POWER ELECTRONIC CONTROL OF INDUCTION MOTORS ppt

ENGNG3070 Power Electronics Devices, Circuits and Applications
 E Levi, Liverpool John Moores University, 2002
74
6. POWER ELECTRONIC CONTROL OF INDUCTION MOTORS
6.1 Introduction
Three-phase induction motors are the most frequently utilised electric machines in
industry. They are characterised with low cost, high reliability, high efficiency, simple
construction and, in the case of squirrel-cage induction motors, with virtually maintenance-free
operation. If operated with stator three-phase voltage supply of fixed frequency and fixed rms
value, induction motors will run at a speed that very slightly depends on loading.
In contrast to DC machines, where choice of methods of speed control and associated
power electronic converters that are nowadays in use is rather limited, there exists a variety of
both speed control techniques and appropriate power electronic converters that are used in
conjunction with three-phase induction motor drives. A three-phase induction machine requires
three-phase AC supply at stator side. In a squirrel-cage type of induction machines this is
simultaneously the only approachable winding. However, in slip-ring induction machines the
three-phase rotor winding may be approached as well. Thus the speed of an induction machine
may be controlled by controlling the stator AC supply for both types of induction machines;
additionally, speed may be controlled in slip-ring machines from the rotor winding side as well.
Squirrel-cage induction machines are by far the most frequently used machines. It is for this
reason that the following discussion will be predominantly devoted to speed control methods
associated with alteration of the stator supply, that are equally applicable for both types of
induction machines. Only one method, specifically aimed at slip ring machines, will be looked
at.
If an induction machine is supplied with a voltage of frequency f then the so-called
synchronous speed is determined with the frequency and the number of pole pairs P and is
expressed in rpm as
n
s
=60f/P (6.1)
However, an induction motor will run at a speed n that differs from the synchronous. The

difference between actual speed of rotation and the synchronous speed is characterised by the
quantity called slip. Slip s is expressed in per unit as ratio of the speed difference normalised
with respect to the synchronous speed, i.e.,
s=(n
s
−
n) / n
s
(6.2)
Thus zero speed of rotation indicates unity slip and synchronous speed of rotation corresponds
to zero slip. Torque-speed characteristic of an induction machine can be derived for steady-
state operation with sinusoidal supply from the per-phase equivalent circuit, given in Fig. 6.1.
I
s
R
s
jX
σs
jX
σr
I
r
V
n
jX
m
I
m
R
r

/s
Fig. 6.1: Steady-state per-phase equivalent circuit of an induction machine for purely sinuso-
idal supply voltage.
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All the parameters of the rotor winding in Fig. 6.1 are referred to the stator winding, by
means of the transformation ratio. Symbols in bold denote phasors. Variable resistor in the
rotor circuit represents both rotor copper loss and power converted into mechanical. Recall
that stator winding of the machine can be connected in either star or delta; the equivalent
circuit is valid for phase rather than line values, regardless of the winding connection. Thus V
n
stands for rated phase to neutral voltage of the stator. All the reactances are given at fixed,
rated frequency. Torque-slip characteristic of a three-phase induction machine follows from
power flow considerations in Fig. 6.1, in the form
()
()
()
Ts
P
f
V
Rs
RRs X X
P
f
V
Rs
RRs X
en

r
sr s r
n
r
sr
()=
+++
=
++
3
2
3
2
2
22
2
2
2
ππ
σσ
(6.3)
while stator current phasor can be expressed directly from Fig. 6.1 as
()
IVZ
ZRjX
jX R s jX
Rs jX X
sne
es s
mr r

rrm
=
=+ +
+
++
σ
σ
σ
()
(6.4)
Torque is a rather complicated function of motor parameters, supply voltage and slip and its
typical appearance is given in Fig. 6.2 for rated supply conditions. The operating region is
restricted to slips up to typically 10%, indicating that speed of rotation changes with load but
remains within rather narrow boundaries from zero load up to rated load. Maximum (pull-out)
torque, rated torque and starting torque, as well as corresponding slips, are indicated in Fig.
6.2 and can be calculated from the following expressions:
T
e
maximum (pull-out) torque
T
en
T
est
operating region
slip
1s
m
s
n
0s

0n
s
speed
Fig. 6.2: Torque-speed characteristic of a three-phase induction machine for rated supply
conditions.
()
()
()
s
R
RX
TTss
P
f
V
Rs
RRs X
P
f
V
RRX
TTs
P
f
V
R
RR X
TTss
P
f

V
Rs
RRs X
m
r
s
em e m n
rm
srm
n
ss
est e n
r
sr
en e n n
rn
srn
=
+
===
++
=
++
===
++
===
++
22
2
2

2
2
22
2
2
2
2
2
2
3
2
3
4
1
1
3
2
3
2
()
()
()
ππ
π
π
(6.5)
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Equations (6.1)-(6.5) enable discussion of all the relevant methods of speed control,

applicable to an induction machine. It follows from (6.1) that synchronous speed can be altered
by changing the number of pole pairs. Assuming that load torque is constant, if pole pair
number is doubled during operation of the machine, synchronous speed will be halved, leading
to operation at essentially one half of the rated speed. This method of speed control is used in
drives that typically require operation at two distinctly different operating speed (say, a
washing machine; spinning is done at high speed, while normal washing cycle takes place at
low speed). Speed control by pole pair changing requires special construction of the stator
winding. It is usually realised for two different speeds of operation and pole pair changing is
performed by mechanical reconnection of the stator winding from one pole pair number to
another. Illustration of torque-speed characteristics is shown in Fig. 6.3 for change-over from
one pole pair to two pole pairs. Power electronics converters are not involved in this speed
control method, and its applicability is restricted to the cases when two speeds, rather than
continuous speed variation, are needed. Therefore speed control by pole pair changing will not
be considered further on.
T
e
load torque
AB
1500 3000 speed (rpm)
Fig. 6.3: Speed control by change of pole pair number: drive operates either in point A or in B.
The two methods of speed control, that are universally applicable to all the three-phase
induction machines and that will be elaborated, are the speed control by stator voltage
variation and speed control by simultaneous stator voltage and frequency variation. The
former, although very simple, has restricted applicability for the reasons that will be explained;
the latter is the most widely used method of speed control of induction machines. Finally, a
method valid for slip-ring machines only, insertion of a resistance in the rotor circuit, will be
considered as well.
6.2 Speed Control by Stator Voltage Variation
Equation (6.3) shows that electromagnetic torque developed by an induction machine is
proportional to the square of the applied rms stator phase voltage. Thus, given the load torque

to be, say, a constant, reduction of voltage will lead to operation with increased slip, i.e., with
decreased speed. As voltage is not allowed to exceed rated value, this method of speed control
can be utilised only for reducing the speed below rated. Torque-speed (slip) characteristics for
this speed control technique are shown in Fig. 6.4. Note that, according to (6.5), pull-out slip
is not function of the applied voltage. Hence the motor develops maximum torque at constant
slip (speed), determined with (6.5), regardless of the applied voltage. However, both maximum
(pull-out) and starting torque are functions of voltage squared. Hence, when voltage is
reduced, maximum torque and starting torque reduce as well, proportionally to the voltage
reduction squared. This is one of the major drawbacks of this speed control method: reduction
in starting torque means that the motor will be able to start only loads that are of small torque
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at low speeds; reduction in maximum torque means that overloading capability of the motor
reduces with reduction in voltage.
T
e
rated voltage
reducing voltage
operating region
slip
1s
m
0s
Fig. 6.4: Torque-slip characteristics of an induction machine with speed control by stator
voltage variation.
Additional drawback of this method is that, when voltage is reduced and speed
therefore reduces as well, additional copper loss in rotor winding takes place. Regardless of
these two serious shortcomings, this method of speed control is widely used in two distinct
cases. When the load torque is proportional to the speed squared (pumps, ventilators,

compressors, etc.) then even a small reduction in speed means significant reduction in the
output power, which is proportional to the cube of the speed. For a number of applications
with load torque of this type it is sufficient to vary the speed in this narrow region. The second
application is in drives that run for prolonged periods of time with very light loads. In such a
situation it is advantageous to reduce the voltage for light load operation as this improves the
efficiency of the drive. In other words, considerable saving in electricity consumption may be
achieved in this way.
Example:
A three-phase squirrel-cage induction motor drives a load of rated torque, with rated
slip of 3%. Stator and rotor resistance (referred to stator) are both equal to 0.015 Ω.
Sum of stator and rotor leakage reactance is X =0.09Ω. Calculate the necessary
reduction in stator supply voltage if the induction motor is to drive the same load with
slip equal to 15%.
Solution:
Pull-out slip of the motor is, from given parameters, equal to
s
R
RX
m
r
s
=
+
=+==
22
22
0 015 0 015 0 09 0164 164% .
The motor is required to operate at slip of 15%. As load torque is constant, this indicates that in new
operating point motor torque will be very close to maximum torque, so that overloading capability will
be almost non-existent. The necessary reduction of the voltage, that will yield operation with 15% slip,

can be calculated as follows:
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78
()
()
()
()
()
()
TT
T
P
f
V
Rs
RRs X
TT
P
f
V
Rs
RRs X
ss
T
T
P
f
V
Rs

RRs X
P
f
V
Rs
RRs X
V
Rs
RRs X
V
Rs
RRs X
V
V
Len
en n
rn
srn
een
r
sr
n
e
en
r
sr
n
rn
srn
r

sr
n
rn
srn
n
=
=
++
==
++
==
==
++
++
=
++
++
at all speeds; hence
3
2
3
2
003 015
1
3
2
3
2
2
2

2
11
2
1
1
2
2
1
1
1
2
1
1
2
2
2
2
2
1
2
1
1
2
2
2
2
2
1
π
π

π
π

()
()
()
()
=
++
++
=
++
++
=
=
2
1
1
2
2
2
2
2
2
2
2
1
015
003
0 015 0 015 015 009

0015 0015 003 009
039
0 624
s
s
RRs X
RRs X
VV
n
sr
srn
n
.
.
/. .
/. .
.
.
Necessary voltage reduction is 37.6%.
Situation is illustrated in accompanying Figure.
T
e
rated voltage
T
L
=T
en
62.4% of
rated voltage
0.16 0.03 slip

0.15
Let us examine, using this example, increase in rotor losses that takes place with this speed control
method. Taking power transferred from stator to rotor to be P
sr
, for these two operating conditions one
has
()
()
()
PsP PsP P
P
s
PP P
PT T s T
s
s
P
PsP P
P
s
PPP
PsP xP P
n n srn curn n srn srn
n
n
ncurn n
en en s en n
n
n
sr sr n n srn

cur sr n n
=− = =
−
==
==−=
−
−
=
=−
=
−
==≡
== =
1
1
103 0031
1
1
1
0876
1
1
0876 085 103
015 103 01545
11 1
1
111 1
1
1
111

.
./
.
ωωω
This consideration shows that power transferred from stator to rotor is the same for the two cases.
Hence reduction in output power reflects itself directly as an increase in rotor copper loss, which goes
up from 3% of the rated power to more than 15% of the rated power. As this loss takes place in the
motor, it will essentially cause overheating. Needless to say, efficiency is sharply reduced.
Starting problem with reduced voltage and this increase in loss are the two major reasons why this
speed control method is not used with constant load torques. Situation is much improved in both
respects when load torque is proportional to the square of the speed. Speed control by stator voltage
variation is therefore applied in conjunction with this type of load in practice.
Speed control by stator voltage variation is realised by using AC-AC voltage controller
in each stator phase of the machine. Voltage controller is of the same structure as in Chapter 4
on reactive power compensation. Figure 6.5 illustrates the connection of the power electronic
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converters for the case when only forward motoring is required. If the machine is required to
run in both directions, two additional sets of back-to-back (anti-parallel) thyristors are
required, in order to enable phase reversal of the stator supply. This part of the drive is shown
in Fig. 6.5 as well, in dotted lines.
It should be noted that although the principle of operation of an AC-AC controller is
very simple, analysis of the system of Fig. 6.5 is extremely tedious even for steady-state
operation. This is so because of the inductive nature of the machine, which makes the instant of
cessation of the current flow through each of the thyristors essentially unknown. As the
voltage exists as long as there is current flow, then it is actually very difficult to evaluate the
actual voltage applied across the machine under given operating conditions. Note that the
voltage value calculated in the previous example for reduced speed operation is the required

rms value of the fundamental harmonic of the output phase to neutral voltage of the AC-AC
voltage controller.
Induction
machine
Fig. 6.5: Speed control of an induction machine by stator voltage variation.
6.3 Speed Control of Slip Ring Machines by Addition of Resistance in the Rotor
As rotor winding of a slip ring machine can be approached from the outside world, it is
possible to add a resistance in each of the three rotor phases. Let R
add
denote per-phase value
of the added resistance in the rotor circuit, referred to the stator winding. From (6.3) and (6.5)
it follows that
()
()
()
()
()
()
s
RR
RX
fR R
TTss
P
f
V
RR s
RRR s X
T
P

f
V
RRX
fR R
TTs
P
f
V
RR
RRR X
m
radd
s
r add
em e m n
raddm
s r add m
em n
ss
r add
est e n
radd
s r add
=
+
+
=+
===
+
++ +

=
++
≠+
===
+
++ +
22
2
2
2
2
22
2
2
2
3
2
3
4
1
1
3
2
()
()
π
π
π
(6.6)
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Equation (6.6) shows that pull-out slip is proportional to the added resistance. Hence the speed
at which maximum torque occurs varies with the amount of added resistance. On the other
hand, maximum torque is not affected by addition of resistance, indicating that overloading
capability of the machine is not affected. Compared with stator voltage variation method,
addition of resistance in rotor is to be preferred, for the following reasons: a) any speed of
operation between zero and rated can be obtained (with stator voltage variation speed control
region is confined to speeds higher than pull-out speed); maximum torque is not affected (with
stator voltage variation it reduces proportionally to the stator voltage reduction squared);
additional copper loss is now developed in added resistance, which is external to the machine
and therefore the problem of overheating does not take place. Note however, that problem of
low efficiency remains to be present: as speed is reduced, larger and larger portion of the total
input power is dissipated in the additional resistors. Principle of this speed control method and
resulting torque speed curves are shown in Fig. 6.6.
Stator T
e
R
add
=0
IM
Load torque
Rotor
Increasing R
add
R
add
1s
1
s

2
s
n
0 slip
Fig. 6.6: Speed control by addition of resistance in rotor winding - principle and torque - slip
curves.
Three possible operating points are shown in Fig. 6.6 for assumed constant load
torque. As added resistance increases slip increases as well (i.e., speed decreases). Of special
interest is the curve with operating slip given as s
1
. Note that for this curve the amount of
added resistance is assumed to be exactly such as to yield development of maximum torque at
zero speed (i.e., slip of one). By doing so, it is possible to start the motor with starting torque
equal to maximum torque: this indeed is one of frequent applications of addition of resistance
in the rotor circuit, even when speed control is not required. If the motor parameters are
known the value of added resistance that gives starting with maximum torque can easily be
determined from (6.6), where it is only necessary to equate slip s
m
to unity.
Example:
A slip ring induction machine is loaded with constant load torque equal to 5 Nm (rated
value). Machine is star connected, its stator is supplied with rated 380 V, 50 Hz
voltage, and the machine has two pairs of poles. Stator and rotor resistance are 10 Ω
and 6.3 Ω, respectively, while leakage reactances of stator and rotor are 12 Ω each.
Magnetising reactance can be neglected. Rotor parameters are referred to stator.
a) Determine slip for rated operating conditions;
b) Calculate added resistance that is needed to reduce the speed to 1000 rpm;
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c) Calculate added resistance that will enable starting of the motor with starting torque
equal to maximum torque.
Solution:
a) Calculation of rated slip:
()
()
()
()
()
TT
T
P
f
V
Rs
RRs X X
V
ss
s
nfP n sn
s
R
RXX
T
P
f
V
R
RR X
Len

en n
rn
srn s r
n
nn
n
snnn
m
r
ssr
est n
r
sr
==
=
+++
=
==
−+=
==
==
=− =
=
++
==
=
++
5
3
2

5
380 3 220
676 1039 5 39 7 0
0039 39%
60 1500 1 14415
0 242 24 2%
3
2
1
1
2
22
2
2
2
2
2
Nm = constant
Nm
Note that due to star connection V
Substituting all the known values in rated torque expression, one gets
rpm rpm
Further,
π
π
σσ
σσ


/.

()
()
()
σσ
σσ
π
sr
em n
rm
srm s r
X
T
P
f
V
Rs
RRs X X
+
=
=
+++
=
2
2
2
2
692
3
2

12 84
.
.
Nm
Nm
b) For operation at 1000 rpm with the same load torque the following resistance is needed:
()
s
nn
n
TT
P
f
V
RR s
R
RR
s
X
RR R
RR
R
R
s
s
een n
radd
s
radd
radd

add
1
1
1
2
1
1
2
2
2
1500 1000 1500 0 3333
5
3
2
0 016 0 9 12168 0
54 7
54 7 6 3 48 4
=
−
=− =
===
+
+
+
+
=+
−+ =
=
=−=
()/.

.
.
π
Let . Substituting all the known values in torque equation
Ω
Ω
c) If machine is to start with starting torque equal to maximum torque, then
TT s
RR
RX
RRXR
em est m
radd
s
add s r
= =
=
+
+
=+−=
1
1197
22
22
. Ω
All the three torque-slip curves are shown in Figure. Note that for operation with 1000 rpm (part b)
slip at which maximum torque occurs is 2.1. For 19.7 Ω machine would operate at 0.16 slip.
T
e

48.4 Ω 19.7 Ω R
add
=0
5Nm
2.1 1 0.33 0.16 0.039 s
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Example:
A three-phase 220 kW, 60 Hz slip ring induction machine has 16 poles and rated slip of
2.5%. It drives a pump and operates at rated speed with rated torque. Pump’s torque is
proportional to the speed squared. Rotor resistance is 0.0175 Ω. Determine the value
of the added resistance that needs to be inserted in rotor phases if the required speed of
rotation is 300 rpm.
Solution:
Only rotor resistance is given out of motor’s parameters. It is therefore not possible to use full
expression for torque in calculations. However, torque of the motor in the operating region (i.e., for
slips between rated and zero) can be approximated with a straight line. Thus T
e
=ksand calculations
in this example will be based on this expression.
()
()
nfPx
nsn
TP x
TK T K K
TK x
snnn
TksTks

s
nns
en n n
LLnn
L
ss
een
== =
=− =− =
== =
=
== =
== =
=− = − =
== ==
60 60 60 8 450
1 1 0 025 450 439
220000 30 439 4785
4785
2 26 300 30 22305
450 300 450 0 333
22305 4785
22
11
22
11
22
//
(.)
/( )

() /
.( /) .
/( )/ .
.
rpm
rpm
Nm
Nm = 2.26 Nms rad
In new steady - state at 300 rpm
Nm
At natural characteristic operation with torque of 2230.5 Nm would result in slip
2
ωπ
ωωωω
ωπ
nn
e
en
en
r
s
r
en
radd
s
radd
r r add
add r r
ss
T

T
x
n
T
P
f
V
Rs
R
R
s
X
T
P
f
V
RR s
R
RR
s
X
R
s
RR
s
R
s
s
RR
rpm However, required speed with this torque is 300 rpm. Hence

natural characteristic
at curve with added resistance
Thus
== =
=
==
+ +
==
+
+
+
+
=
+
=−=
2
2
2
2
2
2
2
2
1
2
1
1
2
2
21

1
2
0 025 22305 4785 0 01165
444 75
22305
3
2
22305
3
2
0 0175 0333
/.
.
.
.
()
.(./
π
π
0 01165 1 048.).−= Ω
Torque-slip curves are illustrated in Figure.
T
e
T
L
0.48 Ω
4785 Nm
2230.5 Nm
0 300 444.75 n(rpm)
439

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If slip ring induction motor speed is to be continuously varied, it is necessary to provide
a method of continuous additional resistance variation. Such a continuous resistance variation
is possible if power electronic converters are used.
Figure 6.7 illustrates electronic additional resistance variation. A three-phase bridge
diode rectifier is connected to the rotor winding, via slip rings. The output of the rectifier is
connected to a chopper, whose circuit contains a resistor R. When chopper switch S is closed,
resistor R is short circuited; when switch is open, resistor R is connected to the output of the
rectifier. Inductor between rectifier and chopper is large and its purpose is to provide almost
level DC current at the rectifier output, regardless of the state of the chopper switch S.
Three-phase diode
bridge rectifier L
DC
stator rotor
IM S R
Fig. 6.7: Speed control of a slip ring induction machine by addition of an electronically con-
trolled variable resistance in the rotor circuit.
What now has to be considered is the correlation between resistance R of Fig. 6.7 and
per phase added resistance R
add
, used previously in all the calculations. If switch S is open all
the time (duty cycle equal to zero), resistance seen by the rectifier will be R.Ifswitchisclosed
all the time (duty cycle equal to one), resistance seen by the rectifier is zero. Hence resistance
presented to the rectifier by the chopper equals
()
RR tT
eon
=− =1

δδ
(6.7)
The input rectifier current (i.e., phase rotor current) is, due to large inductance at the DC side,
of quasi-square waveform, discussed in Chapter 4, with 120 degrees of non-zero value in each
half-period. For level DC current, of value I
DC
, total rms of the rotor current (i.e., rectifier
input current) is
IIdI
rDCDC
==
1
2
2
2
3
2
0
23
π
θ
π
/
(6.8)
Required equivalent per-phase value of the added resistance, R
add
, is determined from the
power balance at the input and at the output of the rectifier. Assuming that there are not any
power losses in the rectifier, chopper and the DC side inductor, real power at the input of the
rectifier must equal real power delivered to the resistance of (6.7). Hence, using (6.7) and

(6.8),
()
()
() ()
()
PRI RI
P per phase RI
II
P per phase R I RI
P per phase R I
RR
DC e DC DC
rDC
DC r
rrr
raddr
add
==−
−=−
=
−=−
=−
−=
=−
22
2
2
2
2
1

1
3
1
3
2
1
3
1
3
2
051
051
δ
δ
δδ
δ
()
() .
()
.
(6.9)
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84
Note that the value of added resistance per rotor phase in (6.9) is actual value and it has to be
referred to the stator winding by means of transformation ratio in order to enable its use in
torque calculations.
Example:
Additional resistance is inserted in the rotor winding of the slip ring induction machine,
analysed in the previous example, using scheme of Fig. 6.7. Determine the chopper

duty cycle that will enable operation with the per phase added resistance calculated in
the previous example, if R =2Ω.
Solution:
Required value of per phase added resistance was calculated as 0.48 Ω. From (6.9)
()
()
RR
xx
add
=−
=−
−= =
051
048 05 2 1 1 048 052
.
. .
δ
δδ δ
6.4 Speed Control by Simultaneous Stator Voltage and Frequency Variation
Both of the speed control methods, elaborated so far, are characterised with poor
efficiency. Additionally, stator voltage variation method leads to a decrease in maximum
torque when stator voltage is reduced. This reduction in maximum torque is a consequence of
the fact that, when voltage is reduced while frequency is kept constant, flux in the machine
reduces as well. From equivalent circuit of Fig. 6.1 it follows that induced emf is
()
EjXI VRjXI
Ekf
mm s s s
==−+
=

σ
Φ
(6.10)
Induced emf is determined with the product of flux and frequency. In order to maintain flux
constant, it is necessary to simultaneously alter both frequency and induced emf. Wide speed
control range can be realised with an induction machine only if stator frequency is made
variable. Additionally, when frequency is varied, it follows from (6.10) that it is necessary to
simultaneously vary the rms value of the supply voltage. Ideally, the change of supply voltage
and frequency should be done in such a way that induced counter electromotive force in the
induction machine is kept constant. In this case flux in the machine will be kept constant,
enabling the maximum torque to be kept constant as well at all operating frequencies. If the
induced counter electromotive force is E then the control law that enables constant flux
operation at all frequencies is
E/f = E
n
/f
n
(6.11)
A set of torque-speed curves that result from application of this control law is shown in Fig.
6.8. Operation above rated speed is achieved by increasing frequency above rated. This region
of operation is again called field-weakening region and once more in this region supply voltage
will be kept at rated value.
In reality however it is not possible to control the machine by using the previously
stated control law as the internal emf cannot be measured. Instead, control is done in such a
way that voltage to frequency ratio is held constant, i.e.
V/f = V
n
/f
n
(6.12)

The consequence of this is that at low operating frequencies stator resistance voltage drop
becomes dominant and flux in the machine significantly reduces leading to the reduction of the
maximum torque at low frequencies. Torque speed characteristics obtainable by this control
law are given in Fig. 6.8 as well. The compromise is found usually by applying a modified
voltage control law that contains so-called voltage boost, i.e.
V=k(V
n
/f
n
)f+V
0
(6.13)
ENGNG3070 Power Electronics Devices, Circuits and Applications
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85
The resulting torque-speed characteristics in this case approach those obtainable with true
constant flux operation. The two possible control laws are illustrated in Fig. 6.9.
20 Hz 40 Hz 60 Hz 80 Hz 60 Hz, two-pole machine
T
e
speed
[rpm]
E/f = E
n
/f
n
1200 2400 3600 4800
Base speed region Field-weakening region
T
e

Rated voltage, rated frequency
speed
V/f = V
n
/f
n
Fig. 6.8: Torque-speed characteristics of an induction machine for the two control laws:
E/f = E
n
/f
n
and V/f = V
n
/f
n
.
VV
n
VV
n
V
0
0f
n
f0f
n
f
Fig. 6.9: Illustration of control laws V/f = V
n
/f

n
and V = k (V
n
/f
n
)f + V
0
.
ENGNG3070 Power Electronics Devices, Circuits and Applications
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86
Let the ratio of operating frequency to rated frequency be a=f/f
n
. The equivalent
circuit of Fig. 6.1 at operating frequency f becomes as shown in Fig. 6.10. All the reactances
are given for rated frequency; hence at any other frequency they become aX. Input stator
voltage is assumed to be determined with V/f = V
n
/f
n
law of Fig. 6.9.
I
s
R
s
jaX
σs
jaX
σr
I

r
V =aV
n
jaX
m
I
m
R
r
/s
Fig. 6.10: Steady-state per-phase equivalent circuit of an induction machine at a frequency
other than rated.
Note that slip s in Fig. 6.10 is a per-unit value. If n
s
= 60f
n
/P is synchronous speed at
rated frequency, then for operation at any other frequency
s=(an
s
-n)/an
s
(6.14)
where n denotes operating speed at this other frequency.
Calculations related to V/f method of speed control are illustrated by means of
examples.
Example:
A three-phase star connected 60 Hz four-pole induction motor has the following
parameters: stator and rotor resistance 0.024 Ω each, stator and rotor leakage
reactance 0.12 Ω each. The motor is controlled according to the law V/f=const. For an

operating frequency of 12 Hz calculate:
a) The maximum torque as a ratio of its value at the rated frequency for both motoring
and generating;
b) The starting torque and stator starting current in terms of their values at the rated
frequency.
Solution:
a) The purpose of this part of the example is to show that with V/f=constant law flux in the motor is
not held constant; hence maximum torque at 12 Hz will be reduced in motoring with respect to the
value at rated frequency.
aff
Tf f
P
f
V
RRX
Tfaf
P
af
aV
RRaX
P
f
aV
aR a R a X
Tfaf
P
f
V
Ra R a X
n

em n
n
n
ss
em n
n
n
ss
n
n
ss
em n
n
n
ss
== =
==
±+
==
±+
=
±+
==
±+
//.
()
()
[/ / ]
()
//

12 60 0 2
3
4
1
3
4
13
4
1
3
4
1
2
22
22
222
2
22 2
2
22 2
π
ππ
π
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87
()
Tfaf
Tff
RRX

Ra Ra X
Tf f
Tff
Tf f
Tff
Tf f
Tff
em n
em n
ss
ss
em n
em n
em n
em n
em n
em n
()
()
(.)
()

./. (./.) .
(.)
()
.
(.)
()
.
=

=
=
±+
±+
=
=
=
±+
±+
=
=
=
=
=
=
22
2
2
22
22
0 2 0 024 0024 0 24
0024 02 0024 02 024
02
068
02
146
Plus sign applies to motoring, minus to generation.
For motoring
For generation
b) This part of the example shows some additional benefits of the V/f speed control. As will be shown,

starting at low frequency gives increased starting torque compared to starting at rated frequency. Even
better, this increased starting torque is obtained with reduced starting current.
()
() ()
()
()
Tf f
P
f
V
R
RR X
Tfaf
P
af
aV
R
RR aX
P
f
V
aR
RR aX
Tfaf
Tf f
aR
RR aX
R
RR X
x

x
Iff
V
R
est n
n
n
r
sr
est n
n
n
r
sr
n
n
r
sr
est n
est n
r
sr
r
sr
sst n
n
()
()
()
()

.

.
()
()
==
++
==
++
=
++
=
=
=
++
++
=
+
+
=
==
3
2
3
2
3
2
02 0024

0048 02 024
0024
0 048 0 24
26
2
2
2
22
2
22
2
2
22
2
22
2
2
222
22
π
ππ
()
()
()
()
sr
sst n
n
sr
sst n

sst n
sr
sr
RX
Ifaf
aV
RR aX
Ifaf
Iff
a
RR X
RR aX
x
++
==
++
=
=
=
++
++
=
+
+
=
2
2
2
22
2

2
2
22
22
222
02
0 048 0 24
0048 02 024
072
()
()
()
()
()
()
.


.
Thus V/f control undoubtedly provides high starting torque with reduced starting current.
Example:
If the motor of the previous example drives a load whose torque is constant and equal
to rated, calculate the motor speed if frequency is 30 Hz. Rated slip (at rated
frequency) is 0.04.
Solution:
()
()
aff
Tf f
P

f
V
Rs
RRs X
Tf af
P
f
V
aR s
RRs aX
n
en n
n
n
rn
srn
en n
n
n
r
sr
== =
==
++
==
++
//.
()
()
30 60 05

3
2
3
2
2
2
2
2
1
1
2
22
π
π
Equating the right hand sides of these two equations gives
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88
()()
()
Rs
RRs X
aR s
RRs aX
a
ss
s
faf n fP
nsn
rn

srn
r
sr
ns
s
++
=
++
=
−+=
==
==
==
=− =
2
2
1
1
2
22
1
2
1
1
111
05
26 04 098 1 0
0 089 89%
30 60 900
1 820

.


/Hz rpm
rpm
Note that at rated frequency synchronous speed and operating speed are 1800 rpm and 1728 rpm; the
difference of the two is 72 rpm. At 30 Hz synchronous speed and operating speed are 900 rpm and 820
rpm; the difference of the two is 80 rpm. Indeed, for constant load torque, difference between
synchronous speed and operating speed in rpm is constant and independent of frequency if true E/f
control is implemented. In this example V/f control is used instead, so that the difference between
synchronous and operating speed slightly varies. Nevertheless, one notes that slip in per unit is
substantially different (4% and 8.9%); this is a consequence of the change in synchronous speed when
frequency varies.
Example:
A 400 V, 50 Hz, 6-pole, 960 rpm, star connected induction motor has the following
parameters: stator resistance and rotor resistance are 0.4 Ω and 0.2 Ω, respectively,
stator and rotor leakage reactance are 1.5 Ω each, and magnetising reactance is 30 Ω.
Motor is controlled by simultaneous stator voltage and frequency variation, using
E/f=constant law. Calculate stator current for rated torque, rated frequency operation,
and speed of the motor at one half of the rated torque and 25 Hz. Then repeat the same
speed calculation assuming that torque-speed curves are straight lines in the operating
region.
Solution:
The exact solution requires application of the motor equivalent circuit. For rated operating conditions
()
nfP s
ZRsjX
ZjXj
Zs s R jX
jX R s jX

R s jX jX
j
IVZ
I
I
Z
ZZ
Ix
EIZ x
sn n
rrn r
mm
ens s
mrn r
rn r m
sn n e
sn
r
m
mr
sn
nrr
== =− =
=+=∠
==
==+ +
+
++
=+ =∠
==

∠
=
=
+
==
== =
60 1000 1000 960 1000 0 04
522 167
30
482 363 6 37
400 3
637
385
30 385 3189 36 22
36 22 522 189
/()/.

() . .
/
.
./ . .

rpm
A
A
σ
σ
σ
σ
Ω

Ω
Ω
V
NmT
P
f
R
s
I
en
n
r
n
r
==
3
2
188
2
π
At 25 Hz motor operates with one half of the rated torque. Hence
()
()
()
aff
TT
P
af
aE R s
Rs aX

x
x
xxs
sx
n
een
n
nr
rr
== =
==
+
=
+
//.
/
/

(. ) . /

25 50 05
2
3
2
188 2
33
05 314159
05 189 02
02 05 15
2

1
2
1
2
2
2
1
1
2
2
22
π
σ
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89
Solution is found as slip equal to 0.0374. Hence operating speed is 481.3 rpm.
Next, it is necessary to examine how accurate linear approximation of the torque-speed curve is in the
operating region. Difference between synchronous speed at rated frequency and rated speed is for rated
torque operation 1000 - 960 = 40 rpm. If the torque-speed curve is straight line then for half of the
rated torque speed difference is 20 rpm. At 25 Hz synchronous speed is 500 rpm. As load torque
remains half of the rated, then difference between synchronous speed and speed of rotation remains
constant and equal to 20 rpm. The operating speed is then 500 - 20 =480 rpm. This compares well
with the exact value of 481.3 rpm.
6.5 Inverters and Inverter Control of Induction Machines
6.5.1 Six-step (Quasi Square-wave) Voltage Source Inverter
Variable voltage, variable frequency operation of induction machines is realised
utilising autonomous inverters, in conjunction with a rectifier and a DC link circuit. There are
two generic structures of autonomous inverters. The first one, voltage source inverter (VSI),
has already been introduced in Chapter 4 in its single-phase form and is the most frequently

applied power supply source for V/f control of induction motors. The second one, current
source inverter (CSI), is beyond the scope of interest here. Principle of operation of a three-
phase VSI is explained next.
Three-phase version of a VSI contains three inverter legs (rather than two, as the case
is in single-phase version). Input voltage for a three-phase VSI is provided by a three-phase (or
single-phase) bridge rectifier with capacitor placed at the output. As shown in Chapter 5,
capacitor provides smoothing of the DC voltage and, for sufficiently large capacitance, DC
voltage at the rectifier output approaches a constant value (for given firing angle). It will
therefore be assumed that inverter input voltage is constant in all the subsequent analysis.
Power circuit of a six-step voltage source inverter is shown in Fig. 6.11. As the inverter
itself controls only the frequency of the output voltage, a controllable rectifier must be used in
order to provide control of the output voltage magnitude (output voltage magnitude is
proportional to the input DC voltage). Each switch in the inverter circuit is again composed of
two back-to-back connected semiconductor devices. One of these two is a controllable switch
(say, BJT), while the other one is a diode. Diode is essential for correct operation of the VSI
as output voltage and current are out of phase due to inductive nature of the machine. It
enables current flow when one switch in given inverter leg is turned off and the other one
turned on, while the current still flows in the previous direction. The three inverter legs are
controlled in such a way that leg voltages constitute three-phase system of square-wave
voltages. This means that, assuming that upper transistor in leg A is fired at time instant zero,
firing of upper transistor in leg B will take place after 120 degrees, while firing of the upper
transistor in leg C will be delayed for another 120 degrees. The conduction of each of the six
semiconductor switches is again 180 degrees so that at any time three out of six switches are
on and the remaining three switches are off. The resulting output voltage waveforms for line-
to-line voltages are quasi-square waves, with two 60 degrees zero intervals and two 120
degrees intervals in which line-to-line voltage equals plus and minus DC voltage, respectively.
VSI operated in the 180 degrees conduction mode is therefore usually called six-step inverter.
Leg voltages of the inverter are given in Fig. 6.12 with respect to the negative pole of the DC
link. Line-to-line voltages applied to the induction machine are obtained directly from leg
voltages as

v
AB
=v
An
-v
Bn
v
BC
=v
Bn
-v
Cn
v
CA
=v
Cn
-v
An
. (6.15)
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90
Line-to-line voltages are shown in Fig. 6.12 as well. Finally, if the machine is star connected, it
can be shown that in the system of Fig. 6.11 phase to neutral voltages of the machine (included
in the Fig. 6.12) are determined with the following expressions:
p
C
V
DC
AB C

n
Rectifier and inverter control: IM
rectifier controls V
DC
while inverter controls
output frequency so that V/f = V
n
/f
n
or
V=k(V
n
/f
n
)f + V
0
.
Fig. 6.11: Three-phase voltage source inverter (VSI) fed induction motor drive.
()
()
()
vv vv
vv vv
vv vv
AAnBnCn
BBnAnCn
CCnBnAn
=−+
=−+
=−+

23 13
23 13
23 13
//
//
//
(6.16)
As is obvious from Fig. 6.12, waveforms of line to line and phase to neutral voltages
arenotpuresinewaves.Linetolinevoltagesareofquasi-squarewaveformofthetype
discussed in Chapter 4 (with 120 degrees non-zero intervals) and they can be represented with
appropriate Fourier series of the type developed in one of the examples of Chapter 4. For
example, line to line voltage AB can be represented with
vVt
tt t t
VV V
VVV
AB DC
DC DC
DC DC
= −+−+−+
==
==
23 5
5
7
7
11
11
13
13

6
078
2
3
0816
1
π
ω
ωω ω ω
π
cos
cos cos cos cos

.
() .
fundamental rms
total rms
(6.17)
Rms values of the fundamental and of the whole waveform are given in (6.17) as well.
Selection of appropriate DC voltage for V/f control is done on the basis of the rms of the
fundamental component, as illustrated later on. Note that line to line voltage does not contain
harmonics divisible by three, although these are present in leg voltages. Indeed, if leg voltages
are referred to the mid-point of the DC supply rather than to negative rail of the DC supply
(Fig. 6.13), they are square waves of amplitude V
DC
/2 and their Fourier series is given with
v
V
t
tt t

Ao
DC
=−+−+−+
4
2
3
3
5
5
77
7
9
9
π
ω
ωωω ω
cos
cos cos cos cos
(6.18)
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As line to line voltages are determined with difference of the two appropriate leg voltages, then
harmonics divisible by three mutually cancel because they have the same phase in all the three
leg voltages.
v
An
V
DC
v

AB
V
DC
v
Bn
v
BC
v
Cn
v
CA
0 60 120 180 240 300 360 ωt[°]
Leg voltages
Line-to-line voltages
2/3 V
DC
v
A
1/3 V
DC
v
B
Phase to neutral voltages
v
C
Fig. 6.12: Leg, line-to-line and phase to neutral voltages in VSI fed induction machine.
Phase voltages, given in Fig. 6.12, can be looked at as being composed of two
waveforms: the first one is a square-wave of amplitude V
DC
/3, while the second one is a quasi-

square wave of amplitude V
DC
/3 and 60 degrees duration of non-zero value. Hence harmonic
content of phase to neutral voltages can be determined as sum of Fourier series of a square-
wave and Fourier series of a quasi-square wave with β = 60 degrees (section 3.6). Phase to
neutral voltage is given with
ENGNG3070 Power Electronics Devices, Circuits and Applications
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92
vV t
tt t
VV
ADC
phase DC
=+++++
=
25
5
7
7
11
11
2
1
π
ω
ωω ω
π
sin
sin sin sin

()
rms of fundamental component
(6.19)
Phase to neutral voltage contains the same harmonic components as the line to line voltage,
(6.17). Note that rms value of the fundamental component of the phase to neutral voltage, as
well as rms values of all the higher harmonic components in the phase to neutral voltage, are
square root from three times smaller than appropriate rms values in the line to line voltage.
Difference in phase is consequence of different selection of zero time instant for Fourier
analysis (line to line voltage is taken as even function, phase to neutral voltage is taken as odd
function).
Harmonic components divisible by three and present in the leg voltages are suppressed
and they appear in voltage between motor’s isolated neutral point (N) and mid point of the DC
supply (o). This is illustrated in Fig. 6.13.
v
Ao
V
DC
/2
v
Bo
v
Co
v
No
V
DC
/6
Fig. 6.13: Leg voltages referred to the mid point of the DC supply and voltage between neutral
point of the motor and the mid point of the DC supply.

From Figs. 6.11 and 6.13 it follows that
()
vvv
vvv
vvv
vvv v vv v
vvvv
Ao AN No
Bo BN No
Co CN No
Ao Bo Co AN BN CN No
No Ao Bo Co
=+
=+
=+
++= + + +
=++
()3
3
(6.20)
The square-wave voltage between point ‘N’ and ‘o’ is of three times higher frequency than the
phase and line to line voltages. Hence its series is given with
()
()
() ()
v
V
t
ttt
No

DC
=+++++
4
6
3
33
3
53
5
73
7
π
ω
ωωω
sin
sin sin sin

(6.21)
Analysis of the motor behaviour under non-sinusoidal supply conditions can again be
done using equivalent circuit approach. However, equivalent circuit of Fig. 6.1 is valid only for
the fundamental component of the supply voltage. Thus, if speed of rotation is known (and all
the parameters are known as well), fundamental component of the motor current can be
calculated using Fig. 6.1, where the voltage at the input is fundamental component of the phase
to neutral inverter voltage given in (6.19).
In order to calculate higher harmonic components of the stator current, equivalent
circuit of Fig. 6.1 has to be modified. For harmonic component of the order k circuit takes the
ENGNG3070 Power Electronics Devices, Circuits and Applications
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93
form given in Fig. 6.14 (operation with fundamental frequency equal to rated is assumed). This

circuit can however be greatly simplified. First of all, stator current harmonic of the order k
creates a rotating field that rotates at speed k times greater than the speed of the fundamental
field. As rotor still rotates at speed n, then slip for the harmonic k can be found as
s
nn
n
kn n
kn
k
k
k
k
s
k
sk
sk
s
s
k
=
−
=
−
≈
=
=
=
=
≈
()

()
.
.
.
.
12 5
086 7
11 11
092 13
1
(6.22)
Approximate calculation in (6.22) assumes that actual speed of rotation n can be taken as equal
to n
s
, and accounts for the fact that the 5th, 11th, etc. harmonics create rotating field that
rotates in opposite direction of the rotor rotation, while the 1st, 7th, 13th, etc. harmonics
create rotating fields that rotate in the same direction as the rotor. Equivalent circuit for higher
harmonics can be further simplified by neglecting at first stator and rotor resistance, which are
much smaller then reactances for the harmonic of the order k. Further, magnetising reactance
can be taken as infinite, leaving the final form of the harmonic equivalent circuit in which only
two leakage reactances are present. Thus rms value of the stator current harmonic of the order
k can be found as
()
[]
IVkXX
sk k s r
=+
σσ
(6.23)
where rms voltage harmonic values are contained within (6.19).

I
sk
R
s
jkX
σs
jkX
σr
I
rk
V
k
jkX
m
I
mk
R
r
/s
k
I
sk
R
s
jkX
σs
jkX
σr
I
rk

V
k
jkX
m
I
mk
R
r
I
sk
jkX
σs
jkX
σr
I
rk
jkX
σs
jkX
σr
I
sk
V
k
jkX
m
I
mk
V
k

Fig. 6.14: Induction motor equivalent circuit for higher harmonics.
ENGNG3070 Power Electronics Devices, Circuits and Applications
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94
Example:
A 400 V, 50 Hz, 6-pole, 960 rpm, star connected induction motor has the following
parameters: stator resistance and rotor resistance are 0.4 Ω and 0.2 Ω, respectively,
stator and rotor leakage reactance are 1.5 Ω each, and magnetising reactance is 30 Ω.
Motor is controlled by simultaneous stator voltage and frequency variation, using
V/f=constant law and is supplied from a three-phase VSI.
a) Determine value of the DC voltage required for operation at rated frequency.
Calculate fundamental component of the stator current for rated torque, rated
frequency operation. Determine rms values of the fifth and the seventh harmonic of the
stator current when the motor operates at rated frequency.
b) Calculate DC voltage value required for operation at 25 Hz. Determine the fifth and
the seventh harmonic of the stator current at this frequency.
Solution:
a) As motor is star connected, rms of the rated stator phase voltage is 400/√ 3 = 231 V. When the
motor is supplied from a VSI, input DC voltage must be such that at rated frequency (50 Hz)
fundamental component of the phase to neutral inverter output voltage equals 231 V. Hence from
(6.19)
VVVV
phase DC DC phase11
2
22
231 513
() ()
= ===
π
ππ

V
In order to calculate fundamental component of the stator current, the motor equivalent circuit of Fig.
6.1 has to be used. For rated operating conditions
()
nfP s
ZRsjX
ZjXj
Zs s R jX
jX R s jX
R s jX jX
j
IV Z
I
sn n
rrn r
mm
ens s
mrn r
rn r m
s phase e
s
== =− =
=+=∠
==
==+ +
+
++
=+ =∠
==
∠

=
60 1000 1000 960 1000 0 04
522 167
30
482 363 6 37
231
637
385
11
1
/()/.

() . .
.
()
rpm
A
σ
σ
σ
σ
Ω
Ω
Ω
Higher harmonics of the stator current can be obtained using approximate equivalent circuit of Fig.
6.14. Rms values of the higher voltage harmonics are given in (6.19). Hence
()
[]
()
[]

()
[]
()
[]
()
[]
VV
k
IVZVkXX
VV VV
IV XX
IV XX
kDC
sk k k k s r
DC DC
ssr
ssr
=
== +
== ==
=+=+=
=+=+=
21
21
5
46
21
7
33
54651515307

7 33 7 15 15 157
57
55
77
π
ππ
σσ
σσ
σσ
VV
A
A
.
.
Note that the fifth voltage harmonic is 20% of the fundamental; however, the fifth harmonic of the
stator current is, due to filtering action of inductances, only 8% of the fundamental stator current
component. Note as well that these harmonic currents are independent of the machine loading;
however, fundamental component of the stator current is obviously load dependent.
b) For operation at 25 Hz, using V/f=constant law, fundamental component of the stator phase voltage
has to be one half of the value at 50 Hz, i.e., 231/2=115.5 V. As fundamental component of the stator
voltage is directly proportional to DC voltage, then required DC voltage value is 513/2=256.5 V. The
fifth and the seventh harmonic of the stator current are
ENGNG3070 Power Electronics Devices, Circuits and Applications
 E Levi, Liverpool John Moores University, 2002
95
()
[]
()
[]
()

[]
()
[]
()
[]
aff
VV
k
IVZVkXX
VV VV
IVaXX xx
IVaXX xx
n
kDC
sk k k k s r
DC DC
ssr
ssr
== =
=
== +
== ==
=+= +=
=+= +=
//.
.
.

25 50 05
21

21
5
23
21
7
165
5 23 5 05 15 15 307
71657051515157
57
55
77
π
ππ
σσ
σσ
σσ
VV
A
A
Stator current higher harmonics therefore have the same value regardless of the operating frequency.
This is so due to the fact that DC voltage and hence rms values of all the voltage harmonics change
proportionally to operating frequency. As leakage reactances are proportional to fundamental
frequency as well, ratio of phase voltage harmonic rms value to sum of leakage reactances for the
given harmonic remains constant regardless of the fundamental frequency.
6.5.2 PWM Operation of a Voltage Source Inverter
In all the analysis so far it was assumed that the inverter operates in 180 degrees
conduction mode and that variation of the inverter output voltage magnitude is achieved by
rectifier control. Such a solution is nowadays rarely applied. Instead, the inverter is operated in
pulse width modulated (PWM) mode and is supplied from a diode bridge rectifier (single-phase
or three-phase), so that input DC voltage of the inverter is constant. Variation of both inverter

output voltage and output frequency is now achieved by the inverter, due to operation in the
PWM mode. The idea of PWM is explained using single-phase VSI circuit of Fig. 3.9. The
most frequently utilised method of PWM is the so-called sinusoidal PWM technique, in which
a reference signal - a sine wave of desired amplitude and frequency - is compared with a
triangular carrier wave of constant amplitude and frequency. The instants for turn-on and turn-
off of semiconductors in Fig. 3.9 are then determined with intersections of the reference signal
and the carrier wave. Switches are turned on and off in pairs: S1 and S2 are always together
either on or off and similarly, S3 and S4 are always together either on or off. The advantage of
this approach is two-fold. First of all, inverter now becomes capable of controlling both the
frequency and the first harmonic magnitude, so that there is no need for application of a
controllable rectifier. Instead, a diode rectifier is used. Secondly, switching now occurs at high
frequency determined with the carrier wave frequency. This enables faster control and
simultaneously greatly improves harmonic spectrum of the load current. Inverter output
voltage now does not contain low order harmonics (the fifth, seventh, etc.). Instead, harmonics
are situated around multiples of the switching frequency (i.e., triangular carrier wave
frequency). Thus, if switching frequency is 5 kHz (typical value nowadays), then the inverter
output voltage will contain, apart from fundamental, higher harmonics whose frequencies are
around 5 kHz, 10 kHz, 15 kHz etc. If the motor operates with 50 Hz fundamental frequency,
then 5 kHz means that the order of the harmonic is around 100 (rather than 5, as it is in the
simple VSI discussed previously). As reactance of the motor (Fig. 6.14) is 100 times greater at
5 kHz than at 50 Hz (rather than only five times greater, as the case is for harmonic frequency
of 250 Hz), harmonic currents in the motor will be very small.
The principle of sinusoidal PWM, as applied in the single-phase bridge inverter of Fig.
3.9, is illustrated in Fig. 6.15. The impact of variation of reference signal amplitude on output
voltage waveform is evident from Fig. 6.15. Widths of the pulses in the two cases shown differ
although output frequency and triangular carrier wave frequency are the same. The variation in
pulse widths leads to subsequent difference in the values of the first harmonic of the output
ENGNG3070 Power Electronics Devices, Circuits and Applications
 E Levi, Liverpool John Moores University, 2002
96

voltage. It can be shown that the fundamental harmonic of the output voltage waveforms
shown in Fig. 6.15 equals in frequency and in amplitude the reference signal.
Reference and carrier signals
High amplitude of reference
v signal
V
Reference and carrier signals
Medium amplitude of reference
signal
v
V
Fig. 6.15: Sinusoidal PWM in single-phase bridge inverter for high and medium amplitudes of
the reference signal at same output frequencies.
Extension of the principle of PWM from single-phase to three-phase voltage source
inverter is rather straightforward. The reference sinusoidal voltage (which equals desired
fundamental voltage at the inverter output) is formed on the basis of the control law for given
operating point (say, V/f = V
n
/f
n
law). Thus the amplitude and the frequency of the reference
sinusoidal signal are those that want to be obtained at machine terminals. The reference
sinusoidal signal is further compared with the carrier signal (again, high frequency triangular
waveform). The instants of semiconductor switching in inverter legs are determined with
intersections of sinusoidal reference and triangular carrier wave. The three-phase inverter
requires three reference sinusoidal signals in order to achieve operation with three-phase
system of output voltages. The three sinusoidal reference signals have mutual displacement of
120 degrees. If triangular carrier frequency is sufficiently high, one carrier may be utilised for
all the three phases. It can be shown that although the output voltage is again composed of a
series of rectangular pulses, fundamental component of the output voltage is of the same

frequency and magnitude as the reference sinusoidal signal is.
Analysis of the PWM inverter output voltages is always conducted using the notion of
the ‘modulation index’. Modulation index m is defined as the ratio of the sinusoidal signal
ENGNG3070 Power Electronics Devices, Circuits and Applications
 E Levi, Liverpool John Moores University, 2002
97
reference amplitude to the amplitude of the triangular carrier wave. Another frequently utilised
term is the frequency ratio F, which is defined as the ratio of the triangular carrier frequency to
the frequency of the reference signal. Hence
mV V F f f
ref tri tri ref
==
−−sin sin
and (6.24)
If the drive is controlled using V/f = const. law, than the amplitude and the frequency of the
sinusoidal reference signal are varied to satisfy this law. Full PWM operation is possible only
for modulation index values between zero and one. If modulation index exceeds the value of
one, so called pulse-dropping region is entered. In this region reference signal is of higher
amplitude than the carrier signal and some of the intersections between the reference and the
carrier do not take place any more (i.e. some of the pulses are dropped - hence the name pulse
dropping). Eventually, if the modulation index is sufficiently high, the inverter essentially
reverts to six-step operation. In general, depending on the application, it is possible to realise
the PWM inverter fed drive in two ways: full PWM operation is maintained in the whole base
speed region (i.e. from zero to rated frequency) or full PWM operation takes place in part of
the base speed region only, while at rated frequency the inverter operates either with six-step
output voltage or with partial PWM (i.e. some pulses are dropped). This region of operation is
frequently called over-modulation. Note that the beneficial feature of PWM inverter, non-
existence of low order harmonics, holds true only so long as the full PWM operation is
preserved. The described two ways of operating the PWM inverter will require different input
DC voltage at rated output frequency. The amplitude of the fundamental output line-to-neutral

voltage for operation in the full PWM mode is given with
()
()
VmV
VmV
peak
DC
DC
1
1
2
22
=
=
(6.25)
while the corresponding output line-to-line fundamental rms voltage value for the six-step
operation is given in equation (6.19).
Example:
A three -phase 415 V, 50 Hz star-connected induction motor is to be supplied from a
three-phase voltage source inverter that is controlled using sinusoidal PWM. Calculate
the required inverter input DC voltage if: a) the inverter operates in the whole base
speed region with full PWM; b) inverter reverts to six-step operation at rated output
frequency.
Solution:
Fundamental component of the inverter line-to-line voltage has to be at 50 Hz equal to 415 V, rms,
regardless of the applied method of control, so that required phase to neutral fundamental component
has to be of 240 V rms. If full PWM operation takes place at 50 Hz, then from (6.25) one has
()
m
VmV V Vm

DC DC
=
=
===
1
2 2 2 2 2 2240 679
11
V
If the operation at 50 Hz is in six-step mode, then from (6.19)
VV VV
DC DC11
2
2 240 2 533=
== =
π
ππ
V/
Considering that the normal three-phase rectifier input voltage is 415 V line-to-line, then, assuming
almost infinite capacitance in the DC link, the DC voltage could at most be equal to the peak of the
input line-to-line voltage, 587 V. The standard practice is therefore to operate the inverter at rated
frequency in the pulse dropping mode, close to six-step waveform. If full PWM operation is needed at
rated frequency, then specially designed motors have to be used, whose rated voltage is below 415 V
but whose insulation is of higher rating.
ENGNG3070 Power Electronics Devices, Circuits and Applications
 E Levi, Liverpool John Moores University, 2002
98
It is usually said that DC voltage utilisation in the PWM mode is poorer than in the six-step mode.
Given the DC voltage, maximum fundamental line-to-line voltage rms values are 61% and 78% of the
applied DC voltage for sinusoidal PWM (with m = 1) and six-step voltage, respectively.
In the past, when available switching frequencies of semiconductors were rather low, it

was customary to keep the frequency ratio F constant, so that good feature of the sinusoidal
PWM with regard to harmonic content was preserved at all operating frequencies. Nowadays,
semiconductor switching frequencies are for major part of the power region in the kHz region.
Typically, for small to medium powers, inverter switching frequency is of the order of 2 to 20
kHz. Consequently, carrier frequency is nowadays kept normally constant and only one carrier
is used for all the three reference signals.
In summary, it can be stated that operation of a VSI in PWM mode yields two
substantial benefits, when compared to operation in 180 degrees conduction mode. A diode
rectifier can be used instead of a controllable rectifier, since the inverter is now capable of
controlling both the frequency and the rms value of the fundamental component of the output
voltage. Additionally, higher harmonics of the voltage are now of substantially higher
frequencies, meaning that current is much closer to a true sine waveform.
One special type of PWM, that is nowadays extremely frequently applied, is the so-
called ‘voltage space vector modulation’. For reasons that are beyond the scope here, this
PWM method is the prevailing one in closed-loop control of induction motors fed from PWM
inverters. Explanation of this method however requires at first the introduction of the notion of
the space vector.
Let us at first suppose that a three-phase supply is purely sinusoidal and balanced, so
that the system of phase voltages can be given with
vVt
vVt
vVt
a
b
c
=
=−
=−
2
223

243
cos
cos( )
cos( )
ω
ωπ
ωπ
(6.26)
Space vector of phase voltages is defined as
()
vvavav a j a j
s
s
Ab c
=++ = =
2
3
23 43
22
,exp( ) exp( )
ππ
(6.27)
and is obviously a complex quantity that simultaneously represents all the three voltages of the
three-phase supply applied to the machine.
Hence, for the case of the sinusoidal supply
()()
()
vVtat at
vVe
s

s
s
s
jt
=+−+−
=
2
3
22343
2
2
cos cos cos
ωωπ ωπ
ω
(6.28)
The result is obtained after relatively simple trigonometric manipulations. This is an equation of
a circle in the complex plane. It describes a complex number of constant amplitude (for given
V) whose phase continuously changes in time. Space vector is therefore a complex number that
is time-dependent. Space vector of stator voltages, for constant V value and constant
frequency, travels uniformly along a circle in the complex plane. One revolution of the space
vector corresponds to one period of the supply frequency. Complex plane and the space vector
of phase voltages are illustrated in Fig. 6.16.
In the previously discussed sinusoidal PWM, the three reference signals were three sine
waves of appropriate amplitude and frequency. Hence the calculated space vector corresponds
to what one wishes to impose to the machine terminals: three-phase system of sinusoidal