SEC. 2.1
CHAPTER
Heinrich Hertz, who was the first to demonstrate electromagnetic wave
propagation. One Hz = 1 cps, 1 kilohertz (kHz) = 103 Hz, 1 megahertz
(MHz) = 106 Hz, 1 gigahertz (GHz) = 109 Hz. The period T of the wave is
the time for one complete cycle; therefore T = l/f. For example, a wave of
frequency 2000 cps or 2 kHz has a period of 0.0005 s. The "60-cycle"
(really 60 cps) line voltage available in most laboratories has a period of
1/60 s = 0.0167 s. Often, the angular frequency {U = 27Tf = 27rIT is used; {U
has units of radians per second (rad/s). (27rrad 360°; 1 rad 57°.) Note
that a radian is really a pure number, so radians per second = S-I. This
follows from the definition of an angle 6 (Fig. 2.2) in radians: 6 == sIr,
where s is the arc length and r is the radius.
2
Alternating Current
Circuits
2.1
PERIOOIC WAVEFORMS
In Chapter 1 we briefly mentioned alternating current (ac) and described it
as current in which the electric charge "sloshed" back and forth along the
wire. The most common type of ac is that in which the back-and-forth
motion is sinusoidal; that is, if we plot a graph of the instantaneous value of
the current i(t) against time t, we get a sine wave as in Fig. 2.l(a). The
instantaneous voltage v(t) can also be sinusoidal, as shown in Fig. 2.l(b).
Such a sine wave is completely determined by specifying three things:
(1) the frequency f of the wave, (2) the amplitude of the wave, and (3) the
phase of the wave. The frequency f of the wave is the number of complete
cycles that occur in one second and is expressed in cycles per second (cps).
The hertz (Hz) is a unit of frequency now widely used. It is named after
+10
T
.......--------- .....
,.
8= !
s curved arc length
radius
FIGURE 2.2
The amplitude of the wave is a measure of how "large" or strong the
wave is. A common way to specify the amplitude is to give the change in
current or voltage from the most positive value to zero; this is 10 in Fig .
2.1(a) and Vo in Fig. 2.1(b). The peak-to-peak amplitude is the change in
current or voltage from the most positive value to the most negative value:
210 in Fig. 2.1(a) and 2Vo in Fig. 2.l(b). Another common measure of the
amplitude is the root-mean-square (rms) amplitude, defined as
v(t)
i(t) +Vo
- Vo
j(t) =
f
period
(a) sinu~oidal current Sinusoidal waveforms.
IT]
v (t) dt
2
1/2 0
(2.1)
Vo sin {Ut and
If v(t) is sinusoidal, v(t)
I
T=l
w" 21rf
1
[T
Vo sin wI
vtl)
frequency
w = angular frequency
FIGURE 2.1
Vnns
10 sin wt
T
54
55
Periodic Wa".rorms
2.".
T
(b) sinusoidal voltage
Vrms
[TlIT Vijsin
0
2
{Utdt
]lf2 = ~
V
= 0. 707Vo
(2.2)
Rms values are useful because they occur in the expression for the power in
ac ,circuits that will be explicitly shown later in this chapter. The 1l0-V,
SEC. 2.1
56
CHAP. 2
57
Perlof6c Weveforms
Alternating Current CircuiD
VI (t)
VI (1)
v,(t)
v1 (r)
60-Hz line voltage common~ available in laboratories has an rms voltage
of 110 V; therefore, Vo =.J2 Vm'I$ = 156 V, and the peak-to-peak voltage
is 312 volts! This line voltage is shown in Fig. 2.3 and can vary from about
lOS V rms to 120 V rms depending on the time of day and the demands
made upon the electric power company.
vet)
+ 156 Y
!lOY
IIV
I
I
I I
f I
'\:
I
II
1
I
f
I
I
phase difference
(b) two waves out of phase
(a) Iwo waves in phase
v(l)
FIGURE 2.3
=
Vo sin 21(/1
= (154 Y)
sin 211'(60)1
FIGURE 2.4 Phase relationships.
110-Y-rms 6O-"cycle" (60·Hz) line voltage.
The phase of a wave is a more subtle concept and has meaning only
when specified relative to another wave of the same frequency. The phase
tells uS when the wave reaches its maximum value compared to the time of
the maximum for the other wave. Two waves of the same frequency are in
phase if they reach their maximum values at exactly the same time, as
shown in Fig. 2.4(a). It makes no sense to compare the phase of two waves
of different frequencies. The two waves of Fig. 2.4(b) are out of phase
because their peaks do not occur at the same time. Their phase difference is
seen to be iTs or 45°, because one complete period T corresponds to 360°.
Mathematically, the initial phase is expressed as the angle I/> in the
equation v(t) = Vosin(21T/t + 1/» = Vo sin(21!1/T + 1/» which is graphed in
Fig. 2.S(a), and the phase is the entire argument (21!1/T + 1/». The
equivalence between phase in degrees (or radians) and time can be seen
from the fact that a change in t by one period T or a change in I/> by 3600
or 2 'IT rad does not change v(t). Expressed in words, the phase determines
the wave amplitude at the arbitrary time t = 0, because at t = 0, V =
Vo sin 1/>. Phase is measured in degrees or radians and does not depend
upon the amplitude or the frequency.
The voltage Vl(t) = VOl sin(wt + I/>} has a phase of I/> radians relative
to the voltage V2(t) = V 02 sin wt; in other words VI(t) leads V2(t) by I/>
radians, or, less precisely, Vt(t) is out of phase with respect to Vz(t) by I/>
radians. If two waves (of the same frequency) are out of phase by 'IT radians
or 180°, then they appear as shown in Fig. 2.5(b). Thus two waves of equal
amplitude and exactly 1800 out of phase add to give exactly zero. From
vdt)
vet)
v2 (t)
(a) vet)
Vo sin (wI
+ ,p)
(b) two waves 18f1' oul
0/ phase
FIGURE 2.5 More phase relationships.
trigonometry any two sine waves of the same frequency and arbitrary
phases will add together to give a sine wave of the same frequency wand a
definite phase 1/>. That is,
VOt sin(wt + 1/>1) + Vozsin(wt + 4>2) = Vosin(wt + 1/»
SEC. 2.2
58
CHAP. 2
where VOl is the amplitude of the first wave, V 02 is the amplitude of the
second wave, 0/1 is the phase of the first wave, <1>2 is the phase of the second
wave, Vo is the amplitude of the resultant wave, and 0/ is the phase of the
resultant wave.
It is often convenient to use cosine waves rather than sine waves. A
cosine wave is merely a sine wave shifted in phase by 90° or Tr/2 rad.
Vo COS wt = Vo sin( wt +
The identity sin(a + b)
=
AC PoWer
59
Aft.",.Ung Cummt Circuits
f)
Notice that the current and the voltage are exactly in phase. The
instantaneous power developed in the resistance is still given by p vi (the
de formula), but now v is the instantaneous value of the voltage drop across
the resistance, and i is the instantaneous value of the current through the
resistance. Hence, p(t) = l5R sin2 wt; the instantaneous power p(t) thus
varies with time, as is shown in Fig. 2.6.
Usually, however, it is much more useful to consider the average
power, that is the power averaged over an integral number of cycles of
voltages and current. The average power can be written by averaging the
instantaneous power over one (or many) periods of time:
sin a cos b + sin b cos a has been used here.
Pay
2.2 AC POWER
=
Consider a sinusoidal voltage v(t) = Vo sin wt applied across a resistance
R. Ohm's law holds at every instant of time, so
;(t) =
v(t)
R=
Vo sin wt = Vo sin wt = 10 sin wt
R
R
(2.3)
Thus the current through the resistance also varies sinusoidally with time at
the same frequency as the voltage. The amplitude 10 of the current equals
VoIR (see Fig. 2.6).
i(t)
.(t)
V. " ••'
~
v(t)
~
P(tl
10
T
FIGURE 2.6
IT
Pay -- ~
T V5
R
But
0
(Vo sin wt)
IT sin
0
2
(~o sin wt) dt
wtdt = 2R
(2.4)
(2.5)
Vrms = Vol.fi
V,;,.S
R
so
Pay =
And
Inns
= lo/.fi =
so
Pay
=
(2.6)
Vol.fiR
=
Vrms/ R ,
(2.7)
l';"sR
p=
T
10
0
The result, then, for sinusoidal current flowing through a resistance is that
the average power dissipated in the resistance equals I';"'R or V';"s/R. The
occurrence of the rms values in the power formula is one reason rms values
are useful. In words, a dc current I = 1m.. = 101.fi would have the same
heating effect as the ac current of amplitude 10 , and similarly for a dc
voltage V = V nns = Vo/.fi.
If the current through a two-terminal element is sinusoidal and the
voltage across it is also sinusoidal of the same frequency but differing in
phase by 0/ rad, then the instantaneous power dissipated in the element is
R
- Vo
-10
IT p(t) dt = T1 IT v(t)i(t) dt
Vo
i(t)
i(t)
= T1
Vo
Ii
Alternatiog curreot and voltage through a resistance.
=
Vo sin(wt + 0/)10 sin wt
(2.8)
T
and the average power dissipated can be shown to be
Pay
=
Volo
-2- cos 0/ = Vrms l.".,. cos 0/
(2.9)
60
2.3
CHAP. 2
Altemating Current ClrculU
Capacitance is very important in ac circuits. A circuit element that has
capacitance is called a capacitor or a condenser; it is represented in circuit
diagrams by the symbol in Fig. 2.7. Capacitance is defined as the charge
stored on one plate divided by the voltage difference between the two
plates:
CEQ
V
(2.10)
The total net charge on the two plates is always zero. (If + Q is on one
plate, then - Q is on the other plate because the presence of positive charge
(a) schemotic symhol
61
Capacimnce
There are two general types of capacitors: polarized and unpolarized.
CAPACITANCE
yp
SEC. 2.3
Q
C;a
~ Val
(b) deftnilion
FIGURE 2.7 Symbol of capacitance.
on one plate repels an equal amount of positive charge from the other
plate.) The larger the capacitance, the more charge is stored on the plates
for a given voltage difference; capacitance is the "amount of charge stored
per volt." A capacitor has two terminals or leads, one going to each plate.
Capacitance is measured in coulombs per volt; one coulomb per volt is
called a farad (F). A I-F capacitor would be physically huge, so other units
are used for capacitance: the microfarad (/LF), which equals 10-6 F, and the
picofarad (pF), which equals 10- 12 F. A picofarad is often called a "puff" in
informal conversation. The reason for the capacitance symbol in Fig. 2.7 is
that a capacitor is actually made by placing two metal plates or foils paranel
to but insulated from each other; the two terminals are connected to the
two plates. Notice that if the dielectric (i.e., the insulating material) between
the two plates does not break down from too high a voltage being applied
across the plates, then there is no way direct cu"ent can flow through the
capacitor. In other words, a capacitor has an infinite dc resistance in the
steady state. The thicker the insulating material, the higher the voltage
rating of the capacitor.
However, alternating current can pass through a capacitor. As a
positive charge surges onto the left-hand plate, the positive charge on the
right-hand plate is repelled toward the right. Thus a surge of current into
one terminal results in a surge of current out of the other terminal, and,
one-half cycle later, charge surges into the right-hand terminal and out of
the left-hand terminal. In other words, the capacitor does pass alternating
current.
Polarized capacitors function properly only when a definite dc voltage
polarity is maintained across the plates. One terminal is labeled + and the
other -; the dc voltage on the + side must always be maintained positive
with respect to the side regardless of the ac voltage present. The most
common type of polarized capacitor is the electrolytic capacitor in which
the dielectric film between the two conducting plates is formed by an
electrochemical reaction. The film is extremely thin; thus, large capaci­
tances can be obtained at relatively low voltage ratings in a small volume.
For example, an electrolytic capacitor as large as half a cigarette may have
100-/LF capacitance with a maximum voltage rating of 25 V dc. If the dc
voltage polarity across an electrolytic capacitor is reversed, the dielectric
film is usually punctured, and the capacitor is permanently shorted out.
Such a situation can usually be detected by measuring the dc resistance of
the capacitor; a good electrolytic capacitor should yield a dc resistance of
several hundred kO or more. In making this resistance measurement, place
the red or positive ohmmeter probe on the + lead of the electrolytic
capacitor.
Unpolarized capacitors can function properly with either polarity of dc
voltage between the plates. There are many types. The tubular capacitor
shown in Fig. 2.8 consists of a thin sheet of insulating material (mylar,
,ubular
disc ceramic
mica
FIGURE 2.8 Various types of capacitors.
paper, etc.) rolled up in a cylinder between two thin metal foils. One wire
lead is connected to each metal foil, and the entire assembly is potted in
wax or some insulating plastic. Tubular capacitors are normally used for
audio frequencies, and typical capacitance values range from 0.001 /LF to
l/LF.
The disc ceramic capacitor, shown in Fig. 2.8, has a thin layer of
ceramic as the insulator between the electrodes and, unfortunately, a rather
large temperature coefficient; that is, the capacitance changes relatively
rapidly with changing temperature. Hence, they are usually used for bypass
applications where only a certain minimum capacitance is required, rather
than a specific value of capacitance. They are used at frequencies up to
several hundred MHz. A special type of disc ceramic, type NPO, has a zero
temperature coefficient at room temperature and is widely used.
62
CHAP. 2
Alternating Current Circuits
The mica capacitor, shown in Fig. 2.8, has a thin sheet of mica as the
insulator and provides the most stable, precise value of capacitance of all
capacitor types. Mica capacitors are useful up to hundreds of megahertz
and are also rather expensive. Another type of expensive capacitor is the
tantalum capacitor, which has low inductance, high capacitance per unit
volume, and provides relatively high capacitance values, up to hundreds of
/-LF. A summary of the properties of different types of capacitors is given in
Appendix A.
Variable capacitors, where the capacitance is adjusted by turning a
shaft, are also available. As the shaft turns, two sets of parallel metal plates
mesh, without touching [Fig. 2.9(b)): the larger the overlap, the higher the
capacitance. The stationary set of plates is called the stator, the rotating set
the rotor. The rotor is connected electrically and mechanically to the shaft,
which is usually connected to the chassis and thus is at ground. The curved
line in the schematic symbol [Fig. 2.9(a)] is always the rotor and is usually
grounded.
#f (al schematic symbols
(b I actual variable capacitor
FIGURE 2.9 Variable capacitor.
A useful way to think of a capacitor in terms of the water-flow current
analogy is to regard the capacitor as an enlargement in the water pipe with
a flexible membrane stretched across the enlargement, as shown in Fig.
2.10. As water surges in from the left, the membrane stretches toward the
right, and water surges out of the right-hand pipe. No water actually passes
completely through, from left to right, but the surge of water does flow out
of the right-hand pipe. In the case of the capacitor, no direct current flows
through, but alternating current does because ac is really a back-and-forth
surging of electrons. Note that as the water pressure between the two sides
increases, the membrane stretches and water flows out of the low-pressure
side. This is analogous to a capacitor, which passes some current every time
the voltage difference across the capacitor plates changes. A very stiff
membrane corresponds to a small capacitance; a very flexible one to a
large capacitance.
SEC. 2.3
63
Capacitance
thin flexible membrane
water _
surge in
water
surge out
-
FIGURE 2.10 Water-pipe analogy of capacitor.
Capacitance depends on the area of the plates, the separation of the
plates, and the material between the plates. It does not depend on current
or voltage. For two parallel plates each of area A and separation d, the
capacitance can be shown to be C = EoA/d, for a vacuum between the
plates, and C = kEoAI d for a material of dielectric constant k (a pure
number) between the plates. If A is in square meters, d in meters, and Eo in
farads per meter, then C is in farads. The constant Eo 8.85 x 10-12 F/m is
the permittivity of free space, which occurs in so many formulae in mks
units. Because C varies inversely with d, we see that if we obtain a higher
voltage rating by using a thicker insulating material (larger d) between the
plates, we must decrease the capacitance for a given volume capacitor. In
other words, we can have a high voltage rating and a low capacitance, or a
low voltage rating and a high capacitance, for a fixed physical size.
If the capacitors of capacitance C 1 and C 2 are connected in parallel as
in Fig. 2.11(a), then the resultant can be regarded as one new capacitor. We
would expect the new capacitor to have a capacitance larger than C 1 or C 2
because the area of the left-hand plate of the new resultant capacitor is the
sum of the areas of the left-hand plates of C 1 and C 2 , and similarly for the
C1
~J+
C",,,,
c2
C1
C2
+11+ -
+11­
+­
+0, -0,
+02 -02
AO
­
C1 t
c2
~oral
+ _I
Cz
(a) parallel connection
FIGURE 2.11
DB
Combinations of two capacitors.
=
C 1 C2
-­
C 1 + C2
(b) series connection
SEC. 2.4
64
CHAP. 2
65
Capacitive Reeetan",
Alternating Current Circuits
for two resistors in parallel. For N capacitors in series
right-hand plates. If we do not change the plate separation of each capaci­
tor when we connect them in parallel, we might, since the areas add, then
expect simply that Clolal = C 1 + C 2 , which is the case, as can be seen by the
following derivation. By definition
Clolal =
QIOlal
(2.11)
V
where Qlolal is the total charge on either plate and V is the voltage
difference across the plates. Note that Qlolal = Ql + Q 2 where Q 1 = the
charge on C 1 and Q 2 = the charge on C 2 . Therefore,
CIOlal = Q 1
+ Q2
(2.12)
V
From the definition of capacitance Q 1 = C 1 VI and Q 2 = C 2 V 2 , and
because C 1 and C 2 are in parallel, VI = V 2 • Therefore
1
Clolal = C V
+ C2 V
=
C
1
+ C2
(2.13)
For example, a O.01-/-LF capacitor and a 0.05-/-LF capacitor in parallel act
effectively like one 0.06- /-LF capacitor. For N capacitors in parallel it is
easily shown that
CIOlal = C 1
+ C 2 + ... + CN
(2.14)
If two capacitors C 1 and C 2 are connected in series as in Fig. 2.1l(b),
then the resultant can be regarded as one new capacitor. If a voltage
difference V is applied between terminals A and B, with A positive with
respect to B, then positive charge from the right-hand plate of C 1 will be
repelled to the left-hand plate of C 2, where it must stop. Thus, Q 1 = Q2,
because the center wire, the right-hand plate of C b and the left-hand plate
of C 2 must remain electrically neutral. Thus the charge on C 1 must exactly
equal the charge on C 2 • Let this charge be denoted by Q. By definition
Clolal = Q / V = Q /( VI + V 2 ). But VI = Qd C 1 and V 2 = Q2/ C 2 and
0 1 = Q 2 = Q, so
Q
Clolal = Q
1
C 1 C2
Q = -1--1 = C + C
-+C1
C2
-+C1
1
(2.15)
1
Clolal = 1
1
1
C1
C2
CN
(2.16)
-+-+ ... + ­
A O.OI-/-LF capacitor and a 0.05-/-LF capacitor in series act like a 0.0083-/-LF
capacitor.
In addition to manufactured capacitors intentionally wired in a circuit,
there is always some "stray" capacitance between any two elements in the
circuit: for example, between a wire and the metal chassis, or between the
grid and plate of a vacuum tube, or between the base and collector of a
transistor. Such stray capacitance is rarely shown on circuit diagrams but,
nevertheless, can often be extremely important, particularly at high
frequencies-for example, at tens or hundreds of MHz. The reason is that at
high frequencies capacitors tend to act as "low resistances." In the limit as
the frequency increases the capacitance acts like a short circuit. We will
show this in Section 2.4
When a capacitor is charged with a voltage difference V between the
plates, there is an electric field present in the region between the plates.
Energy is stored in this electric field, and work must be done to create this
field-that is, work must be done to charge up the capacitor. Conversely,
when the capacitor discharges, the energy stored in its electric field must go
somewhere; it must be dissipated as heat, or it must be stored in some other
form, such as the energy of a magnetic field. The energy stored in a
capacitance C with a voltage difference V between the plates is W =
tCV 2 • The derivation is short:
W =
v V dQ
Io
=
IV V d(CV) =
0
C
IV V dV = -1 CV
0
2
Because Q, the charge on either plate, is related to V by Q
be rewritten as W = QV/2 or W = Q2/2c.
=
2
(2.17)
CV, W can
2.4 CAPACITIVE REACTANCE
Suppose a sinusoidal voltage v = Vo cos wt is applied across a capacitor as
shown in Fig. 2.12(a). Then the current through the capacitor can be
obtained by taking the first derivative with respect to time of the equation
Q = Cv:
2
C2
Thus the total capacitance of two capacitors in series is the reciprocal of the
sum of the reciprocals of the capacitances. This result is analogous to that
.
1=
dQ = -d
d (Cv) = -(CVocos
d
. wt = wCVocos (7T)
_d
wt) = -wCVosm
wt +­
t
t··
dt
2
(2.18)
66
CHAP. 2
i(t)
Alternating Current Circuits
i(t) = wCVOCOS{wt+;)
i{t)
SEC. 2.4
Now we would like to be able to write an ac version of Ohm's law that
takes into account the phase as well as the amplitude of the current through
the capacitor. The current can be written in a form similar to Ohm's law
(i = vIR) by writing
i(t)
C
where we have used
i{t)
(a) circuit
COS
wt
(b) waveforms
FIGURE 2.12 Current and voltage in a capacitance.
where we have used the trigonometric identity
cos(wt + 'lr12) = -sin wt
Thus, we see that the current i varies sinusoidally with time at the same
frequency w as does the applied voltage but is 90° or 1T/2 rad out of phase
with respect to the voltage. The current through the capacitor leads the
voltage across the capacitor by 90°, or, equivalently. the voltage lags the
current by 90° or 'lr12 rad, as shown in Fig. 2.12(b). Notice also that the
higher the frequency w of the applied voltage, the larger the current for a
given voltage; that is, at higher frequencies the capacitor presents less
opposition to the current flow.
All these features can be consolidated very neatly by introducing
complex numbers. For a brief summary of complex numbers see Appendix
E. We now regard the applied voltage Vocos wt as the real part of Voe i"",
which we can call the complex voltage (F equals -1, and e i6 = cos 8 +
j sin 8):
v(t) = Vo cos wt = real part of [Voe i""]
(2.19)
The current can also be written as the real part of a complex number:
::=
wCVocos( wt
Re[wCVoeiw'ei"IZ]
ei"12 ""
+
where Re stands for "real part oL"
m=
Re[wCVoe i (Wlh/2)J
(2.20)
= Re[jwCVoe i""]
(2.21)
j. Then we can write
i(t)
lI(t) = Vo
i(t)
67
Capacitive Reactance
= Re [
Voei""j
_.1_
}wC
(2.22)
The expression says that the current through the capacitor equals the
voltage divided by I/jwC, which depends only on the capacitance and the
frequency of the applied voltage. In other words, the current equals the real
part of the ratio of the complex voltage to the complex number 1/jwC.
Thus, 11 jWC plays the role of the "effective resistance" or "impedance" of
the capacitor to current flow: the larger II jwC, the smaller the current for a
given voltage. The value I/jwC is called the capacitive reactance, usually
denoted by Xc, and is measured in ohms. The capacitive reactance can be
thought of as the ac frequency-dependent resistance of a capacitor. By
convention we usually omit writing "real part of":
i
i(t) = vet) = Voe ""
Xc -1­
(2.23)
jWC
It is important to realize that once we define capacitance by C = QI V,
we are forced to the conclusions that the current and voltage are 90° out of
phase, and that the current through a capacitor equals the voltage across it
divided by II jwc. The presence of the complex number j simply takes into
account the phase of the current relative to the voltage. Notice that because
the current through and voltage across a capacitor differ in phase by
<p = 90°, the power dissipated in the capacitor is zero because P =
Irms V nn. cos <p
[rm. V nnS cos 90° = O. This is in complete contrast to a
resistance where the current and voltage are in phase, and electrical power
is dissipated as heat. Notice also that in the limiting case of zero frequency,
which is direct current, the capacitive reactance goes to infinity; that is, a
capacitor can pass no direct current.
It is often desired to block out the dc component of a voltage while re­
taining the ac component. A series capacitor called a blocking capacitor or
a coupling capacitor does the trick, as shown in Fig. 2.13. If any dc current
Ide flowed through C, there would be a dc voltage drop across R equal to
68
CHAP. 2
'""r
c
V
,
I
,
Uin:
Vt.b -
I
I
I
1 Vout
I
I
I
I
t
Alterneting Current Circuits
1\
..
\"/V t
1\
Vt.b --­
'.~--
t
...t
FIGURE 2.13
Ide R
Coupling or blocking capacitor.
. But the capacitor cannot pass any dc current; hence
Ide
=0
and
Vdeout
=0
2.5 INDUCTANCE
Inductance is also important in ac circuits. A circuit element that has
inductance is called an inductor, a choke, or sometimes an inductance. It is
represented in circuit diagrams as a coil (see Fig. 2.14).
Actual inductors are usually made by winding wire as a coil around
some kind of core; the two terminals are the two ends of the coil. The
definition of inductance is
L=~
(2.24)
I
the ratio of magnetic flux ¢ with respect to current I. (We recall that
magnetic flux equals the magnetic field induction times area, ¢ = BA.) The
current I flowing through the coil produces a magnetic flux ¢ through the
cross-sectional area of the coil. The larger I is, the larger ¢ is, so L is
----'000'­
(a) schematic symbol
FIGURE 2.14
Circuit symbol for inductance.
SEC. 2.5
Inductance
69
positive. A more easily understood definition of inductance is in terms of
the voltage drop v across the inductor; L is defined from the equation
v = - L dI /dt. These two definitions are, of course, equivalent, as can be
seen from the Faraday law expression for the voltage produced by a
changing magnetic flux. Starting with Faraday's law, v = -d¢/dt =
-d/dt(LI) = -LdI/dt. If the current changes through the inductor at a
rate of one ampere per second, and the voltage produced across the
inductor is one volt, then the inductor has an inductance of one henry (H).
Typical units used for inductance are the millihenry (mH), which equals
10- 3 H, and the micro henry (pH), which equals 10-6 H.
The physical reason for the inductance is that the current flowing in the
coil produces a magnetic flux through the coil. As the current changes
(dI/dtjO), the magnetic flux through the coil also changes d¢/dtjO;
hence a voltage is induced by Faraday's law (induced voltage or emf =
-d¢/dt) between the two ends of the coil. In other words, whenever the
magnetic flux lines "cut" the wire turns of the coil, a voltage is induced in
the coil. The minus sign means that the voltage induced is of a polarity to
oppose the change in current that produced the voltage. We can see from
Faraday's law that a direct current (I = constant and dI / dt = 0) will
produce no voltage across the coil terminals; the only voltage drop
produced by direct current I through the coil will be the Ohm's law drop
V = IR, where R is the resistance of the wire in the coil, which is usually
negligibly small. For example, a low-power, 1-H iron-core inductance may
have a dc resistance of about 500; hence 10 rnA dc flowing through the
inductance will produce a steady voltage drop across it of only ! V.
Inductance ideally depends upon the number of turns on the coil, the size of
the coil, and the material inside the coil. It is usually independent of current
or voltage.
Solid or powdered iron is often placed inside the coil to increase the
inductance. The iron produces more magnetic flux ¢ through the coil for a
given number of ampere-turns. However, eddy currents always flow in the
iron core, which has some resistance; thus I2 R power is dissipated as heat
in the core. This power comes from the ac signal in the coil, so that ac
signal is attenuated or reduced. The higher the frequency of the ac, the
larger the power loss. It can be shown that the power loss is linear with
frequency-doubling the frequency exactly doubles the power loss to the
core. This power loss can be reduced by using a powdered iron core in
which very small particles of iron are cemented in an insulating glue to
reduce the eddy currents. Powdered-iron-core inductances can be used at
frequencies up to several tens of MHz. For higher frequencies air-core coils
are usually used. Note that iron-core inductors often change their in­
ductance if the dc current through the coil is changed. This is due to
magnetic saturation t of the core (see Fig. 2.15). L = ¢/I, and if saturation
occurs, L clearly decreases.
(b) actual inductor
tSaturation means a flattening or leveling off of the <l>-vs.-I curve.
70
CHAP. 2
Alternating Current Circuits
4>
through
core
SEC. 2.5
71
If two inductors are connected in parallel (this is almost never done in
actual practice because of magnetic coupling between the two inductors) as
shown in Fig. 2.16(b), the resultant can be regarded as a single inductor of
inductance L = 1/(11 Ll + 1/ ~). This result follows from I = II + I z.
Therefore,
---
lowerL~
dI
dt
dI
dt
dI
dt
- = -I + ­z
By the definition of inductance dIldt v/L, dI1 1dt vdLb dIzldt =
vz/~, and, because Ll and ~ are in parallel, VI = Vz
v. Therefore,
I through coil
L=~
I
FIGURE 2.15
Inductance
V
L
Inductance changes with changing dc current.
V
V
Ll
~
-+-
or
1
-
L
1
1
+­
Ll ~
or
1
L=--­
I
Ll
If two inductors of inductance L, and ~ are connected in series, as
shown in Fig. 2.16(a), then we can show that the resultant can be regarded
as a single inductor with an inductance equal to LI + ~. This result follows
from the argument
v=L
dI,
Ll dt
+ Vz
VI
Hence,
(2.25)
Vl+VZ
dI
dI
dI
+ ~ - z = (L I + ~) - = L
dt
dt
dt
L = Ll + Lz
(2.26)
VI
VI
V2
l----·'\
/1 "
l-----'\
I
-. L,
' ....
_------­
y
L2
L"".,
L,
+
L
L,
+~
Lz
(2.27)
It can be shown that an air coil of wire of length I, radius R, and N turns
dI
dt
But VI = L 1 (dIddt) and V2 ~(dIzldt). And because Ll and ~ are in
series, II = 12 = I and dId dt dIzl dt = dI I dt. Therefore
v =
LI~
or
+
L,
(a) series
FIGURE 2.16 Series and parallel inductances.
L"Hal
=
I
I
-+Lt
L2
LIL2
Ll + L2
(b) parallel
has an inductance of
L
=
N2
J..to-1TR 2
1
It can be shown that for a given length and cross-sectional area, a wire
of rectangular cross section (a flat strip) has much less inductance than a
wire of circular cross section. Hence, flat metal strips are often used in very
high-frequency circuits instead of ordinary round wire, particularly for
ground connections. Even a perfectly straight piece of wire has an in­
ductance because of the finite cross-sectional area of the wire. For example,
a 4-in. (== 10-cm) long piece of No. 18 (0.04-in. diameter) wire has an
inductance of 0.1 1JoH. This inductance can be very important in determin­
ing circuit behavior at high frequencies in the tens or hundreds of MHz.
Current flowing in one part of the wire creates a magnetic flux <t> in other
parts of the wire, and this l1ux changing in time affects the current flow in
the wire. The inductance also will change with frequency, because at higher
frequencies the current tends to be concentrated near the surface of the
wire and not in the interior; this phenomenon is called the skin effect.
Electromagnetic fields do not penetrate into a good conductor. They
decrease according to e- x / o, where x is the depth of the conductor and l5 is
the skin depth of the conductor. It can be shown that l5 is given by
l5 == (1T/alJo)-1!2, where / is the frequency of the electromagnetic field, a is
the conductivity, and IJo is the permeability of the conductor. The skin depth
in copper at 60 Hz is 0.85 mm, and at 1 MHz it is only 0.066 mm.
SEC. 2.6
72
CHAP. 2
When an inductance L has a certain current I flowing through it, there
is a magnetic field present in the region around the inductance. Energy is
stored in this magnetic field, and work must be done to create the field; that
is, work must be done to increase the current flowing through the in­
ductance from zero to I. Conversely, when the current through the in­
ductance falls to zero, the magnetic field decreases to zero, and the energy
must go somewhere. It must be dissipated as heat or stored in some other
form, such as the electric field energy in a charged capacitor. The energy
stored in an inductance through which a current I is flowing is given by
W = tLI2. The derivation is short:
w=
f
V dQ =
f
(L
~~) (I de) =
L
r
I dI =
i
LJ2
In the previous discussion of inductors in series and parallel, we have tacitly
assumed that the magnetic flux lines, which cut the wires of an inductor and
induce a voltage, come from the inductor itself and not from any other coil.
But magnetic flux lines from any source will induce a voltage in a wire if
they cut the wire. Thus we expect a changing current in one conductor to
produce a voltage in a second conductor if the magnetic flux lines from the
first conductor cut through the second.
This concept of changing magnetic flux in one conductor being
produced by changing current in a different conductor leads us to the
concept of mueyal inductance between the two conductors. Consider the
circuit of Fig. 2.17 where we have two coils of inductances LI and ~ with
NI and N2 turns, respectively. Suppose a time-varying current II is flowing
in L I · Let <1>12 be the magnetic flux through coil 2 caused by the current II
flowing in coil 1. The voltage V12 induced in ~ will be given by Faraday's
law:
=-
_ ...
N 2 d<l>12
magnetic flux
FIGURE 2.11
coil2
Mutual inductance between two coils.
N2K dII
de
(2.30)
V12 = -
We define -N2 K as M 12 , the mutual inductance between coil Ll and
terms of M 12 ,
VI2
induced
voltage
due to
1
M 1 2dI
-dt
~.
In
(2.31)
M12 clearly has units of henrys, just like self-inductance. The mutual
inductance depends on the number of magnetic flux lines from Ll that pass
through ~ (i.e., on the geometry-the distance between the coils, the
relative orientation of the two coils, and the medium between LI and ~). A
little thought will show that the further apart the two coils are, the smaller
the mutual inductance is, and if the coils are oriented at right angles, the
mutual inductance will be minimum. If the coil axes are parallel, the mutual
inductance will be maximum. Any substance between the coils with a high
magnetic permeability, such as iron, increases the mutual inductance.
The self-inductance ~ also creates an induced voltage in coil 2 due to
the changing 12 • The total voltage V2 across coil 2 will therefore by given by
V2
A changing current in
a similar argument
=
dII
de
dI2
de
M12--~-
(2.32)
produces magnetic flux lines that cut L 1 ; hence by
VI
/
coil I "'_)
magnetism that the flux in ~
KIl> where K is a constant
coils, the number of turns in
coils. Therefore the voltage
(2.29)
de
~fE]v"
But we know from elementary electricity and
must be proportional to II' That is, <1>21 =
depending on the distance between the two
coil 1, and the relative orientation of the
induced in ~ is
(2.28)
2.6 MUTUAL INDUCTANCE
V12
73
Mutual Inductance
Alternating Current C/rcult.ll
=
dI2
M21
dt
dII
Ll dt
(2.33)
The mutual inductance M21 can be shown to equal M 12 • Two coils wound
close together so that most of the flux lines from one pass through the other
produces a transformer, the circuit symbol for which is shown in Fig. 2.18. If
both coils are wound on a common iron core, then almost all the magnetic
flux lines produced by a current in one coil will pass through the other coil,
as shown in Fig. 2.18(d). The practical details of transformers are discussed
in Appendix A.
74
CHAI'. Z
:JC Alternating CulTftnt Clrcuib
:JllC (a) air core
SEC.
:.'!
defines inductance:
v
f magnetic flux
(..---L-.. ___\
:J11C
I
I
L_--~;
=iBm=:
(c) solid (laminated)
iron core
Now
So
and
;(t) - ;(0)
If we choose t
= -1
;(t) - i(O) = 0 when
di
L- so
dt
= ;(t) -
d;
(d) actllal constructiol1
AGURE 2.18 Transformers.
=
r
(b) powered irOI1 core
core""""
75
Inductive Reactance
L
I'
i{t)
= Vo cos (wt _ .!: )
wL
2
wL
i{t)
Vo sin
wL
wL iW
i(t)
I'
Vo cos wtdt
0
wI
(2.35)
-Vo cos ( wt - -7T)
wL
2
0 and
(2.36)
where we have used the trigonometric identity cos(wt - 7T/2) = sin wt. The
current; varies sinusoidally with time at the same frequency w as does the
applied voltage and is 90° or 7T/2 rad out of phase with respect to the
voltage. The current in the case of the inductance lags the voltage by 7T/2
rad or 90° as shown in Fig. 2.19. Thus, the power dissipated in an
inductance as heat is zero because of the 90° phase difference between the
current and the voltage. Notice that the higher the frequency of the
applied voltage, the smaller the current; that is, at higher frequencies the
inductance presents more "opposition" to the current flow. The use of
complex numbers can be used to consolidate these features. Again, we
regard the applied voltage Vo cos wt as the real part of Voe iwr and write
t. =
+ Vo
L
the current equals zero, then ;(0)
VO COS ( wt v(t)
= -1
v dt
0
.( t) = -Vo.
sm wt
wL
Suppose a sinusoidal voltage v = Vo cos wt is applied across an inductance,
as shown in Fig. 2.19. Then the current; through the inductance can be
obtained by integrating with respect to time the v = L(di / dt) equation that
(2.34)
;(0)
l
2.7 INDUCTIVE REACTANCE
~ dt di =
f) =Re (Voe (W'-1T/2»)
L
i = Re(- jVoe ')
wL
i
w
using e -j'lT/2
- -}
-
•
- Va
v{t)
(a) circuit
FIGURE 2.19
Current and voltage in an inductance.
Va cos wt
; = Re(~oejw') JwL
(b) waveforms
where we have used - j
= 1/ j.
(2.37)
CHAP. Z
76
Alternating Current Cin:uits
This expression for the current is in a form similar to Ohm's law; the
current equals the real part of the complex voltage divided by the complex
number jwL. Thus jwL plays the role of the effective resistance of the
inductance to current flow-the larger jwL, the smaller the current for a
given voltage. The inductive reactance XL is defined as jwL and can be
thought of as the ac frequency-dependent resistance of an inductance. The
presence of the complex number j simply takes into account the phase of
the current relative to the voltage. By convention, we usually omit writing
Re and simply write
Voe i'" = vet)
jwL
XL
SEC.
Z.'
plane by a point that moves counterclockwise in a circle of radius Vo with
an angular frequency w. At any time t the actual voltage Vo cos wt is the
projection of the complex voltage Voe i'" on the real voltage axis. Such a
diagram is often referred to as a rotating vector diagram or phasor diagram
(see Fig. 2.20). The vector drawn from the origin out to the point Voe i'" is
called the voltage vector or phasor. The voltage found in the actual circuit is
the projection of the rotating complex voltage vector on the real axis.
imaginary
voltage axis 1
v
(2.38)
Notice that in the limiting case of zero frequency, which is direct current,
the inductive reactance goes to zero, which means that inductance offers no
opposition to the flow of direct current. Direct current flowing through an
inductance is limited only by the resistance of the wire in the coil.
The reciprocal of reactance is sometimes called susceptance and is
measured in siemens or ohms-I. The susceptance of a capacitance Cis
Vo sin
~
wI
/
real voltage axis
Vo cos wi
IJ
B = jwC
77
The Comp'e. Vo'tage Plane
= Voe jwt = Va (cos wi
+j
sin
wI)
(2.39)
FIGURE 2.20 Basic rotating vector diagram.
The susceptance of an inductance is
B =
_1
(2.40)
jwL
Thus the current
exists is
a capacitor C across which a voltage difference v
i
= Bev
= jwCv
(2.41)
and the current through an inductance L is
v
i = BLv = jwL
One of the main advantages of this kind of diagram is that the phase
difference between two voltages (of the same frequency) is simply the
geometrical angle between their two rotating vectors. This will be parti­
cularly useful in analyzing phase-shifter circuits in which the output voltage
has been shifted in phase with respect to the input voltage. That is, if we
have two voltages
(2.42)
2.8 THE COMPLEX VOLTAGE PLANE
Let us now consider a few points about the representation of sinusoidal
voltages by complex numbers. The voltage Vo cos wt can certainly be
written as the real part of the complex exponential Voe;'" because Voe iw' =
Vo cos wt + jVo sin wi. We can represent Voeiu>t in the complex voltage
VI
VOl cos(wt + .pl) and
V2 =
V02 cos(wt +
.p2)
then at 1= 0 they would be represented by the diagram in Fig. 2.21. The
phase difference between V2 and VI is.p2 .ph the angle between VI and Vz;
V2 leads VI by .pz - .pl. VI + V2 can be calculated very easily from the
rotating vector diagram by adding VI and V2 using the standard paral­
lelogram method. The phase and amplitude of VI + V2 can be read off the
diagram immediately.
An example will be useful in analyzing RC and RL circuits. First,
consider two sinusoidal voltages VI and Vz 90° out of phase, with VI leading
V2. They could be drawn as in Fig. 2.22(a), and their sum would be drawn as
the hypotenuse. Notice that the sum VI + V2 leads V2 by 8 and lags VI by
(90
8). They could also be drawn as in Fig. 2.22(b) with VI + V2 along the
real voltage axis, or as in Fig. 2.22(c) with VI along the real voltage axis.
0
_
SEC. 2.9
imaginary
voltage
axis
~~-,
~
~------;/l
I/
I
III
79
ftC High-Pass Filter
One voltage obviously can be drawn at any angle in any quadrant, but the
voltage triangle and the phases between any two voltages are the same. In
all cases the relative phase of the voltages is the same: V2 lags VI + V2 by 8,
and VI leads VI + V2 by 900 - 8.
+ II,
/
2.9
/ 111
real voltage axis
FIGURE 2.21 Two sinusoidal voltages with different phases and their sum at a certain
instant of time.
VI
VI
(a)
(VI
+ U2)
RC HIGH-PASS FILTER
Consider the circuit of Fig. 2.23 in which the output is taken across the
resistor. The gain of any circuit is defined as the output divided by the
input. The attenuation of any circuit is defined as the reciprocal of the gain.
The voltage gain of the RC high-pass filter equals ~/vl and clearly
depends on the frequency of the input. At very low frequencies the
capacitor presents a very high reactance, thus giving a small output; and at
very high frequencies the capacitor is essentially a short circuit, thus
making the output nearly equal to the input. In other words, low frequen­
cies will be attenuated, and high frequencies will be passed without much
loss in amplitude, with the gain approaching 1.0 as the frequency increases
without limit (w ~ 00). For this reason the circuit is called a high-pass
circuit or filter. The gain is zero for dc (zero frequency) because the
capacitor passes no direct current. If a dc voltage is applied at the input,
once the transients have died out all the dc input voltage will appear across
C and none across R, thus giving zero output.
Let us calculate the gain, assuming that a negligible amount of current
is drawn from the output terminals and that the input voltage is VI
VOl cos wt Re( VOl eiwt ). Then the ac current i will be as shown in Fig.
2.24(a). For this sinusoidal input of angular frequency w we need only
replace the capacitor by its complex capacitive reactance Xc = 11jwC and
treat the circuit as follows [see Fig. 2.24(b)].
iR
VI
i(Xc + R)
(b)
i
1
R
iR
Vl
(_.1
+ R)
JWC
1
jwC+ R
1+
1
(2.43)
c
~
+I
I
I
input
(e)
FIGURE 2.22
78
Voltage addition of v, and
V2
90" out of phase.
1 + jwRC
jwRC
FIGURE 2.23
I
I
II,
i,
I
R
I
I
I u~ =
I
I
I
I
I
+
f
RC high-pass filter.
output
SEC. 2.9
80
CHAP. 2
C
20
jwC
•,
•
I
I
I
I
I
I
I
R
: Vz
R
,,
I
I
t
logl~lt
20
log
We
Ws
OdB~---+--------~
log
:IJ~
I
I
-3 dB ,.-------
W
o
-3 dB
+
(b) equivalent circuit
(a) actual circuit
FIGURE 2.24
81
RC High-Fa., Filter
A'temating Current Circuits
RC high-pass filter with source.
6 dB/octave slope
(20 dB/decade)
The fact that the voltage gain is complex merely means that the output
differs in phase from the input. The magnitude or absolute value of the gain
will tell us the magnitude of the output divided by the magnitude of the
input:
==
VI
(V2)*(V2)
Vi
VI
= ( -jwRC)(
1 - jwRC
2
2
2
jwRC ) == w R C
1 + jwRC
1+
C2
(2.44)
FIGURE 2.25 Gain vs. frequency for RC high-pass filter.
filter with R == 10 kfl and C
Wa
wRC
v'l+~2R2c2
(2.45)
Notice that this expression for the gain approaches 1.0 as the angular
frequency w of the input approaches infinity, which agrees with our in­
tuitive feeling that the capacitor acts like a short circuit at very high
frequencies.
Also notice from (2.45) that for very "low" frequencies (wRC <t 1) the
gain decreases as the first power of the frequency or by a factor of 2 each
time the frequency decreases by a factor of 2.
I~I == wRCoc w
(2.46)
If we plot a graph of the magnitude of the gain versus frequency on
Jog-log paper, we obtain Fig. 2.25. The curved graph [Fig. 2.25(a)] of the
actual gain can be approximated quite well by two straight lines [Fig.
2.25(b)], giving a sharp break or "knee" in the graph. The frequency at
which this break occurs is given by Wa = 1/ RC. This is called the break­
point frequency or usually simply the breakpoint. For example, a high-pass
·lndicates complex conjugate.
(b) straight-fine approximation
(a) actual graph
lIRC =
= 0.1
f-LF will have a breakpoint at
0)(10- 7 F)
rad/s, or fa
Wa/27T
= 160 Hz
At the breakpoint the voltage gain IV21vII = 0.707, as can be seen by
substituting w = 11 RC in the expression for \v2Ivll· As a gross sim­
plification, the RC high-pass filter can be thought of as passing frequencies
above Wa = 11 RC and attenuating those below Wa·
lt is common practice to express gain (or attenuation) in decibels (dB).
The voltage gain v21vi in decibels is defined as
== 20Iog lO (V2)
( V2)
Vi indB
VI
(2.47)
The phase difference (if any) between V2 and VI does not affect the gain
expressed in decibels. Note that a negative gain in decibels merely means the
output is less than the input; a positive gain means, of course, that the
output is greater than the input.
I~:I
in dB
\~l
in dB
0.01
0.1
1.0
10.0
100.0
-40
-20
0
+20
+40
0.5
0.707
1.414
2.0
4.0
-6
-3
+3
+6
+12
82
Alterneting Cu"ent Circuits
-c
Often the power gain is expressed in decibels according to
== 1010g lO (Pz)
(Pz)
PI indB
PI
~
(2.48)
---...,.
imaginary
voltage
axis
UR =
\
U1 ,
If the impedance levels are the same at the input and output, then
Pzlpi = dlvT and the voltage and power gains in dB are equivalent.
Pz) = 1010glO(Pz) = 10 log 10 (V2)2 = 2010g lO (V2)
(PI
dB
PI
VI
VI
•
83
RC Low-Pen Filter
SEC. 2.10
CHAP. 2
I,
.
real voltage axis
-.
.( I) = I.( wC
-1)
Uc = iXc = I iwC
v;+vc
~
(2.49)
Thus a power gain of 20 dB means the output power is 100 times the input
power and the output voltage is 10 times the input voltage.
Notice that because the log of the product of two gains is equal to the
sum of the logs of the individual gains, the total dB gain of two filters,
amplifiers, or whatever in series is equal to the sum of the individual dB
gains. A lO-dB amplifier driving a 30-dB amplifier has a net gain of 40 dB,
provided the second amplifier does not "load" the first one. In other words,
the second amplifier should not draw too much current from the output of
the first amplifier. More on this when input and output impedance are
discussed in Chapter 5.
Expressed in decibels, the voltage gain or power gain at the breakpoint
for an RC high-pass filter is -3 dB or 3 dB "down," relative to the gain of
1.0 at high frequencies. If the gain is down 3 dB at the breakpoint, the
voltage gain is down by a factor of 0.707, and the power gain is down by a
factor of 2. At lower frequencies the voltage gain continues to fall off at a
constant rate of 6 dB per octave frequency change or, equivalently, 20 dB
per decade (x 10) frequency change according to the straight-line ap­
proximation to the gain curve. For example, if the gain at 1000 Hz for an
RC high-pass filter is 10 dB down, then the gain at 500 Hz will be 16 dB
down; at 250 Hz it will be 22 dB down; and so forth.
The phase of the output is different from the phase of the input. This
can be seen most easily from a rotating vector diagram. Choose the phase
of the current as the reference phase, and recall that the current leads the
voltage in a capacitor by 90°. Therefore, the voltage VR = iR must be 90°
more counterclockwise (ahead of vd. Thus, we have the diagram of Fig.
2.26 because VI = VR + Vc. Thus, the output voltage across the resistor
leads the input voltage VI by (J, which is given by (J = tan-l(ivci/ivRi) =
tan- l (l/wRC). Notice that as the frequency increases, the phase shift (J goes
to zero, and that as (J increases, the output voltage decreases. Sometimes
this circuit is used to shift the phase of a sinusoidal signal; the amount of
phase shift can be varied by using a variable resistor for R. When R =
1/wC, (J = 45° and iV21vd = 1/.J2 = 0.707. However, as a phase shifter this
circuit has two disadvantages: The amplitude of the output voltage changes
as R is changed, and also the maximum phase shift possible with thi~ circuit
is 90° in the limit as the output voltage goes to zero.
iR
(b) voltage vector diagram
(a) circuit
l~\
(J
1.0
0.707
90°
45°
w
w
Ws
Ws
= RC
= RC
o
(d) gain vs. frequency
(e) phase shift vs. frequency
FIGURE 2.26
2.10
RC phase shifter.
RC LOW-PASS FILTER
Consider the circuit of Fig. 2.27, in which the output is taken across the
capacitor. The voltage gain v21vI clearly will go to zero at very high
frequencies because the capacitor acts like a short circuit at high frequen­
cies. At zero frequency (dc) the gain will be unity if we assume no current is
drawn from the output. The circuit is therefore called a low-pass filter
because it passes the low frequencies and attenuates the high frequencies.
For a sinusoidal input of angular frequency w, the voltage gain (assuming
·
I
I n
I
j
,
I
R
+I
I
I
I
input
~ .,
I
C
o
fiGURE 2.27
I
I
RC low-pass filter.
\ •• ~ output
0
84
CHAP. 2
Alternating CUrTent Circuits
no output current) is
V2
Vt
iXc
i(R + Xc)
-~
i(ic)
wC
1 )
R
i (R + jwC
j
1
1 + jwRC
(2.50)
wC
The complex gain means that the output has been shifted in phase relative
to the input. The magnitude of the gain is
1::1
[(:r C:)f [c _j~RC)C
n
=
+
SEC. 2.1D
1~lt
20 log
1 ~\
WB
OdB
3 dB ...
log w
imaginary voltage aKis
R
(2.51)
20 log
85
or tube. Thus, we can see that if we are "stuck" with a certain minimum
capacitance between a signal-carrying wire and ground, and if we wish to
maximize the high-frequency response of the circuit, then we should take
care that the effective series resistance R is as small as possible. We will
discuss this in more detail in Chapter 7.
The difference in phase between the input and the output can be
calculated most easily from a rotating vector diagram. Choose the phase of
the current as the reference phase and recall that the current leads the volt­
age in a capacitor by 900 • Thus we have, assuming iou, = 0, the diagrams of
Fig. 2.29. The output is seen to lag the input in phase by an angle
j~RC)r2 =~2R2C2
which goes to unity as w ~ 0, and which goes to zero as w ~ 00. If we plot
the magnitude of the gain versus frequency on log-log paper, we obtain Fig.
2.28. The curved graph of the actual gain can be approximated by two
straight lines, giving a sharp break or knee in the curve at the breakpoint
frequency w l/RC £<.Ia. At the breakpoint the voltage gain is down by
0.707 or 3 dB down relative to the gain of unity at zero frequency. You see
this by substituting w == 1/ RC in the expression for the gain. The slope of
the straight line is 6 dB per octave. That is, if the voltage gain at 1000 Hz is
down 12 dB, then the voltage gain at 2 kHz is down 18 dB.
The RC circuit basically accounts for the decrease in gain with
increasing frequency for all amplifier circuits, so it is worth studying in
detail. The capacitance C is often the stray capacity between the circuit
wiring and the chassis, or it may be an inherent capacity built in a transistor
RC I.ow""a.. RItM
·
i
l
J
I
I
VI
I
I
tI
C
UR
I
I
I i
i
o
= iR
real voltage aKis
V2
V2
-i(~~)
V2=iXC
0
->
VI
VR
+
-.
V2
(b) vollage veclor diagram
(a) circuit
¢
I?I 1.0 -\----..::.--
90°
0.707
45°
I
I~I
Ws =
RC
(e) phase shifl vs. frequency
w
I
(Us
w
RC
(d) R"ain vs. frequency
Wn
o dB
FIGURE 2.29
3 dB J..
RC phase shifter.
0
cf> = tan-t(vRlvd
tan- t wRC. Notice that as the frequency increases, the
phase shift cf> goes to 90" and the output voltage goes to zero in amplitude.
If either the resistance or the capacitance is made variable, the phase shift
(a) actual graph
(b) straight-line approximation
FIGURE 2.28 Gain vs. frequency for RC low-pass filter.
can be varied, but the amplitude of the output varies with the phase shift.
The maximum phase shift is 90°.
Most four-terminal networks composed of simple combinations of R,
L, and C with a resistive source and load have gains that vary with
frequency and contribute a frequency-dependent phase shift; that is, the
phase difference between lhe output and the input is a function of the
86
CHAP. 2
Alternating Current Circuits
frequency. If the network contributes the minimum possible phase shift for
a given gain-versus-frequency behavior, the network is termed a minimum
phase-shift network, and for such networks the gain characteristics and the
phase characteristics are not independent. Knowing one characteristic
enables you to calculate the other. Fortunately, most simple four-terminal
networks fall into this class. Examples are the simple RC and LR low- and
high-pass filters.
The exact mathematical theory of minimum phase-shift networks is
beyond the scope of this book, and the interested reader is referred to
Bode's theory in the Radio Engineer's Handbook (first edition) by Fred. E.
Terman, McGraw-Hill (1943, p. 218). The essence of the theory is that the
phase shift produced by the network at a frequency f depends on the rate of
change of the network gain with respect to frequency evaluated at that
frequency f. In other words, the network introduces the greatest phase shift
in a frequency range where the network gain is rapidly changing. In the
simple RC low-pass filter, for example, at high frequencies (f ~ 1I21TRC)
the gain falls off linearly with frequency (6 dB per octave or 20 dB per
decade), and this corresponds to a phase shift of 90°. For an LC network
with the gain varying at 12 dB per octave, the phase shift is 180°. Gain and
phase shift characteristics are shown for various networks in Fig. 2.30.
It takes considerable ingenuity to produce a simple network that is not
a minimum phase-shift network. One such network is the phase shifter in
Fig. 2.31, in which the gain is absolutely constant from dc to infinite
frequency, and the phase shift varies from 0° to 180°
UJ
o dB
c
.;;;
In general, practical electronic circuits consist of combinations of resis­
tance, inductance, and capacitance as well as tubes and transistors. By a
"brute-force" solution of the Kirchhoff voltage equation for a simple series
combination of R, L, and C, we will now show that the total impedance
(Hnet opposition to the current flow") offered by R, L, and C is given by
the magnitude of the complex sum of the resistance R, the inductive
reactance jwL, and the capacitive reactance I/jwC. The impedance Z of
any collection of R, L, and C is the complex algebraic sum of R, jwL, and
11jwC, treating each term as a "resistance" and combining terms in series
or parallel as the circuit implies. Z is measured in ohms but is complex.
Thus, the impedance of Rand L in series is Z
R + jwL, and the
impedance of Rand C in parallel is R x (l/jwC)/(R + I/jwC). The mag­
nitude 10 of the current through R, L, and C in series is given by
10 = Vo/IZI, where Z
R + jwL + IljwC is the total impedance, Izi
..JR2 + (wL - l/wC)2, and Vo equals the magnitude of the applied voltage.
Consider the series RLC circuit of Fig. 2.32. The applied voltage is
Vo cos wt; we are given R, L, and C and are asked to calculate the current
log w
3 dB
R
"~'P"' I
OIl
E
UJ
0°
'-
.,'"
'"
.c
'0."
I
RC
= lL>o
\1
I
I
:2
..
c
o
0
log w
_45°
_90 0
(a) RC low pass
UJ
=
UJe
=
..
RC
i -~::t--ml log w
C
0---11
t
0
,"P~'P"' '­
..,'"
RLC CIRCUITS
RC
QO
:2
2.11
1
= lL>o
'"
.c
'"
0.
90° I
,!
45°
0° .,---­
log w
I
UJ
= "'"
RC
(b) RC high pass
c
;;;
OIl
1~
I
.:::
:2
.,'"
'"
'"
.c
0.
+I~
-\7
log
·
w
.
;'pIT'""' lllgw
(e) general network
FIGURE Z.30 Gain and phase shift for three circuits.
87
88
CHAP. 2
Ahern.ting Current Circuits
SEC. 2.11
89
RLC Circuits
differential equation
c
in:J
L
d2 j
output Vo
di
+R d +
t
j
.
e = - £1.1 Vo sm wt
R
-.L
(a) circuil
Vin
(b) voltage vector diagram
q,
90·
w=
w
As we have seen, the current has the same frequency as the applied voltage
for the capacitor alone and for the inductance alone; it is therefore a
reasonable guess in the RLe circuit in Fig. 2.32 that the current has the
same frequency as the applied voltage. However, the phase of the current
may be different from the phase of the voltage. To take this possibility into
account, we will assume a solution of the form i = 10 cos(wt - 8) and solve
for 10 and 8. 10 and e are both assumed to be constants. 10 is the magnitude
of the current; 8 is the phase of the current relative to the voltage. If e is
positive, then the current lags the voltage. Of course, the final justification
for the form of the assumed solution lies in our being able to find the
solution that satisfies the differential equation and the boundary conditions.
10 cos(wt e) in the second­
Now we substitute the assumed solution i
order differential equation in i and try to solve for 10 and 8. We obtain one
cos wt term and one sin wt term after a little algebra:
(c) phase shifl vs. frequency
FIGURe 2.31
Phase shifter.
10
wle) cos 8 Rsin 8] cos wt
+ I In (WL
R
1
Vo sin wt
=0
(2.53)
Because the functions cos wt and sin wt are orthogonal, the coefficient of
each iIlUSt equal zero. Another way of seeing this is to note that equation
(2.53) must hold for al\ values of t. In particular, when t 0, sin wt 0, so
the coefficient of cos wI must equal zero; and when t = 7[/2£1.1 T/4,
cos wt ;; cos( 7[/2) is zero, so the coefficient of sin wt must also equal zero.
By either reasoning we obtain two equations that can be solved for 10 and 8.
L
c
FIGURe 2.32
wle) sin e + loR cos (J -
(WL - ~e) cos 8
Series RLC circuit.
R sin 8
0
i. The Kirchhoff voltage law is
and
Vo cos wt ­
VR
VL -
Vc
0
or
~7
'R
I' 0 cos wt ;; I
10 (wL
wle) sin e + loR cos e-
Vo
0
+ L -di + Q
dt
e
(2.52)
which when differentiated once with respect to time yields a second-order
Hence
1
wL- we
tan e =
R
(2.54)
CHAP. 2
90 L
Vo
o
AItern'fing Current Circuits
1
SEC. 2.11
91
RLC Circuits
R
(2.55)
R cos 6 + ( wL - wC) sin 6
-------......
1
From equation (2.54) we find 6 = tan- (wL - l/wC/R). That is, if
wL> l/wC, then the current lags behind the applied voltage by 6, or,
equivalently, the applied voltage leads the current by 6. From equation
(2.55) and the impedance triangle shown in Fig. 2.33, we obtain
Vo
10 =
)2]1/2
1
[R2+(WL- wC
(2.56)
L
jwC
C
(b) equivalent circuit
(a) actual circuit
FIGURE 2.34
which is very interesting. 10 , the magnitude of the current, equals V o, the
magnitude of the voltage, divided by [R2 + (wL 1/wC)2]lI2, which is
precisely the magnitude of the complex number [R + jwL + l/jwC). Thus, by
jwL
Series RLC circuit and equivalent.
The reciprocal of the impedance Z is called the admittance ¥,
1
(2.59)
¥= Z
and is measured in siemens or ohms-lor mhos.
The current flowing in such a series circuit, Fig. 2.34(b), can be written
down from Ohm's law:
V oe jw' Reos 8
(2.60)
1
i =
R + jwL + jWC
R
FIGURE 2.33 Impedance triangle for series RLC circuits.
treating the problem as a series connection of three complex reactances R,
jwL, and 1/ jwC, we can get the magnitude of the current without even
writing down the differential equation, much less solving it! In other words,
using the complex inductive and capacitive reactances provides us with a
quick shortcut to solving the differential equation for the current amplitude.
Let us show explicitly how the use of complex impedance simplifies the
job of finding the amplitude of the current. We replace the voltage
Vo cos wt by Voe i"", L by its inductive reactance jwL, and C by its
capacitive reactance 1/ jwC, as shown in Fig. 2.34. The total impedance is
thus
1
and Z
R + jwL + jwC
(2.57)
=
~ R2 + ( wL - w~r
(2.58)
The magnitude of the current is obtained merely by taking the magnitude
or absolute value of both sides of the above equation. If we let 10 be the
magnitude of the current, then we see that
1 _
\Voei"'l
0-
\R + jwL +
_
Vo
_ Vo
j~cl-l R2 + (wL - ~cr1112 -\Z\
(2.61)
which is the same result as was obtained by the relatively laborious
procedure of solving the differential equation.
To obtain the phase of the current, we will draw a rotating voltage
vector diagram and start by picking a reference phase. Draw VR
iR along
the real voltage axis. Because R, L, and C are in series, the current
through each of them is the same in magnitUde and phase. Thus the voltage
VL across the inductance is given by VL = iXL = ijwL which leads the
current by 900 • The voltage across the capacitance is given by Ve = iXe =
i/jwC. The rotating voltage vector Ve will then be Ve = -ji/wC which lags
92
CHAP. 2
Alternating Current Circuits
the current by 90". The rotating voltage vector diagram is shown in Fig.
2.35.
Notice that the voltage across the resistance is exactly in phase with the
current through the resistance. We now note that, at any instant of time, by
Kirchhoff's voltage law the voltage Vin must exactly equal the instantaneous
sum of VR, VL, and Vc. Thus we have Fig. 2.35(c). From this diagram the
SEC. 2. 12
Series and Parallel Resonance
93
2.12 SERIES AND PARALLEL RESONANCE
In the series RLC circuit 10 V o /[R2 + (wL - l/wC)2rl2; the current
magnitude 10 is maximum when wL 11 wC, and the maximum value of i is
given by lomax = VoIR. A graph of 10 as a function of angular frequency is
shown in Fig. 2.36. The frequency Wo at which wL l/wC is called the
resonant frequency and is given by Wo = I/JLC in angular frequency
(rad/s), or by to = 1I(2~) in Hz or cps. Also, VL and Vc are equal in
R
10
Voe i "'/( rv
c
i(t) =
10 COS
(wt - 0)
i(t)
L
c
(a) circuit
w
Wo =
(a) circuit
10
vs. frequency
/J
III,
(ilL -
IIR
lie
(b) current
real voltage axis
lie)
+ 90° rl-------::::.::=:==
real voltage axis
w
(b) phases of IIR. ilL and Ile
-90"
(c) voltage addition
FIGURE 2.35 Phases of voltages in a series RLC circuit.
(c) phase vs. frequency
phase 8 of the current relative to the applied voltage is given by () =
tan-1(wL - l/wC)/R. If wL is greater than l/wC, the current lags the
applied voltage by 6; if wL is less than 11 wC, the current leads the applied
voltage by 8. For later times (t > 0) all the voltage vectors rotate counter­
clockwise at the constant angular frequency w, with () remaining constant.
And, as usual, the actual voltages measured in the circuit will be the real
parts of the complex voltages we have diagrammed, that is, the projection
of the voltage on the horizontal real voltage axis.
To sum up, complex numbers provide a shortcut to finding the currents
in circuits without having to solve differential equations, and they auto­
matically take care of the phase differences among the various voltages and
currents.
FIGURE 2.36
Series RLC circuit.
magnitude and 180" out of phase at resonance, and Z = R at resonance,
which means that the only limiting agent to the current at resonance is the
resistance. Thus, if R is very small, the current at resonance will be very
large. The voltage at resonance across the inductance VL =: IOmax(wL) may
be extremely high (even greater than V o), perhaps high enough to cause
arcing in the inductance. Similarly, Vc = Iomax(1/wC) at resonance may be
large enough to destroy the capacitor. Notice also that the phase of the
current relative to the applied voltage changes from a hi.g to a lead as w
passes through Wo = I/JLC, because tan () = (wL - l/wC)/R, as shown in
Fig. 2.36(a).
94
CHAP. 2
Alternating Current Circuits
One application of the series RLC circuit is to attenuate a narrow band
of frequencies around Wo by the circuit shown in Fig. 2.37. The voltage gain
versus frequency has a relative minimum at Wo because the impedance
between A and B is a minimum at Wo from equation (2.58). At Wo the gain
equals R /(R + R,) = 1/(1 + R, / R), which can be much less than one if
R,~R.
SEC. 2.12
95
Series and Parallel Resonance
Therefore, at resonance, the current i = v / Z goes to zero, and the voltage
Vab goes through a maximum. Thus we have just the opposite behavior from
the series circuit.
Notice that the impedance changes from inductive to capacitive as w
passes through woo For w < Wo, Z = jwLfy, and for w> wo, Z = -jwL/y,
where y = 1(1 - w2 /w6)1. Because the voltage across the parallel LC circuit
is given by v = iZ, the phase of the output voltage relative to the input
voltage Voe iw1 changes by 180 as w passes through wo, as shown in Fig.
2.39(c).
0
i(l)
R,
A
R
•:
llout
I
I
i
I
lu....,
I
I
II
C
i(l)
FIGURE 2.37
R
R+ R,
I
I
t
~
B
Z
V ab -
Woo
Z
=
;;c
1
jwL+jwC
-
'C
jwL
jwL
1 - w 2 LC
1
wL-wC
Voe
lW1
.
-'>
Voe
iw1
IUabl
10
as
Zab -'>
00
=---;;;
1-­
I
(2.62)
I
Wo
w6
q,
V'+A
L
=
+ z..b·
The parallel LC circuit is sometimes called a tank circuit and is most
often used to select one desired frequency from a signal containing many
Consider the RLC circuit in Fig. 2.38 in which Land C are in parallel;
this circuit behaves rather differently from the series RLC circuit. Let us
calculate the current as a function of frequency. The parallel combination
of Land C has an impedance of
1 )
R+CL-~~)
Zab
R
(2.63)
III
"'0
Series RLC circuit to attenuate frequencies near
jwL (
Voe iw1
v
i=-=
lI'a
=
./LC
W
~
I ",- k"
(?) current vs. frequency
/~/
"(
_90 0
which approaches infinity as w approaches the resonant frequency woo
i (I)
j
+~
FIGURE 2.39
(b) voltage vs.frequency
W
I
(c) phase vs.frequency
Current and voltage as functions of frequency in a parallel LC circuit.
II)'
Voe
iwt
different frequencies. This selection is accomplished by the basic circuit in
Fig. 2.40 in which the voltage gain (Vou,/Vin) is maximum (= 1) at Wo =
l/JLC. This circuit is said to be tuned to the frequency woo Often either L
or C is made variable so that different frequencies may be selected or
tuned. This is precisely what you do when tuning in a radio station; the
tuning knob is usually a variable capacitor in a parallel LC circuit.
The parallel LC circuit will tend to oscillate. This can be seen by the
following physical argument. Suppose at t = 0 the capacitor C is fully
i(l) = 10 cos (wI - q,)
( "--'
c
i(t)
L
}-
\ ,'I\...­
\ / :)(1.\1
FIGURE 2.38
Parallel LC circuit.
()<\""'l,
"
I.)-J
\) (1 ft
\
.
\,
'\')
CHAP. 2
96
--0
97
Series and Parallel Resonance
SEC. 2.12
i = 0
Vout 1
r
c:=
Alternating Current Circuits
vin
~L 1.0"
)\
+Q
C
+ + 1+ +
r 1 I F-
"---"j"---'\...\ magnetic
field
:'
electric
field ';:jL
t,
C
\
-Q
t
L
,
,
I
{I
'<""--,,,: ~, __ ",/I
J
I
Wo
,Jf7;
0, W
(a) I
QC
Cv
2
(b) 0
2
2
FIGURE 2.40 Parallel LC circuit used to select frequencies near Wo.
<I <
-f
i max
charged and the current is zero. The energy of the circuit is then all in the
electric field between the capacitor plates. The capacitor C will start to
discharge through the inductance, as shown in Fig. 2.41(b). The current
flowing through L creates a magnetic field around L, and energy is stored
in this magnetic field. This increase in energy will be exactly balanced by
the decrease in the energy stored in the electric field between the plates of
C, the total energy of the LC circuit remaining constant. But by Lenz's law
the magnetic field lines cutting the turns in L induce a voltage across the
terminals of L of such a polarity as to oppose the increasing current.
Eventually the current will reverse direction because of this induced vol­
tage and will recharge C with the opposite polarity, as in Fig. 2.41(f). The
capacitor C will now discharge through L in the opposite direction, as in
2.41(f), eventually returning the circuit to the state of Fig. 2.4l(a) when the
entire cycle starts over again. This back-and-forth flow of current is called
oscillation, and the oscillations for an LC circuit will in principle last
forever with a constant peak-to-peak amplitude. If, however, there is any
resistance at all in the circuit, there will be a conversion of electrical energy
to heat energy in the resistance at a rate i 2 R. This loss of electrical energy
occurs regardless of the direction of the current flow. Thus the electrical
energy of the circuit gradually decreases, and the oscillations die out. In
actual practice, of course, there is always some resistance associated with
any LC circuit, mainly due to the resistance of the wire in the inductance.
The only exception would be a circuit constructed entirely from super­
conducting material, which, at this writing, must be kept at liquid helium
temperatures (=4 K) and, hence, is somewhat impractical. Thus, the oscil­
lations in a real circuit die out eventually. If constant amplitude oscillations
are desired, then some circuit must be devised to continuously replenish the
energy lost to heat in each cycle. This is precisely what one does in building
an oscillator, which will be covered in Chapter 7.
It can be shown that the current in a pure LC circuit (no resistance) is
exactly sinusoidal, constant in amplitude, and of angular frequency
Wo
=
1
...rrc
it" f"~,\ f...... "
~l
!I/ t
." !\ \\
f
Q = 0
'-",---'t '
/
f,
tt
C
t
\
\
\
\
(
I
Ii'
..,--;.'
-:::::_/
\
I
\
I
C
;
L
,,
/:
I
I
I
/----1'----' +
i
\
L
;
\
"'",
\
\
/
,,
I
\.
. . ~- ...... ~
I
,
I
'~--~I'
/
i = i max
(c) I
=
I. W _
2
Li m.,
I.
< I < I.
4
2
(d)
----y-
4'
o
" ......... ~ ..... ~ /"'--"'-,
I
Q
I
I
I
1
C
1
L
C
-'-r-"--
L
~
\\
,
'",-",'
"
"-~"
\,,
•
:
//
o
(el
I
T w
2'
FIGURE 2.41
QC
(f)
2=
f<
I
Oscillation of current in parallel LC circuit.
To show this, we simply write the Kirchhoff voltage law for the closed loop
containing Land C.
Vc -
Vc
+ L
VL
= 0
di
=0
dt
Q
+ L di = 0
C
dt
(2.64)
SEC. 2.13
98
CHAP. 2
Alternllting Current Circuits
Differentiating once with respect to time gives us an equation in the
current:
d2 i
1
-+
-i=O
dt 2 LC
i
d2 i
-+L-=O
dt 2
C
(2.65)
where we have used i = dQ I dt. This is the familiar wave equation or
oscillator equation. The solution is
i = Iocos(wot
+ <1»
i = 10 sin(Wot + 8)
or
(2.66)
where Wo = l/.../LC. Thus, the current flowing in a pure LC circuit IS
sinusoidal, of constant amplitude 10 and of angular frequency Wo = l/.../LC.
Precisely the same differential equation is obtained for the motion of a mass
attached to a perfect, Hooke's law spring: m(d 2Xldr) = -kX where m is
the mass, X is the displacement, and k is the spring constant. Thus,
d 2X Idt 2 + (k Im)X = 0, which is also the familiar wave equation with
solution X = Xosin(wot + 8) = Xocos(wot + <1», where Wo = ,)klm. L is
seen to be analogous to m, and C to 11k.
2.13
Q (QUALITY FACTOR)
The Q or quality factor of any circuit can be defined as 27T times the energy
stored per cycle divided by the energy lost or dissipated (usually as heat) per
cycle.
Q
= 27T Ws
W
99
Q (Qulllity FlIctor/
cycle and WH is the energy converted to heat in one cycle. Assume
i = 10 cos wt is the current flowing in the circuit. The energy stored in the
2
magnetic field of L when a current i is flowing is W = 1Li • W will
therefore be maximum when i is maximum. Because i and Vc are 90° out of
phase, when i is maximum, Vc = O. Therefore when i = i max = 10 , all the
energy is stored in the magnetic field. W = 1Li~ax, and none is stored in
the electric field of the capacitor because Vc = O. Ws = 1LI6 and WH =
Jl pdt, where p is the instantaneous power converted into heat, p = i2(t)R,
and the period T = 27Tlw. Thus,
Ws
~LI6
Q = 27T W = 27T JoT i2(t)R dt
H
7TL
1
R JOTcos2 wtdt
7TL
R JOT
1
TI2
=R·
7TLI6
15 cos2 wt dt
(2.68)
wL
=R
Note that at the resonance frequency Wo, Q can be written as Q =
l/R,)LIC. Thus the higher the ratio of L to C, the higher the Q at
resonance. However, it is very difficult to make a large inductance
without also increasing R. And, as expected, the larger R is, the lower the
Q, because a larger R will dissipate more energy as heat.
For the parallel RLC circuit of Fig. 2.42, we show that the Q is also
given by Q = wL I R. Here we explicitly draw the resistance of the wire in
the inductance by including R in series with L. But we can always write
il(t) in the form il max cos(wt - 8), so then Q becomes
·2
7TL Ilmax
Q=
T
f
(2.67)
o
L
The energy stored refers to the energy stored in the electric and
magnetic fields. The energy converted from electrical energy to heat energy
is "lost" in the sense that it is no longer available in the electrical current.
The higher the Q, the less energy is converted into heat per cycle. An ideal
circuit that dissipates no energy would have an infinite Q.
Notice that the energy has been converted from a more ordered form
(i.e., the coherent average drift of electrons in the current flow in the wire)
to the more disordered random thermal motion of the molecules. This
conversion corresponds to an increase in entropy and is consistent with the
second law of thermodynamics, which says that the reverse process will
never occur with 100% efficiency-that is, the thermal energy can never be
completely converted back into the more ordered electrical energy. Thus,
some (and usually all in most circuits) of the energy converted into heat is
permanently lost from the electrical current.
We will now show from the above definition that the Q of a series RLC
circuit can be expressed as Q = wLI R. From the definition, equation
(2.67), we can write Q = 27TWs l WH , where Ws is the energy stored in one
=
7TL
=R
.2
2
II max COS (wt
t
T
- 8) R dt
1
cos2 ( wt - 8) dt
R
c=
Q
wL
R
Ws
Q=­
WH
Lj2mi.lx
2
Q = 21r
~L
calculation for a parallel RLC circuit.
(2.69)
2
i,
i2
FIGURE 2.42
7TL 1
R T
=-_.-
[T;i
(t)Rdt
wL
R
SEC. 2.13
CHAP. 2
100
Alternating Current Circuits
Ii
which is the same as for the series RLC circuit. Note that we could have
solved explicitly for il(t) by writing
J• • :: ...
.JR2 + w 2L2
R
t
----br h'lf-po~;
I
I
I
I
c
I
I
I
WI
0.W
Q
Parallel LC circuit with capacitor leakage resistance R.
then an ideal circuit with zero power dissipated as heat would have an
infinite R. Thus we would expect Q to be proportional to R. A calculation
from the definition of Q yields the result Q = R IwL or Q WoRC or
Q = R../C/L at resonance.
We will now show that for a series RLC circuit, the sharpness of the
current-versus-frequency curve depends on the Q: A high Q corresponds to
a narrow, sharp curve and a low Q corresponds to a broad curve. The
current as a function of frequency is given by
=
R
Vo
= Ioe '
[R2
+ (wL
_
wlcr]
Vo
1/2
-
WI
Wo
I .
..;Ie
Wo
FIGURE 2.44 Frequency response of series LC circuit.
~ R2 + (wlL w~cr = ~ R2 + (W2L w~cr =..fiR
(2.70)
= IZ!
Clearly 10 reaches a maximum when wL = l/wC (i.e., at resonance), and
lomax = VoIR, as shown in Fig. 2.44.
Let us now define the width tl.w of the Io-versus-w curve as the width
(2.71)
Thus, WI and W2 are the two roots of (wL I/wC)2 = R2. If we assume a
symmetrical curve about w = Wo and also that tl.w ~ wo, then substituting
W2 = Wo + tl.w/2 and Wl = Wo - tl.w/2, 'we obtain tl.w
wo/Q· In other
words,
Q= tl.w
'(...-8)
+ jwL + jWC
10
= W2
W
Wz
llw
Wo
i
Wo
from WI to W2, where i(wI) = i(W2) Iom•• I..fi. That is, tl.w is the width of
the curve where the current has fallen to 1/..fi = 0.707 of its value at
resonance. At WI and W2 the power dissipated in R is one-half of the power
at resonance. Hence, the points WI and W2 are sometimes called the
half-power points. To find tl.w, we note that lomax = Vol Rand i has fallen
to Iomax/..fi when Z ..fiR. Therefore at W = WI and at W = W2, Z ..fiR
or !Z(Wl)! = !Z(w2)1 = ..fiR.
L
-"----=I
pmo,"
I
L
where (J = tan-l(wLIR). Then i 1ma• = V olJ1f1+-,,?I}, but this procedure
is not necessary to calculate the Q because i lma• cancels out in the
expression for Q.
The resistance R in an actual RLC circuit is usually the resistance of
the wire in the inductance L. Practically speaking, we usually can neglect
any power dissipated as heat in the capacitor, which we have done by not
drawing any resistance associated with C. If we draw the parallel LC circuit
as in Fig. 2.43, with R representing the leakage resistance of the capacitor,
FIGURE 2.43
R
Voei (wr-6)
Voe iwr
il(t) = R + jwL
101
0 {Ouality Factor}
(2.72)
Thus a high-Q circuit has a narrow width tl.w for a given frequency wo, and
a low-Q circuit has a broad width tl.w. A high-Q circuit is said to be sharp
or to have a high selectivity because it can be used to select a narrow band
of frequencies around Wo and to reject other frequencies (see Fig. 2.45).
The best intuitive way to think of Q is with the relationship Q = wo/tl.w.
For the parallel RLC circuit we can also show that, if woL i» R (the
resistance associated with L), Q == wo/tl.w = WoLIR, where Wo is the
angular resonant frequency and tl.w is the full width between the half-power
CHAP. 2
102 CHAP. 2
1
i
103
Problems
Alternating Current Circuits
III
R
"v
L
11< •
c
/
"
•
fIGURE 2.46 Ringing excited by step input.
By inspection Q o CVo is a constant particular solution of (2.73). The
solution to (2.74) can be shown to be
fiGURE 2.45 Frequency response of high·Q and low·Q Series RLC circuit.
points. The impedance of the parallel tank at resonance is Zo = QWoL
QlwoC, and at the half-power points the impedance is Z = Zol..fi. Notice
that at resonance the Q effectively multiplies the impedance of the in­
ductance or of the capacitance.
If an RLC circuit (either series or parallel) is excited with a step
function voltage (or indeed with any rapidly changing voltage), the current
will tend to oscillate or "ring" with an exponentially decreasing amplitude,
as shown in Fig. 2.46. The differential equation from the KVL for a
step-function input is
Vin
= Vo =
L di
dt
or
v:
o
+ V R + V c for
t
~
C
0 at t
Q = CVo(1
w5
or 4Q
~ 2
}>
l.
(2.76)
Thus we see that for a high-Q circuit (4Q 2 }> 1), the charge on the
capacitor oscillates at approximately Wo, and the amplitUde of the oscil­
lations decreases exponentially according to
(2.77)
= O. The solution of
(2.73)
PROBLEMS
(2.73) can be
1. Calculate the period in ~econds of a 400-Hz sinusoidal voltage. If the zero-to­
Q
= Qh + Qp
where Qp is any particular solution and Q h is the solution to the homo­
geneous equation.
o
where A is a constant, and provided R2/4L 2 <is 11 LC
Using Q = WaLl R, we can write the solution as
(2.7S)
In other words, the envelope of the oscillations or ringing decreases accord­
to (2.77) as shown in Fig. 2.46(b). The current and voltage also
decrease according to (2.77). Thus a measurement of the voltage across the
RLC circuit versus time can yield the circuit Q.
L d Q + R dQ + Q dt 2
dt
C
with the boundary condition Q
written as
Ae - R,{2 L e ±iw ,,' e'"wo<l2Q 2
=
Qh
0
+ R' + Q
I
(b) voltage acrossc
(a) circuit
wL
R
Q
e
'" "
a
W
Wo
e
2
L d Qh + R dQh + Q.
dr
dt
C
(2.74)
peak amplitude is 5 V, calculate the maximum instantaneous rate of change of
the voltage in volts per second.
2. Prove that the root-me an-square (rms) value of
V(t) = Vo cos wI
is equal to Vo/J2. What is the rms value of the voltage of Problem I?
\ .
104
.\
105
CHAP. 2 Problema
(
CHAP. 2
Ahern.tin" Current Circuifa
,
3,. Calculate the average power (in watts) dissipated as heat in a device passing a
"current of i(t) 4 cos wt if the voltage across the device is (a) v(t) =:
10 cos(wt + 30"), (b) v(t) = 10 cos wt.
4. Calculate the capacitance in microfarads between two l-cm 2 conducting plates
1 mm apart in a vacuum. Repeat if the space between the plates is filled with a
plastic dielectric with a dielectric constant of 8.
5. Calculate the capacitive reactance in ohms of a O.Ol-j.tF capacitor at (a)
100 Hz, (b) 1 kHz, (c) 100 kHz, (d) 1 MHz.
6. Calculate the inductive reactance in ohms of a 2.5-mH choke at (a) 100 Hz, (b)
1 kHz, (c) 100 kHz, (d) 1 MHz.
7. Express L dimensionally in terms of ohms and seconds. Repeat for C.
8. Calculate the energy in jOllies stored in a 2000-j.tF capacitor charged to 5 V.
Physically. how is the energy stored?
I-H inductance carries a current of 500 mA. The wire breaks, and in 10- 3 s
the current drops to zero. What would happen?
10. Calculate the impedance ZAB in the form a + jb and Izle i8 for
\(~ Design a low-pass RC filter that will attenuate a 60-Hz sinusoidal voltage by
/{ 12 dB relative to the dc gain. Use a 100-fl resistance. Explain in words why the
/
low-pass RC filter attenuates the high frequencies.
15. For a low-pass RC filteryrove that (a) at the frequency w == 1/ RC the voltage
gain equals 0.707 It-J2; (b) the rise time of the output pulse equals 2.2RC for
a zero rise time input pulse.
16. Calculate the slope of the gain versus frequency curve for a low-pass RC filter
in dB per octave and also in dB per decade for high frequencies (w ~ 1/ RC).
[One decade frequency change is a factor of ten (e.g., from 10 to 100 Hz, or
from 50 to 500 kHz).)
Carefully sketch the voltage vector diagram
a high-pass RC filter. and
, calculate the phase of the output voltage relatIve to the phase of the mput
voltage.
18. Derive an expression for the voltage
and phase shift for the following LR
circuit.
\t\
/~
f~r
;",=
R
R
C
A~I
oB
AO>----.t
f--~QB
C
Calculate the impedance
ZAB
in the form a + jb and
Izle!8 for
R
A~I--oB
L
R
C
!
'j~. Calculate the impedance
A
B
19. Design a parallel LC resonant circuit or tank to resonate at 1 MHz. Assume the
inductance L 100 j.tH and has a dc resistance of 10 fl. What is the Q of this
circuit at resonance?
20. Carefully sketch the rotating voltage vector diagram for a series RLC circuit
/at resonance. If the circuit has a Q of 100, calculate the voltage ratings Land
>(. C must have.
/ it: Sketch a graph of the magnitude of the impedance versus frequency for series
'lind parallel RLC circuits. State the change in phase of the impedance as the
,. frequency passes through resonance.
~2) Sketch the approximate gain-versus-frequency curve for the following circuit.
\
You may treat the circuit as being composed of two independent RC filters.
L
ZAB
in the form a + jb and
for
c
A
tR :""ul
~M~'
R
L
13. Design a high-pass RC filter with a breakpoint at 100 kHz. Use a 1-kfl
resistance. Explain in words why the high-pass filter attenuates the low
frequencies.
IJ1.F
IOOkn
~I ~ ~
;"'u:
I kO
f:
~"i
~utput
23. For a high-Q parallel RLC circuit prove that Q'" Wolt:..w, where Wo is the
(angular) resonant frequency and t:..w is the width at the half-power points.
24. For a high-Q parallel RLC circuit prove that at resonance the impedance
equals the Q times the inductive reactance at resonance. Calculate the im­
106
CHAP. 2
Alternating Current Circuits
pedance at resonance for
CHAPTER
L == 100 #LH
C
O.OOl#LF
R
3
50.
25. For a series RLC circuit show that the angular frequency for the charge On C
(or the current) is approximately
Fourier Analysis and Pulses
W=wo(1-8~2)
26. Derive (2.75) for the series RLC circuit excited by a step-function voltage.
[Hint: Solve (2.74) by assuming a solution of the form Q~ = eAt where A is an
unknown constant independe,1t of time but depending upon R, L, and C.
Assume R'/4L" <{ lILC.]
27. A general ac bridge can be made from the Wheatstone bridge of Fig. 1.35 by
replacing each of the four resistances by a general impedance Z, that is,
RI - ZI, etc. The battery of Fig. 1.35 is also replaced by a sinusoidal ac source
of angular frequency w, and the null meter is replaced by a sensitive ac
voltmeter such as an oscilloscope. Show that the general balance condition is
Z:,
Z.
z,.
28. With Z, = R" z,. = R, + jwL" Z:, = R" and Z.
two balance conditions are
RI
R2
R,
R.
and
R. + jwL., show that the
R, '" R,
L,
L.
3.1
INTRODUCTION
In many situations we encounter voltage changes that are not sinusoidal in
shape. Any relatively rapid change in voltage is usually referred to as a
pulse. For example, every time an ionizing particle such as a proton or
electron passes through a Geiger Mueller tube, a small voltage pulse several
microseconds long will be produced at the output of the tube. This pulse
must then be amplified and recorded in some way to count the protons or
electrons. And every time an ionizing particle passes through a scintillation
crystal such as NaI (sodium iodide), a light flash of very short duration is
produced in the crystal or scintillator. This light flash is converted to a
voltage pulse lasting about 10-'6 s by a photomultiplier tube, and this
voltage pulse must then be amplified and counted. Voltage pulses may also
be used to indicate the beginning and the end of a time interval and,
therefore, must be amplified and recorded. The electronic signals in modern
computers also consist of voltage pulses.
This is an inductance bridge. Also sketch the bridge.
29. With Z, "" R" z,. == R2 + 1/ jwC2 , Z:, = R" and Z. = R4 + 1/ jwC., show that
the two balance conditions are
R,
R2
i
R.
and
R,C,
R 3 C.
.?\\ This is a capacitance bridge. Also sketch the bridge.
. 30. \With ZI
R z,.
R211 C2 , Z:,
"
jba\ance conditions
are
R" and Z4
.R. + C2
if:2
R3
-C = -R
4
1
and
R.
+ 1/ jwC4, show that the two
w'R 2 C,R.C4
= 1
This is the Wien bridge. If R2 R. = Rand C2 = C. = C, find the two
balance conditions. Also sketch the bridge.
3.2 DESCRIPTION Of A PULSE
An ideal single pulse is rectangular in shape, as shown in Fig. 3.1 (a). This
pulse can be described by giving its amplitude Vo and its width or duration
'T. An actual pulse encountered in the laboratory will always appear some­
what rounded if observed with a sufficiently fast oscilloscope, as shown in
Fig. 3.1(b). The steepness of the sides of this pulse is described by giving
the rise time, which is defined as the time for the pulse to rise from 10% to
90% of its maximum value. The fall time of the pulse is defined as the time
for the pulse to fall from 90% to 10% of its maximum value. An ideal pulse
has zero rise and fall times.
If a series of pulses occurs regularly in time as shown in Fig. 3.2, we
can describe the pulses completely by giving the number of pulses per
second (called the frequency or repetition rate), the pulse width 'T, the pulse
amplitude Yo, and the phase. In Fig. 3.2 the frequency or repetition rate is
200 kHz, the pulse width is 1 j.Ls, and the amplitude is 10 V. Notice that if
107