Electrical Machines and Systems
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Course in
Electrical Machines and Systems
Year 2
©2004 J D Edwards
ELECTRICAL MACHINES AND SYSTEMS COURSE NOTES
These notes were prepared for a Year 2 course module in the Department of Engineering and
Design at the University of Sussex. The module runs for ten weeks, and absorbs 25% of student
time.
This introductory course in electrical machines follows on from the Year 1 course
Electromechanics, with a similar philosophy. It avoids the traditional mathematical derivation of
the theory of AC and DC machines, and makes extensive use of field plots generated with MagNet
to explain the principles. With this physical background, the theory is developed in terms of circuit
models and phasor diagrams.
Field plots are particularly useful for explaining armature reaction and compensating windings in
DC machines, and for demonstrating the action of variable-reluctance and hybrid stepper motors.
They also show the essential unity of the conventional rotating machines; there are similar plots for
the stator and rotor field components and the resultant field in DC, synchronous and induction
machines.
Electrical Machines and Systems Course Notes Copyright © 2004 J D Edwards
CONTENTS
1
INTRODUCTION
1
2
TRANSFORMERS
2
2.1
Introduction
2
2.2
Types of power transformer
3
2.3
Ideal transformer properties
4
2.4
Circuit model of a transformer
6
2.5
Parameter determination
7
2.6
Transformer performance
9
8
2.7
Current transformers
11
8.1
AC/DC Converters
66
2.8
Transformer design
12
8.2
DC motor control
68
8.3
DC/AC Inverters
68
8.4
AC motor control
70
3
DC MACHINES
15
3.1
Introduction
15
3.2
DC machines in practice
16
3.3
Characteristics and control
18
3.4
Series motors
21
4
INTRODUCTION TO AC MACHINES
24
4.1
Review of 3-phase systems
24
4.2
Rotating magnetic field
25
4.3
Multi-pole fields
28
5
SYNCHRONOUS MACHINES
30
5.1
Introduction
30
5.2
Characteristics
31
5.3
Salient-pole machines
35
5.4
Linear synchronous motors
37
6
INDUCTION MACHINES
39
6.1
Introduction
39
6.2
Characteristics
40
6.3
Losses and efficiency
46
6.4
Parameter determination
48
6.5
Single-phase induction motors
50
6.6
Dynamic conditions
52
6.7
Linear induction motors
54
7
STEPPER MOTORS
56
7.1
Introduction
56
7.2
Variable-reluctance principle
56
7.3
Variable-reluctance stepper motors
58
7.4
Hybrid stepper motors
59
7.5
Stepper motor characteristics
61
7.6
Stepper motor control
64
9
POWER ELECTRONIC CONTROL
66
REFERENCES
76
10 APPENDICES
77
10.1
Induction motor 2-axis equations
77
10.2
List of formulae
80
1 INTRODUCTION
This course follows on from the Term 3 course
Electromechanics. Its purpose is to explore in
greater depth the AC and DC machines that were
introduced in the earlier course. The approach is
that of the application engineer rather than the
machine designer, concentrating on the basic
principles, characteristics, and control. Since
induction motors account for more than 90 per cent
of the motors used in industry, the course gives
particular emphasis to these machines.
Magnetic field plots
A magnetic field plot is often a useful way of
picturing the operation of an electromagnetic
device. Numerous plots have been specially
prepared for these notes, using the MagNet
electromagnetic simulation software, to develop
the basic concepts with a minimum of
mathematics.
References
Course components
The course has three closely linked components:
lectures, problem sheets and laboratories. In
addition, there is a design assignment, which
introduces some of the basic ideas and problems of
design by considering a very simple device: an
electromagnet.
Lectures will use video presentation and
practical demonstrations. These notes provide
support material for the lectures, but they are not a
substitute. Regular lecture attendance is essential.
Problem solving is a vital part of the course.
Problem sheets will be issued at the first lecture
each week, and methods of solving the problems
will be discussed in each lecture.
The laboratory runs from week 4 to week 9,
with three 3-hour experiments:
EMS1: Speed control of induction motors.
EMS2: Characteristics of a power transformer.
EMS3: Control of a stepper motor.
Experiment EMS1 is a sequel to the simple DC
motor-control experiment in Electromechanics.
EMS2 introduces some important electrical
measurement techniques as well as exploring the
properties of a transformer. EMS3 explores a
stepper motor and controller of the kind widely
used in industry.
References to books are listed in section 10, and
cited in the text of the notes with the reference
number in square brackets.
Background material
The course assumes a familiarity with the contents
of the Term 3 course Electromechanics, so the
basic principles covered in that course will not be
repeated. Students are expected to have a copy of
the printed notes for Electromechanics [1], and
further information will be found in references [2]
to [4].
Introduction 1
2
Sinusoidal operation
TRANSFORMERS
If the voltage source is sinusoidal, then the core
flux will also be sinusoidal, so we may put:
m sin t
Introduction
Basic transformer principles were covered in
Electromechanics [1], and the main results are
given below. Figure 2-1 is a schematic
representation of a single-phase transformer with
two coils on a magnetic core, where the magnetic
coupling is assumed to be perfect: the same flux
passes through each turn of each coil.
i1
+
v1
~
+
e1
i2
R1
+
R2
e2
+
Z v2
v1 N 1
V1m N 1 m 2 fN 1 m 2 fN 1 ABm
Figure 2-1: Transformer with source and load
Voltage relationships
Kirchhoff’s voltage law applied to the two
windings gives:
(2-1)
d
R2 i 2
dt
(2-2)
v 2 e 2 R2 i 2 N 2
2
4
6
8 10 12 14 16 18 20
Magnetic intensity H, kA/m
Figure 2-2: Silicon transformer steel.
Current relationships
If the resistances R1 and R2 are negligible, then
equations 2-1 and 2-2 become:
v1 N 1
d
dt
(2-3)
v2 N 2
d
dt
(2-4)
Dividing these equations gives the important result:
v1 N 1
v2 N 2
The relationship between the primary and
secondary currents can be found by considering the
magnetic circuit of the transformer. From the basic
magnetic circuit equation, we have:
F N 1i1 N 2 i2 R
(2-9)
N 1i1 N 2 i2 0
(2-10)
In a well-designed transformer, the reluctance R is
small, so equation 2-9 becomes:
(2-5)
This gives the counterpart of equation 2-5 for
voltage:
i1 N 2
i2 N 1
2
(2-8)
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
d
R1i1
dt
(2-7)
where A is the cross-sectional area of the core and
Bm is the maximum flux density in the core. A
typical value for Bm is 1.4 T for the silicon steel
characteristic in figure 2-2.
N2 turns
v1 e1 R1i1 N 1
d
N 1 m cos t V1m cos t
dt
Thus the maximum primary voltage is:
secondary
primary
N1 turns
(2-6)
Substituting this expression in equation 2-3 gives:
Flux density B, T
2.1
Electrical Machines and Systems Course Notes
(2-11)
If sinusoidal voltages and currents are represented
by phasors, the corresponding forms for the basic
voltage and current equations are:
V1 N 1
V2 N 2
(2-12)
I1 N 2
I 2 N1
(2-13)
Transformer rating
The maximum voltage at the primary terminals of a
transformer is determined by equation 2-8, and is
independent of the current. The maximum primary
current is determined by the I2R power loss in the
resistance of the transformer windings, which
generates heat in the transformer. This power loss
is independent of the applied voltage.
Consequently, for a given design of transformer,
there is a maximum value for the product V1I1 at
the primary terminals. To a first approximation,
this is also equal to the product V2I2 at the
secondary terminals. This maximum value does not
depend on the phase angle between the voltage and
the current. Transformer ratings therefore specify
the apparent power VI (volt-amperes, VA) rather
than the real power VI cos (watts, W).
2.2
Types of power transformer
In addition to the ordinary single-phase power
transformer, two other types are in common use:
auto-wound transformers, and 3-phase
transformers.
Auto-wound transformers
A transformer can have a single coil with an output
taken from a portion of the coil, as shown in figure
2-3. This is known as an auto-wound transformer
or auto-transformer.
Unlike the normal transformer with two windings,
known as a double-wound transformer, the autowound transformer does not provide electrical
isolation between the primary and the secondary,
However, an auto-wound transformer can have a
much larger apparent power rating than a doublewound transformer of the same physical size.
Let N1 be the number of turns on the upper part
of the winding in figure 2-3, and N2 the number of
turns on the lower part. The conventional
transformer equations 2-12 and 2-13 apply to these
parts of the winding, since they are equivalent to
two separate windings with a common connection.
Applying Kirchhoff’s law to this circuit gives:
VS V1 V2
(2-14)
I L I1 I 2
(2-15)
As an example, suppose that N1 = N2. If the
transformer is regarded as ideal (see section 2.3),
then I1 = I2 and V1 = V2. Equations 2-14 and 2-15
give:
VS V1 V2 2V2 2VL
(2-16)
I L I1 I 2 2I1 2I S
(2-17)
where VL is the voltage across the load and IS is the
current supplied by the source. This auto-wound
transformer behaves as a step-down transformer
with a ratio of 2:1, and the current in each winding
is equal to half of the load current.
An elegant application of the auto-wound
transformer principle is the variable transformer,
which has a single-layer coil wound on a toroidal
core. The output is taken from a carbon brush that
makes contact with the surface of the coil; the
brush can be moved smoothly from one end of the
coil to the other, thus varying the output voltage.
Examples of variable transformers are shown in
figure 2-4.
I1
+
VS
ZL V1
IL
I2 V2
+
+
Figure 2-3: Auto-wound transformer.
Transformers 3
a m cos t
b m cos( t 120)
c m cos( t 240) m cos( t 120)
(2-18)
Figure 2-7 shows flux plots for the transformer at
the instants when t = 0, 120º and 240º.
(a)
Figure 2-4: Variable transformers.
(RS Components Ltd)
3-phase transformers
In 3-phase systems (see section 4.1), it is common
practice to use sets of three single-phase
transformers. It is also possible, however, to make
3-phase transformers with three sets of windings
on three limbs of a core, as shown in figure 2-5.
(b)
(c)
Figure 2-5: 3-phase transformer model.
Figure 2-7: 3-phase transformer flux plots:
(a) 0º, (b) 120º, (c) 240º.
The corresponding fluxes are shown in figure 2-6.
There is no requirement for another limb to form a
flux return path, because the fluxes a, b and c
sum to zero in a balanced 3-phase system. The
proof is as follows. From equation 2-18, the sum is
given by:
a
b
c
Figure 2-6: 3-phase transformer flux.
In a balanced system with sinusoidal phase
voltages, the fluxes will be given by:
4
a b c
m
cos t cos( t 120) cos( t 120)
cos t 2 cos t cos120
cos t cos t 0
(2-19)
Because the fluxes in the three limbs sum to zero at
all instants of time, there is no leakage of flux from
the core, as the flux plots in figure 2-7
demonstrate.
Electrical Machines and Systems Course Notes
2.3
Ideal transformer properties
I1
If the primary and secondary windings have zero
resistance, and the magnetic core has zero
reluctance, then the approximate equalities in
equations 2-12 and 2-13 become exact equalities.
This leads to the concept of an ideal transformer
element, to accompany the other ideal elements of
circuit theory. Figure 2-8 shows a circuit symbol
for the ideal transformer element.
i1
+
V1
ZL
Figure 2-9: Ideal transformer with a load.
The secondary impedance is given by:
ZL
+
v1
+
V2
N1:N2
i2
+
I2
v2
V2
I2
(2-22)
At the primary terminals, the circuit presents an
impedance given by:
Figure 2-8: Ideal transformer element
The voltage and current relationships in the time
and frequency domains are given in table 2-1.
Table 2-1: Ideal transformer relationships.
Time domain
Frequency domain
v2 N 2
n
v1 N 1
V2 N 2
n
V1 N 1
(2-20)
i1 N 2
n
i2 N 1
I1 N 2
n
I 2 N1
(2-21)
N 1 V2
V
N2
Z in 1
N
I1
2I2
N1
N
1
N2
(2-23)
V2 N 1
I2 N 2
2
Z
Z L 2L
n
2
Thus the combination of an ideal transformer of
ratio n and an impedance ZL can be replaced by an
equivalent impedance ZL / n2.
The following properties of the ideal transformer
may be deduced from equations 2-16 and 2-17:
The voltage transformation is independent of
the current, and vice versa.
If the secondary is short-circuited, so that
v2 = 0, the primary terminals appear to be
short-circuited since v1 = 0.
If the secondary is open-circuited, so that
i2 = 0, the primary terminals appear to be opencircuited since i1 = 0.
The output power is equal to the input power,
so there is no power loss in the element.
Referred impedances
Impedance transformation
The two expressions for Zin will be identical if:
The ideal transformer has the important property of
transforming impedance values in a circuit.
Consider an ideal transformer with an impedance
ZL connected to its secondary terminals, as shown
in figure 2-9.
Figure 2-10(a) shows an ideal transformer with a
load impedance ZL connected to the secondary.
Another impedance Z2 is in series with ZL. The
input impedance of this circuit is:
Z in
Z2 ZL
n2
(2-24)
The input impedance of the circuit in figure 210(b) is:
Z in Z2
Z2
Z2
n2
ZL
n2
(2-25)
(2-26)
Transformers 5
2.4
Z2
Zin
ZL
N1:N2
(a)
In a practical transformer, the winding resistances
and the core reluctance are not zero. In addition,
there will be some power loss in the core because
of eddy currents and hysteresis in the magnetic
material. All of these effects can be represented by
the equivalent circuit [3, 4] shown in figure 2-12.
I1
Z'2
Circuit model of a transformer
R1
jx1
I2 = nI2
jx2
R2
I2
I0
Zin
ZL
N1:N2
+
I0m
I0c
+
V1
jXm
Rc
V2
(b)
Figure 2-10: Referred impedance – 1.
N1:N2
The impedance Z'2 is termed the secondary
impedance Z2 referred to the primary.
In a similar way, a primary impedance Z1 can
be referred to the secondary, as shown in figure
2-11. In this case, the referred impedance is given
by:
Z1 n 2 Z1
(2-27)
Z1
Zin
ZL
N1:N2
(a)
Z''1
Zin
Figure 2-12: Transformer equivalent circuit.
This circuit is based on the ideal transformer
element, with additional circuit elements to
represent the imperfections. The resistances R1 and
R2 represent the physical resistances of the
windings, and Rc represents the power lost in the
core. The reactance Xm, known as the magnetising
reactance, allows for the current required to
magnetise the core when the reluctance is not zero.
Reactances x1 and x2, known as leakage
reactances, represent the leakage flux that exists
when the magnetic coupling between the primary
and the secondary is not perfect. Figure 2-13 shows
the leakage flux when the core has an artificially
low relative permeability of 10, and one winding at
a time is energised. In practice, the leakage is much
less than this, so the leakage reactances are
normally very much smaller than the magnetising
reactance Xm.
ZL
N1:N2
(b)
Figure 2-11: Referred impedance – 2.
The concept of referred impedance is often a useful
device for simplifying circuits containing
transformers, as will be shown in the next section.
It is conventional to use a single prime (') to denote
quantities referred to the primary side, and a
double prime (") to denote quantities referred to
the secondary side.
6
Electrical Machines and Systems Course Notes
(a)
R1
jXm
jx1
jx2
R2
Rc
(b)
N1:N2
Figure 2-13: Transformer leakage flux:
(a) left coil energised, (b) right coil energised.
In the circuit of figure 2-12, the current I2 is
the effective value of the secondary current as seen
from the primary side of the transformer. It is also
known as the secondary current referred to the
primary.
The current I0 is the no-load current, which is
the current taken by the primary when there is no
load connected to the secondary. It has a
component I0m, known as the magnetising current,
which represents the current required to set up the
magnetic flux in the core. The current I0c is the
core loss component of the no-load current.
(a)
R1
jXm
jx1
R2
jx2
Rc
N1:N2
(b)
Re
jxe
Approximate equivalent circuit
For power transformers with ratings above
100 VA, the values of the series elements R1 and x1
are generally much smaller than the shunt elements
Rc and Xm. Under normal working conditions, the
voltage drop in R1 + jx1 is much smaller than the
applied voltage V1. Similarly, the no-load current I0
is much smaller than the load current I1. It follows
that the shunt elements can be moved to the input
terminals, as shown in figure 2-14(a), with very
little loss of accuracy.
The secondary elements R2 and jx2 can be
replaced by equivalent elements R2 = R2 / n2 and
x2 = x2 / n2 on the primary side (see section 2.3),
giving the circuit shown in figure 2-14(b). Finally,
the series elements can be combined to give an
effective resistance Re and leakage reactance xe, as
shown in figure 2-14(c), where the values are:
Re R1
R2
x
, x e x1 22
2
n
n
(2-28)
jXm
Rc
N1:N2
(c)
Figure 2-14: Approximate equivalent circuit.
2.5
Parameter determination
The parameters of the approximate equivalent
circuit (figure 2-14) can be determined
experimentally from two tests:
An open-circuit test, where the secondary is
left unconnected and the normal rated voltage
is applied to the primary.
A short-circuit test, where the secondary
terminals are short-circuited and a low voltage
is applied to the primary, sufficient to circulate
the normal full-load current.
Transformers 7
Open-circuit test
Short-circuit test
With the secondary unconnected, I2 = 0, so the
equivalent circuit reduces to the form shown in
figure 2-15.
If the secondary terminals are short-circuited, the
ideal transformer in figure 2-14 can be replaced by
a short circuit, so the equivalent circuit takes the
form shown in figure 2-16(a). In a typical power
transformer, the shunt elements Rc and Xm are at
least 100 times larger than the series elements Re
and xe. Consequently, the shunt elements can be
neglected, and the circuit reduces to the form
shown in figure 2-16(b).
I1oc
+
V1oc
jXm
Rc
Re
jxe
Figure 2-15: Open-circuit test.
jXm
To a very close approximation, the current I1oc
supplied to the primary is equal to the no-load
current I0 in figure 2-12. The values of the
elements Rc and Xm can be determined from
measurements of the input voltage V1oc, current I1oc
and power P1oc as follows. The input power is
entirely dissipated in the resistance Rc, giving:
Rc
(a)
I1sc
jxe
Re
+
V2
Rc 1oc
P1oc
(2-29)
V1sc
The input impedance of the circuit is given by:
Z1oc
V1oc
1
1
1
I1oc
Rc jX m
Figure 2-16: Short-circuit test.
In terms of magnitudes, equation 2-30 becomes:
Z1oc
V1oc
I 1oc
1
1
Rc2
1
(b)
(2-30)
(2-31)
X m2
The values of the elements Re and xe can be
determined from measurements of the input voltage
V1sc, current I1sc and power P1sc as follows. The
input power is entirely dissipated in the resistance
Re, giving:
Re
Re-arranging equation 2-31 gives the value of Xm:
Xm
1
I1oc
V1oc
1
2
Rc
2
(2-32)
V
N
n 2 2oc
N 1 V1oc
8
(2-34)
The input impedance of the circuit is given by:
Z1sc
From figure 2-14(c), it follows that the turns ratio
is given by:
P1sc
I12sc
V1sc
Re jx e
I1sc
(2-35)
In terms of magnitudes, equation 2-35 becomes:
(2-33)
Electrical Machines and Systems Course Notes
Z1sc
V1sc
Re2 x e2
I 1sc
(2-36)
Re-arranging equation 2-36 gives the value of xe:
V
X e 1sc
I 1sc
Re2
2
(2-37)
In practice, the open-circuit test is usually
made on the low-voltage side of the transformer to
minimise the value of Voc, and the short-circuit test
is made on the high-voltage side to minimise the
value of Isc. The resulting parameter values are then
referred to the primary side of the transformer.
2.6
Transformer performance
Consider a transformer with an impedance ZL
connected to the secondary. With the approximate
equivalent circuit, this may be represented by the
circuit diagram of figure 2-17.
ZL =
RL + jXL
Rc
Figure 2-17: Transformer with a load.
The load impedance can be referred to the primary
side of the ideal transformer element, giving the
circuit shown in figure 2-18.
Re
jxe
I2 = nI2
I0
+
V1
jXm
Rc
ZL =
ZL / n2
V2 nl V2 fl
V2 nl
(2-38)
where V2nl is the magnitude of the no-load
secondary terminal voltage, and V2fl is the
corresponding full-load voltage.
Power is lost as heat in the windings and core
of the transformer, represented by the resistance
elements Re and Rc in the equivalent circuit. The
efficiency is defined in the usual way as:
Pout
Pin
(2-39)
where Pin is the power input to the primary and Pout
is the power output from the secondary. The power
lost as heat in the transformer is:
Ploss Pin Pout
(2-40)
so we have the following alternative forms of
equation 2-39:
N1:N2
I1
When a load is connected to the secondary of a
transformer, there will be a voltage drop in the
series elements Re and xe, so the secondary terminal
voltage will change. The voltage regulation is
defined as:
jxe
Re
jXm
Voltage regulation and efficiency
+
V2 = V2 / n
Figure 2-18: Circuit with referred impedance.
This circuit is easily solved for the currents I0 and
I2. The referred secondary voltage is given by
V2 = ZLI2, and the secondary terminal quantities
are given by V2 = nV2, I2 = I2 / n.
Pout Pin Ploss
P
1 loss
Pin
Pin
Pin
Pout
Ploss
1
Pout Ploss
Pout Ploss
(2-41)
Large transformers are very efficient. Even a
2 kVA transformer can have an efficiency of about
95%. Above 25 kVA, the efficiency usually
exceeds 99%. It is very difficult to make an
accurate measurement of efficiency by direct
measurement of Pout and Pin, since this would
require an accuracy of measurement of the order of
0.01%. Instead, the normal practice is to determine
the losses from measurements, and calculate the
efficiency from one of the alternative expressions
in equation 2-41. The losses can be calculated with
high accuracy from the equivalent-circuit
parameters determined from tests on the
transformer.
Maximum efficiency
The power loss in a transformer has two
components: the core loss, given by V12 / Rc, and
Transformers 9
the I2R loss, given by I'2Re. The core loss will be
constant if the primary voltage V1 is constant, but
the I2R loss will vary with the secondary current.
When the current is low, the output power will be
low, but the core loss remains at the normal value.
Consequently, the efficiency of the transformer
will be low under these conditions. It may be
shown that the efficiency is a maximum when the
secondary current is such that the variable I2R loss
is equal to the fixed core loss. This result also
applies to other devices where the losses have
fixed and variable components. Transformers are
usually designed to have maximum efficiency at
the normal operating value of secondary current,
which may be less than the maximum rated current.
(b)
(c)
(d)
(e)
the secondary terminal voltage magnitude,
the primary current magnitude,
the voltage regulation,
the efficiency of the transformer.
(a) Secondary current
The load impedance referred to the primary is:
Z L
Z L 6 .0 j 2 .5
26.9 j11.2
n2
( 0.472) 2
The secondary current referred to the primary is:
I2
V1
Z e ZL
230 j 0
( 0.682 j 0.173) ( 26.9 j11.2)
230 j 0
7.12 j 2.94 A
27.6 j11.4
Power relationships
When calculating the transformer performance, the
following power relationships can be useful. The
complex power S is given by [2]:
S = P + jQ = VI*
(2-42)
where V is the voltage phasor, I* is the complex
conjugate of the current, P is the real power, and Q
is the reactive power. If is the phase angle, then:
I 2 7.12 j 2.94 7.70 A
The secondary current magnitude is:
I2
I 2
7.70
16.3 A
n 0.472
P S cos VI cos Re( VI*) (2-43)
(b) Secondary voltage
Q S sin VI sin Im( VI*) (2-44)
The secondary voltage magnitude is:
S VI P Q
2
2
V2 I 2 Z L 16.3 6.0 j 2.5
16.3 6.5 106.0 V
(2-45)
If the voltage phasor V is chosen as the reference
quantity, and defined to be purely real (V = V + j0),
then the power relationships take a simple form:
P VI cos V Re( I )
(2-46)
Q VI sin V Im(I )
(2-47)
(c) Primary current
The no-load current is:
I0
V1
V
230 j 0 230 j 0
1
Rc jX m
1080
j 657
0.213 j 0.350 A
The primary current is:
Worked example 2-1
A 2 kVA, 50 Hz, power transformer has the
following equivalent-circuit parameter values
referred to the primary:
Re = 0.682 , xe = 0.173, Rc = 1080,
Xm = 657, N2 / N1 = 0.472.
If the primary is connected to 230 V 50 Hz supply,
and a load impedance (6.0 + j2.5) is connected
to the secondary, determine:
(a) the secondary current magnitude,
I1 I 0 I2
(0.213 j 0.350) (7.12 j 2.94)
7.33 j 3.29 A
I1 7.33 j 3.29 8.03 A
(d) Voltage regulation
On no load, the secondary voltage is:
V2 nV1 0.472 230 108.6 V
10 Electrical Machines and Systems Course Notes
The voltage regulation is therefore:
V2 nl V2 fl
V2 nl
108.6 106.0
2.33%
108.6
I1
R1
jx1
(e) Efficiency
jx2
I2 = nI2
R2
I0
I2
jXm
ZL
The output power is:
Pout I 22 R L (16.31) 2 6.0 1597 W
N1:N2
The power loss is:
Figure 2-19: Current transformer circuit model.
V12
I 22 Re
Rc
Only the relationship between currents is of
interest, so the primary impedance (R1 + jx1) can be
disregarded. It is now convenient to refer the
primary quantities to the secondary side, giving the
circuit model shown in figure 2-20.
Ploss
(230.0) 2
(7.699) 2 0.682
1080
89.4 W
The input power is therefore:
I1
Pin Pout Ploss 1597 89.4 1686 W
jx2
R2
I2
I0 = I0 / n
Alternatively, from equation 2-46 the input power
is:
jXm = jn2Xm
ZL
Pin V1 Re(I1 ) 230.0 7.330 1686 W
The efficiency is therefore:
Pout 1597
94.70%
Pin 1686
Figure 2-20: Modified circuit model.
The circuit acts as a current divider, where the
current in the secondary branch is given by:
I2
2.7
Current transformers
The use of transformers for measuring current has
been introduced in Electromechanics [1], where
the danger of open-circuiting the secondary has
been explained. This section introduces the
important topic of measurement errors.
Current transformer errors
In a well-designed current transformer, the core
flux density is low and the core is made from a
high-quality magnetic material. Under normal
operating conditions, the core loss will be
negligibly small, so the core loss resistance Rc can
be omitted from the equivalent circuit. A circuit
model for a current transformer connected to a load
therefore takes the form shown in figure 2-19.
jX m I1
Z 2 Z L jX m
(2-48)
where I1 is the primary current referred to the
secondary, and Z2 = R2 + jx2. In a well-designed
transformer, the secondary impedance Z2 is very
small in comparison with the referred magnetising
reactance Xm, so this term introduces very little
error. Equation 2-48 shows that it is desirable to
keep the load impedance ZL as small as possible if
the error is to be minimised.
In practice, current transformers are designed
for a specified maximum secondary voltage at the
rated secondary current. This defines a maximum
apparent power for the secondary load, or burden.
Typically, a small current transformer will have a
rated secondary burden of 5 VA. With the usual
secondary current rating of 5 A, this implies that
the maximum secondary voltage is 1 V, and the
maximum impedance magnitude is 0.2 .
Transformers 11
Worked example 2-2
2.8
A current transformer has 10 turns on the primary
and 100 turns on the secondary. It has a rated
secondary current of 5 A, the magnetising
reactance referred to the secondary is 10 , and the
maximum burden is 5 VA. If the primary current is
50 A, determine the secondary current, and hence
the percentage error in the current measurement, if
the secondary load is (a) purely resistive, (b) purely
inductive. The transformer secondary impedance
may be neglected.
(a) Resistive load
Since the burden is 5 VA and the secondary current
is 5 A, the secondary voltage is 1 V, and the
resistance is 1 / 5 = 0.2 . The turns ratio is n =
100 / 10 = 10, so the primary current referred to the
secondary is 50 / 10 = 5.0 A. The secondary load
current is:
The majority of single-phase transformers use the
shell type of construction shown in figure 2-21.
I2
jX m I1
j10.0 5.0
RL jX m 0.2 j10.0
The magnitude is given by:
I2
10.0 5.0
10.0 5.0
4.999 A
0.2 j10.0
10.002
Transformer design
Figure 2-21: Shell-type transformers.
(RS Components Ltd)
Normally the core laminations are made in two
parts, termed E and I laminations, as shown in
figure 2-22.
The percentage error is thus:
e
5.0 4.999
0.02%
5 .0
(b) Inductive load
Since the impedance magnitude is the same as
before, the load reactance is 0.2 . The secondary
current is now:
jX m I1
j10.0 5.0
I2
jX L jX m
j 0.2 j10.0
The magnitude is given by:
I2
Figure 2-22: E and I laminations.
The centre limb is twice the width of the outer
limbs because it carries twice the flux, as shown by
the flux plot in figure 2-23.
10.0 5.0
4.902 A
0.2 10.0
The percentage error is thus:
e
5.0 4.902
1.96%
5 .0
Figure 2-23: Flux plot: shell-type transformer.
12 Electrical Machines and Systems Course Notes
The coils are wound on a bobbin that fits the centre
limb of the core, and the core is assembled by
inserting E laminations alternately from each side
and adding matching I laminations. Dimensions are
chosen so that two E and two I laminations can be
punched from a rectangular steel sheet without any
waste, as shown in figure 2-24.
T
k c Pl
As
(2-49)
where T is the temperature rise above ambient, Pl
is the total power loss in the windings, As is the
exposed surface area, and kc is a cooling coefficient
with a typical value of 0.04 Km2/W.
Figure 2-25 shows the side view and top view
of a shell-type transformer.
b
2a
mean turn
6a
a
Figure 2-24: Punching E and I shapes.
Current flowing in the resistance of the
transformer windings will produce heat, which
must escape through the surface of the windings. In
addition, there will be power loss in the core,
which also appears as heat.
The power output from a given size of
transformer is governed by the rate at which heat
can be removed. Large transformers are usually
cooled by circulating oil, but small transformers
rely on natural convection cooling in air. A simple
design approach for small transformers is given
below.
3a
a
5a
a
2a
a
a
Figure 2-25: Shell-type transformer dimensions
It is assumed that the core is made from
laminations punched as shown in figure 2-24. If a
is the width of each outer limb of the core, then the
width of the centre limb is 2a, and the other
dimensions are as shown in figure 2-25.
Winding resistance
Thermal model
The rate of cooling depends on the exposed surface
area of the transformer and the temperature rise
above ambient. An exact calculation is complex,
since it needs to take account of temperature
gradients within the transformer as well as the
cooling conditions on different surfaces.
A simple thermal model ignores temperature
gradients, and the power loss in the core. It just
considers the I2R loss in the windings, and assumes
that this heat escapes through the exposed surfaces
of the windings. It is assumed that the temperature
rise is proportional to the power loss per unit area:
The total cross-sectional area of the two windings
is the window area of height 3a and width a. Each
winding occupies half of this area, so the conductor
cross-sectional area for each winding is:
Ac 1.5k s a 2
(2-50)
where ks is the conductor space factor, which
allows for insulation and space between the turns.
For simplicity, it will be assumed that the
primary and secondary windings are placed sideby-side on the core, and that they have the same
number of turns N. From figure 2-25, the mean turn
length of each winding is:
l m a 4a 2b ( 4)a 2b
(2-51)
Transformers 13
If the winding has N turns, then the total length of
wire is Nlm, and the cross-sectional area of the wire
is Ac / N. The winding resistance is therefore:
R
l Nl m N {( 4)a 2b}
A Ac / N
1.5k s a 2
Equation 2-55 gives:
I2
2
(2-52)
I
Temperature rise
If the RMS current in one winding is I, the power
loss is I2R. The cooling surface area of the
winding, from figure 2-25, is:
As 1.5a ( 4a 2 a ) 4a 2 a 2
a 2 (10 4 )
(2-53)
3k s a 4 (5 2 )T
k c N 2 {( 4)a 2b}
3k s a 4 (5 2 ) T
k c N 2 {( 4) a 2b}
3 0.4 (10 10 3 ) 4 (5 2 )(90 30)
0.04 21.9 10 9 (1230 ) 2
{( 4) 10 10 3 2 30 10 3 }
0.216 A
Thus, I = 216 mA, so the transformer rating is
230 0.216 VA = 49.7 VA.
Substituting in equation 2-49 gives:
k c Pl k c I 2 R k c N 2 I 2 {( 4)a 2b}
(2-54)
A relationship between the apparent power rating
As
As
3k s a 4 (5 2 )
of the transformer and the dimensions can be
Thus, the current is given by:
obtained by substituting for N from equation 2-56
in equation 2-55:
3k s a 4 (5 2 ) T
2
(2-55)
I
2 Bm ba 3 3k s (5 2 ) T
k c N 2 {( 4)a 2b}
VI
(2-57)
k c {( 4)a 2b}
From equation 2-10, the number of turns is:
Rating and size
T
N
Vm
2V
2fABm 4fabBm
(2-56)
If it is assumed that the core depth b is proportional
to the dimension a, and other quantities remain
constant, then equation 2-57 gives the following
relationship:
VI a 3.5
Worked example 2-3
A transformer has a primary wound for 230 V, and
the core measures 60 50 30 mm. The maximum
winding temperature is 90ºC, the ambient
temperature is 30ºC, the winding space factor is
0.4, the cooling coefficient is 0.04 Km2/W, and the
resistivity of copper at 90ºC is 21.9 nm. If the
maximum flux density in the core is 1.4 T and the
frequency is 50 Hz, determine (a) the number of
turns on the primary, (b) the maximum current in
the primary.
From equation 2-58, if the dimensions of the
transformer are doubled, the rating will increase by
a factor of 11.3. A similar result is obtained for the
increase in the power output of a DC machine
when the dimensions are doubled – see section 3.2.
Solution
From figure 2-26, a = 10 mm and b = 30 mm. From
equation 2-56, we have:
N
2V
4fabBm
2 230
4 50 10 10 3 30 10 3 1.4
(2-58)
1230
14 Electrical Machines and Systems Course Notes
3 DC MACHINES
Introduction
Basic DC machine principles were covered in
Electromechanics [1], and the main results are
given below. Brushless DC machines, which were
mentioned briefly in [1], are beyond the scope of
this course but are covered in the Year 3 course
Electrical Machine Drives.
Armature voltage (V)
3.1
350
300
250
200
150
100
50
0
0
0.2 0.4 0.6 0.8
1.2
Field current (A)
Basic equations
The generated voltage and the developed torque
are given by:
ea K f ùr
Td K f ia
1
[V]
(3-1)
[Nm]
(3-2)
where K is the armature constant, f is the field
flux, and ia is the armature current. Note that the
rotor angular velocity r must be in radians per
second (rad/s) and not in rev/min. If the rotational
speed is nr rev/s or Nr rev/min, then:
2ð N r
ùr 2ð n r
60
[rad/s]
Figure 3-1: Magnetisation characteristic.
In this example, the relationship is almost
linear up to the rated field current of 0.5 A, but
there is significant non-linearity above this value,
when parts of the magnetic circuit saturate. There
is also a small residual flux when the field current
is zero, giving a corresponding generated voltage.
For the initial part of the magnetisation
characteristic, it is approximately true that f if.
Equations 3-1 and 3-2 then become:
(3-3)
In a permanent-magnet machine, the field flux
f is fixed, but in a wound-field machine, it is a
function of the field current if.
Magnetisation characteristic
From equation 3-1, if the speed r is held constant,
the flux is proportional to the armature generated
voltage ea. A graph of ea against if is known as the
magnetisation characteristic of the machine, and a
typical curve is shown in figure 3-1 for a 3 kW
motor.
ea K i f ùr
(3-4)
Td K i f ia
(3-5)
Armature equation
Figure 3-2 shows a symbolic representation of
a DC machine, with the armature connected to a
voltage source va.
r
if
+
ea
ia
+
va
Figure 3-2: DC machine with voltage source.
If the armature has a resistance Ra, then
Kirchhoff’s voltage law gives the armature voltage
equation:
v a Ra ia ea Ra ia K f ùr
(3-6)
DC Machines 15
3.2
DC machines in practice
Slotted armature
The elementary theory of DC machines assumes
that conductors are on the surface of the armature,
so that the simple expressions e = Blu and f = Bli
are applicable. In practice, the armature conductors
are placed in slots, as shown in figure 3-3.
Figure 3-5: DC machine field flux.
Figure 3-3: DC motor armature.
Figure 3-4 shows a simple model of a machine
with a slotted armature, and figure 3-5 shows the
corresponding flux plot when there is no armature
current.
This is similar to the flux plot when conductors are
on the surface, but with one important difference:
most of the flux lines pass between the conductors,
indicating that the flux density in the slots is very
low. Consequently the force on the conductor,
given by f = Bli, is very small. Most of the force is
exerted on the armature iron and not on the
conductors. The basic equations 3-1 and 3-2 are
not affected by the location of the conductors. In
particular, the generated voltage is not affected,
although a direct application of e = Blu appears to
contradict this. See reference [1] for a discussion
of induced voltage in such situations.
Armature reaction
Current flowing in the armature conductors will
also create a magnetic field in the machine, known
as the armature reaction field. Figure 3-6 shows a
flux plot of this field when the main field flux is
absent.
Figure 3-4: DC machine model.
Figure 3-6: Armature reaction flux plot.
16 Electrical Machines and Systems Course Notes
Note that the axis of the armature reaction field is
at right angles to the axis of the main field. When
currents flow in the field and armature conductors,
the two component magnetic fields combine to
give a resultant field of the form shown in figure 36. The bending of the field lines indicates that the
stator exerts a counter-clockwise torque on the
rotor. Another way of picturing the
electromagnetic action is to consider the magnetic
poles representing the field and armature flux
components, as shown in figure 3-7.
field flux. This has important consequences for the
motor characteristics – see section 3.3.
Efficiency
The efficiency of a motor is defined in the usual
way:
Pout
Pin
(3-7)
where Pin is total electrical power input to the
motor terminals, and Pout is the mechanical power
output from the motor shaft. The power lost as heat
in the motor is:
Ploss Pin Pout
N
S
N
S
so we have the following alternative forms of
equation 3-7:
Pout Pin Ploss
P
1 loss
Pin
Pin
Pin
Pout
Ploss
1
Pout Ploss
Pout Ploss
Figure 3-7: Total flux plot.
An important effect of armature reaction, which is
evident in figure 3-7, is to increase the flux density
at one side of a field pole and decrease it at the
other side. Figure 3-8 shows a shaded plot of the
flux density magnitude, where the colour range
from blue to red represents the flux density range
from minimum to maximum.
(3-8)
(3-9)
As with transformers (see section 2-5), the
efficiency of a large motor is not usually
determined from equation 3-7 by direct
measurement of the input and output power.
Instead, losses are determined from several tests,
and the efficiency calculated from equation 3-9.
The output power is given by
Pout r T r (Td Tl )
(3-10)
where Tl is the rotational loss torque. It follows that
the rotational power loss is rTl. The rotational loss
has two components: mechanical loss, which is
also known as the windage and friction loss, and
core loss in the armature and the field poles
resulting from the rotation of the armature.
The total power loss includes in addition the
I2R loss in the field and armature windings, and the
brush contact loss, which results from the voltage
drop between the brushes and the commutator
segments. For further information, see
references [3, 4].
Figure 3-8: Flux density magnitude.
High values of flux density may result in local
saturation of the steel, increasing the reluctance of
the magnetic circuit, and reducing the value of the
DC Machines 17
P r Td 2 r Bav AV
Power output and size
The developed torque may be calculated from the
equation f = Bli, even though the armature
conductors are in slots. For this purpose, the
machine will be represented by a simple model
with the conductors on the surface, as shown in the
flux plot of figure 3-9.
Thus, the maximum power output of a DC motor is
roughly proportional to the product of the armature
volume and the armature speed.
In practice, the current loading A also increases
with size, so the power output of large machines is
further increased. To quantify this, assume that the
power dissipation per unit surface area is constant.
Let d be the radial depth of a conducting layer
representing the armature conductors. The
resistance of an element ds of this layer is:
dR
ds
r
dP ( di ) dR
dS
l ds
Figure 3-9: Model for torque calculation.
It is useful to define a current loading A as the
current in amperes per metre length of
circumference on the surface of the armature. The
current in an element of length ds is then di = A ds.
The force on this element is:
(3-11)
where l is the axial length of the armature. The
corresponding contribution to the torque is:
dTd r df rBlA ds
(3-15)
( A ds ) 2
l ds
l
d ds
A2
(3-16)
d
If dP / dS is constant, then A d, and if d r
then A r. For geometrically similar machines,
the axial length l is proportional to the radius r. If
the rotational speed r is constant, it follows from
equation 3-13 that the power output is proportional
to r3.5. Thus, if the dimensions are doubled, the
power output will increase by a factor of 11.3. A
similar result was obtained for transformers
(section 2.8), and it holds for other types of
machine.
3.3
Characteristics and control
(3-12)
Speed control
where r is the radius of the armature. If Bav is the
average value of the flux density at the armature
surface, then the total torque is just:
Td rBav lA.2 r 2 Bav Alr 2 2 Bav AV
l
d ds
The cooling surface area of the element is dS =
l ds, so the power dissipation per unit area is:
2
df Bl di BlA ds
(3-14)
(3-13)
In a large motor, the total power loss is small.
Since the armature power loss Raia2 is only a part of
the total loss, this term must be small in
comparison with the armature input power vaia. It
follows that:
where V is the volume of the armature.
Ra ia v a
(3-17)
The maximum value of Bav is limited by the
saturation of the magnetic material of the armature,
So Raia may be neglected in equation 3-6, giving:
and the value of the current loading A is limited by
va K f ùr
(3-18)
the I2R heating of the armature conductors. If it is
assumed that the maximum value of A is
independent of the machine size, equation 3-13
shows that the developed torque is proportional to
the rotor volume. The treatment of the effects of
scale in reference [1] also gives this result.
The gross power output is given by:
18 Electrical Machines and Systems Course Notes
ùr 0
The rotational speed is therefore:
v
ùr a
K f
(3-19)
Equation 3-19 shows that the speed is independent
of the torque, provided inequality 3-17 holds.
Equation 3-19 is the basis of speed control. If
the field flux f is constant, the rotational speed is
proportional to the applied voltage. The normal
values of va and f define the base speed r0, and
the speed can be reduced to zero by varying va.
Electronic controllers for DC motors deliver an
adjustable voltage to the motor armature by phasecontrolled rectification of the AC mains supply,
using thyristors: see section 8. This is the basis of
DC variable-speed drive systems, which are widely
used in industry.
In a wound-field motor, it is possible to
increase the speed above the base speed by
reducing f – known as field weakening. Only a
limited speed increase is possible, for the following
reason. The torque is related to the armature
current through equation 3-2:
Td K f ia
[3-2]
If f is reduced, there will be a compensating
increase in ia to maintain the torque, and there is a
risk of exceeding the current rating of the machine.
Small motors
In small motors, with power ratings below 1 kW,
inequality 3-17 does not hold, so it is not
permissible to neglect the Raia term. The rotational
speed then depends on the developed torque, as
may be seen by substituting for ia in terms of Td in
equation 3-6:
v a Ra ia K f ùr
Ra Td
K f ùr
K f
(3-20)
Thus, the speed is given by:
ùr
va
R T
a d2
K f ( K f )
(3-21)
A graph of speed against torque is a straight line,
as shown in figure 3-10. The no-load speed, which
is the speed when the torque is zero, is given by:
va
K f
(3-22)
r
r
slope Ra
Td
Figure 3-10: Speed-torque characteristic.
Effect of armature reaction
In section 3.2, it was noted that the armature
reaction field could cause local saturation of the
field poles, thereby reducing the value of the field
flux f. This effect increases with the armature
current ia, and therefore with the developed torque
Td. From equation 3-19, a decrease in f will cause
the speed r to rise. Armature reaction therefore
has the opposite effect to armature resistance,
which causes r to fall with increasing torque load.
There is an important difference, however. The
effect of resistance is linear, as illustrated in figure
3-8, but the effect of armature reaction is nonlinear. At low values of armature current, the
uneven distribution of flux density in the field pole
is insufficient to cause saturation, so there is hardly
any reduction in the field flux. At high values of
current, on the other hand, there may be a
significant reduction.
It is possible, therefore, for the speed of a
motor to fall with increasing load when the
armature current is low, but to increase with load
when the current is high. This increase in speed
can be very undesirable, leading to instability with
some kinds of load. A large DC motor generally
includes some form of compensation for armature
reaction.
A simple method of compensation is to provide
a second winding on the field poles, connected in
series with the armature (figure 3-11), to increase
the field MMF when the armature current
increases. The resulting motor is known as a
compound motor [3, 4]. However, this cannot
DC Machines 19
compensate for the non-linear nature of the
armature reaction effect.
N
S
N
S
Figure 3-11: DC compound motor.
A better method, which is frequently used in highpower DC drives, is to use a compensating winding
[4]. This takes the form of conductors embedded in
slots in the field pole faces, connected in series
with the armature. These conductors carry current
in the opposite direction to the armature
conductors, thereby cancelling the armature
reaction flux.
Figure 3-13: Total flux plot.
Compensating winding
Figure 3-12 shows a model of a DC machine with a
compensating winding, figure 3-13 shows the
resulting flux plot, and figure 3-14 shows a shaded
plot of the flux density magnitude. These plots
should be compared with figures 3-7 and 3-8 for
the machine without a compensating winding.
Figure 3-14: Flux density magnitude.
With a compensating winding, the flux distribution
in each field pole is symmetrical and uniform, so
the effect of the armature reaction has been
cancelled in this part of the machine. The torque
exerted on the armature is not affected, however,
because this depends on the interaction of the
armature currents and the field flux.
Figure 3-12: Compensating winding model.
Starting of DC motors
When the speed is zero, the armature generated
voltage is also zero. From equation 3-6, the
corresponding armature current is given by:
ia 0
20 Electrical Machines and Systems Course Notes
va
Ra
(3-23)
Transient conditions
When the voltage applied to a DC motor is
changed, the speed will not change instantly
because of the inertia of the rotating system. If J is
the moment of inertia and TL is the mechanical load
torque, Newton’s second law gives:
d r
J
Td TL
dt
(3-24)
Mechanical loss torque is assumed included with
TL. Substituting for Td in terms of ia from equation
3-2, and ia from equation 3-6, gives:
J
d r K f ( v a K f r )
TL
dt
Ra
(3-25)
The solution of equation 3-25 gives the speed as a
function of time after a change in the applied
voltage. It should be noted that the derivation of
equation 3-25 has neglected the inductance of the
armature circuit. See reference [10] for the
difference this makes to the transient behaviour.
As an example, suppose that TL = 0 and the
motor starts from rest with a suddenly applied
voltage va = V. Equation 3-25 becomes:
Ra J
( K f )
2
d r
V
r
dt
K f
The solution of equation 3-26 is:
r
V
1 e t / r 0 1 e t /
K f
(3-27)
where r0 is the no-load speed and the time
constant is given by
Ra J
(3-28)
( K f ) 2
The armature current can be determined by
substituting for r in equation 3-6:
V Ra ia K f ùr
(3-29)
Ra ia V (1 e t / )
ia
V t /
e
ia 0 e t /
Ra
(3-30)
Figure 3-15 shows graphs of the normalised current
ia / ia0 and the normalised speed r / r0 against the
normalised time t / .
1
Current, Speed
This is the stalled armature current, which is much
larger than the normal running current. In a small
motor, it is permissible to connect the armature
directly to a constant-voltage supply. The armature
can withstand the stalled current for a short time,
and it will accelerate rapidly. As it does so, the
generated voltage will rise and the current will fall
to its normal value.
However, a large motor must not be started in
this way because the armature resistance Ra is very
low, and the stalled current would be large enough
to cause serious damage. An electronic controller
for speed control will limit the starting current to a
safe value. If this is not available, a variable
resistance must be connected in series with the
armature and the value progressively reduced to
zero as the armature accelerates.
Speed r / r0
0.8
0.6
0.4
Current ia / ia0
0.2
0
0
1
2
3
Time t /
4
5
Figure 3-15: Starting performance.
(3-26)
DC Machines 21
3.4
Shunt and series motors
So far, it has been assumed that the field flux f is
independent of the conditions in the armature. This
is the case in a permanent-magnet motor, and in a
wound-field motor where the field winding is
supplied from a separate voltage source. The latter
is sometimes termed a separately excited motor.
There are two ways of introducing constraints
between the armature and field of a wound-field
motor. In a shunt motor, the field winding is
connected in parallel with the armature. In a series
motor, the field winding is designed to carry the
full armature current, and it is connected in series
with the armature.
Series motor
Figure 3-16 shows the connection of a series motor
to a voltage source. The characteristics are readily
deduced if the following assumptions are made:
The field flux is proportional to the current.
The resistance of the windings is negligible.
+
i
Shunt motor
r
+
v
ea
Figure 3-16: Series motor.
If the field winding is connected to the same supply
voltage as the armature, the field current will be
proportional to the armature voltage. Usually, a
variable resistance is connected in series with the
field winding to provide some measure of control
for the field current [3, 4, 13].
The characteristics of a shunt motor may be
deduced from equation 3-19:
va
K f
f k f if
ùr
[3-19]
Provided there is a linear relationship between the
field flux and the field current, we have:
k f va
Rf
(3-33)
v ea K iùr
(3-34)
v
K i
v
T
K d
K
v
K Td
(3-35)
A graph plotted from equation 3-33 for a small
motor is shown in figure 3-17.
(3-31)
200
where Rf is the total resistance of the field circuit
and kf is a constant. Substituting in equation 3-19
gives:
Rf
v
ùr a
K f Kk f
Td K i 2
Eliminating i gives the relationship between speed
and torque:
(3-32)
Thus, the speed of a shunt motor is, to a first
approximation, independent of the applied voltage.
Speed, rad/s
ùr
Equations 3-4 and 3-5 become:
150
100
50
0
0
5
10
15
20
25
Developed torque, Nm
Figure 3-17: Speed/torque characteristic.
22 Electrical Machines and Systems Course Notes
Two features of the graph should be noted:
The motor can develop a very large torque at
low speeds.
When the developed torque is low, the speed is
very high.
The high torque at low speeds is a useful feature,
but the high speed associated with low torque can
be dangerous.
Consider what happens if the external load
torque is removed. The developed torque then just
supplies the mechanical losses in the motor. In a
small motor, the mechanical losses are usually high
enough to limit the no-load speed to a safe value.
However, in motors with power ratings above
about 1 kW, the losses are proportionately smaller,
and the no-load speed would be dangerously high.
Such motors must never be run without a load.
Effect of saturation
From equations 3-2 and 3-5, the torque developed
by a series motor is given by:
Td K f i K i 2
(3-36)
Applications
A correctly designed series motor will work on AC
as well as DC [1, 3, 4]. Such motors are termed
universal. They are widely used in products such
as portable power drills, and in domestic
appliances such as food processors and vacuum
cleaners. In applications such as power tools, the
speed/torque characteristic of the series motor is
advantageous.
An important reason for their use is the high
speeds that are possible with series motors: speeds
of up to 10 000 rev/min (1050 rad/s) are common
in small motors. It was shown in section 3.2 that
the maximum power output of a motor is roughly
proportional to the product of the rotational speed
and the armature (or rotor) volume. Thus, a small
motor can deliver a large amount of power if the
speed is high. Induction motors (see section 6) are
limited to speeds below 3000 rev/min (314 rad/s)
when operated from the 50 Hz mains supply. The
power per unit volume for a series motor is
therefore about three times the value for an
induction motor. Disadvantages include the higher
manufacturing cost, and the need for maintenance
of the commutator and brushes.
This equation is valid only when the current is low
enough to avoid saturation of the magnetic circuit,
so that the flux is proportional to current. When the
current is very high, the field flux will tend
towards a constant value fm, and the torque is then
given by
Td K fm i
(3-37)
Under these conditions, the torque is proportional
to i instead of i2. The performance of motors such
as car starter motors must therefore be measured
under actual operating conditions, where equation
3-36 would give very inaccurate results.
DC Machines 23
4 INTRODUCTION TO AC MACHINES
Va
4.1
1
+
+
Review of 3-phase systems
Industrial AC motors use 3-phase alternating
current to generate a rotating magnetic field from
stationary windings. The 3-phase supply may be
taken from the AC mains, or it may be generated
electronically with an inverter. In either case, the
requirement is a symmetrical set of sinusoidal
currents with relative phase displacements of 120º.
Formally, 3-phase sets of currents and voltages
may be defined as follows in the time domain:
ic I m cos( t 240)
Figure 4-1: 3-phase star connection.
The corresponding voltage phasor diagram is
shown in figure 4-2.
V31 Va
Vc
V23
Figure 4-2: Star connection: voltage phasors.
(4-2)
Vm cos( t 120)
Figure 4-3 shows the delta connection of a
3-phase AC source to three lines, and figure 4-4
shows the corresponding current phasor diagram.
The quantities in equations 4-1 and 4-2 correspond
to a positive phase sequence, in which the
quantities reach their maximum values in the
sequence a b c. If the connections to any two
phases are interchanged, the effect is to reverse the
phase sequence. For example, if b and c are
interchanged, and primes denote the new phase
labels, we have:
v c vb Vm cos( t 120)
1
Ia
Ib
2
Ic
v a va Vm cos( t )
v b vc Vm cos( t 120)
V12
Vb
(4-1)
v a Vm cos( t )
v c Vm cos( t 240)
2
3
I m cos( t 120)
vb Vm cos( t 120)
+
+ Vc
ia I m cos( t )
ib I m cos( t 120)
V12
Vb
3
Figure 4-3: 3-phase delta connection.
(4-3)
These quantities reach their maximum values in the
sequence c b a, corresponding to a negative
phase sequence.
Figure 4-1 shows the star connection of a 3phase AC source to three lines, where the timevarying quantities of equation 4-1 are represented
by complex (phasor) quantities in the usual way.
I3
I1
Ia
Ib
Ic
I2
Figure 4-4: Delta connection: current phasors.
24 Electrical Machines and Systems Course Notes
From these diagrams, we have the relationships
between the magnitudes of line and phase
quantities given in table 4-1:
Table 4-1
4.2
Star
Delta
Vline 3V phase
Vline V phase
I line I phase
I line 3I phase
Rotating magnetic field
The groups labelled a, b, c carry currents in the
positive direction, into the plane of the diagram,
and the groups labelled a', b', c' carry currents in
the negative direction.
All coils have the same shape, with one coil
side at the bottom of a slot and the other side at the
top of a slot, as shown by the pair of circles in
figure 4-6.
If phase a is energised on its own with positive
current, the resulting magnetic field pattern is
shown in figure 4-7(a). The corresponding patterns
for phases b and c are shown in figures 4-7(b) and
4-7(c).
In AC motors, the stationary part is termed the
stator, and the rotating part is the rotor. Figure 4-5
shows the stator of a small AC motor with a
3-phase winding.
(a)
Figure 4-5: 3-phase AC motor winding.
Coils are arranged in slots in a laminated steel
core, rather like the armature coils in a DC motor.
The function of the winding is to route currents to
slots from the three phases, as shown in the
simplified diagram of figure 4-6. This diagram also
shows a steel cylinder in the centre, representing
the rotor of the motor.
(b)
a
c'
b'
b
c
a'
(c)
Figure 4-7: AC machine flux plots:
Figure 4-6: 3-phase conductor groups.
(a) phase a, (b) phase b, (c) phase c.
Introduction to AC machines 25
When all three phases are energised, the individual
phase fields combine to give a resultant field.
Consider the instant when t = 0. From equation
4-1, the currents in the three phases are:
ia I m ,
ib 1 2 I m ,
ic 1 2 I m
Since the currents in phases b and c are negative,
the component fields are in the directions shown in
figure 4-8. They combine to give a resultant in the
same direction as for phase a, so a flux plot of the
resultant field is the same as for phase a alone. As
time advances, the currents in the phases change,
and the resultant pattern changes. Figure 4-9 shows
the patterns when t = 30º, 60º and 90º.
(b)
b
(c)
Figure 4-9: Flux plots for values of t:
(a) 30º, (b) 60º, (c) 90º.
a
c
Figure 4-8: Component fields when t = 0.
Notice that the magnetic field pattern has rotated
by the same angle as the time phase of the currents.
We have a rotating magnetic field, which makes
one revolution for each cycle of the alternating
current. This field is capable of exerting a torque
on a suitably designed rotor.
Two kinds of AC motor exploit the rotating
field effect. In synchronous motors, the rotor has
magnetised poles, which lock in with the rotating
field, so that the rotor moves in synchronism with
the field. With induction motors, the rotating field
induces currents in conductors on the rotor, so the
rotor must move more slowly than the rotating
field.
(a)
26 Electrical Machines and Systems Course Notes
Sinusoidal fields
The rotating magnetic field will induce voltages in
the phase windings. These voltages should be as
nearly sinusoidal as possible, which implies that
the flux density should vary sinusoidally with
angular position. This in turn requires that the
currents producing the field should be distributed
sinusoidally. A sinusoidal distribution is also
required if the resultant field is to rotate at a
uniform speed. The conductor distribution for each
phase shown in figure 4-6 approximates to a
sinusoidal distribution. It is a better winding than it
appears, because harmonics that are multiples of 3
are missing from the Fourier series for the
waveform. This is one of the many advantages of
using three phases. Practical windings generally
use more slots than this simple model, and give a
better approach to the ideal sinusoidal distribution.
Assume that the three phases produce
component fields in the airgap with magnitudes Ba,
Bb and Bc as follows:
Ba kia cos
Bb kib cos( 120)
Bc kic cos( 120)
(4-4)
where k is a constant and is the angle from a
reference axis, as shown in figure 4-10 for the
component field Ba.
Equation 4-5 represents a sinusoidal field with a
constant maximum value, rotating in a counterclockwise direction with a constant angular
velocity r, illustrated in figure 4-10. The axis of
the field makes an angle t with the horizontal
axis. At an angle , the displacement from the field
axis is – t, so the magnitude of the flux density
is Bm cos( – t).
t
Figure 4-11: Rotating magnetic field.
The rotating field represented by equation 4-5
will induce a sinusoidal voltage in any coil on the
stator. Consider a single-turn coil on the stator,
with one side at = 0 and the other side at
= 180º, as shown in figure 4-12.
Figure 4-12: Single-turn coil.
Figure 4-10: Angular position definition.
The total field is the sum of the component fields.
If the currents are given by equation 4-1, then
substituting in equation 4-4 gives the result:
B Ba Bb Bc 23 kI m cos( t )
Bm cos( t )
(4-5)
If the direction of the flux density in the airgap is
assumed radial, it may be shown that the magnetic
flux through this coil is:
2lrBm sin t
(4-6)
where r is the radius and l is the axial length. The
induced voltage in the coil is thus:
e
d
2 lrBm cos t
dt
(4-7)
See references [3, 4] for further information about
sinusoidal fields.
Introduction to AC machines 27
Speed of the rotating field
At any instant, there are two effective magnetic
poles for the field shown in figure 4-8, so this is
termed a 2-pole field. It makes one revolution in
one AC cycle, giving the relationships between the
field and the current shown in table 4-2.
Table 4-2: 2-pole field properties.
Current waveform
Frequency f
2-pole field
[Hz]
Rotational speed ns = f [rev/s]
Rotational speed Ns = 60f [rev/min]
Angular frequency Angular velocity s = [rad/s]
As an example, if f = 50 Hz, ns = 50 rev/s,
Ns = 3000 rev/min and s = 2f 314 rad/s
4.3
Multi-pole fields
Coils in an AC machine winding can be arranged
to produce fields with more than two poles. For
example, figure 4-13 shows the simple winding of
figure 4-6 re-arranged to give a 4-pole field, with
the resulting field pattern shown in figure 4-14 at
the instant when t = 0.
c'
Figure 4-14: 4-pole flux plot.
For every 30º advance in the time phase of the
currents, the field pattern rotates 15º. In one cycle
of the supply, the field pattern moves through two
pole pitches – an angle corresponding to two poles.
With any number of poles, the field moves
through two pole pitches in one cycle of the
supply. If there are p pairs of poles, this
corresponds to a fraction 1 / p of a revolution.
Thus, the general expressions for the speed of the
rotating field are:
s
a
b'
b
ns
c
Ns
a'
a'
b
c
b'
a
2f
p
p
60 f
p
(4-9)
[rev/s]
(4-10)
[rev/min]
Table 4-3 gives the field speeds at 50 Hz for
different numbers of poles.
c'
Figure 4-13: 4-pole winding.
f
p
(4-8)
[rad/s]
Table 4-3: Field speeds at 50 Hz.
Poles
p
2
4
6
8
10
1
2
3
4
5
28 Electrical Machines and Systems Course Notes
s
rad/s
314
157
105
78.5
62.8
ns
rev/s
50
25
16.7
12.5
10
Ns
rev/min
3000
1500
1000
750
600
Reversing the direction of rotation
To reverse the direction of rotation, all that is
required is to interchange the connections to any
two of the three phases. For example, suppose that
the connections to phases b and c are interchanged.
Figure 4-15(a) shows the original current pattern
and figure 4-15(b) shows the new pattern.
c'
a
b'
b
c
a'
a'
b
c
b'
a
c'
(a)
b'
a
c'
c
b
a'
a'
c
b
c'
a
b'
(b)
Figure 4-15: Interchange of connections:
(a) original, (b) b and c interchanged.
The new pattern is a mirror image of the old,
so the resultant magnetic field will progress in the
opposite direction as time advances. It follows that
the direction of rotation depends on the phase
sequence of the 3-phase supply, since
interchanging any pair of phases has the effect of
reversing the phase sequence (see section 4-1).
Therefore, it is important to know the phase
sequence of the supply before connecting the
motor.
Introduction to AC machines 29
5 SYNCHRONOUS MACHINES
5.1
Introduction
the two axes. The flux plots in figure 5-2 show (a)
the field produced by stator currents at a particular
instant of time, (b) the field produced by the rotor
magnets, (c) the resultant field when both sources
are active.
In a synchronous machine, the rotor is magnetised
and it runs at the same speed as the rotating
magnetic field. Permanent-magnet rotors are
common in small machines, so the machine
structure is similar to that of the brushless DC
motor shown in figure 3-4. Figure 5-1 is a
simplified model of the structure of this type of
machine, where the rotor has surface-mounted
segments of permanent-magnet material. Other
forms of rotor with embedded magnets are also
possible.
(a)
Figure 5-1: 2-pole PM synchronous machine.
(b)
In larger sizes, a synchronous machine has a field
winding on the rotor instead of permanent magnets.
Direct current for the rotor excitation can be
supplied through sliprings and brushes, but large
machines normally have a brushless excitation
system [3, 4].
The AC stator of a synchronous machine is
termed the armature. It handles the main electrical
power, so it has a similar function to the rotating
armature of a DC machine.
(c)
Motors and generators
As with DC machines, there is no fundamental
difference between a synchronous motor and a
synchronous generator. In a motor, the magnetic
axis of the rotating magnetic field is ahead of the
magnetic axis of the rotor, resulting in a positive
torque that depends on the displacement between
Figure 5-2: Synchronous motor flux plots:
(a) stator field, (b) rotor field, (c) resultant field.
30 Electrical Machines and Systems Course Notes
In this example, the displacement angle is 90º,
which gives the maximum torque for a given
current. The current has been chosen so that the
axis of the resultant field is at approximately 45º to
the magnetic axis of the rotor.
In a synchronous generator, the displacement is
reversed: the magnetic axis of the rotor is ahead of
the magnetic axis of the rotating field, so the
torque is negative. This is illustrated in figure 5-3.
(a)
Most of the AC generators in electric power
systems are synchronous machines. High-speed
turbine generators normally have two poles. The
rotor is made from a cylindrical steel forging, with
the field winding embedded in slots machined in
the steel. Apart from the slots for conductors, the
active surfaces of the stator and rotor are
cylindrical, so these are uniform-airgap or nonsalient machines.
Low-speed hydro generators have many poles.
These are salient-pole machines, where the poles
radiate like spokes from a central hub.
Synchronous generators will not be considered in
detail in this course, but will be studied in the
context of power systems in year 3.
Large synchronous motors are widely used as
high-efficiency constant-speed industrial drives,
where they can also be used for plant power factor
correction: see section 5.2. These are normally
salient-pole machines, and they generally have four
or more poles.
The main features of synchronous machine
operation can be deduced from the simple theory of
non-salient machines. An introduction to salientpole machines is given in section 5.3.
5.2
Characteristics
Circuit model
(b)
A non-salient synchronous machine can be
represented by a simple equivalent circuit [3, 4],
shown in figure 5-4.
jXs
I
Ra
+
+
V
V
+
E
Figure 5-4: Equivalent circuit.
(c)
Figure 5-3: Synchronous generator flux plots:
(a) stator field, (b) rotor field, (c) resultant field.
This circuit represents one phase of a 3-phase
machine. The voltage V is the phase voltage at the
machine terminals, and the current I is the
corresponding phase current. Other elements in the
circuit have the following significance.
Synchronous Machines 31
The voltage E is termed the excitation voltage.
It represents the voltage induced in one phase
by the rotation of the magnetised rotor, so it
corresponds to the magnetic field of the rotor
shown in figure 5-2(b) or 5-3(b).
The reactance Xs is termed the synchronous
reactance. It represents the magnetic field of
the stator current in the following way: the
voltage jXsI is the voltage induced in one phase
by the stator current. This voltage corresponds
to the magnetic field of the stator shown in
figure 5-2(a) or 5-3(a).
The voltage V = E + jXsI represents the
voltage induced in one phase by the total
magnetic field shown in figure 5-2(c) or 5-3(c).
The resistance Ra is the resistance of one phase
of the stator, or armature, winding.
The resistance Ra is usually small in comparison
with the reactance Xs, so it may be neglected in
most calculations from the equivalent circuit. The
resulting approximate equivalent circuit is shown
in figure 5-5, described by the equation:
V jX s I E
I
+
E
I
jXsI
V
I
Figure 5-6: Phasor diagram: generator.
In power system studies, it is customary to reverse
the reference direction for the current, so the
current is then represented by the phasor I in
figure 5-6.
For operation as a motor, V leads E by an
angle , as shown in figure 5-7. The phase angle
is now less than 90º, indicating a flow of electrical
power into the machine.
(5-1)
jXs
E
I
V
N
jXsI
M
+
E
V
Figure 5-7: Phasor diagram: motor.
Figure 5-5: Approximate equivalent circuit.
Torque characteristic
Phasor diagram
Figure 5-6 shows a phasor diagram corresponding
to the equivalent circuit of figure 5-5, for the
machine operating as a generator. The terminal
voltage V lags the excitation voltage E by an angle
. This is known as the load angle, because it
varies with the torque load. In terms of the
magnetic field, represents the angle between the
axis of the rotor field in figure 5-3(b) and the axis
of the resultant field in figure 5-3(c). The phase
angle is greater than 90º, so the power given by
VI cos is negative, indicating a flow of electrical
power out of the machine.
Consider the line MN in figure 5-7, which is
perpendicular to V. The length of MN can be
expressed in terms of E and from the left-hand
triangle, and in terms of XsI and from the righthand triangle. We have:
MN E sin X s I cos
(5-2)
Multiplying both sides of equation 5-2 by the
voltage magnitude V and dividing by Xs gives:
32 Electrical Machines and Systems Course Notes
VE sin
VI cos
Xs
(5-3)
The right-hand side of equation 5-3 is the electrical
power input to one phase of the motor. Since
energy is conserved, and there are no electrical
losses – the resistance has been neglected – the
total electrical power input must equal the work
done by the rotating magnetic field. If s is the
synchronous speed in rad/s and Td is the developed
torque, then:
3VE sin
3VI cos Td s
Xs
(5-4)
The developed torque is thus given by:
Td
3VEsin
s X s
[Nm]
(5-5)
This has a maximum value when = 90º, given by:
Td max
3VE
s X s
(5-6)
Equation 5-5 may therefore be written in
normalised form:
Td
Td max
Excitation control
In a wound-field synchronous machine, the
excitation voltage E can be controlled by varying
the excitation current in the rotor field winding.
This has consequences for the machine
characteristics, as will now be shown for the case
of a synchronous motor.
If the torque load on a synchronous motor is
held constant, the power output will be constant,
and if losses are neglected, there will be a constant
electrical power input per phase given by:
P VI cos
sin
(5-7)
A graph of this normalised torque is shown in
figure 5-8. Positive values of represent motor
operation; negative values represent generator
operation.
1
Normalised torque
loses synchronism with the rotating field, the load
angle will change continuously. The torque will be
alternately positive and negative, with a mean
value of zero. Synchronous motors are therefore
not inherently self-starting. Induction machines,
considered in section 6, do not have this limitation,
and the induction principle is generally used for
starting synchronous motors.
In normal operation the terminal voltage V is
constant, so the quantity I cos must be constant.
In figure 5-9, the line AB is drawn parallel to the
imaginary axis, and its distance from the axis is
I cos. This line is the locus of the current phasor
I.
The length of the line MN is XsI cos, which is
also constant, so the line CD, parallel to the real
axis, is the locus of the excitation voltage phasor E.
A
0.5
motor
0
generator
-0.5
C
-1
-180
(5-8)
-90
0
90
I
V
N
E
jXsI
M
D
180
Load angle, degrees
Figure 5-8: Synchronous machine torque.
If the mechanical load on a synchronous motor
exceeds Tdmax, the rotor will be pulled out of
synchronism with the rotating field, and the motor
will stall. Tdmax is therefore known as the pullout
torque.
The torque characteristic shown in figure 5-8
has an important practical consequence. If the rotor
B
Figure 5-9: Locus diagram for constant load.
Synchronous Machines 33
specially designed for the higher values of rotor
and stator current.
If the value of E is progressively increased, by
increasing the current in the rotor field winding,
the point M will move towards D, and the angle
will decrease. Figure 5-10 shows the condition
where = 0. This is the unity power factor
condition, which gives the minimum value for I.
Let E0 be the value of E that gives this condition,
and 0 the corresponding value of .
A
I
A
I
V
N
N
jXsI
E
jXsI
C
E0
C
V
M
D
M
D
B
Figure 5-12: Leading power factor condition.
B
Figure 5-10: Unity power factor condition.
When E < E0, is negative, and > 0. The
machine takes a lagging current, as shown in figure
5-9. In this condition, the machine is said to be
under-excited. As the value of E is progressively
reduced, the values of I and increase, until the
limit of stability is reached when = 90º. This
condition is shown in figure 5-11.
If the mechanical load is removed from an
over-excited synchronous motor, then = 0, and
the phasor diagram takes the form shown in figure
5-13. The phase angle is now 90º, so the machine
behaves as a 3-phase capacitor. In this condition, it
is known as a synchronous compensator. The
magnitude of the current, and hence the effective
value of the capacitance, depends on the difference
between E and V.
I
A
V
jXsI
E
V
E
jXsI
C
D
Figure 5-13: Synchronous compensator.
Power factor correction
I
B
Figure 5-11: Stability limit.
When E > E0, is positive, and < 0. The
machine now takes a leading current, as shown in
figure 5-12. In this condition, the machine is overexcited. If continuous operation at a leading power
factor is required, as in power factor correction
(discussed below), then the machine must be
The ability of a synchronous motor to operate at a
leading power factor is extremely useful. It will be
shown in section 7 that induction motors always
operate at a lagging power factor. Many industrial
processes use large numbers of induction motors,
with the result that the total load current is lagging.
It is possible to compensate for this by installing an
over-excited synchronous motor. This may be used
to drive a large load such as an air compressor, or it
may be used without a load as a synchronous
compensator purely for power factor correction.
34 Electrical Machines and Systems Course Notes
When a synchronous motor is used in this way,
the rotor excitation is normally controlled
automatically to keep the total current nearly in
phase with the voltage, so that the plant as a whole
operates at a power factor close to unity. It may not
be economic to give full correction, if this would
require a very large synchronous machine.
Figure 5-14 shows a phasor diagram for the
general case of power factor correction. The
synchronous machine current Is combines with the
load current Il from the rest of the plant to give a
total current It which is more nearly in phase with
the voltage V.
5.3
Salient-pole machines
The theory given in section 5.2 is only valid for
non-salient machines. This section gives a brief
introduction to salient-pole machines; for further
information, see references [3, 4].
When the rotor has salient poles, the airgap is
not uniform, and there is a preferred direction for
magnetic flux in the machine. Figure 5-15 shows a
simple model of a salient-pole machine, where part
of the rotor steel has been removed to
accommodate the field winding.
d-axis
Is
V
q-axis
It
Il
Figure 5-14: Power factor correction.
The relationships between the currents can be
expressed in terms of power, as follows. We have:
It Il I s
(5-9)
The complex power is given by:
S t VI *t V ( I l I s ) * VI *l VI *s S l S s (5-10)
Equating the real and imaginary parts of equation
5-10 gives:
Pt Pl Ps
(5-11)
Qt Ql Q s
(5-12)
Figure 5-15: Salient-pole machine model.
The rotor has a path of easy magnetisation,
known as the direct-axis or d-axis, where the
reluctance is low. This is also the axis of the
magnetic flux produced by the rotor field current.
An axis at right angles, known as the quadratureaxis or q-axis, has a higher reluctance because of
the larger airgap.
Figure 5-16 shows the armature flux in a nonsalient machine produced by a particular set of
currents in the armature (stator). The flux axis is at
45º to the rotor d-axis.
For an over-excited synchronous machine, Qs will
be negative, so equation 5-12 shows that the
overall reactive power Qt will be less than the
reactive power Ql of the original plant. For
complete power factor correction, we require Qt =
0, and therefore Qs = –Ql. If the synchronous
machine is run as a synchronous compensator,
without a mechanical load, then Ps 0, and the real
power demand of the plant is unchanged.
Figure 5-16: Non-salient armature flux.
Synchronous Machines 35
Figure 5-17 is a flux plot for same current pattern
in the salient-pole model. Saliency has the effect of
moving the axis of the armature flux towards the
d-axis.
However, if the windings are connected in delta
(figure 4-2), the third harmonic voltages will
accumulate round the delta:
v va vb v c
V1m cos t V3m cos 3 t ...
V1m cos( t 120) V3m cos 3 t ...
V1m cos( t 120) V3m cos 3 t ...
(5-15)
3V3m cos 3 t ...
Figure 5-17: Salient-pole armature flux.
Voltage harmonics
In a non-salient machine, the flux distribution
shown in figure 5-16 is approximately sinusoidal
(see section 4.2). In a salient-pole machine,
however, figure 5-17 shows that the flux density
distribution is non-sinusoidal. It follows that the
voltages induced in the armature phases will also
be non-sinusoidal. The voltages may be
represented by Fourier series:
v a V1m cos t V3m cos 3 t V5m cos 5 t ...
v b V1m cos( t 120) V3m cos 3( t 120) ...
V1m cos( t 120) V3m cos 3 t ...
v c V1m cos( t 120) V3m cos 3( t 120) ...
V1m cos( t 120) V3m cos 3 t ...
(5-13)
If the windings are connected in star (figure 4-1),
the third harmonics will cancel. For example, the
voltage between lines 1 and 2 is:
v12 va vb
V1m cos t V3m cos 3 t ...
V1m cos( t 120) V3m cos 3 t ...
V1m cos t V1m cos( t 120) ...
(5-14)
This can result in a large third-harmonic circulating
current. Similar considerations apply to higher
harmonics where the harmonic number is a
multiple of 3; these are known as triplen
harmonics. Therefore, it is normal practice to
connect the windings of a synchronous machine in
star.
It is possible to eliminate any harmonic from
the voltage waveform by using short-pitched coils
in the stator winding. Short-pitched coils span less
than a pole pitch; the coils shown in figure 4-6 are
of this form. It may be shown that, if the coil is
short-pitched by a fraction 1/n of the pole pitch,
then the nth harmonic will be eliminated from the
voltage waveform. For example, in a 2-pole
machine, if the coils span 120º instead of 180º, so
that they are short-pitched by 1/3, the third
harmonic will be eliminated. In this particular case,
it is permissible to connect the windings in delta. It
is more usual, however, to short-pitch by 1/6, as
shown in figure 4-6, since this reduces the fifth and
seventh harmonics to low levels. If the windings
are star connected, all the triplen harmonics are
eliminated, so the resulting voltage waveform is a
good approximation to a sine wave.
Salient-pole machine theory
It is possible to analyse the salient-pole machine by
resolving the armature flux into two components: a
component acting along the rotor d-axis, and a
component acting along the q-axis [3, 4]. This is
represented in the phasor diagram by resolving the
current I into components Id and Iq. The
component Id acts on the d-axis, where the low
reluctance gives a correspondingly high reactance
Xd. The component Iq acts on the q-axis, where the
36 Electrical Machines and Systems Course Notes
high reluctance gives a low reactance Xq. The
resulting phasor diagram is shown in figure 5-18,
and the phasor equation is:
V E Ra I jX d I d jX q I q
(5-16)
V
jXqIq
Iq
Id
Reluctance motors
Since the reluctance torque is independent of the
field current, this torque component does not
require the presence of a field winding. It is
therefore possible to make a form of synchronous
motor without a field winding. This is termed a
reluctance motor, or a synchronous reluctance
motor. With no field winding, equation 5-8
becomes:
Td
E
Ra I
I
jXdId
1
1
sin 2
2 X q X d
3V
Xd
1 sin 2
2X d X q
3V
Figure 5-18: Salient-pole phasor diagram.
If the armature resistance Ra is neglected, the
torque developed by a salient-pole synchronous
machine is given by [4]:
Td
3
s
VEsin V 2 1
1
sin 2
2 X q X d
X d
(5-17)
(a)
(b)
Term (a) in this equation represents the normal
synchronous torque, which may be identified with
equation 5-4. Term (b) represents a component of
torque due to the saliency of the rotor. This
component is termed the reluctance torque, which
vanishes when the rotor is cylindrical, for Xd is
then equal to Xq. See section 7.1 for a discussion of
reluctance torque. Figure 5-19 shows graphs of the
two torque components and the resultant torque.
1.25
Normalised torque
2
(c)
1
0.75
(a)
2
(5-18)
from which it follows that the ratio Xd / Xq should
be large for a good design. Say [5] describes
several different design techniques that have been
used to give high values of q-axis reluctance – and
therefore low values of Xq – to maximise the
reluctance torque.
Reluctance motors have many of the
advantages of induction motors, with the added
property that the speed is locked to the frequency
of the AC supply. However, they are prone to
instability under certain operating conditions,
particularly at low frequencies when operated from
variable-frequency supplies for speed control [12].
5.4
Linear synchronous motors
Many applications require motion in a straight line
rather than rotary motion, for example in
automation systems. Linear motors provide this. In
concept, a linear motor is a rotary motor opened
out flat. Figure 5-20 shows the linear counterpart
of the rotating field model of figure 4-6, and figure
5-21(a)–(e) show the field patterns when t = 0º,
30º, 60º, 90º and 120º respectively.
0.5
0.25
0
(b)
-0.25
0
30
60
90
120 150 180
Figure 5-20: Travelling field model.
Load angle, degrees
Figure 5-19: Salient-pole machine torque:
(a) synchronous, (b) reluctance, (c) resultant
(a)
Synchronous Machines 37
(b)
A tubular linear motor can be derived from the
structure in figure 5-22 by ‘rolling it up’ around the
longitudinal axis. Figure 5-24 shows an industrial
motor of this kind. Tubular motors have a
particularly simple structure, since the primary
windings are circular coils that encircle the
secondary magnets.
(c)
(d)
Figure 5-21: Flux plots for travelling field:
(a) 0º, (b) 30º, (c) 60º, (d) 90º.
Figure 5-22 shows a linear synchronous motor
derived from the rotary model of figure 5-1. The
lower part with the windings is termed the primary,
and the upper part with the permanent magnets is
termed the secondary. Figure 5-23 shows the
corresponding flux plots for (a) the field produced
by primary currents at a particular instant of time,
(b) the field produced by the secondary magnets,
(c) the resultant field when both sources are active.
Figure 5-24: Tubular linear synchronous motor.
(Copley Motion Systems)
In addition to applications in industry, linear
synchronous motors are used in advanced ground
transport systems. The German Transrapid system,
for example, which was recently installed in
Shanghai, employs controlled electromagnets for
levitation and linear synchronous motors for
propulsion.
Figure 5-22: Linear synchronous motor.
(a)
(b)
(c)
Figure 5-23: Linear motor flux plots:
(a) primary, (b) secondary, (c) resultant field.
38 Electrical Machines and Systems Course Notes
6 INDUCTION MACHINES
6.1
Introduction
Induction motors exploit the rotating magnetic
field in quite a different way from synchronous
motors. Instead of a magnetised rotor, an induction
motor has a rotor with short-circuited conductors.
The motion of the rotating field induces currents in
the conductors, which in turn interact with the field
to develop a torque on the rotor.
Figure 6-1 shows a typical rotor for a small
induction motor, where conductors in the form of
aluminium bars are placed in slots in the laminated
steel core. At each end, the bars are connected to
short-circuiting aluminium rings known as endrings. In this kind of rotor, the bars and end-rings
are made in a single operation by die-casting.
same speed as the stator field. The flux plots in
figure 6-3 show (a) the field produced by stator
currents at a particular instant of time, (b) the field
produced by the rotor current, (c) the resultant field
when both sources are active.
Figure 6-2: 2-pole induction motor model.
(a)
Figure 6-1: Induction motor cage rotor.
Currents will be induced in the rotor
conductors if there is a speed difference between
the rotor and the rotating magnetic field. In a
motor, the rotor runs more slowly than the rotating
field, and the resulting currents produce a torque
that tends to accelerate the rotor. Unlike
synchronous motors, induction motors are
inherently self-starting. The rotor will accelerate
from rest until it is running just below the
synchronous speed, which is the speed of the
rotating field.
Figure 6-2 is a simplified model of the
structure of a 2-pole induction motor with 12 slots
in the rotor. In a practical motor, the airgap
between the stator and the rotor is made as small as
possible, but in the model the airgap has been
exaggerated for clarity. The stator rotating
magnetic field induces currents in the rotor
conductors, which effectively form a 6-phase
winding. These currents in turn produce a 2-pole
magnetic field that rotates in space at exactly the
(b)
(c)
Figure 6-3: Induction motor flux plots:
(a) stator field, (b) rotor field, (c) resultant field.
Induction Machines 39
The magnetic field produced by the rotor
current in an induction motor is very similar to the
corresponding field in a synchronous motor, but
the underlying mechanism is completely different.
Consequently, the motor characteristics are also
completely different.
If the rotor of an induction motor is driven
mechanically so that it runs faster than the rotating
field, the direction of power flow reverses and
machine functions as a generator. Induction
generators are not widely used, but they have found
application in wind-powered generators.
Induction motors are simple and robust, and
their self-starting capability is a particular
advantage. At least 90 per cent of industrial drives
are induction motors. A typical small motor is
shown in figure 6-4.
It is useful to define a quantity s, known as the
fractional slip, as follows:
s
N N r s r
slip speed
(6-2)
s
synchronous speed
Ns
s
It follows that the slip varies in the opposite way to
the rotor speed. When the rotor is running at the
synchronous speed, so that Nr = Ns, the fractional
slip is s = 0. When the rotor is stationary, so that
Nr = 0, the fractional slip is s = 1. From equation
6-2, the slip speed and the rotor speed in terms of s
are given by:
slip s r s s
(6-3)
r s s s (1 s ) s
(6-4)
The frequency of the rotor currents is
proportional to the slip speed, and therefore
proportional to s. When the rotor is stationary,
s = 1, and the rotor frequency must be equal to the
stator supply frequency, so we have the important
result:
f r sf s
(6-5)
where fr is the frequency of rotor currents and fs is
the stator supply frequency.
Figure 6-4: Small induction motor.
Rotor power relationships
(Brook Crompton)
6.2
Characteristics
An essential feature of induction motors is the
speed difference between the rotor and the rotating
magnetic field, which is known as slip. There must
be some slip for currents to be induced in the rotor
conductors, and the current magnitude increases
with the slip. It follows that the developed torque
varies with the slip, and therefore with the rotor
speed.
Slip speed and fractional slip
The slip speed is given by:
N slip N s N r ,
slip s r
(6-1)
The rotating magnetic field exerts a torque Td on
the rotor, and does work at the rate sTd. This
represents an input of power to the rotor. The rotor
revolves at a speed r, and therefore does work at
the rate rTd, which represents the mechanical
output power of the rotor. The difference between
these two powers represents power lost in the
resistance of the rotor. As in the DC motor, there
will be mechanical losses in the motor, so the shaft
torque T is less than Td. We have the following
rotor power relationships:
Rotor electromagnetic input: Pem s Td
(6-6)
Rotor output: Pmech r Td (1 s ) s Td
(6-7)
Rotor loss: Ploss ( s r )Td s s Td
(6-8)
40 Electrical Machines and Systems Course Notes
Thus, a fraction (1 – s) of the rotor electromagnetic
input power is converted into mechanical power,
and a fraction s is lost has heat in the rotor
conductors. The quantity (1 – s) is termed the rotor
efficiency. Since there are other losses in the motor,
the overall efficiency must be less than the rotor
efficiency.
For high efficiency, the fractional slip s should
be as small as possible. In large motors, with power
ratings of 100 kW or more, the value of s at full
load is about 2%. For small motors, with power
ratings below about 10 kW, the corresponding
value is about 5%. It follows that the full-load
speed of induction motor is close to the
synchronous speed.
When the rotor is stationary, the induction motor
behaves as a 3-phase transformer with a shortcircuited secondary. The equivalent-circuit model
of the transformer in figure 2-14 applies to this
condition. When the rotor moves, the voltage
induced in the rotor will depend on the relative
motion. It may be shown [3, 4] that this can be
represented by a simple change to the equivalent
circuit: the secondary resistance is not constant, but
depends on the fractional slip s. The circuit for one
phase takes the form shown in figure 6-5.
Rs
Performance calculation
The developed torque is obtained by equating the
power absorbed in the resistance Rr / s to the rotor
input power from equation 6-6, giving the result:
Induction motor model
Is
With a transformer, it is possible to simplify
the equivalent circuit by moving the shunt elements
to the input terminals – see section 2.3. This is a
poor approximation with an induction motor,
however, because the magnetising reactance Xm is
much smaller in comparison with the leakage
reactances xs and xr. The reason for this is the
presence of an airgap between the stator and the
rotor, which increases the reluctance of the
magnetic circuit. Consequently, the no-load current
I0 in an induction motor is much larger than the noload current in a transformer of similar rating.
jxs
jxr
Td
Ir
Is
(6-9)
Rs
jxs
jxr
Ir
I0
+
jXm
[Nm]
It is necessary to solve the equations of the
equivalent circuit (figure 6-5) for the currents Is
and Ir, and hence determine the torque from
equation 6-9. This process is simplified by
transforming the equivalent circuit to the form
shown in figure 6-6. Here, the parallel combination
of Xm and Rc has been replaced by the series
combination of xm and rc.
I0
Vs
3 2 Rr
Ir
s
s
Rc
Rr
s
+
rc
Vs
Rr
s
jxm
Figure 6-5: Induction motor equivalent circuit.
This is the conventional form of the equivalent
circuit, first introduced by Steinmetz [6]. The
parameters have the same significance as in a
transformer:
Rs is the stator winding resistance.
xs is the stator leakage reactance, representing
stator flux that fails to link with the rotor.
Rr is the rotor resistance referred to the stator.
xr is the rotor leakage reactance referred to the
stator.
Rc represents core loss, mainly in the stator.
Xm is the magnetising reactance.
Figure 6-6: Modified equivalent circuit.
The series elements in figure 6-6 are related to the
parallel elements in figure 6-5 by the following
equations:
rc
X m2
Rc
Rc2 X m2
(6-10)
xm
Rc2
Xm
Rc2 X m2
(6-11)
Induction Machines 41
The corresponding inverse relationships are:
and the rotor current is:
Rc
rc2 x m2
rc
(6-12)
Xm
rc2 x m2
xm
(6-13)
If the core loss is small, so that Rc >> Xm, then
rc << xm, and we have the following approximate
relationships:
xm X m
(6-14)
rc
X m2
Rc
(6-15)
Rc
x m2
rc
(6-16)
The value of rc depends on Xm, and therefore on the
frequency. When the speed of an induction motor
is controlled by varying the frequency (see section
8-4), the resistance Rc is approximately constant.
Under these conditions, rc is proportional to the
square of the frequency, so the modified equivalent
circuit is less useful. In this section, however, the
frequency is assumed constant, so the modified
circuit will be used.
Values of the stator and rotor currents are
easily determined by first defining impedances as
follows:
Z s Rs jx s
(6-17)
Zr
Rr
jx r
s
(6-18)
Z m rc jxm
(6-19)
The parallel combination of the magnetising
branch and the rotor branch is:
Zp
ZmZr
Zm Z r
Ir
(6-22)
A 4-pole 3 kW star-connected induction motor
operates from a 50 Hz supply with a line voltage of
400 V. The equivalent-circuit parameters per phase
are as follows:
Rs = 2.27 , Rr = 2.28 , xs = xr = 2.83 ,
Xm = 74.8 , rc = 3.95 .
If the full-load slip is 5%, determine:
(a) the no-load current,
(b) the full-load stator current,
(c) the full-load rotor current,
(d) the full-load speed in rev/min,
(e) the full-load developed torque.
Solution
The phase voltage is:
Vs
Vline 400
231 V
3
3
The impedances are:
Z s Rs jx s 2.27 j 2.83
Zr
Rr
2.28
jx r
j 2.83 45.6 j 2.83
s
0.05
Z m rc xm 3.95 j 74.8
Zp
Zm Zr
(3.95 j 74.8)( 45.6 j 2.83)
Z m Z r ( 3.95 j 74.8) ( 45.6 j 2.83)
31.1 j 20.3
(a) The no-load current is:
(6-20)
I0
Vs
Zs Z p
Zr
Worked example 6-1
so the stator current is:
Is
Z pIs
(6-21)
42 Electrical Machines and Systems Course Notes
Vs
231
Zs Zm
( 2.27 j 2.83) (3.95 74.8)
231
231
2.97 A
6.22 j 77.6 77.9
Is
Vs
Vs
Z s Z p ( 2.27 j 2.83) (31.1 j 20.3)
231
4.68 j 3.23 A
33.4 j 23.1
I s I s 4.68 j 3.23 5.69 A
(c) The full-load rotor current is:
Ir
Z pIs
Zr
( 31.1 j 20.3)( 4.68 j 3.23)
45.6 j 2.83
211 j5.89
45.6 j 2.83
I r Ir
211 j 5.89
45.6 j 2.83
211
4.62 A
45.7
(d) The synchronous speed is:
Ns
60 f 60 50
1500 rev/min
p
2
so the full-load speed is:
N r (1 s ) N s (1 0.05) 1500
1425 rev/min
(e) The full-load developed torque is:
Td
3I r2 Rr
ss
3 ( 4.62 ) 2.28
18.6 Nm
2 50
0.05
2
2
Developed torque, Nm
Figure 6-7 shows the torque/speed characteristic
for the 3 kW motor used in the worked example.
breakdown
starting
full load
0
35
30
25
20
15
10
full load
5
0
0
300
600
900 1200 1500
Rotor speed, rev/min
Typical characteristics
70
60
50
40
30
20
10
0
The maximum torque is known as the breakdown
torque. If a mechanical load torque greater than
this is applied to the motor, it will stall. Also
shown in figure 6-7 is the full-load operating point,
and the starting torque when the speed is zero. The
torque is zero at the synchronous speed of 1500
rev/min.
When the rotor is stationary, the fractional slip
is s = 1. When the rotor is running at the
synchronous speed, the fractional slip is s = 0.
Since the frequency of the rotor currents is sfs
where fs is the supply frequency (equation 6-6), the
frequency varies from 50 Hz when the rotor is
stationary to 0 at the synchronous speed. At the
full-load slip of 5%, the frequency is 0.0550 =
2.5 Hz.
Figure 6-8 shows the input current as a
function of slip for the same motor. Observe that
the starting current, when the rotor is stationary, is
about six times the full-load current of 5.69 A. This
ratio of starting current to full-load current is a
typical figure for induction motors.
Stator phase current, A
(b) The full-load stator current is:
300 600 900 120 150
0
0
Rotor speed, rev/min
Figure 6-7: Torque/speed characteristic.
Figure 6-8: Current/speed characteristic.
Induction motor starting
Most induction motors are started by connecting
them straight to the AC mains supply. This is
known as direct on line starting, and as figure 6-8
shows, it will result in a large starting current.
Direct on line starting may be unacceptable, either
because the supply system cannot support such a
large current, or because the transient torque (see
section 6.6) could damage the mechanical system.
A simple method of reducing the starting current is
star-delta starting. With this method, the motor is
designed for delta connection in normal running,
but for starting, the windings are connected in star.
Induction Machines 43
Motor driving a load
When a motor drives a mechanical load, the
operating point is given by the intersection of the
motor torque/speed characteristic with the
corresponding load characteristic, as shown in
figure 6-9.
60
Developed torque, Nm
This reduces the starting current by a factor of 3,
with a corresponding reduction in the torque. After
the motor has reached its full speed, the windings
are re-connected in delta. This method has been
superseded by electronic soft starting (see section
8.4), where the applied voltage is gradually
increased.
30
0
brake
region
motor
region
generator
region
-30
-60
-90
-120
-1500
0
1500
3000
Rotor speed, rev/min
Figure 6-10: Extended torque characteristic.
70
Torque, Nm
60
50
motor
40
The rotor speed is negative in the brake region, so
we may put r = –|r |. From equation 6-2, the
fractional slip is given by:
30
load
20
Brake region of the characteristic
10
s
0
0
300
600
900 1200 1500
Rotor speed, rev/min
Figure 6-9: Motor driving a load.
Extended torque characteristic
If the rotor of an induction motor is driven
mechanically in the reverse direction, the
developed torque opposes the motion, so the
machine acts as a brake. Alternatively, if the rotor
is driven in the forward direction faster than the
rotating field, the torque will reverse, and the
machine acts as a generator. Figure 6-10 shows the
extended torque/speed characteristic that includes
these conditions.
s r s r
s
s
(6-23)
so s > 1. The gross mechanical power output given
by equation 6-7 is therefore negative:
Pmech (1 s ) s Td
[6-7]
A negative mechanical output signifies an input of
mechanical power to the rotor:
Pin ( s 1) s Td
(6-24)
There is also an electromagnetic power input from
the stator, given by equation 6-6:
Pem s Td
[6-6]
The power lost in the rotor resistance, given by
equation 6-8, is very large:
Ploss s s Td
[6-8]
This is equal to the sum of the mechanical power
input (equation 6-24) and the electromagnetic
power input (equation 6-6).
The machine can operate in the brake mode
only for very short periods, or the rotor will
overheat. This mode is sometimes used for
stopping an induction motor rapidly by reversing
the phase sequence of the supply, which merely
requires interchanging two of the connections to
44 Electrical Machines and Systems Course Notes
the motor terminals. The process is called plug
braking or plugging, illustrated in figure 6-11.
Thus the direction of power flow has reversed, and
the rotor efficiency for generator operation is:
r
motor
curve
30
A
0
brake
curve
B
-30
C
-60
-1500
-750
0
750
1500
Rotor speed, rev/min
Figure 6-11: Induction motor plug braking.
Suppose that the machine is operating at point A on
the blue motor curve when plug braking is
initiated. The direction of the rotating field
reverses immediately, giving a new torque/speed
characteristic shown as the red brake curve in
figure 6-11. Since the rotor speed cannot change
instantaneously, the operating point moves to B on
the brake curve, vertically below A. The resulting
negative torque will decelerate the rotor until it
reaches zero speed at C. At this point the supply
must be disconnected promptly, otherwise the rotor
will continue to accelerate in the reverse direction.
A better method of stopping the motor is by DC
dynamic braking, discussed in section 6.6.
Generating region of the characteristic
In the generating region of figure 6-10, the
machine torque is negative, so we may put Td = –
|Td|. The fractional slip is also negative: s = –|s|,
and the resistance Rr / s is negative in the rotor
branch of the equivalent circuit. It follows that the
electromagnetic and mechanical power values,
given by equations 6-6 and 6-7, both become
negative, but the power loss (equation 6-8) remains
positive:
Pem s Td s Td
(6-25)
Pmech (1 s ) s Td (1 s ) s Td
(6-26)
Ploss s s Td s s Td
(6-27)
Pem
1
Pmech 1 s
(6-28)
It follows that an induction machine can operate as
a high-efficiency generator if the slip s is small.
This is the same condition for high efficiency in a
motor.
An induction machine must be connected to a
suitable AC source for it to function as a generator,
since a source of reactive power is required for the
magnetising current that creates the rotating
magnetic field. If it is disconnected from any AC
source, an induction machine will not generate.
Usually induction generators are connected to an
AC supply system, but it is also possible to make
induction machines run as self-excited generators
by connecting capacitors to the terminals in a 3phase configuration [5]. Each capacitor effectively
forms a resonant circuit with the magnetising
reactance, thereby acting as an AC source to
supply the magnetising current.
Effect of parameter values
The stator resistance Rs is responsible for the lack
of symmetry between the motoring and generating
regions of the torque curve in figure 6-10. Figure
6-12 shows the result of setting Rs to zero.
90
Developed torque, Nm
Developed torque, Nm
60
60
30
0
-30
-60
-90
0
750
1500
2250
3000
Rotor speed, rev/min
Figure 6-12: Torque characteristic: Rs = 0.
In large machines, the stator resistance is a smaller
part of the total impedance, so the characteristic is
more symmetrical.
Induction Machines 45
Developed torque, Nm
Figure 6-13 shows the effect on the motor
torque of halving the values of the leakage
reactances xs and xr. This has little effect on the
characteristic near to the synchronous speed, but it
increases the starting torque and the breakdown
torque significantly.
90
80
70
60
50
40
30
20
10
0
(b)
(a)
0
300
600
900 1200 1500
Rotor speed, rev/min
Figure 6-13: Torque/speed characteristics:
(a) original, (b) leakage reactance halved.
Figure 6-14 shows the effect of halving the rotor
resistance Rr. This has the effect of compressing
the characteristic towards the synchronous speed
end, because the torque is a function of Rr / s, so
the value of s must change in the same ratio as Rr
to give the same torque. Thus the value of the
breakdown torque is unchanged, but the slip at
which it occurs is halved.
6.3
70
Developed torque, Nm
With this change in Rs, the full-load slip is
approximately halved, and therefore the full-load
rotor loss is halved. The penalty for this increased
efficiency is that the starting torque is much lower.
For operation from a fixed-frequency supply,
the design of an induction motor is a compromise.
A low value of Rs is desirable to give high
efficiency, but this also results in a low value of
starting torque.
An ingenious solution to this conflict between
efficiency and starting torque is to make the rotor
with very deep slots, or even with two concentric
rotor cage windings [4, 5, 6]. These designs exploit
the fact that the frequency of the rotor currents is
sfs (equation 6-5), which is very low when the rotor
is running normally, but is equal to the supply
frequency when the rotor is stationary. Skin effect
in AC conductors [3, 4] causes current to flow near
the surface at high frequencies, with a consequent
increase in the effective resistance. The special
rotor windings have a low resistance at the fullload slip frequency, when the current is uniformly
distributed, but a high resistance when the slip
frequency is high during starting.
When the motor is supplied from a variablefrequency inverter (see section 8.4), the starting
condition is handled automatically without the
need for a high-resistance rotor, so a low resistance
can be used for optimum efficiency.
60
50
(a)
40
(b)
Losses and efficiency
The efficiency of an induction motor is of great
importance to the user. It is defined in the normal
way as the ratio of useful mechanical output power
Pout to the total electrical input power Pin:
30
20
10
0
0
300
600
900 1200 1500
Rotor speed, rev/min
Pout
Pin
(6-29)
The difference between Pin and Pout is the total
motor power loss Ploss. We therefore have
alternative expressions for efficiency in terms of
the power loss:
Figure 6-14: Torque/speed characteristics:
(a) original, (b) rotor resistance halved.
46 Electrical Machines and Systems Course Notes
Pin Ploss
Pout
Pin
Pout Ploss
(6-30)
Table 6-1 shows typical efficiency values for 4pole induction motors ranging from 0.25 kW to
250 kW from one manufacturer, and the
corresponding power loss as a percentage of the
input power.
Table 6-1: Induction motor efficiency
Output power
0.25 kW
2.2 kW
22 kW
250 kW
Efficiency
66%
77%
91%
95%
Power loss
34%
23%
9%
5%
The importance of efficiency to the user can be
demonstrated from the following calculation.
Suppose that a 250 kW motor runs continuously
for 1 year, and that electrical energy costs 7p per
kWh.
Input power:
250 / 0.95 = 263 kW
Power loss:
0.05 263 = 13.2 kW
Running cost: 263 24 365 0.07 = £161 000
Cost of loss:
13.2 24 365 0.07 = £8050
If the efficiency could be improved by a mere
0.1%, the cost saving per year would be £162.
Users comparing different motors need to know the
efficiency with high accuracy.
Loss components
The motor losses are considered to have five
components as follows:
(1) Stator I2R loss: PRs 3 I s Rs
(6-31)
(2) Rotor I2R loss: PRr 3 I r Rr
(6-32)
Pcore 3 I m rc
(6-33)
2
2
(3) Core loss:
2
(4) Friction and windage loss: Pfw
the other four categories. It may be attributed to
departures from a purely sinusoidal winding
distribution, and to effects of the stator and rotor
slot openings on the magnetic field distribution in
the machine [4, 5, 6].
The efficiency of an induction motor can be
determined by the following methods.
Direct determination: measure Pin and Pout, and
use equation 6-29. This is the preferred method
for small machines, below about 10 kW, but it
is difficult to measure the powers with
sufficient accuracy in large high-efficiency
machines.
Calorimetric: measure Pin and Ploss, and use
equation 6-30. The difficulty with this method
is accurate measurement of the loss from the
heat dissipated by the motor, but it can be an
effective method for large machines with
closed cooling systems.
Loss segregation: measure Pin, and calculate
Ploss from its components. This is the most
widely used method.
Loss segregation method
The loss segregation method of determining
efficiency is given in several national and
international test specifications [7, 8]. It requires
the following steps to determine all the losses
under specified load conditions (usually full load).
1. Determine the stator I2R loss PRs from equation
6-31, using the measured values of the stator
phase resistance Rs and phase current Is.
2. Determine the core loss Pcore and the friction
and windage loss Pfw from a no-load test (see
below).
3. Determine the rotor I2R loss PRr as follows,
based on equation 6-32, using measured values
of the input power Pin and the slip s:
PRr 3 | I r |2 Rr 3s | I r |2
Rr
sPem
s
(6-34)
Pstray
Pem Pin PRs Pcore
The total loss is the sum of items 1 to 5. Core loss
is the eddy-current and hysteresis loss in the
magnetic core of the machine, mostly in the stator,
which is represented by the resistance rc. Friction
and windage loss is the total mechanical power loss
within the motor, from bearing friction and
aerodynamic drag on the rotor. Stray load loss is an
additional loss under load, which is not included in
where Pem is the electromagnetic power
transferred to the rotor.
4. Determine a value for the stray load loss Pstray.
This is the least satisfactory part of the
procedure, because of the uncertainty in this
quantity. As yet, there appears to be no
satisfactory test method for stray load loss. The
(5) Stray load loss:
(6-35)
Induction Machines 47
American standard [8] specifies a special test,
known as the reverse-rotation test, but this has
been shown to give inaccurate results in many
cases. The IEC standard [7] avoids this
difficulty by assuming that Pstray is 0.5% of the
input power Pin, but this is too low for small
motors, and it fails to penalise badly designed
motors with a high value for Pstray.
No-load test
In the no-load test, the induction motor is run
without any mechanical load attached to the shaft.
The input power then just supplies the losses in the
motor, and the slip is negligibly small. Under these
conditions, the rotor current is very small, so the
rotor I2R loss is negligible, and we have:
Pin PRs Pcore Pfw
(6-36)
By definition, the stray load loss is not included in
the no-load test. Stray no-load losses, if they are
significant, are considered included in Pcore and Pfw.
Re-arranging equation 6-36 gives
Pin Pin PRs Pcore Pfw
(6-37)
If the stator terminal voltage Vs is reduced, the
rotor speed will be virtually unchanged because the
slip remains small, so Pfw will remain constant.
Since Pcore varies approximately as Vs2, a graph of
P'in against |Vs|2 will approximate to a straight line
as shown in figure 6-15. Extrapolating this line to
zero volts gives the value of Pfw, and the core loss
is then given by
Pcore Pin PRs Pfw
(6-38)
6.4
Parameter determination
The parameters of the equivalent circuit in figure
6-6 are usually determined from three tests:
A DC measurement of the stator phase
resistance
A no-load test, essentially as described above
for efficiency determination
A locked-rotor (or blocked-rotor) test, where
the rotor is prevented from revolving
These tests resemble the open-circuit and shortcircuit tests for determining the equivalent-circuit
parameters of the transformer.
DC resistance test
In small or medium-sized machines, up to about
100 kW, a DC measurement of the primary
winding phase resistance will give an accurate
value for Rs in the equivalent circuit. If the
machine is star connected, then a measurement
between any pair of line terminals will give 2Rs. If
it is delta connected, the corresponding
measurement will give 2Rs/3. The measurement
should be made for all three pairs of terminals and
the average taken.
In large machines, above about 100 kW, the
value of Rs at the normal operating frequency may
be significantly larger than the DC value because
of skin effect in the stator conductors.
No-load test continued
Since the rotor current is small in the no-load test,
the rotor branch may be removed from the
equivalent circuit of figure 6-6, giving the
simplified form shown in figure 6-16.
P'in
Is
Rs
jxs
+
rc
Vs
Pfw
jxm
|Vs|
Figure 6-16: No-load equivalent circuit.
Figure 6-15: No-load test graph.
48 Electrical Machines and Systems Course Notes
This circuit will give accurate results if the input
power is corrected by subtracting the friction and
windage loss power Pfw, since this power is almost
equal to the power in the rotor branch. We then
have:
Pin Pfw | I s |2 ( Rs rc ) 2
(6-39)
where Pin is the input power per phase. Since the
stator phase resistance Rs is known, equation 6-39
gives the value of the core loss resistance rc. The
magnetising reactance xm is obtained from the
terminal voltage and current as follows:
Vs
Is
| Z | ( Rs rc ) 2 ( x s x m ) 2 (6-40)
This requires an estimate of the stator leakage
reactance xs, which can be obtained from the
locked-rotor test described below.
Locked-rotor (blocked rotor) test
If the rotor is locked (or blocked) to prevent
rotation, then r = 0 and s = 1. The impedance of
the rotor branch is low, and to a first
approximation the magnetising branch may be
ignored, giving the approximate equivalent circuit
of figure 6-17:
Is
Rs
jxs
This test can only determine the sum of the
leakage reactances, xs + xr, so it is necessary to
make an assumption about the ratio xs / xr. For most
purposes, it is sufficient to take this ratio to be 1,
but sometimes a different value is used based on
motor design values. In practice the ratio is
unimportant, because it may be shown that the
performance calculated from the equivalent circuit
is not affected by the ratio, if the parameters are
determined by test, using the same ratio. Therefore
the simplest assumption is generally used, namely
xs = xr .
A problem with the locked-rotor test is that the
rotor resistance Rr varies with the frequency of the
secondary current, due to skin effect. The practical
consequence is that the test should not be carried
out at the normal supply frequency, except in very
small (fractional kilowatt) motors. The two test
codes already cited [7, 8] specify a frequency of
not more than 25% of the normal supply frequency.
The value of xs determined from equation 6-42
should be used in equation 6-40 for the
magnetising reactance.
A more accurate calculation of the parameters
is possible by using the full equivalent circuit of
figure 6-6 for each test, and solving numerically for
the unknowns.
Worked example 6-2
jxr
Tests on a 3 kW 50 Hz star-connected cage
induction motor gave the following results:
DC stator resistance between lines: 4.54 .
+
Vs
No-load test: line current = 2.97 A,
line voltage = 400 V, input power = 214 W,
friction and windage loss = 50 W.
Rr
Locked-rotor test at 12.5 Hz: line current = 5.70 A,
line voltage = 44.5 V, input power = 422 W.
Figure 6-17: Locked-rotor equivalent circuit.
Only a low voltage is required to give the normal
full-load value of Is. As in the no-load test, the
resistance and reactance values are determined as
follows from the measured values of voltage,
current and power:
Pin | I s |2 ( Rs Rr ) 2
(6-41)
Load test: line current = 5.69 A,
line voltage = 400 V, input power = 3.29 kW,
fractional slip = 5.0%.
Determine:
(a) the parameters of the equivalent circuit,
(b) the efficiency of the motor at full load, using a
stray-load loss allowance of 0.5%.
Solution
Vs
Is
| Z | ( Rs Rr ) 2 ( x s x r ) 2 (6-42)
(a) Equivalent circuit parameters.
Induction Machines 49
DC resistance test:
Rs
(2)
RDC 4.54
2.27
2
2
Vs
3
3
Pin
Plr 422
141 W
3
3
Rlr
Pin
I lr2
3290 221 104 2965 W
25.7 V
44.5
141
(5.70) 2
4.33
V
25.7
2
xlr s Rlr2
( 4.33)
I
5
.
70
lr
1.25
2
xlr f 0 1.25 50
2.51
2 f lr
2 12.5
Vline
3
Pin
Rnl
400
3
Pnl Pfw
3
Pin
I nl2
PRr sPem 0.05 2965 148 W
(4)
Pstray 0.005Pfl 0.005 3290 16 W
(5)
Ploss PRs PRr Pcore Pfw Pstray
2
No-load test:
Vs
(3)
231 V
214 50
54.7 W
3
54.7
( 2.97 )
2
6.5
X nl
Pfl Ploss
Pfl
3290 540
83.59%
3290
Single-phase induction motors
Single-phase operation of a 3-phase motor
When one line is disconnected from a 3-phase
induction motor, the result is a single-phase
system. If the machine is a star-connected, as
shown in figure 6-18, two of the three phases are
connected in series to the single-phase supply, and
the third phase carries no current.
Is
6.20
sc
Vs
rc Rnl Rs 6.20 2.27 3.93
V
231
2
s Rnl2
(6.20)
I
2
.
97
nl
77.5
2
221 148 104 50 16 540 W
The efficiency is thus:
Rr Rlr Rs 4.33 2.27 2.06
xr
214 50 3 2.97 2.27 104 W
Pem Pfl PRs Pcore
Locked-rotor test:
Vline
Pcore Pnl Pfw 3I nl2 Rs
2
x c X nl x s 77.5 2.51 75.0
(b) Efficiency.
(1) PRs 3I s2 Rs 3(5.70) 2 2.27 221 W
50 Electrical Machines and Systems Course Notes
+
sb
sa
Figure 6-18: Single-phase connection.
At standstill, the currents in the stator and the rotor
will be sinusoidal alternating quantities if the
supply voltage is sinusoidal. The result is a
magnetic field that pulsates instead of rotating, and
the rotor remains stationary. However, a pulsating
field can be regarded as the resultant of two contrarotating fields: one positive and the other negative,
as shown by the red and blue phasors in figure
6-19.
The rotor current has two components: a positivesequence component Irp, associated with the
forward rotating field, and a negative-sequence
component Irn, associated with the backward
rotating field. Relative to the forward component,
the fractional slip s is given by equation 6-2:
s
Ns Nr
Ns
[6-2]
Relative to the backward component, the fractional
slip is:
s
Ns Nr
N
1 r 1 (1 s ) 2 s
Ns
Ns
(6-43)
These two current components give corresponding
positive and negative torque components, so the
developed torque is:
Figure 6-19: Contra-rotating fields.
When the rotor is stationary, the two rotating field
components will produce equal and opposite
torques. If the rotor is made to revolve by any
means, the two torque components are no longer
balanced, and there is a net torque tending to
accelerate the rotor.
It can be shown [9] that the motor in figure
6-18 may be represented by the equivalent circuit
of figure 6-20.
Is
Rs
jxs
Td
jxr
30
20
Rr
s
jxm
Vs
Torque, Nm
rc
Positive
sequence
10
0
Total
torque
-10
Negative
sequence
-20
jxm
Rr
2s
rc
Rs
jxs
(6-44)
When the rotor is stationary, s = 1 = 2 – s, so the
two torque components cancel out. If the rotor
revolves in the positive direction, the positive
torque term predominates; if it revolves in the
negative direction, the negative term predominates.
Figure 6-21 shows the two torque terms and the
resultant torque for the 3 kW motor used in
previous examples.
Irp
+
1 2 Rr
1 2 Rr
I rp
I rn
s
s s
2s
-30
-1500
-750
0
750
1500
Speed, rev/min
Irn
jxr
Figure 6-21: Single-phase torque characteristic.
Figure 6-20: Single-phase equivalent circuit.
Induction Machines 51
Losses
The I2R loss in the stator is given by:
PRs 2 Rs I s2
Vs
(6-45)
If skin effect in the rotor is negligible for the
negative-sequence component of rotor current, the
I2R loss in the rotor is given by:
PRr
2
Rr ( I rp
2
I rn
)
+
(6-46)
Danger of single-phase operation
If one stator line is accidentally disconnected from
a 3-phase motor, it will continue to run as a singlephase motor. The motor slip will increase so that
the developed torque continues to match the load
torque, but the stator and rotor currents will be
abnormally high. Table 6-2 shows the computed
full-load performance of the example 3 kW motor,
(a) with a normal 3-phase supply, (b) with one line
disconnected.
Figure 6-22: Single-phase induction motor.
The phases of a 2-phase winding are displaced in
space by 90º (for a 2-pole machine), corresponding
to the 90º time-phase displacement between the
currents. As with a 3-phase winding, it may be
shown [3, 4] that a 2-phase winding with 2-phase
currents can produce a rotating magnetic field.
Once the motor has started, it is possible to
disconnect the capacitor, and the motor will
continue to run with a single winding, but the
performance is not as good.
The shaded-pole principle [3, 4] is used in very
small single-phase induction motors such as the
one shown in figure 6-23. They are widely used for
driving small cooling fans.
Table 6-2: 3-phase and 1-phase operation.
Full-load slip
Stator current magnitude
Rotor current magnitude
3-phase
0.05
5.56 A
4.63 A
Stator I2R loss
Rotor I2R loss
Total I2R loss
211 W
147 W
358 W
1-phase
0.082
11.6 A
10.5 A +ve,
11.2 A –ve
611 W
538 W
1150 W
This large increase in the I2R loss will cause the
motor to overheat very rapidly, resulting in serious
damage unless it is disconnected from the supply.
For this reason, it is essential for motor control
gear to detect single-phase operation and
disconnect the supply immediately.
Figure 6-23: Shaded-pole induction motor.
(RS Components Ltd)
This type of motor has a single winding on the
stator, energising a pair of poles. A part of each
pole is enclosed by a ring of copper, known as a
shading coil. Currents are induced in the rings,
giving the effect of a rudimentary 2-phase winding.
Single-phase induction motors
For domestic and light industrial applications,
induction motors need to operate from a singlephase supply. Instead of a 3-phase winding, many
single-phase motors have a 2-phase winding and a
capacitor to give a phase shift of approximately 90º
between the currents, as shown in figure 6-22
[3, 4].
52 Electrical Machines and Systems Course Notes
Dynamic conditions
Dynamic model
The equivalent-circuit model of the induction
motor introduced in section 6.2 is valid only for
steady-state conditions, where the induction
machine runs at a constant speed. There are two
common situations where this does not give an
accurate representation of the machine
performance:
Starting, when the rotor is stationary and the
AC supply is suddenly switched on.
Sudden stopping, either by plugging or by DC
dynamic braking (discussed below).
For both of these conditions, a dynamic model is
required. The 2-axis theory of the induction motor
[4, 9, 10] treats the machine as a system of coupled
circuits described by differential equations. This
model is valid for all conditions, transient as well
as steady state. Although the theory is beyond the
scope of this course, it is useful to consider some
of the results from this model obtained by
numerical solution of the differential equations.
The equations of the 2-axis model are given in
Appendix 10.1.
The value of the peak current at switch-on
depends on the point in the AC cycle when the
switch is closed, and it will be different for each of
the three phases. Figure 6-24(a) shows the worst
case, where the peak current is a maximum.
The starting torque in figure 6-24(b) contains a
large oscillatory component at the supply
frequency of 50 Hz. This may have undesirable
effects on a mechanical system coupled to the
motor shaft. Electronic control, described in
section 8.4, can reduce this effect.
50
25
Current, A
6.6
-25
-50
0
100
200
300
Time, ms
(a)
120
Starting transient
90
Torque, Nm
The torque/speed characteristics in section 6.2
show that the steady-state starting current is about
six times the full-load current in the example 3 kW
motor. This is typical of the ratio of currents in
most induction motors. When the motor is
suddenly switched on to the AC supply, the initial
current may be very high for a few cycles, and it is
accompanied by a large oscillatory component of
torque.
Figure 6-24 shows the stator phase current and
the developed torque when the motor is switched
on. The nature of the transient response will
depend on the inertia of the rotating system. Here,
the motor has a moment of inertia of 0.011 kg m2,
and it is assumed that the load has an inertia of four
times this value, giving a total of 0.055 kg m2.
0
60
30
0
-30
0
100
200
300
Time, ms
(b)
Figure 6-24: Induction motor starting:
(a) phase current, (b) torque.
Induction Machines 53
DC braking
(a)
An induction motor can be stopped rapidly by
disconnecting the AC supply from the stator and
connecting a DC supply. The resulting stationary
magnetic field induces currents in the rotor that
oppose the rotation – a process termed DC braking
or DC injection braking. A common
implementation with a star-connected induction
motor is to link two line terminals together, and
connect the DC source between this point and the
third terminal, as shown in figure 6-25.
Torque, Nm
20
0
-20
-40
-60
0
200
400
Time, ms
(b)
Figure 6-26: Induction motor DC braking:
(a) rotor speed, (b) torque.
+
V
Figure 6-25: DC braking connection.
When the speed of an induction motor is controlled
by an inverter (see section 9), DC braking can be
achieved without altering the connections to the
motor.
Figure 6-26 shows the rotor speed and the
developed torque for the example 3 kW motor
when DC braking is initiated from the full-load
speed of 1425 rev/min. The DC source voltage is
120 V, and total moment of inertia is 0.011 kg
Speed, rev/min
1500
Observe that the negative torque is maintained
for a short time after the rotor reaches zero speed,
because the rotor inductance forces current to
continue flowing, resulting in a brief speed
reversal. This in turn causes the rotor current to
change sign, and the torque goes positive, bringing
the rotor to rest.
6.7
Linear induction motors
As with synchronous motors, a linear form of the
induction motor can be derived by ‘unrolling’ a
rotary motor, so the travelling-field model of figure
5-20 is applicable. Figure 6-27 shows the primary
of a small linear induction motor.
1000
500
0
-500
0
200
400
Time, ms
2
m.
Figure 6-27: Linear induction motor primary.
54 Electrical Machines and Systems Course Notes
Figure 6-28 shows models of two forms of linear
induction motor. The primary has a 3-phase
winding, as in the linear synchronous motor, and
the secondary comprises a conducting plate on a
steel backing. The plate, which is normally made
from a good conductor such as copper or
aluminium, takes the place of the rotor conductors
of a rotary machine, and the steel backing provides
the flux return path.
(a)
(b)
(c)
(a)
(d)
Figure 6-30: Linear induction motor flux and
current plots: (a) 0º, (b) 30º, (c) 60º, (d) 90º.
(b)
Figure 6-28: Linear induction motor models:
(a) short primary, (b) short secondary.
Like rotary induction motors, linear induction
motors are robust and self-starting. They have been
applied in airport baggage handling systems and
theme park rides, and in advanced transport
systems such as the magnetically levitated vehicle
shown in figure 6-31.
A model of the active part of a linear induction
motor is shown in figure 6-29.
Figure 6-29: Linear induction motor model.
Figure 6-30(a)–(e) show plots of the magnetic
flux in the motor and the currents in the conductors
when t = 0º, 30º, 60º, 90º and 120º respectively.
The current magnitude is represented by a colour
that ranges from blue for minimum to red for
maximum.
Figure 6-31: Maglev vehicle with linear
induction motor propulsion. (HSST Corporation)
The behaviour of a linear induction motor can
differ significantly from that of an equivalent
rotary motor [3, 4]. Linear motors usually have
large airgaps, so the magnetising current is larger
than in a rotary motor. In addition, a linear motor
has two ends that have no counterpart in a rotary
motor. In a short-primary machine, transient
currents can be generated as secondary conducting
material enters the active region at one end and
leaves it at the other end. These transient currents
can reduce the force on the secondary and increase
the losses, particularly at high speeds. Shortsecondary machines do not suffer from this
problem, because the secondary conducting
material is always in the active region.
Induction Machines 55
7 STEPPER MOTORS
7.1
Introduction
The conventional AC and DC machines are
designed for continuous rotation of a shaft. It is
often desirable to control the speed, but most
applications do not require precise control of the
angular position.
Specialised applications such as automation
systems do require this kind of control. Often it is
achieved with a feedback control system
incorporating a position sensor, a drive motor, and
a controller. The position measured by the sensor is
compared with the desired value, and the
difference used to control the motor until the error
is acceptably small. AC and DC motors used in this
way are termed servomotors, but they will not be
considered further in this course.
Another way of achieving position control is to
use a special type of motor known as a stepping
motor or stepper motor. These motors are not
designed primarily for continuous rotation,
although they can work in this way if required.
Instead, they are designed to turn the output shaft
through a precise angle whenever the currents in
the windings are switched in a particular way by an
electronic driver. Each switching action results in
one increment or step in the rotor position.
There are three basic types of stepper motor:
simple permanent-magnet, variable-reluctance, and
hybrid. Simple permanent-magnet stepper motors
are similar to permanent-magnet synchronous
motors, except that the stator currents are switched
instead of varying continuously. They are no
longer widely used, and they will not be considered
further.
Variable-reluctance stepper motors are simple
applications of the variable-reluctance principle
introduced in section 7.2. They do not contain
magnets, and they are easy to control, but they
have relatively large step angles. A variant of the
variable-reluctance stepper motor is the switched
reluctance motor, which is designed for continuous
rotation.
Hybrid stepper motors make use of permanent
magnets to enhance the reluctance effect. They can
be made with very small step angles – typically
1.8º – and they can develop large values of torque.
Although they require a more complex type of
electronic driver, hybrid stepper motors are the
most popular variety.
7.2
Variable-reluctance principle
Consider an electromagnetic system of any kind,
energised by a coil carrying a current i. If the
displacement of any part of the system causes a
change in the reluctance R of the magnetic circuit,
the inductance L will also change. In the
Electromechanics course notes [1] it is shown that
the x-component of force on that part is:
fx
1
2i
2
L
R
12 2
x
x
(7-1)
L
R
12 2
(7-2)
where is the magnetic flux through the coil.
Similarly, if rotation through an angle causes a
change in the reluctance and the inductance, then
the corresponding torque is:
T
1
2i
2
Equations 7-1 and 7-2 can be very useful for
determining forces and torques in devices where it
is possible to calculate the inductance or the
reluctance. These equations also explain the
principle of a variety of practical devices that
develop a force or a torque because the reluctance
of the magnetic circuit can change.
Alignment torque
Figure 7-1 shows a simple variable-reluctance
actuator, where a short steel bar can rotate between
the poles of an electromagnet.
Figure 7-1: Variable-reluctance actuator.
56 Electrical Machines and Systems Course Notes
When the electromagnet is energised, there is a
torque on the bar, acting in a direction that would
rotate it into alignment with the poles, as can be
seen from the flux plot in figure 7-2.
From equation 7-2, the torque is:
Figure 7-2: Flux plot for actuator.
Figure 7-3 shows a graph of the alignment torque
as a function of the angular position of the rotor,
computed from the field solution. Note that the
torque opposes the rotation, so a positive
displacement results in a negative torque.
Torque, Nm
5
2.5
0
R
1 2 ( Bdr ) 2
T 1 2 2
1
2 ( Bdr )
2
2l g
0 dr 2
2l g
0 dr
l g drB 2
(7-5)
0
Note that the torque is independent of the angle .
This implies that the torque will be the same for
any angle of displacement from the aligned
position, whereas figure 7-3 shows that the torque
is considerably reduced as the displacement
approaches zero. The difference arises from the
simplifying assumption that the flux is entirely
confined to the overlap region.
Worked example 7-1
The actuator in figure 7-1 has an airgap length of
5 mm, a mean airgap radius of 42.5 mm, and a
depth of 100 mm. The coil has 1000 turns and
carries a current of 5.0 A. If the reluctance of the
steel can be neglected, determine:
(a) the value of the flux density in the airgap,
(b) the maximum alignment torque.
Solution
-2.5
From Ampère’s circuital law, we have:
B 0 H
-5
-50
-25
0
25
50
Angle, degrees
4 10 7 1000 5.0
Figure 7-3: Alignment torque.
The maximum torque can be calculated from
equation 7-2 as follows. Let r be the mean radius
of the airgap, lg the radial length of the airgap,
the angle of overlap between the stator and the
rotor, d the depth of the device perpendicular to the
plane of the diagram, and B the flux density in the
airgap.
If it is assumed that the flux is entirely
confined to the overlapping portion of the stator
and the rotor, then the flux is:
BA Bdr
(7-3)
The reluctance of each airgap is:
Rg
l
ì0 A
lg
ì0 dr
0 Ni
2l g
2 5.0 10 3
0.628 T
From equation 7-11, the maximum torque is:
T
l g drB 2
0
6.68 Nm
5.0 100 42.5 10 9 (0.628) 2
4 10 7
Comment
The maximum torque from the graph in figure 7-3
is only 4.72 Nm. Equation 7-11 over-estimates the
torque because it is based on the assumption that
the flux is confined to the overlap region. The flux
plot in figure 7-2 shows this to be a very rough
approximation. If the airgap is made much smaller
in comparison with the other dimensions, equation
7-5 gives a more accurate result.
(7-4)
Stepper Motors 57
7.3
Variable-reluctance stepper motors
Figure 7-4 shows a 3D model of a simple variablereluctance stepper motor with a 4-pole rotor and a
6-pole stator.
(a)
Figure 7-4: Stepper motor model.
Figure 7-5 shows a cross-section of the model.
Coils on opposite pairs of poles are connected
in series to form one phase, so this is a 3-phase
motor.
(b)
a
c'
b'
b
c
a'
Figure 7-5: Model cross-section.
Figure 7-6(a) shows the flux plot when phase a is
energised, so that the rotor is held in alignment
with phase a. When current is switched from
phase a to phase b, the resulting flux plot is shown
in figure 7-6(b). Magnetic forces act on the
misaligned poles, and if the rotor is free to move it
will take up a new position of alignment with
phase b as shown in figure 7-6(c).
(c)
Figure 7-6: Stepper motor flux plots:
(a) phase a energised, (b) phase b energised,
(c) new rotor position.
Note that the field pattern has rotated 60º
clockwise, but the rotor has moved 30º counterclockwise.
There is a similar effect when current is
switched from phase b to phase c, and then from
phase c to phase a. Each switching transition
causes the rotor to turn through the step angle of
30º.
Variable-reluctance stepper motors are easy to
control, since the current in each phase merely has
to be turned on and off with semiconductor
switches such as MOSFETs [3, 4]. Hybrid stepper
motors (section 7-4) require more complex drive
circuits because the current has to be reversed.
58 Electrical Machines and Systems Course Notes
Torque production
7.4
Equation 7-5 for the alignment torque is applicable
to the variable-reluctance stepper motor:
The hybrid stepper motor is the most important and
widely used type. It is essentially a variablereluctance motor with the addition of a permanent
magnet to give a higher torque for a given current.
Figure 7-7 shows the rotor and stator of a typical
commercial hybrid stepper motor.
Figure 7-8 shows the structure of a simple
model of a 2-phase motor with a 10º step angle.
The rotor comprises two toothed discs joined by a
cylindrical permanent magnet, which gives the
discs opposite magnetic polarities. The stator has
four poles, each bridging the two rotor discs.
Cross-sections of the rotor discs and stator
poles are shown in figure 7-9.
The phases are energised according to the
pattern in table 7-1 for successive steps
T
l g drB 2
0
[7-5]
This is the maximum torque when the rotor and
stator poles are misaligned. The torque will vary
with the displacement angle, as shown in figure
7-3. Equation 7-5 may not be an accurate estimate
of the maximum value – see worked example 7-1 –
but it shows that the torque depends on B2. If the
steel parts of the magnetic circuit are unsaturated,
B will be proportional to the coil current, and the
torque therefore varies as i2. In practice, there is
usually significant saturation of the steel parts, so
the torque cannot be calculated by elementary
methods.
Switched reluctance motors
Although variable-reluctance stepper motors are
not very widely used, the same principle is
exploited in a type of motor known as a switched
reluctance motor [3, 4]. These are designed for
continuous rotation, with the electronic driver
controlled by a position sensor on the motor shaft
instead of using externally generated pulses.
Switched reluctance motors are very simple in
structure, and can be competitive with invertercontrolled induction motors in some applications.
They suffer from the disadvantages of torque
pulsation at low speeds, and noise. Switched
reluctance motors are beyond the scope of this
course but are covered in the Year 3 course
Electrical Machine Drives.
Hybrid stepper motors
Table 7-1: Motor phase currents.
phase A
+I
0
–I
phase B
0
+I
0
0
+I
–I
0
As with the variable-reluctance motor, one phase at
a time is energised, but the current in each phase
has to change direction.
Figure 7-7: Hybrid stepper motor
Stepper Motors 59
(a)
(b)
Figure 7-8: Hybrid stepper motor model:
(a) complete, (b) frame and two coils removed.
Figure 7-9 shows the situation when phase A is
energised. Stator pole A1 has S polarity, so it aligns
with the N rotor disc; stator pole A2 has N polarity,
so it aligns with the S rotor disc.
B1
A2
N
A1
B2
(a)
B1
A2
S
Consider the situation when phase A is
switched off and phase B is switched on. Poles B1
and B2 are symmetrically positioned with respect to
the rotor teeth, but the flux densities in the two
gaps are very different. Pole B1 has S polarity, so
the flux density in the gap between this pole and
the rotor N disc is high. Since pole B2 has N
polarity, the flux density in this gap is low. The
converse holds for the S rotor disc. The result is
that the rotor is pulled into a new alignment
position shown in figure 7-10.
It is not possible to show 2D flux plots for the
hybrid stepper motor because the field is
essentially 3D. Magnetic flux passes along the axis
from the permanent magnet to one rotor disc,
crosses the gap to the stator poles, then along the
poles to the other rotor disc, and returns to the
other end of the magnet. Figure 7-11 shows shaded
plots of the flux density magnitude in the two discs
and the poles when phase B is energised, but the
rotor has not yet moved to the new alignment
position. This demonstrates the effect of the
permanent magnet in concentrating the field in the
required gaps.
After one complete cycle of excitation, or four
steps, the rotor will have moved by one tooth pitch
from its original position. If there are n teeth on the
rotor, there will be 4n steps per revolution, so the
step angle is 90 / n degrees. Practical hybrid
stepper motors usually have a large number of
teeth on the rotor, resulting in a small step angle. A
typical step angle is 1.8º, giving 200 steps per
revolution, with 50 rotor teeth.
The maximum torque can be estimated from
equation 7-5 as follows. Let Bm be the flux density
in a gap due to the permanent magnet, and Be the
flux density due an energised pole. If the magnetic
circuit is unsaturated, the total flux density in one
gap will be Bm + Be, and in the other gap it will be
Bm – Be. If p is the number of active pairs of teeth,
the resultant torque is thus:
Tm
A1
B2
(b)
Figure 7-9: Phase A energised:
(a) N polarity end, (b) S polarity end.
60 Electrical Machines and Systems Course Notes
pl g drB12
pl g drB22
pl g dr
( B12 B22 )
0
0
0
pl g dr
{( Bm Be ) 2 ( Bm Be ) 2 }
0
4 pl g drBm Be
0
(7-6)
B1
A2
N
A1
B2
(a)
B1
A2
S
A1
Equation 7-6 shows that the torque is directly
proportional to Be, and therefore to the current in
the coil, provided that the magnetic circuit is
unsaturated. In contrast, the torque in a variablereluctance motor is proportional to the square of
the current. As with the variable-reluctance motor,
however, there is usually significant magnetic
saturation in practical devices, so equation 7-6 may
not be a good approximation.
In equation 7-6, the term Bm shows the effect of
the permanent magnet. It is desirable to make this
term large, and to avoid conditions that would
weaken the magnet. If Bs is the value of the flux
density at which the steel parts of the magnetic
circuit saturate, it may be shown that the greatest
torque is obtained when the following relationship
holds between Be, Bm and Bs:
Be Bm
B2
(b)
Figure 7-10: Phase B energised:
(a) N polarity end, (b) S polarity end.
(a)
1
2
Bs
(7-7)
It is possible to energise both phases at the
same time, giving the pattern of phase currents in
table 7-2.
Table 7-2: Motor phase currents.
phase A
+I
+I
phase B
–I
–I
+I
+I
–I
–I
+I
–I
This sequence gives a larger torque with the
same step angle, and is the normal mode of
operation. The current in each phase is a square
wave, and there is a phase shift of 90º between the
two currents. A combination of the patterns in table
7-1 and table 7-2 will give eight steps for each
cycle of excitation, which is termed the half-step
mode. The normal mode is the full-step mode.
7.5
Stepper motor characteristics
Static torque characteristics
(b)
Figure 7-11: Phase b energised:
(a) N polarity end, (b) S polarity end.
Consider a stepper motor with a constant current
passed through one or more phase windings. The
rotor will be pulled into alignment, and any attempt
to move it will be opposed by a torque. To a first
approximation, the restoring torque is given by
[11]:
T Tm sin n
(7-8)
where is the angular displacement of the rotor, n
is the number of teeth on the rotor, and Tm is the
peak static torque. This is a spring-like
Stepper Motors 61
characteristic, where the restoring force increases
with displacement.
The maximum restoring torque T = –Tm occurs
when n = 90º, corresponding to a displacement of
a quarter of a tooth pitch. If an external torque
exceeding Tm is applied, then the rotor will slip a
tooth – it will move to the next equilibrium
position one tooth pitch away. The maximum
position error, just before tooth slipping occurs, is
therefore a quarter of a tooth pitch. This is equal to
the step angle in a hybrid stepper motor.
Multi-step operation
With each switching of the motor phase currents,
the rotor of a stepper motor will move to a new
step position. Since the rotor has inertia and the
restoring force has a spring-like characteristic, the
rotor will oscillate about its new equilibrium
position. If the currents are switched at a low rate,
the rotor will move in steps, but it will oscillate at
each step position. If the oscillation has not
decayed to a low level before the next switching
instant, resonance is possible, where the oscillation
amplitude increases at each step until the rotor
motion becomes erratic [11]. This can happen
when the time interval between steps is an integer
multiple of the oscillation period. If the stepping
rate is Sr steps per second, and the oscillation
natural frequency is fn hertz, the condition for
resonance is therefore:
1
k
,
Sr
fn
k 1, 2, 3,
fn
,
k
k 1, 2, 3,
Start/stop rate
If the stepping rate is not too high, a stepper motor
will start from rest and run in the slewing mode
when the winding currents are switched at a
constant rate. Similarly, if the rate is not too high,
the motor will stop suddenly when the current
switching stops. Above a critical stepping rate –
the start/stop rate – the motor may lose steps when
it starts, and over-run or gain steps when it stops.
As long as the start/stop rate is not exceeded, the
rotor moves by a number of steps equal to the
number of switching steps. The start/stop rate
depends on the total inertia of the rotating system
as well as the properties of the motor and driver.
(7-9)
Pullout torque
The critical stepping rates for resonance are
therefore given by:
Sr
The nature of the currents in the motor phases
will depend on the type of electronic driver – see
section 7.6 – as well as the speed and the properties
of the motor. At low speeds, a typical driver will
maintain constant currents in the motor phases for
the duration of each step, so the current waveform
approximates to a square wave.
At high speeds, however, the waveform may be
very different from a square wave, because the rate
of rise of current is limited by the phase
inductance. If the speed is high enough, the current
will not reach its desired value before the end of
the step period. The waveform is then
approximately triangular, decreasing in amplitude
as the stepping rate increases. There will be a
corresponding reduction in the motor torque.
(7-10)
High-speed operation
When the stepping rate is higher than the
oscillation natural frequency, rotor motion is
continuous, and the small speed variation between
steps can be ignored. This mode of operation is
known as slewing. It resembles the steady-state
operation of a synchronous motor.
Like a synchronous motor, a slewing stepper motor
will stall when the applied torque exceeds the
pullout torque. The value of this torque depends on
the phase current waveform, so it varies with the
stepping rate. At low stepping rates, when the
phase current waveform is almost a square wave,
the pullout torque is substantially constant. At
higher rates, as noted above, the mean current will
fall, and the pullout torque will be reduced.
62 Electrical Machines and Systems Course Notes
Acceleration and deceleration
1400
Stepping rate, steps/s
Most stepper motors can run at a much higher
speed than the start/stop rate if the stepping rate is
gradually increased. Similarly, a motor running at
high speed can be stopped without gaining or
losing steps if the stepping rate is gradually
reduced.
Ideally, the acceleration or deceleration should
be related to the motor pullout torque
characteristic, since this indicates the maximum
torque available at any given speed. In practice,
however, a simple pre-defined speed ramp is
generally used. There are three kinds of ramp in
common use:
Linear ramp: the stepping rate is increased at a
uniform rate for acceleration, and decreased at
a uniform rate for deceleration.
S-curve or cosine ramp: the linear ramp is
modified to give a smooth change from zero to
maximum acceleration and back to zero, to
avoid transients at the beginning and end of the
ramp.
Inverse ramp: a curve resulting from a linear
ramp of the time interval between steps.
These ramps are illustrated in figure 7-12 for
acceleration, and figure 7-13 for deceleration.
1200
b
1000
800
600
a
400
c
200
0
0
0.2 0.4 0.6 0.8
1
1.2 1.4
Time, s
Figure 7-13: Typical deceleration ramps:
(a) linear, (b) cosine, (c) inverse.
Most stepper motor drivers offer a linear speed
ramp, and the better ones give the option of an
S-curve or cosine ramp. Some low-cost drivers use
an inverse ramp, because it is easy to implement
with a counter. This is the least desirable form,
because the acceleration or deceleration is greatest
at high stepping rates when the available torque is
at its lowest. However, it is better than no ramp at
all.
Stepping rate, steps/s
1400
1200
b
1000
800
600
a
400
c
200
0
0
0.2 0.4 0.6 0.8
1
1.2 1.4
Time, s
Figure 7-12: Typical acceleration ramps:
(a) linear, (b) cosine, (c) inverse.
Stepper Motors 63
Stepper motor control
Electronic drive circuits
A driver for a stepper motor usually comprises
three sub-systems:
A power drive that supplies current to the
motor windings.
A logic sequencer that generates the required
control signals for the power drive, in response
to step demand pulses.
A controller that generates the required
sequence of step demand pulses.
These sub-systems may be combined in a single
unit. More commonly, the controller is a separate
item that may take the form of a plug-in card for a
computer, and the logic sequencer is combined
with the power drive to form a drive module.
Most hybrid stepper motors require a bipolar
power drive, which can reverse the direction of the
current through the winding. Some types of motor
can use a simpler unipolar drive, which turns the
current on and off without reversing its direction,
but this requires two coils on each pole. Bipolar
drives are the preferred choice. They usually
employ chopper action for controlling the current
magnitude, as described on the next page.
Figure 7-14 shows the essential part of a
bipolar power drive for one phase, with the gate
control circuits for the MOSFETs omitted. Similar
drive circuits are used for the other motor phases.
The circuit operates as follows:
To supply positive current to the motor phase,
transistors Q1 and Q4 are turned on, connecting
terminal T1 to +½Vd and terminal T2 to –½Vd.
When Q1 and Q4 are turned off, current can
continue to flow in diodes D2 and D3, returning
energy to the supply, until the current falls to
zero.
To supply negative current, transistors Q2 and
Q3 are turned on, connecting terminal T1 to
–½Vd and terminal T2 to +½Vd.
When Q2 and Q3 are turned off, current can
continue to flow in diodes D1 and D4, returning
energy to the supply, until the current falls to
zero.
If transistors Q2 and Q3 are turned on before Q1
and Q4 cease to conduct, both arms of the bridge
will short-circuit the power supply. This
undesirable condition is known as shoot-through.
To ensure that it cannot happen, a delay is often
introduced between the positive and negative halfcycles of the current waveform, as shown in figure
7-15.
+½Vd
Q1
100
Current, %
7.6
0
-100
0
90
180
270
360
Time phase, degrees
Q3
D1
T1
Figure 7-15: Phase current waveform.
D3
motor
phase
Q2
T2
Q4
D2
D4
Rc
–½Vd
Figure 7-14: Bipolar power drive.
64 Electrical Machines and Systems Course Notes
Effect of winding inductance
100
0
-100
0
90
180
270
360
Time phase, degrees
Phase current, %
(a)
100
0
-100
0
90
180
270
360
Time phase, degrees
Phase current, %
(b)
100
0
Chopper control
With chopper control, the motor is supplied from a
high voltage source, but without a correspondingly
high phase resistance. The initial rate of rise of
current is given by Vd / L, so a high value of Vd
gives a rapid rise of current. To limit the final
value of the motor current, chopper control works
as follows for positive current:
Transistors Q1 and Q4 are turned on.
Current begins to rise.
When the current reaches an upper threshold
value, Q1 and Q4 are turned off.
Current begins to fall.
When the current reaches a lower threshold
value, Q1 and Q4 are turned on.
The cycle repeats.
A similar sequence applies for negative current,
using transistors Q2 and Q3. The voltage drop
across the current-sensing resistor Rc is used for
controlling the transistors, and the resulting phase
current waveform takes the form shown in figure
7-17. Here it is assumed that the transistors are
switched off when the current reaches 110% of the
desired value, and switched on again when the
current falls to 90%.
-100
0
90
180
270
360
Time phase, degrees
(c)
Figure 7-16: Phase current waveforms at
different stepping rates:
(a) low, (b) medium, (c) high.
Phase current, %
Phase current, %
The rise and fall of current is approximately
exponential, governed by the inductance and
resistance of the motor phase, but modified by the
voltage induced in the winding when the rotor
moves. At higher stepping rates, the time constant
represents a larger fraction of the waveform period,
so the current in one phase takes the form shown in
figure 7-16. In deriving these graphs, the motional
voltage has been ignored, and it is assumed that the
driver does not use chopper control. The mean
current falls progressively, with a corresponding
reduction in the motor torque.
Let R be the resistance of the motor phase plus
any external resistance, L the motor phase
inductance, and Vd the supply voltage. The
maximum phase current is then Vd / R, and the time
constant is L / R. To improve the shape of the
waveform at high stepping rates, it is possible to
increase both Vd and R. This will reduce the time
constant but leave the maximum current
unchanged. However, it is very inefficient because
of the power loss in R. A better method is chopper
control.
110
0
-110
0
90
180
270
360
Time phase, degrees
Figure 7-17: Waveform with chopper control.
Stepper Motors 65
8 POWER ELECTRONIC CONTROL
8.1
AC/DC Converters
AC/DC converters convert alternating current or
voltage into direct current or voltage [3, 4]. In most
applications, the power flow is from the AC source
to the DC load, and the process is called
rectification. If the power flow reverses, the
process is called inversion.
If the diodes are replaced by thyristors, as
shown in figure 8-3, the output voltage can be
controlled by varying the firing angle – the point
on the cycle where the thyristors are turned on by
applying a gate drive signal. The gate drive circuits
are not shown in figure 8-3.
U1
U3
R
+
+
vL
vs
L
U2
Single-phase AC/DC converters
D1
D3
R
+
vs
+
Figure 8-3: Single-phase thyristor bridge.
Figure 8-4 shows the resulting waveform for a
firing angle of 30º. The output goes negative for a
portion of each cycle before the next thyristor is
fired, so the mean value of the waveform falls.
Voltage, %
The simplest AC/DC converter is just a rectifier
bridge circuit using diodes. Figure 8-1 shows a
single-phase bridge supplying a passive load.
vL
L
D2
D4
100
50
0
-50
-100
0
180 360 540 720
Time phase, degrees
Figure 8-1: Single-phase diode bridge.
When the source voltage vs is a sinusoidal
alternating quantity, the output voltage vL is
unidirectional. Figure 8-2 shows the output for two
cycles of the input voltage, as a percentage of the
maximum value of vs.
Figure 8-4: Thyristor bridge output voltage.
The mean output voltage is now given by:
Vd Vd 0 cos
(8-2)
where Vd0 is the diode bridge output given by
equation 8-1 and is the firing angle. Thus, the
output voltage can be varied from Vd0 to zero by
varying the firing angle from 0 to 90º. Figure 8-5
shows the waveform when = 90º, where the
mean output voltage is zero.
100
50
0
0
180
360
540
720
Time phase, degrees
Figure 8-2: Diode bridge output voltage.
The mean output voltage is given by:
Vd 0
2 2V
0.900V
(8-1)
where V is the RMS value of the AC input.
66 Electrical Machines and Systems Course Notes
Voltage, %
Voltage, %
U4
100
50
0
-50
-100
0
180
360
540
720
Time phase, degrees
Figure 8-5: Output voltage when = 90º.
Inverter action
Vd 0
3 2V
1.35V
(8-3)
where V is the RMS line voltage of the AC supply.
100
50
0
-50
-100
0
180
360
540
Voltage, %
Voltage, %
Equation 8-2 indicates that the output voltage will
reverse if the firing angle exceeds 90º. Figure 8-6
shows the output waveform when = 150º, which
is an inversion of figure 8-4.
The output waveform is shown in figure 8-9. The
output is much smoother than for single phase, and
the mean value is given by:
720
100
50
0
Time phase, degrees
0
Figure 8-6: Output voltage when = 90º.
Current cannot flow in the reverse direction
through the thyristors, so the circuit will not work
with a passive load if > 90º. However, current
can continue to flow in the forward direction if the
load contains a voltage source, such as a battery or
a DC motor armature, with the polarity shown in
figure 8-6. The DC source eL then supplies power
to the converter, which returns power to the AC
supply. In this mode, the converter is operating as
an inverter.
U1
U3
540
720
Time phase, degrees
Figure 8-9: 3-phase diode bridge output.
Figure 8-10 shows a 3-phase thyristor bridge, and
figure 8-11 shows the corresponding output voltage
waveform when the firing angle is 30º.
U1
U3
U5
a
R
+
b
vL
L
U2
+
vs
360
c
eL
+
180
U4
U6
R
Figure 8-10: 3-phase thyristor bridge.
U4
Voltage, %
U2
L
Figure 8-7: Inverter action.
3-phase AC/DC converters
A 3-phase diode bridge can be formed from a
single-phase bridge by adding just one extra pair of
diodes, as shown in figure 8-8.
D1
D3
R
b
+
vL
c
L
D2
D4
50
0
0
180
360
540
720
Time phase, degrees
Figure 8-11: 3-phase thyristor bridge output.
D5
a
100
D6
The mean output voltage is given by:
Vd Vd 0 cos
(8-4)
where Vd0 is the 3-phase diode bridge output given
by equation 8-3 and is the firing angle.
Figure 8-8: 3-phase diode bridge.
Power Electronic Control 67
8.2
DC motor control
In section 3.3 it was shown that the speed of a DC
motor could be controlled by varying the voltage
applied to the armature. A thyristor AC/DC
converter is a simple method of generating a
variable DC voltage from the AC mains supply, so
it provides an effective way of controlling the
speed of a DC motor [3, 4]. The combination of a
converter and a DC motor is the basis of variablespeed drive systems that have dominated the
market for many years.
DC motors have several disadvantages in
comparison with induction motors:
High manufacturing cost
Maintenance of the commutator and brushes
Protection required in hostile environments
Until recently, the high cost of inverters for
variable-speed AC drives (see section 8.3) has
offset the disadvantages of the DC motor. This is
no longer the case. AC drives are now the preferred
choice in most applications. However, there is still
a market for large DC drives in specialised
applications such steel rolling mills and mine
winders.
When a DC motor is supplied from a thyristor
converter, the armature generated voltage can
affect the converter voltage waveform, so that the
mean output voltage varies with the load current
[13]. This results in poor speed regulation unless
some form of feedback control is used.
A particular problem with thyristor converters
is that the current cannot reverse. Special measures
are therefore required if the DC motor torque is
required to reverse, as can happen during braking
[3, 13].
An important function of a DC motor
controller is to provide a soft start by gradually
increasing the voltage applied to the armature.
In section 8.2, it was observed that the power
flow through a DC/AC inverter could be reversed
if the firing angle exceeded 90º and a DC voltage
source of the correct polarity was connected to the
output. This is a known as a line-commutated
inverter, since the switching of current from one
device to another is determined by the alternating
voltages applied to the 3-phase input.
It is possible to use a line-commutated inverter
to drive a synchronous motor, since this machine
will generate the required voltage waveforms. This
approach is sometimes used with large
synchronous motors [12]. In general, however,
variable-frequency inverters for motor control use
forced commutation, where the switching of
current from one device to another is
predetermined, and not dependent on an AC source
connected to the output. These are more
complicated than line-commutated inverters, but
are more versatile.
The most common form of inverter for motor
control is the 3-phase bridge inverter [3, 4] shown
in figure 8-12, with the base drive components
omitted. The DC supply voltage Vd is connected to
the terminals labelled +½Vd, –½Vd, and the AC
output is taken from the terminals labelled a, b, c.
Although the switching devices are shown as
transistors, other power electronic switching
devices are frequently used [3, 4]. The most
popular choice is the IGBT (insulated gate bipolar
transistor) for power ratings up to about 500 kW.
The circuit is effectively a 3-phase version of the
bipolar power drive for stepper motors, described
in section 7.6.
+½Vd
Q1
D1
Q3
D3
Q5
D5
a
b
8.3
DC/AC Inverters
The only way to control the speed of a
synchronous motor is to vary the frequency of the
AC supply. This is also the most effective way of
controlling the speed of an induction motor, so
there is a general requirement for a variablefrequency AC source. An electronic system that
converts DC to variable-frequency AC is known as
an inverter.
c
Q2
D2
Q4
D4
Q6
D6
–½Vd
68 Electrical Machines and Systems Course Notes
Figure 8-12: 3-phase bridge inverter.
Six-step inverter
Va, %
Vab, %
0
180
360
540
720
Time phase, degrees
100
0
0
180
360
540
720
Time phase, degrees
Vca, %
100
0
-100
180
360
540
720
Time phase, degrees
0
Figure 8-14: Six-step line-to-line voltages.
-50
180
360
540
720
Time phase, degrees
Pulse-width modulation
50
Vb, %
-100
0
50
0
0
-50
180
360
540
720
Time phase, degrees
50
Vc, %
0
-100
The simplest mode of operation is to control the
switches so that the voltage waveform at each
output terminal is a square wave. Figure 8-13
shows the voltages with respect to the mid-point of
the DC supply, as a percentage of the DC supply
voltage Vd. The line-to-line voltages are then quasisquare waves, as shown in figure 8-14. Because
there are six switching transitions per cycle, this is
known as a six-step inverter.
0
100
Vbc, %
This circuit is switched in a similar way to the
stepper motor drive circuit. In each leg of the
bridge, one transistor at a time is switched on,
connecting an output terminal to either +½Vd or –
½Vd. As in the stepper motor drive, the diodes
provide paths where load current can continue to
flow when the transistors are turned off. It is
essential that one transistor has ceased to conduct
before the second transistor in the same leg is
turned on, otherwise the DC supply will be shortcircuited. This is the same problem of shootthrough that was mentioned in section 7.6.
0
The six-step inverter waveform is rich in
harmonics, which can have an adverse effect on
motor performance [12]. This type of inverter is
generally used in high power drives, typically
above 1 MW, but for lower powers a technique
known as pulse-width modulation (PWM) is used
to generate a better approximation to a sine wave
[3, 4]. Instead of switching the devices at a low
frequency to generate the required output wave
shape, in PWM they are switched at a high
frequency and the mark/space ratio is modulated
sinusoidally. The result is a waveform with an
average value that varies sinusoidally, as shown in
figure 8-15.
-50
0
180
360
540
720
Time phase, degrees
Figure 8-13: Six-step output voltages.
Power Electronic Control 69
100
Vab, %
Dynamic braking
0
-100
0
90
180
270
360
Time phase, degrees
Figure 8-15: PWM inverter voltage waveform
For clarity, the carrier frequency in figure 8-15
is only 12 times the modulation frequency, so the
approximation to a sine wave is not very good. In
practical inverters, the frequency ratio is typically
100 at the highest modulation frequency.
With a PWM inverter, it is a simple matter to
control the amplitude of the equivalent sine-wave
output by varying the modulation index. In figure
8-13, for example, the amplitude is about 70% of
the maximum possible output. For AC motor
control, it is necessary to change the output voltage
amplitude when the frequency changes: see section
8.4.
A problem with the DC link inverter is that the DC
link current cannot reverse, because a simple
AC/DC converter cannot accept current flow from
the DC link. If a motor connected to the output of
the inverter regenerates, the power flow will
reverse, and the current flow from the inverter to
the link will cause the voltage on the capacitor to
rise. Control circuits should shut down the inverter
if this happens.
If the inverter is required to handle reverse
power, which can happen when a motor is braked
by rapidly reducing the inverter output frequency, a
resistor must be connected across the DC link to
absorb the power. This is termed a dynamic
braking resistor, which is usually connected by a
semiconductor switch as shown in figure 8-17. The
switch is turned on only when the DC link voltage
rises above a preset level, so that the resistor is not
connected during normal operation of the inverter.
DC link inverter
Normally the DC supply for the inverter is
obtained from the AC mains via a converter of the
kind described in section 8.1. The DC link between
the converter and the inverter includes inductance
and capacitance, as shown in figure 8-16, to give a
smooth DC input to the inverter. This arrangement
is known as a DC link inverter.
converter
DC link
Figure 8-17: Dynamic braking resistor.
In applications requiring frequent power
reversal, the power loss in a dynamic braking
resistor may be unacceptable. A more complex and
costly alternative is to provide a second AC/DC
converter that operates as an inverter when the
direction of the DC link current reverses, thereby
returning energy to the AC supply [12].
inverter
Figure 8-16: DC link inverter.
In many cases, the converter is simply a diode
bridge, which may be single-phase for low-power
applications.
70 Electrical Machines and Systems Course Notes
8.4
AC motor control
V
f
V0 kV0
f0
(8-6)
E
f
E 0 kE 0
f0
(8-7)
Relationship between voltage and frequency
All AC machines and transformers are limited by
the maximum permissible flux density in the steel
parts of the magnetic circuit. This, in turn, imposes
a relationship between the magnitude of the
applied voltage and the frequency. In a
transformer, we have equation 2-10 for the
magnitude of the primary winding voltage:
V1m 2 fN 1 ABm
V1
R12 (L1 ) 2
s k s 0
(8-5)
where L1 is the self-inductance of the primary. At
very low frequencies, the current will be
determined by the winding resistance R1 rather than
the reactance L1. The constant volts per hertz rule
therefore ceases to apply at very low frequencies,
where a constant voltage is required.
(8-8)
and the reactances are:
X d kX d 0 ,
[2-10]
If the core flux density Bm is held constant, this
equation shows that the magnitude of the voltage is
proportional to frequency. The same principle
applies to AC machines, which gives the constant
volts per hertz rule for variable-frequency
operation.
Equation 2-10 was derived on the assumption
that the resistance of the winding is negligible.
This is usually a good approximation at the normal
AC mains frequency of 50 Hz, but it does not hold
at very low frequencies. If the transformer
secondary is open-circuited, the flux density in the
core is determined by the current in the primary
winding, which in turn depends on the input
impedance:
V
I1 1
Z1
Similarly, the synchronous speed is given by:
X q kX q 0
(8-9)
With these definitions, the torque equation 6.9
becomes:
Td
3
s
3
s0
VEsin V 2 1
1
sin 2
2 X q X d
X d
V E sin V 2 1
1
0 0
0
sin
2
2 X q 0 X d 0
X d 0
(8-10)
Thus the torque depends on the load angle but is
independent of the frequency.
Induction motor soft start
In section 6.2, it was noted that the starting current
of an induction motor is typically six times the fullload current. In addition, as shown in section 6.6,
there is a large transient torque when the full
supply voltage is suddenly applied to the motor.
These problems can be reduced by soft starting –
gradually increasing the applied voltage. This is
easily accomplished with an AC phase controller
of the form shown in figure 8-18, connected
between the 3-phase supply and the load [12].
a1
a2
b1
b2
c1
c2
Synchronous motor speed control
The speed of a synchronous motor can be
controlled by varying the frequency of the applied
voltage. If the armature resistance can be
neglected, the phase voltage V will be proportional
to the frequency. Since the excitation voltage E is
generated by rotation of the magnetised rotor, it
will be proportional to the speed, and therefore
proportional to the frequency. Let V0 and E0 be the
values of E and V at the base frequency f0. If we
define a frequency ratio k = f / f0, the values at any
other frequency f are:
Figure 8-18: Phase controller
Power Electronic Control 71
Voltage, %
Figure 8-19 shows the output voltage waveform for
one phase when the load is purely resistive, and the
firing angle of the thyristors is 60º.
100
Is
+
I0
Vs
jXm
Ir
Rr
s
Figure 8-20: Induction motor equivalent circuit
-100
180
360
540
720
Time phase, degrees
The torque is given by equation 6-9, which
may be re-arranged as follows:
Td
Figure 8-19: Phase controller output voltage
The RMS value of the output voltage falls as the
firing angle increases, reaching zero when
= 180º.
Induction motor speed control
As with the synchronous motor, the most effective
way of controlling the speed of an induction motor
is to vary the supply frequency. For optimum
performance, the supply voltage must also be
varied to maintain a constant magnetic flux density
in the machine. The principles of variablefrequency operation may be deduced from the
equivalent circuit of figure 6-6, reproduced below:
Rs
jxs
jxr
rc
(8-11)
s s s0 s 0
(8-12)
The fractional slip s at frequency f is thus:
s
Ir
Rr
s
Vs
3 2 Rr
3 2
Ir
I r Rr
s
s
s s
The quantity ss is the slip angular frequency, so
the developed torque will be constant if the rotor
current and the slip frequency are held constant.
Constant slip frequency is the preferred operating
mode, so it is necessary to explore the
consequences of this constraint when the frequency
is varied. As with synchronous machines, we
introduce a frequency ratio k = f / f0, where f is the
operating frequency and f0 is the base frequency. If
s0 is the fractional slip at the base frequency f0,
corresponding to a synchronous speed s0, then:
I0
+
jxr
0
0
Is
jxs
Rs
s0 s 0 s0 s 0 s0
s
k s 0
k
(8-13)
The value of the rotor resistance element in the
equivalent circuit is:
Rr
R
kR
r r
s
s0 / k
s0
jxm
(8-14)
and the values of the reactances are:
The core loss resistance is given by equation 6-15:
rc
X m2
Rc
[6-15]
Since Rc is approximately constant under variablefrequency conditions [3, 4], and Xm is proportional
to frequency, the value of rc will be insignificant at
low frequencies. Therefore, this element may be
ignored, giving the circuit shown in figure 8-20.
x s kx s 0 ,
x r kx r 0 ,
X m kX m 0
(8-15)
With these substitutions, the equivalent circuit
takes the form shown in figure 8-21:
Is
Rs
jkxs0
jkxr0
+
I0
Vs
jkXm0
Ir
Figure 8-21: Variable-frequency circuit
72 Electrical Machines and Systems Course Notes
kRr
s0
Torque, Nm
100
If the correct value of stator resistance Rs is
included in the model, it has an increasing effect
on the machine performance at low frequencies
because all the other elements in the equivalent
circuit are getting smaller. Figure 8-23 shows the
resulting torque/speed curves for the example 3
kW motor, when the stator voltage is directly
proportional to frequency.
Torque, Nm
Apart from the resistance Rs, the values of all
the elements in this circuit are proportional to the
frequency ratio k. If we set Rs = 0, and make the
stator voltage Vs proportional to k, then the currents
Is, I0 and Ir will be constant. Since the magnetic
flux density in the machine is proportional to I0,
this constraint ensures that the flux density remains
constant. It is therefore the normal mode for
variable-frequency operation.
Figure 8-22 shows theoretical torque/speed
curves for the example 3 kW motor from section 7,
with Rs = 0 in the equivalent circuit and the voltage
proportional to frequency. The slip is 5% at 50 Hz,
which is close to the full-load value. This gives a
theoretical torque of 20.3 Nm, shown by the blue
horizontal line in figure 8-20. All of the
torque/speed curves have the same shape, with the
same value of breakdown torque, and the full-load
slip speed is 75 rev/min in each case.
70
60
50
40
30
20
10
0
20
30
40
50
10
2.5
0
300 600 900 120 150
0
0
Rotor speed, rev/min
Figure 8-23: Variable-frequency operation:
2.5, 10, 20, 30, 40 and 50 Hz, constant V / f.
80
60
10
40
20
20
30
40
50
2.5
0
0
300 600 900 120 150
0
0
At 50 Hz and a fractional slip of 5%, the
developed torque is 18.5 Nm, shown by the blue
horizontal line in figure 8-21. This is slightly less
than the nominal full-load torque, giving an output
power of 2.76 kW. Table 8-1 shows the
performance for the frequency range 2.5 – 50 Hz.
Table 8-1: Constant V / f
Rotor speed, rev/min
Frequency, Hz
Figure 8-22: Variable-frequency operation:
2.5, 10, 20, 30, 40 and 50 Hz, Rs = 0.
At a supply frequency of 2.5 Hz, the full-load
torque is obtained at standstill. If the motor is
started at this frequency, it will deliver the full
rated torque but only take the normal full-load
current. In contrast, a motor started at 50 Hz will
take about six times the full-load current. Inverter
control therefore gives optimum starting
performance as well as speed control.
50
Phase voltage, V 230
40
30
20
10
2.5
184
138
92.0 46.0 11.5
Phase current, A 5.55 5.49 5.39 5.20 4.68 2.86
Fractional slip, % 5.00 6.25 8.33 12.5 25.0 100
Torque, Nm
18.5 18.1 17.5 16.2 13.2 4.90
Speed, rev/min
1425 1125 825
525
225
75
Power Electronic Control 73
To compensate for the effect of the stator
resistance, it is necessary to boost the voltage at
low frequencies. Table 8-2 shows the
corresponding results when the voltage is increased
to maintain a constant full-load torque. Figure 8-22
shows the corresponding set of torque/speed
graphs.
Table 8-2: Variable V / f
Frequency, Hz
50
Phase voltage, V 230
40
30
20
186
142
98.2 54.5 22.4
10
2.5
Phase current, A 5.55 5.55 5.55 5.55 5.55 5.55
Inverter voltage-frequency characteristic
The results given above for a 3 kW motor show
that it is necessary to modify the simple linear
relationship between voltage and frequency.
Inverter manufacturers normally enable the user to
adjust the volt relationship to suit a particular
motor. As an illustration, the red curve in figure
8-25 shows the required voltage-frequency
relationship for the example 3 kW motor, which
deviates considerably from the simple linear
relationship shown by the blue line.
Fractional slip, % 5.00 6.25 8.33 12.5 25.0 100
Torque, Nm
18.5 18.5 18.5 18.5 18.5 18.5
Speed, rev/min
1425 1125 825
225
75
Phase voltage, V
525
240
70
Torque, Nm
60
50
40
20
10
30
30
40
180
120
60
50
0
20
0
2.5
10
10
20
30
40
50
Frequency, Hz
0
0
300
600
900 1200 1500
Rotor speed, rev/min
Figure 8-24: Variable-frequency operation:
2.5, 10, 20, 30, 40 and 50 Hz, variable V / f.
Although the performance at full-load slip is
now constant throughout the speed range, the
breakdown torque and the starting torque are still
significantly reduced at low frequencies. This is
another effect of the stator resistance Rs. In larger
motors, Rs is a smaller proportion of the motor
impedance, so the characteristics approach the
ideal more closely.
Figure 8-25: Voltage-frequency relationship for
a 3 kW induction motor.
The required characteristic is very nearly a
straight line that does not pass through the origin.
Many inverter manufacturers provide an adjustable
characteristic of the form shown in figure 8-26,
where the boost voltage Vb, break-point voltage Vp,
and break-point frequency fp can be set as required.
74 Electrical Machines and Systems Course Notes
This shape is easily adjusted to give a good
approximation to the characteristic in figure 8-25.
voltage
V0
Vp
Vb
0
fp
f0
frequency
Figure 8-26: Inverter user-defined voltagefrequency characteristic.
When the concept was first introduced, vector
control required the direct measurement of the flux
in the motor with field sensors. Since this was
often impracticable, alternative methods have been
developed based on measurement of the stator
voltage and current and the rotor position. The
latest development is sensorless vector control,
where the rotor position measurement is not
required. This method uses measurements of the
stator voltage and current alone, together with a
mathematical model of the motor.
Drives that implement vector control, with or
without a rotor position sensor, can determine the
parameters of the motor model automatically from
measurements of the stator quantities. They are
more costly than simple inverters with frequency
and voltage control, but they offer improved
dynamic response to changes in the motor speed
and the torque load.
Vector control of induction motors
In a DC motor, the commutator forces the magnetic
axis of the armature flux to be at right angles to the
axis of the field flux. This gives the maximum
torque for a given current in the machine windings,
and it gives the DC machine simple control
properties: the torque is determined by the product
of the armature current and the field flux.
In an induction motor, the corresponding angle
between the flux components depends on the
operating conditions, and is not predetermined by
the design of the machine. However, for any given
rotor speed and load, it is possible to adjust the
stator voltage and frequency so that the magnetic
conditions resemble those of a DC motor, with the
two flux components at right angles. With this
constraint, an induction motor can be controlled in
a similar way to a DC motor, to achieve good
dynamic performance. This method of controlling
the motor is known as vector control or fieldoriented control. A theoretical treatment is beyond
the scope of this course, but an introduction will be
found in references [3, 4, 12]. This topic is covered
in the Year 3 course Electrical Machine Drives.
Power Electronic Control 75
9 REFERENCES
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Edwards, J.D.: Electromechanics Course Notes, 2004.
Smith, R.J. and Dorf, R.C.: Circuits, Devices and Systems (5th edition, Wiley, 1992)
Edwards, J.D.: Electrical Machines and Drives (Macmillan, 1991)
Fitzgerald, A.E., Kingsley, C. Jr., and Umans, S.D.: Electric Machinery (6th edition, McGraw-Hill,
2003).
Say, M.G.: Alternating Current Machines (5th edition, Pitman, 1983).
Alger, P.L.: Induction Machines (2nd edition, Gordon and Breach, 1970).
BS 4999 (IEC 34-2): 1977: Methods for determining losses and efficiency of rotating electrical
machines from tests (British Standards Institution / International Electrotechnical Commission).
ANSI/IEEE Std 112 – 1978, IEEE standard test procedure for polyphase induction motors and
generators (American National Standards Institute).
Hancock, N.N.: Matrix Analysis of Electrical Machinery (2nd edition, Pergamon, 1974).
Edwards, J.D.: Electrical Machines (2nd edition, Macmillan, 1986).
Acarnley, P.P.: Stepping Motors: a Guide to Modern Theory and Practice (4th edition, Institution of
Electrical Engineers, 2002).
Murphy, J.M.D. and Turnbull, F.G.: Power Electronic Control of AC Motors (Pergamon, 1990).
Say, M.G. and Taylor E.O.: Direct Current Machines (2nd edition, Pitman, 1986).
76 Electrical Machines and Systems Course Notes
10 APPENDICES
10.1 Induction motor 2-axis equations
In matrix form, the 2-axis differential equations of a 2-pole induction motor are as follows [9, 10]:
v sd Rs Ls p
v
0
sq
vrd Mp
vrq M r
0
Rs Ls p
M r
Mp
0
Rr Lr p
Lr r
Mp
isd
Mp isq
Lr r ird
Rr Lr p irq
0
(10-1)
In equation 10-1, p d / dt and r is the rotor angular velocity in rad/s. The resistances Rs and Rr are the
normal stator and rotor resistances in the equivalent circuit, and the inductances are related to the
equivalent-circuit reactances as follows:
X m M
x s ( Ls M )
x r ( Lr M )
(10-2)
The variables vsd, vsq, isd and isq are the 2-axis stator voltages and currents. These are related to the 3-phase
terminal voltages and currents by the equations:
isd
i
sq
v sd
v
sq
1 12
2
3
3
0 2
i
12 sa
isb
23
isc
(10-3)
1 12
3
0 2
v
12 sa
v sb
23
v sc
(10-4)
2
3
Provided there is a 3-wire supply to the motor (no neutral connection), there is an inverse transformation
giving the 3-phase variable in terms of the 2-axis variables:
isa
i
sb
isc
1
2 1
3 2
12
0
isd
3
2
isq
23
(10-5)
v sa
v
sb
v sc
1
2 1
3 2
12
0
v sd
3
2
v sq
23
(10-6)
Appendices 77
The variables vrd, vrq, ird and irq are the corresponding 2-axis rotor voltages and currents. They are related to
the physical currents in an equivalent 2-phase rotor winding by the equations:
ird cos
i
rq sin
sin ir
cos ir
(10-7)
v rd cos
v
rq sin
sin v r
cos v r
(10-8)
ir cos
i
r sin
sin ird
cos irq
(10-9)
v r cos
v
r sin
sin vrd
cos v rq
(10-10)
where is the angle between the axis of the phase of the rotor and the a phase of the stator. The inverse
form of these equations is:
Equation 10-1 can be expressed in the form:
v ( R L p G r ) i Z i
(10-11)
where v and i are the column vectors of voltage and current variables, and the impedance matrix Z is given
by:
Z R L p G r
(10-12)
The resistance matrix R, the inductance matrix L and the torque matrix G all have constant elements:
Rs
0
R
0
0
Ls
0
L
M
0
0
0
G
0
M
0
Rs
0
0
0
0
Rr
0
0
Ls
M
0
0
M
Lr
0
0
0
0
0
M
0
0
Lr
78 Electrical Machines and Systems Course Notes
0
0
0
Rr
(10-13)
0
M
0
Lr
(10-14)
0
0
Lr
0
(10-15)
The developed torque is given by [9, 10]:
Td i T G i M (ird isq isd irq )
(10-16)
To solve problems such as starting or DC braking, equation 10-11 must be re-arranged in a suitable
form for numerical solution with all the derivatives on one side:
di
pi L1{v ( R G r ) i}
dt
(10-17)
In addition, there is the mechanical equation of motion:
J
d r
Jp r Td Tl M (ird isq isd irq ) Tl
dt
(10-18)
where J is the moment of inertia of the rotating system, and Tl is the load torque (which may be a function
of the rotor speed r).
Equations 10-17 and 10-18 represent five simultaneous first-order non-linear differential equations,
which can be solved by standard numerical methods such as Runge-Kutta to determine the stator currents
and the rotor speed as functions of time. If the motor has P pairs of poles, two changes are required to the
equations: (a) the left-hand side of the torque equation 10-16 is multiplied by P; (b) in equation 10-1, r is
replaced by Pr.
Appendices 79
10.2 List of formulae
General principles
Pout Pin Ploss
P
1 loss
Pin
Pin
Pin
Pout
Ploss
1
Pout Ploss
Pout Ploss
Efficiency:
Magnetic force on a current element: df idl B
Conductor in a magnetic field: e Blu , f Bli
Ampère’s circuital law:
H.dl i Ni
Materials:
B B ( H ) r 0 H , 0 4 10 7 H/m
Magnetic flux and flux linkage: BA, N
Faraday’s law: e
Inductance: L
d
d d
N
(Li )
dt
dt dt
, M 21 12
i
i1
i2
Stored energy: W ½ L1i12 ½ L2 i22 Mi1i2
Hl R
l
l
Resistance: R
A A
Sinusoidal operation: V1m 2 fN 1 ABm
Ideal transformer:
v2 i1 N 2
n
v1 i2 N1
Voltage regulation:
ZL
n2
V2 nl V2 fl
V2 nl
AC power
Complex power: S = P + jQ = VI*
Real power: P S cos VI cos Re( VI * )
Basic equations: ea K f r , Td K f ia
L
R
(linear system)
½ 2
2
Magnetic force per unit area:
Current: N 1i1 N 2i2 R 0
DC machines
L
R
(linear system)
½ 2
x
x
Torque: T ½i 2
d
d
, v2 N 2
dt
dt
Reactive power: Q S sin VI sin Im( VI * )
Ni
l
Reluctance: R
r 0 A
Force: f x ½i 2
Voltage: v1 N 1
Impedance transformation: Zin
di
di
v1 R1i1 L1 1 M 2
dt
dt
Coupled coils:
di
di
v 2 R 2 i 2 L2 2 M 1
dt
dt
MMF: F Ni
Transformers
f
B
½
A
0
Linear approximation: ea K i f r , Td K i f ia
Armature equation: v a Ra ia ea Ra ia K f r
Ra negligible: ùr
va
K f
Small motors: ùr
va
R T
a d2
K f ( K f )
Series motor: i f ia i
80 Electrical Machines and Systems Course Notes
3-phase systems and rotating field
va Vm cos( t )
3-phase voltages: vb Vm cos( t 120)
vc Vm cos( t 240)
Star connection: Vline 3 V phase
Delta connection: I line 3 I phase
Synchronous speed (2p poles):
s
p
[rad/s], n s
60 f
N s 60n s
p
f
p
[rev/s],
[rev/min].
Induction machines
Slip speed: N s N r
Fractional slip: s
N s N r s r
Ns
s
Rotor input: Pem s Td
Rotor output: Pmech r Td (1 s ) s Td
Rotor loss: Ploss ( s r )Td s s Td
3-phase torque: Td
3 2 Rr
Ir
s
s
1-phase torque: Td
1 2 Rr
1 2 Rr
I rp
I rn
s
s s
2s
Equivalent series/parallel elements:
Synchronous machines
Non-salient model: V jX s I E
Non-salient torque: Td
3VEsin
s X s
Salient-pole torque:
3 VEsin V 2 1
1
Td
sin 2
s X d
2 X q X d
Xd
1 sin 2
2X d X q
Speed control frequency ratio k f / f 0
Reluctance motor: Td
3V
2
V kV0 , E kE 0 , s k s 0
X d kX d 0 ,
Td
3
s0
X q kX q 0
V E sin V 2 1
1
0 0
0
sin 2
2 X q 0 X d 0
X d 0
rc
Rc
X m2
Rc
Rc2 X m2
rc2 x m2
rc
xm
Xm
Rc2
Xm
Rc2 X m2
rc2 x m2
xm
Loss components:
P.oss PRs PRr Pcore Pfw Pstray
PRs 3 I s Rs , PRr 3 I r Rr , Pcore 3 I m rc
2
PRr 3s | I r |2
2
2
Rr
sPem s( Pin PRs Pcore )
s
No-load test
Pin Pin PRs Pcore Pfw
Pin Pfw | I s |2 ( Rs rc ) 2
Appendices 81
Speed control
Td
3 2 Rr
3 2
Ir
I r Rr
s
s
s s
s s s0 s 0
s
s0 s 0 s0 s 0 s0
s
k s 0
k
Rr
R
kR
r r
s
s0 / k
s0
x s kx s 0 ,
x r kx r 0 ,
X m kX m 0
Stepper motors
Alignment torque: Tm
Hybrid stepper: Tm
l g drB 2
0
4 pl g drBm Be
0
Restoring torque: T Tm sin n
Stepper resonance: S r
fn
,
k
k 1, 2, 3,
AC/DC converters
1-phase diode bridge: Vd 0
2 2V
0.900V
3-phase diode bridge: Vd 0
3 2V
1.35V
Thyristor bridge: Vd Vd 0 cos
82 Electrical Machines and Systems Course Notes
