07-Dec-11
Summary
The Basic Capacitor
Basics of electrical systems
Capacitors are one of the fundamental passive
components. In its most basic form, it is composed of
two conductive plates separated by an insulating
dielectric.
The ability to store charge is the definition of
capacitance.
Dielectric
Conductors
Summary
Capacitance
The Basic Capacitor
Dielec tric
VVSS
The charging
process…
Leads
−
Initially
Source
Fully
Charging
charged
removed
uncharged
−
−
−
++
−++
+
+++−
−++
+++
+++
−
− −++
+++
+−
+
−+
+
++
AA +
A
−
− +
−
−−
+−+
Plates
−−−
−
+−
+
−
− −
−
+−
−
− − Electrons
−−−+
−
+−−− B
−
BB
−
−
−
−
A capacitor with stored charge can act as a temporary battery.
Capacitance
Capacitance is the ratio of charge to voltage
C=
Q
V
Rearranging, the amount of charge on a
capacitor is determined by the size of the
capacitor (C) and the voltage (V).
Q = CV
If a 22 µF capacitor is connected to
a 10 V source, the charge is 220 µC
Capacitance
An analogy:
Imagine you store rubber bands in a
bottle that is nearly full.
You could store more rubber bands
(like charge or Q) in a bigger bottle
(capacitance or C) or if you push
them in more (voltage or V). Thus,
Q = CV
A capacitor stores energy in the form of an electric field
that is established by the opposite charges on the two
plates. The energy of a charged capacitor is given by the
equation
W=
1
CV 2
2
where
W = the energy in joules
C = the capacitance in farads
V = the voltage in volts
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07-Dec-11
Summary
Summary
Capacitance
Capacitance
The capacitance of a capacitor depends on
three physical characteristics.
Find the capacitance of a 4.0 cm diameter
sensor immersed in oil if the plates are
separated by 0.25 mm. ( ε r = 4.0 for oil )
ε A
C = 8.85 ×10−12 F/m  r 
 d 
ε A
C = 8.85 × 10 −12 F/m  r 
 d 
2
2
−3
2
The plate area is A = πr = π 0.02 m = 1.26 × 10 m
C is directly proportional to
the relative dielectric constant
and the plate area.
C is inversely proportional to
the distance between the plates
(
)
The distance between the plates is 0.25 × 10−3 m
 ( 4.0 ) (1.26 × 10 −3 m 2 )
C = 8.85 × 10−12 F/m 

0.25 ×10 −3 m

Summary

 = 178 pF


Summary
Capacitor types
Capacitor types
Mica
Mica capacitors are small with high working voltage.
The working voltage is the voltage limit that cannot
be exceeded.
Ceramic disk
Ceramic disks are small nonpolarized capacitors They
have relatively high capacitance due to high εr.
Lead wire soldered
to silver elec trode
Foil
Mica
Foil
Mica
Foil
Mica
Foil
Solder
Ceramic
dielectric
Summary
Dipped phenolic c oating
Silv er elec trodes deposited on
top and bottom of c eramic disk
Summary
Capacitor types
Capacitor types
Plastic Film
Plastic film capacitors are small and nonpolarized. They
have relatively high capacitance due to larger plate area.
Electrolytic (two types)
Electrolytic capacitors have very high capacitance but
they are not as precise as other types and tend to have
more leakage current. Electrolytic types are polarized.
High-purity
foil electrodes
Plastic film
dielec tric
+
Outer wrap of
polyester film
Capacitor section
(alternate strips of
film dielectric and
Lead wire
foil electrodes)
Solder coated end
Al electrolytic
_
Ta electrolytic
Symbol for any electrolytic capacitor
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07-Dec-11
Summary
Capacitors use several labeling methods. Small
capacitors values are frequently stamped on them such
as .001 or .01, which have implied units of microfarads.
Electrolytic capacitors have larger
values, so are read as µF. The unit is usually
stamped as µF, but some older ones may be
shown as MF or MMF).
++++
Capacitor labeling
Variable
Variable capacitors typically have small capacitance
values and are usually adjusted manually.
A solid-state device that is used as a variable
capacitor is the varactor diode; it is adjusted with an
electrical signal.
47VTTMFVTT
Capacitor types
.022
Summary
Capacitor labeling
Series capacitors
A label such as 103 or 104 is read as 10x103
(10,000 pF) or 10x104 (100,000 pF)
respectively. (Third digit is the multiplier.)
When values are marked as 330 or 6800, the
units are picofarads.
222
2 20 0
When capacitors are connected in series, the total
capacitance is smaller than the smallest one. The
general equation for capacitors in series is
CT =
1
1
1
1
1
+
+
+ ... +
C1 C2 C3
CT
The total capacitance of two capacitors is
What is the value of
each capacitor? Both are 2200 pF.
Summary
Series capacitors
0.001 µF
1
1
1
+
C1 C2
…or you can use the product-over-sum rule
Summary
Parallel capacitors
If a 0.001 µF capacitor is connected
in series with an 800 pF capacitor,
the total capacitance is 444 pF
C1
CT =
C2
800 pF
When capacitors are connected in parallel, the total
capacitance is the sum of the individual capacitors.
The general equation for capacitors in parallel is
CT = C1 + C2 + C3 + ...Cn
If a 0.001 µF capacitor is
connected in parallel with
an 800 pF capacitor, the
total capacitance is 1800 pF
C1
C2
0.001 µF
800 pF
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07-Dec-11
Summary
Summary
Vfinal
The RC time constant
When a capacitor is charged
through a series resistor and
dc source, the charging curve
is exponential.
0
t
(a) Capac itor c harging voltage
When a capacitor is discharged
through a resistor, the
discharge curve is also an
exponential. (Note that the
current is negative.)
Iinitial
R
0
t
The RC time constant
Summary
Universal exponential curves
VS
VR
Specific values for
current and voltage
can be read from a
universal curve. For
an RC circuit, the
time constant is
τ = RC
100%
95%
Rising exponential
40%
37%
Falling exponential
14%
5%
0
0
The current has the same shape as VR.
Universal exponential curves
The universal curves can be applied to general formulas for
the voltage (or current) curves for RC circuits. The general
voltage formula is
v =VF + (Vi − VF)e−t/RC
VF = final value of voltage
Vi = initial value of voltage
v = instantaneous value of voltage
The final capacitor voltage is greater than the initial
voltage when the capacitor is charging, or less that the
initial voltage when it is discharging.
99%
63%
60%
20%
Summary
98%
86%
80%
Percent of final value
VC
R
C
t
(b) Discharging current
Summary
VS
−Iinitial
C
0
(b) Charging current
What is the shape of the
current curve?
t
0
(a) Capacitor discharging voltage
R
C
The same shape curves are
seen if a square wave is
used for the source.
Vinitial
The RC time constant
1τ
2%
2τ
3τ
4τ
Number of time constants
1%
5τ
Summary
Capacitive reactance
Capacitive reactance is the opposition to ac by a
capacitor. The equation for capacitive reactance is
XC =
1
2πfC
The reactance of a 0.047 µF capacitor when a
frequency of 15 kHz is applied is 226 Ω
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07-Dec-11
Summary
Summary
Capacitive reactance
Capacitive reactance
When capacitors are in series, the total reactance is the sum of the
individual reactances. That is,
When capacitors are in parallel, the total reactance is the reciprocal
of the sum of the reciprocals of the individual reactances. That is,
X C(tot ) = X C1 + X C2 + X C3 + ⋅⋅⋅ + X Cn
X C(tot ) =
Assume three 0.033 µF capacitors are in series with a 2.5 kHz
ac source. What is the total reactance?
If the three 0.033 µF capacitors from the last example are
placed in parallel with the 2.5 kHz ac source, what is the total
reactance?
The reactance of each capacitor is
XC =
1
1
1
1
1
+
+
+ ⋅⋅⋅ +
X C1 X C2 X C3
X Cn
1
1
=
= 1.93 kΩ
2πfC 2π ( 2.5 kHz )( 0.033 µF )
The reactance of each capacitor is 1.93 kΩ
X C(tot ) = X C1 + X C2 + X C3
X C(tot ) =
= 1.93 kΩ + 1.93 kΩ + 1.93 kΩ = 5.79 kΩ
Summary
1
1
=
= 643 Ω
1
1
1
1
1
1
+
+
+
+
X C1 X C2 X C3 1.93 kΩ 1.93 kΩ 1.93 kΩ
Summary
Capacitive Voltage Divider
Capacitive Voltage Divider
Two capacitors in series are commonly used as a capacitive voltage
divider. The capacitors split the output voltage in proportion to their
reactance (and inversely proportional to their capacitance).
What is the output voltage for the capacitive voltage divider?
Instead of using a ratio of reactances in the capacitor voltage divider
equation, you can use a ratio of the total series capacitance to the
output capacitance (multiplied by the input voltage). The result is
the same. For the problem presented in the last slide,
X C2
1
1
X C1 =
=
= 4.82 kΩ
2πfC1 2π ( 33 kHz )(1000 pF )
1
1
=
=
= 482 Ω
2πfC2 2π ( 33 kHz )( 0.01 µF )
X C ( tot ) = X C1 + X C 2
1.0 V
f = 33 kHz
= 4.82 kΩ + 482 Ω = 5.30 kΩ
Vout
 X

 482 Ω 
=  C 2 Vs = 
1.0 V = 91 mV


 5.30 kΩ 
 X C ( tot ) 
Summary
Capacitive phase shift
When a sine wave
is applied to a
capacitor, there is a
phase shift between
voltage and current
such that current
always leads the
voltage by 90o.
VC
0
0
C1
1000 pF
(1000 pF)( 0.01 µF ) = 909 pF
C1C2
=
C1 + C2
1000 pF + 0.01 µF
C 
 909 pF 
Vout =  ( tot )  Vs = 
 1.0 V = 91 mV
 0.01 µF 
 C2 
C2
Vout
0.01 µF
C1
1000 pF
1.0 V
f = 33 kHz
C2
Vout
0.01 µF
Summary
Power in a capacitor
90o
I
C( tot ) =
Energy is stored by the capacitor during a portion of the ac
cycle and returned to the source during another portion of
the cycle.
Voltage and current are always 90o out of phase.
For this reason, no true power is dissipated by a capacitor,
because stored energy is returned to the circuit.
The rate at which a capacitor stores or returns
energy is called reactive power. The unit for reactive
power is the VAR (volt-ampere reactive).
5
07-Dec-11
Summary
Summary
Power supply filtering
Coupling capacitors
There are many applications for capacitors. One is in
filters, such as the power supply filter shown here.
Coupling capacitors are used to pass an ac signal
from one stage to another while blocking dc.
Rectifier
3V
C
60 Hz ac
Load
resistance
0V
0V
+V
The filter smoothes the
pulsating dc from the
rectifier.
The capacitor isolates dc
Input
between the amplifier stages,
preventing dc in one stage from
affecting the other stage.
Summary
C
R1
Amplifier
stage 1
Amplifier
stage 2
Output
R2
Selected Key Terms
Bypass capacitors
Capacitor An electrical device consisting of two conductive
plates separated by an insulating material and
possessing the property of capacitance.
Another application is to bypass an ac signal to ground
but retain a dc value. This is widely done to affect gain
in amplifiers.
Dielectric The insulating material between the conductive
plates of a capacitor.
dc plus
ac
0V
dc only
0V
The bypass capacitor places
point A at ac ground, keeping
only a dc value at point A.
R1
A
Point in circuit where R
2
only dc is required
C
Selected Key Terms
Farad The unit of capacitance.
RC time A fixed time interval set by the R and C values,
constant that determine the time response of a series RC
circuit. It equals the product of the resistance
and the capacitance.
Quiz
Capacitive The opposition of a capacitor to sinusoidal
reactance current. The unit is the ohm.
Instantaneous The value of power in a circuit at a given
power (p) instant of time.
1. The capacitance of a capacitor will be larger if
a. the spacing between the plates is increased.
True power The power that is dissipated in a circuit
(Ptrue) usually in the form of heat.
b. air replaces oil as the dielectric.
Reactive The rate at which energy is alternately stored
power (Pr ) and returned to the source by a capacitor. The
unit is the VAR.
d. all of the above.
c. the area of the plates is increased.
VAR The unit of reactive power.
(volt-ampere
reactive)
6
07-Dec-11
Quiz
2. The major advantage of a mica capacitor over other
types is
a. they have the largest available capacitances.
b. their voltage rating is very high
c. they are polarized.
Quiz
3. Electrolytic capacitors are useful in applications where
a. a precise value of capacitance is required.
b. low leakage current is required.
c. large capacitance is required.
d. all of the above.
d. all of the above.
Quiz
4. If a 0.015 µF capacitor is in series with a 6800 pF
capacitor, the total capacitance is
a. 1568 pF.
Quiz
5. Two capacitors that are initially uncharged are connected
in series with a dc source. Compared to the larger
capacitor, the smaller capacitor will have
a. the same charge.
b. 4678 pF.
b. more charge.
c. 6815 pF.
c. less voltage.
d. 0.022 µF.
d. the same voltage.
Quiz
6. When a capacitor is connected through a resistor to a dc
voltage source, the charge on the capacitor will reach 50%
of its final charge in
Quiz
7. When a capacitor is connected through a series resistor
and switch to a dc voltage source, the voltage across the
resistor after the switch is closed has the shape of
a. less than one time constant.
a. a straight line.
b. exactly one time constant.
b. a rising exponential.
c. greater than one time constant.
c. a falling exponential.
d. answer depends on the amount of voltage.
d. none of the above.
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07-Dec-11
Quiz
Quiz
8. The capacitive reactance of a 100 µF capacitor to 60 Hz
is
9. If an sine wave from a function generator is applied to a
capacitor, the current will
a. 6.14 kΩ.
a. lag voltage by 90o.
b. 265 Ω.
b. lag voltage by 45o.
c. 37.7 Ω.
c. be in phase with the voltage.
d. 26.5 Ω
d. none of the above.
Quiz
Quiz
10. A switched capacitor emulates a
Answers:
a. smaller capacitor.
b. larger capacitor.
c. battery.
d. resistor.
1. c
6. a
2. b
7. c
3. c
8. d
4. b
9. d
5. a
10. d
Summary
Summary
Sinusoidal response of RC circuits
Impedance of series RC circuits
When both resistance and capacitance are in a series
circuit, the phase angle between the applied voltage and
total current is between 0° and 90°, depending on the
values of resistance and reactance.
In a series RC circuit, the total impedance is the phasor
sum of R and XC.
VR
VC
V R leads VS
V C lags VS
R is plotted along the positive x-axis.
XC is plotted along the negative y-axis.
 XC 

 R 
θ = tan −1 
R
R
θ
θ
Z is the diagonal
R
C
VS
I
I leads V S
XC
XC
Z
Z
It is convenient to reposition the
phasors into the impedance triangle.
8
07-Dec-11
Summary
Summary
Impedance of series RC circuits
Analysis of series RC circuits
Ohm’s law is applied to series RC circuits using Z,
V, and I.
Sketch the impedance triangle and show the
values for R = 1.2 kΩ and XC = 960 Ω.
(1.2 kΩ )
Z=
2
+ ( 0.96 kΩ )
θ
XC =
960 Ω
Z = 1.33 kΩ
Summary
Phasor diagrams that have reactance phasors can only
be drawn for a single frequency because X is a
function of frequency.
R
As frequency changes,
Increasing f
θ
θ
θ
the impedance triangle
Z
X
f
for an RC circuit changes
Z
as illustrated here
because XC decreases
f
X
Z
with increasing f. This
determines the frequency
f
X
response of RC circuits.
Z = 1.33 kΩ
3
θ
1
C3
3
C2
2
C1
1
2
39o
1
VC =
9.6 V
VS = 13.3 V
Summary
Summary
Applications
Applications
For a given frequency, a series RC circuit can be used to
produce a phase lag by a specific amount between an
input voltage and an output by taking the output across
the capacitor. This circuit is also a basic low-pass filter, a
circuit that passes low frequencies and rejects all others.
Reversing the components in the previous circuit produces
a circuit that is a basic lead network. This circuit is also a
basic high-pass filter, a circuit that passes high frequencies
and rejects all others. This filter passes high frequencies
down to a frequency called the cutoff frequency.
V
C
R
θ
Vin
2
3
VR = 12 V
x 10 mA
=
XC =
960 Ω
V
I
Variation of phase angle with frequency
Assume the current in the previous example is 10 mArms.
Sketch the voltage phasor diagram. The impedance
triangle from the previous example is shown for reference.
The voltage phasor diagram can be found from Ohm’s
law. Multiply each impedance phasor by 10 mA.
39o
Z=
Summary
Analysis of series RC circuits
θ
V
Z
Because I is the same everywhere in a series circuit,
you can obtain the voltages across different
components by multiplying the impedance of that
component by the current as shown in the following
example.
39o
−1
R = 1.2 kΩ
I=
R = 1.2 kΩ
= 1.33 kΩ
0.96 kΩ
θ = tan
1.2 kΩ
= 39°
V = IZ
2
C
Vout
(phase lag)
Vout
θ
φ
Vin
(phase lag)
Vin
(phase lead)
Vin
φ
V
Vout
Vout
VR
R
Vout
Vout
Vin
VC
Vin
θ
(phase lead)
9
07-Dec-11
Summary
Summary
Applications
Sinusoidal response of parallel RC circuits
An application showing how the phase-shift network is
useful is the phase-shift oscillator, which uses a
combination of RC networks to produce the required 180o
phase shift for the oscillator.
For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance) and to review
conductance.
G=
Conductance is the reciprocal of resistance.
Amplifier
Rf
Phase-shift network
C
C
C
R
R
BC =
Capacitive susceptance is the reciprocal
of capacitive reactance.
R
Admittance is the reciprocal of impedance. Y =
1
R
1
XC
1
Z
Summary
Summary
Sinusoidal response of parallel RC circuits
Sinusoidal response of parallel RC circuits
In a parallel RC circuit, the admittance phasor is the sum
of the conductance and capacitive susceptance phasors.
The magnitude can be expressed as Y = G 2 + BC 2
 BC 

G 
From the diagram, the phase angle is θ = tan −1 
BC
Some important points to notice are:
G is plotted along the positive x-axis.
BC is plotted along the positive y-axis.
 BC 

G 
θ = tan −1 
Y is the diagonal
BC
Y
Y
VS
BC
G
VS
θ
G
θ
G
Summary
Analysis of parallel RC circuits
Ohm’s law is applied to parallel RC circuits using
Y, V, and I.
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
1
1
=
= 1.0 mS
R 1.0 kΩ
Y = G 2 + BC 2 =
BC = 2π (10 kHz )( 0.01 µ F ) = 0.628 mS
(1.0 mS)
2
G
Summary
Sinusoidal response of parallel RC circuits
G=
BC
+ ( 0.628 mS ) = 1.18 mS
V=
I
Y
I = VY Y =
I
V
2
BC = 0.628 mS
VS
R
C
f = 10 kHz
1.0 kΩ
0.01 µF
Y=
1.18 mS
G = 1.0 mS
Because V is the same across all components in a
parallel circuit, you can obtain the current in a given
component by simply multiplying the admittance of
the component by the voltage as illustrated in the
following example.
10
07-Dec-11
Summary
Summary
Analysis of parallel RC circuits
Phase angle of parallel RC circuits
If the voltage in the previous example is 10 V, sketch the
current phasor diagram. The admittance diagram from the
previous example is shown for reference.
The current phasor diagram can be found from
Ohm’s law. Multiply each admittance phasor by 10 V.
BC = 0.628 mS
Y=
1.18 mS
x 10 V
=
IC = 6.28 mA
IS =
11.8 mA
Notice that the formula for capacitive susceptance is
the reciprocal of capacitive reactance. Thus BC and IC
are directly proportional to f: BC = 2π fC
As frequency increases, BC
and IC must also increase,
so the angle between IR and
IS must increase.
IC
IS
θ
IR
IR = 10 mA
G = 1.0 mS
Summary
Summary
Equivalent series and parallel RC circuits
For every parallel RC circuit there is an equivalent
series RC circuit at a given frequency.
The equivalent resistance and capacitive
reactance are shown on the impedance triangle:
Req = Z cos θ
θ
Z
XC(eq) = Z sin θ
Series-Parallel RC circuits
Series-parallel RC circuits are combinations of both series and
parallel elements. These circuits can be solved by methods from
series and parallel circuits.
Z1
Z2
For example, the
R
C1
1
components in the
R2
C2
green box are in
series: Z1 = R12 + X C21
The components in
the yellow box are
R2 X C 2
in parallel: Z 2 =
R22 + X C2 2
Summary
Measuring Phase Angle
An oscilloscope is commonly used to measure phase angle in
reactive circuits. The easiest way to measure phase angle is to
set up the two signals to have the same apparent amplitude and
measure the period. An example of a Multisim simulation is
shown, but the technique is the same in lab.
Set up the oscilloscope so that
two waves appear to have the
same amplitude as shown.
Determine the period. For the
wave shown, the period is
The total impedance can be found by
converting the parallel components to an
equivalent series combination, then
adding the result to R1 and XC1 to get the
total reactance.
Summary
Measuring Phase Angle
Next, spread the waves out using the SEC/DIV control in order
to make an accurate measurement of the time difference between
the waves. In the case illustrated, the time difference is
 5 µs 
∆t = 4.9 div 
 = 24.5 µs
 div 
The phase shift is calculated from
 24.5 µs 
 ∆t 
o
 360° =  160 µs  360° = 55
T 


θ =
 20 µs 
T = 8.0 div 
 = 160 µs
 div 
11
07-Dec-11
Summary
Summary
The power triangle
The power triangle
Multiplying the voltage phasors by Irms gives the power triangle
(equivalent to multiplying the impedance phasors by I2). Apparent
power is the product of the magnitude of the current and magnitude of
the voltage and is plotted along the hypotenuse of the power triangle.
Recall that in a series RC circuit, you could multiply
the impedance phasors by the current to obtain the
voltage phasors. The earlier example is shown for
review:
R = 1.2 kΩ
θ
x 10 mA
=
39o
Z = 1.33 kΩ
XC =
960 Ω
The rms current in the earlier example was 10 mA.
Show the power triangle.
VR = 12 V
θ
VR = 12 V
θ
39o
VC =
9.6 V
VS = 13.3 V
VS = 13.3 V
Summary
39o
Pr = 96
mVAR
Pa = 133 mVA
Apparent power
The power factor is the relationship between the
apparent power in volt-amperes and true power in
watts. Volt-amperes multiplied by the power factor
equals true power.
Power factor is defined mathematically as
Apparent power consists of two components; a true
power component, that does the work, and a
reactive power component, that is simply power
shuttled back and forth between source and load.
PF = cos θ
Some components such
as transformers, motors,
and generators are rated
in VA rather than watts.
The power factor can vary from 0 for a purely reactive
circuit to 1 for a purely resistive circuit.
Summary
V in
10 V dc
Vin
V 8.46
rms V rms
10 V dc 1.57
0.79V
rms
0
11 µµµFFF
10 V dc
0
Vout
10V
V rms
rms
10
10 V rms
10 V dc
1µ
F
1
µF
F
1 µµ
F
ƒ =
= 1100
Hz
ƒƒ
= 10kHz
kHz
9.87
9
Plotting the response:
8
7
6
1.57
0.79
5.32
5
4
3
2
1
0.1
0 V dc
Vout (V)
9.98
8.46
9.87V rms
5.32 V rms
0.63 V rms
100
Ω
Ω
100
100
100 Ω
Ω
Vout (V)
Plotting the response:
Pr (VAR)
Pa (VA)
Reversing the components, and taking the output across the resistor as
shown, the circuit acts as a high-pass filter.
As the frequency increases, the output amplitude also increases.
Vout
100
Ω
100
100Ω
Ω
Ω
100
20
kHz
ƒƒƒ === 110
kHz
kHz
θ
Frequency Response of RC Circuits
When a signal is applied to an RC circuit, and the output is taken
across the capacitor as shown, the circuit acts as a low-pass filter.
As the frequency increases, the output amplitude decreases.
10
10VVrms
rms
10 V rms
Ptrue (W)
Summary
Frequency Response of RC Circuits
0
VC =
9.6 V
θ
Summary
Power factor
10 V dc
Ptrue = 120 mW
x 10 mA
=
39o
0.63
1
10 20
100
f (kHz)
10
9
8
7
6
5
4
3
2
1
0
0.01
0.1
1
10
f (kHz)
12
07-Dec-11
Selected Key Terms
Impedance The total opposition to sinusoidal current
expressed in ohms.
Phase angle The angle between the source voltage and the
total current in a reactive circuit.
Capacitive The ability of a capacitor to permit current;
suceptance (BC) the reciprocal of capacitive reactance. The
unit is the siemens (S).
Admittance (Y) A measure of the ability of a reactive circuit to
permit current; the reciprocal of impedance.
The unit is the siemens (S).
Quiz
1. If you know what the impedance phasor diagram looks
like in a series RC circuit, you can find the voltage phasor
diagram by
Selected Key Terms
Power factor The relationship between volt-amperes and
true power or watts. Volt-amperes multiplied
by the power factor equals true power.
Frequency In electric circuits, the variation of the output
response voltage (or current) over a specified range of
frequencies.
Cutoff The frequency at which the output voltage of
frequency a filter is 70.7% of the maximum output
voltage.
Quiz
2. A series RC circuit is driven with a sine wave. If the
output voltage is taken across the resistor, the output
will
a. multiplying each phasor by the current
a. be in phase with the input.
b. multiplying each phasor by the source voltage
b. lead the input voltage.
c. dividing each phasor by the source voltage
c. lag the input voltage.
d. dividing each phasor by the current
d. none of the above
Quiz
3. A series RC circuit is driven with a sine wave. If you
measure 7.07 V across the capacitor and 7.07 V across the
resistor, the voltage across both components is
a. 0 V
b. 5 V
c. 10 V
Quiz
4. If you increase the frequency in a series RC circuit,
a. the total impedance will increase
b. the reactance will not change
c. the phase angle will decrease
d. none of the above
d. 14.1 V
13
07-Dec-11
Quiz
5. Admittance is the reciprocal of
a. reactance
Quiz
6. Given the admittance phasor diagram of a parallel RC
circuit, you could obtain the current phasor diagram by
a. multiplying each phasor by the voltage
b. resistance
b. multiplying each phasor by the total current
c. conductance
c. dividing each phasor by the voltage
d. impedance
d. dividing each phasor by the total current
Quiz
7. If you increase the frequency in a parallel RC circuit,
a. the total admittance will decrease
b. the total current will not change
c. the phase angle between IR and IS will decrease
d. none of the above
Quiz
8. The magnitude of the admittance in a parallel RC circuit
will be larger if
a. the resistance is larger
b. the capacitance is larger
c. both a and b
d. none of the above
Quiz
9. The maximum power factor occurs when the phase
angle is
Quiz
10. When power is calculated from voltage and current for an
ac circuit, the voltage and current should be expressed as
a. 0o
a. average values
b. 30o
b. rms values
c. 45o
c. peak values
d. 90o
d. peak-to-peak values
14
07-Dec-11
Quiz
Summary
The Basic Inductor
Answers:
1. a
6. a
2. b
7. d
3. c
8. d
4. c
9. a
5. d
10. b
When a length of wire is formed into a coil., it becomes a
basic inductor. When there is current in the inductor, a threedimensional magnetic field is created.
A change in current causes
the magnetic field to change.
This in turn induces a
voltage across the inductor
that opposes the original
change in current.
Summary
The Basic Inductor
S
N
Summary
Faraday’s law
One henry is the inductance of a coil when a current, changing at a
rate of one ampere per second, induces one volt across the coil. Most
coils are much smaller than 1 H.
Faraday’s law was introduced in Chapter 7 and repeated here
because of its importance to inductors.
The effect of inductance is greatly
magnified by adding turns and winding
them on a magnetic material. Large
inductors and transformers are wound
on a core to increase the inductance.
The amount of voltage induced in a coil is directly proportional to
the rate of change of the magnetic field with respect to the coil.
Magnetic core
Summary
Summary
Lenz’s law
Lenz’s law was also introduced in Chapter 7 and is an extension of
Faraday’s law, defining the direction of the induced voltage:
Lenz’s law
A basic circuit to demonstrate Lenz’s law is shown.
Initially, the SW is open and there is a small current in
the circuit through L and R1.
L
When the current through a coil changes and an induced voltage
is created as a result of the changing magnetic field, the direction
of the induced voltage is such that it always opposes the change
in the current.
VS
SW
+
R1
R2
−
−
+
15
07-Dec-11
Summary
Summary
Lenz’s law
Lenz’s law
SW closes and immediately a voltage appears across L that
tends to oppose any change in current.
L
VS
+
L
SW
−
+
After a time, the current stabilizes at a higher level (due to I2)
as the voltage decays across the coil.
R1
VS
R2
−
−
R1
−
−
+
SW
+
R2
+
Initially, the meter
reads same current
as before the switch
was closed.
Later, the meter
reads a higher
current because of
the load change.
Summary
Summary
Types of inductors
Practical inductors
In addition to inductance, actual inductors have winding
resistance (RW) due to the resistance of the wire and winding
capacitance (CW) between turns. An equivalent circuit for a
practical inductor including these effects is shown:
There are a variety of inductors, depending on the amount of
inductance required and the application. Some, with fine
wires, are encapsulated and may appear like a resistor.
CW
Notice that the winding resistance
is in series with the coil and the
winding capacitance is in parallel
with both.
Common symbols for inductors (coils) are
RW
L
Air core
Iron core
Summary
Factors affecting inductance
Four factors affect the amount of inductance for a coil. The
equation for the inductance of a coil is
L=
N 2µ A
l
where
L = inductance in henries
N = number of turns of wire
µ = permeability in H/m (same as Wb/At-m)
l = coil length on meters
Ferrite core
Variable
Summary
What is the inductance of a 2 cm long, 150 turn coil
wrapped on an low carbon steel core that is 0.5 cm diameter?
The permeability of low carbon steel is 2.5 x10−4 H/m
(Wb/At-m).
A = πr 2 = π ( 0.0025 m ) = 7.85 ×10−5 m 2
2
L=
=
N µA
l
2
(150 t ) ( 2.5 ×10−4 Wb/At-m )( 7.85 ×10−5 m2 )
2
0.02 m
= 22 mH
16
07-Dec-11
Summary
Summary
Practical inductors
Series inductors
When inductors are connected in series, the total inductance is
the sum of the individual inductors. The general equation for
inductors in series is
Inductors come in a variety of sizes. A few common
ones are shown here.
LT = L1 + L2 + L3 + ...Ln
Torroid coil
Encapsulated
If a 1.5 mH inductor is
connected in series with an 680
µH inductor, the total
inductance is
2.18 mH
Variable
Summary
L1
1.5 mH
L2
680 µH
Summary
Parallel inductors
Parallel inductors
When inductors are connected in parallel, the total inductance is
smaller than the smallest one. The general equation for
inductors in parallel is
If a 1.5 mH inductor is connected in
parallel with an 680 µH inductor, the total
inductance is
468 µH
1
LT =
1 1 1
1
+ + + ... +
L1 L2 L3
LT
The total inductance of two inductors is
LT =
L1
1.5 mH
1
1 1
+
L1 L2
L2
680 µH
…or you can use the product-over-sum rule.
Summary
Inductors in dc circuits
Summary
Vinitial
Inductors in dc circuits
VS
When an inductor is connected in
series with a resistor and dc source,
the current change is exponential.
t
0
Inductor voltage after switch closure
The same shape curves are seen if
a square wave is used for the
source. Pulse response is covered
further in Chapter 20.
VL
Ifinal
R
R
L
VS
0
Current after switch closure
L
VR
t
17
07-Dec-11
Summary
Summary
Universal exponential curves
L
R
95%
99%
98%
86%
80%
100%
95%
60%
In a series RL circuit, when
is VR > 2VL?
40%
37%
Falling exponential
20%
14%
5%
0
0
1τ
2%
2τ
3τ
4τ
Number of time constants
1%
5τ
Read the rising
exponential at the 67%
level.
After 1.1 τ
Summary
Universal exponential curves
The universal curves can be applied to general formulas for the current
(or voltage) curves for RL circuits. The general current formula is
i =IF + (Ii − IF)e−Rt/L
When inductors are in series, the total reactance is the sum of the
individual reactances. That is,
X L(tot ) = X L1 + X L2 + X L3 + ⋅⋅⋅ + X Ln
Assume three 220 µH inductors are in series with a 455 kHz
ac source. What is the total reactance?
The reactance of each inductor is
X L = 2π fL = 2π ( 455 kHz )( 220 µH ) = 629 Ω
X L(tot ) = X L1 + X L2 + X L3
= 629 Ω + 629 Ω + 629 Ω = 1.89 kΩ
37%
20%
14%
5%
1τ
2%
2τ
3τ
4τ
Number of time constants
1%
5τ
Inductive reactance
Inductive reactance is the opposition to ac by
an inductor. The equation for inductive
reactance is
X L = 2πfL
The reactance of a 33 µH inductor when a frequency of
550 kHz is applied is
114 Ω
Summary
Inductive reactance
40%
Summary
IF = final value of current
Ii = initial value of current
i = instantaneous value of current
The final current is greater than the initial current when the
inductive field is building, or less than the initial current when the field
is collapsing.
99%
63%
60%
0
0
98%
86%
80%
Rising exponential
63%
Percent of final value
τ=
100%
Percent of final value
Specific values for
current and voltage can
be read from a universal
curve. For an RL circuit,
the time constant is
Universal exponential curves
The curves can give
specific information
about an RL circuit.
Summary
Inductive reactance
When inductors are in parallel, the total reactance is the reciprocal of
the sum of the reciprocals of the individual reactances. That is,
X L(tot ) =
1
1
1
1
1
+
+
+ ⋅⋅⋅ +
X L1 X L2 X L3
X Ln
If the three 220 µH inductors from the last example are placed
in parallel with the 455 kHz ac source, what is the total
reactance?
The reactance of each inductor is 629 Ω
X L(tot ) =
1
1
=
= 210 Ω
1
1
1
1
1
1
+
+
+
+
X L1 X L2 X L3 629 Ω 629 Ω 629 Ω
18
07-Dec-11
Summary
Inductive phase shift
Power in an inductor
True Power: Ideally, inductors do not dissipate power. However, a
small amount of power is dissipated in winding resistance given by
the equation:
When a sine wave is
applied to an inductor,
there is a phase shift
between voltage and
current such that
voltage always leads the
current by 90o.
VL 0
Ptrue = (Irms)2RW
90°
I 0
Reactive Power: Reactive power is a measure of the rate at which
the inductor stores and returns energy. One form of the reactive
power equation is:
Pr=VrmsIrms
The unit for reactive power is the VAR.
Key Terms
Q of a coil
Inductor An electrical device formed by a wire wound around a core
The quality factor (Q) of a coil is given by the ratio of reactive
power to true power.
Q=
I2XL
I 2 RW
For a series circuit, I cancels, leaving
Q=
XL
RW
having the property of inductance; also known as a coil.
Winding The loops or turns of wire in an inductor.
Induced Voltage produced as a result of a changing magnetic
voltage field.
Inductance The property of an inductor whereby a change in current
causes the inductor to produce a voltage that opposes the
change in current.
Key Terms
Quiz
Henry (H) The unit of inductance.
RL time A fixed time interval set by the L and R values, that
constant determines the time response of a circuit. It equals the
ratio of L/R.
Inductive The opposition of an inductor to sinusoidal current. The
reactance unit is the ohm.
1. Assuming all other factors are the same, the inductance of an
inductor will be larger if
a. more turns are added
b. the area is made larger
c. the length is shorter
d. all of the above
Quality factor The ratio of reactive power to true power for an inductor.
19
07-Dec-11
Quiz
2. The henry is defined as the inductance of a coil when
Quiz
3. The symbol for a ferrite core inductor is
a.
a constant current of one amp develops one volt.
a.
b.
one volt is induced due to a change in current of one amp
per second.
b.
c.
one amp is induced due to a change in voltage of one volt.
c.
d.
the opposition to current is one ohm.
d.
Quiz
4. The symbol for a variable inductor is
Quiz
5. The total inductance of a 270 µH inductor connected in series with a
1.2 mH inductor is
a.
a. 220 µH
b.
b. 271 µH
c.
c. 599 µH
d. 1.47 mH
d.
Quiz
6. The total inductance of a 270 µH inductor connected in parallel with
a 1.2 mH inductor is
a. 220 µH
Quiz
7. When an inductor is connected through a series resistor and switch
to a dc voltage source, the voltage across the resistor after the switch
closes has the shape of
b. 271 µH
a. a straight line
c. 599 µH
b. a rising exponential
d. 1.47 mH
c. a falling exponential
d. none of the above
20
07-Dec-11
Quiz
Quiz
8. For circuit shown, the time constant is
9. For circuit shown, assume the period of the square wave is 10 times
longer than the time constant. The shape of the voltage across L is
L
a. 270 ns
b. 270 µs
270 µH
c. 270 ms
VS
10 V
d. 3.70 s
R
1.0 kΩ
a.
L
b.
VS
c.
R
d.
Quiz
Quiz
10. If a sine wave from a function generator is applied to an inductor,
the current will
Answers:
a. lag voltage by 90o
b. lag voltage by
45o
c. be in phase with the voltage
d. none of the above
1. d
6. a
2. b
7. b
3. d
8. a
4. c
9. c
5. d
10. a
Summary
Summary
Sinusoidal response of RL circuits
Impedance of series RL circuits
When both resistance and inductance are in a series circuit, the
phase angle between the applied voltage and total current is
between 0° and 90°, depending on the values of resistance and
reactance.
In a series RL circuit, the total impedance is the phasor sum of R
and XL.
VL
VR
V R lags VS
L
VS
I
I lags VS
 XL 

 R 
θ = tan −1 
V L lead s VS
R
R is plotted along the positive x-axis.
XL is plotted along the positive y-axis.
Z
Z is the diagonal
Z
XL
XL
θ
R
θ
R
It is convenient to reposition the
phasors into the impedance triangle.
21
07-Dec-11
Summary
Summary
Impedance of series RL circuits
Analysis of series RL circuits
Ohm’s law is applied to series RL circuits using quantities of
Z, V, and I.
Sketch the impedance triangle and show the values for R
= 1.2 kΩ and XL = 960 Ω.
(1.2 kΩ )
Z=
2
+ ( 0.96 kΩ )
V = IZ
= 1.33 kΩ
θ = tan −1
= 39°
θ
XL =
960 Ω
39o
Z=
V
I
R = 1.2 kΩ
Summary
Analysis of series RL circuits
Variation of phase angle with frequency
Phasor diagrams that have reactance phasors can only be drawn
for a single frequency because X is a function of frequency.
Assume the current in the previous example is 10 mArms. Sketch the
voltage phasors. The impedance triangle from the previous example is
shown for reference.
x 10 mA
=
Z = 1.33 kΩ
θ
39o
VS = 13.3 V
XL =
960 Ω
θ
39o
Increasing f
As frequency changes, the
impedance triangle for an RL
circuit changes as illustrated
here because XL increases with
increasing f. This determines
the frequency response of RL
circuits.
The voltage phasors can be found from Ohm’s law.
Multiply each impedance phasor by 10 mA.
VL =
9.6 V
Z3
Z2
Z1
XL 3
XL 2
XL 1
θ3
θ1 θ2
R
VR = 12 V
R = 1.2 kΩ
Summary
Summary
Phase shift
Phase shift
For a given frequency, a series RL circuit can be used to produce a
phase lead by a specific amount between an input voltage and an
output by taking the output across the inductor. This circuit is also a
basic high-pass filter, a circuit that passes high frequencies and rejects
all others.
R
Vout
Reversing the components in the previous circuit produces a circuit
that is a basic lag network. This circuit is also a basic low-pass filter,
a circuit that passes low frequencies and rejects all others.
L
Vout
VL
Vin
Vout
φ
L
V
Z
Because I is the same everywhere in a series circuit, you can
obtain the voltage phasors by simply multiplying the
impedance phasors by the current.
Z = 1.33 kΩ
0.96 kΩ
1.2 kΩ
Summary
Vin
I=
2
(phase lead)
φ
θ
VR
Vin
Vin
R
Vin
Vin
Vout
φ (phase lag)
Vout
φ
Vout
22
07-Dec-11
Summary
Summary
Sinusoidal response of parallel RL circuits
Sinusoidal response of parallel RL circuits
For parallel circuits, it is useful to review conductance,
susceptance and admittance, introduced in Chapter 10.
In a parallel RL circuit, the admittance phasor is the sum of the
conductance and inductive susceptance phasors.
The magnitude of the susceptance is
2
2
G=
Conductance is the reciprocal of resistance.
Inductive susceptance is the reciprocal
of inductive reactance.
BL =
1
R
G
1
XL
VS
Admittance is the reciprocal of impedance. Y =
Y = G + BL
 BL 

G
θ = tan −1 
The magnitude of the phase angle is
G
BL
1
Z
BL
Summary
Y
Summary
Sinusoidal response of parallel RL circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BL is plotted along the negative y-axis.
Sinusoidal response of parallel RL circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
B 
θ = tan −1  L 
G
G=
Y is the diagonal
1
1
1
= 0.629 mS
=
= 1.0 mS BL =
2π (10 kHz )( 25.3 mH )
R 1.0 kΩ
Y = G 2 + BL 2 =
G
(1.0 mS)
2
+ ( 0.629 mS ) = 1.18 mS
2
G = 1.0 mS
VS
G
VS
BL
BL
Y
f = 10 kHz
Summary
Analysis of parallel RL circuits
Ohm’s law is applied to parallel RL circuits using quantities of
Y, V, and I.
I
Y=
V
I
V=
Y
I = VY
Because V is the same across all components in a parallel
circuit, you can obtain the current in a given component by
simply multiplying the admittance of the component by the
voltage as illustrated in the following example.
R
1.0 kΩ
L
25.3 mH
BL =
0.629 mS
Y=
1.18 mS
Summary
Analysis of parallel RL circuits
Assume the voltage in the previous example is 10 V. Sketch the
current phasors. The admittance diagram from the previous
example is shown for reference.
The current phasors can be found from Ohm’s law.
Multiply each admittance phasor by 10 V.
G = 1.0 mS
BL =
0.629 mS
Y=
1.18 mS
x 10 V
=
IL =
6.29 mA
IR = 10 mA
IS =
11.8 mA
23
07-Dec-11
Summary
Summary
Phase angle of parallel RL circuits
Series-Parallel RL circuits
Notice that the formula for inductive susceptance is the
reciprocal of inductive reactance. Thus BL and IL are inversely
proportional to f:
1
BL =
As frequency increases, BL and IL
decrease, so the angle between IR
and IS must decrease as well.
2π fL
IR
θ
Series-parallel RL circuits are combinations of both series and
parallel elements. The solution of these circuits is similar to
resistive combinational circuits but you need to combine reactive
elements using phasors.
The components in the
R2
R1
yellow box are in series and
Z1
Z2
those in the green box are
L1
L2
also in series.
Z1 = R12 + X L21
and
IS
IL
The two boxes are in parallel. You can
find the branch currents by applying
Ohm’s law to the source voltage and
the branch impedance.
Z 2 = R22 + X L22
Summary
Summary
The power triangle
The power triangle
Recall that in a series RC or RL circuit, you could multiply the
impedance phasors by the current to obtain the voltage phasors.
The earlier example from this chapter is shown for review:
Multiplying the voltage phasors by Irms gives the power triangle
(equivalent to multiplying the impedance phasors by I2). Apparent
power is the product of the magnitude of the current and magnitude of
the voltage and is plotted along the hypotenuse of the power triangle.
The rms current in the earlier example was 10 mA.
Show the power triangle.
Z = 1.33 kΩ
39o
x 10 mA
=
XL =
960 Ω
R = 1.2 kΩ
VS = 13.3 V
39o
VR = 12 V
VL =
9.6 V
x 10 mA =
VS = 13.3 V
VL =
9.6 V
39o
VR = 12 V
Summary
Power factor
Pr =
96 mVAR
Pa = 133 mVA
39o
Ptrue = 120 mW
Summary
Apparent power
The power factor was discussed in Chapter 15 and applies to
RL circuits as well as RC circuits. Recall that it is the
relationship between the apparent power in volt-amperes and
true power in watts. Volt-amperes multiplied by the power
factor equals true power.
Power factor is defined as
PF = cos θ
Apparent power consists of two components; a true power
component, that does the work, and a reactive power
component, that is simply power shuttled back and forth
between source and load.
Power factor corrections for
an inductive load (motors,
generators, etc.) are done by
adding a parallel capacitor,
which has a canceling effect.
Ptrue (W)
Pr (VAR)
Pa (VA)
24
07-Dec-11
Summary
Summary
Frequency Response of RL Circuits
Frequency Response of RL Circuits
Series RL circuits have a frequency response similar to series RC
circuits. In the case of the low-pass response shown here, the output is
taken across the resistor.
Vin
Vin
10 V dc
10 V rms
10 V dc
0
10
10mH
mH
10
mH
10
mH
ƒ ==
1
ƒƒ
kHz
= 20
10kHz
kHz
8.46
10 V dc 0.79
1.57 V
V rms
rms V rms
100
Ω
100Ω
Ω
100
100
Ω
Vout
10VV rms
rms
10
Vout
10
10 V
V rms
rms
10 V dc
Reversing the position of the R and L components, produces the
high-pass response. The output is taken across the inductor.
100
100 Ω
Ω
100
Ω
100
Hz
kHz
ƒƒ ==101
kHz
10 V rms
10 V dc
0
10
10
10mH
mH
mH
10
mH
Vout (V)
Vout (V)
8.46
9.87
9
Plotting the response:
8
7
6
1.57
0.79
5
4
3
2
5.32
1
0.63
0.1
1
10 20
100
f (kHz)
Key Terms
Inductive
susceptance (BL)
0 V dc
0
9.98
Plotting the response:
9.87 V rms
5.32 V rms
0.63 V rms
The ability of an inductor to permit current; the
reciprocal of inductive reactance. The unit is the
siemens (S).
10
9
8
7
6
5
4
3
2
1
0
0.01
0.1
1
10
f (kHz)
Quiz
1. If the frequency is increased in a series RL circuit, the phase angle
will
a. increase
b. decrease
c. be unchanged
Quiz
2. If you multiply each of the impedance phasors in a series RL circuit
by the current, the result is the
a. voltage phasors
b. power phasors
c. admittance phasors
Quiz
3. For the circuit shown, the output voltage
a. is in phase with the input voltage
b. leads the input voltage
c. lags the input voltage
d. none of the above
Vin
Vout
d. none of the above
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07-Dec-11
Quiz
4. In a series RL circuit, the phase angle can be found from the equation
a.
 XL
 R
V
θ = tan −1  L
 VR
θ = tan −1 



Quiz
5.
In a series RL circuit, if the inductive reactance is equal to the
resistance, the source current will lag the source voltage by



c. both of the above are correct
a. 0o
d. none of the above is correct
c. 45o
b.
b. 30o
d. 90o
Quiz
6. Susceptance is the reciprocal of
a. resistance
Quiz
7. In a parallel RL circuit, the magnitude of the admittance can be
expressed as
a.
b. reactance
c. admittance
b.
d. impedance
Y=
1
1 1
+
G BL
Y = G 2 − BL 2
c. Y = G + BL
d. Y = G 2 + B 2
L
Quiz
8. If you increase the frequency in a parallel RL circuit,
Quiz
9. The unit used for measuring true power is the
a. the total admittance will increase
a. volt-ampere
b. the total current will increase
b. watt
c. both a and b
c. volt-ampere-reactive (VAR)
d. none of the above
d. kilowatt-hour
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07-Dec-11
Quiz
10. A power factor of zero implies that the
a. circuit is entirely reactive
b. reactive and true power are equal
c. circuit is entirely resistive
d. maximum power is delivered to the load
Quiz
Answers:
1. a
6. b
2. a
7. d
3. c
8. d
4. a
9. b
5. c
10. a
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