Year2 Electrical Machines and Systems.pdf

Course in Electrical Machines and Systems

Year 2

©2004 J D Edwards

ELECTRICAL MACHINES AND SYSTEMS COURSE NOTES These notes were prepared for a Year 2 course module in the Department of Engineering and Design at the University of Sussex. The module runs for ten weeks, and absorbs 25% of student time. This introductory course in electrical machines follows on from the Year 1 course Electromechanics, with a similar philosophy. It avoids the traditional mathematical derivation of the theory of AC and DC machines, and makes extensive use of field plots generated with MagNet to explain the principles. With this physical background, the theory is developed in terms of circuit models and phasor diagrams. Field plots are particularly useful for explaining armature reaction and compensating windings in DC machines, and for demonstrating the action of variable-reluctance and hybrid stepper motors. They also show the essential unity of the conventional rotating machines; there are similar plots for the stator and rotor field components and the resultant field in DC, synchronous and induction machines.

Electrical Machines and Systems Course Notes Copyright © 2004 J D Edwards

CONTENTS

1 INTRODUCTION 1

2 TRANSFORMERS 2

2.1 Introduction 2

2.2 Types of power transformer 3

2.3 Ideal transformer properties 4

2.4 Circuit model of a transformer 6

2.5 Parameter determination 7

2.6 Transformer performance 9

2.7 Current transformers 11

2.8 Transformer design 12

3 DC MACHINES 15

3.1 Introduction 15

3.2 DC machines in practice 16

3.3 Characteristics and control 18

3.4 Series motors 21

4 INTRODUCTION TO AC MACHINES 24

4.1 Review of 3-phase systems 24

4.2 Rotating magnetic field 25

4.3 Multi-pole fields 28

5 SYNCHRONOUS MACHINES 30

5.1 Introduction 30

5.2 Characteristics 31

5.3 Salient-pole machines 35

5.4 Linear synchronous motors 37

6 INDUCTION MACHINES 39

6.1 Introduction 39

6.2 Characteristics 40

6.3 Losses and efficiency 46

6.4 Parameter determination 48

6.5 Single-phase induction motors 50

6.6 Dynamic conditions 52

6.7 Linear induction motors 54

7 STEPPER MOTORS 56

7.1 Introduction 56

7.2 Variable-reluctance principle 56

7.3 Variable-reluctance stepper motors 58

7.4 Hybrid stepper motors 59

7.5 Stepper motor characteristics 61

7.6 Stepper motor control 64

8 POWER ELECTRONIC CONTROL 66

8.1 AC/DC Converters 66

8.2 DC motor control 68

8.3 DC/AC Inverters 68

8.4 AC motor control 70

9 REFERENCES 76

10 APPENDICES 77

10.1 Induction motor 2-axis equations 77

10.2 List of formulae 80

Introduction 1

1 INTRODUCTION This course follows on from the Term 3 course Electromechanics. Its purpose is to explore in greater depth the AC and DC machines that were introduced in the earlier course. The approach is that of the application engineer rather than the machine designer, concentrating on the basic principles, characteristics, and control. Since induction motors account for more than 90 per cent of the motors used in industry, the course gives particular emphasis to these machines.

Course components

The course has three closely linked components: lectures, problem sheets and laboratories. In addition, there is a design assignment, which introduces some of the basic ideas and problems of design by considering a very simple device: an electromagnet. Lectures will use video presentation and practical demonstrations. These notes provide support material for the lectures, but they are not a substitute. Regular lecture attendance is essential. Problem solving is a vital part of the course. Problem sheets will be issued at the first lecture each week, and methods of solving the problems will be discussed in each lecture. The laboratory runs from week 4 to week 9, with three 3-hour experiments: EMS1: Speed control of induction motors. EMS2: Characteristics of a power transformer. EMS3: Control of a stepper motor. Experiment EMS1 is a sequel to the simple DC motor-control experiment in Electromechanics. EMS2 introduces some important electrical measurement techniques as well as exploring the properties of a transformer. EMS3 explores a stepper motor and controller of the kind widely used in industry.

Magnetic field plots

A magnetic field plot is often a useful way of picturing the operation of an electromagnetic device. Numerous plots have been specially prepared for these notes, using the MagNet electromagnetic simulation software, to develop the basic concepts with a minimum of mathematics.

References

References to books are listed in section 10, and cited in the text of the notes with the reference number in square brackets.

Background material

The course assumes a familiarity with the contents of the Term 3 course Electromechanics, so the basic principles covered in that course will not be repeated. Students are expected to have a copy of the printed notes for Electromechanics [1], and further information will be found in references [2] to [4].

2 Electrical Machines and Systems Course Notes

2 TRANSFORMERS

2.1 Introduction

Basic transformer principles were covered in Electromechanics [1], and the main results are given below. Figure 2-1 is a schematic representation of a single-phase transformer with two coils on a magnetic core, where the magnetic coupling is assumed to be perfect: the same flux passes through each turn of each coil.

~ Zv1

+ + +

v2e1 e2

i1 i2

R1 R2

primaryN1 turns

secondaryN2 turns

+

Figure 2-1: Transformer with source and load

Voltage relationships

Kirchhoff’s voltage law applied to the two windings gives:

1111111 iRdt

dNiRev

(2-1)

2222222 iRdt

dNiRev

(2-2)

If the resistances R1 and R2 are negligible, then equations 2-1 and 2-2 become:

dt

dNv

11 (2-3)

dt

dNv

22 (2-4)

Dividing these equations gives the important result:

2

1

2

1

N

N

v

v (2-5)

Sinusoidal operation

If the voltage source is sinusoidal, then the core flux will also be sinusoidal, so we may put:

tm sin (2-6)

Substituting this expression in equation 2-3 gives:

tVtNdt

dNv mm

coscos 1111 (2-7)

Thus the maximum primary voltage is:

mmmm ABfNfNNV 1111 22 (2-8)

where A is the cross-sectional area of the core and Bm is the maximum flux density in the core. A typical value for Bm is 1.4 T for the silicon steel characteristic in figure 2-2.

00.20.40.60.8

11.21.41.61.8

2

0 2 4 6 8 10 12 14 16 18 20

Magnetic intensity H, kA/m

Flu

x de

nsity

B, T

Figure 2-2: Silicon transformer steel.

Current relationships

The relationship between the primary and secondary currents can be found by considering the magnetic circuit of the transformer. From the basic magnetic circuit equation, we have:

RF 2211 iNiN (2-9)

In a well-designed transformer, the reluctance R is small, so equation 2-9 becomes:

02211 iNiN (2-10)

This gives the counterpart of equation 2-5 for voltage:

1

2

2

1

N

N

i

i (2-11)

Transformers 3

If sinusoidal voltages and currents are represented by phasors, the corresponding forms for the basic voltage and current equations are:

2

1

2

1

N

N

VV

(2-12)

1

2

2

1

N

N

II

(2-13)

Transformer rating

The maximum voltage at the primary terminals of a transformer is determined by equation 2-8, and is independent of the current. The maximum primary current is determined by the I2R power loss in the resistance of the transformer windings, which generates heat in the transformer. This power loss is independent of the applied voltage. Consequently, for a given design of transformer, there is a maximum value for the product V1I1 at the primary terminals. To a first approximation, this is also equal to the product V2I2 at the secondary terminals. This maximum value does not depend on the phase angle between the voltage and the current. Transformer ratings therefore specify the apparent power VI (volt-amperes, VA) rather than the real power VI cos (watts, W).

2.2 Types of power transformer

In addition to the ordinary single-phase power transformer, two other types are in common use: auto-wound transformers, and 3-phase transformers.

Auto-wound transformers

A transformer can have a single coil with an output taken from a portion of the coil, as shown in figure 2-3. This is known as an auto-wound transformer or auto-transformer.

VS V2

I1

IL

ZL +

+ I2

V1 +

Figure 2-3: Auto-wound transformer.

Unlike the normal transformer with two windings, known as a double-wound transformer, the auto-wound transformer does not provide electrical isolation between the primary and the secondary, However, an auto-wound transformer can have a much larger apparent power rating than a double-wound transformer of the same physical size. Let N1 be the number of turns on the upper part of the winding in figure 2-3, and N2 the number of turns on the lower part. The conventional transformer equations 2-12 and 2-13 apply to these parts of the winding, since they are equivalent to two separate windings with a common connection. Applying Kirchhoff’s law to this circuit gives:

1 2S V V V (2-14)

1 2L I I I (2-15)

As an example, suppose that N1 = N2. If the transformer is regarded as ideal (see section 2.3), then I1 = I2 and V1 = V2. Equations 2-14 and 2-15 give:

1 2 22 2S L V V V V V (2-16)

1 2 12 2L S I I I I I (2-17)

where VL is the voltage across the load and IS is the current supplied by the source. This auto-wound transformer behaves as a step-down transformer with a ratio of 2:1, and the current in each winding is equal to half of the load current. An elegant application of the auto-wound transformer principle is the variable transformer, which has a single-layer coil wound on a toroidal core. The output is taken from a carbon brush that makes contact with the surface of the coil; the brush can be moved smoothly from one end of the coil to the other, thus varying the output voltage. Examples of variable transformers are shown in figure 2-4.


Figure 2-4: Variable transformers. (RS Components Ltd)

3-phase transformers

In 3-phase systems (see section 4.1), it is common practice to use sets of three single-phase transformers. It is also possible, however, to make 3-phase transformers with three sets of windings on three limbs of a core, as shown in figure 2-5.

Figure 2-5: 3-phase transformer model.

The corresponding fluxes are shown in figure 2-6.

a b c

Figure 2-6: 3-phase transformer flux. In a balanced system with sinusoidal phase voltages, the fluxes will be given by:

)120cos()240cos(

)120cos(

cos

tt

t

t

mmc

mb

ma

(2-18)

Figure 2-7 shows flux plots for the transformer at the instants when t = 0, 120º and 240º.

Figure 2-7: 3-phase transformer flux plots: (a) 0º, (b) 120º, (c) 240º.

There is no requirement for another limb to form a flux return path, because the fluxes a, b and c sum to zero in a balanced 3-phase system. The proof is as follows. From equation 2-18, the sum is given by:

0coscos

120coscos2cos

)120cos()120cos(cos

tt

tt

tttm

cba

(2-19)

Because the fluxes in the three limbs sum to zero at all instants of time, there is no leakage of flux from the core, as the flux plots in figure 2-7 demonstrate.

(a)

(b)

(c)

Transformers 5

2.3 Ideal transformer properties

If the primary and secondary windings have zero resistance, and the magnetic core has zero reluctance, then the approximate equalities in equations 2-12 and 2-13 become exact equalities. This leads to the concept of an ideal transformer element, to accompany the other ideal elements of circuit theory. Figure 2-8 shows a circuit symbol for the ideal transformer element.

+ +v1 v2

i1 i2

Figure 2-8: Ideal transformer element

The voltage and current relationships in the time and frequency domains are given in table 2-1.

Table 2-1: Ideal transformer relationships.

Time domain Frequency domain

nN

N

v

v

1

2

1

2 nN

N

1

2

1

2

VV

(2-20)

nN

N

i

i

1

2

2

1 nN

N

1

2

2

1

II

(2-21)

The following properties of the ideal transformer may be deduced from equations 2-16 and 2-17:

The voltage transformation is independent of the current, and vice versa.

If the secondary is short-circuited, so that v2 = 0, the primary terminals appear to be short-circuited since v1 = 0.

If the secondary is open-circuited, so that i2 = 0, the primary terminals appear to be open-circuited since i1 = 0.

The output power is equal to the input power, so there is no power loss in the element.

Impedance transformation

The ideal transformer has the important property of transforming impedance values in a circuit. Consider an ideal transformer with an impedance ZL connected to its secondary terminals, as shown in figure 2-9.

V1 V2

I1 I2

ZL

N1:N2

+ +

Figure 2-9: Ideal transformer with a load.

The secondary impedance is given by:

2

2

IV

Z L (2-22)

At the primary terminals, the circuit presents an impedance given by:

2

2

2

1

2

22

2

1

1

22

2

21

1

1

nN

N

N

N

N

NN

N

LL

in

ZZ

IV

I

V

IV

Z

(2-23)

Thus the combination of an ideal transformer of ratio n and an impedance ZL can be replaced by an equivalent impedance ZL / n

2.

Referred impedances

Figure 2-10(a) shows an ideal transformer with a load impedance ZL connected to the secondary. Another impedance Z2 is in series with ZL. The input impedance of this circuit is:

2

2

nL

inZZ

Z

(2-24)

The input impedance of the circuit in figure 2-10(b) is:

22

nL

inZ

ZZ (2-25)

The two expressions for Zin will be identical if:

22

2n

ZZ (2-26)


ZL

N1:N2

Z2

Zin

(a)

ZL

N1:N2

Z'2

Zin

(b)

Figure 2-10: Referred impedance – 1.

The impedance Z'2 is termed the secondary impedance Z2 referred to the primary. In a similar way, a primary impedance Z1 can be referred to the secondary, as shown in figure 2-11. In this case, the referred impedance is given by:

12

1 ZZ n (2-27)

ZL

N1:N2

Z1

Zin

(a)

ZL

N1:N2

Zin

Z''1

(b)

Figure 2-11: Referred impedance – 2.

The concept of referred impedance is often a useful device for simplifying circuits containing transformers, as will be shown in the next section. It is conventional to use a single prime (') to denote quantities referred to the primary side, and a double prime (") to denote quantities referred to the secondary side.

2.4 Circuit model of a transformer

In a practical transformer, the winding resistances and the core reluctance are not zero. In addition, there will be some power loss in the core because of eddy currents and hysteresis in the magnetic material. All of these effects can be represented by the equivalent circuit [3, 4] shown in figure 2-12.

N1:N2

R1R2jx1 jx2

jXm RcV1 V2

+ +

I1 I2I2 = nI2

I0

I0m I0c

Figure 2-12: Transformer equivalent circuit.

This circuit is based on the ideal transformer element, with additional circuit elements to represent the imperfections. The resistances R1 and R2 represent the physical resistances of the windings, and Rc represents the power lost in the core. The reactance Xm, known as the magnetising reactance, allows for the current required to magnetise the core when the reluctance is not zero. Reactances x1 and x2, known as leakage reactances, represent the leakage flux that exists when the magnetic coupling between the primary and the secondary is not perfect. Figure 2-13 shows the leakage flux when the core has an artificially low relative permeability of 10, and one winding at a time is energised. In practice, the leakage is much less than this, so the leakage reactances are normally very much smaller than the magnetising reactance Xm.

(a)

Transformers 7

(b)

Figure 2-13: Transformer leakage flux: (a) left coil energised, (b) right coil energised.

In the circuit of figure 2-12, the current I2 is the effective value of the secondary current as seen from the primary side of the transformer. It is also known as the secondary current referred to the primary. The current I0 is the no-load current, which is the current taken by the primary when there is no load connected to the secondary. It has a component I0m, known as the magnetising current, which represents the current required to set up the magnetic flux in the core. The current I0c is the core loss component of the no-load current.

Approximate equivalent circuit

For power transformers with ratings above 100 VA, the values of the series elements R1 and x1 are generally much smaller than the shunt elements Rc and Xm. Under normal working conditions, the voltage drop in R1 + jx1 is much smaller than the applied voltage V1. Similarly, the no-load current I0 is much smaller than the load current I1. It follows that the shunt elements can be moved to the input terminals, as shown in figure 2-14(a), with very little loss of accuracy. The secondary elements R2 and jx2 can be replaced by equivalent elements R2 = R2 / n

2 and x2 = x2 / n

2 on the primary side (see section 2.3), giving the circuit shown in figure 2-14(b). Finally, the series elements can be combined to give an effective resistance Re and leakage reactance xe, as shown in figure 2-14(c), where the values are:

22

1n

RRRe ,

22

1n

xxxe (2-28)

N1:N2

R1 R2jx1 jx2

jXm Rc

(a)

N1:N2

R1 R2jx1 jx2

jXm Rc

(b)

N1:N2

Rejxe

jXm Rc

(c)

Figure 2-14: Approximate equivalent circuit.

2.5 Parameter determination

The parameters of the approximate equivalent circuit (figure 2-14) can be determined experimentally from two tests: An open-circuit test, where the secondary is

left unconnected and the normal rated voltage is applied to the primary.

A short-circuit test, where the secondary terminals are short-circuited and a low voltage is applied to the primary, sufficient to circulate the normal full-load current.


Open-circuit test

With the secondary unconnected, I2 = 0, so the equivalent circuit reduces to the form shown in figure 2-15.

jXm Rc

I1oc

V1oc

+

Figure 2-15: Open-circuit test.

To a very close approximation, the current I1oc supplied to the primary is equal to the no-load current I0 in figure 2-12. The values of the elements Rc and Xm can be determined from measurements of the input voltage V1oc, current I1oc and power P1oc as follows. The input power is entirely dissipated in the resistance Rc, giving:

oc

occ P

VR

1

21 (2-29)

The input impedance of the circuit is given by:

mc

oc

ococ

jXR

111

1

11

IV

Z (2-30)

In terms of magnitudes, equation 2-30 becomes:

221

11

11

1

mc

oc

ococ

XR

I

VZ

(2-31)

Re-arranging equation 2-31 gives the value of Xm:

2

2

1

1 1

1

coc

oc

m

RV

I

X

(2-32)

From figure 2-14(c), it follows that the turns ratio is given by:

oc

oc

V

V

N

Nn

1

2

1

2 (2-33)

Short-circuit test

If the secondary terminals are short-circuited, the ideal transformer in figure 2-14 can be replaced by a short circuit, so the equivalent circuit takes the form shown in figure 2-16(a). In a typical power transformer, the shunt elements Rc and Xm are at least 100 times larger than the series elements Re and xe. Consequently, the shunt elements can be neglected, and the circuit reduces to the form shown in figure 2-16(b).

jXm Rc

jxeRe

(a)

jxeReI1sc

V1sc

+

(b)

Figure 2-16: Short-circuit test.

The values of the elements Re and xe can be determined from measurements of the input voltage V1sc, current I1sc and power P1sc as follows. The input power is entirely dissipated in the resistance Re, giving:

21

1

sc

sce

I

PR (2-34)

The input impedance of the circuit is given by:

eesc

scsc jxR

1

11 I

VZ (2-35)

In terms of magnitudes, equation 2-35 becomes:

22

1

11 ee

sc

scsc xR

I

VZ (2-36)

Transformers 9

Re-arranging equation 2-36 gives the value of xe:

22

1

1e

sc

sce R

I

VX

(2-37)

In practice, the open-circuit test is usually made on the low-voltage side of the transformer to minimise the value of Voc, and the short-circuit test is made on the high-voltage side to minimise the value of Isc. The resulting parameter values are then referred to the primary side of the transformer.

2.6 Transformer performance

Consider a transformer with an impedance ZL connected to the secondary. With the approximate equivalent circuit, this may be represented by the circuit diagram of figure 2-17.

N1:N2

Rejxe

jXm RcZL =

RL + jXL

Figure 2-17: Transformer with a load.

The load impedance can be referred to the primary side of the ideal transformer element, giving the circuit shown in figure 2-18.

Rejxe

jXm Rc

ZL =ZL / n2

V2 = V2 / n

+V1

+

I1 I2 = nI2

I0

Figure 2-18: Circuit with referred impedance.

This circuit is easily solved for the currents I0 and I2. The referred secondary voltage is given by V2 = ZLI2, and the secondary terminal quantities are given by V2 = nV2, I2 = I2 / n.

Voltage regulation and efficiency

When a load is connected to the secondary of a transformer, there will be a voltage drop in the series elements Re and xe, so the secondary terminal voltage will change. The voltage regulation is defined as:

nl

flnl

V

VV

2

22 (2-38)

where V2nl is the magnitude of the no-load secondary terminal voltage, and V2fl is the corresponding full-load voltage. Power is lost as heat in the windings and core of the transformer, represented by the resistance elements Re and Rc in the equivalent circuit. The efficiency is defined in the usual way as:

in

out

P

P (2-39)

where Pin is the power input to the primary and Pout is the power output from the secondary. The power lost as heat in the transformer is:

outinloss PPP (2-40)

so we have the following alternative forms of equation 2-39:

lossout

loss

lossout

out

in

loss

in

lossin

in

out

PP

P

PP

P

P

P

P

PP

P

P

1

1 (2-41)

Large transformers are very efficient. Even a 2 kVA transformer can have an efficiency of about 95%. Above 25 kVA, the efficiency usually exceeds 99%. It is very difficult to make an accurate measurement of efficiency by direct measurement of Pout and Pin, since this would require an accuracy of measurement of the order of 0.01%. Instead, the normal practice is to determine the losses from measurements, and calculate the efficiency from one of the alternative expressions in equation 2-41. The losses can be calculated with high accuracy from the equivalent-circuit parameters determined from tests on the transformer.

Maximum efficiency

The power loss in a transformer has two components: the core loss, given by V1

2 / Rc, and


the I2R loss, given by I'2Re. The core loss will be constant if the primary voltage V1 is constant, but the I2R loss will vary with the secondary current. When the current is low, the output power will be low, but the core loss remains at the normal value. Consequently, the efficiency of the transformer will be low under these conditions. It may be shown that the efficiency is a maximum when the secondary current is such that the variable I2R loss is equal to the fixed core loss. This result also applies to other devices where the losses have fixed and variable components. Transformers are usually designed to have maximum efficiency at the normal operating value of secondary current, which may be less than the maximum rated current.

Power relationships

When calculating the transformer performance, the following power relationships can be useful. The complex power S is given by [2]:

S = P + jQ = VI* (2-42)

where V is the voltage phasor, I* is the complex conjugate of the current, P is the real power, and Q is the reactive power. If is the phase angle, then:

*)Re(coscos VI VISP (2-43)

*)Im(sinsin VI VISQ (2-44)

22 QPVIS (2-45)

If the voltage phasor V is chosen as the reference quantity, and defined to be purely real (V = V + j0), then the power relationships take a simple form:

)Re(cos IVVIP (2-46)

)Im(sin IVVIQ (2-47)

Worked example 2-1

A 2 kVA, 50 Hz, power transformer has the following equivalent-circuit parameter values referred to the primary:

Re = 0.682 , xe = 0.173, Rc = 1080, Xm = 657, N2 / N1 = 0.472. If the primary is connected to 230 V 50 Hz supply, and a load impedance (6.0 + j2.5) is connected to the secondary, determine: (a) the secondary current magnitude,

(b) the secondary terminal voltage magnitude, (c) the primary current magnitude, (d) the voltage regulation, (e) the efficiency of the transformer.

(a) Secondary current

The load impedance referred to the primary is:

2.119.26)472.0(

5.20.622

jj

nL

LZ

Z

The secondary current referred to the primary is:

A 70.794.212.7

A 94.212.74.116.27

0230

)2.119.26()173.0682.0(

0230

2

12

jI

jj

j

jj

jLe ZZ

VI

The secondary current magnitude is:

A 3.16472.0

70.722

n

II

(b) Secondary voltage

The secondary voltage magnitude is:

V 0.1065.63.16

5.20.63.1622

jZIV L

(c) Primary current

The no-load current is:

A 350.0213.0

657

0230

1080

0230110

j

j

jj

jXR mc

VVI

The primary current is:

A 03.829.333.7

A 29.333.7

)94.212.7()350.0213.0(

1

201

jI

j

jj

III

(d) Voltage regulation

On no load, the secondary voltage is:

V 6.108230472.012 nVV

Transformers 11

The voltage regulation is therefore:

%33.26.108

0.1066.108

2

22

nl

flnl

V

VV

(e) Efficiency

The output power is:

W 15970.6)31.16( 222 Lout RIP

The power loss is:

W 4.89

682.0)699.7(1080

)0.230( 22

22

21

ec

loss RIR

VP

The input power is therefore:

W 16864.891597 lossoutin PPP

Alternatively, from equation 2-46 the input power is:

W 1686330.70.230)Re( 11 IVPin

The efficiency is therefore:

%70.941686

1597

in

out

P

P

2.7 Current transformers

The use of transformers for measuring current has been introduced in Electromechanics [1], where the danger of open-circuiting the secondary has been explained. This section introduces the important topic of measurement errors.

Current transformer errors

In a well-designed current transformer, the core flux density is low and the core is made from a high-quality magnetic material. Under normal operating conditions, the core loss will be negligibly small, so the core loss resistance Rc can be omitted from the equivalent circuit. A circuit model for a current transformer connected to a load therefore takes the form shown in figure 2-19.

N1:N2

R1 R2jx1 jx2I1

I2

I2 = nI2

I0

jXm ZL

Figure 2-19: Current transformer circuit model.

Only the relationship between currents is of interest, so the primary impedance (R1 + jx1) can be disregarded. It is now convenient to refer the primary quantities to the secondary side, giving the circuit model shown in figure 2-20.

R2jx2I1

I2I0 = I0 / n

jXm = jn2Xm ZL

Figure 2-20: Modified circuit model.

The circuit acts as a current divider, where the current in the secondary branch is given by:

mL

m

Xj

Xj

ZZ

II

2

12 (2-48)

where I1 is the primary current referred to the secondary, and Z2 = R2 + jx2. In a well-designed transformer, the secondary impedance Z2 is very small in comparison with the referred magnetising reactance Xm, so this term introduces very little error. Equation 2-48 shows that it is desirable to keep the load impedance ZL as small as possible if the error is to be minimised. In practice, current transformers are designed for a specified maximum secondary voltage at the rated secondary current. This defines a maximum apparent power for the secondary load, or burden. Typically, a small current transformer will have a rated secondary burden of 5 VA. With the usual secondary current rating of 5 A, this implies that the maximum secondary voltage is 1 V, and the maximum impedance magnitude is 0.2 .


Worked example 2-2

A current transformer has 10 turns on the primary and 100 turns on the secondary. It has a rated secondary current of 5 A, the magnetising reactance referred to the secondary is 10 , and the maximum burden is 5 VA. If the primary current is 50 A, determine the secondary current, and hence the percentage error in the current measurement, if the secondary load is (a) purely resistive, (b) purely inductive. The transformer secondary impedance may be neglected. (a) Resistive load Since the burden is 5 VA and the secondary current is 5 A, the secondary voltage is 1 V, and the resistance is 1 / 5 = 0.2 . The turns ratio is n = 100 / 10 = 10, so the primary current referred to the secondary is 50 / 10 = 5.0 A. The secondary load current is:

0.102.0

0.50.1012 j

j

XjR

Xj

mL

m

I

I

The magnitude is given by:

A 999.4002.10

0.50.10

0.102.0

0.50.102

j

I

The percentage error is thus:

%02.00.5

999.40.5

e

(b) Inductive load Since the impedance magnitude is the same as before, the load reactance is 0.2 . The secondary current is now:

0.102.0

0.50.1012 jj

j

XjjX

Xj

mL

m

II

The magnitude is given by:

A 902.40.102.0

0.50.102

I

The percentage error is thus:

%96.10.5

902.40.5

e

2.8 Transformer design

The majority of single-phase transformers use the shell type of construction shown in figure 2-21.

Figure 2-21: Shell-type transformers. (RS Components Ltd)

Normally the core laminations are made in two parts, termed E and I laminations, as shown in figure 2-22.

Figure 2-22: E and I laminations.

The centre limb is twice the width of the outer limbs because it carries twice the flux, as shown by the flux plot in figure 2-23.

Figure 2-23: Flux plot: shell-type transformer.

Transformers 13

The coils are wound on a bobbin that fits the centre limb of the core, and the core is assembled by inserting E laminations alternately from each side and adding matching I laminations. Dimensions are chosen so that two E and two I laminations can be punched from a rectangular steel sheet without any waste, as shown in figure 2-24.

Figure 2-24: Punching E and I shapes.

Current flowing in the resistance of the transformer windings will produce heat, which must escape through the surface of the windings. In addition, there will be power loss in the core, which also appears as heat. The power output from a given size of transformer is governed by the rate at which heat can be removed. Large transformers are usually cooled by circulating oil, but small transformers rely on natural convection cooling in air. A simple design approach for small transformers is given below.

Thermal model

The rate of cooling depends on the exposed surface area of the transformer and the temperature rise above ambient. An exact calculation is complex, since it needs to take account of temperature gradients within the transformer as well as the cooling conditions on different surfaces. A simple thermal model ignores temperature gradients, and the power loss in the core. It just considers the I2R loss in the windings, and assumes that this heat escapes through the exposed surfaces of the windings. It is assumed that the temperature rise is proportional to the power loss per unit area:

s

lc

A

PkT (2-49)

where T is the temperature rise above ambient, Pl is the total power loss in the windings, As is the exposed surface area, and kc is a cooling coefficient with a typical value of 0.04 Km2/W. Figure 2-25 shows the side view and top view of a shell-type transformer.

2a

b

mean turn

aa 2a a

a

6a

3a 5a

a

Figure 2-25: Shell-type transformer dimensions It is assumed that the core is made from laminations punched as shown in figure 2-24. If a is the width of each outer limb of the core, then the width of the centre limb is 2a, and the other dimensions are as shown in figure 2-25.

Winding resistance

The total cross-sectional area of the two windings is the window area of height 3a and width a. Each winding occupies half of this area, so the conductor cross-sectional area for each winding is:

25.1 akA sc (2-50)

where ks is the conductor space factor, which allows for insulation and space between the turns. For simplicity, it will be assumed that the primary and secondary windings are placed side-by-side on the core, and that they have the same number of turns N. From figure 2-25, the mean turn length of each winding is:

babaalm 2)4(24 (2-51)


If the winding has N turns, then the total length of wire is Nlm, and the cross-sectional area of the wire is Ac / N. The winding resistance is therefore:

2

2

5.1

}2)4{(

/ ak

baN

NA

Nl

A

lR

sc

m

(2-52)

Temperature rise

If the RMS current in one winding is I, the power loss is I2R. The cooling surface area of the winding, from figure 2-25, is:

)410(

4)24(5.12

22

a

aaaaaAs (2-53)

Substituting in equation 2-49 gives:

)25(3

}2)4{(4

222

ak

baINk

A

RIk

A

PkT

s

c

s

c

s

lc (2-54)

Thus, the current is given by:

}2)4{(

)25(32

42

baNk

TakI

c

s

(2-55)

From equation 2-10, the number of turns is:

mm

m

fabB

V

fAB

VN

4

2

2 (2-56)

Worked example 2-3

A transformer has a primary wound for 230 V, and the core measures 60 50 30 mm. The maximum winding temperature is 90ºC, the ambient temperature is 30ºC, the winding space factor is 0.4, the cooling coefficient is 0.04 Km2/W, and the resistivity of copper at 90ºC is 21.9 nm. If the maximum flux density in the core is 1.4 T and the frequency is 50 Hz, determine (a) the number of turns on the primary, (b) the maximum current in the primary.

Solution

From figure 2-26, a = 10 mm and b = 30 mm. From equation 2-56, we have:

12304.110301010504

2302

4

2

33

mfabB

VN

Equation 2-55 gives:

}2)4{(

)25(32

42

baNk

TakI

c

s

A 216.0

}103021010)4{(

)1230(109.2104.0

)3090)(25()1010(4.03

}2)4{(

)25(3

33

29

43

2

4

baNk

TakI

c

s

Thus, I = 216 mA, so the transformer rating is 230 0.216 VA = 49.7 VA.

Rating and size

A relationship between the apparent power rating of the transformer and the dimensions can be obtained by substituting for N from equation 2-56 in equation 2-55:

}2)4{(

)25(32 3

bak

TkbaBVI

c

sm

(2-57)

If it is assumed that the core depth b is proportional to the dimension a, and other quantities remain constant, then equation 2-57 gives the following relationship:

5.3aVI (2-58)

From equation 2-58, if the dimensions of the transformer are doubled, the rating will increase by a factor of 11.3. A similar result is obtained for the increase in the power output of a DC machine when the dimensions are doubled – see section 3.2.

DC Machines 15

3 DC MACHINES

3.1 Introduction

Basic DC machine principles were covered in Electromechanics [1], and the main results are given below. Brushless DC machines, which were mentioned briefly in [1], are beyond the scope of this course but are covered in the Year 3 course Electrical Machine Drives.

Basic equations

The generated voltage and the developed torque are given by:

[V] rfa ùKe (3-1)

[Nm] afd iKT (3-2)

where K is the armature constant, f is the field flux, and ia is the armature current. Note that the rotor angular velocity r must be in radians per second (rad/s) and not in rev/min. If the rotational speed is nr rev/s or Nr rev/min, then:

[rad/s] 60

22 r

rrNð

nðù (3-3)

In a permanent-magnet machine, the field flux f is fixed, but in a wound-field machine, it is a function of the field current if.

Magnetisation characteristic

From equation 3-1, if the speed r is held constant, the flux is proportional to the armature generated voltage ea. A graph of ea against if is known as the magnetisation characteristic of the machine, and a typical curve is shown in figure 3-1 for a 3 kW motor.

0

50

100

150

200

250

300

350

0 0.2 0.4 0.6 0.8 1 1.2

Field current (A)

Arm

atur

e vo

ltage

(V

)

Figure 3-1: Magnetisation characteristic.

In this example, the relationship is almost linear up to the rated field current of 0.5 A, but there is significant non-linearity above this value, when parts of the magnetic circuit saturate. There is also a small residual flux when the field current is zero, giving a corresponding generated voltage. For the initial part of the magnetisation characteristic, it is approximately true that f if. Equations 3-1 and 3-2 then become:

rfa ùiKe (3-4)

afd iiKT (3-5)

Armature equation

Figure 3-2 shows a symbolic representation of a DC machine, with the armature connected to a voltage source va.

if

iar

va+ea

+

Figure 3-2: DC machine with voltage source.

If the armature has a resistance Ra, then Kirchhoff’s voltage law gives the armature voltage equation:

rfaaaaaa ùKiReiRv (3-6)


3.2 DC machines in practice

Slotted armature

The elementary theory of DC machines assumes that conductors are on the surface of the armature, so that the simple expressions e = Blu and f = Bli are applicable. In practice, the armature conductors are placed in slots, as shown in figure 3-3.

Figure 3-3: DC motor armature.

Figure 3-4 shows a simple model of a machine with a slotted armature, and figure 3-5 shows the corresponding flux plot when there is no armature current.

Figure 3-4: DC machine model.

Figure 3-5: DC machine field flux.

This is similar to the flux plot when conductors are on the surface, but with one important difference: most of the flux lines pass between the conductors, indicating that the flux density in the slots is very low. Consequently the force on the conductor, given by f = Bli, is very small. Most of the force is exerted on the armature iron and not on the conductors. The basic equations 3-1 and 3-2 are not affected by the location of the conductors. In particular, the generated voltage is not affected, although a direct application of e = Blu appears to contradict this. See reference [1] for a discussion of induced voltage in such situations.

Armature reaction

Current flowing in the armature conductors will also create a magnetic field in the machine, known as the armature reaction field. Figure 3-6 shows a flux plot of this field when the main field flux is absent.

Figure 3-6: Armature reaction flux plot.

DC Machines 17

Note that the axis of the armature reaction field is at right angles to the axis of the main field. When currents flow in the field and armature conductors, the two component magnetic fields combine to give a resultant field of the form shown in figure 3-6. The bending of the field lines indicates that the stator exerts a counter-clockwise torque on the rotor. Another way of picturing the electromagnetic action is to consider the magnetic poles representing the field and armature flux components, as shown in figure 3-7.

Figure 3-7: Total flux plot.

An important effect of armature reaction, which is evident in figure 3-7, is to increase the flux density at one side of a field pole and decrease it at the other side. Figure 3-8 shows a shaded plot of the flux density magnitude, where the colour range from blue to red represents the flux density range from minimum to maximum.

Figure 3-8: Flux density magnitude.

High values of flux density may result in local saturation of the steel, increasing the reluctance of the magnetic circuit, and reducing the value of the

field flux. This has important consequences for the motor characteristics – see section 3.3.

Efficiency

The efficiency of a motor is defined in the usual way:

in

out

P

P (3-7)

where Pin is total electrical power input to the motor terminals, and Pout is the mechanical power output from the motor shaft. The power lost as heat in the motor is:

outinloss PPP (3-8)

so we have the following alternative forms of equation 3-7:

lossout

loss

lossout

out

in

loss

in

lossin

in

out

PP

P

PP

P

P

P

P

PP

P

P

1

1 (3-9)

As with transformers (see section 2-5), the efficiency of a large motor is not usually determined from equation 3-7 by direct measurement of the input and output power. Instead, losses are determined from several tests, and the efficiency calculated from equation 3-9. The output power is given by

)( ldrrout TTTP (3-10)

where Tl is the rotational loss torque. It follows that the rotational power loss is rTl. The rotational loss has two components: mechanical loss, which is also known as the windage and friction loss, and core loss in the armature and the field poles resulting from the rotation of the armature. The total power loss includes in addition the I2R loss in the field and armature windings, and the brush contact loss, which results from the voltage drop between the brushes and the commutator segments. For further information, see references [3, 4].

S

S

N

N


Power output and size

The developed torque may be calculated from the equation f = Bli, even though the armature conductors are in slots. For this purpose, the machine will be represented by a simple model with the conductors on the surface, as shown in the flux plot of figure 3-9.

Figure 3-9: Model for torque calculation.

It is useful to define a current loading A as the current in amperes per metre length of circumference on the surface of the armature. The current in an element of length ds is then di = A ds. The force on this element is:

dsBlAdiBldf (3-11)

where l is the axial length of the armature. The corresponding contribution to the torque is:

dsrBlAdfrdTd (3-12)

where r is the radius of the armature. If Bav is the average value of the flux density at the armature surface, then the total torque is just:

AVBAlrBrlArBT avavavd 222. 2 (3-13)

where V is the volume of the armature. The maximum value of Bav is limited by the saturation of the magnetic material of the armature, and the value of the current loading A is limited by the I2R heating of the armature conductors. If it is assumed that the maximum value of A is independent of the machine size, equation 3-13 shows that the developed torque is proportional to the rotor volume. The treatment of the effects of scale in reference [1] also gives this result. The gross power output is given by:

AVBTP avrdr 2 (3-14)

Thus, the maximum power output of a DC motor is roughly proportional to the product of the armature volume and the armature speed. In practice, the current loading A also increases with size, so the power output of large machines is further increased. To quantify this, assume that the power dissipation per unit surface area is constant. Let d be the radial depth of a conducting layer representing the armature conductors. The resistance of an element ds of this layer is:

dsd

ldR

(3-15)

The cooling surface area of the element is dS = l ds, so the power dissipation per unit area is:

d

A

dsl

dsd

ldsA

dsl

dRdi

dS

dP 22

2 )()(

(3-16)

If dP / dS is constant, then A d, and if d r then A r. For geometrically similar machines, the axial length l is proportional to the radius r. If the rotational speed r is constant, it follows from equation 3-13 that the power output is proportional to r3.5. Thus, if the dimensions are doubled, the power output will increase by a factor of 11.3. A similar result was obtained for transformers (section 2.8), and it holds for other types of machine.

3.3 Characteristics and control

Speed control

In a large motor, the total power loss is small. Since the armature power loss Raia

2 is only a part of the total loss, this term must be small in comparison with the armature input power vaia. It follows that:

aaa viR (3-17)

So Raia may be neglected in equation 3-6, giving:

rfa ùKv (3-18)

r ds

DC Machines 19

The rotational speed is therefore:

f

ar K

vù

(3-19)

Equation 3-19 shows that the speed is independent of the torque, provided inequality 3-17 holds. Equation 3-19 is the basis of speed control. If the field flux f is constant, the rotational speed is proportional to the applied voltage. The normal values of va and f define the base speed r0, and the speed can be reduced to zero by varying va. Electronic controllers for DC motors deliver an adjustable voltage to the motor armature by phase-controlled rectification of the AC mains supply, using thyristors: see section 8. This is the basis of DC variable-speed drive systems, which are widely used in industry. In a wound-field motor, it is possible to increase the speed above the base speed by reducing f – known as field weakening. Only a limited speed increase is possible, for the following reason. The torque is related to the armature current through equation 3-2:

afd iKT [3-2]

If f is reduced, there will be a compensating increase in ia to maintain the torque, and there is a risk of exceeding the current rating of the machine.

Small motors

In small motors, with power ratings below 1 kW, inequality 3-17 does not hold, so it is not permissible to neglect the Raia term. The rotational speed then depends on the developed torque, as may be seen by substituting for ia in terms of Td in equation 3-6:

rf

f

da

rfaaa

ùKK

TR

ùKiRv

(3-20)

Thus, the speed is given by:

2)( f

da

f

ar

K

TR

K

vù

(3-21)

A graph of speed against torque is a straight line, as shown in figure 3-10. The no-load speed, which is the speed when the torque is zero, is given by:

f

ar K

vù

0 (3-22)

r

Td

slope Ra

r

Figure 3-10: Speed-torque characteristic.

Effect of armature reaction

In section 3.2, it was noted that the armature reaction field could cause local saturation of the field poles, thereby reducing the value of the field flux f. This effect increases with the armature current ia, and therefore with the developed torque Td. From equation 3-19, a decrease in f will cause the speed r to rise. Armature reaction therefore has the opposite effect to armature resistance, which causes r to fall with increasing torque load. There is an important difference, however. The effect of resistance is linear, as illustrated in figure 3-8, but the effect of armature reaction is non-linear. At low values of armature current, the uneven distribution of flux density in the field pole is insufficient to cause saturation, so there is hardly any reduction in the field flux. At high values of current, on the other hand, there may be a significant reduction. It is possible, therefore, for the speed of a motor to fall with increasing load when the armature current is low, but to increase with load when the current is high. This increase in speed can be very undesirable, leading to instability with some kinds of load. A large DC motor generally includes some form of compensation for armature reaction. A simple method of compensation is to provide a second winding on the field poles, connected in series with the armature (figure 3-11), to increase the field MMF when the armature current increases. The resulting motor is known as a compound motor [3, 4]. However, this cannot


compensate for the non-linear nature of the armature reaction effect.

Figure 3-11: DC compound motor.

A better method, which is frequently used in high-power DC drives, is to use a compensating winding [4]. This takes the form of conductors embedded in slots in the field pole faces, connected in series with the armature. These conductors carry current in the opposite direction to the armature conductors, thereby cancelling the armature reaction flux.

Compensating winding

Figure 3-12 shows a model of a DC machine with a compensating winding, figure 3-13 shows the resulting flux plot, and figure 3-14 shows a shaded plot of the flux density magnitude. These plots should be compared with figures 3-7 and 3-8 for the machine without a compensating winding.

Figure 3-12: Compensating winding model.

Figure 3-13: Total flux plot.

Figure 3-14: Flux density magnitude.

With a compensating winding, the flux distribution in each field pole is symmetrical and uniform, so the effect of the armature reaction has been cancelled in this part of the machine. The torque exerted on the armature is not affected, however, because this depends on the interaction of the armature currents and the field flux.

Starting of DC motors

When the speed is zero, the armature generated voltage is also zero. From equation 3-6, the corresponding armature current is given by:

a

aa R

vi 0 (3-23)

S

S

N

N

DC Machines 21

This is the stalled armature current, which is much larger than the normal running current. In a small motor, it is permissible to connect the armature directly to a constant-voltage supply. The armature can withstand the stalled current for a short time, and it will accelerate rapidly. As it does so, the generated voltage will rise and the current will fall to its normal value. However, a large motor must not be started in this way because the armature resistance Ra is very low, and the stalled current would be large enough to cause serious damage. An electronic controller for speed control will limit the starting current to a safe value. If this is not available, a variable resistance must be connected in series with the armature and the value progressively reduced to zero as the armature accelerates.

Transient conditions

When the voltage applied to a DC motor is changed, the speed will not change instantly because of the inertia of the rotating system. If J is the moment of inertia and TL is the mechanical load torque, Newton’s second law gives:

Ldr TT

dt

dJ

(3-24)

Mechanical loss torque is assumed included with TL. Substituting for Td in terms of ia from equation 3-2, and ia from equation 3-6, gives:

La

rfafr TR

KvK

dt

dJ

)( (3-25)

The solution of equation 3-25 gives the speed as a function of time after a change in the applied voltage. It should be noted that the derivation of equation 3-25 has neglected the inductance of the armature circuit. See reference [10] for the difference this makes to the transient behaviour. As an example, suppose that TL = 0 and the motor starts from rest with a suddenly applied voltage va = V. Equation 3-25 becomes:

f

rr

f

a

K

V

dt

d

K

JR

2)(

(3-26)

The solution of equation 3-26 is:

/0

/ 11 tr

t

fr ee

K

V (3-27)

where r0 is the no-load speed and the time constant is given by

2)( f

a

K

JR

(3-28)

The armature current can be determined by substituting for r in equation 3-6:

)1( /

t

aa

rfaa

eViR

ùKiRV

(3-29)

/0

/ ta

t

aa eie

R

Vi (3-30)

Figure 3-15 shows graphs of the normalised current ia / ia0 and the normalised speed r / r0 against the normalised time t / .

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

Time

Cur

rent

, Spe

ed

Figure 3-15: Starting performance.

Speed r / r0

Current ia / ia0

t /


3.4 Shunt and series motors

So far, it has been assumed that the field flux f is independent of the conditions in the armature. This is the case in a permanent-magnet motor, and in a wound-field motor where the field winding is supplied from a separate voltage source. The latter is sometimes termed a separately excited motor. There are two ways of introducing constraints between the armature and field of a wound-field motor. In a shunt motor, the field winding is connected in parallel with the armature. In a series motor, the field winding is designed to carry the full armature current, and it is connected in series with the armature.

Shunt motor

If the field winding is connected to the same supply voltage as the armature, the field current will be proportional to the armature voltage. Usually, a variable resistance is connected in series with the field winding to provide some measure of control for the field current [3, 4, 13]. The characteristics of a shunt motor may be deduced from equation 3-19:

f

ar K

vù

[3-19]

Provided there is a linear relationship between the field flux and the field current, we have:

f af f f

f

k vk i

R (3-31)

where Rf is the total resistance of the field circuit and kf is a constant. Substituting in equation 3-19 gives:

far

f f

RvùK Kk

(3-32)

Thus, the speed of a shunt motor is, to a first approximation, independent of the applied voltage.

Series motor

Figure 3-16 shows the connection of a series motor to a voltage source. The characteristics are readily deduced if the following assumptions are made:

The field flux is proportional to the current.

The resistance of the windings is negligible.

v

iea

+

r+

Figure 3-16: Series motor.

Equations 3-4 and 3-5 become:

2iKTd (3-33)

ra iùKev (3-34)

Eliminating i gives the relationship between speed and torque:

dd

rTK

v

K

TK

v

iK

vù

(3-35)

A graph plotted from equation 3-33 for a small motor is shown in figure 3-17.

0

50

100

150

200

0 5 10 15 20 25

Developed torque, Nm

Spe

ed, r

ad/s

Figure 3-17: Speed/torque characteristic.

DC Machines 23

Two features of the graph should be noted:

The motor can develop a very large torque at low speeds.

When the developed torque is low, the speed is very high.

The high torque at low speeds is a useful feature, but the high speed associated with low torque can be dangerous. Consider what happens if the external load torque is removed. The developed torque then just supplies the mechanical losses in the motor. In a small motor, the mechanical losses are usually high enough to limit the no-load speed to a safe value. However, in motors with power ratings above about 1 kW, the losses are proportionately smaller, and the no-load speed would be dangerously high. Such motors must never be run without a load.

Effect of saturation

From equations 3-2 and 3-5, the torque developed by a series motor is given by:

2iKiKT fd (3-36)

This equation is valid only when the current is low enough to avoid saturation of the magnetic circuit, so that the flux is proportional to current. When the current is very high, the field flux will tend towards a constant value fm, and the torque is then given by

iKT fmd (3-37)

Under these conditions, the torque is proportional to i instead of i2. The performance of motors such as car starter motors must therefore be measured under actual operating conditions, where equation 3-36 would give very inaccurate results.

Applications

A correctly designed series motor will work on AC as well as DC [1, 3, 4]. Such motors are termed universal. They are widely used in products such as portable power drills, and in domestic appliances such as food processors and vacuum cleaners. In applications such as power tools, the speed/torque characteristic of the series motor is advantageous. An important reason for their use is the high speeds that are possible with series motors: speeds of up to 10 000 rev/min (1050 rad/s) are common in small motors. It was shown in section 3.2 that the maximum power output of a motor is roughly proportional to the product of the rotational speed and the armature (or rotor) volume. Thus, a small motor can deliver a large amount of power if the speed is high. Induction motors (see section 6) are limited to speeds below 3000 rev/min (314 rad/s) when operated from the 50 Hz mains supply. The power per unit volume for a series motor is therefore about three times the value for an induction motor. Disadvantages include the higher manufacturing cost, and the need for maintenance of the commutator and brushes.


4 INTRODUCTION TO AC MACHINES

4.1 Review of 3-phase systems

Industrial AC motors use 3-phase alternating current to generate a rotating magnetic field from stationary windings. The 3-phase supply may be taken from the AC mains, or it may be generated electronically with an inverter. In either case, the requirement is a symmetrical set of sinusoidal currents with relative phase displacements of 120º. Formally, 3-phase sets of currents and voltages may be defined as follows in the time domain:

)120cos(

)240cos(

)120cos(

)cos(

tI

tIi

tIi

tIi

m

mc

mb

ma

(4-1)

)120cos(

)240cos(

)120cos(

)cos(

tV

tVv

tVv

tVv

m

mc

mb

ma

(4-2)

The quantities in equations 4-1 and 4-2 correspond to a positive phase sequence, in which the quantities reach their maximum values in the sequence a b c. If the connections to any two phases are interchanged, the effect is to reverse the phase sequence. For example, if b and c are interchanged, and primes denote the new phase labels, we have:

)120cos(

)120cos(

)cos(

tVvv

tVvv

tVvv

mbc

mcb

maa

(4-3)

These quantities reach their maximum values in the sequence c b a, corresponding to a negative phase sequence. Figure 4-1 shows the star connection of a 3-phase AC source to three lines, where the time-varying quantities of equation 4-1 are represented by complex (phasor) quantities in the usual way.

Va

Vb

Vc

1

2

3

+

++

V12

+

Figure 4-1: 3-phase star connection.

The corresponding voltage phasor diagram is shown in figure 4-2.

Va

Vb

Vc

V12

V23

V31

Figure 4-2: Star connection: voltage phasors.

Figure 4-3 shows the delta connection of a 3-phase AC source to three lines, and figure 4-4 shows the corresponding current phasor diagram.

1

2

3Ic

Ia

Ib

Figure 4-3: 3-phase delta connection.

Ia

Ib

Ic

I1

I2

I3

Figure 4-4: Delta connection: current phasors.

Introduction to AC machines 25

From these diagrams, we have the relationships between the magnitudes of line and phase quantities given in table 4-1:

Table 4-1

Star Delta

phaseline VV 3 phaseline VV

phaseline II phaseline II 3

4.2 Rotating magnetic field

In AC motors, the stationary part is termed the stator, and the rotating part is the rotor. Figure 4-5 shows the stator of a small AC motor with a 3-phase winding.

Figure 4-5: 3-phase AC motor winding.

Coils are arranged in slots in a laminated steel core, rather like the armature coils in a DC motor. The function of the winding is to route currents to slots from the three phases, as shown in the simplified diagram of figure 4-6. This diagram also shows a steel cylinder in the centre, representing the rotor of the motor.

Figure 4-6: 3-phase conductor groups.

The groups labelled a, b, c carry currents in the positive direction, into the plane of the diagram, and the groups labelled a', b', c' carry currents in the negative direction. All coils have the same shape, with one coil side at the bottom of a slot and the other side at the top of a slot, as shown by the pair of circles in figure 4-6. If phase a is energised on its own with positive current, the resulting magnetic field pattern is shown in figure 4-7(a). The corresponding patterns for phases b and c are shown in figures 4-7(b) and 4-7(c).

(a)

(b)

(c)

Figure 4-7: AC machine flux plots:

(a) phase a, (b) phase b, (c) phase c.

a

a'

c

c'

b

b'


When all three phases are energised, the individual phase fields combine to give a resultant field. Consider the instant when t = 0. From equation 4-1, the currents in the three phases are:

mcmbma IiIiIi 21

21 ,,

Since the currents in phases b and c are negative, the component fields are in the directions shown in figure 4-8. They combine to give a resultant in the same direction as for phase a, so a flux plot of the resultant field is the same as for phase a alone. As time advances, the currents in the phases change, and the resultant pattern changes. Figure 4-9 shows the patterns when t = 30º, 60º and 90º.

Figure 4-8: Component fields when t = 0.

(a)

(b)

(c)

Figure 4-9: Flux plots for values of t:

(a) 30º, (b) 60º, (c) 90º.

Notice that the magnetic field pattern has rotated by the same angle as the time phase of the currents. We have a rotating magnetic field, which makes one revolution for each cycle of the alternating current. This field is capable of exerting a torque on a suitably designed rotor. Two kinds of AC motor exploit the rotating field effect. In synchronous motors, the rotor has magnetised poles, which lock in with the rotating field, so that the rotor moves in synchronism with the field. With induction motors, the rotating field induces currents in conductors on the rotor, so the rotor must move more slowly than the rotating field.

a

c

b


Sinusoidal fields

The rotating magnetic field will induce voltages in the phase windings. These voltages should be as nearly sinusoidal as possible, which implies that the flux density should vary sinusoidally with angular position. This in turn requires that the currents producing the field should be distributed sinusoidally. A sinusoidal distribution is also required if the resultant field is to rotate at a uniform speed. The conductor distribution for each phase shown in figure 4-6 approximates to a sinusoidal distribution. It is a better winding than it appears, because harmonics that are multiples of 3 are missing from the Fourier series for the waveform. This is one of the many advantages of using three phases. Practical windings generally use more slots than this simple model, and give a better approach to the ideal sinusoidal distribution. Assume that the three phases produce component fields in the airgap with magnitudes Ba, Bb and Bc as follows:

)120cos(

)120cos(

cos

cc

bb

aa

kiB

kiB

kiB

(4-4)

where k is a constant and is the angle from a reference axis, as shown in figure 4-10 for the component field Ba.

Figure 4-10: Angular position definition.

The total field is the sum of the component fields. If the currents are given by equation 4-1, then substituting in equation 4-4 gives the result:

)cos(

)cos(23

tB

tkIBBBB

m

mcba

(4-5)

Equation 4-5 represents a sinusoidal field with a constant maximum value, rotating in a counter-clockwise direction with a constant angular velocity r, illustrated in figure 4-10. The axis of the field makes an angle t with the horizontal axis. At an angle , the displacement from the field axis is – t, so the magnitude of the flux density is Bm cos( – t).

Figure 4-11: Rotating magnetic field.

The rotating field represented by equation 4-5 will induce a sinusoidal voltage in any coil on the stator. Consider a single-turn coil on the stator, with one side at = 0 and the other side at = 180º, as shown in figure 4-12.

Figure 4-12: Single-turn coil.

If the direction of the flux density in the airgap is assumed radial, it may be shown that the magnetic flux through this coil is:

tlrBm sin2 (4-6)

where r is the radius and l is the axial length. The induced voltage in the coil is thus:

tlrBdt

de m

cos2 (4-7)

See references [3, 4] for further information about sinusoidal fields.

t


Speed of the rotating field

At any instant, there are two effective magnetic poles for the field shown in figure 4-8, so this is termed a 2-pole field. It makes one revolution in one AC cycle, giving the relationships between the field and the current shown in table 4-2.

Table 4-2: 2-pole field properties.

Current waveform 2-pole field Frequency f [Hz] Rotational speed ns = f [rev/s] Rotational speed Ns = 60f [rev/min]

Angular frequency Angular velocity s = [rad/s]

As an example, if f = 50 Hz, ns = 50 rev/s, Ns = 3000 rev/min and s = 2f 314 rad/s

4.3 Multi-pole fields

Coils in an AC machine winding can be arranged to produce fields with more than two poles. For example, figure 4-13 shows the simple winding of figure 4-6 re-arranged to give a 4-pole field, with the resulting field pattern shown in figure 4-14 at the instant when t = 0.

Figure 4-13: 4-pole winding.

Figure 4-14: 4-pole flux plot.

For every 30º advance in the time phase of the currents, the field pattern rotates 15º. In one cycle of the supply, the field pattern moves through two pole pitches – an angle corresponding to two poles. With any number of poles, the field moves through two pole pitches in one cycle of the supply. If there are p pairs of poles, this corresponds to a fraction 1 / p of a revolution. Thus, the general expressions for the speed of the rotating field are:

[rad/s] 2

p

f

ps (4-8)

[rev/s] p

fns (4-9)

[rev/min] 60

p

fN s (4-10)

Table 4-3 gives the field speeds at 50 Hz for different numbers of poles.

Table 4-3: Field speeds at 50 Hz.

Poles p s ns Ns

rad/s rev/s rev/min

2 1 314 50 3000

4 2 157 25 1500

6 3 105 16.7 1000

8 4 78.5 12.5 750

10 5 62.8 10 600

a

a'

c

c'

b

b'

b'

b

c

a

c'

a'


Reversing the direction of rotation

To reverse the direction of rotation, all that is required is to interchange the connections to any two of the three phases. For example, suppose that the connections to phases b and c are interchanged. Figure 4-15(a) shows the original current pattern and figure 4-15(b) shows the new pattern.

(a)

(b)

Figure 4-15: Interchange of connections: (a) original, (b) b and c interchanged.

The new pattern is a mirror image of the old, so the resultant magnetic field will progress in the opposite direction as time advances. It follows that the direction of rotation depends on the phase sequence of the 3-phase supply, since interchanging any pair of phases has the effect of reversing the phase sequence (see section 4-1). Therefore, it is important to know the phase sequence of the supply before connecting the motor.

a

a'

c

c'

b

b'

b'

b

c

a

c'

a'

a

a'

b

b'

c

c'

c'

c

b

a

b'

a'


5 SYNCHRONOUS MACHINES

5.1 Introduction

In a synchronous machine, the rotor is magnetised and it runs at the same speed as the rotating magnetic field. Permanent-magnet rotors are common in small machines, so the machine structure is similar to that of the brushless DC motor shown in figure 3-4. Figure 5-1 is a simplified model of the structure of this type of machine, where the rotor has surface-mounted segments of permanent-magnet material. Other forms of rotor with embedded magnets are also possible.

Figure 5-1: 2-pole PM synchronous machine.

In larger sizes, a synchronous machine has a field winding on the rotor instead of permanent magnets. Direct current for the rotor excitation can be supplied through sliprings and brushes, but large machines normally have a brushless excitation system [3, 4]. The AC stator of a synchronous machine is termed the armature. It handles the main electrical power, so it has a similar function to the rotating armature of a DC machine.

Motors and generators

As with DC machines, there is no fundamental difference between a synchronous motor and a synchronous generator. In a motor, the magnetic axis of the rotating magnetic field is ahead of the magnetic axis of the rotor, resulting in a positive torque that depends on the displacement between

the two axes. The flux plots in figure 5-2 show (a) the field produced by stator currents at a particular instant of time, (b) the field produced by the rotor magnets, (c) the resultant field when both sources are active.

(a)

(b)

(c)

Figure 5-2: Synchronous motor flux plots: (a) stator field, (b) rotor field, (c) resultant field.

Synchronous Machines 31

In this example, the displacement angle is 90º, which gives the maximum torque for a given current. The current has been chosen so that the axis of the resultant field is at approximately 45º to the magnetic axis of the rotor. In a synchronous generator, the displacement is reversed: the magnetic axis of the rotor is ahead of the magnetic axis of the rotating field, so the torque is negative. This is illustrated in figure 5-3.

(a)

(b)

(c)

Figure 5-3: Synchronous generator flux plots: (a) stator field, (b) rotor field, (c) resultant field.

Most of the AC generators in electric power systems are synchronous machines. High-speed turbine generators normally have two poles. The rotor is made from a cylindrical steel forging, with the field winding embedded in slots machined in the steel. Apart from the slots for conductors, the active surfaces of the stator and rotor are cylindrical, so these are uniform-airgap or non-salient machines. Low-speed hydro generators have many poles. These are salient-pole machines, where the poles radiate like spokes from a central hub. Synchronous generators will not be considered in detail in this course, but will be studied in the context of power systems in year 3. Large synchronous motors are widely used as high-efficiency constant-speed industrial drives, where they can also be used for plant power factor correction: see section 5.2. These are normally salient-pole machines, and they generally have four or more poles. The main features of synchronous machine operation can be deduced from the simple theory of non-salient machines. An introduction to salient-pole machines is given in section 5.3.

5.2 Characteristics

Circuit model

A non-salient synchronous machine can be represented by a simple equivalent circuit [3, 4], shown in figure 5-4.

jXsRaI

V

+

E+

V

+

Figure 5-4: Equivalent circuit. This circuit represents one phase of a 3-phase machine. The voltage V is the phase voltage at the machine terminals, and the current I is the corresponding phase current. Other elements in the circuit have the following significance.


The voltage E is termed the excitation voltage. It represents the voltage induced in one phase by the rotation of the magnetised rotor, so it corresponds to the magnetic field of the rotor shown in figure 5-2(b) or 5-3(b).

The reactance Xs is termed the synchronous reactance. It represents the magnetic field of the stator current in the following way: the voltage jXsI is the voltage induced in one phase by the stator current. This voltage corresponds to the magnetic field of the stator shown in figure 5-2(a) or 5-3(a).

The voltage V = E + jXsI represents the voltage induced in one phase by the total magnetic field shown in figure 5-2(c) or 5-3(c).

The resistance Ra is the resistance of one phase of the stator, or armature, winding.

The resistance Ra is usually small in comparison with the reactance Xs, so it may be neglected in most calculations from the equivalent circuit. The resulting approximate equivalent circuit is shown in figure 5-5, described by the equation:

EIV sjX (5-1)

jXsI

V

+E

+

Figure 5-5: Approximate equivalent circuit.

Phasor diagram

Figure 5-6 shows a phasor diagram corresponding to the equivalent circuit of figure 5-5, for the machine operating as a generator. The terminal voltage V lags the excitation voltage E by an angle . This is known as the load angle, because it varies with the torque load. In terms of the magnetic field, represents the angle between the axis of the rotor field in figure 5-3(b) and the axis of the resultant field in figure 5-3(c). The phase angle is greater than 90º, so the power given by VI cos is negative, indicating a flow of electrical power out of the machine.

I

E

jXsI

V

I

Figure 5-6: Phasor diagram: generator.

In power system studies, it is customary to reverse the reference direction for the current, so the current is then represented by the phasor I in figure 5-6. For operation as a motor, V leads E by an angle , as shown in figure 5-7. The phase angle is now less than 90º, indicating a flow of electrical power into the machine.

I

EjXsI

V

M

N

Figure 5-7: Phasor diagram: motor.

Torque characteristic

Consider the line MN in figure 5-7, which is perpendicular to V. The length of MN can be expressed in terms of E and from the left-hand triangle, and in terms of XsI and from the right-hand triangle. We have:

cossin IXEMN s (5-2)

Multiplying both sides of equation 5-2 by the voltage magnitude V and dividing by Xs gives:

cos

sinVI

X

VE

s

(5-3)


The right-hand side of equation 5-3 is the electrical power input to one phase of the motor. Since energy is conserved, and there are no electrical losses – the resistance has been neglected – the total electrical power input must equal the work done by the rotating magnetic field. If s is the synchronous speed in rad/s and Td is the developed torque, then:

sds

TVIX

VE cos3

sin3 (5-4)

The developed torque is thus given by:

[Nm] sin3

ssd X

VET

(5-5)

This has a maximum value when = 90º, given by:

ss

d X

VET

3

max (5-6)

Equation 5-5 may therefore be written in normalised form:

sinmax

d

d

T

T (5-7)

A graph of this normalised torque is shown in figure 5-8. Positive values of represent motor operation; negative values represent generator operation.

-1

-0.5

0

0.5

1

-180 -90 0 90 180

Load angle, degrees

Nor

mal

ised

tor

que

motor

generator

Figure 5-8: Synchronous machine torque.

If the mechanical load on a synchronous motor exceeds Tdmax, the rotor will be pulled out of synchronism with the rotating field, and the motor will stall. Tdmax is therefore known as the pullout torque. The torque characteristic shown in figure 5-8 has an important practical consequence. If the rotor

loses synchronism with the rotating field, the load angle will change continuously. The torque will be alternately positive and negative, with a mean value of zero. Synchronous motors are therefore not inherently self-starting. Induction machines, considered in section 6, do not have this limitation, and the induction principle is generally used for starting synchronous motors.

Excitation control

In a wound-field synchronous machine, the excitation voltage E can be controlled by varying the excitation current in the rotor field winding. This has consequences for the machine characteristics, as will now be shown for the case of a synchronous motor. If the torque load on a synchronous motor is held constant, the power output will be constant, and if losses are neglected, there will be a constant electrical power input per phase given by:

cosVIP (5-8)

In normal operation the terminal voltage V is constant, so the quantity I cos must be constant. In figure 5-9, the line AB is drawn parallel to the imaginary axis, and its distance from the axis is I cos. This line is the locus of the current phasor I.

The length of the line MN is XsI cos, which is also constant, so the line CD, parallel to the real axis, is the locus of the excitation voltage phasor E.

I

EjXsI

V

M

N

A

B

C D

Figure 5-9: Locus diagram for constant load.


If the value of E is progressively increased, by increasing the current in the rotor field winding, the point M will move towards D, and the angle will decrease. Figure 5-10 shows the condition where = 0. This is the unity power factor condition, which gives the minimum value for I. Let E0 be the value of E that gives this condition, and 0 the corresponding value of .

I

E0

jXsI

V

M

N

A

B

C D

Figure 5-10: Unity power factor condition.

When E < E0, is negative, and > 0. The machine takes a lagging current, as shown in figure 5-9. In this condition, the machine is said to be under-excited. As the value of E is progressively reduced, the values of I and increase, until the limit of stability is reached when = 90º. This condition is shown in figure 5-11.

I

E jXsI

V

A

B

C D

Figure 5-11: Stability limit.

When E > E0, is positive, and < 0. The machine now takes a leading current, as shown in figure 5-12. In this condition, the machine is over-excited. If continuous operation at a leading power factor is required, as in power factor correction (discussed below), then the machine must be

specially designed for the higher values of rotor and stator current.

I

E

jXsI

V

M

N

A

B

C D

Figure 5-12: Leading power factor condition.

If the mechanical load is removed from an over-excited synchronous motor, then = 0, and the phasor diagram takes the form shown in figure 5-13. The phase angle is now 90º, so the machine behaves as a 3-phase capacitor. In this condition, it is known as a synchronous compensator. The magnitude of the current, and hence the effective value of the capacitance, depends on the difference between E and V.

I

EjXsIV

Figure 5-13: Synchronous compensator.

Power factor correction

The ability of a synchronous motor to operate at a leading power factor is extremely useful. It will be shown in section 7 that induction motors always operate at a lagging power factor. Many industrial processes use large numbers of induction motors, with the result that the total load current is lagging. It is possible to compensate for this by installing an over-excited synchronous motor. This may be used to drive a large load such as an air compressor, or it may be used without a load as a synchronous compensator purely for power factor correction.


When a synchronous motor is used in this way, the rotor excitation is normally controlled automatically to keep the total current nearly in phase with the voltage, so that the plant as a whole operates at a power factor close to unity. It may not be economic to give full correction, if this would require a very large synchronous machine. Figure 5-14 shows a phasor diagram for the general case of power factor correction. The synchronous machine current Is combines with the load current Il from the rest of the plant to give a total current It which is more nearly in phase with the voltage V.

Is

V

Il

It

Figure 5-14: Power factor correction.

The relationships between the currents can be expressed in terms of power, as follows. We have:

slt III (5-9)

The complex power is given by:

slslsltt SSVIVIIIVVIS **** )( (5-10)

Equating the real and imaginary parts of equation 5-10 gives:

slt PPP (5-11)

slt QQQ (5-12)

For an over-excited synchronous machine, Qs will be negative, so equation 5-12 shows that the overall reactive power Qt will be less than the reactive power Ql of the original plant. For complete power factor correction, we require Qt = 0, and therefore Qs = –Ql. If the synchronous machine is run as a synchronous compensator, without a mechanical load, then Ps 0, and the real power demand of the plant is unchanged.

5.3 Salient-pole machines

The theory given in section 5.2 is only valid for non-salient machines. This section gives a brief introduction to salient-pole machines; for further information, see references [3, 4]. When the rotor has salient poles, the airgap is not uniform, and there is a preferred direction for magnetic flux in the machine. Figure 5-15 shows a simple model of a salient-pole machine, where part of the rotor steel has been removed to accommodate the field winding.

Figure 5-15: Salient-pole machine model.

The rotor has a path of easy magnetisation, known as the direct-axis or d-axis, where the reluctance is low. This is also the axis of the magnetic flux produced by the rotor field current. An axis at right angles, known as the quadrature-axis or q-axis, has a higher reluctance because of the larger airgap. Figure 5-16 shows the armature flux in a non-salient machine produced by a particular set of currents in the armature (stator). The flux axis is at 45º to the rotor d-axis.

Figure 5-16: Non-salient armature flux.

d-axis

q-axis


Figure 5-17 is a flux plot for same current pattern in the salient-pole model. Saliency has the effect of moving the axis of the armature flux towards the d-axis.

Figure 5-17: Salient-pole armature flux.

Voltage harmonics

In a non-salient machine, the flux distribution shown in figure 5-16 is approximately sinusoidal (see section 4.2). In a salient-pole machine, however, figure 5-17 shows that the flux density distribution is non-sinusoidal. It follows that the voltages induced in the armature phases will also be non-sinusoidal. The voltages may be represented by Fourier series:

...3cos)120cos(

...)120(3cos)120cos(

...3cos)120cos(

...)120(3cos)120cos(

...5cos3coscos

31

31

31

31

531

tVtV

tVtVv

tVtV

tVtVv

tVtVtVv

mm

mmc

mm

mmb

mmma

(5-13)

If the windings are connected in star (figure 4-1), the third harmonics will cancel. For example, the voltage between lines 1 and 2 is:

...)120cos(cos

...3cos)120cos(

...3coscos

11

31

31

12

tVtV

tVtV

tVtV

vvv

mm

mm

mm

ba

(5-14)

However, if the windings are connected in delta (figure 4-2), the third harmonic voltages will accumulate round the delta:

...3cos3

...3cos)120cos(

...3cos)120cos(

...3coscos

3

31

31

31

tV

tVtV

tVtV

tVtV

vvvv

m

mm

mm

mm

cba

(5-15)

This can result in a large third-harmonic circulating current. Similar considerations apply to higher harmonics where the harmonic number is a multiple of 3; these are known as triplen harmonics. Therefore, it is normal practice to connect the windings of a synchronous machine in star. It is possible to eliminate any harmonic from the voltage waveform by using short-pitched coils in the stator winding. Short-pitched coils span less than a pole pitch; the coils shown in figure 4-6 are of this form. It may be shown that, if the coil is short-pitched by a fraction 1/n of the pole pitch, then the nth harmonic will be eliminated from the voltage waveform. For example, in a 2-pole machine, if the coils span 120º instead of 180º, so that they are short-pitched by 1/3, the third harmonic will be eliminated. In this particular case, it is permissible to connect the windings in delta. It is more usual, however, to short-pitch by 1/6, as shown in figure 4-6, since this reduces the fifth and seventh harmonics to low levels. If the windings are star connected, all the triplen harmonics are eliminated, so the resulting voltage waveform is a good approximation to a sine wave.

Salient-pole machine theory

It is possible to analyse the salient-pole machine by resolving the armature flux into two components: a component acting along the rotor d-axis, and a component acting along the q-axis [3, 4]. This is represented in the phasor diagram by resolving the current I into components Id and Iq. The component Id acts on the d-axis, where the low reluctance gives a correspondingly high reactance Xd. The component Iq acts on the q-axis, where the


high reluctance gives a low reactance Xq. The resulting phasor diagram is shown in figure 5-18, and the phasor equation is:

qqdda jXjXR IIIEV (5-16)

RaI

E

V

Iq

jXqIq

IIdjXdId

Figure 5-18: Salient-pole phasor diagram.

If the armature resistance Ra is neglected, the torque developed by a salient-pole synchronous machine is given by [4]:

(b) (a)

2sin11

2

sin3 2

dqdsd XX

V

X

VET

(5-17)

Term (a) in this equation represents the normal synchronous torque, which may be identified with equation 5-4. Term (b) represents a component of torque due to the saliency of the rotor. This component is termed the reluctance torque, which vanishes when the rotor is cylindrical, for Xd is then equal to Xq. See section 7.1 for a discussion of reluctance torque. Figure 5-19 shows graphs of the two torque components and the resultant torque.

-0.25

0

0.25

0.5

0.75

1

1.25

0 30 60 90 120 150 180

Load angle, degrees

Nor

mal

ised

torq

ue (c)

(a)

(b)

Figure 5-19: Salient-pole machine torque: (a) synchronous, (b) reluctance, (c) resultant

Reluctance motors

Since the reluctance torque is independent of the field current, this torque component does not require the presence of a field winding. It is therefore possible to make a form of synchronous motor without a field winding. This is termed a reluctance motor, or a synchronous reluctance motor. With no field winding, equation 5-8 becomes:

2sin12

3

2sin11

2

3

2

2

q

d

d

dqd

X

X

X

V

XX

VT

(5-18)

from which it follows that the ratio Xd / Xq should be large for a good design. Say [5] describes several different design techniques that have been used to give high values of q-axis reluctance – and therefore low values of Xq – to maximise the reluctance torque. Reluctance motors have many of the advantages of induction motors, with the added property that the speed is locked to the frequency of the AC supply. However, they are prone to instability under certain operating conditions, particularly at low frequencies when operated from variable-frequency supplies for speed control [12]. 5.4 Linear synchronous motors

Many applications require motion in a straight line rather than rotary motion, for example in automation systems. Linear motors provide this. In concept, a linear motor is a rotary motor opened out flat. Figure 5-20 shows the linear counterpart of the rotating field model of figure 4-6, and figure 5-21(a)–(e) show the field patterns when t = 0º, 30º, 60º, 90º and 120º respectively.

Figure 5-20: Travelling field model.

(a)


(b)

(c)

(d)

Figure 5-21: Flux plots for travelling field: (a) 0º, (b) 30º, (c) 60º, (d) 90º.

Figure 5-22 shows a linear synchronous motor derived from the rotary model of figure 5-1. The lower part with the windings is termed the primary, and the upper part with the permanent magnets is termed the secondary. Figure 5-23 shows the corresponding flux plots for (a) the field produced by primary currents at a particular instant of time, (b) the field produced by the secondary magnets, (c) the resultant field when both sources are active.

Figure 5-22: Linear synchronous motor.

(a)

(b)

(c)

Figure 5-23: Linear motor flux plots: (a) primary, (b) secondary, (c) resultant field.

A tubular linear motor can be derived from the structure in figure 5-22 by ‘rolling it up’ around the longitudinal axis. Figure 5-24 shows an industrial motor of this kind. Tubular motors have a particularly simple structure, since the primary windings are circular coils that encircle the secondary magnets.

Figure 5-24: Tubular linear synchronous motor. (Copley Motion Systems)

In addition to applications in industry, linear synchronous motors are used in advanced ground transport systems. The German Transrapid system, for example, which was recently installed in Shanghai, employs controlled electromagnets for levitation and linear synchronous motors for propulsion.

Induction Machines 39

6 INDUCTION MACHINES

6.1 Introduction

Induction motors exploit the rotating magnetic field in quite a different way from synchronous motors. Instead of a magnetised rotor, an induction motor has a rotor with short-circuited conductors. The motion of the rotating field induces currents in the conductors, which in turn interact with the field to develop a torque on the rotor. Figure 6-1 shows a typical rotor for a small induction motor, where conductors in the form of aluminium bars are placed in slots in the laminated steel core. At each end, the bars are connected to short-circuiting aluminium rings known as end-rings. In this kind of rotor, the bars and end-rings are made in a single operation by die-casting.

Figure 6-1: Induction motor cage rotor.

Currents will be induced in the rotor conductors if there is a speed difference between the rotor and the rotating magnetic field. In a motor, the rotor runs more slowly than the rotating field, and the resulting currents produce a torque that tends to accelerate the rotor. Unlike synchronous motors, induction motors are inherently self-starting. The rotor will accelerate from rest until it is running just below the synchronous speed, which is the speed of the rotating field. Figure 6-2 is a simplified model of the structure of a 2-pole induction motor with 12 slots in the rotor. In a practical motor, the airgap between the stator and the rotor is made as small as possible, but in the model the airgap has been exaggerated for clarity. The stator rotating magnetic field induces currents in the rotor conductors, which effectively form a 6-phase winding. These currents in turn produce a 2-pole magnetic field that rotates in space at exactly the

same speed as the stator field. The flux plots in figure 6-3 show (a) the field produced by stator currents at a particular instant of time, (b) the field produced by the rotor current, (c) the resultant field when both sources are active.

Figure 6-2: 2-pole induction motor model.

(a)

(b)

(c)

Figure 6-3: Induction motor flux plots: (a) stator field, (b) rotor field, (c) resultant field.


The magnetic field produced by the rotor current in an induction motor is very similar to the corresponding field in a synchronous motor, but the underlying mechanism is completely different. Consequently, the motor characteristics are also completely different. If the rotor of an induction motor is driven mechanically so that it runs faster than the rotating field, the direction of power flow reverses and machine functions as a generator. Induction generators are not widely used, but they have found application in wind-powered generators. Induction motors are simple and robust, and their self-starting capability is a particular advantage. At least 90 per cent of industrial drives are induction motors. A typical small motor is shown in figure 6-4.

Figure 6-4: Small induction motor. (Brook Crompton)

6.2 Characteristics

An essential feature of induction motors is the speed difference between the rotor and the rotating magnetic field, which is known as slip. There must be some slip for currents to be induced in the rotor conductors, and the current magnitude increases with the slip. It follows that the developed torque varies with the slip, and therefore with the rotor speed.

Slip speed and fractional slip

The slip speed is given by:

rssliprsslip NNN , (6-1)

It is useful to define a quantity s, known as the fractional slip, as follows:

s

rs

s

rs

N

NNs

speed ssynchronou

speed slip (6-2)

It follows that the slip varies in the opposite way to the rotor speed. When the rotor is running at the synchronous speed, so that Nr = Ns, the fractional slip is s = 0. When the rotor is stationary, so that Nr = 0, the fractional slip is s = 1. From equation 6-2, the slip speed and the rotor speed in terms of s are given by:

srsslip s (6-3)

sssr ss )1( (6-4)

The frequency of the rotor currents is proportional to the slip speed, and therefore proportional to s. When the rotor is stationary, s = 1, and the rotor frequency must be equal to the stator supply frequency, so we have the important result:

sr sff (6-5)

where fr is the frequency of rotor currents and fs is the stator supply frequency.

Rotor power relationships

The rotating magnetic field exerts a torque Td on the rotor, and does work at the rate sTd. This represents an input of power to the rotor. The rotor revolves at a speed r, and therefore does work at the rate rTd, which represents the mechanical output power of the rotor. The difference between these two powers represents power lost in the resistance of the rotor. As in the DC motor, there will be mechanical losses in the motor, so the shaft torque T is less than Td. We have the following rotor power relationships:

Rotor electromagnetic input: dsem TP (6-6)

Rotor output: dsdrmech TsTP )1( (6-7)

Rotor loss: dsdrsloss TsTP )( (6-8)


Thus, a fraction (1 – s) of the rotor electromagnetic input power is converted into mechanical power, and a fraction s is lost has heat in the rotor conductors. The quantity (1 – s) is termed the rotor efficiency. Since there are other losses in the motor, the overall efficiency must be less than the rotor efficiency. For high efficiency, the fractional slip s should be as small as possible. In large motors, with power ratings of 100 kW or more, the value of s at full load is about 2%. For small motors, with power ratings below about 10 kW, the corresponding value is about 5%. It follows that the full-load speed of induction motor is close to the synchronous speed.

Induction motor model

When the rotor is stationary, the induction motor behaves as a 3-phase transformer with a short-circuited secondary. The equivalent-circuit model of the transformer in figure 2-14 applies to this condition. When the rotor moves, the voltage induced in the rotor will depend on the relative motion. It may be shown [3, 4] that this can be represented by a simple change to the equivalent circuit: the secondary resistance is not constant, but depends on the fractional slip s. The circuit for one phase takes the form shown in figure 6-5.

Rs

sRr

jxs jxr

jXm RcVs

+

Is Ir

I0

Figure 6-5: Induction motor equivalent circuit.

This is the conventional form of the equivalent circuit, first introduced by Steinmetz [6]. The parameters have the same significance as in a transformer: Rs is the stator winding resistance. xs is the stator leakage reactance, representing

stator flux that fails to link with the rotor. Rr is the rotor resistance referred to the stator. xr is the rotor leakage reactance referred to the

stator. Rc represents core loss, mainly in the stator. Xm is the magnetising reactance.

With a transformer, it is possible to simplify the equivalent circuit by moving the shunt elements to the input terminals – see section 2.3. This is a poor approximation with an induction motor, however, because the magnetising reactance Xm is much smaller in comparison with the leakage reactances xs and xr. The reason for this is the presence of an airgap between the stator and the rotor, which increases the reluctance of the magnetic circuit. Consequently, the no-load current I0 in an induction motor is much larger than the no-load current in a transformer of similar rating.

Performance calculation

The developed torque is obtained by equating the power absorbed in the resistance Rr / s to the rotor input power from equation 6-6, giving the result:

[Nm] 3 2

s

RIT r

rs

d (6-9)

It is necessary to solve the equations of the equivalent circuit (figure 6-5) for the currents Is and Ir, and hence determine the torque from equation 6-9. This process is simplified by transforming the equivalent circuit to the form shown in figure 6-6. Here, the parallel combination of Xm and Rc has been replaced by the series combination of xm and rc.

Rs

sRr

jxs jxr

jxm

rc

Vs

+

Is Ir

I0

Figure 6-6: Modified equivalent circuit.

The series elements in figure 6-6 are related to the parallel elements in figure 6-5 by the following equations:

cmc

mc R

XR

Xr

22

2

(6-10)

mmc

cm X

XR

Rx

22

2

(6-11)


The corresponding inverse relationships are:

c

mcc r

xrR

22 (6-12)

m

mcm x

xrX

22 (6-13)

If the core loss is small, so that Rc >> Xm, then rc << xm, and we have the following approximate relationships:

mm Xx (6-14)

c

mc R

Xr

2

(6-15)

c

mc r

xR

2

(6-16)

The value of rc depends on Xm, and therefore on the frequency. When the speed of an induction motor is controlled by varying the frequency (see section 8-4), the resistance Rc is approximately constant. Under these conditions, rc is proportional to the square of the frequency, so the modified equivalent circuit is less useful. In this section, however, the frequency is assumed constant, so the modified circuit will be used. Values of the stator and rotor currents are easily determined by first defining impedances as follows:

sss jxR Z (6-17)

rr

r jxs

RZ (6-18)

mcm jxr Z (6-19)

The parallel combination of the magnetising branch and the rotor branch is:

rm

rmp ZZ

ZZZ

(6-20)

so the stator current is:

ps

ss ZZ

VI

(6-21)

and the rotor current is:

r

spr Z

IZI (6-22)

Worked example 6-1

A 4-pole 3 kW star-connected induction motor operates from a 50 Hz supply with a line voltage of 400 V. The equivalent-circuit parameters per phase are as follows:

Rs = 2.27 , Rr = 2.28 , xs = xr = 2.83 , Xm = 74.8 , rc = 3.95 .

If the full-load slip is 5%, determine: (a) the no-load current, (b) the full-load stator current, (c) the full-load rotor current, (d) the full-load speed in rev/min, (e) the full-load developed torque.

Solution

The phase voltage is:

V 2313

400

3 line

sV

V

The impedances are:

83.227.2 jjxR sssZ

83.26.4583.205.0

28.2jjjx

s

Rr

rrZ

8.7495.3 jxr mcmZ

3.201.31

)83.26.45()8.7495.3(

)83.26.45)(8.7495.3(

j

jj

jj

rm

rmp ZZ

ZZZ

(a) The no-load current is:

A 97.29.77

231

6.7722.6

231

)8.7495.3()83.227.2(

2310

j

j

VI

ms

s

ZZ


(b) The full-load stator current is:

A 69.523.368.4

A 23.368.41.234.33

231

)3.201.31()83.227.2(

jI

jj

jj

VV

ss

s

ps

ss

I

ZZI

(c) The full-load rotor current is:

A 62.47.45

211

83.26.45

89.5211

83.26.45

89.5211

83.26.45

)23.368.4)(3.201.31(

j

jI

j

j

j

jj

rr

r

spr

I

Z

IZI

(d) The synchronous speed is:

rev/min 15002

506060

p

fN s

so the full-load speed is:

rev/min 1425

1500)05.01()1(

sr NsN

(e) The full-load developed torque is:

Nm 6.1805.0

2

50228.2)62.4(33 22

s

RIT

s

rrd

Typical characteristics

Figure 6-7 shows the torque/speed characteristic for the 3 kW motor used in the worked example.

010203040506070

0 300 600 900 1200

1500

Rotor speed, rev/min

Dev

elop

ed to

rque

, Nm

breakdown

full load

starting

Figure 6-7: Torque/speed characteristic.

The maximum torque is known as the breakdown torque. If a mechanical load torque greater than this is applied to the motor, it will stall. Also shown in figure 6-7 is the full-load operating point, and the starting torque when the speed is zero. The torque is zero at the synchronous speed of 1500 rev/min. When the rotor is stationary, the fractional slip is s = 1. When the rotor is running at the synchronous speed, the fractional slip is s = 0. Since the frequency of the rotor currents is sfs where fs is the supply frequency (equation 6-6), the frequency varies from 50 Hz when the rotor is stationary to 0 at the synchronous speed. At the full-load slip of 5%, the frequency is 0.0550 = 2.5 Hz. Figure 6-8 shows the input current as a function of slip for the same motor. Observe that the starting current, when the rotor is stationary, is about six times the full-load current of 5.69 A. This ratio of starting current to full-load current is a typical figure for induction motors.

0

5

10

15

20

25

30

35

0 300 600 900 1200 1500


Sta

tor

phas

e cu

rren

t, A

full load

Figure 6-8: Current/speed characteristic.

Induction motor starting

Most induction motors are started by connecting them straight to the AC mains supply. This is known as direct on line starting, and as figure 6-8 shows, it will result in a large starting current. Direct on line starting may be unacceptable, either because the supply system cannot support such a large current, or because the transient torque (see section 6.6) could damage the mechanical system. A simple method of reducing the starting current is star-delta starting. With this method, the motor is designed for delta connection in normal running, but for starting, the windings are connected in star.


This reduces the starting current by a factor of 3, with a corresponding reduction in the torque. After the motor has reached its full speed, the windings are re-connected in delta. This method has been superseded by electronic soft starting (see section 8.4), where the applied voltage is gradually increased.

Motor driving a load

When a motor drives a mechanical load, the operating point is given by the intersection of the motor torque/speed characteristic with the corresponding load characteristic, as shown in figure 6-9.

0

10

20

30

40

50

60

70

0 300 600 900 1200 1500


Tor

que,

Nm

load

motor

Figure 6-9: Motor driving a load.

Extended torque characteristic

If the rotor of an induction motor is driven mechanically in the reverse direction, the developed torque opposes the motion, so the machine acts as a brake. Alternatively, if the rotor is driven in the forward direction faster than the rotating field, the torque will reverse, and the machine acts as a generator. Figure 6-10 shows the extended torque/speed characteristic that includes these conditions.

-120

-90

-60

-30

0

30

60

-1500 0 1500 3000


Dev

elop

ed to

rque

, Nm

motorregion

generatorregion

brakeregion

Figure 6-10: Extended torque characteristic.

Brake region of the characteristic

The rotor speed is negative in the brake region, so we may put r = –|r |. From equation 6-2, the fractional slip is given by:

s

rs

s

rss

(6-23)

so s > 1. The gross mechanical power output given by equation 6-7 is therefore negative:

dsmech TsP )1( [6-7]

A negative mechanical output signifies an input of mechanical power to the rotor:

dsin TsP )1( (6-24)

There is also an electromagnetic power input from the stator, given by equation 6-6:

dsem TP [6-6]

The power lost in the rotor resistance, given by equation 6-8, is very large:

dsloss TsP [6-8]

This is equal to the sum of the mechanical power input (equation 6-24) and the electromagnetic power input (equation 6-6). The machine can operate in the brake mode only for very short periods, or the rotor will overheat. This mode is sometimes used for stopping an induction motor rapidly by reversing the phase sequence of the supply, which merely requires interchanging two of the connections to


the motor terminals. The process is called plug braking or plugging, illustrated in figure 6-11.

-60

-30

0

30

60

-1500 -750 0 750 1500


Dev

elop

ed to

rque

, Nm motor

curve

brakecurve

A

B

C

Figure 6-11: Induction motor plug braking.

Suppose that the machine is operating at point A on the blue motor curve when plug braking is initiated. The direction of the rotating field reverses immediately, giving a new torque/speed characteristic shown as the red brake curve in figure 6-11. Since the rotor speed cannot change instantaneously, the operating point moves to B on the brake curve, vertically below A. The resulting negative torque will decelerate the rotor until it reaches zero speed at C. At this point the supply must be disconnected promptly, otherwise the rotor will continue to accelerate in the reverse direction. A better method of stopping the motor is by DC dynamic braking, discussed in section 6.6.

Generating region of the characteristic

In the generating region of figure 6-10, the machine torque is negative, so we may put Td = –|Td|. The fractional slip is also negative: s = –|s|, and the resistance Rr / s is negative in the rotor branch of the equivalent circuit. It follows that the electromagnetic and mechanical power values, given by equations 6-6 and 6-7, both become negative, but the power loss (equation 6-8) remains positive:

dsdsem TTP (6-25)

dsdsmech TsTsP )1()1( (6-26)

dsdsloss TsTsP (6-27)

Thus the direction of power flow has reversed, and the rotor efficiency for generator operation is:

sP

P

mech

emr

1

1 (6-28)

It follows that an induction machine can operate as a high-efficiency generator if the slip s is small. This is the same condition for high efficiency in a motor. An induction machine must be connected to a suitable AC source for it to function as a generator, since a source of reactive power is required for the magnetising current that creates the rotating magnetic field. If it is disconnected from any AC source, an induction machine will not generate. Usually induction generators are connected to an AC supply system, but it is also possible to make induction machines run as self-excited generators by connecting capacitors to the terminals in a 3-phase configuration [5]. Each capacitor effectively forms a resonant circuit with the magnetising reactance, thereby acting as an AC source to supply the magnetising current.

Effect of parameter values

The stator resistance Rs is responsible for the lack of symmetry between the motoring and generating regions of the torque curve in figure 6-10. Figure 6-12 shows the result of setting Rs to zero.

-90

-60

-30

0

30

60

90

0 750 1500 2250 3000


Dev

elop

ed to

rque

, Nm

Figure 6-12: Torque characteristic: Rs = 0.

In large machines, the stator resistance is a smaller part of the total impedance, so the characteristic is more symmetrical.


Figure 6-13 shows the effect on the motor torque of halving the values of the leakage reactances xs and xr. This has little effect on the characteristic near to the synchronous speed, but it increases the starting torque and the breakdown torque significantly.

0102030405060708090

0 300 600 900 1200 1500


Dev

elop

ed to

rque

, Nm

(a)

(b)

Figure 6-13: Torque/speed characteristics: (a) original, (b) leakage reactance halved.

Figure 6-14 shows the effect of halving the rotor resistance Rr. This has the effect of compressing the characteristic towards the synchronous speed end, because the torque is a function of Rr / s, so the value of s must change in the same ratio as Rr to give the same torque. Thus the value of the breakdown torque is unchanged, but the slip at which it occurs is halved.

0

10

20

30

40

50

60

70

0 300 600 900 1200 1500


Dev

elop

ed to

rque

, Nm

(a) (b)

Figure 6-14: Torque/speed characteristics:

(a) original, (b) rotor resistance halved.

With this change in Rs, the full-load slip is approximately halved, and therefore the full-load rotor loss is halved. The penalty for this increased efficiency is that the starting torque is much lower. For operation from a fixed-frequency supply, the design of an induction motor is a compromise. A low value of Rs is desirable to give high efficiency, but this also results in a low value of starting torque. An ingenious solution to this conflict between efficiency and starting torque is to make the rotor with very deep slots, or even with two concentric rotor cage windings [4, 5, 6]. These designs exploit the fact that the frequency of the rotor currents is sfs (equation 6-5), which is very low when the rotor is running normally, but is equal to the supply frequency when the rotor is stationary. Skin effect in AC conductors [3, 4] causes current to flow near the surface at high frequencies, with a consequent increase in the effective resistance. The special rotor windings have a low resistance at the full-load slip frequency, when the current is uniformly distributed, but a high resistance when the slip frequency is high during starting. When the motor is supplied from a variable-frequency inverter (see section 8.4), the starting condition is handled automatically without the need for a high-resistance rotor, so a low resistance can be used for optimum efficiency.

6.3 Losses and efficiency

The efficiency of an induction motor is of great importance to the user. It is defined in the normal way as the ratio of useful mechanical output power Pout to the total electrical input power Pin:

in

out

P

P (6-29)

The difference between Pin and Pout is the total motor power loss Ploss. We therefore have alternative expressions for efficiency in terms of the power loss:

lossout

out

in

lossin

PP

P

P

PP

(6-30)


Table 6-1 shows typical efficiency values for 4-pole induction motors ranging from 0.25 kW to 250 kW from one manufacturer, and the corresponding power loss as a percentage of the input power.

Table 6-1: Induction motor efficiency

Output power Efficiency Power loss 0.25 kW 66% 34% 2.2 kW 77% 23% 22 kW 91% 9% 250 kW 95% 5%

The importance of efficiency to the user can be demonstrated from the following calculation. Suppose that a 250 kW motor runs continuously for 1 year, and that electrical energy costs 7p per kWh.

Input power: 250 / 0.95 = 263 kW

Power loss: 0.05 263 = 13.2 kW

Running cost: 263 24 365 0.07 = £161 000

Cost of loss: 13.2 24 365 0.07 = £8050

If the efficiency could be improved by a mere 0.1%, the cost saving per year would be £162. Users comparing different motors need to know the efficiency with high accuracy.

Loss components

The motor losses are considered to have five components as follows:

(1) Stator I2R loss: ssRs RIP2

3 (6-31)

(2) Rotor I2R loss: rrRr RIP2

3 (6-32)

(3) Core loss: cmcore rIP2

3 (6-33)

(4) Friction and windage loss: fwP

(5) Stray load loss: strayP

The total loss is the sum of items 1 to 5. Core loss is the eddy-current and hysteresis loss in the magnetic core of the machine, mostly in the stator, which is represented by the resistance rc. Friction and windage loss is the total mechanical power loss within the motor, from bearing friction and aerodynamic drag on the rotor. Stray load loss is an additional loss under load, which is not included in

the other four categories. It may be attributed to departures from a purely sinusoidal winding distribution, and to effects of the stator and rotor slot openings on the magnetic field distribution in the machine [4, 5, 6]. The efficiency of an induction motor can be determined by the following methods.

Direct determination: measure Pin and Pout, and use equation 6-29. This is the preferred method for small machines, below about 10 kW, but it is difficult to measure the powers with sufficient accuracy in large high-efficiency machines.

Calorimetric: measure Pin and Ploss, and use equation 6-30. The difficulty with this method is accurate measurement of the loss from the heat dissipated by the motor, but it can be an effective method for large machines with closed cooling systems.

Loss segregation: measure Pin, and calculate Ploss from its components. This is the most widely used method.

Loss segregation method

The loss segregation method of determining efficiency is given in several national and international test specifications [7, 8]. It requires the following steps to determine all the losses under specified load conditions (usually full load). 1. Determine the stator I2R loss PRs from equation

6-31, using the measured values of the stator phase resistance Rs and phase current Is.

2. Determine the core loss Pcore and the friction and windage loss Pfw from a no-load test (see below).

3. Determine the rotor I2R loss PRr as follows, based on equation 6-32, using measured values of the input power Pin and the slip s:

emr

rrrRr sPs

RIsRIP 22 ||3||3 (6-34)

coreRsinem PPPP (6-35)

where Pem is the electromagnetic power transferred to the rotor.

4. Determine a value for the stray load loss Pstray. This is the least satisfactory part of the procedure, because of the uncertainty in this quantity. As yet, there appears to be no satisfactory test method for stray load loss. The


American standard [8] specifies a special test, known as the reverse-rotation test, but this has been shown to give inaccurate results in many cases. The IEC standard [7] avoids this difficulty by assuming that Pstray is 0.5% of the input power Pin, but this is too low for small motors, and it fails to penalise badly designed motors with a high value for Pstray.

No-load test

In the no-load test, the induction motor is run without any mechanical load attached to the shaft. The input power then just supplies the losses in the motor, and the slip is negligibly small. Under these conditions, the rotor current is very small, so the rotor I2R loss is negligible, and we have:

fwcoreRsin PPPP (6-36)

By definition, the stray load loss is not included in the no-load test. Stray no-load losses, if they are significant, are considered included in Pcore and Pfw. Re-arranging equation 6-36 gives

fwcoreRsinin PPPPP (6-37)

If the stator terminal voltage Vs is reduced, the rotor speed will be virtually unchanged because the slip remains small, so Pfw will remain constant. Since Pcore varies approximately as Vs

2, a graph of P'in against |Vs|

2 will approximate to a straight line as shown in figure 6-15. Extrapolating this line to zero volts gives the value of Pfw, and the core loss is then given by

fwRsincore PPPP (6-38)

|Vs|

P'in

Pfw

Figure 6-15: No-load test graph.

6.4 Parameter determination

The parameters of the equivalent circuit in figure 6-6 are usually determined from three tests:

A DC measurement of the stator phase resistance

A no-load test, essentially as described above for efficiency determination

A locked-rotor (or blocked-rotor) test, where the rotor is prevented from revolving

These tests resemble the open-circuit and short-circuit tests for determining the equivalent-circuit parameters of the transformer.

DC resistance test

In small or medium-sized machines, up to about 100 kW, a DC measurement of the primary winding phase resistance will give an accurate value for Rs in the equivalent circuit. If the machine is star connected, then a measurement between any pair of line terminals will give 2Rs. If it is delta connected, the corresponding measurement will give 2Rs/3. The measurement should be made for all three pairs of terminals and the average taken. In large machines, above about 100 kW, the value of Rs at the normal operating frequency may be significantly larger than the DC value because of skin effect in the stator conductors.

No-load test continued

Since the rotor current is small in the no-load test, the rotor branch may be removed from the equivalent circuit of figure 6-6, giving the simplified form shown in figure 6-16.

Rsjxs

jxm

rc

Vs

+

Is

Figure 6-16: No-load equivalent circuit.


This circuit will give accurate results if the input power is corrected by subtracting the friction and windage loss power Pfw, since this power is almost equal to the power in the rotor branch. We then have:

22 )(|| cssfwin rRIPP (6-39)

where Pin is the input power per phase. Since the stator phase resistance Rs is known, equation 6-39 gives the value of the core loss resistance rc. The magnetising reactance xm is obtained from the terminal voltage and current as follows:

22 )()(|| mscss

s xxrRZI

V (6-40)

This requires an estimate of the stator leakage reactance xs, which can be obtained from the locked-rotor test described below.

Locked-rotor (blocked rotor) test

If the rotor is locked (or blocked) to prevent rotation, then r = 0 and s = 1. The impedance of the rotor branch is low, and to a first approximation the magnetising branch may be ignored, giving the approximate equivalent circuit of figure 6-17:

Rsjxs

Vs

+

Isjxr

Rr

Figure 6-17: Locked-rotor equivalent circuit.

Only a low voltage is required to give the normal full-load value of Is. As in the no-load test, the resistance and reactance values are determined as follows from the measured values of voltage, current and power:

22 )(|| rssin RRIP (6-41)

22 )()(|| rsrss

s xxRRZI

V (6-42)

This test can only determine the sum of the leakage reactances, xs + xr, so it is necessary to make an assumption about the ratio xs / xr. For most purposes, it is sufficient to take this ratio to be 1, but sometimes a different value is used based on motor design values. In practice the ratio is unimportant, because it may be shown that the performance calculated from the equivalent circuit is not affected by the ratio, if the parameters are determined by test, using the same ratio. Therefore the simplest assumption is generally used, namely xs = xr. A problem with the locked-rotor test is that the rotor resistance Rr varies with the frequency of the secondary current, due to skin effect. The practical consequence is that the test should not be carried out at the normal supply frequency, except in very small (fractional kilowatt) motors. The two test codes already cited [7, 8] specify a frequency of not more than 25% of the normal supply frequency. The value of xs determined from equation 6-42 should be used in equation 6-40 for the magnetising reactance. A more accurate calculation of the parameters is possible by using the full equivalent circuit of figure 6-6 for each test, and solving numerically for the unknowns.

Worked example 6-2

Tests on a 3 kW 50 Hz star-connected cage induction motor gave the following results:

DC stator resistance between lines: 4.54 .

No-load test: line current = 2.97 A, line voltage = 400 V, input power = 214 W, friction and windage loss = 50 W.

Locked-rotor test at 12.5 Hz: line current = 5.70 A, line voltage = 44.5 V, input power = 422 W.

Load test: line current = 5.69 A, line voltage = 400 V, input power = 3.29 kW, fractional slip = 5.0%.

Determine: (a) the parameters of the equivalent circuit, (b) the efficiency of the motor at full load, using a

stray-load loss allowance of 0.5%.

Solution

(a) Equivalent circuit parameters.


DC resistance test:

27.22

54.4

2DC

sR

R

Locked-rotor test:

V 7.253

5.44

3 line

sV

V

W 1413

422

3 lr

inP

P

33.4)70.5(

14122

lr

inlr

I

PR

06.227.233.4slrr RRR

25.1

)33.4(70.5

7.25 22

22

lrlr

slr R

I

Vx

51.25.12

50

2

25.1

20

lr

lrr f

fxx

No-load test:

V 2313

400

3 line

sV

V

W 7.543

50214

3

fwnl

in

PPP

20.6)97.2(

7.5422

nl

innl

I

PR

93.327.220.6snlc RRr

5.77

)20.6(97.2

231 22

22

nlnl

snl R

I

VX

0.7551.25.77snlc xXx

(b) Efficiency.

(1) W 22127.2)70.5(33 22 ssRs RIP

(2) W 10427.297.2350214

3 2

snlfwnlcore RIPPP

W 29651042213290

coreRsflem PPPP

(3) W 148296505.0 emRr sPP

(4) W 163290005.0005.0 flstray PP

(5) W 5401650104148221

strayfwcoreRrRsloss PPPPPP

The efficiency is thus:

%59.833290

5403290

fl

lossfl

P

PP

6.5 Single-phase induction motors

Single-phase operation of a 3-phase motor

When one line is disconnected from a 3-phase induction motor, the result is a single-phase system. If the machine is a star-connected, as shown in figure 6-18, two of the three phases are connected in series to the single-phase supply, and the third phase carries no current.

sasb

sc

Vs+

Is

Figure 6-18: Single-phase connection.


At standstill, the currents in the stator and the rotor will be sinusoidal alternating quantities if the supply voltage is sinusoidal. The result is a magnetic field that pulsates instead of rotating, and the rotor remains stationary. However, a pulsating field can be regarded as the resultant of two contra-rotating fields: one positive and the other negative, as shown by the red and blue phasors in figure 6-19.

Figure 6-19: Contra-rotating fields.

When the rotor is stationary, the two rotating field components will produce equal and opposite torques. If the rotor is made to revolve by any means, the two torque components are no longer balanced, and there is a net torque tending to accelerate the rotor. It can be shown [9] that the motor in figure 6-18 may be represented by the equivalent circuit of figure 6-20.

Rsjxr

jxm

rc

Vs

+

Is

Irn

s

Rr

2

s

Rr

jxm

rc

jxs

Rs jxrjxs

Irp

Figure 6-20: Single-phase equivalent circuit.

The rotor current has two components: a positive-sequence component Irp, associated with the forward rotating field, and a negative-sequence component Irn, associated with the backward rotating field. Relative to the forward component, the fractional slip s is given by equation 6-2:

s

rs

N

NNs

[6-2]

Relative to the backward component, the fractional slip is:

ssN

N

N

NNs

s

r

s

rs

2)1(11 (6-43)

These two current components give corresponding positive and negative torque components, so the developed torque is:

s

RI

s

RIT r

rns

rrp

sd

2

11 22

(6-44)

When the rotor is stationary, s = 1 = 2 – s, so the two torque components cancel out. If the rotor revolves in the positive direction, the positive torque term predominates; if it revolves in the negative direction, the negative term predominates. Figure 6-21 shows the two torque terms and the resultant torque for the 3 kW motor used in previous examples.

-30

-20

-10

0

10

20

30

-1500 -750 0 750 1500

Speed, rev/min

Tor

que,

Nm

Positive sequence

Negative sequence

Total torque

Figure 6-21: Single-phase torque characteristic.


Losses

The I2R loss in the stator is given by:

22 ssRs IRP (6-45)

If skin effect in the rotor is negligible for the negative-sequence component of rotor current, the I2R loss in the rotor is given by:

)( 22rnrprRr IIRP (6-46)

Danger of single-phase operation

If one stator line is accidentally disconnected from a 3-phase motor, it will continue to run as a single-phase motor. The motor slip will increase so that the developed torque continues to match the load torque, but the stator and rotor currents will be abnormally high. Table 6-2 shows the computed full-load performance of the example 3 kW motor, (a) with a normal 3-phase supply, (b) with one line disconnected.

Table 6-2: 3-phase and 1-phase operation.

3-phase 1-phase Full-load slip 0.05 0.082 Stator current magnitude 5.56 A 11.6 A Rotor current magnitude 4.63 A 10.5 A +ve,

11.2 A –ve Stator I2R loss 211 W 611 W Rotor I2R loss 147 W 538 W Total I2R loss 358 W 1150 W

This large increase in the I2R loss will cause the motor to overheat very rapidly, resulting in serious damage unless it is disconnected from the supply. For this reason, it is essential for motor control gear to detect single-phase operation and disconnect the supply immediately.

Single-phase induction motors

For domestic and light industrial applications, induction motors need to operate from a single-phase supply. Instead of a 3-phase winding, many single-phase motors have a 2-phase winding and a capacitor to give a phase shift of approximately 90º between the currents, as shown in figure 6-22 [3, 4].

Vs+

Figure 6-22: Single-phase induction motor.

The phases of a 2-phase winding are displaced in space by 90º (for a 2-pole machine), corresponding to the 90º time-phase displacement between the currents. As with a 3-phase winding, it may be shown [3, 4] that a 2-phase winding with 2-phase currents can produce a rotating magnetic field. Once the motor has started, it is possible to disconnect the capacitor, and the motor will continue to run with a single winding, but the performance is not as good. The shaded-pole principle [3, 4] is used in very small single-phase induction motors such as the one shown in figure 6-23. They are widely used for driving small cooling fans.

Figure 6-23: Shaded-pole induction motor. (RS Components Ltd)

This type of motor has a single winding on the stator, energising a pair of poles. A part of each pole is enclosed by a ring of copper, known as a shading coil. Currents are induced in the rings, giving the effect of a rudimentary 2-phase winding.


6.6 Dynamic conditions

Dynamic model

The equivalent-circuit model of the induction motor introduced in section 6.2 is valid only for steady-state conditions, where the induction machine runs at a constant speed. There are two common situations where this does not give an accurate representation of the machine performance: Starting, when the rotor is stationary and the

AC supply is suddenly switched on. Sudden stopping, either by plugging or by DC

dynamic braking (discussed below). For both of these conditions, a dynamic model is required. The 2-axis theory of the induction motor [4, 9, 10] treats the machine as a system of coupled circuits described by differential equations. This model is valid for all conditions, transient as well as steady state. Although the theory is beyond the scope of this course, it is useful to consider some of the results from this model obtained by numerical solution of the differential equations. The equations of the 2-axis model are given in Appendix 10.1.

Starting transient

The torque/speed characteristics in section 6.2 show that the steady-state starting current is about six times the full-load current in the example 3 kW motor. This is typical of the ratio of currents in most induction motors. When the motor is suddenly switched on to the AC supply, the initial current may be very high for a few cycles, and it is accompanied by a large oscillatory component of torque. Figure 6-24 shows the stator phase current and the developed torque when the motor is switched on. The nature of the transient response will depend on the inertia of the rotating system. Here, the motor has a moment of inertia of 0.011 kg m2, and it is assumed that the load has an inertia of four times this value, giving a total of 0.055 kg m2.

The value of the peak current at switch-on depends on the point in the AC cycle when the switch is closed, and it will be different for each of the three phases. Figure 6-24(a) shows the worst case, where the peak current is a maximum. The starting torque in figure 6-24(b) contains a large oscillatory component at the supply frequency of 50 Hz. This may have undesirable effects on a mechanical system coupled to the motor shaft. Electronic control, described in section 8.4, can reduce this effect.

-50

-25

0

25

50

0 100 200 300

Time, msC

urre

nt, A

(a)

-30

0

30

60

90

120

0 100 200 300

Time, ms

Tor

que,

Nm

(b)

Figure 6-24: Induction motor starting: (a) phase current, (b) torque.


DC braking

An induction motor can be stopped rapidly by disconnecting the AC supply from the stator and connecting a DC supply. The resulting stationary magnetic field induces currents in the rotor that oppose the rotation – a process termed DC braking or DC injection braking. A common implementation with a star-connected induction motor is to link two line terminals together, and connect the DC source between this point and the third terminal, as shown in figure 6-25.

V+

Figure 6-25: DC braking connection.

When the speed of an induction motor is controlled by an inverter (see section 9), DC braking can be achieved without altering the connections to the motor. Figure 6-26 shows the rotor speed and the developed torque for the example 3 kW motor when DC braking is initiated from the full-load speed of 1425 rev/min. The DC source voltage is 120 V, and total moment of inertia is 0.011 kg

m2.

-500

0

500

1000

1500

0 200 400

Time, ms

Spe

ed, r

ev/m

in

(a)

-60

-40

-20

0

20

0 200 400

Time, ms

Tor

que,

Nm

(b)

Figure 6-26: Induction motor DC braking: (a) rotor speed, (b) torque.

Observe that the negative torque is maintained for a short time after the rotor reaches zero speed, because the rotor inductance forces current to continue flowing, resulting in a brief speed reversal. This in turn causes the rotor current to change sign, and the torque goes positive, bringing the rotor to rest.

6.7 Linear induction motors

As with synchronous motors, a linear form of the induction motor can be derived by ‘unrolling’ a rotary motor, so the travelling-field model of figure 5-20 is applicable. Figure 6-27 shows the primary of a small linear induction motor.

Figure 6-27: Linear induction motor primary.


Figure 6-28 shows models of two forms of linear induction motor. The primary has a 3-phase winding, as in the linear synchronous motor, and the secondary comprises a conducting plate on a steel backing. The plate, which is normally made from a good conductor such as copper or aluminium, takes the place of the rotor conductors of a rotary machine, and the steel backing provides the flux return path.

(a)

(b)

Figure 6-28: Linear induction motor models: (a) short primary, (b) short secondary.

A model of the active part of a linear induction motor is shown in figure 6-29.

Figure 6-29: Linear induction motor model.

Figure 6-30(a)–(e) show plots of the magnetic flux in the motor and the currents in the conductors when t = 0º, 30º, 60º, 90º and 120º respectively. The current magnitude is represented by a colour that ranges from blue for minimum to red for maximum.

(a)

(b)

(c)

(d)

Figure 6-30: Linear induction motor flux and current plots: (a) 0º, (b) 30º, (c) 60º, (d) 90º.

Like rotary induction motors, linear induction motors are robust and self-starting. They have been applied in airport baggage handling systems and theme park rides, and in advanced transport systems such as the magnetically levitated vehicle shown in figure 6-31.

Figure 6-31: Maglev vehicle with linear induction motor propulsion. (HSST Corporation)

The behaviour of a linear induction motor can differ significantly from that of an equivalent rotary motor [3, 4]. Linear motors usually have large airgaps, so the magnetising current is larger than in a rotary motor. In addition, a linear motor has two ends that have no counterpart in a rotary motor. In a short-primary machine, transient currents can be generated as secondary conducting material enters the active region at one end and leaves it at the other end. These transient currents can reduce the force on the secondary and increase the losses, particularly at high speeds. Short-secondary machines do not suffer from this problem, because the secondary conducting material is always in the active region.


7 STEPPER MOTORS

7.1 Introduction

The conventional AC and DC machines are designed for continuous rotation of a shaft. It is often desirable to control the speed, but most applications do not require precise control of the angular position. Specialised applications such as automation systems do require this kind of control. Often it is achieved with a feedback control system incorporating a position sensor, a drive motor, and a controller. The position measured by the sensor is compared with the desired value, and the difference used to control the motor until the error is acceptably small. AC and DC motors used in this way are termed servomotors, but they will not be considered further in this course. Another way of achieving position control is to use a special type of motor known as a stepping motor or stepper motor. These motors are not designed primarily for continuous rotation, although they can work in this way if required. Instead, they are designed to turn the output shaft through a precise angle whenever the currents in the windings are switched in a particular way by an electronic driver. Each switching action results in one increment or step in the rotor position. There are three basic types of stepper motor: simple permanent-magnet, variable-reluctance, and hybrid. Simple permanent-magnet stepper motors are similar to permanent-magnet synchronous motors, except that the stator currents are switched instead of varying continuously. They are no longer widely used, and they will not be considered further. Variable-reluctance stepper motors are simple applications of the variable-reluctance principle introduced in section 7.2. They do not contain magnets, and they are easy to control, but they have relatively large step angles. A variant of the variable-reluctance stepper motor is the switched reluctance motor, which is designed for continuous rotation.

Hybrid stepper motors make use of permanent magnets to enhance the reluctance effect. They can be made with very small step angles – typically 1.8º – and they can develop large values of torque. Although they require a more complex type of electronic driver, hybrid stepper motors are the most popular variety.

7.2 Variable-reluctance principle

Consider an electromagnetic system of any kind, energised by a coil carrying a current i. If the displacement of any part of the system causes a change in the reluctance R of the magnetic circuit, the inductance L will also change. In the Electromechanics course notes [1] it is shown that the x-component of force on that part is:

xx

Lif x

R2

212

21 (7-1)

where is the magnetic flux through the coil. Similarly, if rotation through an angle causes a change in the reluctance and the inductance, then the corresponding torque is:

R2

212

21 L

iT (7-2)

Equations 7-1 and 7-2 can be very useful for determining forces and torques in devices where it is possible to calculate the inductance or the reluctance. These equations also explain the principle of a variety of practical devices that develop a force or a torque because the reluctance of the magnetic circuit can change.

Alignment torque

Figure 7-1 shows a simple variable-reluctance actuator, where a short steel bar can rotate between the poles of an electromagnet.

Figure 7-1: Variable-reluctance actuator.

Stepper Motors 57

When the electromagnet is energised, there is a torque on the bar, acting in a direction that would rotate it into alignment with the poles, as can be seen from the flux plot in figure 7-2.

Figure 7-2: Flux plot for actuator.

Figure 7-3 shows a graph of the alignment torque as a function of the angular position of the rotor, computed from the field solution. Note that the torque opposes the rotation, so a positive displacement results in a negative torque.

-5

-2.5

0

2.5

5

-50 -25 0 25 50

Angle, degrees

Tor

que,

Nm

Figure 7-3: Alignment torque.

The maximum torque can be calculated from equation 7-2 as follows. Let r be the mean radius of the airgap, lg the radial length of the airgap, the angle of overlap between the stator and the rotor, d the depth of the device perpendicular to the plane of the diagram, and B the flux density in the airgap. If it is assumed that the flux is entirely confined to the overlapping portion of the stator and the rotor, then the flux is:

BdrBA (7-3)

The reluctance of each airgap is:

drì

l

Aìg

00

l

Rg (7-4)

From equation 7-2, the torque is:

0

2

20

22

1

0

22

122

1

2)(

2)(

drBl

dr

lBdr

dr

lBdrT

gg

g

R

(7-5)

Note that the torque is independent of the angle . This implies that the torque will be the same for any angle of displacement from the aligned position, whereas figure 7-3 shows that the torque is considerably reduced as the displacement approaches zero. The difference arises from the simplifying assumption that the flux is entirely confined to the overlap region.

Worked example 7-1

The actuator in figure 7-1 has an airgap length of 5 mm, a mean airgap radius of 42.5 mm, and a depth of 100 mm. The coil has 1000 turns and carries a current of 5.0 A. If the reluctance of the steel can be neglected, determine: (a) the value of the flux density in the airgap, (b) the maximum alignment torque.

Solution

From Ampère’s circuital law, we have:

T 628.0100.52

0.51000104

2

3

7

00

gl

NiHB

From equation 7-11, the maximum torque is:

Nm 68.6

104

)628.0(105.421000.57

29

0

2

drBl

T g

Comment

The maximum torque from the graph in figure 7-3 is only 4.72 Nm. Equation 7-11 over-estimates the torque because it is based on the assumption that the flux is confined to the overlap region. The flux plot in figure 7-2 shows this to be a very rough approximation. If the airgap is made much smaller in comparison with the other dimensions, equation 7-5 gives a more accurate result.


7.3 Variable-reluctance stepper motors

Figure 7-4 shows a 3D model of a simple variable-reluctance stepper motor with a 4-pole rotor and a 6-pole stator.

Figure 7-4: Stepper motor model.

Figure 7-5 shows a cross-section of the model. Coils on opposite pairs of poles are connected in series to form one phase, so this is a 3-phase motor.

Figure 7-5: Model cross-section.

Figure 7-6(a) shows the flux plot when phase a is energised, so that the rotor is held in alignment with phase a. When current is switched from phase a to phase b, the resulting flux plot is shown in figure 7-6(b). Magnetic forces act on the misaligned poles, and if the rotor is free to move it will take up a new position of alignment with phase b as shown in figure 7-6(c).

(a)

(b)

(c)

Figure 7-6: Stepper motor flux plots: (a) phase a energised, (b) phase b energised,

(c) new rotor position.

Note that the field pattern has rotated 60º clockwise, but the rotor has moved 30º counter-clockwise. There is a similar effect when current is switched from phase b to phase c, and then from phase c to phase a. Each switching transition causes the rotor to turn through the step angle of 30º. Variable-reluctance stepper motors are easy to control, since the current in each phase merely has to be turned on and off with semiconductor switches such as MOSFETs [3, 4]. Hybrid stepper motors (section 7-4) require more complex drive circuits because the current has to be reversed.

a

a'

c

c'

b

b'

Stepper Motors 59

Torque production

Equation 7-5 for the alignment torque is applicable to the variable-reluctance stepper motor:

0

2

drBl

T g [7-5]

This is the maximum torque when the rotor and stator poles are misaligned. The torque will vary with the displacement angle, as shown in figure 7-3. Equation 7-5 may not be an accurate estimate of the maximum value – see worked example 7-1 – but it shows that the torque depends on B2. If the steel parts of the magnetic circuit are unsaturated, B will be proportional to the coil current, and the torque therefore varies as i2. In practice, there is usually significant saturation of the steel parts, so the torque cannot be calculated by elementary methods.

Switched reluctance motors

Although variable-reluctance stepper motors are not very widely used, the same principle is exploited in a type of motor known as a switched reluctance motor [3, 4]. These are designed for continuous rotation, with the electronic driver controlled by a position sensor on the motor shaft instead of using externally generated pulses. Switched reluctance motors are very simple in structure, and can be competitive with inverter-controlled induction motors in some applications. They suffer from the disadvantages of torque pulsation at low speeds, and noise. Switched reluctance motors are beyond the scope of this course but are covered in the Year 3 course Electrical Machine Drives.

7.4 Hybrid stepper motors

The hybrid stepper motor is the most important and widely used type. It is essentially a variable-reluctance motor with the addition of a permanent magnet to give a higher torque for a given current. Figure 7-7 shows the rotor and stator of a typical commercial hybrid stepper motor. Figure 7-8 shows the structure of a simple model of a 2-phase motor with a 10º step angle. The rotor comprises two toothed discs joined by a cylindrical permanent magnet, which gives the discs opposite magnetic polarities. The stator has four poles, each bridging the two rotor discs. Cross-sections of the rotor discs and stator poles are shown in figure 7-9. The phases are energised according to the pattern in table 7-1 for successive steps

Table 7-1: Motor phase currents.

phase A +I 0 –I 0 +I

phase B 0 +I 0 –I 0

As with the variable-reluctance motor, one phase at a time is energised, but the current in each phase has to change direction.

Figure 7-7: Hybrid stepper motor


(a)

(b)

Figure 7-8: Hybrid stepper motor model: (a) complete, (b) frame and two coils removed.

Figure 7-9 shows the situation when phase A is energised. Stator pole A1 has S polarity, so it aligns with the N rotor disc; stator pole A2 has N polarity, so it aligns with the S rotor disc.

(a)

(b)

Figure 7-9: Phase A energised: (a) N polarity end, (b) S polarity end.

Consider the situation when phase A is switched off and phase B is switched on. Poles B1 and B2 are symmetrically positioned with respect to the rotor teeth, but the flux densities in the two gaps are very different. Pole B1 has S polarity, so the flux density in the gap between this pole and the rotor N disc is high. Since pole B2 has N polarity, the flux density in this gap is low. The converse holds for the S rotor disc. The result is that the rotor is pulled into a new alignment position shown in figure 7-10. It is not possible to show 2D flux plots for the hybrid stepper motor because the field is essentially 3D. Magnetic flux passes along the axis from the permanent magnet to one rotor disc, crosses the gap to the stator poles, then along the poles to the other rotor disc, and returns to the other end of the magnet. Figure 7-11 shows shaded plots of the flux density magnitude in the two discs and the poles when phase B is energised, but the rotor has not yet moved to the new alignment position. This demonstrates the effect of the permanent magnet in concentrating the field in the required gaps. After one complete cycle of excitation, or four steps, the rotor will have moved by one tooth pitch from its original position. If there are n teeth on the rotor, there will be 4n steps per revolution, so the step angle is 90 / n degrees. Practical hybrid stepper motors usually have a large number of teeth on the rotor, resulting in a small step angle. A typical step angle is 1.8º, giving 200 steps per revolution, with 50 rotor teeth. The maximum torque can be estimated from equation 7-5 as follows. Let Bm be the flux density in a gap due to the permanent magnet, and Be the flux density due an energised pole. If the magnetic circuit is unsaturated, the total flux density in one gap will be Bm + Be, and in the other gap it will be Bm – Be. If p is the number of active pairs of teeth, the resultant torque is thus:

0

22

0

22

21

00

22

0

21

4

})(){(

)(

emg

ememg

gggm

BdrBpl

BBBBdrpl

BBdrpldrBpldrBpl

T

(7-6)

N

S

A1

B1

A2

B2

A1

B1

A2

B2

Stepper Motors 61

(a)

(b)

Figure 7-10: Phase B energised: (a) N polarity end, (b) S polarity end.

(a)

(b)

Figure 7-11: Phase b energised: (a) N polarity end, (b) S polarity end.

Equation 7-6 shows that the torque is directly proportional to Be, and therefore to the current in the coil, provided that the magnetic circuit is unsaturated. In contrast, the torque in a variable-reluctance motor is proportional to the square of the current. As with the variable-reluctance motor, however, there is usually significant magnetic saturation in practical devices, so equation 7-6 may not be a good approximation. In equation 7-6, the term Bm shows the effect of the permanent magnet. It is desirable to make this term large, and to avoid conditions that would weaken the magnet. If Bs is the value of the flux density at which the steel parts of the magnetic circuit saturate, it may be shown that the greatest torque is obtained when the following relationship holds between Be, Bm and Bs:

sme BBB 21 (7-7)

It is possible to energise both phases at the same time, giving the pattern of phase currents in table 7-2.

Table 7-2: Motor phase currents.

phase A +I +I –I –I +I

phase B –I +I +I –I –I

This sequence gives a larger torque with the same step angle, and is the normal mode of operation. The current in each phase is a square wave, and there is a phase shift of 90º between the two currents. A combination of the patterns in table 7-1 and table 7-2 will give eight steps for each cycle of excitation, which is termed the half-step mode. The normal mode is the full-step mode.

7.5 Stepper motor characteristics

Static torque characteristics

Consider a stepper motor with a constant current passed through one or more phase windings. The rotor will be pulled into alignment, and any attempt to move it will be opposed by a torque. To a first approximation, the restoring torque is given by [11]:

nTT m sin (7-8)

where is the angular displacement of the rotor, n is the number of teeth on the rotor, and Tm is the peak static torque. This is a spring-like

N

S

A1

B1

A2

B2

A1

B1

A2

B2


characteristic, where the restoring force increases with displacement. The maximum restoring torque T = –Tm occurs when n = 90º, corresponding to a displacement of a quarter of a tooth pitch. If an external torque exceeding Tm is applied, then the rotor will slip a tooth – it will move to the next equilibrium position one tooth pitch away. The maximum position error, just before tooth slipping occurs, is therefore a quarter of a tooth pitch. This is equal to the step angle in a hybrid stepper motor.

Multi-step operation

With each switching of the motor phase currents, the rotor of a stepper motor will move to a new step position. Since the rotor has inertia and the restoring force has a spring-like characteristic, the rotor will oscillate about its new equilibrium position. If the currents are switched at a low rate, the rotor will move in steps, but it will oscillate at each step position. If the oscillation has not decayed to a low level before the next switching instant, resonance is possible, where the oscillation amplitude increases at each step until the rotor motion becomes erratic [11]. This can happen when the time interval between steps is an integer multiple of the oscillation period. If the stepping rate is Sr steps per second, and the oscillation natural frequency is fn hertz, the condition for resonance is therefore:

,3,2,1,1

kf

k

S nr

(7-9)

The critical stepping rates for resonance are therefore given by:

,3,2,1, kk

fS n

r (7-10)

High-speed operation

When the stepping rate is higher than the oscillation natural frequency, rotor motion is continuous, and the small speed variation between steps can be ignored. This mode of operation is known as slewing. It resembles the steady-state operation of a synchronous motor.

The nature of the currents in the motor phases will depend on the type of electronic driver – see section 7.6 – as well as the speed and the properties of the motor. At low speeds, a typical driver will maintain constant currents in the motor phases for the duration of each step, so the current waveform approximates to a square wave. At high speeds, however, the waveform may be very different from a square wave, because the rate of rise of current is limited by the phase inductance. If the speed is high enough, the current will not reach its desired value before the end of the step period. The waveform is then approximately triangular, decreasing in amplitude as the stepping rate increases. There will be a corresponding reduction in the motor torque.

Start/stop rate

If the stepping rate is not too high, a stepper motor will start from rest and run in the slewing mode when the winding currents are switched at a constant rate. Similarly, if the rate is not too high, the motor will stop suddenly when the current switching stops. Above a critical stepping rate – the start/stop rate – the motor may lose steps when it starts, and over-run or gain steps when it stops. As long as the start/stop rate is not exceeded, the rotor moves by a number of steps equal to the number of switching steps. The start/stop rate depends on the total inertia of the rotating system as well as the properties of the motor and driver.

Pullout torque

Like a synchronous motor, a slewing stepper motor will stall when the applied torque exceeds the pullout torque. The value of this torque depends on the phase current waveform, so it varies with the stepping rate. At low stepping rates, when the phase current waveform is almost a square wave, the pullout torque is substantially constant. At higher rates, as noted above, the mean current will fall, and the pullout torque will be reduced.

Stepper Motors 63

Acceleration and deceleration

Most stepper motors can run at a much higher speed than the start/stop rate if the stepping rate is gradually increased. Similarly, a motor running at high speed can be stopped without gaining or losing steps if the stepping rate is gradually reduced. Ideally, the acceleration or deceleration should be related to the motor pullout torque characteristic, since this indicates the maximum torque available at any given speed. In practice, however, a simple pre-defined speed ramp is generally used. There are three kinds of ramp in common use: Linear ramp: the stepping rate is increased at a

uniform rate for acceleration, and decreased at a uniform rate for deceleration.

S-curve or cosine ramp: the linear ramp is modified to give a smooth change from zero to maximum acceleration and back to zero, to avoid transients at the beginning and end of the ramp.

Inverse ramp: a curve resulting from a linear ramp of the time interval between steps.

These ramps are illustrated in figure 7-12 for acceleration, and figure 7-13 for deceleration.

0

200

400

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Time, s

Ste

ppin

g ra

te, s

teps

/s

a

b

c

Figure 7-12: Typical acceleration ramps:

(a) linear, (b) cosine, (c) inverse.

0

200

400

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Time, s

Ste

ppin

g ra

te, s

teps

/s

a

b

c

Figure 7-13: Typical deceleration ramps:

(a) linear, (b) cosine, (c) inverse.

Most stepper motor drivers offer a linear speed ramp, and the better ones give the option of an S-curve or cosine ramp. Some low-cost drivers use an inverse ramp, because it is easy to implement with a counter. This is the least desirable form, because the acceleration or deceleration is greatest at high stepping rates when the available torque is at its lowest. However, it is better than no ramp at all.


7.6 Stepper motor control

Electronic drive circuits

A driver for a stepper motor usually comprises three sub-systems: A power drive that supplies current to the

motor windings. A logic sequencer that generates the required

control signals for the power drive, in response to step demand pulses.

A controller that generates the required sequence of step demand pulses.

These sub-systems may be combined in a single unit. More commonly, the controller is a separate item that may take the form of a plug-in card for a computer, and the logic sequencer is combined with the power drive to form a drive module. Most hybrid stepper motors require a bipolar power drive, which can reverse the direction of the current through the winding. Some types of motor can use a simpler unipolar drive, which turns the current on and off without reversing its direction, but this requires two coils on each pole. Bipolar drives are the preferred choice. They usually employ chopper action for controlling the current magnitude, as described on the next page. Figure 7-14 shows the essential part of a bipolar power drive for one phase, with the gate control circuits for the MOSFETs omitted. Similar drive circuits are used for the other motor phases.

D1

D2

D3

D4

Q1

Q2

Q3

Q4

motorphase

+½Vd

Rc

T1 T2

–½Vd

Figure 7-14: Bipolar power drive.

The circuit operates as follows: To supply positive current to the motor phase,

transistors Q1 and Q4 are turned on, connecting terminal T1 to +½Vd and terminal T2 to –½Vd.

When Q1 and Q4 are turned off, current can continue to flow in diodes D2 and D3, returning energy to the supply, until the current falls to zero.

To supply negative current, transistors Q2 and Q3 are turned on, connecting terminal T1 to –½Vd and terminal T2 to +½Vd.

When Q2 and Q3 are turned off, current can continue to flow in diodes D1 and D4, returning energy to the supply, until the current falls to zero.

If transistors Q2 and Q3 are turned on before Q1 and Q4 cease to conduct, both arms of the bridge will short-circuit the power supply. This undesirable condition is known as shoot-through. To ensure that it cannot happen, a delay is often introduced between the positive and negative half-cycles of the current waveform, as shown in figure 7-15.

-100

0

100

0 90 180 270 360

Time phase, degrees

Cur

rent

, %

Figure 7-15: Phase current waveform.

Stepper Motors 65

Effect of winding inductance

The rise and fall of current is approximately exponential, governed by the inductance and resistance of the motor phase, but modified by the voltage induced in the winding when the rotor moves. At higher stepping rates, the time constant represents a larger fraction of the waveform period, so the current in one phase takes the form shown in figure 7-16. In deriving these graphs, the motional voltage has been ignored, and it is assumed that the driver does not use chopper control. The mean current falls progressively, with a corresponding reduction in the motor torque.

-100

0

100

0 90 180 270 360

Time phase, degrees

Pha

se c

urre

nt, %

(a)

-100

0

100

0 90 180 270 360

Time phase, degrees

Pha

se c

urre

nt, %

(b)

-100

0

100

0 90 180 270 360

Time phase, degrees

Pha

se c

urre

nt, %

(c)

Figure 7-16: Phase current waveforms at different stepping rates:

(a) low, (b) medium, (c) high.

Let R be the resistance of the motor phase plus any external resistance, L the motor phase inductance, and Vd the supply voltage. The maximum phase current is then Vd / R, and the time constant is L / R. To improve the shape of the waveform at high stepping rates, it is possible to increase both Vd and R. This will reduce the time constant but leave the maximum current unchanged. However, it is very inefficient because of the power loss in R. A better method is chopper control.

Chopper control

With chopper control, the motor is supplied from a high voltage source, but without a correspondingly high phase resistance. The initial rate of rise of current is given by Vd / L, so a high value of Vd gives a rapid rise of current. To limit the final value of the motor current, chopper control works as follows for positive current: Transistors Q1 and Q4 are turned on. Current begins to rise. When the current reaches an upper threshold

value, Q1 and Q4 are turned off. Current begins to fall. When the current reaches a lower threshold

value, Q1 and Q4 are turned on. The cycle repeats. A similar sequence applies for negative current, using transistors Q2 and Q3. The voltage drop across the current-sensing resistor Rc is used for controlling the transistors, and the resulting phase current waveform takes the form shown in figure 7-17. Here it is assumed that the transistors are switched off when the current reaches 110% of the desired value, and switched on again when the current falls to 90%.

-110

0

110

0 90 180 270 360

Time phase, degrees

Ph

ase

curr

ent,

%

Figure 7-17: Waveform with chopper control.


8 POWER ELECTRONIC CONTROL

8.1 AC/DC Converters

AC/DC converters convert alternating current or voltage into direct current or voltage [3, 4]. In most applications, the power flow is from the AC source to the DC load, and the process is called rectification. If the power flow reverses, the process is called inversion.

Single-phase AC/DC converters

The simplest AC/DC converter is just a rectifier bridge circuit using diodes. Figure 8-1 shows a single-phase bridge supplying a passive load.

D1 D3

D2 D4

R

L

vs vL

+ +

Figure 8-1: Single-phase diode bridge.

When the source voltage vs is a sinusoidal alternating quantity, the output voltage vL is unidirectional. Figure 8-2 shows the output for two cycles of the input voltage, as a percentage of the maximum value of vs.

0

50

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-2: Diode bridge output voltage.

The mean output voltage is given by:

VV

Vd 900.022

0

(8-1)

where V is the RMS value of the AC input.

If the diodes are replaced by thyristors, as shown in figure 8-3, the output voltage can be controlled by varying the firing angle – the point on the cycle where the thyristors are turned on by applying a gate drive signal. The gate drive circuits are not shown in figure 8-3.

U1 U3

U2 U4

R

L

vs vL

+ +

Figure 8-3: Single-phase thyristor bridge.

Figure 8-4 shows the resulting waveform for a firing angle of 30º. The output goes negative for a portion of each cycle before the next thyristor is fired, so the mean value of the waveform falls.

-100-50

050

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-4: Thyristor bridge output voltage.

The mean output voltage is now given by:

cos0dd VV (8-2)

where Vd0 is the diode bridge output given by equation 8-1 and is the firing angle. Thus, the output voltage can be varied from Vd0 to zero by varying the firing angle from 0 to 90º. Figure 8-5 shows the waveform when = 90º, where the mean output voltage is zero.

-100

-500

50

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-5: Output voltage when = 90º.

Power Electronic Control 67

Inverter action

Equation 8-2 indicates that the output voltage will reverse if the firing angle exceeds 90º. Figure 8-6 shows the output waveform when = 150º, which is an inversion of figure 8-4.

-100-50

050

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-6: Output voltage when = 90º.

Current cannot flow in the reverse direction through the thyristors, so the circuit will not work with a passive load if > 90º. However, current can continue to flow in the forward direction if the load contains a voltage source, such as a battery or a DC motor armature, with the polarity shown in figure 8-6. The DC source eL then supplies power to the converter, which returns power to the AC supply. In this mode, the converter is operating as an inverter.

U1 U3

U2 U4

R

L

eL

+

vs

+

Figure 8-7: Inverter action.

3-phase AC/DC converters

A 3-phase diode bridge can be formed from a single-phase bridge by adding just one extra pair of diodes, as shown in figure 8-8.

D1 D3

D2 D4

R

L

vL

+

D5

D6

a

b

c

Figure 8-8: 3-phase diode bridge.

The output waveform is shown in figure 8-9. The output is much smoother than for single phase, and the mean value is given by:

VV

Vd 35.123

0

(8-3)

where V is the RMS line voltage of the AC supply.

0

50

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-9: 3-phase diode bridge output.

Figure 8-10 shows a 3-phase thyristor bridge, and figure 8-11 shows the corresponding output voltage waveform when the firing angle is 30º.

U1 U3

U2 U4

R

L

vL

+

U5

U6

a

b

c

Figure 8-10: 3-phase thyristor bridge.

0

50

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-11: 3-phase thyristor bridge output.

The mean output voltage is given by:

cos0dd VV (8-4)

where Vd0 is the 3-phase diode bridge output given by equation 8-3 and is the firing angle.


8.2 DC motor control

In section 3.3 it was shown that the speed of a DC motor could be controlled by varying the voltage applied to the armature. A thyristor AC/DC converter is a simple method of generating a variable DC voltage from the AC mains supply, so it provides an effective way of controlling the speed of a DC motor [3, 4]. The combination of a converter and a DC motor is the basis of variable-speed drive systems that have dominated the market for many years. DC motors have several disadvantages in comparison with induction motors: High manufacturing cost Maintenance of the commutator and brushes Protection required in hostile environments Until recently, the high cost of inverters for variable-speed AC drives (see section 8.3) has offset the disadvantages of the DC motor. This is no longer the case. AC drives are now the preferred choice in most applications. However, there is still a market for large DC drives in specialised applications such steel rolling mills and mine winders. When a DC motor is supplied from a thyristor converter, the armature generated voltage can affect the converter voltage waveform, so that the mean output voltage varies with the load current [13]. This results in poor speed regulation unless some form of feedback control is used. A particular problem with thyristor converters is that the current cannot reverse. Special measures are therefore required if the DC motor torque is required to reverse, as can happen during braking [3, 13]. An important function of a DC motor controller is to provide a soft start by gradually increasing the voltage applied to the armature.

8.3 DC/AC Inverters

The only way to control the speed of a synchronous motor is to vary the frequency of the AC supply. This is also the most effective way of controlling the speed of an induction motor, so there is a general requirement for a variable-frequency AC source. An electronic system that converts DC to variable-frequency AC is known as an inverter.

In section 8.2, it was observed that the power flow through a DC/AC inverter could be reversed if the firing angle exceeded 90º and a DC voltage source of the correct polarity was connected to the output. This is a known as a line-commutated inverter, since the switching of current from one device to another is determined by the alternating voltages applied to the 3-phase input. It is possible to use a line-commutated inverter to drive a synchronous motor, since this machine will generate the required voltage waveforms. This approach is sometimes used with large synchronous motors [12]. In general, however, variable-frequency inverters for motor control use forced commutation, where the switching of current from one device to another is predetermined, and not dependent on an AC source connected to the output. These are more complicated than line-commutated inverters, but are more versatile. The most common form of inverter for motor control is the 3-phase bridge inverter [3, 4] shown in figure 8-12, with the base drive components omitted. The DC supply voltage Vd is connected to the terminals labelled +½Vd, –½Vd, and the AC output is taken from the terminals labelled a, b, c. Although the switching devices are shown as transistors, other power electronic switching devices are frequently used [3, 4]. The most popular choice is the IGBT (insulated gate bipolar transistor) for power ratings up to about 500 kW. The circuit is effectively a 3-phase version of the bipolar power drive for stepper motors, described in section 7.6.

D1 D3

D2 D4

+½Vd

D5

D6

a

b

c

Q1 Q3 Q5

Q2 Q4 Q6

–½Vd

Figure 8-12: 3-phase bridge inverter.


This circuit is switched in a similar way to the stepper motor drive circuit. In each leg of the bridge, one transistor at a time is switched on, connecting an output terminal to either +½Vd or –½Vd. As in the stepper motor drive, the diodes provide paths where load current can continue to flow when the transistors are turned off. It is essential that one transistor has ceased to conduct before the second transistor in the same leg is turned on, otherwise the DC supply will be short-circuited. This is the same problem of shoot-through that was mentioned in section 7.6.

Six-step inverter

The simplest mode of operation is to control the switches so that the voltage waveform at each output terminal is a square wave. Figure 8-13 shows the voltages with respect to the mid-point of the DC supply, as a percentage of the DC supply voltage Vd. The line-to-line voltages are then quasi-square waves, as shown in figure 8-14. Because there are six switching transitions per cycle, this is known as a six-step inverter.

-50

0

50

0 180 360 540 720

Time phase, degrees

Va,

%

-50

0

50

0 180 360 540 720

Time phase, degrees

Vb,

%

-50

0

50

0 180 360 540 720

Time phase, degrees

Vc,

%

Figure 8-13: Six-step output voltages.

-100

0

100

0 180 360 540 720

Time phase, degrees

Vab

, %

-100

0

100

0 180 360 540 720

Time phase, degrees

Vbc

, %

-100

0

100

0 180 360 540 720

Time phase, degrees

Vca

, %

Figure 8-14: Six-step line-to-line voltages.

Pulse-width modulation

The six-step inverter waveform is rich in harmonics, which can have an adverse effect on motor performance [12]. This type of inverter is generally used in high power drives, typically above 1 MW, but for lower powers a technique known as pulse-width modulation (PWM) is used to generate a better approximation to a sine wave [3, 4]. Instead of switching the devices at a low frequency to generate the required output wave shape, in PWM they are switched at a high frequency and the mark/space ratio is modulated sinusoidally. The result is a waveform with an average value that varies sinusoidally, as shown in figure 8-15.


-100

0

100

0 90 180 270 360

Time phase, degrees

Vab

, %

Figure 8-15: PWM inverter voltage waveform

For clarity, the carrier frequency in figure 8-15 is only 12 times the modulation frequency, so the approximation to a sine wave is not very good. In practical inverters, the frequency ratio is typically 100 at the highest modulation frequency. With a PWM inverter, it is a simple matter to control the amplitude of the equivalent sine-wave output by varying the modulation index. In figure 8-13, for example, the amplitude is about 70% of the maximum possible output. For AC motor control, it is necessary to change the output voltage amplitude when the frequency changes: see section 8.4.

DC link inverter

Normally the DC supply for the inverter is obtained from the AC mains via a converter of the kind described in section 8.1. The DC link between the converter and the inverter includes inductance and capacitance, as shown in figure 8-16, to give a smooth DC input to the inverter. This arrangement is known as a DC link inverter.

converter inverterDC link

Figure 8-16: DC link inverter.

In many cases, the converter is simply a diode bridge, which may be single-phase for low-power applications.

Dynamic braking

A problem with the DC link inverter is that the DC link current cannot reverse, because a simple AC/DC converter cannot accept current flow from the DC link. If a motor connected to the output of the inverter regenerates, the power flow will reverse, and the current flow from the inverter to the link will cause the voltage on the capacitor to rise. Control circuits should shut down the inverter if this happens. If the inverter is required to handle reverse power, which can happen when a motor is braked by rapidly reducing the inverter output frequency, a resistor must be connected across the DC link to absorb the power. This is termed a dynamic braking resistor, which is usually connected by a semiconductor switch as shown in figure 8-17. The switch is turned on only when the DC link voltage rises above a preset level, so that the resistor is not connected during normal operation of the inverter.

Figure 8-17: Dynamic braking resistor.

In applications requiring frequent power reversal, the power loss in a dynamic braking resistor may be unacceptable. A more complex and costly alternative is to provide a second AC/DC converter that operates as an inverter when the direction of the DC link current reverses, thereby returning energy to the AC supply [12].


8.4 AC motor control

Relationship between voltage and frequency

All AC machines and transformers are limited by the maximum permissible flux density in the steel parts of the magnetic circuit. This, in turn, imposes a relationship between the magnitude of the applied voltage and the frequency. In a transformer, we have equation 2-10 for the magnitude of the primary winding voltage:

mm ABfNV 11 2 [2-10]

If the core flux density Bm is held constant, this equation shows that the magnitude of the voltage is proportional to frequency. The same principle applies to AC machines, which gives the constant volts per hertz rule for variable-frequency operation. Equation 2-10 was derived on the assumption that the resistance of the winding is negligible. This is usually a good approximation at the normal AC mains frequency of 50 Hz, but it does not hold at very low frequencies. If the transformer secondary is open-circuited, the flux density in the core is determined by the current in the primary winding, which in turn depends on the input impedance:

2

121

1

1

11

)( LR

V

Z

VI

(8-5)

where L1 is the self-inductance of the primary. At very low frequencies, the current will be determined by the winding resistance R1 rather than the reactance L1. The constant volts per hertz rule therefore ceases to apply at very low frequencies, where a constant voltage is required.

Synchronous motor speed control

The speed of a synchronous motor can be controlled by varying the frequency of the applied voltage. If the armature resistance can be neglected, the phase voltage V will be proportional to the frequency. Since the excitation voltage E is generated by rotation of the magnetised rotor, it will be proportional to the speed, and therefore proportional to the frequency. Let V0 and E0 be the values of E and V at the base frequency f0. If we define a frequency ratio k = f / f0, the values at any other frequency f are:

000

kVVf

fV (8-6)

000

kEEf

fE (8-7)

Similarly, the synchronous speed is given by:

0ss k (8-8)

and the reactances are:

00 , qqdd kXXkXX (8-9)

With these definitions, the torque equation 6.9 becomes:

2sin11

2

sin3

2sin11

2

sin3

00

20

0

00

0

2

dqds

dqdsd

XX

V

X

EV

XX

V

X

VET

(8-10)

Thus the torque depends on the load angle but is independent of the frequency.

Induction motor soft start

In section 6.2, it was noted that the starting current of an induction motor is typically six times the full-load current. In addition, as shown in section 6.6, there is a large transient torque when the full supply voltage is suddenly applied to the motor. These problems can be reduced by soft starting – gradually increasing the applied voltage. This is easily accomplished with an AC phase controller of the form shown in figure 8-18, connected between the 3-phase supply and the load [12].

a1

b1

c1

a2

b2

c2

Figure 8-18: Phase controller


Figure 8-19 shows the output voltage waveform for one phase when the load is purely resistive, and the firing angle of the thyristors is 60º.

-100

0

100

0 180 360 540 720

Time phase, degrees

Vol

tage

, %

Figure 8-19: Phase controller output voltage

The RMS value of the output voltage falls as the firing angle increases, reaching zero when = 180º.

Induction motor speed control

As with the synchronous motor, the most effective way of controlling the speed of an induction motor is to vary the supply frequency. For optimum performance, the supply voltage must also be varied to maintain a constant magnetic flux density in the machine. The principles of variable-frequency operation may be deduced from the equivalent circuit of figure 6-6, reproduced below:

Rs

sRr

jxs jxr

jxm

rc

Vs

+

Is Ir

I0

The core loss resistance is given by equation 6-15:

c

mc R

Xr

2

[6-15]

Since Rc is approximately constant under variable-frequency conditions [3, 4], and Xm is proportional to frequency, the value of rc will be insignificant at low frequencies. Therefore, this element may be ignored, giving the circuit shown in figure 8-20.

s

Rr

jxs jxr

jXmVs

+

Is Ir

I0

Rs

Figure 8-20: Induction motor equivalent circuit

The torque is given by equation 6-9, which may be re-arranged as follows:

rrs

rr

sd RI

ss

RIT 22 33

(8-11)

The quantity ss is the slip angular frequency, so the developed torque will be constant if the rotor current and the slip frequency are held constant. Constant slip frequency is the preferred operating mode, so it is necessary to explore the consequences of this constraint when the frequency is varied. As with synchronous machines, we introduce a frequency ratio k = f / f0, where f is the operating frequency and f0 is the base frequency. If s0 is the fractional slip at the base frequency f0, corresponding to a synchronous speed s0, then:

00 ss ss (8-12)

The fractional slip s at frequency f is thus:

k

s

k

sss

s

s

s

s 0

0

0000

(8-13)

The value of the rotor resistance element in the equivalent circuit is:

00 / s

kR

ks

R

s

R rrr (8-14)

and the values of the reactances are:

000 ,, mmrrss kXXkxxkxx (8-15)

With these substitutions, the equivalent circuit takes the form shown in figure 8-21:

0s

kRr

jkxs0 jkxr0

jkXm0Vs

+

Is Ir

I0

Rs

Figure 8-21: Variable-frequency circuit


Apart from the resistance Rs, the values of all the elements in this circuit are proportional to the frequency ratio k. If we set Rs = 0, and make the stator voltage Vs proportional to k, then the currents Is, I0 and Ir will be constant. Since the magnetic flux density in the machine is proportional to I0, this constraint ensures that the flux density remains constant. It is therefore the normal mode for variable-frequency operation. Figure 8-22 shows theoretical torque/speed curves for the example 3 kW motor from section 7, with Rs = 0 in the equivalent circuit and the voltage proportional to frequency. The slip is 5% at 50 Hz, which is close to the full-load value. This gives a theoretical torque of 20.3 Nm, shown by the blue horizontal line in figure 8-20. All of the torque/speed curves have the same shape, with the same value of breakdown torque, and the full-load slip speed is 75 rev/min in each case.

0

20

40

60

80

100

0 300 600 900 1200

1500


Tor

que,

Nm

5040302010

2.5

Figure 8-22: Variable-frequency operation:

2.5, 10, 20, 30, 40 and 50 Hz, Rs = 0.

At a supply frequency of 2.5 Hz, the full-load torque is obtained at standstill. If the motor is started at this frequency, it will deliver the full rated torque but only take the normal full-load current. In contrast, a motor started at 50 Hz will take about six times the full-load current. Inverter control therefore gives optimum starting performance as well as speed control.

If the correct value of stator resistance Rs is included in the model, it has an increasing effect on the machine performance at low frequencies because all the other elements in the equivalent circuit are getting smaller. Figure 8-23 shows the resulting torque/speed curves for the example 3 kW motor, when the stator voltage is directly proportional to frequency.

010203040506070

0 300 600 900 1200

1500


Tor

que,

Nm

5040302010

2.5

Figure 8-23: Variable-frequency operation: 2.5, 10, 20, 30, 40 and 50 Hz, constant V / f.

At 50 Hz and a fractional slip of 5%, the developed torque is 18.5 Nm, shown by the blue horizontal line in figure 8-21. This is slightly less than the nominal full-load torque, giving an output power of 2.76 kW. Table 8-1 shows the performance for the frequency range 2.5 – 50 Hz.

Table 8-1: Constant V / f

Frequency, Hz 50 40 30 20 10 2.5

Phase voltage, V 230 184 138 92.0 46.0 11.5

Phase current, A 5.55 5.49 5.39 5.20 4.68 2.86

Fractional slip, % 5.00 6.25 8.33 12.5 25.0 100

Torque, Nm 18.5 18.1 17.5 16.2 13.2 4.90

Speed, rev/min 1425 1125 825 525 225 75


To compensate for the effect of the stator resistance, it is necessary to boost the voltage at low frequencies. Table 8-2 shows the corresponding results when the voltage is increased to maintain a constant full-load torque. Figure 8-22 shows the corresponding set of torque/speed graphs.

Table 8-2: Variable V / f

Frequency, Hz 50 40 30 20 10 2.5

Phase voltage, V 230 186 142 98.2 54.5 22.4

Phase current, A 5.55 5.55 5.55 5.55 5.55 5.55

Fractional slip, % 5.00 6.25 8.33 12.5 25.0 100

Torque, Nm 18.5 18.5 18.5 18.5 18.5 18.5

Speed, rev/min 1425 1125 825 525 225 75

0

10

20

30

40

50

60

70

0 300 600 900 1200 1500


Tor

que,

Nm

5040302010

2.5

Figure 8-24: Variable-frequency operation: 2.5, 10, 20, 30, 40 and 50 Hz, variable V / f.

Although the performance at full-load slip is now constant throughout the speed range, the breakdown torque and the starting torque are still significantly reduced at low frequencies. This is another effect of the stator resistance Rs. In larger motors, Rs is a smaller proportion of the motor impedance, so the characteristics approach the ideal more closely.

Inverter voltage-frequency characteristic

The results given above for a 3 kW motor show that it is necessary to modify the simple linear relationship between voltage and frequency. Inverter manufacturers normally enable the user to adjust the volt relationship to suit a particular motor. As an illustration, the red curve in figure 8-25 shows the required voltage-frequency relationship for the example 3 kW motor, which deviates considerably from the simple linear relationship shown by the blue line.

0

60

120

180

240

0 10 20 30 40 50

Frequency, Hz

Pha

se v

olta

ge, V

Figure 8-25: Voltage-frequency relationship for

a 3 kW induction motor.

The required characteristic is very nearly a straight line that does not pass through the origin. Many inverter manufacturers provide an adjustable characteristic of the form shown in figure 8-26, where the boost voltage Vb, break-point voltage Vp, and break-point frequency fp can be set as required.


This shape is easily adjusted to give a good approximation to the characteristic in figure 8-25.

frequencyf0

voltage

V0

Vp

Vb

0 fp

Figure 8-26: Inverter user-defined voltage-frequency characteristic.

Vector control of induction motors

In a DC motor, the commutator forces the magnetic axis of the armature flux to be at right angles to the axis of the field flux. This gives the maximum torque for a given current in the machine windings, and it gives the DC machine simple control properties: the torque is determined by the product of the armature current and the field flux. In an induction motor, the corresponding angle between the flux components depends on the operating conditions, and is not predetermined by the design of the machine. However, for any given rotor speed and load, it is possible to adjust the stator voltage and frequency so that the magnetic conditions resemble those of a DC motor, with the two flux components at right angles. With this constraint, an induction motor can be controlled in a similar way to a DC motor, to achieve good dynamic performance. This method of controlling the motor is known as vector control or field-oriented control. A theoretical treatment is beyond the scope of this course, but an introduction will be found in references [3, 4, 12]. This topic is covered in the Year 3 course Electrical Machine Drives.

When the concept was first introduced, vector control required the direct measurement of the flux in the motor with field sensors. Since this was often impracticable, alternative methods have been developed based on measurement of the stator voltage and current and the rotor position. The latest development is sensorless vector control, where the rotor position measurement is not required. This method uses measurements of the stator voltage and current alone, together with a mathematical model of the motor. Drives that implement vector control, with or without a rotor position sensor, can determine the parameters of the motor model automatically from measurements of the stator quantities. They are more costly than simple inverters with frequency and voltage control, but they offer improved dynamic response to changes in the motor speed and the torque load.


9 REFERENCES 1. Edwards, J.D.: Electromechanics Course Notes, 2004. 2. Smith, R.J. and Dorf, R.C.: Circuits, Devices and Systems (5th edition, Wiley, 1992) 3. Edwards, J.D.: Electrical Machines and Drives (Macmillan, 1991) 4. Fitzgerald, A.E., Kingsley, C. Jr., and Umans, S.D.: Electric Machinery (6th edition, McGraw-Hill,

2003). 5. Say, M.G.: Alternating Current Machines (5th edition, Pitman, 1983). 6. Alger, P.L.: Induction Machines (2nd edition, Gordon and Breach, 1970). 7. BS 4999 (IEC 34-2): 1977: Methods for determining losses and efficiency of rotating electrical

machines from tests (British Standards Institution / International Electrotechnical Commission). 8. ANSI/IEEE Std 112 – 1978, IEEE standard test procedure for polyphase induction motors and

generators (American National Standards Institute). 9. Hancock, N.N.: Matrix Analysis of Electrical Machinery (2nd edition, Pergamon, 1974). 10. Edwards, J.D.: Electrical Machines (2nd edition, Macmillan, 1986). 11. Acarnley, P.P.: Stepping Motors: a Guide to Modern Theory and Practice (4th edition, Institution of

Electrical Engineers, 2002). 12. Murphy, J.M.D. and Turnbull, F.G.: Power Electronic Control of AC Motors (Pergamon, 1990). 13. Say, M.G. and Taylor E.O.: Direct Current Machines (2nd edition, Pitman, 1986).

Appendices 77

10 APPENDICES

10.1 Induction motor 2-axis equations

In matrix form, the 2-axis differential equations of a 2-pole induction motor are as follows [9, 10]:

rq

rd

sq

sd

rrrrr

rrrrr

ss

ss

rq

rd

sq

sd

i

i

i

i

pLRLMpM

LpLRMMp

MppLR

MppLR

v

v

v

v

00

00

(10-1)

In equation 10-1, p d / dt and r is the rotor angular velocity in rad/s. The resistances Rs and Rr are the normal stator and rotor resistances in the equivalent circuit, and the inductances are related to the equivalent-circuit reactances as follows:

)(

)(

MLx

MLx

MX

rr

ss

m

(10-2)

The variables vsd, vsq, isd and isq are the 2-axis stator voltages and currents. These are related to the 3-phase terminal voltages and currents by the equations:

sc

sb

sa

sq

sd

i

i

i

i

i

23

23

21

21

32

0

1 (10-3)

sc

sb

sa

sq

sd

v

v

v

v

v

23

23

21

21

32

0

1 (10-4)

Provided there is a 3-wire supply to the motor (no neutral connection), there is an inverse transformation giving the 3-phase variable in terms of the 2-axis variables:

sq

sd

sc

sb

sa

i

i

i

i

i

23

21

23

21

32

01

(10-5)

sq

sd

sc

sb

sa

v

v

v

v

v

23

21

23

21

32

01

(10-6)


The variables vrd, vrq, ird and irq are the corresponding 2-axis rotor voltages and currents. They are related to the physical currents in an equivalent 2-phase rotor winding by the equations:

r

r

rq

rd

i

i

i

i

cossin

sincos (10-7)

r

r

rq

rd

v

v

v

v

cossin

sincos (10-8)

where is the angle between the axis of the phase of the rotor and the a phase of the stator. The inverse form of these equations is:

rq

rd

r

r

i

i

i

i

cossin

sincos (10-9)

rq

rd

r

r

v

v

v

v

cossin

sincos (10-10)

Equation 10-1 can be expressed in the form:

iZiGLRv )( rp (10-11)

where v and i are the column vectors of voltage and current variables, and the impedance matrix Z is given by:

rp GLRZ (10-12)

The resistance matrix R, the inductance matrix L and the torque matrix G all have constant elements:

r

r

s

s

R

R

R

R

000

000

000

000

R (10-13)

r

r

s

s

LM

LM

ML

ML

00

00

00

00

L (10-14)

00

00

0000

0000

r

r

LM

LMG (10-15)

Appendices 79

The developed torque is given by [9, 10]:

)( rqsdsqrdT

d iiiiMT iGi (10-16)

To solve problems such as starting or DC braking, equation 10-11 must be re-arranged in a suitable form for numerical solution with all the derivatives on one side:

})({1 iGRvLii

rpdt

d (10-17)

In addition, there is the mechanical equation of motion:

lrqsdsqrdldrr TiiiiMTTJp

dt

dJ )(

(10-18)

where J is the moment of inertia of the rotating system, and Tl is the load torque (which may be a function of the rotor speed r). Equations 10-17 and 10-18 represent five simultaneous first-order non-linear differential equations, which can be solved by standard numerical methods such as Runge-Kutta to determine the stator currents and the rotor speed as functions of time. If the motor has P pairs of poles, two changes are required to the equations: (a) the left-hand side of the torque equation 10-16 is multiplied by P; (b) in equation 10-1, r is replaced by Pr.


10.2 List of formulae

General principles

Magnetic force on a current element: Bddf li

Conductor in a magnetic field: BlifBlue ,

Ampère’s circuital law: NiildH.

Materials: 7

00 104,)( HHBB r H/m

Magnetic flux and flux linkage: NBA ,

Faraday’s law: )(Lidt

d

dt

dN

dt

de

Inductance: 2

12

1

21,ii

Mi

L

Coupled coils:

dt

diM

dt

diLiRv

dt

diM

dt

diLiRv

122222

211111

Stored energy: 21222

211 ½½ iMiiLiLW

MMF: RF HlNi

Reluctance: A

lNi

r 0R

Resistance: A

l

A

lR

Force: xx

Lif x

R22 ½½ (linear system)

Torque:

R22 ½½ L

iT (linear system)

Magnetic force per unit area: 0

2

½B

A

f

Efficiency:

lossout

loss

lossout

out

in

loss

in

lossin

in

out

PP

P

PP

P

P

P

P

PP

P

P

1

1

Transformers

Voltage: dt

dNv

dt

dNv

2211 ,

Current: 02211 RiNiN

Sinusoidal operation: mm ABfNV 11 2

Ideal transformer: 2 1 2

1 2 1

v i Nn

v i N

Impedance transformation: 2L

inn

Z

Z

Voltage regulation: nl

flnl

V

VV

2

22

AC power

Complex power: S = P + jQ = VI*

Real power: )Re(coscos *VI VISP

Reactive power: )Im(sinsin *VI VISQ

DC machines

Basic equations: rfa Ke , afd iKT

Linear approximation: rfa iKe , afd iiKT

Armature equation: rfaaaaaa KiReiRv

Ra negligible: f

ar K

vù

Small motors: 2)( f

da

f

ar

K

TR

K

vù

Series motor: iii af

Appendices 81

3-phase systems and rotating field

3-phase voltages:

)240cos(

)120cos(

)cos(

tVv

tVv

tVv

mc

mb

ma

Star connection: phaseline VV 3

Delta connection: phaseline II 3

Synchronous speed (2p poles):

ps [rad/s],

p

fns [rev/s],

p

fnN ss

6060 [rev/min].

Synchronous machines

Non-salient model: EIV sjX

Non-salient torque: ss

d X

VET

sin3

Salient-pole torque:

2sin

11

2

sin3 2

dqdsd XX

V

X

VET

Reluctance motor:

2sin12

32

q

d

dd X

X

X

VT

Speed control frequency ratio 0/ ffk

0kVV , 0kEE , 0ss k

00 , qqdd kXXkXX

2sin11

2

sin3

00

20

0

00

0 dqdsd XX

V

X

EVT

Induction machines

Slip speed: rs NN

Fractional slip: s

rs

s

rs

N

NNs

Rotor input: dsem TP

Rotor output: dsdrmech TsTP )1(

Rotor loss: dsdrsloss TsTP )(

3-phase torque: s

RIT r

rs

d23

1-phase torque: s

RI

s

RIT r

rns

rrp

sd

2

11 22

Equivalent series/parallel elements:

cmc

mc R

XR

Xr

22

2

m

mc

cm X

XR

Rx

22

2

c

mcc r

xrR

22

m

mcm x

xrX

22

Loss components:

strayfwcoreRrRsoss PPPPPP .

ssRs RIP2

3 , rrRr RIP2

3 , cmcore rIP2

3

)(||3 2coreRsinem

rrRr PPPssP

s

RIsP

No-load test

fwcoreRsinin PPPPP

22 )(|| cssfwin rRIPP


Speed control

rrs

rr

sd RI

ss

RIT 22 33

00 ss ss

k

s

k

sss

s

s

s

s 0

0

0000

00 / s

kR

ks

R

s

R rrr

000 ,, mmrrss kXXkxxkxx

Stepper motors

Alignment torque: 0

2

drBl

T gm

Hybrid stepper: 0

4

emg

m

BdrBplT

Restoring torque: nTT m sin

Stepper resonance: ,3,2,1, kk

fS n

r

AC/DC converters

1-phase diode bridge: VV

Vd 900.022

0

3-phase diode bridge: VV

Vd 35.123

0

Thyristor bridge: cos0dd VV

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