)> L
'" \z\
v
VF
h ='- </>s -k
(7.87
)
where
coL
(7.88)
~R'
By substitutin
g (7.86
) an
d (7.87
) int
o (7.81)
, we obtai
n th
e particula
r solution
:
n
<$> =-■ arcta
Vm
,
cos((ot + <f>s- <j>).
2
y/R + w2L2
Ut) =
(7.89
)
Combinin
g thi
s expressio
n wit
h th
e "homogeneous
" solutio
n yield
s th
e expressio
n
for th
e complet
e solutio
n o
t th
e differentia
l equation
:
.
cos(a)t + <f>s -(/>
) +Ae~*f.
(7.90
)
2
2 2
VR + w L
Now , th
e proble
m s
i nearl
y don
e an
d onl
y th
e unknow
n constan
t A remain
s o
t be
found
. Thi
s s
i accomplishe
d by usin
g th
e initia
l conditio
n (th
e same one as n
i th
e
previou
s RL circuit)
:
iL(t) =
iL (0+) =
A
. Vm
cos(cj>s - 4>) +A = 0,
\/R
+ a)2L2
(7.91
)
2
Vm
= 2
+ OJ2L2
VR
- <$>).
c o s(^
(7.92
)
B y substitutin
g thi
s valu
e fo
rA int
o (7.90)
, we arriv
e at th
e fina
l solution
:
iL(t) =
Vme~lf
Vm
^/R2
cos(a)t + <f>s - <\>) \/R2 + a>2L2
+ o)2L2
cos(</>
, - <^>)
.
(7.93
)
Tw o instructiv
e observation
s ca
n be made concernin
g th
e solutio
n (7.93)
.
Conside
r suc
h an initia
l phas
e <f>s of th
e voltag
e sourc
e tha
t
4 > s - 4> = <f>s - arcta
n —
= -.
(7.94
)
Then, t
is
i clea
r fro
m (7.92
) tha
t
c o s ^( - 4>) = - . Vm
co
s ^ = 0.
2
VR + o L
VR2 + (o2L2
From (7.93)
, (7.94)
, an
d (7.95
) we find:
A = -
V,n
2
2 2
hit) =
,
\/R2
Vm
+ w2L2
cos(co
r +%) = ip(t).
2
(7.95
)
(7.96
)
223
7.2. First-Order Circuits
This mean
s tha
t fi th
e initia
l phas
e cj>s s
i properl
y adjuste
d [se
e formul
a (7.94)
] ther
e
wil
l be no transien
t response
. The reaso
n fo
r thi
ss
i that
:
ip(0) = 0 = L/(0+).
(7.97
)
In othe
r words
, ther
e s
i no transien
t respons
e becaus
e th
e initia
l conditio
n fo
r th
e
inducto
r coincide
s wit
h it
s ne
w steady-stat
e condition
.
The abov
e discussio
n clearl
y reveal
s tha
t th
e physica
l caus
e fo
r transient
s s
i
a mismatc
h betwee
n initia
l condition
s fo
r energ
y storag
e element
s an
d thei
r ne
w
steady-stat
e conditions
. Thi
s mismatc
h drive
s transients
.
Anothe
r observatio
ns
i relate
do
t th
e interpretatio
n of th
e complet
e solutio
n (7.90
)
in term
s of impedanc
e Z(s) = sL +R a
s a functio
n of "complex
" frequenc
y s. t
Is
i
clea
r fro
m (7.90
) (a
s wel
la
s fro
m (7.86)-(7.89)
) tha
t th
e valu
e of thi
s impedanc
e a
t
the frequenc
y of excitatio
n s = jco completel
y determine
s th
e steady-stat
e respons
e
of th
e circuit
.t
Is
i als
o clea
r fro
m (7.44
) tha
t zero
s of thi
s impedanc
e coincid
e wit
h
the exponent
s of th
e transien
t (free
) response
. Thus
, th
e impedanc
e Z(s) provide
s th
e
complet
e informatio
n concernin
g bot
h steady-stat
e an
d transien
t responses
. Fo
r thi
s
reason
, th
e impedanc
e a
s a functio
n of comple
x frequenc
y s
i extensivel
y use
d n
i th
e
advance
d theor
y of electri
c circuits
.
■
EXAMPLE 7.6 Conside
r th
e specifi
c exampl
e of a
n RL circui
t excite
d by a
n a
c
source
.n
I thi
s cas
e R = 1/2(1,
L = 1/
2 H, an
d vs(t) — 2cos(\/30
- Fin
d th
e
inducto
r current
.
First
, we find
th
e impedanc
e magnitud
e
\Z\ = VR2 + o>2L2 = \j{\/2)2
2
+ (\/3/2)
- 1 a
For th
e impedanc
e phas
e we us
e equatio
n (7.88)
:4> = arctan(coL/7?
) = arctan(
y/3)
= 60°
. Pluggin
g thes
e value
s int
o equatio
n (7.89)
, we find
th
e particula
r solutio
n o
t
be ip(t) = 2cos(y3
f ~ 60°)
. The unknow
n constan
tn
i th
e homogeneou
s solutio
n
is foun
d fro
m equatio
n (7.92)
:A = - 2 c o s ( - )
6 0=° - 1. The tim
e constant
, L/R,
for th
e homogeneou
s solutio
n s
i unity
. Combinin
g th
e tw
o solution
s an
d performin
g
some algebra
, we find
th
e final
answer
:iL(f) = cos(y3
0 + y3 sin(y3
0 — e~f A.
A s a usefu
l exercise
, th
e reade
r may wis
h o
t tr
y o
t recove
r th
e sam
e resul
t by usin
g
the metho
d of undetermine
d coefficients
.
Now , we conside
r th
e excitatio
n ofRC circuit
s by sources
. We first
examin
e th
e
excitatio
n of th
e circui
t show
n n
i Figur
e 7.
8 by th
e dc curren
t source
:
is(t) = I0 = constant
.
(7.98
)
First
, we find
th
e initia
l condition
.n
I thi
s case
, we nee
d o
t determin
e th
e initia
l
voltag
e acros
s th
e capacitor
. Sinc
e th
e circui
ts
i no
t excite
d befor
e th
e switc
h closes
,
w e conclud
e tha
t th
e initia
l voltag
e acros
s th
e capacito
r must be zero
. Therefore
, th
e
initia
l conditio
n is
:
v(0+) = 0.
(7.99
)
224
Chapter 7. Transient Analysis
t=0
i(t
)
i—°^r°
d>
->—
+
(t
)
inW
v(t);
Figur
e 7.8
: RC circui
t excite
d by a curren
t source
.
Now , we analyz
e th
e circui
t fo
r t > 0. Applyin
g KCL yields
:
is(t) = kit) + iR(t),
(7.100
)
where ic(t) an
d iR(t) ar
e th
e current
s throug
h th
e capacito
r an
d th
e resistor
, respec
tively
. We nee
d o
t expres
s th
e KCL equatio
n ni term
s of th
e voltag
e acros
s th
e
capacitor
, becaus
e thi
s voltag
e s
i a physica
l quantit
y fo
r whic
h we hav
e th
e initia
l
condition
. To thi
s end
, we evok
e th
e termina
l relationships
:
ic(t) = C
iR{t)
dv(t)
dt
(7.101
)
(7.102
)
R '
Substitutin
g thes
e expression
s int
o (7.100
) give
s th
e differentia
l equatio
n n
i th
e for
m
w e need
. Thi
s equatio
n togethe
r wit
h initia
l conditio
n (7.99
) lead
so
t th
e followin
g
initia
l valu
e problem
:
c
dv(t)
~dT
+
v(r
)
.
lM
(7.103
)
v(0+
) = 0.
(7.104
)
T
=
This initia
l valu
e proble
m ca
n be solve
d n
i th
e standar
d manner
.
W e kno
w tha
t th
e complet
e solutio
n of differentia
l equatio
n (7.103
)s
i a su
m of
the homogeneou
s solutio
n an
d th
e particula
r solution
:
v(t) = vh(t) + vpit).
(7.105
)
The homogeneou
s solutio
n s
i ni th
e for
m
vh(t) = Aest,
(7.106
)
where s s
i th
e solutio
n of th
e followin
g characteristi
c equation
:
sC + - = 0,
R
(7.107
)
225
7.2. First-Order Circuits
which ca
n als
o be rewritte
n a
s
Y(s) = - +sC
R
=0.
(7.108
)
The abov
e equatio
n lead
s to
:
s= - ±
(7.109
)
vh{t) = Ae~t/RC.
(7.110
)
Therefore
,
Now , th
e particula
r solutio
n s
i required
. Fo
rt > 0, th
e right-han
d sid
e of (7.103
)
is aconstant
. Usin
g th
e metho
d of undetermine
d coefficients
, we loo
k fo
r th
e particula
r
solutio
n ni th
e sam
e for
m a
s th
e right-han
d sid
e of th
e differentia
l equation
:
vP(t) =VQ = constant
.
(7.111
)
Substitutin
g vp(t) int
o (7.103
) we find
0 + ^ = /0 ,
(7.112
)
Vb=/o*.
(7.H3)
So, a
s shoul
d be expected
, th
e particula
r solutio
n coincide
s wit
h th
e dc steady-stat
e
respons
e of th
e circuit
. Thus
, th
e complet
e solutio
n s
i th
e su
m of th
e homogeneou
s
and particula
r solutions
:
v(t) = RI0 +Ae~t/RC.
(7.114
)
W e nex
t find
th
e constan
tA by usin
g initia
l conditio
n (7.104)
:
v(0+
) = RI0+A
= 0,
(7.115
)
A = -RIQ,
(7.116
)
v(t) = RIo-RIoe~t/RC.
(7.117
)
which lead
so
t th
e final
answer
:
Now tha
t we hav
e th
e expressio
n fo
r th
e voltag
e acros
s th
e capacitor
, we ca
n
easil
y find
expression
s fo
r th
e current
s throug
h th
e capacito
r an
d th
e resistor
:
ic(t) = C^=I0e-"RC,
at
(7.118
)
iR(t)=V-^=I0-I0e-t/RC.
(7.119
)
K
The graph
s of thes
e tw
o function
s (Figur
e 7.9
) revea
l some interestin
g propertie
s
of th
e capacitor
. At tim
e t = 0, ther
es
i no voltag
e dro
p acros
s th
e capacitor
,s
ot
i act
s
lik
e a shor
t circui
t an
d shunts th
e curren
t source
.n
I othe
r words
, al
l th
e curren
t flows
Chapter 7
. Transient Analysis
226
Time (t)
Time (t)
Figur
e 7.9
: Graph
s o
f th
e curren
t throug
h th
e resisto
r an
d capacitor
.
throug
h th
e branc
h wit
h th
e capacito
r an
d non
e flowsthroug
h th
e branc
h wit
h th
e
resistor
. However
,a
t dc stead
y stat
e (t = sc
) th
e capacito
r act
s lik
e a
n ope
n circuit
.
None of th
e curren
t flows
throug
h t
i becaus
e ther
es
i no tim
e variatio
n of th
e voltage
.
All th
e curren
t flows
throug
h th
e branc
h wit
h th
e resistor
.
The propert
y of a capacito
ro
t ac
ta
s a shor
t circui
t an
d o
t absor
b most of th
e
transien
t curren
t durin
g th
e initia
l stag
e of transient
ss
i utilize
d n
i practice
. Namely
,
capacitor
s ar
e ofte
n connecte
d ni paralle
l wit
h expensiv
e device
s (equipmen
t suc
h
as computers
, etc.
) o
t protec
t thes
e device
s fro
m larg
e (an
d destructive
) transien
t
current
s whic
h usuall
y occu
r durin
g initia
l stage
s of transients
.
EXAMPLE 7.7 Conside
r th
e cas
e of a
n a
c curren
t source
:
is(t) = Imscos((ot + <l>s)
(7.120
)
drivin
g th
e circui
t ni Figur
e 7.8
. Fin
d th
e expressio
n fo
r v(t).
In thi
s case
, differentia
l equatio
n (7.103
) ha
s th
e form
:
dv(t)
v(t)
C— — + — - = Imscos(o)t + </>,)
.
at
R
n nn
(7.121
)
The homogeneou
s solutio
n fo
r thi
s equatio
n si th
e sam
e a
s befor
e (se
e (7.110))
. To
findth
e particula
r solutio
n o
t equatio
n (7.121)
, we tak
e advantag
e of th
e fac
t tha
t thi
s
solutio
n ha
s th
e physica
l meanin
g of th
e a
c steady-stat
e respons
e of th
e RC circuit
.
For thi
s reason
, we shal
l emplo
y th
e phaso
r techniqu
e o
t find
th
e particula
r solution
.
In th
e tim
e domain
, thi
s particula
r solutio
n ha
s th
e form
:
vP{t) = Vmcos(cot + <j>v).
(7.122
)
The phaso
r for
m of thi
s solutio
n is
:
V = Vmej4)v.
(7.123)
7.2. First-Order Circuits
227
Similarly
, th
e phaso
r of th
e curren
t sourc
e is
:
l = Imse^.
(7.124
)
It s
i clea
r tha
t th
e voltag
e phasor
,V, an
d th
e curren
t sourc
e phasor
,Is, ar
e relate
d o
t
one anothe
r throug
h th
e admittance
,Y:
YV,
(7.125
)
Y = - + jcoC.
(7.126
)
V = -,
Y
(7.127
)
ms
(7.128
)
<j>v = <f>s - cf>,
(7.129
)
where
From (7.125)
, we find
:
which accordin
g o
t (7.126
) lead
s to
:
Vm =
where
(/
> = arcta
n o>#C
.
(7.130
)
Thus, th
e particula
r solutio
n (7.122
) is
:
j
vJt) =
ms
,
n
cos(cot + <f>s - 4>).
(7.131
)
y/l + oi2C2R2
Combinin
g (7.131
) wit
h (7.110)
, we fin
d th
e complet
e solutio
n of differentia
l equatio
n
(7.121)
:
= cos(co
f + <j)s-<}>) + Ae~t/RC.
(7.132
)
y/l + (o C R
To fin
d th
e constant
,A, we shal
l invok
e th
e initia
l conditio
n (7.104)
, whic
h lead
s to
:
v(f
) =
,
ImsR
2
2 2
ImsR
\ /l
+ OJ2C2R2
cos(4>
, - <f>) + A = 0.
(7.133
)
Consequently
:
. ImsR
=cos(
^ - $).
(7.134
)
V I + w2C2R2
By substitutin
g (7.134
) int
o (7.132)
, we en
d up wit
h th
e fina
l expressio
n fo
r th
e
respons
e of th
e RC circui
t excite
d by a
c curren
t source
:
A= -
v(t) =
ImsR
ms
V I + OJ2C2R2
cos(at
f + 4>s — 4>) -
I sRe~^RC
,ms
y/l + (o2C2R2
cos((/).
y - 4>),
(7.135
)
Chapter 7. Transient Analysis
228
where
, accordin
g o
t (7.130)
:
<f> = arctancotfC
.
(7.136
)
Formula
s (7.135
) an
d (7.136
) explicitl
y expres
s th
e circui
t respons
e n
i term
s of a
c
excitatio
n an
d circui
t parameters
.
Two instructiv
e observation
s ca
n be made concernin
g th
e solutio
n (7.135)
.t
I ca
n
be show
n tha
t th
e initia
l phase
, ^s, of th
e curren
t sourc
e ca
n be chose
n n
i suc
h a way
tha
t ther
e wil
l be no transient
. Indeed
, fi
<j)s = <f> + j = arctano>/?
C + y,
(7.137
)
the
n fro
m (7.134
) an
d (7.135
) we find:
A = 0,
v(t) =
(7.138
)
J
D
, ms
cos(cot + <f>s-(l)) = vJt).
y/\ + co2C2R2
(7.139
)
It s
i clea
r fro
m (7.139
) (a
s wel
l a
s fro
m (7.131)
) tha
t fi th
e initia
l phas
e of th
e a
c
curren
t sourc
es
i chose
n accordin
g o
t (7.137
) the
n th
e initia
l conditio
n fo
r th
e voltag
e
acros
s th
e capacito
r coincide
s wit
h th
e a
c steady-stat
e conditio
n fo
r thi
s voltage
:
v,(0
) = v(0
.
+) = 0
(7.140
)
The abov
e discussio
n agai
n underline
s th
e fac
t tha
t th
e physica
l caus
e fo
r transient
s
is a mismatc
h betwee
n initia
l condition
s fo
r energ
y storag
e element
s an
d thei
r ne
w
steady-stat
e conditions
. Thi
s mismatc
h drive
s transients
.
Anothe
r observation
, whic
h s
i appropriat
e here
,s
i tha
t th
e admittance
,Y(s), a
s
a functio
n of comple
x frequenc
y full
y determine
s th
e for
m of th
e complet
e solutio
n
(7.132)
. Indeed
, th
e valu
e of thi
s admittanc
e a
t th
e frequenc
y of excitation
,s = jo),
completel
y determine
s th
e steady-stat
e respons
e of th
e circuit
, whil
e zero
s of thi
s
admittanc
e [se
e (7.108)-(7.109)
] ar
e th
e exponent
s of th
e transien
t (free
) response
.
W e recal
l tha
t we reache
d a simila
r conclusio
n wit
h respec
to
t impedanc
e Z{s)
when we analyze
d th
e RL circuit
. The natura
l questio
n s
i why ni on
e cas
e we dea
l
wit
h impedanc
e Z(s) whil
en
i th
e othe
r cas
e we dea
l wit
h admittanc
e Y(s). A relate
d
questio
n is
: "I
st
i possibl
e o
t achiev
e some uniformit
y n
i thi
s matte
r s
o tha
t we shal
l
deal wit
h th
e sam
e quantit
y regardles
s of th
e typ
e of electri
c circuit?
"
The answer
so
t th
e pose
d question
s ar
e base
d on th
e notio
n oftransfer functions,
which ar
e discusse
d n
i detai
ln
i Sectio
n 7.4
.
7.2.3
Circuits Excited b y Initial Conditions and Sources
W e no
w conside
r th
e cas
e of a circui
t excite
d by bot
h initia
l condition
s an
d sources
.
EXAMPLE 7.8 Conside
r a circui
t show
n n
i Figur
e 7.10
, wher
e vs(t) s
i a dc source
:
vs(t) = V0 = constant
.
(7.141
)
7.2. First-Order Circuits
229
FL
i(t)
A/WV
R<
v s (t)Q
t<0
Figur
e 7.10
: Circui
t excite
d by initia
l condition
s an
d sources
.
A t tim
e t < 0 th
e circui
ts
ia
t dc stead
y state
. The
n a
t tim
e t = 0 th
e switc
h s
i throw
n
and th
e circui
t change
s it
s configuration—specifically
, a ne
w resisto
rR2 s
i connected
.
A s a result
, some transient
s wil
l be generate
d an
dt
i wil
l tak
e some tim
e fo
r th
e circui
t
to readjus
to
t ne
w steady-stat
e condition
s correspondin
g o
t th
e ne
w configuration
.
W e approac
h th
e analysi
sn
i th
e sam
e way a
s wit
h an
y proble
m involvin
g tran
sients
. First
, th
e initia
l conditio
n s
io
t be found
.n
I thi
s case
, we nee
d th
e initia
l valu
e
of th
e curren
t throug
h th
e inductor
. Examinin
g th
e circui
t befor
e switching
, we se
e
tha
t th
e curren
ts
i give
n by th
e expression
:
V,
m =fl, +/?q
(7.142
)
This s
i becaus
e immediatel
y befor
e switching
, th
e circui
ts
ia
t dc steady-stat
e condi
tions
. Thi
s mean
s tha
t th
e inducto
r act
sa
s a shor
t circui
t becaus
e it
s dro
p ni voltag
es
i
zero
, and
, consequently
, th
e curren
t throug
h th
e inducto
rs
i give
n by equatio
n (7.142)
.
Therefore
, th
e initia
l conditio
n is
:
/(0+) = *(0_) =
Vn
R\ +R3
(7.143)
Afte
r switching
, th
e circui
t ca
n be represente
d ni th
e ne
w for
m show
n n
i Figur
e
7.11
. Here
,o
t simplif
y th
e calculations
, th
e resistor
s ar
e combine
d int
o on
e equivalen
t
resistor
:
R,
R^ +
RiR
1^2
Ri +R2
(7.144)
W e ca
n no
w writ
e th
e differentia
l equatio
n fo
r thi
s circuit
, whic
h s
i simila
ro
t thos
e
for th
e RL circuit
s we hav
e considered
. By combinin
g thi
s differentia
l equatio
n wit
h
the initia
l condition
, we en
d up wit
h th
e followin
g initia
l valu
e problem
:
230
Chapter 7. Transient Analysis
v s wQ
t>0
Figure 7.11: The circui
t afte
r switchin
g
di(t)
L-^+ Rei{t) = V0,
at
/«M = ^ .
(7.145
)
(7.146)
This proble
m ca
n be solve
d by usin
g th
e method
s outline
d n
i th
e previou
s sections
.
The complet
e solutio
n wil
l be a su
m of th
e homogeneou
s an
d particula
r solutions
:
i(
0 = ih{t) + ip(t).
(7.147
)
The homogeneou
s solutio
n wil
l hav
e th
e for
m
ih(t) = Aes\
(7.148
)
where A s
io
t be foun
d fro
m th
e initia
l conditio
n an
ds s
i th
e solutio
n o
t th
e charac
teristi
c equation
. The characteristi
c equatio
n fo
r thi
s differentia
l equatio
n is
:
Z(s) = sL + Re = 0,
(7.149
)
and t
is
i clea
r tha
t
( 71 5 0)
s = ~ -
'
Therefore
, th
e homogeneou
s solutio
n s
i
ih(t) = Ae~{Re/L)t.
(7.151
)
Now , th
e particula
r solutio
n must be found
. Sinc
e th
e right-han
d sid
e of (7.145
)
is a constant
, we assum
e th
e particula
r solutio
n o
t be a constan
ta
s well
:
ip(t) = o/ = constant
.
(7.152
)
By substitutin
g ip(t) int
o equatio
n (7.145)
, we find:
0+ /W> = V0,
(7.153
)
k - -£•
(7.154
)
Ke
As expected
, th
e particula
r solutio
n coincide
s wit
h th
e ne
w steady-stat
e response
.
7.2
. First-Order Circuits
231
Now , th
e complet
e solutio
n ca
n be writte
n a
s follows
:
V
/L
0
i(t)=^+Ae-^
\
(7.155
)
RP
The constan
tA s
i foun
d by usin
g th
e initia
l conditio
n (7.146)
:
'■<0+)-§ + A = *Ti^
(7156)
A = — ^ ° - - ^.
Ri + R3 Re
Substitutin
g A bac
k int
o (7.155
) give
s th
e final
solutio
n o
t th
e problem
:
*>-£+(*V£)'^-
(7.157
)
ai58)
A s before
, thi
s solutio
n ha
s tw
o components
: th
e particula
r solutio
n whic
h corre
spond
so
t th
e steady-stat
e respons
e an
d th
e homogeneou
s solutio
n whic
h correspond
s
to th
e transien
t response
.
EXAMPL E 7.
9 Conside
r th
e sam
e circui
t excite
d by a
n a
c voltag
e sourc
e (se
e Figur
e
7.10)
:
vs(t) = Vmscos(a)t + <k)
.
(7.159
)
To find
th
e initia
l condition
, we hav
eo
t first
analyz
e th
e a
c steady-stat
e respons
e
which existe
d befor
e switching
. Thi
s tas
k ca
n be accomplishe
d by usin
g th
e phaso
r
technique
. The impedanc
e of th
e circui
t befor
e switchin
g s
i equa
l to
:
Z_ = (Rl + R3) + j<oL,
(7.160
)
where th
e subscript"—
" indicate
s th
e impedanc
e of th
e circui
t fo
r negativ
e times
.
Now , by usin
g th
e voltag
e sourc
e phaso
r
Vs = Vmsej(t)\
(7.161
)
= Im-e^-^,
(7.162
)
w e find
th
e curren
t phasor
:
/- =—
where a
s befor
e th
e subscrip
t "—" mean
s th
e a
c steady-stat
e curren
t valu
e befor
e
switching
. Fro
m (7.160)
, (7.161)
, an
d (7.162)
, we derive
:
—
ms
2
V ^ i +R3) + u2L?
<f>- = arcta
n
R\ + #3
.
(7.163
)
Thus, n
i th
e tim
e domain
, th
e a
c steady-stat
e curren
t prio
ro
t switchin
g ca
n be writte
n
as follows
:
232
Chapter 7
. Transient Analysis
From expressio
n (7.164
) we ca
n find
th
e initia
l conditio
n fo
r th
e curren
t throug
h th
e
inductor
:
Vm
i(0
) = ,
+ ) = /(0V^i
\
= == cos(<f
e - (/>-)
.
+ R3)2
(7.165
)
+ CD2L2
Now , th
e initia
l valu
e proble
m (7.145)—(7.146
) ca
n be reformulate
d a
s follows
:
+Rei(t) = Vms cos(cot + <fc)
,
L ^
(7.166
)
Vm
i'(0
\
===== cos(<f
c - (/>_)
.
+) = ,
V (* i + *3 )2 + u2L2
(7.167
)
The homogeneou
s solutio
n fo
r equatio
n (7.166
)s
i th
e sam
e a
s befor
e (se
e (7.151))
.
The particula
r solutio
n fo
r equatio
n (7.166
) ha
s th
e physica
l meanin
g of th
e a
c steady
stat
e respons
e of th
e circui
t show
n ni Figur
e 7.1
1 an
d t
i ca
n be foun
d by usin
g agai
n
the phaso
r technique
. The final
resul
ts
ia
s follows
:
ip{t) =
ms
fJ
1T1 cos(a>t
2
y/R e + o)2L2
+ <f
c -
tf>),
(7.168
)
where
(oL
<f> = arctan—
.
(7.169
)
Thus, th
e complet
e solutio
n o
t differentia
l equatio
n (7.166
) is
:
i(t) =
Vna
cos(iot + &-<*>
) +Ae-<*</L».
2 2
y/R] + Oi L
(7.170
)
To find
th
e unknow
n constan
tA, we invok
e th
e initia
l conditio
n (7.167)
, whic
h lead
s
to:
Vmscos((j)s - (f>-) _
2
Vmscos(<
k -<f>)
2 2
JR\ + a,2L2
y/(Rx + R3) + <o L
This result
s in
:
A =
V
'ns COS(<t>s -
(f)-)
2
_
Vms COS(cfry - <f>)
2 2
y/(R{ + R3) + oo L
y/R2e + co2L2
By substitutin
g (7.172
) int
o (7.170)
, we find
th
e final
expression
:
ms
i(t) = —
= cos(w
? + (bs — <b)
y/R\ + 0>2L2
+
I
Vmscos((j)s-
WiRx+RiP
(f)-)
+ aW
_ Vmscos(<t>5 - <f>)\ e-(Re/L)t
y/R\ + a>2L2
V
(1
1 7 3)
233
73. Second-Order Circuits
7.3
Second-Orde
r Circuit
s
All th
e circuit
s tha
t we hav
e discusse
d thu
s fa
r dealin
g wit
h transient
s hav
e bee
n
first-order
circuits
, becaus
e the
y containe
d onl
y on
e energ
y storag
e element
. Fo
r thi
s
reason
, the
y le
d o
t first-order
differentia
l equations
. We no
w conside
r circuit
s whic
h
are describe
d by second-orde
r differentia
l equations
. Thes
e circuit
s generall
y featur
e
two energ
y storag
e elements
. We conside
r circuit
s excite
d by initia
l condition
s befor
e
examinin
g circuit
s drive
n by sources
.
7.3,1
Circuits Excited b y Initial Conditions
Conside
r th
e circui
t show
n on th
e lef
tn
i Figur
e 7.12
. Notice
tha
tt
i contain
s bot
h a
n
inducto
r an
d a capacitor
. The approac
h o
t th
e analysi
s of thi
s circui
ts
i basicall
y th
e
same a
so
t th
e analysi
s of first-order
circuits
: namely
, we hav
eo
t first find
th
e initia
l
conditio
n an
d the
n analyz
e th
e circui
t afte
r switching
. However
, when ther
e ar
e tw
o
energ
y storag
e element
s n
i th
e circuit
, tw
o initia
l condition
s must be found
.n
I thi
s
case
, we nee
d o
t findth
e initia
l curren
t throug
h th
e inducto
r an
d th
e initia
l voltag
e
acros
s th
e capacitor
.
Usin
g KVL fo
r th
e circui
t befor
e switching
, we ca
n se
e tha
t th
e voltag
e acros
s
the capacito
r si equa
lo
t th
e sourc
e voltage
:
vs(t) - vcit)
= 0,
(7.174
)
vc(!) = vs(t).
(7.175
)
The initia
l curren
t throug
h th
e inducto
rs
i zero
, becaus
e befor
e switching
, th
e inducto
r
is n
i a
n ope
n branc
h wit
h no curren
t flowing
. Thus
, th
e initia
l condition
s ar
e
vc(0
) = vv(0) = Vb,
+ ) = vc(O-
(7.176
)
/L(0+
) = I'L(O) = 0.
(7.177
)
Afte
r switching
, th
e circui
t ha
s th
e configuratio
n show
n on th
e righ
t ni Figur
e
7.12
. We no
w us
e KVL o
t se
t up th
e differentia
l equatio
n fo
r thi
s circuit
:
vcit) +vL{t) + vR(t) = 0.
(7.178
)
*.«
© o
t<o
Figur
e 7.12
: Second-orde
r circuit
.
t>0
Chapter 7. Transient Analysis
234
Ultimately
, we want o
t ge
t th
e differentia
l equatio
n onl
yn
i term
s of th
e voltag
e acros
s
the capacitor
. To thi
s en
d we first
us
e th
e termina
l relationship
s fo
r th
e inducto
r an
d
the resistor
. Thi
s lead
so
t th
e followin
g for
m of equatio
n (7.178)
:
di(t)
vc(t) + L—~ + Ri(t) = 0.
at
(7.179
)
Sinc
e al
l thre
e element
s ar
e connecte
d n
i series
, th
e sam
e curren
t flowsthroug
h al
l
thes
e elements
. Thi
s curren
t ca
n be expresse
d ni term
s of th
e voltag
e acros
s th
e
capacitor
:
,-(»
) = C ^ >.
at
By substitutin
g (7.180
) int
o (7.179)
, we obtain
:
LC
d2vr(t)
^
dtl
(7.180
)
dvr(t)
+RC-^
+ vc(0 = 0.
dt
(7.181
)
Now th
e initia
l conditio
n (7.177
) must be expresse
d n
i term
s of vc(t) a
s well
. Thi
s
can be accomplishe
d by combinin
g (7.177
) wit
h (7.180)
, whic
h lead
s to
:
i(0
,
+) = C ~ ( 0
+) = 0
dt
(7.182
)
^ ( 0 +) = 0.
(7.183
)
dt
Thus, we arriv
e a
t th
e followin
g initia
l valu
e problem
: find
th
e solutio
n of th
e equation
:
d2Vr(t)
dVr(t)
LC—^r1 +RC^r1
dt1
dt
subjec
to
t th
e initia
l conditions
:
+ vc(0
= 0
vc(0+ ) = %
(7.184
)
(7.185
)
^n(0+
) = 0.
(7.186
)
dt
, homogeneou
s second-orde
r differentia
l equa
Differentia
l equatio
n (7.184
)s
i a linear
tio
n wit
h constan
t coefficients
. As wit
h an
y linea
r homogeneou
s differentia
l equatio
n
wit
h constan
t coefficients
, we loo
k fo
r a solutio
n n
i th
e form
:
vc(t) = Aest.
(7.187
)
By substitutin
g (7.187
) int
o equatio
n (7.184)
, we find:
s2LCAest +sRCAest + Aest = 0.
(7.188
)
By factorin
g ou
tAes\ we obtain
:
Aest(s2LC
+sRC + 1
) - 0.
(7.189
)
7.3. Second-Order Circuits
235
From (7.189
) we conclud
e tha
t th
e functio
n n
i (7.187
) wil
l be a solutio
n of th
e
differentia
l equatio
n onl
y fi th
e exponen
ts satisfie
s th
e characteristi
c equation
:
s2LC + sRC + 1 = 0.
(7.190
)
The las
t equatio
n ca
n be writte
n n
i th
e equivalen
t form
:
(7.191
)
Z(s) =R + sL+^~=0
sC
which agai
n show
s tha
t th
e zero
s of th
e characteristi
c equatio
n ar
e th
e zero
s of
the impedanc
e a
s a functio
n of comple
x frequency
. The characteristi
c equatio
n s
ia
quadrati
c equatio
n whic
h ha
s tw
o root
s (solutions
) give
n by th
e formula
:
A s fa
r a
s th
e root
s of th
e characteristi
c equatio
n ar
e concerned
, thre
e differen
t
case
s ca
n be clearl
y distinguished
. Thes
e case
s depen
d on th
e discriminant, whic
h s
i
the quantit
y unde
r th
e radical
:
d
(7193)
=&-h
If th
e discriminan
ts
i greate
r tha
n zero
, the
n s\ an
d s2 ar
e bot
h rea
l an
d different
.f
It
i
e rea
l an
d equal
.f
I th
e discriminan
ts
i les
s tha
n zero
,
is equa
lo
t zero
, the
n s\ an
d s2 ar
the
n s\ an
d s2 ar
e comple
x conjugates
. Not
e tha
t sinc
e LC > 0, th
e rea
l par
t of bot
h
root
s wil
l alway
s be negativ
e (o
r zer
o if/
? = 0)
.t
Is
i wel
l know
n fro
m th
e theor
y of
differentia
l equation
s tha
t th
e abov
e thre
e case
s correspon
d o
t thre
e distinc
t form
s of
the solution
s of equatio
n (7.184)
. Thes
e thre
e form
s of th
e solutio
n ar
e give
n below
:
{
Aies*f +A2eS2\
fid > 0, (sl <0,s2<
0)
,
Axest + A2test,
fid - 0, (s{ = s2 = s\
e~m{A\ co
s cot + A2 si
n cot), fid < 0,
where or an
d co n
i (7.194
) ar
e determine
d by th
e expressions
:
«r = £,
(7.194
)
w =
(7196
)
Vzz-|?
-
(7.195
)
Sinc
en
i ou
r proble
m we do no
t assum
e specifi
c value
s fo
r th
e circui
t elements
,
w e shal
l discus
s th
e abov
e thre
e case
s separately
.
Case a)
:d > 0.
In thi
s case
, th
e solutio
n o
t equatio
n (7.184
) an
d it
s derivativ
e hav
e th
e forms
:
vc(t) = AxeS{t + A2eS2t,
(7.197
)
dvc(t)
= sxA{es'1 + s2A2eS21.
dt
(7.198
)
Chapter 7. Transient Analysis
236
By settin
g t = 0n
i thes
e expression
s an
d invokin
g th
e initia
l condition
s (7.185
) an
d
(7.186)
, we obtain
:
A]+A2
s]Al+
= V 0,
(7.199
)
s2A2 = 0
.
(7.200
)
By solvin
g thi
s simpl
e syste
m o
f equations,
we find
th
e value
s fo
r Ai an
d A2.
A, =J2¥±_)
(7.201
)
A2 = JlY°_.
7.202
)
(
s2 — s\
S) ~ S2
By substitutin
g (7.201
) an
d (7.202
) int
o (7.197)
, we arriv
e a
t th
e final
solution
:
vc(t) =V0 (-^—e^'
+ — '-^eA .
(7.203
)
\s2 -s\
si-s2
)
Next conside
r th
e cas
e b)
:d = 0.
In thi
s cas
e th
e solutio
n an
d it
s derivativ
e tak
e o
n th
e forms
:
vc{t) = Axes1 + A2test,
(7.204
)
dvc(t)
sAxesr + A2esl + sA2tesl.
(7.205
)
dt
g t = 0 an
d invokin
g th
e initia
l condition
s (7.185
) an
d (7.186)
, we
Now , by settin
obtain
:
Ai=Vo,
sAx + A2 = 0
,
(7.206
)
(7.207
)
A2 = -sV0.
(7.208
)
vc(0 = V0est(l -st).
(7.209
)
Therefore
, th
e final
solutio
n is
:
Finally
, conside
r th
e cas
e c)
:d < 0
For thi
s case
, th
e solutio
n an
d it
s derivativ
e hav
e th
e forms
:
vc(t) = e~at(Ax co
s(ot + A2 si
n cat),
dt
(7.210
)
CTf
?
= —cre~
(Ai co
scot + A2 si
n cot) + e~°"
( — coA\ si
n cot + coA2 co
scot).
(7.211
)
By settin
g t = 0 an
d by invokin
g agai
n th
e initia
l conditions
, we get
:
V 0 =Ah
(7.212
)
-<JA; + coA2 = 0
(7.213
)
237
7.3. Second-Order Circuits
CO
Thus, th
e final
solutio
n is
:
vc(0 = V0e~at fco
s co
r + — si
n coA .
(7.215
)
It s
i helpfu
lo
t examin
e th
e plot
s of th
e solutio
n fo
r th
e abov
e thre
e differen
t case
s
in orde
ro
t gai
n some feelin
g fo
r it
s behavior
. The plo
tn
i Figur
e 7.13
a correspond
s
to positiv
e discriminant
; an
d t
is
i calle
d th
e overdamped case
. Thi
s mean
s tha
t ohmi
c
losse
s du
eo
t th
e resisto
r do no
t allo
w fo
r an
y oscillatio
n n
i th
e solution
. The plo
tn
i
Figur
e 7.13
b correspond
so
t zer
o discriminant
. Thi
ss
i th
e borderlin
e case
, sometime
s
calle
d critically damped. Thi
s mean
s tha
t losse
s du
e o
t th
e resisto
r ar
e jus
t enoug
h
to preven
t an
y oscillation
. Not
e tha
tn
i genera
l criticall
y dampe
d solution
s deca
y
more rapidl
y tha
n overdampe
d solutions
. The final
plo
t (Figur
e 7.13c
) correspond
so
t
negativ
e discriminant
;t
is
i calle
d a
n underdamped solution
, an
d t
i exhibit
s damped
oscillations. Thi
s mean
s tha
t th
e solutio
n oscillate
s whil
et
i decays
.
Note tha
tn
i eac
h of th
e cases
, th
e respons
e alway
s decays
. Thi
ss
i th
e resul
t of
irreversibl
e losse
s of energ
y n
i th
e resistor
.
EXAMPLE 7.10 Conside
r th
e specifi
c cas
e of th
e circui
t show
nn
i Figur
e 7.1
2 when
L = 1H,/
? = 2(1
, an
d VQ = 10 V. Fin
d th
e expressio
n fo
r th
e voltag
e acros
s th
e
d (c
) when C = 4/
5 F. Compar
e
capacito
r (a
) when C = 4/
3 F, (b
) when C = 1 F, an
the tim
e evolutio
n of th
e capacito
r voltag
e fo
r eac
h of th
e thre
e case
s above
.
From equatio
n (7.193)
, th
e discriminan
t fo
r cas
e (a
)s
i foun
d o
t be
:d = 22/4 —
3/4 = 1/
4 > 0. Thus
, th
e circui
ts
i overdampe
d an
d th
e tw
o root
s ar
e give
n by
equatio
n (7.192)
:
1
si = -1 + 1/
2 = - 12/ s"
;
1
s2 = -1 - 1/
2 = - 32/ s"
.
(7.216
)
The expressio
n fo
r th
e capacito
r voltag
e come
s fro
m equatio
n (7.203)
:
vc(t)
= \0[3/2e~t/2
3r/2
- l/2^~
] V.
(7.217
)
2
The discriminan
t fo
r cas
e (b
) s
i d = 2 /4 — 1 =0 an
d th
e circui
ts
i criticall
y
damped. Bot
h root
s ar
e the
n equa
lo
t negativ
e on
e an
d th
e capacito
r voltag
e ca
n be
foun
d fro
m equatio
n (7.209)
:
vc(t) = 10(
1 + f)e~% V.
(7.218
)
Finally
, fo
r cas
e (c
) we hav
e d = 22/4 - 5/
4 = - 14/ < 0 an
d th
e circui
ts
i
underdamped
. Fro
m equation
s (7.195
) an
d (7.196
) we se
e tha
ta = 1 an
d co = 1/2
.
The expressio
n fo
r th
e capacito
r voltag
e come
s fro
m equatio
n (7.215)
:
vc{t) = 10e~'[cos(f/2
) + 2sin(f/2)
] V.
(7.219
)
Sample value
s of th
e capacito
r voltage
s a
t some selecte
d time
s ar
e give
n n
i Tabl
e
7.1. You ca
n se
e tha
t th
e overdampe
d cas
e decay
s considerabl
y mor
e slowl
y tha
n
the othe
r tw
o cases
. Tha
t deca
ys
i limite
d by th
e smalles
t of th
e tw
o roots
.n
I fact
, when
Chapter 7. Transient Analysis
238
v(t)
v(t)
v(t)
-V,
Time (t)
Figure 7.13: Plot
s for
: (a
) overdampe
d (d > 0)
, (b
) criticall
y dampe
d (d = 0)
, an
d (c
)
underdampe
d (d < 0) solutions
.
7.3. Second-Order Circuits
239
Tabl
e 7.1
: The capacito
r voltage
s vc(0 V a
t specifie
d time
s fo
r th
e abov
e thre
e cases
.
f(s
)
cas
e (a
)
cas
e (b
)
cas
e (c
)
0
1
2
5
10
20
10.0
0
7.9
8
5.2
7
1.2
3
0.10
1
-3
0.68X10
10.0
0
7.3
6
4.0
6
0.40
4
0.00
5
6
0.4
3 X10"
10.0
0
6.7
6
3.0
1
.02
7
3
-.7
4 X 10"
7
-.40X10-
t >5 s
, th
e secon
d ter
m n
i th
e expressio
n fo
r th
e overdampe
d cas
e ca
n essentiall
y be
ignored
. The underdampe
d circui
t clearl
y oscillate
s (goe
s negative
) an
d decay
s most
rapidl
y of al
l th
e cases
.
EXAMPL E 7.1
1 Althoug
h ther
e ca
n neve
r be areal-worl
d circui
t tha
t ha
sno energ
y
losses
, thes
e inevitabl
e losse
s ca
n be reduce
d o
t a ver
y small
, negligibl
e amount
. Thi
s
justifie
s th
e consideratio
n of a circui
t whic
h contain
s onl
y a
n inducto
r an
d a capacito
r
(se
e Figur
e 7.14)
.
This circui
ts
i a particula
r cas
e of th
e circui
t we considere
d earlie
r (R = 0) an
d
is calle
d a
n undampe
d circuit
. Fro
m equatio
n (7.192)
, we ca
n se
e that
:
S\,2
=
J
(7.220)
e comple
x
The discriminan
tn
i thi
s cas
e s
i les
s tha
n zer
o s
o th
e root
s s\ an
d S2 ar
conjugates
. Becaus
e ther
es
i no resistor
, thei
r rea
l part
s ar
e zero
:
(7.221
)
( 7 - 0,
and:
(7.222)
t = 0 /r
^ <
I—°~r°~
i(t
)
vs(t)(j
t <0
t >0
Figur
e 7.14
: An LC circui
t wit
h no resistors
.
240
Chapter 7
. Transient Analysis
From (7.215)
, we find
that
:
v
c(0
(7.223)
— VQ COS cot
From equatio
n (7.223)
, we se
e tha
t th
e respons
e of thi
s circui
t neve
r decays
,t
i
jus
t continuousl
y oscillate
s betwee
n V0 an
d — Vb- The frequenc
y of th
e oscillation
,
,s
i calle
d th
e natural frequency of th
e circuit
. Thi
s s
i becaus
e thi
s
co = 1/yLC
frequenc
y s
i no
t cause
d by a sourc
e (indeed
, ther
e s
i no source)
, bu
t rathe
rt
is
i du
e
to th
e structur
e of th
e circui
t itself
. Thus
, a simpl
e LC circui
t ca
n be considere
d a
s
a generato
r of harmoni
c oscillations
, an
d th
e frequenc
y of thes
e oscillation
s ca
n be
chose
n by th
e appropriat
e selectio
n ofL an
d C. n
I practice
, however
, ther
e ar
e alway
s
some losses
. To maintai
n th
e oscillations
, th
e los
t energ
y shoul
d be continuousl
y
replenishe
d withou
t disruptin
g th
e oscillations
. Thi
s ca
n be achieve
d by usin
g LC
circuit
sn
i combinatio
n wit
h activ
e elements—transistors
.
The expressio
n fo
r th
e curren
t throug
h th
e circui
t ca
n be foun
d fro
m th
e previou
s
expressio
n fo
r th
e voltag
e acros
s th
e capacitor
:
KO
=
C—;—
at
(7.224)
= — COCVQ si
n cot
By usin
g th
e abov
e expression
s fo
r voltag
e an
d current
, we ca
n make a
n inter
estin
g observatio
n concernin
g th
e energ
y store
d n
i th
e capacito
r an
d th
e inductor
. For
the capacito
r we hav
e
2
cot
_ Cv2c(t) _ CVl cos
(7.225
)
We(f) =
whil
e fo
r th
e inducto
r we find:
wm(t) =
Li2(t)
2
CV Q sin
cot
(7.226
)
The voltag
e acros
s th
e capacito
r an
d th
e curren
t throug
h th
e inducto
r a
s wel
la
s
the energ
y store
d n
i th
e capacito
r an
d th
e inducto
r ar
e represente
d by th
e graph
sn
i
Figur
e 7.15
.t
Is
i apparen
t fro
m thes
e graph
s tha
ta
s th
e curren
t increases
, th
e voltag
e
time
time
Figur
e 7.15
: Graph
s o
f voltage
, current
, an
d energ
y n
i th
e LC circuit
.
7.3. Second-Order Circuits
241
acros
s th
e capacito
r decreases
, an
d th
e energ
y transfer
s fro
m th
e electri
c field
of th
e
capacito
r o
t th
e magneti
c field
of th
e inductor
. On th
e othe
r hand
, a
s th
e curren
t
decreases
, th
e voltag
e acros
s th
e capacito
r increases
, an
d th
e energ
y transfer
s fro
m
the magneti
c field
of th
e inducto
ro
t th
e electri
c field
of th
e capacitor
. Thi
s energ
y
exchang
e betwee
n th
e capacito
r an
d th
e inducto
rs
i th
e physica
l mechanis
m of th
e
oscillations
.
The LC circuit
s foun
d practica
l applicatio
n year
s ag
o n
i variou
s electri
c device
s
generatin
g electromagneti
c oscillation
s a
t variou
s frequencies
. The
y performe
d thi
s
dut
y well
, excep
ta
t hig
h frequencie
s when energ
y losse
s du
e o
t radiatio
n an
d some
othe
r imperfection
s appeared
. Nowadays
, thei
r rol
e ha
s bee
n filled
by moder
n circuit
s
which avoi
d th
e us
e of inductor
s (difficult-to-work-wit
h circui
t element
s tha
t ar
e bot
h
expensiv
e an
d har
d o
t miniaturize)
.
7.3.
2
Circuit
s Excite
d by Source
s
Now , we conside
r a second-orde
r circui
t tha
ts
i excite
d by a sourc
e (Figur
e 7.16)
. We
begi
n a
s usua
l wit
h th
e initia
l conditions
, whic
h n
i thi
s cas
e ar
e obvious
:
v(0+
) = v(0_
) = 0,
(7.227
)
i(0+
) = i(O) = 0,
(7.228
)
where v(i) stand
s fo
r th
e voltag
e acros
s th
e capacito
r an
d i(t) s
i th
e curren
t throug
h
all of th
e elements
. By usin
g KVL an
d th
e termina
l relationship
s fo
r th
e inducto
r an
d
resistor
, we obtain
:
diit)
v(t) + L — ^ +Ri(t) = vs(t).
at
Then, by invokin
g th
e expressio
n
(7.229
)
dv(t)
i(t) = C — ^ ,
(7.230
)
at
w e arriv
e a
t th
e followin
g initia
l valu
e problem
: find
th
e solutio
n of th
e equation
:
d2v(t)
LC—Y-l
dt
dv(t)
+RC—-1- + v(t) = vs(t)
dt
Figur
e 7.16
: Second-orde
r circui
t wit
h source
.
(7.231
)
242
Chapter 7. Transient Analysis
subjec
to
t th
e initia
l conditions
:
v(0+
) - 0,
(7.232
)
^ ( 0 +) = 0.
(7.233
)
at
Sinc
e we dea
l her
e wit
h anonhomogeneous differentia
l equation
, it
s complet
e solutio
n
is a su
m of th
e homogeneou
s an
d particula
r solutions
:
v(t) = vh(t) + vp(t\
(7.234
)
W e alread
y kno
w ho
w o
t findth
e homogeneou
s solution
. First
, we must solv
e th
e
characteristi
c equation
:
s2LC + sRC + 1 = 0.
(7.235
)
For th
e sak
e of bein
g specific
, we wil
l assum
e tha
t thi
s characteristi
c equatio
n ha
s
two rea
l an
d differen
t roots
. Thi
s mean
s th
e homogeneou
s solutio
n ha
s th
e form
:
vh(t) = AxeSlt + A2eS2t.
(7.236
)
Now , th
e particula
r solutio
n must be found
. The for
m of thi
s solutio
n depend
s
on th
e natur
e of th
e voltag
e source
. We first
conside
r a dc voltag
e source
:
vs(t) = V0 = constant
.
(7.237
)
In thi
s case
, th
e particula
r solutio
n ha
s th
e physica
l meanin
g of th
e dc steady-stat
e
voltag
e acros
s th
e capacitor
. Therefore
,t
is
i clea
r that
:
vp(t) = Vb-
(7.238
)
Now , th
e complet
e solutio
n o
t differentia
l equatio
n (7.231
) is
:
v(f
) = V0 + Axes'x
+A2eS2t.
(7.239
)
Constant
s A\ an
d A2 remai
n o
t be found
. By differentiatin
g (7.239
) an
d settin
g t = 0,
fro
m initia
l condition
s (7.232
) an
d (7.233
) we find:
V0 + Ai + A2 = 0,
(7.240
)
sxAi + s2A2 = 0.
(7.241
)
These tw
o equation
s ca
n be easil
y solve
d o
t find
A\ an
d A2:
^
A
Ax =
A2 =
,
si
^
-si
.
(7.242
)
(7.243
)
243
73. Second-Order Circuits
Thus, th
e final
solutio
n is
:
S l
J\t -L
„S t
1 + —eSxt
+ ^ —eS2t
.
(7.244
)
V
^ - s2
s2- s{
/
From th
e expressio
n fo
r th
e voltage
, we ca
n easil
y findth
e expressio
n fo
r th
e
curren
t throug
h th
e circui
t (se
e equatio
n (7.230))
:
v(t) = V0
2
/(,
) =CV0 (-12^-e*
\si -s2
+^ ^ e s A .
s2-si
J
(7.245
)
EXAMPLE 7.12 Conside
r th
e sam
e circui
t excite
d b
y a
n a
c voltag
e sourc
e (se
e
Figur
e 7.16)
:
vs(t) = Vms cos(a)t + cj)s).
(7.246
)
In thi
s case
, differentia
l equatio
n (7.231
) ha
s th
e form
:
d2v(t)
dv(t)
LC—rt1
+RC—-1 + v(
0 -Vmscos(a)t +
fa).
(7.247
)
at1
at
The homogeneou
s solutio
n fo
r thi
s equatio
n s
i th
e sam
e a
s befor
e (se
e expressio
n
(7.194))
. The particula
r solutio
n fo
r equatio
n (7.247
) ha
s th
e physica
l meanin
g of th
e
e capacito
rn
i th
e circui
t show
n n
i Figur
e 7.16
. Fo
r
ac steady-stat
e voltag
e acros
s th
thi
s reason
, we shal
l emplo
y th
e phaso
r techniqu
eo
t find
th
e particula
r solution
.
First
, we transfor
m th
e voltag
e (7.246
) o
f th
e a
c sourc
e int
o phaso
r form
:
V5 = Vmsej^.
(7.248
)
f th
e curren
t throug
h th
e circuit
:
Next, we ca
n find
th
e phaso
r /o
Vs
Vs
/ =— =
j —.
Z
R + j(coL~^)
Finally
, we ca
n fin
d th
e phaso
r o
f th
e voltag
e acros
s th
e capacitor
:
/ . .
-x
Vc = - ^/ =Vs s
*
£
_
coC
R + j(<oL-^)
(7.249
)
(7.250
)
From (7.250
) we obtain
:
Vc =Vmce^\
Vmc =Vms
(7.251
)
_L
,
"c = ,
& + (o,
L -o>C'
-L)
2
4>c = 4>s-4>- \ ,
(7.252
)
(7.253
)
where
ioL — -~
<f> = arcta
n
R
^.
(7.254
)
244
Chapter 7. Transient Analysis
By transformin
g th
e phaso
r Vc int
o th
e tim
e domain
, we en
d up wit
h th
e particula
r
solution
:
vp(t) = Vms
coC
yj&
^ =
+ (toL - ±?
I
.
.
co
s lot + fa - $ - V
*
2J
•
(7.255
)
To be specific
, we assum
e tha
t th
e discriminan
ts
i positiv
e and
, consequently
, th
e
homogeneou
s solutio
n ha
s th
e form
:
vh(t) = A{eSit + A2eS2t.
(7.256
)
Then, th
e genera
l solutio
n of equatio
n (7.247
) is
:
v(t) = A^'
+ A2eSlt + Vms
yj&
°c
co
s (cot + fa - <p - ^) .
V
2J
+ (<o
L - X)2
(7.257
)
From (7.257)
, we obtain
:
Z
.
si
n (a>t + fa - <P - ^) .
2
V
2/
l& + (co
L - -L)
(7.258
)
By settin
g t = 0n
i (7.257
) an
d (7.258
) an
d by usin
g initia
l condition
s (7.232
) an
d
(7.233)
, we arriv
e a
t th
e followin
g equation
s fo
r Ai an
d A2:
^Pdt
= sxA,e^
+s2A2es'-t - Vms
-7
k —y>
</ —2 ?)
7; cos(<
A , + A2 = -V^**
,
—
\
*2 + (
^ - i2)
5 ] Al
<f2— ^f
)
+ ,2 A 2 = Vms c4 sin(<f>
vy
»? — ^>
^
+ ( w L_ _ L)2
(7.259
)
( 7 # 2 6)0
From (7.259
) an
d (7.260)
, we deriv
e
- $2
Sin
Vmsi-^r
(^ ~ </>
) ~ T COS(^ - <£)
]
coC
2
(*, - W / ?
+ (WL - i )
V ^ [^ sin(<
k -(/> )£ cos(<
k - (/>)
]
fe-.O^
(7.261
)
2
(7.262
)
+ ^ L - ^)
By substitutin
g (7.261
) an
d (7.262
) int
o (7.257
) we obtai
n th
e explici
t analytica
l for
m
of th
e solution
.
EXAMPL E 7.1
3 Conside
r th
e circui
t show
nn
i Figur
e 7.17
, wher
e th
e sourc
e curren
t
is give
n by
:
is(t) = 10cos(f)A
.
(7.263
)
Assumin
g zer
o initia
l condition
s fo
r th
e curren
t throug
h th
e inducto
r an
d th
e voltag
e
acros
s th
e capacitor
, deriv
e th
e expressio
n fo
r th
e curren
t throug
h th
e inductor
.
245
7.3. Second-Order Circuits
t=0
+
(T)i 8 (t)
U(t)
v(t)jiH
Figure 7.17: A second-orde
r circui
t example
.
B y usin
g KCL, we find:
HO + kit)
+ iR{t) = is{t).
(7.264
)
The voltag
e v(t) acros
s th
e inductor
, capacitor
, an
d resisto
r ca
n be expresse
d n
i term
s
of th
e curren
t throug
h th
e inducto
r as follows
:
v(0 = L
diL(t)
dt
(7.265
)
B y usin
g thi
s expressio
n fo
r th
e voltag
e an
d th
e termina
l relationship
s fo
r capacitor
s
and resistors
, we expres
s ic(t) an
d z#(
0 n
i term
s of kit):
= CL
ic(t) = C— :—
dt
(7.266
)
dt2
v(t) = L diL{t)
R
R dt
iit(t)
(7.267
)
Now , by substitutin
g (7.266
) an
d (7.267
) int
o (7.264
) an
d rearrangin
g th
e terms
, we
deriv
e th
e differentia
l equatio
n fo
r //.(?)•
'
CL
d2iL(t)
dt
2
L diL{t)
R
dt
+ iL(t)
= is(t).
(7.268
)
B y substitutin
g th
e expressio
n fo
r is(t) an
d th
e value
s fo
r R, L, an
d C, we obtai
n
d2idt)
dt2
diL{t)
+ 2 —dt:—
.
+ iL(t) = 10 cos?
.
(7.269
)
By usin
g zer
o initia
l condition
s fo
r th
e curren
t throug
h th
e inducto
r an
d th
e voltag
e
acros
s th
e capacito
r alon
g wit
h (7.265)
, we obtain
:
IL(0+
) = 0,
diL
dt
(0+) = 0.
(7.270
)
(7.271
)
Chapter 7. Transient Analysis
246
Expression
s (7.269)
, (7.270)
, an
d (7.271
) constitut
e th
e initia
l valu
e proble
m fo
r iL(t).
To solv
e thi
s proble
m we first
conside
r th
e characteristi
c equatio
n fo
r (7.269)
:
s2 + 2s + 1 = 0.
(7.272
)
si = s2 = - I-s1.
(7.273
)
From thi
s equatio
n we find:
Consequently
, th
e transien
t (free
) respons
e ha
s th
e form
:
iLh(t) = Ale-'+A2te-t.
(7.274
)
Next, we shal
l us
e th
e phaso
r techniqu
e o
t findth
e particula
r solutio
n of equatio
n
(7.269)
.t
Is
i clea
r that
:
/, = 10 A,
(7.275
)
Yin = -+j + 2 = 2U.
(7.276
)
j
From (7.275
) an
d (7.276
) we find
th
e phaso
r V" of th
e voltag
e acros
s th
e inductor
:
V = —
= 5 V.
(7.277
)
* in
Now , we ca
n find
th
e phaso
rli of th
e curren
t throug
h th
e inductor
:
iL = ^-f
= -. = ~j5 A.
(7.278
)
By usin
g (7.278)
, we ca
n writ
e th
e particula
r solutio
n iLp(t) fo
r (7.269
)a
s follows
:
kp(0
= 5 co
s (t - j)
= 5 si
n t A.
(7.279
)
By combinin
g (7.274
) an
d (7.279)
, we obtai
n th
e tota
l response
:
iL(t) = 5 sin
? +Axe~l + A2te~''.
(7.280
)
To find
A\ an
d A2, we us
e th
e initia
l condition
s (7.270
) an
d (7.271)
, whic
h yield
:
iL(0+) = Ax = 0,
(7.281
)
^ ( 0+ ) = 5 + A2 - A] = 0.
at
(7.282
)
By takin
g int
o accoun
t (7.281
) ni (7.282)
, we obtain
:
A2 = - 5
.
(7.283
)
By substitutin
g (7.281
) an
d (7.283
) int
o (7.280)
, we ge
t th
e final
resul
t
iL{t) = 5 si
n t - 5te~' A.
(7.284
)
247
7A. Transfer Functions and Their Applications
7.4
Transfe
r Function
s an
d Thei
r Application
s
It s
i clea
r fro
m th
e previou
s discussio
n tha
t th
e most involve
d par
t of transien
t analysi
s
is th
e derivatio
n of th
e differentia
l equatio
n fo
r th
e appropriat
e circui
t variable
. Thi
s
difficult
y ca
n be circumvente
d by usin
g th
e notio
n of a transfe
r function
. Thi
s notio
n
is centra
lo
t th
e syste
m an
d signa
l area
s an
d t
i als
o permeate
s many differen
t branche
s
of electrica
l engineering
. As fa
r as transien
t analysi
s s
i concerned
, th
e machiner
y of
the transfe
r functio
n allow
s one o
t avoi
d completel
y th
e derivatio
n of differentia
l
equation
s an
d o
t solv
e transien
t problem
s by usin
g mostl
y algebrai
c manipulation
s
on phasors
.
B y definition
, transfe
r function
s ar
e th
e ratio
s of outpu
t o
t inpu
t signals
.I
n th
e
case of electri
c circuits
, th
e inpu
t ca
n be define
d as a sourc
e whic
h drive
s an electri
c
circuit
, whil
e th
e outpu
ts
i th
e circui
t variabl
e whic
h we ar
e intereste
d n
i calculating
.
For example
, n
i th
e cas
e of th
e RL circui
t show
n n
i Figur
e 7.18a
, th
e inpu
t s
i th
e
voltag
e source
, whil
e th
e outpu
t s
i th
e curren
t throug
h th
e circuit
. So, th
e transfe
r
functio
n is
:
H(t) =
(7.285)
vs(t)'
In th
e cas
e of th
e RC circui
t show
n n
i Figur
e 7.18b
, th
e inpu
ts
i th
e curren
t source
,
is(t), whil
e th
e outpu
t s
i th
e voltage
, v0(t), acros
s th
e capacitor
. Thus
, th
e transfe
r
functio
n is
:
Voit)
(7.286
)
H(t)
hit) *
Transfe
r function
s ar
e generall
y use
d no
tn
i th
e tim
e domai
n bu
t rathe
rn
i th
e frequenc
y
domain
.f
I we conside
r ac excitation
, the
n we ca
n defin
e th
e transfe
r functio
n as th
e
rati
o of th
e outpu
t phaso
r o
t th
e inpu
t phasor
. I
n thi
s case
, expression
s (7.285
) an
d
(7.286
) become
, respectively
:
H(j*o) =
H(J*»)
(a)
Vo(jo>)
1
1
Z(jco)
R + jwL'
1
Y(Jv)
^+j(oC'
(b)
Figur
e 7.18
: Two first-orde
r circuits
.
(7.287
)
(7.288
)
Chapter 7. Transient Analysis
248
If th
e abov
e circuit
s ar
e drive
n by source
s of comple
x frequenc
y s
, the
n th
e transfe
r
functio
n H wil
l be a functio
n ofs a
s wel
l an
d expression
s (7.287
) an
d (7.288
) ca
n be
writte
n a
s follows
:
1
n{S) - —-
!
-
Z(s) ' R + sL'
1
- +sC
*'> - h -
(7.289
)
(7.290
)
The circuit
s show
n n
i Figur
e 7.18
a an
d b hav
e bee
n discusse
d n
i Example
s 7.
5 an
d
7.7
, respectively
, an
d we hav
e reache
d th
e followin
g conclusion
: zero
s of Z(s) an
d
Y(s) ar
e th
e exponent
s of transien
t response
s whil
e th
e value
s of Z(s
) an
d Y(s) a
t th
e
frequenc
y of excitatio
n determin
e th
e steady-stat
e responses
. By usin
g expression
s
(7.289
) an
d (7.290)
, thi
s conclusio
n ca
n be rephrase
d n
i th
e followin
g way: pole
s of
the transfe
r function
s ar
e exponent
s of fre
e (transient
) responses
, whil
e th
e value
s of
the transfe
r function
s a
t th
e excitatio
n frequenc
y determin
e steady-stat
e responses
.
This conclusio
n applie
so
t circuit
s wit
h mor
e tha
n on
e energ
y storag
e element
.
Indeed
, le
t us recal
l th
eRLC serie
s circui
t discusse
dn
i Exampl
e 7.12
. Fro
m expressio
n
(7.250)
, we ca
n find
th
e transfe
r function
:
H(jco) = £ =
,f
*
,
Vs
R+
ja)L+-c
(7.291
)
If we conside
r thi
s transfe
r functio
n a
s th
e functio
n of comple
x frequenc
y s, we arriv
e
at:
"(s) =irtrr
=TLC^RCTI-
(7292)
-
Now , we agai
n observ
e that
:
The values of the transferfunctions at the frequency of excitation, s = ja>, completely
determine the ac steady- state response of linear electric circuits, while the poles of the
transfer function coincide with the roots of the characteristic equation (see (7.235))
and, consequently, they determine the form of the transient (free) response.
The state
d fac
ts
i ver
y genera
l ni natur
e an
d s
i extensivel
y use
d n
i linea
r syste
m
theory
. The abov
e discussio
n suggest
s tha
t th
e machiner
y of th
e transfe
r functio
n ca
n
be ver
y powerfu
ln
i th
e analysi
s of transient
s n
i electri
c circuits
. Thi
s analysi
s ca
n
procee
d a
s follows
. A circui
t variable
, whic
h we ar
e intereste
d in
, ca
n be identifie
d
as th
e output
. By usin
g th
e phaso
r technique
, we ca
n find
th
e transfe
r functio
n H(jco)
for thi
s variabl
e and
, consequently
, th
e a
c steady-stat
e response
. Then
, we ca
n find
the pole
s of th
e transfe
r functio
n H(s) whic
h determin
e th
e for
m of th
e transien
t
(free
) response
. The unknow
n constant
s n
i thi
s fre
e respons
e shoul
d be determine
d
fro
m th
e initia
l condition
s fo
r th
e chose
n circui
t variable
. The describe
d approac
h s
i
algebrai
c n
i nature
;t
i completel
y avoid
s th
e derivatio
n an
d solutio
n of differentia
l
equation
s an
d t
i full
y exploit
s th
e machiner
y of th
e phaso
r technique
.
7.4. Transfer Functions and Their Applications
249
The assertio
n tha
t th
e pole
s of th
e transfe
r functio
n coincid
e wit
h comple
x
frequencie
s of th
e fre
e (transient
) respons
e ca
n be n
i genera
l explaine
d by usin
g th
e
followin
g reasoning
.
Conside
r a circui
t whic
h s
i excite
d by a voltag
e sourc
e vs(t) wit
h comple
x
frequenc
y s. We ar
e intereste
d n
i a voltag
e vk(t) acros
s th
e kth branch
. Accordin
g o
t
the definitio
n of th
e transfe
r function
, we have
:
Vk = Hk(s)Vs,
(7.293
)
where Vs an
d Vk ar
e th
e phasor
s of th
e voltag
e sourc
e an
d th
e voltag
e acros
s th
e kth
branch
, respectively
.
Now , we shal
l reduc
e th
e pea
k valu
e of th
e voltag
e sourc
e (and
, consequently
,Vs)
to zero
. Accordin
g o
t th
e las
t expression
, Vk ca
n be nonzer
o onl
y fo
r suc
h comple
x
.t
I si clea
r tha
t suc
h comple
x frequencie
s ar
e th
e
frequencie
s s tha
t Hk(s) = oo
pole
s of th
e transfe
r function
.t
Is
i als
o clea
r tha
t fo
r thes
e comple
x frequencies
, th
e
"sourceless
" voltag
e vk(t) acros
s branc
h numbe
rk ca
n exist
. Sinc
e thi
ss
i a sourceles
s
voltage
,t
i ha
s th
e physica
l meanin
g of fre
e (transient
) response
.
It s
i interestin
g o
t not
e tha
t by usin
g th
e transfe
r functio
n we ca
n als
o find
th
e dc
steady-stat
e response
. For example
, conside
r th
e RLC serie
s circui
t show
n n
i Figur
e
7.16
.n
I th
e cas
e of vs(t) = V0 = constant
, fro
m (7.292
) we derive
:
vp(t) = H(0)vs(t) = V0,
(7.294
)
which correspond
so
t th
e resul
t give
n ni (7.238)
.
Next, we shal
l demonstrat
e th
e transfe
r functio
n approac
h by considerin
g th
e
followin
g examples
.
EXAMPLE 7.14 Recal
l th
eRLC paralle
l circui
t considere
d n
i Exampl
e 7.1
3 (show
n
in Figur
e 7.17)
, fo
r whic
h we hav
e foun
d th
e inducto
r curren
t by derivin
g an
d solvin
g
a second-orde
r differentia
l equation
. Now we sho
w that
, by usin
g th
e transfe
r functio
n
approach
, we ca
n arriv
e a
t th
e sam
e resul
t withou
t havin
g o
t perfor
m an
y calculus
.
From th
e definitio
n of th
e transfe
r functio
n an
d th
e curren
t divide
r rule
, we
obtain
:
H(s) ='-£=
/,
„ ,:; ,„ = , ,vi , ,
YL + YC + YR
-L+sC+l
(7.295
)
By substitutin
g th
e value
s R = 0.
5 Cl, L = 1 H an
d C = 1 F an
d makin
g simpl
e
transformations
, we derive
:
H(s) = -.
1
s
-
± +s + 2
= -02—
s
1
.
s + 2s+
l
(7.296
)
Next, we fin
d th
e pole
s of th
e transfe
r function
:
s2 + 2s + 1 = 0,
(7.297
)
l
5, = s2 = -1 s~.
(7.298
)
250
Chapter 7
. Transient Analysis
Consequently
, th
e fre
e (transient
) respons
e ha
s th
e form
:
iLh(t) = Axe~f
+A2te~f.
(7.299
)
To find
th
e force
d (a
c stead
y state
) response
, we evaluat
e H(s) a
ts = j • 1
:
#0" • i)
1
_ ! _ . ;,
/2 + 2; + l 2j
2"
(7.300)
From (7.275
) an
d (7.300)
, we obtain
:
IL=H(j-l)Is
J
= - ^ -01 = -j5
A.
(7.301
)
By transformin
g phaso
rII int
o th
e tim
e domain
, we en
d up wit
h th
e force
d response
:
*L;,(
0 = 5 co
s (
t — — j = 5 si
n t A.
(7.302
)
By combinin
g (7.299
) an
d (7.302)
, we obtai
n th
e tota
l response
:
iL(t) = 5sh
U +A\e~l + A2te~',
(7.303)
which coincide
s wit
h (7.280
) a
st
i must
. What s
i lef
ts
io
t us
e th
e initia
l condition
s
(7.270
) an
d (7.271
)n
i orde
ro
t find
A\ an
d A2. However
, thi
s par
t of th
e solutio
n
proces
ss
i literall
y th
e sam
e a
s before
. Tha
ts
i why t
is
i omitte
d here
.
It s
i clea
r fro
m th
e abov
e exampl
e tha
t th
e transfe
r functio
n approac
h s
i purel
y
algebrai
cn
i natur
e an
d completel
y avoid
s th
e derivatio
n of an
y differentia
l equations
.
EXAMPL E 7.1
5 Conside
r th
e electri
c circui
t show
n n
i Figur
e 7.19
. Thi
s circui
t
contain
s tw
o energ
y storag
e element
s (capacitor
s C\ an
d C2); tha
ts
i why thi
s s
ia
s th
e capacito
rC2second-orde
r circuit
. We want o
t fin
d th
e voltag
e v2(t) acros
First
, we shal
l fin
d th
e transfe
r functio
n H(s), whic
h n
i ou
r cas
es
i define
d a
s th
e
ratio
of th
e phaso
r V2 of th
e voltag
e v2(t) o
t th
e phaso
r Vs of th
e sourc
e voltage
:
V2
H(s) = -±.
t°
6vs (t)
R
K)
t
R
1<
(7.304)
\,(t)y\2(\)
R
2
:v + 9 (t)
Figur
e 7.19
: A second-orde
r circui
t example
.
7A. Transfer Functions and Their Applications
251
To find
V2, we characteriz
e eac
h branc
h o
f th
e circui
t by impedance
. Ther
e ar
e thre
e
i th
e circui
t an
d thei
r impedance
s a
s function
s o
f comple
x frequenc
y s ar
e
branche
sn
give
n by th
e followin
g expressions
:
Z = R,
zx=Rx
+ -^,
Z2=R2 + -U
(7.305
)
__
zxz.2
2
—zxlz—^= -zxz + z-2z + —
l\ +Z2
Z\ +Z2
(7.306
)
sCx
sL2
By usin
g th
e abov
e impedances
, we deriv
e th
e followin
g expressio
n fo
r th
e equivalen
t
inpu
t impedance
:
Zec,=Z+
W e nex
t find
th
e phaso
rI o
f th
e tota
l curren
ti(t):
z, + z
v
2
/ - —s =Vs
.
Z eq
Z\Z +Z2Z iZ\Z2
(7.307
)
By usin
g th
e curren
t divide
r rule
, we derive
:
I2 = I———
-Vs
.
(7.308
)
V2=I2—
= Vs
—
.
sC2
sC2(Z\Z + Z2Z + Z\Z2)
From (7.304
) an
d (7.309)
, we conclude
:
(7.309
)
1
zx+z2
Now , we ca
n obtai
n th
e expressio
n fo
rV2:
zxz + z2z + zxz2
H(S) = r s y y ^ ' + y y y
(7310
)
st2(Z\Z + Z2Z + Z\Z2)
By substitutin
g (7.305
) int
o (7.310
) an
d b
y usin
g simpl
e algebrai
c transformations
,
w e derive
:
KS)
s2ClC2(RRl
+RR2 +R{R2) + s(C2R + CXR + CXRX +C2R2) + 1
'
(7.311
)
Now , we ca
n find
th
e pole
s of th
e transfe
r function
. The
y ar
e th
e root
s of th
e equation
:
s2C{C2(RRi + RR2 + RYR2) + s{C2R + CXR + CXRX + C2R2) + 1 = 0.
(7.312
)
To b
e specific
, we wil
l assum
e tha
t thi
s characteristi
c equatio
n ha
s tw
o rea
l an
d
distinc
t root
s sx an
d s2. Thi
s mean
s tha
t th
e fre
e (transient
) respons
e ha
s th
e form
:
v2h(t) = Axe^+A2es^.
(7.313
)
In th
e case
s when th
e root
s o
f th
e characteristi
c equatio
n ar
e rea
l an
d identica
lo
r
comple
x an
d conjugate
, th
e fre
e respons
e wil
l b
e give
n b
y expression
s simila
ro
t
(7.204
) an
d (7.210)
, respectively
.
252
Chapter 7. Transient Analysis
Next, we assum
e tha
t th
e circui
ts
i excite
d by a
n a
c voltag
e source
:
v.v(0 = Vmscos(cot + <J>S).
(7.314
)
This mean
s tha
t
V, = Vmse^.
(7.315
)
To find
th
e force
d (a
c stead
y state
) response
, we evaluat
e H(s) a
ts — jco:
H(ja>) = \H(ja>)\ej4>".
(7.316
)
From (7.304)
, (7.315)
, an
d (7.316
) we find:
V2 = Vms\ft(j°>)\eJ{,h++H).
(7.317
)
By transformin
g phaso
r V2 int
o th
e tim
e domain
, we en
d up wit
h th
e force
d response
:
v2p(t) = Vms\H(jco)\ cos((ot + <f>s + <j>H).
(7.318
)
By combinin
g (7.313
) an
d (7.318)
, we obtai
n th
e tota
l response
:
r + <f>s + <f>H).
v2(t) = A{es,t + A2eS2t + Vms\H(joj)\ cos(w
(7.319
)
The previou
s discussio
n clearl
y illustrate
s ho
w th
e machiner
y of th
e transfe
r functio
n
can be use
d ni orde
ro
t find
th
e ful
l respons
e of th
e electri
c circuit
.
The unknow
n constant
sn
i (7.319
) shoul
d be determine
d fro
m th
e initia
l condi
tion
s fo
r v2(t). We assum
e tha
t befor
e switchin
g (t < 0) th
e capacitor
s C\ an
d C2
were no
t charged
. Thi
s mean
s tha
t
v.(0
+)
(7.320
)
v2(0+)
(7.321
)
Expressio
n (7.321
) give
s us on
e initia
l conditio
n fo
r v2(t). n
I orde
ro
t fin
d th
e secon
d
initia
l conditio
n fo
r v2(t), we conside
r th
e circui
t show
n ni Figur
e 7.1
9 at the initial
instant of time t = 0.
By usin
g (7.320
) an
d (7.321)
, thi
s circui
t ca
n be redraw
n ni th
e way show
n n
i
Figur
e 7.20
.
R
i(0
+)
rWWf—
W t j2(0+)t
(0+)6
R,
Figure 7.20
: The circui
ta
tt = 0.
253
7A. Transfer Functions and Their Applications
This s
i becaus
e a
t tim
e t = 0, th
e uncharge
d capacitor
s ca
n be replace
d by shor
t
circuits
. The circui
t show
n n
i Figur
e 7.2
0 ca
n be interprete
d a
s a dc resistiv
e circuit
,
which make
s it
s analysi
s quit
e simple
.t
I si clea
r fro
m th
e abov
e circui
t tha
t
i(0+) = 4 ^ *
(7.322)
where
^
^
Rea
eq = R+
R\Ri
R\R +RoR +R1R7
— L - 4r = —
„
——■
Ri+R2
R\+R2
,m ^^^
x
(7.323
)
By usin
g th
e curren
t divide
r rule
, we fin
d
i2(0+) = i (+0) - 4 | —
.
(7.324
)
K\ -r K2
From (7.322)
, (7.323)
, an
d (7.324)
, we obtain
:
—
.
/2(0+) = vs(0+)
2
}
v
R]R + R2R + RiR2
Now , accordin
g o
t th
e circui
t show
n n
i Figur
e 7.19
, we have
:
(7.325
)
i2(t) = C2——.
at
From (7.325
) an
d (7.326)
, we find:
(7.326
)
^ ( 0 +) =
^ ±W
.
(7.327)
dt
C2{R\R + R2R + R1R2)
Thus, we hav
e determine
d th
e secon
d initia
l conditio
n fo
r v2(t). The way we hav
e
achieve
d thi
ss
i quit
e genera
ln
i natur
e an
d ca
n be summarize
d a
s follows
.
To find the initial values for electric currents in a circuit, we redraw this circuit
for the initial instant of time t = 0by replacing uncharged capacitors by short
circuits and "currentless" inductors by open circuits. As a result, we obtain a dc
resistive circuit which can be used to find initial values for all currents and voltages.
Then, by employing terminal relationshipsfor capacitors and inductors, we determine
additional initial conditions for energy storage elements.
In the case of nonzero initial conditions, capacitors and inductors should be
"replaced" by dc voltage and current sources, respectively. As a result, we arrive
at dc resistive circuits with several sources. Analysis of these circuits yields initial
values for all currents and voltages.
To finish
th
e solutio
n of ou
r problem
, we shal
l us
e initia
l condition
s (7.321
) an
d
(7.327
)o
t find
th
e unknow
n constant
s Ax an
d A2 n
i (7.319)
. Fro
m (7.319)
, (7.321)
,
and (7.327
) we obtain
:
A{+A2
= -Vms\H(j(o)\
fa),
cos(<f
c +
.A
,
,
,
M i +s2A2 = coVms\H(jco)\ sm(4>s + 4>H) + r
(7.328
)
VmvR\ co
s d>c
,
)
p p 1 ' p ^ p p , (7.329
\^2\K\K
~r K2K ~r K\K2)
254
Chapter 7
. Transient Analysis
where we hav
e use
d th
e fac
t tha
t accordin
g o
t (7.314
) Vy(0
) = Vms co
s <j)s. Simulta
neous equation
s (7.328
) an
d (7.329
) ca
n be solve
d fo
rA\ an
d A2. By substitutin
g th
e
foun
d value
s fo
r A! an
d A2 int
o (7.319)
, we en
d up wit
h th
e final
expressio
n fo
r v2(t).
This complete
s th
e solutio
n of th
e problem
.
■
EXAMPLE 7.16 Conside
r th
e circui
t show
n n
i Figur
e 7.21
.t
Is
i assume
d tha
t th
e
initia
l voltage
s acros
s th
e capacitor
s C\ an
d C2 ar
e equa
lo
t zero
. We inten
d o
t find
the voltag
e vi(
0 acros
s resisto
rR\ fo
r th
e followin
g value
s of circui
t parameters
:
6
6
6
6
Ci = 1(T
F, R\ = 10
ft,C2 - 10~
F, R2 = 0.
5 X 10
ft.
By usin
g th
e definitio
n of transfe
r functio
n an
d th
e voltag
e divide
r rule
, we
derive
:
R^
H(s) =
^ 2' Tr7-
1
(7.330)
^ic;
By usin
g simpl
e algebrai
c transformations
, we find:
H(s) =
(7.331
)
R, + 4+ sR C + ]
sC\
2
2
sRlCl(sR2C2 + 1
)
s2RlR2ClC2 + s(RiCi + R2C2 + R2CX) + 1'
By substitutin
g th
e value
s of C\, C2, R\, an
d R2 int
o (7.332)
, we obtain
:
H(s) =
&(\
H{S)
s(s
=
f+As
+ 2)
+ 2-
(7.332
)
(7.333)
Next, we fin
d th
e pole
s of th
e transfe
r function
:
s2 + As + 2 = 0,
(7.334
)
-0.5
9 s- 1,
(7.335
)
1
s2 = - 2- \fl = -3.4
1 s"
.
(7.336
)
j, = -2 + V ^=
t=0
I H W W t
+ Mt)( T ) v s (t) = 50cos(t)
c2
Figur
e 7.21
: A second-orde
r circui
t example
.
7.4. Transfer Functions and Their Applications
255
Consequently
, th
e fre
e transien
t respons
e ha
s th
e form
:
vh(t) = Ale-°-59t+A2e-3AU.
(7.337)
To find
th
e force
d (a
c stead
y state
) response
, we evaluat
e H(s) a
ts = j • 1:
H(J
' l) = p+4j + 2 = TT4
7 = " 1 7 '"
41
H(j) = 0.54^'
°.
( 7 3 3)8
(7.339
)
, we obtain
:
From (7.330)
, (7.339)
, an
d th
e fac
t tha
tVs = 50
41
Vi = 0.54^'
° • 50 = 21ej4l° V.
(7.340
)
By transformin
g phaso
r Vx int
o th
e tim
e domain
, we en
d up wit
h th
e force
d response
:
vlp(t) = 21cos(t + 41°) V.
(7.341)
By combinin
g (7.337
) an
d (7.341)
, we obtai
n th
e tota
l response
:
vi(
0 = 27cos(
r + 41°
) +A{e~Q-59t + A2e~3Alt.
(7.342
)
To find
A\ an
d A2, we nee
d th
e initia
l condition
s vi(0+
) an
d ^-(0+)
. To thi
s end
, we
conside
r th
e circui
t show
n n
i Figur
e 7.2
1 a
t th
e initia
l instan
t of tim
e t = 0. Sinc
e a
t
d C2 ar
thi
s instan
t of tim
e voltage
s acros
s th
e capacitor
s C\ an
e equa
lo
t zero
, thes
e
capacitor
s ca
n be replace
d by short-circui
t branches
. Thi
s lead
so
t th
e circui
t show
n
in Figur
e 7.22
. Fro
m thi
s figure
, we find:
V!(0
,
+) = 50V
(7.343
)
6
i(0+
) - 50/10A.
(7.344
)
,
It s
i clea
r tha
t ?'(0+
) = «c,(0+
) = ic2(0+). Consequently
6
*c, (0+
) = ic
) = 50 •10~
.
2(0+
(7.345
)
Next, we conside
r agai
n th
e circui
t show
n ni Figur
e 7.2
1 an
d writ
e KVL fo
r th
e loo
p
consistin
g of th
e source
, capacito
r C\, resisto
rR\, an
d capacito
rC2:
50 co
st = vcXt) + vM) + vcXt).
i(0)
50
6
R
^AAA^
+ v1 (0) -
Figur
e 7.22
: The previou
s circui
ta
tt = 0.
(7.346
)
Chapter 7
. Transient Analysis
256
By differentiatin
g th
e las
t equation
, we obtain
:
- 5 0 s i,n = * £W + * ^>
a/
a
?
The las
t equatio
n ca
n be rewritte
n a
s follows
:
+ * S «.
a/
__ icx(t) , rfvj(0 i'c
(0
2,
- 50sin
/ - - ^^ + — ^ + - ~ .
^
Ci
at
C2
By settin
g r = 0 an
d usin
g (7.345)
, we derive
:
.347
)
(7
(7.348
)
^ - ( 0)
+ = -lOOV/s
.
(7.349
)
at
Havin
g foun
d th
e initia
l condition
s (7.343
) an
d (7.349)
, we ca
n determin
e A{ an
d A2
in (7.342)
. Indeed
, th
e abov
e initia
l condition
s lea
d o
t th
e equations
:
50 = 27 co
s 41
° +Ax + A2,
(7.350
)
- 10
0 = -7
2 sin41
° - 0.59A
i - 3.41A
2.
(7.351
)
By solvin
g thes
e equation
s fo
r A\ an
d A2, we find:
Ax
= 6.64
,
A2 = 22.98
.
(7.352
)
From (7.352
) an
d (7.342)
, we finally
obtain
:
a59r
341r
v{(t) = 27cos(
f + 41°
) + 6.64e"
+ 22.98e~
V.
(7.353
)
To bette
r appreciat
e th
e benefit
s of th
e transfe
r functio
n approach
,t
is
i suggeste
d tha
t
the reade
r tr
y o
t solv
e thi
s proble
m by usin
g th
e differentia
l equatio
n approach
.
7.5
Impuls
e Respons
e an
d Convolutio
n Integra
l
U p unti
l now, we hav
e discusse
d electri
c circuit
s whic
h contai
n onl
y dc or a
c sources
.
Usin
g th
e metho
d of undetermine
d coefficients
, we ca
n als
o analyz
e many circuit
s tha
t
have source
s describe
d by othe
r mathematica
l functions
. But
, ho
w ca
n we calculat
e
a respons
e of a
n electri
c circui
to
t a
n arbitrary source
? Thi
s ca
n be accomplishe
d by
usin
g a specia
l techniqu
e calle
d th
e convolution integral technique. The convolutio
n
integra
l give
s th
e respons
e of a circui
to
t an
y voltag
e or curren
t source
.n
I thi
s case
,
ther
e s
i no separatio
n of th
e respons
e int
o transien
t an
d steady-stat
e component
s
becaus
e wit
h a
n arbitrar
y sourc
e ther
e may no
t be an
y steady-stat
e respons
eo
t spea
k
of.
To introduc
e th
e convolutio
n integral
, we shal
l first
conside
r a specifi
c circuit
,
namel
y th
e RL circui
t use
d n
i ou
r earlie
r discussions
. Late
r we wil
l sho
w ho
w o
t
deriv
e th
e convolutio
n integra
l fo
r an
y linea
r circuit
.
257
7.5. Impulse Response and Convolution Integral
7.5.1
Convolution Integral for an RL Circuit
The differentia
l equatio
n fo
r th
e RL circui
t show
n n
i Figur
e 7.
6 was give
n n
i (7.62
)
as:
L ~ +Ri(t) = vs(t),
(7.354
)
dt
where we hav
e omitte
d th
e subscrip
t "L
" fo
r th
e sak
e of convenience
. Multiplyin
g
(7.354
) by e^f an
d dividin
g by L yields
:
Rtdi(t)
R
/?,.
,N
eTJ
^^r-r,
ezt-jr
+ je-L'iit) = —v,(0
(7.355
)
dt
L
L
By usin
g th
e rul
e fo
r th
e differentiatio
n of th
e produc
t of tw
o functions
, thi
s equatio
n
can be rewritte
n as
:
-[ei%)]
= ^-vs(t).
(7.356
)
dt
L
By integratin
g th
e las
t equatio
n fro
m 0o
t ,f an
d by usin
g initia
l conditio
n (7.58)
, we
obtain
:
rt
et'iit) =
-r
/ —vs(r)dr.
Jo L
(7.357
)
Next, we multipl
y bot
h side
s of (7.357
) by e~^t whic
h yields
:
i(t) =
/
Jo
L
vs(T)dr.
(7.358
)
Expressio
n (7.358
)s
i th
e convolutio
n integra
l fo
r ou
r particula
r circuit
.t
I wil
l giv
e
the respons
e (current
) of th
e circui
t fo
r an
y voltag
e source
. Indeed
, sinc
e vs(t) s
i
nowher
e define
d ni thi
s problem
, we ca
n se
e tha
t th
e convolutio
n integra
l wil
l wor
k
for arbitrary sources.
In general
, integral
s whic
h fit
th
e followin
g for
m
i(t) =
h(t- T)vs(r)dr
(7.359
)
Jo
are calle
d convolutio
n integrals
.
In equatio
n (7.359)
,t s
i know
n a
s observation time, r s
i integration time, an
d
h{t — T) s
i calle
d th
e kernel of convolution. n
I th
e expressio
n (7.358)
, th
e kerne
l of
convolutio
n s
i
h(t -
T) = —
.
(7.360
)
Expressio
n (7.358
) was derive
d by usin
g th
e integratin
g facto
r techniqu
e whic
h s
i
applicabl
e o
t solvin
g first-order
differentia
l equations
.
258
Chapter 7
. Transient Analysis
EXAMPLE 7.17 Conside
r th
e RL circui
tn
i Figur
e 7.
6 whic
h s
i excite
d by th
e non
periodi
c sourc
e
v , W = |2 V >
(7.361
)
f > ls
l an
d L = 2 H. Use th
e convolutio
n integra
lo
t find
a
n expressio
n fo
r th
e
LetR = 1 f
time dependenc
e of th
e curren
t throug
h th
e inductor
.
Sinc
e th
e voltag
e sourc
es
i specifie
d differentl
y fo
r tw
o differen
t tim
e intervals
,
w e must conside
r separatel
y thes
e tw
o intervals
. The kerne
l of convolutio
n s
i th
e sam
e
for bot
h tim
e intervals
, sinc
e t
i depend
s onl
y on th
e circui
t structur
e and
, accordin
g
to equatio
n (7.360)
,t
is
i equa
lo
t
h(t-r)
= e~^~r)/2.
(7.362
)
For time
s betwee
n zer
o an
d on
e second
, we find
th
e convolutio
n integra
l (7.358
)o
t
yield
:
ft e-kt~r)
ft
= /
2rdr = e~t/2 / er/2T^T.
Jo
2
J0
This integra
l ca
n be evaluate
d by usin
g integratio
n by parts
, whic
h lead
so
t
kit)
iL{i) = 2
f - 4 +4e~t/2.
(7.363
)
(7.364
)
For time
s greate
r tha
n on
e second
, th
e convolutio
n integra
l becomes
:
ft e-\Sf-r)
rx
k(t) = / —
2dr + e~t/2 / eT/2Tdr.
(7.365
)
The first
integra
l ca
n be foun
d n
i a straightforwar
d way
, whil
e th
e secon
d integra
l ca
n
be calculate
d n
i th
e sam
e way a
s fo
r th
e first
tim
e interval
. Thus
, th
e final
answe
r ca
n
be expresse
d as
:
. ,.
lL(t)
=
f
2/ - 4 +4e~^2 A,
\ 2(
1 -2e^2)
+4e~^2 A,
0 <t < 1 ,
s
/> 1.
s
„,
.
n
( ? 3 6 )6
Ther
e ar
e tw
o importan
t point
s o
t not
e here
. First
, J'L(I) = /L(
1 +>) a
s t
i m u st be
accordin
g o
t th
e continuit
y of electri
c curren
t throug
h a
n inductor
. Second
,ii(t) -^
2A a
s t —> cot Thi
s als
o shoul
d be expecte
d becaus
e th
e inducto
r wil
l ac
t lik
e a shor
t
circui
t afte
r th
e transien
t die
s awa
y (tha
t is
,a
t dc steady-state)
.
Later
, we shal
l sho
w tha
t th
e convolutio
n integra
l ca
n be derive
d fo
r an
y linea
r
circuit
. To achiev
e this
, we shal
l nee
d some fact
s concernin
g th
e uni
t ste
p functio
n
and th
e uni
t impuls
e function
. Thes
e fact
s wil
l als
o be instrumenta
l ni arrivin
g a
t th
e
physica
l interpretatio
n of th
e convolutio
n kerne
l (7.360)
.
The unit step function, denote
d a
s u{t), s
i define
d a
s follows
:
_
U(t)=
{
/ 0, if
f < 0,
1
, i f r >.0
n „ „
( 7 3 6 )7
This expressio
n mean
s tha
t when th
e argumen
t (expressio
n ni parentheses
) of th
e uni
t
ste
p functio
n s
i les
s tha
n zero
, thi
s functio
n s
i equa
lo
t zero
, an
d when it
s argumen
t
is greate
r tha
n zero
, th
e uni
t ste
p functio
n s
i equa
lo
t 1 (se
e Figur
e 7.23)
. The uni
t
7.5. Impulse Response and Convolution Integral
259
u(t
)
1
Figur
e 7.23
: Grap
h o
f th
e uni
t ste
p function
.
ste
p functio
n s
i ver
y usefu
l fo
r mathematica
l description
s of switchin
g of sources
.
For example
, fi a sourc
es
i define
d a
s
vs{t)u{t) =
0,
vs(t),
fit < 0,
fit > 0,
(7.368
)
thi
s mean
s tha
t th
e sourc
e s
i "off
" when t < 0 an
d th
e sourc
e si "on
" when t > 0
(se
e Figur
e 7.24)
. The uni
t ste
p functio
n ca
n als
o be use
d o
t describ
e th
e switchin
g
off of sources
. Fro
m th
e ver
y definitio
n of th
e uni
t ste
p function
, we fin
d that
:
n w \
(7-369
)
/ 1
, fi r < 0,
« ( - 0 =| Q> i f, > a
This versio
n of th
e uni
t ste
p functio
n ca
n be use
d o
t tur
n source
s of
fa
tt = 0. Indeed
,
v , ( 0 « ( - 0 a= |
(7-370
)
i f / a>
which mean
s tha
t th
e sourc
es
i on when t < 0 an
d th
e sourc
es
i of
f whe
n t > 0. Next
,
w e introduc
e th
e shifte
d uni
t ste
p functio
n u(t — to). t
Is
i clea
r fro
m th
e definitio
n of
the uni
t ste
p functio
n that
:
u(t - t0)
0, fit < t0,
1, i f r 0>.r
(7.371
)
It s
i eas
y o
t se
e that
:
vs(t)u{t - t0) =
\r^\
v,(t)u(t
)
0,
vs(t),
fit < t0,
fit > t0.
(7.372)
v(t)u(-t
)
Figur
e 7.24
: Descriptio
n o
f a sourc
e switche
d on an
d of
f by usin
g th
e uni
t ste
p func
tion
.
Chapter 7
. Transient Analysis
260
U(t -t 0)
1
d uni
t ste
p function
.
Figure 7.25: Shifte
Consequently
,vs(t)u(t — to) ca
n be interprete
d a
s a voltag
e sourc
e whic
h s
i turne
d on
at tim
e t0 (se
e Figur
e 7.25)
.
One of th
e importan
t application
s of th
e uni
t ste
p functio
n s
i th
e staircas
e approx
imatio
n of continuou
s functions
. First
, we not
e tha
t staircas
e (o
r stepwise
) function
s
can be represente
d a
s sums of shifte
d ste
p functions
:
(7.373)
fit) = ^T aku(t - tk\
where th
e ^'
s for
m a
n ordere
d sequenc
e (f*
> f*)
. Thes
e function
s loo
k lik
e th
e
+1
one show
n ni Figur
e 7.26
.t
Is
i clea
r fro
m thi
s figure
tha
t th
e ak hav
e th
e meanin
g of
the increment
s off(t) a
t time
s tk {k = 1
, 2,...)
. Thes
e stepwis
e function
s ca
n n
i tur
n
i a continuou
s sourc
e function
,
be use
d o
t approximat
e continuou
s functions
.f
Ivs(t) s
the
n t
i ca
n be approximate
d wit
h ste
p function
s by usin
g th
e expression
:
vs(t) « vs(P)u(t) + ^Av^uit
- tk)
(7.374)
where Av^ ar
e th
e increment
s of th
e sourc
e functio
n on successiv
e tim
e intervals
.A
grap
h of a sourc
e functio
n an
d it
s stepwis
e approximatio
n s
i show
n n
i Figur
e 7.27
.
The secon
d importan
t functio
n whic
h must be covere
d o
t full
y understan
d th
e
convolutio
n integra
l s
i th
e unit impulse function. Thi
s function
, denote
d 8(t), s
i
formall
y define
d a
s th
e derivativ
e of th
e uni
t ste
p function
:
8(0
0, fit < 0,
oo, fit = 0,
0, fit > 0.
du(i)
dt
(7.375
)
<x,
+ a
2+ a
3+ a
4
a,
+ a2+a3
a,
+ a2
a,
i — i —
t,
-1
t2
1
t3
1
t4
t5
r
Figure 7.26: Staircas
e functio
n represente
d by ste
p functions
.
7.5. Impulse Response and Convolution Integral
261
v,(t)
1
i
t, t t3
2
1
t4
t5
Figur
e 7.27
: Approximatin
g a continuou
s functio
n wit
h ste
p functions
.
This mean
s tha
t th
e uni
t impuls
e functio
n s
i infinit
e fo
r t = 0 an
d zer
o fo
r al
l othe
r
value
s oft. We ca
n als
o evaluat
e th
e integra
l of th
e uni
t impuls
e functio
n by usin
g it
s
definition
:
a
8(t)dt
du{t)
dt = u(a) — u(b).
dt
(7.376
)
From thi
s expressio
n we ca
n se
e tha
t th
e integra
l of th
e uni
t impuls
e functio
n s
i equa
l
to eithe
r on
e or zero
, dependin
g on whethe
r or no
t zer
o fall
s insid
e or outsid
e th
e
limit
s of integration
:
8(0* -
1, i f O e ( M,)
0, i f O g ( M.)
(7.377)
Equatio
n (7.377)
, alon
g wit
h th
e expressio
n 8(
0 = 0 fit £
= 0, s
i sometime
s use
d a
s
anothe
r definitio
n of th
e uni
t impuls
e function
.
Now tha
t we kno
w th
e uni
t ste
p functio
n an
d th
e uni
t impuls
e function
, we ca
n
elucidat
e th
e physica
l meanin
g of th
e convolutio
n integral
. Conside
r agai
n th
e simpl
e
RL circui
t excite
d by a voltag
e sourc
e show
n n
i Figur
e 7.28
. Thi
s circui
ts
i basicall
y
identica
lo
t th
e othe
r RL circuit
s considere
d previously
; however
,t
i ha
s a uni
t ste
p
voltag
e source
. The curren
t produce
d by a uni
t ste
p voltag
e source
,iu(t), s
i calle
d th
e
uni
t ste
p response
. We ca
n writ
e th
e differentia
l equatio
n fo
r thi
s circui
t a
s befor
e
(th
e onl
y differenc
e s
i th
e sourc
e term)
:
at
u
«0 O R
Figur
e 7.28
: RL circui
t wit
h uni
t ste
p voltag
e source
.
(7.378
)
Chapter 7. Transient Analysis
262
The solutio
n o
t thi
s equatio
n s
i th
e sam
e a
s th
e on
e fo
r th
e constan
t dc voltag
e source
.
Indeed
, fo
r t < 0 th
e circui
ts
i switche
d off
. Fort > 0 th
e sourc
es
i switche
d on
, an
d
w e jus
t trea
t th
e uni
t ste
p voltag
e sourc
e a
s a constan
t voltag
e sourc
e wit
h a valu
e
of 1. Therefore
, accordin
g o
t (7.75)
, th
e solutio
n o
t thi
s differentia
l equatio
n s
i
l
• ^
Ut)
= -~-e
l
-R-t
*'
.
(7.379
)
W e no
w retur
n o
t equatio
n (7.378
) an
d differentiat
e bot
h side
s wit
h respec
to
t time
:
du{t)
djuit)
d_ diu{t)
(7.380)
+ R dt
dt
dt
' dt
t be th
e derivativ
e of th
e
Now , we defin
e th
e uni
t impuls
e response
, denote
d is(t), o
uni
t ste
p functio
n response
:
diu(t)
dt
Substitutin
g (7.381
) int
o (7.380
) an
d recallin
g (7.375
) yields
:
is(t) =
(7.381
)
di8(t)
(7.382)
+ Rid(t) = 8(t).
dt
This equatio
n ha
s th
e sam
e mathematica
l for
m a
s th
e origina
l differentia
l equatio
n
(7.378
) an
d ca
n be interprete
d a
s th
e equatio
n describin
g a curren
tn
i th
e circui
t
excite
d by a uni
t impuls
e voltag
e sourc
e (se
e Figur
e 7.29)
. Thus
, th
e uni
t impuls
e
respons
e ca
n be als
o define
d a
s a respons
e produce
d by a uni
t impuls
e source
.
W e nex
t want o
t fin
d i§(t). Sinc
e iu(t) s
i alread
y know
n (equatio
n (7.379))
, cal
culatin
g th
e uni
t impuls
e functio
n respons
es
i straightforward
. Accordin
g o
t (7.381)
,
w e simpl
y tak
e th
e derivativ
e of (7.379)
:
T
diuit)
(7.383)
dt
L
Notic
e tha
t th
e expressio
n fo
r is(t — r) s
i identica
lo
t th
e kerne
l of convolutio
n we
derive
d earlie
r fo
r thi
s circui
t (se
e formul
a (7.360))
. Thus
, th
e convolutio
n integra
l
can be writte
n a
s follows
:
hit) =
i(t) =
/ is(t - T)vs(r)dT.
Figur
e 7.29
: RL circui
t wit
h uni
t impuls
e functio
n source
.
(7.384)
7.5. Impulse Response and Convolution Integral
263
Therefore
,we have established that the kernel of the convolution integral has the
physical meaning of the unit impulse response. Thus
, fi th
e uni
t impuls
e respons
e
is known
, the
n th
e respons
e o
t an
y sourc
e ca
n be foun
d by usin
g th
e convolutio
n
integral.
n
I thi
s sense
, th
e uni
t impuls
e respons
e give
s a complet
e characterizatio
n of
a linea
r circuit
. Of course
, we hav
e discusse
d onl
y th
e RL circui
t an
d thi
s doe
s no
t
prov
e ou
r genera
l statement
. Tha
t tas
k s
i lef
to
t th
e nex
t section
.
7.5.2
Convolution Integral for Arbitrary Linear Circuits
W e want o
t prov
e tha
t fi th
e uni
t ste
p functio
n respons
e (an
d consequentl
y th
e uni
t
impuls
e response
) of a circui
ts
i known
, the
n th
e respons
e fo
r an
y sourc
e ca
n be
found
. We begi
n wit
h a
n arbitrar
y linea
r circui
t drive
n by a
n arbitrar
y voltag
e source
,
shown on th
e lef
tn
i Figur
e 7.30
. The sam
e circui
t excite
d by th
e uni
t ste
p functio
n
sourc
e s
i show
n on th
e righ
tn
i th
e sam
e figure.
We kno
w fro
m th
e previou
s sectio
n
tha
t an
y continuou
s functio
n ca
n be approximate
d by usin
g linea
r combination
s of
shifte
d ste
p functions
:
vs(t) ~ vs(0)u(t) + V Av?>ii(
f -tk).
(7.385
)
k
Each ter
m n
i (7.385
) ca
n be interprete
d a
s a voltag
e sourc
e whic
h s
i turne
d on a
t
time ffc
. Consequently
, th
e las
t expressio
n ha
s a physica
l interpretation
, illustrate
d n
i
Figur
e 7.31
. Al
l thes
e voltag
e source
s ar
e connecte
d n
i serie
s an
d contribut
e o
t th
e
respons
e current
. Sinc
e th
e circui
ts
i linear (an
d tim
e invariant)
, fi a voltag
e sourc
e
u(t) result
sn
i a respons
e curren
tiu(t)9 a sourc
e Avf ^u(t — t^) wil
l resul
tn
i a respons
e
g o
t th
e superposition principle
, we ca
n ad
d thes
e separat
e
Avf ^ f
( — tk). Accordin
component
s of th
e respons
e curren
t togethe
ro
t find
th
e tota
l current
. Thi
s curren
ts
i
give
n by th
e expressio
n
Kt) « vs(0)h(t) + £
(7.386
)
A v ^ Cf - tk).
)
Usin
g elementar
y calculus
, Avf
ca
n be replace
d with
:
Av w
Kt)
!
D
v,(t)
«
dvs(t)
dt t=t
(7.387)
Af*.
k
'„(*)
Passive
Linear
Circuit
p
Passive
Linear
Circuit
Figur
e 7.30
: Arbitrar
y circui
t an
d it
s uni
t ste
p response
.
Chapter 7. Transient Analysis
264
v.{t)Q
-»•
+A
vs(0)u(t
)
+A
Av( 1 )u(t-t,
)
+/ A v > ( t - t
)
2
!"(J)Av s <k, u(t-t k )
Figur
e 7.31
: A serie
s connectio
n o
f voltag
e sources
.
Substitutio
n of (7.387
) int
o (7.386
) yields
:
i(
0 « v,(0)/„(r
) + Y^ M ~ ^
dvs{t)
dt
Af*.
(7.388)
A s th
e tim
e interval
s A.tk ge
t smalle
r an
d smalle
r th
e las
t expressio
n get
s mor
e an
d
e las
t expressio
n become
s a
n exac
t equality
,
more accurate
.n
I th
e limi
t of Af
y — 0, th
whil
e th
e su
m ni thi
s expressio
n become
s a
n integral
. Thus
, we obtain
:
T
)^p-dT-
i(t) = vs(0)iu(t) + I iu(t ~
(7.389
)
This integra
ls
i calle
d th
e superposition integral
, an
d t
is
i interestin
g ni it
s own right.
c circuit
,
It show
s tha
t fi we kno
w th
e uni
t ste
p functio
n respons
e iu{t — r) of an electri
the
n we ca
n fin
d th
e respons
e of thi
s circui
to
t any voltag
e source
.
The las
t integra
l ca
n be transforme
d throug
h integratio
n by parts
:
/(
0 = vMUt)
dlu(t
+ iu(t - T)VS(T)|;:J
> -J
T)
vs(r)dr.
dr
(7.390
)
By substitutin
g th
e limit
s of integration
, we find:
m = v,(o)iH(o + /«(o)v,(o - wov/o) - /
Jo
dlu(t T)
~ vs{T)dr.
(7.391
)
dr
By cancelin
g th
e identica
l terms
, we derive
:
I(0 = »«(0)v
5(0 - /
dT
vs(r)dr.
(7.392
)
Next, we recal
l (se
e (7.381)
) that
:
diu(t dT
T)
diu(t - T)
dt
-i8(t -
T)
.
(7.393
)
7.5. Impulse Response and Convolution Integral
265
By substitutin
g th
e las
t expressio
n int
o (7392)
, we en
d up wit
h th
e convolutio
n
integral
:
i(t) = iu(0)vs(t) + / i8(t ~ r)vs(r)dr.
Jo
(7394
)
For many circuit
s tha
t we wil
l encounter
, we wil
l findtha
t iu(0) = 0. Fo
r thos
e
circuit
s (7394
) reduce
s to
:
i(
0
i8(t - r)vs(T)dr.
(7395
)
This s
i th
e resul
t we sough
to
t prove
.
Previously
, we hav
e stresse
d th
e importanc
e of th
e convolutio
n an
d superpositio
n
integral
s fro
m th
e computationa
l poin
t of view
. Namely
, we hav
e emphasize
d that
, by
computin
g th
e uni
t impuls
e respons
e or th
e uni
t ste
p functio
n response
, we ca
n the
n
findth
e respons
e o
t a
n arbitrar
y sourc
e jus
t by performin
g certai
n integrations
. The
calculatio
n of th
e uni
t ste
p functio
n respons
e (a
s wel
la
s th
e uni
t impuls
e response
)
require
s th
e analysi
s of a
n electri
c circui
t fo
r th
e simples
t sourc
e excitation
. Thi
s
underline
s th
e computationa
l efficienc
y of convolutio
n an
d superpositio
n integrals
.
These integral
s ar
e als
o importan
t fro
m th
e experimenta
l poin
t of view
. The
reaso
n s
i tha
t th
e uni
t ste
p functio
n respons
e an
d th
e uni
t impuls
e respons
e ca
n be
measure
d by usin
g relativel
y simpl
e tests
. Then
, thes
e measure
d response
s ca
n be
utilize
d o
t predic
t response
s of a
n electri
c circui
to
t arbitrar
y sourc
e excitations
.
7.5.
3 Application
s of th
e Convolutio
n Integra
l
In thi
s sectio
n we wil
l demonstrat
e th
e usefulnes
s of th
e convolutio
n integra
l tech
niqu
e by applyin
g t
io
t a numbe
r of circui
t problems
.
EXAMPL E 7.1
8 n
I thi
s example
, we wil
l us
e th
e convolutio
n integra
l techniqu
e o
t
solv
e th
e proble
m considere
d previousl
y n
i Exampl
e 7.7
. The circui
t unde
r consid
eratio
n s
i redraw
n ni Figur
e 7.32
. Not
e tha
t no explici
t switche
s ar
e draw
n n
i thi
s
circuit
. Instead
, th
e turnin
g "on
" of th
e sourc
e s
i describe
d by usin
g th
e uni
t ste
p
Figur
e 7.32
: RC circui
t excite
d by a curren
t source
.
Chapter 7
. Transient Analysis
266
functio
n as
:
is{t)u(t).
(7396
)
In orde
r o
t findth
e circui
t respons
e by usin
g th
e convolutio
n techniques
, we
must first find
th
e uni
t ste
p respons
e of th
e circuit
. Jus
t prio
ro
t Exampl
e 7.7
, th
e
expression
s (7.117)
, (7.118)
, an
d (7.119
) wer
e derive
d fo
r th
e cas
e of excitatio
n of
, we obtai
n th
e
the RC circui
t by a dc curren
t source
. By puttin
g 70 = 1ni (7.117)
uni
t ste
p respons
e vu(t):
vu(t) = R - Re~t/RC.
(7.397
)
From th
e las
t expression
, we ca
n find
th
e uni
t impuls
e response
, vs(t):
vg(0
dvuii)
dt
-t/RC
(7.398
)
By usin
g thi
s uni
t impuls
e response
, we ca
n emplo
y th
e convolutio
n integra
l an
d find
the respons
e o
t a
n arbitrar
y curren
t source
:
v(0
o
v8(t - r)is(r)dr =
e
Jo
RC
C
is{T)dr.
(7.399
)
In Exampl
e 7.7
, we considere
d a
n a
c curren
t source
:
is{t) = Ims cos(cot + <k)
.
(7.400)
By substitutin
g thi
s expressio
n fo
r th
e curren
t sourc
e int
o th
e convolutio
n integra
l
(7.399)
, we obtain
:
1 f*
v(0 = 7; / e~{t~T)/RCImscos(cor+ <j>s)dr.
(7.401
)
C Jo
. Afte
r usin
g thi
s
W e ca
n simplif
y th
e integratio
n by recallin
g tha
tRe(ejx) = COS(JC)
fac
t an
d rearrangin
g term
s we arriv
e at
:
v(t) - Re \l^e-t/RC
['e^e^+Mdrl.
(7.402)
Now we ca
n easil
y perfor
m th
e integratio
n an
d rearrang
e term
so
t get
:
v(f
) = Re
1 -jcoRC
RI„
2
y/\ + (0)RC)
y/l + ((ORC)2
Jo
eJHe *
~ e~,/RC
(7.403)
The brackete
d ter
m s
i jus
te ^ wher
e 4> = arctan(o>/?C)
. Thu
s we ca
n tak
e th
e rea
l
par
to
t arriv
e a
t th
e final
answer
:
v(0 =
RIn
^ -</>)]
.
:[cos(to
? + 4>s-—A \4>) „~t/RC
— e cos(<
y/1 + ((0RC)2
This agree
s wit
h equatio
n (7.135
)a
st
i must
.
(7.404)
7.5. Impulse Response and Convolution Integral
267
EXAMPL E 7.1
9 Conside
r th
e RLC paralle
l circui
t of Example
s 7.1
3 an
d 7.14
,
which s
i redraw
n n
i Figur
e 7.33
. Use th
e convolutio
n integra
l techniqu
e o
t find
th
e
curren
t throug
h th
e inductor
.
W e ca
n find
th
e require
d differentia
l equatio
n by replacin
g th
e sourc
e curren
tn
i
(7.268
) wit
h th
e uni
t ste
p functio
n an
d pluggin
g n
i th
e componen
t values
:
d ir
dij
(7.405)
1
at
at
The particula
r solutio
n s
i simply
:
Kit) = .I
(7.406
)
The homogeneou
s solutio
n was foun
d ni (7.274
)o
t be
:
h„(t) = A\e~l
+ A2te~'.
(7.407
)
iL (0+) = 0 = ^ ( 0 +,)
at
(7.408
)
From (7.270
) an
d (7.271
) we have
:
and a fe
w algebrai
c manipulation
s yield
s th
e uni
t ste
p response
:
iu{t) = 1 -e~l - te~x.
(7.409
)
The uni
t impuls
e respons
es
i give
n by th
e derivativ
e of (7.409)
:
(7.410
)
The convolutio
n integra
ln
i ou
r cas
e (whe
n th
e sourc
e an
d outpu
t quantitie
s ar
e bot
h
currents
) take
s on th
e form
:
i(t) =
(7.411
)
/ hit - T)isir)dT
Jo
which fo
r thi
s exampl
e becomes
:
i(t) = 10 Re | /(t - r)e~{t~T)ejTdr\
iit) = We'* Re { \
7dr
A,
\ - 5te~l A,
L(t) = 10cos(t)u(t)
Figur
e 7.33
: RLC paralle
l circui
t excite
d by a curren
t source
.
(7.412
)
(7.413
)
Chapter 7
. Transient Analysis
268
i(
0 = 5 sin(f
) -5te~f A.
(7.414
)
Equatio
n (7.414
)s
i identica
lo
t (7.284)
,a
st
i must be
. Thi
s conclude
s thi
s example
.
EXAMPLE 7,20 Conside
r th
e circui
t show
n ni Figur
e 7.34
. Fin
d th
e voltag
e v(t)
acros
s th
e 2 F capacitor
.
For th
e abov
e circuit
, th
e differentia
l equatio
n whic
h describe
s th
e tim
e evolu
tio
n of v(t) s
i quit
e difficul
t o
t deriv
e directl
y fro
m KVL, KCL, an
d th
e termina
l
relationships
. However
, thi
s equatio
n s
i neede
d n
i orde
ro
t find
th
e particula
r solutio
n
for th
e give
n voltag
e source
:
vs(t) -
fcT'.
(7.415
)
The abov
e difficult
y ca
n be circumvente
d by combinin
g th
e convolutio
n integra
l an
d
transfe
r functio
n techniques
. The convolutio
n integra
l techniqu
e require
s us o
t find
the uni
t impuls
e response
, whic
h ca
n be don
e fi we kno
w th
e uni
t ste
p response
.
Becaus
e dc source
s ar
e a limitin
g cas
e of a
c sources
, th
e metho
d of th
e transfe
r
functio
n ca
n be use
d o
t findth
e uni
t ste
p respons
e an
d we ca
n procee
d wit
h th
e
convolutio
n techniqu
e fro
m there
.
The transfe
r functio
n approac
h fo
r thi
s circui
t was considere
d ni Exampl
e 7.15
.
For th
e parameter
s give
n ni Figur
e 7.34
, th
e transfe
r functio
n (7.311
) becomes
:
s+ 1
6s +6s + T
The fre
e respons
e ca
n be foun
d fro
m th
e pole
s of (7.416)
:
H(s)
2
1
sl2
= —U
= ± 1 A / 3S)
\
(7.416)
(7.417
)
The force
d respons
e ca
n be foun
d fro
m (7.294
) an
d (7.416
)o
t be
:
(7.418
)
.
vp(t) = H(0)u(t) = 1
Thus, th
e complet
e respons
e is
:
vu(t)=l+Ale-°-2Ut+A2e-°jm.
(7.419
)
rAAA/v
1 Q
te( u
( T) " (t)
1 f l \
1 Q
1 F
2F
+
v(t)
Figure 7.34: A second-orde
r circui
t excite
d by a voltag
e source
.
7.6. Circuits with Diodes (Rectifiers)
269
The initia
l condition
s ar
e foun
d fro
m (7.321
) an
d (7.327
)o
t be
:
vw (0+) = 0
(7.420
)
^ ( 0 +) = i
(7.421
)
at
6
By usin
g thes
e tw
o initia
l condition
s an
d (7.419)
, we ca
n solv
e fo
r A\ an
d A 2 an
d
arriv
e a
t th
e expressio
n fo
r th
e uni
t ste
p response
:
a21u
a789r
vu(t) = 1 - 1.077e"
+ 0.077^"
.
(7.422
)
The uni
t impuls
e respons
en
i foun
d n
i th
e usua
l way by differentiatin
g (7.422)
:
a211
a789r
v8(t) = 0.227e~
' - 0.061^"
.
(7.423
)
When th
e sourc
e an
d outpu
t signal
s ar
e bot
h voltages
, th
e convolutio
n integra
l take
s
the form
:
/ v8(t - T)vs(T)dr.
(7.424
)
Jo
Pluggin
g th
e sourc
e voltag
e (7.415
) an
d th
e uni
t impuls
e respons
e (7.423
) int
o th
e
convolutio
n integra
l (7.424
) yield
s th
e solution
:
v ( 0=
0 211
a789
v(t) = e~f + 0.3666T
' ' - 1.366<T
'A
(7.425
)
afte
r integratin
g by parts
. Thi
s conclude
s th
e example
.
7.6
Circuit
s wit
h Diode
s (Rectifiers
)
In thi
s section
, we shal
l appl
y th
e transien
t analysi
s technique
s develope
d ni previ
ous section
s o
t th
e analysi
s of steady-stat
e response
s of specia
l electri
c circuits—
rectifiers
.
Many electri
c device
s requir
e dc powe
r supplies
. Utilit
y companie
s suppl
y a
c
power
. Thus
, ther
e s
i a nee
d fo
r specia
l circuit
s whic
h conver
t a
c voltag
e source
s
int
o dc voltag
e sources
. Suc
h circuit
s ar
e calle
d rectifier
s an
d th
e mai
n elemen
t of
thes
e circuit
ss
i a diode
. The diod
e ca
n be define
d a
s a two-termina
l elemen
t whos
e
resistanc
e depend
s on th
e polarit
y of applie
d voltage
. The circui
t notatio
n fo
r th
e
diod
es
i show
n n
i Figur
e 7.35
.
i(t
)
D
O—
+
v(t
)
Figur
e 7.35
: The circui
t notatio
n fo
r a diode
.
270
Chapter 7. Transient Analysis
l
k
Figur
e 7.36
: The i-v curv
e o
fa
n idea
l diode
.
W e shal
l conside
r onl
y idea
l diodes
. Suc
h diode
s ar
e characterize
d by th
e i-v
curv
e show
n n
i Figur
e 7.36
. Accordin
g o
t thi
s figure,
th
e diod
e resistanc
es
i equa
lo
t
zer
o when th
e voltag
e acros
s th
e diod
e s
i positive
, an
d th
e diod
e resistanc
e s
i equa
l
to infinit
y fi th
e voltag
e acros
s th
e diod
e s
i negative
.n
I othe
r words
, th
e diod
e act
s
as a shor
t circui
t fi th
e applie
d voltag
es
i positive
, an
d t
i act
s a
sa
n ope
n circui
t fi th
e
applie
d voltag
es
i negative
. Thi
s mean
s tha
ta
n electri
c curren
t throug
h th
e diod
e ca
n
flowonl
y n
i on
e direction
, an
d thi
s si implie
d by th
e circui
t notatio
n fo
r th
e diode
.
Actua
l (real-world
) diode
s deviat
e somewha
t fro
m th
e idea
l propertie
s describe
d
above
. However
, a
s a first
approximation
, thes
e deviation
s ca
n be neglected
. Thi
s
help
so
t understan
d th
e effec
t of diode
s on th
e operatio
n of electri
c circuits
.
The diode
s ar
e usuall
y constructe
d by usin
g semiconductors
. One of th
e most
frequentl
y use
d diode
ss
i th
ep-n junctio
n diode
, whic
h s
i studie
d n
i detai
ln
i course
s
on semiconducto
r devices
.
The fac
t tha
t th
e diod
es
i a polarity-dependen
t elemen
t suggest
s tha
t diode
s ca
n
be use
d fo
r rectificatio
n of a
c powe
r sources
. As a
n exampl
e of suc
h rectification
, le
t
us first
conside
r th
e circui
t show
n n
i Figur
e 7.37
. Here
, th
e circui
ts
i excite
d by a
c
voltag
e sourc
e vs(t):
vs(t) = Vms sin cot,
the grap
h of whic
h s
i depicte
d n
i Figur
e 7.38
.
i(t
)
v
s«>0
Figur
e 7.37
: A half-wav
e rectifie
r circuit
.
(7.426
)
7.6. Circuits with Diodes (Rectifiers)
271
v.(t
)
Figure 7.38
: The inpu
t voltag
e source
.
Durin
g th
e first
half-perio
d (
0 < cot < 77)
, a voltag
e of positiv
e polarit
y s
i
applie
d acros
s th
e diod
e D. Thus
, thi
s diod
e s
i "closed
" an
d th
e tota
l sourc
e voltag
e s
i
applie
d acros
s th
e resistiv
e loa
d R. Durin
g th
e nex
t half-period
, a voltag
e of negativ
e
polarit
y s
i applie
d acros
s th
e diode
. As a result
, th
e diod
e s
i "open
" an
d no voltag
e s
i
applie
d acros
s th
e resistiv
e load
.I
n th
e subsequen
t half-periods
, th
e situatio
n repeat
s
itself
. Thus
,t
i ca
n be conclude
d tha
t th
e resistiv
e loa
d s
i subjec
to
t th
e rectifie
d voltag
e
)
v( r (0
show
n n
i Figur
e 7.39
.
Accordin
g o
t Ohm's la
w i(t) = v(t)/R an
d th
e shap
e of th
e loa
d curren
t mimic
s
the shap
e of th
e rectifie
d voltag
e (se
e Figur
e 7.40)
.
Thus, we hav
e achieve
d rectification
. However
, n
i th
e circui
t show
n n
i Figur
e
7.3
7 th
e curren
ts
i bein
g conducte
d durin
g onl
y positiv
e half-periods
. For thi
s reason
,
v(r'(t) I
ft
27C
3TC
47i
5TC
6 TI
co
t
Figure 7.39: Voltag
e acros
s th
e resistor
.
i(t
) A
R
7C
27C
37
1
47
C
57
1
Figure 7.40: Curren
t throug
h th
e resistor
.
67
E COt
272
Chapter 7. Transient Analysis
vs(t)
Figure 7.41: A full-wav
e rectifie
r circuit
.
thi
s circui
ts
i calle
d ahalf-wave rectifier
. To achiev
e th
e curren
t conductio
n durin
g
s ar
e designed
. Thes
e rectifier
s
positiv
e an
d negativ
e half-periods
,full-wave rectifier
usuall
y emplo
y th
e diod
e bridg
e circui
t show
n n
i Figur
e 7.41
.n
I thi
s circuit
, durin
g
the positiv
e half-periods
, a voltag
e of positiv
e polarit
y s
i applie
d acros
s diode
s Dx
and £>
,
whil
e
a
voltag
e
o
f
negativ
e
polarit
y
s
i
applie
d
acros
s
diode
s
D
an
d
Z)
.
r
3
2
4 Fo
thi
s reason
, diode
s D\ an
an
d
D
ar
e
open
.
Thus
,
th
e
d £>
ar
e
closed
,
whil
e
diode
s
D
4
3
2
tota
l voltag
e sourc
es
i applie
d acros
s th
e resistiv
e branc
h wit
h positiv
e polarit
y bein
g
applie
d o
t th
e uppe
r termina
l of R an
d negativ
e polarit
y bein
g applie
d o
t th
e lowe
r
termina
l of R.
Durin
g th
e negativ
e half-periods
, th
e polarit
y of th
e a
c voltag
e sourc
e vs(t) s
i
reversed
. Fo
r thi
s reason
, a voltag
e of positiv
e polarit
y s
i applie
d acros
s diode
s D2
and D4, whil
e a voltag
e of negativ
e polarit
y s
i applie
d acros
s diode
s D\ an
d Z)
3. As
a result
, diode
s D2 an
d D4 ar
e closed
, whil
e diode
s D\ an
d D3 ar
e open
. Again
, th
e
tota
l voltag
e sourc
es
i applie
d acros
s th
e resistiv
e branc
h wit
h positiv
e polarit
y bein
g
applie
d o
t th
e uppe
r termina
l of R an
d negativ
e polarit
y bein
g applie
d o
t th
e lowe
r
termina
l of R.
Thus, t
i ca
n be conclude
d tha
t th
e resistiv
e branc
h s
i subjec
t o
t th
e rectifie
d
voltag
e show
n n
i Figur
e 7.42
.t
Is
i clea
r tha
t th
e shap
e of th
e electri
c curren
t mimic
s
the shap
e of th
e rectifie
d voltage
.n
I thi
s way
, full-wav
e rectificatio
n s
i achieved
.
vs (t
)
V
L -,
^
Figure 7.42: Voltag
e acros
s th
e diagona
l branch
.
co
t
7.6. Circuits with Diodes (Rectifiers)
vs (t)
Figur
e 7.43
: A diod
e bridg
e circui
t wit
h a
n inductor
.
However
, th
e rectifie
d voltag
e an
d curren
t hav
e a substantia
l amoun
t of ripple
.To
reduce the ripple level, diod
e circuit
s wit
h energ
y storag
e element
s (inductor
s an
d
capacitors
) ar
e employed
. One of thes
e circuit
s s
i show
n n
i Figur
e 7.4
3 an
d wil
l
be analyze
d below
.t
Is
i clea
r fro
m th
e previou
s discussio
n tha
t th
e rectifie
d voltag
e
(show
n n
i Figur
e 7.42
) s
i applie
d acros
s th
e LR branch
. Consequently
, th
e circui
t
shown n
i Figur
e 7.4
3 ca
n be replace
d by th
e equivalen
t circui
t show
n n
i Figur
e 7.44
.
This circui
ts
i equivalen
to
t th
e on
e show
n n
i Figur
e 7.4
3 a
s fa
ra
s th
e curren
t throug
h
the LR branc
h a
s wel
la
s th
e voltage
s acros
s L an
d R ar
e concerned
.
As s
i eviden
t fro
m Figur
e 7.42
, th
e rectifie
d voltag
e source
, v ^ (,0 s
i periodi
c
s implie
s tha
t th
e steady-state current
,i(t), si periodi
c
wit
h perio
d equa
lo
t TT/CO. Thi
as wel
l an
d ha
s th
e sam
e period
. Fo
r thi
s reason
,t
i suffice
s o
t conside
r th
e circui
t
shown n
i Figur
e 7.4
4 durin
g on
e period
:
(7.427)
v
s(t)
Q
Figur
e 7.44
: Equivalen
t circui
t fo
r th
e full-wav
e rectifie
r wit
h inductor
.
274
Chapter 7. Transient Analysis
During this period, th
e current
,i(t), satisfie
s th
e differentia
l equation
:
di(i)
L~^
+ Ri(t) = Vms si
n cot
at
(7.428
)
and th
e followin
g periodi
c boundar
y condition
:
i(0
) =i ( ^
) .
(7.429
)
(O
A s fa
ra
s th
e differentia
l equatio
n s
i concerned
,t
i jus
t follow
s fro
m KVL. Indeed
,
the first
ter
m n
i thi
s equatio
n s
i th
e voltag
e dro
p acros
s th
e inductor
, th
e secon
d ter
m
is th
e voltag
e dro
p acros
s th
e resistor
, an
d th
e right-han
d sid
es
i th
e expressio
n fo
r th
e
rectifie
d voltag
e sourc
e durin
g th
e first
perio
d specifie
d by (7.427)
. (Pleas
e not
e tha
t
thi
s expressio
n s
i no
t vali
d fo
r th
e secon
d period.
) As fa
r a
s th
e periodi
c boundar
y
conditio
n (7.429
) s
i concerned
,t
is
i jus
t a consequenc
e of th
e periodicit
y of th
e
steady-stat
e curren
ti(t) n
i ou
r circuit
.
Differentia
l equatio
n (7.428
)s
i identica
lo
t th
e differentia
l equatio
n fo
r th
e RL
circui
t excite
d by a
n a
c voltag
e source
. Consequently
, th
e solutio
n o
t equatio
n (7.428
)
can be foun
d by usin
g th
e sam
e techniqu
e a
s before
. (Thi
ss
i th
e mai
n reaso
n why
w e conside
r circuit
s wit
h diode
s ni thi
s chapter.
) The genera
l solutio
n o
t equatio
n
(7.428
) ha
s th
e form
:
i(t) =
. Vms
sin(a)t -<£>
) +Ae'^\
V/?2 +co2L2
(7.430
)
coL
= arctan—
.
R
(7.431
)
where
6v
The first
ter
m s
i th
e particula
r solutio
n of equatio
n (7.428)
, whic
h s
i foun
d by usin
g
the phaso
r technique
. The secon
d ter
m s
i th
e homogeneou
s solutio
n wit
h unknow
n
constan
t A. When we discusse
d transient
s fo
r th
e RL circuit
, thi
s constan
t was foun
d
fro
m initia
l conditions
.n
I ou
r problem
, an
d thi
ss
i th
e mai
n mathematica
l difference
,
thi
s constan
t shoul
d be foun
d fro
m th
e periodi
c boundar
y conditio
n (7.429)
. By
i (7.430
) an
d by usin
g (7.429)
, we obtai
n th
e followin
g
settin
g t = 0 an
d t = TT/OJ n
equatio
n fo
r A:
~Vmssin(l)
2
^R
2 2 +
+ o) L
,n AI~>\
(7.432
)
Vm , sin(7
r - (/)
)
M
A = —VR
. 2 + o>2L—2 +Ae *L.
By solvin
g (7.432
) fo
r A, we arriv
e at
:
A=
2Vms si
n (b
™ ,y
2 2
(1 - £ - 5 r ) 2V #+ o)
L
(7.433
)
By substitutin
g (7.433
) int
o (7.430)
, we find
th
e final
solution
:
i(t) =
Vme
ms
VR2 + o>2L2
sin(ot
f -4>)+
2 Vm v s i n )
d
Rf
y
J"
e~i'.
2
1
( - e-5E)Vtf+<o2L2
(7.434
)
275
7.6. Circuits with Diodes (Rectifiers)
From (7.434
) we obtain
:
vR(t) =
sm(cot - <\>) +
—
e S.
(7.435
)
2
V 7/?
+ co2L2
(l-e~^)VR2
+ co2L2
Let us examin
e th
e las
t equation.
First
, le
t us evaluat
e th
e averag
e valu
e of vR(t):
v S 0=
- /"
' vR(t)dt.
(7.436
)
By substitutin
g (7.435
) int
o (7.436
) an
d by performin
g th
e integration
, we derive
:
2VmsR co
s 4>
2VmsR si
n§
a>L
,
+
;
: ,"
— (1 - * " *.) (7.437
)
2 2
TTVR2
+ co
L
77(
1 -e'^)y/R2
+ co2L2 R
By performin
g some cancellation
s ni th
e secon
d ter
m of (7.437
) an
d by recallin
g
(7.431)
, we transfor
m (7.437
)a
s follows
:
vR(t) =
vR(t) =
2VmsR
TT\/R
2
(
2 2
+ (0 L
\
, coL
co
s <j) + — si
n >
c/
R
2Vm.sR
TT^R
2
,
.,
, . .x
:(co
s
(p
+
ta
n
q>
si
n q>)
+ OJ L
2 2
2VmsR
7T\/R
2
2VmsR
2 2
+ C0L COS(/>
\JR2 + o)2L2
2Vm
R
I*
2 2
2
7 \ / / ? + C0 L
(7.438
)
So, we arriv
e a
t th
e followin
g remarkabl
e result
:
v*(0 = -Vms.
(7.439
)
77
It s
i remarkabl
e becaus
e the average voltage across the resistor does not depend on
the value of inductance or resistance. t
Is
i th
e sam
e fo
r an
y valu
e of th
e inductance
,
L, an
d resistance
,R. However
, th
e valu
e of L wil
l affec
t ripple
s of th
e voltag
e acros
s
the resistor
. To demonstrat
e this
, we conside
r tw
o limitin
g cases
: L —» 0 an
d L — °°
.
In th
e first
case
, fro
m (7.435
) an
d (7.431
) we find:
r)
v^(0 = V^sinot
f = vi
(0.
(7.440
)
Thus, fo
r ver
y smal
l L, th
e voltag
e acros
s th
e resistanc
e ha
s th
e sam
e (substantial
)
amount of rippl
e a
s th
e rectifie
d voltag
e sourc
e show
n ni Figur
e 7.42
.
For ver
y larg
e L, we have
:
\-e-%~—,
VR2 + OJ2L2
(oL
« o)L,
(7.441
)
(7.442
)
Chapter 7. Transient Analysis
276
By usin
g (7.441)
, (7.442)
, an
d (7.443
) ni (7.435)
,n
i th
e secon
d limitin
g cas
e we find:
v*(0 = -Vms
7T
= vR(t\
(7.444
)
Thus, fo
r ver
y larg
e L, th
e voltag
e acros
s th
e resisto
r ha
s a ver
y smal
l amoun
t of
ripple
. Physically
, thi
s ca
n be explaine
d a
s follows
. Fo
r larg
e L, th
e secon
d ter
m n
i
(7.435
)s
i almos
t constan
t an
d equa
lo
t th
e averag
e value
,vR(t), whil
e th
e first
ter
m
in (7.435
)s
i quit
e smal
l an
d mostl
y responsibl
e fo
r th
e ripple
.
The voltag
e vR(t) acros
s th
e resisto
r ni th
e circui
t show
n ni Figur
e 7.4
3 ca
n be
viewe
d a
s th
e outpu
t voltage
.t
I si clea
r fro
m th
e abov
e discussio
n tha
t fo
r larg
e L thi
s
outpu
t voltag
e ca
n be considere
d a
s a high-qualit
y dc voltag
e source
.t
Is
i importan
t
to not
e tha
t th
e voltag
e of thi
s dc sourc
e doe
s no
t depen
d on th
e resistiv
e load
. Thi
s
follow
s fro
m expressio
n (7.439)
, whic
h show
s tha
t th
e averag
e voltag
e acros
s th
e
resisto
r doe
s no
t depen
d on th
e valu
e of/?
. Al
l thi
s suggest
s tha
t fo
r larg
e L th
e circui
t
shown ni Figur
e 7.4
3 s
i a goo
d rectifie
r circuit
. The proble
m s
i tha
tt
is
i technicall
y
difficul
to
t realiz
e a sufficientl
y larg
e inductance
. Thi
s inductanc
e als
o result
s ni ver
y
lon
g transients
. Fo
r thi
s reason
, many differen
t rectifie
r circuit
s whic
h avoi
d th
e us
e
of inductor
s hav
e bee
n developed
.
One exampl
e of suc
h a circui
ts
i show
n ni Figur
e 7.45
. Fo
r th
e sak
e of analysis
,
r)
thi
s circui
t ca
n be replace
d by th
e equivalen
t on
e show
n ni Figur
e 7.46
. Here
, v^
(/
)
is th
e rectifie
d voltag
e sourc
e show
n ni Figur
e 7.42
.
The circui
t show
n ni Figur
e 7.4
6 s
i equivalen
to
t th
e circui
t show
n ni Figur
e 7.4
5
onl
y when i(t) > 0, tha
t is
, when th
e curren
t i(t) flowsfro
m nod
e 1o
t nod
e 2. The
opposit
e flowof th
e curren
ts
i prohibite
d by th
e diod
e bridg
e circuit
. Le
t us conside
r
the tim
e interva
l 0 < t < TT/CO an
d findwhen th
e conditio
n i(t) > 0 s
i violated
.
v 8 (0
Figur
e 7.45
: Full-wav
e rectifie
r wit
h capacitor
.
7.6. Circuits with Diodes (Rectifiers)
277
i(t
) {1
' (t)
iR (t) t
R 7
r l
2
Figur
e 7.46
: Equivalen
t circui
t fo
r th
e full-wav
e rectifie
r wit
h capacitor
.
From th
e circui
t show
n ni Figur
e 7.4
6 we infer
:
i{t) = iR(t) + ic{i),
(7.445
)
Vn
si
n coty
R
(7.446
)
lR(t)
dv[r)
ic(t) = C
= VmscoC co
s cot.
at
(7.447
)
By substitutin
g (7.446
) an
d (7.447
) int
o (7.445)
, we obtain
:
i(t) = —— si
n cot + Vmsa)C co
s cot,
R
(7.448
)
Let us find
th
e instan
t of tim
e t\ when th
e curren
ti(t) reache
s zero
:
i(t\) = —^ si
n o)?
i + Vms(oCcos cot\ = 0.
(7.449
)
The las
t expressio
n yield
s
tan coti =—(oRC.
(7.450
)
d
Ther
es
i onl
y on
e solutio
n of equatio
n (7.450
)n
i th
e tim
e interva
l 0 < t < TT/CO an
thi
s solutio
n s
i give
n by
:
7T
1
CO
CO
- - -taxTl((oCR).
(7.451
)
It s
i clea
r fro
m (7.448
) tha
t fit\ < t < TT/CO the
n i(t) < 0. However
, th
e negativ
e
curren
t value
s ar
e prohibite
d by th
e diod
e bridg
e circuit
. Thi
s mean
s tha
t th
e curren
t
i(t) shoul
d be equa
lo
t zer
o fo
r t\ < t < TT/CO. As soo
n a
s th
e curren
t i(t) cease
s
to flow,
th
e capacito
r C start
s o
t discharg
e throug
h th
e resisto
r R. Thi
s mean
s tha
t
the equivalen
t circui
t take
s th
e for
m show
n n
i Figur
e 7.47
. Thi
ss
i exactl
y th
e circui
t
tha
t we discusse
d a
t th
e ver
y beginnin
g of th
e chapter
. We foun
d tha
t th
e capacito
r
discharg
e s
i describe
d by th
e equation
:
vc(t) = Ae *c,
(7.452
)
278
Chapter 7
. Transient Analysis
i(t
)
Figur
e 7.47
: Equivalen
t circui
t fo
r th
e capacito
r discharge
.
where th
e constan
t A shoul
d be foun
d fro
m th
e initia
l condition
.n
I ou
r case
, thi
s
conditio
n is
:
vc(*i
) = Vmssino)th
(7.453
)
which s
i immediatel
y clea
r fro
m th
e circui
t show
n n
i Figur
e 7.46
.
From (7.452
) an
d (7.453)
, we find
:
A = Vmse«c si
n wf .
i
(7.454)
By substitutin
g (7.454
) int
o (7.452)
, we obtain
:
vc(0 = Vmse^
sincofi
.
(7.455
)
This capacito
r discharg
e wil
l continu
e unti
l th
e voltag
e acros
s th
e capacito
r become
s
equa
lo
t th
e rectifie
d voltag
e sourc
e v^fe
) a
t some instan
t of tim
e ti n
i th
e interva
l
ir/co < t < 2TT/CO (se
e Figur
e 7.48)
. Thi
s lead
so
t th
e followin
g equation
:
Vms si
n (otie'^
= v{p{t2).
(7.456
)
Sinc
e th
e voltag
e vc(t) s
i periodi
c wit
h th
e perio
d equa
lo
t TT/O), we find
that
:
h = t0 + - ,
(7.457
)
CO
where tim
e to s
i indicate
d on Figur
e 7.48
.t
Is
i als
o clea
r tha
t
v{P(h) = v(sr\to) = Vms si
n cof
0.
(7.458
)
By substitutin
g (7.457
) an
d (7.458
) int
o (7.456)
, we en
d up wit
h th
e followin
g
equatio
n fo
rt$\
e
«RC
si
n (oti
si
n o)to,
(7.459
)
eRC sinatf
o = e "RC sin a)t\.
(7.460
)
which ca
n be reduce
d to
:
Note tha
tt\ s
i give
n by expressio
n (7.451)
. Consequently
, expressio
n (7.460
) ca
n be
construe
d a
s a nonlinea
r algebrai
c equatio
n whic
h ca
n be solve
d (b
y usin
g graphica
l
technique
s or a computer
) fo
r 0f
- Afte
r ?
s
i
found
,
th
e
expressio
n
fo
r
vc(t)
s
i
give
n
0
7.7. MicroSim PSpice Simulations
279
Normalized time (cot)
Figure 7.48: Voltag
e acros
s th
e capacitor
.
by:
vc(0 =
if 0f < f < t\,
Vms si
n o)t,
Vms smcot^e RC ,iff
! < t < t0 + -.
t]->
(7.461
)
This voltag
e s
i show
n by th
e continuou
s lin
e n
i Figur
e 7.48
, whil
e th
e full-wav
e
rectifie
d sourc
e voltag
e s
i presente
d by a dashe
d line
.t
Is
i clea
r fro
m thi
s figure
tha
t th
e rippl
e leve
l fo
r th
e voltag
e vc(t) (whic
h by th
e way s
i equa
lo
t vR(t)) s
i
substantiall
y smalle
r tha
n th
e rippl
e leve
l fo
r th
e full-wav
e rectifie
d sourc
e voltage
.
e variatio
n of th
e voltag
e
This figure
als
o clearl
y reveal
s th
e physica
l mechanis
m of th
vc(0
- Durin
g th
e tim
e interva
l t0 < t < t\, th
e capacito
rs
i bein
g charge
d by th
e
rectifie
d voltag
e source
, whil
e durin
g th
e tim
e interva
l t\ < t < to + IT/CO th
e
capacito
rs
i bein
g discharge
d throug
h th
e resisto
rR. Thi
s patter
n periodicall
y repeat
s
itself
.
7.7
MicroSi
m PSpic
e Simulation
s
The first
PSpic
e simulatio
n tha
t we wil
l conside
rn
i thi
s sectio
ns
i fo
r th
e source-drive
n
first-order
RL circui
tn
i Exampl
e 7.5
. The circuit
, redraw
n by th
e PSpic
e schemati
c
generator
,s
i show
nn
i Figur
e 7.49
. The value
s of th
e passiv
e element
s ar
e indicate
d n
i
the figure
an
d th
e voltag
e sourc
e ha
s bee
n give
n th
e tim
e dependenc
e vi
a th
e "VSIN
"
dialo
g box
.
The par
t name fo
r th
e switc
h s
i "Sw_tClose,
" whic
h ca
n be accesse
d by typin
g
thi
s name n
i th
e "Ad
d Part
" dialo
g box
.f
I on
e s
i uncertai
n of th
e name of a circui
t
model or eve
n uncertai
n fi a mode
l exists
, on
e ca
n pres
s th
e "Browse...
" butto
n
280
Chapter 7. Transient Analysis
tClose=0
Figur
e 7.49
: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 7.5
.
in th
e "Ad
d Part
" dialo
g bo
x fo
r help
. Tha
t butto
n bring
s up th
e "Ge
t Part
" dialo
g
box tha
t list
s th
e variou
s librarie
s tha
t contai
n th
e par
t model
s an
d list
s th
e part
s
containe
d n
i whicheve
r librar
y s
i highlighted
. The switc
h require
d fo
r thi
s exampl
e
can be foun
d by clickin
g on th
e "eval.slb
" librar
y (usin
g th
e scrol
l bar
s fi necessary
)
and scrollin
g down th
e part
s lis
t unti
l th
e name "Sw.tClose
"s
i visible
. Clic
k on th
e
name o
t highligh
t it
. Thi
s name wil
l appea
rn
i th
e "Par
t Name" bo
x a
t th
e to
p of th
e
windo
w an
d th
e descriptio
n "Switch
: close
s a
t tClose=?
" wil
l appea
r directl
y belo
w
the par
t name box
. Pressin
g th
e "OK" butto
n wil
l selec
t th
e switc
h an
d t
i ca
n the
n be
place
d on th
e schemati
cn
i th
e usua
l way
.
The propertie
s of th
e switc
h ca
n be foun
d by double-clickin
g on th
e switc
h
afte
r t
i ha
s bee
n place
d ni th
e circuit
. Ther
e ar
e five
propertie
s of thi
s mode
l tha
t
can be adjusted
. The first
propert
y s
i th
e Packag
e Referenc
e Designato
r "PKGREF"
(i.e.
, name
) whic
h uniquel
y identifie
s eac
h par
t tha
t si place
d n
i th
e circuit
. An idea
l
switc
h s
i modele
d by infinit
e ope
n resistanc
e an
d zer
o close
d resistance
. However
,
for numerica
l reason
s an
d becaus
e some switche
s ar
e nonideal
, th
e switc
h mode
l ha
s
the parameter
s Rope
n an
d Rclose
d o
t mode
l th
e nonidea
l ope
n an
d close
d resistance
s
of th
e switch
, respectfully
. The defaul
t value
s ar
e give
n n
i th
e dialo
g bo
x an
d ca
n be
change
d fi necessary
.n
I particular
, car
e shoul
d be take
n fi eithe
r of thes
e value
ss
i
the sam
e orde
r of magnitud
e a
s othe
r resistance
sn
i th
e circuit
. The "tClose
" variabl
e
indicate
s th
e tim
e when th
e switc
h begin
so
t chang
e fro
m it
s ope
n stat
e o
t it
s close
d
state
. The final
paramete
rs
i "ttran,
" whic
h si th
e tim
e ti take
so
t make th
e transitio
n
fro
m th
e high-impedanc
e stat
eo
t th
e low-impedanc
e state
. Al
l of th
e switc
h defaul
t
value
s wer
e use
d fo
r thi
s simulation
.
A transien
t analysi
ss
i calle
d fo
r wit
h a final
tim
e of 10 s an
d a prin
t ste
p of 50
ms. The simulate
d curren
t throug
h th
e inducto
rs
i plotte
d ni Figur
e 7.50
. The analyti
c
solutio
n s
i als
o plotte
d n
i th
e figure.
For th
e circui
t parameter
s state
d above
, equatio
n
(7.93
) give
s th
e analyti
c solutio
n as
:
iL](t) = 2cos(\/3
' - 60°
) -e'f
A.
(7.462
)
A s expected
, th
e tw
o curve
s ar
e virtuall
y identica
l fo
r th
e entir
e simulatio
n period
.
The difference
s ste
m fro
m th
e resistanc
e of th
e switc
h an
d th
e inaccuracie
s of th
e
PSpic
e numerica
l methods
.
7.7. MicroSim PSpice Simulations
281
4
5
Time (s)
6
10
Figure 7.50: The simulate
d (soli
d line
) an
d analyti
c (dashe
d line
) solution
s fo
r th
e
inducto
r curren
tn
i th
e circui
to
f Figur
e 7.49
.
The nex
t PSpic
e simulatio
n revolve
s aroun
d th
e second-orde
r circui
t of Exampl
e
7.1
0 whic
h s
i excite
d by initia
l conditions
. The numerica
l value
s n
i th
e schemati
c
shown n
i Figur
e 7.5
1 ar
e identica
lo
t th
e value
s give
n ni th
e example
. A parametri
c
sweep of th
e capacitanc
e s
i selecte
d (wit
h Lis
t Values
: C = 0.1
, 0.8
, 1.0
, an
d 1.33
3
F) a
s describe
d n
i Sectio
n 4.7
. The onl
y ne
w par
t typ
e s
i th
e openin
g switch
, whic
h
has th
e name "Sw_tOpen
" an
d ca
n be foun
d n
i th
e "eval.slb
" library
.
The result
s of th
e simulatio
n fo
r th
e voltag
e acros
s th
e capacito
r ar
e plotte
d n
i
Figur
e 7.5
2 fo
r al
l fou
r cases
. As expected
, th
e voltag
e decay
s mor
e slowl
y fo
r th
e
overdampe
d cas
e C
( = 1.33
3 F) tha
n fo
r th
e criticall
y dampe
d cas
e C
( = IF)
. The
underdampe
d case
sC
( = 0.1
, 0.
8 F) deca
y th
e most rapidly
. The voltag
e oscillation
s
tOpen=0
szQ
tClose=0
PARAMETERS:
0.8
cv
Figure 7.51: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 7.10
.
Chapter 7
. Transient Analysis
282
10
i V
3
c == 4/
c == 1
—
5
c == 4/
~~_ c =
= 1/10
. _.
CD
"o
>
o
^
•Bo
CO
Q.
CO
O
\ \ s X
1
I
\\ V
\ r
\
\ i
\ i
N
x
*»»
V^ ^ Lr^
, _
\j
i
i
i
4
5
Time (s)
6
7
10
Figure 7.52: The simulate
d capacito
r voltag
en
i th
e circui
to
f Figur
e 7.51
.
for th
e C = 0.
1 F cas
e ar
e quit
e evident
, bu
t th
e capacito
r voltag
e fo
r th
e C = 0.
8
F cas
e goe
s onl
y slightl
y negative
. Thi
ss
i difficul
t o
t se
e on th
e figure
becaus
e th
e
dampin
g s
is
o strong
, bu
t ca
n be see
n ni Prob
e by expandin
g th
e y-axis
. A compariso
n
of th
e figure
wit
h th
e numerica
l value
s liste
d ni Tabl
e 7.
1 show
s goo
d agreement
.
The thir
d PSpic
e simulatio
n demonstrate
s th
e powe
r an
d utilit
y of PSpic
e by
showin
g th
e eas
e wit
h whic
h we generat
e th
e solutio
n o
t a difficul
t second-orde
r
transien
t problem
. The schemati
cs
i show
nn
i Figur
e 7.5
3 an
d use
s th
e source-excite
d
dual capacito
r circui
t of Exampl
e 7.15
. The parameter
s fo
r th
e passiv
e element
s ar
e
indicate
d n
i th
e figure.
The voltag
e sourc
es
i a dampe
d sinusoida
l wit
h a
n amplitud
e
tClose=
0
U1
R0
AA/Vv
1k
R1 >1k
R2>3k
V1
C1
1u C2
1u
Figure 7.53: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 7.15
.
7.7. MicroSim PSpice Simulations
283
4
6
Time (ms)
Figure 7.54: The simulate
d capacito
r voltag
en
i th
e circui
to
f Figur
e 7.53
.
of 10 V, a frequenc
y of 200
0 Hz, a
n offse
t voltag
e of 0 V, a phas
e of 0 (i.e.
, a sin(cot
)
dependence)
, an
d a dampin
g facto
r (DF
) of 150
. The dampin
g facto
r si simpl
y relate
d
to th
e comple
x frequency
, whic
h fo
r thi
s cas
e becomes
:
s = a + j(o = -150 + y27r(2000
)"
s
(7.463
)
The voltag
e acros
s th
e capacito
r Cl s
i plotte
d a
s a functio
n of tim
e n
i Figur
e
7.54
. The reade
r who stil
l doubt
s th
e usefulnes
s of acircui
t simulato
rs
i encourage
d o
t
reproduc
e th
e analyti
c resul
t fo
r thi
s circui
t an
d compar
et
io
t th
e PSpic
e simulation
.
The schemati
c show
n ni Figur
e 7.5
5 s
i th
e PSpic
e realizatio
n of th
e second-orde
r
transien
t proble
m tha
t was analyze
d ni Exampl
e 7.16
. The numerica
l value
s indicate
d
in th
e figure
ar
e th
e sam
e a
s ni th
e example
. Likewise
, th
e sinusoida
l voltag
e sourc
e
(VSIN) ha
s a magnitud
e of 50 V an
d a
n angula
r frequenc
y of 1 rad/s
. Remember tha
t
you must us
e "MEG" fo
r th
e resistance
so
t ge
t th
e prope
r values
.
C1
1u
V1
R1
AA/W
1MEG
C2
U1
1u
R2
.5MEG
Figure 7.55: The PSpic
e schemati
c fo
r the circui
tn
i Exampl
e 7.16
.
Chapter 7
. Transient Analysis
284
A transien
t analysi
s si selecte
d wit
h a "Prin
t Step
" of 50 ms an
d a "Fina
l Time
"
of 20 s
.n
I th
e sam
e "Transient
" dialo
g bo
x tha
t si use
d o
t selec
t thes
e times
, ther
e
are tw
o additiona
l option
s a
t th
e botto
m of th
e "Transien
t Analysis
" sectio
n of th
e
window. The lef
t optio
n call
s fo
r a "Detaile
d Bia
s Pt.
" an
d th
e righ
t optio
n read
s "Us
e
Init
. Conditions.
" To ge
t th
e desire
d simulatio
n results
, plac
e a
n "x
" ni th
e bo
x o
t th
e
immediat
e lef
t of th
e righ
t optio
n by clickin
g on eithe
rt
i or th
e text
. The "bia
s poin
t
detail
" si normall
y selecte
d by defaul
t ni th
e "Analysi
s Setup
" dialo
g box
,s
o ther
e si
no nee
d o
t activat
e th
e lef
t option
. The analyti
c expressio
n fo
r th
e voltag
e acros
s Rl
,
which s
i give
n ni (7.353)
, si indicate
d by th
e uppe
r trac
e ni Figur
e 7.56
. The lowe
r
trac
e ni tha
t figur
e display
s th
e resul
t fo
r th
e simulation
. Once again
, th
e result
s ar
e
virtuall
y identical
.
The final
simulatio
n use
s PSpic
eo
t analyz
e th
e filtered,
full-wav
e diod
e rectifie
r
circui
t considere
d ni th
e previou
s sectio
n (se
e Figur
e 7.45)
. As see
n ni th
e text
, thi
s
proble
m s
i quit
e challengin
g an
d require
s th
e solutio
n of a transcendenta
l equation
.
However
, th
e PSpic
e solutio
n s
i no mor
e difficul
t tha
n an
y of th
e previou
s problems
.
The schemati
c si draw
n ni Figur
e 7.5
7 an
d ha
s a 1 kH resistiv
e load
. The capacito
r
has bee
n parameterize
d C
( = 30
0 pF
, 3 /xF
, 30 /xF
, an
d 30
0 /xF
)o
t demonstrat
e th
e
□ 27*cos(Time+.71558)+6.64*exp(-.59*Time)+22.98*exp(-3.41 *Time)
20 s
4s
□ v(M:1)-v(M:2)
Time
Figur
e 7.56
: The analyti
c expressio
n (upper
trace
) an
d simulate
d resul
t (lowe
r trace
)
for th
e voltag
e acros
s Rl n
i th
e circui
to
f Figur
e 7.55
.
7.7. MicroSim PSpice Simulations
285
D1N4148
D1N4148
PARAMETERS:
cv
30u
Figur
e 7.57
: The PSpic
e schemati
c fo
r th
e filtere
d full-wav
e diod
e rectifier
.
effec
t of it
s magnitud
e on filtering.
We assum
e a househol
d voltag
e sourc
e wit
h a
n
amplitud
e of 169.
7 V, a phas
e of 0°
, an
d a frequenc
y of 60 Hz.
The onl
y element
s n
i thi
s circui
t tha
t we hav
e no
t ye
t considere
d ni previou
s
example
s ar
e th
e fou
r identica
l diodes
, whic
h hav
e th
e par
t name "D1N4148
" an
d
can be foun
d n
i th
e "eval.slb
" library
. The modelin
g of thi
s elemen
t si a subjec
t fo
r
an electronic
s course
, bu
t we nee
d o
t make on
e simpl
e modificatio
n nonetheless
.t
I
suffice
so
t sa
y tha
t whil
e a
n idea
l diod
e ha
s zer
o curren
t fo
r an
y voltag
e applie
d acros
s
it n
i th
e "wrong
" direction
, a rea
l diod
e ha
s a maximu
m revers
e breakdow
n voltag
e
tha
t shoul
d no
t be exceeded
. Fo
r thi
s diode
, th
e mode
l ha
s thi
s breakdow
n voltag
e
set o
t Bv = 10
0 V. Sinc
e we ar
e usin
g a voltag
e sourc
e wit
h a maximu
m voltag
e of
169.
7 V, we wil
l chang
e th
e breakdow
n voltag
e of th
e mode
lo
t 20
0 V.(Note: if we
were to use real DlN4148's in this bridge, exceeding the reverse breakdown voltage
would most likely cause them to burn out with potentially dire consequences!) Clic
k
on a diod
e o
t selec
tt
i an
d fro
m th
e "Edit
" drop-dow
n menu selec
t "Model.
" n
I th
e
"Edi
t Model
" dialo
g bo
x tha
t appears
, pus
h th
e "Edi
t Instanc
e Model
" button
.A
"Model Editor
" dialo
g bo
x wil
l appea
r tha
t wil
l lis
t al
l th
e parameter
s of th
e model
.
Change th
e "Bv
" variabl
e fro
m 10
0 o
t 20
0 an
d the
n pus
h th
e "OK" button
. Repea
t
thi
s procedur
e fo
r th
e othe
r thre
e diodes
. (Th
e procedur
e require
d o
t creat
e entirel
y
new model
ss
i beyon
d th
e scop
e of thi
s tex
t bu
ts
i describe
d n
i some of th
e reference
s
liste
d n
i Appendi
x C.
)
The voltag
e acros
s th
e resistor/capacito
r loa
d si show
nn
i Figur
e 7.5
8 fo
r th
e fou
r
differen
t cases
. The lowes
t capacitanc
e doe
s virtuall
y nothin
g an
d th
e voltag
e s
i th
e
same a
s fo
ra
n unfiltere
d full-wav
e rectifier
. The tim
e constan
t fo
r th
e 3 /x
F capacito
r
is a fe
w time
s smalle
r tha
n th
e sourc
e perio
d an
d th
e rippl
es
i quit
e large
. The rippl
e
is much reduce
d fo
r th
e 30 /x
F capacitor
, whic
h ha
s a tim
e constan
t nearl
y twic
e tha
t
of th
e sourc
e period
. The ripple
fo
r th
e 30
0 JULF capacito
rs
i th
e lowest
, of course
,
but th
e tim
e constan
ts
is
o lon
g tha
t th
e voltag
e ha
s no
t ye
t reache
d stead
y stat
e afte
r
thre
e period
s of th
e sourc
e (si
x period
s of th
e effectiv
e rectifie
d source)
.
Chapter 7
. Transient Analysis
286
200
160
•#
: 1
0 120
- 1
"o
v
o 80
:|
>
. f
\ NJ'
*
*
/
I
) i
.\ /
y
l>
/ ^
° 40 J' ,
"coc
a.
cc
.1
20
30
Time (ms)
Figur
e 7.58
: The simulate
d capacito
r voltag
e n
i th
e circui
to
f Figur
e 7.57
. The soli
d
0 /x
F case
, th
e dashe
d lin
es
i fo
rC— 3
0 fiF, th
e dot-das
h
lin
e correspond
so
t th
e C = 30
lin
e si fo
r C = 3 jiiF
, an
d th
e dotte
d lin
es
i fo
r C = 30
0 pF
7.8
Summary
In thi
s chapter
, we systematicall
y develope
d variou
s technique
s fo
r th
e analysi
s of
transient
s n
i firstan
d second-orde
r electri
c circuits
. The most importan
t fact
s an
d
result
s discusse
d n
i th
e chapte
r ca
n be summarize
d a
s follows
:
• Transient
s n
i electri
c circuit
s occu
r du
e o
t th
e presenc
e of energ
y storag
e
element
s (i.e.
, inductor
s an
d capacitors)
.
• Transient
sn
i electri
c circuit
s ca
n be excite
d by initia
l conditions
, by sources
,
or by both
.
• Analysi
s of transient
s ca
n be broke
n down int
o tw
o majo
r steps
:
1. Determinatio
n of initia
l condition
s fo
r th
e energ
y storag
e element
s by
usin
g th
e continuit
y of voltag
e acros
s a capacito
r an
d th
e continuit
y of
curren
t throug
h a
n inductor
.
2. Analysi
s of electri
c circuit
s afte
r switching
. Thi
s ste
p normall
y involve
s
the solutio
n of initia
l valu
e problem
s fo
r ordinar
y differentia
l equations
.
• Analysi
s of transient
s excite
d by initia
l condition
s require
s th
e solutio
n of
homogeneous differentia
l equation
s subjec
to
t nonzer
o initia
l conditions
.
• Analysi
s of transient
s excite
d by source
s require
s th
e solutio
n ofnonhomogeneous differentia
l equation
s subjec
to
t zero initia
l conditions
.
7.8. Summary
287
• Analysi
s of transient
s excite
d by initia
l condition
s an
d source
s require
s th
e
solutio
n ofnonhomogeneous equation
s subjec
to
t nonzero initia
l conditions
.
• A complet
e solutio
n of a nonhomogeneou
s linea
r differentia
l equatio
n ca
n be
represente
d a
s a su
m of a particula
r solutio
n of th
e nonhomogeneou
s equatio
n
and a genera
l solutio
n of th
e correspondin
g homogeneou
s equation
.
• n
I th
e cas
e of excitatio
n by a
c sources
, th
e particula
r solutio
n of th
e nonhomo
geneou
s equatio
n ca
n be foun
d by usin
g th
e phaso
r techniqu
e fo
r th
e calculatio
n
of th
e steady-stat
e response
. The particula
r solutio
n ha
s th
e physica
l meanin
g
offorced response.
• A genera
l solutio
n of th
e correspondin
g homogeneou
s equatio
n ha
s th
e physica
l
meanin
g of free (transient) response. It
s calculatio
n require
s th
e solutio
n of
characteristi
c equation
s an
d determinatio
n of unknow
n constant
s fro
m initia
l
conditions
.
• The natur
e of th
e transien
t (free
) respons
e of second-orde
r circuit
ss
i determine
d
by th
e root
s of th
e characteristi
c equation
. Ther
e ar
e fou
r distinc
t cases
:
a) overdampe
d respons
e when th
e tw
o root
s ar
e real
, negative
, an
d distinct
;
b) criticall
y dampe
d respons
e when th
e tw
o root
s ar
e real
, negative
, an
d
identical
;
c) underdampe
d respons
e when th
e tw
o root
s ar
e comple
x an
d conjugate
;
d) undampe
d respons
e when th
e tw
o root
s ar
e imaginar
y an
d conjugate
.
• The machiner
y of transfe
r function
s s
i a ver
y powerfu
l too
ln
i th
e analysi
s
of transient
s n
i electri
c circuits
. Thi
s analysi
s proceed
s a
s follows
. A circui
t
variable
, whic
h we ar
e intereste
d in
,s
i identifie
d a
s th
e output
. The transfe
r
functio
n s
i define
d a
s th
e rati
o of th
e outpu
t phaso
ro
t th
e inpu
t phasor
, i.e.
,
excitatio
n source
. By usin
g th
e phaso
r technique
, th
e transfe
r functio
n H(s) s
i
foun
d a
s th
e functio
n of comple
x frequenc
y s. The valu
e of thi
s functio
n a
t
the excitatio
n frequenc
y (s = jo)) full
y determine
s th
e a
c steady-stat
e (forced
)
response
, whil
e th
e pole
s of th
e transfe
r functio
n determin
e th
e exponent
s
and, consequently
, th
e for
m of th
e fre
e response
. The describe
d approac
h
is algebrai
c n
i nature
;t
i completel
y avoid
s th
e derivatio
n an
d solutio
n of
differentia
l equation
s an
d full
y exploit
s th
e machiner
y of th
e phaso
r technique
.
• To calculat
e th
e respons
e of a
n electri
c circui
t o
t a
n arbitrar
y source
, th
e
convolution integral ca
n be used
. The convolutio
n integra
l ha
s th
e form
:
i(t) = [ i8(t ~ T)VS(T)CIT,
(7.464
)
Jo
where i(t) s
i th
e respons
e (current
) cause
d by th
e excitatio
n by th
e voltag
e
sourc
e vs(t), whil
e i$(t) s
i th
e uni
t impuls
e respons
e cause
d by th
e uni
t impuls
e
sourc
e excitation
. The uni
t impuls
e respons
e ca
n be foun
d a
s th
e tim
e derivativ
e
Chapter 7. Transient Analysis
288
of uni
t ste
p response
, whic
h ni tur
n ca
n be foun
d fro
m th
e transien
t analysi
s
of th
e electri
c circui
t excite
d by a uni
t dc source
. Thus
, th
e analysi
s of a
n
electri
c circui
t by usin
g th
e convolutio
n integra
l consist
s of tw
o majo
r steps
:
a) calculatio
n of uni
t impuls
e response
; b) evaluatio
n of convolutio
n integra
l
s simila
ro
t (7.464
)
for a
n arbitrar
y (bu
t given
) voltag
e sourc
e vs(t). Expression
hol
d fo
r th
e convolutio
n integral
s when th
e desire
d circui
t respons
es
i a voltag
e
and/o
r th
e circui
t excitatio
n s
i a curren
t source
.
• A diod
e s
i a two-termina
l elemen
t whos
e resistanc
e depend
s on th
e polarit
y
of th
e applie
d voltage
. Idea
l diode
s ac
ta
s shor
t circuit
s fi applie
d voltage
s ar
e
positive
, an
d the
y ac
ta
s ope
n circuit
s fi applie
d voltage
s ar
e negative
. Diode
s
are use
d n
i rectifie
r circuit
s o
t conver
t a
c voltag
e source
s int
o dc voltag
e
sources
. A singl
e diod
e ca
n be use
d o
t construc
t ahalf-wave rectifier
. A diod
e
bridg
e circui
t ca
n be use
d o
t construc
t afull-wave rectifier. Energ
y storag
e
element
s ar
e employe
d ni rectifie
r circuit
so
t reduc
e th
e leve
l of ripples
.The
techniques for transient analysis of electric circuits can be used for the
steady-state analysis of rectifiers.
7.9
Problem
s
1. A nonidea
l capacitor
, afte
r bein
g charge
d o
t a
n initia
l voltag
e V0, wil
l slowl
y discharge
.
For th
e capacitances
, conductances
, an
d initia
l voltage
s give
n ni eac
h ro
w of th
e followin
g
table
, find
th
e tim
e ti take
s fo
r th
e capacito
ro
t discharg
e o
t 1 V.
Circuit
2.
3.
V„ (V
) C(ixF)
Gc (mU)
a
2
.00
1
100.
b
5
2.2
10.
c
10
470
1.
An initia
l current
,I0, ni a shorte
d nonidea
l inducto
r wil
l slowl
y deca
y wit
h time
. Fo
r th
e
inductances
, resistances
, an
d initia
l current
s give
n ni eac
h ro
w of th
e followin
g table
,
findth
e tim
e ti take
s fo
r th
e inducto
r curren
to
t reduc
e o
t 1
0 mA.
Circuit
h (A)
L(JJLH)
RL W
a
.02
1
1.
b
5
10
10,000
.
c
.06
470
100.
Plo
t th
e curren
t throug
h th
e 1
0 f
l resisto
r ni th
e circui
t show
n ni Figur
e P73 fo
r t > 0.
289
7.9. Problems
Plo
t th
e voltag
e acros
s th
e 1
0 /x
H inducto
r ni th
e circui
t show
n ni Figur
e P74 fo
r t > 0.
4.
^
.=0
^°
W>
CKA-O-J
o^o
5 mA (T) 10 |aH£v 10
Figure P7-3
Figure P7-4
Fo
r th
e circui
t show
n ni Figur
e P7-5
, find
th
e voltag
e acros
s R3 fo
r t > 0.
5.
6. Fo
r th
e circui
t show
n n
i Figur
e P7-6
, find
th
e voltag
e acros
s C2 fo
r t > 0.
7. Fo
r th
e circui
t show
n ni Figur
e P7-6
, le
t V0 = 20 V, /?
, = 6 kft
, /?
, an
d
2 = 2 kft
Ci = C2 = 47
0 /x
F Fin
d th
e valu
e of R3 require
d s
o tha
t th
e voltag
e acros
s C2 decay
s
i 1 second
.
to 1 V n
8. For th
e circui
t show
n ni Figur
e P7-5
, le
t V0 = 1
0 V,R{ = 10
0 fl
, R2 - 25 fl
, an
d
L\ = L2 = 10 mH. Fin
d th
e valu
e ofR3 require
d s
o tha
t th
e curren
t throug
h L2 decay
s
to 2 mA n
i 2 ms.
1
v
^°
W)
0 ^ 0 "#
o>j
>
o
Figure P7-5
9.
Figure P7-6
Fin
d th
e curren
t throug
h th
e inducto
r ni th
e circui
t show
n n
i Figur
e P79 (a
) by th
e
metho
d of phasor
s an
d (b
) by th
e metho
d of undetermine
d coefficients
. Assum
e vs(t) =
3 cos(2
/ + TT/8
) V an
d L/(0_) = 0.
10. Fin
d th
e voltag
e acros
s th
e capacito
r n
i th
e circui
t show
n ni Figur
e P7-1
0 (a
) by th
e
metho
d of phasor
s an
d (b
) by th
e metho
d of undetermine
d coefficients
. Assum
e is(t) =
6
10sin(10
0 mA an
d vc(0) = 0.
11.
Assum
e tha
tis(t) = 3t mA when t > 0 n
i th
e circui
t show
n ni Figur
e P7-1
1 an
d plo
t th
e
curren
t throug
h th
e inducto
r fo
r t > 0. Al
l quantitie
s ar
e zer
o fo
r t < 0.
Chapter 7. Transient Analysis
290
2 Q
rAAA/VH
dxw
v
s(t
) O
Figure P7-9
Figure P7-10
L
R,
Figure P7-11
12.
Assum
e tha
tvs(t) = 2e2t/t° V when t > 0 ni th
e circui
t show
n ni Figur
e P7-1
2 an
d plo
t
the voltag
e acros
s th
e capacito
r fo
r t > 0. Al
l quantitie
s ar
e zer
o fo
r t < 0.
13.
Fin
d th
e curren
t throug
h th
e resisto
r ni th
e circui
t show
n ni Figur
e P7-1
0 fi is(t)
2(t/ta)2u(t - t0) - {t/t())u{t - 3f„)/x
A an
d ta = 1 juts
.
=
14. Fin
d th
e voltag
e acros
s th
e resisto
r ni th
e circui
t show
n ni Figur
e P79 fi vs(t) =
e{t~X)u{t-
15.
l ) k .V
Use th
e convolutio
n integra
lo
t find
th
e voltag
e acros
s th
e resisto
rR\ ni th
e circui
t show
n
in Figur
e P7-1
2 fi th
e sourc
e voltag
e sivs(t) = cos(t) V an
d vc(0_) = 0.
16. Use th
e convolutio
n integra
lo
t findth
e curren
t throug
h th
e resisto
r R2 n
i th
e circui
t
shown ni Figur
e P7-1
1 fi th
e sourc
e curren
t siis(t) = sin(f
) A an
d //,(()) = 0.
17.
Fin
d th
e uni
t ste
p an
d impuls
e response
s fo
r th
e curren
t throug
h th
e lf
t resisto
r ni th
e
circui
t show
n ni Figur
e P7-17
.
fAAA/W
0Jv(t) a
s
Figure P7-12
Figure P7-17
18.
Fin
d th
e uni
t ste
p an
d impuls
e response
s fo
r th
e voltag
e acros
s th
e capacito
r ni th
e circui
t
shown n
i Figur
e P7-18
.
19.
Fin
d th
e curren
t throug
h th
e resisto
r /?
, ni th
e circui
t show
n ni Figur
e P7-18
. Assum
e
the initia
l voltag
e acros
s th
e capacito
rs
i vc(0) = 2 V,vs(t) = cos(
0 V, C = IF
, an
d
R2 = 2R\ = 2 H.
20.
Fin
d th
e voltag
e acros
s th
e 1 H resisto
r ni th
e circui
t show
n ni Figur
e P7-17
. Assum
e th
e
initia
l curren
t ni th
e inducto
rs
i I'L(0_
) = 1 mA an
d vs(t) = 3u(t) mV.
21.
Fin
d th
e voltag
e acros
s th
e resisto
r ni th
e circui
t show
n ni Figur
e P7-2
1 fo
r t > 0.
8
8
Assume tha
tv\(t) = cos(10
f) V an
d v2(f
) = cos(
2 X 10
f) V.
7.9. Problems
291
t=0
rAAAVi
R
2
fA/VW+
FT
v.(t)u(t
) 0
t=0
-o—X-o+<
v S 3 kQ
6 »*
«
©*,<*>
C
40 |iH
Figure P7-21
Figure P7-18
22.
Fin
d th
e voltag
e acros
s th
e 1
0 Ci resisto
r ni th
e circui
t show
n ni Figur
e P7-2
2 fo
r t > 0.
23.
Conside
r th
e circui
t show
n ni Figur
e P7-23
. Dra
w th
e circui
t fo
r t < 0 an
d fin
d th
e
relevan
t initia
l conditions
. Dra
w th
e circui
t fo
r t > 0 an
d fin
d th
e differentia
l equatio
n
for th
e curren
t throug
h th
e inductor
. Fin
d th
e solutio
n fo
r th
e inducto
r curren
t fo
r / > 0.
Plot th
e powe
r throug
h th
e inductor
. (B
e carefu
lo
t us
e a
n appropriat
e tim
e scale.
)
t=0
5 Q.
t=0
t=0
10 a
rA/VW
6
39 (iF^
p Vl
O Q(T)2sin(10
t)
p 6V
25.
W W i
20 Q
-o—V-o-
(T)lO
V
Figure P7-22
24.
t=0
1 |iH(* 10 Q
Figure P7-23
Conside
r th
e circui
t show
n ni Figur
e P7-24
. Fin
d th
e voltag
e acros
s th
e capacito
r fo
r
t >0.
Conside
r th
e circui
t show
n ni Figur
e P7-25
. Fo
r th
e capacitances
, resistances
, induc
tances
, an
d initia
l voltage
s give
n ni eac
h ro
w of th
e followin
g table
, fin
d th
e tim
et
i take
s
for th
e capacito
ro
t discharg
e o
t 1 V.
Circuit
V0 (V
)
C(F)
L(H)
/?(fl
)
a
2
10"8
10"3
470
b
10
0.2
1
20
c
100
0.12
5
2
8
292
Chapter 7. Transient Analysis
t=0
t=o
^°
.=
0
9.1 kQ ^
5V
<5cos(10 3 t+45°) v ( T )
0.1 uF
Figure P7-24
Figure P7-25
26.
Conside
r th
e circui
t show
n ni Figur
e P7-25
.f
I Z, = 1 mH, C = 47
0 mF,R = 2 kO, an
d
V() = 5 V, plo
t th
e curren
t throug
h th
e inducto
r whe
n t > 0.
27.
Conside
r th
e circui
t show
n ni Figur
e P7-27
.f
IL = 1/JLH, C = I /xF
, ?/ = 1
0 D, an
d
/0 = 1 A, plo
t th
e curren
t throug
h th
e resisto
r when t > 0.
t=0
t=0
6
Figure P7-27
28.
29.
Conside
r th
e circui
t show
n ni Figur
e P7-27
. Fo
r th
e capacitances
, resistances
, induc
tances
, an
d initia
l current
s give
n ni eac
h ro
w of th
e followin
g table
, fin
d th
e tim
e ti take
s
for th
e inducto
r curren
to
t reduc
e o
t 1
0 mA.
Circuit
h (A)
L(H)
C(F)
R{Q)
a
.02
10"3
10"7
200
b
0.2
10"4
6
2 X 10~
20
c
2.
.01
io- 7
2000
Plo
t th
e curren
t throug
h th
e resisto
r ni th
e circui
t show
n n
i Figur
e P7-2
9 fo
r t > 0.
Assume R = 1/
2 fl, =C 1 F,L = 1 H, an
d V0 = 5 V.
30. Plo
t th
e voltag
e acros
s th
e inducto
r ni th
e circui
t show
n ni Figur
e P7-3
0 fo
r t > 0.
Assume R = 1/
2 fi,
C = 1 F,L = 1 H, an
d „/ = 5 A.
7.9. Problems
293
,u(-.
> 6
Figure P7-29
31.
Figure P7-30
Fo
r th
e circui
t show
n n
i Figur
e P7-31
, us
e th
e convolutio
n integra
lo
t findth
e voltag
e
e resisto
r fo
rt > 0
. Assum
e tha
t R = 4 ft,
L = 1H, C = 0.2
5 F, an
d
acros
s th
v,(
0 -{It - 3t2)u(t) V.
Q
V
s(t
)
Figure P7-31
32.
33.
Fo
r th
e circui
t show
n n
i Figur
e P7-31
, us
e th
e convolutio
n integra
lo
t findth
e voltag
e
acros
s th
e capacito
r fo
rt > 0
. Assum
e tha
tR — 1
0 ft, =L 0.
1 H, C = 0.
1 F, an
d
vs(t) = 1
(
-e~2t)u(t)\.
Fin
d th
e curren
t throug
h th
e inducto
r n
i th
e circui
t show
n n
i Figur
e P7-31
. Assum
e
vs(t) - 2cos(20
r + TT/6)M(
0 V,L = 0.1H
, C - 0.0
2 F, an
d R = 5 ft.
34.
Fin
d th
e voltag
e acros
s th
e capacito
r n
i th
e circui
t show
n n
i Figur
e P7-31
. Assum
e
v,(
0 = 5 sinQO
f + 30°)K(
0 kV, L = 0.
5 H,C = 0.
2 F, an
d R = ^/w
35.
ft.
Fin
d th
e uni
t ste
p respons
e an
d th
e uni
t impuls
e respons
e o
f th
e curren
t throug
h th
e
inducto
rn
i th
e circui
t show
n n
i Figur
e P7-35
. Le
t L — 1.
2 H, C = 0.
3 F, an
d R = 1 ft.
Figure P7-35
36.
Fin
d th
e uni
t ste
p respons
e an
d th
e uni
t impuls
e respons
e of th
e voltag
e acros
s th
e resisto
r
in th
e circui
t show
n n
i Figur
e P7-35
. Assum
e tha
t(1/2RC)2 - l/(LC) < 0
.
Chapter 7. Transient Analysis
294
37.
Fin
d th
e voltag
e acros
s th
e resisto
r n
i th
e circui
t show
n n
i Figur
e P7-3
5 fi is(t)
Im cos(at
f + $)u(t) A. Assum
e tha
t(1/2RC)2 > 1/(LC)
.
38.
Conside
r tha
tis(t) = ltu(t)mA n
i th
e circui
t show
n n
i Figur
e P7-3
5 an
d plo
t th
e curren
t
throug
h th
e inducto
r fo
r t > 0. LetL = 1 H, C = 1 F, an
d R = 3 ft.
39.
Plo
t th
e powe
r dissipate
d n
i th
e capacito
r an
d inducto
rn
i th
e circui
t show
n ni Figur
e
P7-2
5 a
s a functio
n of tim
e fiR = 0 a, L = 1 mH, C = l m F, an
d V0 = 1 kV.
40.
Plo
t th
e powe
r dissipate
d ni th
e capacito
r an
d inducto
rn
i th
e circui
t show
n ni Figur
e
, C = 22 juF
, an
d 7
P7-3
0 a
s a functio
n of tim
e fiR -+ c
o n, l = 91 /xF
0 = 1 mA.
41.
Plo
t th
e curren
t throug
h th
e sourc
e (ys(t) — Vms cos(co
^ +4>s)u(t)) n
i th
e circui
t show
n n
i
Figur
e P7-4
1 ove
r on
e period
.
42.
Fin
d th
e steady-stat
e voltag
e acros
s th
e capacito
r ni th
e circui
t show
n n
i Figur
e P7-4
2
when is(t) = Ims\ sin(cof)|
.
v.(t)C
) R>
Figure P7-41
=
ns
Figure P7-42
43.
Use th
e convolutio
n integra
lo
t find
th
e outpu
t curren
t indicate
d ni th
e circui
t show
n ni
Figur
e P7-4
3 when vs(t) = 1
( +2t)u(t). Clearl
y explai
n al
l majo
r step
s of th
e analysis
.
44.
Use th
e convolutio
n integra
lo
t find
th
e outpu
t curren
t indicate
d n
i th
e circui
t show
n ni
[ + cos(t)]u(t) A. Clearl
y explai
n al
l majo
r step
s of th
e
Figur
e P7-4
4 when is(t) = 1
analysis
.
Figure P7-43
45.
Figure P7-44
Conside
r th
e circui
t show
n ni Figur
e P7-45
. Fin
d th
e equatio
n whic
h describe
s th
e
evolutio
n of v0(t). f
I al
l paramete
r value
s ar
e unit
y an
d vs(t) = cos(3t)u(t), find
va{t).
(Hint
: yo
u may want o
t conside
r usin
g transfe
r functions.
)
7.9. Problems
295
46.
Conside
r th
e circui
t show
n ni Figur
e P7-45
.f
Ivs(t) = 10w(-f)V
,C\ = 1/xF
, C2 = 3/xF
,
R\ = 2 kft
, an
d /?
, findth
e curren
t throug
h C2 a
s a functio
n of time
.
2 = 1 kft
47.
Conside
r th
e circui
t show
n n
i Figur
e P7-47
.f
I C = 47
0 jmF
,L = 1
0 JUH, /?
! = 10 kft
,
and R2 = 2 kft
, findth
e uni
t ste
p an
d impuls
e response
s of vG(t).
rAA/Wl-AA/W
6".«
C
^
Rr
Figure P7-47
Figure P7-45
48.
Conside
r th
e circui
t show
n n
i Figur
e P7-47
.f
I al
l paramete
r value
s ar
e unit
y an
d vs(t) =
cos(4t)u(t), find
va(t). (Hint
: yo
u may want o
t conside
r usin
g transfe
r functions.
)
49.
Conside
r th
e circui
t show
nn
i Figur
e P7- 49
. Fin
d th
e differentia
l equatio
n whic
h describe
s
the evolutio
n of i0(t). f
I Ri = 1/
2 ft,
R2 = 1 ft,
L = 2 H, C = I F
, an
d is(t) =
cos(2
£ + 45°)w(0
, find
i0(t).
Figure P7-49
50.
Conside
r th
e circui
t show
n ni Figur
e P7-49
.f
IL = 1 mH, C = 1 nF
, an
d R{ = 1 kft
,
findth
e valu
e ofR2 tha
t wil
l resul
t ni critica
l dampin
g of i0(t).
In Problem
s 51-60
, us
e PSpic
e o
t plo
t th
e circui
t parameter
s indicated
.
51.
Plo
t th
e curren
t throug
h th
e 1
0 ftresisto
r ni th
e circui
t show
n n
i Figur
e P73 fo
r t > 0.
52.
Plo
t th
e voltag
e acros
s th
e 1
0 /x
H inducto
r ni th
e circui
t show
n ni Figur
e P74 fo
r t > 0.
53.
Plo
t th
e voltag
e acros
s th
e resisto
r ni th
e circui
t show
n ni Figur
e P7-2
1 fo
rt > 0. Assum
e
8
8
tha
t vi(
0 = cos(10
0 V an
d v2(0 = cos(
2 X 10
0 V.
54.
Plo
t th
e curren
t throug
h th
e resisto
r n
i th
e circui
t show
n n
i Figur
e P7-2
9 fo
r t > 0.
Assume R = 1/
2 ft,
C = 1 F,L = 1 H, an
d V0 = 5 V.
55.
Plo
t th
e voltag
e acros
s th
e inducto
r ni th
e circui
t show
n ni Figur
e P7-3
0 fo
r t > 0.
Assume R = 1/
2 ft,
C = 1 F,L = 1 H, an
d I0 = 5 A.
296
Chapter 7. Transient Analysis
56.
Plo
t th
e powe
r dissipate
d ni th
e capacito
r an
d inducto
r ni th
e circui
t show
n n
i Figur
e
d V0 = 1 kV.
P7-2
5 a
s a functio
n of tim
e fiR = 0Ct, L = 1 mH,C = 1 mF, an
57.
Plo
t th
e powe
r dissipate
d n
i th
e capacito
r an
d inducto
r ni th
e circui
t show
n ni Figur
e
P7-3
0 a
s a functio
n of tim
e fiR +
- o ft,
L = 91 /xF
, C = 22 juF
, an
d 1Q = 1 mA.
58.
Plo
t th
e curren
t throug
h th
e sourc
e (ys(t) = Vms cos((t + (f)s)u(t)) n
i th
e circui
t show
n ni
Figur
e P7-4
1 ove
r on
e period
.
59.
Conside
r th
e circui
t show
n n
i Figur
e P7-45
.f
I al
l paramete
r value
s ar
e unit
y an
d vs(t) =
cos(3t)u(t), plo
t v0(t).
60.
Conside
r th
e circui
t show
n ni Figur
e P7-49
.f
Iis(t) = lOu(-t) mA, C = 1 /xF
,L = 1 H,
R\ = 2 kfl
, an
d 7?
, plo
t th
e curren
t throug
h th
e inducto
r a
s a functio
n of time
.
2 ~ 1 kfl
Chapte
r 8
Dependent Sources and
Operational Amplifiers
8.1
Introductio
n
In previou
s chapters
, we hav
e considere
d tw
o basi
c sources
: curren
t an
d voltag
e
sources
. Thes
e source
s wer
e alway
s assume
d o
t be independent. Tha
t mean
t tha
t
thei
r value
s di
d no
t depen
d on othe
r quantitie
s ni circuit
s the
y wer
e connecte
d to
.
In thi
s chapter
, we stud
y dependent sources
.These are the sources whose values do
depend on other quantities in the circuit The
y usuall
y appea
r a
s linea
r model
s fo
r
transistor
s an
d othe
r semiconducto
r devices
.
Afte
r definin
g th
e fou
r differen
t type
s of dependen
t sources
, we giv
e a brie
f
. We introduc
e a dependen
t
introductio
n o
t th
e tw
o most importan
t type
s of transistors
sourc
e mode
l fo
r th
e MOSFET transisto
r an
d demonstrat
e it
s validity
. We the
n revisi
t
the formalism
s fo
r genera
l noda
l analysis
, genera
l mes
h analysis
, an
d Thevenin'
s
theore
m an
d sho
w ho
w th
e presenc
e of dependen
t source
s affect
s th
e procedure
s tha
t
w e hav
e previousl
y develope
d o
t analyz
e electri
c circuits
. We demonstrat
e tha
t fo
r
each techniqu
e onl
y mino
r modification
s ar
e require
d o
t accommodat
e th
e presenc
e
of dependen
t sources
.
In th
e secon
d par
t of thi
s chapte
r we discus
s a
n importan
t typ
e of dependen
t
voltag
e sourc
e wit
h ver
y specia
l properties
. Thi
s dependen
t voltag
e sourc
e s
i calle
d
an operational amplifier (als
o calle
d a
n op-amp fo
r short
) an
d s
i realize
d vi
a a
n
integrate
d circui
t whic
h contain
s many transistors
, resistors
, an
d capacitors
. We wil
l
not ge
t int
o th
e detail
s of th
e interna
l structure-tha
ts
i lef
t fo
r a futur
e cours
e on
electronics
. Instead
, we characteriz
e th
e op-am
p n
i term
s of it
s termina
l propertie
s
and us
e the
m o
t analyz
e many circuit
s containin
g op-amps
. Thes
e circuit
s ca
n be
used o
t isolat
e load
s fro
m sources
, amplif
y voltag
e or curren
t signals
, inver
t signals
,
add signal
s together
, integrat
e signals
, or differentiat
e them
. The chapte
rs
i conclude
d
wit
h a demonstratio
n tha
t op-am
p circuit
s ca
n ac
ta
s analo
g computer
s an
d sin
e wav
e
generators
.
297
Chapter 8. Dependent Sources and Operational Amplifiers
298
8.
2
Dependen
t Source
s asLinea
r Model
s
for Transistor
s
Ther
e ar
e fou
r type
s of dependen
t sources
: voltage-controlle
d voltag
e source
s
(VCVS), current-controlle
d voltag
e source
s (ICVS)
, voltage-controlle
d curren
t
source
s (VCIS)
, an
d current-controlle
d curren
t source
s (ICIS)
. The dependen
t
source
s ar
e represente
d by diamon
d notation
s instea
d of circle
s a
s wit
h independen
t
sources
. The circui
t notation
s fo
r th
e fou
r source
s ar
e show
n n
i Figur
e 8.1
.n
I thi
s
figure,
v
^ an
d ix ar
e controllin
g voltage
s an
d currents
, respectively
.n
I thes
e notations
,
controllin
g branche
s ar
e show
n adjacen
to
t dependen
t sources
. Thi
s may no
t be th
e
cas
e fo
r actua
l circuits
! The constant
s A, G, an
d R ar
e factor
s whic
h coupl
e control
lin
g quantitie
s wit
h sources
. Usually
,A ha
s th
e meanin
g of amplificatio
n facto
r an
d
is dimensionless
, whil
e G an
d R hav
e th
e dimension
s of conductanc
e an
d resistance
,
respectively
. (The
y ar
e ofte
n calle
d transconductance
s an
d transresistances.
) Thi
s
explain
s th
e notation
s fo
r th
e couplin
g factors
. n
I summary
,n
i orde
ro
t specif
y a
dependen
t sourc
e we hav
eo
t specif
y a controllin
g branc
h an
d a couplin
g factor
.
VCV S
ICVS
VCIS
ICI
S
Figur
e 8.1
: Circui
t notatio
n fo
r variou
s dependen
t sources
.
Where do dependen
t source
s come from
? The answe
r is
: dependent sources
usually appear in electric circuits as linear models for transistors. To understan
d
how the
y fulfil
l thi
s role
, we must briefl
y examin
e transistor
s themselves
.
Ther
e ar
e tw
o most importan
t type
s of transistors
. The first
kin
d si th
e bipola
r
junctio
n transistor
, or BJT. A diagra
m of th
e BJT s
i show
n n
i Figur
e 8.2
. The BJT si
made ou
t of a semiconducto
r whic
h ha
s thre
e distinc
t region
s (n+, p, n~), eac
h on
e
connecte
d o
t a
n ohmi
c contact
. The thre
e ohmi
c contact
s ar
e calle
d th
e emitter
, th
e
collector
, an
d th
e bas
e an
d the
y ar
e specifie
d ni Figur
e 8.2
.
emitter
collector
base o
Figur
e 8.2
: Bipola
r junctio
n transistor
.
8.2. Dependent Sources as Linear Models for Transistors.
299
To understan
d thes
e notations
, we recal
l tha
tn
i semiconductor
s ther
e ar
e tw
o
type
s of curren
t carriers
: electron
s an
d (positively-charged
) holes
. Usually
, a semi
conducto
rs
i no
t ver
y usefu
l fi th
e amount
s of electron
s an
d hole
s ar
e balanced
. Fo
r
thi
s reason
, thi
s balanc
es
i deliberatel
y upse
t by introducin
g impuritie
s int
o semicon
ductors
. Thi
s proces
s s
i calle
d doping. Afte
r doping
, fi th
e amoun
t of electron
s s
i
greate
r tha
n th
e amoun
t of holes
, th
e semiconducto
rs
i calle
d a
n n type
.f
I th
e opposit
e
is true
, the
n th
e semiconducto
rs
i ap type
. The semiconducto
r of th
e BJ
T s
i dope
d o
t
creat
e th
e thre
e distinc
t regions
. The emitte
r regio
n s
iheavily dope
d and
,a
s a result
,
the densit
y of electron
ss
imuch large
r tha
n densit
y of holes
. Thi
ss
i indicate
d by th
e
superscrip
t "+.
" The bas
e regio
n s
i dope
d n
i suc
h a way tha
t th
e densit
y of hole
ss
i
large
r tha
n th
e densit
y of electrons
. Fo
r thi
s reason
, th
e bas
es
i ap region
. Finally
, th
e
collecto
r regio
n s
i lightly n dope
d an
d thi
ss
i indicate
d by a superscrip
t "-.
" When
two oppositel
y dope
d region
s ar
e adjacen
to
t on
e anothe
r the
y for
m what s
i calle
d a
p—n junction
. Two suc
h junction
s ar
e show
n n
i Figur
e 8.2
. Thi
s explain
s why thi
s
transisto
rs
i calle
d th
e bipola
r junctio
n transistor
. The middl
e area
, th
e base
,s
i made
very narrow o
t allo
w fo
r interaction
s betwee
n th
e tw
o junctions
. Thi
ss
i essentia
l fo
r
the prope
r operatio
n of th
e B JT
.
The BJT s
i a current-controlle
d device
. Indeed
,t
i ca
n be show
n tha
t injectio
n
of hole
s int
o th
e bas
e cause
s larg
e number
s of electron
so
t move fro
m th
e emitte
ro
t
the collector
. Thus
, th
e bas
e curren
t control
s th
e emitter-collecto
r current
. Fo
r thi
s
reason
, th
e BJT ca
n be modele
d a
s a current-controlle
d source
.
A t on
e time
, th
e BJT was th
e most importan
t devic
e n
i electronics
, bu
t no
w t
i
has bee
n somewha
t eclipse
d by th
e MOSFET transistor
. The MOSFET transisto
r
is th
e mai
n buildin
g bloc
k of VLSI (ver
y larg
e scal
e integrated
) circuits
, whic
h ar
e
essentia
l component
s of moder
n computers
. The acrony
m MOSFET stand
s fo
r meta
l
oxid
e semiconducto
r field
effec
t transistor
.t
Is
i picture
d n
i Figur
e 8.3
. The MOSFET
transisto
r s
i als
o made of a semiconducto
r wit
h thre
e distinctl
y dope
d regions
.t
I
has fou
r ohmi
c contacts
: th
e source
, th
e drain
, th
e gate
, an
d th
e substrate
. The first
thre
e letter
s of MOSFET describ
e th
e compositio
n of th
e device
. Indeed
, th
e word
s
metal oxid
e semiconducto
r refe
r o
t th
e metalli
c gate
, th
e Si0
n an
d th
e bul
k
2 regio
semiconducto
r region
. The las
t words
, field
effec
t transistor
, ar
e relate
do
t th
e principl
e
of operatio
n of th
e device
. Unlik
e th
e BJT, th
e MOSFET transisto
r s
i a voltage
controlle
d device
. When apositiv
e voltag
es
i applie
do
t th
e gate
, avertica
l electri
c field
is created
. Thi
s field
s
i indicate
d n
i Figur
e 8.
3 by arrow
s directe
d fro
m th
e gat
e towar
d
the p region
. The positivel
y charge
d hole
sn
i th
e p regio
n ar
e repelle
d by thi
s electri
c
e gat
e voltag
e and
, consequently
, th
e field
field.
Thi
s proces
ss
i calle
d depletion. As th
are furthe
r increased
, electron
s ar
e introduced
. Thi
ss
i calle
d inversion becaus
e th
e
applie
d gat
e voltag
e ha
s inverte
d th
e region
, makin
g electron
s mor
e numerou
s tha
n
holes
. Thi
s inversio
n create
s a
n n channe
l betwee
n th
e sourc
e an
d th
e drain
, allowin
g
curren
to
t flowbetwee
n the
m when some source-to-drai
n voltag
es
i applied
.
The curren
t id betwee
n th
e sourc
e an
d drai
n (whic
h s
i usuall
y calle
d a drai
n
current
) depend
s on th
e voltage
, v«/
, betwee
n sourc
e an
d drai
n a
s wel
l a
s on th
e
voltage
,vsg, betwee
n sourc
e an
d gate
. Mathematically
,t
i ca
n be writte
n a
s follows
:
id = f(vsd,
Vsgl
(8.1
)
300
Chapter 8. Dependent Sources and Operational Amplifiers
source
gate
drain
SiO
substrate
Figur
e 8.3
: MOSFET transistor
.
This dependenc
e ca
n be experimentall
y measure
d an
d plotte
d a
s a famil
y of
e give
n fo
r
curve
s show
n ni Figur
e 8.4
. n
I thi
s figure
, th
e plot
s of i^ vs.
vsci ar
monotonicall
y increasin
g value
s of gat
e voltag
e vsg:
„0)
- V„(2
(3) </ V„(4
)
K ) ^ lt
,
<*
< V\,
sg •
(8.2
)
It s
i clea
r fro
m th
e abov
e figure
tha
t fo
r th
e sam
e fixed
drai
n voltage
,vsd, th
e drai
n
current
,id, s
i increase
d a
s th
e gat
e voltage
,vsg, s
i increased
. The las
t fac
ts
i transparen
t
fro
m th
e physica
l poin
t of view
: th
e large
r th
e gat
e voltage
, th
e large
r th
e cros
s sectio
n
of th
e inversio
n channel
, th
e smalle
r it
s resistance,
an
d th
e large
r th
e drai
n curren
t
for th
e sam
e drai
n voltage
. Thi
s clearl
y suggest
s tha
t th
e drai
n curren
ts
i controlle
d
by th
e gat
e voltage
. Fo
r thi
s reason
, th
e MOSFET transisto
rs
i a voltage-controlle
d
devic
e an
d ca
n be modele
d a
s a voltage-controlle
d source
.
In orde
ro
t arriv
e a
t suc
h models
, we conside
r th
e cas
en
i whic
h th
e applie
d drai
n
and gat
e voltage
s hav
e tw
o distinc
t components
:
Figur
e 8.4
: Drai
n curren
ta
s a functio
n o
f source-to-drai
n voltag
e fo
r differen
t value
s
of gat
e voltag
e vsg.
8.2
. Dependent Sources as Linear Models for Transistors.
301
vsd =vfj + AvS£/,
(8.3
)
^
(8.4
)
= <> + Avv,.
Here,vfj an
d v^ ar
e constan
t (quiescent
) component
s of th
e drai
n an
d gat
e voltages
,
which ar
e usuall
y calle
d bias voltages
, whil
e Lvsd an
d kvsg ar
e smal
l time-varyin
g
component
s calle
d signals
.
The bia
s voltage
s determin
e th
e bia
s curren
t throug
h th
e transistor
, whic
h ac
cordin
g o
t (8.1
)s
i give
n by
:
f ,
(0 ) =
( 0))
(0
)
(8.5
)
The "signal
" voltage
s caus
e th
e time-varyin
g (signal
) componen
t of th
e drai
n current
,
which accordin
g o
t (8.1
) ca
n be define
d a
s follows
:
Md = /(vf
f + Av„, vg
> + Av„) -f(vfj,
vg>
>
(8.6
)
Sinc
e th
e signa
l voltage
s ar
e assume
d o
t be small
, we ca
n us
e n
i (8.6
) th
e Taylo
r
expansio
n an
d retai
n onl
y th
e first
tw
o terms
. Thi
s lead
s to
:
Md
dvsd bias
Av^ +
dVco
»sg.
bia
s
(8.7
)
Notatio
n "Ibias
" mean
s tha
t th
e correspondin
g derivative
s n
i (8.7
) ar
e evaluate
d a
t
bia
s voltage
s v^ an
d v ^. Expressio
n (8.7
) s
i a loca
l linearization of transisto
r
characteristic
s (8.1
) aroun
d a
n operatin
g poin
t of th
e transisto
r whic
h s
i determine
d
by th
e bia
s conditions
. Thi
s linearizatio
n lead
s o
t linear model
s fo
r th
e MOSFET
transistor
.
Now , we introduc
e ne
w notation
s fo
r signa
l components
:
Md = Id,
kvsd = Vsd,
Avsg
v«,
(8.8
)
as wel
la
s th
e followin
g notation
s fo
r th
e derivatives
:
1
dvsd bia
s
R
(8.9
)
dv
*g bia
s
In th
e ne
w notations
, expressio
n (8.7
) ca
n be writte
n n
i th
e form
:
h = VR~
+ GV,gg.
(8.10
)
The las
t expressio
n clearl
y suggest
s tha
ta
s fa
ra
s th
e relationshi
p betwee
n th
e signa
l
component
s of th
e drai
n current
, drai
n voltage
, an
d gat
e voltag
e ar
e concerned
,
the MOSFET transisto
r ca
n be modele
d a
s th
e nonidea
l voltage-controlle
d curren
t
sourc
e show
n n
i Figur
e 8.5
.n
I thi
s figure,
source
, drain
, an
d gat
e terminal
s ar
e
marke
d by "s,""d,
" an
d "g,
" respectively
, an
d gat
e an
d drai
n voltage
s ar
e identified
.
It s
i clea
r fro
m thi
s figure
tha
t th
e drai
n current
,Id, s
i th
e su
m of tw
o currents
: th
e
curren
t throug
h th
e resisto
r /?
, whic
h s
i equa
lo
t Vsd/R, an
d th
e curren
t GVsg of th
e
dependen
t curren
t source
. Consequently
, th
e drai
n current
,Id, s
i give
n by (8.10)
. Thi
s
302
Chapter 8. Dependent Sources and Operational Amplifiers
™£l
R
v
Figur
e 8,5
: Small-signa
l MOSFET model
.
prove
s ou
r assertio
n tha
t th
e voltage-controlle
d curren
t sourc
e show
n n
i Figur
e 8.
5 s
i
the appropriat
e mode
l fo
r th
e MOSFET transistor
.
By usin
g th
e equivalen
t transformatio
n of a nonidea
l curren
t sourc
e int
o a non
idea
l voltage-source
, we en
d up wit
h th
e nonidea
l voltage-controlle
d voltag
e sourc
e
model fo
r th
e MOSFET transistor
. Thi
s mode
ls
i show
n ni Figur
e 8.6
. Here
, th
e
followin
g expressio
n s
i vali
d fo
r th
e amplificatio
n facto
r A:
A=RG.
(8.11
)
Thus, we hav
e demonstrated that dependent sources appear as linear models for
transistors.
It s
i importan
to
t not
e tha
t th
e techniqu
e whic
h we use
d n
i orde
ro
t arriv
e a
t th
e
voltage-controlle
d sourc
e model
s fo
r th
e MOSFET s
i calle
d small-signal analysis.
This techniqu
es
i extensivel
y use
d fo
r th
e analysi
s of electri
c circuit
s an
ds
i discusse
d
in detai
ln
i course
s on electronics
.
It s
i als
o importan
to
t not
e tha
t ou
r discussio
n of bipola
r an
d MOSFET transistor
s
has bee
n ver
y superficial
. We intende
d onl
y o
t giv
e some ver
y genera
l idea
s concern
-
Figur
e 8.6
: VCVS mode
lo
f aMOSFET.
8.3. Analysis of Circuits with Dependent Sources
303
ing th
e structure
s an
d principle
s o
f operatio
n o
f thes
e device
s an
d o
t demonstrat
e
tha
t ther
e ar
e real-worl
d device
s behin
d th
e notio
n o
f dependen
t sources
. Extensiv
e
1
f thes
e transistor
s ca
n be foun
d n
i book
s o
n semiconducto
r devices.
discussion
s o
83
Analysi
s o
f Circuit
s wit
h Dependen
t Source
s
Analysi
s o
f circuit
s wit
h dependen
t source
s s
i simila
r o
t th
e analysi
s o
f circuit
s
wit
h independen
t sources
. Thi
s s
i tru
e fo
r noda
l analysis
, mes
h curren
t analysis
,
and Thevenin'
s theorem
. Ther
e are
, however
, some sligh
t differences
, whic
h wil
lb
e
discusse
d below
.
8-3.1
Nodal Analysis
The applicatio
n o
f noda
l analysi
s o
t electri
c circuit
s wit
h dependen
t source
s wil
lb
e
illustrate
d by th
e followin
g examples
.
EXAMPLE 8.1 Conside
r th
e circui
t show
n n
i Figur
e 8.7
.t
Is
i assume
d tha
tY\, Y2,
F3,Is\ ar
e given
. The dependen
t source
s ar
e specifie
d a
s follows
:
= G2V23,
(8.12
)
%> = G3Vl3.
(8.13
)
If
Thus, th
e first
dependen
t sourc
e depend
s o
n th
e voltag
e betwee
n node
s 2 an
d 3,
and th
e secon
d sourc
e depend
s on th
e voltag
e betwee
n node
s 1 an
d 3. Not
e tha
t th
e
controllin
g branche
s ar
e no
t adjacen
to
t th
e dependen
t sources
.
M
^h
.0 U
^
Figure 8.7
: Circui
t wit
h dependen
t sources
.
*See
, fo
r example
,Electronic Design b
y C. .
J Savant
, M. S. Roden
, an
d G. L. Carpente
r
(Redwoo
d City
: Benjami
n Cummings
, 1991
) orMicroelectronic Circuits b
y A. S. Sedr
a an
d
K. C. Smit
h (Ne
w York
: Holt
, Rinehar
t an
d Winston
, 1987)
.
304
Chapter 8. Dependent Sources and Operational Amplifiers
The first
ste
p of usin
g noda
l analysi
s wit
h dependen
t source
s s
io
t trea
t al
l
the circui
t source
s a
sf
i the
y wer
e independen
t an
d writ
e th
e correspondin
g matri
x
equations
. For ou
r circui
t thes
e equation
s are
:
y, +Y3
-Y3
(d)
-Y3
vi
/,
i +.
/
Y2 +Y3
h
l
'(d)
s2
j(d)
(8.14
)
s3
where nod
e 3 ha
s bee
n chose
n a
s areferenc
e node
.
The secon
d ste
p s
io
t expres
s th
e dependen
t source
s n
i term
s of th
e nod
e poten
tials
. I
n thi
s example
, th
e dependen
t source
s ar
e voltage-controlle
d s
o the
y ca
n be
rewritte
n as
/£> =G2(v2 - v3) = G2v2,
(8.15
)
/.S
° = G3(vi -h) = G3vh
(8.16
)
Afte
r th
e source
s hav
e bee
n expresse
d n
i term
s of th
e nod
e potentials
, th
e thir
d
and fina
l ste
p si o
t substitut
e th
e ne
w expression
s int
o th
e matri
x equation
s an
d modif
y
them accordingly
. Afte
r substitutio
n we have
:
Yi+Y3
~Y3
~Y3
Y2 + Y3
vi
hi
v2
G2v2
+ G2v2
— G3V]
(8.17
)
Now , al
l unknown
s must be moved o
t th
e left-han
d side
. Thi
s lead
s o
t th
e followin
g
fina
l matri
x equation
:
Y\ + Y3
-F3 + G3
— Y3 — G2
Y2 + Y3 + G2
vi
Isl
v2
0
(8.18
)
The las
t matri
x equatio
n ca
n be solve
d fo
r vi an
d v2, an
d afterwar
d al
l current
s ca
n
be easil
y found
.
EXAMPL E 8.
2 Conside
r a circui
t wit
h more complicate
d dependen
t source
s (se
e
e give
n a
s wel
l a
s th
e
Figur
e 8.8)
.t
Is
i assume
d tha
t Yu Y2, F3, F4,Y5, 76,Vs\, ar
followin
g specification
s fo
r th
e dependen
t sources
:
id)
V[s2
Rih*
(8.19
)
Ad)
V%'
= A3V{
(8.20
)
A s before
, th
e firs
t ste
p s
io
t writ
e th
e nod
e equation
s a
sf
i al
l source
s wer
e
independent
:
Yx + Y2 + Y3 + Y4
~Y3-Y4
-Y3-Y4
Y3 + Y4 + Y5 + Y6
YiVsi + Y4Vld2] '
-Y4V^ + Y6P%\
(8.21
)
where agai
n nod
e 3s
i chose
n a
s areferenc
e node
.
8.3. Analysis of Circuits with Dependent Sources
305
Figur
e 8.8
: Exampl
e circui
t wit
h dependen
t sources
.
Next, we expres
s th
e dependen
t source
sn
i term
s of th
e unknowns
:
Vf
= R2I2 = R2Y2Vl3
V^ = A3Vl2 = A3(v,
= R2Y2(vi ~ h)
= R2Y2K
-v2\
(8.22)
(8.23)
Finally
, we substitut
e thes
e expression
s int
o th
e previou
s nod
e equation
s an
d
modif
y th
e matri
x of thes
e equation
s by movin
g al
l unknown
so
t th
e left-han
d side
:
Y, + Y2 + Y3 + Y4- Y4Y2R2
- F3 - Y4 + R2Y2Y4 - Y6A3
-Y3 - Y4
Y3 + Y4 + Y5 + Y6 +A3Y6
YiVsl
0
(8.24)
Now , thes
e equation
s ca
n be solve
d fo
r i>
i an
d v2 an
d the
n al
l circui
t variable
s ca
n be
found
.
8.3.
2 Mesh Curren
t Analysi
s
The majo
r step
s fo
r usin
g mes
h curren
t analysi
s of circuit
s wit
h dependen
t source
s
are simila
ro
t thos
e use
d ni noda
l analysis
. Thes
e step
s ar
e a
s follows
.
1. Writ
e th
e matri
x equation
s a
s fi al
l th
e source
s wer
e independent
.
2. Expres
s th
e dependen
t source
sn
i term
s of th
e unknown
sn
i th
e matri
x equations
.
For mes
h curren
t analysis
, thi
s mean
s expressin
g th
e dependen
t source
s ni term
s
of mes
h currents
.
3. Substitut
e th
e expression
s fo
r th
e dependen
t source
s int
o th
e matri
x equations
.
Then modif
y th
e coefficien
t matri
x by movin
g expression
s fo
r dependen
t source
s
in term
s of mes
h current
so
t th
e left-han
d sid
e of th
e matri
x equation
.
Chapter 8. Dependent Sources and Operational Amplifiers
306
M
4
->
*„©
#
6
&*:
Figur
e 8.9
: Exampl
e circui
t wit
h dependen
t sources
.
EXAMPL E 8,
3 Conside
r th
e circui
t show
n n
i Figur
e 8.9
. Our goa
ls
io
t writ
e th
e
mesh curren
t equations
.t
Is
i assume
d tha
tZ]y Z2, Z3, Z4, Z5, Vs\, Vs2 ar
e give
n a
s
well a
s th
e followin
g specification
s fo
r th
e dependen
t sources
:
Vf
= AV2,
(8.25
)
(8.26
)
where V2 s
i th
e voltag
e acros
s th
e impedanc
e Z2, whil
e 74 s
i th
e curren
t throug
h th
e
impedanc
e Z4.
First
, we writ
e th
e mes
h curren
t equation
s treatin
g al
l source
s a
s independent
:
Z2+Z4
- z2
-Z4
-z 2
z{+z2 + z3
1 fti
- z4
-z3
^2
z3 + z4 + z5
- v(52f +1>,
=
lAJ
v,3 + v;
(8.27)
Now , we expres
s th
e dependen
t sourc
e ni term
s of th
e mes
h currents
:
Vg } = AV2
A/2Z2 = A Z2 ( i2 - J1 ),
(8.28
)
W*>=Rl4=R&-%).
(8.29
)
Substitutin
g th
e expression
s fo
r th
e dependen
t source
sn
i term
s of mes
h current
s
int
o th
e matri
x equation
s an
d movin
g the
m o
t th
e left-han
d sid
e yield
s th
e fina
l se
t
of equations
:
Z2 + Z4 — R
~Z2
- Z2 -AZ2 Zl+Z2+Z3+
AZ2
- Z4 +AZ2
- Z3 - AZ2
- Z4 +R
"Zj
z3 + z4 + z 5
Xl
12 =
L^J
0
+vsl
[-Vs3
. (8.30)
307
8.3. Analysis of Circuits with Dependent Sources
By solvin
g th
e las
t matri
x equation
, we find
th
e mes
h currents
, whic
h ca
n the
n
be use
d o
t calculat
e th
e actua
l current
s ni th
e circuit
.
■
8.3.3
Thevenin , s Theorem
W e first
recal
l tha
t th
e us
e of Thevenin'
s theore
m fo
r analysi
s of electri
c circuit
s wit
h
independen
t source
s require
s th
e followin
g thre
e steps
:
1. Remove th
e branc
h ni questio
n an
d determin
e th
e open-circui
t voltage
.
2. Se
t al
l source
s n
i th
e activ
e circui
to
t zer
o an
d findth
e inpu
t impedanc
e wit
h
respec
to
t th
e ope
n terminals
. When we se
t source
s o
t zero
, we replac
e curren
t
source
s by ope
n branche
s an
d voltag
e source
s by short-circui
t branches
.
3. Use th
e equatio
n
I =
Vn
(8.31
)
to find
th
e curren
t throug
h th
e branc
h n
i question
.
The primar
y obstacl
e n
i th
e us
e of Thevenin'
s theore
m fo
r circuit
s wit
h dependen
t
source
s s
i th
e secon
d step
. Thi
s s
i becaus
e th
e value
s of dependen
t source
s ar
e
determine
d by controllin
g voltage
s an
d current
sn
i othe
r branche
s and
, fo
r thi
s reason
,
dependen
t source
s canno
t independentl
y be se
to
t zer
o withou
t affectin
g controllin
g
branches
. However
, thi
s obstacl
e ca
n be remove
d by usin
g a
n alternativ
e approach
.
W e recal
l tha
t th
e inpu
t impedanc
e ca
n als
o be define
d as
:
7 = Y^
(8.32
)
Therefore
,o
t find
th
e inpu
t impedanc
e we nee
d o
t find
th
e open-circui
t voltag
e an
d
the short-circui
t current
, bot
h of whic
h ca
n be foun
d fo
r circuit
s containin
g dependen
t
sources
. The Theveni
n theore
m s
i especiall
y effectiv
e when th
e branc
h ni questio
n s
i
a controlling branch.
EXAMPLE 8.4 Conside
r th
e circui
t show
n ni Figur
e 8.10
.t
Is
i assume
d tha
tZ\, Z2,
Z 3,Vsi ar
e give
n a
s wel
la
s th
e followin
g specification
s fo
r th
e dependen
t sources
:
id
o
M
(d)
Figure 8.10: Circui
t containin
g dependen
t sources
.
Chapter 8. Dependent Sources and Operational Amplifiers
308
vf = AV3,
(8.33
)
j(d)
(8.34
)
GV-x.
In thi
s circuit
, branc
h 3s
i th
e controllin
g branc
h becaus
e al
l of th
e dependen
t source
s
depen
d on th
e voltag
e acros
s thi
s branch
. We want o
t solv
e fo
r th
e curren
t 3/ throug
h
thi
s branch
.
W e begi
n by removin
g thi
s branc
h an
d analyzin
g th
e resultin
g circui
t ni orde
ro
t
findth
e open-circui
t voltage
. Figur
e 8.1
1 show
s th
e circui
t wit
h th
e branc
h removed
.
Now we make th
e followin
g observations
. Firs
t of all
, th
e curren
t throug
h th
e onl
y
loo
p s
i equa
lo
t th
e sourc
e current
. Second
, voltag
e V3 whic
h control
s th
e curren
t
sourc
e s
i equa
lo
t th
e voltag
e acros
s th
e curren
t source
. Thi
ss
i becaus
e ther
e s
i no
curren
t throug
h Z2 an
d henc
e no voltag
e dro
p acros
s it
. Wit
h thes
e observation
s n
i
mind, we se
t up th
e KVL fo
r th
e loop
:
id) _ j(d)v
Vx] + V, + V\
_
(8.35)
= 0.
Substitutin
g ni th
e expression
s fo
r th
e dependen
t source
s ni term
s of V3 yields
:
- Vsl +%+
AV3 - GZX % = 0.
(8.36
)
Now , we ca
n solv
e fo
r V3, whic
h s
i by th
e way equa
lo
t th
e open-circui
t voltage
:
V, = Vnr =
V,S\
1 +A -GZ
(8.37
)
Next, we short-circui
t th
e branc
h n
i questio
n n
i orde
ro
t findth
e short-circui
t
current
. Her
e we must make anothe
r importan
t observation
. Short-circuitin
g branc
h
3 mean
s tha
t th
e voltag
e dro
p acros
s thi
s branc
h s
i zero
, V3 = 0. Thus
, al
l th
e
source
s controlle
d by thi
s branc
h ar
e als
o zero
. Thi
s s
i th
e tric
k whic
h simplifie
s
the calculatio
n of short-circui
t current
s throug
h controllin
g branches
. We no
w hav
e
the circui
t show
n ni Figur
e 8.12
.t
Is
i trivia
lo
t solv
e thi
s circui
t fo
r th
e short-circui
t
current
. Thi
s curren
ts
i give
n by
:
/,
K
Zi +Z7
(8.38
)
o
(d)
9
O "
' ^"
V
3 = VQ C
Figur
e 8.11
: Branc
h s
i removed
, creatin
g ope
n circuit
.
8.3. Analysis of Circuits with Dependent Sources
309
v,6
Figure 8.12: Branc
h s
i short-circuited—dependen
t source
s shu
t off
.
With th
e open-circui
t voltag
e an
d th
e short-circui
t curren
t n
i hand
, we ca
n
calculat
e th
e inpu
t impedance
:
"'
Isc
Zx + Z2
1 +A - GZX
(8.39
)
Usin
g th
e las
t formul
a n
i (8.31)
, we find
th
e curren
t throug
h branc
h 3o
t be
h
Vac
VSX
Zin + Z3
Zx+Z2+
(8.40)
Z3(l +A - GZX)
Anothe
r method
, base
d on th
e principl
e of equivalen
t transformations
, ca
n be
used o
t find
bot
h parameter
s of a Theveni
n or Norto
n equivalen
t circui
t wit
h a singl
e
circui
t calculation
.n
I thi
s method
, we attac
h a "probing
" independen
t voltag
e or
curren
t sourc
e acros
s th
e circui
t terminal
s fo
r whic
h we ar
e seekin
g th
e equivalen
t
circuit
.
When usin
g a probin
g voltag
e source
,a
s show
nn
i Figur
e 8.13a
, acircui
t analysi
s
l hav
e
is performe
d o
t findth
e curren
t throug
h tha
t source
. Thi
s current
, Ip, wil
two components
, on
e constan
t an
d on
e proportiona
lo
t Vp:Ip = A — BVp. Fo
r th
e
equivalen
t circui
t show
n n
i Figur
e 8.13b
, we ca
n us
e KVL o
t find:
Ip = VTH/ZTH ~
Vp/ZTH. Becaus
e thes
e circuit
s ar
e equivalent
, th
e expression
s fo
rIp must be identica
l
} +^
- ^ - n
"TH
VPO
vTH Q
vp O
f
-O- 1
(a)
(b)
(c)
Figure 8.13: Applicatio
n o
f th
e probin
g voltag
e metho
d o
t fin
d a
n equivalen
t circuit
.
310
Chapter 8. Dependent Sources and Operational Amplifiers
CZJo-i
•TH
'P vT H U
(a)
A
V
(b)
cbt \jb
Y NO
,0
1
(c)
Figure 8.14: Applicatio
n o
f th
e probin
g curren
t metho
d o
t fin
d an equivalen
t nonidea
l
source
.
s wil
l be tru
e onl
y fiZTH = 1/
5 an
d VTH = A/B. Therefore
,
for an
y valu
e of Vp. Thi
by solvin
g fo
r Ip n
i term
s of Vp an
d th
e othe
r circui
t parameters
, we ca
n rea
d of
f
the Theveni
n equivalen
t voltag
e an
d impedance
. Similarly
, usin
g a probin
g voltag
e
sourc
e o
t find Norto
a n equivalen
t circuit
,a
s show
n n
i Figur
e 8.13c
, woul
d resul
tn
i
the expressions
:I^o = A an
d YNo = B.
If a circui
t appear
so
t be easie
ro
t analyz
e wit
h genera
l noda
l analysis
, a probin
g
curren
t sourc
e shoul
d be use
d an
d th
e voltag
e acros
s thi
s sourc
es
i th
e circui
t variabl
e
which must be obtained
. The applicatio
n of KVL o
t th
e Theveni
n equivalen
t circui
t
in Figur
e 8.14
b yield
s Vp = VTH + ZTHIp. Noda
l analysi
s of th
e circui
t show
n n
i
Figur
e 8.14
a yield
s Vp = A + BIp. By followin
g th
e sam
e lin
e of reasonin
g a
sn
i th
e
above paragraph
, we conclud
e tha
tVTH = A an
d ZTH = 5. An applicatio
n of KCL o
t
the circui
t of Figur
e 8.14
c wil
l resul
tn
i simila
r expression
s fo
r th
e Norto
n equivalen
t
circuit
.
EXAMPLE 8.5 Let'
s us
e a probin
g sourc
e o
t findth
e Theveni
n equivalen
t circui
t
fro
m th
e previou
s exampl
e depicte
d n
i Figur
e 8.11
.f
I we conver
t th
e serie
s combi
natio
n of Vs an
d Z\ o
t a nonidea
l curren
t sourc
e an
d attac
h a prob
e curren
t source
,
w e arriv
e a
t th
e circui
t show
n n
i Figur
e 8.15
, wher
e Y\ = l/Z\ an
d Y2 = 1/Z
.
Th
e
2
Figure 8.15: The circui
t fo
r th
e previou
s exampl
e wit
h a probin
g curren
t source
.
8.3. Analysis of Circuits with Dependent Sources
311
genera
l noda
l analysi
s matri
x equatio
n is
:
Yi
0
0
0
Y2
~Y2
° \ (h h
lx + Gv3
~Y2
(8.41
)
Y2 J \ h
By addin
g th
e first
tw
o equation
s an
d usin
g th
e fac
t tha
tv\ = Av3 + v2, we reduc
e
the orde
ro
f th
e matri
x equatio
n by one
:
7, +Y2
-Y2
Y.A-GY2
Y2
v2
VSiYi
(8.42
)
Z, + Z,
\l+A-ZiG
h.
(8.43)
h
The nod
e voltag
e v3 s
i the
n foun
d o
t be
v3 =
V,
\+A-Z{G
+
so
VTH —
V,
1 + A - Z jG
and ZTH =
Z, + Z9
1 + A - Z Gi
(8.44)
These result
s ar
e identica
lo
t th
e previou
s solutio
n give
n by equation
s (8.37
) an
d
(8.39)
,a
s the
y must be
.
■
EXAMPLE 8.6 Le
t us fin
d th
e Norto
n equivalen
t nonidea
l sourc
e fo
r th
e circui
t
shown n
i Figur
e 8.
9 a
t th
e terminal
s forme
d by th
e remova
lo
f Z2. The ne
w circuit
,
i show
n n
i Figur
e 8.16
. Thi
s sourc
e voltag
e
complet
e wit
h a probin
g voltag
e source
,s
Figur
e 8.16
: Exampl
e circui
t wit
h probin
g voltag
e source
.
Chapter 8. Dependent Sources and Operational Amplifiers
312
is equa
lo
t th
e controllin
g voltag
e V2 as reflecte
d n
i th
e figur
e (se
e equatio
n (8.25))
.
The genera
l mesh equation
s ar
e give
n by
:
Z4
0
- Z4
Z{+Z3
lx
0
- Z3
?2
Z4
Z3
Vp-Rds-Ti)
Vsl -VpAVP
.
AVP - Vs3
=
Z3 + Z4 ~"f Z5
(8.45
)
B y movin
g th
e unknown
s o
t th
e left-han
d sid
e of th
e equatio
n we arriv
e at
Z4-R
0
0
Z{+Z3
R - Z4
- Z3
—Z 4
—Z3
Z3 + Z4 + Z5
1\
—
%J
[
Vsl - 1
( + A)%
AVP - Vs3
'(8.46
)
The curren
t throug
h th
e prob
e sourc
e s
i Ip = T2 — %, s
o we must us
e Gaussia
n
eliminatio
n an
d back-substitutio
n o
t findal
l th
e mesh currents
. The equivalen
t upper
triangula
r matri
x s
i foun
d o
t be
:
Z4-R
0
0
R-Z4
- Z3
Z5 + z>z3
0
Z,+Z3
0
^3J
z, +z
,
V„ - 1( + A)fc
,
Z^fzl^
l ~ ^3 + \ z4-/
?+
(8.47)
Z|A-Z3
Z, +Z3
V„
Solvin
g fo
r Ip — T2 ~ ^1
» afte
r some algebrai
c manipulation
s we arriv
e at
:
Z5t>9l + ZlVs3
h = (Z, + Z3)(Z5 + Z,)
1
Z4-i?
1
-
+
ZPR
Z3(Z5 + Ze)
+
Z e( l + A)
Z1Z3
Z\ZP
1 +
Z3(Z5 + Ze)
Vn
(8.48)
where Ze
= ZiZ3/(Zi
+ Z3) s
i th
e paralle
l combinatio
n of Z\ an
d Z3. Thus
,
Z5VsX+ZxVs3
INO
(8.49
)
(Zi + Z3XZ5 + Ze)
and
i^v
o
=
r ! [11 Z (Z *+Z,)1_
I Z i -?i
3
Ze
Z e d + A)
5
ZXZ3
This conclude
s th
e analysi
s of thi
s problem
.
1 +
Z l ze
Z 3 ( Z5 + Z e)_
(8.50
)
8.4. Operational Amplifiers
8-4
313
Operational Amplifiers
A n operationa
l amplifie
r (op-amp
)s
i a voltage-controlle
d voltag
e sourc
e wit
h ver
y
specia
l propertie
s whic
h make thi
s devic
e ver
y usefu
ln
i numerou
s applications
. The
circui
t notatio
n fo
r a
n operationa
l amplifie
rs
i show
n n
i Figur
e 8.17
.
It s
i clea
r fro
m thi
s figure
tha
t th
e operationa
l amplifie
rs
i a four-termina
l device
.
One termina
l (show
n a
t th
e botto
m of th
e figure)
s
i common o
t bot
h inpu
t voltage
s
d th
e outpu
t voltag
e VQ. Thi
s termina
ls
i commonl
y calle
d th
e groun
d
vi an
d V2 an
termina
l (groun
d lead
) an
d t
i ca
n be use
d a
s a natura
l zer
o referenc
e node
. The outpu
t
voltag
es
i relate
d o
t inpu
t voltage
s v\ an
d v>i by th
e expression
:
v0 = A ( v2 - v ,,)
(8.51
)
whereA s
ia
n amplificatio
n facto
r (sometime
s calle
d th
e open-loo
p gain)
.t
Is
i apparen
t
fro
m (8.51
) tha
t th
e outpu
t voltag
es
i controlle
d by th
edifference of tw
o inpu
t voltages
.
For thi
s reason
, thi
s amplifie
r s
i als
o calle
d adifferential amplifier
. The controllin
g
, whic
h explain
s why v
is
i calle
d th
e "inverting
" input
, whil
e v2 s
i
voltag
e s
i V2 — vi
calle
d th
e "noninverting
" input
.n
I th
e abov
e figure,
th
e invertin
g termina
l 1s
i marke
d
by a "-,
" whil
e th
e noninvertin
g termina
l 2s
i marke
d by a "+.
"
Operationa
l amplifier
s ar
e characterize
d by th
e followin
g thre
e properties
:
1. A ver
y hig
h amplificatio
n facto
r A whic
h typicall
y ha
s th
e followin
g rang
e of
values
:
7
104 < A < 10
.
(8.52
)
2. A ver
y larg
e inpu
t resistanc
e R[ (i.e.
, th
e resistanc
e betwee
n terminal
s 1 an
d 2)
:
15
io5n</?
; < io
a
(8.53
)
3. A relativel
y smal
l outpu
t resistanc
e R0 (tha
ts
i th
e resistanc
e betwee
n outpu
t an
d
groun
d terminals)
:
3
1 I < R0 < 10
0
ft.
1
I' '
M
Figur
e 8.17
: A circui
t notatio
n fo
r operationa
l amplifier
.
(8.54
)
314
Chapter 8. Dependent Sources and Operational Amplifiers
It s
i customar
y o
t conside
r a
n idea
l operationa
l amplifie
r wit
h th
e followin
g
idealize
d properties
:
A = oo
, R.
= oo
, RQ = 0.
(8.55
)
It s
i thes
e thre
e propertie
s whic
h make th
e operationa
l amplifie
r a ver
y valuabl
e cir
cui
t device
. Internally
,a
n operationa
l amplifie
rs
i a complicate
d activ
e circui
t whic
h
contain
s many transistors
, resistors
, an
d capacitor
s an
d must be connecte
d o
t on
e
or tw
o powe
r sources
. However
, detail
s of th
e interna
l structur
e of th
e operationa
l
amplifie
r ar
e immateria
la
s fa
ra
s th
e effec
t of th
e operationa
l amplifie
r on a
n externa
l
circui
ts
i concerned
. This effect is completely determined by the above three terminal properties. Fo
r thi
s reason
,n
i th
e analysi
s of electri
c circuit
s wit
h operationa
l
amplifier
s we ca
n completel
y ignor
e th
e interna
l structur
e of thes
e amplifier
s an
d
use onl
y thei
r termina
l propertie
s describe
d above
. Thes
e termina
l propertie
s ca
n be
used a
s th
e definition of th
e operationa
l amplifie
r a
s acircuit element. Thi
s axiomatic
approac
h s
i als
o justifie
d by th
e extremel
y wid
e us
e of operationa
l amplifiers
. n
I
fact
, thes
e amplifier
s ar
e encountere
d n
i moder
n circuit
s almos
ta
s ofte
n a
s resistor
s
and capacitors
. Thi
s justifie
s th
e elevatio
n of th
e operationa
l amplifie
r o
t th
e leve
l
of abasic circuit element, th
e elemen
t whic
h s
i define
d by th
e termina
l propertie
s
(8.55)
.
Sinc
e th
e amplificatio
n facto
r si assume
d o
t be infinite
, no signa
l (n
o voltage
)s
i
applie
d betwee
n terminal
s 1 an
d 2. Usually
, onl
y on
e of thes
e terminal
s s
i directl
y
biased
, whil
e th
e potentia
l of th
e othe
r termina
l adjust
s itsel
f n
i orde
ro
t kee
p th
e
outpu
t voltage
, v0, finite.
Accordin
g o
t expressio
n (8.51)
, th
e infinit
e gai
n A an
d
y that
:
finite
outpu
t voltag
e VQ impl
vx = v2.
(8.56
)
The las
t formul
a s
i th
e first
consequenc
e of th
e termina
l propertie
s of th
e idea
l
operationa
l amplifier
. Thi
s formul
a significantl
y simplifie
s th
e analysi
s of circuit
s
wit
h operationa
l amplifier
s an
d t
i wil
l be use
d tim
e an
d tim
e agai
n n
i ou
r subse
quent discussions
. For actua
l operationa
l amplifier
s equalit
y (8.56
)s
i fulfille
d onl
y
approximately
. However
, deviation
s fro
m thi
s equalit
y ar
e almos
t alway
s negligi
ble
.
Operationa
l amplifier
s ar
e use
d wit
h some circuitr
y aroun
d the
m n
i orde
r o
t
achiev
e desire
d effects
. One constan
t elemen
t of thi
s circuitr
y s
i a feedback branch
which connect
s th
e outpu
t termina
l wit
h th
e invertin
g inpu
t terminal
. Dependin
g on
the compositio
n of thi
s feedbac
k branch
, th
e operationa
l amplifier
s ca
n perfor
m many
functions
, whic
h we procee
d o
t discus
sn
i th
e followin
g sections
.
8-4.
1 Voltag
e Follower-Buffe
r Amplifie
r
An electri
c circui
t of a voltag
e followe
r si show
n ni Figur
e 8.18
. A voltag
e followe
r
has a short-circui
t feedbac
k branc
h whic
h connect
s th
e outpu
t termina
l wit
h th
e
invertin
g inpu
t terminal
. Consequently
:
vi(0 = v0(t).
(8.57
)
8.4. Operational Amplifiers
315
I
X
Figure 8.18: Voltag
e follower
.
On th
e othe
r hand
, th
e voltag
e sourc
e vs(t) s
i connecte
d betwee
n th
e groun
d termina
l
and th
e noninvertin
g terminal
. Thi
s mean
s that
:
v2(0 = vs(t).
(8.58
)
By substitutin
g (8.57
) an
d (8.58
) int
o (8.56)
, we obtain
:
v0(t) = vs{t\
(8.59
)
which explain
s why th
e abov
e circui
ts
i calle
d a voltag
e follower
. The practica
l utilit
y
of th
e voltag
e followe
rs
i tha
tt
i provide
s isolatio
n of a voltag
e sourc
e fro
m a load
.
Indeed
, fi some loa
d s
i connecte
d betwee
n th
e outpu
t an
d groun
d terminals
, thi
s loa
d
wil
l "feel
" th
e sourc
e voltage
. At th
e sam
e time
, th
e actua
l voltag
e sourc
e wil
l no
t fee
l
the load
, becaus
e th
e voltag
e sourc
e face
s th
e infinit
e inpu
t resistanc
e (impedance
)
of th
e amplifier
. Thus
, th
e loa
d wil
l no
t affec
t th
e curren
t throug
h th
e voltag
e sourc
e
and, consequently
,t
i wil
l no
t affec
t th
e voltag
e acros
s th
e inpu
t terminals
. Thi
s mean
s
tha
t any loa
d wil
l experienc
e th
e same sourc
e voltag
e appearin
g acros
s th
e outpu
t
and groun
d terminals
.n
I othe
r words
, anonideal voltag
e sourc
e connecte
d betwee
n
the noninvertin
g an
d groun
d terminal
s wil
l ac
ta
sa
n ideal voltag
e sourc
e betwee
n th
e
outpu
t an
d groun
d terminals
. Due o
t th
e describe
d isolatio
n of a voltag
e sourc
e fro
m
a load
, th
e voltag
e followe
rs
i als
o calle
d a buffe
r amplifier
.
It s
i instructiv
eo
t examin
e what effec
t finite
value
s of A,Ri9 an
d R0 wil
l hav
e on
the performanc
e of th
e voltag
e follower
.n
I othe
r words
, we woul
d lik
eo
t investigat
e
how th
e voltag
e followe
r wil
l perfor
m fi th
e operationa
l amplifie
rs
i no
t ideal
.f
I th
e
operationa
l amplifie
rs
i no
t ideal
, the
n th
e voltag
e followe
r ca
n be represente
d by a
n
electri
c circui
t show
n n
i Figur
e 8.19
.n
I thi
s figure,
Rt s
ia
n inpu
t resistanc
e of th
e
amplifie
r (tha
ts
i resistanc
e betwee
n invertin
g an
d noninvertin
g terminals)
,R0 s
ia
n
outpu
t resistanc
e (thi
ss
i th
e resistanc
e whic
h ca
n be measure
d betwee
n th
e outpu
t an
d
groun
d terminal
s when no signa
ls
i applie
d o
t th
e amplifie
r an
d no loa
d s
i connecte
d
betwee
n th
e groun
d an
d outpu
t terminals)
, an
d RL s
i a loa
d resistance
. The amplifie
r
itsel
fs
i represente
d a
s a voltage-controlle
d voltag
e sourc
e wit
h amplificatio
n facto
r
(gain
)A an
d controllin
g voltag
e vx{t):
vx{t) = vi(f)-v
2 (0.
(8.60
)
Chapter 8. Dependent Sources and Operational Amplifiers
316
v
x«
rWWfo
FL
Av
vs(t)Q
x<L>
Figur
e 8.19
: Electri
c circui
to
f a voltag
e followe
rn
i th
e cas
e o
f a nonidea
l op-amp
.
Accordin
g o
t expression
s (8.51
) an
d (8.60)
, th
e voltag
e of th
e controlle
d voltag
e
sourc
es
i equa
l to
:
Avx(t) = A ( v2 ( 0" vi(r))
,
(8.61
)
which si reflecte
d n
i polarit
y of thi
s voltag
e source
. Finally
, th
e short-circui
t feedbac
k
branc
h of th
e voltag
e followe
r si represente
dn
i Figur
e 8.1
9 by th
e short-circui
t branc
h
which connect
s th
e outpu
t an
d invertin
g terminals
.
W e shal
l appl
y th
e noda
l analysi
so
t th
e circui
tn
i Figur
e 8.19
, namely
, th
e versio
n
of thi
s analysi
s whic
h deal
s wit
h circuit
s containin
g voltag
e source
s ni serie
s wit
h
admittances
. Accordin
g o
t thi
s versio
n of noda
l analysis
, we obtai
n th
e followin
g
e outpu
t terminal
:
equatio
n fo
r potentia
lv0 of th
1
Ri
1
R0
1\
1
Ri
RLJ
-Avv
(8.62
)
From Figur
e 8.19
, we als
o find:
Vj =
V2 =
V0,
(8.63
)
Vs
By substitutin
g (8.63
) int
o (8.60)
, we obtain
:
Vr
=
Vn -
(8.64
)
Vc.
By substitutin
g (8.64
) int
o noda
l equatio
n (8.62)
, we deriv
e th
e followin
g equatio
n
for v0:
1 1
1
A ,
— + — +
+ — )Vr,
Ri
R0
RL
RO
By solvin
g th
e las
t equatio
n fo
r v0, we arriv
e at
:
1
Rt
A
R0
(8.65
)
± + A
Ri
Rt)
+
4r + 1+A
lk
R„
(8.66
)
8.4. Operational Amplifiers
317
4
5
Conside
r th
e "worst
" cas
e (se
e (8.52)
, (8.53)
, an
d (8.54)
) when A = 10
,/?
/ = 10
ft,
3
3
RQ = 10 ft,
an
d RL = 10 ft.
By substitutin
g thes
e value
s int
o (8.66)
, we find:
va{t) = 0.9998v
.
v(0
(8.67
)
By comparin
g (8.67
) wit
h (8.59
) we ca
n se
e tha
t th
e "nonidea
l nature
" of th
e oper
ationa
l amplifie
r result
s ni a ver
y smal
l deviatio
n n
i th
e performanc
e of th
e voltag
e
follower
. Thi
s conclusio
n s
i of genera
l natur
e an
d applicabl
e o
t al
l othe
r circuit
s
discusse
d below
.
8.4.2
Noninverting Amplifier
A n electri
c circui
t fo
r thi
s amplifie
r si show
n n
i Figur
e 8.20
. The noninvertin
g
k branc
h wit
h resistanc
e Rf whic
h connect
s th
e outpu
t termina
l
amplifie
r ha
s a feedbac
wit
h th
e invertin
g terminal
. The current
,,
/ whic
h flows
throug
h thi
s branc
h s
i give
n
by:
I
=
R
(8.68
)
The invertin
g termina
ls
i connecte
d o
t th
e groun
d termina
l by th
e branc
h wit
h re
sistanc
e R\. Sinc
e th
e inpu
t resistanc
e of th
e operationa
l amplifie
rs
i assume
d o
t be
infinite
, th
e sam
e curren
t flows
throug
h R\. Thi
s curren
t ca
n be expresse
d a
s follows
:
vi
(8.69
)
Finally
, th
e voltag
e sourc
e vs(t) s
i connecte
d betwee
n th
e noninvertin
g an
d groun
d
terminals
. Consequently
:
v2
(8.70)
Thus, we hav
e expresse
d al
l electri
c connection
sn
i th
e circui
t show
n ni Figur
e 8.2
0
in term
s of mathematica
l equation
s (8.68)
, (8.69)
, an
d (8.70)
. We shal
l nex
t combin
e
rW</V^
Figur
e 8.20
: Noninvertin
g amplifier
.
318
Chapter 8. Dependent Sources and Operational Amplifiers
thes
e equation
s wit
h (8.56
)n
i orde
ro
t find
th
e expressio
n fo
r v0 n
i term
s ofvs. Fro
m
(8.56
) an
d (8.70)
, we find:
(8.71)
vi = v5
By substitutin
g (8.71
) int
o (8.69
) an
d (8.68
) an
d by excludin
g i fro
m th
e las
t tw
o
equations,
we obtain
:
Vn ~ V.
(8.72
)
Rf
By solvin
g th
e las
t equatio
n fo
r v0, we obtain
:
1 + ^ ) v,(0.
v0(t)
(8.73)
It s
i clea
r fro
m (8.73
) tha
t a
t an
y instan
t of tim
e th
e outpu
t voltag
e ha
s th
e sam
e
polarit
y a
s th
e sourc
e voltag
e an
d t
is
i amplifie
d by th
e factor
:
i
vs(t)
Ri
(8.74
)
That s
i th
e reaso
n why th
e circui
t show
n n
i Figur
e 8.2
0 s
i calle
d a noninvertin
g
amplifier
. The utilit
y of thi
s circui
ts
i that
, by varyin
g th
e resistance
s Rf an
d R\, we
can contro
l th
e amplificatio
n facto
r (gain
) of th
e amplifier
.
8.4.3
Inverting Amplifier
A n electri
c circui
t fo
r thi
s amplifie
rs
i show
n n
i Figur
e 8.21
. As before
, th
e invertin
g
amplifie
r ha
s afeedbac
k branc
h wit
h resistanc
eRf whic
h connect
s th
e outpu
t termina
l
wit
h th
e invertin
g terminal
. The current
,,
/ throug
h thi
s branc
h si give
n by
:
i —
Rf
(8.75
)
Voltag
e sourc
e vs(t) s
i connecte
d o
t th
e invertin
g termina
l throug
h resistanc
e Ri.
Sinc
e th
e inpu
t resistanc
e (impedance
) of th
e operationa
l amplifie
rs
i assume
d o
t be
Figur
e 8.21
: Invertin
g amplifier
.
8.4. Operational Amplifiers
319
infinite
, th
e sam
e current
,,
/flows
throug
h resistanc
e R\. Thi
s curren
t ca
n be expresse
d
as follows
:
Vv — V]
i = ^ -i .
(8.76
)
Sinc
e th
e noninvertin
g termina
l si connecte
d o
t th
e groun
d terminal
, we find:
v2 = 0.
(8.77
)
W e hav
e expresse
d al
l electrica
l connection
s n
i th
e circui
t show
n n
i Figur
e 8.2
1 n
i
term
s of mathematica
l equation
s (8.75)
, (8.76)
, an
d (8.77)
. We shal
l combin
e thes
e
equation
s wit
h (8.56
)n
i orde
ro
t find
th
e expressio
n fo
r v0 n
i term
s of vs.
From (8.77
) an
d (8.56)
, we find:
V! - 0.
(8.78
)
By substitutin
g (8.78
) int
o (8.75
) an
d (8.76
) an
d by excludin
g th
e curren
t /fro
m th
e
las
t tw
o equations,
we obtain
:
Vn
V,
Rf
Ri
(8.79
)
which lead
so
t
v0(t) = ~^vs(t).
(8.80
)
It s
i clea
r fro
m (8.80
) tha
t a
t an
y instan
t of tim
e th
e outpu
t voltag
e ha
s a polarit
y
opposit
e o
t th
e polarit
y of th
e sourc
e voltage
.t
Is
i als
o clea
r tha
t th
e amplificatio
n
facto
r (sometime
s calle
d th
e closed-loo
p gain
) of th
e amplifie
rs
i give
n by
:
Af
= -§.
£
(8.81
)
That s
i why thi
s amplifie
rs
i calle
d a
n invertin
g amplifie
r or inverter
. The gai
n of thi
s
amplifie
r ca
n be controlle
d by varyin
g resistance
s Rf an
d R\.
8.4.4
Adder (Summer) Circuit
This circui
ts
i show
n ni Figur
e 8.22
. As before
, ther
e s
i a feedbac
k branc
h wit
h
resistanc
e Rf whic
h connect
s th
e outpu
t termina
l wit
h th
e invertin
g terminal
. The
current
,,
/ throug
h thi
s branc
h s
i give
n by
:
i = V^-
< 8 - 82 >
= Y1f*.
<8-83)
K
f
Ther
es
i agrou
p ofn paralle
l branche
s an
d thi
s grou
ps
i connecte
d betwee
n th
e invert
ing an
d groun
d terminals
. Sinc
e th
e inpu
t resistanc
e (impedance
) of th
e operationa
l
amplifie
rs
i assume
d o
t be infinite
, th
e overal
l curren
t of th
e abov
e grou
p of paralle
l
branche
ss
i equa
lo
t th
e current
,i, throug
h th
e feedbac
k branch
. Thus
:
* =i
320
Chapter 8. Dependent Sources and Operational Amplifiers
Figure 8.22: Adde
r circuit
.
Each paralle
l branc
h contain
s a resistanc
e Rk n
i serie
s wit
h a voltag
e sourc
e vsk(t).
Consequently
:
Vsk ~
Vi
-, (k = 1,2,...,«)
.
(8.84)
Rk
Finally
, th
e noninvertin
g termina
ls
i connecte
d o
t th
e groun
d terminal
, whic
h mean
s
that
:
Ik =
(8.85)
v2 = 0.
W e hav
e expresse
d al
l electrica
l connection
s ni th
e circui
t show
n ni Figur
e 8.2
2 ni
term
s of mathematica
l equation
s (8.82)
, (8.83)
, (8.84)
, an
d (8.85)
. We shal
l nex
t
combin
e thes
e equation
s wit
h expressio
n (8.56
)n
i orde
ro
t findth
e expressio
n fo
r
the outpu
t voltag
e v0 ni term
s of th
e sourc
e voltage
s v^
. Fro
m (8.56
) an
d (8.85)
, we
find:
(8.86
)
Vj = 0.
W e the
n substitut
e (8.86
) int
o (8.84
) an
d (8.82)
, whic
h result
s in
:
k = TT>
K-k
(k = 1
, 2,...
,n),
(8.87)
(8.88)
Ri
From (8.87)
, (8.88)
, an
d (8.83)
, we obtain
:
n
Rf
Vsk_
^ Rk
From (8.89)
, we arriv
e at
:
vo(0 =
-^2akvsk(t),
k=\
(8.90
)
8.4. Operational Amplifiers
321
where
ak = §*.
(8.91
)
It s
i clea
r fro
m (8.91
) tha
t th
e outpu
t si a "weighted
" su
m of sourc
e voltages
. For thi
s
reason
, th
e abov
e circui
ts
i calle
d a
n adde
r circuit
.t
Is
i als
o clea
r tha
t th
e weighte
d
, 2,...
,n).
coefficient
s ca
n be controlle
d by varyin
g resistance
s Rf an
d Rk (k = 1
The right-han
d sid
e of equatio
n (8.90
) ca
n be writte
n n
i th
e for
m of a
n inne
r
produc
t of tw
o vectors
. Indeed
, by introducin
g th
e vecto
rvs of voltag
e source
s
V* = (v
vb vs2, ...vsn),
(8.92
)
the vecto
ra of weighte
d coefficient
s
a = (a\, ci2> ...an),
(8.93
)
and th
e inne
r produc
t
n
< ay vs >= ^2<2kvsk,
(8.94
)
k=\
w e ca
n represen
t (8.90
)a
s follows
:
v0(t) = - <a,vs>.
(8.95
)
For thi
s reason
, th
e circui
t show
n ni Figur
e 8.2
2 ca
n als
o be calle
d a
n inne
r produc
t
circuit
.
EXAMPLE 8.7 We woul
d lik
e o
t desig
n a four-stag
e (binary
) digital-to-analo
g
(D/A) converte
r usin
g tw
o op-amp
s an
d an
y numbe
r of resistor
s wit
h resistanc
e
value
s up o
t lOkfl
.
In Chapte
r 4 we sa
w tha
t a D/
A converte
rs
i jus
t a voltag
e summer wher
e th
e
weighte
d coefficient
s ar
e power
s of 2. Fo
r a four-stag
e syste
m we nee
d a
n outpu
t
voltag
e tha
t satisfie
s vout = 8v
e summer circui
t describe
d abov
e
4 + 4v
3 + 2v
2 +v\. Th
yield
s a negativ
e outpu
t voltag
e (fo
r positiv
e inputs)
,s
o we wil
l hav
eo
t connec
tt
i wit
h
a simpl
e inverte
r op-am
p circui
to
t restor
e th
e positiv
e sign
. The circui
t schemati
c s
i
shown ni Figur
e 8.2
3 on pag
e 322
.
From equatio
n (8.81)
, we se
e tha
tR/Ri = at = T~x fo
r 1 < / < 4. f
I we tak
e
R = 1
0 HI, the
n we wil
l nee
d o
t hav
e R{ = 10 kft
,R2 = 5 kfl
,R3 = 2.
5 kfl
, an
d
R4 = 1.2
5 kfl
. The outpu
t of th
e first
stag
e wil
l be jus
t th
e negativ
e of th
e desire
d
result
. So, we se
e fro
m equatio
n (8.70
) tha
t we must hav
e RA = RB. We migh
ta
s
well tak
e bot
h o
t be equa
lo
t 10 kfl
.
■
8.4.5
Integrator
A n integrato
r circui
ts
i show
nn
i Figur
e 8.24
. The operatio
n of thi
s circui
t begin
s wit
h
the simultaneou
s closin
g of switc
h SW1 an
d openin
g of switc
h SW2 a
t tim
e t = 0.
The dc voltag
e sourc
e Vb
> whic
h fo
r tim
e t < 0s
i connecte
d n
i paralle
l wit
h feedbac
k
capacito
r C/
, wil
l ensur
e th
e followin
g initia
l conditio
n fo
r th
e voltage
,vc(t\ acros
s
322
Chapter 8. Dependent Sources and Operational Amplifiers
WW
r^A/V\An
FL
FL
Vout
Figur
e 8.23
: An op-am
p D/A converter
.
the capacitor
:
vc( 0+ )=
-V0
(8.96
)
The feedbac
k capacito
r Cf s
i connecte
d betwee
n th
e outpu
t an
d invertin
g terminals
.
Consequently
:
vc(0 = v , ( 0 - v „
-( 0
(8.97
)
The noninvertin
g termina
l si connecte
d o
t th
e groun
d terminal
, whic
h mean
s that
:
v2 = 0.
(8.98
)
SW 2
SW 1
vs(t
)
Figur
e 8.24
: Integrator
.
8.4. Operational Amplifiers
323
From (8.56
) an
d (8.98)
, we find:
vi(r
) = 0.
(8.99
)
By substitutin
g (8.99
) int
o (8.97)
, we have
:
vc(0 = -v„(0
.
(8-100
)
From (8.100
) an
d (8.96
) we find
th
e initia
l conditio
n fo
r v0(t):
vo(0+) = V0.
(8.101
)
For th
e current
,i{i), throug
h th
e feedbac
k branc
h we have
:
dVr(t)
i(t) = Cf-^-,
at
(8.102
)
which accordin
g o
t (8.100
) lead
s to
:
dv0(t)
i(t) = —Cf—-—.
(8.103
)
at
Sinc
e th
e inpu
t resistanc
e (impedance
) of th
e operationa
l amplifie
r s
i assume
d o
t
be infinite
, th
e sam
e current
, /(/)
, flowsthroug
h resistanc
e R. Thi
s curren
t ca
n be
expresse
d a
s follows
:
i(,) = V'(l) ~ V | ( ,,)
R
which accordin
g o
t (8.99
)s
i tantamoun
t to
:
i(t) = - ^.
K
(8.104)
(8.105
)
From (8.103
) an
d (8.105)
, we obtain
:
— T — = " - ^ - ^ -( 0
(8.106
)
at
RCf
By integratin
g (8.106
) an
d takin
g int
o accoun
t initia
l condition
s (8.101)
, we arriv
e
at:
Vo(t) = - - L
- / vs(r)dr + V0.
K<
-f
JO
(8.107
)
It s
i clea
r fro
m (8.107
) tha
t th
e outpu
t voltag
e s
i a
n integra
l (weighte
d integral
) of
the sourc
e voltage
. For thi
s reason
, th
e circui
t show
n n
i Figur
e 8.2
4 s
i calle
d a
n
integrator
.t
Is
i importan
to
t not
e tha
t th
e abov
e circui
t als
o ensure
s th
e appropriat
e
initia
l conditio
n fo
r th
e outpu
t voltage
.
8A6
Differentiator
A differentiato
r circui
ts
i show
n n
i Figur
e 8.25
. Here
, ther
es
i a feedbac
k branc
h wit
h
resistanc
e Rf whic
h connect
s th
e outpu
t termina
l wit
h th
e invertin
g terminal
. The
Chapter 8. Dependent Sources and Operational Amplifiers
324
rAAAAn
1/rt
1
* T
->— K
R
—* -
vs «6
v0w
Figur
e 8.25
: Differentiator
.
curren
ti(t) throug
h thi
s feedbac
k branc
h s
i give
n by
:
v\(t)-v0(t)
m=
Rf
(8.108
)
Sinc
e th
e inpu
t resistanc
e (impedance
) of th
e operationa
l amplifie
rs
i assume
d o
t be
infinite
, th
e sam
e current
, i(t), flow
s throug
h capacito
r C. The curren
t throug
h th
e
capacito
r ca
n be expresse
d a
s follows
:
/(f) = C
d{vs(t) - vi(Q
)
dt
(8.109
)
Sinc
e th
e noninvertin
g termina
ls
i connecte
d o
t th
e groun
d terminal
, we have
:
v2(0 = 0.
(8.110
)
From (8.56
) an
d (8.110)
, we find:
V!(0 = 0.
(8.111
)
By substitutin
g (8.111
) int
o (8.108
) an
d (8.109
) an
d by eliminatin
g th
e curren
t i(t)
fro
m thes
e tw
o equations
, we obtain
:
v0(t)
cdvs{t)
Ri
dt
v0(t) = -CRf
dvs(t)
dt
(8.112
)
From (8.112)
, we arriv
e at
:
(8.113
)
Thus, th
e outpu
t voltag
e s
i proportiona
lo
t th
e tim
e derivativ
e of th
e voltag
e source
.
For thi
s reason
, th
e circui
t show
n ni Figur
e 8.2
5 s
i calle
d a differentiator
.
325
8.4. Operational Amplifiers
8.4.7
Application of Operational Amplifiers to the Integration of
Differential Equations (Analog Computer)
Conside
r th
e followin
g initia
l valu
e problem
: find
th
e solutio
n o
t th
e equatio
n
d2v(t)
a 2 ^ ^ +
a
dv(t)
^ +
a
ov(0 = fit)
(8.114
)
subjec
to
t th
e initia
l condition
s
v(0+
) = V0,
(8.115
)
^ ( 0+ ) = Vj
.
(8.116
)
dt
, a0,f(t), VQ, an
d V\ ar
e known
.
In th
e abov
e expression
s a2, ai
Befor
e constructin
g a
n electri
c circui
t wit
h operationa
l amplifier
s whic
h inte
grate
s th
e initia
l valu
e proble
m (8.114)—(8.116)
, we rewrit
e differentia
l equatio
n
(8.114
)a
s follows
:
d2v(t)
— j^ar
=
ax dv(t)
a0
f(t)
— - —v(t) +
.
a2 at
a2
a2
,Q i n ,
(8.117
)
Now , le
t us assum
e fo
r th
e tim
e bein
g tha
t we hav
e a voltag
e equa
lo
t d2v(t)/dt2.
This assumptio
n s
i made jus
tn
i orde
ro
t star
t th
e desig
n of ou
r circuit
. At th
e en
d
of ou
r design
, we shal
l se
e ho
w thi
s voltag
e ca
n be obtaine
d and
, consequently
, ou
r
s th
e inpu
to
t th
e integrato
r
assumptio
n wil
l be removed
. We appl
y voltag
e d2v(t)/dt2 a
n n
i Figur
e 8.26
.f
IR\ an
d Cj\ ar
e chose
n ni suc
h a
circui
t containin
g op-am
p 1 show
way tha
tR\ Cf\ = 1
, the
n th
e outpu
t voltag
e of thi
s integrato
rs
i equa
lo
t —dv(t)/dt.
W e als
o choos
e th
e dc voltag
e sourc
e Vs\ (i
n paralle
l wit
h Cf\) equa
lo
t — V\, whic
h
guarantee
s initia
l conditio
n (8.116
) fo
r dv(t)/dt. The first-stage
outpu
t voltag
e of
—dv(t)/dt serve
s a
s th
e inpu
t fo
r th
e secon
d integrato
r show
n n
i th
e sam
e figure.
f
I
R2 an
d C/
2 ar
e chose
n suc
h tha
tR2Cf2 = 1
, the
n th
e outpu
t voltag
e of thi
s integrato
r
is equa
lo
t v(t). We als
o choos
e th
e dc voltag
e sourc
e Vs2 (i
n paralle
l wit
h Cf2) equa
l
s guarantee
s initia
l conditio
n (8.115
) fo
r v(t). The circui
t voltage
s equa
lo
t
to VQ. Thi
—dv(t)/dt an
d f(t) serv
e a
s th
e input
s fo
r th
e adde
r circui
t containin
g op-am
p 3. We
choos
e 7?3
, 7?
d Rf3 suc
h tha
t
4, an
* £ = !,
K3
*£ = f!
.
a2
K4
(8.118
)
a2
Then, th
e outpu
t voltage
, v3(f)
, of thi
s adde
r circui
ts
i equa
l to
:
a\ dv(t)
fit)
v3(0 = — — r 1 - —•
(8.119
)
a2 dt
a2
This voltag
e an
d th
e voltag
e equa
lo
t v(t) serv
e a
s th
e input
s fo
r th
e adde
r circui
t
containin
g op-am
p 4. We choos
e 7?
,
7?
,
an
d
Rf
suc
h
tha
t
5
6
4
|l
K5
= ,
l f
K6
=
a2
"Ji.
(8.120
)
326
Chapter 8. Dependent Sources and Operational Amplifiers
Then, th
e outpu
t voltage
,v4(t), of thi
s adde
r circui
ts
i equa
l to
:
a0
(8.121
)
v4(0 = -v
)
-v(t).
3 (r
a2
From (8.121
) an
d (8.119)
, we find:
a\ dv{t)
(8.122
)
v 4 (0 = a2
a2
a2 dt
Now , we connec
t th
e outpu
t of adde
r circui
t 4 wit
h th
e inpu
t of integrato
r 1. n
I thi
s
way we forc
e th
e equalit
y
d2v{t)
(8.123
)
dt2 = v4(0,
which accordin
go
t (8.122
)s
i equivalen
to
t differentia
l equatio
n (8.117)
. Thus
, we ca
n
conclud
e tha
t th
e voltag
e v(t) n
i th
e circui
t show
n n
i Figur
e 8.2
6 satisfie
s differentia
l
equatio
n (8.114
) an
d initia
l condition
s (8.115
) an
d (8.116)
. Consequently
, fi a
t tim
e
t = 0 we simultaneousl
y ope
n switche
s SW1 an
d SW2 an
d clos
e switc
h SW3 an
d
the
n measur
e th
e output
,v(t), of integrato
r 2a
s afunctio
n of time
, we find
th
e solutio
n
of initia
l valu
e proble
m (8.114)—(8,116)
.n
I othe
r words
, th
e electri
c circui
t show
n n
i
Figur
e 8.2
6 ca
n be considere
d a
s a
n analo
g compute
r whic
h solve
s th
e initia
l valu
e
proble
m (8.114)-(8.116)
.
v4(t
)
Figur
e 8.26
: Circui
t whic
h integrate
s th
e initia
l valu
e problem
.
8.4. Operational Amplifiers
327
EXAMPL E 8.
8 We shal
l desig
n a
n electri
c circui
t wit
h operationa
l amplifier
s whic
h
solve
s th
e followin
g initia
l valu
e problem
:
1
d2v{t)
dt2
~v(t),
(8.124
)
v(0+) = V0,
(8.125
)
dv
(0+) = 0.
dt
A s before
, we assum
e fo
r th
e tim
e bein
g tha
t we hav
e a voltag
e equa
lo
t
(8.126
)
_1_\ d2v(t)
~^J dt2 '
This assumptio
n wil
l be remove
d a
t th
e en
d of ou
r design
. We appl
y th
e voltag
e
(l/o)2)d2v(t)/dt2
a
s th
e inpu
to
t integrato
r 1 show
n n
i Figur
e 8.27
. We choos
e R\
and Cf\ suc
h tha
tR\Cf\ = 1/w
, the
n th
e outpu
t voltag
e of thi
s integrato
rs
i equa
l
to —( 1/V
) dv(t)/dt. We als
o us
e a short-circui
t branc
h n
i paralle
l wit
h feedbac
k
capacito
r C/
i ni orde
ro
t guarante
e zer
o initia
l conditio
n (8.126
) fo
r dv{t)/dt. The
voltag
e of (—l/a))dv/dt serve
s a
s th
e inpu
t fo
r integrato
r 2. We choos
e R2 an
d C/
2
such tha
tRiCf2 = 1 A>, the
n th
e outpu
t voltag
e of thi
s integrato
rs
i equa
lo
t v(t).
W e als
o us
e th
e dc voltag
e sourc
e VQ ni paralle
l wit
h feedbac
k capacito
r C/
2 ni orde
r
to guarante
e initia
l conditio
n (8.125
) fo
r th
e voltag
e v(t). Thi
s voltag
e serve
s a
s th
e
e R3 an
d Rf?, suc
h tha
t
inpu
t fo
r invertin
g amplifie
r 3. We choos
R/3 _
(8.127
)
= 1
.
R^
Then, th
e outpu
t voltag
e of invertin
g amplifie
r 3s
i equa
lo
t — v(t). Now, we connec
t
h th
e inpu
t of integrato
r 1
.n
I thi
s way
, we forc
e
the outpu
t of invertin
g amplifie
r 3 wit
the equalit
y (8.124)
. Thus
, we ca
n conclud
e tha
t fi a
t tim
e t = 0 we simultaneousl
y
open switche
s SW1 an
d SW2 an
d clos
e switc
h SW3, the
n th
e voltag
e v(t) n
i th
e
^
SW 1
Oi
1 d2 v(t
)
2
(a2 dt
r-AA/Wn
rAA/WH
1
w
W
VWHdv(t
)
dt
<X
*-WWH-
SW 3
O
Figur
e 8.27
: Generato
r circui
t fo
r harmoni
c electri
c oscillations
.
328
Chapter 8. Dependent Sources and Operational Amplifiers
circui
t wil
l satisf
y differentia
l equatio
n (8.124
) an
d initia
l condition
s (8.125
) an
d
(8.126)
.t
I si eas
y o
t se
e tha
t th
e solutio
n of initia
l valu
e proble
m (8.124)—(8.126
) ha
s
the form
:
KO = VQ COS Q)t.
(8.128
)
Consequently
, fi we measur
e th
e voltag
e v(t) betwee
n th
e outpu
t termina
l of integrato
r
2 an
d th
e groun
d terminal
, we find
ou
t tha
t thi
s voltag
e exhibit
s undampe
d harmoni
c
oscillations
.n
I othe
r words
, th
e circui
t show
n ni Figur
e 8.2
7 ca
n be considere
d a
s a
generato
r of harmoni
c electri
c oscillations
. The frequenc
y of thes
e oscillation
s ca
n
be controlle
d by varyin
g th
e resistanc
e of resistor
s R\ or R2.
The questio
n ca
n be aske
d ho
w th
e circui
t show
n n
i Figur
e 8.2
7 ca
n generat
e
undampe
d oscillation
s when ther
e ar
e resistor
s an
d henc
e energ
y losse
sn
i th
e circuit
.
i tha
t thes
e energ
y losse
s ar
e replenishe
d by th
e dc powe
r supplie
s whic
h
The answe
rs
are connecte
d o
t th
e operationa
l amplifiers
.
A remarkabl
e propert
y of th
e abov
e circui
ts
i tha
tt
i generate
s undampe
d har
monic oscillation
s withou
t usin
g inductor
s (compar
e wit
h th
e LC circui
t discusse
d n
i
the previou
s chapter)
. Thi
s featur
e s
i typica
l fo
r moder
n activ
e RC circuit
s whic
h ca
n
perfor
m divers
e function
s withou
t utilizin
g expensiv
e an
d bulk
y inductors
. As fa
r a
s
the cos
t of operationa
l amplifier
s s
i concerned
,t
is
i quit
e lo
w du
eo
t th
e remarkabl
e
progres
sn
i semiconducto
r technology
.
■
EXAMPLE 8.9 We want o
t desig
n a
n op-am
p circui
t tha
t generate
s th
e dampe
d
oscillatio
n v{t) = e~f sin(0
- Use a
s fe
w op-amp
s a
s possible
; limi
t th
e resistor
s o
t
100 kf
l an
d make th
e capacitor
s a
s smal
la
s possible
.
First
, we nee
d o
t determin
e th
e differentia
l equatio
n tha
t describe
s thi
s function
.
By takin
g tw
o derivative
s of v(t) an
d performin
g a fe
w algebrai
c manipulations
, we
quickl
y findtha
td2v/dt2 = —2{dv/dt) — 2v
. We wil
l nee
d tw
o op-am
p integrato
r
circuits
, on
e summer circuit
, an
d possibl
y on
e inverte
r circuit
. Let'
s se
e fi we ca
n
avoi
d th
e inverte
r by bein
g a bi
t clever
. Conside
r th
e circui
t show
n n
i Figur
e 8.28
.
The ke
y o
t avoidin
g th
e inverte
r s
in
i th
e rightmos
t op-am
p circuit
. At first
glance
,t
i appear
s o
t be a circui
t tha
t we hav
e no
t analyze
d before
. However
, fi
dzv
dt 2
Figure 8.28
: Circui
t whic
h generate
s th
e functio
n v{t) = e * sin(0
.
8.5. MicroSim PSpice Simulations
329
the voltag
e a
t th
e noninvertin
g termina
l wer
e zero
,t
i woul
d be a simpl
e invertin
g
amplifier
. Likewise
, fiv(t) wer
e se
to
t zero
, th
e op-am
p circui
t woul
d be a simpl
e
noninvertin
g amplifier
. Fro
m th
e superpositio
n principle
, we se
e tha
t th
e outpu
t of
thi
s circui
ts
i jus
tvx(l +R5/R4) — v(t)R5/R4. The differentia
l equatio
n require
s tha
t
R5/R4 = 2, s
o we ca
n tak
e R5 = 10 kf
l an
d R4 = 5 kfl
. Thi
s mean
s tha
t we must
have vx = — \{dv/di). Equatio
n (8.97
) tell
s us tha
tR\ C\ must be 3/2
,s
o we ca
n tak
e
R\ = 10
0 kl
l an
d Ci = 1
5 JWF. The sam
e equatio
n place
s th
e conditio
n R2C2 = 2/3
,
so we ca
n tr
y R2 = 10
0 kl
l an
d C2 = 6.6
6 /JLF. Ther
es
i no direc
t expressio
n fo
r R3
sinc
e no curren
t flow
s throug
h it
, bu
tn
i practic
e t
is
i usuall
y bes
to
t make t
i equa
lo
t
R4. The determinatio
n of V\ s
i lef
ta
s a
n exercis
e fo
r th
e reader
.
8.5
MicroSi
m PSpic
e Simulation
s
In thi
s section
, we wil
l us
e PSpic
e simulation
s o
t examin
e th
e performanc
e of si
x
circuit
s whic
h contai
n dependen
t source
s an
d operationa
l amplifiers
. The first
circui
t
tha
t we wil
l conside
r contain
s tw
o dependen
t source
s an
d was originall
y draw
n n
i
Figure
s 8.1
0 an
d 8.15
.n
I th
e latte
r figure,
on
e of th
e impedance
s (Z
s replace
d
3) wa
by a probin
g curren
t sourc
e o
t find
th
e Theveni
n equivalen
t nonidea
l voltag
e sourc
e
at th
e ope
n terminals
. We wil
l us
e PSpic
eo
t find
th
e Theveni
n voltag
e an
d impedanc
e
for th
e specifi
c cas
e of a dc circui
t wit
h Z{ = Rl = 5 fl
,Z2:= R2 = 10 H, Vvl =5
V, A = 2, an
d G = 0.
5 (thes
e variable
s ar
e define
d n
i Exampl
e 8.5)
. Fo
r thes
e values
,
d a
n impedanc
e ofZTH
(8.44
) yield
s a Theveni
n equivalen
t voltag
e ofWTH = 10 V an
= 30 a.
The circuit
, a
s draw
n by th
e schematic
s editor
,s
i show
n ni Figur
e 8.29
. The
independen
t voltag
e an
d curren
t source
s us
e "VSRC" an
d "ISRC" models
, respec
tively
, an
d ar
e give
n dc value
s a
s indicate
d above
. The voltage-controlle
d voltag
e
Figure 8.29: The schemati
c drawin
g o
f th
e circui
to
f Exampl
e 8.5
.
330
Chapter 8. Dependent Sources and Operational Amplifiers
sourc
e mode
ls
i containe
d ni th
e "analog.sib
" librar
y an
d ca
n be selecte
d by enterin
g
an "E" n
i th
e "Ad
d Part
" dialo
g box
. Not
e tha
t th
e VCVS par
t doesn'
t utiliz
e th
e
standar
d symbo
l bu
ts
i give
n instea
d by a four-termina
l box
. Thi
s s
i don
e s
o tha
t
the controllin
g voltag
e ca
n be specifie
d simpl
y by connectin
g th
e plu
s an
d minu
s
"sensing
" terminal
s o
t th
e appropriat
e circui
t node
s vi
a wires
. The tw
o terminal
s
which ar
e connecte
d o
t th
e usua
l voltag
e sourc
e symbo
l containe
d withi
n th
e bo
x ar
e
the terminal
s of th
e dependen
t source
. Double-clickin
g on th
e VCVS symbo
l bring
s
up th
e mode
l attribute
s dialo
g box
. The gai
n shoul
d be se
to
t 2 vi
a th
e "Sav
e Attr
"
button
. The par
t name fo
r th
e voltage-controlle
d curren
t sourc
es
i "G" an
d th
e mode
l
is als
o a four-por
t devic
e tha
ts
i containe
d n
i th
e "analog.sib
" library
. The gai
n fo
r
the VCIS shoul
d be se
to
t 0.
5 usin
g th
e procedur
e outline
d fo
r th
e othe
r dependen
t
source
.
Under "Setup...
"n
i th
e Analysi
s drop-dow
n menu we selec
t a "DC Sweep.
"n
I
the "DC Sweep
" dialo
g bo
x we selec
t "Curren
t source
" fo
r th
e "Swep
t Var
. Type
" an
d
"Linear
" fo
r th
e "Swee
p Type.
" We ente
r "Ip
" unde
r "Name
" an
d selec
t a star
t valu
e
of zero
,a
n en
d valu
e of two
, an
d a
n incremen
t of 0.1
. The result
s of th
e simulatio
n
are indicate
d n
i Figur
e 8.30
, wher
e th
e voltag
e acros
s th
e probin
g sourc
e s
i plotte
d
as a functio
n current
. The zer
o intercep
t give
s th
e Theveni
n equivalen
t voltag
e an
d
the slop
e of th
e lin
e s
i equa
lo
t th
e Theveni
n impedance
. As expected
, thes
e value
s
agre
e wit
h th
e result
s of (8.44)
.
The nex
t circui
t tha
t we examin
e come
s fro
m Exampl
e 8.
6 an
d s
i redraw
n wit
h
the schematic
s edito
rn
i Figur
e 8.31
. Thi
s tim
e we wil
l us
e a probin
g voltag
e sourc
e
and a dc swee
p analysi
s wit
h PSpic
e o
t determin
e th
e Norto
n equivalen
t nonidea
l
curren
t sourc
e fo
r th
e resistance
s give
n n
i th
e figure.
The dc sourc
e parameter
s fo
r
thi
s exampl
e ar
e Vsi = 2 V, Vs3 = 5 V, A = 3, an
d R = 10 £
1 (thes
e value
s ar
e define
d
ou
.
70
^
^60
>
0) 50
O)
2 40
o
>
■5 30
^ y ^
Q.
=5 20
0
^
^—
^ y ^
^^^
^y^
^^^
^ s ^
^^^^
^^^
^^^
10
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Probing current (A)
1.4
1.6
1.8
2.0
Figure 8.30: The simulate
d outpu
t voltag
e a
s a functio
n o
f probin
g curren
t fo
r th
e
circui
to
f Exampl
e 8.5
.
8.5. MicroSim PSpice Simulations
331
R5<T15
Figur
e 8.31
: The schemati
c drawin
g o
f th
e circui
to
f Exampl
e 8.6
.
in Figur
e 8.16)
. The par
t name fo
r th
e current-controlle
d voltag
e sourc
e s
i "H" an
d
the model
, containe
d n
i th
e "analog.sib
" library
,s
i onc
e agai
n a four-por
t device
. The
directio
n of th
e controllin
g curren
ts
i indicate
d by th
e directio
n of th
e arro
w whic
h s
i
connecte
d betwee
n th
e tw
o "sensing
" terminals
. (Note
: th
e current-controlle
d curren
t
source
, whic
h s
i no
t considere
d n
i eithe
r of thes
e examples
, ha
s th
e par
t name "F
"
and s
in
i th
e sam
e librar
y a
s th
e othe
r dependen
t sources.
)
In th
e "DC Sweep
" dialo
g bo
x we selec
t "Voltag
e source
" fo
r th
e "Swep
t Var
.
Type" an
d "Linear
" fo
r th
e "Swee
p Type.
" We ente
r "Vp" unde
r "Name
" an
d selec
ta
star
t valu
e of zero
,a
n en
d valu
e of ten
, an
d a
n incremen
t of 0.2
. The curren
t throug
h
the probin
g source
,a
s calculate
d by PSpice
,s
i plotte
d n
i Figur
e 8.3
2 a
s a functio
n
prob
e voltage
. Accordin
g o
t equatio
n (8.48)
, th
e zer
o intercep
t give
s th
e Norto
n
equivalen
t curren
t an
d th
e negativ
e of th
e slop
e of th
e lin
e s
i equa
lo
t th
e Norto
n
admittance
. The circui
t value
s give
n above
, when inserte
d n
i equation
s (8.49
) an
d
9 A an
d YN0 = 0.4
3 U. Again
, thes
e
(8.50)
, resul
tn
i th
e Norto
n parameter
s INO = 0.2
value
s ar
e virtuall
y identica
lo
t thos
e indicate
d on th
e PSpic
e outpu
t graph
.
The schemati
c fo
r a noninvertin
g amplifie
r circuit
, wit
h th
e nonidea
l op-am
p
circui
t show
n n
i Figur
e 8.19
,s
i indicate
d n
i Figur
e 8.33a
. The mode
l fo
r th
e op-am
p
include
s th
e voltage-controlle
d voltag
e sourc
e Av wit
h a gai
n of 10,00
0 an
d th
e tw
o
resistor
s Ri an
d Ro. An a
c voltag
e sourc
e (par
t name = "VSRC"
) wit
h a
n amplitud
e
of 1 V s
i th
e input
. The gai
n of a
n idea
l noninvertin
g amplifie
rs
i Ay = 1 + Rf_
a /Rl _
a
= 11. An "AC Sweep
" analysi
ss
i selecte
d wit
h 21 point
s pe
r decad
e calculate
d fro
m
100 Hz o
t 10
0 kHz. The outpu
t voltag
e magnitud
e ove
r tha
t rang
e of frequencies
,
indicate
d by th
e soli
d lin
e n
i Figur
e 8.34
, si abou
t 0.12
% belo
w th
e idea
l valu
e of
332
Chapter 8. Dependent Sources and Operational Amplifiers
1
^-—^
<
0
c 1-
CD
i^_
&_
3 -2
•+-*
D
a.
O
3 3
-4
3
4
5
6
Probing voltage (V)
7
10
Figure 8.32: The simulate
d outpu
t curren
t a
s a functio
n o
f probin
g voltag
e fo
r th
e
circui
to
f Exampl
e 8.6
.
11 V, Our simpl
e mode
l of a
n op-am
p s
i independen
t of frequency
, of course
,s
o th
e
voltag
e curv
es
i jus
t a lin
e wit
h zer
o slope
.
Sinc
e practica
l circuit
s ofte
n contai
n many op-amps
,t
i woul
d be usefu
l fi th
e se
t
of element
s tha
t make up a
n op-am
p mode
l (Av
, Ri
, an
d Ro fo
r ou
r simpl
e model
)
coul
d be combine
d int
o on
e bloc
k tha
t coul
d be inserte
d int
o a schemati
c a
s a singl
e
entity
. PSpic
e ha
s th
e mean
s o
t do thi
s by definin
g a "subcircuit.
" n
I th
e window
s
versio
n of PSpice
, subcircuit
s ca
n be create
d by th
e "Creat
e Subcircuit
" lin
e ni th
e
circuit (a)
0
Vi_a
Rf a > 10k
Vi b
Figure 8.33: Noninvertin
g amplifie
r circuits
: (a
) nonidea
l op-am
p mode
l an
d (b
) LM
324 op-am
p model
.
\
CO
Output voltage (V)
8.5. MicroSim PSpice Simulations
circuit (a)
" " ■ circuit (b)
\
\
\
\
\
\
\
\
\
\
1
\
10'
10°
^0A
10'
Frequenc
y (Hz
)
Figur
e 8.34
: The frequenc
y dependenc
e o
f th
e outpu
t voltage
.
"Tools
" drop-dow
n menu
. Subcircui
t generatio
n s
i beyon
d th
e scop
e of thi
s text
,
however
, an
d th
e intereste
d reade
rs
i referre
d o
t th
e reference
s give
n n
i Appendi
x C.
W e wil
l instea
d us
e on
e of th
e op-am
p model
s tha
ts
i provide
d n
i th
e "eval.slb
"
library
. The par
t name fo
r thi
s commonl
y use
d op-am
p s
i "LM324
" an
d t
i ca
n be
obtaine
d fro
m th
e "Ge
t New Part
" lin
e of th
e "Draw
" drop-dow
n menu n
i th
e usua
l
way. Thi
s mode
ls
i considerabl
y mor
e sophisticate
d tha
n th
e nonidea
l mode
l picture
d
in Figur
e 8.33
a an
d s
i base
d on th
e actua
l makeu
p of th
e devic
e (transistors
, resistors
,
etc.
, an
d thei
r interconnections)
. A thoroug
h analysi
s of th
e operatio
n of thi
s op
amp s
i a subjec
t fo
r a cours
e on electronics
; herewe wil
l jus
t mentio
n tw
o genera
l
propertie
s of rea
l op-amp
s tha
t shoul
d be understoo
d prio
ro
t thei
r introductio
n int
o
PSpic
e schematics
. First
, the
y requir
e connection
so
t externa
l powe
r supplie
s whic
h
provide
, amon
g othe
r things
, th
e energ
y necessar
y o
t amplif
y inpu
t signals
. Often
,
two powe
r supplie
s ar
e use
d wit
h equa
l voltag
e magnitude
s bu
t opposit
e sign
s (±1
2
V ar
e widel
y use
d fo
r many applications)
. Thi
ss
i required
, fo
r example
,o
t produc
e
an amplifie
d sinusoida
l signa
l wit
h zer
o offse
t voltage
, becaus
e th
e rang
e of possibl
e
outpu
t voltage
ss
i necessaril
y containe
d betwee
n th
e tw
o sourc
e voltages
. Second
, op
amps exhibi
t frequenc
y dependence
s du
e o
t eithe
r capacitor
s whic
h ar
e deliberatel
y
place
d n
i th
e op-am
p circui
t or capacitance
s tha
t ar
e a consequenc
e of transisto
r
structure
. Typically
, thes
e capacitance
s resul
tn
i a decreas
e n
i th
e open-loo
p gai
n a
s
the driv
e frequenc
y s
i increased
.
The noninvertin
g circui
t wit
h th
e LM324 op-am
p s
i show
n ni Figur
e 8.33b
.
The op-am
p require
s five
connections
. Thre
e of th
e connection
s ar
e th
e sam
e a
s fo
r
our simpl
e model
: th
e invertin
g termina
l ( -,) th
e noninvertin
g termina
l (+)
, an
d
the outpu
t termina
l (no
t marked)
. The othe
r tw
o terminal
s ar
e fo
r th
e powe
r suppl
y
connections
. Vp an
d Vm ar
e dc powe
r supplie
s of +12 an
d —12 V, respectively
. The
y
are connecte
d o
t th
e V+ an
d V— terminal
s of th
e op-am
p vi
a "bus
" lines
. Bus line
s
334
Chapter 8. Dependent Sources and Operational Amplifiers
are typicall
y use
d n
i a circui
t when many connection
s ar
e require
d o
t th
e sam
e poin
t
and ca
n be generate
d by th
e "Bus
" lin
e of th
e "Draw
" drop-dow
n menu
. The outpu
t
voltag
e terminal
s of powe
r supplie
s ar
e excellen
t example
s of bu
s lines
. The gai
n of
thi
s circuit
, whic
h ideall
y shoul
d be th
e sam
e a
s tha
tn
i th
e circui
t of Figur
e 8.33a
,
is give
n by th
e dashe
d lin
e n
i Figur
e 8.34
. The low-frequenc
y gai
n of th
e op-am
p
circui
ts
i slightl
y highe
r tha
n n
i th
e previou
s case
, bu
t th
e gai
n drop
s of
f sharpl
y a
s
the frequenc
y goe
s abov
e 10 kHz. Not
e tha
t th
e tw
o circuit
s ar
e place
d n
i th
e sam
e
schemati
c drawin
g an
d ar
e simulate
d a
t th
e sam
e tim
e (henc
e wit
h th
e sam
e analysi
s
specifications
)n
i PSpice
.
The nex
t op-am
p circui
t tha
t we wil
l investigat
es
i th
e integratin
g circui
t show
n
in Figur
e 8.35
. The suppl
y voltage
s ar
e ±12 V an
d th
e resistanc
e an
d capacitanc
e
are take
n s
o tha
t RICf = 1
. The voltag
e sourc
e o
t be integrate
d use
s a piecewis
e
linea
r mode
l an
d ha
s th
e par
t name "VPWL" n
i th
e "source.slb
" library
. The tim
e
dependenc
e of th
e inpu
t voltag
e s
i indicate
d by th
e soli
d lin
e n
i Figur
e 8.36
. Thi
s
voltag
e rise
s fro
m zer
o o
t 6 volt
sn
i 1 second
, remain
s a
t 6 volt
s fo
r 2 seconds
, fall
s
back o
t zer
o n
i 1 second
, an
d the
n decrease
so
t -1
2 Vn
i th
e final
second
. Two case
s
are simulate
d vi
a a parametri
c analysis
: Vo = 0 V an
d Vo = 5 V.
A transien
t analysi
s was conducte
d fo
r 5 second
s an
d th
e result
s fo
r bot
h initia
l
condition
s ar
e plotte
dn
i Figur
e 8.36
. The nonzer
o initia
l conditio
n cas
es
i plotte
d wit
h
the dot-dashe
d line
. As expected
, th
e outpu
ts
i paraboli
c when th
e inpu
ts
i changin
g
linearl
y an
d varie
s linearl
y when th
e inpu
ts
i constant
. The outpu
t decrease
s fo
r
positiv
e inpu
t voltage
s du
e o
t th
e invertin
g natur
e of th
e integrator
. The zer
o initia
l
conditio
n cas
e outpu
ts
i indicate
d by th
e dashe
d line
. The circui
t clearl
y doe
s no
t
tOpen=0
Figur
e 8.35
: An integrato
r circui
t wit
h a
n LM324 op-amp
.
335
8.5. MicroSim PSpice Simulations
1U
5
> n ^ - .* ^
CD
x
*
N
N
CO
?"5
\
>
-10
Source
"
v
V
Vo = 0 V
x
/
A
N
^ • ^ _—
*
•
/
\
Vo = 6 V
i
i
0
1
i
2
Time (s)
i
i
3
4
i
Figur
e 8.36
: Inpu
t an
d outpu
t voltage
s o
f th
e integrato
r circuit
.
integrat
e th
e inpu
t signa
l afte
r abou
t 2.
5 seconds
. The proble
m arise
s becaus
e th
e
outpu
t voltag
e attempt
s o
t go belo
w th
e minimu
m allowabl
e valu
e of Vm = —12
V. We sa
y tha
t th
e outpu
t ha
s "saturated
" durin
g thi
s time
.t
Is
i interestin
g o
t not
e
tha
t ther
e s
i a dela
y afte
r th
e inpu
t goe
s negativ
e befor
e th
e outpu
t agai
n track
s
the (negativ
e of the
) integra
l of th
e inpu
t signal
. The intereste
d reade
r ca
n finda
n
explanatio
n fo
r thi
s phenomeno
n by plottin
g th
e voltag
e a
t th
e invertin
g termina
l
Afte
r th
e outpu
t saturates
,t
is
i no longe
r necessaril
y tru
e tha
t V+ = V— an
d th
e
outpu
t doe
s no
t begi
n o
t functio
n properl
y agai
n unti
l thi
s conditio
n s
i restored
.
The final
PSpic
e simulatio
n presente
d n
i thi
s sectio
n involve
s th
e harmoni
c
generato
r op-am
p circui
t of Exampl
e 8.8
. Thi
s circuit
, wit
h parameter
s selecte
d o
t
generat
e a
n angula
r frequenc
y of 1
0 rad/s
,s
i show
nn
i Figur
e 8.37
. The powe
r supplie
s
Figur
e 8.37
: The schemati
c drawin
g o
f th
e circui
to
f Exampl
e 8.8
.
Chapter 8. Dependent Sources and Operational Amplifiers
336
0.8
1.
0
1.
2
2.0
Time (s
)
Figur
e 8.38
: The outpu
t voltag
e o
f th
e circui
to
f Exampl
e 8.8
.
are agai
n ±12 V an
d th
e initia
l voltag
e s
i Vo = 5 V. A transien
t analysi
ss
i selecte
d
wit
h a tim
e ste
p of 5 ms an
d a final
tim
e of 2 s
. The tim
e dependenc
e of th
e outpu
t
voltag
es
i give
nn
i Figur
e 8.3
8 an
d clearl
y indicate
s th
e sinusoida
l natur
e of th
e outpu
t
wit
h th
e prope
r initia
l conditio
n an
d a frequenc
y of f = 10/(277
) HZ.
8.6
Summary
The mai
n result
s of thi
s chapte
r ca
n be summarize
d a
s follows
:
• Fou
r type
s of dependen
t source
s hav
e bee
n introduced
. The
y ar
e th
e voltage
controlle
d voltag
e sourc
e (VCVS)
, current-controlle
d voltag
e sourc
e (ICVS)
,
voltage-controlle
d curren
t sourc
e (VCIS)
, an
d current-controlle
d curren
t
sourc
e (ICIS)
.t
I ha
s bee
n demonstrate
d tha
t dependen
t source
s usuall
y appea
r
as linear models for transistors an
d semiconducto
r devices
.
• The noda
l analysi
s an
d mes
h analysi
s technique
s fo
r circuit
s wit
h dependen
t
source
s hav
e bee
n presented
. Thes
e technique
s involv
e th
e followin
g steps
:
(a) Writ
e th
e noda
l potentia
l or mes
h curren
t equation
s a
s fi al
l th
e source
s
were independent
, (b
) Expres
s th
e dependen
t source
sn
i term
s of th
e unknow
n
variables
. Fo
r noda
l analysis
, thi
s mean
s o
t expres
s th
e dependen
t source
sn
i
term
s of noda
l potentials
. Fo
r mes
h curren
t analysis
, th
e dependen
t source
s
shoul
d be expresse
d n
i term
s of mes
h currents
, (c
) Substitut
e th
e abov
e expres
sion
s fo
r th
e dependen
t source
sn
i term
s of unknow
n variable
s int
o th
e origina
l
equations
. Then
, modif
y th
e coefficien
t matri
x by movin
g thes
e expression
so
t
the left-han
d sid
e of th
e equations
.
337
8.7. Problems
• The applicatio
n of Thevenin'
s theore
m o
t th
e analysi
s of electri
c circuit
s wit
h
dependen
t source
s ha
s bee
n discussed
.t
I ha
s bee
n pointe
d ou
t tha
t th
e calcu
latio
n of th
e open-circui
t voltag
e an
d th
e short-circui
t curren
t wit
h respec
to
t
the terminal
s of th
e branc
h n
i questio
n s
i require
d n
i orde
ro
t find
th
e param
eter
s of th
e Theveni
n equivalen
t circuit
. Thes
e calculation
s ar
e substantiall
y
simplifie
d when th
e branc
h n
i questio
n s
i acontrolling branch. The "probing
"
techniqu
e fo
r th
e calculatio
n of Theveni
n an
d Norto
n equivalen
t circuit
s ha
s
been presente
d a
s well
.
• An idea
l operationa
l amplifie
r ha
s bee
n define
d a
s a voltage-controlle
d voltag
e
sourc
e wit
h infinit
e gain
, infinit
e inpu
t resistance,
, an
d zer
o outpu
t resistance
.
It ha
s bee
n demonstrate
d tha
t operationa
l amplifier
s ca
n be use
d o
t desig
n
variou
s circuit
s suc
h a
s buffe
r amplifier
s (voltag
e follower)
, noninvertin
g an
d
invertin
g amplifiers
, adde
r circuits
, integrators
, an
d differentiators
.t
I ha
s bee
n
shown tha
t th
e abov
e operationa
l amplifie
r circuit
s ca
n be utilize
d o
t desig
n
analo
g computer
s an
d sin
e wave generators
.
8.7
1.
Problem
s
Write
, bu
t do no
t solve
, th
e noda
l equation
s fo
r th
e circui
t show
n n
i Figur
e P8-1
.
Figure P8-1
2.
Fin
d th
e voltage
s of al
l th
e node
s ni th
e circui
t show
n ni Figur
e P8-2
.
rAAA/Vi
1 Q
2Q,
-e-k
+ 2A
't>2
ld 12 Q<M 1 13
/Q <
> 2
V Of)
Figure P8-2
338
3.
Chapter 8. Dependent Sources and Operational Amplifiers
Determin
e th
e valu
e of th
e resistanc
e R require
d o
t achiev
e a 2 V potentia
l differenc
e
rn
i th
e circui
t show
n n
i Figur
e P8-3
.
acros
s th
e 1 O resisto
rA/lX/V
n
v
V
+
21,
2 fl(T)2 RAQ S I /12 /i
Figure P8-3
In Problem
s 4 an
d 5, assum
e th
e source
s al
l hav
e a
n angula
r frequenc
y of o»
.
4.
Write
, bu
t do no
t solve
, th
e noda
l equation
s fo
r th
e circui
t show
n ni Figur
e P8-4
.
CD', c,
Figure P8-4
5.
Write
, bu
t do no
t solve
, th
e mes
h curren
t equation
s fo
r th
e circui
t show
n n
i Figur
e P8-5
.
Figure P8-5
6.
g fro
m
Determin
e th
e valu
e of th
e resistanc
e R whic
h result
sn
i a curren
t of 1 A emergin
the positiv
e termina
l of th
e dependen
t voltag
e sourc
en
i th
e circui
t show
n n
i Figur
e P8-6
.
8.7. Problems
339
3 D. + <
>
1 A
2Q
1
Q )2 V R &
VR S l /
2 Q
3 Q. ' <
1 Q
2V,
2a
Figur
e P86
7. Fin
d al
l of th
e mes
h current
sn
i th
e circui
t show
n ni Figur
e P8-7
.
8. Write
, bu
t do no
t solve
, th
e mes
h curren
t equation
s fo
r th
e circui
t show
n n
i Figur
e P8-8
.
O
5cos(3t
•5 H 3V
10 Q
fAA/Vv
3Q
V 1 J 2 H -2V1
sin(3t)
-O-
AAA/V
Figure P8-7
7 ft
AAAA^
Figure P8-8
9.
Remove th
e 1/
3 ftresisto
r fro
m th
e circui
t show
n ni Figur
e P82 an
d find
th
e Theveni
n
equivalen
t circui
t wit
h respec
to
t th
e ope
n terminals
.
10
Remove th
e 2 H resisto
r fro
m th
e circui
t show
n ni Figur
e P83 an
d findth
e Norto
n
equivalen
t circui
t wit
h respec
to
t th
e ope
n terminals
. Le
t fi = 1(1
.
11
Remove th
e 1/
2 ftresisto
r fro
m th
e circui
t show
n ni Figur
e P86 an
d findth
e Norto
n
equivalen
t circui
t wit
h respec
to
t th
e ope
n terminals
. Le
tR = 1 ft.
12.
Remove th
e 5 ftresisto
r fro
m th
e circui
t show
n ni Figur
e P87 an
d findth
e Theveni
n
equivalen
t circui
t wit
h respec
to
t th
e ope
n terminals
.
13.
Conside
r th
e circui
t show
n ni Figur
e P8-13
. Solv
e fo
r al
l th
e nod
e voltages
.
Figure P8-13
340
14.
Chapter 8. Dependent Sources and Operational Amplifiers
Conside
r th
e circui
t show
n ni Figur
e P8-14
. Solv
e fo
r al
l th
e nod
e voltages
.
—K-
JM1
-
1/2 F
1Q
ww—
1 H
1/2 n
<0>
2
\2Q
i
K - ^ cos(3t
-G )
2F
2sin(3t) (T)
-N3
2£> <>
I <f
Figure P8-14
15.
Conside
r th
e circui
t show
n n
i Figur
e P8-15
. Identif
y an
d labe
l clearl
y al
l th
e meshe
s ni
the circui
t an
d solv
e fo
r al
l th
e correspondin
g mes
h currents
.
16. Conside
r th
e circui
t show
n ni Figur
e P8-16
. Solv
e fo
r al
l th
e phaso
r mes
h currents
.
-2jQ
21(3
(P-3JA
wvw
4\3Q
'
-4jQ
5jQ
Figure P8-16
Figure P8-15
17.
(
s~M^
Give
n th
e circui
t show
n ni Figur
e P8-17
, fin
d th
e Norto
n equivalen
t circui
ta
t th
e ope
n
terminals
.
vwv
Qcos(t
)V
1Q
MH
fAAAA/-fAA/VV1Q
Figure P8-17
+
341
8.7. Problems
18.
Desig
n a
n idea
l op-am
p circui
t whic
h produce
s a
n outpu
t voltag
e of vout(
0 = 1000v
3 —
100v2 + 10v
i + v0, wher
e v09 v\, v2, v3 ar
e inpu
t voltages
. Use onl
y resistor
sn
i th
e rang
e
100 1
2 o
t 1
0 kf
l an
d minimiz
e th
e numbe
r of op-amps
.
19.
Conside
r th
e circui
t show
n n
i Figur
e P8-19
. Fin
d th
e outpu
t voltag
e ni term
s of th
e inpu
t
voltag
e a
s a functio
n of vi
, v2.
R
2R
V2 o A A A / V t A A A A n
Figure P8-19
20.
Analyz
e th
e circui
t show
n ni Figur
e P8-2
0 o
t findth
e outpu
t voltag
e a
s a functio
n of
Vi,V2,...Vn.
Figure P8-20
21.
Desig
n a
n idea
l op-am
p circui
t whic
h produce
s a
n outpu
t voltag
e of vout(0 = 22v
i +
49v2 — 17v
e vi
, v2, an
d v3 ar
e inpu
t voltages
. The maximu
m resistanc
e valu
e yo
u
3, wher
may us
es
i lOkfl
.
4
2
4
2
n a
n op-am
p circui
t tha
t produce
s a
n outpu
t voltag
e of v0(t) = 5t + 12f + 6
22. Desig
give
n a
n inpu
t voltag
e sourc
e ofvs(t) = t6/6 + t4 +It. You ca
n us
e an
y combinatio
n of
capacitors
, batteries
, an
d switches
, bu
t yo
u may onl
y us
e 1
0 kO resistors
.
n a
n op-am
p circui
t tha
t produce
s a
n outpu
t voltag
e of v0(t) = 5t + 12f + 6
23. Desig
give
n a
n inpu
t voltag
e sourc
e of vs(t) = \0t3 + I2t. You ca
n us
e an
y combinatio
n of
capacitors
, batteries
, an
d switches
, bu
t yo
u may onl
y us
e 10 kf
l resistors
.
24.
Desig
n a
n op-am
p circui
t tha
t solve
s th
e differentia
l equation
:
d2v
dv
3-7T1 +4— +2v(
0 - It.
at
at
342
Chapter 8. Dependent Sources and Operational Amplifiers
Assume v(0
) = 0 an
d v'(0
) = 2 V. You ca
n us
e an
y combinatio
n of capacitors
, batteries
,
switches
, an
d resistor
s (u
p o
t 10 kft)
.
25.
Desig
n a
n op-am
p circui
t tha
t ca
n generat
e th
e functio
n v(t) = e"'(co
s It + 1/
2 si
n 2i).
You ca
n us
e an
y combinatio
n of capacitors
, batteries
, switches
, an
d resistor
s (u
p o
t 1
0
kfl)
.
26. Usin
g onl
y switches
, 10 /x
F capacitors
, resistor
s wit
h 10
0 ft R< < 10 kfl
, an
d fou
r
op-amps
, desig
n (an
d draw
) a circui
to
t realiz
e th
e followin
g equation
, wher
e vj
, v2, v3,
V4 ar
e inpu
t voltag
e sources
:
dv3
10vi - 100v
2 + 4
dt
f
v'4(T) dr.
Jo
27. Desig
n a
n op-am
p circui
t tha
t ca
n generat
e th
e functio
n v(t) = te2r
w(/)
« You ca
n us
e
any combinatio
n of capacitors
, batteries
, switches
, an
d resistor
s (u
p o
t 10 kfl)
.
28.
Analyz
e th
e op-am
p circui
t show
n ni Figur
e P8-2
8 o
t determin
e th
e tim
e dependenc
e of
the outpu
t voltag
e v0 on th
e inpu
t voltage
s v\ an
d v2.
29.
Analyz
e th
e op-am
p circui
t show
n ni Figur
e P8-2
9 o
t determin
e th
e tim
e dependenc
e of
the outpu
t voltag
e v0 on th
e inpu
t voltage
s v
j an
d v2.
v1 o - V W V
NAAA/S
Figure P8-29
Figure P8-28
30.
Analyz
e th
e op-am
p circui
t show
n ni Figur
e P8-3
0 o
t determin
e th
e tim
e dependenc
e of
the outpu
t voltag
e v0 on th
e inpu
t voltage
s v\, v2, an
d v3.
3 Q
1 Q,
2Q
r
r
7Q
11 Q.
5Q
4Q
6 Q.
AA/vV- 1
JD>
Figure P8-30
343
8.7. Problems
31.
The
p circui
t show
n n
i Figur
e P8-3
1 (fo
r t > 0) ha
s bee
n constructe
d o
t solv
e a
The op-am
specifi
c differentia
l equation
. Fin
d th
e differentia
l equatio
n whic
h th
e outpu
t voltag
e va
satisfies
. (Th
e circui
ts
i show
n fo
r tim
e / > 0; do no
t attemp
to
t find
th
e initia
l conditions.
)
Figure P8-31
Use PSpic
e o
t solv
e Problem
s 32-41
.
32.
Fin
d th
e voltage
s of al
l th
e node
s ni th
e circui
t show
n n
i Figur
e P8-2
.
33.
Determin
e th
e valu
e of th
e resistanc
e ?
7 whic
h result
s ni a curren
t of 1 A emergin
g fro
m
the positiv
e termina
l of th
e dependen
t voltag
e sourc
en
i th
e circui
t show
n n
i Figur
e P8-6
.
34.
Remove th
e 2 ftresisto
r fro
m th
e circui
t show
n n
i Figur
e P83 an
d findth
e Norto
n
equivalen
t circui
t wit
h respec
to
t th
e ope
n terminals
. LetR = 1 ft.
35.
Remove th
e 5 ftresisto
r fro
m th
e circui
t show
n n
i Figur
e P87 an
d findth
e Theveni
n
equivalen
t circui
t wit
h respec
to
t th
e ope
n terminals
.
36.
Conside
r th
e circui
t show
n ni Figur
e P8-14
. Plo
t th
e nod
e voltage
s Nl, N2, an
d N3 a
sa
functio
n of time
.
37.
Desig
n a
n op-am
p circui
t tha
t solve
s th
e differentia
l equation
:
d2v
3 —2
dt
dv
+ 4— + 2v(t) - 2t.
dt
You ca
n us
e an
y combinatio
n of capacitors
, batteries
, switches
, an
d resistor
s (u
p o
t 10
kft)
. Plo
t th
e outpu
t voltag
e an
d th
e analyti
c resul
t fo
r v(
0 ove
r a
n appropriat
e tim
e
interval
.
38.
f
Desig
n a
n op-am
p circui
t tha
t ca
n generat
e th
e functio
n v(t) = e~
(co
s It + 1/
2 si
n It).
You ca
n us
e an
y combinatio
n of capacitors
, batteries
, switches
, an
d resistor
s (u
p o
t 1
0
344
Chapter 8. Dependent Sources and Operational Amplifiers
kft)
. Plo
t th
e outpu
t voltag
e an
d th
e analyti
c resul
t fo
r v{t) ove
r a
n appropriat
e tim
e
interval
.
t th
e outpu
t
39. Desig
n a
n op-am
p circui
t tha
t ca
n generat
e th
e functio
n v(t) = te~2tu(t). Plo
voltag
e an
d th
e analyti
c resul
t fo
r v(t) fro
m t — 0o
tt = 5 .
s
40.
Plo
t th
e outpu
t voltag
e of th
e op-am
p circui
t show
n ni Figur
e P8-2
9 when v
i =
cos(lOf
) V, v2 = 2 sin(5f
) V,Rf - 1
0 kH,R = 5 kO, an
d C - 10
0 JUR
41.
Plo
t th
e outpu
t voltag
e v0{i) fo
r th
e op-am
p circui
t show
n n
i Figur
e P8-3
1 ove
r a
n
appropriat
e tim
e interval
.
Chapte
r9
Frequency Characteristics
of Electric Circuits
9.1
Introductio
n
In earlie
r chapters
, we discusse
d electri
c circuit
s tha
t wer
e drive
n by sinusoida
l
voltag
e an
d curren
t sources
. t
I was alway
s assume
d n
i ou
r previou
s discussion
s
tha
t th
e frequencie
s of thes
e source
s wer
e give
n an
d fixed.
Now, we conside
r ho
w
response
s of electri
c circuit
s var
y wit
h changin
g frequencies
.
First
, we focu
s on a simpl
e circui
t wit
h a
n a
c voltag
e source
, resistor
, inductor
,
and capacito
r al
l connecte
d ni series
. We demonstrat
e tha
t thi
s circui
t may exhibi
ta
n
interestin
g phenomeno
n calle
d resonance
. The resonanc
e conditio
n (tha
t is
, when th
e
curren
ts
i ni phas
e wit
h th
e sourc
e voltage
)s
i discusse
d ni detail
. We the
n introduc
e
the notio
n of afilter an
d demonstrat
e th
e propertie
s of fou
r differen
t filter
type
s by
evaluatin
g th
e voltag
e signal
s acros
s th
e variou
s passiv
e component
s a
s a functio
n
of frequency
. An approximat
e metho
d enablin
g th
e quic
k plottin
g of th
e transfe
r
functio
n amplitud
e an
d phas
e a
s function
s of frequenc
y s
i presented
. Thi
s metho
d s
i
base
d on th
e logarithmi
c graph
s tha
t ar
e know
n a
s Bode plots
.
Activ
e filters,
whic
h utiliz
e op-am
p circuit
s tha
t contai
n onl
y resistor
s an
d ca
pacitors
, ar
e introduced
. We demonstrat
e tha
t eac
h of th
e fou
r filter
type
s ca
n be
realize
d by some combinatio
n of op-am
p circuits
. Her
e th
e fac
t tha
t th
e inpu
t an
d
outpu
t stage
s of op-am
p circuit
s ar
e decouple
d fro
m eac
h othe
rs
i exploite
d o
t desig
n
cascade
d circuits
. The chapte
rs
i close
d wit
h th
e developmen
t of a procedur
e fo
r th
e
synthesi
s of transfe
r function
s by usin
g active-/?
C circuits
. The procedur
e utilize
s
pole
s an
d zero
s of transfe
r functions
. Generi
c op-am
p circuit
s fo
r th
e realizatio
n of
pol
e factor
s an
d zero-factor
s ar
e introduced
, an
d th
e synthesi
s of th
e transfe
r function
s
is accomplishe
d by cascadin
g thes
e generi
c circuits
.
345
Chapter 9. Frequency Characteristics of Electric Circuits
346
9.2
Resonanc
e
W e shal
l stud
y th
e simpl
e RLC circui
t show
n ni Figur
e 9.1
. We assum
e tha
tvs(t) s
i
an a
c voltag
e source
:
v.v(
0 = Vms cos(ar
t + <j>s),
(9.1
)
and we shal
l be intereste
d n
i th
e respons
e of thi
s circui
ta
s a functio
n of frequenc
y co.
The curren
ti(t) produce
d by th
e abov
e voltag
e sourc
es
i sinusoida
la
s well
:
(9.2
)
/(?
) = lm COS(ftr
f + (j)[).
However
, th
e pea
k value
,Im, an
d initia
l phase
, (/>/
, of th
e curren
t depen
d on th
e valu
e
of co. The relationshi
p betwee
n thes
e quantitie
s an
d ws
i easil
y foun
d by usin
g th
e
phaso
r technique
:
v,(r
) - Vs = Vmsej^,
(9.3
)
j,)> (o,
i(t) — I((o) = Im(oo)e ' \
(9.4
)
Z((o) = R + j(a>L--^)=R
+ jX(a>).
(9.5
)
By usin
g th
e impedance-typ
e relationshi
p betwee
n voltag
e an
d current
, we find
that
:
(9.6
)
/(<»
) =
Im(oo) =
v„
V ms
|Z(o>)
|
(9.7
)
-)
\JV + (^ - ^ „
c
<^/(w
) = <k - <j>(<o),
2
X(a>)
ta
n <J>(<o
)
/?
(9.8
)
d 4)
/s
i clea
r fro
m equation
s (9.7
) an
d (9.8)
.
The frequenc
y dependenc
e oflm an
W e ar
e intereste
d ni findin
g th
e particula
r frequenc
y o)0 whic
h cause
s th
e pola
r
angl
e of th
e impedanc
e o
t be equa
lo
t zero
. Mathematically
, we see
k suc
h value
s of
^V^
.6
Figur
e 9.1
: A simpl
e RLC circuit
.
347
9.2. Resonance
CDQ tha
t satisfy
:
</>(*>o
) =0.
(9.9
)
Equation
s (9.9
) an
d (9.8
) impl
y tha
t th
e reactanc
e must be equa
lo
t zero
:
X(co0) =0.
(9.10
)
a)0L--^-=0,
(x)0C
(9.11
)
From (9.5
) an
d (9.10)
, we find:
coo = - ^ =
.
\/LC
(9.12
)
Thus, a
t th
e frequenc
y co
e circui
t behave
s a
s fi t
i contain
s onl
y a resistor
. The
0, th
contribution
s of th
e inductanc
e an
d th
e capacitanc
e ar
e cancele
d out
. As a result
, th
e
initia
l phas
e of th
e curren
ts
i equa
lo
t th
e initia
l phas
e of th
e voltag
e source
:
In othe
r words
, th
e voltag
e an
d curren
t ar
e ni phase
. Thi
s phenomeno
n si calle
d resonance. The frequenc
y COQ a
t whic
h resonanc
e occur
ss
i calle
d th
e resonant frequency.
A circui
t possesse
s a fe
w interestin
g propertie
s when a
t resonance
. First
, th
e
impedanc
e achieve
s it
s minimu
m absolut
e valu
e a
t th
e resonan
t frequency
:
|Z(OJO)
| -minJZ(a>)
| = R.
(9.14
)
A direc
t resul
t of equatio
n (9.14
)s
i tha
t th
e pea
k valu
e of th
e curren
t achieve
s it
s
maximum valu
e a
t th
e resonan
t frequency
:
Im((oo) =maxjm(a)) = ~ .
K
(9.15
)
Second
, th
e voltage
s acros
s th
e inducto
r an
d th
e capacito
r hav
e a
n interestin
g
relationshi
p a
t resonance
. To se
e this
, we recal
l th
e followin
g expression
s fo
r thes
e
voltages
:
VL((o) =jioLI (a),
(9.16
)
Vc(u)=
(9.17
)
-^Hu).
By combinin
g (9.11
) wit
h (9.16
) an
d (9.17)
, we arriv
e a
t th
e followin
g relationship
:
V L(co
j(o0Lhco0)
0) =
yL(co
0) = -VC(<OQ).
= -^-/(co
0) = -y
c(o)
0),
o)0C
(9.18
)
(9.19
)
This mean
s tha
ta
t resonanc
e th
e phas
e differenc
e betwee
n thes
e voltage
s s
i 180°
,
whil
e th
e pea
k value
s ar
e th
e same
. As a result
, th
e su
m of th
e voltage
s acros
s th
e
348
Chapter 9. Frequency Characteristics of Electric Circuits
inducto
r an
d th
e capacito
r s
i equa
l o
t zer
o a
t an
y instan
t of time
. Thus
, we ca
n
conclud
e tha
t th
e voltag
e acros
s th
e resisto
r si equa
lo
t th
e sourc
e voltage
:
VR(OJO) = Vs = Rkool
(9.20
)
This lead
so
t some interestin
g consequences
.n
I particular
, we examin
e th
e rati
o of
the pea
k value
s of th
e voltage
s acros
s th
e energ
y storag
e element
so
t th
e pea
k valu
e
of th
e voltag
e source
:
Vlmi^o) = Vcm(teQ)
Vms
=
Q)QLIm((Oo) = <*>oL
Vms
RIm((Oo)
R
"
Thus, we come o
t what may see
m o
t be a startlin
g conclusion
: fi co
n
0L > R, the
I othe
r words
, the voltage across the energy storage elements
VL/W(^O) ^ Vms. n
could attain peak values that are far greater than the source voltage itself. Thi
s
"amplification
" depend
s solel
y on th
e value
s of th
e circui
t element
s an
d th
e frequenc
y
of th
e voltag
e source
. Thi
s resul
t suggest
s tha
tt
is
i unwis
eo
t touc
h expose
d a
c circuit
s
even fi th
e voltag
e sourc
e s
i know
n o
t be relativel
y small
. Voltage
s acros
s energ
y
storag
e element
s ni th
e circui
t may be much large
r tha
n th
e sourc
e voltag
e an
d may
caus
e seriou
s harm
.
Anothe
r propert
y of a circui
ta
t resonanc
e si tha
t al
l supplie
d powe
rs
i consumed
;
none of ti s
i stored
. Thi
s si becaus
e th
e circui
t behave
s a
s a pur
e resistor
. As fa
r a
s
the inducto
r an
d capacito
r ar
e concerned
,a
t resonanc
e the
y continuousl
y exchange
the electromagneti
c energ
y withou
t involvin
g th
e sourc
e ni thi
s process
.
When a circui
t si no
t operatin
g a
t it
s resonan
t frequenc
y a>
e reactiv
e par
t of
0, th
the impedanc
e wil
l ac
ta
s a capacitanc
e or a
s a
n inductanc
e dependin
g on whethe
r
the operatin
g frequenc
y co s
i les
s tha
n or greate
r tha
n co0.
If CD < (X)Q,
toL - — < 0,
o)C
X((o) < 0,
(9.22
)
(9.23
)
and th
e curren
t lead
s th
e voltag
e an
d we dea
l wit
h "capacitive
" reactance
.
If o
c > COQ,
a)L - ~
> 0
,
(9.24
)
X(co) > 0,
(9.25
)
(X)C
and th
e curren
t lag
s behin
d th
e voltag
e an
d we dea
l wit
h "inductive
" reactance.
EXAMPLE 9.1 For th
e circui
t show
n ni Figur
e 9.1
, we assum
eR = 1
0 Cl,L =
1 mH,C = 47
0 nF
,Vms = 1
0 V, an
d 4>s = 0. We inten
d o
t calculat
e th
e circuit'
s
resonan
t frequency
, th
e maximu
m voltag
e acros
s th
e inducto
r a
t resonance
, an
d th
e
magnitud
e an
d phas
e of th
e curren
ta
ta)0/2, co
d 3co
0, an
0.
9.3. Passive Filters
349
Tabl
e 9.1
: Curren
t magnitud
e an
d phas
e values
.
Frequenc
y cu
Curren
t magnitud
e Im{oi)
Curren
t phas
e </>/(<»>
)
wo/2
o>o
3a)o
143 mA
1 A
81 mA
+ 81.78
°
0°
- 8 5 . 3°4
The resonan
t frequenc
y s
i give
n by equatio
n (9.12
) as
:
7
4
co0 = 1/V(4.
7 X 10)(1 X 1CT3) = 4.6
1 X 10
rad/s
,
(9.26
)
The maximu
m voltag
e acros
s th
e inducto
rs
i give
n by expressio
n (9.21
) as
:
4
3
VLn(wo) = Vmsa)0L/R = 1
0 X 4.6
1 X 10
X 1(T
/10 = 46.
1 V!
(9.27
)
W e ca
n rewrit
e th
e expression
s fo
r th
e magnitud
e an
d phas
e o
f th
e curren
t vi
a
equation
s (9.7)
, (9.8)
, an
d (9.12
) a
s follows
:
/m (a>)
Vms/R
yj\ +
[ooL/R]2[l-(Mo/a>)2}:
(9.28
)
and
1
|^
<M*>) = - tan"
1
[ ~im/cof]
|•
(9.29
)
o equation
s we ca
n quickl
y fill
ou
t th
e entrie
s n
i Tabl
e 9.1
, whic
h
With thes
e tw
complete
s thi
s problem
.
9.3
Passiv
e Filter
s
The behavio
r o
f a circui
t ca
n be change
d appreciabl
y by varyin
g th
e frequenc
y o
f
excitation
. By exploitin
g thi
s fact
, a circui
t ca
n be use
d a
sa filter. n
I general
, afilter
is adevic
e designe
d o
t selectivel
y le
t signal
s a
t some frequencie
s pas
s throug
h whil
e
suppressin
g signal
s a
t othe
r frequencies
. When applie
d o
t electri
c circuits
, afilter
s
ia
circui
t tha
t let
s some voltag
e or curren
t signal
s pas
s throug
h whil
e attenuatin
g othe
r
signal
s base
d on th
e frequenc
y o
f thos
e signals
. We conside
r belo
w fou
r type
s o
f
filters:
high-pas
s filters,
low-pas
s filters,
band-pas
s filters,
an
d band-notc
h filters.
A
high-pas
s filter
allow
s onl
y high-frequenc
y signal
so
t pas
s through
, whil
e alow-pas
s
filter
permit
s th
e passag
e o
f onl
y low-frequenc
y signals
. A band-pas
s filter
limit
s th
e
passag
e o
f signal
s o
t a somewha
t narro
w ban
d o
f frequencie
s aroun
d th
e resonan
t
frequency
, whil
e a band-notc
h filter
attenuate
s onl
y th
e signal
s nea
r th
e resonan
t
frequency
. Eac
h o
f thes
e filters
ca
n be realize
d by th
e sam
e circui
t show
n n
i Figur
e
9.1.
350
9.3.1
Chapter 9. Frequency Characteristics of Electric Circuits
High-Pass Filter
To demonstrat
e th
e high-pas
s filtering
propertie
s of th
e abov
e circuit
, we conside
r
specificall
y th
e voltag
e acros
s th
e terminal
s of th
e inductor
. By usin
g th
e familia
r
impedance-typ
e relationshi
p betwee
n voltag
e an
d curren
t phasors
, we ca
n expres
s
thi
s voltag
en
i phaso
r for
m a
s follows
:
VL(co) = jo>LI(a>) =
R+
j(oLVs
j{a)L-±)
(9.30
)
W e no
w defin
e atransfer function Hi(co) a
s th
e rati
o of Vi o
t Vs:
HL(OJ)
=
Vs
(9.31
)
R + jfrL-^Ly
Equatio
n (9.31
) define
s a comple
x transfe
r functio
n whos
e absolut
e valu
e is
:
ioL
\£r ( \\
\HL(co)\ = ^BP- + (coL -
JL)
*
<^LC
V" R Ci + {^LC
2 2
- 1)
2
(9.32
)
It s
i clea
r fro
m (9.32
) tha
t \Hi{o))\ s
i equa
lo
t zer
o when co = 0. As co s
i increased
,
\HL(co)\ s
i increase
d a
s wel
l an
d reache
s it
s maximu
m value
. Wit
h furthe
r increas
e
of co
,\HL(co)\ s
i somewha
t decrease
d an
d asymptoticall
y reache
s th
e valu
e tha
ts
i
equa
lo
t 1
. The plo
t of \HL(CO)\ s
i show
n n
i Figur
e 9.2
.t
Is
i clea
r fro
m thi
s plo
t tha
t
if we measur
e th
e voltag
e acros
s inducto
r terminals
, the
n th
e outpu
t signal
s wit
h
low frequencie
s wil
l be suppressed
, whil
e th
e outpu
t signal
s a
t hig
h frequencie
s wil
l
have almos
t th
e sam
e pea
k value
s a
s th
e inpu
t signals
. Fo
r thi
s reason
, fi th
e inducto
r
terminal
s ar
e use
d a
s th
e outpu
t terminals
, the
n th
e circui
t show
nn
i Figur
e 9.
1 act
sa
s
We shal
l furthe
r investigat
e th
e functio
n \HL(O))\. Namely
, we shal
l
a high-pas
s filter.
findth
e answe
ro
t th
e followin
g question
. What s
i th
e maximu
m valu
e of \HL(CO)\
and a
t what frequenc
y (relativ
eo
t CJQ) doe
s ti occur
?
.<o.
Figur
e 9.2
: Plo
to
f \HL(<o)\ vs
93. Passive Filters
351
If we inser
t equatio
n (9.12
) int
o (9.32
) an
d defin
e th
e dampin
g facto
r £ by
2£/(o0 = RC, the
n we obtai
n (wit
h th
e abbreviatio
n x = (O)/Q)Q)2):
\HL(x)\ =
,
*
(9.33)
=.
2
JMpx +(x - l)
W e find
th
e maximu
m by solvin
g fo
r th
e zer
o of th
e derivativ
e of \HL(x)\ wit
h respec
t
to x:
owAo o = 1 /
0
2
2
"2£
= 1A/
1 " (*>o#C)
/2
(9.34
)
and
I^L(co
max)| =
1
= =.
v W ? 0 2 ~ (cwoi?C)V
4
(9.35
)
Ther
e ar
e tw
o thing
so
t not
e here
. First
, th
e maximu
m occur
sa
t th
e resonan
t frequenc
y
onl
y when R = 0, n
i whic
h cas
e th
e maximu
m s
i infinite
. Otherwise
, th
e maximu
m
frequenc
y exceed
s th
e resonan
t frequency
. Second
, fi th
e resistanc
e s
is
o larg
e tha
t
n ther
es
i no loca
l maximu
m an
d th
e transfe
r functio
n magnitud
es
i
R > \jlL/C, the
a monotonicall
y increasin
g functio
n of frequenc
y whic
h asymptoticall
y approache
s
1. Fo
r th
e parameter
s of th
e previou
s exampl
e (R = 10 ft, =L 1 mH, C = 47
0 nF)
,
w e find
tha
t £ = 0.108
, tOmax/^
o = 1.012
, an
d \HL(comSiX)\ = 4.64
. Thi
s represent
s
a pea
k voltag
e of 46.
4 V, whic
h s
i onl
y slightl
y highe
r tha
n th
e calculate
d voltag
e at
resonance
.
9.3.2
Low-Pass Filter
To arriv
e at th
e low-pas
s filter,
we examin
e th
e voltag
e acros
s th
e terminal
s of th
e
capacitor
:
Vc(<o) = ^
J(oC
=
fc———i—.
jooC[R + j(coL - ^ ) ]
(9.36
)
The transfe
r function
, Hc(to), define
d as th
e rati
o ofVc o
t Vs, is
:
Hc(co) = - 4^ =
r—.
Vs
j(oC[R + j(a>L - JL)]
(9.37
)
The absolut
e valu
e ofHc(o>) s
i give
n by:
\Hc(io)\ =
1
O * ^
+ ( wL _ J^)
2
=
*
v/0)2/?2C
2 + ( ^L C
(9.38
)
- 1)
2
It s
i clea
r fro
m (9.38
) tha
t \Hc((o)\ s
i equa
l o
t 1 when (o = 0. As co s
i increased
,
|//c(co)
|s
i increase
d as wel
l an
d reache
s it
s maximu
m value
. Wit
h furthe
r increas
e
352
Chapter 9
. Frequency Characteristics of Electric Circuits
Figur
e 9.3
: A plo
to
f \Hc(co)\ vs
. o>
.
of co
,\Hc(co)\ monotonicall
y decrease
s an
d asymptoticall
y reache
s zero
. A typica
l
i show
n n
i Figur
e 9.3
.t
Is
i eviden
t fro
m thi
s plo
t tha
t fi th
e capacito
r
plo
t of \Hc(co)\ s
terminal
s ar
e viewe
d a
s outpu
t terminals
, the
n th
e outpu
t signa
la
t hig
h frequencie
s
wil
l be suppressed
, whil
e th
e outpu
t signa
l a
t lo
w frequencie
s wil
l hav
e th
e sam
e
(or eve
n somewha
t larger
) pea
k value
s a
s th
e inpu
t signal
. Thus
, fi th
e capacito
r
terminal
s ar
e use
d a
s th
e outpu
t terminals
, the
n th
e circui
t show
n n
i Figur
e 9.
1 act
s
as a low-pas
s filter.
Next, we conside
r th
e followin
g question
. What s
i th
e maximu
m valu
e of\Hc(co)\
and a
t what frequenc
y (relativ
eo
t co
)
doe
s
t
i
occur
?
0
2
If we inser
t (9.12
) int
o (9.38
) an
d defin
e 2£/co
d x = (o>/a)
n
0 = RC an
0) , the
w e obtain
:
\Hc(x)\ =
1
y/4?X +
(x-\)2
(9.39
)
W e find
th
e maximu
m by solvin
g fo
r th
e zer
o of th
e derivativ
e of \Hc{x)\ wit
h respec
t
tox:
0 )n
J<oo = x/
l " U1 = y/\ ~ ((o0RC)2/2
(9.40
)
and
l # c O m a xl
) =
^(OJQRC)2
4
- (a>
0#C) /4
(9.41
)
Again
, th
e maximu
m occur
sa
t th
e resonan
t frequenc
y onl
y when R = 0, n
i whic
h cas
e
the maximu
m s
i infinite
. Otherwise
, th
e maximu
m frequenc
y s
i belo
w th
e resonan
t
frequency
. Furthermore
, fi th
e resistanc
e s
is
o larg
e tha
tR > y 2L/C
, the
n ther
es
i
no loca
l maximu
m an
d th
e transfe
r functio
n magnitud
es
i a monotonicall
y decreasin
g
functio
n of frequency
. Fo
r th
e parameter
s of th
e previou
s exampl
e (R = 1
0 ft, =L
1 mH,C = 47
0 nF)
, we stil
l hav
e £ = 0.10
8 an
d | £ c ( ^lw )= 4.64
, bu
t no
w
^maxM) = 0.988
.
353
9.3. Passive Filters
9.3.3 Band-Pass Filter
To arriv
e a
t th
e band-pas
s filter,
we conside
r th
e voltag
e acros
s th
e terminal
s of th
e
resistor
:
VR(a>) = RI(o)) = VV
R
(9.42
)
The transfe
r function
,HR(CO), define
d a
s th
e ratio
of VR o
t Vs, is
:
HR(a>)
VR((O)
R
Vs
R + j(a>L - 0)C'^)
(9.43
)
The absolut
e valu
e of \HR(a))\ s
i give
n by
:
\HR((o)\ =
R
Ri + (co
L - ^)
2
coRC
y/^&C*
+ (u2LC ~ \f
(9.44
)
It s
i clea
r fro
m (9.44
) tha
t \HR(a))\ s
i equa
lo
t zer
o when >
o = 0. As o
c s
i increased
,
\HR((x))\ s
i increase
d a
s wel
l an
d reache
s it
s maximu
m valu
e of on
e a
t th
e resonanc
e
frequenc
y co
n by (9.12)
. As th
e frequenc
y >
o s
i furthe
r increased
, \HR(co)\ s
i
0 give
i
montonicall
y decrease
d an
d asymptoticall
y approache
s zero
. The plo
t of \HR(<o)\ s
shown n
i Figur
e 9.4
.t
Is
i eviden
t fro
m thi
s plo
t tha
t fi th
e resisto
r terminal
s ar
e
viewe
d a
s th
e outpu
t terminals
, the
n th
e outpu
t signal
s of lo
w an
d hig
h frequencie
s
wil
l be suppressed
, whil
e th
e outpu
t signal
s a
t frequencie
s centere
d aroun
d a>
l
0 wil
have approximatel
y th
e sam
e pea
k value
s a
s th
e inpu
t signals
. Thus
, fi th
e resisto
r
terminal
s ar
e use
d a
s th
e outpu
t terminals
, th
e circui
t show
n n
i Figur
e 9.
1 act
s a
s a
band-pas
s filter.
IHf©)l
0
co,
Figure 9.4: A plo
to
f \HR(to)\ vs
.(o.
354
9.3.4
Chapter 9. Frequency Characteristics of Electric Circuits
Band-Notch Filter
This filter
come
s fro
m considerin
g th
e ne
t outpu
t voltag
e acros
s th
e inducto
r an
d th
e
capacitor
. Accordin
g o
t KVL, thi
s voltag
e s
i equa
lo
t th
e sourc
e voltag
e minu
s th
e
resisto
r voltage
. Thus
, fro
m equatio
n (9.42
) we get
:
HLC(O>) = VLC((o)/Vs =
j(o)L — 1/coC)
R +j(o)L - l/o)C)
(9.45
)
which result
s ni th
e followin
g expressio
n fo
r th
e magnitud
e of th
e transfe
r function
:
I#LCM| =
\orLC -
1|
y/(a>RC)2 + (a>2LC-
2
l)
'
(9.46
)
Comparin
g thi
s equatio
n o
t (9.44)
, we se
e tha
t \HLc(oo)\ = A/1 — \HR(a))\2. Conse
quently
, thi
s transfe
r functio
n ha
s a minimu
m of zer
o a
t th
e resonan
t frequenc
y an
d
tend
s towar
d a maximu
m of one a
s th
e frequenc
y head
s towar
d eithe
r zer
o or infinity
.
This circui
t arrangemen
ts
i calle
d a band-notc
h filte
r becaus
et
i allow
s signal
so
t pas
s
at al
l frequencie
s excep
t fo
r a smal
l ban
d of frequencie
s nea
r th
e circui
t resonance
.
9.4
Bode Plots
Bode plot
s ar
e piecewis
e straight-lin
e approximation
s fo
r th
e frequenc
y dependenc
e
r function
. The
y wil
l be demonstrate
d throug
h
of th
e magnitud
e an
d phas
e of a transfe
the discussio
n of severa
l examples
.
EXAMPLE 9.2 Thi
s exampl
e deal
s wit
h th
e circui
t show
n ni Figur
e 9.5
. Fro
m th
e
voltag
e divide
r rule
, we get
,
H(co) = 5
WjcoC
R + 1/ j coC
1
1 +jcoRC
1
1 +jo)/O)Q
(9.47
)
where co
Thi
s transfe
r functio
n ha
s a simpl
e pol
e n
i th
e comple
x plane
.
0 = l/RC.
2
The magnitud
e of th
e transfe
r functio
n si \H(co)\ = l / l\ /+ (o)/co
e th
e
0) , whil
l
phas
es
i give
n by 4> = — tan~
((D/CJQ). We conver
t th
e magnitud
eo
t th
e decibe
l (dB
)
scal
e by takin
g th
e logarith
m of th
e transfe
r functio
n magnitud
e an
d multiplyin
g th
e
rA/WVn
R
O v.
Figure 9.5
: Exampl
e o
f a low-pas
s filter
.
355
9A. Bode Plots
resul
t by 20
:
|£(eo)|(dB
) = 201og
10
=
y/\
+(CO/C0 0 ) 2
7
2
= -201og
10 V ! +(co/a>o)
2
-101og
1 0(l+(a)/o)
0) )
(9.48
)
The las
t tw
o equalitie
s resul
t fro
m standar
d logarithmi
c manipulation
s whic
h ar
e
used her
e o
t ge
t th
e answe
r ni a simpl
e form
. The facto
r of 20 s
i use
d (rathe
r tha
n
10) becaus
e decibel
s ar
e usuall
y take
n a
s a
n indicatio
n of relativ
e powe
r rathe
r tha
n
relativ
e voltage
.
Notic
e tha
tn
i th
e low-frequenc
y limi
t (co «
too)
, we get tha
t |//|(dB
) —
— 101og
(l
)
=
0
dB
.
n
I
contrast
,
w
e
ca
n
neglec
t
th
e
constan
t
n
i
th
e
high-frequenc
y
10
limi
t (c
o >> co
)
o
t
approximat
e
th
e
transfe
r
functio
n
magnitud
e
a
s
0
2
|#|(dB
) - -101og
10((co/co
0) ) = -201og
10(co/co
0)dB.
On a plo
t of \H\ (dB
) versu
s frequency
, wher
e co s
i plotte
d on a logarithmi
c scale
,
thi
s high-frequenc
y approximatio
n represent
s a straigh
t lin
e whic
h decrease
s by 20
dB fo
r ever
y orde
r of magnitud
e increas
en
i co
. We approximat
e th
e actua
l magnitud
e
by combinin
g th
e tw
o straigh
t line
s correspondin
g o
t th
e low
- an
d high-frequenc
y
h s
i sometime
s
approximations
. Thes
e tw
o straigh
t line
s intersec
t a
to
c = co
0, whic
referre
d o
t a
s th
e breakpoin
t (o
r corner
) frequency
. A plo
t of th
e actua
l magnitud
e
and it
s approximatio
n s
i show
n n
i Figur
e 9.6
. We ca
n se
e tha
t th
e greates
t discrep
ancy betwee
n th
e tw
o curve
s occur
s a
t th
e breakpoin
t frequency
. The approximat
e
curv
e estimate
s th
e transfe
r functio
n magnitud
e o
t be 0 dB bu
t th
e exac
t resul
ts
i
|//|(dB)
= —10 log
(2
)
~
—3
dB
.
Therefore
,
th
e
tw
o
curve
s
agre
e
o
t
withi
n
3
d
B
10
Breakpoint frequency
0
ST
D
^
(D
Q
-5
1.-10
E
<
-15
Actual solution
Approximate solution
-20
cob/10
COo
IOCOQ
Frequency
Figure 9.6: Bod
e plo
to
f th
e magnitud
e o
f th
e transfe
r function
.
Chapter 9. Frequency Characteristics of Electric Circuits
356
— ■ Approximate solution
-100
co0/100
co0/10
co0
10co 0
lOOcOo
Frequency
Figur
e 9.7
: Bod
e plo
to
f th
e phas
e o
f th
e transfe
r function
.
everywhere
. Becaus
e of this
, th
e breakpoin
t (corner
) frequenc
y si als
o ofte
n referre
d
to a
s th
e 3 dB point
.
A plo
t of th
e phas
e ofH(GJ) a
s a functio
n of frequenc
y s
i give
n ni Figur
e 9.7
. The
curv
e reache
s 0° a
s th
e frequenc
y approache
s zer
o an
d asymptoticall
y approache
s
—90 ° a
s th
e frequenc
y tend
s o
t infinity
. Thi
s curv
e s
i usuall
y approximate
d by a
piecewis
e linea
r functio
n tha
ts
i equa
lo
t 0° fo
r frequencie
s up o
t a decad
e belo
w
the corne
r frequenc
y an
d s
i equa
lo
t —90° fo
r al
l frequencie
s exceedin
g th
e corne
r
frequenc
y by a
n orde
r of magnitude
. Thes
e tw
o line
s ar
e connecte
d by a thir
d lin
e
wit
h a slop
e of —45° pe
r decad
e whic
h goe
s throug
h th
e breakpoin
t frequenc
y a
t
—45°. The rational
e fo
r th
e abov
e approximat
e plo
ts
i no
ta
s firmly
establishe
d a
s
for th
e transfe
r functio
n magnitud
e approximation
; however
,t
is
i a fairl
y simpl
e an
d
reasonabl
y accurat
e approximatio
n nonetheless
.
EXAMPL E 9.
3 We revisi
t th
e second-orde
r high-pas
s transfe
r functio
n define
d by
equatio
n (9.31)
. Recal
l tha
t afte
r we define
d th
e resonan
t frequenc
y co
0 = 1/yLC
and th
e dampin
g facto
r 2£/co
t th
e transfe
r functio
n magnitud
e n
i th
e
0 = RC, we go
form:
i^)i =
(0)/Mo)2
V^SWwo)
2
.
2
2
+ ((CO/COQ) - l)
When we conver
t thi
s magnitud
e o
t th
e dB scal
e we obtai
n tw
o distinc
t terms
:
2
2
2
2
\H(a>)\(dB) = +401og
}.
10(o)/cu
0) - 101og
10{4£(a>Ao
0) + l(co/co0) - l]
This clearl
y reveal
s th
e advantag
e of usin
g a logarithmi
c scale
. Namely
, we ca
n con
struc
ta
n approximatio
n fo
r th
e transfe
r functio
n magnitud
e by graphicall
y combinin
g
the tw
o individua
l terms
. The firs
t ter
m need
s no approximatio
n a
t all
.t
I generate
s
357
9A. Bode Plots
a simpl
e straigh
t lin
e wit
h a slop
e of + 40 dB/decad
e whic
h passe
s throug
h 0 dB at
co0. The secon
d ter
m ca
n be approximate
d on th
e basi
s of it
s behavio
r n
i th
e high
and low-frequenc
y limits
.n
I th
e low-frequenc
y limi
t (c
o << co
th
e secon
d
0), we find
) = 0 dB. I
n th
e high-frequenc
y limi
t (c
o >> ti>o) we
ter
m s
i equa
l o
t - 1 0 1 o1g
0( l
4
approximat
e th
e secon
d ter
m as — 101og
) = —401og10(co/co
e
10((co/coo)
0) dB. Th
two approximation
s intersec
t atco0. I
n Figur
e 9.
8 we plo
t th
e first
term
, th
e piecewis
e
straight-lin
e approximatio
n of th
e secon
d term
, thei
r su
m (dashe
d line)
, an
d th
e actua
l
functio
n (soli
d line)
. When o
c < co
e secon
d ter
m s
i approximate
d as zero
,s
o th
e
0, th
e tw
o addin
g term
s
net resul
ts
i jus
t th
e lin
e wit
h th
e + 40 dB slope
. When o
c > co
0, th
cance
l on
e another
, resultin
g n
i a flat
lin
e at 0 dB. Not
e tha
t thi
s high-pas
s filter
roll
s
off at twic
e th
e rat
e of th
e previou
s example
. Thi
s s
i becaus
e n
i th
e previou
s exampl
e
w e deal
t wit
h a first-order
pole
, whil
e n
i thi
s exampl
e we hav
e a second-orde
r pole
. To
appreciat
e th
e leve
l of discrepanc
y betwee
n th
e piecewis
e straight-lin
e approximatio
n
and actua
l curve
, we not
e tha
t at resonanc
e (c
o = co
e approximatio
n estimate
s a
0) th
. Exac
t
gai
n of 0 dB, whil
e th
e actua
l valu
e depend
s on £: | # | ()d B= -201og
1 0(2£)
e give
n n
i th
e figure.
curve
s fo
r £ = 1 an
d £ = 1 /4 ar
The frequenc
y dependenc
e of th
e phas
e of th
e transfe
r functio
n s
i give
n by
4> = tan
"
2£co/co0
[(co/coo)2 - 1]
and s
i plotte
d n
i Figur
e 9.9
. The curv
e reache
s 180
° as th
e frequenc
y approache
s zer
o
and t
i asymptoticall
y approache
s 0° as th
e frequenc
y tend
s o
t infinity
. A straight
lin
e approximatio
n s
i give
n by a piecewis
e linea
r functio
n tha
ts
i equa
l o
t 180
° fo
r
frequencie
s up o
t a decad
e belo
w th
e breakpoin
t an
d s
i equa
lo
t 0° fo
r al
l frequencie
s
exceedin
g th
e breakpoin
t by an orde
r of magnitude
. Thes
e tw
o line
s ar
e connecte
d
1 s
t ter
m
2nd ter
m
co 0
10co
0
Frequency
Figur
e 9.8
: Bode plo
t magnitud
e of a second-orde
r high-pas
s filter
.
Chapter 9. Frequency Characteristics of Electric Circuits
358
COn/10
100co
0
co0
Frequency
Figure 9.9
: Bode plo
t phas
e of a second-orde
r high-pas
s filter
.
by a thir
d lin
e wit
h a slop
e of —90° pe
r decad
e whic
h goe
s throug
h th
e breakpoin
t
frequenc
y at 90°
. Exac
t curve
s fo
r £ = 1 an
d £ = 1/
4 ar
e give
n n
i th
e
figure.■
EXAMPLE 9.4 Conside
r th
e circui
t show
n n
i Figur
e 9.10
. To findth
e transfe
r
functio
n H((o) = VQ/VS9 we us
e th
e standar
d mesh analysi
s wit
h th
e thre
e meshe
s
indicate
d n
i th
e figure.
The matri
x equatio
n s
i foun
d o
t be
:
j/co
0
2j(co - I/w)
j/a>
j/<o
1 - j/<o
r % ] r vs i
x2 = 0
0
L^3 J
Becaus
e th
e outpu
ts
i th
e voltag
e acros
s a 1Cl resistor
, we find:
ff(a>) =
i3
Vs
rA/WV
l fl
^Oft
1/
2
221]1
( + j(o)[l + jco + (jo))
2H
1 F 7k ( T ) Fi
3
)IQ.
Figure 9.10: Exampl
e of a Butterwort
h circuit
.
(9.49
)
9.4. Bode Plots
359
0
-10
m
-20
a)
-3
0
E
<
-50
-60
Actua
l solutio
n
Approximate solution
-70
co0/10
co 0
10co
0
Frequency
Figur
e 9.11: Bode plo
t magnitud
e o
f a third-orde
r Butterwort
h filter
.
The las
t resul
t s
i obtaine
d by applyin
g Gaussia
n eliminatio
n o
t th
e abov
e matri
x
equation
. We wil
l discus
s th
e Bode plo
t onl
y fo
r th
e magnitud
e an
d leav
e th
e Bode
plot fo
r th
e phas
e a
sa
n exercis
e fo
r th
e reader
. The magnitud
e of th
e transfe
r functio
n
is: \H(co)\ = 1/(
2 vl + a)6). Thi
s functio
n s
i athird-orde
r low-pas
s filter.
The circui
t
is calle
d athird-orde
r Butterwort
h filter
an
d ha
s th
e propert
y tha
t \H(co)\ s
i maximall
y
flat,
i.e.
, ha
s many derivative
s equa
l o
t zer
o a
tco = 0. The approximat
e an
d exac
t
plot
s ar
e give
n n
i Figur
e 9.11
. The constan
t lin
e s
ia
t —6 dB a
s a resul
t of th
e facto
r
of 1/
2 n
i th
e numerator
. The breakpoin
t frequenc
y co0 s
i 1 rad/s
, wher
e th
e tw
o curve
s
diffe
r by 3 dB. The actua
l plo
ts
i asymptoti
c o
t th
e approximat
e lin
e wit
h a slop
e of
- 60 dB/decade
.
■
EXAMPLE 9.5 Assum
e we hav
e a circui
t whos
e transfe
r functio
n s
i give
n by
:
H(co) =
> (1 +jw/100)
_ HI(<O)H2(<D)
2
2
[1 + jco/10 + (jco/10)
](
l + jco/1000) H3(w)H4(coy
(9.50
)
Let us construc
t Bod
e plot
s fo
r bot
h th
e magnitud
e an
d th
e phas
e o
f thi
s function
.
Becaus
e o
f th
e propertie
s o
f logarithms
, we ca
n write
:
\H(co)\(dB) = l / f ^ K d)B + |#
) - \H3(co)\(dB) - |#
.
2(e»)|(dB
4(fi>)|(dB)
Likewise
, fo
r th
e phas
e we have
:
arg (H) = ar
g (H{) + arg(£
g (H3) 2) - ar
ar
g (#
4).
Thus, we ca
n approximat
e eac
h individua
l ter
m separatel
y an
d ad
d the
m u
p a
t th
e
end. Figur
e 9.1
2 show
s th
e individua
l magnitud
e contribution
s an
d Figur
e 9.1
3 give
s
the phas
e contributions
. H\(co) = jco s
i simila
r o
t a facto
r we considere
d whil
e
Chapter 9. Frequency Characteristics of Electric Circuits
360
80
60
40
CO
T^
20
CD
D
3
+~L
0
Q.
-20
£
<
-40
-60
-80
0.1
Actual solution
Approximate solution
Individual sections
1
10
100
1000
10000
Frequency
Figur
e 9.12
: Bod
e plo
t magnitud
e o
f th
e transfe
r functio
n give
n n
i equatio
n (9.50)
.
investigatin
g th
e high-pas
s filte
r an
d result
sn
i a straigh
t lin
e of slop
e 20 dB/decad
e
which goe
s throug
h 0 dB a
t 1 rad/s
. The phas
es
i aconstan
t 90°
. /^(w
) = 1 + y'w/lO
O
has a simpl
e zer
o wit
h a breakpoin
t of COQ = 10
0 rad/s
. The magnitud
e Bode plo
t
is equa
lo
t 0 dB unti
l th
e breakpoint
, the
n increase
s by 20 dB/decade
. The phas
e
begin
s a
t 0° an
d the
n increase
s o
t 90°
. The tw
o fla
t section
s ar
e linke
d by a lin
e
wit
h a slop
e of 45
° whic
h passe
s throug
h 45
° a
t th
e breakpoint
. i¥
(o>
)
s
i
simila
r
n
i
3
for
m o
t th
e "second-orde
r pole
" transfe
r functio
n we discusse
d n
i Exampl
e 9.
3 wit
h
100
-100
Actual
Approximate
Individual
-200
0.1
1
10
100
1000
10000
Frequency
Figur
e 9.13
: Bod
e plo
t phas
e o
f th
e transfe
r functio
n give
n n
i equatio
n (9.50)
.
9.5. Active-RC Filters
36
1
£ = 1/
2 an
d o>
o = 10
. Thus
, —//
s a magnitud
e Bode plo
t tha
ts
i fla
t a
t0d
B
3 ha
unti
l 10 rad/s
, the
n decrease
s a
t —40 dB/decade
. The phas
e (—arg(H
s a
t 0°
,
3)) start
the
n decrease
so
t —180° vi
a a lin
e wit
h aslop
e o
f —90° whic
h passe
s throug
h —90°
when o
c = 1
0 rad/s
. The final
part
, — H4, s
i reall
y th
e produc
t o
f tw
o "first-orde
r
pole
" transfe
r functions
. Therefore
, th
e magnitud
e Bode plo
ts
i flat
a
t 0d
B unti
l
the breakpoin
t frequenc
y o
f 100
0 rad/s
. Afterward
s t
i decrease
s a
t a rat
e o
f —40
dB/decade
. The phas
e stay
s a
t0
° unti
lo
c = 10
0 rad/
s an
d decrease
s o
t —180° b
y
the tim
e th
e frequenc
y reache
s 10,00
0 rad/s
. The resultin
g Bode plot
s ar
e als
o show
n
in th
e figure,
a
s ar
e th
e actua
l plots
.t
Is
i quit
e clea
r tha
t th
e approximatio
n s
i fairl
y
i considerabl
y easie
ro
t reproduc
e tha
n th
e actua
l function
.
accurat
e an
d s
The followin
g conclusio
n ca
n be draw
n fro
m th
e abov
e discussion
. Bode plot
s
are piecewis
e linea
r approximation
s fo
r transfe
r functio
n magnitude
s an
d phase
s a
s
function
s o
f frequenc
y o
n logarithmi
c scale
.The logarithmic scale allows one to
represent Bode plots for complicated transfer functions as sums of very simple
piecewise straight-line plots which are determined by the zeros and poles of the
transfer functions.
9.5
Active-jR
C Filter
s
The filters
discusse
d n
i th
e previou
s section
s hav
e bee
n realize
d b
y usin
g circuit
s
d capacitors)
. Fo
r thi
s
tha
t contai
n onl
y passiv
e element
s (resistors
, inductors
, an
reason
, thes
e circuit
s ar
e calle
d passiv
e filters.
The mai
n deficienc
y o
f passiv
e filters
e us
e o
f inductors
. Inductor
s ar
e "bulky
" circui
t element
s whic
h ar
e difficul
to
t
is th
miniaturiz
e an
d whic
h ar
e no
t compatibl
e wit
h integrate
d circui
t technology
. Thus
,t
i
can be conclude
d tha
tt
is
i desirabl
eo
t replac
e passiv
e filters
wit
h some othe
r circuit
s
which wil
l perfor
m th
e sam
e function
s withou
t usin
g inductors
.t
Is
i als
o desirabl
e
to combin
e filtering
propertie
s o
f electri
c circuit
s wit
h amplification
.t
I turn
s ou
t tha
t
thi
s ca
n be accomplishe
d b
y usin
g resistors
, capacitors
, an
d operationa
l amplifiers
.
f electri
c circuit
s calle
d active-/?
C filters
emerges
.The main
A s a result
, a ne
w clas
s o
idea of the design of these filters is to come up with active-RC circuits which realize
the transfer functions with the same frequency dependence as in the case of transfer
functions HL(CD), HC(O>), HR((O), an
d Hic(o)).
For th
e sak
e o
f generalit
y (an
d notationa
l simplicity
) we agai
n introduc
e th
e
comple
x frequencies
. Fo
r tim
e harmoni
c signals
, th
e relationshi
p s
i give
n b
y
s = jco.
(9.51
)
Now , expression
s (9.31)
, (9.37)
, (9.43)
, an
d (9.45
) ca
n be writte
n a
s follows
:
M^
HAs) =
sL
s2]LC
r =—
,
R + sL + ±
s2LC + sRC + 1
sC(R + sL+ ±)
s2LC +
sRC+l'
(9.52
)
(9.53
)
Chapter 9. Frequency Characteristics of Electric Circuits
362
HR(s)
R
R + sL+ ±
sRC
s LC + sRC+
2
sC
1'
(9.54
)
s2LC + 1
(9.55
)
HLC(s) = 2
s LC + sRC + 1'
Transfe
r function
s fo
r active-/?
C high-pass
, low-pass
, band-pass
, an
d band-notc
h fil
ter
s shoul
d mimi
c th
e transfe
r function
s (9.52)
, (9.53)
, (9.54)
, an
d (9.55)
, respectively
.
This mean
s tha
t the
y shoul
d hav
e th
e followin
g functiona
l dependenc
e on s
:
2
HhP(s)
sa
s + sb +c
= 2
d
s + sb + c
Hhp(s) = 2 sg
s + sb +c
Hlp(s) =
2
(9.56
)
(9.57
)
(9.58
)
s2e + f
—
s +st + c
^
where th
e subscript
s "hp,
" "lp,
" "bp,
" an
d "bn
" stan
d fo
r "high-pass,
" "low-pass,
"
"band-pass,
" an
d "band-notch,
" respectively
. Now, we demonstrat
e tha
t th
e low-pas
s
transfe
r functio
n (9.57
) ca
n be realize
d by th
e active-R
C circui
t show
n ni Figur
e 9.14
.
This demonstratio
n s
i als
o usefu
l a
s a
n exampl
e of th
e analysi
s of electri
c circuit
s
wit
h operationa
l amplifiers
.
W e begi
n ou
r analysi
s of th
e circui
t show
n ni Figur
e 9.1
4 by writin
g th
e noda
l
equatio
n fo
r nod
e numbe
r 3, an
d we shal
l us
e th
e versio
n of noda
l equation
s develope
d
for th
e circuit
s whic
h contai
n voltag
e source
s ni serie
s wit
h admittances
. Accordin
g
to thi
s versio
n we have
:
14, (v\ —
2
sCi + -IT + ^ + -!r) V3 /?i
-!-V
, - — V0 = —Vin.
Ro
rA/VW
FL
r^\AA/V-fA/VVV
vin6
Figur
e 9.14
: Active-RC low-pas
s filter
.
(9.60
)
9.5. Active-RC Filters
363
Sinc
e th
e inpu
t impedanc
e of operationa
l amplifie
r s
i assume
d o
t be infinite
, th
e
same curren
t / flow
s throug
h th
e resistanc
e R2 an
d capacitanc
e C2. Thi
s result
sn
i th
e
followin
g equation
:
Sinc
e th
e gai
n of th
e operationa
l amplifie
r s
i assume
d o
t be infinite
, node
s 1 an
d 2
have equa
l potentials
. Takin
g int
o accoun
t tha
t noninvertin
g termina
l 2s
i grounded
,
w e obtain
:
Vi = V2 = 0.
(9.62
)
By substitutin
g (9.62
) int
o (9.61)
, we obtain
:
= sC2V0,
(9.63
)
% = sR2C2V0.
(9.64
)
^
which s
i tantamoun
t to
:
By substitutin
g (9.62
) an
d (9.64
) int
o (9.60)
, we arriv
e a
t th
e followin
g equatio
n fo
r
- sR,R2C2 (sCi + -i
- + J- + - i) v 0 ~ ^-Vo = Vin\
K\
K3)
K2
(9.65
)
K3
From (9.65)
, we deriv
e th
e followin
g expressio
n fo
r th
e transfe
r function
:
H(s) = X^ =
—.
r
,
Vin siR{R2C{C2 + sR{R2C2 {i + Jr2+i) + %
(9.66
)
which ca
n be furthe
r transforme
d a
s follows
:
l
H(s)
=
7
S2 + S
V /v
j C]
^ Q G
^ 2 ^1
(9
^ 3 ^1 /
^ 2 ^3
6?)
cxc2
Now , we easil
y observ
e tha
t th
e transfe
r functio
n (9.67
) ha
s th
e sam
e functiona
l
dependenc
e on s a
s th
e transfe
r functio
n (9.57
) fo
r a low-pas
s filter.
Thi
s prove
s tha
t
the circui
t show
n n
i Figur
e 9.1
4 s
ia
n active-i?
C low-pas
s filter.
By usin
g thi
s circuit
,
w e ca
n als
o desig
n active-7?
C band-pas
s an
d high-pas
s filters.
An active-/?
C circui
t
which accomplishe
s thi
s tas
k s
i show
n n
i Figur
e 9.15
.t
I si clea
r tha
t th
e lef
t bloc
k
of thi
s circuit
, whic
h contain
s th
e operationa
l amplifie
r 1
,s
i identica
lo
t th
e circui
t
shown n
i Figur
e 9.14
. Fo
r thi
s reason
:
V{1)
-gVin
-RRlrr
-
= Hlp{s) =
l/?2ClC
2
s2 + s (-1-
*
,
+ - L- + -L
_ )+
.
!
(9.68
)
Chapter 9. Frequency Characteristics of Electric Circuits
364
rAMAR
3
A
>
2
A
1
r RI
3
R
If
r -
2
2
*c,
0 (1)
v
4
1
pVWVi
"
c3
' 3,
? ^ ^1
+ J ^i
5 ^
o
A (2
)
*
5
^AAA/Vn
A
i
R
*
K
C
4
,6
^
7
.0
Figur
e 9.15
: Active-RC filte
r circuit
.
Next, we fin
d th
e relatio
n betwee
n V^ an
d Vj?\ Sinc
e th
e inpu
t impedanc
e of
operationa
l amplifie
r 2s
i assume
d o
t be infinite
, we fin
d tha
t th
e sam
e curren
t 72
, we have
:
flow
s throug
h capacito
r C3 an
d resistanc
e R4. Consequently
% - w>
h = sC3(V(0l) - %)
(9.69
)
RA
Next, we recal
l tha
t
V4 = V5 = 0.
(9.70
)
By substitutin
g (9.70
) int
o (9.69)
, we obtain
:
sCVil) =
1/(2)
(9.71
)
RA'
which s
i tantamoun
t to
:
(9.72
)
From (9.68
) an
d (9.72)
, we derive
:
fttQ
vh
R\RiCxC-,
Hbp{s)
^
A x
R\C\
R2Ci
(9.73
)
R}Ci I
R2RiC1C2
Now , we easil
y observ
e tha
t th
e transfe
r functio
n (9.73
) ha
s th
e sam
e functiona
l
dependenc
e on s a
s th
e transfe
r functio
n (9.58
) fo
r a band-pas
s filter
. Thi
s prove
s tha
t
if voltag
e V£2) s
i take
n a
s th
e outpu
t voltage
, the
n th
e circui
t show
n n
i Figur
e 9.1
5
act
s a
s a band-pas
s filter
.
Next, we relat
e V£2) o
t V£3). By usin
g th
e sam
e lin
e of reasonin
g a
s before
, we
find
:
(2) _
h = *c 4 w - v6)
R,
'
(9.74
)
9.6. Synthesis of Transfer Functions with Active-RC Circuits
365
and
% = V7 = 0.
(9.75
)
By substitutin
g (9.75
) int
o (9.74)
, we derive
:
y(3
)
-sR5C4.
(2)
n
(9.76
)
By multiplyin
g (9.73
) by (9.76)
, we obtain
:
— p2 R4R5C3C4
T / ( 3)
2
*|/?2ClC
,
-£- = Hhp(s) =
R\C\
^2^
1
R3C1 /
.
(9.77
)
^2^3^1 W
Now , we easil
y observ
e tha
t th
e transfe
r functio
n (9.77
) ha
s th
e sam
e functiona
l
dependenc
e on s a
s th
e transfe
r functio
n (9.56
) fo
r a high-pas
s filter.
Thi
s prove
s tha
t
if voltag
e V£3) s
i take
n a
s th
e outpu
t voltage
, the
n th
e circui
t show
n n
i Figur
e 9.1
5
act
s a
s a high-pas
s filter.
The constructio
n of a
n active-/?
C band-notc
h filter
s
i somewha
t mor
e difficul
t
tha
n th
e previou
s cases
. Fo
r thi
s reason
, we outlin
e onl
y th
e genera
l idea
. Recal
l tha
t
the band-notc
h filter
aros
e fro
m th
e RLC circui
t by takin
g th
e differenc
e betwee
n th
e
sourc
e voltag
e an
d th
e voltag
e acros
s th
e resistance
.n
I othe
r words
, th
e band-notc
h
filter
was constructe
d by "subtracting
" a band-pas
s filter
fro
m th
e sourc
e voltage
.n
I
a simila
r manner
, we ca
n generat
e a band-notc
h filter
by introducin
g a
n inverte
r an
d
a summer circui
to
t th
e en
d of a band-pas
s filter.
We leav
e th
e tas
k of determinin
g
the resistanc
e value
s necessar
y o
t achiev
e th
e prope
r cancellatio
n o
t th
e reader
.
9.6
Synthesi
s of Transfe
r Function
s wit
h
Active-R
C Circuit
s
In th
e previou
s sectio
n we exploite
d th
e ide
a of cascading op-am
p circuits
. Namely
,
w e hav
e constructe
d th
e active-/?
C circui
t show
n ni Figur
e 9.1
5 a
s a cascad
e of th
e
active-/
? C low-pas
s filter
circui
t show
n n
i Figur
e 9.1
4 an
d tw
o activ
e circuit
s wit
h
simila
r transfe
r function
s whic
h ca
n be genericall
y describe
d as
:
H(s) = sd.
(9.78
)
Now , we shal
l furthe
r exploi
t th
e ide
a of cascading
. Let us suppos
e tha
t we want o
t
desig
n a circui
t realizatio
n fo
r a give
n transfe
r functio
n H{s). Let us als
o suppos
e tha
t
thi
s functio
n ca
n be represente
d a
s a produc
t of severa
l (o
r many
) presumabl
y simpl
e
function
s Hk(s) (k = 1
, 2,...
,n) :
H(s) = H{(s)H2(s)---Hn(s).
(9.79
)
Such a representatio
n s
i calle
d factorization
.
Now , fi we ca
n find circui
a t realizatio
n fo
r eac
h facto
rn
i (9.79)
, the
n a circui
t
realizatio
n fo
rH(s) ca
n be designe
d a
s th
e cascad
e of circuit
s whos
e transfe
r function
s
Chapter 9. Frequency Characteristics of Electric Circuits
366
Figur
e 9.16
: Notatio
n fo
r acircui
t wit
h transfe
r functio
n //^O)
.
are th
e factor
s Hk(s) n
i (9.79)
. Thi
s statemen
t ca
n be prove
d a
s follows
. Figur
e 9.1
6
present
s th
e notatio
n fo
r a circuit
, whic
h provide
s a realizatio
n o
f th
e facto
r Hk(s),
This mean
s tha
t th
e inpu
t V^i an
d outpu
t% o
f thi
s circui
t ar
e relate
d o
t on
e anothe
r
as follows
:
Vk(s)
(9.80
)
= Hk(s).
Conside
r th
e cascad
e o
f th
e abov
e circuit
s (se
e Figur
e 9.17)
. Accordin
g o
t (9.80
) we
have
:
Vl(5
)
.
= Hi(s),
V2(S)
..
Vn-xjs)
= H2(s), • • ■
~
= H„-\(s),
V0(S)
-
Vin(s)
Vi(s)
Vn-2(s)
By multiplyin
g th
e las
t equalities,
we en
d up with
:
V^is)
Hn(s).
(9.81
)
(9.82
)
Vin(s)Vi(s) • • • ^ - 2 ( 5 ) ^ - 1 ( 5 )
Afte
r obviou
s cancellation
sn
i (9.82)
, we obtain
:
Vo(s)
Vin(s)
H(s) =
Hl(s)H2(s)---Hn^(s)Hn(s).
(9.83
)
This prove
s tha
t th
e cascad
e show
n n
i Figur
e 9.1
7 indee
d provide
s th
e realizatio
n fo
r
the give
n transfe
r functio
n H(s).
The abov
e discussio
n underline
s th
e practica
l utilit
y o
f factorizatio
n (9.79)
.t
I
shows tha
tf
i acomplicate
d transfe
r function
,H(s), ca
n be represente
d a
s a produc
t of
Figur
e 9.17
: Cascad
e o
f circuit
s wit
h transfe
r function
s Hk(s).
9.6. Synthesis of Transfer Functions with Active-RC Circuits
367
simpl
e transfe
r function
s Hk(s) whic
h ca
n be separatel
y realize
d by simpl
e circuits
,
the
n H(s) ca
n be realize
d a
s th
e cascad
e of thes
e simpl
e circuits
.Thus, cascading is
a way to build complicated circuits by using simple ones as the main building blocks.
To utiliz
e th
e ide
a of cascading
, we hav
e o
t find wa
ay o
t factoriz
e transfe
r
functions
. Here
, we not
e tha
t atransfe
r functio
n ca
n be represente
d (o
r approximated
)
as a rati
o of tw
o polynomials
:
H(s) = Nn^
= GnSn + an-isn~l + * - + a\s + ap
Dm(s)
bmsm +bm^s™-1 + ---+blS + bom
W e nex
t conside
r th
e cas
e when "numerator
" polynomia
lNm(s) an
d "denominator"
polynomia
l Dm(s) hav
e onl
y rea
l an
d negativ
e roots
. Root
s of Nn(s) ar
e zero
s of
transfe
r function
s an
d we shal
l us
e th
e followin
g notatio
n fo
r them
:
zuzir-zn.
(9.85
)
Roots ofDm(s) ar
e pole
s of transfe
r function
s an
d the
y wil
l be denote
d as
:
Pi,P2> --Pm-
(9.86
)
It s
i wel
l know
n tha
t polynomial
s ca
n be factorize
d by usin
g thei
r roots
. Thes
e
factorization
s fo
rNn(s) an
d Dm(s) ca
n be writte
n a
s follows
:
Nn(s) = an(s -zi)(s -z2)'"(s-
Zn),
(9.87
)
Dm(s) = bm(s - pi)(s - p2) • • (s• - pm).
(9.88
)
By substitutin
g (9.87
) an
d (9.88
) int
o (9.84)
, we obtain
:
H(s) = d C » - * ) ( » - * ) - ( » - * . )
(S ~ P\)(S ~ p2) ' • *(S-
9
m
pm)
where d = an/bm. Now, th
e factorizatio
n of transfe
r function
s s
i readil
y available
.
Indeed
, expressio
n (9.89
) ca
n be rewritte
n a
s follows
:
H(s) = (S- zi)(s -z2)--(s-
Zn)-^—
S - pi
•- ^- • • • - ^ ,
S - p2
S ~ pm
(9.90
)
where d = d\d2 • •- dm.
Thus, th
e whol
e proble
m of circui
t realizatio
n of transfe
r function
s s
i reduce
d o
t
the desig
n of generi
c circuit
s whic
h wil
l realiz
e th
e factor
s (s - Zk) an
d dk/{s — Pk).
A s soo
n a
s th
e desig
n of thes
e circuit
ss
i accomplished
, th
e transfe
r function
, H(s),
can be realize
d a
s a cascad
e of thes
e generi
c circuits
.
To arriv
e a
t th
e abov
e generi
c circuits
, le
t us conside
r th
e circui
t show
n n
i Figur
e
9.18
. Sinc
e th
e inpu
t impedanc
e of th
e operationa
l amplifie
rs
i assume
d o
t be infinite
,
w e find:
1 = %=£ - ^ f
Z(s)
Zf{s)
(9.9.)
Chapter 9. Frequency Characteristics of Electric Circuits
A
Z f( s)
1
^
A
Z(s)
L
__|
0
1
1
I
1
I
I
[\
1
r
"
^ ^
A
2
V.
in
^
V
o—
0
Figur
e 9.18
: Generi
c circuit
.
It s
i als
o clea
r fro
m th
e abov
e circui
t that
:
Vi=V2
(9.92
)
= 0.
By substitutin
g (9.92
) int
o (9.91)
, we obtain
:
*,) = £ = -?*).
(9.93
)
Vin
Z(s)
Now , conside
r th
e circui
t show
n n
i Figur
e 9.19
. Thi
s circui
ts
i a particula
r cas
e of th
e
circui
t show
n n
i Figur
e 9.18
. Fo
r thi
s circuit
, we have
:
Z(s) = R,
7
r \
(9.94
)
sC
*
f
(9.95
)
S + RfCt
By substitutin
g (9.94
) an
d (9.95
) int
o (9.93)
, we derive
:
RCf
H(s)
(9.96
)
S + RfCr
rAAAAn
\
1
1
oA/VW'j
0
<> —
\(
K
I
f
'
*
r\
0
Figur
e 9.19
: Active-RC circui
t wit
h a prescribe
d pole
.
9.6. Synthesis of Transfer Functions with Active-RC Circuits
vvvv-
r-AA/Wn
u
l^
'
^v^, i
If
11
c
1—
0
P
n
VJ
0
Figur
e 9.20
: Active-RC circui
t wit
h a prescribe
d zero
.
From (9.96)
, we conclud
e tha
t th
e circui
t show
nn
i Figur
e 9.1
9 ha
s a transfe
r functio
n
wit
h a prescribe
d pole
:
1
Rff^f
C
(9.97
)
This mean
s tha
t thi
s circui
t ca
n be use
d fo
r realizatio
n of factor
s d^ /(s —pk)m (9.90)
.
Next, conside
r th
e circui
t show
n n
i Figur
e 9.20
. Thi
s circui
ts
i als
o a particula
r
cas
e of th
e circui
t show
n ni Figur
e 9.18
. For thi
s circuit
, we have
:
Z(s)
R-±
(9.98
)
sC
R+ ±
Zf(s) = Rf.
*
sC
RC
(9.99
)
By substitutin
g (9.98
) an
d (9.99
) int
o (9.93)
, we derive
:
H(s)
RfC
(s +
1
RC
(9.100
)
From (9.100)
, we observ
e tha
t th
e circui
t show
n n
i Figur
e 9.2
0 ha
s atransfe
r functio
n
wit
h a prescribe
d zero
:
z=
RC'
(9.101
)
This mean
s tha
t thi
ss
i a generi
c circui
t fo
r realizatio
n of factor
s (s — Zk) ni (9.90)
.
Thus, we can conclude that the circuit realization of a transfer function (9.90)
with real and negative poles and zeros can be achieved by cascading generic circuits
shown in Figures 9.19 and 9.20.
W e shal
l illustrat
e th
e previou
s discussio
n by th
e followin
g examples
.
EXAMPL E 9.
6 We woul
d lik
eo
t desig
n a
n active-i?
C circui
t wit
h a transfe
r functio
n
give
n by th
e followin
g expression
:
370
Chapter 9. Frequency Characteristics of Electric Circuits
Figure 9.21: Active-RC
circui
t realizatio
n of th
e exampl
e transfe
r function
.
H(s)
s+ 1
(s + 2)(,
s + 3)(.
s + 4)
(9.102)
This transfe
r functio
n ha
s on
e zer
o
(9.103)
Z\
and thre
e pole
s
P\ — —2, p2 — —3,
and /?3 = —4.
(9.104)
By usin
g th
e techniqu
e describe
d above
, we ca
n realiz
e thi
s transfe
r functio
n by
cascadin
g on
e circui
t of th
e typ
e show
n ni Figur
e 9.2
0 an
d thre
e circuit
s of th
e typ
e
shown n
i Figur
e 9.19
. The resultin
g circui
ts
i show
n ni Figur
e 9.21
.
To guarante
e th
e correc
t pole
s an
d zer
o of th
e transfe
r functio
n (9.102)
, resis
tance
s an
d capacitance
s of th
e abov
e circui
t ca
n be chose
n a
s follows
:
RiQ
= 1
,
(9.105
)
RlCf2 = 1
,
(9.106
)
= 1
, RfiQ
Rf2Cf2 = ^
>
1
R?>CfT, = 1
,
3'
1
Rf4.CfA
—, RAC
1.
U*-/4
1/4^/4 — 4'
Expression
s (9.105)-(9.108
) follo
w fro
m (9.100
) an
d (9.96)
.
R.pC/3
-
(9.107
)
(9.108
)
EXAMPL E 9.
7 We wil
l synthesiz
e a normalized
, third-orde
r Butterwort
h filte
r by
usin
g active-/?
C filte
r circuits
. The transfe
r functio
n fo
r thi
s filte
rs
i give
n by
:
H(s)
1/2
(s2 + s + l)(s + 1)'
(9.109
)
Therefore
, we wil
l nee
d o
t us
e op-am
p circuit
s cascade
d n
i th
e arrangemen
t show
n ni
Figur
e 9.22
. The leftmos
t op-am
p circui
t wil
l generat
e th
e second-orde
r polynomia
l
in th
e denominato
r an
d th
e rightmos
t circui
t wil
l generat
e th
e first-orde
r pole
. Ther
e
are a fe
w extr
a degree
s of freedo
m ni thi
s problem
,s
o le
t us make a fe
w arbitrar
y
9.7. MicroSim PSpice Simulations
371
i^WW+AAA/V
R.
R,
6vs
Figur
e 9.22
: Activ
e circui
t realizatio
n o
f a third-orde
r Butterwort
h filter
.
Comparin
g equatio
n
assignments
. First
, we wil
l assum
e tha
tR{ = R2 = R3 = 1 ft.
(9.67
) wit
h th
e Butterwort
h transfe
r function
, we se
e tha
t we wil
l nee
d o
t se
tC\ = 3
F an
d C2 = 1/
3 F. Thi
s give
s us th
e prope
r for
m fo
r th
e denominato
r of th
e first
par
t of th
e transfe
r functio
n an
d leave
s th
e numerato
r a
t — 1. The secon
d op-am
p
circui
t wil
l the
n hav
e o
t yiel
d th
e prope
r pol
e an
d hav
e a numerato
r equa
lo
t -1/2
.
Again ther
es
i some flexibility,
s
o we arbitraril
y pic
k C3 = 1 F. Equatio
n (9.96
) the
n
require
s tha
tR$ = 1 ftan
d 7?
Thi
s completel
y specifie
s on
e possibl
e desig
n
4 = 2 ft.
of th
e third-orde
r Butterwort
h
filter.
In conclusion
,t
is
i importan
to
t mentio
n tha
t cascadin
g s
i possibl
e du
e o
t ver
y
smal
l (almos
t zero
) outpu
t impedance
s of operationa
l amplifiers
. Becaus
e of tha
t
property
, an
y loa
d (circuit
) whic
h s
i place
d acros
s th
e outpu
t terminal
s of operationa
l
amplifie
r doe
s no
t affec
t th
e outpu
t voltag
e of thi
s amplifier
, and
, consequently
,t
i
does no
t affec
t th
e transfe
r functio
n of th
e circui
to
t whic
h thi
s amplifie
r belongs
.
9.7
MicroSi
m PSpic
e Simulation
s
The first
circui
t tha
t we wil
l analyz
e wit
h PSpic
e s
i th
e resonan
t circui
t show
n n
i
Figur
e 9.
1 an
d draw
n by th
e PSpic
e schemati
c generato
rn
i Figur
e 9.23
. The value
s
of th
e passiv
e element
s ar
e indicate
d n
i th
e figure
an
d correspon
d o
t th
e parameter
s of
Example 9.1
. The voltag
e sourc
e use
s th
e "VSRC" mode
l an
d ha
sa
n a
c magnitud
e of
10 V. An "AC Sweep
"s
i selecte
d n
i th
e "Analysi
s Setup
" dialo
g box
.n
I Exampl
e 9.
1
the resonan
t frequenc
y was calculate
d o
t be o
f = 733
7 Hz, s
o we swee
pn
i frequenc
y
fro
m 10
0 Hz o
t 10
0 kHz.
The result
s of th
e simulatio
n fo
r th
e voltage
s acros
s eac
h of th
e passiv
e element
s
is plotte
d on alog-lo
g scal
en
i Figur
e 9.24
. The voltag
e acros
s th
e inducto
rs
i indicate
d
by th
e soli
d line
. Fo
r frequencie
s fa
r abov
e th
e resonanc
e frequenc
y th
e voltag
e s
i
nearl
y constan
t an
d equa
lo
t th
e sourc
e voltag
e of 10 V. Fo
r frequencie
s fa
r belo
w
372
Chapter 9. Frequency Characteristics of Electric Circuits
R1
L1
10
1m
r^VW V
0
C1 i . 4 7 u
V1
Figur
e 9.23
: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 9.1
.
resonance
, th
e voltag
e decrease
s a
t abou
t arat
e of 20 dB pe
r decad
e n
i voltag
e (4
0
dB/decad
e n
i power)
. Thes
e ar
e exactl
y th
e propertie
s expecte
d of a second-orde
r
high-pas
s filter.
At resonance
, th
e analyti
c calculatio
n predicte
d th
e inducto
r voltag
e
to excee
d th
e sourc
e voltag
e by a facto
r of 4.6
4 = 6.6
7 dB. Thi
ss
i indee
d th
e case
,
as indicate
d n
i th
e figure.
The voltag
e acros
s th
e capacito
rs
i indicate
d n
i th
e figure
by th
e dot-dashe
d line
.
r low-pas
s filter
a
s expected
.
This curv
e display
s th
e classi
c behavio
r of a second-orde
The overshoo
t of th
e voltag
e a
t resonanc
e matche
s tha
t of th
e inducto
r (recal
l tha
t
at resonanc
e th
e inducto
r an
d capacito
r voltage
s ar
e equa
ln
i magnitud
e bu
t ou
t of
phas
e by 180°)
. The dashe
d lin
e trace
s th
e voltag
e acros
s th
e resisto
r a
s a functio
n
of frequency
. The resisto
r act
s a
s a band-pas
s filter,
wit
h a maximu
m voltag
e a
t
10
10 J
Frequency (Hz)
10'
10^
Figur
e 9.24
: The simulate
d voltage
s acros
s th
e passiv
e element
s fo
r th
e circui
t show
n
in Figur
e 9.23
. The inducto
r voltag
es
i give
n by th
e soli
d line
, th
e resisto
r voltag
e by
the dashe
d line
, an
d th
e capacito
r voltag
e by th
e dot-dashe
d line
.
373
9.7. MicroSim PSpice Simulations
1
R 2< 1
Figur
e 9.25
: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 9.4
.
resonance
equa
lo
t th
e sourc
e voltag
e an
d a drop-of
f awa
y fro
m resonanc
e of abou
t
10 dB pe
r decad
e (i
n voltage
) on bot
h side
s of th
e maximum.
For th
e secon
d PSpic
e simulatio
n we investigat
e th
e third-order
, low-pas
s But
terwort
h filter
circui
t of Exampl
e 9.4
. The circui
ts
i redraw
n by th
e schemati
c edito
r
in Figur
e 9.25
. The passiv
e componen
t value
s ar
e th
e sam
e a
sn
i th
e exampl
e an
d ar
e
shown n
i th
e figure.
Agai
n we us
e th
e "VSRC" par
to
t mode
l th
e source
, bu
t we se
t
the a
c magnitud
eo
t 5 V. The breakpoin
t frequenc
y s
i calculate
d n
i be o
c = 1 rad/
sf
(
- 0.15
9 Hz)
,s
o th
e "AC Sweep
"s
i take
n fro
m 0.00
1 Hz o
t 10
0 Hz.
The result
s of th
e simulatio
n fo
r th
e outpu
t voltag
e acros
s th
e resisto
r ar
e plotte
d
in Figur
e 9.26
. The voltag
e a
t lo
w frequencie
s s
i one-hal
f th
e sourc
e voltag
e a
s
predicte
d n
i th
e example
. The effec
t of th
e "maximall
y flat"
propert
y of thi
s filter
is eviden
t n
i thi
s figure
by th
e extremel
y flat
outpu
t voltag
e n
i th
e passban
d of
10
10"1
10°
Frequency (Hz)
101
Figur
e 9.26
: The simulate
d outpu
t voltag
e o
f th
e third-orde
r Butterwort
h filter
.
374
Chapter 9. Frequency Characteristics of Electric Circuits
vminus
R5
A/VW
1k
R4
R1
1k
1k
rAA/W+A/VW^
Q vs C3 i 3m
C1
0.333m
LM324
R2
U1A
2k
vplus
Figure 9.27: The PSpic
e schemati
c fo
r th
e circui
tn
i Exampl
e 9.7
.
the circuit
. The outpu
t voltag
e drop
s of
f fa
r abov
e th
e breakpoin
t frequenc
y a
t th
e
expecte
d third-orde
r rat
e of 30 dB/decade
.
The final
PSpic
e simulatio
n examine
s th
e activ
e versio
n of th
e third-orde
r low
pass Butterwort
h filter.
Thi
s circui
t was analyze
d ni Exampl
e 9.
7 an
d s
i redraw
n by
the circui
t simulato
r ni Figur
e 9.27
. The resistor
s hav
e bee
n scale
d by afacto
r of 100
0
(O —
kfl
) an
d th
e capacitor
s hav
e bee
n reduce
d by a facto
r of 100
0 F
( — mF) o
t
yiel
d somewha
t mor
e realisti
c componen
t values
. However
, sinc
e an
y resistanc
e n
i
the transfe
r functio
n s
i alway
s multiplie
d by a capacitance
, th
e ne
t transfe
r functio
n
10"
10"1
10°
Frequency (Hz)
101
Figur
e 9.28
: The simulate
d voltag
e a
t th
e outpu
t o
f eac
h stag
e o
f th
e third-orde
r
Butterwort
h filter
. The dashe
d lin
e give
s th
e outpu
to
f th
e firs
t stag
e an
d th
e soli
d lin
e
indicate
s th
e outpu
to
f th
e fina
l stage
.
9.8. Summary
375
of thi
s circui
ts
i identica
lo
t th
e on
en
i th
e example
. The tw
o op-amp
s us
e th
e LM324
five-connection
subcircui
t mode
l fro
m th
e "eval.slb
" librar
y wit
h al
l th
e standar
d
defaul
t values
. Thi
s mode
l was describe
d previousl
y n
i Sectio
n 8.5
. VI an
d V2 ar
e
the op-am
p dc suppl
y voltage
s of minu
s an
d plu
s 1
2 V, respectively
. Vs s
i th
e sourc
e
voltage
, whic
h ha
s a
n a
c magnitud
e of 5 V o
t matc
h tha
t of th
e previou
s example
.
All source
s ar
e "VSRC" parts
.
The voltag
e a
t th
e outpu
t of eac
h of th
e op-am
p stage
ss
i show
n n
i Figur
e 9.28
.
Recal
l tha
t thi
s filter
was synthesize
d by combinin
g a second-orde
r low-pas
s filter
an
d
a first-order
low-pas
s filter.
The dashe
d lin
e indicate
s th
e outpu
t of th
e second-orde
r
filter.
The low-frequenc
y gai
n s
i unit
y an
d th
e voltag
e drop-of
f a
t hig
h frequenc
y s
i
20 dB/decade
. Not
e tha
t ther
e s
i a sligh
t overshoo
tn
i th
e voltag
e a
t th
e resonanc
e
frequency
. Thi
s overshoo
t help
so
t compensat
e fo
r th
e kne
e of th
e first-order
low-pas
s
filter
of th
e secon
d stag
e o
t produc
e th
e "maximall
y flat"
characteristi
c of th
e filter.
The secon
d stag
e ha
s a low-frequenc
y gai
n of 1/
2 an
d th
e sam
e breakpoin
t frequenc
y
as th
e first
stage
. The ne
t outpu
t voltag
e of th
e filter
s
i indicate
d n
i th
e figure
by th
e
soli
d line
. Thi
s curv
e s
i virtuall
y identica
lo
t th
e outpu
t dependenc
e of th
e passiv
e
versio
n of thi
s filter,
a
s expected
.
9.8
Summar
y
In thi
s chapter
, we hav
e discusse
d th
e frequenc
y characteristic
s of electri
c circuits
.
The mai
n result
s of thi
s chapte
r ca
n be briefl
y summarize
d a
s follows
:
• t
I ha
s bee
n demonstrate
d tha
t a simpl
e RLC circui
t exhibit
s a
n interestin
g
phenomeno
n calle
d resonance
. At resonance
, th
e curren
t throug
h th
e circui
t
is in phase wit
h th
e sourc
e voltag
e n
i spit
e of th
e presenc
e of energ
y storag
e
elements
. Thi
s occur
s a
t a resonan
t frequenc
y co
o = 1/y/LC. At resonance
,
the impedanc
e a
s a functio
n of frequenc
y assume
s it
s minimu
m value
, whil
e
the curren
t a
s a functio
n of frequenc
y assume
s it
s maximu
m value
. Voltage
s
acros
s th
e capacito
r an
d inducto
r hav
e a phas
e differenc
e of 180°
, whil
e thei
r
peak value
s ar
e th
e same
. Thes
e pea
k value
s may significantl
y excee
d th
e pea
k
valu
e of th
e voltag
e source
.
• t
I ha
s bee
n demonstrate
d tha
t th
e RLC circui
t ca
n be utilize
d a
s a filter.
f
I th
e
inducto
r terminal
s ar
e use
d a
s th
e outpu
t terminals
, the
n th
e circui
t act
s a
s a
high-pass filter.
Thi
s mean
s tha
t outpu
t signal
s wit
h lo
w frequencie
s wil
l be
suppressed
, whil
e th
e outpu
t signal
s a
t hig
h frequencie
s wil
l hav
e almos
t th
e
same pea
k value
s a
s th
e inpu
t signals
.f
I th
e capacito
r terminal
s ar
e use
d a
s
the outpu
t terminals
, the
n th
e circui
t act
sa
s alow-pass filter.
Thi
s mean
s tha
t th
e
outpu
t signal
s a
t hig
h frequencie
s wil
l be suppressed
, whil
e th
e outpu
t signal
s
at lo
w frequencie
s wil
l hav
e almos
t th
e sam
e pea
k value
s a
s th
e inpu
t signals
.
If th
e resisto
r terminal
s ar
e use
d a
s th
e outpu
t terminals
, the
n th
e circui
t act
sa
s
a band-pass filter.
Thi
s mean
s tha
t th
e outpu
t signal
s wil
l be suppresse
d a
t lo
w
and hig
h frequencies
, whil
e th
e outpu
t signal
s a
t frequencie
s centere
d aroun
d
the resonan
t frequenc
y co
l hav
e approximatel
y th
e sam
e pea
k value
s a
s
0 wil
Chapter 9
. Frequency Characteristics of Electric Circuits
376
the inpu
t signals
.f
I th
e ne
t voltag
e acros
s th
e inducto
r an
d capacito
rs
i use
d
as th
e outpu
t voltage
, the
n th
e circui
t act
s a
s aband-notch filter.
Thi
s mean
s
tha
t th
e outpu
t signal
s a
t frequencie
s centere
d aroun
d o)0 wil
l be suppressed
,
whil
e th
e outpu
t signal
s a
t lo
w an
d hig
h frequencie
s wil
l hav
e almos
t th
e sam
e
peak value
s a
s th
e inpu
t signals
.
• Bode plot
s fo
r transfe
r function
s hav
e bee
n presented
.These plots are piecewise straight-line approximations for transfer function magnitudes and
phases as functions of frequency on a logarithmic scale. The logarithmi
c
scal
e allow
s on
eo
t represen
t Bode plot
s fo
r complicate
d transfe
r function
s a
s
sums of ver
y simpl
e straight-lin
e plot
s whic
h ar
e determine
d by th
e zero
s an
d
pole
s of th
e transfe
r functions
.
• Active-/?
C filters
hav
e bee
n discussed
. Thes
e filters
contai
n resistors
, capaci
tors
, an
d operationa
l amplifiers
, bu
t the
y do no
t contai
n inductor
s (whic
h ar
e
incompatibl
e wit
h integrate
d circui
t technology)
. The mai
n ide
a of th
e desig
n of
an active-/?
C filter
s
io
t come up wit
h active-7?
C circuit
s whic
h hav
e th
e sam
e
transfe
r function
s a
s high-pass
, low-pass
, band-pass
, an
d band-notc
h filters.
Example
s of th
e desig
n an
d analysi
s of active-/
? C filters
hav
e bee
n given
.
• The synthesi
s of transfe
r function
s by usin
g active-jR
C circuit
s ha
s bee
n pre
sented
. The mai
n ide
a of thi
s synthesi
s procedur
es
io
t represen
t a complicate
d
transfe
r functio
n a
s a produc
t of simpl
e transfe
r function
s whic
h ca
n be sepa
ratel
y realize
d wit
h simpl
e active-/?
C circuits
. Then
, th
e origina
l transfe
r func
tio
n ca
n be realize
d a
s th
e cascad
e of thes
e simpl
e circuits
.Thus, cascading is
a way to build complicated circuits by using simple ones as the main building blocks. n
I th
e chapter
, pole
s an
d zero
s hav
e bee
n use
d n
i th
e factorizatio
n
of transfe
r functions
. Generi
c op-am
p circuit
s fo
r realizatio
n of pole-factor
s
and zero-factor
s hav
e bee
n designe
d an
d th
e synthesi
s of th
e transfe
r function
s
has bee
n accomplishe
d by cascadin
g thes
e generi
c circuits
.
9.9
Problem
s
In Problem
s 1-6
, findth
e transfe
r functio
n indicate
d ni th
e appropriat
e figure
an
d identif
y th
e
typ
e of filter
t
i represent
s (low-pass
, high-pass
, band-pass
, or band-notch)
. Sketc
h on a grap
h
the magnitud
e of th
e transfe
r functio
n fro
m o)
0 o
t 10to
0/1
01. H(s) = v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-1
.
2. H(s) = va/vs
fo
r th
e circui
t show
n n
i Figur
e P9-2
.
r A / VW
ex
R
Figure P9-1
Figure P9-2
377
9.9. Problems
3.
H(s) =i0/is fo
r th
e circui
t show
n n
i Figur
e P9-3
.
4. H(s) = v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-4
.
vo
+
rA/WV
-
R
CK
Figure P9-4
Figure P9-3
5. H(s) = v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-5
.
6. H(s) =v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-6
.
C
4i H
1Q
V2 F
Qv s
R
r-AA/Wi
R
He—w w
1a
Figure P9-5
Figure P9-6
In Problem
s 7-12
, fin
d th
e indicate
d transfe
r functio
n an
d construc
t Bode plot
s fo
r th
e magni
tud
e an
d phas
e dependenc
e o
n frequency
.
7. H{s) =v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-7
.
8. H(s) = v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-8
.
1 Q
0.
5 it
F
rAAAAr? \$—
(T)v s 1 (xH ] 1 uH 5 1 &
Figure P9-7
Figure P9-8
378
Chapter 9. Frequency Characteristics of Electric Circuits
9. H(s) — v0/vs
fo
r th
e circui
t show
n ni Figur
e P9-9
.
oorY
rAA/Vv—1(
1Q
ex
0.05 F
"
20 H
+
0.05 H < 20 F
1 Q^>Vc
Figure P9-9
10. H(s) — v0/vs
fo
r th
e circui
t show
n ni Figur
e P9-10
.
11. H(s) = v0/vs fo
r th
e circui
t show
n ni Figur
e P9-11
.
rA/WV
5 mH
50 Q
1 u.F
50 Q >
v
o
Figure P9-11
Figure P9-10
12. H(s) = v0/vs
5 u.F
fo
r th
e circui
t show
n ni Figur
e P9-12
.
i
rAAA/VWWVrA/WVi
100 kQ
10 kQ
100 kQ
1 MQ
-)h
10 ixF ,_J
' W W
1 MQ
~\\
)\
3 U.
F
1 MQ
AA/W
-1/3 u.F
i—
1
|
iT
I
-^\ 1
V
.
Figure P9-12
In Problem
s 13-18
, construc
t Bode plot
s fo
r th
e magnitud
e an
d phas
e dependenc
e on frequenc
y
of th
e transfe
r function
s give
n n
i Tabl
e 9.2
.
9.9. Problems
379
Table 9.2: Sampl
e transfe
r functions
.
13.
H\{s)
1+^/1
0
(1 +s)(\ + 5/IOO
)
H2(s)
_ (s + \0)(s + 1000
)
2
(5 + l)(s + 100)
H3(s) =
100^(
1 +5/100
)
(1 +5/10)(
l +5/1000
)
H4(s) =
10s
(1 +55 + 52)(
l +5/10
)
H5(s) =
(5 + 5)
5
2
(52 + 3
5 + 1)
(5 + 50
)
H6(s)
3
2000(
5 + 1)5
2
2
(5 + 100)
(5 + 10)
Transfe
r functio
n H\.
14. Transfe
r functio
n H2.
15.
Transfe
r functio
n H3.
16. Transfe
r functio
n H4.
17. Transfe
r functio
n H5 .
18.
Transfe
r functio
n H 6.
In Problem
s 19-24
, desig
n activ
e RC circuit
s whic
h realiz
e th
e transfe
r function
s give
n n
i
F capacitors
. Use resistanc
e value
s R < 1 Mfl an
d a
s fe
w op-amp
s
Tabl
e 9.2
. Use onl
y 1
0 JLL
as possible
.
19. Transfe
r functio
n H\.
20.
Transfe
r functio
n H 2.
21.
Transfe
r functio
n //
3.
22.
Transfe
r functio
n H4.
23.
Transfe
r functio
n H5.
24.
Transfe
r functio
n H6.
In Problem
s 25-34
, us
e a frequenc
y analysi
s ni PSpic
e o
t plo
t th
e frequenc
y respons
e of th
e
transfe
r functio
n magnitude
s indicated
. Plo
ta
t leas
t on
e decad
e abov
e an
d belo
w an
y corne
r
frequencies
.
25. H(o)) = i0/is fo
r th
e circui
t show
n n
i Figur
e P9-3
. Le
tR = 2 kfl
, C = 1 mF, an
d Lf
mH .
= 1
26. H(co) — v0/vs
fo
r th
e circui
t show
n ni Figur
e P9-5
.
27. H(co) = v0/vs
fo
r th
e circui
t show
n n
i Figur
e P9-6
. LetR = 3Rf = 24 kf
i an
d C = 47
Chapter 9. Frequency Characteristics of Electric Circuits
380
28. H(OJ) = v0/vs
graph
.
fo
r th
e circui
t show
n n
i Figur
e P9-8
. Plo
t th
e Bode estimat
e on th
e sam
e
29. H(co) = v0/vs
graph
.
fo
r th
e circui
t show
n ni Figur
e P9-9
. Plo
t th
e Bode estimat
e on th
e sam
e
30. H{co) = v0/vs fo
r th
e circui
t show
n ni Figur
e P9-10
. Plo
t th
e Bode estimat
e on th
e sam
e
graph
.
31. H((o) = va/vs fo
r th
e circui
t show
n ni Figur
e P9-12
. Plo
t th
e Bode estimat
e on th
e sam
e
graph
.
32. H(o)) — v0/vs fo
r th
e circui
t show
n n
i Figur
e P9-11
. Le
t2R = Rf
mH .
= 1
0 kf
l an
d Lf = 1
33.
Desig
n a
n activ
e RC circui
t wit
h th
e transfe
r functio
n give
n by H\ ni Tabl
e 9.2
. Plo
t th
e
simulate
d transfe
r functio
n magnitud
e an
d th
e Bode plo
t estimat
e on th
e sam
e graph
.
34.
Desig
n a
n activ
e RC circui
t wit
h th
e transfe
r functio
n give
n by H5 n
i Tabl
e 9.2
. Plo
t th
e
simulate
d transfe
r functio
n magnitud
e an
d th
e Bode plo
t estimat
e on th
e sam
e graph
.
Chapte
r 10
Magnetically Coupled
Circuits and Two-Port
Elements
10.
1
Introductio
n
In ou
r previou
s discussions
, we hav
e deal
t wit
h circuit
s n
i whic
h circui
t element
s
have bee
n couple
d electrically
. Thi
s ha
s bee
n realize
d by usin
g electri
c wires
.t
I
turn
s ou
t tha
t circui
t element
s (an
d circuits
) ca
n als
o be couple
d magnetically
, tha
t is
,
withou
t direc
t electri
c contacts
. Thi
s couplin
g occur
s becaus
e time-varyin
g electri
c
current
s ni on
e circui
t produc
e time-varyin
g magneti
c fields
an
d magneti
c field
lines
.
These field
line
s may lin
k a
n adjacen
t electri
c circui
t and
, accordin
g o
t Faraday'
s law
,
induc
e a voltag
en
i thi
s circuit
.n
I thi
s way
, th
e time-varyin
g electri
c current
s ni on
e
circui
t may affec
t (ma
y be couple
d with
) th
e current
sn
i th
e adjacen
t circuit
.
In some cases
, magneti
c couplin
g ca
n be usefu
l and
, fo
r thi
s reason
,t
i ca
n be
utilize
d an
d eve
n deliberatel
y enhance
d ni th
e desig
n of certai
n electri
c devices
.
Example
s includ
e transformer
s an
d inductio
n motors
.n
I othe
r cases
, magneti
c cou
plin
g ca
n be detrimenta
lo
t desig
n purpose
s an
d shoul
d be suppressed
. An importan
t
exampl
e of thi
ss
i high-densit
y interconnect
s of VLSI circuit
s wher
e "cross-talk
"s
i
becomin
g prohibitivel
y larg
e du
eo
t wirin
g proliferatio
n on silico
n chips
.
In thi
s chapter
, we conside
r magneticall
y couple
d circuits
. We begi
n ou
r discus
sio
n wit
h th
e definitio
n of mutua
l inductanc
e an
d establis
h a ver
y importan
t inequalit
y
betwee
n th
e mutua
l inductanc
e an
d th
e produc
t of self-inductances
. We the
n us
e th
e
concep
t of mutua
l inductanc
eo
t deriv
e th
e couple
d circui
t equations
. Thes
e equation
s
constitut
e th
e foundatio
n fo
r th
e discussio
n of a transformer
. We first
conside
r th
e
theor
y of a
n idea
l transforme
r when many smal
l parameter
s an
d imperfection
s ar
e
neglected
. The presentatio
n of th
e theor
y of a nonidea
l transforme
r the
n follows
. The
importanc
e of suc
h smal
l parameter
s a
s leakag
e reactance
ss
i expose
d an
d stressed
,
381
382
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
and th
e couple
d circui
t equation
s ar
e modifie
d ni orde
ro
t accoun
t explicitl
y fo
r thes
e
leakag
e reactances
. Thi
s modificatio
n lead
s directl
y o
t th
e equivalen
t circui
t fo
r th
e
transformer
.
The secon
d par
t of th
e chapte
r deal
s wit
h th
e theor
y of two-por
t elements
.t
Is
i
emphasize
d tha
t th
e two-por
t element
s serv
e a twofol
d purpose
: the
y ca
n be use
d
as equivalen
t replacement
s fo
r complicate
d circuit
s and
, mor
e importantly
, the
y ca
n
be employe
d a
s circui
t model
s fo
r rea
l device
s suc
h a
s transistors
, transformers
, an
d
transmissio
n lines
. The linea
r an
d homogeneou
s equation
s whic
h describ
e termina
l
voltage-curren
t relation
s fo
r two-por
t element
s ca
n be writte
n ni differen
t form
s
which resul
tn
i differen
t representation
s of two-por
t elements
.n
I th
e chapter
, thes
e
variou
s representation
s ar
e studie
d n
i detai
l alon
g wit
h thei
r experimenta
l identifica
tion
s an
d circui
t realizations
.
10.
2
Mutua
l Inductanc
e an
d Couple
d
Circui
t Equation
s
In circui
t theory
, magneti
c couplin
g betwee
n electri
c circuit
s ca
n be describe
d by
usin
g th
e concep
t of mutua
l inductance
. To defin
e th
e mutua
l inductance
, conside
r
the tw
o coil
s show
n n
i Figur
e 10.1
. The first
coi
l ha
s N\ turns
, whil
e th
e secon
d coi
l
hasN2 turns
. First
, we assum
e tha
t th
e first
coi
ls
i energize
d whil
e th
e curren
t throug
h
the secon
d coi
ls
i equa
lo
t zero
:
= 0,
i\ £
*2
(10.1
)
0.
The curren
t throug
h th
e first
coi
l create
s th
e magneti
c field
aroun
d thi
s coil
, whic
h
can be represente
d by magneti
c field
line
s (se
e Figur
e 10.1a)
. Some of thes
e field
line
s may lin
k th
e secon
d coi
l resultin
g ni th
e flu
x linkages
, i/^i
- Her
e th
e orde
r of
subscript
s indicate
s tha
t i/r
i th
e flux
linkage
s of th
e secon
d coi
l du
eo
t th
e curren
t
2i s
throug
h th
e first
coil
. The mutua
l inductanc
e M12 betwee
n th
e first
an
d th
e secon
d
coil
ss
i define
d a
s th
e followin
g ratio
:
(10.2
)
M„. = *Z.
(b)
Figure 10.1: Two magneticall
y couple
d coils
.
10.2. Mutual Inductance and Coupled Circuit Equations
383
Next, conside
r th
e cas
e when th
e secon
d coi
ls
i energize
d whil
e th
e curren
t throug
h
the first
coi
ls
i equa
lo
t zero
:
i{ = 0,
i2 £
= 0.
(10.3
)
In thi
s case
, th
e curren
t throug
h th
e secon
d coi
l create
s th
e magneti
c field
aroun
d
thi
s coi
l whic
h ca
n be represente
d by magneti
c field
line
s (se
e Figur
e 10.1b)
. Some
of thes
e field
line
s may lin
k th
e first
coi
l resultin
g n
i th
e flux
linkage
s \\f\2. Here
, th
e
i th
e fluxlinkage
s of th
e first
coi
l du
e o
t th
e
orde
r of subscript
s indicate
s tha
t \\f\2 s
curren
t throug
h th
e secon
d coil
. The mutua
l inductanc
e M2\ betwee
n th
e secon
d an
d
the first
coil
ss
i define
d a
s
M 21 = ^.
(10.4
)
It turn
s ou
t tha
t mutua
l inductance
s M12 an
d M2\ ar
e th
e same
:
M 2i = Mu = M.
(10.5
)
This fac
ts
i rigorousl
y prove
d ni electromagneti
c field
theor
y an
d t
is
i know
n a
s th
e
reciprocit
y principle
. The physica
l meanin
g of th
e reciprocit
y principl
e s
i apparen
t
fro
m th
e definition
s of M12 an
d M2\ give
n above
. Namely
, thi
s principl
e state
s tha
t
the interchange of excitation and response doe
s no
t affec
t th
e result
. Indeed
,n
i th
e
e first
coi
ls
i excite
d whil
e we conside
r th
e respons
e (flu
x linkages
)
cas
e of Mi2, th
of th
e secon
d coil
.n
I th
e cas
e of M2i, th
e secon
d coi
ls
i excite
d whil
e we conside
r
the respons
e (flu
x linkages
) of th
e first
coil
.t
Is
i clea
r fro
m (10.2)
, (10.4)
, an
d (10.5
)
tha
t fi th
e excitation
s ar
e th
e same
, the
n th
e response
s wil
l be th
e same
, whic
h s
in
i
complianc
e wit
h th
e meanin
g of th
e reciprocit
y principle
. Reciprocit
y wil
l be furthe
r
discusse
d n
i thi
s chapte
r when we stud
y th
e theor
y of two-por
t elements
.
By usin
g equalit
y (10.5
)n
i formula
s (10.2
) an
d (10.4)
, we ca
n rewrit
e the
m a
s
follows
:
M
= -^
= —.
(10.6
)
ifci(
0 = Mi1(0,
(10.7
)
<M 0 = Mi2{t).
(10.8
)
h
12
The las
t expressio
n s
i tantamoun
t to
:
It s
i clea
r fro
m expression
s (10.7
) an
d (10.8
) tha
t th
e mutua
l inductance
, M, ha
s th
e
physica
l meanin
g of th
e measur
e of magneti
c couplin
g betwee
n tw
o coils
. Indeed
,
the mor
e th
e mutua
l inductanc
e betwee
n tw
o coils
, th
e mor
e th
e flux
linkage
s of th
e
secon
d coi
l fo
r th
e sam
e curren
t throug
h th
e first
coi
l and
, consequently
, th
e stronge
r
the magneti
c couplin
g betwee
n tw
o coils
.
The mutua
l inductanc
e was define
d a
s th
e rati
o of fluxlinkage
s o
t currents
.
However
, th
e mutua
l inductanc
e doe
s no
t depen
d on fluxlinkage
s or currents
.t
Is
i
rathe
r th
e coefficien
t of proportionalit
y betwee
n flux
linkage
s an
d current
s a
s clearl
y
suggeste
d by formula
s (10.7
) an
d (10.8)
. The mutua
l inductanc
e depend
s on relativ
e
location
s of tw
o coil
s wit
h respec
to
t on
e another
, thei
r geometri
c dimensions
, th
e
384
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
physica
l propertie
s of th
e medi
a n
i th
e vicinit
y of th
e tw
o coils
, an
d th
e number
s of
turn
s of th
e coils
. Actually
, th
e mutua
l inductanc
e s
i proportiona
lo
t th
e produc
t of
the number
s of turns
:
M - N{N2.
(10.9
)
The calculatio
n of mutua
l inductanc
e s
i a difficul
t proble
m whic
h belong
s o
t
electromagneti
c field
theory
.n
I electri
c circui
t theory
,t
is
i alway
s assume
d tha
t th
e
mutua
l inductanc
es
i known
. The MKS A uni
t of mutua
l inductanc
es
i th
e henry
, tha
t
t cal
l self-inductance)
.
is th
e sam
e a
s tha
t of inductanc
e (whic
ht
is
i no
w mor
e prope
ro
The give
n definitio
n of mutua
l inductanc
e may creat
e th
e impressio
n tha
t th
e
mutua
l inductanc
e s
i alway
s smalle
r tha
n th
e self-inductance
. However
, thi
s impres
sio
n s
i erroneous
. To elucidat
e thi
s issue
, le
t us recal
l th
e definitio
n of self-inductanc
e
give
nn
i Chapte
r 1. The self-inductance
s of th
e first
an
d secon
d coil
s show
n n
i Figur
e
10.
1 ca
n be define
d a
s follows
:
U
= —,
(10.10
)
U = —.
(10.11
)
h
Here, \\t\ \ s
i th
e fluxlinkage
s of th
e first
coi
l when onl
y thi
s coi
ls
i energized
, an
d
ha
s
a
simila
r
physica
l
meaning
.
t
I
s
i
clea
r
tha
t
ip\
\
an
d
i//
ca
n
b
e
calle
d
self-flu
x
i//
22
22
linkages
.
i th
e flux
linkage
s du
e o
t al
l magneti
c field
line
s an
d \\s2\ s
i th
e flux
Sinc
e \\jn s
linkage
s du
eo
t onl
y some par
t of thes
e field
lines
, th
e impressio
n may be create
d tha
t
\}/u s
i alway
s large
r tha
n «//
d thus
, accordin
g o
t (10.6
) an
d (10.10)
,L\ s
i alway
s
2i an
large
r tha
n M. However
, thi
ss
i no
t tru
e becaus
e th
e flux
linkage
s ar
e determine
d no
t
onl
y by th
e numbe
r of field
line
s whic
h lin
k th
e coi
l bu
t by th
e numbe
r of turn
sn
i
the coi
la
s well
. Althoug
h onl
y a smal
l par
t of magneti
c field
line
s may lin
k th
e turn
s
of th
e secon
d coil
, th
e flux
linkage
s if/2\ ca
n be made arbitraril
y larg
e by increasin
g
the numbe
rN2 of turn
s of thi
s coil
.n
I thi
s way
, we ca
n make M large
r tha
n L\. Thi
s
suggest
s tha
t ther
es
i no direc
t an
d universall
y tru
e inequalit
y betwee
n M an
d L\ (o
r
betwee
n M an
d L2, fo
r tha
t matter)
. However
, ther
e s
ia
n importan
t an
d universall
y
tru
e inequalit
y betwee
n M an
d th
e produc
t ofL\ an
d L2, whic
h s
i give
n below
:
M<^[L^L2.
(10.12
)
W e shal
l prov
e thi
s inequalit
y fo
r th
e practicall
y importan
t cas
e when th
e turn
s of
each of th
e coil
s ar
e closel
y spaced
.n
I thi
s case
, al
l turn
s of th
e sam
e coi
l ar
e linke
d
by th
e sam
e numbe
r of magneti
c field
line
s and
, consequently
, by th
e sam
e flux.
By
usin
g thi
s fact
, we ca
n write
:
ife
i =N2<jy2h
(10.13
)
where <£
i th
e flux
whic
h s
i create
d by th
e curren
t throug
h th
e first
coi
l an
d whic
h
2i s
link
s al
l th
e turn
s of th
e secon
d coi
l (Figur
e 10.1a)
.
It s
i clea
r tha
t
<foi ~ 4>\u
(10.14
)
385
20.2
. Mutual Inductance and Coupled Circuit Equations
where fa \ s
i th
e flux
whic
h s
i create
d by th
e curren
t throug
h th
e first
coi
l an
d whic
h
link
s al
l th
e turn
s of th
e sam
e coil
.
By substitutin
g (10.14
) int
o (10.13)
, afte
r simpl
e transformation
s we find:
<fc
i ^N2fax
= ^Nlfal.
(10.15
)
fax =#!<*>!!
.
(10.16
)
It s
i clea
r that
:
By usin
g formul
a (10.16
) ni (10.15)
, we obtain
:
fci ^ ^ n.
(10.17
)
By dividin
g bot
h side
s of (10.17
) by i\ an
d by usin
g th
e definition
s (10.6
) an
d (10.10)
,
w e derive
:
M<^L n.
(10.18
)
Next, conside
r th
e cas
e show
n n
i Figur
e 10.1b
.t
Is
i clea
r that
:
fai=Nxfa2,
(10.19
)
where fa2 s
i th
e fluxwhic
h s
i create
d by th
e curren
t throug
h th
e secon
d coi
l an
d
which link
s al
l th
e turn
s of th
e first
coil
.
It s
i als
o clea
r tha
t fa2 s
i onl
y some par
t of th
e flux
fa2 whic
h s
i create
d by i2
and link
s al
l th
e turn
s of th
e secon
d coil
. Thus
:
fa2 <
fa2.
(10.20
)
By usin
g inequalit
y (10.20
) ni expressio
n (10.19)
, afte
r simpl
e transformation
s we
find:
fa2<NXfa2
= ^ 2 < f e-2
(10.21
)
Sinc
e
\\f22 = N2fa2,
inequalit
y (10.21
) ca
n be writte
n a
s follows
:
*12 ^ ^ f e
By dividin
g bot
h side
s of th
e las
t formul
a by i2 an
d by recallin
g th
e definition
s (10.6
)
and (10.11)
, we derive
:
M < ^ L2 .
N2
(10.22
)
386
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
Now , by multiplyin
g inequalitie
s (10.18
) an
d (10.22)
, we obtain
:
M2 <LXL2,
(10.23
)
which s
i tantamoun
to
t (10.12)
.
Inequalit
y (10.12
) suggest
s th
e followin
g definitio
n of a ver
y importan
t quantit
y
k calle
d th
e couplin
g coefficient
:
M
(10.24
)
V^Li
It s
i clea
r fro
m inequalit
y (10.12
) tha
t
0<k<
1
.
(10.25
)
s of th
e first
It s
i als
o clea
r tha
tk doe
s no
t depen
d on th
e number
s N\ an
d N2 of turn
and secon
d coils
. Thi
ss
i becaus
e th
e self-inductance
s L\ an
d L2 ar
e proportiona
lo
t
the square
s of th
e number
s of turn
s of thei
r coils
:
Lx - Nt
nl
(10.26
)
whil
e fo
r th
e mutua
l inductanc
e M th
e expressio
n (10.9
) si valid
. By usin
g formula
s
(10.26
) an
d (10.9
) ni th
e definitio
n (10.24)
, we conclud
e tha
tk doe
s no
t depen
d on
N\ an
d N2. Thi
s assertio
n suggest
s tha
t th
e couplin
g coefficien
t ca
n be use
d a
s a ver
y
importan
t measur
e of magneti
c couplin
g betwee
n tw
o coils
, a measur
e whic
h doe
s
not depen
d on th
e number
s of turn
s of thes
e coils
. The couplin
g coefficien
t depend
s
onl
y on relativ
e location
s of tw
o coil
s an
d th
e physica
l propertie
s of th
e medi
a ni th
e
vicinit
y of thes
e coils
. The latte
r poin
ts
i ver
y importan
t an
d t
is
i use
d n
i practic
e ni
orde
ro
t enhanc
e th
e magneti
c couplin
g betwee
n tw
o coils
. To achiev
e this
, th
e coil
s
are usuall
y place
d on th
e sam
e iro
n (o
r ferrite
) cor
e of hig
h magneti
c permeabilit
y (se
e
Figur
e 10.2)
. Becaus
e of it
s hig
h magneti
c permeability
, th
e iro
n cor
es
ia
n excellen
t
e o
fa
n iro
n cor
eo
t enhanc
e th
e magneti
c couplin
g betwee
n tw
o
Figur
e 10.2: The us
coils
.
20.2
. Mutual Inductance and Coupled Circuit Equations
387
guid
e fo
r magneti
c field
lines
. Thi
s mean
s tha
t most of th
e magneti
c field
line
s li
e
entirel
y withi
n th
e iro
n core
, whil
e onl
y a ver
y smal
l fractio
n of the
m (attribute
d o
t
leakage
) partiall
y go throug
h air
. As a result
, th
e first
an
d secon
d coil
s ar
e alway
s
linke
d by almos
t th
e sam
e flux.
Thi
s suggest
s tha
t inequalitie
s (10.14
) an
d (10.20
)
are ver
y clos
e o
t equalities,
which
,n
i turn
, suggest
s tha
t th
e couplin
g coefficien
ts
i
ver
y clos
eo
t unity
.
Havin
g define
d an
d discusse
d th
e concep
t of mutua
l inductance
, we shal
l nex
t
conside
r ho
w thi
s concep
t ca
n be use
d n
i electri
c circui
t theory
. Our ultimat
e goa
l
is o
t find
th
e relationshi
p betwee
n th
e termina
l voltage
s v\{t) an
d v2(t) an
d termina
l
o magneticall
y couple
d coils
. To thi
s end
, conside
r th
e
current
s i\(t) an
d i2(t) fortw
cas
e when th
e tw
o coil
s show
nn
i Figur
e 10.
1 ar
e simultaneousl
y energized
. By usin
g
the superpositio
n principle
, we ca
n clai
m tha
t th
e tota
l fluxlinkag
e of th
e first
coi
l
is th
e algebrai
c su
m of self-flu
x linkag
e if/\\(t), whic
h ar
e du
e o
t th
e curren
t i\(t)
throug
h th
e first
coil
, an
d mutua
l fluxlinkag
e \p\2(t), whic
h ar
e du
e o
t th
e curren
t
i2(t) throug
h th
e secon
d coil
:
<M 0 = <M0 ± <M0.
(10.27
)
Similarly
, fo
r th
e tota
l flux
linkag
e of th
e secon
d coi
l we have
:
W e remar
k tha
t th
e " ±" sign
sn
i equation
s (10.27
) an
d (10.28
) indicat
e tha
t mutua
l
fluxlinkage
s may ad
d o
t or subtrac
t fro
m self-flu
x linkages
. Whic
h of thes
e tw
o
eventualitie
s occur
s depend
s on th
e relativ
e direction
s of tw
o current
s a
s wel
la
s th
e
relativ
e windin
g direction
s of tw
o coils
. Thi
s matte
r wil
l be discusse
d ni detai
l a bi
t
later
.
By usin
g expression
s (10.7)
, (10.8)
, (10.10)
, an
d (10.11
) n
i formula
s (10.27
)
and (10.28)
, we ca
n expres
s th
e tota
l fluxlinkage
sn
i term
s of termina
l current
s a
s
follows
:
Mt) = LMt)
± Mi2(t\
(10.29
)
fo(t) = L2i2(t) ± Mh(t).
(10.30
)
Accordin
g o
t Faraday'
s law
, termina
l voltage
s ca
n be relate
d o
t tota
l flux
linkage
sn
i
the manne
r specifie
d below
:
#i(
Q
#2 (0
,
n n a i
y
~J7~>
2W = —1^(10.31
)
at
at
By substitutin
g expression
s (10.29
) an
d (10.30
) int
o equation
s (10.31
) an
d by assum
ing tha
t self-inductanc
e an
d mutua
l inductanc
e ar
e tim
e invariant
, we obtain
:
di\(t)
dii(t)
v1(r
) = L1 - ^ ± A f - ,
^
(10.32
)
at
at
v
i^
=
v2(t) = L 2 ^ ± M - ^ .
(10.33
)
at
at
Now , we shal
l addres
s th
e issu
e of ± sign
s n
i equation
s (10.32
) an
d (10.33
) (a
s
well a
sn
i previou
s equations)
.t
I ha
s bee
n remarke
d befor
e tha
t th
e sig
n ambiguit
y
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
388
appear
s becaus
e tw
o factor
s ar
e simultaneousl
y involved
: relativ
e curren
t direction
s
and relativ
e coi
l windin
g directions
. To illustrat
e th
e secon
d factor
, tw
o set
s of
e tw
o set
s of couple
d
magnicall
y couple
d coil
s ar
e show
n n
i Figure
s 10.3
a an
d b. Thes
coil
s ar
e identica
ln
i al
l respect
s excep
t tha
t th
e secondar
y coil
s hav
e opposit
e windin
g
directions
. As a consequence
, th
e same direction
s of secondar
y current
s wil
l resul
tn
i
opposit
e direction
s of magneti
c field
line
s an
d n
i opposit
e sign
s n
i equation
s (10.32
)
and (10.33)
. Thi
s sig
n ambiguit
y ca
n be remove
d by th
e preliminar
y calibratio
n of
two magneticall
y couple
d coil
s an
d by establishin
g th
e dot convention on th
e basi
s
of thi
s calibration
. The calibratio
n ca
n be accomplishe
d experimentall
y withou
t an
y
prio
r knowledg
e of th
e relativ
e coi
l windin
g direction
s or t
i ca
n be accomplishe
d
theoreticall
y by usin
g th
e right-hand rule whic
h relate
s th
e direction
s of current
s o
t
the direction
s of magneti
c field
line
s produce
d by thes
e currents
. I
n an
y case
,t
is
i
assume
d n
i circui
t theor
y tha
t thi
s calibratio
n ha
s bee
n performe
d an
d it
s result
s ar
e
represente
d by th
e do
t convention
. The essenc
e of th
e do
t conventio
n ca
n be state
d
as follows
: an increasin
g (i
n time
) curren
t enterin
g th
e dotte
d termina
l of on
e coi
l
result
s n
i suc
h an open-circui
t voltag
e at th
e terminal
s of th
e othe
r coi
l tha
t th
e dotte
d
termina
ls
i positive
.
The do
t conventio
n s
i indicate
d n
i Figure
s 10.4
a an
d b alon
g wit
h relativ
e
(reference
) direction
s of electri
c currents
. The couple
d circui
t equation
s (10.32
) an
d
(10.33
) fo
r th
e abov
e tw
o case
s ca
n be written
, respectively
, as follows
. For cas
e (a)
:
vi(t)
= L] — —
at
M——,
at
+
di2(t)
(10.34
)
„dW)
(10.35
)
whil
e fo
r cas
e (b)
:
di\(t)
vi(
0 = Li
dt
M-
diiit)
dt
M
v2(t) = L2
©
(
c
c
di2{t)
dt
di\(t)
i2(t
)
c
c
(a)
)
(
(10.37
)
dt
Q
)
)
)
)
(10.36
)
c
c
c
)
)
)
)
c
(
i2(t
)
(
4
0
)
(b)
Figure 10.3: The cas
e when th
e secondar
y coil
s hav
e opposit
e windin
g directions
.
20.2
. Mutual Inductance and Coupled Circuit Equations
389
(a
)
(b)
Figur
e 10.4: Dot convention
.
The couple
d circui
t equation
s n
i th
e for
m (10.36)—(10.37
) ar
e sometime
s preferabl
e
fro
m th
e physica
l poin
t of view
. Thi
s s
i th
e cas
e when th
e secon
d coi
l doe
s no
t hav
e
an independen
t sourc
e of excitatio
n an
d s
i excite
d onl
y du
e o
t th
e magneti
c couplin
g
with th
e first
coil
. Then
, accordin
g o
t Lenz'
s law
, th
e curren
t 12(f), n
i th
e secon
d
coi
l s
i alway
s induce
d n
i suc
h a way as o
t counterac
t th
e caus
e of induction
. The
primar
y caus
e of inductio
n of 12(f) s
i th
e voltag
e v\(t), an
d th
e minu
s sig
n n
i (10.36
)
reflect
s th
e counte
r actio
n of thi
s caus
e of induction
. The describe
d situatio
n s
i typica
l
for transformer
s an
d inductio
n motors
. For thi
s reason
, couple
d circui
t equation
s n
i
the for
m (10.36)( 10.37
) wil
l be mostl
y use
d n
i ou
r subsequen
t discussion
. Thes
e
equation
s ca
n be easil
y generalize
d n
i orde
r o
t accoun
t fo
r finite
resistance
s of th
e
couple
d coils
. Thes
e resistance
s resul
tn
i additiona
l drop
s of voltage
s whic
h lea
d o
t
the followin
g modificatio
n of couple
d circui
t equations
:
vx(t) = RM0 + L{ dt
di\(t)
di2(t)
dt
(10.38
)
di2(t)
dt
di\(t)
dt
(10.39
)
v2(t) = R2i2(t) + L2
In th
e cas
e of ac stead
y state
, th
e abov
e equation
s ca
n be writte
n n
i th
e phaso
r for
m
as follows
:
V{ = R{I{
+ y x ni/ -7*12/2
,
(10.40
)
V2 = Rih
+ 7*22/
2 - 7*12/1
,
(10.41
)
where x\ \, x22, an
d x{2 ar
e th
e self-reactance
s an
d mutua
l reactance
, whic
h ar
e relate
d
to self-inductance
s an
d mutua
l inductanc
e by th
e expressions
:
x\\ = coLj
,
X\2 — (OM.
X22 = G)L2,
(10.42
)
(10.43
)
The abov
e couple
d circui
t equation
s ca
n be als
o writte
n n
i th
e impedanc
e form
:
390
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
V{ = znii + znh,
(10.44
)
V2 = znh +z22h
(10.45
)
where
:
zn = R\ + jxn,
z22 = Ri + j*22,
z\2 = -jxn-
(10.46
)
It s
i appropriat
e o
t stres
s her
e tha
t th
e mutua
l inductanc
e s
i associate
d wit
h tw
o
pair
s of terminal
s or wit
h tw
o ports
.n
I thi
s respect
, tw
o inductivel
y couple
d coil
s ca
n
be treate
d a
s a two-por
t elemen
t (se
e Sectio
n 10.4)
. The couple
d circui
t equation
s
(10.44)( 10.45
) provid
e a complet
e termina
l descriptio
n of thi
s two-por
t element
.
These equation
s ca
n be use
d o
t complemen
t th
e KVL an
d KCL equation
s writte
n
for circuit
s connecte
d o
t th
e first
an
d secon
d coils
, respectively
.n
I thi
s way
, we ca
n
obtai
n th
e complet
e se
t of equation
s fo
r magneticall
y couple
d circuits
.
It s
i clea
r fro
m th
e previou
s discussio
n tha
t magneti
c couplin
g betwee
n tw
o
coil
ss
i realize
d fo
r time-varyin
g current
sn
i thes
e coils
. Thi
s follow
s fro
m Faraday'
s
law (se
e formul
a (10.31))
, whic
h lead
so
t induce
d voltage
s onl
y fo
r time-varyin
g flu
x
linkages
.t
Is
i instructiv
eo
t not
e tha
t ther
es
ia
n interestin
g an
d importan
t exceptio
n o
t
the abov
e assertion
. Thi
s exceptio
n hold
s fo
r superconductin
g coils
. Thi
ss
i becaus
e
for superconductin
g coil
s thei
r tota
l fluxlinkage
s ar
e "frozen.
" The
y do no
t chang
e
wit
h tim
e an
d the
y remai
n equa
lo
t thei
r initia
l values
, tha
t is
, th
e value
s a
t th
e
instanc
e of superconductin
g transition
. Accordin
g o
t equation
s (10.29
) an
d (10.30)
,
thi
s lead
so
t th
e followin
g constraint
s impose
d on current
sn
i th
e magneticall
y couple
d
superconductin
g coils
:
i//i(
0 = Lih(t) ± Mi2(t) = constant
,
(10.47
)
ip2(t) = L2i2(t) ± Mh(t) = constant
.
(10.48
)
It s
i apparen
t fro
m equation
s (10.47
) an
d (10.48
) tha
t eve
n ver
y slo
w variation
s n
i
one curren
t resul
tn
i immediat
e adjustment
s ni anothe
r curren
tn
i orde
ro
t kee
p th
e
tota
l flux
linkage
s th
e same
. Thi
s suggest
s tha
t ther
es
i "static
" (dc
) couplin
g betwee
n
superconductin
g coils
.n
I practice
, thi
s couplin
g prohibit
s independen
t (separate
)
excitatio
n of superconductin
g coils
.
10.
3
Transformer
s
In thi
s section
, we shal
l us
e th
e couple
d circui
t equation
s n
i orde
r o
t discus
s th
e
operatio
n an
d performanc
e of a ver
y importan
t electri
c devic
e know
n a
s a transformer
.
The transforme
r s
i use
d o
t ste
p up or ste
p down a
c voltages
. Fo
r thi
s reason
, th
e
transforme
rs
ia
n indispensabl
e lin
k ni electri
c powe
r systems
. The transforme
r ca
n
als
o be use
d o
t electricall
y isolat
e a sourc
e fro
m a loa
d or fo
r impedanc
e matchin
g
purposes
. Thi
s explain
s why th
e transforme
r s
i als
o a vita
l componen
t n
i many
low-powe
r applications
.
The transforme
r ca
n be define
d a
s a devic
e n
i whic
h tw
o (o
r more
) stationar
y
coil
s (calle
d windings
) ar
e couple
d magnetically
. One of th
e windings
, know
n a
s th
e
10.3. Transformers
391
primary
, receive
s powe
r a
t a specifie
d voltag
e an
d frequenc
y fro
m th
e source
, whil
e
the othe
r winding
, know
n a
s th
e secondary
, deliver
s powe
ro
t th
e loa
d a
t a differen
t
voltag
e bu
t th
e sam
e frequency
. To enhanc
e electromagneti
c couplin
g betwee
n th
e
primar
y an
d secondar
y windings
, the
y ar
e place
d on th
e sam
e iro
n cor
e (se
e Figur
e
10.5)
. Thi
s iro
n cor
es
i subjec
to
t atime-varyin
g magneti
c flux
whic
h link
s th
e primar
y
and secondar
y windings
. Sinc
e th
e iro
n cor
e ha
s a finite
(nonzero
) conductivity
, thi
s
time-varyin
g fluxinduce
s edd
y current
s n
i th
e iro
n core
. Thes
e edd
y current
s may
produc
e substantia
l powe
r losse
s calle
d edd
y curren
t losses
. To reduc
e edd
y curren
t
losses
, th
e iro
n cor
e s
i laminated
. Thi
s mean
s tha
t th
e iro
n cor
e s
i assemble
d fro
m
many ver
y thi
n stee
l lamination
s whic
h ar
e electricall
y isolate
d fro
m on
e anothe
r by
ver
y thi
n oxidatio
n (o
r varnish
) layers
. The stee
l use
d fo
r transforme
r lamination
ss
i
calle
d transforme
r steel
.t
Is
i usuall
y deliberatel
y dope
d wit
h silico
nn
i orde
ro
t reduc
e
it
s intrinsi
c conductivit
y withou
t appreciabl
y affectin
g it
s hig
h magneti
c permeability
.
This reductio
n n
i stee
l conductivit
y furthe
r diminishe
s edd
y curren
t losses
.
To bette
r understan
d th
e principl
e of operatio
n of th
e transformer
, we shal
l first
conside
r th
e idea
l transformer
. n
I th
e cas
e of th
e idea
l transformer
, we neglec
t th
e
smal
l resistance
s R\ an
d R2 of th
e primar
y an
d secondar
y windings
, respectively
. We
als
o neglec
t leakag
e fluxi//g
.n
I othe
r words
, we assum
e tha
t al
l magneti
c field
line
s
are entirel
y confine
d o
t th
e iro
n core
. We als
o assum
e tha
t th
e magneti
c permeability
,
jii
,
o
f
th
e
iro
n
cor
e
s
i
infinite
,
whil
e
th
e
conductivity
,
cr
,
o
f
th
e
sam
e
cor
e
s
i
equa
l
o
t
c
c
zero
. Al
l th
e mentione
d assumption
s ar
e summarize
d below
:
fli = R2 = 0,
ifj£ = 0,
ov = 0.
He
(10.49
)
Sinc
e we neglec
t th
e leakag
e flux,
t
i mean
s tha
t we assum
e tha
t al
l turn
s of th
e
<f>(t) whic
h s
i
primar
y an
d th
e secondar
y winding
s ar
e linke
d wit
h th
e sam
e flux
forme
d by th
e magneti
c field
line
s entirel
y confine
d o
t th
e iro
n core
. Thi
s mean
s tha
t
the flux
linkage
s of th
e primar
y an
d secondar
y winding
s ca
n be expresse
d a
s follows
:
M) = Nrfit),
©
0—
tMO =N24>(t).
1(t
)
u(t
)
v
1«
©
v*(t
)
<\
- 6—
Figure 10.5: Transformer
.
(10.50
)
392
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
Now , by usin
g Faraday'
s la
w an
d expression
s (10.50)
, we find:
Ml) = !m = Nlm,
at
at
v2(t) = — — = N2——.
at
at
By dividin
g expressio
n (10.51
) by expressio
n (10.52)
, we derive
:
(10 . 5 i)
(10.52
)
vi(
0
v2(t)
(10.53
)
a = g.
(10.54)
where a stand
s fo
r th
e turn
s ratio
:
If th
e applie
d (source
) voltage
, v\{t), of th
e primar
y windin
g s
i sinusoidal
, then
,
accordin
g o
t (10.53)
, th
e secondar
y (induced
) voltage
,v2(t), s
i sinusoida
la
s well
.t
I
is als
o clea
r fro
m (10.53
) tha
t th
e phasor
s of th
e primar
y an
d secondar
y voltage
s ar
e
relate
d o
t on
e anothe
ra
s follows
:
^ = a.
V2
(10.55
)
Expression
s (10.53
) an
d (10.55
) clearl
y elucidat
e th
e principl
e of operatio
n of th
e
transformer
. The
y sugges
t tha
t by manipulatio
n of th
e turn
s rati
o we ca
n easil
y achiev
e
the desire
d secondar
y voltage
.
Next, we shal
l deriv
e th
e expressio
n fo
r th
e rati
o of primar
y an
d secondar
y
currents
. To thi
s end
, we shal
l exploi
t ou
r assumption
s tha
tR\ = R2 = 0 an
d crc = 0.
These assumption
s mean tha
t ther
e ar
e no powe
r losse
s n
i th
e idea
l transformer
.
Consequently
, th
e primar
y power
,p\{t), delivere
d o
t th
e terminal
s of th
e primar
y
d o
t th
e load
:
windin
g s
i equa
lo
t th
e secondar
y power
,p2(t), delivere
P\(t) = p2(t).
(10.56
)
Recallin
g th
e expression
s fo
r instantaneou
s power
, we have
:
Pi(t) = vx(t)i{{t\
p2(t) - v2(t)i2(t).
(10.57
)
By substitutin
g expression
s (10.57
) int
o equalit
y (10.56)
, we en
d up with
:
VimiO
= V2(t)i2(t).
(10.58
)
By dividin
g bot
h part
s of expressio
n (10.58
) by th
e produc
tv\(t)i2(t), we find:
^>
hit)
= *H.
v,(
0
(1059
)
20.3
. Transformers
393
Now , by recallin
g expressio
n (10.53)
, we finally
derive
:
ii(t)
1
.fA
a
i2(t)
The las
t formul
a ca
n be writte
n n
i th
e phaso
r for
m a
s follows
:
£
= -.
(10.60
)
(10.61)
Expression
s (10.55
) an
d (10.61
) completel
y defin
e th
e idea
l transforme
r a
s a two-por
t
elemen
to
f electri
c circuits
.t
Is
i sometime
s mor
e convenien
to
t writ
e thes
e equation
s
in th
e form
:
Vi = aV2,
(10.62
)
(10.63
)
/, = -h
a
which provid
e complet
e "terminal
" characterizatio
n o
f thi
s two-por
t element
. Thes
e
d wit
h KVL an
d KCL equation
s fo
r th
e res
to
f
termina
l relation
s shoul
d be combine
electri
c circuit
sn
i orde
ro
t analyz
e th
e electri
c circuit
s wit
h idea
l transformers
.
To conclud
e ou
r discussio
n of th
e idea
l transformer
, conside
r it
s inpu
t impedance
.
W e recal
l tha
t th
e inpu
t impedanc
e s
i define
d as
:
.
Zin = y-
(10.64
)
By usin
g expression
s (10.62
) an
d (10.63
)n
i formul
a (10.64)
, we obtain
:
Zin =a2^.
(10.65
)
e rati
o o
f th
e secondar
y voltag
e V2 o
t th
e secondar
y curren
th s
i th
e loa
d
However
, th
impedance
:
ZL = ^.
h
(10.66
)
By combinin
g expression
s (10.65
) an
d (10.66)
, we derive
:
Zin =a2ZL.
(10.67
)
Thus, th
e loa
d impedanc
e ZL viewe
d fro
m th
e primar
y terminal
s s
i equa
lo
t a2ZL.
This fac
t suggest
s tha
t th
e idea
l (an
d real
) transforme
r ca
n be use
d fo
r impedanc
e
matchin
g purposes
, an
d thi
s s
i actuall
y th
e cas
e n
i some applications
. t
Is
i als
o
clea
r fro
m expressio
n (10.67
) that
, wit
h respec
to
t th
e primar
y terminals
, th
e idea
l
transforme
r ca
n be represente
d by th
e equivalen
t circui
t show
n n
i Figur
e 10.6
. The
equivalenc
e her
e s
i understoo
d n
i th
e sens
e that
,a
s fa
r a
s th
e relationshi
p betwee
n
the primar
y voltag
e V\ an
d primar
y curren
tI\ s
i concerned
, th
e idea
l transforme
r an
d
the circui
t show
n n
i Figur
e 10.
6 ar
e indistinguishable.
t
Is
i a ver
y powerfu
l ide
a o
t
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
394
a2Z,
Figur
e 10.6: Equivalen
t circui
t fo
r th
e idea
l transformer
.
replac
e a complicate
d devic
e by a simpl
e equivalen
t electri
c circui
t whic
h provide
s
the sam
e relationshi
p fo
r termina
l voltage
s an
d current
s a
s we hav
e fo
r th
e actua
l
device
. Thi
s ide
a permeate
s many differen
t area
s of electrica
l engineering
.
Next, we procee
d o
t th
e discussio
n of th
e transforme
r theor
y by removin
g th
e
first
thre
e assumption
s n
i (10.49)
. The onl
y assumptio
n tha
t stil
l wil
l be n
i plac
e s
i
tha
t <TC = 0, whic
h s
i tantamoun
to
t th
e neglec
t of edd
y curren
t losses
. Thes
e losse
s
wil
l be take
n int
o accoun
ta
t th
e ver
y en
d of ou
r discussion
, albei
tn
i a somewha
ta
d
hoc manner
.
By removin
g th
e first
thre
e assumption
s ni (10.49)
, we ca
n no
w bas
e ou
r analysi
s
of th
e transforme
r on th
e couple
d circui
t equation
s (10.40)( 10.41)
. Al
l th
e quantitie
s
in thes
e equation
s marke
d by subscrip
t 1 ar
e relate
d o
t th
e primar
y winding
, whil
e
all th
e quantitie
s marke
d by subscrip
t 2 ar
e relate
d o
t th
e secondar
y winding
, an
d x\2
is th
e mutua
l reactanc
e betwee
n th
e primar
y an
d secondar
y windings
.
First
, we us
e equatio
n (10.40
) an
d solv
et
i fo
r I\:
Vi
7*12
(10.68
)
R{ + jxn + R{ + jxu
Next, we substitut
e expressio
n (10.68
) int
o equatio
n (10.41
) and
, afte
r simpl
e trans
formations
, we obtain
:
I\ =
Vn
=
R\
"7*1
2
Vx + (R2 + jx22)
+7*1
1
(jxnf
(*! + 7*n)(*2 + 7*22)
h,
(10.69
)
Expressio
n (10.69
) fo
r V2 ha
s tw
o distinc
t terms
. The first
ter
m
^in
d
"7*1
2
'2
= R\ +7*n•Vi
vr
(10.70
)
is th
e secondar
y voltag
e n
i th
e cas
e when I2 = 0. n
I othe
r words
, thi
s s
i a
n open
circui
t secondar
y voltag
e whic
h ca
n be physicall
y interprete
d a
s th
e voltag
e induce
d
in th
e secondar
y windin
g by th
e magneti
c flux
create
d by th
e curren
tn
i th
e primar
y
winding
. Thi
s explain
s why we us
e superscrip
t "ind
" fo
r V2 n
i (10.70)
.
The secon
d ter
m
(jxn)7
V? o p = (R2 + jx22) 1 (R{ + jxn)(R2 + jx22)
(10.71
)
10.3. Transformers
395
has th
e physica
l meanin
g of th
e dro
p of th
e voltag
e du
e o
t th
e finite
(nonzero
) loa
d
curren
tI2. t
Is
i highl
y desirabl
eo
t hav
e thi
s ter
m a
s smal
la
s possibl
en
i orde
ro
t kee
p
the secondar
y voltag
e V2 (whic
h s
i th
e voltag
e acros
s th
e loa
d terminals
)a
s constan
t
as possibl
en
i th
e fac
e of continuousl
y (an
d quit
e ofte
n unpredictably
) changin
g load
.
It s
i clea
r tha
t th
e smalle
r V2rop th
e bette
r th
e qualit
y of th
e transformer
.
Next, we conside
r anothe
r importan
t characteristi
c of th
e transformer
, namely
, it
s
c
secondar
y short-circui
t curren
t 7|
. Thi
s curren
t occur
s when th
e secondar
y windin
g
is accidentl
y short-circuited
, tha
t is
, when
:
V2 = 0.
From th
e las
t expressio
n an
d (10.69)
, we find:
JX\2
2
1
+ jx22) 1 " (Rl+jX (M2)
)(R +jX22)\
(/?
! + jxn)(R2
n
(10.72
)
2
W e ca
n se
e tha
t th
e sam
e quantit
y
D = 1
^^
(/?
, + jxnXRi + jXTi)
(10.73
)
appear
s ni th
e expression
s (10.71
) an
d (10.72
) fo
r V2rop an
d 22c, respectively
. The
smalle
r D, th
e bette
r th
e qualit
y of th
e transforme
r wit
h respec
to
t it
s abilit
yo
t maintai
n
more or les
s constan
t voltag
e acros
s th
e loa
d terminal
sn
i th
e fac
e of changin
g load
.
vc
However
, smal
lD result
sn
i alarg
e short-circui
t curren
t7
h s
i undesirable
. Thi
s
2 whic
suggest
s tha
t th
e successfu
l desig
n of th
e transforme
r shoul
d carefull
y balanc
e ou
t
thes
e competin
g requirements
.
The previou
s discussio
n clearl
y reveal
s th
e importanc
e of th
e quantit
y D. Fo
r thi
s
reason
, we shal
l carefull
y analyz
e thi
s quantity
. Usually
, th
e primar
y an
d secondar
y
resistance
s ar
e much smalle
r tha
n th
e primar
y an
d secondar
y reactances
:
Rx <xn,
R2<x22.
(10.74
)
For thi
s reason
, expressio
n (10.73
) fo
rD ca
n be simplifie
d a
s follows
:
D
x2
« i - —!2_a
(10.75
)
XUX22
By recallin
g formula
s (10.42)
, (10.43)
, an
d (10.24)
, expressio
n fo
r D ca
n be furthe
r
transforme
d a
s follows
:
M2
9
D = 1 - — - = 1 -k2.
L\L2
(10.76
)
It ha
s bee
n pointe
d ou
tn
i th
e previou
s sectio
n that
, fo
r strongl
y couple
d coil
s (whic
hs
i
certainl
y th
e cas
e fo
r th
e primar
y an
d secondar
y transforme
r windings)
, th
e couplin
g
coefficien
tk s
i ver
y clos
eo
t unity
. Thi
s suggest
s tha
tD s
i a smal
l quantity
. We shal
l
next sho
w tha
t thi
s quantit
y s
i full
y determine
d by leakag
e parameters
.
Conside
r th
e self-flu
x linkage
s \\t\ \ of th
e primar
y winding
. Thes
e flux
linkage
s
are equa
lo
t th
e su
m of flux
linkage
s du
eo
t th
e magneti
c fluxc/>
, whic
h s
i forme
d by
396
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
the magneti
c field
line
s entirel
y confine
d o
t th
e iro
n core
, an
d th
e flux
linkage
s if/f,
which ar
e du
e o
t leakag
e field
line
s whic
h partiall
y or entirel
y li
e outsid
e th
e cor
e
(se
e Figur
e 10.2)
. Thus
, th
e flux
linkage
s ifju ca
n be writte
n a
s follows
:
i//n = A ^ + i//f.
(10.77)
By recallin
g expressio
n (10.10
) fo
r self-inductance
, fro
m formul
a (10.77
) we find:
Lx = -^
+ ^.
(10.78
)
Thus, we ca
n se
e tha
t th
e self-inductanc
e consist
s of tw
o distinc
t components
. The
first
componen
t
U[l = ~
(10.79)
h
can be interprete
d a
s th
e mai
n self-inductance
. The ter
m "main
" emphasize
s th
e fac
t
7
tha
t L"
s
i th
e predominan
t par
t of L\.
The secon
d ter
m
L\ = -9-
(10.80)
'i
can be interprete
d a
s th
e "leakage
" self-inductance
.
Thus:
Lx = L}[1 +Lf
.
(10.81
)
Next, we shal
l establis
h th
e connectio
n betwee
n th
e mutua
l inductanc
e M an
d th
e
main self-inductanc
e L™. First
, we recal
l that
:
M = —.
(10.82
)
ifc
i - iV
.
2</>
(10.83
)
h
It si als
o clea
r that
:
By substitutin
g expressio
n (10.83
) int
o formul
a (10.82)
, afte
r simpl
e transformatio
n
w e obtain
:
N2d>
NiN](b
M = -^
= ^ ^ ^.
/
N[ i\
By usin
g expressio
n (10.79
)n
i (10.84)
, we derive
:
(10.84
)
M = ^L™.
(10.85
)
N\
By literall
y repeatin
g th
e sam
e lin
e of reasonin
g a
s was use
d n
i derivatio
n of (10.81
)
and (10.85)
, we ca
n establis
h that
:
L2 = L%+l4,
(10.86
)
10.3. Transformers
397
and
M = — U£.
(10.87
)
N2
By substitutin
g expression
s (10.85
) an
d (10.87
) int
o formula
s (10.81
) an
d (10.86)
,
respectively
, we find:
— M = Lx ~ 1%,
N2
(10.88
)
2
M = L2N-.
(10.89
)
LJ.
By multiplyin
g th
e expression
s (10.88
) an
d (10.89)
, we obtain
:
M2 = (Li -L\)(L
2 -LJ) = LXL2 - LiLJ - L2h\ + L\L\.
(10.90
)
By dividin
g bot
h side
s of expressio
n (10.90
) by th
e produc
tL\L2, we arriv
e at
:
M2
M
Ll
L£
LlLl
^ L kL2 + L[L
V l.
.9i
)
(10
L\L2 = i - L[
2
By substitutin
g th
e las
t expressio
n int
o formul
a (10.76)
, we finall
y derive
:
L£
L£
L£L£
D = ±- + ^ - ^±±-.
L\
L2
L\L2
(10.92
)
The las
t expressio
n clearl
y reveal
s th
e importanc
e of leakag
e inductances
. Thes
e
inductance
s determin
e th
e valu
e of Z)
, which
,n
i turn
, determine
s th
e abilit
y of th
e
transforme
ro
t maintai
n mor
e or les
s constan
t voltag
e acros
s th
e loa
d terminal
sa
s wel
l
as th
e vulnerabilit
y of th
e transforme
r wit
h respec
to
t th
e occurrenc
e of accidenta
l
short
s of th
e loa
d terminals
.
The importanc
e of th
e leakag
e inductance
s ca
n als
o be demonstrate
d fro
m th
e
purel
y mathematica
l poin
t of view
. Conside
r couple
d circui
t equation
s (10.40
) an
d
(10.41
) a
s a se
t of tw
o linea
r simultaneou
s equation
s wit
h respec
to
t I\ an
d I2^ t
Is
i
eas
y o
t se
e tha
t th
e determinant
, A, of thes
e equation
ss
i equa
l to
:
A = (/?
, +jxn)(R2
+ jx22) - {jxnf
-
(/?
, + jxn)(R2
+jx22)D.
(10.93
)
It s
i eviden
t fro
m expression
s (10.93
) an
d (10.92
) tha
t thi
s determinan
ts
i smal
l an
d
it become
s equa
lo
t zer
o when we neglec
t th
e leakag
e inductance
s (alon
g wit
h smal
l
resistance
s R] an
d R2). n
I othe
r words
, th
e se
t of couple
d circui
t equation
s (10.40)
(10.41
) become
s degenerat
e (singular
) fi we neglec
t th
e leakag
e inductances
.n
I
mathematics
, th
e problem
s n
i whic
h th
e neglec
t of smal
l parameter
s lead
s o
t de
generac
y (singularity
) ar
e calle
d singularly perturbed problems
. Thus
, th
e couple
d
circui
t equation
s (10.40)( 10.41)
, whic
h describ
e th
e operatio
n of th
e transformer
,
are singularl
y perturbed
.t
I ha
s lon
g bee
n understoo
d n
i mathematic
s tha
t smal
l pa
rameter
s ni singularl
y perturbe
d problem
s ar
e ver
y important
. Thi
s s
i becaus
e t
is
i
the smal
l parameter
s whic
h make th
e singularl
y pertrube
d problem
s wel
l define
d
(nonsingular)
. The presente
d discussio
n bring
s othe
r (mathematical
) evidenc
e fo
r th
e
importanc
e of th
e leakag
e inductances
.
398
Chapter 10. Magnetically Coupled Circuits and Two-Port
Elements
Thus, we hav
e establishe
d tha
t th
e leakag
e inductance
s ar
e ver
y important
.
However
, th
e couple
d circui
t equation
s (10.40)( 10.41
) do no
t contai
n th
e leakag
e
reactance
s explicitly
. Thes
e reactance
s ar
e absorbe
d by th
e tota
l reactance
s xxx an
d
x22 an
d thei
r significanc
es
i maske
d an
d lost
. Thi
s suggest
s tha
tt
isi highl
y desirabl
eo
t
modif
y th
e couple
d circui
t equation
s (10.40)( 10.41
)n
i suc
h a way tha
t th
e leakag
e
reactance
s wil
l be explicitly accounted for. As s
i typica
ln
i singularl
y perturbe
d
problems
, th
e smal
l parameters—leakag
e reactances—ca
n be expose
d a
s a resul
t
of th
e appropriat
e scaling. Fo
r th
e couple
d circui
t equation
s (10.40)( 10.41)
, thi
s
scalin
g s
i accomplishe
d by introducin
g scale
d (o
r primed
) secondar
y voltag
e an
d
secondar
y current
:
% = aV2,
V2 = -,
(10.94
)
where
,a
s before
,a s
i th
e turn
s ratio
.
The couple
d circui
t equation
s (10.40)( 10.41
) ca
n be writte
n n
i term
s o
fV]2 an
d
I{ a
s follows
:
Vx = RXIX +jxuh
- jaxl2li,
V^ = a2R2I2 + ja2x22I2
(10.95
)
-jaxnh-
(10.96
)
Now , we introduc
e th
e scale
d secondar
y resistanc
e an
d secondar
y reactance
:
R'2 =a2R2,
x'22 = a2x22,
(10.97
)
e previou
s equation
sn
i th
e for
m give
n below
:
and rewrit
e th
h = RXIX + jxuh - jaxl2li,
(10.98
)
% =R& + jx!22U -jaxx2K
(10.99
)
Next, we shal
l make th
e followin
g equivalen
t transformation
s o
f th
e las
t tw
o equa
tions
: we shal
l ad
d an
d subtrac
t th
e term
s jax\2I\ an
d jax^I^ n
i equation
s (10.98
)
and (10.99)
, respectively
:
Vx = RXIX + j(xxx
~ axX2)Ix + jaxX2(Ix
- l[\
(10.100
)
Vi = R2I2 + j(*2
2 -axn)H + jaxnOi ~ hi
(10.101
)
Now , conside
r th
e physica
l meanin
g o
f th
e coefficient
s n
i equation
s (10.100)
(10.101)
. First
, by usin
g formula
s (10.43)
, (10.54)
, an
d (10.85)
, we derive
:
Nx
Nx N7
ax]2 = —u)M = — co—Lf
Af2
N2 Nx
= coL
f - jtft
,
(10.102
)
where x™x has th
e physica
l meanin
g of th
e main reactanc
e of th
e primar
y winding
.
Next, by usin
g formula
s (10.42)
, (10.43)
, and (10.81)
, we find:
xxx ~ axX2 = (oLx - coL'
f = coLf
= jcf
,
where x\ stand
s fo
r th
e leakag
e reactanc
e o
f th
e primar
y winding
.
(10.103
)
103. Transformers
399
Finally
, by usin
g formula
s (10.42)
, (10.43)
, (10.86)
, (10.87)
, an
d (10.97)
, we
obtain
:
/
_
2
_
2 /
^
x22 ~"ctx\2 — a X2
2 ~ ax\2 — a
I
*22 _ ~X\2
= <^2
N2
- T T^
~ L™)
^(^
2
= <^2
(10.104
)
0
=
L
2
(10.105
)
x
(4)'.
2
A22
=
X2
2 ~ -X\2
a
a xi
^
X l2 =— "
mi\2
-^
2 — V^2
(10.106
)
where x| stand
s fo
r th
e leakag
e reactanc
e of th
e secondar
y windin
g an
d {x\)f s
i th
e
scale
d valu
e of thi
s reactance
.
By substitutin
g expression
s (10.102)
, (10.103)
, an
d (10.106
) int
o equation
s
(lO.lOO)-(lO.lOl)
, we derive
:
Vx = Rjx + jxpx + jx^ih - fy,
(10.107
)
v> = R^
(10.108
)
+jx4yii
+jx^iH - hi
Thus, by mean
s of equivalen
t mathematica
l transformation
s we hav
e represente
d
the couple
d circui
t equation
s (10.40
) an
d (10.41
)n
i th
e for
m (10.107)-(10.108
)n
i
which th
e leakag
e reactance
s ar
e explicitl
y accounte
d for
. The usefu
l by-produc
t of
the abov
e equivalen
t transformation
s s
i th
e fac
t tha
t equation
s (10.107
) an
d (10.108
)
can be interprete
d a
s KVL equation
s fo
r th
e electri
c circui
t show
n n
i Figur
e 10.7
.
This circui
t ca
n be considere
d th
e equivalen
t electri
c circui
t fo
r th
e transformer
.
In othe
r words
, thi
s circui
t an
d th
e transforme
r ar
e describe
d by mathematicall
y
identica
l set
s of equation
s and
, consequently
, th
e transforme
r an
d thi
s circui
t ar
e
indistinguishabl
e a
s fa
r a
s th
e relationshi
p betwee
n termina
l voltage
s an
d termina
l
current
s ar
e concerned
. The loa
d impedanc
e of th
e transforme
r s
i modele
d n
i th
e
equivalen
t circui
t by it
s scale
d valu
e a2ZL. Thi
s immediatel
y follow
s fro
m (10.94
)
and th
e followin
g brie
f derivation
:
^2
'1
R
i
_
, t2> _
(x?V
,
(10.109
)
A '
L
A/WV^> -
a2Z,
Figur
e 10.7
: Equivalen
t circui
t fo
r th
e transforme
r withou
t edd
y curren
t losses
.
400
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
In ou
r discussions
, we hav
e s
o fa
r neglecte
d edd
y curren
t losse
s n
i th
e transforme
r
core
. Now, we shal
l tr
y o
t tak
e the
m int
o accoun
t ni th
e equivalen
t circui
t fo
r th
e
transformer
. t
I ca
n be show
n (an
d thi
s s
i beyon
d th
e scop
e of thi
s text
) tha
t edd
y
curren
t losse
s ar
e directl
y proportiona
lo
t th
e squar
e of th
e magneti
c flux
throug
h th
e
iro
n core
. By usin
g Faraday'
s law
,t
is
i eas
y o
t se
e tha
t th
e rms valu
e of th
e voltag
e
induce
d n
i th
e primar
y windin
g by th
e cor
e flux
s
i directl
y proportiona
lo
t thi
s flux.
Thus, we ca
n conclud
e tha
t th
e edd
y curren
t losse
s ar
e proportiona
lo
t th
e squar
e of
the voltag
e induce
d n
i th
e primar
y windin
g by th
e cor
e flux.
n
I th
e equivalen
t circuit
,
thi
s voltag
e s
i th
e voltage
, V\2, acros
s th
e terminal
s of th
e reactanc
e x™y Thus
, we
can writ
e tha
t
V \2>
(10.110
)
where Pe s
i th
e edd
y curren
t loss
.
Expressio
n (10.110
) lead
so
t th
e ide
a of modelin
g th
e actua
l edd
y curren
t losse
s
Pe n
i th
e iro
n cor
e of th
e transforme
r by th
e ohmi
c losse
s PR€ n
i resisto
rRe connecte
d
acros
s th
e terminal
s of J™C (se
e Figur
e 10.8)
. The rational
e behin
d thi
s ide
as
i th
e fac
t
tha
t th
e ohmi
c losse
sn
i Re ar
e als
o proportiona
lo
t th
e squar
e of th
e sam
e voltage
:
p
PRe
_V?2
~ ~RZ
(10.111
)
The resistor
,Re, ca
n be chose
n fro
m th
e conditio
n tha
t losse
s Pe an
d PRS shoul
d be
the same
:
(10.112
)
Pe — PRe-
By usin
g expressio
n (10.112
)n
i formul
a (10.111)
, we easil
y derive
:
Re = - £.
P„
(10.113
)
The electri
c circui
t show
n ni Figur
e 10.
8 s
i th
e complet
e equivalen
t circui
t fo
r
the transformer
. One of th
e importan
t feature
s of thi
s equivalen
t circui
ts
i tha
t th
e
importan
t smal
l parameter
s suc
h a
s th
e leakag
e reactance
s ar
e explicitl
y accounte
d
i./
A
^AAAAr
i
1
AAAAro-
+
a2Z,
Figure 10.8: Complet
e equivalen
t circui
t fo
r th
e transformer
.
20.3
. Transformers
401
for
. I
ts
i interestin
g o
t not
e that
,n
i th
e cas
e of th
e idea
l transformer
, th
e equivalen
t
circui
t show
n n
i Figur
e 10.
8 s
i reduce
d o
t th
e equivalen
t circui
t show
n n
i Figur
e 10.6
.
Indeed
,n
i th
e cas
e of th
e idea
l transforme
r we assum
e tha
t primar
y an
d secondar
y
r
resistance
s R\ an
d R2 a ^
equa
lo
t zero
. We als
o assum
e tha
t leakag
e fluxlinkage
s ar
e
equal o
t zero
, whic
h s
i tantamoun
t o
t th
e assumptio
n tha
t leakag
e reactance
s xf an
d
{x\y ar
e zeros
. The abov
e assumption
s mean tha
t we shoul
d replac
e th
e correspondin
g
circui
t element
s by short-circui
t branches
. I
n th
e cas
e of th
e idea
l transformer
, we
, consequently
, Pe = 0, whic
h accordin
g o
t (10.113
)
als
o assum
e tha
t crc = 0 and
lead
s o
t th
e infinit
e valu
e of Re. I
ts
i als
o assume
d tha
t [ic = oo
, whic
h lead
s o
t th
e
infinit
e valu
e of JC™. Thus
, th
e las
t tw
o assumption
s sugges
t tha
t th
e correspondin
g
circui
t element
s shoul
d be replace
d by open-circui
t branches
. By introducin
g th
e
above-mentione
d short-circui
t an
d open-circui
t branches
, we reduc
e th
e equivalen
t
circui
t show
n n
i Figur
e 10.
8 o
t th
e equivalen
t circui
t show
n n
i Figur
e 10.6
.
EXAMPLE 10.1 Conside
r th
e circui
t show
n n
i Figur
e 10.9a
. The parameter
s of th
e
,LJ = 1 mH,R} = 0.0
4 H,
transforme
r ar
e L™ = 1 mH,L™ = 25 mH,h\ = 40 /iH
R2 = 1.
0 H, an
d a = 0.2
. The voltag
e sourc
e s
i give
n by v\(t) = 100sin(2500f
) V.
Find th
e voltag
e acros
s th
e loa
d resistor
.
Firs
t not
e tha
tf
i th
e transforme
r wer
e ideal
, th
e outpu
t voltag
e woul
d be v0(t) =
500 sin(2500?
) V. To solv
e th
e proble
m wit
h th
e nonidea
l transformer
, we first
nee
d o
t
findth
e equivalen
t transforme
r parameter
s indicate
d n
i Figur
e 10.7
. The
n we ca
n us
e
mesh analysi
s o
t comput
e th
e scale
d outpu
t voltage
. Finally
, we ca
n us
e th
e scalin
g
relatio
n (10.94
) o
t conver
t th
e scale
d resul
to
t th
e actua
l outpu
t voltage
.
From (10.102)—(10.106
) an
d th
e parameter
s liste
d above
, we find
x7(\ = 2.
5 12
,
;
xf = 0 . i n , a n d ( 4=)0.
1 H. Fro
m (10.97
) we find
R' = 0.0
4 I an
d fro
m (10.67
)
w e calculat
e th
e scale
d loa
d impedanc
e o
t be C^RL
4 £1. The equivalen
t circui
ts
i
shown n
i Figur
e 10.9b
.
The generalize
d mesh equation
s are
:
0.0
4 + 2.6 j
-2.5 j
- 25.j
4.0
4 + 2.6.
/
100/
0
1\
V
fro
m whic
h we ca
n findth
e scale
d secondar
y voltag
e
V' = 4 1' =
v,(t
) r
-1000.
/
= - 9 4 . 2 j *a 0-3'28' V.
10.60
8 + 0.3484
/
zL=iooQSv
)
o(t
(a)
ioo
v r
(b)
Figure 10.9: (a
) Transforme
r circui
t an
d (b
) equivalen
t circui
t wit
h scale
d parameters
.
402
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
The actua
l voltag
e s
i V2 = V^/a, whic
h ca
n be represente
d n
i th
e tim
e domai
n a
s
follows
:
v2(t) = 47
1 sin(2500
r - 1.88°
) V.
In th
e conclusio
n of thi
s section
,t
is
i appropriat
e o
t poin
t ou
t tha
t th
e theor
y
develope
d her
e fo
r th
e transforme
rs
i quit
e genera
ln
i natur
e an
ds
i applicabl
eo
t many
electri
c device
s whic
h contai
n strongly couple
d coil
s (windings)
. Fo
r thes
e devices
,
the correspondin
g couple
d circui
t equation
s ar
e singularl
y perturbe
d an
d th
e leakag
e
inductance
s (o
r leakag
e reactances
) pla
y a ver
y importan
t rol
e whic
h suggest
s tha
t
the
y shoul
d be explicitl
y accounte
d for
. This
, fo
r instance
, explain
s why th
e theor
y
of inductio
n motor
s ha
s many feature
s simila
ro
t th
e theor
y of th
e transformer
.
10.
4
Theor
y of Two-Por
t Element
s
A two-por
t elemen
t si a
n elemen
t of a
n electri
c circui
t whic
h s
i associate
d wit
h
two pair
s of terminal
s (se
e Figur
e 10.10)
. Two lef
t terminal
s wil
l be calle
d th
e first
port an
d th
e voltag
e an
d curren
t associate
d wit
h thes
e terminal
s wil
l be marke
d by
subscrip
t "1.
" Similarly
, tw
o righ
t terminal
s wil
l be calle
d th
e secon
d por
t an
d th
e
voltag
e an
d curren
t associate
d wit
h thes
e terminal
s wil
l be marke
d by subscrip
t "2.
"
In circui
t theory
, two-por
t element
s serv
e a twofol
d purpose
. First
, the
y ca
n be
used a
s equivalen
t replacement
s fo
r complicate
d electri
c circuits
.n
I thi
s way
, th
e us
e
of th
e two-por
t element
s ca
n simplif
y (an
d facilitate
) th
e analysi
s of electri
c circuits
.
Secon
d (an
d mor
e important)
, th
e two-por
t element
s ca
n be employe
d a
s circui
t
models fo
r complicate
d device
s suc
h a
s transistors
, transformers
, an
d transmissio
n
lines
. The interio
r dynamic
s of thes
e device
s s
i usuall
y to
o comple
x o
t be full
y
describe
d by electri
c circui
t theory
. However
,n
i th
e analysi
s of electri
c circuit
s wit
h
thes
e devices
, quit
e ofte
n on
e doe
s no
t nee
d o
t kno
w al
l th
e detail
s of thei
r behavior
;
the mathematica
l relationship
s betwee
n devic
e termina
l voltage
s an
d current
s ar
e
quit
e sufficient
. Thi
s s
i exactl
y th
e purpos
e tha
t th
e theor
y of two-por
t element
s
fulfills
. The two-por
t theor
y provide
s termina
l voltage-curren
t characterization
s of
the devices
. Thes
e termina
l voltage-curren
t relation
s ca
n the
n be couple
d wit
h KVL
A
'1
+
O
A
" o
Figure 10.10: The two-por
t element
.
20.4
. Theory of Two-Port Elements
403
and KCL equation
s an
d use
d fo
r th
e analysi
s of electri
c circuit
s wit
h complicate
d
devices
.
In thi
s section
, we shal
l conside
r linea
r an
d passiv
e two-por
t elements
. The
termina
l voltage-curren
t relation
s fo
r thes
e two-por
t element
s ar
e describe
d by linear
and homogeneous equations
. Thes
e equation
s ca
n be writte
n n
i many differen
t form
s
which resul
tn
i differen
t representation
s of two-por
t elements
.
W e begi
n wit
h th
e ^-representatio
n of two-por
t elements
.n
I thi
s representation
,
the termina
l voltage
s ar
e expresse
d a
s linea
r an
d homogeneou
s function
s of termina
l
currents
:
Vi =ZiJi+Zi2h,
(10.114
)
Vl = Z2l/
l + Z22/2
-
(10.115
)
The linearit
y of equation
s (10.114
) an
d (10.115
)s
i a consequenc
e of linearit
y of th
e
two-por
t element
. Indeed
, th
e two-por
t elemen
t ca
n be viewe
d a
s bein
g excite
d by
two curren
t source
s Is\ = I\ an
d Is2 = h connecte
d o
t th
e first
an
d secon
d ports
,
respectively
. Then
, accordin
g o
t th
e superpositio
n principle
, th
e termina
l voltage
s
shoul
d be linea
r combination
s of thes
e curren
t sources
. Thi
s s
i exactl
y what s
i ex
presse
d by equation
s (10.114
) an
d (10.115)
. The homogeneit
y of thes
e equation
ss
i
a consequenc
e of th
e passivit
y of th
e two-por
t element
.n
I othe
r words
, th
e homogeneit
y of th
e abov
e equation
s come
s fro
m th
e fac
t tha
t th
e two-por
t elemen
t doe
s
not contai
n an
y independen
t sources
;a
s a result
, termina
l voltage
s V\ an
d V2 shoul
d
e se
to
t zero
.
be equa
lo
t zer
o when termina
l sourc
e current
s I\ an
d I2 ar
Equation
s (10.114
) an
d (10.115
) ca
n be writte
n n
i th
e matri
x form
:
V = ZI,
(10.116
)
where vector
s V, I an
d matri
x Z ar
e define
d a
s follows
:
V = (I1
V
I=(ll),
Z=(
Zn
Zu
).
(10.117
)
Next, we sho
w tha
t th
e z-parameter
sn
i th
e abov
e equatio
n ca
n be determine
d by
usin
g th
e tw
o open-circui
t test
s show
n ni Figur
e 10.11
.n
I tes
t (a)
, th
e curren
t sourc
e
+
I
CD
v
I
o
A
i
(a)
+
+
AW
A ^
1
V
(a)
\
1
(b
)
Figur
e 10.11: Two open-circui
t test
s fo
r identificatio
n o
f th
e ^-representation
.
404
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
l[a) s
i connecte
d o
t th
e terminal
s o
f th
e first
por
t whil
e th
e terminal
s o
f th
e secon
d
port ar
e kep
t open
. The latte
r mean
s that
:
I(2a) = 0.
d V^
In thi
s test
, voltage
s V\a) an
and (10.118)
, we find:
(10.118
)
ar
e measured
. Fro
m equation
s (10.114)
, (10.115)
,
V\a) = znt[a\
V(2l)=z{2i[a)-
(10.119
)
Consequently
,
yUi)
zn
= yfe,
y(a)
Z2
. = jU-
dO.120
)
The las
t tw
o expression
s ar
e quit
e ofte
n writte
n n
i th
e forms
:
zn = y-k=o
'
Z2
i = ^ %= 0 ,
(10.121
)
which explicitl
y reflec
t th
e conditio
n o
f th
e tes
t (a)
.
In th
e cas
e o
f tes
t (b)
, th
e curren
t sourc
e I^] s
i connecte
d acros
s th
e secon
d port
,
whil
e th
e terminal
s o
f th
e first
por
t ar
e kep
t open
. The latte
r mean
s tha
t
l\b) = 0
.
±11 Liii.5 ic;at, vuiiagt;^ v ,
aiiu v^
<nc m c *
(10.122
)
L U l l l CVJUC
and (10.122)
, we find:
V\b) = zjf, Of == Z22W'.
(10.123
)
From expression
s (10.123)
, we derive
:
y(b)
zn = j ^ 9
l
£22 = j(b)>
l
2
(10.124
)
2
which ca
n als
o be writte
n as
:
zn -
^•i
y-|/,=o
.
^22 =
(10.125
)
The test
s (a
) an
d (b
) ca
n be use
d o
t find
ou
tf
i th
e two-por
t elemen
ts
i reciprocal
.n
I
the cas
e o
f th
e reciproca
l two-por
t element
, th
e equalit
y o
f excitation
s
l\a) =/<*
>
(10.126
)
result
sn
i th
e followin
g equalit
y o
f responses
:
V\b) = %a)
(10.127
)
In othe
r words
, reciprocit
y mean
s tha
t measuremen
t result
s ar
e invarian
t wit
h respec
t
to th
e interchang
e o
f excitatio
n an
d respons
e terminals
.n
I th
e cas
e o
f th
e reciproca
l
10.4. Theory of Two-Port Elements
405
two-por
t element
, fro
m expression
s (10.120)
, (10.124)
, (10.126)
, an
d (10.127
) we
derive
:
Z\2
(10.128
)
— ?>2\-
This mean
s tha
t fo
r reciproca
l two-por
t element
s Z-matrice
s ar
e symmetric
.
It ca
n be show
n that
, when two-por
t element
s ar
e use
d a
s equivalen
t replacement
s
for passiv
e electri
c circuit
s withou
t dependen
t sources
, the
n thes
e two-por
t element
s
are alway
s reciprocal
. The forma
l proo
f s
i base
d on th
e symmetr
y of matrice
s fo
r
mesh curren
t equation
s an
d Cramer'
s rule
. Thi
s proo
fs
i omitted
.
When two-por
t element
s ar
e use
d a
s circui
t model
s fo
r devices
, the
y ar
e no
t
necessaril
y reciprocal
. Fo
r instance
, th
e two-por
t element
s use
d a
s th
e model
s fo
r
transformer
s ar
e reciproca
l (se
e th
e previou
s section)
, whil
e th
e two-por
t element
s
used a
s th
e model
s fo
r transistor
s ar
e no
t reciprocal
.
The test
s (a
) an
d (b
) describe
d abov
e ca
n be use
d fo
r experimenta
l determinatio
n
of z-parameters
.n
I thi
s sense
, thes
e tw
o test
s constitut
e th
e experimenta
l procedur
e fo
r
the identificatio
n of th
e two-por
t element
sa
s circui
t model
s fo
r rea
l devices
. The abov
e
two test
s ca
n be als
o employe
d fo
r computationa
l determinatio
n of z-parameter
s of th
e
two-por
t element
s when thes
e two-por
t element
s ar
e use
d a
s equivalen
t replacement
s
for electri
c circuits
. We illustrat
e thi
s statemen
t by th
e followin
g example
.
EXAMPLE 10.2 Conside
r th
e two-por
t circui
t show
n n
i Figur
e 10.12
. We want o
t
findz-parameter
s of th
e equivalen
t two-por
t element
.
It s
i clea
r fro
m Figur
e 10.1
1 an
d formula
s (10.120
) an
d (10.124
) tha
t zn an
d Z22
can be construe
d a
s th
e inpu
t impedance
s of circuit
s show
n n
i Figure
s 10.13
a an
d b,
respectively
.n
I th
e cas
e of th
e circui
t show
n n
i Figur
e 10.13a
, we ca
n clearl
y se
e tha
t
impedance
s Z4 an
d Z5 ar
e connecte
d ni serie
s an
d togethe
r the
y connecte
d ni paralle
l
wit
h Z3. Thi
s grou
p of thre
e impedance
ss
i connecte
d ni serie
s wit
h Z2 an
d th
e grou
p
of th
e abov
e fou
r impedance
ss
i connecte
d n
i paralle
l wit
h Z5. Thus
, zn a
s th
e inpu
t
impedanc
e of th
e circui
t show
n n
i Figur
e 10.13
as
i give
n by
:
Z (7
Z\\
=
-4- ^3(^4+^5) \
Zi +Z>
,
Z,(Z4+Z5)
^ z3+z4+z5
Figure 10.12: A two-por
t electri
c circuit
.
(10.129
)
406
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
(a)
(b)
Figure 10.13: Two circuit
s use
d fo
r th
e determinatio
n o
f z-parameters
.
By usin
g Figur
e 10.13
b an
d th
e sam
e lin
e of reasoning
, we find:
7A7.
+ Z3(Z,+Z
2K
^ 5 1 ^4 T
Z3+Zi+Z2)
Z22
7
_L 7, _i
_ Z3(Zi +Z2) '
5 ^ ^4 ^ Z3+Z,+Z2
Ar
(10.130
)
To find
zi\ we shal
l us
e th
e secon
d formul
a give
n n
i (10.119)
.t
Is
i clea
r tha
tn
i ou
r
7(0) s
case
, voltag
e V^ * equa
lo
t th
e voltage
, V5, acros
s impedanc
e Z5:
%a) = V5.
(10.131
)
By consecutivel
y usin
g th
e curren
t divide
r rul
e fo
r th
e circui
t show
nn
i Figur
e 10.13a
,
w e find
:
h = l\a)
Z
1
7
z
i
/5 = /
2~
_|
_7
+ z
'2 f Z + zr
(10.132
)
_
L Z3(Z
4+Z5)
2 -+
- ZrT^+z
7
T T — ^ --
(10.133
)
From formul
a (10.131)
, we conclud
e that
:
%a)=i5Z5.
(10.134
)
By combinin
g expression
s (10.119)
, (10.132)
, (10.133)
, an
d (10.134)
, we obtain
:
Z21 =
Z1Z3Z5
(Z3 + Z4 + ^)(Z
! + Z2 + m ± § :)
(10.135
)
z3+z4+z5-
i identica
lo
t tha
t forz
en
i ou
r proble
m th
e equivalen
t
The expressio
n fo
rz
12 s
2i becaus
two-por
t elemen
ts
i reciprocal
. However
,t
is
i suggeste
d tha
t th
e reade
r verif
y th
e
reciprocit
y by th
e direc
t analysi
s of th
e circui
t show
n n
i Figur
e 10.13b
.
10.4. Theory of Two-Port Elements
407
Thus, we hav
e see
n tha
t two-por
t element
s ca
n be use
d a
s equivalen
t replacement
s
for electri
c circuits
.t
Is
i interestin
g o
t conside
r th
e invers
e problem
. Give
n a two
port elemen
t define
d by equation
s (10.114)( 10.115)
, we want o
t fin
d a simpl
e circui
t
realizatio
n fo
r thi
s two-por
t element
.n
I othe
r words
, we want o
t fin
d asimpl
e two-por
t
electri
c circui
t wit
h termina
l voltage-curren
t relation
s identica
lo
t thos
e describe
d by
formula
s (10.114
) an
d (10.115)
.
l conside
r th
e cas
e o
f reciproca
l two-por
t elements
.n
I thi
s case
,
First
, we shal
it s
i natura
lo
t loo
k fo
r circui
t realization
s withou
t dependen
t sources
. To thi
s end
,
conside
r th
e circui
t show
nn
i Figur
e 10.14
. By writin
g KVL equation
s fo
r loop
s Ian
d
II
, we find:
Vx =ZJX + Zc(h + hi
(10.136
)
V2 = ZbI2 +Zc(h +hi
(10.137
)
By combinin
g th
e simila
r term
sn
i th
e abov
e equations,
we obtain
:
V{ = (Za + Zc)h + ZCI2,
(10.138
)
V2 =ZCIX +(Zb + Zc)h-
(10.139
)
e las
t tw
o equation
s wit
h equation
s (10.114
) an
d (10.115)
, we
By comparin
g th
conclud
e tha
t thes
e pair
s o
f equation
s wil
l be identica
l unde
r th
e conditions
:
Za+Zc
= Z\\,
Zb + Zc = Z
22>
Zc = Zn = Z21
-
(10.140
)
By solvin
g equation
s (10.140
) fo
rZa, Zb, an
d Zc, we find:
%a = Zn - Z\2,
Zh =Z22 ~ Z\2>
%c = Zl2
-
(10.141
)
Thus, fo
r impedance
s define
d by expressio
n (10.141)
, th
e electri
c circui
t show
n n
i
Figur
e 10.1
4 wil
l hav
e termina
l voltage-curren
t relation
s identica
lo
t thos
e o
f th
e
two-por
t elemen
t describe
d by equation
s (10.114)—(10.115)
.n
I thi
s sense
, we hav
e
arrive
d a
t th
e circui
t realizatio
n o
f th
e two-por
t element
.t
Is
i importan
to
t kee
p n
i
mind tha
t impedance
s Za,Zb, an
d Zc may no
t alway
s be physicall
y realizable
. Thi
s
wil
l be th
e cas
e when rea
l part
s o
f thes
e impedance
s ar
e negative
. To circumven
t thi
s
difficult
y a
s wel
la
so
t includ
e nonreciproca
l two-por
t elements
, circui
t realization
s
wit
h dependen
t source
s ca
n be used
. One o
f thes
e realization
s whic
h contain
s tw
o
0-^-
3hI «J2 O
„
Figure 10.14: A circui
t realizatio
n o
f th
e z-representation
.
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
-^0
0-VA
/+
z 1 2 i2
- / Z 2 1 I,
Figure 10,15: A circui
t realizatio
n o
f th
e ^-representatio
n wit
h tw
o dependen
t sources
.
dependen
t voltag
e source
ss
i show
n n
i Figur
e 10.15
.t
Is
i apparen
t tha
t th
e tw
o KVL
equation
s fo
r thi
s circui
t coincid
e wit
h equation
s (10.114)—(10.115)
, whic
h mean
s
t realizatio
n fo
r th
e two-por
t element
. Anothe
r
tha
t Figur
e 10.1
5 provide
s acircui
circui
t realizatio
n fo
r th
e two-por
t element
ss
i give
n n
i th
e followin
g example
.
EXAMPLE 10.3 Conside
r th
e circui
t show
n ni Figur
e 10.16
. We want o
t sho
w tha
t
it provide
s a circui
t realizatio
n fo
r th
e two-por
t element
.
A
I
O - *-
Z Z
-
^11 ^ 1 2
( Z 21" Z 12) 'l
— O
Z -Z
'
cr ^ o
Figure 10,16: A circui
t realizatio
n o
f th
e z-representatio
n wit
h on
e dependen
t source
.
First
, we writ
e tw
o KVL equations
:
Vi = (z
n - zi
i +ZuOi + hi
2)/
% = (z22 - zn)h + (Z2
i -znih
(10.142
)
+zi2(h + hi
(10.143
)
Be performin
g cancellatio
n of th
e identica
l term
sn
i equation
s (10.142
) an
d (10.143)
,
w e en
d up with
:
Vi =zn/
i +znh,
(10.144
)
% =Z2l/
l +Z22/2
-
(10.145
)
The las
t tw
o equation
s ar
e th
e sam
e a
s equation
s (10.114
) an
d (10.115)
.
The z-representatio
n of two-por
t element
ss
i ver
y convenien
t fo
r th
e analysi
s of
serie
s connectio
n o
f two-por
t elements
. Thi
s connectio
n s
i show
n n
i Figur
e 10.17
.
10.4. Theory of Two-Port Elements
409
A/
A'
J l.
!
2
'4
i
+
+
/
v
i
A '
V.J
A
VJ
+
A
A
V,
"
/
V.
A>
A"
A "
V\
) *
]
2 J
//
/
^
+
™
T
A "
2
ou
-c
\J
I
Figur
e 10.17
: Serie
s connectio
n o
f two-por
t elements
.
Our inten
ts
io
t find
th
e z-representatio
n fo
r th
e resultin
g two-por
t element
. We star
t
our derivatio
n by writin
g ^-representation
s fo
r th
e "primed
" an
d "double-primed
"
two-por
t elements
:
n
z'n
4
v?
vZ
z7"
711
u
7"
Z
n
Z22
(10.146
)
(10.147
)
^22
21
Due o
t th
e serie
s connectio
n of th
e abov
e two-por
t elements
, we have
:
V\ = V[ + V"
V2 = % + V",
(10.148
)
h
(10.149
)
V = J"
i±
1"
By addin
g equation
s (10.146
) an
d (10.147
) an
d the
n by takin
g int
o accoun
t expres
sion
s (10.148
) an
d (10.149)
, we derive
:
V2
z
l\
^ zll
z
Z
+ Z
Z
2\
22
12 ^ z 12
22 "^ z 22
(10.150)
which s
i th
e z-representatio
n of th
e resultin
g two-por
t element
. The las
t resul
t ca
n be
writte
n ni concis
e matri
x for
m a
s follows
:
Z = Z' + Z".
(10.151
)
It s
i ver
y convenien
to
t coupl
e th
e z-representatio
n of th
e two-por
t elemen
t wit
h
the mes
h curren
t analysi
s technique
. Indeed
, th
e mes
h current
s ca
n be introduce
d n
i
such a way tha
t branc
h current
s]
/ an
d I2 (se
e Figur
e 10.10
) wil
l coincid
e wit
h mes
h
currents
. Then
, two-por
t elemen
t equation
s (10.114
) an
d (10.115
) ca
n be directl
y
combine
d wit
h mes
h curren
t equation
s fo
r th
e circuit
s connecte
d o
t th
e firs
t an
d
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
410
secon
d ports
.n
I thi
s way
, we en
d up wit
h a complet
e se
t of mes
h curren
t equation
s
for th
e overal
l circuit
.
In th
e cas
e of th
e noda
l analysi
s technique
, th
e y-representation
of th
e two-por
t
element
ss
i preferable
.n
I thi
s representation
, th
e termina
l current
s ar
e expresse
d a
s
linea
r an
d homogeneou
s function
s of termina
l voltages
:
h = y\\V\ +V12V2,
(10.152
)
h
(10.153
)
= ^21^
1 +^22^2
.
These equation
s ca
n be writte
n n
i th
e matri
x for
m a
s follows
:
7 = YV,
(10.154
)
i give
n
where vector
s I an
d V ar
e define
d ni formul
a (10.117)
, whil
e th
e matri
xY s
by:
Y
= ( yn
yn
J21
J2
2
(10.155
)
By comparin
g expression
s (10.116
) an
d (10.154)
, we find
th
e followin
g connectio
n
betwee
n Z an
d Y matrices
:
Y -1
Y =Z -1
(10.156
)
To find
y-parameters
, th
e tw
o short-circui
t test
s show
n n
i Figur
e 10.1
8 ca
n be used
.
In tes
t (a)
, th
e voltag
e sourc
es
i connecte
d o
t th
e terminal
s of th
e first
port
, whil
e th
e
terminal
s of th
e secon
d por
t ar
e short-circuited
. The latte
r mean
s that
:
V2 = 0.
(10.157
)
By usin
g conditio
n (10.157
) ni equation
s (10.152
) an
d (10.153)
, we derive
:
y2\ -
A
A
i
U
1
K
+
Li
0
O
(a)
+
(10.158
)
^lft=o
-
L
>1
i
V,
^
(b)
Figure 10.18: Two short-circui
t test
s use
d fo
r identificatio
n o
f th
e ^-representation
.
20.4
. Theory of Two-Port Elements
411
In tes
t (b)
, th
e voltag
e sourc
es
i connecte
d o
t th
e terminal
s of th
e secon
d port
, whil
e
the terminal
s of th
e first
por
t ar
e short-circuited
, whic
h mean
s that
:
Vx = 0.
(10.159
)
By usin
g conditio
n (10.159
)n
i equation
s (10.152
) an
d (10.153)
, we derive
:
-
7l
l
y22 _ ^ H ^= 0 .
V2
(10.160
)
V2
It s
i clea
r tha
t fo
r reciproca
l two-por
t element
s we have
:
yn
(10.161
)
= J2i
-
Next, we conside
r circui
t realization
s of two-por
t element
s define
d by equation
s
(10.152)—(10.153)
. As before
, we want o
t find simpl
a e two-por
t electri
c circui
t wit
h
termina
l voltage-curren
t relation
s identica
lo
t thos
e describe
d by equation
s (10.152)
(10.153)
. First
, we discus
s th
e cas
e of reciproca
l two-por
t elements
.n
I thi
s case
,
it s
i natura
lo
t loo
k fo
r circui
t realization
s withou
t dependen
t sources
. To thi
s end
,
conside
r th
e circui
t show
nn
i Figur
e 10.19
. By usin
g KCL equation
s fo
r th
e tw
o uppe
r
nodes
, we find:
/, = YaV{ + YC(VX - V2),
(10.162
)
h = YbV2 - YC{VX - V2).
(10.163
)
By rearrangin
g th
e term
sn
i th
e las
t tw
o equations
, we obtain
:
h = (Ya + Ycm - YCV2,
(10.164
)
h = -Y&
(10.165
)
+ (Yb + YC)V2.
By comparin
g th
e las
t tw
o equation
s wit
h equation
s (10.152)—(10.153)
, we conclud
e
tha
t thes
e pair
s of equation
s wil
l be identica
l unde
r th
e followin
g conditions
:
Ya + Yc = yu,
Yb + Yc = y22,
Yc = -yl2.
(10.166
)
By solvin
g equation
s (10.166
) fo
r Ya, Yb an
d Yc, we obtain
:
Ya=y n+ yn,
Yb = y22 +yl2,
Yc = -yl2.
(10.167
)
* i -0
Figur
e 10.19
: A circui
t realizatio
n o
f th
e ^-representation
.
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
412
Thus, fo
r admittance
s define
d by expression
s (10.167)
, th
e electri
c circui
t show
n
in Figur
e 10.1
9 ha
s th
e sam
e termina
l voltage-curren
t relation
s a
s th
e two-por
t ele
ment describe
d by equation
s (10.152)—(10.153)
. The admittance
s give
n by formul
a
(10.167
) may no
t be physicall
y realizabl
e fi thei
r rea
l part
s ar
e negative
. To cir
cumvent thi
s difficult
y an
d o
t exten
d circui
t realization
s o
t nonreciproca
l two-por
t
elements
, electri
c circuit
s wit
h dependen
t source
s ca
n be used
. Thi
ss
i demonstrate
d
in th
e followin
g tw
o examples
.
EXAMPLE 10-4 Conside
r th
e circui
t show
n ni Figur
e 10.2
0 an
d demonstrat
e tha
tt
i
is a circui
t realizatio
n fo
r th
e two-por
t elemen
t describe
d by equation
s (10.152
) an
d
(10.153)
.
-+—+
v, y„
-yi 2 v 2 (T
-y^v,
Figure 10.20: A circui
t realizatio
n o
f th
e y-representatio
n wit
h tw
o dependen
t sources
.
By usin
g KCL equations
, we easil
y deriv
e th
e followin
g expression
s fo
r I\ an
d
h = y\\V\ + ynV2,
(10.168
)
h = yi\Vi + y22V2,
(10.169
)
which ar
e identica
lo
t (10.152)-(10.153)
.
EXAMPLE 10.5 Conside
r th
e circui
t show
n n
i Figur
e 10.21
. Thi
s circui
t contain
s
onl
y on
e dependen
t curren
t source
. We want o
t sho
w tha
tt
i provide
s a circui
t real
izatio
n fo
r th
e two-por
t elemen
t define
d by equation
s (10.152)—(10.153)
.
-Yi2
■^-0
O— ■*-
v
Vn + yi2
(yi2-y 2 i)v/t
y?,+y1
Figure 10.21: A circui
t realizatio
n o
f th
e y-representatio
n wit
h on
e dependen
t source
.
10.4. Theory of Two-Port Elements
413
By usin
g KCL equation
s fo
r tw
o uppe
r node
s of th
e abov
e circuit
, we obtain
:
h = Cvn +ynWx - yxi{V\ - v2),
i
(o.no
)
h = {yii + ynW2 - (yn - y2\)Vi + yX2{Vx - v2).
(10.171
)
By performin
g cancellatio
n of th
e identica
l term
sn
i th
e las
t tw
o equations
, we derive
:
h =yuVi+ynK
(10.172
)
h = yi\Vx + y22V2.
(10.173
)
The equation
s obtaine
d ar
e th
e sam
e a
s equation
s (10.152)—(10.153)
.
W e conclud
e ou
r discussio
n of th
e y-representation
wit
h th
e analysi
s of paralle
l
connectio
n of two-por
t elements
. Thi
s connectio
n s
i show
n n
i Figur
e 10.22
. Our
inten
ts
io
t find
th
e y-representation
fo
r th
e resultin
g two-por
t element
. We begi
n ou
r
reasonin
g by writin
g y -representation
s fo
r th
e prime
d an
d double-prime
d two-por
t
elements
:
=
yii
y'n
y'n
n
(10.174
)
/ y'n
yii
v{'
y'n
y'n j \ v"
(10.175
)
The paralle
l connectio
n of th
e two-por
t element
s result
sn
i th
e followin
g equa
tion
s fo
r termina
l voltage
s an
d termina
l currents
:
v[ = v,
" = vh
(10.176
)
h = U + V'
A'
A'
\
«'2 c
+
A '
/
A
Figur
e 10.22
: Paralle
l connectio
n o
f two-por
t elements
.
(10.177
)
414
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
By addin
g equation
s (10.174
) an
d (10.175
) an
d the
n by takin
g int
o accoun
t expres
sion
s (10.176
) an
d (10.177)
, afte
r simpl
e transformation
s we derive
:
h\
h
=
( y'n+yk
y'n + yb \(
vy'n + y'lx y!2i + y"2
^
\%
(10.178
)
The las
t resul
t ca
n be writte
n n
i concis
e matri
x for
m a
s follows
:
Y = ?' + Y".
(10.179
)
It turn
s ou
t tha
t no
t al
l two-por
t element
s hav
e z- or y-representations.
Fo
r
instance
, th
e idea
l transforme
r discusse
d n
i th
e previou
s sectio
n s
i th
e two-por
t
1
elemen
t wit
h th
e followin
g termina
l relations
:
V2 =aVu
(10.180
)
h = ~-h
(10.181
)
a
which canno
t be frame
d n
i term
s o
fz- or ^-representations
.
r th
e two-por
t element
s
This suggest
s th
e importanc
e of hybri
d representation
s fo
and suc
h representation
s ar
e indee
d usefu
ln
i variou
s applications
. Ther
e ar
e tw
o hy
bri
d representation
s fo
r th
e two-por
t elements
: /z-representatio
n an
d g-representation
.
In th
e cas
e o
f //-representation
, th
e termina
l voltag
e V\ an
d th
e termina
l curren
tI2 ar
e
expresse
d a
s linea
r an
d homogeneou
s function
s o
fI\ an
d VVV{ =hnIi
h
+hl2V2y
(10.182
)
=h2\h +h22V2.
(10.183
)
The abov
e equation
s ca
n als
o be writte
n n
i th
e matri
x form
:
I)=* ( \ ) ■
(,ai84)
where th
e matri
xH s
i define
d a
s follows
:
H=(
11
f
12
f V
(10.185
)
To find
/z-parameters
, th
e tw
o test
s show
n n
i Figur
e 10.2
3 ca
n be used
.n
I tes
t (a)
, th
e
curren
t sourc
e s
i connecte
d o
t th
e terminal
s o
f th
e first
port
, whil
e th
e terminal
s o
f
the secon
d por
t ar
e short-circuited
. The latte
r mean
s tha
t
V2 = 0
.
1
(10.186
)
Pleas
e not
e th
e appearanc
eo
f minu
s sig
n ni equatio
n (10.181
) ni compariso
n wit
h equatio
n
(10.63)
. Thi
s si becaus
e th
e referenc
e directio
n fo
rI2 ni th
e two-por
t elemen
t (se
e Figur
e 10.10
)
is opposit
e ot th
e referenc
e directio
n o
fI2 ni th
e transforme
r (se
e Figur
e 10.5)
.
20.4
. Theory of Two-Port Elements
^
1 °
415
A
A
J2
-,'
* o
+
A
P:
v1
(a)
I
6*
(b)
Figur
e 10.23
: Two test
s use
d fo
r identificatio
n o
f th
e /^-representation
.
By usin
g conditio
n (10.186
)n
i equation
s (10.182
) an
d (10.183)
, we derive
:
&ii =
— V?=0'
h21
'2
'^2=0'
(10.187
)
In tes
t (b)
, th
e voltag
e sourc
es
i connecte
d o
t th
e terminal
s of th
e secon
d port
, whil
e
the terminal
s of th
e first
por
t ar
e kep
t open
. The latte
r mean
s that
:
h = 0.
(10.188
)
By usin
g th
e las
t conditio
n ni equation
s (10.182
) an
d (10.183)
, we derive
:
h\r = — \r/i=0>
V,
h22 =
V7 >/,=o-
(10.189
)
It s
i instructiv
eo
t not
e tha
t th
e /^-representatio
n s
i extensivel
y use
d fo
r th
e character
izatio
n of bipola
r junctio
n transistor
s an
d th
e followin
g terminolog
y s
i adopte
d fo
r
/z-parameters
:
^n = hi — shor
t circui
t inpu
t impedance
,
h\2 = hr — ope
n circui
t revers
e voltag
e gain
,
h2\ ~ hf — shor
t circui
t forwar
d curren
t gain
,
^22 = h0 — ope
n circui
t outpu
t admittance
.
The abov
e terminolog
y s
i consisten
t wit
h expression
s (10.187
) an
d (10.189)
.
In th
e cas
e of ^-representation
, th
e termina
l curren
tI\ an
d termina
l voltag
e V2
are expresse
d a
s linea
r an
d homogeneou
s function
s of V\ an
d I2:
h = gii^
i + gnh,
(10.190
)
V2 = g2\V{ + g2ih-
(10.191
)
The abov
e equation
s ca
n als
o be writte
n n
i th
e matri
x form
:
h
v2
h
(10.192
)
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
416
where th
e matri
xG s
i define
d a
s follows
:
8\i
#21
G =
812
822
(10.193
)
By comparin
g equation
s (10.184
) an
d (10.192)
, we conclud
e tha
t matrice
s H an
dG
are relate
d o
t on
e anothe
r by th
e expressions
:
H = G~\
G=H~K
(10.194
)
The identificatio
n of g-parameter
s ca
n be accomplishe
d by usin
g th
e tw
o test
s show
n
in Figur
e 10.24
.t
Is
i clea
r tha
t ni tes
t (a)
:
h = 0,
(10.195
)
which
, accordin
g o
t equation
s (10.190
) an
d (10.191)
, lead
so
t th
e followin
g formula
s
forg
n an
d g2i
:
= lL|
'1
1
=Yl\
- l/
2=o>
#2
1
(10.196
)
- l/
2=o-
In th
e cas
e of tes
t (b)
, we have
:
(10.197
)
V\ = 0,
which
, accordin
g o
t equation
s (10.190
) an
d (10.191)
, lead
so
t th
e followin
g expres
sion
s fo
r gn an
d g22:
-7 ll
ri
2 - j - |1 =y0'
§22 -
-y 2 ,
"5-1^=0
-
(10.198
)
It s
i interestin
g o
t establis
h connection
s betwee
n th
e y-representatio
n an
d grepresentation
. To thi
s end
, we solv
e equatio
n (10.153
) fo
r V2:
—y?i -
V2 = -^-V
722 x
1 -
(10.199
)
+ J22—I2.
By substitutin
g expressio
n (10.199
) int
o equatio
n (10.152)
, afte
r simpl
e transforma
tion
s we find
:
^22
1
A
1
'l
+
91
A
V
(10.200
)
J2
2
J
O
>
O
o
(a
)
H
6
1
'
(b)
Figur
e 10.24
: Two test
s use
d fo
r th
e identificatio
n o
f g-representation
.
20.4
. Theory of Two-Port Elements
417
By comparin
g equation
s (10.190
) an
d (10.191
) wit
h equation
s (10.200
) an
d (10.199)
,
respectively
, we derive
:
dety
J22
'12
_ Jl
2
J2
J2
2
.
-J2
1
#21
J2
2
1
822 =
J2
2
(10.201
)
From th
e abov
e formula
s an
d symmetr
y conditio
n (10.161)
, we fin
d tha
t fo
r th
e
reciproca
l two-por
t element
s th
e followin
g equalit
y holds
:
821
(10.202
)
■gi2-
From th
e las
t expressio
n an
d formul
a (10.194)
, itca
n be derive
d tha
t reciprocit
y
impose
s th
e followin
g constrain
t on ^-parameters
:
hi
(10.203
)
= ~h\2-
Next, we conside
r some example
s relate
d o
t h- an
d ^-representations
.
EXAMPLE 10.6 We conside
r ho
w h- an
d g-representation
s ca
n be applie
d o
t th
e
idea
l transformer
.
By comparin
g equation
s (10.180)-(10.181
) wit
h equation
s (10.182)-(10.183)
,
w e fin
d tha
t matri
x H ha
s th
e form
:
H =
0
a
I
-i
o
0
a
-a
0
(10.204
)
Similarly
, by comparin
g equation
s (10.180)—(10.181
) wit
h equation
s (10.190)
(10.191)
, we conclud
e tha
t matri
x G ca
n be writte
n a
s follows
:
G =
(10.205
)
By usin
g expression
s (10.204
) an
d (10.205)
, we easil
y verif
y that
:
HG = GH =
1
0
0
1
(10.206
)
which s
i consisten
t wit
h expression
s (10.194)
.
EXAMPLE 10.7 Conside
r th
e circui
t show
n n
i Figur
e 10.2
5 an
d demonstrat
e tha
tt
i
give
s a circui
t realizatio
n fo
r /z-representation
.
O-V
+
A
- ^O
+
A /+
h1 2v2V
h
2 l l'
h2
V
,
Figur
e 10.25
: A circui
t realizatio
n o
f th
e /^-representation
.
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
418
By applyin
g KVL o
t th
e lef
t par
to
f th
e circui
t an
d KCL o
t th
e righ
t par
to
f th
e
same circuit
, we find:
Vx = huh +hnV2,
(10.207
)
h
(10.208
)
=h2Ji + h22V2.
The las
t "terminal
" equation
s ar
e identica
lo
t thos
e whic
h defin
e th
e /i-representatio
n
of th
e two-por
t element
.
■
EXAMPLE 10,8 Conside
r th
e circui
t show
n n
i Figur
e 10.2
6 an
d demonstrat
e tha
tt
i
give
s acircui
t realizatio
n fo
r ^-representation
.
- ^O
+
A
- < U? ';
92i V,
Figure 10.26: A circui
t realizatio
n o
f th
e ^-representation
.
r th
e lef
t par
to
f th
e circui
t an
d KVL fo
r th
e righ
t par
t o
f th
e
By usin
g KCL fo
same circuit
, we derive
:
h = gii$i + 8nk
(10.209
)
V2 = g2iV{ + g22V2.
(10.210
)
t thos
e whic
h defin
e th
e ^-representatio
n
The las
t "terminal
" equation
s ar
e identica
lo
of th
e two-por
t element
.
W e Sliclll
shal
lUUIIUIUUC
conclud
e Uthi
s Ssectio
th
e U1SUUSS1U11
discussio
n Uof
fl L11C
th
e
llS
eCUUn
i l Wwit
l l lh
l lllC
l / lA5C£)-representatio
Z 3 L ^ - l C p i C S C i l l < U l Un
l l Uo
; elements
. Thi
s representatio
n s
i als
o calle
d th
e transmissio
n representatio
n
ana t
is
i ver
y convenien
t fo
r th
e analysi
s o
f cascadin
g o
f two-por
t elements
.n
I th
e
cas
e o
f ASCD-representation
, V\ an
d I\ ar
e expresse
d a
s linea
r an
d homogeneou
s
fV2 an
d I2\
function
s o
V{ = AV2 -BI2,
(10.2H
)
/, = CV2 - DI2.
(10.212
)
The las
t equation
s ca
n be writte
n n
i th
e matri
x for
m
(10.213
)
where th
e matri
xt s
i define
d a
s follow
s
A B
T =
C D
(l 0.214
)
20.4
. Theory of Two-Port Elements
419
Figure 10.27: Test
s use
d fo
r identificatio
n o
f th
e A5CD-representation
.
The identificatio
n of th
e parameter
s A, B, C, D ca
n be performe
d by usin
g th
e fou
r
test
s show
n n
i Figur
e 10.27
. By employin
g th
e same lin
e of reasonin
g a
s before
,t
is
i
easy o
t sho
w tha
t th
e followin
g expression
s ar
e valid
.
A
-
!
/2 =0'
B =
Vy
{
v2h
v2=o'
D
= -rk=o-
(10-215)
EXAMPLE 10.9 Fin
d th
e transmissio
n representatio
n fo
r th
e idea
l transformer
.
B y comparin
g equation
s (10.180)-(10.181
) wit
h equation
s (10.211)-(10.212)
,
w e conclud
e tha
t matri
x t ha
s th
e form
:
T =
i
0
a
(10.216
)
0 a
It s
i eas
y o
t se
e fro
m expressio
n (10.216
) tha
t
det
f = 1
.
(10.217
)
It ca
n be show
n tha
t formul
a (10.217
)s
i generall
y tru
e fo
r an
y reciproca
l two-por
t
element
.
Next
, conside
r cascadin
g o
f two-por
t element
s show
n n
i Figur
e 10.28
. Our goa
l
is o
t find
th
e f -matri
x fo
r th
e resultin
g two-por
t element
. To thi
s end
, we shal
l first
remar
k tha
t th
e followin
g termina
l relation
s ar
e vali
d du
e o
t th
e ver
y natur
e o
f th
e
cascadin
g connection
:
-u,
vi1 = n.
(10.218
)
Now , we shal
l writ
e th
e transmissio
n representation
s fo
r th
e prime
d an
d double
-
420
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
Figure 10.28: Cascadin
g o
f two-por
t elements
.
prime
d two-por
t element
s and
, ni doin
g so
, we shal
l tak
e int
o accoun
t formula
s
(10.218):
V,
A'
c
B1
D'
vi
A"
C"
B"
D"
A1
B'
C D'
v2
(10.219
)
(10.220
)
By substitutin
g (10.220
) int
o (10.219)
, we derive
:
Vx \
/, )
A1' B"
C" D"
( A1 B' \(
\C> B' ){
Vi
(10.221
)
The las
t resul
t ca
n be expresse
d succinctl
y ni th
e matri
x for
m a
s follows
:
nr
__
nrlnhll
(10.222
)
Finally
, we shal
l remar
k tha
t th
e sam
e formalis
m of AfiCD-matrice
s s
i extensivel
y
used (albei
t wit
h a differen
t physica
l connotation
) ni lase
r course
s fo
r ra
y tracin
g an
d
transformatio
n of Gaussia
n beams
.
10.
5
MicroSi
m PSpic
e Simulation
s
Conside
r th
e transforme
r circui
t whic
h was analyze
d n
i Exampl
e 10.
1 an
d s
i picture
d
as circui
t (a
) n
i Figur
e 10.2
9 (a
s draw
n by th
e schematic
s editor)
. The voltag
e
sourc
e s
i th
e "VSIN
" typ
e wit
h a
n amplitud
e of "VAMPL=100,
" a frequenc
y of
Circuit (a
Via
Figure 10.29: Circuit
s fro
m Exampl
e 10.1
.
20.5
. MicroSim PSpice Simulations
421
"FREQ=397.887,
" a phas
e of "PHASE=0,
" an
d a
n offse
t voltag
e of "VOFF"=0
.
Two ground
s ar
e require
d becaus
e th
e tw
o coil
s of th
e transforme
r ar
e electricall
y
isolated
. The par
t name of th
e transforme
r s
i "XFRMJLINEAR" an
d t
i reside
s n
i
the "analog
" library
. Double-clickin
g on th
e transforme
r bring
s up th
e dialo
g bo
x
where th
e primar
y inductance
, secondar
y inductance
, an
d couplin
g coefficien
t ca
n
be defined
. Fro
m th
e example
, we hav
e "Ll_value=lm,
" "L2_value=25m,
" an
d
"COUPLING=0.9583.
" The latte
r valu
e come
s fro
m (10.91)
, whic
h ca
n be rewritten
:
M
k = - ==
^JUL2
/
=
\
Lf
Lf
L f L f 72
V
l - -1 - -1 - - Jl.
U
L2
LXL2)
(10.223
)
The windin
g resistance
s ar
e no
t include
d n
i th
e PSpic
e model
,s
o the
y ar
e explicitl
y
include
d a
s R\a an
d R2a.
The scale
d equivalen
t circui
t si show
n n
i circui
t (b
) of Figur
e 10.29
. Not
e tha
t
xll, Rib, an
d RLb hav
e bee
n multiplie
d by a2 = 0.04
, bu
t al
l othe
r quantitie
s ar
e
unchanged
. The printe
r symbol
s come fro
m th
e "VPRINT1
" devic
en
i th
e "special
"
library
. Attachin
g the
m o
t th
e node
s indicate
d wil
l caus
e PSpic
e o
t prin
t th
e corre
spondin
g nod
e voltage
s n
i th
e outpu
t file.
The defaul
t printin
g occur
s fo
r transien
t
analyses
, bu
t double-clickin
g on th
e printe
r symbo
l bring
s up a dialo
g bo
x tha
t ca
n
be use
d o
t selec
t othe
r type
s of analyses
.
First
, we chos
e a transien
t analysi
s o
t compar
e th
e tim
e dependence
s o
t th
e
analyti
c result
.n
I th
e "Transient
" dialo
g bo
x selec
t a "Prin
t Step
" of 20 us an
d a
"Fina
l Time
" of 4 ms. The voltag
e acros
s th
e loa
d resisto
rs
i give
n n
i Figur
e 10.30
.
The soli
d lin
e give
s th
e outpu
t voltag
e of circui
t (a)
. The dashe
d lin
e indicate
s th
e
actua
l voltag
e acros
s th
e loa
d resisto
r a
s compute
d fo
r circui
t (b)
. Thi
s voltag
e s
i
fivetime
s ( = \/a) th
e voltag
e acros
s th
e scale
d 4 O resistor
. The dotte
d lin
e give
s th
e
500
250 h
O)
Of
CO
>
-250 \
-500 '
0
'
1
'
2
Time (ms)
'
3
Figure 10.30: Tim
e dependenc
e o
f th
e outpu
t voltages
.
'
4
422
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
10
100
Frequency (Hz)
1000
10000
Figure 10.31: Frequenc
y dependenc
e o
f circui
t voltages
.
analyti
c resul
ta
s compute
d n
i Exampl
e 10.1
. Al
l thre
e result
s ar
e virtuall
y identical
,
as expected
.
It s
i instructiv
eo
t analyz
e th
e outpu
t voltag
e a
s a functio
n of frequency
. Thi
s ca
n
be easil
y don
en
i PSpic
e by makin
g tw
o change
so
t th
e previou
s example
. First
, edi
t
"n
I th
e "Analysi
s Setup
" dialo
g
bot
h voltag
e source
s an
d ad
d th
ea
c voltag
e "AC = 100.
box, switc
h o
t a
n a
c sweep
. Se
t th
e "AC Swee
p Type
"o
t "Decade,
" th
e "Pts/Decade
"
to 51, th
e "Star
t Freq.
"o
t 1
, an
d th
e "En
d Freq.
"o
t 10k
. The result
s of th
e simulatio
n
are show
n n
i Figur
e 10.31
. The scale
d voltag
e acros
s th
e loa
d resisto
rs
i give
n by th
e
t fall
s of
f a
t lowe
r an
d
soli
d line
. The respons
e s
i quit
e fla
t fro
m 10
0 Hz o
t 1 kHz bu
highe
r frequencies
. At lo
w frequencies
, th
e primar
y windin
g draw
s to
o much curren
t
and a larg
e voltag
e dro
p occur
s du
e o
t th
e primar
y windin
g resistance
. The voltag
e
acros
sRib s
i plotte
d wit
h a dashe
d lin
e ni th
e figure.
At hig
h frequencies
, th
e dro
ps
i
cause
d by th
e leakag
e inductance
. The scale
d voltag
e acros
s th
e secondar
y leakag
e
inductanc
e s
i give
n by th
e dotte
d line
.
10.
6
Summar
y
The mai
n result
s of thi
s chapte
r ca
n be briefl
y summarize
d a
s follows
:
The magneti
c couplin
g of tw
o coil
s ca
n be characterize
d by th
e mutua
l induc
tanc
e M. By definition
,M s
i equa
lo
t th
e rati
o of th
e flux
linkage
s of on
e coi
l
to th
e curren
t flowing
throug
h th
e othe
r coi
l (assumin
g zer
o curren
tn
i th
e first
coil)
:
M= ifci(0Ai(
0 = tyn(i)/i2(t).
(10.224
)
423
10.6. Summary
• The mutua
l inductanc
e s
i alway
s les
s tha
n th
e geometri
c mean o
f th
e self
inductance
s o
f th
e tw
o coils
:
M < y/LxQ.
(10.225
)
g coefficien
tk s
i define
d a
s
• The couplin
k =M/y/Zih
(10.226
)
and t
i depend
s o
n th
e relativ
e location
s o
f th
e tw
o coils
, thei
r geometri
c dimen
sions
, an
d th
e propertie
s o
f th
e medi
an
i th
e vicinit
y o
f th
e tw
o coils
.
• The couple
d circui
t equation
s fo
r tw
o magneticall
y couple
d coil
s ca
n be writte
n
as follows
:
V{ = Rji + jxnh - jxnh,
(10.227
)
V2 = R2I2 + jx22I2 - jx{2Ih
(10.228
)
• An idea
l transforme
r ha
s acouplin
g coefficien
t of unity
, zer
o resistance
s o
f th
e
primar
y an
d secondar
y windings
, an
d no edd
y current
sn
i th
e iro
n core
.
• Fo
ra
n idea
l transformer
, th
e rati
o o
f secondar
y voltag
e (current
)o
t th
e primar
y
voltag
e (current
)s
i directl
y (inversely
) proportiona
lo
t th
e turn
s ratio
. Thus
,
s a
n effectiv
e valu
e o
f a2ZL when viewe
d fro
m th
e
a loa
d impedanc
e ZL ha
primar
y terminals
. Fo
r thi
s reason
, transformer
s ar
e frequentl
y use
d o
t matc
h
t sourc
e impedances
.
loa
d impedance
so
• Leakag
e inductance
s resul
t fro
m imperfectl
y couple
d (k < 1
) transforme
r
d the
y diminis
h th
e transformer'
s abilit
y o
t maintai
n a constan
t
winding
s an
voltag
e acros
s th
e secondar
y terminal
sn
i th
e fac
e o
f varyin
g loads
.
• Leakag
e reactance
s ar
e ver
y importan
t an
d shoul
d be explicitl
y accounte
d fo
r
in transforme
r models
.
l transforme
r ca
n be modele
d a
s aT -network
.
• A nonidea
• Two-por
t element
s ar
e ofte
n use
d a
s model
s fo
r complicate
d device
s an
d als
o
as equivalen
t replacement
s fo
r comple
x circuits
.
" an
d "output
" variable
s fo
r th
e variou
s two-por
t representation
s ar
e
• The "input
give
n n
i Tabl
e 10.1
.
• Simpl
e open
- an
d short-circui
t test
s ca
n be performe
d o
t determin
e th
e matri
x
element
s o
f th
e variou
s representations
.
• f
I a two-por
t elemen
t contain
s onl
y passiv
e components
, the
n th
e Z an
dY
matrice
s ar
e symmetric
.
• z-parameter
s ar
e usefu
l fo
r modelin
g serie
s connection
s o
f two-por
t elements
.
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
424
Table 10.1: Summary of two-por
t representations
.
Representation
Matrix
Impedanc
e
Admittanc
e
Hybri
d
Hybri
d
Transmissio
n
Z
Y
H
G
T
"Input" variables
/.
,
V,,
/.
,
h,
-h,
"Output" variables
h
Vi,
V2
h,
h,
Vi
/l,
v2
/l,
v2
v2
h
Vx
v2
V.
• y-parameter
s ar
e usefu
l fo
r modelin
g paralle
l connection
s of two-por
t elements
.
• AZ?CZ)-parameter
s ar
e usefu
l fo
r th
e analysi
s of cascadin
g of two-por
t elements
.
10.7
Problems
1.
Two inductor
sn
i a circui
t ar
e magneticall
y couple
d s
o tha
t when 30 mA s
i flowing
n
i th
e
first
inductor
,L{, ane
t flux
of 20 /xWb link
s th
e secon
d coil
, L2, when ti s
i open-circuited
.
(a) What s
i th
e mutua
l inductanc
e of thi
s arrangement
? (b
) How much flux
wil
l lin
k a
n
open-circuite
d L\ fi 20
0 /x
A s
i made o
t flowni L2?
2.
Two set
s of measurement
s wer
e made on a pai
r of magneticall
y couple
d coil
s an
d th
e
result
s ar
e indicate
d ni th
e followin
g table
. Calculat
e th
e self-inductance
s of th
e tw
o
coils
, th
e mutua
l inductance
, an
d th
e couplin
g coefficient
.
Test#
h (A
)
h (A
)
«fo (j*Wb)
«fe (A*Wb)
1
1
2
25
3
2
1
3
20
7
3.
Two identica
l inductor
s ar
e magneticall
y couple
d s
o tha
t when 1 A flowsthroug
h th
e
first
coi
l an
d 2 A flows
throug
h th
e secon
d coil
, a ne
t flux
of 50 mW b link
s th
e first
coi
l
whil
e 20
0 mW b link
s th
e secon
d coil
. Calculat
e th
e mutua
l an
d self-inductance
s of thi
s
arrangemen
ta
s wel
la
s th
e couplin
g coefficient
.
4.
Two identica
l inductor
s ar
e magneticall
y couple
d s
o tha
t when 3 mA flows
throug
h th
e
first
coi
l an
d 5 mA flows
throug
h th
e secon
d coil
, a ne
t flux
of 12
5 ptWb link
s th
e first
coi
l whil
e 15
0 jnWb link
s th
e secon
d coil
. Calculat
e th
e mutua
l an
d self-inductance
s of
thi
s arrangemen
ta
s wel
la
s th
e couplin
g coefficient
.
5.
Two inductor
s ar
e magneticall
y coupled
. When th
e 5 /x
H inducto
rs
i energize
d wit
h 2 A,
it produce
s a flux
linkag
e of 6.
2 ju,W
b ni th
e open-circuite
d 2 jit
H inductor
. Calculat
e th
e
couplin
g coefficient
.
425
20.7. Problems
6. Two nonidea
l inductor
s ar
e magneticall
y coupled
. Fin
d th
e voltage
s acros
s th
e inductor
s
if th
e drivin
g current
s ar
e i{(t) = 20cos(377
r - 45°
) mA an
d i2(t) = 30 sin(377
f + 45°
)
m A an
d th
e inducto
r parameter
s ar
e L\. — 9 mH,LI — 4 mH, an
d M = 6 mH.
7. Two magneticall
y couple
d nonidea
l inductor
s hav
e inductance
s an
d windin
g resistance
s
of Li =2 mH,R{ = 0.
1 ft,
an
d L2 = 5 mH, /?
2 ft.
The couplin
g coefficien
t s
i
2 = 0.
k = 0.
5 an
d th
e voltage
s acros
s th
e coil
s ar
e foun
d o
t be v\(t) = 5 cos(377
f + 30°
) an
d
r - 45°)
. Fin
d th
e tim
e dependence
s of th
e current
s flowing
throug
h
v2(t) = 4cos(377
the tw
o coils
.
8. A transforme
r ha
s a primar
y self-inductanc
e ofL\ = 2 mH an
d a leakag
e inductanc
e of
27.
7 /xH
. The secondar
y windin
g ha
s a primar
y self-inductanc
e of L2 = 72 mH an
d a
leakag
e inductanc
e of 1 mH. The turn
s rati
o s
ia = 1/6
. Calculat
e th
e scale
d inductance
s
of th
e transformer'
s equivalen
t circuit
,xf, x\ ,an
d JC^, fi th
e transforme
r s
i drive
n a
t
a> = 37
7 rad/s
.
9. Conside
r th
e circui
t show
n n
i Figur
e PI 0-9
. The parameter
s of th
e nonidea
l transforme
r
are a
s follows
: L'
f = 10
0 juH
, Lf = 2 JUH,I™ = 1
0 mH,Lj = 20
0 /xH
, /?
, = 0.0
5 ft,
and 7?
2= 1
0 ft.
Fin
d a
n expressio
n fo
r th
e voltag
e acros
s th
e loa
d resisto
rRL = 2 kf
t fi
6
the sourc
e voltag
e s
i give
n by
:vs(t) = 25 cos(10
£) kV. The turn
s rati
o s
ia = 0.1
.
Figure P10-9
10.
A nonidea
l transforme
r ha
s th
e followin
g properties
:L™ — 25/xH
, Lf = 1/xH
,™
L = 1
0
mh,L\ = 40
0 JLLH, 7?
I = 0.0
5 ft,an
d R2 = 2.5 ft.Fin
d th
e ^-representatio
n of th
e
transformer'
s scale
d equivalen
t circui
ta
t a frequenc
y of co = 200
0 rad/s
. The turn
s rati
o
is a = 0.05
.
In Problem
s 11-34
, th
e tw
o leftmos
t terminal
s ni th
e circui
t indicate
d constitut
e th
e first
por
t
and th
e tw
o rightmos
t terminal
s constitut
e por
t 2.
In Problem
s 11-16
, calculat
e th
e z-parameter
s of th
e circui
t show
n n
i th
e figure
indicated
.
11.
Figur
e P10-11
.
12.
Figur
e PI0-12
.
Figure P10-11
Figure P10-12
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
426
13.
Figur
e P10-13
,>
o = 500
0 rad/s
.
14.
6
Figur
e PI0-14
,>
a = 2 X 10
rad/s
.
1 kQ
22 |iH
Figure P10-13
Figur
e PI0-15
.
16.
Figur
e PI0-16
.
100 pF
1
o
Figure P10-14
o » t VW V
AA/VVtAAMH)
2 kQ
FL
R„
Figure P10-16
Figure P10-15
17.
0
^ 5 0 pF
2|i
F
o
15.
)
|
O - ^ ^ ^
20 mH
Conside
r th
e serie
s combinatio
n of tw
o two-por
t network
s ni Figur
e PI0-1
7 wher
e th
e
z-parameter
s of networ
k 1 ar
e give
n by
:
/l5
2
50
I 5
0
10
0
Find th
e z-parameter
s of th
e tota
l network
.
18.
Conside
r th
e circui
t show
n ni Figur
e PI0-1
8 wher
e th
e z-parameter
s of networ
k 1 ar
e
give
n by
:
'69
22
22
"
64 / "
Use mes
h analysi
so
t fin
d th
e curren
t ni th
e 1
0 O resistor
.
427
10.7. Problems
1
50 Q. >
25 Q,
> 25 Q.
Figure P10-17
Ql2V
10 a
AAA/V
Figure P10-18
19.
Desig
n a circui
t whic
h realize
s eac
h se
t of z-parameter
s give
n ni th
e followin
g table
. Use
dependen
t source
sn
i th
e desig
n onl
y fi necessary
.
Case
20.
zxi (H)
z.i(ft
)
Z21 (H)
Z22 (ft
)
a) (rad/s
)
a
75
50
50
125
0
b
25
50
50
75
0
c
50 - /7
5
50
75
50 + y'5
0
400
Desig
n a circui
t whic
h realize
s eac
h se
t of y-parameter
s give
n ni th
e followin
g table
. Use
dependen
t source
sn
i th
e desig
n onl
y fi necessary
.
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
428
21.
a; (rad/s
)
Case
vii ( U)
yn (U
)
V21 ( 0
)
J22 ( U)
a
75
- 50
- 50
100
0
b
50 - 7I
O
y25
7'2
5
-yio
o
377
c
20 + j'2
0
7'2
5
15 + j'2
5 10 - y'4
0
5000
Conside
r th
e serie
s combinatio
n of tw
o two-por
t network
s ni Figur
e PlO-2
1 wher
e th
e
e give
n by
:
^-parameter
s of networ
k 1 ar
0.3
-0.
1
-0.
1
0.
2
Find th
e y-parameter
s of th
e tota
l network
.
Figure PlO-21
In Problem
s 22-2
4 calculat
e th
e _y-parameter
s of th
e circui
t show
n ni th
e figur
e indicated
.
22.
Figur
e P10-11
.
23.
Figur
e P10-13
,o - 700
0 rad/s
.
24.
Figur
e P10-24
.
In Problem
s 25-2
7 calculat
e th
e g-parameter
s of th
e circui
t show
n ni th
e figur
e indicated
.
25.
Figur
e PI0-12
.
26.
6
Figur
e PI0-14
,co = 3 X 10
rad/s
.
27.
Figur
e PI0-27
.
20.7
. Problems
429
220 Q
-500
j a
470j Q. >
90
j Q> 170 Q'
^AAAM—le-
Figure P10-27
In Problem
s 28-3
0 calculat
e th
e /z-parameter
s of th
e circui
t show
n ni th
e figure
indicated
. The
notatio
n N\ :N2 fo
r a
n idea
l transforme
r indicate
s tha
t th
e turn
s rati
o si a — N] /N2 an
d th
e
secondar
y windin
g s
i th
e rightmos
t inductor
.
28.
Figur
e P10-15
.
29.
Figur
e PI0-24
.
30.
Figur
e PI0-30
.
ideal
transformer
1440
Q'
1:6
Figure P10-30
e circui
t show
n ni th
e figure
indicated
.
In Problem
s 31-3
3 calculat
e th
e ABCD-parameter
s of th
31. Figure P10-16.
32. Figure P10-27.
33. Figure P10-30.
34.
Conside
r th
e cascade
d combinatio
n of two-por
t network
s ni Figur
e PI0-3
4 wher
e th
e
z-parameter
s of networ
k 1 ar
e give
n by
:
6
3
3
7
Find th
e ABCD-parameter
s of th
e tota
l network
.
430
Chapter 10. Magnetically Coupled Circuits and Two-Port Elements
rsrw-\
r
-
*> 5j Q T
1
2 QK >
-3j Q ^
Figure P10-34
35.
The y-representation of a two-port network s
i give
n by:
( -'
- . ' )
Find th
e g- an
d A5CD-representation
s of th
e sam
e network
.
36.
The z-representatio
n of a two-por
t networ
k s
i give
n by
:
Z\\
Zi\
Z\i\
Zll)
Find th
e /z-representatio
n of th
e sam
e network
.
37.
The /z-representatio
n of a two-por
t networ
k s
i give
n by
:
GO Find th
e g- an
d AZ?CD-representation
s of th
e sam
e network
.
38.
39.
A transforme
r ha
s a primar
y self-inductanc
e of Lj =2 mH an
d a leakag
e inductanc
e of
27.
7 jaH
. The secondar
y windin
g ha
s a primar
y self-inductanc
e of L2 = 72 mH an
d a
leakag
e inductanc
e of 1 mH. The transforme
r s
i drive
n wit
h a1
2 vol
t sinusoida
l signa
l
at o
c = 37
7 rad/
s an
d s
i connecte
d o
t a loa
d resistanc
e of 1 kf
t (se
e Figur
e P10-9)
. Use
PSpic
eo
t find
th
e tim
e dependenc
e of th
e voltag
e acros
s th
e load
.
Conside
r th
e circui
t show
n ni Figur
e PI0-9
. The parameter
s of th
e nonidea
l transforme
r
are a
s follows
: Lf
= 10
0 JLLH
,i
f
= 2 JUH, L£
= 1
0 mH, h\
= 20
0 JLLH
,RX = 0.0
5 fl
,
and R2 - 10ft
. Use PSpic
eo
t plo
t th
e expressio
n fo
r th
e voltag
e acros
s th
e loa
d resisto
r
6
RL = 2 kf
t fi th
e sourc
e voltag
e s
i give
n by
: vs(£) = 25 cos(10
£) kV.
40.
A nonidea
l transforme
r ha
s th
e followin
g properties
:U[ = 25 /xH
,h\ = 1 jutH
,L™ = 1
0
mH ,h\ = 40
0 /xH
,R\ = 0.0
5 ft,
an
d R2 = 2.
5 ft.
The transforme
r ha
s a
n equivalen
t
cor
e los
s of 1 ftan
d si connecte
d o
t a 1 kf
t loa
d a
s ni Figur
e PI0-9
. Use PSpic
e o
t plo
t
the rati
o of th
e voltag
e acros
s th
e loa
d o
t th
e inpu
t voltag
e fro
m 1
0 Hz o
t 1
0 kHz.
Appendix A
Complex Numbers
To becom
e fluent
wit
h phasor
s an
d phaso
r manipulations
, on
e shoul
d hav
e a soli
d
backgroun
d ni th
e are
a of comple
x numbers
. Ther
e ar
e many goo
d introductor
y
textbook
s on comple
x numbers
; on
e suc
h boo
ks
iComplex Variables and Applications
by R. V. Churchill
,J
. W. Brown
, an
d R. F. Verhe
y (McGraw-Hill
, 1976)
. Here
, we
briefl
y summariz
e a fe
w importan
t fact
s concernin
g th
e comple
x number
s tha
t ar
e
relevan
to
t th
e comple
x algebr
a require
d fo
r basi
c a
c circui
t analysis
.
The cornerston
e of th
e concep
t of a
n imaginar
y numbe
r come
s fro
m th
e equatio
n
j 2 = ~l
(A.l
)
for whic
h ther
e s
i no rea
l numbe
r solution
. By acceptin
g equatio
n (A.l
) a
s th
e
definitio
n of th
e numbe
r j, we ca
n for
m a
n imaginar
y numbe
r lin
e by multiplyin
g
j by an
y rea
l number
. A genera
l comple
x numbe
rz s
i compose
d of bot
h rea
l an
d
imaginar
y parts
:z = x + jy, wher
e x an
d y ar
e bot
h rea
l numbers
. Thi
s for
m of
comple
x numbe
rs
i referre
d o
t a
s th
e algebrai
c or Cartesia
n form
.t
Is
i a convenien
t
for
m when we nee
d o
t ad
d or subtrac
t comple
x numbers
.
It s
i ofte
n usefu
lo
t plo
t comple
x number
s on atwo-dimensiona
l plan
ea
s indicate
d
in Figur
e A. 1. Thi
s graphica
l representatio
n lead
s us o
t conside
r a pola
r for
m fo
rz
.
The magnitud
e of z s
i clearl
y \z\ = \Jx2 + y2 an
d th
e pola
r angl
e of z s
i give
n by
tan 6 = y/x. The graphica
l representatio
n of comple
x number
ss
i extremel
y usefu
l
for ensurin
g th
e selectio
n of th
e prope
r quadran
t fo
r 0.
Ther
e s
i a thir
d way o
t expres
s a comple
x numbe
r know
n a
s th
e exponentia
l or
exponential-pola
r form
. Thi
s for
m s
i th
e most relevan
to
t th
e phaso
r techniqu
e an
d t
i
is define
d a
s z = \z\ej0, wher
e \z\ an
d 0 ar
e th
e sam
e magnitud
e an
d phas
e a
s ni th
e
pola
r form
, respectively
. To understan
d th
e basi
s fo
r thi
s representation
, conside
r th
e
Taylo
r serie
s expansio
n fo
r ex\
ex = 1 + x + x2/2 + x3/3\ + • • • + xn/n\ + • • •.
431
(A.2)
Appendix A. Complex Numbers
Imaginary Axis
jy
Figur
e A.l
: The comple
x plane
.
o tha
tj 3 = —j an
df = 1
.f
I we replac
e x wit
h j6 ni (A.2
)
Recal
l tha
tj 2 = — 1, s
and grou
p th
e term
s int
o rea
l an
d imaginar
y parts
, we arriv
e at
:
eje = [l-d2/2
+ d4/4l-d6/6\
+ - -] + j[6-63/3\
+ 65/5\-67/1\
+ • • •]
. (A.3
)
The first
ter
m ni bracket
s s
i jus
t th
e Taylo
r serie
s expansio
n fo
r co
s d an
d th
e
secon
d brackete
d ter
m s
i jus
t si
n 6. The relatio
n we hav
e jus
t derive
d s
i know
n a
s
Euler'
s equation
:
eje = co
s 6 + j si
n 0
.
(A.4
)
The exponentia
l for
m s
i quit
e convenien
t fo
r multiplicatio
n an
d divisio
n of comple
x
numbers
. Indeed
:
z = ZiZ2 = \zi\ejei\z2\eJ0i
=
\zi\\z2\e™+ei\
z1\_e^=\z}\
Z\
JWi-h)
z= — =
zi
|z
\Z2\
2le
(A.5
)
(A.6
)
Ther
e ar
e severa
l usefu
l manipulation
s on comple
x number
s tha
t we wil
l us
e
in thi
s text
. We wil
l ofte
n tak
e th
e rea
l an
d imaginar
y part
s of a comple
x number
:
x = Re (z
) an
dy = m
I (z). Car
e must be use
d wit
h thes
e tw
o operations
.f
I
Z\ = x\ + jyx an
d Z2 — x2 + jyi the
n ni genera
l
: fc
Re(z,)Re(z
2)? Re(z
1z2).
(A.7
)
433
Appendix A. Complex Numbers
The proo
f come
s directl
y fro
m th
e multiplicatio
n result
:
ziz2 = (x{x2 - y\y2) + j(x\y2 + x2y{).
(A.8
)
Thus,
Re(ziz
2) = xxx2 ~y{y2,
=
(A.9
)
Re (
I (
I (z
Z l) Re (z2) - m
Z l) m
2).
(A
. 10
)
By examinin
g th
e imaginar
y par
t of (A.8)
, on
e ca
n als
o easil
y se
e tha
t
Im(z1z2)^Im(z
1)Im(z
2).
(A.11
)
A direc
t consequenc
e of (A.9
) s
i tha
t yo
u ca
n "pull
" rea
l number
s n
i an
d ou
t of
d Re (x\z2) = x\X2 =
the Re( ) operator
.f
Iy\ = 0, fo
r example
, the
n z\ = x\ an
x{ Re (z
2).
W e wil
l denot
e th
e comple
x conjugat
e of z a
s *
z an
d defin
e t
ia
s a numbe
r
equa
ln
i magnitud
e o
t z bu
t possessin
g th
e polar
angl
e of opposit
e sign
. Thus
, fi
J
z — \z\eje = x +jy, the
n *
z = |zk~
° = x — jy. One us
e of thi
s definitio
n si o
t not
e
tha
t
zz^\zW°\z\e^e
= \z\2 = x2 + y2,
(A.12
)
Let us conside
r a concret
e exampl
e o
t gai
n some familiarit
y wit
h th
e abov
e
j45
concepts
. Le
t z\ = ~3 — j4 an
d z
a
s indicate
d ni Figur
e A.2
. The
n
2 = 2e °
i
jy
42 4
1
1
1
1
/
1
^ 1 1 1 1 1 /
-4
-2
/
/
i
i
i
i
2
i
i
i
i
4
x
>.
*■
/ -2_4_
Figur
e A.2
: Graphica
l representatio
n o
f tw
o comple
x numbers
.
Appendix A. Complex Numbers
434
Re (z\) = —3 an
d m
I (z\) = —4. The comple
x conjugat
e z\ = — 3 + y'4
. To conver
t
int
o exponentia
l form
, we findIz
J = J x\ +y2 = 5 an
d 6 = tan~](y\/xi)
=
-1
tan (4/3)
. We ca
n us
e a calculator
, an
d tak
e int
o accoun
t th
e fac
t tha
t z\ s
in
i
the thir
d quadrant
,o
t ge
t 6 = —126.9°
. Fro
m Euler'
s equatio
n we se
e tha
tz
2 =
2(cos45
° +j si
n 45°
) = y/2 +jy2. We subtrac
t by usin
g th
e Cartesia
n form
:z\ —
zi = -3 + j4-(v^2 + j^)
- -3( + \/2
) + y ( 4 - y)^ = - 4 441 + 72.586.Finally
,
we divide by using the polar form: z\/zi = (5/2)ej[~l26-9~45°] = 2.5e~jnL9\
Dril
l Exercise
s
1. Conver
t th
e followin
g number
so
t exponentia
l form
:
(a) 5 + A (b
) 1
2 - A (c
) -3 + y/lj9 (d
) -1 - 5j.
2.
Conver
t th
e followin
g number
so
t Cartesia
n form
:
(a) 3ejir/69 (b) 7eJl22\(c) 25e4-l7TJ, (d) -e~j/\
3.
Plo
t th
e followin
g number
s on th
e comple
x plane
:
jn5
(a)6y/le \
(b
) 6 - 5y
, (c
) 1
0 + 24y
\ (d
) l3e~jU7r.
4.
Perfor
m th
e specifie
d operation
s an
d expres
s th
e answe
r ni Cartesia
n form
:
(a) (5 + 3j) * 2eJ45°, (b) 3ej7r/6 + 4e~jir/\ (c) le]m° / ( l + ;), (d) (7 + 5j) - 5ejl20°
5.
Perfor
m th
e specifie
d operation
s an
d expres
s th
e answe
r ni pola
r form
:
(a) 1
( +j) * (2- + 3j), (b
)Ij + 4e*, (c
) (1- + ;)/(
3 + 4;)
, (d
) 4-
e^\
Appendix B
Gaussian Elimination
The stud
y of linea
r algebrai
c equation
s belong
s o
t linea
r algebra
, whic
h s
i a broa
d
are
a of mathematic
s tha
ts
i fa
r beyon
d th
e scop
e of thi
s text
. Ther
e ar
e many goo
d
reference
s on th
e subject
,a
n exampl
e of whic
hs
iLinear Algebra and Its Applications,
by G. Stran
g (Academi
c Press
, 1976)
.
In thi
s text
, th
e reade
rs
i expose
d o
t a ver
y specifi
c subse
t of matri
x operation
s
relate
d o
t th
e solutio
n of genera
l noda
l an
d mes
h equations
.n
I thes
e problems
, we
x
are alway
s equatin
g a know
n sourc
e vecto
r wit
h th
e produc
t of a know
n n X n matri
and a
n unknow
n colum
n vecto
r (a
n n X 1 matrix)
:
/
M il
M12
M2\
'•
•
Mi
V MiNX
Mn
\
(
^ \
) \ U n )
'!
\S„
\
(B.l
)
)
W e ca
n writ
e thi
s mor
e compactl
y a
s M •U = S. Circui
t analysi
s usuall
y require
s
tha
t we find
on
e or mor
e of th
e element
sn
i th
e unknow
n vector
.
The complet
e solutio
n fo
rU s
i give
n symbolicall
y by
U = M~l -S
(B.2
)
where M~l s
i th
e invers
e matri
x of M, define
d by
M-M~l
= M~x -M = 1.
(B.3
)
Here ,
/ know
n a
s th
e identit
y matrix
,s
i th
e matri
x equivalen
t of unity
.t
Is
ia
n n Xn
matri
x whos
e diagona
l element
s ar
e on
e an
d whos
e off-diagona
l element
s ar
e zero
:
°\
/'
(B.4
)
/ =
\o
1 /
435
Appendix
436
B. Gaussian
Elimination
The produc
to
fa
n identit
y matri
x wit
h an
y matri
xM o
f th
e sam
e dimension
ss
i jus
t
equa
lo
t M itself
. Thi
s fact
, couple
d wit
h equation
s (B.l
) an
d (B.3)
, ca
n be use
d
to prov
e tha
t (B.2
)s
i th
e solution
. Consequently
,f
i we ca
n findth
e invers
e matri
x
for th
e circui
t proble
m we ar
e attemptin
g o
t solve
, simpl
e matri
x multiplicatio
n wil
l
yiel
d th
e unknowns
. Many analyti
c an
d numeri
c technique
s fo
r invertin
g matrice
s ar
e
describe
d n
i th
e relevan
t literature
. One particularl
y powerfu
l metho
d s
i know
n a
s
Gaussian elimination with back-substitution. Fo
r thi
s reason
, Gaussia
n eliminatio
n
is th
e onl
y techniqu
e describe
d n
i thi
s appendix
. Reader
s who ar
e intereste
d n
i th
e
othe
r method
s ar
e referre
d o
t th
e literature
.
The basi
c ide
a behin
d Gaussia
n eliminatio
n s
i describe
d below
. Recal
l tha
t we
want o
t solv
e th
e matri
x equatio
n (B.l
) fo
r th
e element
s o
f th
e unknow
n colum
n
vector
. To do this
, we wil
l transfor
m th
e give
n matri
x equatio
n int
o a
n equivalen
t
proble
m wher
e th
e n Xn matri
x s
i upper-triangula
r (i.e.
, a matri
x fo
r whic
h /> /
implie
s tha
t M// = 0)
:
0
M{ 2
M<22
V 0
Kn
0
( s[ \
u2
M'm j
Si
\ Un )
(B.5
)
\ S> )
If we ca
n fin
d thi
s equivalen
t problem
, a simpl
e algorith
m ca
n be use
d ot comput
e
the unknowns
. Thi
s procedur
e s
i know
n a
s back-substitutio
n an
d work
s a
s follows
.
The final
ro
w o
f th
e equivalen
t matri
x equatio
n (B.5
) is
:
(B.6
)
M'nnUn = S'n,
so tha
t th
e final
unknow
n elemen
ts
i give
n by
S'n/M'nn.
(B.7
)
K-U-lUn-\+K-l,nUn=S'a
(B.8
)
UR =
The second-to-las
t ro
w ca
n be written
:
which ca
n be solve
d fo
r Un-\\
Un-X^{S'n^-M'n_uUn)/M'n_u_x.
(B.9
)
All th
e term
s o
n th
e right-han
d sid
e o
f equatio
n (B.9
) ar
e known
, s
o Un-\ ca
n be
calculated
. Thi
s procedur
e ca
n be repeate
d o
t fin
d al
lo
f th
e unknowns
. The genera
l
expressio
n fo
r £/
s
i
1
(
<
j
<
n
1)
:
y
U,
£
M'jM
/M'JJ
(B.10
)
< = /+!
Equation
s (B.7
) an
d (B.10
) full
y describ
e th
e back-substitutio
n technique
.
To validat
e thi
s technique
, we retur
n o
t th
e questio
n o
f producin
g th
e equiva
len
t proble
m wit
h a
n upper-triangula
r matrix
. Thi
s procedur
e s
i know
n a
s Gaussia
n
Appendix B. Gaussian
437
Elimination
eliminatio
n an
d s
i describe
d below
. Let us assum
e tha
t th
e matri
x equatio
n n
i ( B .)l
represent
s a collectio
n of n linearl
y independen
t equations
. I
ts
i known fro
m linea
r
algebr
a tha
t f
i we multipl
y an
y of th
e equation
s by a nonzer
o constan
t an
d ad
d th
e
resul
t t
o an
y othe
r equation
, we stil
l hav
e a se
t of n linearl
y independen
t equation
s
with th
e same solution
.
Let'
s assum
e tha
t M\\ =
£ 0 =
£ M\2. I
f we multipl
y th
e firs
t ro
w n
i ( B .)l by
—Myi/M\\
an
d ad
d th
e resul
t o
t th
e secon
d ro
w we obtain
:
0
M22 -
M2n
Mi2M2l/Mn
W„ ,
MXn
\
-MXnM2X/Mn
1
U2 *
w
Mn
/
5,
V
s
\
» l
( B . l)l
s th
e
If we continu
e n
i thi
s fashio
n (i.e.
, we subtrac
t fro
m th
e jht ro
w Mj\ /M\ \ time
firstrow) we wil
l ge
t
V«
/
M„,
Afi
0
M7
5
,
' U2X
M„
M„
M\nM,
M
rfll
/
\
(B.12
)
\uj
\5 «
5I
M?7/
The firs
t colum
n s
in
i th
e prope
r form
, s
o we tur
n our attentio
n t
o th
e secon
d column
.
B y multiplyin
g th
e secon
d ro
w by appropriat
e factor
s an
d addin
g t
i t
o th
e thir
d
throug
h th
e nt
h row, we ca
n zer
o th
e lowe
r element
s of th
e secon
d column
. Repeatin
g
thi
s procedur
e fo
r th
e thir
d throug
h th
e (f
t — l )tsrow, we ca
n zer
o th
e lowe
r element
s
of th
e thir
d throug
h th
e (n — 1 )s
t columns
, respectively
, an
d arriv
e at th
e desire
d
result
.
Let'
s conside
r th
e specifi
c cas
e of a two-mes
h problem
. T he matri
x equatio
n is
:
Z i,
Z21
(B.13
)
Vs2
Z22
Multiplyin
g th
e firs
t ro
w by — Z2\ /Z\ \ an
d addin
g t
io
t th
e secon
d ro
w yields
:
Z\\
0
Z12
Z22 - Z12Z2i/Z
11
VSl
V:Si
vSlz2l/zn
(B.14
)
The secon
d ro
w s
i use
d o
t fin
d h'.
j _ -Z2iVSl +ZnVS2
n —
ZwZyy — Z\jZ
12-^21
(B.15
)
A n applicatio
n of (B.10
) gives
:
/. =(vSl -zi2i2)/zn
Z Vs
~ Z Vs
t
l2 2
= Z\22 1Z?
? — Z\ 7Z
12^21
(B.16
)
Appendix B. Gaussian Elimination
438
A s a secon
d example
, suppos
e tha
t we hav
e a four-nod
e circui
t tha
t result
s ni th
e
followin
g phaso
r proble
m
1 +j
-j
0
-j
1-j
-1
0
\
/ V,
v2
2 J \ %
-1
(B.17
)
Let us comput
e al
l thre
e nod
e potentials
. To zer
o Y2\, we multipl
y th
e first
ro
w by
jf/(
l + j) = 1
( +j)/2 an
d ad
d t
io
t th
e secon
d row
. The resultin
g equatio
n is
:
+j
0
0
j
0
1(1 - J)
-1
-1
2
(V,
\
/
v2
U
±±i
'
(B.18
)
-J
The first
colum
n s
i no
w zeroe
d becaus
e we ha
d y31 = 0o
t begi
n with
,s
o we move on
to th
e secon
d column
. We nee
d o
t multipl
y th
e secon
d ro
w by 2| /1( — j) = 1
( + j)/3
and ad
d t
io
t th
e thir
d ro
w o
t get
:
(B.19
)
W e us
e equatio
n (B.7
)o
t find
V3:
% = -41
~3
1^-^-
-1
( - 5jf)/13
.
(B.20
)
From (B.10
) we find
that
:
V2 = (^r-
+ ^1 f^
]^ 2
( + 37)71
3
(B.21
)
and
Vi = 1
( + 7V2)/(
l + 7) = 6
( - 4/)/13
.
(B.22
)
While equation
s (B.20)-(B.22
) for
m th
e answe
ro
t thi
s exampl
e an
d th
e proble
m s
i
completed
,t
is
i ofte
n pruden
to
t chec
k fo
r algebrai
c mistake
s by pluggin
g th
e answe
r
back int
o th
e origina
l proble
m (B.17)
. Performin
g tha
t exercis
e wit
h thi
s exampl
e
reveal
s tha
t th
e answe
rs
i consistent
,s
o we conclud
e tha
t we n
i fac
t hav
e foun
d th
e
correc
t solution
.
Drill Problems
In Problem
s 1-5
, us
e Gaussia
n eliminatio
n an
d back-substitutio
n ot find
al
l th
e unknowns
.
■■c>mH',
Appendix B. Gaussian Elimination
1
Mi' 1
j
(2 + j) J \ h
3.
0
4.
i +; -j
-j
2
3 +7 0
(
1
-1/
2
-2
0
3+ j
0
2+ 7
- 12/
2
-1
- 13/
-2
-1
3
-1
439
2 -7
1
Appendix C
MicroSi
m PSpic
e Reference
s
Conant
, Roger
,Engineering Circuit Analysis with PSpice and Probe, McGraw-Hill
,
New York
, 1993
.
Goody, Roy
,PSpice for Windows, Prentic
e Hall
, Englewoo
d Cliffs
, NJ, 1995
.
Herniter
, Marc
,Schematic Capture with MicroSim PSpice, 2n
d Edition
. Prentic
e Hall
,
Englewoo
d Cliffs
, NJ, 1996
.
MicroSi
m Corporation
,Circuit Analysis Users Guide, Irvine
, CA, 1992
.
Nilsson
, James
,Introduction to Spice, Addison-Wesley
, New York
, 1990
.
441
Index
A .See amper
e
ABC£>-parameters
, 41
8
ac, 25
power
, 80
steady-stat
e equations
, 74
Active-RC circui
t
filter.
See filter
"pole
" circuit
, 36
8
"zero
" circuit
, 36
9
Adder circuit
, 31
9
Admittance
, 73
matrix
, 17
7
tabl
e for
, 90
Ampere, 4
Amplifie
r
buffer
, 31
4
inverting
, 31
8
noninverting
, 31
7
Analo
g computer
, 32
5
Angula
r frequency
.See frequenc
y
Averag
e power
, 80
B. See susceptanc
e
Band-notc
h filter.
See filter
Band-pas
s filter.
See filter
Basi
c circui
t quantities
,9
Battery
, 24
Bias voltage
, 30
1
Bode plot
, 35
4
Branch
, 34
Breakpoin
t frequency
, 35
5
Bridg
e circui
t
wit
h linea
r elements
, 15
0
wit
h nonlinea
r resistor
, 16
8
Buffe
r circuit
.See amplifie
r
Butterwort
h filter
passive
, 35
8
active
, 37
0
C.See Capacitanc
e
C.See Coulom
b
Capacitance
,1
9
Capacitiv
e voltag
e divider
, 10
5
Capacito
r
fabrication
, 22
ideal
, 19
initia
l condition
.See initia
l
conditio
n
nonideal
, 14
4
shoc
k hazard
, 14
5
Cartesia
n form
, 43
1
Cascade
d network
.See two-por
t
networ
k
Characteristi
c equatio
n
first-order,
21
1
second-order
, 23
5
Charge
.See electrica
l variable
s
Coil
,8
Complet
e solutio
n of a differentia
l
equation
, 21
8
Complex
frequency
, 87
number
, 63
power
, 83
Conductance
,1
2
tabl
e for
, 90
Conductivity
, 14
Conservatio
n
of charge
,3
of current
,4
443
444
Continuit
y
of capacito
r voltage
, 21
of inducto
r current
,1
7
Convolutio
n integral
, 25
6
for genera
l circuit
, 26
5
forRL circuit
, 25
7
Core loss
.See edd
y curren
t
Corne
r frequency
.See breakpoin
t
frequenc
y
Coulomb, 2
Coulomb'
s law
,2
Couple
d circui
t equations
, 38
8
Couple
d coils
, 38
2
Couplin
g coefficient
, 38
6
Cramer'
s rule
, 40
5
Critica
l damping
, 23
7
Curren
t
displacement
,3
electric
,3
Current-controlle
d source
.See
dependen
t sourc
e
Curren
t divider
, 10
4
Curren
t sourc
e
ideal
, 24
nonideal
, 14
4
Cut set
, 35
Index
Directe
d graph
, 35
Discriminan
t of a second-orde
r circuit
,
235
Dot convention
, 38
8
Dualit
y of inducto
r an
d capacitor
, 26
Eddy current
, 40
0
Electrica
l variable
s
charge
,2
device
,8
energy
,6
field,
2
power
,7
Energ
y store
d
in capacitor
, 21
in inductor
,1
8
Equivalen
t circui
t value
s
calculatio
n wit
h probin
g source
.See
prob
e metho
d
Thevenin
, 14
9
Norton
, 15
6
Equivalen
t transformatio
n
of nonidea
l sources
, 14
5
of passiv
e elements
, 98
Euler'
s equation
, 432
Excitatio
n
by a
c source
s
D/A converter
.See Digital-to-analo
g
first-order
circuit
, 22
1
converte
r
second-orde
r circuit
, 24
3
by genera
l source
s
Damping factor
, 35
1
first-order
circuit
, 21
7
dB.See decibe
l
second-orde
r circuit
, 24
1
dc, 1
7
by initia
l condition
s
Decibel
, 35
4
first-order
circuit
, 20
9
Dependen
t sources
, 29
8
second-orde
r circuit
, 23
3
transisto
r model
, 30
2
by initia
l condition
s an
d sources
, 22
8
Differentia
l equatio
n
first-order
F.See fara
d
homogeneous
, 21
1
Farad
, 20
inhomogeneous
, 21
8
Faraday'
s law
,9
second-orde
r
Filte
r circuit
s
homogeneous
, 23
4
activ
e low-pass
, 36
1
inhomogeneous
, 24
1
passive
, 34
9
Differentiato
r circuit
, 32
3
band-notch
, 35
4
Digital-to-analo
g converter
, 12
2
band-pass
,
35
3
Diode
, 26
9
Index
high-pass
, 35
0
low-pass
, 35
1
First-orde
r
circuit
, 20
8
differentia
l equation
, 21
0
Flux linkages
,8
Force
d response
, 22
0
Free response
, 22
0
Frequenc
y
angular
, 59
corner
.See breakpoin
t frequenc
y
resonant
.See resonan
t frequenc
y
3-dB.See breakpoin
t frequenc
y
Full-wav
e rectifier
.See rectifie
r
G .See conductanc
e
g-parameters
.See hybri
d parameter
s
Gain
closed-loop
, 31
9
open-loop
, 31
3
Gaussia
n elimination
, 43
5
Graph tree
, 35
Graphica
l solutio
n fo
r circui
t wit
h
nonlinea
r resistor
, 16
3
Ground
, 31
3
H .See henr
y
/j-parameters
.See hybri
d parameter
s
Half-wav
e rectifier
.See rectifie
r
Harmoni
c generator
.See sin
e wave
generato
r
Heater
, 14
Henry
, 16
Hertz
, 59
High-pas
s filter.
See filter
Househol
d electricity
, 19
5
Hybri
d parameters
,
^-parameters
, 41
6
/z-parameters
, 41
5
Hz.See hert
z
ICIS.See dependen
t sourc
e
ICVS.See dependen
t sourc
e
Idea
l component
.See individua
l
componen
t name
445
Identit
y matrix
, 43
5
Impedance
, 67
cube
, 11
3
infinit
e grid
, 12
1
LC branch
, 72
matrix
, 18
9
measurement
, 11
4
RC branch
, 72
RL branch
, 72
RLC branch
, 70
tabl
e for
, 90
Impedanc
e parameters
.See
z-parameter
s
Impuls
e
function
, 26
0
response
, 26
2
Independen
t
curren
t source
.See curren
t sourc
e
voltag
e source
.See voltag
e sourc
e
Inductance
, 16
leakage
, 39
6
mutual
, 38
2
self
, 38
4
Inducto
r
ideal
,1
6
initia
l condition
.See initia
l
conditio
n
nonideal
, 14
4
Initia
l conditio
n
capacitor
, 21
inductor
,1
7
in second-orde
r circuits
,
252
Inpu
t
admittance
, 17
0
impedance
, 10
7
Integratin
g factor
, 25
7
Integratio
n by parts
, 25
8
Integrato
r circuit
, 32
1
Interna
l
conductance
, 26
resistance
, 26
Invertin
g
amplifier
.See amplifie
r
terminal
, 31
3
Index
446
iv characteristi
c
diode
, 27
0
linea
r resistor
,1
2
nonlinea
r resistor
, 16
2
J.See joul
e
Joule
,6
KCL .See Kirchhof
f scurren
t la
w
Kerne
l of convolution
, 25
7
Kirchhoff'
s curren
t law
, 36
linearl
y independen
t equations
, 41
Kirchhoff'
s voltag
e law
, 37
linearl
y independen
t equations
, 42
KVL .See Kirchhoff'
s voltag
e la
w
L.See inductanc
e
Ladder networ
k
R-2R,109
R-3R, 14
1
Leakag
e
inductance
.See inductanc
e
resistance
, 14
4
Linea
r system
, 39
Linearize
d mode
l of a MOSFET, 30
1
Load, 84
Loop, 34
Low-pas
s filter.
See filter
LTI system
, 15
9
M .See mutua
l inductanc
e
Magneti
c
energy
,1
8
fluxdensity
,8
Magnitud
e plot
.See Bode plo
t
Matri
x algebra
, 43
5
Mesh, 34
Mesh analysi
s
wit
h dependen
t sources
, 30
5
wit
h idea
l curren
t sources
, 19
1
wit
h nonidea
l curren
t source
, 19
1
wit
h trivia
l meshe
s containin
g a
voltag
e source
, 19
2
wit
h voltag
e sources
, 18
6
Mho, 13
Multiplyin
g factor
,1
Mutual inductance
.See inductanc
e
Natura
l frequency
, 24
0
Network
.See two-por
t
Neutra
l wire
, 18
5
Nodal analysi
s
wit
h curren
t sources
, 17
4
wit
h dependen
t sources
, 30
3
wit
h idea
l voltag
e source
, 18
0
wit
h nonidea
l voltag
e sources
, 18
1
wit
h trivia
l node
s containin
g a curren
t
source
, 18
0
Node, 34
Nonidea
l component
.See individua
l
componen
t name
Noninvertin
g
amplifier
.See amplifie
r
terminal
, 31
3
Nonlinea
r resistor
.See resisto
r
Nonreciproca
l networ
k
^-parameters
, 41
8
/^-parameters
, 41
7
y-parameters,
412
z-parameters
, 40
8
Norton'
s theore
m
application
, 15
6
wit
h dependen
t sources
, 31
1
proof
, 15
5
Ohm , 1
2
Ohm's law
,1
2
Op-amp.See operationa
l amplifie
r
Open circuit
, 11
6
test
, 15
5
Operationa
l amplifie
r
ideal
, 31
4
LM324, 33
3
nonideal
, 31
6
Oscillatio
n n
i a second-orde
r circuit
,
237
Overdampe
d solution
, 23
7
Paralle
l connection
, 10
1
equivalen
t impedance
, 10
2
447
Index
Particula
r solution
, 21
9
Passiv
e
filter
.See filter
sig
n convention
, 11
Peak amplitude
, 59
Period
, 59
Periodi
c source
, 25
Permittivity
, 21
Phas
e
initial
, 59
lag/lead
, 34
8
plot
.See Bode plo
t
relatio
n
for capacitor
, 62
for inductor
, 61
for resistor
, 60
Phasor
, 63
conversion
, 64
in th
e solutio
n of a transien
t probl
<
222
Plana
r circuit
, 34
Pola
r form
, 43
1
Pole of transfe
r function
, 24
8
Port
, 40
2
Power, 8
average
, 80
of capacitor
, 21
of inductor
,1
7
of resistor
, 13
Power factor
, 83
Probemetho
d
wit
h curren
t source
, 31
0
wit
h voltag
e source
, 30
9
PSpic
e
ac sweep
, 12
8
add part
, 12
6
alias
, 13
2
analysi
s setup
, 12
8
damped source
, 28
2
dc sweep
, 33
0
dependen
t source
, 32
9
diode
, 28
5
ground
, 12
7
installation
, 12
5
netlist
, 12
8
op-amp
, 33
1
parametri
c analysis
, 13
3
print
, 42
1
PROBE , 12
9
Schematics
, 12
6
sin
e wave generator
, 33
5
switch
, 28
0
three-phas
e circuit
, 19
5
transformer
, 42
1
transien
t analysis
, 13
1
Puls
e source
, 25
Quadrant
, 43
4
R. See resistanc
e
Ramp source
, 25
Reactance
, 71
tabl
e for
, 90
Real part
, 43
2
Rectifie
r
ful
l wave
resistiv
e load
, 27
2
RC load
, 27
6
RL load
, 27
3
hal
f wave
, 27
0
Referenc
e
direction
, 10
node, 17
5
Regimes fo
r superpositio
n method
, 11
6
Resistance
,1
Resistiv
e circuit
, 47
Resisto
r
fabrication
, 1
4
ideal
,1
nonideal
, 14
5
nonlinear
, 16
0
Resonance
, 34
6
Resonan
t frequency
, 34
7
rms.See root-mean-squar
e
Root-mean-square
, 80
Second-orde
r circuit
, 23
3
Sel
f inductance
.See inductanc
e
Serie
s
definition
, 84
Index
448
equivalen
t impedance
, 10
0
genera
l connection
, 98
Shock hazard
.See capacito
r
Shor
t circuit
, 11
6
test
, 15
5
Shunt
, 22
5
SI, 1
Siemens
,1
3
Sine wave generator
, 32
7
Sinusoida
l
source
, 59
steady-state
, 58
Small signa
l model
.See linearize
d
model
Source
.See specifi
c sourc
e name
SPICE, 12
5
Staircas
e function
, 26
0
Ste
p
function
, 25
8
response
, 26
1
shifting
, 25
9
Summer circuit
.See adde
r circui
t
Superpositio
n
integral
, 26
4
method
, 11
5
wit
h multipl
e frequencies
, 12
0
Susceptance
, 73
tabl
e for
, 90
Switch
, 20
9
Symmetry
, 11
0
Synthesis
.See transfe
r functio
n
Terminal
,7
Thevenin'
s theore
m
application
, 14
9
for circuit
s wit
h on
e nonlinea
r
resistor
, 16
3
wit
h dependen
t sources
, 30
7
proof
, 14
6
3-dB frequency
.See breakpoin
t
frequenc
y
Three-phas
e circui
t sta
r connection
,
Time constan
t
RC, 212
RL, 21
6
Topology
, 34
Transfe
r function
, 24
7
synthesis
, 36
5
Transformer
, 39
0
idea
l
equivalen
t circuit
, 39
4
parameters
, 39
1
two-por
t model
, 41
4
nonidea
l equivalen
t circuit
, 40
0
scale
d equations
, 39
9
turn
s ratio
, 39
2
Transien
t signal
, 20
8
Transien
t respons
e
paralle
lRLC circuit
, 24
4
RC circuit
, 20
9
RL circuit
, 21
4
serie
s RLC circuit
, 24
1
Transisto
r
BJT, 29
8
MOSFET , 29
9
Transmissio
n line
,1
5
Two-por
t elements
, 40
2
cascade
d connection
, 41
9
paralle
l connection
, 41
3
serie
s connection
, 40
9
Two-termina
l element
, 10
Undamped circuit
, 23
9
Underdampe
d circuit
, 23
6
Undetermine
d coefficients
, 21
9
Uniqueness
, 33
Unit impulse
.See impuls
e
Unit step
.See ste
p
Units
.See SI
tabl
e of
,9
V.See vol
t
VCIS.See dependen
t sourc
e
VCVS .See dependen
t sourc
e
Volt
,6
Voltage
,6
Voltage-controlle
d source
.See
dependen
t sourc
e
Voltag
e divider
, 10
3
Voltag
e follower
.See amplifie
r
Index
Voltag
e regulator
, 16
6
Voltag
e sourc
e
ideal
, 24
nonideal
, 14
3
W .See wat
t
Watt, 7
Wb .See Weber
Weber, 9
Work, 6
X, See reactanc
e
Y. See admittanc
e
y-parameters,
41
0
circui
t realization
, 412
computation
, 41
1
Z.See impedanc
e
z-parameters
, 40
3
circui
t realization
, 40
8
computation
, 40
5