CONTENTS vii
8.5 One phase Controlled Rectifiers 104
8.5.1 Inverter Mode 104
8.6 Three-Phase Controlled Converters 106
8.7 *Notes 107
9 Inverters 109
9.1 1-phase Inverter 109
9.2 Three-phase Inverters 111
Preface
The purpose of these notes is be used to introduce Electrical Engineering students to Electrical
Machines, Power Electronics and Electrical Drives. They are primarily to serve our students at
MSU: they come to the course on Energy Conversion and Power Electronics with a solid background
in Electric Circuits and Electromagnetics, and many want to acquire a basic working knowledge
of the material, but plan a career in a different area (venturing as far as computer or mechanical
engineering). Other students are interested in continuing in the study of electrical machines and
drives, power electronics or power systems, and plan to take further courses in the field.
Starting from basic concepts, the student is led to understand how force, torque, induced voltages
and currents are developed in an electrical machine. Then models of the machines are developed, in
terms of both simplified equations and of equivalent circuits, leading to the basic understanding of
modern machines and drives. Power electronics are introduced, at the device and systems level, and
electrical drives are discussed.
Equations are kept to a minimum, and in the examples only the basic equations are used to solve
simple problems.
These notes do not aim to cover completely the subjects of Energy Conversion and Power
Electronics, nor to be used as a reference, not even to be useful for an advanced course. They are
meant only to be an aid for the instructor who is working with intelligent and interested students,
who are taking their first (and perhaps their last) course on the subject. How successful this endeavor
has been will be tested in the class and in practice.
In the present form this text is to be used solely for the purposes of teaching the introductory
course on Energy Conversion and Power Electronics at MSU.
E.G.STRANGAS
E. Lansing, Michigan and Pyrgos, Tinos
ix
A Note on Symbols
Throughout this text an attempt has been made to use symbols in a consistent way. Hence a script
letter, say v denotes a scalar time varying quantity, in this case a voltage. Hence one can see
v = 5 sinωt or v = ˆv sin ωt
The same letter but capitalized denotes the rms value of the variable, assuming it is periodic.
Hence:
v =
√
2V sinωt
The capital letter, but now bold, denotes a phasor:
V = V e
jθ
Finally, the script letter, bold, denotes a space vector, i.e. a time dependent vector resulting from
three time dependent scalars:
v = v
1
+ v
2
e
jγ
+ v
3
e
j2γ
In addition to voltages, currents, and other obvious symbols we have:
B Magnetic flux Density (T)
H Magnetic filed intensity (A/m)
Φ Flux (Wb) (with the problem that a capital letter is used to show a time
dependent scalar)
λ, Λ, λ
λ
λ flux linkages (of a coil, rms, space vector)
ω
s
synchronous speed (in electrical degrees for machines with more than
two-poles)
ω
o
rotor speed(in electricaldegreesfor machines with morethantwo-poles)
ω
m
rotor speed (mechanical speed no matter how many poles)
ω
r
angular frequency of the rotor currents and voltages (in electrical de-
grees)
T Torque (Nm)
(·), (·) Real and Imaginary part of ·
x
1
Three Phase Circuits and Power
Chapter Objectives
In this chapter you will learn the following:
• The concepts of power, (real reactive and apparent) and power factor
• The operation of three-phase systems and the characteristics of balanced loads in Y and in ∆
• How to solve problems for three-phase systems
1.1 ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES
We start from the basic equation for the instantaneous electric power supplied to a load as shown in
figure 1.1
+
v(t)
i(t)
Fig. 1.1 A simple load
p(t) = i(t) ·v(t) (1.1)
1
2 THREE PHASE CIRCUITS AND POWER
where i(t) is the instantaneous value of current through the load and v(t) is the instantaneous value
of the voltage across it.
In quasi-steady state conditions, the current and voltage are both sinusoidal, with corresponding
amplitudes
ˆ
i and ˆv, and initial phases, φ
i
and φ
v
, and the same frequency, ω = 2π/T −2πf:
v(t) = ˆv sin(ωt + φ
v
) (1.2)
i(t) =
ˆ
i sin(ωt + φ
i
) (1.3)
In this case the rms values of the voltage and current are:
V =
1
T
T
0
ˆv [sin(ωt + φ
v
)]
2
dt =
ˆv
√
2
(1.4)
I =
1
T
T
0
ˆ
i [sin(ωt + φ
i
)]
2
dt =
ˆ
i
√
2
(1.5)
and these two quantities can be described by phasors, V = V
φ
v
and I = I
φ
i
.
Instantaneous power becomes in this case:
p(t) = 2V I [sin(ωt + φ
v
) sin(ωt + φ
i
)]
= 2V I
1
2
[cos(φ
v
− φ
i
) + cos(2ωt + φ
v
+ φ
i
)] (1.6)
The first part in the right hand side of equation 1.6 is independent of time, while the second part
varies sinusoidally with twice the power frequency. The average power supplied to the load over
an integer time of periods is the first part, since the second one averages to zero. We define as real
power the first part:
P = V I cos(φ
v
− φ
i
) (1.7)
If we spend a moment looking at this, we see that this power is not only proportional to the rms
voltage and current, but also to cos(φ
v
− φ
i
). The cosine of this angle we define as displacement
factor, DF. At the same time, and in general terms (i.e. for periodic but not necessarily sinusoidal
currents) we define as power factor the ratio:
pf =
P
V I
(1.8)
and that becomes in our case (i.e. sinusoidal current and voltage):
pf = cos(φ
v
− φ
i
) (1.9)
Note that this is not generally the case for non-sinusoidal quantities. Figures 1.2 - 1.5 show the cases
of power at different angles between voltage and current.
We call the power factor leading or lagging, depending on whether the current of the load leads
or lags the voltage across it. It is clear then that for an inductive/resistive load the power factor is
lagging, while for a capacitive/resistive load the power factor is leading. Also for a purely inductive
or capacitive load the power factor is 0, while for a resistive load it is 1.
We define the product of the rms values of voltage and current at a load as apparent power, S:
S = V I (1.10)
ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES 3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
−0.5
0
0.5
1
i(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−2
−1
0
1
2
u(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−0.5
0
0.5
1
1.5
p(t)
Fig. 1.2 Power at pf angle of 0
o
. The dashed line shows average power, in this case maximum
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
−0.5
0
0.5
1
i(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−2
−1
0
1
2
u(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−0.5
0
0.5
1
1.5
p(t)
Fig. 1.3 Power at pf angle of 30
o
. The dashed line shows average power
and as reactive power, Q
Q = V I sin(φ
v
− φ
i
) (1.11)
Reactive power carries more significance than just a mathematical expression. It represents the
energy oscillating in and out of an inductor or a capacitor and a source for this energy must exist.
Since the energy oscillation in an inductor is 180
0
out of phase of the energy oscillating in a capacitor,
4 THREE PHASE CIRCUITS AND POWER
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
−0.5
0
0.5
1
i(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−2
−1
0
1
2
u(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
−0.5
0
0.5
1
p(t)
Fig. 1.4 Power at pf angle of 90
o
. The dashed line shows average power, in this case zero
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
−0.5
0
0.5
1
i(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−2
−1
0
1
2
u(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1.5
−1
−0.5
0
0.5
p(t)
Fig. 1.5 Power at pf angle of 180
o
. The dashed line shows average power, in this case negative, the opposite
of that in figure 1.2
the reactive power of the two have opposite signs by convention positive for an inductor, negative for
a capacitor.
The units for real power are, of course, W , for the apparent power V A and for the reactive power
V Ar.
SOLVING 1-PHASE PROBLEMS 5
Using phasors for the current and voltage allows us to define complex power S as:
S = VI
∗
(1.12)
= V
φ
v
I
−φ
i
(1.13)
and finally
S = P + jQ (1.14)
For example, when
v(t) =
(2 · 120 ·sin(377t +
π
6
)V (1.15)
it =
(2 · 5 ·sin(377t +
π
4
)A (1.16)
then S = V I = 120 · 5 = 600W, while pf = cos(π/6 − π/4) = 0.966 leading. Also:
S = VI
∗
= 120
π /6
5
−π/ 4
= 579.6W −j155.3V Ar (1.17)
Figure 1.6 shows the phasors for laggingand leadingpower factors and thecorresponding complex
power S.
S
S
jQ
jQ
P
P
V
V
I
I
Fig. 1.6 (a) lagging and (b) leading power factor
1.2 SOLVING 1-PHASE PROBLEMS
Based on the discussion earlier we can construct the table below:
Type of load Reactive power Power factor
Reactive Q > 0 lagging
Capacitive Q < 0 leading
Resistive Q = 0 1
6 THREE PHASE CIRCUITS AND POWER
We also notice that if for a load we know any two of the four quantities, S, P, Q, pf, we can
calculate the other two, e.g. if S = 100kV A, pf = 0.8 leading, then:
P = S · pf = 80kW
Q = −S
1 − pf
2
= −60kV Ar , or
sin(φ
v
− φ
i
) = sin [arccos 0.8]
Q = S sin(φ
v
− φ
i
)
Notice that here Q < 0, since the pf is leading, i.e. the load is capacitive.
Generally in a system with more than one loads (or sources) real and reactive power balance, but
not apparent power, i.e. P
total
=
i
P
i
, Q
total
=
i
Q
i
, but S
total
=
i
S
i
.
In the same case, if the load voltage were V
L
= 2000V , the load current would be I
L
= S/V
= 100 · 10
3
/2 · 10
3
= 50A. If we use this voltage as reference, then:
V = 2000
0
V
I = 50
φ
i
= 50
36.9
o
A
S = V I
∗
= 2000
0
· 50
−36.9
o
= P + jQ = 80 ·10
3
W −j60 · 10
3
V Ar
1.3 THREE-PHASE BALANCED SYSTEMS
Compared tosingle phasesystems, three-phase systems offer definite advantages: for thesame power
and voltage there is less copper in the windings, and the total power absorbed remains constant rather
than oscillate around its average value.
Let us take now three sinusoidal-current sources that have the same amplitude and frequency, but
their phase angles differ by 120
0
. They are:
i
1
(t) =
√
2I sin(ωt + φ)
i
2
(t) =
√
2I sin(ωt + φ −
2π
3
) (1.18)
i
3
(t) =
√
2I sin(ωt + φ +
2π
3
)
If these three current sources are connected as shown in figure 1.7, the current returning though node
n is zero, since:
sin(ωt + φ) + sin(ωt − φ +
2π
3
) + sin(ωt + φ +
2π
3
) ≡ 0 (1.19)
Let us also take three voltage sources:
v
a
(t) =
√
2V sin(ωt + φ)
v
b
(t) =
√
2V sin(ωt + φ −
2π
3
) (1.20)
v
c
(t) =
√
2V sin(ωt + φ +
2π
3
)
connected as shown in figure 1.8. If the three impedances at the load are equal, then it is easy
to prove that the current in the branch n − n
is zero as well. Here we have a first reason why
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