IM Starting&Control

8/3/2019 IM Starting&Control

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EN0567Power, machines and renewable energy

Lecture Notes - Dr M Jovanovic Page 1

1 Starting of Cage Induction Motors

There are a number of ways of starting squirrel cage induction motors. Each method has its

own characteristics and place of correct application. The most commonly used methods are

outlined in the following paragraphs.

1.1 Direct-On-Line (DOL) Starting

At standstill, the cage-type induction motor is similar to a transformer with a shorted

secondary winding. Hence, the application of rated voltage to the stator winding causes a

high stator current of around 6 times the rated full-load value, at low power-factor. This

current quickly decreases as the machine accelerates and slip decreases. The high starting

current is generally not a problem for the motor itself, since the machine thermal time

constant is usually much greater than the rotor run-up time. However, it has to be taken

into account when choosing protection devices for the machine, such as fuses andoverloads. The protection devices are normally designed to effectively ignore this starting

characteristic.

An induction motor with a Direct-On-Line (DOL) starter is shown in Fig. 1. Note the fuse

and overload sensor (OLS) used as protection devices. The drawing of large starting

currents, in all but small machines, causes the supply voltage in the vicinity of the machine

to fall, or dip slightly. This is due to the voltage drop across the source impedance and

may be explained with reference to Fig. 1 in which the voltage at the main distribution

board is given by: V = ES - Idol ZS

The system voltage dip may lead to objectionable lamp 'flicker' if excessive or frequent.

Incandescent lamps are particularly sensitive to fluctuations of voltage magnitude in that afluctuation of 0.5% at a frequency of 8-Hz will cause annoying lamp flicker. Thus, frequent

starting could be particularly problematic. In addition to incandescent lamp dimming,

voltage dips can affect other motors running on the same supply, e.g. increase in running

slip or possibly stalling. This simple method is therefore used where system capacity and

stiffness is sufficient to stand the high inrush currents without excessive voltage drop.

The following factors are to be considered while selecting the type of starting:

Size: e.g. DOL starting of pump motors up to 45-kW.

Inertia of Load: e.g. DOL starting of fan motors up to 18.6-kW.

Frequent Start: e.g. DOL starting of pump motors up to 15-kW.

It is good practice to avoid, where possible, simultaneous starting of motors, which may

lead to excessive voltage drops.

I

Z

ES V~

S

Other loads

M

oo

DOLcontactor

starter

OLSFuse

Fig.1: Connections and equivalent circuit for DOL starting


2/36



1.2 Reduced Voltage Starting

- Star-Delta starting

This starting method is employed when the motor windings are connected in delta () for

normal operation and in star (Y) during start-up. Fig.2 shows a / starter in which three

contactors (RLS) per phase are used to connect the stator windings in Y and/or . At

starting, the star (S) contactors are closed while the delta contactors R are open to allow the

stator windings to be Y connected. When the motor picks up speed and current falls below

a set value, the star contactors S are de-energised and after a very short delay (to reduce

electrical transients) the delta () contactors R are energised connecting the stator winding

in delta.

Fig.2: Winding connections for Y-start -run (Y/) method

When the Y-connected motor is started, the stator phase voltage is reduced to 1 3/ , or

58%, of its normal running value (line voltage). The star phase current is, therefore, also

reduced by 1 3/ of its normal running delta phase value.

To consider the effect of star-delta starting on the motor line current, assume an effective

motor impedance of Z /phase, as shown in the figure below.

I Line

VPhase

R

Y

B

VLine

R

Y

B

I PhaseVLine

Z

Z

Z

ZZ

Z

I Line

Y

If the motor is started in delta-connection (i.e. DOL) then:


3/36



IV

ZPhase

Line

i.e.I IV

Z Line Phase

Line

3 3

However, for the star-connected motor the starting line current is significantly reduced as

follows from:

I IV

Z

V

Z

V

Z Line Phase

Phase Line Line

/ 3 1

3I I Line Line

1

3.

The advantage of this scheme is that no equipment other than a / starter is needed.

However, the disadvantages are that all six ends of the stator windings must be accessible

and six 1-core cables or two 3-core cables are required between the starter and the motor.

In addition, both the starting current and torque are greatly reduced, since the

electromagnetic torque developed by the motor (at any slip) is directly proportional to the

square of the applied winding voltage (phase voltage) i.e. T k V 2 where kis a constant.

If the motor is started with delta-connection (i.e. DOL), VLine = VPhase, and therefore

T k V DOL Line2. For Y-connected windings during start-up, V VPhase Line / 3 , so that

T k VLine ( / )32

meaning that:

TT

k V

k VDOL

Line

Line

3 1

3

2

2.

The starting torque is 1/3 of that available with DOL starting (the low torque leads to an

increased run-up time) which makes this method suitable for applications where load

torque during acceleration is low. Note that since the line current for star connection is also

1/3 of the DOL value as shown above, the Torque/Line Ampere for the machine remainsconstant.

- Autotransfomer starting

In this method, an auto-transformer is placed in series with the motor during starting as

shown in Fig.3 and the corresponding single-line diagram in Fig. 4-b. The transformer

action reduces the voltage applied to the motor terminals (by transformer ratio) and the line

current is consequently less than the motor current. For this reason this method is usually

used in applications where complete motor acceleration at reduced amperes is needed.

At starting, the contactor R is open and S & Y are closed so that the stator windings are

supplied via the low-voltage taps of an autotransformer (Fig.4-b). After run-up, during

normal running, the switching states of the contactors are changed (R is closed and S&Y

are open) to allow the machine direct on-line connection.

Fig.3: Motor terminal connections for auto-transformer starting


4/36



MV

IDOL

MV

xV

x IDOL

I = x2 IDOL

(a) (b)

Fig. 4: Single-line diagram for DOL (a) and auto-transformer (b) starting

In Fig. 4-a, a voltage V is directly applied to the motor, which results in a starting current

of IDOL (DOL starting). When starting by an autotransformer, each phase of the motor

windings is supplied with xVvolts which results in a phase current of xIDOL . Consider

now the power-balance of the autotransformer:

V I xV x I DOL i.e. I x I DOL2

From the above expression it is obvious that when starting with an autotransformer atx-pu.

tapping (x is less than 1), the starting current is reduced tox2

of the DOL value.

As with the / starting, due to the lower applied voltage, the starting torque is also

reduced:

TT

xV

VxX

DOL

2

2

2

so that the Torque/Ampere is unchanged.

Three tappings are commonly provided with an autotransformer starter (e.g. 40, 60, 75 %

or 50, 65, 80 % of full voltage) but only one is used during starting. For example, if a 50%

tapping is used, the motor current is reduced by a factor of 2 and the line starting current is

reduced to of that with DOL starting. Note that the starting torque is also reduced by a

factor of 4.

The main advantages of this scheme are that only a 3-core cable is required between the

starter and the motor and that the value of tapping (x) can be chosen to suit the

circumstances. Note that the motor may be either star or delta connected.The main disadvantages are the loss in the starting torque and that the autotransformer

starter (its cost is comparable to or in excess of the / starter) is only utilised during

starting.

2 Effects of Source Impedance on Motor Starting

In the previous section, the effects of source impedance on starting performance of the

induction motor were ignored, i.e. the motor terminal voltage was assumed constant.However, the flow of starting current through the source impedance creates a voltage drop,

which reduces the voltage at the terminals of the motor.


5/36



For DOL starting the induction motor and supply system may be represented by the

equivalent circuit shown in Fig.5, where:

Zm is the motor standstill impedance per phase and

ZS is the source impedance.

Hence IE

Z ZDOL

S

S m

and V I ZZ

Z ZE DOL DOL m

m

S m

S

When starting with a / starter, the starting

line current is reduced to 1/3 of the DOL

value. For the same supply voltage ES, this is

equivalent to increasing the motor effective

impedance by a factor of 3 (i.e. 3Zm). Thusthe equivalent circuit becomes that in Fig.6,

which gives:

IE

Z Z

S

S m

3

and V I ZZ

Z ZEm

m

S m

S

33

3

With autotransformer starting, the effective

standstill impedance as seen at the terminals

of the autotransformer, shown in Fig.7, can beexpressed as:

ZV

I

V

x V

Z

Z

xin

LX

m

m '' ''

''2 2

Hence IE

ZZ

x

LXS

Sm

2

and V I ZE

ZZ

x

Z

x

Z

Z x Z E LX in

S

Sm

m m

m S

S' '

2

2 2

Several important observations can be made from the above expressions:

1/31

/1

3

sm

sm

ms

ms

DOL

Y

ZZ

ZZ

ZZ

ZZ

I

Ii.e. the line current is generally lower for Y/

starting as compared to DOL as expected. For instance, if sm ZZ then

DOLY

II 5.0 and forsm

ZZ 4 this is only 38% of DOL current.

~ E

Z

Zm

I =LXx V2 ''

Zm

V''

xV''

I =mxV ''

Zm

Z in

S

S

Fig.7: Autotransformer starting

IES V~

ZSoo

ZmDOL

DOL

Fig. 5: DOL starting

IES V~

ZSoo

3ZmY

Y

Fig. 6: Y/ starting


6/36



11

/313

22

s

m

sm

ms

ms

Y

LX

Zx

Z

ZZ

x

ZZ

ZZ

I

Iif 3/1

2x i.e. 58.03/1 x .

Therefore, if the autotransformer tap-ratio at start-up is less than 58% then the current

loading is lower than with Y/ starting.

13/1

/1

ms

ms

DOL

Y

ZZ

ZZ

V

Vi.e. the voltage drop on the source impedance is smaller for

star/delta starting method due to lower starting current as shown above.

1/1

3/1''

2

ms

ms

Y ZZx

ZZ

V

Vfor 58.0x . Under this tap-ratio condition, the

autotransformer starting is better than star-delta one as it gives a higher terminal

voltage i.e. lower voltage drop.

Example 1 : Auto-transformer starting

A 3-phase, 93.5-kW, 460-V, 1141-rpm induction motor has the locked-rotor torque at rated

voltage equal to 125% its full load value. If started DOL, the motor inrush current is 1-kA.

Determine starting torque and corresponding stator inrush current if the motor is started at

reduced voltage using an auto-transformer with a 65% tap. What is the inrush line current

under this condition?

Solution:

Given the rated parameters of the motor, its full load torque is :-

Nm-52.7821141

935003030

n

PT

The starting torque is therefore : Nm-15.97825.1 TTT DOLst .

Since the motor input impedance is constant at standstill (n = 0 i.e. s = 1), the stator inrush

current is proportional to the stator applied voltage. Thus, 650100065.0 mI -A.

With the 65% tap, the motor terminal voltage is 29946065.0 mV -V. The starting

(locked-rotor) torque is proportional to the voltage square which means that :-

15.97865.0)( 22 DOLrated

mat T

V

VT 413.27-Nm.

The current calculated above is the secondary current of the auto-transformer. The primary

(line) current can be easily determined from the power balance expression :-

65065.0mline

mlinemmlineline I

V

VIIVIV 422.5-A.

Note that the above current can be actually expressed as (see Fig.4b) : DOLline IxI 2

where x = 0.65 and DOLI =1000-A.


7/36



Example 2 : Star-delta (also called Wye-Delta) starting

A 3-phase, 45-kW, 460-V, 60-Hz, 1750-rpm induction motor has a locked-rotor impedance

of 0.547 /phase while the torque is 140% rated. Assuming the machine is to be started

Wye-Delta (a) sketch the starter-motor connections and explain briefly the starting

procedure; (b) determine the starting current and torque as well as the machine pole-number. What would be the motor terminal current if started delta connected?

Solution:

(a) The winding connections for this starting method can be found in Fig.2. On starting, the

supply is first applied to the motor with its stator windings star-connected (star contactors S

are closed and delta ones R are open). As the motor accelerates, its speed stabilises when

its developed torque become equal to its load torque. This usually happens at about 75% -

80% of nominal speed as shown in Fig 8. The star contactors (S) are then de-energised, and

the delta contactors (R) energised to delta-connect the stator windings. Each winding is

now fed with the full supply voltage, and the motor adopts its normal operating

characteristics. The run-up time with the windings star-connected is controlled by a timer

which, typically, can be adjusted from 0 to 30 seconds. This timer is adjusted during

commissioning to ensure that the star-delta changeover occurs, as closely as possible, at the

point of torque equilibrium (Fig 8).

Fig. 8: Typical starting characteristics of conventional cage induction motors: star-delta vs

DOL method

(b) The motor is star-connected at start-up and its starting current is therefore:-

52.4853547.0

460

3

Z

V

Z

VI LL

phasestart -A

The full load torque of the machine can be calculated using the rated parameters provided

as follows:-


8/36



55.2451750

4500030

60

2

fl

fl

fl

flfl n

PPT -Nm

The locked rotor torque for delta connected stator is given and it is 140% of the above

value (this is in fact the starting torque for delta connected machine i.e. the DOL torque)i.e. 77.3434.1 flDOL TT -Nm. Bearing in mind that the torque is proportional to phase

voltage square the following torque relationship applies:-

59.1143

77.343

3

1)

3/( 2 start

LL

LL

DOL

start TV

V

T

T-Nm.

If started -connected then LLphase VV and the DOL line current would be :-

57.1456547.0

34603

Z

VI LLline -A.

Finally, the machine pole-number can be determined using the fact that the rated speed isvery close to synchronous i.e. about 0.92-0.97 sn . The synchronous speed in rpm can be

expressed aspolespolespoles

fn

plys

720060120120 sup

and the number of poles is therefore 4

as 1800sn -rpm under this condition (this is the first synchronous speed value above the

rated speed of the machine = 1750-rpm).

Example 3 : Series-resistance starting

A 3-phase, Y-connected, 230-V, 78-A, induction motor has a locked-rotor impedance0.0978 + j0.2549 /phase. The motor is to be started using series resistors in each line. (a)

Draw starter-motor connections for this starting method and explain its mechanism; (b)

Determine the resistance of the resistors required to limit the starting current to 3 times

rated current; (c) Calculate the stator phase voltage at locked rotor; (d) If the locked-rotor

torque at rated voltage and frequency is 150% rated value, what is the starting torque (in pu

taking the rated torque as a base) at reduced voltage?

Solution:

(a)Series-impedance starters start the motor at reduced voltage by connecting a resistance(in this case) or reactance bank in series with the motor windings (see figures below) tolimit the inrush current during start-up. The impedance seen by the power system then

is that of the reactance plus that of the motor. The running contacts (R) are open when

starting (S contacts are closed) and once the motor has run up and its speed has

stabilised (Fig.11), the resistance (reactance) bank is shorted out by closing R contacts

(and opening S contacts), and the motor becomes DOL connected. This change-over is

normally controlled by an adjustable timer within the starter. The ohmic value of the

resistor or reactor is generally selected to provide approximately 70-80% rated voltage

at the motors terminals when starting. The series impedance starter is of simple design

and allows smooth acceleration.


9/36



Fig. 9: Series resistorstarting Fig. 10: Series reactorstarting

Fig.11: Typical characteristics for series impedance induction motor starting

(b)The total impedance as seen from the line terminals is obviously (0.0978 + extR ) +j0.2549 /phase. Due to the system Y-connection, the voltage from line to neutral is

79.1323/230 -V, and the locked-rotor current is to be limited to

2347833 ratedstart II -A. The 3-phase stator and series-connected resistors

present a balanced circuit and the current magnitudes in the respective branches are

consequently equal. Applying Ohms law to one phase:-

2342549.0)0978.0(

79.132

22

exttotal

LNstart

RZ

VI -A

Solving the above quadratic for extR one gets 4093.0extR -.

(c)The voltage applied to the machine (across terminals T1-T3 in the above figures) isactually the voltage drop on the motor impedance Z:-

88.63234273.0 startm IZV -V.


10/36



(d)The starting torque at reduced stator voltage of 63.88-V can be calculated from thefollowing torque ratio :-

flstart

start

fl

start

voltsfullTT

T

T

T

T

347.0

88.63

79.1325.1

2

2

The starting torque is therefore 34.7% rated value or 0.347-pu.

Example 4 : Series-reactance starting

A certain 208-V, 24-A, 60-Hz, Y-connected induction motor has a locked-rotor current of

139.5-A at full voltage. The power factor under this condition is 0.342 lagging. The motor

is to be started using the series-impedance method with pure inductors (of negligible

resistance) in each line. Calculate the inductance and voltage rating of each series-

connected inductor required so that the starting current is 200% rated.

Solution:

Following on from the previous example and applying Ohms law to one phase of Y-

connected stator windings one can determine the locked-rotor impedance of the machine :-

-861.035.139

208

lr

LNm

I

VZ

Since the power factor is 0.342 lagging, the impedance phase angle is therefore70342.0cos 1 .

The total impedance of the system is the external reactance plus the motor impedance i.e.

)809.0(294.0)70sin342.0(861.0 extext XjXjj . The starting current needs

to be limited not to exceed 482 ratedI -A. Therefore :-

A-48)809.0(294.0

3/208

22

exttotal

LNstart

XZ

VI

From the above equation, one can solve for -675.12 extext LfX . The added

inductance is thus 44.4extL -mH as f = 60-Hz.

The voltage drop across each of the inductors represents their voltage rating. This is simply

4.80675.148 extstart XI -V.

3 Starting of Wound-Rotor Induction Motors

The DOL starting current of a wound-rotor motor may be reduced by connecting an

external 3-phase variable resistance in series with the rotor windings, via the slip rings. The

value of external resistances is gradually reduced as the rotor speed increases. An interlock

is usually employed to prevent main switch closure unless the starting resistances are attheir maximum values. The circuit arrangement and torque-slip characteristics of a wound-

rotor type induction motor with a four-point starter are shown in Fig.12.


11/36



When the required number of resistance steps is small, the starting resistors may be

mounted on the rotor shaft and cut out using a centrifugal switch. This avoids the

requirement of slip rings and external switchgear. However, this device is complicated if

frequent starting is required, due to heat dissipation requirements, or if more than two

resistance steps are required. A motor fitted with a centrifugal switch will automatically

reset the starting resistors if stalled.

Fig.12: Starting of a wound-rotor induction motor with a four-point starter

As a remainder and for convenience of the following analysis, it would be useful to

reproduce here the steady-state equivalent circuit of an induction machine (Fig.13), the

corresponding equations as well as a typical torque-slip characteristic similar to those in

Fig.12.

Is Rs jXs jXr Ir

Vn Im jXm Rr /s

Fig.13: Steady-state per-phase equivalent circuit of an induction machine for purely

sinusoidal supply voltage (iron loss resistance not shown)

All the rotor parameters in Fig.13 are referred to the stator winding, by means of thetransformation ratio. Symbols in bold denote phasors. The variable resistor in the rotor

circuit represents both rotor copper loss and power converted into mechanical. Recall that

the 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

Vn stands for the rated phase to neutral voltage of the stator. All the reactances are given at

fixed, rated supply frequency.

The general torque-slip characteristic for a 2P-pole machine shown in Fig.14 follows from

power flow considerations in Fig.13, in the form:-

222

22

2

2

3

2

3)(

XsRR

sRV

f

P

XXsRR

sRV

f

PsT

rs

rn

rsrs

rne

(1)


12/36



Temaximum (pull-out) torque

Ten

Test

operating region

slip

1 sm sn 0 s

0 ns speed

Fig.14: Torque-speed curve of an induction machine for rated supply conditions

The operating region is restricted to slips up to typically 10%, indicating that the speed of

rotation changes with load but remains within rather narrow boundaries from no load up to

rated load. The maximum (pull-out) torque, rated torque and starting torque, as well as the

corresponding slips (indicated in Fig.14) can be calculated from the following expressions:

222

22

2

22

2

22

2

22

2

3)(

2

3)1(

1

4

3

2

3

)(

XsRR

sRV

f

PssTT

XRR

RV

f

PsTT

XRRVf

P

XsRR

sR

Vf

P

ssTT

XR

Rs

nrs

nrnneen

rs

rneest

ssn

mrs

mr

nmeem

s

rm

(2)

Referring to the first two expressions for maximum electromagnetic torque )( emT and the

corresponding slip (m

s ) it is apparent that the rotor resistance (r

R ) influencesm

s , but not

the magnitude of the maximum torque. It is obviously possible to obtain maximum torque

at starting by adding the appropriate resistance in the rotor circuit (in series with the rotor

windings) so that 1ms (see Fig. 15). Therefore the value of external resistance per phase

required to achieve maximum torque at start-up can be determined using the first

expression above as follows:-

rsext

s

extrm RXRR

XR

RRs

22

221 (3)

This unique property of a wound rotor (slip ring) induction machine to achieve maximum

starting torque with relatively low supply current is its principal advantage over a cagecounterpart. It also serves as a basis for speed control of this machine (Fig.15) as we shall

see in the following sections.


13/36



Stator Te

Radd = 0

IM

Load torque

Rotor

Increasing Radd

Radd

1 s1 s2 sn 0 slip

Fig.15: Speed control of a WRIM by addition of resistance in the rotor windingbasic

principle and torque-slip curves.

Example 5 : External resistance control for maximum torque at start-up of a WRIM

A 380-V, 50-Hz, 4-pole, Y-connected slip ring induction machine is fully loaded at 5-Nm.

The stator and rotor resistance are 10- and 6.3-, respectively, while the leakage

reactances are 12- each. A magnetising reactance can be neglected.

(a) Determine the slip for rated operating conditions;

(b) Calculate added resistance that will enable starting of the motor with the maximum

starting torque.

Solution:

(a) Calculation of rated slip:

Nm84.12

2

3

Nm92.61

1

2

3

%2.24242.0

Further,

rpm5.14411rpm1500/60

%9.3039.0

07.395.1039676

-:getsone,expressiontorqueratedinesknown valutheallngSubstituti

V2203380connectionstartoduethatNote

Nm52

3

constant=Nm5

22

2

22

2

22

2

22

2

rsmrs

mrnem

rsrs

rnest

rss

rm

nnns

n

nn

n

rsnrs

nrnen

enL

XXsRR

sRV

f

PT

XXRR

RV

f

PT

XXR

Rs

nsnPfn

s

ss

V

XXsRR

sRV

f

PT

TT


14/36



(b) If the machine is to start with starting torque equal to maximum torque, then an

additional rotor resistance per phase required can be calculated as follows:-

7.191

1

22

22 rsadd

s

addr

mestem

RXRRXR

RR

sTT

4 Power Electronic Control of Induction Machines

4.1 Introduction

Three-phase induction motors are the most frequently utilised electric machines in industry

(more than 70%). They are characterised with low cost, high reliability, high efficiency,

simple construction and, in the case of squirrel-cage induction motors, with virtuallymaintenance-free operation. If operated with the stator three-phase voltage supply of fixed

frequency and magnitude, induction motors will run at a speed that very slightly depends

on loading.

In contrast to DC machines, where the choice of speed control methods 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 3-phase induction motor drives. A 3-phase induction machine requires

3-phase AC supply at the stator side. In a squirrel-cage type of induction machines this is

simultaneously the only accessible winding. However, in slip-ring induction machines

there is more degrees of control freedom as the three-phase rotor winding may be

approached as well. Thus the speed of an induction machine may be controlled via the

stator voltage variations for both types of induction machines; additionally, the speed may

be controlled in slip-ring machines from the rotor side as well. Squirrel-cage induction

machines are by far the most frequently used. It is for this reason that the following

discussion will be predominantly devoted to speed control methods associated with the

alteration of the stator supply voltage. Only one method, specifically aimed at slip ring

machines, will be looked at.

If a 2P-pole induction machine is supplied with a voltage of frequencyfthen the so-called

synchronous speed of the stator revolving field can be expressed in rpm as:ns = 60f / P (4)

From this expression it follows that the synchronous speed can be altered by simply

changing the number of pole pairs - P. Assuming that the load torque is constant, ifP is

doubled during operation of the machine, the synchronous speed will be halved, leading to

operation at approximately one half of the rated speed (as slip is small under normal

operating conditions). This method of speed control is used in low performance 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) but where control accuracy is not an imperative.

Speed control by pole pair changing requires a specially wounded stator and is usually

realised by mechanically reconnecting the stator winding from one pole pair number to

another. The linear portions of typical torque-speed characteristics for a change-over from


15/36



2-pole to 4-pole stator are shown in Fig.16. Power electronics converters are obviously not

involved in this speed control method, and its applicability is restricted to the cases where

two discrete speed values, rather than continuous speed variation, are needed. Therefore

speed control by pole pair changing will not be considered further on.

The general torque expression for an induction machine (1), suggests that, for a giventorque, the speed (slip) of an induction machine can be adjusted by changing the stator

applied voltage (V) and/or frequency (f). The two consequent methods of speed control,

universally applicable to all the 3-phase induction machines, that will be elaborated in

more detail are:

(1)the speed control by stator voltage variation and(2)speed control by simultaneous stator voltage and frequency variation.The former, although relatively simple, has restricted applicability for the reasons that will

be explained in the following; the latter is the most widely used method of speed control of

induction machines. Finally, a method valid for slip-ring machines only based on theinsertion of a series resistance in the rotor circuit, will be discussed as well.

Te

load torque

A B

1500 3000 speed (rpm)

Fig. 16: Speed control by changing pole pair number: drive operates either at point A or B.

4.2 Speed Control by Stator Voltage Variation

The equation (1) shows that electromagnetic torque developed by an induction machine is

proportional to the square of the applied rms stator phase voltage (V). Thus, for a given

load torque, a reduction of voltage will lead to the operation with an increased slip, i.e.,

with a decreased speed as shown in Fig. 17. As the supply voltage is not allowed to exceed

its rated value, this method of speed control can be utilised only for reducing the speed

below rated. Note that, according to (3), the pull-out slip is not a function of the applied

voltage. Hence the motor develops maximum torque at constant slip (speed), determined

with (5), regardless of the applied voltage. However, as both maximum (pull-out) and

starting torques are functions of the voltage squared, the reduction in voltage results in the

reduction in starting torque which means that the motor will be able to start only light loads

at low speeds; also a decrease of maximum torque implies that overloading capability of

the motor is compromised. This is one of the major drawbacks of this speed control

method.

Another limitation of the method is that the speed drops with reducing the voltage

incurring additional copper losses in the rotor winding due to increased slip as previously

mentioned. Regardless of these two serious shortcomings, this speed control technique is

widely used in two distinct cases. When the load torque is proportional to the speedsquared (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.


16/36



For a number of applications having 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.

Te

rated voltage

reducing voltage

operating region

slip

1 sm 0 s

Fig.17: Torque-slip characteristics of an induction machine with speed control by stator

voltage variation.

Example 6: Speed control by stator voltage variation

A 3-phase squirrel-cage induction motor drives a full load at rated slip of 3%. A stator and

rotor resistance (referred to stator) are both equal to 0.015- and a total leakage reactance

is X = 0.09-. (a) Calculate the necessary reduction in the stator supply voltage if the

induction motor is to drive the same load but with a 15% slip; (b) Compare the rotor losses

for the two operating points and comment on the results obtained.

Solution:

(a)Pull-out slip of the motor is:%4.16164.009.0015.0015.0 22

22

XR

Rs

s

rm

The reduction of the voltage to allow the operation with 15% slip at unchanged load torquecan be calculated as follows.

37.6%.isreductionvoltageNecessary624.0

39.009.003.0/015.0015.0

09.015.0/015.0015.0

03.0

15.0

2

3

2

3

1

15.0and03.0:pointsoperating

1

22

22

22

2211

21

22

2

221

121

22

2

221

121

1

1

n

nrs

rs

nn

nrs

nrn

rs

r

nrs

nrn

rs

r

en

e

n

VV

XsRR

XsRR

s

s

V

V

XsRR

sRV

XsRR

sRV

XsRR

sRV

f

P

XsRR

sRV

f

P

T

T

ss


17/36



This situation is illustrated in the accompanying figure.

Te

rated voltage

TL = Ten

62.4% of

rated voltage

0.16 0.03 slip

0.15

(b)Let us now examine an increase in rotor losses that takes place with this speed controlmethod. Taking the power transferred from the stator to the rotor (electromagnetic

power) to be Psr, for these two operating conditions one has

nnsrcur

srnnnsr

nn

nensenen

nsrnncurnnn

nsrnsrnnn

n

PPsPP

PPPs

PP

Ps

sTsTTP

s

PPsPPs

PPPsP

s

1545.003.115.0:losscopper

03.185.0/876.01

:powerneticelectromag

876.01

11:powermechanical

:torque(rated)sameat the15.0slipnewfor

031.0:lossescopper;03.11

1

:03.0slipratedfor

This consideration shows that power transferred from stator to rotor is the same for the two

cases. Hence reduction in output power of 12.4% reflects itself directly on an increase in

rotor copper loss from 3.1% 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 problems with reduced voltage and this increase in losses are the two

main reasons why this speed control method is not used with constant load torques. The

situation is much improved in both respects when the load torque is proportional to the

speed square as is the case in pump-like applications. 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 an AC-AC voltage controller

in each stator phase of the machine as shown in Fig.18. The firing delay angle of the

thyristors may be controlled to provide soft-starting as well in order to alleviate the

starting problems of cage-type motors as will be discussed earlier. The same voltage

controller structure is also a basic configuration used in power systems for reactive power

compensation.

It should be noted that although the principle of operation of an AC-AC controller is very

simple (to be reviewed in the following section), an analysis of the system in Fig.18 is

extremely tedious even for steady-state operation. This is due to the inductive nature of themachine, 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


18/36



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.

IM

R

Y

B

T1

T2

T3

Fig.18: Stator voltage variation of an IM using an AC-AC voltage controller

4.3 Speed Control of Slip Ring Induction Machines

From a set of equations (2) it follows that the pull-out slip is proportional to the added

resistance. On the other hand, the corresponding maximum torque is not influenced at all

by addition of resistance, indicating that overloading capability of the machine is

unaffected. Compared with the stator voltage variation method, an addition of rotor

resistance is to be preferred, for the following reasons: (a) any operating speed between

zero and rated can be obtained (with the stator voltage variation speed control region is

confined to speeds higher than pull-out speed); (b) maximum torque is not affected (with

stator voltage variation it reduces proportionally to the stator voltage reduction squared);

(c) additional copper loss is now developed in the added resistance, which is external to the

machine and therefore the problem of overheating does not take place. Note however, that

the problem of low efficiency still remains to be present: as speed is reduced, larger and

larger portions of the total input power are dissipated in the additional resistors. A principle

of this speed control method and resulting torque speed curves are shown in Fig.15.

Example 7. For a wound rotor induction motor from Example 5, calculate an added

resistance that is needed to reduce its speed from the rated value to 1000 rpm.

Solution:

For operation at 1000-rpm with the same load torque the following resistance is needed:

4.483.67.547.5402168.19.0016.0equationin torqueesknown valutheallngSubstituti.Let

2

35

3333.01500/)10001500(

2

2

2

1

121

11

add

addr

addrs

addrnene

s

s

RRRRRRR

Xs

RRR

sRRV

f

PTT

n

nns


19/36



Example 8. A 3-phase 220-kW, 16-pole, 60-Hz slip ring induction machine drives a pump

at rated slip of 2.5% and full load torque. The pumps torque is proportional to the speed

squared. A rotor phase resistance is 0.0175-. Determine the value of an added resistance

that needs to be inserted in series with the rotor phases to achieve the speed of rotation of

300-rpm. Assume a linear torque-speed characteristic around the rated operating point.

Solution:

Motor torque in the operating region (i.e., for slips between rated and zero) can be

approximated with a straight line i.e. Te = ks. Calculations in this example will be therefore

based on this expression.

48.0)101165.0/333.0(0175.0Thus

resistanceaddedwithcurveat)(

2

35.2230

sticcharacterinatural2

35.2230

Hencerpm.300istorquewith thisspeedrequiredHowever,rpm75.444

01165.04785/5.2230025.047855.2230

slipinresultwouldNm2230.5ofewith torquoperationsticcharacterinaturalAt

333.0450/)300450(/

Nm5.2230)30/300(26.2

rpm300atstate-steadynewIn

rad/Nms2.26=Nm4785)(

Nm4785)439/(30220000

rpm439450)025.01(1rpm4508/6060/60

2

1

12

22

1

121

22

2

222

2222

11

2211

222

rraddaddrr

addrs

addrne

rs

rne

en

ennene

ss

L

nnLL

nnen

snns

RRs

sR

s

RR

s

R

Xs

RRR

sRRV

f

PT

Xs

RR

sRVf

PT

n

xT

TssksTksT

nnns

xKT

KKTKT

xPT

nsnxPfn

Torque-slip curves for this example are illustrated below.

Te TL

0.48

4785 Nm

2230.5 Nm

0 300 444.75 n(rpm)

439


20/36



If the speed of a slip ring induction motor is to be continuously varied, it is necessary to

provide a method of continuous additional resistance variation. This is possible if power

electronic converters are used as shown in Fig. 19. A 3-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 the switch S (normally a transistor) is

closed, the resistorR is short circuited; else i.e. with the switch S open, R is connected tothe rectifier output. The purpose of a large inductor DCL in the circuit is to provide almost

level DC load current, regardless of the state of the switch S.

Three-phase diode

bridge rectifier LDC

stator rotor

IM S R

Fig. 19: Continuous speed control of a slip ring induction machine by addition of an

electronically controlled variable resistance in the rotor circuit.

What now has to be considered is the correlation between the resistance R of Fig. 19 and

per phase added resistance Radd in the rotor circuit (Fig.15) used previously in all the

calculations. If the switch S is open all the time (duty cycle equal to zero), the resistance

seen by the rectifier will be R. If the switch is permanently closed (unity duty cycle) then

the equivalent resistance as seen from the rectifier terminals is zero. Hence the resistance

presented to the rectifier by the chopper is:

TtRR one where1 (5)

The input rectifier current (i.e., phase rotor current) is, due to large inductance at the DC

side, of quasi-square waveform with 120 degrees of non-zero value in each half-period. For

a constant DC current, of value IDC, the total rms of the rotor current (i.e., rectifier input

current) is then:

I I d I r DC DC 1

22

2

3

2

0

2 3

/

(6)

The required equivalent value of the added resistance per phase, Radd, is determined from

the power balance at the input/output of the rectifier. Assuming that the rectifier, chopper

and the DC side inductor are ideal (i.e. lossless), real power at the input of the rectifier

must equal that delivered to the resistance of (5). Hence, using (5) and (6), one gets the

following relationship:

RRIRRIRIIRP addraddrDCDCeDC 15.03)1(2

31 2

222(7)

Note that the value of added resistance per rotor phase in (7) is the actual value and it has

to be referred to the stator side by means of transformation ratio in order to enable its use in

torque calculations.


21/36



Example 9: Chopper speed control of a WRIM

For the slip ring induction machine analysed in Example 8, determine the chopper duty

cycle that will enable operation with the calculated additional resistance per phase if the

chopper output resistance isR = 2-.

Solution :

Required value of per phase added resistance was calculated as 0.48 . From (7) it follows:

52.048.01125.048.015.0 xxRRadd

Example 10: A 30-kW, 1170-rpm, 60-Hz slip ring induction motor has the open-circuit

rotor voltage of 400-V. A chopper with a resistive load of 5- and switching frequency of

200-Hz is used for continuous variations of the rotor external resistance. Calculate the

chopper on-time required for the motor to develop a 200-Nm torque at 900-rpm.

Solution:

For a 60-Hz supply frequency and 1170-rpm rated speed the synchronous speed must be

the first possible value above 1170-rpm i.e. sn 1200-rpm this corresponding to a 6-pole

machine. The slip at 900-rpm is therefore: 25.01200/9001 s .

The rotor voltage between the any two slip-rings for the above slip value and the locked-

rotor (open-circuit) voltage of 400-V is: V-10040025.0 ocr EsE .

The DC voltage of the diode bridge rectifier: V-13510035.123

rdc EV .

The power delivered to the rotor from the stator at 200-Nm: kW-13.2560

2 sem

nTP

.

The power dissipated as heat in the rotor circuit: W-62822513025.0 emloss PsP .

The chopper input resistance determined from power balance:

.-9.26282

13522

dc

dce

P

VR

The chopper duty-cycle calculated using (5): 42.05/9.21/1 RRfT eswon .The chopper on-time is therefore: ms-1.2200/42.0 onT .

4.4 Soft-starting of induction motors

Soft starters (also called solid state starters) are used for gentle starting and stopping of 3-

phase induction motors. Pumps, centrifuges, compressors, escalators, belt conveyors, mills,

fans, stone crushers and saws are typical applications. If a motor is not adequately protected

the sudden change in rotation torque and speed, which occurs on starting and stopping will

jolt the equipment linked to it. Over the long-term this can lead to increased mechanical

wear of gearboxes, clutches, transmission and conveyor systems. Abrupt starting and


22/36



stopping can also damage goods being handled by mechanical equipment. For instance, the

filling and distribution of glass bottles and containers loaded onto a conveyor holds the

potential for a minor disaster. One sudden jerk during starting and stopping and the whole

process line could be turned into a mass of broken glass and dripping liquid or sticky

product. With pumps soft starters eliminate pressure shocks in pipes and valves when the

pump starts or stops. This poses a particular safety hazard when the transfer of volatileand/or inflammable liquids is involved.

In all the applications mentioned previously, the back-to-back converter configuration of

Fig.18 with associated electronics (Fig. 20) can be used to chop the sine-wave system

power so that only a portion of the supply voltage is applied to the motor during starting

and/or stopping. This kind of voltage control ensures smooth acceleration and deceleration

thereby eliminating the disruptive effects of sudden starts/stops. The gradual supply of

current to a motor also eliminates unwanted tripping, erratic current supply and motor

overheating. The logic circuits (Fig.20) can be programmed to respond to any of several

sensors to control the voltage: internal time ramp; current sensor feedback, or tachometerfeedback. In addition, most such starters also have provisions for reducing the voltage

when the loading on the motor is low, thus minimizing the no load losses of the motor and

improving its efficiency (issues related to this will be addressed in the following).

Fig.20: Design of an electronically controlled soft starter for induction motors

The starting controls are set so that, initially, the voltage increases rapidly until the motor

just begins to turn, after which the voltage rises linearly with time until full voltage is

reached (Fig.21a). Some of these ramp-up schedules incorporate a short pulse of full (or

nearly full) voltage to overcome the static friction of machinery that has not operated for

some time or that is covered with frost (Fig.21b). In other schedules (Fig.21e) the starting

current can be limited automatically to, say, 4 times the rated current.

During stalling, when power is shut off, the motor may coast to a stop too quickly. In such

cases the slow ramp-down feature of the electronic starter is a significant advantage

(Fig.21d). In this phase the motor terminal voltage is reduced gradually until the motor

comes to rest. The starter can be appropriately programmed to generate the desired ramp

characteristics (Fig.21d) that are best suited to a particular load.


23/36



Fig.21: Typical soft starting/stopping characteristics of cage induction motors

Usually soft starters are only active during the starting and stopping procedure; the rest of

the time the motor is supplied directly from the mains. For improved steady-state

performance it is useful to be able to keep control of the motor and the load at all times so

most soft starters are designed to continuously control or monitor the running of the motor

instead of controlling it only during starting and stopping. The explanation for this can befound in the fact that once a start has been completed the motor operating efficiency

becomes of interest. When working at or near full load, a typical 3-phase induction motor


24/36



is relatively efficient, readily achieving efficiencies of 85% to 95%. However, motor

efficiency falls dramatically when the load falls to less than 50% of rated output. In fact, a

very few motors actually experience consistent fully rated operation, the vast majority

being operated at much lower loads due to either over-sizing (a very frequent situation), or

natural load variations. At light loads and mains voltages, induction motors always have

excess magnetic flux, and suffer efficiency loss and power factor degradation as a result.By detecting the load at any instant, and adjusting the motor terminal voltage accordingly,

it is possible to save some of the excitation energy and reduce core losses, thus improving

the motor efficiency at light loads (as shown earlier in Section 4.2). For this reason, many

soft-starters are often microprocessor controlled which brings them a number of important

advantages. Firstly, all the calculations necessary to find the most optimal firing angle of

the thyristors for any load condition is made by the micro-computer. Secondly, the start

always synchronises with the supply voltage and a special structure of turn-on pulses

virtually eliminates the inrush currents normally associated with motor start-up. Lastly,

there is the absolutely step-less starting process, found only with series impedance electro-

mechanical starters - but without the wasted energy, with improved reliability, salientcontact-less operation and with the opportunity to control the maximum current allowed to

flow during the starting process. Other features such as soft stopping can be also included

to give considerable control over all modes of induction motor operation.

4.5 Speed Control by Simultaneous Stator Voltage and Frequency Variation

All the speed control methods elaborated so far are characterised by relatively 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 field weakening in the machine i.e. the fact that, when the supply voltage is reducedwhile the applied frequency is kept constant, the machine flux reduces as well. From the

equivalent circuit of Fig.13 it follows that the induced emf is:

fkIjXRVIjXE sssmm (8)

i.e. it is determined as the product of the flux and frequency. Clearly, in order to maintain

flux constant, it is necessary to simultaneously alter both f and E. A wide speed control

range with an induction machine can be realised only if the stator frequency f is made

variable and reasonably high. Additionally, when f is varied, it follows from (8) that it is

necessary to simultaneously vary the rms value of the supply voltage V. Ideally, the changeofVandfshould be done in such a way that Eis held constant. In this case both the flux

and maximum torque of the machine will be kept constant at all operating frequencies. The

control law that allows constant flux operation at all frequencies can be therefore expressed

as follows:

E/f = En /fn (9)

A set of resulting torque-speed curves is shown in Fig. 22. The machine operating region

above the rated speed, which is achieved by increasing frequency above rated at full

voltage, is called the field-weakening region as the air-gap flux is then reduced.


25/36



In reality however it is not possible to control the machine by using the previously stated

law as the internal emf cannot be measured. Instead, control is done in such a way that the

supply voltage to frequency ratio is held constant, i.e.

V/f = Vn /fn (10)

The consequence of this is that at low operating frequencies the stator resistance voltage

drop (RI) becomes dominant, this significantly reducing the flux and the maximum torque

of the machine (Fig. 22). A compromise is found usually by applying a modified voltage

control law that contains the so-called voltage boost at low frequencies, i.e.

V = (Vn /fn ) f + V0 (11)

The two possible control laws are illustrated in Fig. 23.

20 Hz 40 Hz 60 Hz 80 Hz 60 Hz, two-pole machine

Te

speed

[rpm]

E/f = En /fn

1200 2400 3600 4800

Base speed region Field-weakening region

Te

Rated voltage, rated frequency

speed

V/f = Vn /fn

Fig. 22: Torque-speed characteristics of an induction machine for two control laws:

E/f = En /fn (top) and V/f = Vn /fn (bottom)


26/36



V Vn V Vn

V0

0 fn f 0 fn f

Fig. 23: Voltage vs frequency relationship for V/f = Vn /fn and V = (Vn /fn )f + V0 control

Let the ratio of operating frequency to rated frequency be a =f/fn . The original equivalent

circuit of Fig.13 at operating frequency f now becomes as shown in Fig. 24. All the

reactances are given for the rated frequency; hence at any other frequency they become aX.The input stator voltage is assumed to be determined using either (10) or (11).

Is Rs jaXs jaXr Ir

V = a Vn Im jaXm Rr /s

Fig. 24: Steady-state equivalent circuit of an induction machine at an arbitrary frequency

Note that slip s in the above figure is defined as per-unit (pu) value. Ifns = 60fn /P is the

synchronous speed of a 2P-pole machine at rated frequency, then for operation at any other

frequency

s = (ans - n)/ans (12)

where n denotes operating speed at this other frequency. Calculations related to V/f methodof speed control are best illustrated by means of examples.

Example 11: V/f control of Induction machines

A 3-phase, 60-Hz, 1728-rpm induction motor has the following parameters: stator

resistance = rotor resistance = 0.024-; stator leakage reactance = locked rotor leakage

reactance = 0.12-. The motor is to be 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;

b) The starting torque and current in terms of their values at the rated frequency.c) If this motor is used to drive a load of constant torque equal to rated, calculate the motor

speed for the supply frequency of 30-Hz.


27/36



Solution:

a) The purpose of this part of the example is to show that with V/f = constant law the air-

gap flux in the motor is not constant; hence the maximum torque at 12-Hz will be

consequently reduced in motoring with respect to the value at rated frequency.

222

2

222

2

222

22

22

2

//

1

4

3

]//[

1

4

31

4

3)(

1

4

3)(

2.060/12/

XaRaR

Vf

P

XaRaRa

aVf

P

XaRR

Vaaf

PafT

XRR

Vf

PfT

ffa

ss

nn

ss

nn

ss

nn

nem

ss

nn

nem

n

68.024.0)2.0/024.0(2.0/024.0

24.0024.0024.0

)(

)2.0(

)(

)(

22

22

22

22

nem

nem

ss

ss

nem

nem

ffT

ffT

XaRaR

XRR

ffT

affT

b) This part of the example demonstrates some additional benefits of the V/f speed control.

As will be shown, starting at low frequency gives an increased starting torque compared to

starting at rated frequency. Even better, this higher starting torque is achieved with reduced

starting current.

72.0

24.02.0048.0

24.0048.02.0

)(

)(

)(

)(

6.2

24.0048.0

024.024.02.0048.0

024.02.0

)(

)(

2

3

2

3)(

2

3)(

222

22

222

22

)(

)(

222)(

22)(

22

222

22

222

222

2

222

22

22

2

xXaRR

XRRa

ffI

affI

XaRR

aVaffI

XRR

VffI

x

x

XRR

R

XaRR

aR

ffT

affT

XaRR

aRV

f

P

XaRR

RVa

af

PaffT

XRR

RV

f

PffT

rs

rs

nsts

nsts

rs

nnsts

rs

nnsts

rs

r

rs

r

nest

nest

rs

rn

nrs

rn

nnest

rs

rn

nnest

Thus V/f control undoubtedly provides high starting torque with reduced starting current.


28/36



c) Since the motor rated speed at 60-Hz is 1728-rpm, the motor has 4 poles and the

corresponding synchronous speed is 6060/2 = 1800-rpm.

The rated slip is therefore: 11728/1800 = 0.04 i.e. 4%.

givesequationstwotheseofsideshandrighttheEquating

2

3)(

2

3)(

5.060/30/

222

2

22

2

XasRR

saRV

f

PaffT

XsRR

sRV

f

PffT

ffa

rs

rn

nnen

nrs

nrn

nnen

n

rpm-8201rpm-900/60Hz30

%9.8089.00198.004.26 2

22222

ssn

rs

r

nrs

nr

nsnPfnaff

sss

XasRR

saR

XsRR

sR

Note that at rated frequency synchronous speed and operating speed are 1800-rpm and

1728-rpm respectively; the difference of the two is 72-rpm. At 30-Hz synchronous speed

and operating speed are 900-rpm and 820-rpm respectively; the difference between the two

is now 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 12: Scalar (V/f = const) control of induction machines

A 3-phase, 60-Hz, 460-V, 1746-rpm cage induction motor operates with constant V/f speed

control. Assume the motor torque-speed characteristic in the range of 0 150% rated

torque to be linear.

a) Calculate the applied voltage and speed of the fully loaded motor at 45-Hz supplyfrequency.

b) If the motor drives a centrifugal pump (of approximately square torque-speedcharacteristic) under rated operating conditions, determine the supply voltage and

frequency at 75% of full-load torque.

Solution:

Using a linear approximation of the torque-slip (speed) curve at small slips

( sr RsR / and lr XsR / ) the motor torque can be written in a form convenient for the

analysis under the conditions of constant V/f control :-

sff

VRps

RfpV

Xs

RR

Vpfs

RsT L

rr

L

lr

s

r

22

22

2

22)(

23)(


29/36



a) The machine is operated with constant V/f ratio, and therefore at 45-Hz supply

frequency the applied voltage is :

line)-to-(lineV-34546060

45 r

rr

r Vf

fV

f

V

f

V

In order to determine the rotation speed at 45-Hz one should take into account that the

machine is fully loaded, which means that the load torque is equal to the rated value. The

motor rated slip is:-

03.01800

174611

s

rr

n

ns

Since the load torque is constant, the slip frequency has to be constant too as follows from

the above torque expression. This allows the speed to be calculated in the following way:-

rpm-1296135096.02

456096.0)1(

04.003.0

45

60

s

rr

rr

nsn

s

f

fssffs

b) The centrifugal load torque is proportional to the speed squared and at 75% rated torque

the corresponding speed value is:-

rpm-08.151275.01746

2

rr

rr T

Tnn

n

n

T

T

The slip frequency under the same loading conditions is now using the initial torque

expression:-

Hz-35.16003.075.075.0 sffs

sf

T

T

rrr

and the supply frequency can be obtained as :-

Hz-753.5160

208.151235.1

60)

601()1(35.1

f

npff

f

npf

n

nsf

s

The applied line voltage required is consequently: V-77.39646060

753.51 r

r

Vf

fV .

4.6 Control of Inverter-fed Induction Machines

Variable voltage, variable frequency operation of induction machines is realised utilising

autonomous inverters, in conjunction with a rectifier and a DC link circuit as shown in

Fig.25. There are two generic structures of autonomous inverters. The first one, voltage

source inverter (VSI), is the most frequently applied power supply source for control of

induction motors. The second one, current source inverter (CSI), is beyond the scope of

interest here and wont be considered. The principle of operation of a three-phase VSI is

explained next.


30/36



C

L

d.c.link

+

-

Filter InverterControlled

RectifierIM

3-phfixedfrequency

supply

3-phvariablefrequency

V f

Fig.25: Variable speed drive system (top) and family of torque-speed curves with

V/f = const control (bottom)

The 3-phase VSI contains three inverter legs with the input voltage being provided by a

three-phase (or single-phase) bridge rectifier with a capacitor placed at the output (Fig.25).

The capacitor provides smoothing of the DC voltage and, for sufficiently large capacitance,

the DC output voltage approaches a constant value for a given firing angle. It will therefore

be assumed that inverter input voltage is constant in the subsequent analysis.

The power circuit of a six-step voltage source inverter is shown in Fig. 26. As the inverter

itself controls only the frequency of the output AC voltage, a controllable rectifier must be

used in order to provide control of the output voltage magnitude via DC voltage variations

(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 normally a transistor switch (most commonly an IGBT), while the

other one is a diode. The diode is essential for the correct operation of the VSI as the output

voltage and current are out of phase due to inductive nature of the machine. It enables

current flow when one switch in a given inverter leg is turned off and the other one turned

on, while the current still flows in the previous direction. This diode is usually called afreewheeling or feedback diode.

The three inverter legs are controlled in such a way that the corresponding voltages

constitute a 3-phase system of square-wave voltages. This means that, assuming that the

upper transistor in the leg A is fired at time instant zero, firing of upper transistor in the leg

B will take place after 120, while firing of the upper transistor of the leg C will be delayed

for another 120. The conduction of each of the six semiconductor switches is again 180

so that at any time three out of six switches are on and the remaining three switches are off.

The resulting output voltage waveforms of line-to-line voltages in Fig.27 are quasi-square

waves, with two 60 zero intervals and two 120 non-zero intervals in which line-to-linevoltage equals plus and minus DC link voltage, respectively. A VSI operated in this 180

conduction mode is therefore usually called a six-step inverter.


31/36



The leg voltages of the inverter are given in Fig.27 with respect to the negative pole of the

DC link. The line-to-line voltages applied to the induction machine, also illustrated in

Fig.27, are obtained directly from the leg voltages as vAB = vAn - vBn, vBC = vBn - vCn and

vCA = vCn - vAn . Finally, if the machine is star connected, it can be shown that in the

system of Fig.26 line to neutral (phase) voltages of the machine (presented in the Fig.27)

are determined with the following expressions:

p

C

VDC A B C

n

Rectifier and inverter control: IM

rectifier controls VDC while inverter controls

output frequency so that V/f = Vn /fn or

V = (Vn /fn )f + V0.

Fig.26: 3-phase voltage source inverter (VSI) fed induction motor.

AnBnCnC

CnAnBnB

CnBnAnA

vvvv

vvvv

vvvv

3/13/2

3/13/2

3/13/2

(13)

As is obvious from Fig.27, waveforms of line to line and line to neutral voltages are notsine waves. Indeed, it can be verified by Fourier analysis that they contain higher odd

harmonics but not those divisible by three. The situation is quite similar with the current

harmonic content.

In the previous analysis it has been assumed that the inverter is operated in 180

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 (not

thyristor) bridge rectifier (single-phase or three-phase), so that the input DC voltage of the

inverter is constant. This operating mode allows variations of both inverter output voltage

and output frequency to be achieved by appropriate inverter switching.


32/36



The idea of PWM can be relatively easily explained using a single-phase VSI circuit of Fig.

28. The most frequently utilised method of PWM is the so-called sinusoidal PWM

technique, in which the 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 are then determined with intersections

of the reference signal and the carrier wave. Switches are turned on and off in pairs: S1 andS2 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 twofold. Firstly, the 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 quite

sufficient. Secondly, switching now occurs not at line frequency, as is the case with a six-

step inverter, but at much higher carrier wave frequency. This enables significantly faster

control and, at the same time, greatly improves harmonic spectrum of the load current. The

inverter output voltage now does not contain low order harmonics (the fifth, seventh, etc.)

present with the six-step inverter. Instead, harmonics are situated around multiples of the

switching frequency (i.e., triangular carrier wave frequency). Thus, if the switchingfrequency is 5-kHz (typical value nowadays), then the inverter output voltage will contain,

apart from fundamental, higher harmonics of frequencies 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 the motor reactance is 100 times greater at 5-kHz than at 50-Hz, harmonic

currents in the motor will be consequently very small.

The principle of sinusoidal PWM applied to the single-phase bridge inverter of Fig. 28, is

illustrated in Fig.29. The impact of amplitude variations of the reference signal on the

output voltage waveform is evident from the same figure. The widths of the pulses in the

two cases shown differ although output frequency and triangular carrier wave frequency arethe same. This change in pulse widths leads to subsequent difference in the values of the

first harmonic of the output voltage. It can be shown that the fundamental harmonic of the

output voltage waveforms shown in Fig.29 equals in frequency and in amplitude the

reference signal.

An extension of the PWM principle from a single-phase to a 3-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 = Vn /fn law). This is further compared with the carrier

signal (again, high frequency triangular waveform) and the intersections points representthe switching instants for the inverter legs. The 3-phase inverter however requires three

120 mutually displaced reference sinusoidal signals in order to achieve operation with

three-phase system of output voltages. If a 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.


33/36



vAn VDC vAB VDC

vBn vBC

vCn vCA

0 60 120 180 240 300 360 t []

Leg voltages

Line-to-line voltages

2/3 VDC

vA 1/3 VDC

vB

Line to neutral voltages

vC

Fig.27: Typical voltage waveforms in a VSI fed induction machine.

In summary, it can be stated that the operation of a VSI in PWM mode yields two

substantial benefits, when compared to operation in 180 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. Due to

high control rate and improved power supply quality, the machine performance is also

generally much better and its operation can be very smooth and free of torque ripples.


34/36



Rectifier i

S1 S3

V C v Load

S4 S2

0


35/36



Example 13: Induction motors with V/f control and PWM inverter supply

A 3-phase, 415-V, 50-Hz, 1440-rpm inverter-fed cage induction motor with constant V/f

speed control drives a centrifugal pump under rated operating conditions.

a) Using the approximate equivalent circuit of the machine, derive the torque expressionstating the meanings of all the parameters used.

b) Show that the torque-slip (speed) characteristic is linear at small slips and that themotor torque is approximately constant providing that the slip frequency is constant.

c) Use the above approximation to estimate the supply voltage and frequency required toreduce the drive speed to 1000-rpm.

Solution:

This example is closely related to the previous and the two should be considered together.

a) The electromagnetic torque expression for a 2p-pole IM is :

sfXsRR

VpR

s

IRsT

rs

r

s

rr

])/[(2

33)(

22

22

where rI is the rotor phase current, srR , are the rotor and stator phase resistances

respectively, V is the machine phase voltage, X is the total leakage reactance (stator plus

rotor leakage), s is the slip andfis the supply frequency.

b) Linear approximation at small slips: XRsR sr ,/

sff

Vcs

f

V

R

ps

sfR

VpRsT

rr

r 222

2

2

2

3

2

3)(

Therefore ifsf= rotor (slip) frequency is constant then the motor torque is constant too.

c) Centrifugal load torque at 1000-rpm and 1440-rpm (torque ratio under constant V/f

control) :

rrrrated

load

fs

sf

n

n

T

T

22

1440

1000

Slip frequency:6060

)1(np

fpf

nfff

n

nsf

s

.

The above torque ratio now becomes (the machine is 4-pole as its rated speed is 1440-rpm

i.e. the synchronous speed at 50-Hz is 1500-rpm) :-

Hz-3.34120

200060

14405060

100060

60

60

1440

10002

f

f

p

pf

pnf

npf

fs

sf

rrrr

3

The supply voltage at 34.3-Hz is thus:

LLV-69.28441550

3.34 rrr

r VffVconst

fV

fV


36/36


Example 14: Constant Volts/Hz control of IMs using a PWM voltage source inverter

A 3-phase, 440-V, 6-pole, 50-Hz, delta-connected induction motor has the following rated

parameters per phase: stator resistance = 0.2-; rotor resistance = 0.18-; stator and rotor

leakage reactance = 0.58- each. The machine operates with constant V/f scalar control

supplied from a PWM voltage source inverter.(a)The machine is subjected in service to a 40% fall in both voltage and frequency. What

total mechanical load torque is it safe to drive so that the machine does not stall under

these conditions?

(b)IfV and f were both halved, what would be the increase in starting torque from thenormal DOL start? Compare the starting currents for these two cases and comment

upon the results obtained.

Solution:

(a)This part of the question asks effectively for the maximum torque with voltage andfrequency reduced to 0.6 of rated values. Substituting for a = 0.6 in the expression for

maximum torque derived earlier (refer to Example 12) one gets:

Nm.-4.1800

)16.16.0(2.02.0

)4406.0(

506.04

33)(

4

3)(

max

22

2

222

2

max

T

XaRR

aV

af

PafT

ss

r

rr

(b)The ratio of starting torque at reduced and full voltage and frequency for a = 0.5 is (seeExample 12) :

88.0

58.038.0

16.138.05.0

)(

)(

55.15.058.038.0

16.138.0

)(

)(

22

22

222

22

22

22

222

22

XaRR

XRRa

ffI

affI

aXaRR

XRR

ffT

affT

rs

rs

rst

rst

rs

rs

rst

rst

Note that due to improved power factor following the reduction of reactance (because of

the lower supply frequency), the starting torque is increased by 55% relative to the DOL

value while the current required to achieve this torque rise is 88% DOL value. Therefore

the starting torque per ampere (TPA) is much higher with reduced voltage and frequency

starting and can be calculated as follows:

rf

f

rst

rst

st

st

rst

st

rst

st

TPA

TPA

fI

fT

fI

fT

fI

fI

fT

fT

)(

)(

)(

)(

76.188.0

55.1

)(

)(

)(

)(

The TPA is increased by 76% compared to DOL starting This ratio is even better at lowerf i

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