Cambridge University Press & Assessment
We are indebted to Sandeep S. Kaler of Toronto Metropolitan
University for creating almost all the figures here, as in the book
itself. This manual builds on and extends the work done by David
A. Torrey and Aleksandar M. Stankovic on the corresponding
manual for the first edition of the book. The authors also thank
Peter Buckles of Cambridge University Press for providing helpful
guidance.
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2 Form and Function: An Overview
PROBLEMS
2.1 The resistor voltage and current are shown in Fig. 2.19. As we go back through the switches,
the current is unfolded to give the source current shown in Fig. 2.20. Because iac and vac
are in phase,
⟨p(t)⟩ = Iacrms Vacrms
Vs Vs V2
⟨p(t)⟩ = √ √ = s
2R 2 2R
2.2 Even though the current through the load resistor is alternating, the current drawn from
the battery is constant. The bridge creates the alternating current in the load, with the
switching operations being transparent to the battery. The battery current is
Figure 2.19 The resistor voltage and current for problem 2.1.
Figure 2.20 The ac source current in problem 2.1.
1
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2 Chapter 2: Form and Function: An Overview
Vdc
Idc =
R
The power delivered to the load can be computed in two ways. The first is to determine
the time-varying current through the resistor and then calculate the power dissipation
2
P = RIR rms
. The second takes advantage of a power balance which recognizes that the
power supplied by the battery equals the power dissipated in the load. We know this to be
true since the ideal switches do not dissipate any power. This gives the power dissipation
in the resistor as
2
Vdc
P =
R
2.3 In periodic steady state operation there is zero average voltage across the load inductor.
The average voltage applied to the load by the converter is
2Vs
⟨vd ⟩ = cos α = Vdc
π
It follows then that
πVdc 75π
α = cos−1 = cos−1 = 0.8052r/s
2Vs 2(170)
2.4 The ac source current, ia , is shown in Fig. 2.21, where it has been assumed that the P
and N switches conduct for equal amounts of time and the P switches are closed a radians
after the source voltage goes through a positive going zero crossing. The ac source voltage
Figure 2.21 The ac source current and voltage for Problem 2.4.
is also shown in Fig. 2.21. The average power delivered to the current source is the average
power supplied by the ac source since the ideal switches do not dissipate any power.
Z
1 π+α 2Vs Idc
⟨p⟩ = Idc Vs sin ωt d(ωt) = cos α
π α π
2.5 Since an inductor cannot support an average voltage in the periodic steady state, we know
that ⟨vR ⟩ = ⟨vd ⟩ . It follows that
Z
1 π+α 2Vs
⟨vR ⟩ = Vs sin ωt d(ωt) = cos α
π α π
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Problems 3
This function is plotted in Fig. 2.22.
Figure 2.22 The average value of vd as a function of α for Problem 2.5.
2.6 When the P switches are conducting vd = V1 . When the N switches are conducting,
vd = −V1 . Since ⟨vd ⟩ = V2 ,
1
V2 = (V1 DT − V1 (1 − D)T ) = V1 (2D − 1)
T
2.7 With the switches operated so ωs = ωo , the amplitude of the nth harmonic of vac is given
by
−ȷnωo C 4
|Vacn | = R 2 2 Vdc
n ωo LC − ȷnωo RC − 1 nπ
This can be reduced to
ωo RC 4
|Vacn | = q Vdc
2 π
(1 − n ) + (nωo RC)
2 2
In this problem ωo has the value
1 1
ωo = √ = p ≈ 1.6 · 105
LC (159 · 10 ) (0.25 · 10−6 )
−6
The fundamental component of vac has an amplitude of
4
|Vac1 | = 100 = 127.3 V
π
The third harmonic has an amplitude of
1.6 · 105 (5) 0.25 · 10−6 4
|Vac3 | = q 100 = 3.15 V
2 π
(1 − 3 ) + [3 (1.6 · 10 ) (5) (0.25 · 10 )]
2 2 5 −6
Therefore,
|Vac3 |
≈ 2.5%
|Vac1 |
If we make the reasonable assumption that the fundamental and the third harmonic con-
tribute essentially all of the power dissipated in the resistor, the average power dissipated
in the resistor is given by
1 |Vac1 |2 |Vac3 |2 1 127.322 3.152
⟨p⟩ = + = + = 1622 W
R 2 2 5 2 2
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4 Chapter 2: Form and Function: An Overview
When the switching frequency is adjusted so that ωs = 3ωo , the relationship between the
harmonic voltages and va becomes
3ωo RC 4
|Vacn | = q Vdc
2 π
(1 − (3n) ) + (3nωo RC)
2 2
We now expect that the fundamental and the third harmonic are significantly smaller than
before. Putting in the numbers gives
3 1.6 · 105 (5) 0.25 · 10−6 4
|Vac1 | = q 100 = 9.4 V
2 π
(1 − (3) ) + [3 (1.6 · 10 ) (5) (0.25 · 10 )]
2 2 5 −6
|Vac3 | = 0.95 V
The ratio of the third to the first harmonic amplitudes is now
|Vac3 |
≈ 10%
|Vac1 |
Note that the third harmonic now makes up a larger percentage of vac . Making the same
approximation regarding the power delivered to the load, we get
1 9.42 0.952
⟨p⟩ = + = 8.9 W
5 2 2
Because the RLC impedance looks inductive above ωo , we expect vac to be an approximate
sinusoid which is lagging behind va , as shown in Fig. 2.23.
Figure 2.23 Voltages va and vac when ωs = 3ωo in Problem 2.7.
2.8 The output waveform for vd using the topologies of Figs. 2.7 and 2.8 in the text are
given in Figs. 2.24 and 2.25. In each case the duty ratio is determined by the constraint
⟨vd ⟩ = 1/2V1 . The switching period T is assumed to be 10−3 s.
With either converter, the ripple in the inductor current can be determined to first order
by assuming the output voltage vo is constant at Vo . Under this assumption, the voltage
across the filter inductor is vd − Vo . When this voltage is positive, the inductor current
is increasing, and when this voltage is negative, the inductor current is decreasing. The
peak to peak ripple in the current is found by considering either the total increase or the
total decrease over one cycle. If the increase is considered, vd = V1 and the peak to peak
current ripple is
1
∆io = (V1 − Vo ) DT
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We are indebted to Sandeep S. Kaler of Toronto Metropolitan
University for creating almost all the figures here, as in the book
itself. This manual builds on and extends the work done by David
A. Torrey and Aleksandar M. Stankovic on the corresponding
manual for the first edition of the book. The authors also thank
Peter Buckles of Cambridge University Press for providing helpful
guidance.
, i i
solns CUP/HE10-mono-design kassakian9-21-2022.sty August 2, 2023 20:34 Page-1
i i
2 Form and Function: An Overview
PROBLEMS
2.1 The resistor voltage and current are shown in Fig. 2.19. As we go back through the switches,
the current is unfolded to give the source current shown in Fig. 2.20. Because iac and vac
are in phase,
⟨p(t)⟩ = Iacrms Vacrms
Vs Vs V2
⟨p(t)⟩ = √ √ = s
2R 2 2R
2.2 Even though the current through the load resistor is alternating, the current drawn from
the battery is constant. The bridge creates the alternating current in the load, with the
switching operations being transparent to the battery. The battery current is
Figure 2.19 The resistor voltage and current for problem 2.1.
Figure 2.20 The ac source current in problem 2.1.
1
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2 Chapter 2: Form and Function: An Overview
Vdc
Idc =
R
The power delivered to the load can be computed in two ways. The first is to determine
the time-varying current through the resistor and then calculate the power dissipation
2
P = RIR rms
. The second takes advantage of a power balance which recognizes that the
power supplied by the battery equals the power dissipated in the load. We know this to be
true since the ideal switches do not dissipate any power. This gives the power dissipation
in the resistor as
2
Vdc
P =
R
2.3 In periodic steady state operation there is zero average voltage across the load inductor.
The average voltage applied to the load by the converter is
2Vs
⟨vd ⟩ = cos α = Vdc
π
It follows then that
πVdc 75π
α = cos−1 = cos−1 = 0.8052r/s
2Vs 2(170)
2.4 The ac source current, ia , is shown in Fig. 2.21, where it has been assumed that the P
and N switches conduct for equal amounts of time and the P switches are closed a radians
after the source voltage goes through a positive going zero crossing. The ac source voltage
Figure 2.21 The ac source current and voltage for Problem 2.4.
is also shown in Fig. 2.21. The average power delivered to the current source is the average
power supplied by the ac source since the ideal switches do not dissipate any power.
Z
1 π+α 2Vs Idc
⟨p⟩ = Idc Vs sin ωt d(ωt) = cos α
π α π
2.5 Since an inductor cannot support an average voltage in the periodic steady state, we know
that ⟨vR ⟩ = ⟨vd ⟩ . It follows that
Z
1 π+α 2Vs
⟨vR ⟩ = Vs sin ωt d(ωt) = cos α
π α π
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Problems 3
This function is plotted in Fig. 2.22.
Figure 2.22 The average value of vd as a function of α for Problem 2.5.
2.6 When the P switches are conducting vd = V1 . When the N switches are conducting,
vd = −V1 . Since ⟨vd ⟩ = V2 ,
1
V2 = (V1 DT − V1 (1 − D)T ) = V1 (2D − 1)
T
2.7 With the switches operated so ωs = ωo , the amplitude of the nth harmonic of vac is given
by
−ȷnωo C 4
|Vacn | = R 2 2 Vdc
n ωo LC − ȷnωo RC − 1 nπ
This can be reduced to
ωo RC 4
|Vacn | = q Vdc
2 π
(1 − n ) + (nωo RC)
2 2
In this problem ωo has the value
1 1
ωo = √ = p ≈ 1.6 · 105
LC (159 · 10 ) (0.25 · 10−6 )
−6
The fundamental component of vac has an amplitude of
4
|Vac1 | = 100 = 127.3 V
π
The third harmonic has an amplitude of
1.6 · 105 (5) 0.25 · 10−6 4
|Vac3 | = q 100 = 3.15 V
2 π
(1 − 3 ) + [3 (1.6 · 10 ) (5) (0.25 · 10 )]
2 2 5 −6
Therefore,
|Vac3 |
≈ 2.5%
|Vac1 |
If we make the reasonable assumption that the fundamental and the third harmonic con-
tribute essentially all of the power dissipated in the resistor, the average power dissipated
in the resistor is given by
1 |Vac1 |2 |Vac3 |2 1 127.322 3.152
⟨p⟩ = + = + = 1622 W
R 2 2 5 2 2
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4 Chapter 2: Form and Function: An Overview
When the switching frequency is adjusted so that ωs = 3ωo , the relationship between the
harmonic voltages and va becomes
3ωo RC 4
|Vacn | = q Vdc
2 π
(1 − (3n) ) + (3nωo RC)
2 2
We now expect that the fundamental and the third harmonic are significantly smaller than
before. Putting in the numbers gives
3 1.6 · 105 (5) 0.25 · 10−6 4
|Vac1 | = q 100 = 9.4 V
2 π
(1 − (3) ) + [3 (1.6 · 10 ) (5) (0.25 · 10 )]
2 2 5 −6
|Vac3 | = 0.95 V
The ratio of the third to the first harmonic amplitudes is now
|Vac3 |
≈ 10%
|Vac1 |
Note that the third harmonic now makes up a larger percentage of vac . Making the same
approximation regarding the power delivered to the load, we get
1 9.42 0.952
⟨p⟩ = + = 8.9 W
5 2 2
Because the RLC impedance looks inductive above ωo , we expect vac to be an approximate
sinusoid which is lagging behind va , as shown in Fig. 2.23.
Figure 2.23 Voltages va and vac when ωs = 3ωo in Problem 2.7.
2.8 The output waveform for vd using the topologies of Figs. 2.7 and 2.8 in the text are
given in Figs. 2.24 and 2.25. In each case the duty ratio is determined by the constraint
⟨vd ⟩ = 1/2V1 . The switching period T is assumed to be 10−3 s.
With either converter, the ripple in the inductor current can be determined to first order
by assuming the output voltage vo is constant at Vo . Under this assumption, the voltage
across the filter inductor is vd − Vo . When this voltage is positive, the inductor current
is increasing, and when this voltage is negative, the inductor current is decreasing. The
peak to peak ripple in the current is found by considering either the total increase or the
total decrease over one cycle. If the increase is considered, vd = V1 and the peak to peak
current ripple is
1
∆io = (V1 − Vo ) DT
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