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electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
chapter 10
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of 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.
VR
VC
V R leads VS
V C lags V S
R
C
VS
I
I leads V S
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Impedance of series RC circuits
In a series RC circuit, the total impedance is the phasor
sum of R and XC.
R is plotted along the positive x-axis.
XC is plotted along the negative y-axis.
 XC 

 R 
  tan 1 
R
R


Z is the diagonal
XC
XC
Z
Z
It is convenient to reposition the
phasors into the impedance triangle.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Impedance of series RC circuits
Sketch the impedance triangle and show the
values for R = 1.2 kW and XC = 960 W.
Z
1.2 kW 
2
+  0.96 kW 
 1.33 kW
  tan 1
0.96 kW
1.2 kW
 39
Electronics Fundamentals 8th edition
Floyd/Buchla
2
R = 1.2 kW

39o
Z = 1.33 kW
XC =
960 W
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Analysis of series RC circuits
Ohm’s law is applied to series RC circuits using Z,
V, and I.
V  IZ
V
I
Z
V
Z
I
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Analysis of series RC circuits
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.
R = 1.2 kW

39o
Z = 1.33 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XC =
960 W
VR = 12 V

39o
VS = 13.3 V
VC =
9.6 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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.
R
As frequency changes,
Increasing f



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.
3
2
1
3
C3
3
C2
2
C1
1
2
1
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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.
V
R

Vin
C
Vout
VR
f
(phase lag)
Vout
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
f
Vin
Vin
(phase lag)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Applications
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.
C
V
Vout

Vin
(phase lead)
Vin
R
Vout
Vout
VC
Electronics Fundamentals 8th edition
Floyd/Buchla
Vin

(phase lead)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Applications
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.
Amplifier
Rf
Phase-shift network
C
C
C
R
Electronics Fundamentals 8th edition
Floyd/Buchla
R
R
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance) and to review
conductance.
Conductance is the reciprocal of resistance.
Capacitive susceptance is the reciprocal
of capacitive reactance.
G
1
R
1
BC 
XC
1
Admittance is the reciprocal of impedance. Y 
Z
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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  G2 + BC 2
 BC 

G 
From the diagram, the phase angle is   tan 1 
BC
Y
VS
G
BC

Electronics Fundamentals 8th edition
Floyd/Buchla
G
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BC is plotted along the positive y-axis.
 BC 

G 
  tan 1 
Y is the diagonal
BC
Y
VS
G
BC

Electronics Fundamentals 8th edition
Floyd/Buchla
G
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
G
1
1

 1.0 mS
R 1.0 kW
Y  G 2 + BC 2 
BC  2 10 kHz  0.01 mF  0.628 mS
1.0 mS +  0.628 mS  1.18 mS
2
2
BC = 0.628 mS
VS
f = 10 kHz
R
1.0 kW
C
0.01 mF
Y=
1.18 mS
G = 1.0 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Analysis of parallel RC circuits
Ohm’s law is applied to parallel RC circuits using
Y, V, and I.
V
I
Y
I  VY Y 
I
V
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Analysis 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
G = 1.0 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 V
=
IC = 6.28 mA
IS =
11.8 mA
IR = 10 mA
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Phase angle of parallel RC circuits
Notice that the formula for capacitive susceptance is
the reciprocal of capacitive reactance. Thus BC and IC
are directly proportional to f: BC  2 fC
As frequency increases, BC
and IC must also increase,
so the angle between IR and
IS must increase.
Electronics Fundamentals 8th edition
Floyd/Buchla
IC
IS

IR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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 

Z
Electronics Fundamentals 8th edition
Floyd/Buchla
XC(eq) = Z sin 
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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
R1
C1
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
Electronics Fundamentals 8th edition
Floyd/Buchla
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.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
 20 μs 
T  8.0 div 
  160 μs
 div 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
 t 
 24.5 μs 
o
    360  
 360  55
T 
 160 μs 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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 kW

39o
Z = 1.33 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XC =
960 W
VR = 12 V

39o
VS = 13.3 V
VC =
9.6 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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.
The rms current in the earlier example was 10 mA.
Show the power triangle.
VR = 12 V
x 10 mA
=

39o
VS = 13.3 V
Electronics Fundamentals 8th edition
Floyd/Buchla
VC =
9.6 V
Ptrue = 120 mW

39o
Pa = 133 mVA
Pr = 96
mVAR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Power factor
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
PF = cos 
The power factor can vary from 0 for a purely reactive
circuit to 1 for a purely resistive circuit.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Apparent power
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.
Some components such
as transformers, motors,
and generators are rated
in VA rather than watts.
Electronics Fundamentals 8th edition
Floyd/Buchla
Ptrue (W)

Pa (VA)
Pr (VAR)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
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.
Vin
10 V dc
Vout
10
10VVrms
rms
10 V rms
0
10 V dc
100
100W
W
W
100
20
kHz
ƒƒƒ == 110
kHz
kHz
V 8.46
rms V rms
10 V dc 1.57
0.79V
rms
0
11 mm
mFF
F
Vout (V)
9.98
Plotting the response:
8.46
1.57
0.79
9
8
7
6
5
4
3
2
1
0.1
Electronics Fundamentals 8th edition
Floyd/Buchla
1
10 20
100
f (kHz)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Frequency Response of RC Circuits
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.
Vin
10 V dc
Vout
10V
V rms
rms
10
10 V rms
0
10 V dc
m
F
11mm
mF
F
1
F
ƒ
=
100
Hz
ƒ
=
1
kHz
ƒ = 10 kHz
9.87V rms
5.32 V rms
0.63 V rms
100 W
W
100
100
100 W
0 V dc
Vout (V)
9.87
Plotting the response:
5.32
0.63
Electronics Fundamentals 8th edition
Floyd/Buchla
10
9
8
7
6
5
4
3
2
1
0
0.01
0.1
1
10
f (kHz)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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).
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
a. multiplying each phasor by the current
b. multiplying each phasor by the source voltage
c. dividing each phasor by the source voltage
d. dividing each phasor by the current
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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. be in phase with the input.
b. lead the input voltage.
c. lag the input voltage.
d. none of the above
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
d. 14.1 V
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
5. Admittance is the reciprocal of
a. reactance
b. resistance
c. conductance
d. impedance
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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. multiplying each phasor by the total current
c. dividing each phasor by the voltage
d. dividing each phasor by the total current
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
9. The maximum power factor occurs when the phase
angle is
a. 0o
b. 30o
c. 45o
d. 90o
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
10. When power is calculated from voltage and current for an
ac circuit, the voltage and current should be expressed as
a. average values
b. rms values
c. peak values
d. peak-to-peak values
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
Answers:
Electronics Fundamentals 8th edition
Floyd/Buchla
1. a
6. a
2. b
7. d
3. c
8. d
4. c
9. a
5. d
10. b
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.