Miller, T.J.E. - SPEED's Electric Motors

SPEED's Electric Motors

An outline of some of the theoryin the SPEED software for electric machine design

with problems and solutions

© TJE Miller, University of Glasgow, 2002


Contents

Chapter 1 Sizing, gearing, cooling, materials and design

Chapter 2 Brushless permanent-magnet machines

Chapter 3 Induction machines

Chapter 4 Switched reluctance machines

Chapter 5 Commutator machines

SPEED's Electromagnetic Primer

Problems and solutions

1. Sizing, gearing, cooling, materials and design

1.1 Motion control systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.2 Why adjustable speed? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.3 Large versus small drives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4

1.4 Structure of drive systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4

1.5 Drive system requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

1.6 New technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

1.7 Which Motor ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8

1.8 Sizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.14

1.9 Gearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.18

1.10 Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.20

1.11 Intermittent operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.27

1.12 Permanent magnet materials and circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.30

1.13 Properties of electrical steels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.39

1.14 Machine and drive design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.42

1.15 Computer-aided design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.43

Fig. 1.1 Flow process controlled by recirculation can produce energy losses in the flow control valve.

1. SIZING, GEARING, COOLING, MATERIALS and DESIGN

1.1 MOTION CONTROL SYSTEMS

Technology is so saturated with developments in microelectronics that it is easy to forget the vitalinterface between electrical and mechanical engineering. This interface is found wherever mechanicalmotion is controlled by electronics, and pervades a vast range of products. A little considerationreveals a large and important area of technology, in which motor drives are fundamental. In Japan theterm 'mechatronics' is applied to this technology, usually with the connotation of small drives. In thewest the term 'motion control system' is often used for small controlled drives such as position orvelocity servos. In the larger industrial range the term 'drive' usually suffices.

Many engineers have the impression that the technology of motors and drives is mature, even static.But there is more development activity in drives today than at any time in the past, and it is notconfined to the control electronics. Two important reasons for the development activity and theincreasing technical variety are:

(1) Increasing use of computers and electronics for motion control. Automation demands driveswith a wide variety of physical and control characteristics.

(2) New technology in power semiconductors, sensors, integrated circuits, and microcontrollers,facilitating the development of nonclassical motors such as brushless DC motors, steppers andswitched reluctance motors in a wide variety of designs.

1.2 WHY ADJUSTABLE SPEED?

Three common reasons for preferring an adjustable-speed drive over a fixed-speed motor are :

(a) energy saving;

(b) velocity or position control; and

(c) amelioration of transients.

(a) Energy saving. In developed economies about one-third of all primary energy is converted intoelectricity, and about two-thirds of that is re-converted in electric motors and drives, mostlyintegral-kW induction motors running essentially at fixed speed. If a constant-speed motor is used todrive a flow process (such as a fan or pump), the only ways to control the flow rate are by throttling orby recirculation, Fig. 1. Since the motor runs at full speed regardless of the flow requirement, therecan be excessive energy losses in the recirculation valve. Similar considerations apply to the controlof airflow by adjustable baffles in air-moving plant.

1. Sizing, gearing, cooling, materials and design Page 1.3

1 Series control of induction motors is inefficient; produces excessive line harmonics; and is not very stable. Thesefactors can usually be tolerated for a few seconds during starting, but they render the soft-starter unsuitable for continuousspeed control.

Fig. 1.2 Flow control efficiency is improved by adjustable-speed drive

In such applications it is often possible to reduce average energy costs by 50 per cent or more by usingadjustable-speed drives, which eliminate the throttling or recirculation loss; see Fig. 1.2. The additionallosses in the adjustable-speed drive are generally much less than the throttling and recirculation losses,since the drive efficiency is usually of the order of 90 per cent or more. The adjustable-speed drive maybe more expensive, but its capital cost can be offset against energy savings and the reduction ofmaintenance requirements on mechanical components.

Recirculation in a flow-control process is analogous to the control of a DC motor by means of anadjustable series resistance. The technique is inherently wasteful, and although it is cheap toimplement, it is increasingly hard to justify in the face of concerns about energy efficiency andpollution, even at low power levels.

(b) Velocity or position control. Obvious examples of speed control are the electric train, portablehand tools, and domestic washing-machine drives. In buildings, elevators or lifts are interestingexamples in which not only position and velocity are controlled, but also acceleration and its derivative(jerk). Countless processes in manufacturing industry require position and velocity control of varyingdegrees of precision. Particularly with the trend towards automation, the technical and commercialgrowth in drives below about 20 kW is very vigorous. Many system-level products incorporate anadjustable-speed drive as a component. A robot, for example, may contain between 3 and 6 independentdrives, one for each axis of movement. Other familiar examples are found in office machinery:positioning mechanisms for paper, printheads, magnetic tape, and read/write heads in disk drives.

(c) Amelioration of transients. The electrical and mechanical stresses caused by direct-on-line motorstarts can be eliminated by adjustable-speed drives with controlled acceleration. A fulladjustable-speed drive is used in this situation only with very large motors or where the start-stopcycles are so frequent that the motor is effectively operating as a variable speed drive. Mostsoft-starting applications are less onerous than this, and usually it is sufficient (with AC motors) toemploy series SCR's (or triacs with smaller motors) which ‘throttle' the starting current to a controlledvalue, and are bypassed by a mechanical switch when the motor reaches full speed.1 The soft-starteris less expensive than a full adjustable-speed drive, which helps to make it economical for short-timeduty during starting.

Page 1.4 SPEED’s Electric Motors

Fig. 1.3 Drive system structure

1.3 LARGE VERSUS SMALL DRIVES

There are marked design differences between large and small drives. Large motors are almost alwayschosen from one of the classical types: DC commutator (with wound field); AC induction; andsynchronous. The main reasons are the need for high efficiency and efficient utilization of materials;and the need for smooth, ripple-free torque. In small drives there is greater variety because of the needfor a wider range of control characteristics. Efficiency and materials utilization are still important, butso are control characteristics such as torque/inertia ratio, dynamic braking, speed range, acousticnoise and torque ripple.

There are also several breakpoints in the technology of power semiconductors. At the highest powerlevels (up to 20MW) SCRs (thyristors) and GTOs (gate turn-off thyristors) are the only devices withsufficient voltage and current capability, but IGBTs (insulated-gate bipolar transistors) also now havevoltage ratings measured in kV and current ratings of hundreds of amps. Naturally-commutated orload-commutated converters are preferred, because of the saving in commutation components and forreliability and efficiency reasons. In the medium power range (up to a few hundred kW) forcedcommutation and PWM are normal, and IGBT’s are very widely used. At low powers (below a few kW)the power MOSFET is attractive because it is easy to switch at high chopping frequencies.

1.4 STRUCTURE OF DRIVE SYSTEMS

The general structure of a drive system is shown in Fig. 3. It comprises the load, the motor, theelectronic drive, and the control.

The range of modern motion-control applications is virtually unlimited. Any random list illustratesthe variety—aerospace actuators; washing machines; computer disk and tape drives; printer plattensand printheads; inertial guidance systems; adjustable-speed pumps, blowers, fans and compressors;locomotive and subway traction; automatic machine tools and machining centers; servo drives andspindle drives; robots; automotive auxiliaries; refrigeration and air-conditioning drives; and manyothers.

Loads have widely differing requirements. The commonest requirement is for speed control, withvarying degrees of precision and accuracy. Position control is of increasing importance, particularlyin automated plants and processes, and in office machinery and computer peripherals. In some casesit is the steady-state operation that is most important, for example in air-conditioning and pump drives.


1. Compliance with national, EC, USA and industry standards2. Maximum continuous power or torque requirements3. Forward/reverse operation4. Motoring/braking operation5. Dynamic or regenerative braking6. Overload rating and duration7. Supply voltage (AC or DC) and frequency8. Type of control: speed, position, etc.9. Precision required in controlled speed or position10. Programmability: speed and/or position profiles, start/stop ramps etc.11. Interface with plant control and communications12. Dynamic requirements: torque/inertia ratio, acceleration/deceleration13. Gearbox or direct drive; gear ratio14. Reliability and redundancy of components15. Protection arrangements, both mechanical and electrical16. Maximum level of acoustic noise; noise spectrum17. Compliance with EMC regulations18. Limitation of harmonics in the supply system19. Maintenance; spare parts; provision for expansion or reconfiguration20. Environment: indoor/outdoor installation; enclosure; temperature; humidity; pollution; wind and seismic factors;

type of coolant.

TABLE 1.1

DRIVE SYSTEM REQUIREMENTS CHECKLIST

In other cases, such as in robots, tape drives, and actuators, dynamic performance is important becauseof the need to minimize the time taken to perform operations or effect control changes. In these casesthe torque/inertia ratio of the motor is an important parameter. In automotive applications the primerequirements are low cost and low noise. Efficiency is important in motors that run continuously, e.g.heater blowers, but not in intermittent-duty motors such as window-winders. Torque ripple isincreasingly important in servo motors and applications such as automotive power steering, where lessthan 1% ripple is typically required.

1.5 DRIVE SYSTEM REQUIREMENTS

Table 1.1 provides a checklist of application requirements. The speed/torque capability diagram isespecially important, Fig. 4. Typically at low speed the drive operates under current limit, and sincetorque is generally proportional to current (or nearly so), the torque is controlled in this mode ofoperation, such that any value can be obtained up to the value corresponding to maximum current. Athigh speed, the back-EMF increases to a level at which the drive cannot maintain maximum current;or even if it can maintain maximum current, it may not have sufficient voltage to maintain the correctphase angle or waveform of the current. Base speed is the maximum speed at which rated torque canbe developed. Above this speed, drives are often evaluated according to the speed range over whichthey can maintain constant power, the torque decreasing as the speed increases. Some drives (such astriac-controlled AC universal motors) have almost no constant-torque region and their characteristicsare said to be “highly inverse”. On the other hand, brushless PM motors tend to have a predominantlyconstant-torque characteristic with limited speed range above base speed.

1.6 NEW TECHNOLOGY

Several new technologies are contributing to the development of motion control systems.

Digital electronics. It would be hard to overstate the importance of microelectronics in motioncontrol. At the 'heavy' end of the spectrum are the multiple drives found in steel rolling mills, papermills, and other heavy process plants, where it is normal to coordinate the motion of all the shafts bymeans of a computer or a network of computers, some of which may be quite large. At the light endof the power range are the drives found in office machinery, small computers, and portable goods suchas cameras and compact-disk players, where custom integrated circuits and gate arrays are common.Between these extremes there are many microprocessor-controlled systems of all levels of complexity.


Fig. 1.4 Speed/torque characteristic

The first functions implemented with microprocessors were low-speed functions such as monitoringand diagnostics, but digital control has penetrated from outer position loops through the intermediatevelocity loop and is now routinely used even in high-bandwidth current regulators. The developmentof field-oriented control for AC induction and synchronous motor drives would not have been practicalwithout the microprocessor. Field-oriented control is based on reference-frame transformation whichmay require rapid computation of trigonometric functions of rotor position. It permits the outercontrol loops of AC and DC drives to be the same, both in hardware and software, and improves thedynamic performance of the AC drive. Its development since the mid 1970's was a key factor enablingthe AC induction motor to compete in precision speed control with the DC motor, which had beenpreferred for speed control for at least 50 years before then.

Such is the sophistication and speed of modern microelectronics that the PWM schemes employed toregulate the voltage and current can be optimized with respect to many attributes, such as efficiency,acoustic noise, dynamic response, and harmonic content. There is increasing use of field-programmablegate arrays for the high-speed functions, often combined with microcontrollers. Digital signalprocessors (DSP) are also used in advanced drives, and many of these operate with “sensorless”control, i.e., without shaft position sensing.

In motion control systems the ultimate objective is true instantaneous control of the torque, preferablywith a minimum of reliance on shaft position sensors and detailed knowledge of motor parameters.In pursuit of this objective the processing power of modern microcomputers and DSPs has beenexploited together with new forms of control based on neural networks or fuzzy logic. At the other endof the drive, communication with computers and other controllers is another area of continuousdevelopment.

Power semiconductor devices. The IGBT is the most popular device in integral-kW drives, and thepower MOSFET in low-power drives, especially at low voltage. GTO thyristors are still used in largedrives especially above 1 MW.

New magnetic materials. The permanent-magnet industry has continued to improve thecharacteristics of the main families of magnet used in electric machines: neodymium-iron boron, whichwas pioneered by Sumitomo and General Motors; rare-earth/cobalt; and ferrite. At room temperatureNdFeB has the highest energy product of all the common magnet materials. In its early days theperformance was rather sensitive to temperature, but it is now widely used even in automotive andindustrial servo applications where high operating temperatures up to and exceeding 100EC arecommon, even in the presence of strong demagnetizing fields.

CAD and numerical analysis in design. Motor design has been computerized since the early daysof computers, initially with the coding of well established design procedures. The last 20 years has seena steady development of commercially available finite-element software capable of analysing fields inmachines in two and three dimensions. The most advanced programs can determine eddy-currents andtransients, but the coupling of finite-element solvers to circuit-analysis and simulation algorithms is


2Others include design programs from Trimerics in Stuttgart, Germany and Yeadon Engineering Services, Wisconsin.

still in the research laboratory. In spite of the technical advances it remains the case that finite-element analysis requires the application of highly skilled personnel, and although machine designersuse it fairly widely, they either follow well-established procedures or rely on specialists to practisecomputations in support of more conventional design calculations.

The finite-element method is still far from being a complete design tool and is too slow for many of theroutine processes. It is fundamentally an analysis tool rather than a design tool: and it suffers fromcertain limitations when applied to motor design. It requires detailed input data and the results needskilled interpretation. It is accurate only in idealized situations where parasitic effects have beenremoved. It is too slow to be cost-effective as part of a synthesizing CAD package, and is likely to remainso for some time. It is most useful in helping to understand a theoretical problem that is too difficultfor conventional analysis, and in this role it has undoubtedly helped to refine many existing motordesigns and improve some new ones.

At the same time the development of special design software for electric machines has produced anumber of products which are much faster and are specificall intended for motor and generatordesign. Such programs are widely used in the industry for initial design, for “what-if” analysis, and(with suitable calibration against test data) for recording the characteristics of entire lines of motorproducts. One example is the SPEED software from the University of Glasgow, which is used in placesto illustrate this book.2

The primary problems in motor and generator design are not simply electromagnetic, but require anintegrated approach to materials utilization and design-for-manufacture. The philosophy needs to bea synthesizing one rather than an analytical one. This multidisciplinary problem includes heattransfer and mechanical design as well as electromagnetics. The situation is more complicated inadjustable-speed drives where the supply waveforms are 'switchmode' chopped waveforms rather thanpure sinewaves or DC. In these cases time-stepping simulation may be necessary to determine theexpected performance of a given design over a wide range of operating conditions.

Because of the fact that finite-element analysis usually requires foreknowledge of the currentwaveform, it could be argued that preoccupation with this class of software tool could hinder thedevelopment of drive systems which escape from the classical DC and sinusoidal waveforms in orderto explore the possibilities of a wider class of drive current waveforms, coupled with new concepts inmotor geometry. The key lies in using the finite-element tool in the correct way to determineparameters that can be used in time-stepping simulations that will be executed by other software.

It is perhaps surprising that design software for electric machines is rarely capable of synthesis in thetrue sense. It is much more common to find “optimization” techniques which rely on the automaticgeneration of large numbers of “feasible” designs, and then rank them according to a more or lesscomplex criterion which includes constraints on particular dimensions and other parameters. Theimaginative element in the design process remains in the human mind, and the computer appears tobe far from taking it over.

For simulation of complete systems, such as an automotive power steering system or an aircraft flightcontrol surface actuation system, there are several software packages, such as Simulink™, Saber™,Simplorer™, Easy5™ and many others. Suitably modified and extended, some of these packages permitthe simulation of detailed motor models and their drives and controls. They may be used for thedevelopment of control algorithms that are subsequently programmed in a microprocessor or gatearray.

Other contributing technologies. Plastics and composite materials find many applications in motors.Fans, slot liners and wedges, end-bells and covers, and winding supports are the commonest, butmoulded slot insulation and encapsulation of rotors are also widely used. In brushless motors designedfor high peripheral speeds, the magnets are often restrained against centrifugal force by banding ortubes made of Kevlar™, glass-reinforced plastics, or carbon fiber.

Motor drives generally require transducers for control and protection, and there has been progressin current-sensor and shaft position sensor technology. In particular the linearity and


temperature-independence of Hall-effect current sensors has improved greatly, and it is common tomount these devices in the same package, or on the same printed-circuit card, as the driver stage of thepower electronics in small drives. For larger drives flux-nulling current sensors can be used withbandwidths of up to several kHz and isolation at least as good as that of a C.T.

In brushless drives the commutation signals are often derived from Hall sensors, activated either bythe rotor magnet or by a separate magnet ring. Alternatively, optical interrupters may be used witha shaft-mounted slotted disk. At high speeds the commutation sensor can be used to generate a speedsignal via a frequency-to-voltage conversion. For motion control systems and servo-quality drivesseparate velocity and position transducers usually have to be used. For such systems the resolver isattractive because of its ruggedness, resolution, and its ability to provide accurate absolute positionand velocity signals from one sensor.

1.7 WHICH MOTOR ?

The proliferation of new ideas, materials and components obviously generates many opportunities butalso complicates the question, what is the best drive for a particular job? We can perhaps address thisby attempting to trace the evolution of the different motor types in such a way as to bring out theirmost important advantages and disadvantages. It is the motor that determines the basic characteristicsof the drive system, and it alsodetermines the requirements on the power semiconductors, theconverter circuit, and the control.

Evolution of motors. The evolution of brushless motors is shown in Fig. 5. Row 1 contains the three'classical' motors : DC commutator (with wound field); AC synchronous; and AC induction. The term'classical' emphasizes the fact that these motors satisfy three important criteria:

(1) they all produce essentially constant instantaneous torque (i.e., very little torque ripple);

(2) they operate from pure DC, or AC sinewave supplies, from which

(3) they can start and run without electronic controllers.

The classical motors of row 1 are readily coupled to electronic controllers to provide adjustable speed;indeed it is with them that most of the technical and commercial development of power electroniccontrol has taken place. Together with the PM DC commutator motor in row 2 and the series-woundAC commutator motor or 'universal' motor, the row 1 motors account for the lion's share of all motormarkets, both fixed-speed and adjustable-speed, even though they represent only a minority of themany different principles of electromechanical energy conversion on which motor designs may bebased. By contrast, the nonclassical motors are essentially confined to specialist markets and untilrecently, few of them except the brushless DC motor have been manufactured in large numbers. Table1.2 is a classification of some common types of motor according to these criteria.

The motors in row 2 are derived from those in row 1 by replacing field windings with permanentmagnets. The synchronous motor immediately becomes brushless, but the DC motor must go throughan additional transformation, from row 2 to row 3 with the inversion of the stator and rotor, before thebrushless version is achieved. The induction motor in row 1 is, of course, already brushless in its 'cage'version, but not in its wound-rotor or slip-ring version. The brushless motors are all those with threeterminals, together with the switched reluctance motor, which cannot be derived from any of the othermotors. Its awkward placement in Fig. 5 reflects the fact that it has various properties in common withall the brushless motors. Obviously the emphasis in this book is on the brushless motors, with onlya relatively superficial treatment of the DC motor and the induction motor. Stepper motors are alsoexcluded.

The DC commutator motor. The traditional DC commutator motor with electronically adjustablevoltage has always been prominent in motion control. It is easy to control, stable, and requiresrelatively few semiconductor devices in the drive. For many years the wound-field DC motor held itsown against the challenge of AC drives—arguably for at least fifty years from the mid-1930's until themid-1980's—but AC field oriented control, manufacturing cost structures, the development of the IGBT,and huge R&D investments finally forced it into a declining role.


DC wound field (brush)AC/DC universal (brush)

[Brushless PM Exterior Rotor]

Brushless PMSquarewave or sinewave

Brushless PM AC

Synchronousreluctance

AC inductionAC synchronouswound-field

Switchedreluctance

DC PM (brush)

Fig. 1.5 Evolution of brushless motors from classical AC and DC motors


Fig. 1.6 Integrated motor/inverter and hand-held controller(courtesy of Grundfos A/S, Denmark)

The main objections to the commutator motor are brush and commutator wear, and the fact that thelosses arise mostly on the rotor, making cooling more difficult than in AC motors where the losses arisemostly on the stator. It is not that brushgear is unreliable—on the contrary, it is reliable, well-proven,and 'forgiving', as is proven by the widespread use of DC motors in railway systems throughout theworld, and in automotive auxiliaries where the life of the brushes is not a serious limitation.

The PM DC commutator motor. In small DC commutator motors, replacing the field winding andpole structure with permanent magnets usually permits a considerable reduction in stator diameter,because of the efficient use of radial space by the magnet and the elimination of field losses. Armaturereaction is reduced and commutation is improved, owing to the low permeability of the magnet. Theloss of field control is not as important as it would be in a larger drive, because it can be overcome bythe controller. In small drives the need for field weakening is less common anyway. The PM DC motoris usually fed from an adjustable voltage supply, either linear or pulse-width modulated.

In automotive applications the PM DC motor is well entrenched because of its low cost and because ofthe low-voltage DC supply. Here it is usually operated at fixed speed or with series-resistance control.For safety-critical and demanding applications such as electric power steering and braking, brushlessmotor drives are more suitable. The development of higher-voltage automotive power supply systems(above 40V) will help to make brushless motors more acceptable by reducing the current levels andtherefore the size and cost of MOSFETs required in the drive.

AC induction motor drives. AC induction or synchronous motors are often preferred because of thelimitations of commutation and rotor speed in DC motors. Slip is essential for torque production inthe induction motor, and it is impossible, even in theory, to achieve zero rotor losses. This is one of thelimitations of the induction motor, since rotor losses are more difficult to remove than stator losses,and it is one main reason to use permanent-magnet and/or reluctance-type synchronous motors.

The efficiency and power factor of induction motors falls off in small sizes because of the natural lawsof scaling, particularly at part load. As a motor of given geometry is scaled down, if all dimensions arescaled at the same rate the MMF required to produce a given flux-density decreases in proportion to thelinear dimension. But the cross-section available for conductors decreases with the square of the lineardimension, as does the area available for heat transfer. This continues down to the size at which themechanical airgap reaches a lower limit determined by manufacturing tolerances. Furtherscaling-down results in a more-or-less constant MMF requirement while the areas continue to decreasewith the square of the linear dimension. There is thus an “excitation penalty” or “magnetizationpenalty” which becomes rapidly more severe as the scale is reduced. It is for this reason thatpermanent magnets are so necessary in small motors. By providing flux without copper losses, theydirectly alleviate the excitation penalty.


Fig. 1.7 The Minas brushless PM motor produced by Matsushita with its stator fabricated from segments. Courtesy ofMatsushita Ltd., Japan

The induction motor is “brushless” and can operate with simple controls without a shaft positiontransducer. The simplest type of inverter is the six-step inverter. With no shaft position feedback, themotor remains stable only as long as the load torque does not exceed the breakdown torque, and thismust be maintained at an adequate level by adjusting the voltage in proportion to the frequency as thespeed changes. At low speeds, oscillatory instabilities may appear. To overcome these limitationsthere have been several improvements including slip control and, ultimately, full field-oriented or“vector” control in which the phase of the stator currents is regulated to control the angle betweenstator MMF and rotor flux. Field orientation usually requires a shaft position encoder and may includean in-built control model whose parameters are specific to the motor, and which must be compensatedfor changes caused by changing load and temperature. Such controls are complex and generally cannotbe justified in small drives, but excellent results have been achieved in larger sizes (above a few kW).

In the fractional and low integral-horsepower range the complexity of the AC drive is a drawback,especially when dynamic performance, high efficiency, and a wide speed range are among the designrequirements. These requirements cannot be met adequately with series- or triac-controlled inductionmotors, which are therefore restricted to applications where low cost is the only criterion. Togetherthese factors favour the use of brushless PM motor drives in the low power range.

The brushless DC PM motor. The smaller the motor, the more sense it makes to use permanentmagnets for excitation. There is no single 'breakpoint' below which PM brushless motors outperforminduction motors, but it is usually in the1!10 kW range. Above this size the induction motor improvesrapidly, while the cost of magnets works against the PM motor. Below it, the PM motor has betterefficiency, torque per ampere, and effective power factor. Moreover, the power winding is on thestator where its heat can be removed more easily, while the rotor losses are extremely small. Thesefactors combine to keep the torque/inertia ratio high in small motors. The brushless DC motor is alsoeasier to control, especially in its ‘squarewave' configuration (Chapter 4). Although the inverter issimilar to that required for induction motors, usually with six transistors for a 3-phase system, thecontrol algorithms are simpler and readily implemented in 'smartpower' or special-purpose ICs.

The brushless PM AC synchronous motor. In Row 2 of Fig. 5 the brushless synchronous machinehas permanent magnets instead of a field winding. Field control is again sacrificed for the eliminationof brushes, sliprings, and field copper losses. This motor is a classical salient-pole synchronous ACmotor with approximately sine-distributed windings, and it can therefore run from a sinewave supplywithout electronic commutation. If a cage winding is included, it can self-start 'across-the-line'.

The magnets can be mounted on the rotor surface (Chapter 5) or they can be internal to the rotor. Theinterior construction simplifies the assembly and relieves the problem of retaining the magnets againstcentrifugal force. It also permits the use of rectangular instead of arc-shaped magnets, and usuallythere is an appreciable reluctance torque which leads to a wide speed range at constant power.


The PM synchronous motor operates as a synchronous reluctance motor if the magnets are left out ordemagnetized. This provides a measure of fault-tolerance in the event of partial or totaldemagnetization through abnormal operating conditions. It may indeed be built as a magnet-freereluctance motor, with or without a cage winding for starting 'across-the-line'. Although the powerfactor and efficiency are not as good as in the PM motor, synchronous reluctance motors can bedesigned with wide speed range and substantial short-term overload capacity..

In larger sizes the brushless synchronous machine is sometimes built with a brushless exciter on thesame shaft, feeding a rotating rectifier which passes DC to a field winding on the main rotor. Thismotor has full field control. It is capable of a high specific torque and high speeds. As a generator, thisconfiguration is popular in high-speed aircraft generators (at 24,000 and 12,000 rpm, 400 Hz) and in awide variety of small industrial applications.

All the motors on the diagonal of Fig. 5 operate with inverters that share the same power circuittopology (three 'totem-pole' phaselegs with the motor windings connected in star or delta to themidpoints). This gives rise to the concept of a family of motor drives providing a choice of motors andmotor characteristics, but with a high degree of commonality in the control and power electronics andall the associated transducers. The trend towards integrated phaselegs, or indeed complete three-phasebridges, with in-built control and protection circuitry makes this concept more attractive. This familyof drives covers a wide range of requirements, the main types being the conventional brushless DC(efficient in small sizes with good dynamics); the interior-magnet synchronous motor (wide speedrange); the synchronous reluctance motor (free from magnets and capable of high speeds orhigh-temperature operation); and the induction motor. It should be noted that all these drives areessentially “smooth-torque” systems with low torque ripple.

Stepper motors represent a major class of motors not included in Fig. 5. Steppers are alwaysbrushless and usually operate without shaft position sensing. Although they have many properties incommon with synchronous and brushless DC motors, they cannot naturally be evolved from the motorsin Fig. 5. By definition they are pulsed-torque machines incapable of achieving ripple-free torque bynormal means. Variable-reluctance (VR) and hybrid steppers can achieve an internal torquemultiplication through the use of multiple teeth per stator pole and through the ‘vernier' effect ofhaving different numbers of rotor and stator poles. Both these effects work by increasing the numberof torque impulses per revolution, and the price paid is an increase in commutation frequency and ironlosses. Steppers therefore have high torque-to-weight and high torque-to-inertia ratios, but are limitedin top speed and power-to-weight ratio. The fine tooth structure requires a small airgap, which addsto the manufacturing cost. Beyond a certain number of teeth per pole the torque gain is “washed out”by scale effects that diminish the inductance variation on which the torque depends. Because of thehigh magnetic frequency and the effect of MMF drop in the iron, such motors require expensivelamination steels to get the best out of them.

Switched reluctance motors are derived from the single-stack VR stepper, in which the currentpulses are phased relative to the rotor position to optimize operation in the 'slewing' (continuousrotation) mode. This usually requires a shaft position transducer similar to that which is required forthe brushless DC motor, and indeed the resulting drive is like a brushless DC drive without magnets.With this form of control the switched reluctance motor is not a stepper motor because it can producecontinuous torque at any rotor position and any speed. There is still an inherent torque ripple,however, which can be compensated only by current waveform profiling and accurate phase controlof the current waveform relative to the shaft position. The switched reluctance motor suffers the same'excitation penalty' as the induction motor and cannot equal the efficiency or power density of the PMmotor in small sizes.

When the classical motors are interfaced to switchmode converters (such as rectifiers, choppers, andinverters) they continue to respond to the average voltage (in the case of DC motors) or thefundamental voltage (in the case of AC motors). The harmonics associated with the switching operationof the converter cause parasitic losses, torque ripple, and other undesirable effects in the motors, sothat de-rating may be necessary. The nonclassical motors are completely dependent on the switchmodeoperation of power electronic converters. In steppers it is acceptable for the torque to be pulsed, butmost brushless drives are designed for smooth torque even though the power is switched.


Motor Drive or supply Typical application

DC commutator motors

(a) Wound-field Constant DC or variable DC from DCgenerator, phase-controlled rectifier,or chopper

Integral-kW industrial drives; railwaytraction

(b) Permanent-magnet Automotive and aircraft auxiliaries;small servo and variable-speed drives

DC homopolar motors with slip-ringsor segmented slip-rings

Ship-propulsion specials

Brushless PM motors

(a) Squarewave Electronically commutated rectangularcurrent waveform from inverter

Disk drives, automotive auxiliaries,small portable goods, servo motors,spindles

(b) Sinewave Sinewave current waveform frominverter

Servo motors, spindles

Universal AC/DC series commutatormotors

Fixed AC; triac-controlled AC; alsovariable or constant DC

Domestic appliances (washingmachines, food processors); powertools

Induction motors

(a) Cage-type, 3-phase Constant voltage AC or inverter-fed Pumps, fans, compressors, industrialmachinery of all types

(b) Cage-type, 1-phase Constant-voltage AC Low-power pumps, fans, machinery

(c) Wound-rotor, 3-phase Constant voltage AC or inverter-fed High-power drives with difficult startingconditions; sometimes used with slip-recovery energy systems

Synchronous motors

(a) Wound rotor with DC winding andslip-rings

Constant voltage AC or inverter-fed Very large drives

(b) Wound-rotor with brushless exciter Constant voltage AC or inverter-fed(motoring)

Most often used as a generator

(c) Permanent magnet Same as Sinewave brushless PM motor

Reluctance motors

(a) Synchronous reluctance (line-start,with rotor cage)

Inverter-fed AC Multi-machine variable-speed driveswith several motors fed from oneinverter

(b) Synchronous reluctance (cageless) Inverter-fed AC Servo/spindle motors

(c) Switched reluctance Electronically regulated andcommutated DC

Washing machine drives, miningmachinery, door openers, automotiveauxiliaries, aircraft actuators, high-speed compressors

(d) Single-phase reluctance Switched DC or AC Small actuators, clocks

Stepper motors

(a) VR, single-stack Switched DC, usually current-limited Motion-contol applications dependenton open-loop stepping for positioncontrol (indexing); office equipment,industrial machinery

(b) VR, multiple-stack

(c) Permanent-magnet

(d) Hybrid

Hysteresis motors

(a) Cylindrical rotor Inverter-fed or constant AC voltage Simple synchronous motor with goodstarting characteristics

(b) Can-type rotor Constant-voltage AC Motorized valves and actuatorsTABLE 1.2

A SELECTION OF MOTORS WITH TYPICAL APPLICATIONS


A 'Total ampere&conductors

Airgap circumference'

2mNph I

BDA/m (1.1)

M1 ' B ×BDLstk

2pWb (1.2)

Fig. 1.8 Definition of tooth pitch and J

1.8 SIZING

When a new electric machine is to be designed from scratch, the requirements usually includes a setof performance specifications and a set of constraints or limitations such as the maximum physical size,the maximum temperature rise, and the supply voltage. This section explains how the basic size of amachine can be determined, starting from the performance specifications and working within thelimits of material properties and temperature rise.

In many cases, new machine designs are evolved from existing ones, by modifying existing laminationsand components to minimize the cost of changes in tooling and components. Even so, the sameprinciples determine how much power and performance can be achieved from a machine of given sizeand temperature rise.

The output equation. The classical output equation applies to (and unifies) all electrical machinesfrom the tiniest micromotors (a few µW) up to the largest AC motors used in process plants or shippropulsion (up to 20MW). Intuitively it comes from the fundamental law of electromagnetic force whichis often loosely stated as “force = flux × current”, according to the left-hand rule. For engineeringpurposes we need to derive a more precise statement of this law. Except in linear motors, we are moreinterested in torque than force. It is convenient to work with flux-density and current-density, becausethese parameters have values which do not change greatly from one machine to another. Further, theflux and current densities are closely related to the power loss density which determines the coolingrequirements and temperature distribution throughout the machine.

Specifically, the output equation relates the torque per rotor volume (TRV) to the electric loading A andthe magnetic loading B. We will define A and B first before deriving a precise form of the outputequation. The definitions are written in a form suitable for AC synchronous and induction machines.For other types of machine the definitions are similar, but with slight variations of multiplyingconstants and interpretation.

The electric loading A is defined as the linear current density around the airgap circumference, thatis, the number of ampere-conductors per metre around the stator surface that faces the airgap.

where I is the RMS phase current, m is the number of phases, Nph is the number of turns in series perphase, and D is the diameter of the airgap. The airgap is assumed to be small compared to the rotordiameter, so that no distinction is made between the rotor diameter and the stator diameter. The RMScurrent is used because it determines the I 2R heating, which is what limits the electric loading.

The magnetic loading B is defined as the average flux-density over the rotor surface. In AC motors theflux-density is distributed sinusoidally so that the fundamental flux/pole is

where p is the number of pole-pairs and Lstk is the stack length, i.e., the axial length of the active partof the machine. In slotted stators and rotors, the peakflux density in the teeth Bt(pk) must be limited toabout 1.6T, otherwise the magnetizing current and/orthe iron losses may become excessive. The peak flux-density Bg(pk) in the airgap is therefore Bg(pk) . JBt(pk),where J is the ratio of tooth width to slot-pitch,measured at a diameter where the tooth flux-densityis maximum; see Fig. 1.8. Typically J is of the order of0.5. Thus B = 2Bg(pk)/B = 2JBt(pk)/B, so B is normallylimited to around 2 × 0.5 × 1.6/B . 0.5T.


3See chapter 2.

4 In Imperial units, if D and Lstk are in inches, then T is in lbf-in. If F = 1 lbf/in2, TRV = 13.8 kNm/m3. 5 In some references the output coefficient K is defined as T/(D

2L), so K = TRV × B/4.

E '2B

2kw1NphM1 f '

B2

2

kw1Nph BDLstk f

pV. (1.3)

TRV 'TVr

'B

2kw1 AB Nm/m 3 . (1.4)

TRV 'TVr

' 2F. (1.5)

Fig. 1.9 Airgap shear stress

Fig. 1.10 Airgap shear stress

The generated EMF per phase is given by the standard equation3

where f is the fundamental frequency, kw1 is the fundamental harmonic winding factor, and the productkw1Nph is the effective number of turns in series per phase. The maximum available electromagneticpower at the airgap is mEI. We consider this as being converted into mechanical power TT/p, whereT/p = 2Bf is the speed in rad/sec. (Note also that T = 2B/60 × rpm). We can obtain the TRV asT/(BD2Lstk/4), and substituting from equations. 1.1, 1.2 and 1.3 we get

This equation reflects the “flux-current product” in the form AB. Themultiplying factor is simply a constant multiplied by the windingfactor kw1, which incidentally casts kw1 in the role of a utilizationfactor—the higher the winding factor, the greater is the utilization offlux and current in producing torque. Since kw1 is usually about0.85!0.95, TRV . 2AB.

The TRV is also related to the airgap shear stess F, which is thetangential (torque-producing) force per unit of swept rotor surfacearea; see Figs. 1.9 and 1.10. For every unit of rotor surface area, theelectromagnetic torque is rF = DF/2, so the total torque is T = BD ×DF/2 = BD2F/2, from which it follows that

The airgap shear stress F is measured in kN/m2. Typical values aregiven in Table 1.3.4 The winding factor kw1 is generally between 0.8and 0.95, so that TRV . 2 BA and F . BA. For example, if the electricloading A = 20 A/mm and the magnetic loading B = 0.5 T, F . 0.5 × 20× 103 = 10 kN/m2. For totally-enclosed motors the lower values of Fand TRV would apply with natural convection, while the higher valueswould apply with forced-air cooling supplied by an external or shaft-mounted fan.5

Class of machine TRV kNm/m3 F lbf/in2

Small totally-enclosed motors (Ferrite magnets) 7 ! 14 0.5 ! 1

Totally-enclosed motors (sintered Rare Earth or NdFeB magnets) 14 ! 42 1 ! 3

Totally-enclosed motors (Bonded NdFeB magnets) 20 1.5

Integral-hp industrial motors 7 ! 30 0.5 ! 2

High-performance servomotors 15 ! 50 2 ! 4

Aerospace machines 30 ! 75 2 ! 5

Large liquid-cooled machines (e.g. turbine-generators) 100 ! 250 7 ! 18

TABLE 1.3TYPICAL VALUES FOR TRV AND F (CONTINUOUS OPERATION)


The coefficient Bkw1//2 in eqn. (1.11) is peculiar to AC machines where the ampere-conductordistribution and the flux-density are sinusoidally distributed in space around the airgap; this can bewritten

The product A(2)B(2) is the force per unit of rotor surface in N/m2, and therefore the torque can beobtained by integrating rA(2)B(2)d2 = rA(pk)B(pk) sin2 2 over the entire rotor surface, where r = D/2,and dividing by the rotor volume BD2Lstk/4. This gives TRV = A(pk)B(pk). But the RMS value of A isA(pk)//2, while the average value of B is BB(pk)/2, giving TRV = BAB//2. The winding factor appearsbecause only the fundamental space-harmonic component of the current distribution produces torque,in conjunction with the fundamental component of flux-density, and the effectiveness of the windingin producing the fundamental component is represented by kw1.

In DC machines the integral rA(2)B(2)d2 has no sin2 2 term, and the result is TRV = 2AB, where B is theaverage value of flux-density over the whole rotor periphery. The RMS value of A is equal to the peakvalue, since the current is uniformly distributed around the rotor, and

where Z is the number of rotor conductors, a is the number of parallel paths, Ia is the armature current,and D is the armature (rotor) diameter.

The electric loading A is limited by the slot fill factor, the depth of slot, and the cooling. It is alsorelated to the current density J in the conductors. Suppose the area of one slot is Aslot. Let d = slot-depth, t = tooth width, w = slot width, and 8 = slot pitch = BD/Ns, where Ns is the number of slots. Alsolet J = t/8. Then t + w = 8 and Aslot = wd = (1 ! t)8d. Now if the slot-fill factor Fslot is defined as the ratioof actual copper cross-section area to the total area of each slot, we can write

For example, if the slot depth is d = 15mm, the slot-fill factor is Fslot = 0.4, the tooth-width/slot-pitchratio is J = 0.5, and the electric loading is A = 20 A/mm, the current density is

Typical values of current density for use in AC or brushless machines for different applications aregiven in Table 1.4. Note that in machines operated from electronic drives there are usually time-harmonics in the current which increase the current-density without increasing the torque-producingvalue of A, and it may be necessary to allow for this by multiplying J by a form factor kf. In ACmachines this will be the ratio of the true RMS current to the RMS value of its fundamental component.In DC machines it will be the ratio of the RMS current to the average current.

Condition A/mm2 A/in2

Totally enclosed 1.5—5 1000—3000

Air-over, fan-cooled 5—10 3000—6000

Liquid cooled 10—30 6000—20000

TABLE 1.4

TYPICAL CURRENT DENSITIES (CONTINUOUS OPERATION)

J 'A8

Fslot Aslot

'A

Fslot d(1 ! J ). (1.8)

J '20

0.4 × 15 × (1 ! 0.5)' 6.7 A/mm 2 . (1.9)

A(2) ' A(pk) sin 2 and B(2) ' B(pk) sin 2 . (1.6)

A 'ZIa /a

BD(1.7)


6Switched reluctance machines have very high local flux-densities but a comparatively low magnetic loading, becausethe high flux-density is limited to a small fraction of the stator periphery.

These current-density values assume that the windings are varnished for good heat transfer. In air-cooled machines, the fan is mounted on the rear of the motor outside the frame with a shroud whichfocuses the air over the outside of the motor. Liquid cooled motors may have a passageway around theoutside of the stator with a cooling fluid circulating to remove the heat. The highest values are obtainedwith hollow conductors with coolant flowing through them (“direct conductor cooling”).

It might seem strange to evaluate the magnetic loading as the average flux-density in the airgap ratherthan the peak or RMS value, but the idea behind this is to indicate how well the entire cylinder of steelis being utilized.6 Its value is limited by the available MMF of the excitation source, and by core losseswhich increase rapidly at high flux-density.

It is interesting to see why it is the rotor volume and not its surface area that primarily determines thetorque capability or 'specific output'. As the diameter is increased, both the current and the fluxincrease if the electric and magnetic loadings are kept the same. Hence the diameter (or radius) appearssquared in any expression for specific output. On the other hand, if the length is increased, only the fluxincreases, not the current. Therefore the length appears linearly in the specific output. Thus thespecific output is proportional to D2Lstk, or rotor volume. In practice as the diameter is increased, theelectric loading can be increased also, because more intense fan-cooling or liquid cooling can be usedwithout reducing the efficiency. Consequently the specific output (TRV) increases faster than the rotorvolume.

Although it is theoretically possible to write one general equation from which the torque of any electricmotor can be calculated, in practice a different torque equation is used for every different type ofmotor. Only in certain cases is it possible to discern in this equation an explicit product of flux andcurrent, or even of quantities directly related to them. For example, in the DC commutator motor theelectromagnetic torque is given by

where N is the flux and k is a constant. Here the flux-current product is obvious. In rotating-field ACmachines the classical torque equations do not contain this product explicitly. However, the recentdevelopment of 'field-oriented' or 'vector' controls has necessitated the transformation of the classicalequations into forms in which the flux and current may appear explicity in a scalar or vector product.In eqn. (1.6) it is tacitly assumed that the flux and current are oriented at such angles as to maximizethe torque, but this is not automatically the case except in field-oriented drives. By contrast, in DCmachines the commutator automatically maintains the optimum relative angle of orientation betweenthe flux and the ampere-conductor distribution. In the case of doubly-salient motors such as theswitched reluctance motor and stepper motors, the torque cannot be expressed as the explicit productof a flux and a current. However, the TRV can still be used for initial sizing provided that A and B canbe meaningfully defined (Miller, [1993]).

So far we have restricted attention to the torque per unit rotor volume, a natural consequence of thefact that the torque appears at the rotor surface. For a very rough estimation of overall size includingthe stator, we can use a typical value of 'split ratio' S (i.e., rotor/stator diameter ratio): thus

A typical value of split ratio for an AC motor is in the range 0.55!0.65. For switched reluctance motorsrather smaller values are found. For DC commutator motors the value is usually somewhat higher.

The best way to acquire typical practical values of F or TRV is by experience. An engineer who isfamiliar with a particular design of motor will have built and tested several, and the test data providesvalues of TRV correlated with temperature rise, electric and magnetic loadings etc. The values quotedin Table 1.3 relate to the continuous rating. Peak ratings may exceed these values by 2!3 times,depending on the duration and other factors.

T ' kNIa (1.10)

Stator volume 'Rotor volume

S 2 (1.11)


7This assertion is valid only in the normal range of sizes. In large intensively-cooled machines such as power-stationgenerators, the electric machine clearly outperforms the steam turbine — it is typical for the generator to be dwarfed by theturbines.

" 'Tmp

n Jm %JL

n 2

(1.14)

The TRV determines the volume of the rotor but not its shape. To estimate the rotor diameter and lengthseparately, a length/diameter ratio should be specified. A value around 1 is common; however, it is alsocommon to design motors of different ratings using the same laminations but with different stacklengths. The length/diameter ratio may then vary over a range of 3:1 or more. Very largelength/diameter ratios are undesirable because of inadequate lateral stiffness, but may be used wherea high torque/inertia ratio is desired, or in special cases where the motor has to fit into a narrow space.

The foregoing discussion concerns the electromagnetic torque, that is, the raw torque produced by theelectromechanical energy conversion process at the airgap. The actual torque available at the shaftcoupling is less than this in motors, or greater in generators, by the amount of the mechanical losseswhich include friction, windage, and certain electromagnetic losses appearing on the rotor. Allowanceshould be made for these losses, which typically amount to less than 5% of the electromagnetic torque,and in larger machines or high-efficiency machines, less than 1%.

1.9 GEARING

Compared to the torque density in mechanical and hydraulic devices, the torque density (TRV) inelectric motors is miserably low in comparison with what engineers would really like to achieve.7 Italways has been low, and it always will be low until someone discovers or invents a material that cancarry ten times as much flux as steel for the same magnetizing force; or a material that has a fractionof the resistivity of copper. Such inventions would not by themselves be enough to increase the flea-power of the electric motor by an order of magnitude, unless they were manufacturable in reasonablequantity at reasonable cost—a test which has been repeatedly failed by laboratory prototypes and“wonder motors” for many decades.

For this reason motors are often used with gearboxes to drive the load. A gearbox is the obvious wayto step up the torque. If the gear ratio is n, and Tm is the motor torque, the torque applied to the loadis nTm. The motor speed Tm is increased over the load speed TL by the same ratio. Thus

In most cases the increased motor speed falls in a standard speed range for 'high-speed' motors, whichmay be typically anywhere from a few hundred rev/min to 30,000 rev/min or more.

If the gearbox efficiency is 100%, the output power of the motor is equal to the power applied to theload. The choice of gear ratio depends on how the drive operates. If the speed is constant it is usuallya simple matter of matching the load torque TL to the rated continuous motor torque Tmc :

If, however, the load has a 'dynamic' requirement which specifies a profile of speed or position as afunction of time, the choice of the gear ratio and the motor parameters is more complicated.

Simple acceleration of pure inertia load. Referring to Fig. 1.11, if the motor torque is its peak ratedtorque Tmp, the acceleration of the load is given by

n 'TL

Tmc

(1.13)

TL ' nTm and Tm ' nTL . (1.12)


" 'Tmp & TL /n

n Jm % JL /n 2.

Fig. 1.11 Gear ratio

n 'TL

Tmp

1 % 1 %JL

JM

.Tmp

2

TL2

.

where the term in brackets is the inertia of the motor combined with the load inertia, referred to themotor shaft. If n is large the gearing makes the load inertia insignificant, but it reduces the load speedand acceleration relative to those of the motor. If n is small the referred load inertia is large, and thislimits the acceleration. Between the extremes of large and small n, there is a value that gives maximumacceleration for fixed values of Tmp and the separate inertias. This 'optimum' value can be determinedby equating the differential coefficient d"/dn to zero, giving

which is a well-known result. This value of n makes the referred load inertia equal to the motor inertia.The maximum acceleration of the load is therefore

The corresponding acceleration of the motor is n times this value. In this analysis, the inertias of thepinions (gearwheels) have been ignored. For a very precise evaluation, in the case of a single-stagegearbox, the pinion inertias can be combined with (added to) the respective motor and load inertias.

Acceleration of inertia with fixed load torque. A slightly more complicated example is where theload has a fixed torque TL in addition to its inertia.

Again there is one value of gear ratio n that produces maximum acceleration, and by the samedifferentiation process it is found to be

If the inertias are unchanged from the previous case, the gear ratio is increased. The expression for theoptimum ratio can be substituted back in the formula for acceleration to find the maximum loadacceleration. The result is the same as eqn. (1.16); the difference is that with a larger ratio n the loadacceleration will be smaller. It is interesting to note that the maximum acceleration of the motor isunchanged, and is equal to one-half the torque/inertia ratio of the motor.

"max '12

Tmp

Jm

1n

(1.16)

n 'JL

JM

(1.15)


Peak/continuous torque ratio of motor. In the constant-speed case, the choice of n maximizes theutilization of the continuous torque rating of the motor, Tmc. In the acceleration case, the choice of nmaximizes the utilization of the motor's peak acceleration capability as expressed by its peaktorque/inertia ratio Tmp /Jm. Consider a load that requires both short periods of acceleration and longperiods at constant speed. Then there is a question, can the two values of n be the same? If so, theutilization of both aspects of motor capability will be maximized at the same time.

This problem can be solved analytically in a few special cases, and one solution is given here as anexample of the kind of analysis that is needed to get a highly optimized system design. Assume that theload torque is constant at all times, but that short bursts of acceleration (or deceleration) are requiredfrom time to time. The peak rated torque of the motor will be used for acceleration, and the continuousrated torque for constant speed. If we equate the two separate values of n from the appropriateformulas given above, and if we write

where k is the ratio of peak motor torque to continuous motor torque, then the following relationshipcan be derived:

The left-hand expression is the ratio of the referred motor inertia to the load inertia, and we can referto it as the 'referred inertia ratio' or just 'inertia ratio'. For a range of values of the inertia ratio, theequation can be solved to find the values of k that simultaneously optimize n for both theconstant-speed and acceleration periods. The most interesting result of this is that a large range ofinertia ratio is encompassed by only a small range of values of k : as the inertia ratio changes frominfinity down to 2, k changes only from 2 to 4. But values of k in this range are extremely common: socommon, in fact, as to appear to be a natural characteristic of electric motors. This implies that formost inertia ratios where the referred motor inertia is more than twice the load inertia, the gear ratiocan be chosen to make good utilization of both the continuous torque and the peak acceleration of themotor, provided k $ 2. If k < 2, the gear ratio must be chosen for constant speed or for acceleration, andcannot be optimal for both. The property of electric motors to provide short bursts of peak torque foracceleration is one of the most important aspects of their use in motion control systems.

General speed and position profiles. The cases considered are all idealised by rather restrictiveassumptions that may be too simple in a complex motion-control system. For detailed work it isdesirable to simulate the performance of the whole system using system-simulation software

1.10 COOLING

The need for cooling

There are two major aspects to the thermal problem in electrical machines:

1. heat removal; and 2. temperature distribution within the motor.

The main reasons for limiting the temperature rise of the windings and frame of a motor are:

1. to preserve the life of the insulation and bearings;2. to prevent excessive heating of the surroundings; and3. to prevent injury caused by touching hot surfaces.

Insulation life. The "life" of electrical insulation is inversely related to the temperature. A sustained10EC increase in temperature reduces the insulation life by approximately 50%. The extent to whichexcessive temperatures can be tolerated depends on the duration and the actual temperatures reached.An interesting example of a motor designed for exceptionally high temperatures is the FUMEX motormanufactured by Invensys Brook Hansen, used to extract fumes via the ventilation systems of public

n 2Jm

JL

'k 2

(k & 1)2& 1

. (1.20)

Tmp ' kTmc (1.19)


Q ' k AdTdx

. k A)T

tW (1.22)

buildings in the event of fire; these motors can operate in an ambient temperature of 300EC for a limitedperiod of 30 minutes. Similar considerations apply to bearings. Grease-lubricated bearings may be filledwith high-temperature grease for hot-running applications, but in aerospace machines the bearingsare usually lubricated by separately-cooled oil or oil mist.

Heating of the surroundings is obviously undesirable especially if the motor is heating the equipmentit is driving. For this reason it is important to minimize rotor losses conducted along the shaft. PMmotors have cooler rotors than DC or induction motors. In some applications such as hermeticcompressors used in air-conditioning, refrigeration, etc., the motor losses are removed by the workingfluid, reducing the thermodynamic efficiency of the system.

To prevent injury or harm from touching, exposed surfaces must be kept below 50EC. In certainapplications (e.g., under car bonnets), this requirement is impossible to meet because the "ambient"temperature under the bonnet may reach 100EC. In industrial applications the ambient temperatureis generally less than 50EC, and NEMA ratings for electrical insulation assume an ambient temperatureof 40EC. In aerospace applications motors and generators may be directly cooled by oil or fuel andcoolant temperatures can be as high as 100EC.

The increase in winding temperature increases the resistivity of the windings: a 50EC rise by 20%, anda 135EC rise by 53%, increasing the I 2R losses by the same amount if the current remains the same. Theresistance increase is used in test procedures to determine the actual temperature rise of the winding,but this is obviously an average temperature; hot-spot temperatures can be 10!20E higher. At anytemperature T EC the resistivity of copper can be calculated as

where " = 0.00393 /EC , is the temperature coefficient of resistivity and D20 is the resistivity at 20EC, thatis, 1.728 ×10!8 ohm-m.

Heat Removal

In most industrial and commercial motors, heat is removed by a combination of

1. conduction to the frame mountings;2. air convection, which may be natural or forced; and 3. radiation.

In highly-rated machines direct cooling methods are used:

1. oil mist, especially in aerospace machines;2. immersion in refrigerant, in "hermetic" motors used in refrigerator compressors;3. direct conductor cooling, with hydrogen, oil, or water forced through hollow conductors,

especially in turbine-generators.

Conduction. The conduction equation for a block of thickness t and area A is

where )T is the temperature difference through the thickness t. The coefficient k is the thermalconductivity, with units (W/m2) per (EC/m), i.e. W/EC-m. The thermal conductivity is a materialproperty, and usually it is a function of temperature. Most metals have high thermal conductivities,especially those which are also good electrical conductors. On the other hand, electrical insulatingmaterials and most fluids have low thermal conductivities.

D ' D20 [1 % " (T ! 20)] ohm&m (1.21)


8The absolute temperature in degrees Kelvin (K) is the temperature in EC plus 273.

dTdx

'Q

kA'

2.7

387 × 64 × 10!6' 108 EC/m. (1.23)

QA

' h )T W/m 2 (1.27)

As an example, consider the flow of heat along a conductor whose cross-section area is A = 64 mm2 andlength 50 mm, when the RMS current-density is 7 A/mm2. The electrical resistivity of copper is 1.7 ×10!8 ohm-m, so heat is produced at the rate of J 2D = (7 × 106)2 × 1.7 × 10!8 = 833,000 W/m3 or 83.3 W/cm3.In one conductor the I 2R loss is therefore 833,000 × 64 × 10!6 × 50 × 10!3 = 2.7 W. To take the mostpessimistic estimate, assume that all of this heat is generated at the mid-point of the coil-side, half-wayalong the motor. The thermal conductivity of copper is 387 (W/m2) per (EC/m). So the temperaturegradient along the coil-side is given by eqn. (1.22) as

Since the heat can flow in both directions, the temperature-gradient is only half this value, and thetemperature rise between the ends of the stack and the centre is therefore 110/2 × 50 × 10!3/2 = 1.4 EC,which is negligible. A more thorough analysis would have to consider the full diffusion equation alongthe length of the coil-side, but this quick calculation reveals that such sophistication is not needed inthe example considered.

Thermal resistance and contact resistance. Eqn. (1.24) can be used to define thermal resistance asthe ratio of temperature difference )T to heat flow rate Q : the symbol used for thermal resistance isR, with units EC/W. Thus

The thermal resistance is a "lumped parameter" that can be used to model the conduction through aregion or interface where the individual values of k, A, and t may be difficult to determine. The contactresistance between two surfaces is usually treated in this way, as, for example, between the frame andthe stator core. The temperature drop across a thermal resistance is given by eqn. (1.24) as )T = QR.For example, if the contact resistance between the motor flange and the mounting plate is 1EC/W, thenwith 40W flowing though it the temperature difference across the interface would be 40EC.

Radiation. Radiation is described by the Stefan-Boltzmann equation

where F is the Stefan-Boltzmann constant, 5.67 × 10!8 W/m2/K4 for a black body, T1 is the absolutesurface temperature of the radiating body in degrees Kelvin, and T2 is the absolute temperature of thesurroundings.8 A black body is a perfect radiator, that is, one which reflects no radiated heat butabsorbs all the heat radiated towards it. Real surfaces are imperfect radiators, and their radiativeeffectiveness relative to that of a black body is called the emissivity e. A black lacquered surface canachieve an emissivity as high as 0.98, but a more practical rule of thumb is to take 0.9 for black-paintedor lacquered surfaces. For example, a surface with an emissivity of 0.9 that is 50EC above thesurroundings at 50EC, has a net heat transfer rate of

which is 432 W/m2 or 0.28 W/in2. A surface 30EC above the surroundings at 20EC has a rate of 186 W/m2

— quite a useful component of the heat-removal capability of the frame.

Convection. Heat removal by convection is governed by Newton's Law:

where )T is the temperature difference between the cooling medium and the surface being cooled, andh is the heat-transfer coefficient. The units of h are W/m2/EC. The value of h depends on the viscosity,

0.9 × 5.67 × 10!8 × ((50%50%273)4& (50%273)4 ) (1.26)

QA

' e F (T14 ! T2

4) W/m 2 (1.25)

R ')TQ

't

kAEC/W. (1.24)


h . 7.5 )TD

1/4

W/m 2/EC (1.28)

h . 125 VL

W/m 2/EC (1.29)

thermal conductivity, specific heat, and other properties of the coolant, and also on its velocity. Innatural convection the flow of coolant is not assisted by fans, blowers, pumps etc. In forced convectionthe flow is assisted by one of these external means.

The heat transfer coefficient for natural convection around a horizontally-mounted unfinnedcylindrical motor can be roughly estimated as

where D is in mm. For example, for an unfinned cylinder of diameter D = 100 mm and a temperaturerise of 50EC, the natural-convection heat-transfer coefficient is calculated as 6.3 W/m2/EC. For a )T of50EC, the heat transfer rate is then given by eqn. (1.28) as 6.3 × 50 = 315 W/m2. As a first approximationthis value can be applied to the whole surface including the ends, but if the motor is flange-mountedthen only one end is available for convective cooling.

Forced convection (with a shaft-mounted fan or an external blower) increases the heat-transfercoefficient by as much as 5!6 times, depending on the air velocity. The increase in heat-transfercoefficient is approximately proportional to the square-root of the air velocity. An approximate formulafor the forced-convection heat-transfer coefficient is

where V is the actual air velocity [m/s] and L is the frame length [mm] (assumed parallel to thedirection of airflow). For a motor of length 100 mm, if the air velocity is 4 m/s, this formula predictsh = 25 W/m2/EC. This is 4 times higher than for natural convection.

Some rules of thumb for "calibration". In a water-immersed wire 1 m long, 1 mm diameter, a powerloss of 22 W (0.022 W per mm length) is sufficient to boil the water at the wire surface. The wire surfacetemperature is 114EC and the heat transfer coefficient (see below) is 5000 W/m2/EC. The heat flow at thewire surface is 0.07 W/mm2 and the current-density in the wire is approximately 35 A/mm2. In normalmotors, the rate at which heat can be abstracted is far less than this, and current-densities over 30A/mm2 are achievable only for short bursts. 35 A/mm2 is sufficient to fuse a copper wire in free air.

The maximum rate of heat removal by natural convection and radiation (with 40EC rise) is only about800 W/m2. With forced air convection the rate increases to about 3000 W/m2, and with direct liquidcooling about 6000 W/m2. A motor that generates more heat than can be removed at these rates mustabsorb the heat in its thermal mass, which permits the output power to be increased for a short time.These rates limit the heat generated per unit volume to about 0.012 W/cm3 for natural convection, 0.3W/cm3 for metallic conduction or forced-air convection, and 0.6 W/cm3 for direct liquid cooling.

The permissible current-density cannot be directly related to the temperature rise of the winding bya simple general equation, because the heat transfer rate depends on the shape of the conductors. Asan example, 1 cm3 of copper can be made into a stubby cylinder of 1 cm diameter and 1.27 cm length,or a long wire of 1 mm diameter and 1.27 m length. The short cylinder has a cylindrical surface areaof 4 cm2 while the long wire has a surface area of 40 cm2. The loss density in the conductor is J 2D whereJ is the current density and D is the resistivity. With ten times the surface area the long wire candissipate ten times the heat, assuming the same heat transfer coefficient in both cases. This suggeststhat the permissible current-density in the long wire can be /10 times that in the short stubby cylinder.

If rated torque is required at very low speed, a shaft-mounted fan may not provide enough coolant flowto keep the motor cool. DC motors often have separate AC-driven fans, because they have to work forprolonged periods at low speed with high torque. Since most of the heat in a DC motor is generated onthe rotor, good internal airflow is essential. In DC motors the external fan is usually mounted to oneside of the motor, where it is easily accessible, and does not increase the overall length. With vector-controlled induction motors a common practice is to mount the fan in line with the motor at the non-drive end, and arrange it to blow air over the outside of the finned frame. The fan may increase theoverall length by up to 60%. Brushless motors have less severe problems because most of the heat atlow speed is generated in the stator windings, and very little on the rotor.


L 2 T %1kMqMt

'1"

MTMt

(1.30)

" 'kDc

m 2/s (1.32)

Internal temperature distribution

The steady-state temperature distribution within the motor is essentially a diffusion problem. The mostimportant aspect of the problem is finding the hottest temperature in the motor, given a certaindistribution of losses and a known rate of heat removal. It is difficult to solve precisely, because ofthree-dimensional effects and because some thermal resistances (such as the resistance between slotconductors and slot liner) are hard to calculate.

The differential equation for three-dimensional conduction of heat is the so-called diffusion equation:

where

and

is the diffusivity in SI units. In SI units, k is the thermal conductivity in W/mEC; c is the specific heatin kJ/kgEC, and D is the density in kg/m3. In a structure as complex as an electric motor the heatconduction equation is a complex boundary-value problem that is best solved by computer-basednumerical methods such as the finite-element method.

In electric motors internal convection and radiation may be as important as conduction, and when thedifferential equation is extended to include them, matters become very complicated, even for steady-state calculations. During transients the temperature distribution can be very different from thesteady-state distribution, and different methods of analysis may be needed for the two cases.

Thermal equivalent circuit. For most purposes it is sufficient to use a thermal equivalent circuit ofthe interior of the motor, Fig. 1.12. This is analogous to an electric circuit, in that heat is generated by"current sources" and temperature is analogous to voltage. The rate of generation of heat in a sourceis measured in Watts. The heat flow rate, which is also measured in Watts, is analogous to current.Resistance is measured in EC/W. The copper losses, core losses, and windage & friction losses arerepresented by individual current sources, and the thermal resistances of the laminations, insulation,frame, etc. are represented as resistances. In the simplest possible model, all the losses are representedtogether as one total source, i.e. the individual sources are taken as being in parallel. The thermalequivalent circuit is really a lumped-parameter model of all the heat-flow processes within the motoras well as the heat removal processes discussed earlier.

The thermal equivalent circuit should ideally take into account the anisotropy effects: for example, theeffective thermal conductivity through a lamination stack is lower in the axial than the radialdirection. A more complex thermal equivalent circuit may include provision for direct cooling of thewinding conductors, or for direct cooling of the rotor shaft. If it also includes the thermal masses orcapacities of the winding, the rotor and stator laminations, the frame, the shaft, and other massivecomponents, then it can be solved for transient as well as steady-state heat transfer. The heat removalroutes by conduction, radiation, and convection are represented by thermal resistances. For convectionthe appropriate resistance Rv is given by

where A is the appropriate surface area for convective heat-transfer and the subscript "v" stands forconvection. If h is a function of the temperature-difference, the equivalent circuit becomes non-linearand requires an iterative solution. For radiation the equivalent thermal resistance Rv is the ratio ofthe temperature difference T1 ! T2 to the radiation heat exchange rate Q in eqn. (1.25). Clearly this isnon-linear. However, the non-linearity is often neglected and a fixed value of Rv is calculated assumingthat the case temperature is known.

Rv '1

hAEC/W (1.33)

L 2 T 'M2T

Mx 2%

M2T

My 2%

M2T

Mz 2(1.31)


Fig. 1.12 Thermal equivalent circuit. S = stator (tooth centre), T = tooth (at airgap), Y= stator yoke, E = end-winding, C = conductors (at central plane), G = airgap,H = shaft, A = ambient. BloCool = heat abstracted by through airflow (W), R =radiation, U = conduction, V = convection. Double letters refer to thermalresistances, e.g., CT = thermal resistance from the conductors to the statorteeth.

Some useful data is provided in the following tables.

Motor type Class B Class F Class H

1.15 Service Factor 90 115 140

1.00 Service Factor 85 110 135

TEFC 80 105 125

TENV 85 110 135

TABLE 1.5TEMPERATURE RISE BY RESISTANCE AND INSULATION

(NEMA Standard MG-1), EC. Assumes 40EC ambient temperature.

Material Emissivity

Polished aluminium 0.04

Polished copper 0.025

Mild steel 0.2-0.3

Grey iron 0.3

Stainless steel 0.5-0.6

Black lacquer 0.9-0.95

Aluminium paint 0.5

TABLE 1.6


SELECTED EMISSIVITIES

Material D (20EC)

ohm-m × 10!8

k

(W/m K)

Sp. Heat

kJ/kg/EC

Density

kg/m3

Copper 1.72 360 0.38 8950

Aluminium 2.8 220 0.90 2700

0.1% Carbon steel 14 52 0.45 7850

Silicon steel 30!50 20!30 0.49 7700

Cast iron 66 45 0.5 7900

Cobalt-iron 40 30 0.42 8000

Ceramic magnet 104 4.5 0.8 4900

Re-Co magnet 50 10 0.37 8300

NdFeB magnet 160 9 0.42 7400

Kapton® 303 V/µm* 0.12 1.1 1420

Teflon 260V/µm* 0.20 1.2 2150

Pressboard/Nomex 10kV/0.22mm* 0.13 — 1000

Epoxy resin 30kV/mm* 0.5 1.7 1400

Water (20EC) 0.0153 4.18 997.4

Freon 0.0019 0.966 1330

Ethylene Glycol 0.0063 2.38 1117

Engine oil 0.0037 1.88 888

TABLE 1.7 SELECTED MATERIAL PROPERTIES

*Dielectric strength


tcy ' tON % tOFF . (1.34)

d 'tON

tcy

'tON

tON % tOFF

. (1.35)

J ' RC (1.36)

Fig. 1.13 Intermittent operation

Fig. 1.14 Simple thermal equivalent circuit for transient calculations

1.11 INTERMITTENT OPERATION

Intermittent operation is normal for brushless PM motors, because most of the applications that usethem are motion-control applications with programmed moves, accelerations, decelerations, stops,starts, and so on. Consequently the temperatures of the windings and magnets are constantly varying.A simple example is shown in Fig. 1.13, where the motor executes a simple on-off sequence: on for tON

and off for tOFF, after which the on/off cycle repeats indefinitely. The cycle time tcy is

The duty-cycle d is defined as

The most efficient use of the thermal capability of the motor will be made if the maximum windingtemperature Tmax just reaches the rated value Tr at the end of each on-time. Because the powerdissipation is interrupted with cool-down intervals tOFF, the power Pd that can be dissipated during theon-times may exceed the steady-state continuous dissipation rating of the motor Pr, and therefore themotor may be permitted to exceed its steady-state output power rating during the on-times. Thesimplified thermal equivalent circuit model in Fig. 1.14 makes it possible to calculate the permissibleoverload factor as a function of the on-time tON and duty-cycle d for a given motor.

The thermal equivalent circuit is a parallel combination of thermal resistance R and thermalcapacitance C. R represents the steady-state thermal resistance between the winding and thesurroundings in EC/W. C represents the thermal capacity of the entire motor in J/EC. The thermal time-constant J is given [in seconds] by eqn. (1.36):

The analysis proceeds by equating the temperature rise during the on-time with the temperature fall


Tr ! T0 ' RPr , (1.39)

k 2'

Pd

Pr

. (1.40)

Tmax ' Tr , (1.41)

(Tr ! T0 ) [1 ! k 2 (1 ! e!tON/J

) ] ' (Tc ! T0 )e!tON/J (1.42)

Tc ! T0 ' (Tr ! T0 )e!tOFF/J

. (1.44)

during the off-time. To do this we need the equations for the temperature rise and the temperature fall.

Temperature rise during ON-time. During the on-time tON, the power dissipation in the motor is Pd

and the temperature rises according to the equation

The temperature rise is expressed relative to the ambient temperature T0. The second term in eqn.(1.37) is due to the initial condition in which the temperature rise is (Tc ! T0) at t = 0. At t = tON,

By definition, the steady-state rated temperature-rise (Tr ! T0) is given by

where Pr is the rated steady-state power dissipation in the motor, i.e., the continuous power dissipationthat produces rated temperature rise. We can use this to "calibrate" Pd in eqns. (1.37) and (1.38), bydefining the dissipation overload factor k2, where

The reason for using k2 instead of k is that in most types of brushless servomotor the losses aredominated by I2R losses while the load torque is proportional to the current I. If the load is increasedby a factor k, it means that the current and torque are increased by the factor k while the lossesincrease by k2. Thus k is the overload factor for torque and current.

Substituting equations 1.39 and 1.40 in eqn. (1.38) and rearranging, and assuming that

we obtain the following equation relating the temperature rise to the overload factor and the on-time:

Temperature fall during OFF-time. When the motor is switched off, the power dissipation falls tozero and the winding temperature falls according to the equation

where t is measured from the end of the on-time, i.e. the beginning of the off-time. At tOFF,

Steady-state : equating the temperature rise and fall. First, multiply eqn. (1.44) by e!tON/J :

The left-hand side of eqn. (1.45) is identical to the right-hand side of eqn. (1.42), so the right-hand sideof eqn. (1.45) can be equated to the left-hand side of eqn. (1.42). With suitable rearrangement, the resultcan be expressed in different ways, all of which are useful for different purposes.

Maximum overload factor. First, we get a solution for the dissipation overload factor k2 in terms ofthe on-time and the duty-cycle: writing tON/d instead of tON + tOFF, i.e., instead of tcy, the expression is

( Tc ! T0)e!tON/J

' ( Tr ! T0)e!(tON % tOFF)/J

. (1.45)

Tmax ! T0 ' RPd(1 ! e!tON/J

) % (Tc ! T0 )e!tON/J

. (1.38)

T ! T0 ' RPd (1 ! e !t/J) % (Tc ! T0 )e !t/J . (1.37)

k 2'

1 ! e!tON/Jd

1 ! e!tON/J

(1.46)

T ! T0 ' (Tr ! T0 )e !t/J (1.43)


k 2'

1 ! e !0.2/0.25

1 ! e !0.2' 3.04, (1.47)

k 2'

1d

. (1.48)

tOFF ' ! 40 × ln[2 ! (2 ! 1) × e 8/40 ] ' 10.0 min. (1.51)

tON ' J ln k 2

k 2 ! 1. (1.53)

k 2'

1

1 ! e!tON/J

. (1.49)

For example, if the duty-cycle is 25% (d = 0.25) and tON = 0.2 × J, the dissipation overload factor is

which means that the dissipation can be increased to 304% of its rated steady-state value for a periodof tON = 0.2J in every cycle of length tcy = tON/d = (0.2/0.25)J = 0.8J. If J = 40 min, the dissipation can beraised to 304% for 8 minutes followed by a cool-down period of 24 minutes. Increasing the dissipationto 304% corresponds to an increase in current and torque to %k = %3.04 = 1.74 times their rated values.

If tcy << J, then eqn. (1.46) simplifies so that

This means that when the on/off cycles are very short compared with the thermal time-constant of themotor, the mean dissipation will be equal to Pr when the peak dissipation Pd = k2Pr is equal to Pr/d. Thissimple result is intuitive.

Maximum overload for a single pulse. Eqn. (1.46) can also be used to calculate the maximumdissipation overload factor for a single pulse, for which d = 0. In this case

For example, if tON = 8 min and J = 40 min, then the maximum dissipation overload factor k2 is 5.5 or550%, and the maximum overload factor k is 2.35 or 235%.

Required cool-down period for a given overload factor and on-time. The second result that arisesfrom equating the temperature rise and temperature fall is an expression for the necessary cool-downtime tOFF as a function of the dissipation factor k2 and the on-time tON. The expression is

Together with eqn. (1.35), this can be used to determine the maximum duty-cycle d that can be usedwith a given dissipation overload factor k2 and a given on-time tON, for a motor of thermal time-constantJ. For example, if the dissipation is 200% of rated, and if tON = 8 min, J = 40 min,

The minimum cycle time is therefore 18 min and the maximum duty-cycle (with 8 minutes' on-time) is8/18 = 0.44 or 44%.

Maximum on-time for a given overload factor and cool-down time. A third result obtained byequating the temperature rise and fall is an expression for the maximum on-time tON as a function ofthe dissipation overload factor k2 and the off-time tOFF. The expression is

Maximum duration of single pulse. This expression can be used to calculate the maximum durationof a single pulse having a given dissipation overload factor k2. For a single pulse, tOFF is infinite and

tON ' J ln k 2 ! e!tOFF/J

k 2 ! 1. (1.52)

tOFF ' ! J ln [k 2 ! (k 2 ! 1)etON/J

] . (1.50)


Fig. 1.15 Intermittent heating curves

For example, if k2 = 5.5 and J = 40 min, then tON =8 min.

Graphical transient heating curves. Fig. 1.15shows the relationship expressed by eqn. (1.53)graphically in terms of the duty-cycle d, the on-time tON as a fraction of the time-constant J, andthe overload factor k.

This graph can be used in a number of ways. Forexample, to determine the maximum permissibleduration of a single pulse with a given overloadfactor k, the duty-cycle d should be set to zero.Thus with k = 1.5 the maximum pulse duration is0.58J. With a time-constant of 40 min this is 23.2min.

The graph shows the maximum duty-cycle thatcan be used with a given overload factor. Forexample, at 200% load the maximum duty-cycle is0.25 or 25%, but in this limiting case the on-timemust be vanishingly small. With an on-time of0.1J at 200% load, the maximum duty-cycle isapproximately 0.2, which means that the cool-down period in each cycle must be (1!d)J = 0.9J.If J is 40 min, this means a maximum operatingtime at 200% load of 4 min, followed by a cool-down period of 36 min before the cycle can berepeated. Operations that need a short on-timewith a high duty-cycle must use a lower overloadfactor.

1.12 PERMANENT MAGNET MATERIALS AND CIRCUITS

The permanent-magnet industry has continually improved the properties of PM materials in the past20!30 years, mainly by painstaking development of the metallurgy of existing materials. Samples of themain families of PM materials used in electric machines are shown in Table 1.8.

Property Units Alnico 5-7 Ceramic Sm2Co17 NdFeB

Remanence Br T 1.35 0.41 1.06 1.2

Coercivity Hc kA/m 60 325 850 1000

Energy product (BH)max kJ/m3

60 30 210 250

Relative recoil permeability µrec 1.9 1.1 1.03 1.1

Specific gravity 7.3 4.8 8.2 7.4

Resistivity µS-cm 47 >104

86 150

Thermal expansion coefficient 10!6

/EC 11.3 13 9 3.4

Temperature coefficient of Br %/EC !0.02 !0.2 !0.025 !0.1

Saturation H kA/m 280 1120 > 3200 > 2400

TABLE 1.8TYPICAL MAGNET PROPERTIES


Fig. 1.16 B-H loop of a hard PM material with electrical steel shown for comparison

The 'strength' of a magnet is sometimes measured by its 'energy product' (see below). At roomtemperature NdFeB has the highest energy product of all commercially available magnets. The highremanence and coercivity permit marked reductions in motor size, compared with motors usingFerrite (ceramic) magnets. However, ceramic magnets are considerably cheaper than Rare Earth orNdFeB.

Both ceramic and NdFeB magnets are sensitive to temperature and special care must be taken if theworking temperature is above 100EC. For very high temperature applications Alnico or RareEarth/Cobalt magnets must be used, for example Sm2Co17 which is useable up to 200 EC or even 250 EC.

NdFeB is produced either by a mill-and-sinter process (Neomax) or by a melt-spin casting processsimilar to that used for amorphous alloys (Magnequench). NdFeB magnets are often made in ringswhich may be sintered or polymer bonded, but they can be formed in a wide variety of other shapes.They are not 100% dense and coatings or electroplating may be necessary to prevent corrosion.

B-H loop and demagnetization characteristics. The starting-point for understanding magnetcharacteristics is the B-H loop or 'hysteresis loop', Fig. 1.16. The x-axis is the magnetizing force or'magnetic field intensity' Hm in the material. The y-axis is the magnetic flux-density Bm in the material.An unmagnetized sample has Bm = 0 and Hm = 0 and therefore starts out at the origin. If it is subjectedto a magnetic field, as for example in a magnetizing fixture, Bm and Hm in the magnet will follow theinitial magnetization curve as the external ampere-turns are increased. If the external ampere-turnsare switched off, the magnet relaxes along the curve shown by the arrows. Its operating point (Hm, Bm)will depend on the shape of the magnet and the permeance of the surrounding 'magnetic circuit'. If themagnet is surrounded by a highly permeable magnetic circuit, that is, if it is 'keepered', then its polesare effectively shorted together so that Hm = 0 and the flux-density is then equal to the remanence Br.This is the maximum flux-density that can be retained by the magnet at a specified temperature afterbeing magnetized to saturation.

External ampere-turns applied in the opposite direction (i.e., Hm < 0) cause the magnet's operating pointto follow the curve through the second and third quadrants until the magnet is saturated in theopposite direction. Again, if the current is switched off the operating point returns towards the point


9 Sometimes eqn. (1.54) is written Bm = µ0(Hm + M) and then the magnetization is measured in kA/m instead of T.

Fig. 1.17 2 n d - q u a d r a n t d e m a g n e t i z a t i o ncharacteristic showing intrinsic curve

Bm ' µ0 Hm % J. (1.54)

(0,!Br), but because of the demagnetizing effect of the external magnet circuit, Bm falls to a (negative)value smaller than Br. It is now magnetized in the opposite direction and the maximum flux-densityit can retain when 'keepered' is !Br.

To bring the flux-density to zero from the original positive remanence point (0,Br), the externalampere-turns must provide within the magnet a negative magnetizing force !Hc, called the coercivity.Likewise, to return the flux-density to zero from the negative remanence point (0,!Br), the field +Hcmust be applied. The entire loop is usually symmetrical and can be measured using instrumentsdesigned specially for magnet testing.

If negative external ampere-turns are applied, starting from the positive remanence point (0,Br), andswitched off at R, the operating point of the magnet 'recoils' and will operate along the lower curve ofa 'minor loop'. For practical purposes the minor loop of high-coercivity magnets is very narrow andcan be taken as a straight line, the recoil line, whose slope is equal to the recoil permeability, µrec. Thisis usually quoted as a relative permeability, so that the actual slope is µrecµ0 H/m. Operation along therecoil line is stable provided that the operating point does not go outside the original hysteresis loop.

A 'hard' PM material is one whose recoil lines are straight throughout all or most of the secondquadrant, which is where the magnet normally operates in service. In very hard magnets that are fullymagnetized, the recoil line is coincident with the second-quadrant section of the hysteresis loop. Thisis characteristic of ceramic, Rare Earth/Cobalt, and NdFeB magnets, which usually have µrec between1.0 and 1.1. 'Soft' PM materials have a 'knee' in the second quadrant, such as Alnico. While Alnicomagnets have very high remanence and excellent mechanical and thermal properties, they have lowcoercivity and are therefore limited in the demagnetizing field they can withstand.

Compared with 'electrical steel' used in laminations, even the 'soft' PM materials are very 'hard' : inother words, the hysteresis loop of a typical nonoriented electrical steel is very narrow and has a lowcoercivity and a high permeability; see Fig. 1.16. The high permeability is desirable in order tominimize the magnetizing MMF (which is supplied by the magnets in PM motors, or by the magnetizingcurrent in induction motors). The narrow loop is desirable because the loop area represents an energyloss or “hysteresis” loss which is dissipated every time the loop is traversed, and in AC motors(including brushless PM motors) the loop is traversed at the fundamental frequency.

The most important part of the B-H loop is the secondquadrant, Fig. 1.17. This is called the demagnetizationcurve. In the absence of externally applied ampere-turns,the magnet operates at the intersection of thedemagnetization curve and the 'load line', whose slope isthe product of µ0 and the 'permeance coefficient' (PC) of theexternal circuit: i.e., at (Hm,Bm), with Hm < 0.

Since Bm and Hm in the magnet both vary according to theexternal circuit permeance, it is natural to ask what it isabout the magnet that is 'permanent'. The relationshipbetween Bm and Hm can be written

The first term is the flux-density that would exist if themagnet were removed and the magnetizing force remainedat the value Hm. Therefore the second term can be regarded as the contribution of the magnet to theflux-density within its own volume; accordingly, J is called the magnetization and it is measured intesla.9 If the demagnetization curve is straight, and if its relative slope µrec = 1, then J is constant. Thisis shown in Fig. 1.17 for negative values of Hm up to the coercivity !Hc. In most hard magnets µrec isslightly greater than 1 and there is a slight decrease of J as the negative magnetizing force increases,but this is reversible down to the 'knee' of the B-H loop (which may be in either the second or the third


Fig. 1.18 Closed and gapped magnetic circuits

quadrant, depending on the material and its grade). Evidently the magnet can recover or recoil backto its original flux-density as long as the magnetization is constant. The coercive force required todemagnetize the magnet permanently is called the intrinsic coercivity and this is shown as Hci.

For engineering purposes we normally represent the recoil line by the equation

which can be related to eqn. (1.54) by expanding it as follows:

which indicates that

where P is the susceptibility, 1 ! µrec.

Another parameter often calculated is the magnet energy product, BmHm. This is not the actual storedmagnet energy but simply indicates how hard the magnet is working against the demagnetizinginfluence of the external circuit. Contours of constant energy product are rectangular hyperbolas BmHm= constant, often drawn on data sheets. The maximum energy product (BH)max occurs where thedemagnetization characteristic is tangent to the hyperbola of its (BH)max value. If the relative recoilpermability is unity, this occurs for a permeance coefficient of unity, with Bm = Br/2, provided thatthere are no externally applied ampere-turns from windings or other magnets.

Calculation of Magnet operating point. Fig. 1.18 shows a simple magnetic circuit in which themagnet is 'keepered' by a material or core of relative permeability µr. The core and magnet togetherform a closed magnetic circuit. Applying Ampere's law, and assuming uniform magnetizing force inboth the magnet and the core,

Hmlm % HFelFe ' 0. (1.58)

Bm ' µ0 Hm % µ0 (1 ! µrec)Hm % Br (1.56)

Bm ' µ0 µrecHm % Br (1.55)

J ' µ0 (1 ! µrec)Hm % Br ' µ0PHm % Br (1.57)


Bm

Hm

' ! µ0

Ag

Am

lm

lg

' ! µ0 × PC (1.63)

Hmlm % HFelFe % Hglg ' 0. (1.60)

Hm ' !lFe

lm

HFe (1.59)

CM 'Am

Ag

. (1.64)

BmHm 'BgHgAglg

Amlm

'2Wg

Vm

(1.65)

Hm . !lg

lm

Hg (1.61)

where Hm is the magnetic field in the magnet, HFe is the magnetic field in the iron core (assumed to beuniform around the core length lFe, and lm is the length of the magnet in the direction of magnetization.This is effectively the line integral of H around the magnetic circuit, and it is zero because there areno externally applied ampere-turns. Hence

which establishes that the magnet works in the second quadrant of the B-H loop. Now consider thegapped magnetic circuit in Fig. 1.18, in which there is an airgap in series with the magnet and the twosections of iron core. Now

where Hg is the magnetic field in the airgap and lg is the airgap length. The permeability of theelectrical steel used in motors is usually several thousand times higher than µ0, so that the term HFelFecan be neglected as a first approximation, even though lFe may be much bigger than lg. Then

Now by Gauss' law, the flux-densities in the magnet and the airgap are related by

so that if we take the ratio of Bm/µ0Hm, recognizing that in the airgap Bg = µ0Hg, we get

where PC is the permeance coefficient. The ratio of magnet pole area to airgap area is sometimes calledthe flux-concentration factor or flux-focussing factor:

In order to minimize the risk of demagnetization we need to operate the magnet fairly close to Br, i.e.,with a high permeance coefficient. On the other hand, the airgap flux-density Bg is increased if we usea high value of the flux-concentration factor Am/Ag. But this reduces the permeance coefficient andeqn. (1.63) shows that this reduces the ratio Bm/Hm, which increases the risk of demagnetizationbecause it moves the operating point further down the recoil line away from Br towards the knee of theB-H curve.

To achieve a high permeance coefficient with a high flux-concentration factor we must increase theratio lm/lg to compensate for the demagnetizing effect of the airgap: in other words, use a magnet thatis long in the direction of magnetization and also long relative to the airgap length. It does not meanlong in relation to the lateral dimensions of the magnet, and indeed most modern magnets exceptAlnico have such high coercivity that the length in the direction of magnetization is thesmallest dimension and is intuitively referred to as the 'thickness'!

The energy product is given by

where Wg is the magnetic energy stored in the airgap volume and Vm is the volume of the magnet. Thisshows that the minimum magnet volume required to magnetize a given working volume of airspace

BmAm ' BgAg (1.62)


u ' Hglg A&t (1.66)

!u . Hmlm A&t . (1.67)

is inversely proportional to the working energy product BmHm . Therefore, in these cases it pays todesign the magnet length and pole area in such proportions relative to the length and area of theairspace, as to cause the magnet to work at (BH)max, which is a property of the particular material ata given temperature. In motors this principle cannot be applied so simply, because the armaturecurrent produces demagnetizing ampere-turns that may be very great under fault conditions. Toeliminate the risk of demagnetization, motors are designed so that on open-circuit or no-load, themagnet operates at a high permeance coefficient with an adequate margin of coercive force to resistthe maximum demagnetizing ampere-turns expected under load or fault conditions.

The lower diagrams in Fig. 1.18 illustrate the relative intensities of Bm and Hm under different workingconditions, in all cases with no externally applied ampere-turns. Note that B is continuous throughoutthe magnetic circuit (because it obeys Gauss' law), but H is not. The discontinuities of H are associatedwith the appearance of magnetic poles at the interfaces between different sections of the magneticcircuit, notably at the 'poles' of the magnet and the working airspace. The polarization of surfaces givesrise to a magnetic potential difference across the airspace which is useful for calculating fluxdistribution in motors. In Fig. 1.18 this potential difference is

If the magnetic potential drop in the steel is neglected, the corresponding magnetic potential differenceacross the magnet is

C.g.s. units are still widely used in the magnet industry, whereas motors are usually designed in metric(SI) units in Europe and Japan, and in metric or Imperial units in the U.S.A. Some conversion factorsare as follows:

1 inch 25.4 mm

1 T 10 kG

1 kA/m 4B Oe

1 kJ/m3 B/25 MGOeTABLE 1.9

CONVERSION FACTORS

Temperature effects; reversible and irreversible losses

High-temperature effects. Exposure to high temperatures for long periods can produce metallurgicalchanges which may impair the ability of the material to be magnetized and may even render itnonmagnetic. There is also a temperature, called the Curie temperature, at which all magnetizationis reduced to zero. After a magnet has been raised above the Curie temperature it can be remagnetizedto its prior condition provided that no metallurgical changes have taken place. The temperature atwhich significant metallurgical changes begin is lower than the Curie temperature in the case of theRare Earth/Cobalt magnets, NdFeB, and Alnico; but in ceramic ferrite magnets it is the other wayround. Therefore ceramic magnets can be safely demagnetized by heating them just above the Curiepoint for a short time. This is useful if it is required to demagnetize them for handling or finishingpurposes. Table 1.10 shows these temperatures for some of the important magnets used in motors.

Metallurgical change EC Curie temperature EC

Alnico 5 550 890

Ceramic 1080 450

Sm2Co17 350 800

NdFeB 200 310TABLE 1.10

METALLURGICAL CHANGE AND CURIE TEMPERATURE


Material Temp. coefft. of Br %/EC

Alnico 5-7 !0.02

Ceramic !0.19

Sm2 Co17 !0.02

NdFeB !0.11TABLE 1.11

REVERSIBLE TEMPERATURE COEFFICIENTS OF BR

Reversible losses. The B-H loop changes shape with temperature. Over a limited range the changesare reversible and approximately linear, so that temperature-coefficients for the remanence andcoercivity can be used. Table 1.11 gives some typical data. Ceramic magnets have a positive coefficientof Hc, whereas the high-energy magnets lose coercivity as temperature increases. In ceramic magnetsthe knee in the demagnetization curve moves down towards the third quadrant, and the permeancecoefficient at the knee decreases. Thus ceramic magnets become better able to resist demagnetizationas the temperature increases up to about 120EC. The greatest risk of demagnetization is at lowtemperatures when the remanent flux-density is high and the coercivity is low; in a motor, this resultsin the highest short-circuit current when the magnet is least able to resist the demagnetizingampere-turns. In high-energy magnets the knee moves the other way, often starting in the thirdquadrant at room temperature and making its way well into the second quadrant at 150 EC. Grades witha high resistance to temperature are more expensive, yet these are often the ones that should be usedin motors, particularly if high temperatures are possible (as they usually are under fault conditions).

All the magnets lose remanence as temperature increases. For a working temperature of 50EC abovean ambient of 20EC, for instance, a ceramic magnet will have lost about 10%. This is spontaneouslyrecovered as the temperature falls back to ambient.

Irreversible losses recoverable by remagnetization. (a) Domain relaxation. Immediately aftermagnetization there is a very slow relaxation, starting with the least stable domains returning to astate of lower potential energy. The relaxation rate depends on the operating point and is worse below(BH)max, i.e. at low permeance coefficients. In modern high-coercivity magnets at normal temperaturesthis process is usually negligible, particularly if the magnets have been stabilized (by temperaturecycling and/or AC flux reduction) immediately after magnetization. Elevated temperatures duringsubsequent operation may, however, cause an increased relaxation rate. This can be prevented bytemperature-cycling in the final assembly over a temperature range slightly wider than the worst-caseoperating range. Subsequent relaxation is reduced to negligible levels by this means. Table 1.12 showsthe stability of different magnet materials at 24 EC.

Material % loss after 10 years (typ.)

Ceramic < 0.01

Rare earth/Cobalt 0.2

Alnico 0.5TABLE 1.12

LONG-TERM STABILITY OF MAGNET MATERIALS

(b) Operating point effect. Temperature alters the B-H loop. If this causes the operating point to 'fall off'the lower end of a recoil line, there will be an irreversible flux loss. This is illustrated in Fig. 1.19.Initial operation is at point a on the load line Oa, which is assumed to remain fixed. The remanencecorresponding to point a is at point A. When the temperature is raised from T1 to T2 the operatingpoint moves from a to b, and the corresponding remanence moves from A to B. Note that because theknee of the curve has risen above point b, the effective remanence at B' is less than that at B, whichis what it would have been if the magnet had been working at a high permeance coefficient.


Fig. 1.19 Reversible and irreversbile loss caused by operating at a high temperature with a low permeance coefficient.

If the temperature is now reduced to T1 the operating point can recover only to a', which lies on therecoil line through A'. The recovery from b to a' is reversible, but there has been an irreversible lossof flux-density )Bm in the magnet, relative to point a. The remanence at T1 has fallen from A to A'. Theloss can be recovered only by remagnetization at the lower temperature.

Manufacturers' data for irreversible loss should be interpreted carefully to distinguish between thelong-term stability and the effects just described. Irreversible loss is usually quoted at a fixedpermeance coefficient. If the magnet is used at a lower permeance coefficient, the irreversible loss overthe same temperature range will be higher.

Mechanical properties, handling, and magnetization

Magnets are often brittle and prone to chipping, but proper handling procedures are straightforwardenough as long as the rules are followed. Modern high-energy magnets are usually shipped in themagnetized condition, and care must be taken in handling to avoid injury that may be caused bytrapping fingers. A further hazard is that when two or more magnets are brought close together theymay flip and jump, with consequent risk to eyes. Table 1.13 summarizes some of the important safetyprecautions.

The best way to 'tame' magnetized magnets is to keeper them. Fixtures for inserting magnets can bedesigned so that the magnets slide along between steel guides which are magnetically short-circuitedtogether. There still remains the problem of entering the magnets between the guides, but usually thereis enough space to provide for this to be done gently. Obviously it is important to keep magnets clearof watches and electronic equipment that is sensitive to magnetic fields. Floppy disks, magnetic tapes,credit cards and key cards are particularly vulnerable, and high-energy magnets can distort the imageon computer terminals and monitors.

Magnets are usually held in place by bonding or compression clips. In motors with magnets on therotor, adhesive bonding is adequate for low peripheral speeds and moderate temperatures, but for highspeeds a kevlar banding or stainless steel retaining shell can be used. In motors it is not advisable tomake the magnet an integral part of the structure. Mechanically, the magnet should be regarded as a'passenger' for which space and fixturing must be provided. The important requirements are that themagnet should not move and that it should be protected from excessive temperatures.


Permanent magnets require strict adherence to safety procedures at all stages of handling and assembly.

Always wear safety glasses when handling magnets. This is particularly important when assemblingmagnets into a motor. When a large pole magnet is being assembled from smaller magnets, the magnetshave a tendency to flip and jump unexpectedly and may fly a considerable distance.

Work behind a plexiglass screen when experimenting or assembling magnet assemblies. Watchout for trapped fingers, especially with large magnets or high-energy magnets.

Avoid chipping by impact with hard materials, tools or other magnets.

Never dry-grind rare-earth magnets – the powder is combustible. In case of fire, use LP argon ornitrogen dry chemical extinguishers – never use water or halogens.

Use suitable warning labels, especially on large machines. PM motors generate voltage when the shaftis rotated, even when disconnected from all power supplies. This may be obvious to an engineer, but isa potential safety hazard for electricians and maintenance personnel.

Never leave magnetized members open or unprotected. When assembling a rotor to a stator, witheither one magnetized, the rotor must be firmly guided and the stator firmly located.

TABLE 1.13MAGNET SAFETY

A wide range of shapes is available, but in motors the most common are arcs and rectangles. Closetolerances of +/!0.1mm can be held in the magnetized direction even for standard magnets. But if thedesign permits a relaxation of the required tolerance, particularly in the dimensions perpendicular tothe magnetic axis, this should be exploited because it reduces the cost of the finished magnets.

Thermal expansion of magnets is usually different in the directions parallel and perpendicular to themagnetic axis. The coefficients in Table 1.8 are along the direction of magnetization. Most magnetshave a high compressive strength but should never be used in tension or bending.

Magnetization of high-energy magnets requires such a high magnetizing force that special fixtures andpower supplies are essential, and this is one reason why high-energy magnets are usually magnetizedbefore shipping. The magnetizing force Hm must be raised at least to the saturation level shown inTable 1.8, and this normally requires ampere-turns beyond the steady-state thermal capability ofcopper coils. Therefore pulse techniques are used, or in some cases superconducting coils. Ceramicand Alnico magnets can sometimes be magnetized in situ in the final assembly, but this is impracticalwith high-energy magnets.

Application of permanent magnets in motors

Permanent magnets provide a motor with life-long excitation. The only cost is the initial cost, whichis buried in the cost of the motor. It ranges from a few pennies for small ferrite motors, to severalpounds for rare-earth motors. Even so, the cost of magnets is typically only a small fraction of the totalcost of the motor. Broadly speaking, the primary determinants of magnet cost are the torque per unitvolume of the motor; the operating temperature range; and the severity of the operational duty.

Power density. We have seen that for maximum power density the product of the electric andmagnetic loadings must be as high as possible. The electric loading is limited not only by thermalfactors, but also by the demagnetizing effect on the magnet. A high electric loading necessitates a longmagnet length in the direction of magnetization, to prevent demagnetization. It also requires a highcoercivity, and this may lead to the more expensive grades of material (such as Sm2Co17, for example),especially if high temperatures will be encountered. The magnetic loading, or airgap flux, is directlyproportional to the remanence, and is nearly proportional to the pole face area of the magnet. A highpower density therefore requires the largest possible magnet volume (length times pole area).

With ceramic magnets the limit on the magnet volume is often the geometrical limit on the volume ofthe rotor itself, and the highest power densities cannot be obtained with these magnets. With rare-earthor other high-energy magnets, the cost of the magnet may be the limiting factor.


(BH)max 'Br

2

4µ0

J/m 3 . (1.68)

Fig. 1.20 DC B-H curve for electrical steels

The airgap flux-density of AC motors is limited by saturation of the stator teeth. Excessive saturationabsorbs too much excitation MMF (requiring a disproportionate increase in magnet volume); or causesexcessive heating due to core losses. For this reason there is an upper limit to the useable energy of apermanent magnet. With a straight demagnetization characteristic throughout the second quadrantand a recoil permeability of unity, the maximum energy-product (BH)max is given by

Assuming that the stator teeth saturate at 1.6T and that the tooth width is half the tooth pitch, themaximum airgap flux-density cannot be much above 0.8T and is usually lower than this. Thereforethere will be little to gain from a magnet with a remanent flux-density above about 1 or 1.2 T, implyingthat the highest useable energy product is about 300 kJ/m3. At 100 EC, such characteristics are justwithin the range of available high-energy magnets. Evidently it is just as important to develop magnetmaterials with 'moderate' properties and low cost, rather than to develop 'super magnets'

Operating temperature range. Because of the degradation in the remanence and coercivity withtemperature, the choice of material and the magnet volume must usually be determined with referenceto the highest operating temperature. Fortunately brushless PM motors have very low rotor losses.The stator is easily cooled because of the fine slot structure and the proximity of the outside air.Consequently the magnet can run fairly cool (often below 100 EC) and it is further protected by its ownthermal mass and that of the rest of the motor. The short-time thermal overload capability of theelectronic controller would normally be less than that of the motor, providing a further margin ofprotection against magnet overtemperature.

Severity of operational duty. Magnets can be demagnetized by fault currents such as short-circuitcurrents produced by inverter faults. In brushless motors with electronic control the problem isgenerally limited by the protective measures taken in the inverter and the control. With anover-running load, or where two motors are coupled to a single load, shorted turns or windings can betroublesome because of drag torque and potential overheating of the stator. But by the same token, thedynamic braking is usually excellent with a short-circuit applied to the motor terminals, and motorsmay well be designed to take advantage of this. As is often the case, characteristics that are desirablefor one application are undesirable for another. The design must accommodate all the factors thatstress the magnet: electromagnetic, thermal, and mechanical.

1.13 PROPERTIES OF ELECTRICAL STEELS

Fig. 1.20 shows the DC B-H curve in the first quadrantfor two steels. The lower curve is a typical electricalmotor steel having 1.5% Silicon to increase theresistivity to limit eddy-current losses. The saturationflux-density of such steels (i.e. the flux-density at whichthe incremental permeability becomes equal to µ0) istypically about 2.1T. The upper curve is for a cobalt-iron alloy with a saturation flux-density of about 2.3T.This material is much more expensive than normalelectrical steel, and is only used in special applicationssuch as highly rated aircraft generators, where lightweight and high power density are at a premium.

The maximum permeability of electrical steels is of theorder of 5,000 µ0, and usually occurs between 1 and 1.5 T.In Fig. 1.20, the total permeability of the electrical steelat 2.0T is about 2.0/3,000 which is approximately 530 µ0.


P ' Ch f Bpkn % Ce f 2 Bpk

2 . (1.69)

P ' Ch f Bpka%bBpk % Ce f 2 Bpk

2 . (1.70)

P ' Ch f Bpka % bBpk % Ce1

dBdt

2

. (1.71)

Fig. 1.21 Typical form of variation of losses in electrical steel,versus frequency and flux-density

Losses. Under AC conditions, a power lossarises in electrical steel as shown in Fig. 1.21,which indicates increasing loss as the frequencyand flux-density increase. The loss is attributedto

(a) hysteresis;(2) eddy-currents; and(3) “anomalous loss”.

The hysteresis component is associated with thechanging magnitude and direction of themagnetization of the domains, while the eddy-current loss is generated by induced currents.Eddy-currents can be inhibited by laminating thesteel, so that the eddy-currents become resistancelimited and the loss is then inverselyproportional to the resistivity. If the eddy-currents are resistance-limited the loss is alsoproportional to 1/t2, where t is the laminationthickness. At higher frequencies the resistance limited condition is lost, and the losses increase rapidlywith frequency. For this reason, very thin laminations, as thin as 0.1 mm, may be used at very highfrequencies (such as 400 Hz in aircraft generators or 3000 Hz in certain specialty machines). The“anomalous loss” is associated with domain wall movement and is not often accounted for in empiricalexpressions of the iron loss.

Characterization of core loss. Core-loss data from steel suppliers is almost always obtained frommeasurements in which a sinusoidal flux waveform is applied to a sample of laminations in the formof a stack of rings or an “Epstein square” made up from strips interleaved at the corners. The loss maybe characterized by the so-called Steinmetz equation with separate terms for hysteresis and eddy-current loss:

The units of P are usually W/kg or W/lb. Bpk is the peak flux-density in T, and f is the frequency in Hz.Ch is the hysteresis loss coefficient and Ce is the eddy-current loss coefficient. The exponent n is oftenassumed to be 1.6!1.8, but it varies to a certain extent with Bpk. To a first approximation we can writen = a + bBpk. With this modification,

The flux-density in motor laminations may be far from sinusoidal, and one approximate way to dealwith this is to modify the Steinmetz equation in the following way, recognizing that the eddy-currentloss component is expected to vary as the square of the EMF driving the eddy-currents, and that this EMF

varies in proportion to dB/dt. Thus

The hysteresis loss component is unchanged, but the eddy-current component is taken to beproportional to the mean squared value of dB/dt over one cycle of the fundamental frequency. Eqn.(1.71) can be applied in the respective sections of the magnetic circuit, after calculating the relevantflux-density waveforms.


Pf

' Ch Bpka % bBpk % Ce f Bpk

2 (1.74)

Pf

' D % Ef . (1.75)

D ' Ch Bpka % bBpk . (1.76)

log D1 ' log Ch % (a % bBpk1 ) log Bpk1 . (1.77)

Ce1 'Ce

2B2. (1.72)

The eddy-current loss coefficient Ce1 in the modified form can be derived from the sinewave coefficientCe if we assume that eqn. (1.71) holds with B = Bpk sin (2B f t). Then dB/dt = 2B f Bpk cos (2B f t) and(dB/dt)2 = 4B2 f 2Bpk

2 cos2 (2B ft), the mean value of which is [dB/dt]2 = 2B2 f 2Bpk2. For sinewave flux-

density, equations 1.70 and 1.71 give the same result if

Extracting the core loss coefficients from test data. Two procedures are used for extracting thecoefficients Ch, Ce1, a and b from sinewave loss data. The more elaborate of these requires a completeset of curves of core loss vs. frequency at different flux-densities. When this data is not available, asimpler procedure is used based on five parameters.

Simple procedure—It is often the case that only a single value of P is available, for example, 8 W/kg at50 Hz, measured with Bpk = 1.5 T. There is not enough data to determine the four loss coefficientsuniquely, so we use an estimate for n in eqn. (1.69); for example, n = 1.7. It is further necessary toestimate the split between hysteresis and eddy-current loss. If h is the fraction of the total lossattributable to hysteresis, then it can be shown that

Then a = n; b = 0, and Ce1 = Ce/2B2.

Procedure used with complete set of core-loss data—The core loss data is usually in the form of graphsof P vs. f at different flux-densities, or P vs. Bpk at different frequencies. The procedure is to try toseparate the hysteresis and eddy-current components of P. First we divide eqn. (1.70) by f :

We then plot graphs of P/f vs. f for three values of Bpk, e.g. 1, 1.5 and 2T with f from 50 to the highestfrequency. The graphs should be straight lines and can be represented by

The intercept D on the vertical (P/f) axis must be equal to

The intercepts D1, D2 and D3 for the three values of Bpk are substituted into the logarithm of eqn. (1.76),giving three simultaneous linear algebraic equations for Ch, a and b of the form

Ce 'P(1 ! h)

f 2 Bpk2

and Ch 'hP

Bpkn f

. (1.73)


10 The ampere-conductor distribution is often loosely termed the MMF (magneto-motive force), or MMF distribution.

These are solved for log Ch, a and b; Ch is then obtained from log Ch. Next, three values of Ce areobtained from the gradients of the three graphs of P/f vs. f , eqn. (1.74). The average or the highest valuecan be taken for Ce. Finally Ce1 = Ce/2B2. The loss curves may be re-plotted from the formula as acheck. Any extrapolation to higher Bpk or f should be checked carefully.

Note that Ce is approximately inversely proportional to t 2 , where t is the lamination thickness. Thiscan be used to modify Ce (or Ce1) for different thicknesses if test data is not available.

1.14 MACHINE AND DRIVE DESIGN

The equation TRV . 2AB reflects the fact that torque is produced by the interaction between flux andcurrent. This simple equation is the cornerstone of electrical machine design. It leads to the moreadvanced work of

machine design which is concerned with producing the torque with a minimum of material andpower loss; and

drive design which is concerned with the control of torque and speed, subject to constraintson the electric and magnetic loadings.

In relation to the machine design, it is always important to minimize power losses and temperaturerise caused by I 2R heating of the conductors, core losses caused by hysteresis and eddy-currents in themagnetic steel, and other losses. But there are many other aspects, such as the need to minimize torquepulsations and acoustic noise, and to use materials economically. There is a huge variety of differenttypes of electrical machine, arising partly from constraints imposed by the available power supply.For example, in automobiles the DC commutator motor is universally used because of the low-voltageDC power supply. But in industry, AC induction motors are used primarily because of the availabilityof polyphase AC power (which has a natural rotation between the phases), and because the inductionmotor has no brushes and therefore requires very little maintenance. In traction applications(railways, transit vehicles etc.), traditionally the DC motor was used because although AC supplieswere available, the control equipment was less expensive for DC drives. Since the 1980's, modern powerelectronics has become so cost-effective that AC drives have steadily taken over from DC drives evenin the most demanding traction applications.

At the same time the variety in types and designs of electrical machines has greatly increased in manyother fields of application because of the advances made in power electronics and microelectroniccontrol. This is clearly evident in such products as tape drives, computers, office machinery, and soon; but there are many others less well known—for example, the use of very high-speed brushlesspermanent-magnet motors in machine tools. These motors can run at several tens of kW and severaltens of thousands of revolutions per minute.

In relation to the drive design, one of the fundamental aspects of electrical machines is the orientationof the flux and the ampere-conductor distribution in relation to one another. The flux and MMF10 mustbe orthogonal in space, i.e. the axes of their spatial distributions must be displaced by B/2p radians,if the electromagnetic torque is to be maximized for a given flux and current. If the displacement angleis zero, there is no torque and the power factor is zero. In DC commutator machines the orientationbetween the flux and the armature MMF is maintained at B/2p radians by the action of the commutator,and therefore if the machine is controlled by a chopper or phase-controlled rectifier the controller isnot concerned with orientation and need do no more than regulate the current. By contrast, in ACinduction motors and synchronous machines, the orientation is not guaranteed to be B/2p radians,even though the machines might be designed to achieve this approximately under normal operation.For this reason, modern AC drives employ field-oriented control, (also called vector control), to orientthe MMF and the flux orthogonally. This is quite complex and typically requires the use ofmicrocontrollers or DSP’s (digital signal processors). The most modern embodiments of field-orientedcontrol are sophisticated enough to include estimators for important parameters such as the flux, thedirection of the flux, the rotor temperature, and the electromagnetic torque itself.


Fig. 1.22 Design loop

1.15 COMPUTER-AIDED DESIGN

When new designs are evolved from old ones, computer-aided design is valuable for

1. calculating and evaluating a large number of options, often characterized by small changes ina large number of parameters; and

2. performing detailed electromagnetic and mechanical analysis to permit the design to be"stretched" to its limit. With accurate computer software, we can reduce the need forprototypes, which are expensive and time-consuming.

Modern computer methods are rapidly reaching the stage where a new prototype can be designed withsuch confidence that it will be "right first time", without the need for reiteration of design and test thatwould otherwise be necessary. Computer-aided design goes hand-in-hand with the modern designengineering environment. Custom designs are often required within a very short space of time, whilecost pressures force the designer ever closer to the limits of materials and design capabilities.

Moreover, customers are becoming more sophisticated in their requirements, and may specify (or askto see) particular parameters that traditionally were part of the "black art" of the motor builder. Oftenthese parameters are required for system simulation purposes long before the motor is actuallymanufactured. Regulatory pressures on matters such as energy efficiency, acoustic noise, and EMC alsotighten the constraints on the motor designer.

No matter how effective the computer software available, it is always important to check the overallparameters of a motor design using common sense and fundamental engineering principles. For thisreason it will always be necessary to be able to perform a single set of design calculations on thecomputer and check the results against manual calculations. The next stage is to repeat designcalculations, modifying the dimensions and parameters until the performance objectives are attained.These processes are illustrated graphically in Fig. 1.22. The SPEED software is designed to be used inthis way.

The synthesis of a design by an optimization process is a much more complex undertaking beyond thescope of this book. However, the development of scripting languages which can run programs such asSPEED motor design programs automatically opens up new opportunities for user-defined designautomation procedures.


Index

Acceleration 3, 5, 18-20Adjustable speed 2, 8Airgap shear stress 15Base speed 5Brushless DC 2, 8, 11, 12Brushless PM 5, 11, 13, 27, 32, 39B-H loop 31CAD 6, 7Commutator 4, 8, 10, 13, 17, 42Computer-aided design 43Conduction 21-25Convection 15, 21-25Cooling 1, 2, 10, 14-17, 20-24Copper losses 10, 11, 24Core loss 40, 41Current density 14, 16, 23DC commutator motor 8, 10, 17, 42Demagnetization 12, 31-36, 38, 39Design 1, 2, 4, 6, 7, 11, 17, 18, 20, 35, 38, 39, 42, 43Digital electronics 5Drive system requirements 5Duty-cycle 27-30eddy-current loss 40Efficiency 3-6, 10-12, 17, 18, 21, 43Electric loading 14-17, 38Electrical steels 39Energy product 6, 30, 31, 33-35, 39Energy saving 2Evolution of motors 8Fan 2, 15-17, 23Finite-element 6, 7, 24Flow process 2Gearing 1, 2, 18, 19Heat removal 20-24Hysteresis 13, 32, 40-42hysteresis loss 40Induction motor 6, 8, 10-12, 42Inertia 4, 5, 11, 12, 18-20Insulation life 20Intermittent operation 27Irreversible losses 35, 36

Magnet operating point 33Magnet safety 38Magnetic loading 14, 15, 17, 38Magnetic materials 6Magnetization 31-38, 40Microelectronics 2, 5, 6Minas 11Motion control systems 2, 5, 6, 8, 20New technology 2, 5Numerical analysis 6Output equation 14Overload 5, 12, 27-30, 39Permanent magnet 13, 30, 39Permeance coefficient 33-37Position control 2-4, 13Power 2-8, 10-15, 18, 23, 27, 28, 38-40, 42Power density 12, 38, 39Power factor 10-12, 42Power semiconductors 2, 4, 8Radiation 21-25Reluctance motors 2, 12, 13, 17Reversible losses 36Sizing 1, 2, 14, 17Slip 8, 10, 11, 13Steinmetz 40Steinmetz equation 40Stepper motors 8, 12, 13, 17Structure of drive systems 4Switched reluctance 2, 8, 12, 13, 17Synchronous motors 10, 13Synchronous reluctance 12, 13Temperature rise 14, 17, 20-23, 25, 27-29, 42Thermal equivalent circuit 24, 25, 27Torque 4-6, 8, 10-12, 14-20, 23, 28, 29, 38, 39, 42Transients 2, 3, 6, 24TRV 14-18, 42

2. Brushless Permanent-Magnet Machines

2.1 What is a brushless machine? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.2 Basic operation of the brushless DC motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3

2.3 Torque/speed characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9

2.4 Magnetic circuit analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10

2.5 Winding inductance of surface-magnet motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17

2.6 Inductances of salient-pole motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.18

2.7 Basics of sinewave operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.23

2.8 The origin and development of AC vector control for brushless PM motors . . . . . 2.25

2.9 History of brushless PM motor drives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.32

2.10 Switching strategy for sinewave drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.35

2.11 Simplified analysis of squarewave drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.37

2.12 Back-emf sensing with squarewave drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.43

2.13 Unipolar drive circuits, and motors with one or two phases . . . . . . . . . . . . . . . . . . . 2.44

2.14 Windings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.48

2.15 Calculation of torque using the finite-element procedure . . . . . . . . . . . . . . . . . . . . . 2.49

2.16 Torque and Torque ripple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.53

2.17 Cogging torque calculations using finite-element analysis . . . . . . . . . . . . . . . . . . . . 2.55

2.18 Getting inductance from finite-element calculations . . . . . . . . . . . . . . . . . . . . . . . . . 2.60

2.19 Torque per ampere and kVA/kW of squarewave and sinewave motors . . . . . . . . . 2.60

2.20 Permanent magnets versus electromagnetic excitation . . . . . . . . . . . . . . . . . . . . . . . 2.62

2.21 Slotless motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.65

2.22 PM generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.69

Notes

1“Salient” means “sticking out” and refers to the projecting poles on wound-rotor synchronous machines. The term “salient-pole” is now used more generally to refer to machines with different permeances in the direct and quadrature axes of the rotor,even when there are no physical salient poles. The difference in d-axis and q-axis permeance leads to a difference in thesynchronous inductances Ld and Lq, and gives rise to reluctance torque.

2. BRUSHLESS PERMANENT-MAGNET MACHINES

2.1 WHAT IS A BRUSHLESS MACHINE?

"Brushless" electrical machines are those in which all components associated with sliding contacts areeliminated. These components include brushes, commutators, slip-rings, etc. used to supply currentto the rotor. Although brushed motors are still widely used, their disadvantages have encouraged thedevelopment of brushless machines: for example,

(1) in computer disk drives because of the undesirability of brush debris;

(2) in blowers and fans because of the need for low noise, high efficiency, and speed control;

(3) in precision servomotors for factory automation, where downtime must be minimized; and

(4) in vehicle traction, because of the cost of brushgear maintenance and the need for highefficiency.

The development of brushless machines would not have been possible without certain enablingtechnologies : in particular, power transistors, microelectronic controls and sensors, and permanent-magnet materials. These components are now so highly developed that they have opened up many newopportunities for electric motor applications, especially where variable speed is required.

There are so many varieties of brushless electrical machine that it is difficult to classify them, but wecan recognize four main categories:

(a) Brushless DC Machines are derived directly from the classical DC machine by replacing thecommutator and brushes with an electronic power supply. The motor is often designed to havea trapezoidal back-EMF waveform and the current waveforms are rectangular, with alternatingpolarity. The current polarity is switched in synchronism with the rotor position, by means ofpower semiconductors which are also used to regulate the current. Fig. 2.1.1 shows a typicalexample. Permanent magnets (mounted on the rotor surface) are generally used for excitation.

(b) AC brushless machines are similar to brushless DC machines, but the back-EMF waveform isdesigned to be sinusoidal and the current waveform is also controlled to be sinusoidal. They areused in servosystems where smooth torque control is required. Resolvers are often used forshaft position feedback. In some cases the permanent magnets are mounted inside the rotor,as shown in Fig. 2.1.3. This tends to introduce saliency and a reluctance torque component.1

The saliency can also be helpful when the motor is to operate at constant power over a widespeed range—as, for example, in drives for electric vehicles. It is even possible to dispense withthe magnets altogether, as in the synchronous reluctance motor, Fig. 2.1.4.

(c) Self-synchronous AC brushless machines in which the DC field winding on the rotor is fedfrom a rotating rectifier, which in turn is fed from a rotating AC exciter mounted on the sameshaft. This category includes some large machines (several MW) and high-speed AC machinessuch as aircraft generators. Fig. 2.1.5 shows an example (not including the rectifier or exciter).

(d) Specialty brushless machines not derived from classical DC or AC machines. This categorycovers a huge range of different designs, including a wide variety of blower motors (Fig. 2.1.2),timing motors, and others. Often the manufacturing volumes of these motors are in millions.

In this introductory theory section we will concentrate on the classical three-phase brushless DCmachine, Fig. 2.1.1. The design principles of machines in category (c) are more in common with thoseof classical synchronous machines. For machines in category (b), the design principles are in commonwith both categories (a) and (c).

Other important types of "brushless" electric machine include the induction motor, the switchedreluctance motor, and the stepper motor.


Fig. 2.1 Types of brushless motor.

Brushless permanent-magnet machines Page 2.3

Fig. 2.2 Generation of back-EMF

ea1 'dR1

dt' Tm

dR1

d2(2.1)

Mg ' m0

B/pB(2)rd2Lstk ' Bg ×

BDLstk

2p(2.2)

R1(2) 'p2

B/2R1max (2.3)

ea1 ' Tm ×2p

B× TcMg (2.4)

Ea ' Tm ×2p

B× TphMg (2.5)

2.2 BASIC OPERATION OF THE BRUSHLESS DC MOTOR

Fig. 2.2 shows a 2-pole motor with a magnet rotatingcounter-clockwise at the instant when the flux-linkagewith coil 1 is at a negative maximum. The magnet isshown with an arc $M = 180Eelec., and the coil-pitch is also180Eelec, so that the flux-linkage R1 of each coil varies witha triangular waveform as the rotor rotates: see Fig. 2.4.

By Faraday’s Law the EMF induced in coil 1 is

where Tm is the angular velocity in mechanical rad/s and2 is the rotor position in mechanical radians. As long asthe flux-linkage is varying linearly with rotor position, theinduced EMF is constant. When the flux-linkage reaches amaximum, it starts to decrease at the same absolute rate,and the EMF changes polarity. The result is a squarewavegenerated EMF, ea1.

A second coil a2A2 is displaced by a certain angle ( from the first coil. It has the same number of turnsas the first coil, Tc. Its EMF waveform is identical to that of the first coil, but retarded in phase by (electrical radians. If the two coils are connected in series, their EMFs add, giving an EMF waveform withtwice the peak value of the individual coil EMF. If each phase is made up of just two coils connected inseries, this is the phase EMF. Its waveform is stepped because of the phase displacement between coils1 and 2. In practice, because of fringing of the flux at the edges of the magnet, the edges of the EMF

waveform are not sharply stepped but appear smoothed, as shown by the dotted line in Fig. 2.4 . Thisis the so-called trapezoidal back-EMF waveform that is characteristic of brushless DC motors.

A 3-phase brushless DC motor is generally designed so that the flat top of the phase back-EMF waveformis just over 120E wide. Then each phase is supplied with a current waveform consisting of blocks ofconstant current 120E wide. During each 120E period, the electromagnetic power conversion is eaia =TeTm, where Te is the electromagnetic torque. If the EMF and current waveforms are sufficiently flatduring this period, and the speed is essentially constant, Te is also constant.

The EMF can be calculated from the airgap flux distribution. If Bg is the average flux-density over onepole-pitch, the airgap flux Mg is given by

where r is the stator bore radius, D is the stator bore diameter (= 2r), Lstk is the stack length, and p isthe number of pole-pairs. The peak flux-linkage of coil 1 is R1 max = TcMg, and if the linear variationof R1 with 2 is

then by eqns. (2.1!2.3) the peak coil EMF is

For a machine with Tph turns in series per phase, the peak EMF/phase can be written


Fig. 2.3 Drive circuit for wye-connected brushless DC motor

Te ' kT I (2.8)

kE '4pTphMg

B. (2.7)

Fig. 2.3 shows the commonly used bridge circuit for a 3-phase brushless permanent-magnet motor. For“squarewave” operation (Fig. 2.4), it has two phases conducting at any time. If the motor is wye-connected they carry the same current I in series, and the line-line EMF during each 60E interval is

where kE is the back-EMF constant in Vs/rad:

The electromagnetic power is ELLI and the electromagnetic torque is Te = ELLI/Tm. This can be written

where k T = kE is the torque constant in Nm/A. When driven this way with "two phases on", the motorbehaves very much like a permanent-magnet DC commutator motor. The torque is produced in blocks60E wide, and there are 6 such blocks every electrical cycle. See Fig. 2.7.

The electronic power supply is called the drive. In low-power drives, MOSFETS are popular because theyare easy to control, and they can be switched at high frequency. This makes it possible to regulate thecurrent by chopping with low acoustic noise. MOSFETS are ideal for low-voltage drives because of theirlow on-state voltage drop. At higher powers and voltages, IGBTS are used.

Current and EMF waveforms are shown in Figs. 2.4 and 2.7 for the wye-connected motor, and in Fig.2.8 for the delta-connected motor, whose drive circuit is shown in Fig. 2.5. The line current waveformsare identical for the two connections, as is the commutation table for the transistors, Table 2.1.

Fig. 2.6 shows the interaction of the "phasebelts" of ampere-conductors with the magnet arcs in wye-and delta-connected motors. In the wye-connected motor, the magnet arc is 180E. With two phasesconducting, the positive and negative phasebelts produce belts of ampere-conductors 120E wide.Therefore the motor can rotate 60E with no change in the overlap between each magnet and the belt ofampere-conductors nearest to it. This ensures constant torque over a 60E angle. At the end of each 60Eperiod, the current is commutated from one phase into the next. In Fig. 2.6, phase 2 will be the next toturn off, and phase 3 will be the next to turn on, if the rotor is rotating CCW.

In the delta-connected motor, three phases are conducting at any time, giving 180E belts of ampere-conductors. To ensure that the overlap is constant for 60E, the magnet arc is reduced to 120E.

ELL ' 2 Ea ' kETm (2.6)


Fig. 2.4 Flux-linkage, EMF and current waveforms for the motor in Fig. 2.2.


Fig. 2.6 Interaction of arc magnets and ampere-conductor phasebelts in wye- and delta-connectedbrushless DC motors.

Fig. 2.5 Drive circuit for delta-connected brushless DC motor.


Fig. 2.7 Ideal waveforms of line currents ia, ib, ic; phase EMFs e1, e2, e3; phase torques T1, T2,T3, and total electromagnetic torque Te in wye-connected brushless DC motor drive.


Fig. 2.8 Ideal waveforms of line currents ia, ib, ic; phase current i1; and phase EMF e1 in delta-connected brushless DC motor and drive. For more detail see Hendershot &Miller[1994].

Rotor position [Eelec]

Line Phaseleg A Phaseleg B Phaseleg C

A B C Q1 Q4 Q3 Q6 Q5 Q2

330!30 0 !1 +1 0 0 0 1 1 0

30!90 +1 !1 0 1 0 0 1 0 0

90!150 +1 0 !1 1 0 0 0 0 1

150!210 0 +1 !1 0 0 1 0 0 1

210!270 !1 +1 0 0 1 1 0 0 0

270!330 !1 0 +1 0 1 0 0 1 0TABLE 2.1

COMMUTATION TABLE FOR BRUSHLESS DC MOTOR DRIVE: 120E SQUAREWAVE LINE CURRENTS


Fig. 2.9 Torque/speed diagram

Vs ' E % RI (2.9)

T

T0

' 1 !T

T0

' 1 !I

I0

(2.10)

T0 'Vs

kE

rad/s (2.11)

T0 ' kT I0 (2.12)

I0 'Vs

R. (2.13)

2.3 TORQUE/SPEED CHARACTERISTIC

The torque/speed characteristic is similar to that of the permanent-magnet DC commutator motor, Fig.2.9. For two-phase-on operation, if E is the line-line EMF and R is the resistance of two phases in series,then during any 60E period we can write

where I is the DC current. Using eqns. (2.6) and (2.8), and neglecting losses, the torque/speedcharacteristic can be derived in the form

where the no-load speed is

and the stall torque is

and the stall current or locked-rotor current is

The stall current given by this expression may be large enough to demagnetize the magnets, but usuallythis current is far beyond the capability of the power transistors in the drive. Therefore, the currentis limited by chopping to a safe value that is normally well below the "demag." current.

The torque/speed characteristic is plotted in Fig. 2.9. As in the case of the DC commutator motor, thespeed is controlled by the voltage and Fig. 2.9 shows the effect of reduced voltage on the torque/speedcharacteristic. The voltage is varied by chopping the power transistors with a certain duty-cycle d; or,alternatively, by regulating with a constant current in such a way that the effective duty-cycle isautomatically maintained at the correct value corresponding to the actual speed of the motor.

Fig. 2.9 also shows the effect of imposing a current limit, which limits the torque. For a short time itmay be permissible to operate at currents higher than the normal limit; accordingly, Fig. 2.9 is dividedinto continuous and intermittent operating regions by the current-limit line.


Pm0 ' µrecµ0

AM

LM

' µrecµ0

$M

2

rMLstk

LM

Wb/At (2.14)

Fig. 2.10 Simple magnetic circuit

2.4 MAGNETIC CIRCUIT ANALYSIS

The most basic magnetic calculation in brushless permanent-magnet motors is to determine the fluxproduced by the magnets. An important result of this calculation is the “operating point” of themagnets and the general saturation level of the iron. It is also important to determine the distributionof the flux around the airgap, because this in turn determines the waveform of the generated EMF. TheEMF waveform is also affected by the winding distribution, but in this section we will concentrate onlyon the calculation of magnet flux, that is, the “open-circuit” condition.

Magnetic circuit calculations in brushless permanent-magnet machines can be classified into threemain classes:

(1) very simple methods which can be executed using a calculator;(2) more complex analytical methods which may require the use of a computer; and(3) finite-element methods which always require the use of a computer.

All three of these classes are important. The first is important for “sanity checks” on the other two,and is useful for roughing out new designs or checking existing ones. The second is generally the oneused for serious design calculations because it can be executed repetitively and very fast when suitablyprogrammed on a computer. Finite-element methods are the most accurate, but also the slowest. Thethree classes also give a rich variety of physical interpretations, but they are not equally appropriatefor all types of motor. For surface-magnet motors it is often sufficient to work with the first two, butfor embedded-magnet motors the finite-element method is essential because of the complexity of therotor geometry, with intense local saturation.

This section describes two methods for the magnetic calculation of surface-magnet motors, the firstbeing from class 1 and the second from class 2.

Simple magnetic circuit analysis

The magnetic circuit calculation begins with a simple estimate of the airgap flux Mg based on themagnetic equivalent circuit in Fig. 2.10. This is used as the starting-point for a nonlinear calculationthat uses the magnetic equivalent circuit in Fig. 2.11. The permeances in both models are associatedwith one half-pole. The magnet permeance is

where $M is the pole-arc of the magnet in mechanical radians, AM is the magnet half-pole area, Lstk isthe stack length, and LM is the magnet length in the direction of magnetization. The radius rM is theeffective radius of the magnet.


Mg 'Pg

Pg % PL % Pm0

Mr . (2.17)

Mg '1

1fLKG

% Pm0 Rg

Mr(2.18)

Rg 'g

µ0Ag

A&t/Wb (2.15)

PC 'Bm

µ0*Hm*. (2.21)

fLKG 'Mg

Mm

'Mg

Mg % ML

< 1. (2.16)

Mg '1

1 % (1 % prl )Pm0 Rg

Mr (2.19)

1 % prlPm0 Rg '1

fLKG

. (2.20)

Similarly the airgap reluctance is

where Ag is the airgap area. The flux Mg is the flux crossing the airgap, Mm is the magnet flux, and MLis the leakage flux, i.e., Mm ! ML.

Leakage factor : We can identify two separate (but related) ways of dealing with leakage. For surface-

magnet motors the leakage factor f LKG is defined as the ratio of gap flux to magnet flux:

For embedded-magnet motors, including “spoke”-type and “IPM” (interior permanent-magnet) motors,it is possible to calculate the rotor leakage permeance PL directly from rotor geometry, and express itas a fraction prl of the magnet permeance Pm0: thus PL = prl Pm0. Any saturating bridges (as for examplein Fig. 2.1.3) are in parallel with the leakage flux path PL.

In Fig. 2.10 the magnet is represented by a Norton equivalent circuit and the half-pole gap flux Mg isexpressed a fraction of the half-pole remanent flux Mr:

For surface-magnet motors Mg is expressed in terms of f LKG as

but for spoke-type and IPM motors it is expressed in terms of prl as

where Rg = 1/Pg. Eqns. (2.18) and (2.19) are equivalent, with

Note that having f LKG < 1 means that Mg is reduced by the effect of the leakage. A typical value for f LKGis 0.9 !0.95 for a wide range of surface-magnet motors. If this value is specified, prl can be calculatedfrom eqn. (2.20). For spoke-type and IPM motors, prl is estimated directly or by finite-element analysis,and f LKG is calculated from eqn. (2.20).

Permeance coefficient : By definition, the permeance coefficient is

The values of Bm and Hm are obtained from the nonlinear magnetic circuit calculation (below). Havingf LKG < 1 means that Mm is greater than it would be with no leakage, and therefore PC is increased byleakage.


Fig. 2.12 Magnet fringing

Fig. 2.11 Nonlinear magnetic circuit

Hm '(Fg % FSY % FRY % FST )

LM

(2.22)

b '12

1 ! e!(2 ! 2a )/a(2.23)

Nonlinear calculation : The magnetic circuit in Fig. 2.11 is drawn for half of one Ampère's Lawcontour, representing the MMF drops associated with one airgap. The magnet is now represented by aThévenin equivalent circuit comprising the "open-circuit" MMF Fma and the reluctance Rm0 = 1/Pm0. Theflux densities in the yoke and teeth sections are calculated from their permeance areas, and theassociated MMF drops F are obtained using the nonlinear BH curve of the steel. The total magnetisingforce in the magnet is calculated as

The circuit is solved iteratively. The basic result of the nonlinear magnetic circuit calculation is theairgap flux Mg and from this the average flux density over the magnet pole arc can be calculated.

Apart from a simple allowance for fringing, the magnetic analysis has so far been described as thoughthe magnet flux had the ideal rectangular distribution around the airgap shown by the dotted line inFig. 2.12. In reality the distribution has rounded corners and there is a fringing zone near the edgesof the magnet. An approximate functional representation of this is the exponential relationship

in the fringing zone to the right of 2a, where b is the normalized value of the flux-density and a is anempirical coefficient given by


2Hague’s work was done at the University of Glasgow in the 1920's. It was adapted for permanent-magnet motors by Boules[1984, 1985] by means of the equivalent current-sheet . Subsequently, new original solutions of the Maxwell equations werepublished by Zhu et al [1993] and Rasmussen et al [1999]. These later solutions relied on a harmonic series representation ofthe magnetization vector and therefore considerably extended the scope of the analysis. 3Several others have been developed for the PC-BDC program.

a '12

g [g % Lm/µrec] . (2.24)

b '12

e(2 ! 2a )/a(2.25)

To the left of 2a the fringing function is simply

and similar functions are used on the right half of the distribution, symmetrical about 2b. This fringingfunction can be modified for skew, and although it is approximate it is extremely fast in computation.

Once the flux and its distribution are known, it is a straightforward matter to calculate thefundamental space-harmonic component B1(2) and from this the fundamental magnet flux/pole MM1and the peak fundamental open-circuit airgap flux density, B1

(oc).

Demagnetizing effect at locked-rotor: The magnetic circuit model can be used to estimate thedemagnetizing field in the magnet under various conditions, for example, if the rotor is locked or inany other condition where high current is liable to flow in the windings. The armature ampere-turnsFa per half-pole are calculated according to the winding type and are assumed to be concentrated at asingle point in the airgap. An example of this calculation is given in Chapter 5 in connection with DCcommutator motors, which have a similar magnetic circuit. See also Hendershot & Miller [1994].

Analytical solution of Laplace/Poisson equation

Another class of analytical methods for calculating the magnet flux distribution is based on the directsolution of Maxwell’s equations, which reduce to the Laplace equation in the air region and the Poissonequation in the magnet. The original basis for this class of methods is the book by Bernard Hague[1929],2 which provides a comprehensive solution for the magnetic field between two concentric smoothiron cylinders, for an arbitrary distribution of current-carrying conductors in the airgap or on thesurfaces of the cylinders. This work was applied by Boules [1984], who replaced the magnet by anequivalent distribution of ampere-conductors and used Hague’s solution to compute the field. The“equivalent distribution of ampere-conductors” can be determined only in special cases: generallywhere there is no irregular iron shape in the rotor, the stator has a smooth bore, and the magnet hasa simple geometric shape and a certain direction of magnetization. Boules developed solutions forcertain basic shapes of magnet including surface magnets with radial and parallel magnetization.3

Equivalent ampere-conductor distributions: The magnet is replaced by a current sheet K = M ×n [A/m], where M is the magnetization vector inside the magnet and n is the unit vector normal to themagnet surface. Since M and n are both always in the x,y plane, transverse to the axis of rotation, Kis always in the z direction along the axis of rotation, i.e. K = (0,0,K). M is the actual magnetizationof the magnet, which includes an induced component due to the demagnetizing field of the externalmagnetic circuit. Unfortunately this is not known a priori. However, if the recoil permeability is near1, the susceptibility Pm of the magnet is nearly zero, and the induced magnetization is small. Boulespoints out that on open-circuit the magnets are normally worked between B r/2 and B r, and he uses theaverage magnetization over this range, i.e., M = kmM0 = km B r/µ 0 where k m = (1 + 0.75 Pm)/(1 + Pm).Note that M is equivalent to the “apparent coercivity” H ca, i.e. the coercivity that the magnet wouldhave if its recoil line was straight throughout the second quadrant with relative permeability µr . Thevalue of the susceptibility Pm and the constant k m can be seen in Table 2.2 for typical values of µ r. Formost magnets µ r does not exceed 1.1, so the maximum error from this approximation is less than 2.5%.


Fig. 2.14 Filamentary coil model

Br ' jn

q (MnHn % NnKn)

µr (q 2 ! 1)[r q!1

% a 2qr !q!1]c !q%1Hn cos q2 (2.27)

µ r Pm k m

1 0 1

1.05 0.05 0.988

1.1 0.1 0.977

1.2 0.2 0.958

TABLE 2.2

Boules derived the ampere-conductor distributions for arc magnets whose edges xy lie along radii. Forradially-magnetized magnets, K = 0 on the curved surfaces; and on the edges, K = M. For parallel-magnetized magnets, on the curved surfaces K = M sin 2 and on the edges K = M cos $M/2, where $M isthe magnet pole arc expressed in mechanical degrees. The equations for magnets with parallel edgesare as follows: along the outer curved surface, K = M sin 2; on the inner curved surface K = !M sin 2;and on the edges K = M. For the “full ring” (solid 2-pole) magnet, on the outer curved surface K = Msin 2; and on the flat chamfers K = M. On the inner circular curved surface K = !M sin 2.

The magnetic field is given by Hague’s solution for the field of coils distributed in the airgap betweentwo concentric cylinders. Fig. 2.13 shows a basic 4-pole distribution of single-turn coils having a radiusc and span 2 >, equivalent to the magnet arrangement in Fig. 2.14. The field produced by the coilset ofFig. 2.13 at the point (r, 2) is given by

where i is the coil current, p is the number of pole-pairs, and the sum is taken over all odd electricalharmonics, i.e., n = (2j ! 1), j = 1,2,3.... The factor k Fn is the n’th harmonic skew factor, which is equalto sin (nF/2) /(nF/2), where F is the skew in electrical radians. When the contributions of all thefilamentary coils are summed, a similar factor arises if the magnetization tapers off from a peak valueto zero over an angle F electrical radians, proving that tapered magnetization and skew are equivalentin terms of their effect on the airgap field.

The methods developed by Rasmussen [1999] and by Zhu et al [1993] go beyond the Hague-Boules methodjust described, by using a direct scalar potential solution that relies on a harmonic seriesrepresentation of the magnetization vector. The airgap field is given by expressions of the form

where q = np, p is the number of pole-pairs, Mn and Nn are the n’th harmonic components of the radialand tangential components of magnetization, and Hn and Kn are functions of q, µr and the various radiigiven in Rasmussen [op cit.]. The sum is taken over all odd electrical harmonics, i.e. n = (2j ! 1), j =1,2,3... and similar expressions are given for B2 and for the field in the magnet itself. The magnetizationis assumed to be invariant with r, i.e., it does not vary through the thickness of the magnet.

Br ' 2pµ0 i

Br j4

n

a n

c n@

c 2n% b 2n

a 2n ! b 2n@

r n

a n%

a n

r n@ kFn sin n> cos n2 (2.26)

Fig. 2.13 Airgap model of surface magnet


Fig. 2.15 Example with typical proportions. Good resultscan be obtained even with the simple magneticcircuit analysis model.

Fig. 2.17 With a small rotor diameter to accommodate athick non-magnetic retaining ring, the simplemagnetic circuit model is inadequate and theHague/Boules or Rasmussen methods givemuch better results.

Fig. 2.16 With very thin magnet and a short airgap, goodresults can be obtained with all methods.

Fig. 2.18 “FullRing” magnet type, a solid 2-pole magnet.In this case the Hague/Boules method is thebest analytical method although Rasmussen’smethod is also appropriate. The simplemagnetic circuit model is inadequate in thiscase.

Figs. 2.15!19 show examples of different permanent-magnet brushless motors for which the abovemethods are appropriate.


Fig. 2.19 Simple finite-element mesh for brushless PMmotor

Fig. 2.20 Magnetic field solution for Fig. 2.19 (opencircuit)

Fig. 2.21 Comparison of finite-element and magnetic circuit model

Finite-element analysis

Because the Hague-Boules and Rasmussen methods are analytical, they are not well adapted to dealwith saturation effects, but the method is nevertheless still useful, either for machines wheresaturation is not important, or where simplified allowances can be made for it. The simple magneticcircuit model can make crude allowances for saturation, but for thorough analysis of the magnetic fieldthe finite-element method is by far the most powerful. It is particularly effective in computing thedetails of local geometric features and the effects of arbitrary distributions of ampere-conductors andmagnetization patterns. These details continually increase in importance, partly because ofcompetitive pressure to improve performance and cost-effectiveness, but also because of the need toreduce torque ripple and acoustic noise.

Figs. 2.19 and 2.20 show typical examples of finite-element computations for a simple brushlesspermanent-magnet motor on open-circuit, and Fig. 2.21 shows the comparison of the airgap flux-densitydistribution obtained by the finite-element and magnetic-circuit methods. The finite-element solutionincludes the effect of the slot-openings, which is absent from the analytical solution.


Fig. 2.22 Calculation of inductance

H 'Ni

2 gN(2.28)

Bga 'µ0 N i

2 gN. (2.31)

Fig. 2.23 Distribution of armature reaction MMF and flux

R ' N × Bga ×BD

2pLstk (2.32)

gN ' kc g %LM

µrec

(2.29)

kc '5 % s

5 % s ! s 2 g/8(2.30)

Lg 'B

4

µ0 Tph2 Lstk D

p 2 gN(2.33)

2.5 WINDING INDUCTANCE OF SURFACE-MAGNET MOTORS

Fig. 2.22 shows the magnetic flux established by a full-pitch coil. In a motor with one slot/pole/phase,this could be a complete phase winding. The total MMF around a complete flux path is equal to N i,where N is the number of turns in the coil and i is the current. If the steel in the rotor and stator isassumed to be infinitely permeable, then the MMF is concentrated across two effective airgaps. Eacheffective airgap includes the thickness of the magnet, LM, which is assumed to have a relative recoilpermeability µrec close to 1. Therefore the MMF drop across each airgap is Ni/2. If the flux is assumedto cross the gap in the radial direction, the magnetizing force in each gap is

where gN is given by

in which g is the actual airgap modified for slotting by Carter’s coefficient kc. An approximation forCarter’s coefficient suitable for surface-magnet motors is

where s is the ratio of slot-opening to the total airgap g, and 8 is the ratio of the slot-pitch to the gap g.The flux-density produced by this magnetizing force at the stator bore diameter D is

The ideal flux distribution around the airgap is plotted in Fig. 2.23. The flux linkage of the coil is

where p is the number of pole-pairs. The self-inductance is given by Lg = R/i, that is,

where Tph is the number of turns in series per phase (= Np/a, where a is the number of parallel pathsin the winding). The subscript g denotes the fact that this is only the "airgap" component of theinductance. The total self inductance is given by


kd '1

kw12 j

n>1

kwn2

n 2

4Classical AC machine theory gives the differential leakage reactance via the differential leakage factor

M ' !Lg

3% Mslot % Mend . (2.36)

R21 ' N × Bga ×D

2pLstk × !

7B6

!B

2%

B

2!

B

6(2.35)

where the subscript "ph" denotes the phase inductance and LF is the leakage inductance per phase.Eqn. (2.34) identifies two separate components of leakage inductance: slot leakage and end-turn leakage,which arise from fluxes that are not included in the airgap flux.

The mutual inductance between phases can be calculated in a similar way, by adding the flux-linkagesof a second coil placed in the field of the first one, Fig. 2.23. The second coil (phase 2) is located in slotsat the angles !90 + 120 = 30E and +90 + 120 = 210E. Its flux-linkage due to current in coil 1 is

from which it can be deduced that the airgap component of mutual inductance is !Lg /3. If Mslot andMend are the mutual inductance components in the slot and end-turns respectively, then

in which Mslot and Mend are separate components of mutual inductance associated with "leakage" fluxesother than the airgap flux.

Hague's method [5] can be used to calculate the airgap components of the phase inductance and themutual inductance between phases, if the ampere-conductors in the slots are represented by currentfilaments at the centres of the slot openings, and the stator bore is assumed to be smooth. The airgaplength is augmented by Carter’s coefficient and the magnet length, according to eqn. (2.29). It ispossible to program such a calculation for any arbitrary distribution of winding conductors, but thecomputation may require a large number of harmonics (of the order of 100) to produce a satisfactoryresult. As in the analytical examples above, the slot and end-winding components of inductance mustbe added separately.

The slot-leakage components Lslot and Mslot can be calculated as in induction motors, using the slotpermeance coefficient calculated from the slot geometry. The calculation of Mslot must take accountof the coincidence of conductors from different phases in each slot. The end-turn inductance Lend canbe calculated from circular-coil formulas. Mutual coupling in the end-windings can often be ignored.

Because Hague's method relies on a Fourier series expansion of the space-harmonics of the windingdistribution, it is possible to isolate the contribution of the fundamental, and then the sum of all higherharmonic terms is called the "differential leakage inductance" Ldiff.

4 The fundamental term is the onethat gives rise to the "magnetizing" or "airgap" component of the synchronous inductance, which istreated in more detail in the next section.

2.6 INDUCTANCES OF SALIENT-POLE MOTORS

In “embedded-magnet” motors, including “interior-magnet” and “inset-magnet” motors, the windinginductances vary as the rotor rotates; this property is known as saliency and such motors are classifiedas salient-pole motors. It is typical in such motors for the inductance to vary also as a function ofcurrent. As a result, it is difficult to perform a time-stepping simulation. The dq-axis transformationhas long been used for the analysis of salient-pole motors, [9].

Lph ' Lg % LF ' Lg % [Lslot % Lend ] (2.34)


Laa (2 ) ' (LF % Lg0 ) % Lg2 cos22Lbb (2 ) ' (LF % Lg0 ) % Lg2 cos(22 % 120E )Lcc(2 ) ' (LF % Lg0 ) % Lg2 cos(22 ! 120E )

(2.37)

Laa[d] ' (LF % Lg0 ) % Lg2

Laa[q] ' (LF % Lg0 ) ! Lg2(2.38)

or (LF % Lg0 ) 'Laa[d] % Laa[q]

2; Lg2 '

Laa[d] ! Laa[q]

2. (2.39)

Lab (2 ) ' MF ! 1

2Lg0 % Lg2 cos(22 ! 120E ) (2.40)

LLL ' Laa % Lbb ! 2Lab (2.41)

Fig. 2.25 Reference axes for dq analysisFig. 2.24 Cross-section of PM motor showing d-axis

The dq-axis transformation renders the machine inductances independent of rotor position by castingthe equations in a frame of reference that rotates in synchronism with the rotor. Under steady-stateAC conditions the currents and voltages in this frame of reference are constant, as though they wereDC quantities. Although the inductances still vary with current, the functional variation is simpler andis not time-dependent.

The dq-axis transformation is valid only if the motor is sinewound: i.e., the windings are sine-distributed (or nearly so). In this case, irrespective of the rotor shape, there is only a 2nd harmonicvariation for the phase self-inductances (Laa, Lbb, Lcc) and mutual inductances (Lab, Lbc, Lca) as the rotorrotates: thus if 2 is the rotor position we can write

The angle 2 and the relative orientation of the phase winding axes are shown in Figs. 2.24 and 2.25. Theconstant term in the phase inductance has a leakage component LF, and two "airgap components" Lg0

and Lg2. Both Lg0 and Lg2 are associated only with the fundamental space-harmonic component of theconductor distribution. The leakage component includes slot-leakage, end-turn leakage, and"differential" leakage —i.e., inductance associated with higher-order space harmonics of the windingdistribution. The self-inductance attains extreme values when the d- and q- axes of the rotor are alignedwith the phase axis: thus, for phase a:

The mutual inductance between two phases is given by

where MF represents mutual coupling mainly in the slots. The line-line inductance is


LLL ' LLL0 ! 3Lg2 sin(22 ! 30E ) with LLL0 ' 2(LF ! MF ) % 3Lg0 (2.42)

LLL[d] ' 2(LF ! MF ) % 3[Lg0 % Lg2 ] ' LLL0 % 3Lg2

LLL[q] ' 2(LF ! MF ) % 3[Lg0 ! Lg2 ] ' LLL0 ! 3Lg2(2.43)

LLL0 'LLL[d] % LLL[q]

2; Lg2 '

LLL[d] ! LLL[q]

6(2.45)

Ld ' LF ! MF % Lmd ; Lq ' LF ! MF % Lmq (2.46)

Lmd '3

2(Lg0 % Lg2 ) ; Lmq '

3

2(Lg0 ! Lg2 ) . (2.47)

Xd ' 2B fLd ' Xmd % XF

Xq ' 2B fLq ' Xmq % XF(2.49)

Xmd ' 2B fLmd ' 'd Xm0

Xmq ' 2B fLmq ' 'q Xm0(2.50)

Xm0 ' TLm0 '3Tµ0 Dl

Bp 2 g )(kw1 Tph )2 . (2.51)

'd 'gN

gdNN; 'q '

gN

gqNN(2.52)

LLL[d,q] ' LLL0 '2

3(LF ! MF ) % Lg0 ± Lg2 . (2.44)

LLL[q] ' 2Lq and LLL[d] ' 2Ld (2.48)

When eqns. (2.37) and (2.40) are substituted in eqn. (2.41), LLL simplifies to

Like the phase inductance, LLL also has a constant and a second-harmonic term, and it attains extremevalues when the d- and q-axes are aligned such that

For delta-connected sinewound machines it can be shown that

The extreme values of LLL could be measured, and then eqns. (2.43) could be solved for LLL0 and Lg2 :

Using the d,q-axis transformation [9], we obtain the synchronous inductances

where

From eqns. (2.43) with (2.46) and (2.47),

suggesting that static measurement of LLL[d] and LLL[q] can be used to measure Ld and Lq, provided thatthe windings are sine-distributed or nearly so.

In most brushless PM motors Lmd< Lmq and Ld < Lq, so Lg2 < 0. The synchronous inductances Ld andLq and the reactances Xd and Xq can be calculated from the following:

where

Xm0 is the airgap component of the synchronous reactance of a nonsalient pole machine having anairgap length equal to gN = kc g, where g is the actual airgap and kc is the Carter coefficient: thus

where kw1 is the fundamental winding factor, Tph is the number of turns in series per phase, and T =2B f. The coefficients 'd and 'q are given by

where gdNN is the effective airgap in the d-axis including the effects of the magnet and the saliency; andgqNN is the same in the q-axis. Formulas for gdNN and gqNN are derived for a number of rotor types in [1,2],using a magnetic field solution directly and thereby avoiding the need to precalculate Lg0 and Lg2.


Lg0 '1

3[Lmd % Lmq ] ; Lg2 '

1

3[Lmd ! Lmq ]. (2.53)

Ldiff ' Lg ! Lg0 ; Mdiff ' Mg ! (!Lg0 /2) . (2.54)

ksg '

2

3Lm0

Lgg

'23

Xm0

TLgg

(2.55)

Lph ' Lg % Lslot % Lend ; Mph ' Mg % Mslot . (2.56)

Once Lmd and Lmq are calculated from eqn. (2.50), Lg0 and Lg2 can be calculated from eqn. (2.47): thus

A sine-distributed winding has an airgap inductance of Lg0 and a phase-phase mutual inductance of!Lg0/2. Therefore if Lg and Mg are the actual airgap self- and mutual inductances (including windingharmonics), the respective components of differential leakage inductance can be estimated as

Criterion for sinewound motors: To help decide whether a motor is sinewound, a parameter

can be calculated. ksg is unity for sinewound motors. Lm0 is the synchronous inductance and Lgg is theactual airgap inductance that would be obtained if the winding was perfectly sine-distributed and therotor was a smooth iron cylinder with the same airgap as in the actual machine. Both Lm0 and Lgg arefree of end-effects and slot-leakage. For 1- or 2-phase motors the 2/3 factor is omitted. If ksg is close to1, the winding is nearly sine-distributed, and calculations with d,q equations should agree with thoseperformed with direct phase variables. On the other hand, if ksg is far from 1, the differential leakagewill not be negligible and calculations with d,q equations will not agree with those performed withdirect phase variables. This applies equally to nonsalient-pole motors, so ksg is calculated for these too.

Summary—phase inductances

In nonsalient-pole motors, Lph and Mph are constant and are calculated from

The airgap components Lg and Mg can be calculated by summing the self- and mutual flux-linkages ofeach coil due to current flowing in every coil, preferably using a method based on Hague’s analysisdiscussed earlier, which makes it possible to account for any winding distribution. This procedureincludes the differential leakage reactance in Lg, that is, the reactance associated with all space-harmonics of the armature-reaction field except the fundamental.

In salient-pole motors Lph and Mph vary with rotor position. For analysis by dq-axis theory thevariation is assumed to follow eqns. (2.37) and (2.40). The leakage inductances LF and MF must includethe differential leakage components Ldiff and Mdiff respectively, and these are estimated using eqn.(2.54).

Summary—synchronous inductances

In general the synchronous inductances are calculated by eqns. (2.49) and (2.50). The associatedreactances X d and X q are used in the phasor diagram. In nonsalient-pole motors, ideally Xd = Xq.However, there may be a slight difference between Xd and Xq, even in surface-magnet motors, if therelative permeability of the magnet is slightly higher than 1 and the magnet is not totally rotationallysymmetrical. In salient-pole motors Xd and Xq differ, usually with Xq > Xd, and in embedded-magnetmotors they can vary significantly as a function of current. The component inductances Lg0 and Lg2 canbe recovered using eqn. (2.53). The maximum and minimum values of the phase inductance can beobtained with eqn. (2.38), and the maximum and minimum values of the line-line inductance with eqn.(2.48).


Classification of motors

Table 2.3 shows a suggested classification of different types of permanent-magnet rotor into “salient-pole” and “nonsalient pole” types.

Rotor type Inductance variation with2222

Salient-pole or nonsalient pole

Surface radial or parallel Constant nonsalient pole

Breadloaf Weakly variable Can be treated as nonsalient pole

Spoke Strongly variable salient pole

Exterior radial orparallel

Constant nonsalient pole

IPM Strongly variable salient-pole

Inset magnet Variable salient-poleTABLE 2.3

VARIATION OF INDUCTANCE WITH COMMON ROTOR TYPES

Classification of inductances

Because salient-pole motors are analyzed using the dq-axis equations, they require the calculation ofthe synchronous inductances Ld and Lq and associated component inductances. Nonsalient pole motors,on the other hand, are analyzed in direct phase variables and they require the actual phaseinductances. To clarify these differences, the inductance parameters can be classified into two groupslabelled “actual” and “synchronous”, as shown in Table 2.4.

Actual (phase) Lph Mph

SP = salient pole Lg (NSP) or Lgg (SP) Lslot Lendt

NSP = nonsalient-pole Mg (NSP) or Mgg (SP) Mslot Ldiff

LLL[d] LLL[q] LLL (NSP) or LLL0 (SP)

Lg0 Lg2 Laa[d]

Synchronous Ld Lq Laa[q]

Xd Xq LF

'd 'q MF

kw1 Xm0 XFTABLE 2.4

CLASSIFICATION OF INDUCTANCEPARAMETERS

With salient-pole motors the inductances Lph, Mph, Lg and Mg are variable, but Lgg and Mgg are theapproximate values that would be obtained with a solid steel rotor of the same radius and airgap g.These can be used in the calculation of ksg.

For time-stepping simulations, nonsalient-pole motors need not be sinewound, because their circuitequations can be expressed in direct phase variables. Salient-pole motors, on the other hand, shouldbe sinewound if at all possible, to facilitate the solution of their circuit equations either by the phasordiagram, or by time-stepping simulation in dq axes.


5 Because of the distribution of conductors the winding links only a fraction kw1 of the fundamental flux. It acts as a filter. Toget a sinusoidal EMF requires low winding factors kwn for the harmonic fluxes (n > 1) and a value of kw1 close to 1.6The phase lead results from the use of the “sink” convention in which e = dR/dt rather than !dR/dt. This is appropriate formotors, but less intuitive for generators. However, permanent magnet motors are vastly more common than generators.

Fig. 2.26 Phasor diagram on open-circuit

Q1Md ' kw1 Tph M1Md 'kw1 Tph

2

B1Md DLstk

p[V&s r.m.s.] (2.57)

E ' jEq1 ' jTQQQQ1Md (2.58)

2.7 BASICS OF SINEWAVE OPERATION

The most fundamental parameters of a brushless permanent-magnet motor drive are the number ofphases and the method of controlling the current. The drive circuit configuration depends primarilyon the number of phases. Broadly speaking there is a division into two groups:

(1) Low-cost brushless DC drives with one or two phases, widely used in blowers and fans; and(2) Polyphase brushless DC drives with smooth torque and high efficiency.

Group 1 drives often use single-ended drive circuits, sometimes with bifilar windings. Group 2 drivesuse bridge circuits for high efficiency and maximum controllability. In some instances, full H-bridgecircuits are used with 1-phase or 2-phase motors to achieve high efficiency. With 1 or 2 phases, it isgenerally not possible to achieve reliable starting in both directions from any position, unless auxiliarymethods are adopted, including tapered airgaps and reluctance poles, [16].

The simplest type of “drive” is a plain AC voltage source with 2 or 3 phases. In the steady state, a motoroperating from such a supply can be represented by its phasor diagram, provided that it is sinewoundor approximately so. Even before the motor is connected to the supply, the open-circuit phasordiagram can be drawn as shown in Fig. 2.26. Since the motor is on open-circuit there is no current, andthe phasor of the generated EMF E leads the phasor of the fundamental flux-linkage QQQQ1Md by 90E. Thefundamental flux-linkage Q1Md is the product of the effective number of series turns/phase kw1Tph andthe fundamental component of magnet flux in the airgap M1Md, where kw1 is the fundamental harmonicwinding factor and Tph is the number of series turns per phase.5 Thus if B1Md is the peak value of thefundamental flux-density produced by the magnet on open-circuit,

and is centred on the d-axis. The subscript 1 refers to the fundamental space-harmonic component. Thephasor relationship between E and QQQQ1Md is

where the subscript 1 emphasizes the fundamental, q the q-axis, d the d-axis, and M the magnet. The90E phase lead comes from Faraday’s law, when the phasors are expressed as complex numbers, andit is normal to consider the magnet flux-linkage phasor to be aligned with the d-axis while the generatedEMF is aligned with the q-axis.6


7This definition of power factor is valid only if the voltage and current are sinusoidal.8Also known as “amortisseur” windings, these are usually in the form of a cast cage similar to the cage of an induction motorrotor. See, for example, Jordan [1983], Miller [1984].

Fig. 2.28 Phasor diagram

Fig. 2.27 Circuit diagram of one phase of nonsalient pole brushless PM motor, with AC voltage supply

The connection of the motor to a balanced AC voltage source is shown inFig. 2.27. If the motor is nonsalient pole it can be represented electricallyby its generated EMF E, its phase resistance R and its synchronousreactance Xs (which is equivalent to Xd in a salient-pole machine). TheAC voltage source is assumed to have a constant voltage V and nointernal impedance. The phasor diagram for the circuit in Fig. 2.27 isshown in Fig. 2.28. It shows that the supply voltage V comprises theseries combination of the volt-drops RI, jXsI, and E, which must be added“vectorially” because they are not in phase with one another. The volt-drop RI is in phase with the current I and its phasor is parallel to thecurrent phasor, but the volt-drop jXsI is at right-angles to the current andleads it in phase because Xs is an inductive impedance.

The angles shown in Fig. 2.28 are important. The angle N between V andI is the power-factor angle such that the power factor is cos N.7 The angle( between E and I is the “torque angle”, which is very important indrives which control the phase and magnitude of the current relative tothe shaft position. The angle * between V and E is called the “load angle”and is also sometimes called the “torque angle” especially when themotor is operating from an AC voltage source (i.e., without currentcontrol).

Operation from an AC voltage source is useful for understanding the basic concepts of the phasordiagram, but in practice it is very rare. Brushless PM motors operated in this way are inherentlyunstable and tend to “hunt” or oscillate about the synchronous speed unless they are fitted withdamper windings on the rotor.8 Moreover, without the damper winding the motor has no means ofstarting. With damper windings, such motors are referred to as “line-start” motors. They are used incertain applications where several motors operate in synchronism from a single inverter, or directfrom the mains supply where, for some reason, exact synchronous speed is required. Several attemptshave been made to develop such machines to compete with induction motors on the basis of highefficiency, even with single-phase capacitor motors, and although some very good technical resultshave been reported, there has been little commercial development because of the cost and certainapplication problems.


9AC induction motors do not have obvious d,q axes because (i) both the stator and the rotor are cylindrical, i.e., rotationallysymmetric; and (ii) there is no distinct excitation winding. However, dq axis theory still applies to the induction motor. See,for example, Fitzgerald and Kingsley, [1961]. In descriptions of field-oriented control of induction motors, since about 1975, thelabels d,q are applied extensively, but several different frames of reference are used and there is probably no single definitionof d,q axes that would work equally well for all variants. In dealing with PM brushless machines it is natural to fix the dq axesto the rotor, as in the wound-field synchronous machine of Fig. 2.29, but this is not the only convention that can be adopted.

Fig. 2.29 Wound-field synchronous machine Fig. 2.30 Surface-magnet motor

Fig. 2.31 Interior-magnet motor

2.8 THE ORIGIN AND DEVELOPMENT OF AC VECTOR CONTROL FOR BRUSHLESS PM MOTORS

The terms direct axis and quadrature axis refer to the two axes of symmetry of the magnetic fieldsystem as defined by the field winding or excitation winding. In the DC machine the excitation windingis on the stator9 and therefore the d.q axes are fixed to the stator. In the AC synchronous machine theexcitation winding is on the rotor and therefore the d.q axes are fixed to the rotor, Fig. 2.29. The d-axisis the axis of symmetry centred on one rotor pole. Sometimes it is called the polar axis or field axis.The q-axis is also an axis of symmetry and it is known as the interpolar axis. Because there are twoaxes of symmetry, the rotor is said to have two-axis symmetry. In electrical terms the d- and q-axes areorthogonal, i.e., separated by 90 “electrical degrees”. Indeed the definition of “electrical degrees” issuch that there are 180 electrical degrees between consecutive d-axes.

In the permanent-magnet machine the wound poles are replaced by permanent magnets, Figs. 2.30 and2.31, but the meaning of the d and q axes is unchanged. The interior-magnet motor in Fig. 2.31 is asalient-pole machine with different inductive properties along the d- and q-axes, but the surface-magnetmotor in Fig. 2.30 is nonsalient-pole, being rotationally symmetric apart from the magnetization of themagnets and the possibility of slight differences in permeability along the d- and q-axes.

To understand vector control we need first to understand the phasor diagram, Fig. 2.32, which isslightly more complex than the one in Fig. 2.28 in that it is drawn for a salient-pole machine with Xddifferent from Xq. The voltage drop XsI is replaced by two separate voltage drops XdId and XqIq.


Fig. 2.32 Phasor diagram and flux-linkage vector diagram

In the steady-state the r.m.s. voltage V in each phase is related to its r.m.s. flux-linkage Q by a simpleequation V = TQ where T is the frequency in electrical rad/sec. We have seen that the phase angle ofthis voltage is 90E ahead of the flux-linkage. For example, in Fig. 2.32, V is represented by an arrow(called a phasor) which is 90E ahead of the arrow representing Q. Mathematically, the phasor valueof V is the complex number V = V ej(* + B/2), the angle (* + B/2) being the phase angle of V relative to thereference axis (the d-axis). Both V and Q are sinusoidal quantities in time, and both are representedin the diagram by their r.m.s. values, while the continual advance of their phase angles is representedby the rotation of all the phasors at the angular velocity T rad/sec. During this rotation the phasedisplacement between Q and V remains at 90E.

The phasor diagram in Fig. 2.32 is split into two parts. On the left are the electrical quantities, i.e.,voltages and currents. On the right are the corresponding magnetic flux-linkages. The separation intotwo parts makes the diagram clearer. The flux-linkage part of the diagram is usually omitted, but it canbe considered to have a physical reality, in that the flux actually rotates in space, at an angular velocityof T elec. rad/s. The flux-linkages can therefore be interpreted as space vectors (if considered asrotating physically in space); or simply as phasors (if considered as time-varying sinusoidal quantities).

We can now examine how the entire phasor diagram is built up. In a permanent-magnet machine thereis a flux due to the magnets which links all the windings in turn, and gives rise to the flux-linkage Q1Mdin each phase, even when there is no current flowing. Corresponding to this flux-linkage is the “open-circuit” voltage E, which leads Q1Md in phase by 90E, just as V leads Q by 90E. In the phasor diagram,the flux Q1Md is along the d-axis, and therefore E is along the q-axis. Note that the dq axes in the phasordiagram on the left-hand side are really fictional axes defined in terms of the time phasor diagram.However, since the time phasor diagram (of voltages and currents) and the space vector diagram (offlux-linkages) both rotate synchronously in their respective coordinate systems, we tend to blur thedistinction and regard the dq axes as being the same for both. In common engineering parlance,everyone takes this for granted and one would be thought pedantic if one continually reiterated thedistinction between them.


10The above theory is called the two-axis theory, two-reaction theory, or dq-axis theory of the synchronous machine. Historicallyit dates back to A. Blondel in the 1890's, but its modern exposition is generally attributed to R.H. Park along with others around1920!1930, notably Doherty, Nickle, and W.V. Lyon. See Park RH, Two-reaction theory of synchronous machinery; I ! Generalizedmethod of analysis, Transactions IEEE, Vol. 48, pp. 716!730, July 1929. One of the most famous accounts of it is given by CharlesConcordia in Synchronous Machines, John Wiley, 1951.

Te ' mp (Qd Iq ! Qq Id ) (2.60)

Id ' ! I sin ( ; Iq ' I cos ( (2.63)

Qd ' Q1Md % Ld Id ; Qq ' Lq Iq . (2.61)

Te ' mp [Q1Md I cos ( ! I 2 sin ( cos ( (Ld ! Lq ) ] (2.64)

Te ' mp [Q1Md Iq % Id Iq (Ld ! Lq ) ] (2.62)

( ' sin!1 14

!Q1Md

)Q%

Q1Md

)Q

2

% 8 (2.65)

I ' Id % jIq (2.59)

When current flows in the stator windings it creates an additional flux which is easier to analyse if wefirst “resolve” the current into two components in the phasor diagram: Id along the d axis and Iq alongthe q axis. In terms of the complex phasor I, this is written

The flux-linkage produced by Id is LdId, where Ld is the d-axis synchronous inductance. It is in phasewith Id and induces a voltage XdId which is 90E ahead of Id (i.e., parallel to the q axis); then Xd = TLdis the d-axis synchronous reactance. Likewise the current Iq produces a flux-linkage LqIq and a voltageXqIq which is parallel to the negative d axis, with Xq = TLq the q-axis synchronous reactance. The totalvoltage at the phase terminals is the sum of the component voltages E, XdId, and XqIq, added together“vectorially” by means of the polygon formed by the respective phasors placed nose-to-tail. Similarlythe total flux-linkage is the vector sum of the component flux-linkages Q1Md, LdId and LqIq.10

The phasor diagram is useful in understanding how the torque is limited by the voltage and currentavailable from the drive. Neglecting losses, the electromagnetic torque is given by

where m is the number of phases, p is the number of pole-pairs, and Qd and Qq are the d- and q- axiscomponents of the r.m.s. flux-linkage per phase. In the d-axis the flux-linkage has two components,Q1Md due to the magnet and LdId due to the stator current component Id. In the q-axis there is nomagnet flux but only the armature-reaction component LqIq. Thus

If we substitute the expressions for Qd and Qq in eqn. (2.60), we get

which shows that there are two components of torque, a permanent-magnet alignment torque Q1MdIq and a reluctance torque IdIq(Ld ! Lq). If there is no saliency Ld = Lq and no reluctance torque. If themagnet flux is constant the torque is proportional to Iq, and the torque constant k T = Te/I is constant.The controller should then maintain I = Iq, which is sometimes called “quadrature control”. If,however, there is saliency and Ld and Lq are unequal, the mix of permanent-magnet alignment torqueand reluctance torque can be adjusted by changing the phase angle of the current (() as well as itsmagnitude. Noting that ( is measured from the q-axis in the positive (CCW) direction, we can write

and if we substitute these expressions in equation (2.60) we get

At any given current level I, we can differentiate this expression with respect to ( to find the value of( which gives maximum torque. The result is

where )Q = (Ld ! Lq)I.


11Vm is the fundamental harmonic component of the actual phase voltage.

Id 'Vq ! E

Xd

'Vm cos * ! E

Xd

; Iq '!Vd

Xq

'Vm sin *

Xq

. (2.66)

Fig. 2.33 Current-limit circle and voltage-limit circle at the change-over speed; non-salient-pole motor with Xd = Xq.

Vm2

' Vd2

% Vq2 , (2.67)

(Xq Iq )2% (E % Xd Id )2

' Vm2 . (2.68)

If there is no saliency )Q = 0 and from eqn. (2.64) the phase angle that gives maximum torque is ( = 0:i.e., the current must be oriented in the q-axis ($ = 90E) in phase with the EMF E. Phase advance (( >0) in a surface-magnet motor reduces the torque constant k T, which is defined as Te/I. This can be seenin eqn. (2.64) (with Ld = Lq). The reduction in k T is in the ratio cos (.

The argument leading to the optimum phase angle given by eqn. (2.65) assumes that the current I hasa certain value. If this is the rated r.m.s. current Im, the tip of the current phasor lies on a circle, centre0, radius Im, Fig. 2.33. In practical terms this corresponds to the case where the drive is operatingunder current limit and the current regulators have complete control of the current waveform.

The EMF E is equal to TQ1Md and at high speed it approaches the maximum available supply voltage Vm.At certain values of ( the drive may not have sufficient voltage to maintain the current Im. We musttherefore examine what happens when the drive is voltage-limited rather than current-limited. Imaginethat the drive is supplying maximum voltage Vm to each phase of the motor, such that the phase anglebetween Vm and E is * (Fig. 2.32).11 From the phasor diagram

and

so that


Te 'mp

T

EVm

Xd

sin * %Vm

2

21

Xd

!1

Xq

sin 2* . (2.69)

* ' cos!1 ! J ± J2% 8

4(2.70)

(Vm/Xd )2' (E/Xd )2

% Im2 (2.72)

J 'E/Vm

1 ! Xd /Xq

. (2.71)

TQ 'Vm

Q1Md2

% (Ld Im)2rad/sec. (2.73)

According to these equations, the tip of the voltage phasor lies on a circle of radius Vm, but the tip ofthe current phasor has an elliptical locus as * and ( vary. Under this type of control, with constant(maximum) voltage Vm but variable phase angle * between Vm and E, the torque can be calculated bysubstituting eqns. (2.66) in eqn. (2.62) to eliminate Id and Iq. The result is

Again we can differentiate this expression to find the phase angle * which maximises the torque. Aftersome simplification the result is

where

If there is no saliency, the angle which gives maximum torque is * = 90E. The phase angle of the currentwill vary in a more complex manner as * varies.

For machines with no saliency (such as surface-magnet motors), Xd = Xq and we can summarise theguidelines for maximum torque as follows:

Low speed Control current with ( = 0

High speed Control voltage with * = 90ETABLE 2.5

The change-over between “low speed” and “high speed” is the maximum speed at which the ratedcurrent Im can be driven into the motor with ( = 0. For a surface-magnet (non-salient-pole) motor withXd = Xq the voltage-limit ellipse becomes a circle, and the change-over frequency can be determinedfrom its intersection Q with the current-limit circle, Fig. 2.33. At this speed the voltage-limit circle isjust large enough to intersect the current-limit circle at point Q, where Iq = Im and Id = 0. Therefore

In this equation Xd and E are both proportional to speed or frequency T, but Im and Vm are fixed. If wesubstitute E = TQ1Md and Xd = TLd we can re-arrange eqn. (2.72) to give the change-over frequency:

The subscript Q identifies this value of frequency as the change-over value, sometimes known as thecorner-point or base value. The corresponding speed in rpm is NQ = TQ /p × 30/B. At speeds higher than NQ it is still possible to drive rated current Im into the motor, but not at theoptimum angle for maximum torque (( = 0 in a non-salient-pole motor). As the frequency increases theradius of the voltage-limit circle decreases and the intersection with the rated-current circle movesalong the arc QD. To illustrate this, Fig. 2.34 shows the conditions at four different speeds, N1 < N2 (=NQ) < N3 < N4.


NQ

ND

' u ! 1 ! u 2 (2.74)

1

2< u < 1 (2.75)

Fig. 2.34 Circle diagram at four different speeds (non-salient-pole motor)

At low speed (circle 1) the voltage-limit circle completely encloses the current-limit circle at Im, whichmeans that this current can be driven into the motor with any phase angle. In fact the current couldbe increased up to the value OL, approximately twice Im, under the control of the current regulator.

As the speed increases the voltage-limit circle shrinks, until at NQ the maximum current that can bedriven along the q-axis is Im, circle 2. At a still higher speed circle 3 shows operation at P with I = Im,but the phase angle ( is advanced as shown, and the torque is reduced by the factor cos (. Eventuallya speed is reached at which the current Im can be driven only along the negative d axis, circle 4. Allthe current is now used to suppress the flux (flux weakening), and none of it is available to producetorque. The intersection is at point D and the torque is zero. To achieve this point the currentregulator must operate with a massive phase advance of 90E and with maximum current reference.

It can be shown that the speed at which point D is reached is related to the change-over speed by

where u = TQQ1Md/Vm = EQ/Vm, EQ being the value of E at the change-over speed. For a solution toexist at a positive speed, we must have

For example, if u = 0.8, ND = 5 NQ, but if u = 0.9, ND = 2.155 NQ. Alternatively suppose that the motormust maintain maximum torque at speeds up to 3,000 rpm and be capable of just reaching 6,000 rpm.Then ND/NQ = 2 and according to equation 15, u must be no higher than 0.911.

The ratio of the speeds ND and NQ can be expressed in terms of the reactance or inductance of themotor. Define the per-unit synchronous inductance as


Te 'mp

TIdN Iq )X (2.78)

) X ' Xd ! Xq and IdN ' Id %E

)X. (2.79)

x 'Ld Im

Q1Md

. (2.76)

ND

NQ

'x 2

% 1x ! 1

. (2.77)

Fig. 2.35 Constant-torque loci (salient-pole motor)

The per-unit synchronous reactance is identical to x, and by equating CD to OP in Fig. 2.34 we get

Even with x = 0.3 (a high value for a surface-magnet motor), ND/NQ = 1.491, which is quite a narrowspeed range above the corner-point speed. For a motor which must maintain constant torque up to3,000 rpm and just be able to reach 6,000 rpm ( ND/NQ = 2), eqn. (2.77) prescribes that x must be at least2.347, which is unrealistically high. This helps to explain why in some instances, additional inductancehas been connected in series with the motor to extend the speed range above the corner-point speed.The additional inductance makes the system more responsive to phase advance in weakening the totalflux, but some of this flux is in the external inductance and not in the motor. Note that eqns. (2.74) and(2.77) are independent and must both be satisfied.

Although eqns. (2.74) and (2.77) are an incomplete account of the variation of torque with speed, andapply only to non-salient-pole (surface-magnet) motors, they help to resolve the question as to whetherphase advance is needed “to overcome the rising EMF” or “to compensate for inductance” as the speedincreases. The answer is both.

A constant-torque locus can be superimposed on the circle diagram, as in Fig. 2.35, by writing thetorque equation as

where

This is a rectangular hyperbola asymptotic to the negative d-axis and to a q-axis which is shifted to theright by E/)X. With high-energy magnets the constant-torque contours are more nearly horizontal,but with low-energy magnets they have more curvature. Fig. 2.35 is drawn for a salient-pole motor (Xd… Xq). For a nonsalient pole motor the torque loci are horizontal straight lines.


12See Strauss F, Synchronous machines with rotating permanent-magnet fields, AIEE Transactions, Vol. 71, Pt. II, pp. 887-893,October 1952. Also Merrill FW, Permanent-magnet excited synchronous motors, Transactions AIEE, Vol. 74, pp. 1754 1760, 1955.13See DD Hershberger, Design considerations of fractional horsepower size permanent-magnet motors and generators, AIEETransactions, pp. 581!584, June 1953.

( ' sin!1 14

!0.91.0

%0.91.0

2

% 8 ' 31.13E . (2.81)

I ' Id % j Iq ' I e j$' I e j (( % 90E) ;

V ' Vd % j Vq ' V e j (* % 90E) .(2.82)

( ' sin!1 14

!u

)x%

u

)x

2

% 8 . (2.80)

The constant-torque loci in Fig. 2.35 are drawn for three torques T1 < T2 < T3. The middle one goesthrough the point Q which we earlier associated with the corner-point for a nonsalient pole motor.With saliency, however, the constant-torque loci show that the torque can be increased by phaseadvance between Q and P, with the current maintained at the rated value Im. The additional torqueis reluctance torque, the second term in eqn. (2.64), which comes from the saliency. As ( increases, thesin ( term increases quickly from zero while cos ( changes only slowly. As we have seen, eqn. (2.65)can be used to determine the phase advance angle which maximises the torque. Having defined theper-unit synchronous reactance x for a non-salient-pole motor, it is a simple matter to extend this toxd and xq, the per-unit synchronous reactances in the d- and q-axes respectively, and put )x = (xd ! xq).If Im is the rated current, its per-unit value can be taken as 1, and if we use the per-unit EMF u asdefined earlier, eqn. (2.65) gives

For example, suppose u = 0.9 and xd = 0.5 and xq = 1.5. Then )x = !1.0 and

The phase advance can be used to achieve the same torque at a lower current, or the same torque ata higher speed as the voltage-limit ellipse shrinks in size.

So far we have discussed the controlled variation of the current or voltage in terms of their magnitudesand phase angles, I and ( or V and *; in other words, we have represented the voltage and current inpolar coordinates. It is equally straightforward to describe the controlled variation of current in termsof the cartesian components Id and Iq, and the controlled variation of voltage in terms of Vd and Vq.Mathematically these components are equivalent to Ig( and Vg*. Using complex numbers,

2.9 HISTORY OF BRUSHLESS PM MOTOR DRIVES

The permanent-magnet synchronous machine was certainly known in the early 1950's.12 Although mostpermanent-magnet machines at that time were generators, motors were also manufactured,13 and theclassical two-axis theory was used to analyze them. The motors were “line-start” motors supplieddirectly from the AC mains, without electronics. In the early 1970's the discovery of high energy cobalt-samarium magnets gave new impetus to the development of permanent-magnet AC motors. Lower-energy ferrite magnets were already used in DC brush-type motors and improvements in these magnetsalso encouraged new work in AC line-start machines, notably by Brown Boveri (Isosyn motor, 1978)and Reliance Electric (1979). Some of these motors were used with inverters, but they were still “line-start” motors without shaft position sensing: therefore they were not self-synchronous in the sensethat modern servo-motors and brushless DC motors are.

The so-called “brushless DC” motor emerged at this time (mid-1970's), notably from Papst. The“brushless DC motor” is strictly speaking a DC machine with electronic commutation. Although themachine is physically similar to the AC brushless permanent-magnet machine, and in many cases


14Lajoie-Mazenc M, Foch H and Villanueva C, Feeding permanent-magnet machines by a transistorized inverter,PCI/MOTORCON, pp. 558!570, September 1983. See also Lajoie-Mazenc et al [1976] in the main references, for an earlierdescription of this system.15Since about 1985 a good many more papers have been published specifically on field-oriented dq-axis control, withdevelopments in sensorless operation (i.e., the elimination of the encoder or resolver), and more sophisticated methods ofdetermining Id and Iq and of controlling the current regulators. See Jahns and Van Nocker [1990], Morimoto et al [1990!1994].

Fig. 2.36 Lajoie-Mazenc’s I!( controller (1983)

identical, the method of driving it is fundamentally different. The brushless DC or “electronicallycommutated” motor, sometimes also known as the “squarewave” or “trapezoidal” motor, does not havea rotating ampere-conductor distribution and since it does not have sine-distributed windings, phasoranalysis and dq-axis theory are not applicable to it. To a purist, the squarewave drive is strictlyappropriate only with surface-magnet motors which have no “saliency” (i.e. Ld = Lq) and no reluctancetorque. However, motors of this type are sometimes used with squarewave drives.

While “brushless DC” and “line-start AC” motors were emerging in the 1970's, many engineersenvisaged the possibility of removing the rotor cage from the “line-start” AC motor and of feeding thismotor with sinewave currents phase-shifted to maximize the torque per ampere. An account of suchan investigation is reported by Lajoie-Mazenc [1976,1983],14 including

— self-synchronization by means of shaft encoder feedback;— the addition of a variable phase-shift to optimize the torque production;— the use of a digital encoder signal to index a sinewave reference for the current waveform; and— the accommodation of saliency (Xd … Xq).

Although Lajoie-Mazenc used dq-axis theory in deriving equations for the optimum phase-shift angle,he did not describe what would now be termed a field-oriented dq controller. The architecture of hiscontroller is reproduced in Fig. 2.36, and it is evident that it is an “I-( controller” in the sense that itprovides for the adjustment or control of the magnitude and phase of the current, as discussed in theprevious section. Since the current magnitude is determined in the rectifier upstream of the inverter,it is not possible for this controller to exercise direct independent control of Id and Iq.

A field-oriented dq controller in the strict sense is one in which the d- and q-axis components of thecurrent are controlled independently, one of them being substantially oriented to control the flux andthe other one being oriented to control the torque. In general this gives rise to a control block diagramin which the separate d- and q-axis components are identifiable, as well as the means of controllingthem. A controller of this type is described by Leonhard [1983].15


16See, for example, Jahns TM, Kliman GB and Neumann TA, Interior permanent magnet motors for adjustable speed drives, IEEETransactions, Vol. IA-22, No. 4, pp. 738!747, July/August 1986.

Fig. 2.37 dq controller described by Jahns et al.

Fig. 2.38 Effect of phase advance on torque/speed characteristic. With no phase advance, the torque falls to zero at 2500 rpm.With phase advance increasing from 0 at 1200 rpm to 60E at 2400 rpm, the torque at 2400 rpm is 0.85 Nm. The motoris of the type shown in Figs. 2.24 and 2.31.

Another example of a true dq controller is shown in Fig. 2.37.16 In the simplest case the d-axiscomponent Id can be held at zero ( fd = 0) while the torque would be varied by controlling Iq via fq. Athigh speed, when the EMF of the machine approaches the supply voltage, the flux can be “weakened”by applying negative current in the d-axis: Id < 0. Alternatively, especially in salient-pole machines,more complex variation with Id and Iq with torque and/or speed can be contemplated.

There is only one choice of reference frame for permanent magnet synchronous machines that alignsthe d-axis with the magnet flux, and in non-salient-pole machines the q-axis is the natural axis for thetorque-producing component of current. This reference frame is the only one which separates the flux-controlling component of current from the torque-controlling component in nonsalient pole machines.With salient-pole machines there is more justification for adopting a different dq axis frame ofreference: e.g., one in which the d-axis is aligned with the total flux rather than just the magnet flux.

Effect of phase advance on torque/speed characteristic: Fig. 2.38 shows an example of atorque/speed characteristic with no phase advance. The torque starts to fall off at about 1400 rpm, andhas fallen to zero when the speed reaches 2500 rpm. Also shown is the torque/speed characteristic witha phase advance angle ( which increases linearly from 1200 rpm to 2400 rpm. The torque does not reachzero. Even at 2400 rpm it is 0.85 Nm. At 2100 rpm it is still more than 1.0 Nm.

If we take base speed as 1200 rpm, the power at the base speed is 1200 rpm × 1.55 Nm = 195 W. With nophase advance, at 1800 rpm the torque is 0.7 Nm and the power is about 132 W and is less than thepower at base speed. With phase advance, the torque at 1800 rpm is 1.3 Nm and the power is 245 W.Phase advance almost doubles the power at 1800 rpm, which is 50% higher than the base speed.


17 The term “switch control strategy” means much the same thing as “pulse-width modulation strategy”. There is a largenumber of different PWM strategies and the subject has been extensively researched, especially in relation to the control offield-oriented AC motor drives. The particular strategy described here is a simple intuitive one. 18The “base segment” is a 60E interval during which PC-BDC computes the current waveforms recursively, reconstructing theentire period from this segment by a substitution process once the solution has converged.

Fig. 2.39 Voltage vectors

Fig. 2.40 Six-step connections

The principle of phase advance is equally applicable to the squarewave drive. But since squarewaveoperation cannot be expressed in terms of phasors, computer simulations are used for analysis instead.

2.10 SWITCHING STRATEGY FOR SINEWAVE DRIVE

The switch control strategy for obtaining sinusoidal currentcan be expressed in terms of voltage vectors, as shown inFig. 2.39.17 This diagram can be explained with the help ofthe connection diagrams in Fig. 2.40, which show thepolarities of the motor line terminals corresponding to thestates of the six transistors in the bridge circuit, Fig. 2.3.Note that the numbering of the transistors in Fig. 2.3reflects the order in which they switch on, as in thesquarewave drive; but in the sinewave drive it is normal tohave three transistors conducting instead of only two, andthe conduction period for each transistor is 180E instead of120E. Such switching schemes are called “3-phase-on” and“2-phase-on” respectively; or, for short, “3Q” and “2Q”.

Each of the six vectors in Fig. 2.39 corresponds to one of the connections in Fig. 2.40. For example, Q612means that transistors 6,1 and 2 are on, so that line A is connected to the positive terminal and linesB and C to the negative terminal. In a simple “six-step” controller, this condition may persist for 60E,and then transistor 6 switches off and transistor 3 switches on, producing the state Q123 in which linesA and B are connected to the positive terminal and line C to the negative terminal. In the motor, theorientation of the stator ampere-conductor distribution (or “MMF vector”) advances 60E, correspondingto the transition from the 1,0,0 to the 1,1,0 state. Over a full 360 electrical degrees of rotation the sixstates follow one another in this way to produce a coarse approximation to a rotating MMF.

Even though there are only six possible states in a 3Q scheme, the current regulator or PWM controllercan switch between states at a much higher frequency than 6 times per cycle in such a way as to makethe current track a sinusoidal reference signal whose amplitude and phase are determined by thetorque demand and the speed. The coding example on the following page describes a simple, intuitivecurrent regulator algorithm to achieve this. Only the four highlighted vectors are used during the basesegment.18 In the regulator algorithm, the currents (ik1 and ik2) in two lines are sensed at each stepof the time-stepping solution, and the decision is made as to which transistors to switch on or off. Thecoding is as follows:


Fig. 2.41 Usage of dq equations and direct phase variables

{Current regulation in Sixstep}beginiCR1 := ISP * Sin(ThRe); {Set reference in line 1}iCR2 := ISP * Sin(ThRe - 2*pi/3); {Set reference in line 2}If (k*j) mod kTs = 0 then {Check switching freq. }If ik1 > iCR1 thenif ik2 > iCR2 thenVoltageVector := Q456

elseVoltageVector := Q234

elseif ik2 > iCR2 thenVoltageVector := Q612

elseVoltageVector := Q123;

end;

The transistors do not always begin to conduct when they are switched on, because there may be afreewheeling current in the antiparallel diode. At high speed the diode currents may be flowing all thetime so that although the transistors receive turn-on signals they may never conduct. Note that it isnot immediately obvious from the current waveforms whether, say, a positive current in phase 1 isflowing in Q1 or D4, etc.

Although this description is written for a wye connection, the same control logic applies with a deltaconnection. Also note that the 3Q strategy uses complementary switching of the upper and lower powertransistors in each phase leg.

Analytical note: Time-stepping simulation in dq-axes requires that the d,q terminal voltages vd and vq be determinate. In asquarewave 2Q drive, normally two transistors are conducting (PERIOD B) and two terminal voltages and one current aredeterminate, instead of the three terminal voltages. Therefore vd and vq are not determinate and the d,q equations cannot beused. During commutation or freewheeling (PERIOD A), however, a diode connects the freewheeling line terminal to thepositive or negative rail and all three terminal voltages are known. Therefore PC-BDC switches from dq-axes in PERIOD A todirect phase variables in PERIOD B. With two phases conducting, the inductance is the line-line inductance of two phases inseries. It varies sinusoidally with rotor position (assuming that the machine is sinewound), and it and its derivative with respectto position are derived from Ld and Lq.

In a 3Q drive the terminal voltages and hence vd and vq are determinate at all times, permitting the dq equations to be usedthroughout. Note that the actual phase self- and mutual inductances Lph and Mph are used when he dynamic simulation isworking in actual phase variables, while the synchronous inductances Ld and Lq. are used when the dynamic simulation is indq axes.

Fig. 2.41 summarizes these considerations. The solution is no faster in dq axes than in direct phase variables, because undertransient conditions the reference-frame transformations to vd, vq from va,vb,vc and to id,iq from ia,ib,ic (and their inverses) mustbe performed every time-step. This neutralizes any saving in computation time which might be expected with the dq solution.


19The theory of space vectors relies only on the assumption of sinusoidally distributed windings, so that the voltages andcurrents need not be sinusoidal. The theory of space vectors could be applied to the squarewave drive, but only if the motorhad sine-distributed windings — in which case it would be a sinewave motor fed with squarewave currents, not a squarewavemotor in the normal sense of the electronically commutated brushless DC motor.20The “dwell” or conduction period of the transistors in a 2Q drive can be shortened below 120E. It can also be extended beyond120E, in which case there are periods with 3 transistors conducting. If the dwell is extended to 180E the 2Q scheme becomes a3Q scheme all the time. All these possibilities are included in the PC-BDC program.

Fig. 2.42 3-phase drive for wye-connected brushless DC motor.

(a) Conduction in lines A and B with transistors 1 and 6 conducting.

(b) Conduction in lines A and B with transistor 1 switched off ("chopped"). Diode 4 freewheels the line A current.

We have already seen that permanent-magnet brushless motors are often driven with squarewavecurrents, and in the next section we will return to the subject of squarewave control and currentregulation. For operation with squarewave drive, ideally the motor should be designed with“concentrated windings” instead of distributed windings, and it should have a trapezoidal EMF

waveform as described earlier. The dq-axis theory, including the phasor diagram, does not apply to thismode of operation, because it relies on the assumption of sinusoidal voltages and currents.19

2.11 SIMPLIFIED ANALYSIS OF SQUAREWAVE DRIVE

Current regulation

We have seen in Fig. 2.7 that the currents are commutated into the appropriate phases in synchronismwith the rotation of the rotor. For 3-phase sinewave drives it is often the practice to have threetransistors conducting at any time, but with squarewave drives normally two transistors areconducting: one “upper” and one “lower” transistor in the inverter bridge, Fig. 2.42. This is a “2Q”switch control strategy.20

The line currents are regulated or controlled by chopping one or more power transistors in the drive.Fig. 2.42(a) shows the normal path of the current in two lines A and B, and in two phases 1 and 2 inseries, during the interval from 30E to 90E in Fig. 2.7. The line currents and the states of the transistorsare identical for wye- and delta-connected motors, but the way in which the current divides betweenthe phases is different in the two cases. (A detailed analysis is given in Hendershot and Miller, [1994]).

With both transistors Q1 and Q6 conducting, the voltage across the series combination of phases 1 and2 is v12 = Vs. When Q1 switches off, v12 = 0, and the current freewheels through diode D4, as shown inFig. 2.42(b). Instead of switching off Q1, we could equally well switch off Q6; then v12 = 0 and the currentwould freewheel through D3. If both transistors Q1 and Q6 are turned off, the current freewheelsthrough both D4 and D3, but now v12 = !Vs. This is summarized in Table 2.6.


Fig. 2.43 Switching duty-cycle

v1 ' e1 % Ri1 % Lp1 % Mp2

v2 ' e2 % Ri2 % Lp2 % Mp1(2.84)

V12 'Vs × tON % 0 × (ts ! tON)

ts

'tON Vs

ts

' dVs. (2.83)

v12 ' v1 ! v2 ' e12 % 2Ri1 % 2(L ! M )p1 . (2.85)

Fig. 2.44 Simplified circuit for chopping Fig. 2.45 Average di/dt

Q1 D4 Q6 D3 v12

1 0 1 0 Vs

0 1 1 0 0

1 0 0 1 0

0 1 0 1 !VsTABLE 2.6

TRUTH TABLE FOR SWITCHES IN FIG. 2.42

If the transistors are switched at sufficiently high frequency, the motor inductance keeps the currentwaveform smooth and the motor responds to the average applied voltage V12. Suppose the circuit isswitched between the states in Fig. 2.42(a) and 2.42(b) at high frequency, by chopping Q1 with a duty-cycle d, where d is the ratio ton/ts with which Q1 is switched: see Fig. 2.43. The switching frequency isfs = 1/ts. Then the average voltage V12 is given by

The effect of the chopping on the motor current can be determined by applying this voltage to theequivalent circuit of phases 1 and 2 in series. The simplified circuit is shown in Fig. 2.44.

We have

where p1 = di1/dt and p2 = di2/dt. R is the phase resistance, L is the phase self-inductance, and M is themutual inductance between phases. Since i3 = 0 during the 60E interval of interest, we have i2 = !i1 andtherefore p2 = !p1. Then if we write e12 = e1 ! e2,

Now if the circuit is switched at high frequency between Fig. 2.42(a) and Fig. 2.42(b), we can replace v12by V12, and p12 by the averaged value of p1 in the sense depicted in Fig. 2.45: this is denoted P1 = )i/)t.The most important practical case is where the chopping is arranged to maintain the current constantand equal to a set-point value Isp, as shown in Fig. 2.46.


Fig. 2.47 Torque/speed curve

d 'e12 % 2RIsp

Vs

. (2.86)

Fig. 2.46 Chopping to maintain constant current

Fig. 2.48 Commutation interval between line C and line A

Tb 'Vs ! 2RIsp

kE

. (2.87)

In this case P1 = 0 and from eqns. (2.83) and (2.85), therequired duty-cycle d is given by

The maximum value of d is 1. When d is 1, the choppingtransistor (Q1) is on all through the 60E interval. Sincee12 (the line-line voltage eLL) is proportional to speed,this condition occurs at a particular speed, called thebase speed Tb. Evidently

The base speed is the maximum speed at which the set-point current Isp can be forced into the motor. If Isp isequal to the rated current, Tb is therefore also themaximum speed at which rated torque can be developed.At higher speeds, even though d = 1, the rising EMF

rapidly makes it impossible to force current into themotor, and the torque decreases quickly to zero as thespeed rises. This is shown in Fig. 2.47, which also showsthe hyperbola of constant power Pmax passing through thebase-speed point. Sometimes this is called the "cornerpoint".

Commutation

At the end of each 60Einterval the current must switch from one pair of lines to another pair. In thepreceding 60E interval the current would have been flowing in lines BC, and at the end of that intervalit is “commutated” into lines AB.

Ideally, the current in line B remains constant while the current in line C switches to line A. Inpractice, however, iC does not fall to zero instantaneously, neither does iA rise from zero to Ispinstantaneously. Rather, there is a brief interval when iC is falling and iA is rising, and there is currentin all three lines at once. This interval is called the freewheeling or commutation interval: Fig. 2.48.

The circuit can be analyzed during the commutation interval by writing the voltage equations for twomeshes: one in which the current is rising or "building up", and the other in which it is falling or"freewheeling". We will apply this to the commutation from lines BC to lines AB, where transistor Q5has reached the end of its conduction period and Q1 is switched on. The current in line C freewheelsthrough D2, and when this current is extinguished the current in lines AB continues ("two-phase-on"operation) for the remainder of the 60E interval. The chopping analysis in the previous section appliesto this condition, i.e., after the extinction of iC.


Vbld ' v12 ' v1 & v2 ' vAB. (2.88)

Vfwh ' v23 ' v2 & v3 ' vBC . (2.89)

Vbld '

***

Vs & 2Vq & 2Rqi1 (sQ1'1)&Vq & Rqi1 & Vd (sQ1'0) (2.90)

Vbld ' sQ1 (Vs & 2Vq & 2Rqi1) % (1 & sQ1) (&Vq & Rqi1 & Vd ) . (2.91)

Vfwh ' Vd % Vq % Rq (&i2) . (2.92)

i1 'Vbld & (e1 & e2 )

2R. (2.93)

d 'e1 & e2 % 2RIsp % Vd % Vq % Rq Isp

Vs & Vq % Vd & Rq Isp

. (2.94)

d 'e12 % 2RIsp

Vs

(2.95)

The voltage Vbld controlling the build-up of current in phases 1 and 2 (lines A and B) is the line-linevoltage across these two phases in series, which can be written

Similarly the voltage Vfwh ("fwh" = "freewheel") controlling the decay of current in phases 3 and 2 (linesC and B) is the line-line voltage across these two phases in series, which can be written

If Q1 is the controlling (chopping) transistor, the value of Vbld is determined by the state of Q1, whichis labelled sQ1 :

Note that each transistor is modelled by a resistance Rq and a voltage-drop Vq ; and each diode by avoltage-drop Vd. These equations can be combined:

The value of Vfwh is not affected by the state of Q1:

If Q1 is chopped at a sufficiently high frequency then sQ1 can be set equal to the duty-cycle d, and Vbld canbe interpreted as the average voltage across the terminals AB, or V12. When we make sQ1 equal to d weare using the principle of "state space averaging", to replace a binary value sQ1 by a real number d, assuggested by eqn. (2.83). d is the effective value of sQ1 under the stated conditions, where the choppingis at high frequency and the inductance is sufficient to keep the current waveform reasonably smooth.

While we are using the state-space average value d in algebraic calculations of the steady-statecurrents, the very same formulas for Vbld and Vfwh can be used in digital simulation, where sQ1 is theactual state of the chopping transistor, either 1 or 0 (on or off).

At the end of the commutation interval the line current iC = i3 reaches zero and the line current iA = i1

arrives close to Isp, the set-point value of the current-regulator. iA is then maintained constant bychopping Q1. The current i1 is essentially a DC value. This is possible only if d < 1. Evidently

If we set i1 = Isp then d can be calculated as

A simplified version of this arises if the voltage drops across the transistors and diodes are negligiblecompared with the supply voltage Vs and the line-line back-EMF of the motor, e1 ! e2 = e12. In this case,writing Vd = Vq = Rq = 0,

which is exactly the same as eqn. (2.86). In 12V systems this approximation is often not admissiblebecause the volt-drop across power transistors and diodes can be appreciable in comparison with Vs.


Fig. 2.49 Accumulation of current and current2

Conduction mode diagrams

Fig. 2.50 (case 1) shows the circuit diagram. For purposes of analysis one cycle is divided into sixsegments and the interval from 30E to 90E is the “base segment” (see Fig. 2.7). The base segment startswhen Q5 switches off and Q1 switches on. Fig. 2.50 (case 2) shows the "build" loop (solid line) via linesA and B, and the "freewheel" loop (dotted line) via lines C and B. The state of the chopping transistor(s)is sQChop = 1: i.e., both transistors Q1 and Q6 are on.

Fig. 2.51 shows the effect of soft chopping the line A current while the freewheeling current in line Cis still flowing: i.e., during the commutation interval shown in Fig. 2.48. Case (4) is where Q1 isswitched off and and case (5) is where Q6 is switched off.

Fig. 2.52 (case 3) shows the effect of hard chopping, where both Q1 and Q6 are switched off. Case (6)shows the over-running condition, where the generated EMF exceeds the supply voltage and the diodesare rectifying the generated voltage. Note that this is an unregulated condition.

Reconstruction of waveforms for 3-phase squarewave motors

The PC-BDC program reconstructs 360E of waveforms from the base segment which is only 60E wide.For "120E" controls segment A [30-90E] and segment B [90-150E] are not the same (Fig. 2.49). In thesecases the base segment is calculated twice, with a different switch control in the two segments. Whenboth passes are complete, the integrals of current and current2 in the lines, transistors, and diodes areconstructed as indicated by the shaded areas in Fig. 2.49.

Accumulation of IIIIi dt and IIIIi 2 dt for mean and r.m.s. currents

During any 60E commutation interval three accumulations are made forIi dt and Ii2 dt:

AQn incoming transistor (e.g. Q1)ADn diode that freewheels the current for the incoming transistor (e.g. D4 for Q1)AQo outgoing transistorADo diode that freewheels the current for the outgoing transistor

In addition, the freewheeling current in the "third line" is accumulated as ADt. This current flows onlyduring the commutation interval. The currents in lines A and B are accumulated as ALA and ALBrespectively. Note that ALA = AQn + ADn, and ALB = AQo + ADo. The label “C60 Q1" in Fig. 2.49refers to one of the standard switching strategies in PC-BDC, in which each transistor has control ofthe chopping for 60E, and during the base segment it is Q1.


Fig. 2.51 Soft chopping with Q1 or Q6

Fig. 2.50 (1) circuit diagram and (2) forward conduction mode in 3-phase squarewave drive

Fig. 2.52 Hard chopping and over-running (rectifying)


edet 'Vs

2! v1 % e3 (2.96)

R1 ' (L1 ! L12 ) i1 . (2.98)

v1 ' e1 % Rph i1 % (Lph ! Mph )di1

dt. (2.99)

L1B ' (LF ! MF %3

2Lg0 ) % Lg2 3 cos (22 % B/6) and

dL1B

d2' !Lg2 2 3 sin (22 % B/6) .

(2.102)

L1 ' LF % Lg0 % Lg2 cos 22

L12 ' MF ! 1

2Lg0 % Lg2 cos (22 ! 2B/3) (2.100)

dR1

dt' L1B

di1

dt% i1

dL1B

dt' L1B

di1

dt% i1Te

dL1B

d2(2.101)

v1 ' e1 % Rph i1 %dR1

dt. (2.97)

Fig. 2.53 Back-EMF sensing

2.12 BACK-EMF SENSING WITH SQUAREWAVE DRIVE

Motors with 120E square-wave currents are oftenoperated with “back-EMF sensing” to eliminate theneed for a shaft position sensor. The schematic inFig. 2.53 produces a waveform of the detectionvoltage e det, whose zero-crossings are used forcommutation. This voltage is obtained by switchingthe detection circuit to the idle line every 60E.

The detection EMF e det is multiplexed from thevoltages e AM, e BM, and e CM according to the states ofthe power transistors. The waveforms of thecurrent and EMF repeat every 60E (commutated intodifferent phases), so it is necessary to analyse onlyone 60E period—the base segment described earlier.At the beginning of this period Q5 has just switchedoff, and i C continues to freewheel through D2,forcing e det = e CM = !Vs/2. During the freewheelingperiod (“PERIOD A”) current flows in all three lines.Since e det=!Vs/2, due to the clamping action of D2,the detection voltage is useless during this period.

When D2 switches off, PERIOD A ends and theremainder of the conduction segment, “PERIOD B”,starts. During PERIOD B, only two lines are conducting, A and B. There is only one mesh current, thatis, the one flowing through phase 1 and negatively through phase 2. The detection voltage is given by

where

The flux-linkage R1 includes a self- and a mutual term, and since i2 = !i1 this is

In a nonsalient pole motor, L1 = Lph and L12 = Mph, and both these inductances are constant so that

In a salient-pole motor, if it is a sinewound motor then

with Lg2 < 0 in most cases. Both i1 and the inductances are functions of time, so that R1 in eqn. (2.97)must be differentiated by the chain rule. If we write L1B = L1 ! L12 for the apparent inductance of phase1 during PERIOD B, then

where


21 i.e., there are no even harmonics in the EMF waveform unless they are deliberately introduced by means of geometric or otherimbalances or asymmetry in the magnetic circuit. Generally such imbalances are acceptable only in very small motors.

LLL ' L1 % L2 ! 2L12

' 2(LF ! MF ) % 3Lg0 ! 3Lg2 sin (22 ! B/6) (2.103)

dLLL

d2' !6Lg2 cos (22 ! B/6) (2.104)

Vbld ' e1 ! e2 % 2Rph i1 % LLL

di1

dt% Tei1

dLLL

d2(2.105)

Fig. 2.54 Bifilar motor windings

By substituting all these equations back into eqn. (2.96), we arrive at an equation for the detectionvoltage e det during PERIOD B. However, we still need the differential voltage equation for the wholecircuit so that the current waveform i1 can be calculated. This is simple enough if we combine all theinductances in the single conducting loop AB into one line-line inductance:

The derivative of LLL with respect to 2 is

and the required voltage equation is

which can be rearranged for integration by Euler's method. The "build" voltage V bld is the voltageapplied to loop A-B or 1-2, to build up the current in phase 1. If both Q1 and Q6 are on, it is equal to thesupply voltage Vs minus the voltage-drops in the power transistors. Eqn. (2.96) is valid only forchopping strategies in which Q1 remains on during PERIOD B, which requires that any chopping mustbe performed by Q6 during the base segment. It is straightforward to program the solution for i1 inPERIOD B using direct phase variables, because there is only one conducting mesh or loop. However,PERIOD A with all three phases conducting is slightly more complicated.

The zero-crossing angles are subject to variation caused by induced speed voltages associated with anysecond-harmonic variation in the self- and mutual inductances of the phases. Other sources ofvariation in the zero-crossings include even-harmonic distortion of the EMF, causing the waveform overone half-pole to differ from the waveform over the second half-pole. This would happen if the magnetswere not centred, or if their magnetization varied over their width, or if the magnetization of magnetswas not uniform under all the poles. These effects can be checked by running the motor open-circuitand recording the line-neutral (phase) voltage waveform.

2.13 UNIPOLAR DRIVE CIRCUITS, AND MOTORS WITH ONE OR TWO PHASES

Fig. 2.55 shows the basic waveforms of EMF, current and torque inmotors with one or two phases. In any phase winding the EMF e1 is analternating quantity in which the positive and negative half-cycles aresimilar in shape.21 This being the case, equal impulses of torque in thepositive and negative half-cycles of EMF require current pulses ofopposite polarity and similar waveshape, as shown by the symmetricalcurrent squarewave i1. The torque produced by the interaction of e1 andi1 is shown as T1. The EMF and the current both require a finite time tochange from positive to negative, and this results in a dip in the torquewaveform every half-cycle, i.e., at twice the fundamental frequency.

The most common means of driving a symmetrical alternating currentwaveform such as i1 is a bridge circuit such as the one labelled “H-bridge” in Fig. 2.55. This circuit has four transistors, two of which arereferenced to the negative supply rail and the other two to the positive rail. The upper transistors arecalled “high-side” transistors and the lower ones “low-side” transistors. The gating and protectioncircuits for the high-side devices are more complicated than for the low-side ones, and for this reasonthe bifilar circuit is sometimes used. In this case the phase winding is split into two equal parallelpaths which are connected within the motor with opposite polarity, as shown in Fig. 2.54. The half-


Fig. 2.55 Single-phase and two-phase motor waveforms

wave or “unipolar” current i1a flows in one path for half a cycle; then it is switched off and the currenti1b flows in the other path. In Fig. 2.54 the paths are shown as separate phases 1 and 2, but this is amatter of convention; the bifilar motor in Fig. 2.54 is often termed “1-phase bifilar”. Complementaryswitching is used with the transistors controlling the two paths or phases: normally each conducts forhalf a cycle in any full cycle. When one switches off, the current ideally transfers immediately to theother, but in practice the mutual coupling between the two paths is imperfect and the resulting leakageinductance retains a fraction of the inductive energy, which must be dissipated. There is also anovervoltage on the outgoing transistor, which may cause avalanching if a snubber is not used.

The torque dips in the T1 waveform in Fig. 2.55 are not a problem in applications such as low-poweredfans or blowers, provided that the motor can start satisfactorily. Sometimes a tapered airgap is usedto “park” the rotor at a position away from the location of zero torque when the motor comes to rest.Alternatively the torque dips can be eliminated by adding a second phase, as shown in Fig. 2.55, withwindings displaced 90Eelec from those of the first phase, and currents 90E out of phase with those of thefirst phase. The current, EMF and torque contribution of a second phase are shown in Fig. 2.55. Thedrive circuit requires a duplication of the first H-bridge (leading to a total of 8 transistors for 2 phases),or a duplication of the bifilar drive circuit (leading to a total of 4 transistors). The bifilar drive withtwo orthogonal sets of windings is often termed “2-phase bifilar” but it can equally well be regardedas a 4-phase motor in which the complementary phases (1&3 and 2&4) are tightly coupled.

The total torque (T1 + T2) in the case shown in Fig. 2.55 will have no zeroes but may have considerableripple because of the particular waveforms of the individual phase torques. This ripple can be reducedby reducing the conduction angle from 180E to 90E in each half-cycle in each phase, which is easilyaccomplished with the H-bridge full-wave circuit. In the bifilar circuit, however, with less than 180Econduction the stored inductive energy in the winding is not transferred to the complementarywinding. Unipolar circuits usually have no means of returning it to the supply, being designed for lowcomponent cost: therefore this energy must be dissipated, with inevitable loss of efficiency.

Note that if the number of phases is increased to 3, the 3-wire connection can be used (with internal wyeor delta connection of the windings), and in this case the 6-transistor bridge can be used with norestriction on the current control, in the sense that positive and negative current and voltage can beapplied to the motor terminals by appropriate switching controls. The three-phase motor has better


22Unipolar or “half-wave” circuits are also sometimes called “single-ended”.

Fig. 2.56 Single-ended circuit with noseparate freewheeling path

Fig. 2.57 Single-ended circuit with Zenerdiode and damping resistor

Fig. 2.58 Waveforms corresponding to Fig. 2.56

ON : v ' Vs ! Vq ! (Rs % Rq ) i (2.106)

OFF : v ' Vs ! Vz ! (Rs % Rd ) i (2.107)

overlap for covering the natural torque dips, as we can see in Fig. 2.7 where each line need conduct foronly 120E in each half-cycle, leaving 60E for the EMF waveform to change polarity. Because of thecontrol flexibility of the three-phase drive, and the fact that it requires only 3 leads and 6 transistors,it is overwhelmingly the most popular choice except where extremely low component cost is required.

A more detailed look at unipolar (half-wave) drive circuits

Fig. 2.56 shows the simplest form of unipolar circuit,22 in which the winding W is controlled by a singletransistor Q. Diode D is not a freewheel diode but is the body diode if Q is a field-effect transistor.There is no separate freewheel diode shown in Fig. 2.56. Positive current i flows from the DC supplyVs when Q is switched on. When Q switches off, a reverse voltage v < 0 must be developed across thewinding in order to reduce the current to zero. The only source of the necessary reverse voltage is theavalanche voltage of the transistor or diode, Vz.

It is possible to protect the transistor and diode by connecting a separate voltage-limiting circuit inparallel with them, with a Zener diode Vz and a damping resistor Rd as shown in Fig. 2.57. To ensurethat v < 0 it is obviously necessary to have Vz + Rd × i > Vs !!!! Rs × i where e is the generated EMF andRs is the supply resistance. The peak forward voltage appearing across the transistor when it turnsoff is Vz + Rd × i.

As we have already seen in Fig. 2.55, the EMF,current and torque waveforms for a single-endedcircuit such as the one in Fig. 2.56 are of the formshown in Fig. 2.58, which shows the additionalpossibility of spurious conduction during thenegative half-cycles of EMF, if the sum of Vs and thepeak EMF epk exceeds the avalanche voltage Vz. Thisuncontrolled current produces a pulse of negativetorque.

The circuit equations for the circuit in Fig. 2.57 are:

where Vq is the forward voltage drop in the transistor during conduction and Rq is the transistorresistance. These equations can be used with the circuit of Fig. 2.56 if Rd = 0.

Fig. 2.59 shows an alternative single-ended circuit in which the suppression circuit is in parallel withthe winding so that the supply voltage does not appear in the freewheeling path. The suppressionvoltage is developed across the damping resistor Rd together with a small additional forward voltagedrop across the freewheel diode. A disadvantage is that the suppression voltage Rd × i decays with i,so that the defluxing of the motor winding is slow.


23This is the one that is sometimes termed “single-phase bifilar”!

Fig. 2.59 Unipolar circuit withplain freewheeling diode

Fig. 2.60 Unipolar circuit withZener suppression diode

Fig. 2.61 Current in bifilar circuit withnegative spike at commutation

vt1 ' e1 % Rph i1 % Lph p1 % Mph p2

vt2 ' e2 % Rph i2 % Mph p1 % Lph p2(2.109)

OFF : v ' !Vd ! Vz ! Rd i (2.108)

p1 '(1 ! k )Vs ! Vz

L (1 ! k 2 ); p2 '

(1 ! k )Vs % kVz

L (1 ! k 2 ). (2.110)

Fig. 2.60 shows a modification with a Zener diode to sustain the suppression voltage at a higher level.As in the circuits of Figs. 2.56 and 2.57, spurious conduction is possible during the negative half-cyclesof EMF, unless Vz is high enough to exceed the peak EMF epk.

When the transistor is on, eqn. (2.106) also describes the operation of the circuits in Figs. 2.59 and 2.60.When it is off, the operation is described by the equation

The circuit in Fig. 2.60 can be used with any number of phases, with or without bifilar windings.However, there is no point in contemplating it for use with 2-phase motors which have phases displacedby 90E, or 4-phase motors with phases uniformly displaced by 45E, because it cannot deliver thealternating (full-wave) currents required in these cases.

It is important to account for mutual coupling between phases during the freewheeling period aftereach phase transistor turns off. The circuit equations for the bifilar motor can be taken in pairs, sincethere is no coupling between the first pair (phases 1&3) and the second pair (phases 2&4) in a 4-phasemotor. With no loss of generality we can consider the 2-phase motor, labelling the phases 1&2.23 Thus

where p = di/dt and vt1 and vt2 are the terminal voltages of the phase windings, which depend on thestates of the transistors and diodes. These equations are rearranged in the canonical Euler form forstepwise integration of p1 and p2 to give new currents i1 and i2, and in a simulation program the statesof the transistors are updated according to the switching algorithm and the values of the currents.

For the Zener diode circuit of Fig. 2.57, which also represents the circuit of Fig. 2.56 when the transistoris avalanching, eqns. (2.109) can be solved algebraically for the turn-off period of the transistor. If EMF

and resistance are neglected and Mph/Lph = k, the coupling coefficient, the solution for p1 and p2 is

If k is nearly !1, these equations show that if Vz # 2Vs, p1 = p2 $ 0, so that commutation requires Vz > 2Vs.For example if Vz = 3Vs, p1 = p2 = !Vs/L (1 ! k2), and since both of these are negative it is evident thatwhen one transistor switches off, a negative current is induced in the complementary phase. Thiscurrent appears as a spike which disappears and is followed by the positive current pulse after a fewdegrees of rotation, Fig. 2.61. If chopping is employed, the negative spike current recurs every time atransistor is switched off, and the inductive energy in the leakage inductance is dissipated. To avoidchopping, phase advance can be used to control the torque, (also at the expense of efficiency). Highvoltage stress on the transistor (>Vz) is inevitable with these unipolar circuits during turn-off.


24The terms “coil span” and “coil pitch” are synonymous. They are usually measured in slot-pitches. The term “throw” is alsooften used with the same meaning. However, in some factories “throw” is one less than “span” : a coil that is said to be woundin slots “1 and 6” has a “throw” of 6 and a span of 5. In SPEED software, however, “throw” means the same as “span”.

Fig. 2.62 Definition of some terms used in motor windings

2.14 WINDINGS

Fig. 2.62 shows some of the terms used in connection with motor windings. Although the subject ofwindings is vast, brushless permanent-magnet motors basically use two types of winding: theconcentrated winding for squarewave drive and the distributed winding for sinewave drive. Becausethe number of slots/pole/phase is usually quite small, the practical differences between these windingsare often not apparent.

The winding shown in Fig. 2.62 has three phases, of which only one is shown. There are four poles(corresponding to the number of magnet poles), and for each pole there is a group of 2 coils which aresaid to be “concentric” because of the way in which the outer coil embraces the inner one. The outercoil in each pole-group in Fig. 2.62 has a span of 5 slots, and the inner one a span of 3 slots.24


25Such windings are suitable for delta connection since there will be no zero sequence EMF to drive current around the delta.

Fig. 2.63 Phasor diagram for salient-pole brushless PM machine

Taking two coil groups together, and recognizing that adjacent coil-groups alternate in polarity , in Fig.2.62 there are four coil-sides in adjacent slots all carrying current in the same direction, and these coil-sides constitute what is known as a phase belt. In this winding the phase belt spans 4 slot-pitches, andsince there are 24 slots covering 4 poles, this is 4/6 or 2/3 of a pole-pitch or 120 electrical degrees. In aperfectly concentrated winding the phase belt would have a span of zero: in other words, all thecoilsides in it would be in the same slot. Thus the winding in Fig. 2.62 is a distributed winding.

In order to make a perfectly concentrated winding, the coils in Fig. 2.62 would all have to have a spanof 6 slot-pitches (“full-pitch” coils). In that case only 4 slots would be occupied by any phase, and halfthe slots in the machine would be empty. As it is, every slot holds two coil-sides, one from each of twodifferent phases (this is called a “double-layer” winding).

The distribution of coil-sides is critical in the determination of the EMF waveform, and also has asignificant influence on the winding inductance and the mutual inductance between phases. A full-pitch concentrated winding produces an EMF time-waveform on open-circuit that is a replica of thespatial distribution of magnet flux-density around the airgap. It is possible to construct the EMF

waveform of any symmetrical winding by adding together the EMF waveforms of an equivalentdistribution of full-pitch coils. The phase-shifts between the coils can be cleverly combined to eliminateselected harmonics from the resulting phase EMF waveform. For example, windings in which all thecoils have 2/3 pitch (i.e. 2/3 of a pole-pitch, or a span of 4 in Fig. 2.62) have no 3rd harmonic in the EMF

waveform.25 To eliminate the 5th harmonic, the coil span should be 5/6 of the pole-pitch (5 in Fig. 2.62).

Formulas for the harmonic winding factors of standard windings are given in Chapter 3.

2.15 CALCULATION OF TORQUE USING THE FINITE-ELEMENT PROCEDURE

(a) Simple, fast procedure for sinewound machines with low saturation level

Fig. 2.24 shows the general cross-section of an interior-magnet brushless AC motor, and Fig. 2.25 showsthe reference axes of phases a, b, and c together with the rotor reference axes d and q. These axes areconsistent with the equations for variation of phase inductances with rotor position in §2.6. The phasordiagram is reproduced in Fig. 2.63 for a normal motoring condition.


T ' m p [Q1d Iq ! Q1q Id ] (2.113)

Q1d 'kw1 Tph

2

B1d DLstk

p; Q1q '

kw1 Tph

2

B1q DLstk

p(2.114)

V ' E % jXd Id % jXq I q % Rph I (2.112)

Id ' I cos $ ' ! I sin ( and Iq ' I sin $ ' I cos ( . (2.111)

The current phasor is defined by its r.m.s. value I and its phase angle ( relative to the open-circuit EMF

E. The phasor values are I = Id + jIq and E = 0 + jEq = jTQ1Md , where Q1Md is the r.m.s. fundamentalflux-linkage per phase on open-circuit, produced by the magnet: see eqn. (2.57).

The angle ( in the phasor diagram is the angle between the current phasor I and the q-axis, and ismeasured positive if I leads E. In Fig. 2.63, ( < 0. The angle $ is the angle between the current phasorand the d-axis, such that $ = B/2 + (. Thus

In the physical cross-section of the machine, $ is the angle between the axis of the stator MMF

distribution and the d-axis. When $ = 0, the stator MMF is aligned with the d-axis in the magnetizingdirection, and Id = I with Iq = 0. When $ = 180E, Id = !I with Iq = 0, and the stator MMF is aligned withthe negative d-axis in the demagnetizing direction. When $ = 90E and ( = 0, the current is "in the q-axis": Iq = I with Id = 0. In surface-magnet sinewave motors at low speed, the current is normallyoriented in the q-axis with ( = 0.

The phasor voltage V = Vd + jVq is the airgap voltage produced by the r.m.s. fundamental airgap flux-linkage QQQQ1 under load: thus V = jTQQQQ1 = Q1d + jQ1q. The phasor diagram is completed by the equation

where Id is written as the phasor Id = Id + j0 and Iq is written as the phasor Iq = 0 + jIq. Ra is the phaseresistance. In this equation, both Xd and Xq should incorporate the slot, end-turn, and differentialleakage components (and any skew-leakage component) in addition to the main airgap componentassociated with the fundamental MMF distribution of the stator winding.

In a sinewound motor we have seen in §2.8 that the torque is computed directly from the fundamentalflux-linkage per phase QQQQ1 and the phase current I using the equation

where p is the number of pole-pairs and m is the number of phases, and Q1d and Q1q are the componentsof QQQQ1 resolved along the d- and q-axes. This is the average torque produced by the fundamentalharmonic components of the airgap flux and the stator MMF distribution. It does not include coggingtorque or any harmonic torques. However, provided that the correct fundamental components Id, Iq,Q1d and Q1q are computed, it is the correct running torque for a sinewave motor, and in a well-designedmotor with a sinusoidal back EMF, it is virtually constant.

Eqn. (2.113) does not require the d-axis flux-linkage to be resolved into the separate componentscontributed by the magnet and the d-axis armature current; neither does it even require a knowledgeof Xd or Xq: these are necessary only when the current has to be computed from the voltage equation(2.64) or (2.110). If the current is known in magnitude and direction, as is normally the case with acurrent-regulated drive, then Xd and Xq are not required.

The fundamental flux-linkages Q1d and Q1q are obtained from the finite-element procedure as follows.First, the d-axis is aligned with the axis of phase a. Then the field solution is obtained for a range ofvalues of I and (, where I = Id + jIq = I ej(B/2 + (), from ( = !90E to +90E (i.e., $ ranges from 0 to 180E). Foreach solution the airgap flux-density waveform is Fourier-analyzed to give B1d and B1q, where B1d isthe peak value of the d-axis component and B1q is the peak value of the q-axis component. Then Qd1 andQq1 are obtained using the same form as eqn. (2.57), i.e.

The leakage components IdLF and IqLF should be added to Q1d and Q1q respectively. If the leakagereactance is saturable, the flux-MMF diagram technique should be used instead (see below). Thesevalues are substituted in eqn. (2.113) to produce a set of torque curves as illustrated in Fig. 2.64.


Ik ' C [a,k] ia % C [b,k] ib % C [c,k] ic . (2.115)

ia ' Isp cos $ib ' Isp cos ($ ! 2B/3)ic ' Isp cos ($ % 2B/3)

(2.118)

Fig. 2.65 Direction of currents and stator MMF around a S magnet for three values of (.

ia ' Id cos 2 ! Iq sin 2ib ' Id cos (2 ! 2B/3) ! Iq sin (2 ! 2B/3)ic ' Id cos (2 % 2B/3) ! Iq sin (2 % 2B/3)

(2.116)

ia ' Id

ib ' Id cos (!2B/3) ! Iq sin (!2B/3) ' (!Id % 3Iq)/2

ic ' Id cos (2B/3) ! Iq sin (2B/3) ' (!Id ! 3Iq)/2 .

(2.117)

Fig. 2.64 Torque vs. torque angle (, for various currents

The values of currents in the individual slotsare obtained from the conductor locationvector (CLV), that is, an array C which has mcolumns, one for each phase, and n rows,where n is the number of slots. The value ofthe (j,k)th element is the number ofconductors belonging to the jth phase in the kth

slot. C[j,k] is signed according to the directionof the conductors. Then for the kth slot thetotal current is

Instantaneous phase currents ia, ib and ic areobtained from the phasor values by Park’stransformation:

where Id and Iq are peak values (i.e., the values taken from the phasor diagram and multiplied by /2).Since the d-axis is aligned with phase a, 2 = 0 and eqns. (2.116) simplify to

Substituting from eqn. (2.111), with $ = ( + 90E the angle between the current and the d-axis,

where Isp = /2*I* is the setpoint current of the current-regulator. In PC-BDC the sign conventions aresuch that the main magnet pole analyzed in the finite-element model is a south pole. Fig. 2.65 showsthe directions of the current for three different values of (: ( = 0 is the normal "q-axis" control, ( = !90Eis "magnetizing" with $ = 0, and +90E is demagnetizing with $ = 180E.

It is not always necessary to compute the entire set of T((,I) curves. If the current is regulated to besinusoidal, both I and ( are controlled and only one finite-element computation is needed at that point.


Fig. 2.66 i-psi loop

Fig. 2.67 i-psi loop

Te 'm

2B× W . (2.119)

(b) Flux-MMF diagram technique

Method (a) is fast because it uses a fixed mesh andsimple post-processing of the finite-element fieldsolution, but it fails to give adequate results if themachine is not sinewound, or if the airgap flux-distribution is not sinusoidal, or if there is significantsaturation.

Moreover. under saturated conditions the values of Eand Xd become indeterminate, because thesuperposition implied by the equation Qd = Q1Md +LdId, (or the equivalent voltage equation Vq = Eq + XdId)no longer applies, so that it is impossible to resolve Qd(or Vq) uniquely into separate components, oneattributed to the magnet and the other to the current.Probably the most realistic physical interpretation is that both Eq and Xd are reduced by saturation.In the q-axis, the question of uniqueness does not appear to arise, because the linear theory ofsinewound machines tacitly assumes that there is no Ed produced by "magnet flux" in the q-axis. Butunder saturated conditions a component of Ed can appear, while Xq is certainly reduced by saturation,often to a severe degree.

Under these circumstances the most powerful method for computing the torque is to calculate asequence of finite-element field solutions at intervals of rotation throughout an electrical cycle. Thecurrent (in all phases) set to the correct value for each rotor position, and the i-R loop is plotted, thatis, the loop formed by a point whose coordinates are the phase current i and the phase flux-linkage R.

With a sinewound machine having ideal sinusoidal current waveforms, the i-R loop or "energy-conversion loop" is elliptical, its major axis being oriented at an angle that depends on the phase anglebetween the current and the d-axis. The average electromagnetic torque is proportional to the area Wof the i-R loop: thus

The phase flux-linkage R is extracted from the finite-element solution. The flux through a single coilis N = Lstk (Ar ! Ag) [Wb], where Ar is the vector potential at the "return" coilside and Ag is the vectorpotential at the "go" coilside, and Lstk is the stack length. The flux-linkage is R = TcN [V-s], where Tcis the number of turns in the coil.

If required, the d! and q-axis components of flux-linkage can be approximated using thetransformation, Rd = 2Ra/3 ! (Rb + Rc)/3 and Rq = (Rb ! Rc)//3. However, this transformation is notof interest unless the winding is sinewound.

The flux-MMF diagram technique works for anymachine with any drive. It is not restricted tosinewound machines or sinewave drives, and worksequally well with machines having non-sinusoidal EMF

waveforms and squarewave drive, regardless of thelevel of saturation.

The main limitation of the method is that the phasecurrents must be known at every step of rotation,because the finite-element computation is "current-driven".

Fig. 2.67 shows an example of the i-R loop computed fora "squarewave" brushless DC motor, with the currentregulated to follow a 120E squarewave by chopping.


26Eqn. (2.120) therefore applies to surface-magnet motors but is incomplete for embedded-magnet motors.

Tavg '2B

0Te (2)d2 (2.121)

Te Tm ' e1 i1 % e2 i2 % e3 i3 % ... % emim (2.120)

e1 ' epk cos Tt i1 ' ipk cos (Tt ! ()e2 ' epk cos (Tt ! 2B/3) i1 ' ipk cos (Tt ! 2B/3 ! ()e3 ' epk cos (Tt % 2B/3) i1 ' ipk cos (Tt % 2B/3 ! ()

(2.122)

Te ' 3EI

Tm

cos ( (2.123)

e1 i1 ' epk ipk cos Tt cos (Tt ! ()

'epk ipk

2[cos ( % cos (2Tt ! () ]

(2.124)

2.16 TORQUE AND TORQUE RIPPLE

Two main sources of torque ripple can be identified in brushless permanent-magnet motors:

(1) electromagnetic torque ripple; and

(2) cogging torque.

Electromagnetic torque in brushless PM machines is the result of the interaction between the phasecurrents i1, i2,... and the EMF’S e1, e2,... generated by the rotation of the magnets. Ideally the energyconversion is described by the equation

where Tm is the speed in rad/sec and m is the number of phases. It is important to recognize Te as theinstantaneous torque and not the average torque. The average torque (averaged over one or morerevolutions) is the main output parameter,

and in most cases Tavg is the parameter that is quoted, with never any mention of the instantaneoustorque. Quite possibly the credit for this must go to the designers of classical DC and AC machines who,for the first century of the history of electric machines, delivered such smooth torque that no-one gavea second thought to the existence of torque ripple, and the whole subject of instantaneous torque andthe theory of electromechanical energy conversion was academic.

Now in the age of electronic drives, torque ripple is an important topic, probably for three reasons:

(a) power electronics has opened up many applications which are extremely sensitive to torqueripple, e.g., automotive power steering, machine tool feed drives, and computer disk drives;

(b) power electronics has made it possible to use motors which do not inherently deliver smoothtorque, except in their most ideal theoretical forms; and

(c) power electronics and associated digital controls can exacerbate the torque ripple problem.

It is worth reviewing the principles of smooth torque production starting from eqn. (2.120), even thoughthis equation includes only the “alignment” torque of the permanent magnets and does not include anyreluctance torque.26 In AC machines the most classical instance of “constant torque” is the three-phase machine wherein the EMF’s and currents are balanced polyphase quantities:

Substituting in eqn. (2.120), and recognizing the r.m.s. values E = epk//2 and I = ipk//2, we get

which is constant. The contributions of the individual phases, however, are not constant: for example,


Te 'm

Tm

[Eq Iq % (Xd ! Xq)Id Iq] (2.127)

Trel 'm

Tm

× 2B f [Ld ! Lq ] id iq 'mp

2[(Ld ! Lq) id iq] (2.128)

id '2

3[ia cos 2 % ib cos (2 ! 2B/3) % ic cos (2 % 2B/3)] ;

iq ' ! 2

3[ia sin 2 % ib sin (2 ! 2B/3) % ic sin (2 % 2B/3)] .

(2.129)

e1 ' epk cos Tt i1 ' ipk cos (Tt ! ()e2 ' epk cos (Tt ! B/2) i1 ' ipk cos (Tt ! B/2 ! () (2.125)

Te ' 2EI

Tm

cos ( . (2.126)

which has an average value (EI/2) cos ( with a double-frequency oscillatory component whoseamplitude is greater than the average value unless ( = 0 (i.e., unless e1 and i1 are in phase). Thisenormous torque ripple is the reason why pure single-phase AC motors are used in only a limitednumber of applications. It also explains the use of auxiliary capacitors with split-phase inductionmotors operating from a single-phase supply. With two phases the phase displacement between phasesis ideally B/2 or 90E instead of the 2B/3 or 120E used in three-phase machines, and if

then the sum e1i1 + e2i2 yields

In balanced polyphase systems it is always the case that as the torque contribution of one phase varies,the variation is exactly compensated by variations in the other phases.

Eqn. (2.120) also suggests the other main principle of attaining constant instantaneous torque, whichis to maintain one or two of the product terms constant for a fixed angle of rotation, and then“commutate” to another product term or pair of terms. For example, Fig. 2.7 shows how this isaccomplished in the 3-phase brushless motor. The angle of rotation over which one of the productterms (such as T1) can be kept constant is limited by the magnet arc and the winding distribution, andalthough the current can be held constant for very nearly 180E of rotation, it is not possible to maintaina flat-topped phase EMF waveform wider than about 175E or so. For this reason it is impossible toachieve constant torque in a two-phase brushless motor, and it is better to settle for three phases and120E flat tops, which can be adjusted to minimize any ripple arising at the commutation points.

Even with a motor which has ideal sinusoidal or flat-topped trapezoidal EMF waveforms, the currentwaveform may depart from the ideal sinusoid or 120E squarewave because of chopping (PWM) andcommutation in the drive. Moreover, at high speeds the ideal sine or 120E squarewave current cannotbe achieved because of the combination of series inductance and the growth in the EMF relative to theavailable supply voltage.

In sinewound motors with sinusoidal supply currents the torque equation has already been expressedin various forms such as eqns. (2.60), (2.62), (2.69), (2.78), and (2.113): all of these equations refer to theaverage torque. Another variant of these equations is

The reluctance torque is identifiable as the second term in this equation, and this is the average valuesince Eq, Iq and Id are all r.m.s. quantities. However, the instantaneous reluctance torque Trel can bedetermined if the instantaneous currents are used instead of the r.m.s. values:

where Tm = 2B f /p and id and iq are the instantaneous d- and q-axis currents, obtained from theinstantaneous phase currents by Park's transformation:

and 2 is the angle between the d-axis and the axis of phase a. The inverse Park transformation is givenby eqn. (2.116). This is an interesting illustration of the fact that Park’s transformation is valid undertransient conditions, or with nonsinusoidal currents, provided only that the windings are sinewound.


Fig. 2.68 Representation of magnet by equivalent air-cored coil

B ' Br % µrec µ0 H (2.131)

Wf '1

2 m BH dV (2.130)

Fig. 2.69 Result of a small perturbation in magnetoperating point from X to Y.

2.17 COGGING TORQUE CALCULATIONS USING FINITE-ELEMENT ANALYSIS

A magnet (or any element thereof) can be represented by an equivalent distribution of linear surfacecurrent density K = M × n, together with a volume distribution of current density J = curl M, whereM is the magnetization and n is the normal to the magnet surface. For isotropic magnets J = 0 and K= Hca, the apparent coercivity Br/µrecµ0. In the following, Hca is written Hc. The total coercive MMF ofthe magnet is Fc = Hc Lm, where Lm is the magnet length in the direction of magnetization.

At first sight it appears that the stored magnetic field energy can be obtained by calculating the integral

in all regions of the motor, and adding the resulting integrals together. In the air regions, B = µ0H. Inthe iron regions, B and H are related by the B/H characteristic of the iron. But within the volumeoccupied by the magnet,

where Br is the remanence and µrec is the relative recoilpermeability. It is not obvious that we can substitutethis equation in eqn. (2.130) and get the correct value forthe energy in the volume occupied by the magnet. Anenergy analysis is required, and this must use theprinciple of virtual work, because a change in storedenergy can be brought about only by an exchange ofmechanical work when the motor is on open-circuit.

Consider a small displacement )2 such that theoperating point of the magnet moves from X to Y, as aresult of the change in the permeance of the magneticcircuit as the magnet moves past the slot openings. Themagnet flux changes by )Mm, and if the change takesplace in a time )t, an induced voltage e = )Mm/)tappears at the terminals of the equivalent coil in Fig.2.68, if we assume that the coil has one turn. The powerat these terminals is just p = e i and the energyexchange during the time interval )t is e i)t = i)Mm.But i = Fc, the coercive MMF of the magnet. Thereforethe energy exchange with the fictitious excitation coilis Fc)Mm. Because the coil and current source arefictitious, there is no external electrical exchange ofenergy via i and e. Nevertheless, the fictitious coil and current source must be capable of “sourcing”or “sinking” energy. The flux change )Mm causes a change )Wmag in the stored field energy in the spaceoccupied by the magnet, and a further change )Wgap in the remainder of the magnetic circuit, (i.e., theairgap and the iron). Let Wmag = ½FmagMmag and Wgap = ½FgapMgap, where Fmag and Fgap are therespective MMF drops across these sections of the magnetic circuit, and Mmag = Mgap = Mm, the total flux.


27 See Deodhar RP, Staton DA, Miller TJE and Jahns TM : Prediction of cogging torque using the flux-mmf diagram technique,IEEE Transactions on Industry Applications, Vol. IAS-32, No. 3, May/June 1996, pp. 569!576.

)Wmag ' mmmVm

1

2 m HdB dVm (2.137)

Fig. 2.70

Tcog '1

2Fc

dMm

d2(2.136)

)Wmag '1

2( Fmag @)Mmag % )Fmag @Mmag )

)Wgap '1

2( Fgap @)Mgap % )Fgap @Mgap )

(2.132)

)Wmag % )Wgap '1

2(Fmag % Fgap))Mm %

1

2( )Fmag % )Fgap )Mm (2.133)

)Wmag % )Wgap '1

2Fc )Mm (2.134)

)Wm ' )Wmag % )Wgap ! Fc)Mm ' ! 1

2Fc)Mm (2.135)

Then

The sum of these is

But )Fmag + )Fgap = )Fc = 0, since Fc is by definition constant and Fmag + Fgap = Fc. Therefore

We can now complete the energy audit using the law of energy conservation. If the movement XY isfrom a position of higher permeance to one of lower permeance, then mechanical work is suppliedto the motor via the shaft. The stored magnetic energies increase while the flux Mm decreases.Therefore

In order to move the rotor and cause the flux change )Mm, mechanical work )Wm must have beenexerted via a torque at the shaft. This torque is the cogging torque Tcog, and if we regard torque exertedby the motor as positive, then Tcog )2 = ! )Wm and in the limit as )2 tends towards zero,

The shaded triangle )Wm in Fig. 2.69 is equal to½Fc)Mm, and therefore represents the work of coggingtorque during the rotation from X to Y. Fig. 2.70 showsthat the area )Wm is also equal to the triangle OXY.27

The change of energy )Wmag in the volume Vmoccupied by the magnet is suggested by classical theoryas

This is represented by the shaded area in Fig. 2.71. Thedifference between )Wm and )Wmag is the energyexchanged with the air + iron regions of the motor,denoted as )Wgap. It is important to recognize that )Wm= ½Fc)Mm includes all the mechanical work, eventhough the expression ½Fc)Mm contains only magnetparameters. Indeed B and H do not need to be known inthe airgap and iron regions in order to calculate )Wm(and from it the cogging torque). All the necessaryinformation about the mechanical energy exchange is included in the expression ½Fc)Mm. Thisexpression also applies to magnetically linear variable-reluctance devices that are supplied from aconstant-current source. The factor ½ reflects a partition of energy between mechanical energyconversion and field storage. Evidently permanent magnets have additional storage capability beyondthe integral in eqn. (2.137). It is associated with the internal current source in Fig. 2.68 and it is, ofcourse, what distinguishes them as “permanent”.


Fig. 2.71 Fig. 2.72

The equivalent-coil model helps remove doubt about the absolute value of the stored energy in themagnet. Consider a change in the external circuit permeance that takes the operating point up to thepoint R, where B = Br, the remanence, and H = 0 in the magnet. The mechanical energy exchange alongXR is shown as Wmag in Fig. 2.72. For the purposes of calculating cogging torque, with the motorwindings open-circuited, the magnet energy at R can be taken to be zero, because any displacementfrom R requires mechanical work to be provided. Of course, in a motor with a finite airgap, the magnetnever operates at R, but approaches closest to it when no external torque is applied.

It follows from Fig. 2.70 that the total energy exchange along XR is OXR = Wm, and therefore the energyexchange with the air and iron regions is Wm ! Wmag , which is shown as Wgap in Fig. 2.72. Fig. 2.72 alsoshows the separate components of the coercive MMF Fc : that is, Fgap across the air + iron part of themagnetic circuit and Fmag across the magnet itself. When the magnet is working near the remanencepoint R, the total stored energy is small and the magnet energy is smaller than the airgap component.If the rotor is displaced so that the magnet operating point moves close to the coercive point C, the totalstored energy is large: an external torque must have been applied to the shaft to make this change, andthe energy stored in the magnet is larger than that stored in the gap. The gap energy is low becausethe flux-density is reduced to a low level, as a result of the reduction in the circuit permeance. Thepotential energy given to the magnet will be recovered if the shaft is allowed to turn back to its originalposition. During continuous rotation the cogging torque is generated as the operating point cyclesbetween points such as X and Y. Point X is a point of maximum permeance and point Y is a point ofminimum permeance. When the rotation is from X to Y, the stored energy is increasing and the coggingtorque is a retarding torque in the direction of rotation. At Y the rotor continues in the same physicaldirection but the operating point moves towards X, which corresponds to a new maximum permeanceposition. The angle of rotation between two X’s depends on the number of slots and poles; in anintegral- slot motor, it is equal to the slot-pitch.

Fig. 2.70 also shows that the area )Wm used to evaluate the cogging torque can be taken as anincrement in the total stored field energy, or as an equal and opposite increment !)Wm in thecoenergy. The total coenergy is represented by triangle OCX and its components WNgap and WNmag areshown in Fig. 2.72. Since there is no external electrical energy exchange, it does not matter which oneis used, and the cogging torque is the total derivative dWm/d2. The PM motor on open-circuit differsfrom the electrically excited actuator or motor, where it is necessary to use partial derivativesMWN(i,2)/M2 or !MWf(M,2)/M2 and keep to the rule of “constant MMF” or “constant flux” respectively. Inthe PM motor on open-circuit there is no distinction between these partial derivatives, except as totheir sign.


Fig. 2.73

Even when the magnetic circuit is linear, the coenergy and energy are in general unequal, i.e. OXR isnot generally equal to OCX. This is a further difference between the PM motor on open-circuit and thenormal electrically-excited device.

The sign of the cogging torque is not particularly important, because it is purely oscillatory when therotor rotates continuously, and since it cannot sustain any continuous energy conversion, its averagevalue must be zero. Suffice it to say that the cogging torque will be such that mechanical work mustbe supplied to the motor when the magnetic stored energy is increasing, and mechanical work will besupplied by the rotor when the magnetic stored energy is decreasing.

Finite-element procedure

In a finite-element analysis we must rotate the rotor in small steps (e.g. 1E) and at each positioncalculate the change in the total stored energy. By evaluating the triangle OCX for each magnetelement, we are automatically including the coenergy stored in the “airgap” (i.e., in the rest of themagnet circuit), and therefore it is not necessary to evaluate Wgap separately by integrating BH/2 inthe air and iron regions. The same is true if we use triangle OXR to evaluate the total stored fieldenergy due to the current magnet element. The required process is as follows: for each magnetelement, per metre of stack length,

Coenergy Triangle OCX ½Hc CB × element area

Stored field energy Triangle OXR ½Br CH × element areaTABLE 2.7

The coenergy and energy calculations are shown in Fig. 2.73. Since Br and Hc are constants, the areasrequire the integration of a linear function over each element. The integral on the right of Fig. 2.73 isthe coenergy evaluation that would be done in air regions using ½B H = ½B2 /µ0 = ½µ0H

2, and it isquadratic. According to the theory, this is not needed for calculating the cogging torque.

We can now use one of the magnetic energy evaluations in Figs. 2.73. Once the element coenergyintegral of Fig. 2.73 or Table 2.7 has been summed over the entire magnet region, it represents thecoenergy of the entire machine in the same way that OCX represents this coenergy for a lumped magnetin Fig. 2.70. When the rotor moves from position X to position Y, the change in this coenergy sum willbe of the form OCX ! OCY, and this will give the average torque through the interval )2 = XY. On theother hand, if we use the alternative magnetic energy integral in Fig. 2.73, the total energy change fromX to Y will be of the form OYR ! OXR. Both methods should produce the same result.

Because of “discretization noise”, the total energy sums should be extracted from the FE analysis andfitted with a cubic spline; then the differentiation to get the cogging torque will be smoother. Thismethod helps avoid the discretization noise which troubles the Maxwell stress method.


Fig. 2.76 Variation of airgap flux-density

Fig. 2.74 Flux calculation

Fig. 2.75

Deodhar’s method: A simple method for evaluating thetriangles OXY treats the entire magnet as a single element.The magnet flux Mm is evaluated the from the difference ofvector potential at the magnet edges (multiplied by the stacklength), and used together with the magnet MMF obtained fromthe relation F = Lm/µrecµ0 × (Mm /Am ! Br) to plot the triangleOXY directly, moving from position to position and calculatingthe torque at each step. The vector potentials can be evaluatedat points shown in Fig. 2.74 on the edges of the magnets, usingthe expression Mm = (A1 ! A2) × Lstk, where Lstk is the stacklength or magnet length in the axial direction. If the centrelineof the magnet is a line of symmetry, generally A = 0 there andA2 = ! A1, so that Mm = 2A1 Lstk. An alternative is shown in Fig.2.75, where Mm is approximately equal to 2(AQ ! AZ) × Lstk. Inmany cases AZ will be zero.

This method could be made more efficient by recognizing thatOXY = ½Fc)Mm where )Mm = MY!MX. Alternatively, OXY =½Mr(FY ! FX) where FY = HY × Lm and FX = HX × Lm, and HX andHY are the values of H in the magnet at the positions X and Yrespectively, evaluated by H = (B ! Br)/µrecµ0. Together withthe integrals in Table 2.7, evidently there are several simpleprocesses for determining the cogging torque once the FEsolution is available.

PC-BDC’s approximate method: The rotor is stepped round in intervals of 1Eelec. At each positionthe airgap flux-density distribution is calculated including the effect of slotting. In the interests of rapidcalculation the slot modulation is very crude. It is estimated using an effective airgap that varies asshown in Fig. 2.76 as the point P sweeps across the slot opening. Mm is evaluated at each rotor positionby means of a ratio function Lm/(Lm + gN), where gN is the effective airgap at each position of the pointP. Then the total airgap flux is evaluated by integrating the airgap flux over the pole pitch, and therotor is stepped to the next position.

Fractional-slot motors can be accommodated by evaluating the cogging torque for the N and S polemagnets independently, and then adding them together. Both methods allow for skew by summing theeffects taken at several axial positions.


Fig. 2.77 Calculation of flux-linkagefrom vector potential

T/I '32

2Br1 Lstk B1Md kw1 Tph

2Nm/A (2.139)

N ' m A @dl (2.138)

T/i ' 4r1 BMgLstk Tph Nm/A (2.140)

2.18 GETTING INDUCTANCE FROM FINITE-ELEMENT CALCULATIONS

Plain inductance (phase self inductance or mutual inductance): Most finite-element programsclaim to be able to calculate stored field energy W, (and coenergy), suggesting the extraction ofinductance from the formula W = ½LI 2, if the current I is known. However, this method is ambiguousif there is any saturation, because the formula W = ½LI 2 relies on the assumption that L (the ratiobetween flux-linkage and current) is constant. Moreover, as we have seen in §2.17, the stored energyin permanent magnets introduces another ambiguity.

A more rigorous approach is to use the vector potential A directly,with the equation

in which N is the flux linking the contour along which A is integrated.In 2-D problems, the flux N linking a coil (per metre of axial length) isgiven by N = Ac1 ! Ac2, where Ac1 and Ac2 are the vector potential valuesat the coilside positions, Fig. 2.77. If there is a complete winding withcoilsides in different locations, the method can be extended bysumming the fluxes with appropriate polarities according to thedirection of the conductors. If the coils are in series carrying currentI, and all have Nc turns, the inductance is L = NcEN /I. Thisinductance is the total inductance, not the incremental inductance.The method is simple to implement because it uses point values of vector potential.

Incremental inductance: For calculations relating to the power electronic circuit, it may be importantto know the incremental inductance dR/dI when the machine is fully fluxed with full current, becausethis is the inductance presented to the current regulator. Using the A method described above, theincremental inductance is given by )R/(I1 ! I2), where )R is the difference in computed flux-linkage attwo slightly different current levels I1, I2.

Synchronous inductance: The extraction of the synchronous inductances Ld and Lq has beendescribed earlier. The vector-potential method calculates total flux linkage, and therefore it is notpossible to resolve the flux-linkage uniquely into separate “magnetizing” and “leakage” components.For a winding which is essentially sinewound, a practical procedure is to get a finite-elementcalculation with two phases connected in series line-line, and determine the total line-line inductanceLLL for the two rotor positions that give the maximum and minimum values LLL(max) and LLL(min). Thenif Lq > Ld, Lq = LLL(max)/2 and Ld = LLL(min)/2. End-winding inductances (not calculated in 2-D FE) mustbe added to these values of Ld and Lq. Since these are generally small, approximate methods ofestimation should suffice.

2.19 TORQUE PER AMPERE AND KVA/KW OF SQUAREWAVE AND SINEWAVE MOTORS

With a torque angle of 90E degrees the torque per r.m.s. ampere of phase current in the three-phasesinewave motor is

where r1 is the stator bore radius and B1Md is the peak fundamental flux-density produced by themagnet in the airgap. The torque per peak ampere is /2 times smaller. In the squarewave motor(assuming 180E magnet arcs, star connection and 120E squarewave phase currents), the torque per peakampere of phase current is

where BMg is the peak flux-density produced by the magnet in the airgap.


28The ratio actually depends on the depth of modulation.

peak kVA/kW ' 2q × is × Vs (2.145)

I ' i 2

3(2.141)

vLL ' V ' 3 vph ' 6Vph (2.146)

r.m.s. kVA/kW ' 2q × Is × Vs (2.144)

T/I ' 4 3

2r1 BMgLstk Tph Nm/A (2.142)

4 3

2

32

× 2 ×B2

' 1.47 (2.143)

The r.m.s. phase current I is derived from the 120E squarewave as:

where i is the peak or flat-top value of the phase current. Therefore the torque per r.m.s. ampere ofphase current is

With these expressions the ratio of torque per ampere in the two motors can be compared. With equalr.m.s. phase currents, the torque of the squarewave motor exceeds that of the sinewave motor by thefactor

With equal peak currents, the factor is 1.27. The squarewave motor appears to have a significantlyhigher torque per ampere. However, the comparison assumes equal peak magnet flux-densities in theairgap, which is likely to require significantly more magnet and a thicker stator yoke in thesquarewave motor. To rectify this imbalance it is perhaps better to compare the motors on the basisof the same flux per pole. For the same peak flux-density the flux per pole of the square-wave motorexceeds that of the sinewave motor by the factor B/2. Now with equal r.m.s. phase currents the torqueratio is only 0.94, and with equal peak currents it is 0.81. This analysis neglects many important effects,such as armature reaction and losses, but it indicates that with equal amounts of copper, iron, andmagnet, the torque per ampere is not greatly different between the two machines.

The comparison is now carried to the volt-ampere requirements of the drive. A simple estimate of thedrive 'rating' can be made in terms of the total kVA rating of its main switches, per kW of power fed tothe motor. The relevant parameters are defined as follows. With respect to the r.m.s. current in eachswitch, if q is the phase number,

where Is is the r.m.s. current in each switch and Vs is the peak voltage across each switch. For thedrives normally used with brushless DC motors the peak device voltage is nominally equal to the DCsupply voltage, because each switch must block this voltage while the other one in the same phaselegis conducting. Obviously there must be a margin for voltage spikes caused by stray inductance andreverse-recovery of diodes. While these effects are not small, they are parasitic and not fundamentalto the operation of the motor. Therefore the DC supply voltage will be used in this comparison.

With respect to the peak current in each switch,

where îs is the peak current in each switch, and the voltage conditions are unchanged.

In the sinewave motor the line currents are assumed to be sinewaves and each switch conducts a halfsinewave for 180E and is then off for 180E. The r.m.s. switch current is therefore 1//2 times the r.m.s.line current, which will be assumed to be the same as the phase current (i.e. the motor isstar-connected). The peak device current is equal to the peak phase current. The peak line-line voltageof the motor is approximately equal to V.28 Thus


r.m.s. switch VA ' 6Vi 2

3;

peak switch VA ' 6Vi ;

drive power output ' Vi

(2.148)

r.m.s. switch VA ' 6VI

2;

peak switch VA ' 6Vi ;

drive power output ' 3VphI ' 3V

6I

(2.147)

We can now write

From the appropriate ratios between these quantities, the r.m.s. switch kVA/kW is 3.5 and the peakkVA/kW is 6.9.

For the squarewave motor the corresponding equations are:

From the appropriate ratios between these quantities, the r.m.s. switch kVA/kW is 4.9 and the peakkVA/kW is 6.0. The result is the same for the star- and delta-connected motors considered earlier.

Thus the squarewave motor has a slightly better utilization of the peak current capability of the driveswitches. Although the sinewave motor appears from this analysis to have a much better utilizationof their r.m.s. current capability, this advantage may be offset by the greater duty cycle in the sinewavemotor, where three drive switches are conducting at any instant, not two. In the analysis it has beenassumed that the ideal waveforms are achieved by pulse-width modulation at a sufficiently highfrequency, and of course this incurs switching losses which tend to lessen the significance of the lowerr.m.s. switch currents required by the sinewave motor. More detailed comparison requires the use ofcomputer simulation.

2.20 PERMANENT MAGNETS VERSUS ELECTROMAGNETIC EXCITATION

Because of the laws of electromagnetic scaling there is an 'excitation penalty' associated with smallmotors. As the size is decreased, the cross-section area available for conductors decreases with thesquare of the linear dimension, but the need for MMF decreases only with the linear dimension, beingprimarily determined by the length of the airgap. As the motor size is further decreased, the airgaplength reaches a minimum manufacturable value (typically of the order of 0.15!0.3 mm). Past this pointthe MMF requirement decreases only slowly while the copper area continues to decrease with thesquared linear dimension. The per-unit copper losses increase even faster, and the efficiency decreasesrapidly. The loss-free excitation provided by permanent magnets therefore increases in relative valueas the motor size is decreased.

In larger motors magnets improve efficiency by eliminating the losses associated with electromagneticfield windings. But in larger motors the relative excitation penalty is small. At the same time therequired volume of magnets with adequate properties increases with motor size to the point where PMexcitation is just too expensive. It is therefore rare to find PM motors of more than a few kilowatts.

Sometimes operational or safety considerations work against the PM motor. For example, in railwayor transit-car traction, a PM motor with a short-circuit winding fault would have a large brakingtorque, possibly with a high ripple component and a definite risk of overheating which coulddemagnetize the magnets or even cause a fire. There is no simple way to protect against this type offault. A conventional synchronous or DC machine can be de-excited under fault conditions, and theinduction motor is self-protecting.


Am 'B1DLstk

p(Br

(2.149)

Fg '2

0i

N

2sin p2 d2 '

iN

2pcos p2 (2.155)

Hmlm ' Hg g ' 0 (2.150)

gB1

µ0

'iN

2p(2.156)

lm 'gBgµrec

(1 & () Br

(2.151)

iN × 2p ' 2piN (2.157)

Vm ' 2pAmlm (2.152)

AcJ ' 2piN (2.158)

Vm '2B 2

1 D 2 gµrec

k1 B 2r ((1 & ()

(2.153)

Fg ' gHg ' gBg

µ0

'g

µ0

B1 cos p2 (2.154)

There is no hard-and-fast power level below which permanent-magnet excitation becomesadvantageous, but it is possible to examine the excitation penalty in ways which indicate roughlywhere the breakpoint lies, and why. For a given level of excitation the choice can be made betweenmagnets or copper windings operating at a current density J (in the copper).

In the following analysis, several gross assumptions are made, and the equations derived should notbe used for detailed design purposes but only for guidance and interpretation.

Using a permanent magnet, the fundamental flux per pole is B1 DLstk/p, where B1 is the peakfundamental flux-density produced by the magnet in the airgap. If rotor leakage is neglected this fluxcan be taken to be approximately equal to the flux through the magnet in a surface-magnet motor, thatis, Bm Am. If the magnet is operated at a fraction ( of its remanent flux-density, Bm = (Br and hence

From Ampere’s law we have

from which

where Bg is the average airgap flux-density. The required volume of magnet is then

If B1 = k1 Bg, and if we assume that l = D, then

Now we can calculate the amount of copper needed to magnetize the airgap to the same level. Assuminga sinusoidal distribution of conductors,

But with N conductors in series per phase,

Hence

and the total ampere-conductors required are

If Ac is the cross-section of the copper winding in the whole motor cross-section and J is the currentdensity,

and hence


Vm

Vc

'D

230(2.163)

Ac '4p 2gB1

µ0J(2.159)

Vm

Vc

'µ0 µrecJB1 D

4k1 B 2r p 2((1 & ()

(2.161)

Vc '8p 2gDB1

µ0J(2.160)

Vm

Vc

'D

488(2.162)

The copper cross-section required is proportional to p2, unlike the magnet volume which is independentof the number of poles. The volume of copper can be estimated by assuming a mean length of conductorequal to twice the stack length (to allow for end-turns); thus

The relative volumes of magnet and copper required can be compared by taking the ratio

where D is the rotor diameter.

Consider two four-pole motors with B1 = 0.7T and k1 = 1.1. The PM motor has rare-earth magnets withBr = 0.8T and µrec = 1.05. The electrically-excited motor has J=4 A/mm2 , giving

where D is measured in mm. This means that for motors less than about 500 mm in rotor diameter themagnet volume is less than the volume of copper needed for excitation by a separate field winding.Unfortunately the cost per unit volume of high-energy magnets at this level is of the order of 25 timesthat of copper. For the magnet cost to be less than the cost of the copper in a separate field winding, therotor diameter must therefore be less than 500/25, i.e. only 20 mm, giving a stator diameter of about 40mm. The technical potential of high-energy magnets is thus offset by their high cost in all but thesmallest motors. In very small motors a smaller value should be used for the current density J; withJ = 2.5 A/mm2 the stator diameter for equal cost would be increased from 40 mm to 64 mm (2.5 in). Ingeneral, high-energy magnets can only be justified where there is a special premium on efficiency orcompactness. Of course this argument is simplistic, ignoring factors such as process and manufacturingcosts and many others, but it provides a basic physical understanding of the application potential ofmagnets, and the effects of scale. Motors magnetized with ceramic magnets must settle for a lowerairgap flux-density. Using values of J = 4 A/mm2 ; B1 = 0.3 T; µrec = 1; Br = 0.35 T, the result is

For motors of less than 460 mm stator diameter the magnet volume indicated is less than the volumeof copper in a separate field winding. Ceramic magnets are much less expensive than high-energymagnets, the cost per unit volume being of the order of 0.6 times that of copper, so that the magnet costwill be less than the cost of field copper in motors of diameter less than 460 × 0.6, i.e. 275 mm. In practicePM motors as large as this are relatively uncommon. With ferrite, the flux-density is too low; withrare-earth magnets the cost is too high.

If running costs are taken into account, the comparison between PM and electrically excited motorschanges significantly. With the present cost of raw materials and present kWh tariffs, the kWh cost ofelectrical excitation would outstrip the raw-material cost of the copper in just a few months, assumingthe motor runs at full excitation 24 h/day. Even when all the manufacturing costs are added up, the PMmotor should eventually pay for itself in this way.


b '2(r/R)p%1

[1 % (r/R)2p](2.164)

2.21 SLOTLESS MOTORS

The availability of very high energy rare-earth and Neodymium-Iron-Boron magnets has re-awakenedinterest in the slotless motor, in which the stator teeth are removed and the resulting space is partiallyfilled with additional copper. At least one such motor is manufactured commercially. The slotlessconstruction permits an increase in rotor diameter within the same frame size, or alternatively anincrease in electric loading without a corresponding increase in current density. The magneticflux-density at the stator winding is inevitably lessened, but the effect is not so drastic as might beexpected.

For a motor with an iron stator yoke and an iron rotor body the magnetic field and its harmoniccomponents can be calculated by Hague’s method described in §2.4, or the methods described by Hughesand Miller [1977]. Considering the fundamental radial component of B, the value is greatest at the rotorsurface (radius r) and falls off with increasing radius to its smallest value just inside the stator yoke(radius R). The ratio between the values of the fundamental radial component at these two radii isgiven by:

Consider a rotor body of 40 mm diameter with a high-energy magnet of remanent flux-density 1.2 T andthickness 5 mm. If the radial thickness of the stator winding is 5 mm (including the airgap), then fora 4-pole magnet b = 0.78. The magnet flux-density will be about half the remanent flux-density withthese proportions, so that the radial flux-density in the stator winding varies from about 0.6 T near thebore to 0.47 T just inside the stator yoke, giving a mean value of fundamental flux-density of about 0.53T. The electric loading may be increased relative to that of a slotted stator, because of the additionalspace available for copper; but the increase may not be much because the close thermal contact withthe teeth is lost, and the cooling of the slotless winding by conduction to the stator steel may not be aseffective. Taking these factors into account, the power density should be roughly the same as that ofthe conventional motor, and possibly a little higher, since the stator tooth iron losses are eliminated.This machine may well accept less expensive grades of lamination steel because of the absence ofslotting and the relatively low flux-density in the stator yoke. The reactance is also lessened by theelimination of slot leakage effects, and the risk of demagnetization is decreased.

In this type of motor the maximum useable magnet energy is obviously higher than in a conventionalslotted motor, because the there are no teeth to limit the flux by saturating; indeed the concept wouldnot be viable at all without magnets of high remanence and coercivity.

Once the stator teeth are removed, the conductors are no longer constrained to lie parallel to the axis.They may be skewed by a small amount to reduce torque ripple (which is already reduced by theelimination of cogging effects against the stator teeth). A further possibility is a completely helicalwinding such as that proposed for superconducting AC generators [Ross, 1971], or as used in very smallPM commutator motors. Because the helical winding has no end-turns its utilization of copper is higherthan the severe skew might suggest, and it might permit the design of a very compact motor.

2.21 SIGN CONVENTIONS

Figs. 2.78 and 2.79 show a single conductor fixed on the stator, and a moving magnet that is fixed on therotor.

Sign conventions follow the mathematical ones: in the radial direction, outwards is positive; in thecircumferential direction, counter-clockwise (CCW) is positive; and in the axial direction, out of thepaper is positive.

LeFt-hand rule (Force on current-carrying conductor)

In Fig. 2.78, by the left-hand rule: First finger = Field (produced by the magnet alone), seCond finger= Current, and thuMb = Motion, the force on the conductor is downwards and to the right; the reactionforce on the magnet is upwards and to the left, producing positive (CCW) torque T.


Fig. 2.78 i > 0, T > 0 producing positive torque.

Fig. 2.80 Armature reaction field of a single coil

Fig. 2.79 T > 0, e < 0

Fig. 2.81 Torque-producing armature current with magnet

Right-hand rule (EMF geneRated in conductor moving in magnetic field)

In Fig. 2.79, by the right-hand rule, First finger = Field (produced by the magnet alone), thuMb =Motion (of conductor relative to magnet), seCond finger = induced Current (which gives the directionof the EMF). Note that e < 0 in Fig. 2.79, i.e., it is negative.

The product ei is the instantaneous electrical output power pelec[out], because e is regarded as agenerated EMF and i is in the direction of positive e. In Figs. 2.78 and 2.79, pelec[out] is negative. Theinstantaneous electrical input power is pelec[in] = ! pelec[out], and this is positive. Therefore theconditions shown in Figs. 2.78 and 2.79 correspond to motoring action.

The product TT is the instantaneous mechanical output power pmech[out], because T and T are in thesame direction. If there were no losses, pelec[in] = pmech[out] and T = ei/T. Note that if the EMF isproportional to speed, e = kET, then T = kE i, where kE is the EMF constant or torque constant kT = kE.

Fig. 2.80 shows the direction of the magnetic field produced by a single armature coil with a plainairgap. In Fig.2.81 two magnets are introduced into the airgap with polarities as shown. Using theresult of Fig. 2.78, the torque is positive. We can deduce the direction of the torque from the polarityof the armature reaction flux, which tends to make the stator surface a south pole, as indicated by thelower-case s. Since like poles sS repel, and opposite poles sN attract, the rotor is pushed in the CCWdirection.

The arrows show that the armature reaction field tends to strengthen the magnet flux at one edge ofeach magnet, and to weaken it at the other edge. If the rotor is rotating in the forward (CCW) direction,the flux is strengthened at the leading edges and weakened at the trailing edges (motoring operation).


Fig. 2.82 Armature reaction with the current advanced 90Ein phase relative to the generated EMF.

With perfect symmetry the net flux through themagnet would be unchanged. But perfect symmetryis not achieved in general, for the following reasons:

(1) The stator slots may not be symmetricallyarranged about the magnet centre-line. Withsquarewave drive, the current pattern remainsfixed for typically 60E intervals of rotation. Insinewave motor drives, the fundamentalcomponent rotates in synchronism with therotor but space-harmonics in the windingdistribution may distort it.

(2) The stator teeth are liable to saturate more onthe side with higher flux-density, that is, wherethe magnet flux and armature reaction flux arein the same direction.

(3) If the demagnetization characteristic of themagnet is not linear, the magnet may sufferirreversible loss of magnetization on the sidewhich has the lowest flux-density.

Fig. 2.82 shows the armature reaction with the current advanced 90E in phase relative to the generatedEMF. The conductors are in the same position, with the same currents as in Figs. 2.80 and 2.81, and thephase advance is represented by a 90E rotation of the rotor in the negative (CW) direction. Thearmature-reaction field is demagnetizing across the entire width of each magnet. We say that thearmature reaction is “in the negative d-axis”. In a sinewave motor drive in the steady-state, thisrelationship is fixed. The MMF axis is aligned with the magnet centre-line (the d-axis), and the armatureMMF polarity ns opposes the magnet flux polarity NS.

In Fig. 2.82 the flux-linkage produced by the magnet in the coil is at a negative maximum, since themagnet flux is radially inwards and it is symmetrically aligned with the MMF axis of the coil. As therotor moves forwards (CCW) the flux-linkage R changes in the positive direction unti l it reaches zeroat the position shown in Fig. 2.81, which is 90 electrical degrees later than Fig. 2.82. The generated EMF

e = !dR/dt is zero in Fig. 2.82 and negative in Fig. 2.81. Also, Fig. 2.81 is consistent with Fig. 2.79 inrelation to the sign of the generated EMF.

If instead of the generated EMF e we consider the back-EMF !e, and substitute e instead of !e, then e isat a positive maximum in Fig. 2.81. The back-EMF e = +dR/dt is convenient in motor theory, because itis has the same sign as the applied voltage and has the same sign as an inductive voltage drop L di/dt.

The demagnetizing armature reaction in Fig. 2.82 is associated with a phase advance of the currentrelative to the back-EMF. This is exploited in the technique known as flux-weakening, which permitsbrushless PM motors to operate at high speeds even when the back-EMF exceeds the supply voltage.

Fig. 2.83 shows the phasor diagram of a sinewave motor in the flux-weakening condition. The phaseadvance angle ( is less than 90E because otherwise there would be no magnet-alignment torque. Thecurrent phasor is resolved into d- and q-axis components I = Id + jIq. The demagnetizing effect of Id isclearly seen in that the voltage generated by its associated flux-linkage is jXdId which is in phaseopposition to E, permitting the motor to operate even though the supply voltage V is less than E.

Fig. 2.84 shows the airgap flux-density distribution in a 24-slot, 4-pole squarewave brushless PM motor,with the armature MMF in two positions: in the q-axis in the top trace, and in the negative d-axis in thebottom trace. In the top trace the (south) d-axis is at 0E, and in the bottom trace it is at 270E or !90E(electrical), retarded 90E relative to the MMF. The EMF in the top trace is in phase with the current,while in the bottom trace it is retarded by 90E. Note that the leading edge of the magnet is further tothe right than the trailing edge, since the x-axis is the azimuth or circumferential coordinate. The toptrace shows the strengthening of the flux at the leading edge and the weakening at the trailing edge.The bottom trace shows the demagnetization symmetrical about the d-axes.


Fig. 2.83 Phasor diagram of sinewave brushless PMmotor drive, in which ( > 0 and thecurrent I has a negative d-axis componentId which is in the demagnetizing direction.

Fig. 2.84 (top) airgap flux-density distribution in 24-slot, 4-pole squarewave brushless PM motor with the armature MMF axisat right-angles to the d-axis. The armature reaction is cross-magnetizing, with the south-pole d-axis at 0E.

(bottom) airgap flux-density distribution with the armature MMF axis in the negative d-axis., with 90E phase advanceof the current relative to the back-EMF. The armature reaction is demagnetizing, with the south-pole d-axis at !90Eor 270E.

In the sinewave motor in the steady-state, the flux pattern remains more or less fixed and rotates atsynchronous speed, but in the squarewave motor the mmf remains fixed intervals typically of 60E, andas the magnet sweeps past the airgap flux-density distribution changes shape. If no phase advance isapplied, the top trace in Fig. 2.84 represents the conditions at the middle of the commutation interval.


M1Md 'B1

[oc] DLstk

p(2.169)

f ' n × p Hz (2.165)

f 'rpm

60×

Poles2

'rpm × Poles

120. (2.166)

T ' 2B f rad/sec (2.167)

Eq1 '2B

2kw1Tph f M1 ' 4@44 kw1Tph f M1 V. (2.168)

2.22 PM GENERATORS

A PM brushless machine can operate as a motor or a generator. When it is motoring, theelectromagnetic torque is in the same direction as the rotation. When it is generating, theelectromagnetic torque opposes the rotation. In either case, there is a generated EMF in each phaseproportional to speed. This EMF is an AC quantity. Its fundamental frequency is given by

where n is the speed in rev/sec, i.e. rpm/60, and p is the number of pole-pairs. Alternatively,

The waveform of the EMF is not necessarily sinusoidal. It depends on

— the profile of the airgap flux-density waveform produced by the magnet

— the winding distribution

— the connection of the winding (wye, delta, etc.)

— the amount of skew (if any).

Sometimes the frequency is expressed in electrical radians per second, with symbol T; thus

and the period is J = 2B/T. The period J is the time taken for the rotor to rotate through 1 cycle, i.e.through two pole-pitches, where the “pole-pitch” is B/p radians or 360/Poles in degrees. A “cycle” isalso known as 2B “electrical radians” or 360 “electrical degrees”. This is equal to 2B/p mechanical oractual radians, or 360/p mechanical degrees.

To begin, we will assume that the EMF is sinusoidal, because this is normally the case for ACgenerators, and it means that we can use the classical methods of analysis to describe the performance,especially the phasor diagram. The generated EMF is proportional to the product of the speed and theflux produced by the magnet, and it obeys the classical equation for AC machines:

Here Tph is the number of “turns in series per phase”. If each phase has a total of T turns and theyare connected in a parallel paths, then Tph = T/a. The factor kw1 is the fundamental harmonic windingfactor. A properly designed winding has the property of filtering out harmonics in the EMF waveform,rendering it more sinusoidal. This filtering is achieved at the expense of a slight loss of EMF comparedto that which would be obtained if all the coils were “fully pitched” and concentrated together. kw1expresses this reduction. Usually it has a value between 0.8 and 1, so the slight loss of EMF is not a highprice to pay for the elimination of unwanted harmonics which would distort the waveform. Sometimesthe product kw1 Tph is termed the “effective series turns per phase”, Tph1.

The quantity M1Md is the “fundamental flux per pole” produced by the magnet when the machine isrunning on open-circuit. Only the fundamental harmonic component of airgap flux contributes to thefundamental component of EMF. M1Md can be obtained from the total open-circuit airgap flux by Fourieranalysis. It is related to the peak open-circuit airgap flux-density B1

[oc] by the equation

where D is the stator bore diameter and Lstk is the stack length.


29 Note that a spinning PM generator is electrically "hot" (i.e. "live") whenever it is spinning, and it is dangerous to assume thatit is safe just because it is disconnected or isolated from its load. PM generators should ideally have warning labels on theterminal box to remind electricians of this fact.30 Occasionally a PM generator may be connected to a cycloconverter, as in one or two VSCF (variable-voltage/constant-frequency) aircraft power generating systems; but this is beyond the scope of these notes.

E ' jTQQQQ1Md (2.170)

Fig. 2.85 Generator connected to infinitebus

Fig. 2.86 Generator connected to R!L load

The subscript q in Eq1 tells us that the EMF lies along the q-axis in the phasor diagram; the flux M1 liesalong the d-axis. In phasor terms, E is written 0 + jEq1, or simply jEq or even just jE. Also

where QQQQ1Md = Q1Md + j0 is the fundamental flux-linkage per phase, and Q1Md = kw1TphM1//2 representsthe product of the magnet flux and the effective series turns/phase. The units of Qd1 are volt-secondsand the /2 converts the peak value into the r.m.s. value, since QQQQ1 represents a flux-linkage that isvarying sinusoidally in time. All the quantities in the phasor diagram are r.m.s. values. Looselyspeaking, eqn. (2.170) means that the generated EMF is proportional to speed times flux, or frequencytimes flux. Strictly speaking, it states that the EMF is proportional to speed times flux-linkage, becauseit is also proportional to the number of effective series turns per phase Tph1 = kw1Tph. The “j” in eqn.(2.170) means that the EMF phasor E leads the flux-linkage phasor QQQQ1 by 90E. The phasor diagram onopen-circuit is shown in Fig. 2.26. With no current the terminal voltage V is just equal to E.29

Load current and impedance: The current that flows from a PM generator depends on the loadconnect to it. Four important kinds of load can be identified as being important in understanding theoperation of the PM generator:

— The "infinite bus".

— A fixed-impedance load.

— Another synchronous machine.

— A rectifier.30

The infinite bus is represented as a voltage source having a fixedvoltage and frequency. Physically it is approximated by a verylarge network such as the U.K. National Grid, which is so large(60,000 MW) that neither the voltage nor the frequency can beperceptibly altered by connecting one small additional generator(or load) to it. Fig. 2.85 shows the connection of a generator to aninfinite bus. This diagram is only a schematic diagram becauseit does not show the internal impedance of the generator; (it alsoomits the circuit-breaker!)

A fixed-impedance load is represented electrically as a passiveelectrical circuit containing resistance R and inductance L ineach phase. This is represented in Fig. 2.86.

Fig. 2.87 shows a generator connected to another synchronousmachine, and Fig. 2.88 shows a generator connected to a rectifier.The rectifier itself is loaded with an impedance comprising aresistance R and inductance L. The complex representation R +jTL is not used on the DC side, since it applies only when thevoltages and currents are sinusoidal AC.

The rectifier load is probably the commonest type of load to which PM generators are applied.Unfortunately, a rectifier is a non-linear load, and even though it is connected to a sinusoidal AC supply(the PM generator), it draws a non-sinusoidal current i. This means that we cannot use phasors tocalculate the current. The DC current id is also not a pure DC current, because it contains harmonics.


V ' E ! (R % jXd ) I . (2.171)

Fig. 2.87 Generator connected to another synchronousmachine

Fig. 2.88 Generator connected to a rectifier with a DC loadhaving resistance R and inductance L.

Fig. 2.89 Equivalent circuit of one phase of nonsalient-pole generator

The computation of rectifier loads requires computer simulation. To get round this difficulty we designthe PM generator in two stages. The first stage involves designing as though the generator was goingto be connected to a linear AC load. The generator can be "rated" to operate with this linear load. Wecall this the sinewave rating, because it is strictly valid only for loads that draw sinusoidal current.

Then, in the second stage, we make an allowance for the nonlinear effects of the rectifier load, leadingto a "rectifier rating" which can go into the catalogue as an addition to the sinewave rating.

Generator equivalent circuits — internal impedance: The windings of the generator haveresistance R, and they also have self-inductance and mutual inductance between phases. The simplesttype of generator is a nonsalient pole generator, in which the phase inductances are unaffected by therotor position. In this case, in the steady state with AC sinusoidal current and EMF, the generator hasa very simple equivalent circuit for each phase, Fig. 2.89.

The total internal impedance of the generator is Z = R + jXd. The reactance Xd is the synchronousreactance and it is equal to TLd where Ld is the synchronous inductance. The synchronous inductanceis not simply the self-inductance per phase, but includes the effect of armature-reaction flux ingenerating an internal voltage jXdI which is in series with E and the resistance voltage-drop RI. Theinternal voltage jXdI is a sinusoidal voltage represented as a phasor, and therefore Xd incorporates thefiltering effect of the winding distribution (and any skew).

When current flows, the terminal voltage V deviates from the open-circuit value E, and according toOhm’s law applied to Fig. 2.89, the relationship between V and E is


Te ' mpQ1Md Iq N m. (2.175)

P ' V I cos N (2.172)

Fig. 2.90 Phasor diagram of nonsalient-pole generator with lagging power factor angle N.

P 'EV

Xd

sin * (2.173)

P ' EI cos ( ! RI 2 (2.174)

This is represented in the phasor diagram, Fig. 2.90, which is drawn to emphasize the relationshipsbetween the phasors, so the resistance R and reactance Xd have been exaggerated relative to theirnormal values in a PM generator. Points to note about the phasor diagram are as follows:

(a) The phasor diagram and all equations derived from it are strictly valid only when the voltageand current are sinusoidal. Two cases where the voltage and current are not sinusoidal are (a)when the output is rectified and (b) when the PM machine has a nonsinusoidal EMF waveform.

(b) With lagging power factor the terminal voltage V is generally less than the open-circuit voltageE. The “armature reaction” (i.e., the internal voltage-drop across Xd) has a demagnetizing effectand the machine is said to be “overexcited” (i.e., with E > V). If the load is disconnected, theterminal voltage will jump up to the higher value E. This condition must be considered inpractice because it might be unsafe if E is too high.

(c) Since E is proportional to speed, V may have to be regulated by an electronic controller (suchas a phase-controlled rectifier) on order to maintain constant output voltage as speed varies.

(d) The output power per phase is given by

in watts, where N is the power-factor angle. If the load is an impedance ZL = ZLejN = RL + jXL,then cos N = RL/ZL = RL//(RL

2 + XL2)

(e) If R is much smaller than Xd, eqn. (2.172) can be expressed in terms of E and V in watts/phase:

where * is the load angle. This equation is commonly used in power systems engineering toexpress the fact that there is a maximum power Pmax which can be generated stably. In powersystems the generator is usually considered to be connected to an infinite bus with V =constant. Pmax is the power obtained when * = 90E and sin * = 1.

(f) If we write ( = * + N, where ( is the angle between the current phasor I and the q-axis, the powerper phase can be written as

in which RI2 represents the resistive power loss per phase. The term EI cos ( represents themechanical power per phase, which is written Pm/m, where m is the number of phases (usually3). Now I cos ( = Iq, the q-axis component of the current, and E = Eq = TQ1Md. If Te is theelectromagnetic torque and Tm is the angular velocity, then Tm = T/p and TmTe = Pm, so


Te ' mp (Qd1 Iq ! Qq1 Id ) N m (2.183)

T ' Te % Tf . (2.176)

Vd ' V sin * ; Vq ' V cos *Id ' I sin ( ; Iq ' I cos ( (2.178)

P 'EVd

Xd

! Vd Vq1

Xd

!1

Xq

(2.181)

T ' Te % Tf %WFe

Tm

. (2.177)

Vd ' !RId % Xq Iq ;Vq ' E ! RIq ! Xd Id . (2.179)

P 'EV

Xd

sin * !V 2

21

Xd

!1

Xq

sin 2* . (2.182)

P ' EIq % (Xq ! Xd )Id Iq ! R (Id2

% Iq2) (2.180)

This equation is commonly used in connection with PM motors. It shows that the torque ismaximized if the current phasor is oriented along the q-axis such that it is “in quadrature withthe flux” (and in phase with the EMF E). [Note: Iq is in r.m.s. amperes, and Q1Md in r.m.s. V-sec].

(g) The phasor diagram includes resistive losses (RI 2) but it does not include mechanical losses oriron losses. The mechanical losses can easily be allowed for by a friction torque Tf which mustbe added to Te to give the shaft torque T. Thus

The simplest way to deal with the iron losses is to treat them as a mechanical loss and includethe corresponding torque along with Tf. Thus the power loss in the core is WFe = TmTFe and

It is also possible to represent the core loss as an electrical loss, but this modifies the phasordiagram in rather a complicated way, requiring an iterative solution.

Salient-pole machines: A salient-pole machine is one in which the rotor has two axes of symmetry.Generally one of these axes is the d-axis which is the axis of magnetization, and the other one is the q-axis or interpolar axis. In a salient-pole machine the phase inductances vary with rotor position, butif the machine has sinusoidally distributed windings the analysis can be simplified by Park’stransformation such that the phasor diagram can be used for sinusoidal operation in the steady state,with a relatively simple modification. This modification is to split the armature-reaction voltage dropinto separate components aligned respectively with the d- and q-axes, as shown in Fig. 2.91.

The power per phase is still given by eqn. (2.172), but eqns. (2.173!2.175) acquire additional “saliency”terms. First we resolve the voltages and currents into their d- and q-axis components

then from the phasor diagram

The electrical power output per phase is P = Re{VI*} = VdId + VqIq, giving

and if R is negligible then

in watts per phase. This is often written

The electromagnetic torque is

where Qd1 is the total fundamental d-axis flux-linkage and Qq1 is the fundamental q-axis flux-linkage.


Fig. 2.91 Phasor diagram of salient-pole generator with a load having a lagging power-factor angle N

3VLL IL cos N ' VDC IDC (2.184)

Q '2/3

( 6/B) cos "'

1cos N

. (2.187)

cos N '3B

cos " (2.185)

IL 'VD ID

3VLL cos NA[rms] (2.186)

Rectifier loads and “rectifier rating”

Rectifiers draw nonsinusoidal current waveforms, and a rigorous analysis would involve complex time-stepping simulation which is a difficult and time-consuming process, especially if there is significantsaturation in the machine. To continue to use the simpler theory of the phasor diagram it is desirableto develop an equivalent linear load with constant impedance. This involves the following steps:

(1) Determine the r.m.s. value of the fundamental AC line current. For a three-phase dioderectifier we can equate the DC power to the AC input power:

where VDC is the mean DC voltage and IDC is the mean DC current, VLL is the r.m.s. AC line-linevoltage at the terminals of the generator, and IL is the AC line current. The input power factorof the rectifier is

where " is the phase-delay angle. In a diode rectifier " = 0 and cos N = 3/B = 0.955. The AC linecurrent is therefore

(2) Determine the form factor Q of the rectifier load current. This is the ratio of the actual r.m.s.current I to the r.m.s. value of its fundamental component I1, and is given by

The generator is designed for a sinewave current of IL and power factor cos N, but the calculated I2Rloss in the generator is increased by the factor Q2 to allow for the additional losses caused by harmonicsin the current waveform.

The process may need to be repeated several times to determine a complete operating chart coveringa range of currents and power factors, especially if the generator is to operate at different load levels.


31 By convention, an overexcited generator is said to be “generating VArs” (reactive power), while an underexcited generatoris “absorbing VArs”. For example, if the load is inductive, the VArs generated by the generator are absorbed by the load.

Control and protection

If the generator is running at a certain speed, the generated EMF is fixed and the power output dependson the load impedance, which determines the current and the power factor. If the load draws morepower, then the prime mover must respond by providing more torque, otherwise the generator willslow down. For isolated generators it is normally necessary to provide a governor to make sure thatthe prime mover does this. The simplest form of governor is one that maintains the speed constant,i.e. a closed-loop speed controller, because this will ensure that the generator will receive whatevertorque is necessary to supply the load (plus the losses in the generator).

The governor must also protect the generator and prime mover against overspeed. Overspeed is mostlikely to occur if the generator loses its load, which can happen quite normally if the load is switchedoff, or because of a malfunction that causes the load to disappear or to be disconnected from thegenerator. A sudden loss of load causes all the prime mover power to be applied in accelerating thegenerator/prime mover, and unless the prime mover is shut down quickly, a dangerous overspeed canresult very quickly. For this reason the governor must be designed with a sufficiently fast responseto any speed error between the actual speed and the set-point speed.

In wound-field synchronous generators connected to a local load, isolated from the grid system, thevoltage can be varied by changing the field current. The power factor, however, is still determined bythe load impedance. If a wound-field generator is connected to an infinite bus, the voltage cannot bechanged by varying the field current. Instead, the power factor changes. Increasing the field currenttends to make the power factor more leading (i.e. "overexcited"), whereas decreasing it makes thepower factor more lagging ("underexcited").31 When a generator is connected to an infinite bus, thepower is controlled by the prime mover torque (and by that alone).

PM generators have no means of excitation control, i.e., no field winding. Therefore the voltage at thegenerator terminals cannot be varied without changing the load impedance or the speed. Changingthe speed also changes the frequency. If a PM generator is connected to an infinite bus, the frequencyand voltage are fixed by the infinite bus, and the only means of control is the prime mover torque,which determines the power. The current and power factor both vary in a manner that depends on theinternal impedance and open-circuit EMF of the generator.

If the generator is connected to an infinite bus, the output power must be limited such that thegenerator remains synchronized with the frequency and phase of the infinite bus. With isolatedgenerators the issue of maintaining synchronism does not arise.

Another hazard with all generators is the over-voltage which occurs when the load is suddenlydisconnected. If the load current was at its maximum value just before disconnection, there may beconsiderable energy stored in the internal reactance of the generator. When the disconnection takesplace, the circuit-breaker must dissipate this energy. When the current has fallen to zero, however,the voltage-drop across the internal impedance disappears, and the generated EMF now appears at theterminals. Generally this will exceed the rated terminal voltage; by how much depends on the internal(synchronous) reactance. The internal reactance of PM generators tends to be fairly low, but on theother hand PM generators are also used at extremely high speeds, so the potential overvoltage mustbe allowed for in the insulation system. There is no possibility of switching it off, since the magnetsare permanently excited. If there is a filter capacitor on the DC side, overvoltage protection isessential. This can be provided by a fast-acting “crowbar” circuit which detects the overvoltage andconnects a dump resistance across the DC terminals.

The overvoltage problem can also arise with a PM motor if it is driven into an overspeed condition bythe mechanical system to which it is connected. In this case there is also a possibility of destroyingthe freewheel diodes which rectify the generated current and feed it to the DC link capacitor. This isone reason for caution in the use of “embedded magnet” motors, which have higher per-unit inductancethan surface-magnet motors, so that if the load is lost at high speed the generator terminal voltagetends to rise more than that of a surface-magnet machine.


With all generators the load current must be monitored, and if the current exceeds the generatorrating, appropriate protective measures must be activated. If the current is only a few percent abovethe rated value, it may be sufficient to do no more than display a warning light. More sophisticatedprotection would use inverse-time overcurrent relays and automatically disconnect the generator byopening a circuit-breaker. Generator protection may also include differential current relays (to detectinternal generator faults); over-temperature relays; and negative-sequence relays to protect againstexcessive unbalance between phase currents. In addition, the overall system will generally require aground fault protection scheme. One of the things to bear in mind with electrical generators is thatthere may be huge amounts of energy stored in the magnetic field and even more in the rotating mass,and the protection system must be designed, in general terms, to protect against its uncontrolledrelease.

REFERENCES

1. SPEED’s Electric Motors, the theory text that is used with the SPEED training courses.

2. Hendershot J.R. and Miller TJE : Design of brushless permanent-magnet motors, Oxford University Press,Monographs in Electrical and Electronic Engineering No. 37, 1994. ISBN No. 0-19-859389-9 (or in USA, 1-881855-03-1)

3. Mohan N, Undeland T.M. and Robbins W.P.: Power electronics: converters, applications, and design, JohnWiley & Sons, 1989, 1995 [2nd edition]. ISBN 0-471-58408-8.

4. Murphy J.M.D. and Turnbull F.G.: Power electronic control of AC motors, Pergamon Press, 1988. ISBN 0-08--22683-3.

5. Hague B : The principles of electromagnetism applied to electrical machines, Dover Publications Inc., N.Y.,1962.

6. Boules N : Prediction of no-load flux-density distribution in permanent magnet machines, IEEE Transactions onIndustry Applications, Vol. IA-21, No. 4, May/June 1985, pp. 633-643.

7. Jahns, T.M.: Torque production in PM synchronous motor drives with rectangular current excitation, IEEETransactions, IA-20, pp 803-813, 1984

8. Demerdash, N.A. with Arkadan, A., Nehl, T., Vaidya, J. and others: papers on brushless DC and AC motorsand drives, published in IEEE Transactions, including EC-3, Sept 88, 722-732; EC-2, March 87, 86-92; PAS-104,Aug 85, 2206-2213; 2214-2222; 2223-2231; PAS-103, July 84, 1829-1836; PAS-102, Jan 83, 104-112; PAS-101, Dec 82,4502-4506; PAS-100, Sept 81, 4125-4135; EC-3, Sept 88, 714-721; EC-3, Dec 88, 880-889; 890-898.

9. Fitzgerald AE and Kingsley C : Electric Machinery, McGraw-Hill (second edition)

10. Kostenko and Piotrovsky, Electric Machines, MIR Publishers

11. Rasmussen KF, Miller TJE, Davies JI, McGilp M and Olaru M : Analytical and numerical computation of airgapmagnetic fields in brushless permanent-magnet motors, to be presented at IEEE Industry Applications SocietyAnnual Meeting, Phoenix, Az, October 1999.

12. Reliance Motion Control Inc.: DC Motors, SPEED Controls, Servo Systems: The Electro-craft EngineeringHandbook, 6th edn.

13. Hughes A. and Miller TJE : Analysis of fields and inductances in air-cored and iron-cored synchronous machines,Proceedings IEE, Vol. 124, No. 2, February 1977, pp. 121-131.

14. Rasmussen KF, Analytical prediction of magnetic field from surface mounted permanent magnet motor, IEEEIEMDC Conference, Seattle, pp. 34!36, May 9!12, 1999.

15. Clayton AE and Hancock NN, The performance and design of direct current machines, 3rd edn., Pitman, London,1959-66, p. 36.

16. Kenjo T and Nagamori S, Permanent magnet and brushless DC motors, Sogo Electronics PublishingCompany, Tokyo, 1984.


Index

2Q 36AC brushless machines 1AC Vector control 25Airgap flux distribution 3Alignment torque 27, 67Armature reaction 17, 61, 66-68Back-emf

trapezoidal 3Back-EMF 1, 3, 4, 40, 43, 67, 68Back-EMF constant 4Back-EMF sensing 43Basic operation 3Bifilar 23, 44, 45, 47Blowers 1, 23, 45Boules 13-16, 76Brushed motors 1Brushless DC machines 1Chopping 4, 9, 37-42, 44, 47, 52, 54Classification of inductances 22Cogging torque 50, 53, 55-59Commutation 4, 8, 32, 36, 39-41, 43, 47, 54, 68Computer disk drives 1, 53Constant-torque locus 31Core loss 73Current regulation 36, 37Delta 4, 6, 8, 20, 36, 37, 45, 49, 62, 69Demagnetization 65, 67Differential leakage inductance 18, 21dq-axis transformation 18, 19Drive 4, 6-9, 23, 27-29, 33, 35-37, 42-46, 48-50, 52, 54, 61, 62,

67, 68Duty-cycle 9, 38-40Effective airgap 17, 20, 59Efficiency 1, 23, 24, 45, 47, 62, 64Electromagnetic excitation 62Electromagnetic torque 3, 4, 7, 27, 52, 53, 69, 72, 73Embedded-magnet 10, 11, 21, 53Exciter 1Fans 1, 23, 45Finite-element 10, 11, 16, 49-52, 55, 58, 60Finite-element analysis 11, 16, 55, 58Flux-MMF diagram 50, 52, 56Fringing 3, 12, 13Generated EMF 3, 10, 23, 24, 41, 46, 66, 67, 69, 70, 75Generators 1, 23, 32, 65, 69, 70, 75, 76Hague 13-16, 76Inductance 17-22, 27, 30, 31, 36, 38, 40, 43-45, 47, 49, 54,

60, 61, 70, 71, 75Infinite bus 70, 72, 75IPM 11, 22Laplace 13Leakage factor 11, 18Line-start 32Locked-rotor 9, 13Magnet 1, 1, 3, 4, 9-18, 20-23, 25-29, 31-34, 37, 48-61, 63-70,

75, 76Magnetic circuit analysis 10, 15Magnetization 10, 13, 14, 16, 25, 44, 55, 67, 73Magnets 1, 6, 9, 10, 13, 14, 25, 26, 31, 32, 44, 53, 55, 56, 59,

60, 62-66, 75

Nonlinear magnetic circuit 11, 12Nonsalient pole 20, 22, 24, 31, 32, 34, 43, 71No-load speed 9Period A 36Permanent magnet 23, 34, 63, 76Permanent magnets 1, 25, 53, 56, 60, 62Permeance coefficient 11, 18Phase advance 28, 30-32, 34, 35, 47, 67, 68Phase inductances 22, 49, 71, 73Phasor diagram 21-28, 37, 49-51, 67-70, 72-74Phasors 23, 26, 27, 35, 70, 72Power 1, 3, 4, 9, 24, 34, 36, 37, 39, 40, 43, 44, 53, 55, 60, 61,

63, 65, 66, 70, 72-76Power factor 24, 72, 74, 75Rasmussen 13-16, 76Rectifier 1, 33, 70-72, 74Reluctance torque 1, 27, 32, 33, 53, 54Saliency 1, 18, 20, 27-29, 32, 33Salient-pole 18, 21, 22, 24, 25, 28, 29, 31, 32, 34, 43, 49, 73,

74Salient-pole motors 18, 21, 22Self-synchronous 1, 32Servo-motors 32Shaft position feedback 1Sign conventions 51, 65Sinewave drive 35, 48Sinewave operation 23Sinewound 19-23, 36, 43, 49, 50, 52, 54, 60Single-ended drive 23Sliding contacts 1Slotless motors 65Slots 18, 19, 48, 49, 51, 57, 67Squarewave drive 33, 35, 37, 42, 43, 48, 52, 67Stall torque 9Stepper motor 1Surface-magnet 10, 11, 17, 21, 25, 28, 29, 31, 33, 50, 53, 75Switching strategy 35Synchronous inductance 18, 21, 27, 30, 60, 71Synchronous reactance 20, 24, 27, 31, 32, 71Time-stepping 18, 22, 35, 36, 74Torque 1, 3, 4, 7, 9, 16, 23, 27-35, 39, 44-47, 49-62, 65-67, 69,

72, 73, 75, 76Torque constant 4, 27, 28, 66Torque control 1Torque per ampere 33, 60, 61Torque ripple 16, 53, 54, 65Torque/speed characteristic 9, 34Traction 1, 62Trapezoidal 1, 3, 37, 54trapezoidal back-emf 3Turns in series per phase 3, 17, 20Unipolar drive 44Vector control 25Winding

inductance 17Windings 13, 18-20, 23, 24, 26, 27, 33, 37, 44, 45, 47-49, 54,

57, 62, 63, 71, 73Wye 4, 6, 7, 36, 37, 45, 69

3. Induction machines

3. 1 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.2 Equivalent circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3

3.3 Equivalent-circuit parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8

3.4 Winding factors and other winding-related matters . . . . . . . . . . . . . . . . . . . . . . . . . 3.10

3.5 Rotor and stator slot numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11

3.6 Slot permeance calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13

3.7 Location of ampere-conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16

3.8 Transients and eigenvalue analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.17

3.9 Split-phase motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.24

3.10 Cross-field theory of tapped-winding capacitor motor . . . . . . . . . . . . . . . . . . . . . . . . 3.30

3.11 Interbar currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.35

3.12 Saturation of Leakage Reactance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.35

Induction machines Page 3.1

Fig. 3.1 Sine-distribution of ampere-conductors Fig. 3.2 2-pole sine-distributed field

K (2,T t) ' K (sinp2 cos T t ! cosp2 sin T t) ' K sin (p2 ! T t) (3.1)

Ns '602B

×T

p'

120 × f

poles[rpm] (3.2)

3. INDUCTION MACHINES

3. 1 BASIC THEORY

Sine-distributed windings

A sine-distributed winding has a sinusoidal distribution of ampere-conductors around the stator bore,Fig. 3.1, and produces a sine-distributed flux in the airgap, Fig. 3.2.

Rotating magnetic field

The flux distribution of a single sine-distributed winding is fixed in space: in particular, its orientationis fixed along the winding axis. In Figs. 3.1 and 3.2 the winding axis is the x-axis, 2 = 0. The flux-densityat any point varies in proportion to the current (neglecting magnetic nonlinearity for the moment).

The effect of the current in a sine-distributed winding is closely approximated by a current-sheetaround the stator bore, K1 = K sin p2 [A/m], such that K d2 is the number of ampere-conductors in anarc d2 at the stator bore. p is the number of pole-pairs. In Figs. 3.1 and 3.2, p = 1. K1 can be interpretedas the product of the phase current i1 and the conductor distribution C sin p2. If the current variessinusoidally in time, i1 = Imax cos Tt, then the current sheet remains fixed in space but its amplitude atany point varies sinusoidally: thus K1(2, Tt) = K sin p2cos Tt.

2-phase winding: If a second "phase" is added, with its axis rotated 90E from that of the first phase, andif it has the same number of turns as the first phase, it will produce an ampere-conductor distributionorthogonal to the first one: thus K2 = K cos p2. The axis of phase 2 is the angle at which cos p2 = 0, andsince cos p2 = sin (p2 + 90E), this is at !90E/p. If phase 2 carries a sinewave current which leads i1 by90E in time, we can write i2 = Imax cos (Tt + 90E) = !Imax sin (Tt), and therefore K2 = !K cos p2 sin(Tt).The total ampere-conductor distribution is the sum of K1 and K2:

which is interpreted as a rotating ampere-conductor distribution. The rotation speed is T/p rad/s: or

since T = 2B f, where f is the supply frequency in [Hz]. This is the synchronous speed. If the polarityof one winding was reversed, the ampere-conductor distribution would rotate in the opposite direction.Similarly, if the current I2 was lagging I1 by 90E, it would reverse the direction of rotation.


K(2,T t) ' K sinp2 cosTt% K sin(p2 ! 2B/3) cos(Tt & 2B/3)% K sin(p2 % 2B/3) cos(Tt % 2B/3)

'32

K sin(p2 & Tt) .

(3.3)

Fig. 3.3 3-phase winding

3-phase windings: Three-phase windings are much morecommon in practice than 2-phase windings. Fig. 3 shows thearrangement of three sine-distributed phase windingsarranged symmetrically so that their axes are displaced by120E from each other. Balanced three-phase currents supplythese windings, so that the total ampere-conductordistribution is given by

This represents a rotating ampere-conductor distribution whose peak value is 3/2 times the peak valueproduced by any one phase. It rotates at T/p rad/s.

Multiple poles: The ampere-conductor distributions shown in Figs. 3.1!3. 3 are two-pole distributionsbecause there is only one North and one South. Windings can have any number of pole-pairs. Forexample, a four-pole winding (p = 2) has NSNS and a six-pole winding (p = 3) has NSNSNS.

Electrical and mechanical radians and degrees: In a P-pole winding the pole-pitch is 2B/P radians,i.e. B/p since p = P/2. These radians are actual or mechanical radians. However, it is logical to thinkof a pole-pitch as spanning B electrical radians, where one electrical radian is equal to 2/P mechanicalradians or 1/p mechanical radians. The synchronous speed is T electrical rad/s, or T/p mechanicalrad/sec. In rpm it is given by eqn. (3.2). Common examples are tabulated below:

Poles Frequency Speed [rpm] Frequency Speed [rpm]

2

50

3,000

60

3,600

4 1,500 1,800

6 1,000 1,200

8 750 900Table 3.1

SYNCHRONOUS SPEED VS. FREQUENCY AND NUMBER OF POLES

In order to achieve a synchronous speed of 24,000 rpm in a 2-pole motor, the frequency would need tobe f = 400 Hz. To rotate an 8-pole motor at 15 rpm, the frequency would need to be 1 Hz.

Rotor speed and frequency

In the induction motor, the rotor winding is usually short-circuited: that is, a closed winding that linksthe flux produced by the stator windings, but is not connected to the supply. The currents flowing inthe rotor are induced, and in order to obtain a driving EMF the rotor must rotate at some speed differentfrom synchronous speed. For if the rotor were rotating at synchronous speed, the flux linkage of therotor winding would not vary, and there would be no induced EMF.

The relative speed between the rotor and the rotating flux is called the slip speed. If the rotor speed isTr elec rad/sec, the slip is defined as

In general, the induction motor rotates slightly slower than synchronous speed and s > 0. The inductiongenerator rotates slightly faster than synchronous speed, and s < 0. In variable-frequency motor drives,negative slip is used for braking purposes because it reverses the torque. The rotor speed is alsoexpressed as Tr = (1 ! s)T or Nr = 120 f /P × (1 ! s) rpm. The relative speed between the rotating fluxand the rotor is T ! Tr = sT elec rad/s. Therefore the frequency of EMF's and currents induced in therotor is equal to s f ; i.e., the slip is the ratio of the rotor frequency to the stator frequency.

s 'T ! Tr

T[p.u.] . (3.4)


Fig. 3.4 Equivalent circuit of one phase of polyphase induction motor, with different frequencies in the primary andsecondary. The ideal transformer has the effective primary:secondary turns ratio n1/n2 × 1/s for voltage and n2/n1

for current. Here the circuit is drawn as though n1/n2 = 1. Rc accounts for the core loss, and Xm for the magnetizingcurrent.

In polyphase induction motors designed for high efficiency, the full-load slip is generally small (< 5%),but in small single-phase machines it can be higher than this. For example, a 4-pole motor running at2.5% slip from a 60-Hz supply is rotating at (1 ! 0.025) × 1,800 = 1,755 rpm. The slip speed is 45 rpm andthe rotor frequency is 0.025 × 60 = 1.5 Hz.

The slip-frequency EMF’s induced in the rotor drive slip-frequency currents in the rotor conductors,producing a secondary ampere-conductor distribution which rotates forwards in synchronism with thestator ampere-conductor distribution, i.e. at synchronous speed. The rotor ampere-conductordistribution therefore rotates forwards relative to the rotor, at the relative speed sT electrical rad/s.Because both ampere-conductor distributions rotate at synchronous speed relative to the stator, anyvoltages induced in the stator windings by will be at the line frequency f. There is only one resultantmagnetic flux, produced by the combined effect of the rotor and stator. Provided that the windings aresine-distributed, it rotates at synchronous speed.

3.2 EQUIVALENT CIRCUIT

The induction motor rotor is similar to the secondary of a transformer, with two major differences:

(a) the rotor is rotating relative to the primary, so the frequency in the secondary is s f ; and

(b) the airgap decreases the coupling between the primary and secondary, so that the leakageinductances are higher than in a typical power transformer. Consequently the magnetizingreactance is lower and the magnetizing current is higher.

In all other respects the equivalent circuit is the same as for the transformer, shown in Fig. 3.4 for onephase. The rotor frequency s f is usually in the range 0.5!5 Hz in normal operation.

To simplify analysis we refer all parameters to the primary (stator) winding. To do this we need thetransformation ratios for the voltage, current, and impedance. By Faraday's Law, the sinusoidalinduced voltage in each winding is proportional to the turns and the frequency. In a transformer thevoltage ratio is n1 f/n2 f = n1 /n2, the turns ratio, where n1 and n2 are the effective numbers of turns inthe stator and rotor windings, respectively: the frequency cancels because it is the same in bothwindings. In the induction motor the voltage ratio is n1 f/n2 sf = n1/sn2.

The current ratio is n2/n1, exactly as in the transformer : current balance is a magnetostatic effectcontrolled by Ampere's Law, and is unaffected by frequency.

For impedance the referral ratio is the quotient of the voltage and current ratios, i.e., n12/sn2

2.


Pgap ' 3I2

2 R2

s(3.6)

Pmech ' Pgap (1 & s ) . (3.7)

Tgap 'Pmech

(1 ! s)T'

p

TPgap [Nm]. (3.8)

Tshaft ' T & Tfric . (3.9)

Fig. 3.5 Equivalent circuit of one phase of polyphase induction motor. The frequency is the line frequency in both primaryand secondary, and the turns ratio is eliminated. The EMF Erb is used to represent leakage flux in saturated rotor slot-bridges, and does not contribute to torque production. Boxed variables appear in the phasor diagram, Fig. 3.9.

R2

s' R2 % R2

1s

! 1 . (3.5)

It is now possible to refer all voltages, currents, and impedances in Fig. 3.4 to the stator winding, as inFig. 3.5. All voltages and currents in Fig. 3.5 are at line frequency f, and all reactances have their line-frequency values. While the secondary EMF and reactance lose their dependence on slip, the rotorresistance becomes R2/s, and this is further divided into two components in Fig. 3.5, using

The first term represents I 2R loss, while the second term represents electromechanical powerconversion. The referred equivalent circuit (Fig. 3.5 ) shows that any increase in rotor current appearsas an increase in the stator current I1 drawn from the supply. It can be used to calculate all theimportant currents and voltages in the induction motor, as well as the torque, power, and efficiency.

Power, torque and efficiency

The current I2 flows in the referred rotor resistance R2/s and causes an apparent power dissipation of

where the 3 accounts for all three phases. This is the electromechanical power or airgap power, i.e. thepower transferred across the airgap from the stator to the rotor. The actual copper loss in the rotor,however, is only 3I2

2R2, and since s < 1 (in motoring operation), the airgap power exceeds the rotorcopper loss by 3I2

2R2/s !3I22R2 = Pgap (1 ! s). The electromechanical power is therefore

Since the actual rotor speed is (1 ! s) T/p rad/s, electromagnetic torque must be

The airgap power Pgap divides into two. A fraction s goes as rotor copper loss (heating the rotor). Theremaining fraction (1 ! s) is converted into mechanical power at a speed that is equal to (1 ! s) timesthe synchronous speed, producing the electromagnetic torque Tgap. The shaft torque Tshaft is less thanthe electromagnetic torque because of friction torque Tfric :


0 'Tm × Tshaft

3V1 I1 cos N1

× 100%, (3.10)

I2 'V1

(R1 % R2 /s ) % jXL

(3.11)

Tgap '3p

T×

R2

s×

V12

(R1 % R2 /s )2% XL

2[Nm] . (3.12)

T 6 3p

T×

V12

R2

× s , (3.13)

T 6 3p

T×

V12 R2

XL2

×1s

, (3.14)

The efficiency is now calculated as

where cos N1 is the power factor and Tm = Tr/p is the speed in mechanical radians/sec. Note that V1 isthe phase voltage, since the equivalent circuit is on a per-phase basis. Likewise, I1 is the phase current.If the motor is connected in wye and the phases are balanced, V1 = VL//3 and I1 = IL, where VL is theline-line voltage and IL is the line current. If it is connected in delta, V1 = VL and I1 = IL//3.

Effect of slip on torque, efficiency and power factor

Induction motors are usually designed to operate with low values of slip, typically 0.01!0.05. Thisincreases the fraction of airgap power that is converted to mechanical power, and decreases thefraction that is dissipated as rotor I 2R loss, helping to maximise the efficiency. A small value of sincreases the ratio (R2/s)/X2, which also improves the power factor.

The speed during normal motoring operation is just below synchronous speed, and it changes onlyslightly as the load torque changes. For this reason the induction motor is often described as a"constant speed" machine, provided that the supply frequency is fixed. From the equivalent circuit (Fig.3.5), if we neglect the no-load current Inl and lump the leakage reactances together as XL = X1 + X2,

If this is substituted into eqn. (3.6) for Pgap, then from eqn. (3.8) the airgap torque is given by

This equation relates the electromagnetic torque to the slip, when the supply voltage and frequencyare fixed. It can therefore be used to plot a graph of torque vs. slip. Since the speed in rev/min is givenby (1 ! s) Ns, the graph also shows the variation of speed with torque. Fig. 3.6 shows a typical example.

At speeds near synchronous speed, s 6 0 and

i.e., the torque is proportional to slip. Since the slip is generally small for torques which are less thanor equal to the rated load torque, the speed remains near synchronous speed as the torque varies. Theactual torque is determined by the load, and the operating point is at the intersection of the motor'storque/speed characteristic with the torque/speed characteristic of the load, Fig. 3.6.

If the speed rises above synchronous speed, the slip becomes negative and the torque changes sign. Thisis called regeneration or braking. If the induction machine is driven (for example, by a wind turbine)at speeds above synchronous speed, it generates and feeds power into the AC supply system.

At low speed, s 6 1 and XL tends to exceed (R1 + R2/s), so that

i.e., the torque is inversely proportional to slip at low speed. The locked-rotor torque is the value atstandstill, with s =1. This is the torque when the motor is first switched on.


sTmax 'R2

R12% XL

2(3.15)

Tmax '12

×3p

T×

V12

R1 % R12% XL

2[Nm]. (3.16)

Fig. 3.6 Torque/speed characteristic

Breakdown torque

Fig. 3.6 shows that there is a maximum torque Tmax, occurring at a value of slip slightly beyond thelinear section of the torque/slip curve. This value of slip can be estimated by differentiating the torqueequation with respect to slip, and setting the derivative to zero. The result is

The corresponding maximum torque is called the breakdown torque:

If the load is gradually increased above the rated load, the slip increases and the motor produces moretorque. The operating point moves up the torque/slip curve until the slip reaches sTmax. Any increaseof load torque then causes the motor to slow down still more, increasing the slip. Beyond sTmax themotor torque decreases, and the motor rapidly decelerates and stalls; this is called breakdown.

Induction motors are normally rated such that at rated voltage and rated frequency the rated torqueis roughly half the breakdown torque. This provides a margin of safety against stalling due to transientchanges of load torque, or undervoltage or underfrequency conditions. Since Tmax is proportional toV1

2, a 10% reduction in voltage produces a 20% reduction in breakdown torque.

Phasor diagram

The phasor diagram (Fig. 3.9) represents the steady-state operation of the equivalent circuit of Fig. 3.5.The equivalent-circuit parameters are normally calculated for AC sinewave operation with constantsupply voltage and balanced conditions. Iterative solution of the phasor diagram makes it possible toallow for nonlinearities such as saturation of the magnetizing reactance and leakage reactance, theeffect of slip on the rotor bar resistance, and the variation of core loss and stray loss with the flux level.However, it should be remembered that the phasor analysis is based on the fundamental space-harmonic MMF and is limited to sinusoidal waveforms of voltage and current.


Fig. 3.7 Varying R2 to achieve high starting torque Fig. 3.8 Varying the supply frequency at constant V/Hz

Speed control

Fixed supply: If the supply voltage and frequency are fixed, the speed can be controlled by varyingthe rotor resistance by means of an external resistor connected into the rotor circuit by slip-rings andbrushes. This technique is used in large wound-rotor induction motors, especially for controlling therate at which they start up. The effect is to change the torque/speed characteristic as shown in Fig. 3.7.A high resistance R2 maximizes the starting torque. As the rotor accelerates the external resistanceis shorted out and the characteristic changes to a "low-slip, high-efficiency" characteristic. Wound-rotormachines are expensive and are relatively less common than cage-rotor machines.

Pole amplitude modulation: With a fixed supply, the speed of a cage-rotor machine can be changedby reconnecting the stator windings in such a way as to change the pole number. For example,reconnecting the stator from 6 poles to 8 poles reduces the synchronous speed by 25%. Otherwise thereis no practical way to control the synchronous speed of cage-rotor machines on a fixed supply.

Variable voltage: Changing the voltage causes the torque to be scaled in proportion to V12, so the

operating point moves to a higher or lower speed as the slope of the torque/slip curve changes. Therange of speed variation is small, unless the motor is designed with a high rotor resistance, but thismakes the motor inefficient. This technique is used with single-phase and inexpensive triac controllers.

Variable frequency: The ideal way to control the speed of an induction motor is by varying the supplyfrequency. This causes the torque/slip curve to be translated along the speed axis, Fig. 3.8. If thevoltage/frequency ratio is kept constant ("constant volts/Hz"), the breakdown torque remains constantover most of the speed range; at lower speeds it tends to fall as the stator resistance begins to becomesignificant compared with the leakage reactance. With this type of drive the slip for a given torque canbe held constant while the speed is varied (almost proportional to frequency).

Modern field-oriented drives are capable of extremely rapid torque response. In principle they operateby orienting the stator MMF distribution at an optimal angle relative to the flux trapped by the rotorcurrents, and under transient conditions they are not limited to sinusoidal current. However, theequivalent circuit model is still the basis of analysis and design of these drives.

Double-cage and deep-bar rotors: Some induction motors are designed with a double cage. The innercage has a high leakage reactance and a low resistance, and the outer cage has a low reactance and ahigh resistance. The resulting torque/speed characteristic is similar to the sum of the high-resistanceand low-resistance curves in Fig. 3.7, providing high starting torque and low operating slip (thereforehigh efficiency) in the one motor. As an alternative to the double cage, skin-effect is used in the deep-barrotor to increase the rotor R/X ratio as the slip (and therefore the rotor frequency) increases.


1The classical references are Alger, Veinott, Richter and others. To understand these calculations it is virtually essential tostudy the original references, but some idea of the procedures can be seen in §3.6 which describes the calculation of slotpermeance. PC-IMD provides a range of optional methods for calculating the most important equivalent-circuit parameters,including most of the classical methods and a number of original ones.

V1

VR1

VX1VZ1

VR2

VX2

VZ2

Erb

E1

I1

I2

ER2

Irc

Imag

Inl

φ

Fig. 3.9 Phasor diagram for one phase of a balanced polyphaseinduction motor. The phasor Erb represents rotor slotbridge leakage.

3.3 EQUIVALENT-CIRCUIT PARAMETERS

Although the equivalent circuit (Fig. 3.5) is simple and elegant, the calculation of its parameters is aformidable and complex task. They are normally calculated from the motor geometry, the windingparameters, and the appropriate material properties. The phasor diagram for one phase of a balancedpolyphase induction machine corresponding to Fig. 3.5 is shown in Fig. 3.9.

To a certain extent it is possible to modify the equivalent circuit parameters to account for saturationof the main magnetic flux path (Xm); saturation of the leakage reactances (X1, X2 and Erb); deep-bar effect(especially X2); inter-bar currents (X2 and R2), and stray-load loss (R1 or Rc); and inverter harmonics(Rc). The equivalent-circuit and its parameters will remain approximately valid for space-vectoranalysis (as used in the analysis of field-oriented drives), provided that the windings are sufficientlysine-distributed. If the voltage and current waveforms are non-sinusoidal in time, then of course thelosses, saturation levels, and frequency-dependent impedances will be less accurate.

Several alternative methods are known forcalculating leakage reactance components,including belt-leakage, zig-zag, skew anddifferential components.1 Saturation ofmagnetizing reactance is calculated usingthe "60E" method (Alger [1970]; Say [1948])or a specially developed method with moredetail, (Ionel et al [1998]). Saturation ofleakage reactance is calculated by means ofErb, or by a modification of Norman’s [1934]method. (See also Boldea et al, [2000]).Deep-bar effect is calculated using a specialmethod that works with slots of any shape,(Boldea [1995]). Alternatively, the classicalrectangular-bar theory can be used,(Kostenko and Piotrovsky [1974]; Schuisky[1960]). For double-cage rotors the rotorbranch containing Erb, X2 and R2 isduplicated in parallel with the one in thecircuit of Fig. 3.5.

Closed rotor slots and Erb: For closedrotor slots Erb can be used to represent theflux through the bridges that close therotor slots. These bridges cannot bemodelled by a fixed component of the rotor slot permeance, because they tend to saturate. It is assumedthat there will be a sine-distributed component of the airgap flux that enters the rotor radially and thentravels circumferentially around the rotor, via the bridges. At a position 90Eelec from the point of peakairgap flux-density, the bridge flux density will be a maximum, and this density is arbitrarily assumedto be 2.1T. When this is multiplied by the cross-section area of the bridge, the resulting flux is acomponent of the fundamental flux in parallel with (and in phase with) the leakage flux through X2.

Alger’s equivalent circuit

The classical equivalent circuit model is based on the fundamental space-harmonic of the airgap MMF.To include the effect of phase belt and slot-permeance harmonics, Alger [1970] proposed an extendedequivalent circuit such as the one in Fig. 3.10, which includes two sets of phase-belt harmonics (5th and7th); the forward and backward slot permeance harmonics of order S1/p±1, and the forward andbackward slot-MMF harmonics of order S1/p±1, where S1 = stator slots and p = pole-pairs.


In a 3-phase motor the MMF wave produced by the 5th harmonic of the winding distribution rotates inthe backward direction at 1/5 synchronous speed, and the MMF wave produced by the 7th harmonicrotates in the forward direction at 1/7 synchronous speed. Likewise the MMF waves of the (6k ! 1)th

winding harmonic rotate backwards, and those of the (6k + 1)th winding harmonic rotate forwards.

In a two-phase motor the MMF wave produced by the 5th harmonic of the winding distribution rotatesin the forward direction at 1/5 synchronous speed, and the MMF wave produced by the 7th harmonicrotates in the backward direction at 1/7 synchronous speed. These directions are opposite to those inthe three-phase motor.

R1 1jX

jXm Rc

jX pm

jX 2m

1 + m(1!s)

R2m

jX2

R2

jX

RjXpn

2n

2n1 ! n(1!s)

1I I 2

I m

b

f

Second cage

harmonics

jX m5

jX 25

R256 ! 5s

jX m7

jX

R

27

277s ! 6

jX mm

jXR2n

1 ! n(1!s)

1 + m(1!s)

R2m

jX 2m

jX2n

b

b

f

f

Phasebeltharmonics

Slot MMF(zigzag)harmonics

1V

mn

Slot permeance

Fig. 3.10 Harmonic equivalent circuit for polyphase machines given by Alger [1970]


kpn ' sinnB"

2(3.17)

kdn '

sinnq(

2

q sinn(

2

(3.18)

ksn '

sinnF

2

nF

2

(3.19)

3.4 WINDING FACTORS AND OTHER WINDING-RELATED MATTERS

For standard concentric and lap windings these are defined as in the classical literature.

Pitch factor

where " is the per-unit coil pitch (i.e., the span in electrical radians divided by B), and n is the order ofthe harmonic. Sometimes this is expressed in terms of the “chording angle”, g = 1 ! ", in which case kpn= cos [gnB/2] for odd non-triplen harmonics.

Distribution factor

where ( is the slot pitch in electrical radians and q is the number of slots per pole per phase. This is usedonly with lap windings.

Skew factor

where F is the skew angle in electrical radians, and p = pole-pairs.

For fractional-slot windings the winding factor is obtained by Fourier analysis of the MMF distributionof the winding. The harmonic coefficients an and bn of the MMF are calculated for each individual coil.Then the an and bn are added together for all the coils, assuming that 1A flows in all the windings inseries. The resultant magnitude of the n'th harmonic MMF coefficient is cn = %(an

2 + bn2). The winding

factor, is the ratio of cn and the n'th harmonic winding factor of a "base" winding with the same numberof series turns distributed equally in full-pitch coils among the 2p poles. The "base" winding is assumedto start in slot 0, so that it has only sine coefficients Bn. Thus kwn = cn/Bn. Note that the phaseinformation in an and bn is lost in this process, so kwn is always positive, even with a winding for whichnegative values of kwn are possible. For example, in a 24-slot 2-pole motor we could wind two coils eachwith a span of 8 slots (i.e., 2/3 pitch) diametrically opposite to each other. (Coil 1 in slots 1!9; coil 2 inslots 21!13). This is a concentric winding for which " = 2/3 and so k5 = sin (5 × 2/3 × B/2) = !0.866. Theabove procedure gives a5/B5 = !0.836 and b5/B5 = !0.224, so that c5/B5 = +0.866. When the winding isskewed, the total winding factor for the n'th harmonic is obtained by multiplying kwn by ksn.

Conductivity of rotor bars

99.75% pure Al has conductivity 60—61% of that of pure electrical grade copper. After casting, the Alabsorbs iron and other impurities that reduce the conductivity to 58!59%. A figure as low as 50% isoften used. For copper cast rotors the conductivity might be typically 80!85%.

Current density in the stator copper

The "recommended" current density depends on the ventilation and cooling system. For a totallyenclosed fan-ventilated motor, like the general purpose 3-phase induction motor, a value of 6-7 A/mm2

is a reasonable starting point. This figure can be increased by 10-15% in drip-proof motors and by 25-30%in forced-ventilated motors. For totally enclosed naturally cooled motors a current density of3-3.5A/mm2 can be used for initial sizing, but the influence of the total surface area is important, as isthe airflow and the emissivity of the paint (for cooling by radiation).


3.5 ROTOR AND STATOR SLOT NUMBERS

For a given stack length, rotor diameter and electromagnetic loading (air-gap flux density and statorcurrent density) the number of stator slots has little influence on the stator resistance or the copperweight. The number of stator slots can then be used to control the value of the leakage reactance: itcontrols the differential component directly, and the slot component indirectly, through the ratio of slotdepth to slot width. Similar considerations are valid for the number of rotor slots.

To minimise the harmonic effects (including the differential leakage reactance) a "golden" rule is to usean integral number of stator slots per pole per phase (with a minimum value of 2), the choice beingmade, if necessary, to allow the use of special types of windings (such as the 5/6 chorded lap winding).

The number of rotor slots must be correlated with the number of stator slots to reduce harmonic effectssuch as parasitic synchronous and asynchronous torques, stray-load losses, vibrations and noise.

Alger [1970] states that the number of rotor slots should be 0.75!0.85 or 1.2!1.35 times the number ofstator slots in order to maximise the secondary zig-zag reactance for the stator slot MMF harmonics andto minimize other harmonic loss components. According to Alger, synchronous crawling torques are"present in 3-phase motors at ±Ns × Poles/Rotor slots, if the difference between stator and rotor slotsis 1,2, or 4 per pole; while standstill locking will occur if the difference is 3 per pole". (Ns = synchronousspeed). The choice of stator and rotor slot numbers is discussed in Heller and Hamata [1977] andKopilov et al [1980].

Two tables from these references with recommended numbers for stator and rotor slots are providedbelow. These values were derived from theory, mainly developed for line start machines, and practicalexperience. It should be noted that on this subject there are divergences between various authors ascan be seen for example by inspecting the two tables. The tabulated slot combinations should not benecessarily regarded as “extremely safe” and also it should be kept in mind that there are known casesof successful motor designs that do not fulfill these recommended combinations.

Poles Stator Slots Rotor Slots (Bars)

2

24 (16), [20], ([22]), (28), [30]

30 (16), [20], (22), [26], [34], [36]

4

36 [24], 26, [28], 30, ([32]), 42, (44), [46]

48 (32), 34, [36], 38, [40], ([44]), (56), 58, [60]

6

36 24, [26], [46]

54 38, 40, [44], [64], 66, [68]

8

48 34, [62]

72 50, 52, 54, [56], 58, 86, 88, [90]

TABLE 3.2SUITABLE COMBINATIONS FOR THE NUMBERS OF STATOR AND ROTOR SLOTS FOR SMALL AND MEDIUM-SIZE SQUIRREL-CAGE MACHINES

WITH 2-8 POLES AND AN OUTER DIAMETER UP TO 300MM

The number of rotor slots in round brackets are not suitable for reversible drives (because of largesynchronous parasitic torques in the braking region). The number of rotor slots in square bracketsmaybe used only if the rotor slots are skewed by one slot pitch. It should be noted that according to [3]:"no general rules exist for the choice of the number of slots which would be universally valid for smallas well as large machines".


Poles Stator Slots Rotor Slots (Bars)

Unskewed Skewed

2 12 91, 151 —

18 111, 121, 151, 211, 221 141, 182, 191, 221, 26, 281, 302, 31, 33, 34,35

24 151, 1612, 171, 19, 32 18, 20, 26, 31, 33, 34, 35

30 22, 38 182, 20, 21, 23, 24, 37, 39, 40

36 26, 28, 44, 46 25, 27, 29, 43, 45, 47

42 32, 33, 34, 50, 52 —

48 38, 40, 56, 58 37, 39, 41, 55, 57, 59

4 12 91 151

18 101, 141 181, 221

24 151, 161, 17, 322 16, 18, 202, 30, 33, 34, 35, 36

36 26, 44, 46 242, 27, 28, 30, 322, 34, 45, 48

42 342, 502, 52, 54 332, 34, 382, 512, 53

48 34, 38, 56, 58, 62, 64 362, 382, 392, 40, 442, 57, 59

60 50, 52, 68, 70, 74 48, 49, 51, 56, 64, 69, 71

72 62, 64, 80, 82, 86 61, 63, 68, 76, 81, 83

6 36 26, 46, 482 281, 33, 47, 49, 50

54 44, 64, 66, 68 42, 43, 51, 65, 67

72 56, 58, 62, 82, 84, 86, 88 57, 59, 60, 61, 83, 85, 87, 90

90 74, 76, 78, 80, 100, 102, 104 75, 77, 79, 101, 103, 105

8 48 342, 36, 44, 62, 64 35, 44, 61, 63, 65

72 56, 58, 86, 88, 90 56, 57, 59, 85, 87, 89

84 66, 682, 70, 98, 100, 102, 104 682, 692, 712, 972, 992, 1012

96 78, 82, 110, 112, 114 79, 80, 81, 83, 109, 111, 113

10 60 44, 46, 74, 76 57, 69, 77, 78, 79

90 68, 72, 74, 76, 104, 106, 108, 110, 112,114

70, 71, 73, 87, 93, 107, 109

120 86, 88, 92, 94, 96, 98, 102, 104, 106, 134,136, 138, 140, 142, 144, 146

99, 101, 103, 117, 123, 137, 139

12 72 56, 64, 80, 88 69, 75, 80, 89, 91, 92

90 68, 70, 74, 88, 98, 106, 108, 110 712, 732, 86, 87, 93, 94, 1072, 1092

108 86, 88, 92, 100, 116, 124, 128, 130, 132 84, 89, 91, 104, 105, 111, 112, 125, 127

144 124, 128, 136, 152, 160, 164, 166, 168,170, 172

125, 127, 141, 147, 161, 163

14 84 74, 94, 102, 104, 106 75, 77, 79, 89, 91, 93, 103

126 106, 108, 116, 136, 144, 146, 148, 150,152, 154, 158

107, 117, 119, 121, 131, 133, 135, 145

16 96 84, 86, 106, 108, 116, 118 90, 102

144 120, 122, 124, 132, 134, 154, 156, 164,166, 168, 170, 172

138, 150

TABLE 3.3RECOMMENDED COMBINATIONS OF STATOR AND ROTOR SLOT NUMBERS

1 used especially for fractional horse power machines. 2 might cause increased motor vibrations.


Fig. 3.11 Slot sections

Lc_slot ' µ0 Lstk N 2 (P1 % P2 ) (3.20)

3.6 SLOT PERMEANCE CALCULATIONS

The slot-leakage inductance of a single coil is given by an equation of the form

where N is the number of turns, Lstk is the stack length, and P1 and P2 are the permeance coefficientsfor the two slots in which lie the two coilsides. The permeance coefficient for each slot permits theinductance to be calculated as though all N conductors linked the same flux. In practice they do not:conductors towards the bottom of the slot link more flux than those towards the top. Therefore incalculating the permeance coefficient the distribution of flux within the slot must be taken into account.

When the slot can be considered to be made up of sections, that are segments of circles or trapezoids,Fig. 3.11, analytical expressions for P can be derived if it is assumed that the flux crosses the slot in thex-direction (that is, B = (Bx,0,0). The contribution )P of any section depends on the variation Bx(y)through the depth of the slot, and on the MMF of all sections lying below that section.

Consider an isolated section as in Fig. 3.11 (b) or (c) with an elementary flux tube dN. The MMF drivingthis flux element is JA(y), where J is the average current-density in the wound part of the slot: i.e., J =NI/Aw, where Aw is the total wound area. Then dN = µ0JA(y)Lstkdy/x = [µ0LstkNI/Aw] A(y)dy/x. The fluxdN is linked by the fraction A(y)/Aw of the total turns N, so the contribution dR to the total flux-linkageis equal to dR = dN.N A(y)/Aw = [µ0LstkN2I/Aw

2] A2(y)dy/x. We can write this as dP = (1/Aw2) A2(y)dy/x,

where dP is the contribution of the flux element to P.

The contribution )P of any whole section is obtained by integrating dP over the height h of the section.If there is another current-carrying section of area U below the current section, then we must integratedP = (1/Aw

2) [U + A(y)]2dy/x over h.

The feasibility of building up P in this way depends on the integrability of the expression for dP. Simple slots can be treated algebraically, using a few sections, but slots with more complicated shapesmay need to be divided into a large number of layers, each of which is calculated with eqn. (3.23).

As a simple example, )P is calculated for a slot bottom that is a circular segment spanning an angle 2$,Fig. 3.11(c). Since the section is at the bottom, U = 0 and dP = (1/Aw

2) [A(y)]2dy/x. It is convenient tointegrate with respect to 2 rather than y, so we write y = r(1 – cos 2); dy = r sin 2 d2; and x = 2r sin 2.A(y) is the sector area r2(22 – sin 22)/2, and Aw is also given by this formula with 2 = $. Making all thesesubstitutions and performing the integration with respect to 2 from 0 to $, we get

)P '$ [4$2 /3 % 1/2 % 2cos2$ ] & (5/4)sin2$

2[2$ & sin 2$ ]2. (3.21)


)P '2k

Aw2

B 2 lnw1

w0

%h

2(w1%w0 ){ B %

k

4(w 2

1 % w 20 )} (3.22)

)P '1

Aw2

U (U % a ) %a 2

3h

w. (3.23)

)P '13

h

w(3.24)

)P 'h

w1 ! w0

lnw1

w0

(3.26)

)P '2h

w0 % w1(3.27)

)P '4h

3w1 % w0

; w1 < w0 (3.28)

When $ = B/2, the slot-bottom is semicircular and )P = 0.1424. With $ = B we get the "classical" valuefor the slot permeance coefficient of a round slot, 0.6231.

Now consider the trapezoidal section, Fig. 3.11 (b). The area A(y) is written in terms of x as k(x2–w02)

where k = h/2(w1 – w0) and x = w0 + (w1 – w0)y/h, so that when dP is integrated with respect to x fromx = w0 to x = w1, we get the following expression (with B = U – kw0

2):

When the trapezium has parallel sides w1 = w0 = w, so k 6 4 and if a = hw, )P simplifies to

For a rectangular section at the bottom of the slot, U = 0; and if this is the only section there are noconductors above it, so Aw = a and

which is the well known formula for a rectangular slot. Another special case arises at the bottom ofa slot if w0 = 0; then the section is triangular and

Empty sections: For a section that is empty of conductor we must integrate dP = (1/Aw2) U 2dy/x over

h. For a trapezoidal section this gives

and if w1 and w0 are nearly equal this becomes

which is commonly quoted in textbooks. Veinott in his VICA-31 program for slot constants uses amodified form

in which w1 is equal to the slot opening and w0 is the width at the bottom of the slot wedge. By givingthree times more weight to w1 than to w0, he increases the value of )P and makes an allowance forfringing in a section of the slot where it is generally most significant. (See table 3.4).

Finally, the contribution of the slot opening region is given by eqn. (3.26) with w1 = w0 = w equal to theslot opening, and h equal to the depth of the tooth-tip: i.e., )P = h/w if there is no conductor in the slotopening. If there is conductor in the slot-opening, eqn. (3.22) is used; it gives a slightly lower result.

Closed slots: For slots closed at the top there is no formula for )P that gives a finite result, because thistheory assumes infinitely permeable iron. Closed slots, and even slots with significant saturation ofthe tooth tips, require a different treatment and their effective permeance depends on the slot current.

)P '1

Aw2

h 3 w1

16(3.25)


Fig. 3.12 Typical finite-element flux plots.

The permeance coefficients are a few percenthigher than those calculated on the assumptionthat the flux crosses the slot in parallel tubes.

Comparison with finite-element calculations

The analysis assumes that the flux crosses the slot in the x-direction with no fringing. In practicefringing increases the permeance, and finite-element studies of all the standard example slots in PC-IMDindicate that the analytical P is typically 10% low. Fig. 3.12 shows a typical flux-plot from this study,and the table summarizes the results. The permeance coefficient is calculated from the expressionE/µ0I

2, where E is the energy in J/m of axial length.

Bar type FE PC-IMD

1 1.98 1.8

2 2.01 1.88

3 2.36 2.2

4 2.95 2.73

5 2.41 2.3

6 2.72 2.57

7 4.44 3.63

8 3.78 3.41

9 2.03 1.85

10 2.57 2.3

Opencustom

2.42 2.16

Rectangularslot

2.166 2.167

Table 3.4

Comparison with VICA-31(Type 13 ER=0.5, DR=1.0, CR=0.455, BR=4.0, A1R=3.0)

VICA-31 FE PC-IMDSlot opening region empty

PC-IMD Slot opening full

Classical formulaKostenko & Piotrovsky

2.85 2.87 2.71 2.58 2.2

TABLE 3.5

Stator slots

Rectangular Round-bottomed

FE PC-IMD FE PC-IMD

2.03 2.03 1.97 2.02

TABLE 3.6

Deep-bar effect

For the deep-bar effect (skin effect in rotor conductors), PC-IMD has two alternative methods: one is theclassical method for a rectangular slot, and the other is an integration of the complex diffusion equationthroughout the slot, using a layered model developed for SPEED by Prof. I. Boldea. This is similar tothe analysis above, except that the integration of contributions from the layers in the slots is complex,to account for the change in phase of the current density throughout the conductor.


2 If the amp-conductor distribution is C cos 2 then the MMF distribution is the space-integral !C sin 2. The axis of the MMF

distribution is also the axis of the corresponding space vector.3 PC-IMD's phasor diagram shows only the current I2N (labelled as I2).

Fig. 3.13 Sine-distributed ampere-conductor distribution

Fig. 3.14 Distribution of ampere-conductors.

3.7 LOCATION OF AMPERE-CONDUCTORS

It is sometimes useful to be able to visualize how the ampere-conductors in the rotor and stator are distributed, and how they areoriented with respect to each other, especially in finite-elementcalculations.

Fig. 3.13 shows the basic model of a sine-distributed amp-conductordistribution. When a 3-phase winding is fed with balanced 3-phasecurrents, the fundamental ampere-conductor distribution rotates atsynchronous speed. The axis of the airgap MMF lags 90E behind the axisof the amp-conductor distribution.2

The current Im in the stator sets up a rotating ampere-conductordistribution ACm whose MMF axis is along the x-axis at the instantshown in Fig. 3.13. This distribution rotates in the positive direction(i.e., anticlockwise). The subscript ‘m’ means ‘magnetizing’.

A rotor conductor experiences a magnetic flux-density B (produced by Im) and rotates clockwise relativeto B at slip speed, so its velocity in the magnetic field is the vector v. The EMF v × B in this conductoris in the positive (dot) direction. It causes current to flow in the same direction. Because of the sine-distribution of B, the resulting induced current in the rotor also follows a sine-distribution, and it islabelled AC2. AC2 lags behind the rotor EMF distribution by a small angle because of the rotor leakagereactance in series with the rotor resistance. Note that the force F = i × B on the rotor conductor is inthe right direction to produce positive torque (counterclockwise).

The principle of ampere-turn balance (i.e., Ampère's law) requires that an equal and opposite amp-conductor distribution appear in the stator windings, to cancel the MMF produced by the rotor MMF I2.This "referred" MMF is I2N = !I2, and the corresponding ampere-conductor distribution is AC2N.

3

The total stator current is the sum of the originally postulated current Im plus the referred rotor currentI2N : thus I1 = Im + I2N. The resulting ampere-conductor distribution in the stator is AC1 = ACm + AC2Nas shown in Fig. 3.14.

Only two of the ampere-conductordistributions physically exist : AC1 and AC2.The other two, ACm and AC2N, are abstractcomponents of the stator current distributionwhich are used to explain the "mechanism" ofinduction. The ACm or Im component is the"magnetizing" component since it alone isresponsible for the mutual flux.

Fig. 3.14 shows that the rotor ampere-conductor distribution is largely in oppositionto the stator ampere-conductor distribution, asthough the rotor tries to cancel the fluxproduced by the stator. This must be so if themagnetizing current is to be small comparedwith the load current. Since the positiveampere-conductors of AC1 and the negativeampere-conductors of AC2 are nearly oppositeeach other across the airgap, they tend to forceflux in the peripheral direction around theairgap, i.e. into the leakage paths.


4The main ones are: 2r instead of 22 (rotor angle); subscripts d,q instead of 1d,1q; and subscripts D,Q instead of 2d,2q. Thephases are labelled a,b,c here, but in the SPEED software they are generally labelled 1,2,3 (lines being labelled a,b,c).

Fig. 3.15 Orientation of axes

3.8 TRANSIENTS AND EIGENVALUE ANALYSIS

PC-IMD can calculate starting transients. To model the capability of soft-starters with point-on-waveswitching, the AC supply voltage can be switched in two stages, so that the DC offset can be suppressed.This practically eliminates the oscillatory torque which occurs when the three lines closesimultaneously, and considerably lightens the duty on the shaft coupling, [Wood, 1965].

PC-IMD can also calculate the eigenvalues of the motor operating from a sinusoidal voltage source atzero slip over a range of frequencies from the normal line frequency down to zero. The resulting root-locus diagram is indicative of the relative stability of the motor when fed from a variable-frequencyinverter without shaft position or velocity feedback.

The theory of these dynamic calculations is presented here.

Reference Frame Transformations

PC-IMD’s dynamic calculations use two sets of d,q axes:

1. synchronously rotating (Kron), for eigenvalue analysis; and

2. d,q axes fixed on the rotor (Park) for transient analysis (starting calculations).

The conventions are the same as in Fitzgerald and Kingsley [1961], with only one or two minor changesof notation: see Fig. 3.15.4 Although this approach is rather old-fashioned it is practical and clear. PC-IMD does not include zero-sequence voltages or currents, so the theory works only for three-wireconnections and the transformation matrices are not square.

The transformation from (a,b,c) variables (i.e., direct phase variables) to dq axes is eqn. (3.29), where2 is the angle between the d-axis and the axis of phase a in elec. rad. The same transformation (and itsinverse, eqn. (3.30)) is used for voltages, currents, and flux-linkages.

vd '23

[va cos 2 % vb cos (2 ! 2B/3) % vc cos (2 % 2B/3)] ;

vq ' !23

[va sin 2 % vb sin (2 ! 2B/3) % vc sin (2 % 2B/3)] .(3.29)


va ia % vb ib % vcic '3

2[vd id % vq iq ] (3.31)

Ra ' Laa ia % Lab ib % Lacic % LaAiA % LaBiB % LaC iC

Rb ' Lba ia % Lbb ib % Lbcic % LbAiA % LbBiB % LbC iC

Rc ' Lca ic % Lcb ib % Lccic % LcAiA % LcBiB % LcCiC

(3.32)

LaA(2 ) ' LaA cos 2r

LaB(2 ) ' LaA cos (2r % 2B/3)LaC(2 ) ' LaA cos (2r ! 2B/3)

(3.33)

Rq ' L11 iq % L12 iQ. (3.35)

Rd '2

3[Ra cos 2 % Rb cos (2!2B/3) % Rc cos (2%2B/3)] ' L11 id % L12 iD (3.34)

RD ' L22 iD % L12 id ;RQ ' L22 iQ % L12 iq . (3.36)

Inverse:

At this stage it is not specified whether the dq axes are fixed to the rotor or to the stator or to the flux,or what. All that is necessary to specialize the transformation to one of the classic frames of referenceis to constrain 2 appropriately.

The total electrical instantaneous power at the terminals is

Flux-linkages and inductances

There are three phases a, b, c on the stator and three on the rotor, A,B,C. For the stator,

The self-inductances are constant and symmetrical: Laa = Lbb = Lcc; LAA = LBB = LCC. The stator-statormutual inductances are constant: Lab = Lbc = Lca = Lba = Lcb = Lac. The rotor-rotor mutual inductancesare constant: LAB = LBC = LCA = LBA = LCB = LAC.

The stator-rotor mutual inductances vary sinusoidally with rotor position 2r: for example,

A similar set of equations applies for LbA(2), LbB(2), LbC(2),LcA(2), LcB(2), and LcC(2), with appropriateangles taken from Fig. 3.15. Also an equation similar to eqn. (3.32) is written for the rotor flux-linkagesRA,RB, and RC.

The three stator phases will be replaced by two fictitious, orthogonal coils d and q fixed to (and rotatingwith) the d- and q- axes respectively. The flux-linkage of the d-coil is obtained by applying the voltagetransformation to Ra, Rb and Rc :

where L11 = Laa ! Lab and L12 = 3/2 LaA and much algebra has been omitted. Similarly in the q-axis,

For the rotor D and Q coils, the transformation is the same but with 2r instead of 2; thus

where L22 = LAA ! LAB, and again L12 = 3/2 LaA.

va ' vd cos 2 ! vq sin 2 ;vb ' vd cos (2 ! 2B/3) ! vq sin (2 ! 2B/3) ;vc ' vd cos (2 % 2B/3) ! vq sin (2 % 2B/3) .

(3.30)


va ' Ra ia % pRa ; vb ' Ra ib % pRb ; vc ' Ra ic % pRc (3.37)

vd ' Rd id % pRd ! p2 .Rq ;vq ' Rq iq % pRq % p2 .Rd . (3.38)

Fig. 3.16 d-axis aligned with phase a

Physical attributes of the fictitious d and q coils

Consider the instant when the d-axis is aligned with the axisof phase a, and ia = I, ib = ic =!½I in the 3-phase winding,Fig. 3.16. According to the reference frame transformationid = I. If the flux-linkage in phase a is Q, the flux-linkages ofphases b and c must be !½Q each, so that by the referenceeframe transformation Rd = 2/3 × Q × [1 !½(!½)!½(!½)] =Q. The flux-linkage per ampere in the d-coil, is Rd/id = Q/I.The d-coil evidently has the same number of turns as the a-coil. If we consider I to be a DC current then the voltagedrop in phase a is va = RaI and the voltage-drops in phasesb and c are vb = vc = !RaI/2. According to the reference frame transformation, vd = 2/3 × RaI × [1!½(!½)!½(!½)] = RaI. The resistance of the d-coil is evidently vd/id = Ra. This is consistent withhaving the same number of turns, provided that the total cross-section of copper in the conductors ofthe d-coil is the same as that of the a-coil.

Consider the stator flux-linkage produced by the stator currents alone at the position shown in Fig. 3.16,2r = 0. Ra is made up of LaaI due to self-flux-linkage, plus (!½I) × Lab mutually coupled from each ofphases b and c, for a total of (Laa ! Lab)I = L11I. The flux-linkage of phase b is Laa × (!½I) due to selfflux-linkage, with Lab × (!½I) mutually coupled from phase c, and LabI mutually coupled from phasea, for a total of !½L11I. The flux-linkage of phase c is the same. So Rd = 2/3 × L11I ×[1+(!½)×(!½)+(!½)×(!½)] = L11I. According to the transformation equation the d-coil current is I, sothe apparent inductance is L11. The d-coil represents all three stator phase coils along the d-axis,including their mutual coupling.

Now consider the stator flux-linkage produced by the rotor currents alone, with current I flowing in coilA and !½I in coils B and C. The flux-linkage of stator coil a is Ra = LaAI from coil A and !½LaA × (!½I)from each of coils B and C, for a total of LaAI × [1 + (!½) × (!½) + (!½) × (!½)] = 3/2 × LaAI. The flux-linkage of stator coil b is Rb comprising !½LaA × I from coil A, LaA × (!½I) from coil B, and !½LaA ×(!½I) from coil C for a total of !¾LaAI. The flux-linkage of coil c is the same, so that according to thereference frame transformation Rd = 2/3 × [3/2 + (!½) × (!¾) + (!½) × (!¾)] × LaAI = 3/2 × LaAI.According to the transformation the current iD is I × 2/3 × [1 + (!½) × (!½)+(!½) × (!½)] = I, so thatthe apparent mutual inductance is 3/2 × LaA.

Voltage equations

For the stator

where p is the operator d/dt, and we have assumed that Ra = Rb = Rc. For the stator d- and q-coils thevoltage equations are derived by incorporating eqns. (3.37) into the reference-frame transformationequations for vd and vq. When p operates on the trigonometric functions of 2, we get the speed voltages.Missing out a lot of algebra,

In synchronously rotating axes p2 = T = 2Bf. In d,q axes fixed to the rotor, p2 = Tr, i.e., the rotor angularvelocity in elec rad/s. Note that Rd = Rq = Ra.

A similar procedure applies to the D- and Q-coils on the rotor, except that they are themselves rotatingat angular velocity Tr, the physical angular velocity of the rotor in elec rad/s. The result is


vD ' 0 ' RD iD % pRD ! p2s .RQ

vQ ' 0 ' RQiQ % pRQ % p2s .RD . (3.39)

Te '32

P

2[Rd iq ! Rq id ] [Nm]. (3.40)

Te ' JdTr

dt×

2P

% DTm (3.41)

pRd ' vd ! Rd id % TRq

pRq ' vq ! Rq iq ! TRd

pRD ' ! RD iD % (T ! Tr )RQ

pRQ ' ! RQiQ ! (T ! Tr )RD

pTr 'P

2J[Te ! DTm]

(3.42)

va ' vpk cos (T t % * ) ;vb ' vpk cos (T t % * ! 2B/3) ;vc ' vpk cos (T t % * % 2B/3) .

(3.43)

vd ' vpk cos (T t ! 2 % * ) ;vq ' vpk sin (T t ! 2 % * ) . (3.44)

vd ' vpk cos * ;vq ' vpk sin * . (3.45)

where p2s is the slip velocity in elec rad/s, i.e. the angular velocity of the d,q axes relative to a pointfixed on the rotor. Thus in synchronously rotating axes p2s = T!Tr, with T = 2B f, but in d,q axes fixedto the rotor, p2s = 0. Note that RD = RQ = RA.

The electromagnetic torque is determined from the power associated with the stator current and thespeed voltages, divided by the angular velocity in mechanical rad/s. If P is the number of poles,

The dynamical equation of motion is

where Tr is in elec rad/s and Tm = Tr × 2/P is the rotor speed in rad/s. D is a viscous damping term.

Transient simulation

Transients are calculated by integrating the differential eqns. (3.38),(3.39) and (3.40) in the form.

where T is the velocity of the rotating reference frame in elec rad/s. At the end of each timestep, newvalues of Rd, Rq, RD, RQ, and Tr become available and the currents must be updated by solving eqns.(3.34!3.36). Likewise the torque must be updated by means of eqn. (3.40). A typical timestep would be0.002 s, but it depends on the time-constants of the particular motor. The system of equations is non-linear because of products like Rdiq in the torque equation and TRd in the voltage equation.

If the d,q axes are fixed to the rotor, the speed voltages in eqns. (3.42) vanish, since T = Tr. Theintegration is "driven" by the applied voltages vd and vq, which are defined in terms of va, vb, and vc.while the flux-linkages and currents are "outputs". For example, suppose the supply is defined by

where T = 2B f and * is an arbitrary phase angle. According to the transformation equations

If 2 = 2r the d,q axes are fixed to the rotor. If 2 = Tt they rotate at synchronous speed, with T = 2B f.

Then vd and vq are constant. In particular, if * = 0, vd = vpk and vq = 0.


vab ' vpk 3 cos (T t % B/6) ;

vbc ' vpk 3 cos (T t ! B/2) ;

vca ' vpk 3 cos (T t % 5B/6) .

(3.46)

LLL ' Laa % Lbb ! 2Lab ' 2L11 . (3.47)

id '2

3ia [cos 2 ! cos (2 ! 2B/3)] ' 2 ia / 3 cos (2 % B/6) ;

iq ' ! 2

3ia [sin 2 ! sin (2 ! 2B/3)] ' !2 ia / 3 sin (2 % B/6)

(3.48)

pRD ' !RD iD . (3.49)

Fig. 3.17 Non-simultaneous switching of the supply phases

Non-simultaneous switching of the phases

Fig. 3.17 shows the connection of the motor to the supply. The contacts in line a are assumed to bealready closed, with no current flowing. Those in line b close at the "point on wave" Tt = 21, and linec closes at the angle Tt = 21 + 22; that is, after a delay 22/T measured from the closure of line b.

With line-neutral voltages given by eqns. (3.43), then with * = 0 the line-line voltages are

During the interval 21 # Tt # 22 vc is undefined, but ic = 0 and ia = !ib. Current flows in the loop ab. Notorque is produced, and the rotor remains stationary. A particular case of interest is to close line b ata peak of vab, that is, when Tt = !B/6, and then close line c after a further delay of 90E (21 = 0; 22 = 90E).This strategy eliminates most of the oscillatory component in the transient torque, [Wood, 1965].

During the delay, stator current flows only in the loop through phases a and b. The line-line resistanceis RLL = 2Ra and the line-line inductance LLL is

Obviously current is induced in the rotor. The rotor circuit can be reduced to a single circuit by fixingthe d-axis to the rotor and aligning it with the axis of the effective stator coil ab that results from theloop current flowing in lines a and b. Then the whole circuit is reduced to that of a single-phasetransformer. With ic = 0 we have

and the required alignment is achieved by setting 2 = 2r = !B/6, with id =2 ia/%3 and iq = 0: that is, allthe stator current is in the d-axis and none in the q-axis.

On the rotor, the Q coil is orthogonal to the axis of the effective stator coil ab and so RQ = iQ = 0, whileeqn. (3.42) holds for RD with T = Tr = 0:

The mutual inductance between the rotor D-coil and the stator d-coil is L12 = 3/2 × LaA. Since the d- anda- coils have the same number of turns, the mutual inductance between the D-coil and phase a is also3/2 × LaA = L12 when their axes are aligned. With 2 = 2r = !B/6 this mutual inductance is decreased to


RLL ' LLL ia % 3L12 iD

RD ' L12 id % LD iD

(3.50)

Rd ' L11 id % L12 iD ;RD ' L12 id % LD iD . (3.52)

pRLL ' vLL !3

2RLL id . (3.53)

Fig. 3.18 Alignment of dq axes with the conducting loop in phases a and b.

RLL ' Rab ' Ra ! Rb ' 3 Rd (3.51)

L12 cos (!B/6) = %3/2 × LaA. The axis of phase b is at an angle 5B/6 relative to the d-axis, so that withcurrent in the stator loop ab and ib = !ia phase b makes an equal contribution to Rd. Therefore the totalmutual inductance MD-LL is 2 × L12 × %3/2 = %3 L12, and

where RLL is the flux-linkage of the series connection of phases ab. Using the inverse reference-frametransformation it can be shown that

and since ia = /3/2 id and LLL = 2 L11, we can write eqns. (3.50) as

This agrees with eqns. (3.34) and (3.36), since LD = L22 . The voltage equation for the stator loop is

The solution during the delay interval can therefore proceed by integrating eqns. (3.49) and (3.53),updating the currents at each timestep using eqns. (3.50); there is no torque and no rotation. After linec closes, all three motor terminal voltages are known: eqn. (3.45) can be used for vd and vq, and thesolution can proceed in d,q axes as before, provided that the final currents at the end of the delay aretransformed into initial values for id and iq.


vd ' Rd id ! TRq ' Rd id ! T (L11 iq % L12 iQ) ;vq ' Rq iq % TRd ' Rq iq % T (L11 id % L12 iD ) . (3.54)

ia ' [id cos 2 ! iq sin 2 ] ' id cos T t % iq cos (T t % B/2) . (3.55)

Vd % jVq ' [R1 % jTL11 ] (Id % jIq ) % jTL12 (ID % jIQ) (3.56)

0 ' [R2 % jsTL22 ] (ID % jIQ) % jsTL12 (Id % jIq ) . (3.57)

a51 '32

(P/2)2

J[ iq0 !

Rq0

LsN] (3.59)

a52 ' !32

(P/2)2

J[ id0 !

Rd0

LsN] (3.60)

Steady-state operation

The theory so far assumes that we know the inductances L11, L12 etc., but we need to relate these to thefamiliar "equivalent circuit" parameters of the induction motor: R1, R2, X1, X2, and Xm. In the steadystate the terms in eqns. (3.42) containing the p operator are zero so that

We can combine these equations by writing V = Vd + jVq where Vd = vd/%2 and Vq = vq/%2 and similarlywith the currents. This can be seen as follows: if 2 = Tt then

From eqn. (3.54) it follows that

where R1 = Ra and L12 = 3/2 × LaA. We can write I2 = ID + j IQ directly for the "rotor current", whilefor the "stator current" I1 = Id + jIq. The construction of the equivalent circuit requires also theequivalent of eqn. (3.56) for the rotor, i.e.

where R2 = RA = RD. Eqns. (3.56) and (3.57) represent the conventional equivalent circuit and so wehave done enough to show that Rd = Rq = Ra; L11 = [X1 + Xm]/T, L22 = [X2 + Xm]/T, and L12 = 3/2 × LaA

= Xm/T. Note the appearance of slip s in eqn. (3.57): by dividing this equation throughout by s, therotor circuit is "referred" to the stator with the same frequency T and the familiar R2/s appears as theonly manifestation of rotation.

Small-signal analysis

For this we assume small perturbations (denoted by )) about a steady operating point (denoted bysubscript 0): vd = vd0 + )vd; vq = vq0 + )vq; id = id0 + )id; iq = iq0 + )iq; iD = iD0 + )iD; iQ = iQ0 + )iQ; Rd =Rd0 + )Rd; Rq = Rq0 + )Rq; RD = RD0 + )RD; RQ = RQ0 + )RQ; Tr = Tr0 + )Tr; and Te = Te0 + )Te. T is missingbecause it is considered fixed and equal to 2Bf, but Tr can vary. If we substitute these into the voltageeqn. (3.42) and the torque eqn. (3.40) and simplify by subtracting the "steady-state" components, whileignoring products of the form )x)y, then after some grinding we get

p)Rd !1/JsN T kr/JsN 0 0 )Rd )vd

p)Rq !Tr0 !1/JsN 0 kr/JsN 0 )Rq )vq

p)RD = ks/JrN 0 !1/JrN T!Tr0 !Rq0 )RD + 0 (73)

p)RQ 0 ks/JrN 0 !1/JrN Rd0 )RQ 0

p)Tr a51 a52 a53 a54 !D/J )Tr 0

where


a53 '32

(P/2)2

J

kr

LsNRq0 (3.61)

a54 ' !32

(P/2)2

J

kr

LsNRd0 (3.62)

vd0 ' Rd id0 ! TRq0 ; Rq0 ' L11 iq0

vq0 ' Rq iq0 % TRd0 ; Rd0 ' L11 id0(3.63)

id0 'TL11 vq0 % Ra vd0

Ra2% T2 L11

2; iq0 ' !

TL11 vd0 ! Ra vq0

Ra2% T2 L11

2 (3.64)

Fig. 3.19 Equivalent circuit of split-phase motor. During start-up, the cut-out switch switches from "start" to "run" at a certainpercentage of the synchronous speed. The OC (open-circuit) position of the cut-out switch simply represents here theelectrical connections of a single-phase motor.

and JsN = LsN/R1, JrN = LrN/R2, LsN = (LsLr ! M2)/Lr, LrN = (LsLr ! M2)/Ls, ks = M/Ls, kr = M/Lr, and P isthe number of poles. The notation has been changed using Ls = L11; Lr = L22, and M = L12 so that thematrix eqn. (3.58) can be more easily compared with Vas' eqn. 2.10-36 [9] which is derived using space-vector theory.

In the calculation of the coefficients a51..a54, the determination of Rd0 and Rq0 generally requires theinversion of a 4 × 4 matrix but this can be avoided if the eigenvalues are calculated for the zero-slipcondition. This makes iD0 = iQ0 = 0 and T!Tr0 = 0. From eqn. (3.57) the steady-state conditions are

for which the solution is

3.9 SPLIT-PHASE MOTORS

Fig. 3.19 shows the equivalent circuit for a split-phase induction motor and Fig. 3.20 shows a simplifiedform of the phasor diagram for balanced operation. Split-phase motors are calculated by the methodof forward and backward rotating fields (Veinott [1959], Morrill [1929]); or by the method of symmetricalcomponents (Fitzgerald and Kingsley [1961]); or by the cross-field method. The harmonics of the MMF

distribution are ignored.


F ' im0Nmcos2cosTt %

% ia0Nacos(2%.)cos(Tt%")

'1

2im0Nm[cos(2%Tt)%cos(2!Tt)]

%1

2ia0Na[cos(2%Tt%.%")%cos(2!Tt%.!")]

(3.65)

F '1

2Fm[cos(2%Tt)%cos(2%Tt%.%")]...[b]

%1

2Fm[cos(2!Tt)%cos(2!Tt%.!")]...[f] (3.66)

" % . ' B (3.67)

a 'Na

Nm

'Im

Ia

'Va

Vm

(3.68)

Fig. 3.20 Phasor diagram for split-phase induction motor. The axes of the main and auxiliary windings are displaced by 90E(elec).

The auxiliary winding is connected in series with an auxiliary phase-shifting impedance. If the speedis below the cut-out speed, this auxiliary impedance is usually that of a start capacitor Cstart. Above thecut-out speed it is usually zero or Crun. The cut-out speed is typically 70% of synchronous speed.

Balancing theory

Total MMF distribution; conditions for balance: The airgap MMF distribution is given by the followingequation in which "m" and "a" refer to the main and auxiliary windings, respectively; . is the anglebetween the winding axes, and " is the phase difference between the currents ia and im :

Let im0Nm = ia0Na = Fm : then

For the backward component b to be zero,

Then if . = B/2, " = B/2, i.e., the auxiliary current must lead the main current by 90E. In this case theMMF reduces to . F ' Fmcos(2!Tt)

Let


I a 'I m

ae j" ; V a ' aV me j" (3.69)

I a ' jI m

a; V a ' jaV m (3.70)

V m ' V s ' V a % V c (3.71)

e !j " ! a ' !jXc

aZm

e !j N (3.72)

cos " ! a ' !Xc

aZm

sin N ; sin " 'Xc

aZm

cos N (3.73)

Xc 'aZmsin.

cos N(3.74)

a ' sin. . tanN ! cos. (3.75)

Xc 'aZm

cos Nand a ' tan N (3.76)

V c ' V m ! V a ' V m[1 ! ae j"] (3.77)

Vc ' Vm 1 % a 2 (3.79)

then

and if . = B/2 then

These relationships are shown in the phasor diagram, Fig. 3.20.

Capacitor reactance and turns ratio: The circuit connection (Fig. 3.19) makes

If we write and , eqn. (3.71) reduces to V m ' ZmI m ' ZmIme jN V c ' !jXcI a

Taking real and imaginary parts,

Taking the squares of these equations. and adding them, and substituting " = B ! ., we get

and if the first of eqns. (3.73) is divided by the second we get

If . = 90E then

The turns ratio can be 1 only if N = 45E, i.e. if the motor power factor is 1/%2.

Capacitor voltage

If . = B/2 then

and

V c ' V m[1 ! ja] (3.78)


Qc 'Im

2 Zm

sin N(3.83)

Is ' Im1 % a 2

a'

Im

sin N(3.84)

Z s ' Zs e j Ns 'V s

I s

'V m

Im(1 % j/a)

'Zme jN

(1 % j/a)

'aZm

a 2% 1

e j[N ! tan!1(1/a)]

(3.85)

Qc ' VcIa ' Vm 1 % a 2 ×Im

a' Im

2 Zm1 % a 2

a(3.80)

Pin ' 2Im2 ZmcosN (3.81)

Qc ' Pin .1 % a 2

2a cos N' Pin .

1 % a 2

2a(3.82)

Capacitor reactive power (with . = B/2)

In the balanced condition the total input power to the motor is

But so anda ' tan N cos N ' 1/ 1 % a 2

This has a minimum value Qc = Pin, if a = 1. The value of Qc is not unduly sensitive to the turns ratio.Another expression for Qc is

Supply current (with . = B/2)

The supply current does not have a minimum value. For all practical cases (N < B/2), the supply currentexceeds the main phase current.

Supply power factor (with . = B/2) :The apparent impedance at the primary terminals is

In general a = tan N , so in the special case when a = 1, N = 45E, then Ns = 0 and the supply power-factoris 1. To minimize the capacitor reactive power and maximize the supply power-factor, we require a =1, i.e., a motor with a power-factor of 1/%2.


5The desirability of this approach was recognized 60 years ago, but it was impractical on desktop computers until recently.6 In principle the models could be derived from one another by reference-frame transformations of generalized-machine theory.

Fig. 3.21 Single-phase equivalent circuit, modified with core-loss resistance

Forward and backward revolving-field theory

The calculation methods available in PC-IMD include the forward- and backward-rotating field methodof Morrill, [1929]; the symmetrical-component method; and the cross-field method. All of these aredescribed by Veinott [1959] in their basic form, but PC-IMD implements them with a core-losscalculation in which the resistance Rc is computed iteratively until the circuit equations converge. Thecore losses are calculated independently from the flux-density waveforms and are represented as anelectrical loss included in the equivalent circuit model.5 Secondary effects such as the deep-bar effect,saturation of the magnetizing reactance, core losses and stray-load losses are not equally easilyincorporated in all three methods, and PC-IMD’s calculations are all independently based on originalanalysis with several ad hoc modifications and improvements.6

In PC-IMD the forward- and backward-rotating field model and the cross-field model are furtherextended to cover tapped-winding motors of various configurations.

Fig. 3.21 shows the circuit representation of the pure single-phase motor using forward and backwardrotating fields. The impedance Zf is the parallel combination (jXm/2 ) 2 (R2/s + jX2)/2, and Zb is theparallel combination (jXm/2 ) 2 (R2/(2 ! s) + jX2)/2, in accordance with the fact that the actual pulsatingairgap flux in a single-phase motor is resolved into two equal counter-revolving fields, such that halfthe voltage induced by the airgap flux is due to the forward field and half to the backward field. AsVeinott puts it, half the mutual reactance is charged to the forward field and half to the backward field.

Fig. 3.22 shows the rotating-field model of Morrill [1929] as used with the capacitor motor. This methodmixes direct phase variables (in the main and auxiliary windings) with the forward and backwardrotating components. The currents Im and Ia are the actual currents in these windings.

Fig. 3.23 shows the symmetrical-component model. The main and auxiliary currents are related to thepositive and negative sequence currents by Im = Ip + In and Ia = j (Ip ! In) /a, where a is the effectiveauxiliary/main turns ratio. Likewise the main and auxiliary winding voltages are given by Vm = Vp +Vn and Va = j (Vp ! Vn) /a.

Most of the analysis used in these circuits has been published in IEEE and/or ICEM papers (see Miller,Gliemann, Rasmussen and Ionel [1998], and Rasmussen and Miller [2000, 2001]). The most importantcase is the cross-field theory of the tapped-winding motor, which is described in the next section.


Z b

I m

E 2m

z 1a

E bm

Rc

E fm Z f

Ic

a2Z f

SplitPh.wpg

VaVm

I az 1m I2

a2Z b

E fa!ja

jaE fm

E ba

!jaE bm

E fa

E baja

Z c

Fig. 3.22 Rotating-field model of capacitor motor, including core-loss resistance

Zb

I p! In

E b

RcfE f

Zd

Icb

Vm

z 1m

Vp

Vn

I n

I fz 1m

Z f

Zc

Zn

Zp

Rcb

Icf

I p

I b

Fig. 3.23 Symmetrical-component model of capacitor motor


7This makes the per-unit speed S equal to (s ! 1) instead of (1 ! s), where s is the slip.

Fig. 3.24 Tapped-winding capacitor motor circuits. (a) Base circuit analyzed; (b) G-tap circuit, [1].

3.10 CROSS-FIELD THEORY OF TAPPED-WINDING CAPACITOR MOTOR

Circuits

Fig. 3.24(a) shows the generic circuit of a tapped-winding capacitor motor, in which the auxiliarywinding is fed from a tapping on the main winding. Positive rotation is in the counter-clockwise (CCW)direction, and the axis of the main winding is retarded 90E relative to the axis of the aux winding, so thatthe aux winding current and voltage normally lead the main winding voltage and current in phase, andthe rotor rotates from the aux towards the main. The directions of positive current (closed arrows) andpositive MMF and flux (open arrows) are shown in Fig. 3.24(a), and the polarities of the winding sectionsare denoted by the customary dots. The capacitor is normally connected in series with the aux winding,but provision is made in Fig. 3.24(a) for a capacitor also in series with the main winding. The tappingis represented by the tap ratio t. When t = 0 the tap is at full voltage, but when 0 < t < 1 the tap is at alower voltage.

Fig. 3.25(a) shows a more flexible circuit that can represent several different configurations of tapped-winding capacitor motor connections. The intention is to have a basic “core” analysis for the circuitof Fig. 3.24(a), and to transform all the other circuits into this circuit for analysis.

As an example, the Grundfos or G-tap connection in Fig. 3.25(b) is reproduced in Fig. 3.24(b). Thiscircuit can be transformed into the base circuit by exchanging the “aux” and “main” labels andreflecting it about a vertical line. Mathematically this reflection is the same as reversing the directionof rotation.7 The exchange of labels means that the impedances must be exchanged before the analysisstarts, and the appropriate currents must be exchanged after it finishes. Note that the “main” capacitorwill be used in the analysis, and the “aux” capacitor will be short-circuited.

Other proprietary configurations can be modelled by setting the tapping parameters x, y and t in Fig.3.25(a), assigning the capacitor to the "main" or "aux" winding, and (if necessary) switching the per-unitspeed from positive to negative, so that the voltages and currents of the "aux" winding always lead thoseof the main winding.


Fig. 3.25 Tapped-winding capacitor motor connections.

(a) General case

(b) G-tap configuration

In all cases, the aux winding axis is assumed to be 90Eelec ahead of the main winding axis.


9 Flux associated with the reactance !t(1 !t)Xm does not link the auxiliary winding or any rotor circuits, and therefore does notgenerate any speed voltages. This reactance is merely an artefact of the T-equivalent circuit and accounts for the couplingbetween the tapped and untapped sections of the main winding.10The sign convention for the rotor currents is opposite to that used in [3]. All coils on the same axis produce MMF and flux inthe same direction, if their currents are in the same direction.

Construction of voltage equations

The tapped winding is decomposed into a T-equivalent as shown in Fig. 3.27. The magnetizing reactanceis separated into two components tXm and (1 ! t)Xm and the tap is connected via a reactance !t(1 ! t)Xm.9

The separation leaves three uncoupled reactances which are assumed to interact independently withthe rotor circuits. The resulting circuit is suitable for analysis by either the cross-field method or thedouble revolving-field method. The dot convention in Fig. 3.27 means that the two sections of the tappedwinding produce MMF’s in the same direction when they both conduct positive current. The directionof positive current is shown by the arrows.

The voltage equations are constructed “graphically” in Fig. 3.27. Fig. 3.26 is drawn to assist in checkingthe signs of the speed voltages.10 The main winding 1m has a self-impedance volt-drop and a mutual volt-drop due to current in the “rotor main” winding 2m. Likewise the auxiliary winding 1a has a self-impedance volt-drop and a mutual volt-drop due to current in the “rotor aux” winding 2a. There areno speed voltages in the stator windings.

The rotor main circuit 2m has a self-impedance volt-drop and a mutual impedance volt-drop due tocurrent in the “stator main” 1a. In addition it has speed voltages of the form SxI induced by rotationat the per-unit speed S through flux-linkages xI established by the orthogonal windings 1a and 2a. Forwinding 1a, reactance x excludes the stator leakage; but the rotor leakage is included in x for rotorwinding 2a: in other words all the flux produced by current in rotor aux 2a generates a speed voltagein rotor main 2m. The direction of the speed voltage in 2a induced by Im1 is shown in Fig. 3.26(c). Thespeed voltage in 2m induced by Ia1 is shown in Fig. 3.26(d). Fig. 3.26(e) summarizes the directions of thespeed voltages the rotor circuits in each axis. With this we have enough to construct the voltageequations directly, as expressed in Fig. 3.27.

Fig. 3.26 Directions of speed voltages. In (a) is shown the direction of positive MMF and flux produced by current in a coil withthe polarity indicated by the cross and the dot. The arrow representing the flux defines the axis of the coil. The axesof all the four windings are shown in (b). Note that the auxiliary axis is ahead of the main axis, since positive rotationis counter-clockwise. Also, the axes of the stator and rotor coils 1m and 2m are in the same direction, as are those of1a and 2a. This means that positive current in coil 1m produces flux in the same direction as positive current in coil2m. In (c), positive current in coil 1m induces a speed voltage in coil 2a in the negative direction. Likewise, positivecurrent in coil 2m will also induce a negative speed voltage in coil 2a. In (d), positive current in coil 1a induces a speedvoltage in coil 2m in the positive direction. Likewise. positive current in coil 2a will induce a positive speed voltagein coil 2m. No speed voltage is induced in coils 1m or 1a.


j x A(I1a ! Ica)

j x A

j (1 ! t) x MI2m

j (1 ! t) x M

j t x MI2m

j t x M

j x AI2a

r 1a + j x 1 a

j x A

V1m

I1m

(1 ! t) ( r 1 m + j x 1 m )

! j t ( 1 ! t ) x M

V 1 a

Zcm Z ca

t ( r 1 m + j x 1 m )

I2m I2a

r 2a + j x 2 a

j (1 ! t) x M(I1m ! I1 a ! Icu)

S a t x M (I 1 m ! Ict)

S a ( x 2 m + x M ) I 2 a

r 2m + jx 2 m

j x M

j t x M(I1m ! Ict)

S a x M (I 1 a ! Ic a)

Sa(1 ! t)xM(I1m ! I1a ! Icu)

S a (xM + x2m)I2m

main

aux

stator

rotor

I1a

Eu

Ea

Ict

Icu

Ica

E t

tapped section

untapped section

Fig. 3.27 Cross-field equivalent circuit for general tapped-winding capacitor motor

Core losses

Core losses are incorporated in Fig. 3.27 by means of conductances connected in parallel with the flux-generating elements in each branch of the circuit. Consider a winding wound on a core in which thecore losses are W. If the EMF in the winding is E, the expression W = GE2 reflects the assumption thatthe core loss is proportional to the square of the flux, which itself is proportional to the EMF. The scalingfactor G is a fictitious conductance, which draws current from the circuit in phase with the EMF E.


10An alternative is to use dq-axis theory. Puchstein & Lloyd [1941] quote a torque formula that is effectively based on dq-axistheory, and Veinott [1959] provides a physical explanation of this method; but in many early works the formal mathematicaldevelopment of the dq theory is not given in detail, greater reliance being placed on the physical arguments.

Note that when torque is quoted in “synchronous watts” TSW, the value in Nm is obtained by dividing by the synchronous speedin rad/s, i.e. Te = TSW × p/T. In balanced polyphase machines TSW = Pgap.

WFe ' GEt

2

t%

Eu2

1 ! t%

Ea2

a 2. (86)

Pgap ' Re[E1I1m(% E2 (I1m ! I1a )( % E3I1a

( ] ' Re[(E1 % E2)I1m(

% (E3 ! E2)I1a( ] (87)

E1 ' v1 % j txMI1m ' j txM(I1m % I2m) , (88)

MMMM f ' MMMMm ! jMMMM a ; MMMM b ' MMMMm % jMMMM a

F f ' Fm ! jFa ; Fb ' Fm % jFa .(92)

E2 ' v2 % j (1 ! t )xM(I1m ! I1a ) ' j (1 ! t )xM(I1m % I2m ! I1a ) , and (89)

Pgap ' Re[v1I1m(% v2 (I1m ! I1a )( % v3I1a

( ] . (91)

If the winding is re-wound with a times as many turns, then W = (G/a2)E2. By this logic the conductanceassociated with the auxiliary winding is G/a2, and it represents that fraction of the core loss that isassumed to be associated with the component of flux in the auxiliary axis.

If there were only one main winding there would be only one conductance G in the main windingcircuit, but where the main winding is tapped it is assumed that a fraction t of the core loss Wmattributable to the component of flux in the main axis is associated with the tapped part of the winding.The remaining fraction (1 ! t) is associated with the untapped part. Thus tWm = t(G/t2)Et

2 = (G/t)Et2, so

that the conductance associated with the tapped part of the winding is G/t. Similarly the conductanceassociated with the untapped part is G/(1 ! t). Note that the relationship between induced EMF and thecorresponding component of core loss is assumed to be the same for all three winding sections.

The total core loss satisfies the equation

The total core losses WFe are calculated separately from an estimate of the flux-density waveforms inthe various parts of the magnetic circuit, and include an allowance for stray load loss.

The solution of the circuit equations representing Fig. 3.27 is straightforward but tedious. In practicea direct algebraic solution is used, and this is recursed until the value of G converges to a steady value.(During this process the iron losses may need to be recalculated from the flux-density waveforms.)

Airgap power

The airgap power Pgap is given by the interaction of the currents and EMF’s in each branch: thus

where the core-loss currents have been omitted for clarity, and

Removing products of voltages and currents which are obviously in quadrature, we get

Torque

In a balanced polyphase motor the electromagnetic torque is equal to Pgap in “synchronous watts”, butin unbalanced motors this is not so, and it is necessary to differentiate between the separate torquescontributed by the forward and backward revolving fields.10 The torque can also be expressed in termsof forward and backward fluxes MMMMf and MMMMb and MMF’s Ff and Fb, related to the main-axis flux MMMMm and MMF

Fm and the auxiliary-axis flux MMMMa and MMF Fa by

E3 ' v3 % jxAI1a ' jxA(I1a % I2a ) . (90)


11Eqns. (3.94)—(3.97) apply to all AC induction machines provided that MMMMm, MMMMa, Fm and Fa are defined as the actual fluxes andMMF’s. Equivalent equations are obtained with symmetrical component analysis except that the factor ½ is replaced by a factorthat depends on the form of the symmetrical components transformation; for capacitor motors the factor is usually 2.

E t ' j txM(I1m ! I ct % I2m) ! j t ( 1 ! t)xM(I1a ! Ict % Icu) (3.99)

E a ' j txA(I1a ! Ica % I2a ) (3.100)

Te 'T

2Re (jMMMM fF

(

f ! jMMMM bF(

b ) . (3.94)

Pgap 'T

2Re (jMMMM fF

(

f % jMMMM bF(

b ) . (3.96)

Tpls 'p

2 / jMMMM fFb ! jMMMM bFf / , (3.95)

MMMMm 'E t

jT tNm

and MMMM a 'E a

jTaNm

(3.98)

E u ' E t ×1 ! t

t. (3.101)

Fm ' tNm(I1m ! Ict ) % (1 ! t )Nm(I1m ! I1a ! Icu) (3.102)

Fa ' aNm(I1a ! Ica ) . (3.103)

WCu R ' Pgap ! (1 ! s )Te . (3.97)

The inverses are

The average electromagnetic torque is then given in synchronous watts by 11

where T = 2Bf. The amplitude of the double-frequency pulsating torque is given in Nm by

and the airgap power is given by

With Te in synchronous watts, the power conversion is (1 ! s)Te, so the rotor copper loss is

The fluxes MMMMm and MMMMa are obtained as

where

and

and Nm and Na are the effective series turns in the main and auxiliary windings respectively. If t = 0,Eu can be used in place of Et in eqn. (3.98), with (1 ! t) instead of t in the denominator, since

Finally the MMF’s Fm and Fa are given by

and

3.11 INTERBAR CURRENTS

See [31].

3.12 SATURATION OF LEAKAGE REACTANCE

See [32].

MMMMm '1

2(MMMM f % MMMM b ); MMMM a ' j 1

2(MMMM f ! MMMM b )

Fm '1

2(F f % Fb ); Fa ' j 1

2(F f ! Fb ) .

(3.93)


REFERENCES

1. SPEED’s Electric Motors, the theory text that is used with the SPEED training courses.

2. Say MG, The performance and design of alternating current machines, Pitman, London, SecondEdition, 1948

3. Alger PL, Induction machines, their behavior and uses, Gordon & Breach Science Publishers,New York, London, Paris, Second Edition, 1970. [Original edition, copyrighted 1965 under thetitle The nature of induction machines]. Library of Congress Catalog Card No. 64-18799

4. Heller B and Hamata V, Harmonic field effects in induction machines, Elsevier, Amsterdam,Oxford, New York, 1977 ISBN0-444-99856-X

5. Veinott CG, Theory and design of small induction motors, McGraw-Hill, New York, 1959

6. Kostenko M and Piotrovsky L, Electrical machines, (two volumes), MIR Publishers, Moscow, 3rdedition, 1974

7. Fitzgerald AE and Kingsley C Jr., Electric machinery, McGraw-Hill, Second Edition, 1961

8. Richter R., Elektrische Maschinen, Springer, 1954

9. Schuisky W, Berechnung Elektrischer Maschinen, Springer, 1960

10. Vas P, Electrical machines and drives: a space-vector theory approach, Clarendon Press, Oxford,1992 ISBN 0-19-859378-3

11. Wood WS, Flynn F and Shanmugasundaram A, Transient torques in induction motors, due toswitching of the supply, Proc. IEE, Vol. 112, No. 7, July 1965, pp. 1348-1354

12. Engelmann RH and Middendorf WH [Eds], Handbook of electric motors, Marcel Dekker, NewYork, Basel, Hong Kong, 1995, ISBN 0-8247-8915-6

13. Morrill, WJ, The revolving-field theory of the capacitor motor, Trans AIEE, April 1929, pp.614!632.

14. Levi E, Polyphase motors: a direct approach to their design, John Wiley & Sons Inc., New York,1984 ISBN 0-471-89866-X

15. Hendershot JR and Miller TJE, Design of brushless permanent-magnet motors, Magna PhysicsPublications/Oxford University Press, 1994 ISBN0-19-859389-9

16. Veinott CG and Martin JE, Fractional and subfractional horsepower electric motors, Fourthedition, McGraw-Hill Book Company 1992, ISBN 0-07-067393-4

17. Boldea I, Deep bar effect for slots of any shape, hand-written notes, SPEED Laboratory, 1995.

18. Ionel DM, Cistelecan MV, Miller TJE and McGilp MI, A new analytical method for thecomputation of airgap reactances in 3-phase induction motors, IEEE Industry ApplicationsSociety, Annual Meeting, St. Louis 12-15 October 1998, pp. 65!72.

19. Miller TJE, Gliemann JH, Rasmussen CB and Ionel DM, Analysis of a tapped-winding capacitormotor, ICEM ‘98, Istanbul, 2-4 September 1998, Vol. I, pp. 581!585.

20. Kopilov IP, Goriainov FA, Klokov BK, Design of Electrical Machines (in Russian: Proektirovanieelektriceskih masin), Moscow, Energhia, 1980.

21. Puchstein AF and Lloyd TC, The cross-field theory of the capacitor motor, Trans. AIEE, Vol. 60,February 1941, pp. 58!63.

22. Puchstein AF and Lloyd TC, Capacitor motors with windings not in quadrature, Trans. AIEE,November 1935, pp. 1235!1239.

23. Kingsley C and Lyon WV, Analysis of unsymmetrical machines, Trans. AIEE, May 1936, pp.471!476.

24. Trickey PH, Performance calculations on capacitor motors; the revolving field theory, Trans.AIEE, Vol. 60, February 1941, pp. 73!76.


25. Trickey PH, Capacitor motor performance calculations by the cross-field theory, Trans.AIEE, Vol.76, February 1957, pp. 1547!1553.

26. Suhr FW, Symmetrical components as applied to the single-phase induction motor, Trans.AIEE,Vol. 64, September 1945, pp. 651!655.

27. McFarland TC, Current loci for the capacitor motor, Trans. AIEE, Vol. 61, March 1942, pp.152!155.

28. Bewley LV, Alternating current machinery, Macmillan, N.Y. 1949.

29. Norman HM, Induction motor locked saturation curves, Trans. AIEE, Vol. 53, 1934, pp. 536!541.

30. Kylander, G, Thermal modelling of small cage induction motors, Technical report No. 265, 1995,PhD dissertation, Chalmers University of Technology.

31. Dorrell DG, Miller TJE, Rasmussen CB [2001] Interbar Currents in Induction Machines, IEEEIndustry Applications Society, IAS 2001, Chicago, USA, 30 Sep – 5 Oct. 2001.

32. Miller TJE, Boldea I, Dorrell DG, Rasmussen CB [2000] Leakage Reactance Saturation inInduction Motors, ICEM 2000, Helsinki, Finland, 28-30 August 2000 pp. 203-207.


Index

2-phase winding 13-phase windings 2Airgap power 4, 5, 34, 35Airgap torque 5Alger 8, 9, 11, 36Balancing theory 25Breakdown torque 6, 7Cage-rotor 7Capacitor motor 28-31, 33, 36, 37

balance theory 25Capacitor reactance 26Capacitor voltage 26Closed rotor slots 8Closed slots 14Conductivity 10Constant Volts/Hz 7Core loss 3, 6, 33, 34Core losses 28, 33, 34Cross-field theory 28, 30, 36, 37Current density 10, 11, 15Deep-bar effect 15Deep-bar 7, 8, 15, 28Differential 8, 11, 20Distribution factor 10Double-cage 7, 8Efficiency 3-5, 7Eigenvalue analysis 17Electromagnetic torque 4, 5, 20, 34, 35Equivalent circuit 3-9, 23, 24, 28, 32, 33Finite-element 15, 16Forward and backward components 25Forward and backward revolving-field theory 28Fractional-slot 10Harmonics 8, 10, 11, 24Interbar currents 35, 37Leakage reactance 6-8, 11, 16, 35, 37Location of ampere-conductors 16Magnetizing reactance 3, 6, 8, 28, 32Non-simultaneous switching 21Park 17

Phase-belt harmonics 8Phasor diagram 4, 6, 8, 16, 24-26Pitch factor 10Pole amplitude modulation 7Power 3-5, 12, 18, 20, 26, 27, 34, 35Power factor 5, 26, 27Rotating magnetic field 1Saturation 6, 8, 14, 28, 35, 37Shaft torque 4Sine-distributed windings 1Skew 8, 10Skew factor 10Slip 2-7, 16, 17, 20, 23, 24, 30Slot designs 25Slot numbers 11, 12

ratio 11Slot permeance 8, 13, 14Slot-MMF harmonics 8Speed control 7Split-phase 24, 25Split-phase motors 24Starting torque 7Stray load loss 6, 34Synchronous speed 1-5, 7, 9, 11, 16, 20, 24, 25, 34Synchronous watts 35Tapped-winding 28, 30, 31, 33, 36Torque 2, 4-7, 16, 17, 20-23, 34, 35Torque/slip curve 6, 7Torque/speed characteristic 5-7Transients 17, 20Winding factors 10Wound-rotor 7Zig-zag 8, 11

4. Switched reluctance machines

4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2

4.2 Linear analysis of the voltage equation and torque production . . . . . . . . . . . . . . . . . 4.4

4.3 Nonlinear analysis of torque production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

4.4 Continuous torque production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.11

4.5 Energy conversion analysis of the saturated machine . . . . . . . . . . . . . . . . . . . . . . . . 4.14

4.6 Obtaining the magnetization curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.17

4.7 Solution of the machine equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.19

4.8 Control principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.20

4.9 Variation of current waveform with torque and speed . . . . . . . . . . . . . . . . . . . . . . . 4.22

4.10 Current regulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.27

4.11 Mathematical description of chopping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.33

4.12 Regulation algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.34

4.13 Optimisation of the control variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.36

Fig. 4.1 Simple reluctance machinewith one phase and two poleson both the stator and rotor.

Fig. 4.2 Aligned position Fig. 4.3 Unaligned position

4. SWITCHED RELUCTANCE MACHINES

4.1 DEFINITIONS

A reluctance machine is an electric machine in which torque isproduced by the tendency of its moveable part to move to a positionwhere the inductance of the excited winding is maximized. As wehave seen in chapter 1, this definition covers both switched andsynchronous reluctance machines. The switched reluctance motorhas salient poles on both the rotor and the stator, and operates likea controlled stepper motor.

A primitive example is shown in Fig. 4.1. This machine is denoted"2/2" because it has two stator poles and two rotor poles. The twocoils wound on opposite stator poles are excited simultaneously,and generate magnetic flux as shown. There is only one phase. Inthe position shown, the resulting torque tends to rotate the rotor inthe counterclockwise direction towards the aligned position, Fig.4.2. This machine can produce torque only over a limited arc ofrotation, roughly corresponding to the stator pole arc $s. However,it is the basic model on which the theory of torque production isbased, so we will analyze it first, and then consider methods ofstarting and the extension to multiple poles and phases.

Aligned and unaligned positions

For the primitive reluctance machine in Figs. 4.1!4.3, the aligned and unaligned positions arecharacterized by the properties summarized in Table 4.1.

Aligned Unaligned

2 = 0, 180E 2 = ±90E

Maximum inductance Minimum inductance

Magnetic circuit liable to saturate Magnetic circuit unlikely to saturate

Zero torque : stable equilibrium Zero torque : unstable equilibriumTABLE 4.1

PROPERTIES OF THE ALIGNED AND UNALIGNED POSITIONS

Variation of inductance with rotor position

In the simple machine shown in Figs. 4.1!4.3 the coil inductance L varies with rotor position 2 as shownin Fig. 4.4. Positive rotation is in the counterclockwise direction. Assume that the coil carries aconstant current. Positive motoring torque is produced only while the inductance is increasing as therotor approaches the aligned position; that is, between positions J and A. At J, the leading edge of therotor pole is aligned with the first edge of a stator pole; at A, the rotor and stator poles are fully aligned.Thus J defines the start of overlap, A the maximum overlap, and K the end of overlap.

Switched reluctance machines Page 4.3

Fig. 4.4 Variation of inductance and torque with rotor position; coil current is constant. Thesmall icons show the relative positions of the rotor and stator poles, with the rotormoving to the right. A = aligned position; U = unaligned position; J = start ofoverlap; K = end of overlap.

Fig. 4.5 Variation of inductance, current, flux-linkage, torque, and EMF with rotor position, with ideal pulsed unidirectionalcurrent

The torque changes direction at the aligned position. If the rotor continues past A, the attractive forcebetween the poles produces a retarding (braking) torque. If the machine rotates with constant currentin the coil, the negative and positive torque impulses cancel, and therefore the average torque over acomplete cycle is zero. To eliminate the negative torque impulses, the current must be switched offwhile the poles are separating, i.e., during the intervals AK, as in Fig. 4.5.


v ' Ri %dR

dt' Ri % Tm

dR

d2

' Ri % Tmd (Li)

d2' Ri % L

di

dt% Tm i

dL

d2

(4.1)

vi ' Ri 2% Li

di

dt% Tmi 2 dL

d2. (4.3)

e ' Tm idL

d2. (4.2)

The ideal current waveform is therefore a series of pulses synchronized with the rising inductanceintervals. The ideal torque waveform has the same waveform as the current. The cycle of torqueproduction associated with one current pulse is called a stroke. Evidently the production of continuousunidirectional torque requires more than one phase, such that the gaps in the torque waveform arefilled in by currents flowing in the other phases. The numbers of phases and poles are discussed in §4.4.Normally there is one stroke per rotor pole-pitch in each phase, and the current in any phase isgenerally flowing for only a fraction of the rotor pole-pitch. Note that the current and inductancewaveforms imply a sawtooth waveform of flux-linkage R = Li. Such a waveform is not practicalbecause the sudden extinction of the flux and current would require an infinite negative voltage dR/dt= !4. Similarly the current cannot be established in step fashion unless the inductance at the beginningof the stroke (J) is zero. In practice the inductance along UJ is very small so the leading-edge di/dt canbe very large, presenting a possible problem for the power semiconductors. The rectangular currentwaveform in Fig. 4.5 can be approximated at low speed by chopping the current along JA, which hasthe effect of reducing the average forward applied voltage along JA to a value Va much lower than thesupply voltage Vs. If there is no chopping after commutation at the end of the stroke, the reversevoltage !Vs makes the current fall to zero over a very small angle of rotation.

4.2 LINEAR ANALYSIS OF THE VOLTAGE EQUATION AND TORQUE PRODUCTION

Linear analysis assumes that the inductance is unaffected by the current: that is, there is no magneticsaturation. For simplicity we also ignore the effect of fringing flux around the pole corners, andassume that all the flux crosses the airgap in the radial direction. Mutual coupling between phases isnormally zero or very small, and is ignored. The voltage equation for one phase is

where v is the terminal voltage, i is the current, R is the flux-linkage in volt-seconds, R is the phaseresistance, L is the phase inductance, 2 is the rotor position, and Tm is the angular velocity in rad/s.The last term is sometimes interpreted as a “back-EMF” e:

It is helpful to visualize the supply voltage as being dropped across the three terms in (4.1): namely, theresistance voltage drop, the L di/dt term, and the back-EMF e. The instantaneous electrical power vi is

The rate of change of magnetic stored energy at any instant is given by

According to the law of conservation of energy, the mechanical power conversion p = TmTe is what isleft after the resistive loss Ri2 and the rate of change of magnetic stored energy are subtracted from thepower input vi , Te being the instantaneous electromagnetic torque. Thus from eqns. (4.2) and (4.3),writing Te = p/Tm = vi ! Ri2 ! d(½Li2)/dt, we get

Note that dL/d2 is the slope of the inductance graph in Fig. 4.5.

d

dt

12

Li 2'

12

i 2 dL

dt% Li

di

dt'

12

i 2TmdL

d2% Li

di

dt(4.4)

Te '12

i 2 dL

d2. (4.5)


Fig. 4.6 Single phaseleg circuit

Fig. 4.7 Alternative single-phaseleg circuits

Drive circuit — unidirectional current, bidirectional voltage

Eqn. (4.5) says that the torque does not depend on the directionof the current, since i2 is always positive. However, the voltagemust be reversed at the end of each stroke, to return the flux-linkage to zero. By Faraday’s law, this requires a negativevoltage applied to the coil, to ensure that dR/dt < 0. Fig. 4.6shows the half-bridge phaseleg circuit that accomplishes this.When Q1 and Q2 are both on, the voltage across the motorwindings is v = Vs, the supply voltage. When Q1 and Q2 are bothoff, v = !Vs while the current freewheels through D1 and D2.

To reduce R and i to zero as quickly as possible, as in Fig. 4.5, thereverse voltage must be much larger than the forward voltage,otherwise the flux-linkage will persist beyond the alignedposition, producing an unwanted negative pulse of torque. At lowspeeds this is achievable by chopping the forward voltage,reducing its effective value compared to the reverse voltage.

The circuit of Fig. 4.6 can operate the machine as a motor or as agenerator, since the electrical power vi can be positive ornegative. If the average power is negative (i.e., generating), the energy supplied during transistorconduction in one stroke must be less than the energy recovered during freewheeling. The transistorconduction period (with positive applied voltage) is still necessary to establish the flux, which is builtup from zero and returned to zero each stroke. The voltage-time integrals during transistor conductionand freewheeling must be approximately equal (apart from resistive volt-drop), regardless of whetherthe machine is motoring or generating. As regards control, the main difference between motoring andgenerating is the phasing of the conduction pulse relative to the rotor position. From Fig. 4.4 it appearsthat generating current pulses must coincide with AK, just as motoring pulses coincide with JA.

Other circuits use only one transistor per phase, and employ various means to produce the"suppression voltage" (i.e., reverse voltage) needed to de-flux the windings at the end of each stroke.The circuit in Fig. 4.7a uses separate voltage sources for “fluxing” and “de-fluxing”. The circuit in Fig.4.7b goes one stage further by having two isolated windings with a common magnetic circuit. The twoparts of the winding could be bifilar-wound or they could have completely different numbers of turnsand wire sizes. Unfortunately the leakage inductances of the two parts of the winding are usually quitelarge, even with a bifilar winding, and this leads to problems with transistor overvoltage.Consequently the circuit of Fig. 4.7b is rarely used, although it is common in stepper-motor drives.


Fig. 4.8 Linear energy conversion diagram

W ' OJA ' Te)2 '1

2im

2 dL

d2)2 '

1

2im

2 )L . (4.6)

ei)t ' ABCJ ' im2Tm

dL

d2×

)2

Tm

' im2 )L (4.7)

Q '8 ! 1

28 ! 1. (4.10)

S ' im2 )L %

12

im2 Lu . (4.8)

Q 'energy convertedenergy supplied

'W

S%

W

W % R(4.9)

Additional phases simply use additional drive circuits of the same form as the first phase, usually witha common voltage source. With four-phase motors it is possible to use a common chopping transistorbetween two phases, so that only six power transistors are required; but the control range ofconduction angles must be limited limited to allow de-fluxing of the “complementary” phase in eachpair.

Limitations of the ideal linear model

At the aligned position A in Fig. 4.5, the currentmust be switched off quickly to avoid theproduction of negative torque after the poles havepassed the aligned position. The magnetic storedenergy ½Lai2 must be returned to the supply. In anonsaturating reluctance machine of this type, themagnetic energy stored at A is large, because bothL and i are at their maximum values. We can getfurther insight from an energy audit taken over onestroke as the rotor moves from J to A. The processis shown in the energy conversion diagram, Fig. 4.8,which plots flux-linkage against current. The slopeof OU is the inductance at the unaligned position,Lu, and the slope of OA is the inductance at thealigned position, La. At intermediate positions theinductance is represented by a line of intermediateslope between Lu and La. At the J position (start of overlap), if fringing is neglected (as in the idealizedinductance variation in Fig. 4.4), LJ = Lu. The complete stroke is represented by the locus OJAO. Inmotoring operation it is traversed in the counterclockwise direction, and in generating operation inthe clockwise direction.

Although the current in Fig. 4.5 has a step from 0 to a maximum value im at the position J, in Fig. 4.8this step is along OJ and energy OJC = ½Luim

2 must be supplied to the magnetic circuit as the currentincreases from 0 to im. The step cannot be accomplished in zero time, since that would require infinitevoltage, but if the angular velocity is low, the angle of rotation along OJ is small. Along JA theelectromechanical energy conversion is W or OJA, given by

Along JA there is a back-EMF e which absorbs energy equal to the area ABCJ:

where )t = )2/Tm is the time taken to rotate through the interval )2 = JA, and )L = La ! Lu is thechange in L. The total energy supplied is the sum of areas OCJ and ABCJ, i.e., area S = W + R = OJAB,and

We can now define the energy ratio as

and if we write 8 = La/Lu we can substitute for W and S from (4.6) and (4.8) to give


tf

tr

' d . (4.11)

In this type of nonsaturating motor, less than half the energy S supplied by the drive is converted intomechanical work in each stroke (even neglecting losses). During the “working” part of the stroke JA,the energy is partitioned equally between mechanical work and stored field energy; this is evident fromthe ratio of eqns. (4.6) and (4.7), or OJA/ABCJ. The energy ratio would have a value of 0.5 but for the“overhead” of stored field energy OJC which must be built up before the torque zone JA. This makesQ < 0.5. The inverse of the energy ratio is the converter volt-ampere-seconds per joule of energyconversion C = S/W = 1 + R/W, an important quantity in understanding the basic requirement for“silicon” in the converter.

The stored field energy reaches a maximum value R at A, and must be returned to the supply at the endof the stroke by commutating the current into the diodes, so that the voltage reverses and forces theflux-linkage to fall to zero. In the ideal locus this fall is along AO. However this is not possible withfinite supply voltage. If the current is chopped with a duty-cycle d along JA, the average forwardapplied voltage along JA is d × Vs. In a circuit of the form of Fig. 4.6, the reverse voltage aftercommutation is !Vs. By integrating Faraday’s law the rise and fall periods of flux-linkage can be shownto be in the ratio

This shows that “instant” suppression of the flux is effectively achieved at very low speed when d issmall. But at higher speed there is no chopping: the forward and reverse voltages are equal inmagnitude, and d =, so tf = tr and the flux-linkage waveform is triangular. The time taken along OA isthe same as the time taken along OJA, The waveforms of flux-linkage and current corresponding toFig. 4.8 are shown in Fig. 4.9, and show a tail in the current waveform extending past the alignedposition so that some negative torque must be produced.

Figs. 4.8 and 4.9 represent an important operating condition where the back-EMF is just sufficient tomaintain a flat-topped current waveform. With full voltage applied, the speed at which this occurs iscalled the base speed. The current im could be called the base current. (It is not necessarily the ratedcurrent, because that depends on the cooling arrangements).

Fig. 4.9 Current and flux-linkage waveforms corresponding to Fig. 4.8


Wf ' m idR ;

Wc ' m Rdi .(4.13)

Fig. 4.10 Equivalent circuit

Fig. 4.11 A magnetization curve at one rotor position

R ' W ×1 ! Q

Q. (4.12)

In practical terms the nonsaturating switched reluctance motor makes poor use of the powersemiconductors, because they have to supply more than 2J of energy in order to get 1 J of mechanicalwork, and they must also provide a means to recover the unconverted energy at the end of the stroke.The DC link filter capacitance is directly related to the value of R, which is a large fraction of theenergy conversion W: in fact

For example, if 8 = 8, Q = 0.467, C = 2.14, and R = 1.14W. This implies that for zero ripple voltage at theDC link, the filter capacitance must be large enough to absorb R joules with negligible change ofvoltage. This requirement may be reduced by overlapping charge/discharge requirements of adjacentphases, but still it is a serious consideration.

Eqns. (4.1) and (4.2) imply the existence of an equivalentcircuit of the form shown in Fig. 4.10, in which there is aback-EMF e = TmidL/d2. Unfortunately e is not anindependent parameter, but depends on the current. Ina normal equivalent circuit we interpret the product e i asthe electromechanical power conversion TmTe, implyingthat Te = i2dL/d2. However, eqn. (4.5) states that thetorque is only ½i2dL/d2. Of the power ei, only half isconverted into mechanical power during the “working”part of the stroke JA. The other half is being stored asmagnetic field energy in the increasing inductance. WithL also varying, the equivalent circuit is misleading and cannot be interpreted in the same way as it can,for example, with permanent-magnet motors. This means that the simulation of switched reluctancemachines and their drives requires the direct solution of eqns. (4.1) and (4.5), even when saturation isignored. An elegant and thorough solution for the nonsaturating motor was presented by Ray andDavis [1979]. Usually saturation cannot be ignored and the full nonlinear equations must be solved.

4.3 NONLINEAR ANALYSIS OF TORQUE PRODUCTION

We have already seen that the nonsaturating (i.e.,magnetically linear) switched reluctance machine hasa low energy ratio and makes poor utilization of thedrive. Practical switched reluctance machines aremore effective but they are far from beingmagnetically linear. To understand theelectromechanical energy conversion properly, weneed a nonlinear analysis that takes account of thesaturation of the magnetic circuit. One such analysisis based on the magnetization curves. A magnetizationcurve is a curve of flux-linkage R versus current i ata particular rotor position, Fig. 4.11.

We also need to define the stored magnetic energy Wfand the coenergy Wc graphically, as in Fig. 4.12.

Mathematically,

In a magnetically linear device with no saturation, the magnetization curve is straight, and Wf = Wc.


)Wm ' ) We ! ) Wf

' ABCD ! (OBC ! OAD )' OABCD ! OBC' OAB

(4.16)

Fig. 4.12 Definition of energy Wf and coenergy Wc)We ' m eidt ' m idR

dtdt ' m idR ' ABCD (4.14)

Fig. 4.13 Determination of electromagnetic torque

Fig. 4.14 Ideal magnetization curves

The effect of saturation is to make Wf < Wc. Inmachines with magnetic circuits similar to the one inFig. 4.1, saturation of a typical magnetization curveoccurs in two stages. When the overlap between rotorand stator pole corners is quite small, theconcentration of flux saturates the pole corners, evenat low current. When the overlapping poles are closerto the aligned position, the yokes saturate at highcurrent, tending to limit the maximum flux-linkage.Magnetization curves near the aligned position mayappear “double-jointed” as in Fig. 4.15, if the airgap issmall and the curve is plotted to high flux levels.

In a displacement )2 or AB at constant current (Fig.4.13), the energy exchanged with the supply is

and the change in magnetic stored energy is

The mechanical work done must be

and this is equated to Te)2, so that when )2 6 0,

Ideal cases: In a motor with no saturation the magnetizationcurves would be straight lines, Fig. 4.14a. At any position, Wf =Wc = ½L(2)i2, and in this case (4.17) reduces to Te = ½i2 dL/d2,which we saw earlier.

In a motor with a very small airgap, and especially if the steelhas a "square" B/H curve, the magnetization curvesapproximate to parallel straight lines with a shallow slope, Fig.4.14b. In this case the stored field energy is small and Wc - Ri,so Te = i dR/d2 = e i. This condition approximates to thepermanent-magnet motor which has an EMF that is independentof the current. Because of the small airgap, only a small fractionof the current is needed to raise the flux-density to thesaturation level in the overlap region, and the flux-linkagevaries linearly with the overlap angle. The energy ratio is 1, andthe utilization of the power semiconductors in the drive is high.Practical reluctance motors, especially highly-rated ones, areoften designed to try to approximate this ideal condition.

)Wf ' OBC ! OAD . (4.15)

Te 'MWc

M2 i'const(4.17)


Fig. 4.15 Calculation of average torque

Fig. 4.16 Current, torque, and flux-linkage waveforms with a naturally-determined flat-toppedcurrent waveform. i = phase current; T = phase electromagnetic torque; o = per-unitoverlap between active stator and rotor poles; R = phase flux-linkage; W = energyconversion loop area; U = unaligned; A = aligned.

Average torque: So far we have seen that the torqueis produced in impulses as the rotor rotates, and wehave determined the instantaneous torque Te in termsof the rate of change of coenergy at constant current.The average torque could be calculated by integratingTe over one cycle, (i.e., one rotor pole-pitch J = 2B/Nr,where Nr is the number of rotor poles), and dividing byJ. However, it is difficult in practice to calculate Teaccurately, and it is better to calculate the averagetorque from the enclosed area W in the energy-conversion diagram, Fig. 4.15.

In one cycle of operation the maximum possible energyconversion at a current I is the area W enclosedbetween the unaligned magnetization curve U, thealigned magnetization curve A, and the vertical line UAat the current I. One cycle of operation, i.e., oneexecution of this loop, is called a stroke. If S is thenumber of strokes per revolution, the averageelectromagnetic torque in Nm is

The drive must switch the current on and off at the correct rotor angles to cause the operating pointto follow this loop as closely as possible. Along UA, the current can be regulated by chopping, but athigh speed this may not be achievable and the loop may be smaller than the maximum loop, Fig. 4.30.

Fig. 4.16 shows an example of a motor operating with a naturally flat-topped current waveform, whichis obtained when the back-EMF is approximately equal to the DC supply voltage. This motor is analyzedin more detail later. The speed in this example is 1015 rpm.

Te[avg] 'SW

2B. (4.18)


Fig. 4.17 3-phase 6/4 switched reluctance motor

Fig. 4.18 Waveforms in 3-phase 6/4 switched reluctance machine

S ' m Nr . (4.19)

f1 'rpm

60× Nr Hz. (4.20)

f ' mf1 Hz. (4.21)

4.4 CONTINUOUS TORQUE PRODUCTION

The motor in Fig. 4.1 is useful for analyzing torqueproduction, and although it can maintain a nonzeroaverage torque when rotating in either direction, thistorque is discontinuous, which means that continuousrotation depends on the momentum or flywheel effect.Moreover it cannot self-start from every rotor position. Forexample, at the unaligned and aligned positions the torqueis zero. Unidirectional torque can be produced only over alimited angle where the overlap angle 8 between the rotorand stator poles is varying. To provide continuousunidirectional torque, with starting capability from anyposition, the motor needs more phases, and this requires a“multiplicity” of stator and rotor poles, as in Fig. 4.17.

The number of strokes per revolution is related to thenumber of rotor poles Nr and the number of phases m, andin general

This can be substituted in eqn. (4.18) to give the average torque including all m phases, provided thatS is the same for all of them. The motor in Fig. 4.17 has m = 3 and Nr = 4, so S = 12. The stroke angle isg = 360/12 = 30E. The three phases are labelled AA', BB ' and CC ,' and the ideal current/torque pulsesare shown in Fig. 4.18. The resultant torque is ideally constant and covers 360E of rotation. In practice,of course, the waveforms are more complex than the ideal ones in Fig. 4.18, and their computationrequires a numerical simulation of the transient electromagnetic behaviour throughout one stroke.

Magnetic frequency: The fundamental frequency f1 of the current in each phase is evidently equalto the rotor pole passing frequency, i.e.,

The number of strokes per second is given by

This frequency and its harmonics appear in the flux waveforms in various parts of the magnetic circuit.


g '2BS

'2B

mNr

, (4.22)

Fig. 4.19 3-phase 6/8 motor. Each phase has two coils onopposite poles.

Stator/rotor pole numbers

To provide a structure for ordering the numbers of stator and rotor poles, we can start by defining aregular switched reluctance motor as one in which the rotor and stator poles are symmetrical abouttheir centre-lines and equally spaced around the rotor and stator respectively. An irregular motor isone which is not regular. This chapter, indeed the whole book, is more concerned with regularmachines, since they usually have the most sophisticated power electronic control requirements; andtherefore it focuses on machines with m = 3 or 4 phases. Machines with m = 1 or 2 are usually irregularand they are discussed in Chapter 11 together with various other irregular and special machines.

The absolute torque zone Ja is defined as the angle through which one phase can produce non-zerotorque in one direction. In a regular motor with Nr rotor poles, the maximum torque zone is Ja(max) =B/Nr. The effective torque zone Je is the angle through which one phase can produce useful torquecomparable to the rated torque. The effective torque zone is comparable to the lesser pole-arc of twooverlapping poles. For example, in Fig. 4.17 the effective torque zone is equal to the stator pole-arc: Je= $s = 30E.

The stroke angle g is given by 2B/(strokes/rev) or

The absolute overlap ratio Da is defined as the ratio of the absolute torque zone to the stroke angle:evidently this is equal to m/2. A value of at least 1 is necessary if the regular motor is to be capable ofproducing torque at all rotor positions. In practice a value of 1 is not sufficient, because one phase cannever provide rated torque throughout the absolute torque zone in both directions. The effectiveoverlap ratio De is defined as the ratio of the effective torque zone to the stroke angle, De = Je/g . Forregular motors with $s < $r this is approximately equal to $s/g. For example, in Fig. 4.17 the effectiveoverlap ratio is 30E/30E = 1. Note that De < Da. A value of De of at least 1 is necessary to achieve goodstarting torque from all rotor positions with only one phase conducting, and it is also a necessary (butnot sufficient) condition for avoiding torque dips.

Three-phase regular motors: With m = 3, Da = 1.5 and De can have values of 1 or more, so regular 3-phase motors can be made for 4-quadrant operation. In the 6/4 motor in Fig. 4.17 , forward rotationcorresponds to negative phase sequence. This is characteristic of vernier motors, in which the rotorpole-pitch is less than B/m. The 3-phase 6/4 motor has S = mNr = 12 strokes/rev, with a stroke angleg = 30E, giving De = $s/g = 30/30 = 1.

With regular vernier motors there is always thechoice of having either Nr = Ns ! 2, as in the 6/4; orNr = Ns + 2, which gives the 6/8 motor shown inFig. 4.19; it has S = 24 strokes/rev and g = 15E. Theadvantage of the larger Nr is a smaller strokeangle, leading possibly to a lower torque ripple; butinevitably the price paid is a lower inductanceratio which may increase the controller volt-amperes and decrease the specific output. Thestator pole arc has to be reduced below that of the6/4 motor and this decreases the alignedinductance, the inductance ratio, and themaximum flux-linkage (although it increases theslot area). The consequent reduction in availableconversion energy tends to offset the increase inthe number of strokes/rev, and the core losses maybe higher than those of the 6/4 motor because ofthe higher switching frequency.


1 i.e., the well-known OULTON motor introduced in 1983 by Tasc Drives Ltd., Lowestoft, England.

Fig. 4.21 4-phase 8/6 motor

Fig. 4.20 3-phase 12/8 motor. Each phase has four coils, andthe magnetic flux-pattern is 4-pole.

The 12/8 three-phase motor is effectively a 6/4 witha "multiplicity" of two. It has S = 24 strokes/rev,with a stroke angle g = 15E and Da = 1.5. In Fig. 4.20,De = 15/15 = 1, the same as for the 6/4 motordiscussed earlier. A high inductance ratio can bemaintained and the end-windings are short: thisminimizes the copper losses, shortens the frame,and decreases the unaligned inductance.Moreover, the magnetic field in this machine hasshort flux-paths because of its four-pole magneticfield configuration, unlike the two-poleconfiguration in the 6/4 (or the 8/6; see below), andthe four-pole magnetic circuit helps to minimizeacoustic noise (see chapter 4). Although the MMF

per pole is reduced along with the slot area, theeffects of long flux-paths through the stator yokeare alleviated. The 12/8 is possibly the mostpopular configuration for three-phase machines.

Four-phase regular motors: The 4-phase regular8/6 motor shown in Fig. 4.21 has 24 strokes/rev anda stroke angle of 15E, giving Da = 2. With $s = 21E, De= 1.33, which is sufficient to ensure starting torquefrom any rotor position, and it implies that therewill be no problem with torque dips. However, it isgenerally impossible to achieve the same flux-density waveform in every section of the statoryoke, because of the polarities of the stator poles(NNNNSSSS, NNSSNNSS, or NSNSNSN). Thisconfiguration was one of the first to be producedcommercially.1

With Ns = Nr + 2 = 10, S = 32 strokes/rev and g =11.25E. The inductance ratio is inevitably lowerthan in the 8/6, and the poles are narrower, whilethe clearance between pole-corners in theunaligned position is smaller, increasing theunaligned inductance. This motor is probably onthe borderline where these effects cancel eachother out; with higher pole-numbers, the loss ofinductance ratio and energy-conversion area tends to dominate the gain in strokes/rev. For thisreason, higher pole-numbers are not considered here.

Table 4.2 gives some examples of stator/rotor pole-number combinations for motors with up to m = 7phases. The parameter NwkPP is the number of working pole-pairs: that is, the number of pole-pairs inthe basic magnetic circuit. For example, the 4-phase 8/6 has NwkPP = 1 (a 2-pole flux pattern), while the3-phase 12/8 has NwkPP = 2 (a 4-pole flux pattern). The unshaded boxes in Table 4.2 are probable the bestchoices, the others having too many poles to achieve a satisfactory inductance ratio or too high amagnetic frequency.


Fig. 4.22 Nonlinear energy conversion analysis, with alignedmagnetization curve represented by a straight line OS anda parabola SA.

(R ! Rs0 )2' 4a ( i ! is0) (4.23)

m Ns Nr NwkPP gE S

2 4 2 1 90.00 4

2 8 4 2 45.00 8

2 4 6 1 30.00 12

2 8 12 2 15.00 24

2 12 18 3 10.00 36

2 16 24 4 7.50 48

m Ns Nr NwkPP gE S

3 6 2 1 60.00 6

3 6 4 1 30.00 12

3 6 8 1 15.00 24

3 12 8 2 15.00 24

3 18 12 3 10.00 36

3 24 16 4 7.50 48

m Ns Nr NwkPP gE S

4 8 6 1 15.00 24

4 16 12 2 7.50 48

4 24 18 3 5.00 72

4 32 24 4 3.75 96

4 8 10 1 9.00 40

m Ns Nr NwkPP gE S

5 10 4 1 18.00 20

5 10 6 1 12.00 30

5 10 8 1 9.00 40

5 10 12 1 6.00 60

m Ns Nr NwkPP gE S

6 12 10 1 6.00 60

6 24 20 2 3.00 120

6 12 14 1 4.29 84

m Ns Nr NwkPP gE S

7 14 10 1 5.14 70

7 14 12 1 4.29 84

TABLE 4.2EXAMPLES OF VALID STATOR/ROTOR POLE NUMBER COMBINATIONS

4.5 ENERGY CONVERSION ANALYSIS OF THE SATURATED MACHINE

Energy ratio and converter volt-ampere requirement

Fig. 4.22 shows a model of the energyconversion process in which the unalignedmagnetization curve is assumed to bestraight, while the aligned curve iscomposed of two sections, a straight line OSand a parabola SA. For a given peakcurrent im the energy conversioncapability W is completely defined by thethree points U,S,A. It is shown in Millerand McGilp [1990] that the parabola sectionSA is represented by

with is0 = is ! a/Lau2, Rs0 = Rs ! 2a/Lau, a =

Rms2/4(ims ! Rms/Lau), Rms = Rm ! Rs, and

ims = im ! is. These relationships ensurethat the first derivatives of the segments OSand SA are equal at S. The area R can becalculated by direct integration:

R ' Rms

Rm 2 % Rs (Rm % Rs)

12a% Rs0

Rs0 ! (Rm % Rs)

4a% is0 %

12

Lau is2 (4.24)


Fig. 4.23 Nonlinear energy-conversion analysis, with alignedmagnetization curve represented by two straightlines.

F 'Rm/ im

Rs/ is

(4.26)

iE 'RE

Lau

. (4.31)

Vs$s

Tm

' imLu (8F ! 1) ' )Rm (4.27)

W ' imRm ! R !12

Lu im2 . (4.25)

Rc 'Vs$s

Tm

( 1 ! c % k ) ' c)Rm % Lu im (4.28)

c '(1 % k ) (8F ! 1) ! 1

2(8F ! 1). (4.29)

RE 'k

1 ! c % kRc (4.30)

and then the energy-conversion area W can be obtained as

The energy ratio Q is defined as W/(W + R), with inverse C = 1 + R/W. For example, in Fig. 4.22 theunsaturated inductance ratio 8 = Lau/Lu = 8, and is drawn with Rm = 0.7 V-s, Rs = 0.5 V-s, Ru = 0.25 V-s,im = 80A, and is = 20A. The resulting values are approximately W = 27.7J, R = 13.3J , Q = 0.67, and C =1.5. The energy ratio is about 45% greater than in the linear nonsaturating machine (Q = 0.467), andthe converter volt-ampere-second requirement of the nonsaturating machine is about 45% higher.

Estimation of the commutation angle

Fig. 4.23 shows a model of the energy conversionprocess in which the unaligned magnetizationcurve is again straight, while the aligned curveis fitted by straight lines OS and SA. For a givenpeak current the energy conversion capability Wis again defined by the three points U,S,A. Thismodel is like the one which J.V. Byrne [1970]used to describe controlled saturation. Thesaturation effect is characterized by the ratio

which is effectively the ratio of the saturated tothe unsaturated inductance in the alignedposition. At the base speed the currentwaveform is naturally flat-topped with peakvalue im, because the back-EMF e of the motorequals the supply voltage (resistance isneglected). If the angular rotation JA isassumed equal to the stator pole-arc $s, then itcan be shown that e = Vs when

Commutation is at point C such that the change of flux-linkage between J and C is c)Rm, where c # 1.After commutation the current continues to flow throughout the angle (1 ! c)$s and for anundetermined interval k$s thereafter. Since the peak flux-linkage Rc is c)Rm + Lui, we can write

which describes the de-fluxing interval with !Vs applied via the diodes. Eqns. (4.27) and (4.28) can beused to solve for c:

When the rotor reaches the aligned position the flux-linkage is

and it follows that the current at the aligned position is

The example in Fig. 4.22 has F = 0.350 and 8 = 8. We can consider two extreme cases:


n 'Tp

Tb

'12

2p

2b

2(8F)2

8F ! 1(4.32)

Ta ' im2 Lu (8F ! 1)

mNr

4Bc (2 ! c/s ) (4.33)

T0

Tb

'2s ! 1

cb (2s ! cb ). (4.34)

(i) Commutation at the aligned position — In this case c = 1 and from eqn. (4.29) k = 1.556, so thecurrent continues to flow for 1.556 times the angle $s after the aligned position. The flux-linkageat the aligned position is Rm and the current is iE = Rm/Lau = 40Rm = 28.0A.

(ii) Commutation such that the current extinguishes at the aligned position — In this casek = 0 and from eqn. (4.29), c = 0.222, and RE = iE = 0. Although all negative torque is completelyeliminated, the energy conversion with c = 0.222 is far below the capability of the machine.

In practice the commutation angle is selected to maximize the torque per ampere, and falls betweenthese extremes, with k > 0 and some negative torque. In the example shown in Fig. 4.16, the negativetorque is very small and c = 2/3, giving k = 0.89 and RE = 0.727Rc. Although resistance accelerates theflux suppression, it can be seen that the current tail in the negative torque zone has little impact on theoverall energy conversion.

Basic torque/speed characteristic

An interesting simplified analysis of the speed range at constant power was given by Byrne andMcMullin [1982], in which they derived a formula for the speed range at constant power. If Tp is themaximum speed at which the power can be developed equal to the maximum power at base speed Tb,then in terms of the parameters used in this chapter,

where 2p and 2b are the dwell angles (transistor conduction angles) at the speeds Tp and Tb respectively.For the example motor 8F = 8 × 0.35 = 2.8 and if we assume 2p = 1.52b, we get n = 4.9. This is probablyoptimistic but it shows the importance of phase advance and the saturated inductance ratio 8F. Atypical speed range at constant power is probably nearer 3.

The parameter c in Fig. 4.23 effectively controls the energy conversion loop area and the averagetorque. At low speeds it is possible to work with c = 1, so that the entire available energy conversionarea between the aligned and unaligned curves, bounded on the right by the peak current, can be used.Miller [1985b] gives this average electromagnetic torque the equation

where s = (8 ! 1)/(8F ! 1). At the base speed c has a value cb < 1 and from (4.33) the ratio of the torqueat zero speed to the torque at base speed is derived as

In the example motor cb = 2/3 and s = (8 ! 1)/(8 × 0.35 ! 1) = 3.89, and T0/Tb = 1.43. The motor canevidently produce 43% more torque at standstill than at the base speed, for the same peak current.(The r.m.s. current must be increased because the dwell is greater). With the same peak current, thepeak torque is the same in both cases, implying that the torque waveform must be peakier at the basespeed than at standstill. Miller [1985b] goes on to compare the volt-ampere requirements of theswitched reluctance motor with those of a comparable high-efficiency induction motor, and concludesthat the switched reluctance motor requires 14% more volt-amperes based on peak current, or 20%more based on r.m.s. current. Such comparisons are, however, extremely difficult to generalize.


Fig. 4.24 Analysis of energy-conversion loop:transistor conduction

Fig. 4.25 Analysis of energy-conversion loop: diodefreewheeling

Fig. 4.26 Analysis of energy-conversion loop: thecombined loop

Analysis of the energy-conversion loop

During a typical motoring stroke the locus of theoperating point (i,R) follows a curve similar to the onein Fig. 4.24, which is drawn with the aligned andunaligned magnetization curves and anothermagnetization curve at the commutation angle C. Atthis point the supply voltage is reversed and thecurrent freewheels through the diode. At C theaccumulated energy from the supply is equal to thetotal area U = Wmt + WfC. The stored magnetic energy isequal to WfC. Therefore the mechanical work donebetween turn-on and commutation is Wmt, during theperiod of transistor conduction. In Fig. 4.24 this isroughly comparable to WfC, meaning that only half ofthe energy supplied has been converted to mechanicalwork. The other half is stored in the field.

After commutation, Fig. 4.25, the supply voltage isreversed and the energy Wd is returned to the supplyvia the diodes. The mechanical work done during thefreewheel interval is Wmd = WfC ! Wd. In Fig. 4.25 thisis less than half of WfC. The energy balance can bededuced from the areas in Figs. 4.24 and 4.25. Supposethat the energy supplied from the controller during the"fluxing" interval (transistor conduction) is U = Wmt +WfC = 10 Joules. At C, 5J has been converted tomechanical work and 5J are stored in the field. Duringthe "de-fluxing" period (diode freewheeling), Wd = 3.5Jis returned to the supply and Wmd = 1.5J is converted towork. The total work is therefore W = Wmt + Wmd = 5 +1.5 = 6.5J or 65% of the energy supplied by thecontroller. The energy returned to the supply is Wd =3.5J or 35% on each "stroke".

The entire stroke is shown in Fig. 4.26, which combinesthe two previous diagrams. The energy conversion isnow shown as the area W, while the energy returned tothe supply is R = Wd. The original energy supplied bythe controller is U = W+R, and the energy ratio is 0.65.

4.6 OBTAINING THE MAGNETIZATION CURVES

Calculation

The aligned magnetization curve can be calculated bylumped-parameter magnetic circuit analysis, with anallowance for the stator slot-leakage which increases athigh flux levels. The unaligned curve is more difficultto calculate because of the complexity of the magneticflux paths in this position, but practical results arereported by Miller and McGilp [1990] using a dual-energy method based on a quadrilateral discretizationof the slotted region. Earlier work by Corda andStephenson [1979] also produces adequate results formany practical or preliminary design calculations.


Fig. 4.27 Finite-element flux-plot in a partial-overlap position

fu 'R1 ! R0 % Lstk

Lstk

(4.36)

Lu ' Lu0 × (eu % fu) (4.35)

The computation should include the “partial linkage” effect, meaning that at certain rotor positionsthe turns of each coil do not all link the same flux. In a finite-element calculation this implies that theflux-linkage should be calculated as the integral of A@dl along the actual conductors, with the coilsidesin the finite-element mesh in their correct positions in the slot.

End effects are important in switched reluctance motors. (See Michaelides and Pollock [1994]; Reeceand Preston, [2000]). When the rotor is at or near the unaligned position, the magnetic flux tends to"bulge out" in the axial direction. The associated increase in the magnetic permeance can raise theunaligned inductance Lu by 20!30%. Since this inductance is critical in the performance calculations,it is important to have a reasonable estimate of it. Unfortunately 2-dimensional finite-element analysiscannot help with this problem, and 3-dimensional finite-element calculations tend to be expensive andslow. When the rotor is at or near the aligned position, the flux is generally higher and the "bulging"of flux outside the core depends on the flux level in the laminations near the ends of the stack. At ornear the aligned position at high flux levels, the stator and rotor poles can be highly saturated and theexternal flux-paths at the ends of the machine can increase the overall flux-linkage by a few percent.

In spite of the complexity of the field problem, good results have been obtained with relatively simpleend-effect factors for Lu. For example, the PC-SRD computer program [Miller, 1999] splits the end-effectcalculation into two parts. For Lu,

where eu = Lend/Lu0 represents the self-inductance Lend of the end-windings (including any extension),expressed as a fraction of the uncorrected 2-dimensional unaligned inductance Lu0. Lend is theinductance of a circular coil whose circumference is equal to the total end-turn length of one pole-coil,including both ends. It is multiplied by the appropriate function of turns/pole and parallel paths beforebeing normalized to Lu0. fu is a factor that accounts for the axial fringing in the end-region. It iscalculated by the approximation

which is derived by analogy with the fringing formula for two opposite teeth or poles. For the alignedposition the procedure is similar, with a factor fa = (g + Lstk)/Lstk.

At positions intermediate between thealigned and unaligned positions, thecalculation of individual magnetizationcurves is not practical by analytical methodsand the finite-element method should beused. Fig. 4.1 shows a simple example of afinite-element flux-plot at a position ofpartial overlap, and Fig. 4.27 a more complexexample. Considerable success has beenachieved with interpolating proceduresespecially in computer programs for rapiddesign, e.g., [Miller and McGilp, 1990],[Miller et al 1998].

Measurement of the magnetization curves isdescribed in [Cossar 1992].


v ' e % Ri (4.37)

e 'MR

Mt' Tm

MR

M2(4.38)

4.7 SOLUTION OF THE MACHINE EQUATIONS

The PC-SRD computer program has been widely used for design and analysis of switched reluctancemachines for many years. Each phase is treated as a variable inductance in which the flux-linkage Ris a nonlinear function of both the current i and the rotor position 2: thus R = R(i, 2). PC-SRD computesthe function R(i, 2) from the geometry, the winding details, and the B/H curve of the steel, and itrepresents the result graphically as a set of static magnetization curves in which R is plotted vs. i atseveral rotor positions between the unaligned and the aligned positions. For a given machine, thecurves are a fixed property. It is not necessary to recalculate them unless changes are made to thegeometry, the windings, or the steel. For precise work PC-SRD can import external magnetiztioncurves which have been obtained either by measurement or by 3D finite-element calculation.

PC-SRD solves the electrical circuit by stepwise integration of eqn. (4.1) for one phase by Euler’smethod. Each integration step produces a new value of flux-linkage R, and PC-SRD computes a newvalue of current i from the function R(i,2), i.e., from the magnetization curves. The method of solvingfor each new current value i depends on whether the curves are internal or external. With internalmagnetization curves PC-SRD uses a fast algebraic interpolation method based on so-called gaugecurves [Miller and McGilp, 1990]. With external curves, PC-SRD fits the curves with a set of cubicsplines, and interpolates them. This is slower than the gauge curve method, but more accurate.

The instantaneous torque is calculated from the rate of change of coenergy MWc(i,2)/M2. With internalmagnetization curves, the derivative is evaluated using approximate algebraic expressions derivedfrom the gauge curve model. With external mag curves it is evaluated from a precalculated set of splinefunctions that represent the coenergy Wc as a function of current and rotor position, Wc(i,2). Theaverage electromagnetic torque is computed from the loop area W in Fig. 4.15, and a typical exampleis given in Fig. 4.16. Several computed examples are given in chapter 5. The average electromagnetictorque is given by (4.18). Since W is evidently an integral quantity, errors in the distribution of themagnetization curves tend to cancel out, provided that the aligned and unaligned curves are accurate.On the other hand, the calculation of the instantaneous torque is sensitive to the precision in theintermediate magnetization curves and the method of representing them mathematically is critical.

PC-SRD’s model is a single-phase model. The currents in phases 2,3,... are determined by phase-shiftingthe current waveform of phase 1, which is calculated as though it were the only current flowing. Thefluxes and torques of the other phases are added to those of phase 1 without taking account ofinteractions in shared magnetic circuits. This is one of the main limitations of PC-SRD, but it alsoexplains the extraordinary speed of computation. A full magnetic model of a polyphase switchedreluctance motor, including all magnetic interactions between phases, requires a multi-dimensionalset of magnetization curves and is a formidably complex thing to contemplate. PC-SRD’s simple modelis successful when the yokes are sufficient to avoid mutual interaction between phases. Underconditions of extreme loading or (e.g., faults), the PC-SRD model cannot be expected to give accurateresults, even with external magnetization curves, since these are valid only for one phase conducting.

A method that uses coenergy and avoids integration

The somewhat convoluted process described in the previous section could in principle be replaced bya more direct method based on a coenergy map. The voltage equation for one phase is

where the back-EMF is given by

and this can be obtained from the R(i,2) curves by differentiating with respect to 2 at constant current.


T *

Fig. 4.28 Nested control loops. T * = torque demand; Tm* = speed demand, Tm = speed; 2* = position demand; 2 = shaftposition. Tacho = tachometer or speed transducer; Enc = encoder or position transducer.

e ' Tm

M2Wc

M2 Mi(4.42)

i 'v ! e

R(4.39)

T 'MWc

M2(4.40)

R 'MWc

Mi(4.41)

The solution proceeds at each timestep by calculating

using the current values of v and e. Once the “new” current is calculated from (4.39), e is re-evaluatedusing (4.38) for the next integration step. The torque is calculated using

which is also evaluated at constant current. It is interesting to note that the flux-linkage can also beevaluated from the coenergy using

evaluated at constant 2. This suggests that the back-EMF can be evaluated using

This suggests that the machine can be represented by a surface Wc (i,2) whose derivatives can be usedat any position 2 and any current i to determine the back-EMF e (4. 38) the flux-linkage R (4.41) and thetorque T (4.40). The finite-element solution of the magnetization curves can therefore be expressed interms of the surface Wc (i,2) without even computing the flux-linkage R, given that the finite-elementmethod can compute Wc directly by means of a global integration. A difficulty with this approach, aswith the previous one, is the representation of the coenergy function by a sufficiently smoothinterpolating function that is differentiable both 2 and i. Also, it does not naturally provide data whichcan be compared with measurements.

4.8 CONTROL PRINCIPLES

Torque in the switched reluctance machine is produced by pulses of phase current synchronized withrotor position. The timing and regulation of these current pulses are controlled by the drive circuitand the torque control scheme. Usually there are also outer feedback loops for controlling speed orshaft position, as shown in Fig. 4.1. The outer loops are generally similar to those used in other typesof motor drive, but the inner torque loop is specific to the switched reluctance machine.

The torque demand signal generated by the outer control loops is translated into individual currentreference signals for each phase, [Bose, 1987]. The torque is controlled by regulating these currents.Usually there is no torque sensor and therefore the torque control loop is not a closed loop.Consequently, if smooth torque is required, any variation in the torque/current or torque/positionrelationships must be compensated in the feed-forward torque control algorithm. This implies that thetorque control algorithm must incorporate some kind of “motor model”.


Unlike the DC or brushless DC motor drive, the switched reluctance motor drive cannot becharacterized by a simple torque contstant kT. (torque/ampere). The drive must be specificallyprogrammed for a particular motor, and even for particular applications. One cannot take a switchedreluctance motor from one source and connect it to a drive from another source, even when the voltageand current ratings are matched. On the contrary, the motor and drive control must be designedtogether, and usually they must be optimized or tuned for a particular application.

The power electronic drive circuit is usually built from phaselegs of the form shown in Fig. 4.6. Thesecircuits can supply current in only one direction, but they can supply positive, negative, or zero voltageat the phase terminals. Each phase in the machine may be connected to a phaseleg of this type, and thephases together with their phaseleg drive circuits are essentially independent. The circuits in Fig. 4.7can be adapted to operate the phases with separate DC supplies of different voltages, although the mostusual case is to connect them all to a common DC supply. Figs. 4.7a and 4.7b also show the possibilityof “fluxing” at one voltage V1 and “de-fluxing” at another voltage !V2.

At lower speeds the torque is limited only by the current, which is regulated either by voltage PWMor current regulation. As the speed increases the back-EMF increases to a level at which there isinsufficient voltage available to regulate the current; the torque can then be controlled only by thetiming of the current pulses. This control mode is called “single-pulse mode” or “firing angle control”,since the firing angles alone are controlled to produce the desired torque. Many applications requirea combination of the high-speed and low-speed control modes. Even at lower speeds with voltage PWMor current regulation, the firing angles must be varied with speed to optimise performance.

This chapter is concerned with control of average torque, i.e., the torque averaged over one stroke ( g= 2B/mNr). The amplitude and phase of the current reference signal (relative to the rotor position) areassumed to remain constant during each stroke. This corresponds to the operation of a “variable-speeddrive”, as distinct from a servo drive which would be expected to control the instantaneous torque.Average torque control requires a lower control-loop bandwidth than instantaneous torque control.

Differences between switched reluctance machines and classical machines: Much of the classicaltheory of torque control in electric drives is based on the DC machine, in which torque is proportionalto flux × current. The flux and current are controlled independently, and the “orientation” of the fluxand the ampere-conductor distribution, both in space and in time, is fixed by the commutator. In ACfield-oriented control, mathematical transformations are used, in effect, to achieve independent controlof flux and current, and the commutator is replaced by a shaft-position sensor which is used by thecontrol processor to adjust the magnitude and phase of the currents to the correct relationship withrespect to the flux. The current can be varied rapidly so that a rapid torque response can be achieved.Generally speaking, in classical DC and AC machines the flux is maintained constant while the currentis varied in response to the torque demand. In both cases the torque control theory is characterizedby the concept of “orthogonality”, which loosely means that the flux and current are “at right angles”.In the architecture of the machine and the drive, this concept has a precise mathematical meaningwhich depends on the particular form or model of the system.

In switched reluctance machines, unfortunately there is no equivalent of field-oriented control. Torqueis produced in impulses and the flux in each phase must usually be built up from zero and returned tozero each stroke. The “orthogonality” of the flux and current is difficult to contemplate, because themachine is “singly excited” and therefore the “armature current” and “field current” areindistinguishable from the actual phase current. Although this appears to be the case also withinduction machines, the induction machine has sine-distributed windings and a smooth airgap, so thatthe theory of space vectors can be used to resolve the instantaneous phase currents into an MMF

distribution which has both direction and magnitude, and the components of this MMF distribution canbe aligned with the flux or orthogonal to it. The switched reluctance machine does not have sine-distributed windings or a smooth airgap, and there is virtually no hope of “field-oriented” control. Toachieve continuous control of the instantaneous torque, the current waveform must be modulatedaccording to a complex mathematical model of the machine.


Fig. 4.29 Low-speed motoring waveforms. i = phase current, R = phase flux-linkage, T = phase torque, and o = overlapbetween stator and rotor poles. Horizontal axis is rotor angle (degrees). Unaligned position U = 45E; aligned positionA = 90E. The position J is the start of overlap between the active rotor poles and the stator poles of this phase.

4.9 VARIATION OF CURRENT WAVEFORM WITH TORQUE AND SPEED

The average electromagnetic torque is given by eqn. (4.14), and the energy-conversion loop area W isshown in Figs. 4.12 and 4.13. The objective of “average torque control” is a simple current pulsewaveform which produces the required value of W corresponding to the torque demand. Even insimple cases, this is more complex than simply determining the required “value of current”, since thetorque/ampere varies with both position and current. The following sections describe the generalproperties of the current waveform at different points in the torque/speed diagram, Fig. 4.35.

Low-speed motoring — At low speed the motor EMF e is low compared to the available supply voltageVs, and the current can be regulated by chopping. If voltage-drops in the semiconductor devices areneglected, the drive can apply three voltage levels +Vs, !Vs or 0 to the winding terminals to raise orlower the flux and current. A simple strategy is to supply constant current throughout the torque zone,i.e., over the angle through which the phase inductance is substantially rising. Fig. 4.29 shows a typicallow-speed motoring current waveform of this type in a 3-phase 6/4 motor at 500 rev/min.

The current i is chopped at about 8 A, starting 5E after the unaligned position (at 45E) and finishing 10Ebefore the aligned position (at 90E). At first no torque is produced because the inductance is low andunchanging, but when the corners of the stator and rotor poles are within a few degrees of conjunctionJ, torque suddenly appears. It is controlled by the regulating the current. When the transistors areswitched off, 10E before the aligned position, the current commutates into the diodes and falls to zero,reaching the “extinction” point a few degrees later, so that virtually no negative torque is produced.

The flux-linkage R grows from zero and falls back to zero every stroke. When the driving transistorsare first switched on, R grows linearly at first because the full supply voltage is applied across thewinding terminals. When the current regulator starts to operate, R is also regulated to a constantvalue at first because the constant current is being forced into an inductance that is still almostconstant at the low value around the unaligned position, before the poles begin to overlap. As soon asthe pole corners approach conjunction J, the inductance starts to increase, so the flux-linkage R alsoincreases as constant current is now being forced into a rising inductance. The flux-linkage continuesto increase until the commutation point. After that, the diodes connect a negative “de-fluxing” voltage!Vs across the winding terminals and therefore R falls to zero very rapidly. In this example theresistive voltage-drop is small, and therefore the rate of fall of flux-linkage is almost linear. At lowspeed the dwell is made approximately equal to $s, since this is “width” of the “torque zone”, and thisangle might typically be a little less than 30E in a typical 6/4 motor. De-fluxing is completed over onlya small angle of rotation since the speed is low, so the entire conduction stroke occupies only about 30E.


Fig. 4.30 High-speed motoring waveforms.

The process is summarized in the energy-conversion loop, which fits neatly between the aligned andunaligned magnetization curves as a result of the selection of the firing angles. It appears that theenergy conversion W could be increased slightly, by retarding the commutation angle to extend theloop up to the aligned magnetization curve. This would not require any increase in peak current, butit would increase the average and r.m.s. values. It is also possible that delayed commutation couldincur a period of negative torque just after the aligned position, which would appear as a re-entrantdistortion of the energy-conversion loop, limiting the available gain in torque. Operation is at pointM1 in the torque/speed characteristic, Fig. 4.35. It is possible to maintain torque constant withessentially the same current waveform as the speed increases up to a much higher value, since themotor EMF is still much lower than the supply voltage.

High-speed motoring — At high speed the motor EMF is increased and the available voltage may beinsufficient for chopping, so that the torque can be controlled only by varying the firing angles of asingle pulse of current. Fig. 4.32 shows a typical example, in which the speed is 1300 rev/min.

The driving transistors are switched on at 50E and off at 80E, the same as in Fig. 4.29. At first theoverlap between poles is small, and the supply voltage forces an almost linear rise of current di/dt =Vs/Lu into the winding. Just before the start of overlap the inductance begins to increase and the back-EMF suddenly appears, with a value that quickly exceeds the supply voltage and forces di/dt to becomenegative, making the current fall. The higher the speed, the faster the current falls in this region.Moreover, for a given motor there is nothing that can be done to increase it, other than increasing thesupply voltage. The torque also falls. Operation is at point M2 in Fig. 4.35.

Operation at much higher speed — At a certain “base speed” the back-EMF rises to a level at whichthe transistors must be kept on throughout the stroke in order to sustain the rated current. Anychopping would reduce the average applied voltage and this would reduce the current and torque. The“base” speed is marked B in Fig. 4.35. If resistance is ignored, the peak flux-linkage achieved duringthe stroke is given by Vs)2/T, where )2 is the “dwell” or conduction angle of the transistors. If the peakflux is to be maintained at higher speeds, the “dwell” must be increased linearly with speed above thebase speed. At high speed the turn-on angle can be advanced at least to the point where the sum of thefluxing and de-fluxing intervals is equal to the rotor pole-pitch, at which point conduction becomescontinuous (i.e. the current never falls to zero). This corresponds to a dwell of 45E and a totalconduction stroke of 90E, neglecting the effect of resistance (which tends to shorten the de-fluxinginterval).


Fig. 4.31 Very high speed motoring

Fig. 4.32 Energy-conversion loops at low and high speed, 1300 and3900 rev/min.

Thus it appears that the dwell or “flux-building angle” can increase from 30E at lowspeed to 45E at high speed, an increase of50% or 1.5:1. Over a speed range of 3:1, thepeak flux-linkage might therefore fall to1.5/3 = 0.5, or one-half its low-speed value.This is illustrated in Fig. 4.31 for a speed of3900 rev/min. The peak current isapproximately unchanged but the loop areaW is only about one-third of its low-speedvalue. The comparison between the loopareas at 1300 and 3900 rev/min is shownmore clearly in Fig. 4.32. The averagetorque is therefore only about one-third ofits low-speed value, but the power remainsalmost unchanged. Operation is at point M3in Fig. 4.35.

Low-speed generating — Low-speed generating is similar to low-speed motoring except that the firingangles are retarded so that the current pulse coincides with a period of falling inductance. Fig. 4.33shows a typical example. The average torque is negative and the energy-conversion loop is traversedin the clockwise direction. At the start of the stroke, there is a slight positive torque because thecurrent is switched on shortly before the aligned position, while the inductance is still rising. In thisexample the torque falls to zero before the current is commutated, indicating that the commutationangle could be advanced slightly without reducing the average torque. The reduction in copper losswould increase the efficiency. During that “tail” period when there is current but no torque, thecurrent is maintained by the drive which is simply exchanging reactive energy with the DC link filtercapacitor. Operation is at G1 in Fig. 4.35.


Fig. 4.33 Low-speed generating waveforms

High-speed generating — High-speed generating is similar to high-speed motoring, except that thefiring angles are retarded so that the current pulse coincides with a period of falling inductance. Fig.4.34 shows a typical example. The torque is negative and the energy-conversion loop is again traversedin the clockwise direction. At the start of the stroke, there is a slight positive torque because thecurrent is switched on a few degrees before the aligned position, while the inductance is still rising.Operation is at G2 in Fig. 4.35.

Operating regions — torque/speed characteristic

For control purposes the torque/speed envelope can be divided into regions as shown in Fig. 4.35.

Constant torque region—The base speed is the maximum speed at which maximum current andrated torque can be achieved at rated voltage. In this region the torque is controlled by regulating thecurrent, with relatively minor adjustments in the firing angles as necessary to alleviate noise orimprove the current or torque waveform, or to improve efficiency.

Fig. 4.34 Energy-conversion loop: high-speed generating


Fig. 4.35 Torque-speed characteristics

Constant power region—As the speed and back-EMF increase, the dwell is increased to maintain thepeak flux-linkage at the highest possible level. If the dwell is equal to half the rotor pole-pitch and thede-fluxing angle is negligible at the base speed, then in principle the dwell can be doubled before theonset of continuous conduction. Therefore if the dwell is increased in proportion to speed, the peak flux-linkage can be maintained up to about twice the base speed. However, constant power can bemaintained to a higher speed than this, because the loss of loop area dW/dT is compensated by theincrease in speed. If power is taken as TT and T % W, then P % TW and for constant power we requirethat )P = T)W + W)T = 0, which says that constant power can be maintained up to the point where)W/W = !T/)T. In other words, the maximum speed at constant power is the speed at which the rateof loss of loop area is balanced by the rate of increase of speed. The rate of increase in back-EMF is lessthan proportional to the speed, because the current decreases with speed and MR/M2 is reduced. (In thelinear analysis e = iTdL/d2, and i is decreasing while T is increasing and dL/d2 remains constant.

Falling power region—Eventually as the speed increases, the turn-on angle can be advanced no more,and the torque falls off more rapidly so that constant power cannot be maintained, even though veryhigh speeds can be attained against a light load. The maximum phase advance depends on the drivecontroller. If the turn-on angle is advanced beyond the point where the dwell becomes equal to abouthalf the rotor pole-pitch, continuous conduction will begin: the phase current never falls to zero andthe energy-conversion loop “floats” away from the origin. As it does so, it moves to a region where theseparation between the aligned and unaligned curves is increased, and the torque per ampere actuallyincreases. For this reason, operation with continuous conduction is a possible means of increasing thepower density, not only at high speeds but even at low speeds. The increase in copper loss is acceptableif there is a greater gain in converted power and the machine can withstand the temperature rise. Asimilar effect can be achieved with a DC bias winding in 3-phase motors, [Horst, 1995].

Reversibility—Fig. 4.35 shows only two quadrants of the torque/speed characteristic, correspondingto motoring and generating (or braking). The direction of rotation is the same in both quadrants.Operation in the opposite direction is symmetrical, provided that the rotor position transducer canprovide the correct reference position and direction sense. The firing angles for motoring in onedirection become generating angles in the reverse direction, at least at low speed. The machine is thusreversible and regenerative, and able to operate in all four quadrants of the torque/speed diagram.

Multiple-phase operation — To produce torque at all rotor positions the entire 360E of rotation mustbe ‘covered’ by segments of rising inductance from different phases, as shown in Fig. 4.16, and the phasecurrents must be sequenced to coincide with the appropriate segments. The total torque averaged overone revolution is usually assumed to be the sum of the torque contributions from each phase. Althoughthe calculation and control of torque are both referred to one phase, some degree of overlap is requiredin practice to minimise notches in the instantaneous torque waveform when the phases arecommutated, and to produce adequate starting torque at all rotor positions.


Fig. 4.36 Conduction modes

4.10 CURRENT REGULATION

Soft chopping, hard chopping, and conduction modes

At high speed the current is controlled solely by the on/off timing of the power transistor switching,but at low and medium speeds it is regulated by chopping. This means that the power transistors areswitched on/off, usually at a high frequency compared with the fundamental frequency of the phasecurrent waveform. The voltage applied to the winding terminals is +Vs if both transistors are on, 0 ifone is on and the other is off, and !Vs if both transistors are off and the phase current is freewheelingthrough both diodes. In the zero-volt state the phase current freewheels through one transistor andone diode. These three conduction modes are shown in Fig. 4.36, and Table 4.3 shows the states of thepower transistors and diodes in the three conduction modes.

Soft chopping is when only one transistor is chopping. The other transistor remains on, and it iscalled the "commutating" transistor because its only function is to steer or commutate the current intoits associated phase winding at the beginning and end of the conduction period. The voltage appliedto the winding switches between +Vs and 0. During the zero-volt period the rate of change of flux-linkage is very small (in fact it is equal to !Ri), and therefore the current falls slowly. This means thatthe chopping frequency and DC link capacitor current can both be greatly reduced for a given currentripple or hysteresis band (see below).

State Q1 Q2 D1 D2 VA 1 1 0 0 Vs

B 1 0 0 1 0C 0 1 1 0 0D 0 0 1 1 -Vs

TABLE 4.3

TRUTH TABLE FOR THE STATES OF THE TRANSISTORS AND DIODES

Hard chopping is when both transistors are switched on/off together. It generally produces moreacoustic and electrical noise, and increases the current ripple and DC link capacitor current for a givencurrent ripple or hysteresis band. It is necessary in certain conditions particularly duringregeneration, to prevent loss of control of the current waveform, and of course the final “chop” at 2cat the end of the conduction period is a hard chop.


2 Note that "r is the angle of rotation between two successive aligned positions.

Rc '1T

2c

20

(Vs ! Ri) d2 % R0 (4.43)

TRc ' Vs(1 & u1 ) . 2D (4.44)

0 ' Rc %1T

2q

2c

(!Vs & Ri) d2 (4.45)

T Rc ' Vs(1 % u2 ) (2q & 2c). (4.46)

2q ! 20 •TRc

Vs

.2 ! u1 % u2

(1 % u2 )(1 ! u1 ). (4.47)

2D max ' "r .1 % u2

2 ! u1 % u2

. (4.48)

2D max ' "r .(1 % u)

2. (4.49)

Single-pulse control at high speed

The flux must be established from zero every stroke. Its build-up is controlled by switching both powertransistors on at the turn-on angle 20 and switching them off at the commutation angle 2c. In motoringoperation the dwell )2 = 2c ! 20 is timed to coincide with a period of rising inductance, and ingenerating operation with a period of falling inductance. At a sufficiently high speed, the waveformsof voltage, flux-linkage, current, and idealised inductance are as shown in Fig. 4.30 and 4.31 (motoring)and Fig. 4.34 (generating). The "idealised" inductance that would be obtained with no fringing and withinfinitely permeable iron has a waveform similar to that of the pole-overlap waveform, and providesa convenient means for relating the waveforms to the rotor position.

At constant angular velocity T the build-up of flux-linkage proceeds according to Faraday's Law:

where R0 is the flux-linkage pre-existing at 20 (ordinarily zero). Vs is the supply voltage, R is the phaseresistance, and i is the instantaneous current. All impedances and volt-drops in the controller and thesupply are ignored at this stage. Eqn. (4.43) can be written as

where 2D = (2c ! 20)is the dwell and v1 = u1Vs is the mean volt-drop in the resistance and transistorsduring 2D. If u1 << 1 the flux-linkage rises linearly. In motoring operation the flux should ideally bereduced to zero before the poles are separating, otherwise the torque changes sign and becomes abraking torque. To accomplish this the terminal voltage must be reversed at 2c, and this is usually doneby the action of the freewheeling diodes when the transistors turn off. The angle taken for the negativevoltage to drive the flux back to zero at the "extinction angle" 2q is again governed by Faraday's Law:

and this can be written as

where v2 = u2Vs is the mean volt-drop in the resistance and diodes in the de-fluxing period (2q - 2c). Ifu2 << 1 the flux-linkage falls linearly, and at constant speed the angle traversed is nearly equal to thedwell angle, both being equal to Rc/Vs. The peak flux-linkage Rc occurs at the commutation angle 2c. Thetotal angle of phase current conduction covers the fluxing and de-fluxing intervals and is equal to

If u1 = u2 = 0 this reduces to 2TRc/Vs. The entire conduction period must be completed within one rotorpole-pitch "r = 2B/Nr, otherwise there will be a ratcheting or pumping effect in which R0 has a seriesof non-zero values increasing from stroke to stroke.2 This condition is also called "continuousconduction". That is, 2q ! 20 # "r. Eqns. (4.44) and (4.47) combine to give the maximum permissible dwellangle,

If the mean volt-drops u1 and u2 are both approximately the same fraction of Vs, so that v1/Vs = v2/Vs= u, then eqn. (4.48) reduces to

For example, in a symmetrical 6/4 motor the pole-pitch is "r = 90E (360 elecE) and if u = 0 the maximum


2D #"r

2.

1 % u

2. (4.50)

dwell angle is 2D = 45E, giving a total angle of conduction in the phase winding of 90E. But if u = 0.2 themaximum dwell angle is 54E. In a regular switched reluctance motor the angle of rising inductance isonly "r/2. Ideally the flux should be zero throughout the period of falling inductance, because currentflowing in that period produces a negative or braking torque. To avoid this completely, the conductionangle must be restricted to "r/2 and the maximum dwell angle is then

In the 6/4 motor, with u = 0.2 this indicates a maximum dwell angle of 27E (108 elecE) and a conductionangle of 54E. In practice, larger dwell angles than this are used because the gain in torque-impulseduring the rising-inductance period exceeds the small braking-torque impulse, which generally occursin a region when the torque/ampere is low (i.e., near the aligned and/or unaligned positions). Thiscondition is shown in Figs. 4.30 and 4.31, where the current has a "tail" extending beyond the alignedposition. The torque is negative during this tail period, but it is small.

The turn-on angle in Fig. 4.30 is just after the unaligned position, and the current rises linearly untilthe poles begin to overlap. The rising inductance generates a back-EMF which consumes an increasingproportion of the supply voltage, until at the peak of the current waveform the back-EMF equals Vs.Subsequently the back-EMF grows greater than Vs because the flux-linkage is still increasing, while thespeed is constant. What was an excess of applied forward voltage now becomes a deficit, and the currentbegins to decrease. At the point of commutation the applied terminal voltage reverses, and there is asharp increase in the rate of change of current. At the aligned position the back-EMF reverses, so thatinstead of augmenting the negative applied terminal voltage, it diminishes it, and the rate of fall ofcurrent decreases. In this period there is a danger that the back-EMF may exceed the supply voltage andcause the current to start increasing again. It is for this reason that in single-pulse operation,commutation must precede the aligned position by several degrees. The commutation angle must beadvanced as the speed increases.

Figs. 4.30 and 4.31 also show the importance of switching the supply voltage on before the poles beginto overlap. This permits the current to grow to an adequate level while the inductance is still low. Foras long as the inductance remains nearly constant, there is no back-EMF and the full supply voltage isavailable to force the increase in current. The turn-on angle may be advanced well ahead of theunaligned position at high speed, even into the previous zone of falling inductance.

Current regulation and voltage-PWM at low and medium speeds

The method of current regulation is a question of the timing of the voltage pulses. Broadly speakingthere are two main methods: current-hysteresis control and voltage-PWM control, but manyvariations exist on these basic schemes. The drive circuit is assumed to be the same for both methods,although several variants of drive circuit have been devised to effect various improvements in thecurrent waveform control or to reduce the cost of the controller. In both cases there is a “flux-building”interval from initial turn-on 20 to commutation at 2c, when the flux is built up from zero to its peakvalue. This interval is called the “dwell” or “transistor conduction angle”. At 2c both transistors areswitched off, and the freewheeling diodes connect the reverse of the supply voltage to the phaseterminals, causing the flux to decay to zero. The “de-fluxing interval” lasts from 2c to 2q, and in generalis shorter then the fluxing interval.

Voltage PWM (Fig. 4.37): In voltage-PWM, at least one of the two transistors in a phaseleg is switchedon and off at a predetermined frequency fchop, with a duty-cycle D which is interpreted as ton × fchop =ton/Tchop, ton being the on-time and Tchop = 1/fchop being the period at the switching frequency. Involtage-PWM there is no closed-loop control of the instantaneous current. The current waveform hasits "natural" shape at all speeds, as though the supply voltage was "chopped down" to the value D × Vs.However, for safety and protection a current-limiting function is provided such that if the currentreaches a predetermined level iHi, the current will be limited by switching off at least one of thephaseleg transistors.


TRc ' 2D (D & u )Vs ' (2q & 2c) (1 % u )Vs . (4.51)

2q & 20 ' 2D1 % D

1 % u. (4.52)

2D < "r .1 % u

1 % D. (4.53)

2D <"r

2.

1 % u

1 % D, (4.54)

Fig. 4.37 Voltage-PWM waveforms with soft chopping

With soft chopping, Q2 remains on throughout the dwell angle, Fig. 4.36. When Q1 is on, voltage Vs isconnected to the phase winding. When it is off, the winding is short-circuited through Q2 and D2. Q1is called the "chopping transistor" and D2 the "chopping diode". Q2 is called the "commutatingtransistor" and D1 the "commutating diode", because they change state only at the commutation angles20 and 2c. During the dwell angle the average voltage applied to the phase winding is D × Vs. Againusing u to represent the averaged per-unit effect of volt-drops in the resistance and the semiconductors,the flux-linkage rise in the dwell period can be equated to the flux-linkage fall in the de-fluxing period:

This can be rearranged to show that the total conduction angle is

To prevent continuous conduction, 2D must be restricted to

For example, in the 6/4 motor, if u = 0.2 and D = 0.5, the maximum dwell is 1.2/1.5 × 90 = 72E. To preventany braking torque, 2D must be restricted to

i.e. one-half of the absolute maximum, or 36E in the example. The dwell can be increased as the duty-cycle is decreased, up to the maximum given by eqn. (4.48) or (4.50).

A similar analysis can be carried out for hard chopping, in which both transistors are switchedtogether at high frequency. In both soft and hard chopping, the flux-linkage waveform increases inregular steps with a more-or-less constant average slope. Before the start of overlap, the average slopeof the current waveform is also nearly constant as the linearly-increasing flux is forced into a constantinductance. Thereafter, the inductance increases more or less linearly while the flux-linkage continuesto rise linearly. Consequently the current tends to become constant or flat-topped. Voltage-PWM tendsto produce quieter operation than current hysteresis control.

The waveforms in Fig.s 4.29 and 4.33 show that at low speed, when chopping is the preferred controlstrategy, the whole of the absolute torque zone can be used. As is evident from eqn.(4.51), the ratio ofthe slopes of the rising and falling parts of the flux-linkage waveform is approximately equal to D, sothat with a low duty-cycle (needed to "throttle" the voltage at low speed), the de-fluxing is accomplishedin a very few degrees, permitting late commutation.

Although the pole arcs do not appear in any of the equations constraining the limiting values of thefiring angles, they are important in determining their optimum values.


Fig. 4.38 Architecture of voltage-PWM controller

Fig. 4.39 Device current waveforms

The duty-cycle is typically set by the outer speed and position control loops, while the firing angles canbe scheduled with speed to optimise efficiency. Fig. 4.38 shows the concept of average torque controlwith voltage PWM, with typical voltage and current waveforms as shown in Fig. 4.37. The torquedemand is represented by the duty-cycle command signal D*, which may vary as the torque demandvaries, with consequent variation in the current. Because no attempt is made to control the currentsinstantaneously, there is no need for current sensors in individual phases. Voltage-PWM schemes maytherefore be designed with only one current sensor at the DC link for over-current protection.

Current hysteresis (Fig. 4.39): In current hysteresis control, at least one of the two transistors in aphaseleg is switched off when the current exceeds a specified set-point value iHi. It is switched on againwhen the current falls below a second level iLo = iHi ! )i, where )i is called the "hysteresis-band".Current hysteresis control maintains a generally flat current waveform, as shown in Fig. 4.29 or 4.33,with ripple determined by )i and the bandwidth of the current-regulator. At high speed, the back-EMF

may prevent the current from ever reaching iHi, and then the current waveform is naturallydetermined by the changing inductance and back-EMF as the rotor rotates (this is sometimes calledsingle-pulse mode). Fig. 4.39 shows the waveforms obtained with a hysteresis-type current regulatorand soft chopping, in which one power transistor is switched off when i > iHi and on again when i < iLo.The instantaneous phase current i is measured using a wide-bandwidth current transducer, and fedback to a summing junction. The error is used directly to control the states of the power transistors.Both soft and hard chopping schemes are possible, but only the soft-chopping waveforms are shownin Fig. 4.39. The waveforms for hard chopping are similar. As in the case of voltage-PWM, soft choppingdecreases the current ripple and the filter requirements, but it may be necessary in braking orgenerating modes of operation.


Fig. 4.40 Architecture of current-hysteresis controller

Fig. 4.41 Zero-volt loop used to achieve an interval of approximately constant flux-linkage.

Delta modulation: A variant of the current hysteresis controller is delta modulation in which thecurrent is sampled at a fixed frequency. If the phase current has risen above the reference current i*the phase voltage is switched off, and if it has fallen below i* it is switched on. The switching frequencyis not fixed but is limited by the sampling rate.

Soft braking: Soft braking is used for regenerative braking and uses the same zero-voltage loopprinciple as soft chopping. Initially both transistors are on and full supply voltage is applied until thephase current in the winding exceeds a predetermined limit, i > iHi. Both transistors are then switchedoff and !Vs is applied until the current falls below iLo. Thereafter one transistor is switched on and offto regulate the current until the end of the conduction period. The state of the switches alternatesbetween B and D, or between C and D, in Table 4.4. During the freewheeling state B or C, the flux-linkage remains approximately constant and at low speed the rate of rise of current is low, so that thisstrategy can be used to limit the switching frequency and the current ripple. At high speed, however,the current rise may be too rapid during the zero-volt periods B and C, and hard chopping may becomenecessary, in which the supply voltage Vs is used to suppress the rate of rise of current and the switchstates alternate between A and D. Generated energy is returned to the DC link during state D.

Zero-volt loop: Just as current-hysteresis control can be used to maintain constant current, zerovoltage can be used to maintain constant flux-linkage during part of the stroke. [Miller et al, 1985],[Gunnar, 1990]. An example is shown in Fig. 4.41. By this means it is possible to reduce the torquewithout current chopping and its associated losses. The peak flux is limited to a lower value. Theenergy conversion loop still makes good utilization of the available energy. This technique has beenused also for noise reduction. [Horst, 1995].


2elec ' (2mech ! U) × Nr . (4.55)

4.11 MATHEMATICAL DESCRIPTION OF CHOPPING

For analysis or computer simulation the firing angles must be defined with respect to a reference valueof rotor position, Fig. 4.42. The rotor position reference is the graph of “per-unit overlap” between thestator poles of a reference phase and any pair of rotor poles. This graph is periodic with a period of "r,the rotor pole-pitch, which defines a range of “principal values” of the rotor position 2. It is convenientto start at the “previous aligned position” and end at the “current aligned position”, A. When 2 isoutside the principal range it can be “reduced” to the principal range by adding or subtracting integralmultiples of "r. In a symmetrical machine the magnetization curves are generally defined only betweenthe unaligned and aligned positions, and it is convenient to divide the principal range into two sectionsAU and UA. If 2 is in AU, it is replaced by A!2 so that the reduced rotor angle is always in the rangeUA. A flag ST is set to !1 to represent the sign of the torque when 2 is in the AU range, and +1 whenit is in the UA range. Otherwise all calculations can proceed as if the rotor was in the range UA. If themachine is not symmetrical, the principal range cannot be divided and must extend over a completerotor pole-pitch; also the magnetization curves must be available over this entire range.

Fig. 4.42 also shows the definition of “electrical degrees”. The origin for electrical degrees is theunaligned position and one electrical cycle is equal to the rotor pole-pitch. Therefore the conversionfrom electrical degrees to mechanical degrees is

The interval between unaligned and aligned positions is always 180E(elec). In a 3-phase 6/4 motor witha pole-arc of 30E, the J position (start of overlap) is at 60E(mech) or (60!45) × 4 = 60E(elec).

Referring to the power circuit in Fig. 4.36, transistors Q1 and Q2 are turned on at 20 and off at 2c. Incurrent hysteresis control, the applied voltage during the conduction interval 2c ! 20 is Vs. In acomputer simulation, at the end of each integration step the phase current i is compared with iHi. If i> iHi, Q1 is switched off; otherwise it is switched on. In hard chopping, Q2 is switched off as well as Q1.When i falls below iLo, the chopping transistor is switched on again. In voltage-PWM, the choppingtransistor is switched on and off at the frequency fchop, with a duty-cycle D. At 2c, Q1 and Q2 areswitched off, and the reverse ("de-fluxing") voltage is !Vs. The voltage equation for one phase is:

Fig. 4.42 Definition of firing angles


Fig. 4.43 Typical variation of torque with current

where d1 is the duty-cycle with Q1 and Q2 on, d2 with Q1 off, and d3 with Q1 and Q2 off. This equationcaters for all combinations of the states of Q1 and Q2. At each integration step d1, d2 and d3 are assignedthe correct values, and d1+d2+d3 = 1 within each integration step. The compact voltage equationexpressed in this form with d1, d2 and d3 embodies all the switching states and logic required for bothcurrent hysteresis control and voltage- PWM control. In current hysteresis control, d1, d2 and d3 arescalar values that multiply the voltage terms in the equation. This is the principle of "state spaceaveraging", and is based on the notion of an upstream chopper controlling the DC source voltage, withinfinite chopping frequency. d1 is either equal to D or 0; d2 is either 1 ! D or 0; and d3 is either 0 or 1.In voltage-PWM, d1, d2 and d3 are binary states having the value 0 or 1. The states are determined bythe combined states of the transistors and diodes. Only one of the three states d1,d2 and d3 can benon-zero during one integration step.

The transistor and diode currents and their squares are accumulated in each integration step usingiQ1 = d1 × i; iQ2 = (d1+d2) × i; iD1 = d3 × i; iD2 = (d2+d3) × i; and iDC = (d1!d3) × i {= iQ1 ! iD1 = iQ2 ! iD2}. Whenthe integration is finished, the mean and mean-squared values are calculated from the accumulationsby dividing by the number of steps. This process is the same for both current hysteresis control andvoltage-PWM. An exception to this calculation is the DC link ripple current. This has to be constructedfrom the phase-shifted sums of the phaseleg currents, which flow in both directions in the DC link. Itcan be constructed from the array of samples of phase current.

4.12 REGULATION ALGORITHMS

Motor control

As already noted at the beginning of this chapter, the regulation algorithm in DC motor drives is basedon the simple linear relationship between torque and current, and the torque demand produced by thevelocity loop is translated directly into a current command by a simple constant of proportionality, thetorque constant kT. A similar principle is implemented in field-oriented AC motor drives. Thetranslation of the torque demand signal T* into a current command signal i* is the function of the“feedforward” part of the torque controller, since there is usually no torque transducer and thereforeno torque feedback. In the switched reluctance drive the relationship between torque and current isnot linear. An example is shown in Fig. 5.16.

TdR

d2' d1[Vs ! Rph i ! 2Rq i ! 2Vq]

% d2 [!Rph i ! Rq i ! Vq ! Vd]

% d3 [!Vs ! Rph i ! 2Vd]

' (d1 ! d3)Vs ! Rph i ! 2(d1 % d2)(Rq i % Vq) ! (d2 % 2d3)Vd

(4.56)


Fig. 4.44 Graphical representation of look-up table for turn-on angle

Fig. 4.45 Architecture of single-pulse controller

The torque also depends on the firing angles 20 and 2c. To complicate matters, 20 and 2c may be requiredto vary for reasons other than torque control — for example, to minimize noise or to compensate forback-EMF at high speed. The result is that the feedforward torque control must usually be implementedas a mapping from T* to i* with, with additional links to the firing angles and possibly also the supplyvoltage. The mapping can be implemented in a digital memory, with interpolation in the processor,or possibly by equations. In either case, the mapping must be computed or determined by experiment;this can be a laborious and time-consuming process. Unfortunately the mapping will be specific to aparticular motor and drive, and usually also specific to a particular application.

Fig. 4.44 shows a graphical representation of a look-up table for the turn-on angle 20 as a function ofspeed and torque demand. With single-pulse control the variation of torque with 20 and 2c is equallycomplex. The architecture of a single-pulse controller is shown in Fig. 4.45.

For closed-loop control of the phase currents both linear and non linear current regulators may be used[Kjaer, Gribble and Miller, 1996]. Linear regulators normally use proportional-integral (PI) control toeliminate the error between the reference and actual currents, and to give a smooth variation of phasevoltage with reduced torque ripple and electromagnetic noise. The main disadvantage stems from thevariation of inductance with rotor position, which can cause the electrical time constant to vary byas much as 10:1, making it difficult to tune for satisfactory transient performance.


Generator control

The switched reluctance machine will regenerate power to the DC supply if the current pulse is timedto coincide with an interval of falling inductance, and typical generator waveforms are shown in Figs.4.33 and 4.34. Excitation power is supplied by the DC source when the transistors are both on, andgenerated power is returned to the DC source when they are both switched off. The power circuits ofFigs. 4.7 and 4.8 may be used, but several variants have been published, e.g. [Radun, 1994].

In the steady state the switched reluctance machine can sustain itself in the generating mode with theDC source disconnected, but the DC link capacitor must be retained to provide excitation power duringthe “fluxing interval” during the first part of each stroke. Generally the load will be connected inparallel with the DC link capacitor, and in general its impedance will be variable and not under thecontrol of the switched reluctance generator controller. Inevitably the DC voltage will decrease duringthe fluxing interval, and increase during the de-fluxing interval when power is being returned throughthe diodes. The variation or ripple in the DC link (capacitor) voltage depends on the energy conversionper stroke, the energy ratio, and the capacitance. The controller must maintain the average DC voltageconstant in much the same way as it must maintain constant average torque in motoring mode. Thisis equivalent to the maintenance of constant torque, averaged over at least one stroke. Therefore thearchitecture of a generator controller can be similar to that of a motor controller. However, the DClink capacitance has an integrating effect such that it requires lower bandwidth to control the DCvoltage than to control the DC current (or torque). At low speed, therefore, the DC link voltage iscontrolled by varying the set-point current i* or the duty-cycle D*, and at high speed by varying thecontrol angles 20 and 2c. See Fig. 4.46. As in motoring operation, a control map is required to determinehow the control variables must vary in response to the voltage error )Vd. Various linearising schemeshave been presented to simplify the control, [Radun, 1993], [Kjaer, Cossar, Gribble, Li and Miller, 1994].

Control mode Control variables

Current hysteresiscontrol

iHi, iLo

Delta-modulation i*

Voltage-PWM control D

Zero-volt-loop mode iHi, iLo, 2z

Single-pulse control 20, 2cTABLE 4.4

CONTROL MODES WITH THEIR CONTROL VARIABLES

4.13 OPTIMISATION OF THE CONTROL VARIABLES

Table 4.4 summarizes the different control schemes for torque (in motoring operation) or DC voltage(in generating operation), and their associated control variables. Average torque can be controlled byvarying any one or indeed all of the control variables in a given mode, but the configuration dependson the performance requirements, the acceptable level of complexity, and the cost. Since there aremany possible combinations of control variables which produce the same torque, a secondary controlobjective is needed to select and define the variation of control variables for optimum performance.Examples of such secondary control objectives are efficiency, acoustic noise, and torque ripple.

4.14 REFERENCES

See Miller TJE (Ed.), Electronic control of switched reluctance machines, Newnes, 2001, ISBN 0 750650737 and Miller TJE, Switched reluctance motors and their control, Oxford University Press/MagnaPhysics Publishing, 1993, ISBN 1-881855-02-3 (available from Motorsoft). All references in brackets arelisted in this book, together with many others.


Index

Acoustic noise 13, 36Aligned 2, 3, 5-7, 9-12, 14-19, 21-26, 28, 29, 33Average torque 3, 10, 11, 16, 21, 24, 31, 36Back-EMF 4, 6-8, 10, 15, 19-21, 23, 26, 29, 35Braking 3, 26, 28-32Brushless DC motor 21Byrne 15, 16Capacitor 24, 27, 36Chopping 4-7, 10, 22, 23, 27, 29-34Classical machines 21Coenergy 8-10, 19, 20Commutation angle 15-17, 23, 24, 28, 29Constant power 16, 26Constant torque 25, 36Control 5, 6, 12, 20, 21, 25-36Converter volt-ampere requirement 14Current hysteresis 30-34, 36Current regulation 21, 27, 29Definition 2, 9, 33Drive circuit 5, 20, 21, 29Electromagnetic torque 4, 9, 10, 16, 19, 22Energy conversion 6-8, 10, 14-17, 23, 32, 36Energy ratio 6-9, 14, 15, 17, 36Energy-conversion diagram 10Energy-conversion loop 17, 22-26Equivalent circuit 8Euler’s method 19Finite-element 18-20Firing angles 21, 23-26, 30, 31, 33, 35Flat-topped current 7, 10Half-bridge phaseleg 5Hard chopping 27, 30-33High-speed generating 25High-speed motoring 23, 25Inductance 2-4, 6, 8, 12, 13, 15, 16, 18, 19, 22-26, 28-31, 35,

36Instantaneous torque 10, 19, 21, 26Linear analysis 4, 26Look-up table 35

Low-speed generating 24, 25Low-speed motoring 22, 24Magnetization curves 8, 9, 17-20, 23, 33Motor 2, 5, 7-13, 15, 16, 19-23, 28-30, 33-36Multiple-phase operation 26Multiplicity 13Nonlinear analysis 8Number of strokes per revolution 10, 11Overlap ratio 12PC-SRD 18, 19Pole arc 2, 12Pole numbers 12Ripple 8, 12, 27, 31, 32, 34-36Saturation 4, 8, 9, 15Single-pulse 28, 29, 35, 36Soft chopping 27, 29-32Stator/rotor pole number 14Stator/rotor pole numbers 12Stepper motor 2Stepper-motor 5Stored field energy 7, 9Stroke 4-8, 10-13, 17, 21-25, 28, 32, 36Stroke angle 11-13Strokes/rev 12, 13Suppression voltage 5Switching frequency 12, 29, 32Synchronous reluctance 2Torque 2-13, 16, 19-26, 28-36Torque zone 7, 12, 16, 22, 30Torque/speed characteristic 16, 23, 25, 26Turn-on angle 23, 26, 28, 29, 35Unaligned 2, 3, 6, 10, 11, 13-19, 22, 23, 26, 29, 33Vernier 12Voltage equation 4, 19, 33, 34Voltage PWM 21, 29, 31Zero-volt 27, 32, 36

5. Commutator machines

5.1 DC Commutator machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

5.1.1 Electric circuit model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

5.1.2 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

5.1.3 Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

5.1.4 Torque/speed characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

5.1.5 Operating characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

5.1.6 Time constants of DC motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

5.1.7 Resistance and inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

5.1.8 Magnetic circuit analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

5.1.9 Demagnetizing effect of armature reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

5.1.10 Torque per rotor volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

5.1.11 Armature windings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5.1.12 Winding examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Commutator machines Page 5.1

1 Large DC machines generally have field windings. They have traditionally been used in rail traction and high-power motioncontrol applications such as steel rolling mills, elevators, excavators, etc. Many of these now use AC motor drives.

Fig. 1 Electrical equivalent circuit model of DC motor

V ' ea % (Ra % Rleads ) ia % La

dia

dt% 2Vb (5.1)

Ea ' kETm (5.2)

kE 'pZMg

aB. (5.3)

5. COMMUTATOR MACHINES

5.1 DC Commutator machines

Permanent-magnet DC commutator machines are widely used in automotive auxiliaries (for example,windscreen wipers, heater blowers, cooling fans), where the supply is low-voltage DC at 12 or 24V.Because of their excellent control characteristics they are also used in low-power motion-controlapplications, where they can be controlled with low-cost electronic drives such as choppers and phase-controlled converters.1 The theory of the DC machine is also fundamental in the understanding of manyother types of motor and drive system.

5.1.1 ELECTRIC CIRCUIT MODEL

The DC motor model comprises an electrical circuit model and a magnetic field model. The electriccircuit model is shown in Fig. 1, in which V is the terminal voltage. It is determined by the supply,which can be a DC voltage source such as a battery, or a chopper, or a phase-controlled rectifier.

The motor model is represented by a single differential equation,

in which ia is the instantaneous armature current, ea is the instantaneous back-EMF, Vb is the volt-dropin each brush, Ra is the armature resistance, Rleads is the resistance of the leads, and La is the armatureinductance. With pure DC, dia/dt = 0 and the instantaneous quantities become constant DC quantities,ia = Ia and ea = Ea. The circuit equation then reduces to Ia = (V ! Ea ! 2Vb)/(Ra + Rleads). With a chopperor phase-controlled rectifier drive, the differential equation (1) must be solved as described in Ref. [6].

The parameters Ea, Ra and La can be calculated from the motor geometry, the winding details, and theproperties of the steel and magnet materials specified via the materials database. Ea is obtained fromthe magnetic circuit model in the classical form

where the back-EMF constant kE is given by

The flux/pole Mg is calculated in the magnetic circuit model. Z is the total number of armatureconductors, p is the number of pole-pairs, and a is the number of parallel paths. Tm is the angularvelocity in rad/sec, equal to 2B × RPM/60.

Page 5.2 SPEED's Electric Motors

T ' kT ia ! Trot Nm. (5.4)

Tm 'Ea

kE

'V ! 2Vb ! Ra Ia

kE

'V ! 2Vb

kE

!Ra

kE2

(T % Trot ) (5.6)

Tm ' T0 !Ra

kE2

T (5.7)

T0 'V ! 2Vb

kE

!Ra

kE2

Trot . (5.8)

5.1.2 TORQUE

In S.I. units the back-EMF constant kE is equal to the torque constant k T, which is used to calculate theelectromagnetic torque Te = kTia. The torque constant is one of the most important parameters of a DCmotor and is usually quoted in catalogue data. It can be measured by running the machine on adynamometer and plotting the torque vs. current, then measuring the slope of the resulting graph. TheEMF constant kE can be measured by driving the machine with an auxiliary motor and measuring thegenerated voltage. Generally this has a linear variation with speed. The measured k T is often less thankE because of friction, and at high current the armature reaction distorts the magnetic flux and reducesk T still further. In spite of the "saturation" of k T at high current, the linearity of Te vs. ia is animportant characteristic of the PM DC motor, because it ensures good behaviour in control systems.

The shaft torque T is equal to the electromagnetic torque minus the rotational loss torque Trot, whichincludes the windage and friction loss. The iron loss is also often lumped in Trot.

5.1.3 EFFICIENCY

The mechanical power output is TmT and the electrical input power is VIa, so the efficiency is

The copper loss is calculated from I2Ra, using the r.m.s. value of Ia. The iron loss WFe can be calculatedusing the modified Steinmetz equation, taking into account the variation of the flux-density in the teethand the rotor core as the rotor rotates.

5.1.4 TORQUE/SPEED CHARACTERISTIC

We can now derive an equation relating the speed Tm to the shaft torque T :

or

where T0 is the no-load speed:

The torque/speed characteristic given by eqn. (8) has the form shown in Fig. 3 in normalized form. With a fixed supply voltage, normal operation at rated load would be at a speed slightly below the bo-load speed, preferably in the neighbourhood of maximum efficiency.

The speed can be varied by introducing series resistance, which steepens the negative slope of thetorque/speed characteristic. This is inherently wasteful. If, for example, the series resistance issufficient to reduce the voltage at the motor terminals to 50% of the supply voltage, then the powerdissipated in the series resistance must be equal to the entire input power to the motor, so the overallefficiency must necessarily be less than 50%.

A better way to control the speed is by varying the supply voltage V, which varies the no-load speed T0and therefore raises or lowers the whole speed curve. DC choppers can be used for this when the supplyis DC; or phase-controlled rectifiers when the supply is AC.

0 'output powerinput power

'Tm T

VIa

× 100%. (5.5)


0 'Tm T

Tm T % W1 % kT 2 (5.9)

Fig. 2 Torque/speed characteristics of PM DC motor

V ' Ra Ia % kETm (5.11)

Te ' kT Ia (5.12)

5.1.5 OPERATING CHARACTERISTICS

Normalized (i.e., per-unit) operating characteristics at constant voltage are shown in Fig. 2. There isa peak efficiency and a peak output power, occurring at different loads. The load torque at which thepeak efficiency occurs depends on the way in which the losses vary with torque and speed.

Maximum efficiency — Suppose the total power loss comprises a fixed loss W1 and a variable loss W2= kT 2, i.e., the variable loss varies with the square of the torque (effectively, with the square of thecurrent since torque is proportional to current). The efficiency 0 is given by

and if this is differentiated and d0/dT set to zero, the condition for maximum efficiency is found as W1= W2 : i.e., the maximum efficiency occurs when the variable loss is equal to the fixed loss.

5.1.6 TIME CONSTANTS OF DC MOTORS

In the steady state, assuming no load and neglecting rotational losses, the speed is given by

This steady-state equation gives no indication as to how the speed varies dynamically if the supplyvoltage V is changed, as it might be by the action of the electronic controller. The dynamic responsecan be determined from the transfer function between speed and voltage, which is derived as followsfrom the equations that describe the system. These equations are:

where J = Jm + JL, is the total inertia of motor + load, and " is the angular acceleration, dTm/dt.

Tm 'VkE

. (5.10)

Te ' J " ' JdTm

dt. (5.13)


V ' La

dIa

dt% Ra Ia % kETm . (5.18)

Te(s ) '(V ! kETm(s ) )kT

La s % Ra

' (Js % D )Tm(s ) . (5.22)

V(s ) ' (La s % Ra )Ia (s ) % kETm(s ). (5.19)

" ' s Tm (5.14)

Te(s ) ' (Js % D )Tm(s ) (5.21)

Te ' JdTm

dt% DTm % Trot % TL . (5.20)

Tm 'Te

Js'

kT Ia

Js'

kT (V & kETm)

Ra Js. (5.15)

Tm (s)

V(s)'

1/kE

1 % sRa J

kT kE

'1/kE

1 % sJm

.(5.16)

Jm 'Ra J

kT kE

. (5.17)

Assuming that the speed is zero at time t = 0, in terms of Laplace transforms,

so that

This can be rearranged to give the transfer function between speed and voltage as

The denominator has only one root, so the transfer function has only one pole at s = !1/Jm, and thesystem is said to be a first-order system. Jm is called the mechanical time constant:

After a disturbance such as a step-change in voltage, the speed will settle exponentially to a final steady-state value, because the transfer function is of the form 1/(s + a), which has an inverse transform thatis a decaying exponential function of time. After an infinite time, in the steady-state all derivatives arezero and the corresponding condition in the frequency domain is s = 0. If this is substituted in eqn. (18)the s-term in the denominator disappears and the transfer function degenerates to the steady-stateexpression 1/kE, as in eqn. (12). This value is called the gain of the system.

Effect of inductance — the electrical time-constant

The simple dynamic motor model can be extended to include the effects of inductance as follows. Theelectrical equation of the motor is

where V is the terminal voltage and brush volt-drops are neglected. By Laplace transformation,

The mechanical equation of the motor is

where DTm is an additional "viscous" torque sometimes described as a damping torque that isproportional to speed. Of most interest is the transfer function between speed and voltage, and thesimplest case is at no load, i.e. with TL = 0. If we also neglect Trot, then by Laplace Transformation,

From equations (5.14),(5.21) and (5.21),


Tm(s)

V'

kE

(La s % Ra ) Js % D % kEkT

. (5.23)

G(s) 'Tm(s)

V(s)'

1/kE

(sJe % 1)(sJm % 1)(5.24)

Je .La

Ra

, Jm .Ra J

kEkT

. (5.25)

La ' Lslot % Lgap % Lend , (5.27)

Ra ' DCu ×Z×LT /2a 2

Nsh ×BdCu2 /4

(5.26)

From this the transfer function between speed and voltage can be extracted as

The poles of the system are the roots of the denominator of eqn. (25), which is quadratic. There aretherefore two roots, and it is common to write eqn. (25) in a form which separates these two roots: thus

where Je is the electrical time constant and Jm is the mechanical time constant. As before, 1/kE is thesystem gain. Assuming no motor damping (D = 0) and low inductances, the time constants are

5.1.7 RESISTANCE AND INDUCTANCE

The armature resistance Ra is calculated from the winding details at the operating temperature:

where Nsh is the number of strands-in-hand in each conductor, dCu is the strand diameter, DCu is theresistivity of copper, and LT is the mean length of turn.

The armature inductance La is also calculated from the winding details and the motor geometry. Thiscalculation is considerably more complex than the resistance calculation, and the theory is given in fullin Ref. [8]. Basically there are three components of armature inductance,

where Lslot is the slot-leakage component, Lgap is the airgap component, and Lend is the end-windingcomponent. The airgap component Lgap is liable to be affected by frame saturation especially in 2-polemotors, or where the frame is very thin. Therefore, for work requiring very accurate values of La itmight be advisable to use finite-element calculations to check the value.

The inductance is important for two particular reasons. One is in the formulation of the electrical time-constant Je = La /Ra, which is sometimes important in determining the dynamic response, especiallyfor servo-type applications. The other is in the calculation of the current waveform when the motor isdriven from a chopper or phase-controlled rectifier. In these cases the current waveform may besensitive to the value of La, and if an accurate value is not used, the current and the torque will becalculated incorrectly.


Fig. 3 No-load flux distribution in PM DC motor

Fig. 4 Flux-density waveform under one pole at no-load

5.1.8 MAGNETIC CIRCUIT ANALYSIS

The flux produced by the magnet at no-load has the general form shown in Fig. 3. The magnet flux Mmcan be separated into two components: the larger of these is the airgap flux Mg = fLKG × Mm, whichcrosses the airgap and links the armature winding. The fraction fLKG is called the leakage factor, anda typical value is 0.9. The leakage flux component (1 ! fLKG)Mm does not link the armature winding.

Fig. 4 shows a graph of the airgap flux-density under one pole at no-load. The peak value is Bg and theairgap flux per pole Mg is taken to be the integral of B over 180Eelec, over the whole armature length.If $M is the magnet arc in Eelec, we can get a very rough estimate of the average airgap flux-density B= Bg × $M/180. Then Mg = BAg, where Ag = Larm × BDarm/2p is the airgap area per pole, Larm being thearmature length and 2p being the number of poles. Darm is the rotor diameter.

A more detailed analysis must include the leakage flux and the iron teeth, the rotor core, and the frame.For this purpose a lumped-parameter equivalent-circuit is used, Fig. 5. The permeances in this modelcorrespond roughly to the main flux-paths in Fig. 3. They are calculated from the geometry and fromthe material characteristics. The iron parts are non-linear, so the solution method is iterative.

Under load conditions the armature current produces a MMF distribution ("armature reaction") thatdistorts the airgap flux-density waveform. This effect is less severe in PM motors than in wound-fieldmachines. For accurate analysis of the magnetic field, it is desirable to use a finite-element analysis.


Fig. 5 Magnetic equivalent circuit

Bm ' Br × PCµrec % PC

. (5.29)

Fig. 6 Magnet operating point P at the intersection of the load line and the recoil line, which is obtained from the B-H curve.The H-axis is scaled by µ0 so that both axes are in [T] and a recoil line with µrec = 1 has a slope of 1.

Magnet operating point

The magnet operates at a point P on the recoil line shown in Fig. 7, where Bm = Mm/Am. The recoil lineis described by the equation

where µrec is the relative recoil permeability and Br is the remanence. The magnetic circuit must bedesigned so that the operating point (Hm,Bm) lies on the straight part of this line, to avoid irreversibleloss of magnetization. Fig. 7 also shows the non-linear effect of saturation on the load line OP.

The slope of the load line is known as the permeance coefficient PC, and

Bm ' Br % µrec µ0 Hm , (5.28)


Fd 'ZIa /a

2p × 2×

$M

180. (5.30)

TRV ' 2BQ Nm/m 3 (5.33)

Hm LM % Hg g ' Fd (5.31)

Q 'Z × Ia /a

BDarm

. (5.34)

Hm '

±Fd !gµ0

Br

LM % µrec g. (5.32)

Fig. 7 Effect of armature reaction on magnet operating point

5.1.9 DEMAGNETIZING EFFECT OF ARMATURE REACTION

The armature ampere-conductors are distributed uniformly around the rotor in blocks of alternatingpolarity, such that the ampere-conductors under one magnet pole are all in the same direction, Fig. 8.Under loaded conditions the trailing pole tips of the magnets are subject to a demagnetizing MMF Fd :

where p is the number of pole-pairs and $M is the magnet arc in Eelec. The effect of Fd on the magnetoperating point can be calculated by combining eqn. (33) with Ampère’s Law,

where g is the effective airgap length. Assuming that Bm . Bg = µ0 Hg,

The positive value of Fd applies at the leading pole-tip and the negative value at the trailing pole-tip. Thedemagnetization current Idemag can be extracted from eqns. (5.30) and (5.32) after setting Hm to thecoercivity, which is usually taken to be the intrinsic coercivity HcJ.

5.1.10 TORQUE PER ROTOR VOLUME

For DC machines the torque per rotor volume TRV is given by the output equation in the form

where B is the average flux-density in the airgap in [T] and Q is the electric loading in [A/m], given by

If J is the RMS current density in the armature conductors, and each conductor has a cross-sectionAcond, then JAcond = kf Ia/a, where kf = Irms/Ia is the current form factor, i.e., the ratio of RMS to averagecurrent. Note that the average magnetic flux-density in the airgap at no-load is called the magneticloading.


Fig. 8 Armature amp-conductor distribution

Fig. 9 Single coil in lap-wound DC armature

5.1.11 ARMATURE WINDINGS

The conductors are laid in slots on the rotor. The ampere-conductors under each magnet pole must allbe in the same direction, so that all the ampere-conductors produce torque in the same direction. Sincethe magnet polarities alternate NSNS..., the polarities of the ampere-conductors must also alternatewith the same "pitch" or "wavelength", Fig. 8.

There are two main forms of winding: the lap winding, Fig. 9, and the wave winding, Fig. 10. In the lapwinding the coil ends are terminated on adjacent commutator segments, but in wave windings they areapproximately 360Eelec apart; see Figs. 11, 13 and 14. In small machines the winding is usually woundin one operation on a winding fixture, but large machines use separate form-wound coils.

Fig. 10 Single coil in a wave-wound DC armature


Fig. 11 Front pitch yC and back pitch yB. (a) Progressive lap; (b) retrogressive lap; (c) progressive wave

CSL 'CNs

. (5.35)

In DC armatures all coils have the same throw, that is, the number of slot-pitches between the "go" and"return" coilsides. In Figs. 9 and 10 the throw is 4 slot-pitches. The throw is also known as the span orback pitch of the coil, yB. The throw must be chosen so that the coil links all the flux of one magnetwhen its axis is aligned with the magnet axis. If there are Ns slots and 2 p poles, the pole-pitch is Ns/2pslots or 2B/Ns radians or 360/Ns × p Eelec. The magnet arc $M is usually rather less than the pole-pitch,perhaps 120Eelec. The coil throw yB should be chosen such that Ns/2 p > yB > $M (in consistent units).

The ends of the coils are connected to commutator segments. The front pitch of a coil is the number ofcommutator segments between the two ends of the coil, yC. Coils can be laid up in two ways,progressive and retrogressive, depending on the value of yC compared to the number of commutatorsegments per pole. Table 1 summarizes the permissible values of yC. C is the number of commutatorsegments, and x is known as the plex or multiplicity of the windings. If x = 1 the winding is simplex; ifx = 2 it is duplex, x = 3 triplex, and so on. The overwhelming majority of windings are simplex (x = 1).Multiplex windings are interleaved and may be independent of each other.

Lap windings Wave windings

yC = +1 progressive yC = (C + x)/p progressive

yC = !1 retrogressive yC = (C ! x)/p retrogressiveTABLE 1

Multiple coilsides per layer

It is common to use multiple coilsides per layer (CSL) in each slot and to connect the ends of thecoilsides in one layer to CSL adjacent commutator segments, Fig. 12. This reduces the leakageinductance of the coil which is being shorted by the brush as it undergoes commutation, that is, thereversal of the direction of its current. It is usually at the position where its flux-linkage is maximumand its induced voltage dN/dt is ideally zero, so that the brush can reverse the polarity of the currentunder zero-voltage conditions, which helps to avoid sparking. However, the coil flux-linkage includesleakage flux components in the end-windings and in the slots. Unlike the main flux, these are directlyproportional to the current in the coil or coils undergoing commutation, and the reversal of currenttherefore induces a voltage (sometimes called the reactance voltage) between the commutator segmentsthat are shorted by the brush. If the reactance voltage exceeds about 2!3 V, sparking will occur. Clearlythis is liable to increase when the armature current increases. Therefore it is always desirable todesign for minimum leakage inductance by using multiple coilsides per slot, and by other means suchas wide slot-openings and shallow slots. In general


Fig. 12 Multiple coilsides/layer; CSL= 3

CSL is also known as the "number of bars per slot". It should not be confused with the multiplicity orplex of the winding.

Numbers of parallel paths and brushes

In wave windings there are always two parallel paths through the winding, a = 2, but in lap windingsthe number of parallel paths is equal to the number of poles: a = 2p. The number of brushes b in lapwindings is also equal to the number of poles: thus a = b = 2p. In wave windings only two brushes areneeded to make the electrical circuit, but in practice it is common to use b = 2p, i.e. p sets in parallel,so that the commutator can be made shorter and the current density per brush can be decreased. Theserelationships are summarized in Table 2.

Lap windings Wave windings

a = 2p a = 2

b = 2p b = 2 or b = 2p

TABLE 2

5.1.12 WINDING EXAMPLES

Fig. 13 shows a progressive lap winding in a 15-slot, 4-pole armature with 2 coilsides/slot. The throwor back pitch is yB = 3 slot-pitches, and since the slot-pitch is 360/15 × 4/2 = 48Eelec, yB = 3 × 144Eelec.In this machine the magnet arc is $M = 120Eelec and yC = 1. Fig. 14 shows an example of a retrogressivewave winding with yC = 7. In both examples CSL = 1.


Fig. 13 Progressive lap winding with 2 coilsides/slot in a 15-slot, 4-pole armature. yB = 3, yC = 1

Fig. 14 Retrogressive wave winding with 1 coilside/slot in a 15-slot, 4-pole armature. yB = 3, yC = 7


REFERENCES

[1] Hamdi, ES [1994] Design of small electrical machines, John Wiley & Sons, Chichester, ISBN 0 47195202 8

[2] Perrine, R [anticipated 1999] DC motor design : contact Motorsoft Inc., P.O. Box 442, 30 E.Mulberry St., Suite 1, Lebanon, OH 45036, USA [e-mail: [email protected]]

[3] Gieras JF and Wing M [1997] Permanent magnet motor technology, Marcel Dekker, New York,ISBN 0-8247-9794-9

[4] Clayton AE and Hancock NN [1966] The performance and design of direct current machines,Pitman & Sons Ltd., London

[5] Kostenko M and Piotrovsky L [1974] Electrical machines, MIR Publishers, Moscow

[6] Sen PC [1981] Thyristor DC drives, John Wiley & Sons Inc, New York, ISBN 0-471-06070-4

[7] Staton DA, McGilp MI, and Miller TJE [1994] Interactive computer-aided design of permanent-magnet DC motors. IEEE Transactions on Industry Applications. Vol. 31, No.4, July/August 1995,pp 933-940

[8] Miller TJE, McGilp MI, Staton DA, Bremner JJ [1999] Calculation of inductance in permanent-magnet DC motors, IEE Proceedings on Electric Power Applications., Vol 146, No. 2, March 1999,pp. 129!137.

[9] Moczala H et al : [1998] Small electric motors, IEE Power and Energy Series No. 26, London, ISBN0 85296 921 X

[10] The Electro-craft engineering handbook, Reliance Motion Control, Inc.


Index

Armature inductance 2, 6Armature resistance 2, 6Armature windings 10Automotive 2Back pitch 11, 12Back-EMF 2, 3Back-EMF constant 2, 3Battery 2Brush 2, 5, 11, 12Brushes 12Chopper 2, 6Demagnetizing 9Duplex 11Efficiency 3, 4Electric circuit 2Electric loading 9Flux/pole 2Form factor 9Front pitch 11Inertia 4Iron loss 3kE 2, 3, 5, 6kT 3Lap winding 10, 12, 13Losses 4Magnet operating point 8, 9Magnetic circuit analysis 7Magnetic loading 9Maximum efficiency 3, 4

Multiplex 11No-load speed 3Operating characteristics 4Parallel paths 2, 12Permeance coefficient 8Phase-controlled rectifier 2, 6Pitch 10-12Plex 11, 12Power 2-4, 15Progressive 11-13Reactance voltage 11Rectifier 2, 6Retrogressive 11-13Saturation 3, 6, 8Series resistance 3Shaft torque 3Simplex 11Slot-leakage 6Span 11Throw 11, 12Time constant 5, 6Torque 3-6, 9, 10Torque constant 3Torque per rotor volume 9Torque/speed characteristic 3Triplex 11Wave winding 10, 12, 13Windings 2, 10-12

SPEED 's ELECTROMAGNETIC PRIMER

1. INTRODUCTION

Electromagnetics can be a bit confusing?

Not to worry! It took the philosophers about nineteen centuries to figure out eventhe rudiments of the subject. So we shouldn’t feel too discouraged.

After the philosophers, the engineers got to work. Working about 100 times fasterthan the philosophers they applied the subject so vigorously that by 1900 many ofthe fundamental torque-producing mechanisms had been developed, including thed.c. commutator machine, and a.c. synchronous and induction machines. Thetransformer, too, came at about this time. Even though it produces no torque, it 'squite a useful thing.

Thanks to the early engineers, the form of all these electromagnetic energyconverters is such as to exploit the fundamental electromagnetic principles intheir simplest form . In other words, electric machines is a good place to begin thestudy of electromagnetics.

All electromagnetic energy converters comprise two fundamental elements :

(a) an electric circuit, and

(b) a magnetic circuit.

With the exception of the transformer, they all also have a third element :

(c) a mechanical system of moving parts.

We’ll forget this and concentrate on the electric and magnetic parts.

Page 2 SPEED's Electromagnetic Primer

1In the United States, Joseph Henry made the discovery of electromagnetic induction atabout the same time as Faraday. Faraday is accorded priority because he was first to publishhis results. Henry 's name is used as the unit of inductance.

2The word "flux" means "flow", and magnetic f lux is mathematical ly analogous to otherquantities which obey Laplace 's equation and are known as "streamlines". Many examples existin aerodynamics, f luid flow, electrostatics, and other fields.

e ' N dNdt

(1)

2. ENGINEERING EFFECTS OF MAGNETIC FIELDS AND ELECTRIC CURRENTS

There were three natural philosophers who did more than any others to get theengineers started in their exploitation of electromagnetic phenomena. They wereOersted, Ampere and Faraday1, and their most important work was done between1820 and 1831. Our modern view of electromagnetics owes much to Clerk Maxwell,who formulated the basic laws mathematically as Maxwell's Equations. Othersmade fundamental contributions in the application of the electromagnetic laws.One of these is Steinmetz, who is credited with some of the basic early work onphasors and the properties of electrical steels.

It is perhaps best if we take the three original discoverers in the wrongchronological order.

Faraday discovered that an electromotive force (EMF), and from it a current, couldbe generated by means of a changing magnetic field. The change in the magneticfield can arise in two ways, corresponding to the transformer and the "dynamo-electric" machine.

A coil of copper wire which is stationary in a magnetic field of varying intensityhas a voltage (EMF) induced in it. This is "transformer action".

A coil which is moving relative to a magnetic field of fixed intensity also has anEMF induced in it. This is "generator action" (sometimes called "flux-cutting").

Faraday formulated this discovery precisely: the EMF induced is proportional tothe number of turns, N, on the coil; and is also proportional to the rate of changeof what he called “magnetic flux”, N . Thus

But what is magnetic flux? What is a magnetic field?

Flux is an abstract concept, justified by its usefulness.2 It is so useful, and we talkabout it so freely, that it is easy to come to think that it really exists. In realitythe only evidence for i ts existence is the very phenomenon it is supposed toexplain — induced voltage!

SPEED's Electromagnetic Primer Page 3

3The l ines are defined by another property of the field which is that they are the pathswhich would be fol lowed by isolated magnetic poles free to move in the field.

N ' B dA. (4)

N '1N

e dt % constant (2)

N ' BA . (5)

i

i

Flux

Fig. 1 Flux l inking a coil

F ' BLi (3)

We usually think of flux in terms of flux lines, or “lines of force” which link thecoil. This is how Faraday thought of it. 3

So far, the only way we can define the flux N numerically is in terms of Faraday’sLaw itself. (We can measure e , but we have no independent way of measuring N).Solving eqn. (1) for N , we get

The concept of magnetic flux is suggested by another (earlier) experiment ofFaraday’s which is of equal important in applications: that there is a physicalforce acting on a conductor which carries current through a magnetic field. Thisis the historical basis of motor technology. Faraday formulated this discoveryprecisely also: the force is proportional to the current, to the length of theconductor and to what we now call the magnetic “flux-density”, B :

The three quantities F , B and i are mutually at right angles in space. The flux-density B can be thought of as the number of flux lines passing through an areaof one square metre. In general the flux passing through any area A is given bythe simple integral,

If the flux-density is uniform (i.e. if B does not vary across the area A), then thisrelationship simplifies to


ŠH .ds ' H × 2Ba ' Ni, so H 'Ni

2Ba. (7)

a H

Ni

Fig. 2 Magnetic field produced by a current

ŠH .ds ' i (6)

3. THE ESTABLISHMENT OF MAGNETIC FIELDS BY ELECTRIC CURRENT

Although Faraday’s discoveries give us a precise way of quantifying flux and flux-density [eqns. (1) and (3)] , they cannot be used for engineering design andanalysis unless we can calculate N or B by some other, independent, method. Forexample, we cannot predict the EMF induced in a coil unless we already know therate of change of the flux linking it.

Up to now we’ve probably been vaguely aware that the flux discussed in Section2 was caused or established by another current not shown in the diagrams, orperhaps by a permanent magnet. This is true. The basic law governing theestablishment of magnetic fields by electric current was formulated by Ampere,following experiments by Oersted which shows that electric currents do indeedestablish magnetic fields.

Ampere’s Law says that there is a magnetizing force, or magnetic field strengthH, encircling any electric current and that the integrated value of H around anyclosed contour surrounding the current is equal to the current itself. Thus,

where ds is an element or the path. The current is sometimes called the total"magnetomotive force" (MMF) in this context. If there are I strands or turns ofconductor, each carrying the current i , then the M M F is F = Ni . The simplestexample is that of a long, straight wire. If we take our closed contour to be anycircle of radius a with the current Ni at the centre, then by symmetry H is thesame all the way around and

H in this case is in the circumferential direction, as shown in the diagram.


Iron

Air

B

H

µ0

Fig. 3 B/H relationship for iron and air

B ' µH. (8)

µo ' 4B× 10! 7 H/m (9)

4. MATERIAL EFFECTS : PERMEABILITY

We now have two apparently independent systems of equations for describing themagnetic field; one in terms of B (derived from Faraday) and one in terms of H(derived from Oersted and Ampere). We have to relate these two systems to oneanother to get a complete set of equations capable of describing the magnetic field.

It turns out that the relationship between B and H is a property of the medium(i.e ., material) in which they are measured. This property is called thepermeability, µ, and is the ratio between B and H:

Since B has the units Wb/m2 and H has the units A/m, it turns out that µ has theunits of inductance per unit length, H/m, and it has sometimes been called the“specific inductivity” of a material.

In air, µ has the value

In iron, µ is not constant and we often represent the relationship between B andH graphically:

For low flux-densities, µ in iron can be hundreds and even thousands of timesgreater than µ0. We say that iron is "highly permeable" and observe that in irononly a small value of H is required to achieve a large value of B .


g

Fluxφ

Ni

Iron ring

Fig. 4 Effect of an airgap in a magnetic circuit

As an example, suppose we enclose the current of a previous diagram by an ironring. H inside the iron will have the same value as before (by symmetry), i.e. H =i/2Ba . But B has the value µH, which is much larger than the value outside thering, or in the absence of the ring.

The high flux-density exists only within the iron. To get access to it (for example,to produce a force on a current carrying conductor), we must cut a gap in the ring,as shown :

What is the flux-density in the gap?

If we assume now that the iron is infinitely permeable then, since Ampere’s Lawstill holds for H, the integral ŠH.ds must be developed entirely across the gap, andis equal to H × g . This is equal to i (or Ni if there are N turns or strands). In otherwords, all the MMF is expended in forcing flux across the gap and none in the iron.The flux density in the gap is just µ0Hg = µ0Ni/g .

If, instead of a long straight wire, we have a complete coil wound around the iron,the topology becomes essentially the same as that of an electric machine ortransformer, except that the moveable parts are still missing.

Another way of looking at the high permeability of iron is in terms of "inducedmagnetization". Permanent magnets acquire this induced magnetization whenthey are magnetized (by a coil with a high current), and they have the property ofretaining this magnetization when the current is switched off, so they tend tosustain the flux. This shows that the permanent magnet has an internal"excitation" similar to that of the original magnetizing current.


B .dS ' 0 (10)

B .dS ' B1 A1 & B2 A2 ' 0

B1 A1 ' B2 A2

B1

B2

'A2

A1

> 1

(11)

It may bother some people as to why we have two parameters (B and H) to describemagnetic fields. What’s the difference between them anyway?

(a) B obeys Faraday’s Law (through N = BA): H does not.

(b) H obeys Ampere’s Law: B does not.

Therefore we have no choice but to use both. To solve magnetic problems we mustrelate Ampere’s Law with Faraday’s Law and this we do through the"constitutive" relationship B = µH.

5. GAUSS’ LAW

Gauss' Law states that magnetic flux always flows in closed loops. Mathematically it is stated as

where dS is an element of the area over which B is being integrated. This is amathematical way of stating that there are no isolated magnetic poles or"charges". If we draw any closed surface, the flux entering must equal the fluxleaving. A simple example is the polepiece of a laboratory magnet that is taperedto concentrate the flux, i.e. to increase the flux-density :

An alternative to the "integral" formulation of Gauss’ Law is the "differential"formation: div B = 0, which is one of Maxwell's equations.

Units — In the SI system of units, A is measured in square metres and N inWebers. One Weber equals one volt-second, which is clearly the correct unit foreqn. (2), if e is measured in volts and t in seconds. Consequently, B is measuredin Webers per square metre (Wb/m2) and this unit is called the Tesla, after theinventor of the induction motor. In the USA, B is often measured in lines persquare inch, the flux being measured in lines (also called "maxwells"). One Weberequals 108 lines.

B A1 1

B A2 2


6. SUMMARY

The basic laws of magnetics are :

(a) Faraday’s Law e = N dN/dt

(b) Ampere’s Law m H.ds = Ni

(c) Material property B = µ H between B and H in any material

(d) Gauss’ Law m B.dS = 0

7. EXTRAS

Permanent magnets behave like sources of flux, but with very low permeabilitycalled the recoil permeability. They are described by the equation

where Br is the remanence and µrec is the relative recoil permeability (i.e., relativeto µ0). This equation describes the so-called recoil line , which is anapproximation to a very narrow minor hysteresis loop along which the magnetoperates.

Flux-linkage is often defined as the product of "flux × turns", NN . In magneticdevices it is always the case that "not all the flux links all the turns", and thisgives rise to a distinction between "leakage flux" and "main flux" or "mutual flux".But flux-linkage is always the integral of the induced voltage e in eqn. (1).

Vector potential is another magnetic field variable like B or H (and like themit is a vector quantity). It is defined such that B can be derived from it bydifferentiation. The two vital things to know about vector potential are

(a) it combines two or more Maxwell's equations into a single equation(Laplace's or Poisson's equation) that can be solved by the finite-element method; and

(b) the flux linkage of a coil can be calculated directly from the differencein vector potential between the two coil sides.

B ' µrec µ0 H % Br (12)


Problems

1. Foundation

Sizing, gearing, permanent magnets, permanent-magnet equivalent circuits, cooling

Answers

1.2 B

1.3 B

1.4 A

1.5 A — stator; B — rotor

1.6 n H@@@@dl ' E Ni

1.7 C

1.8 *BmHm*

1.9 398 J

1.10 C

1.11 Y

1.12 36.4 mm; 51 mm

1.13 14.4 kNm/m3; 7.7 kN/m2 or 1.04 psi

1.14 3.6 A/mm2

1.15 4.49 × 10!3 kg-m2; 3.35; 5,225 rad/s2

1.16 (a) 17@1 A/mm; (b) 18@7 kNm/m3, 9.35 kN/m2; (c) 823 kW/m3.

1.17 (a) 145 kW/m3; (b) 1@92 kW/m2; (c) 215EC; (d) 49@5EC.

1.18 (a) 1.37EC/W; (b) 32 min; (c) 765 W; (d) 51.5 min.

1.19 PC = 2.618; Bm = 0.2894 T; Bg = 1.1056 T; Energy product = 25.5 kJ/m3 (3.2 MGOe)

1.20 PC = 1.8933; Bm = 0.6544 T; Bg = 0.6567 T; Energy product = 180 kJ/m3 (22.6 MGOe)

1.21 PC = 3.9184; Bm = 0.7967 T; Bg = 0.2033 T; Energy product = 130 kJ/m3 (16.2 MGOe)

1.22 (a) 0.55; (b) Bm = 0.4 T, Hm = 111.4 kA/m; (c) 23%; (d) 0.27.

Full solutions are given at the end

1.1 Sketch the complete B-H curve for a typical ‘hard’ permanent magnet. Indicate the remanentflux-density, the coercive force, and the recoil permeability.

1.2 In which of the following electrical machines would you expect to find permanent magnets?

A Induction motorB Brushless DC servomotorC Variable-reluctance stepper motor

1.3 Which of the following motors never has permanent magnets?

A DC commutator motorB induction motor

1.4 Which of the following materials is a 'hard' permanent magnet?

A Samarium CobaltB Stainless steelC Gallium Arsenide

1.5 In the following motors, are the permanent magnets on the rotor or the stator?

A DC commutator motorB Brushless DC motor

1.6 State Ampere's Law.

1.7 Which of the following units are identical to the henry?

A Wb/At (Webers per ampere-turn)B At/Wb (Ampere-turns per Weber)C V-s/A (Volt-seconds per ampere)

1.8 Define the energy product of a permanent magnet.

1.9 If magnetic flux could be bottled, how much energy would be stored in a one-litre bottlecontaining flux at a uniform density of 1.0T ?

1.10 Large magnets are sometimes assembled from smaller magnets in much the same way as a wallis built from bricks. It is necessary to test the polarity of each small magnet before adding it tothe assembly. If the material is high-energy Neodymium-Iron-Boron, which of the followingpolarity tests would you recommend? Give reasons for your choice.

A floating the small magnet on a cork in water and noting which way it points in theearth's field;

B offering the small magnet up to a compass and noting which way the compass needleswings;

C offering the small magnet up to another small magnet whose polarity is known, andnoting whether the poles attract or repel.

1.11 An electric motor contains coils and magnets and the flux is fixed in magnitude. Can theflux-linkage of any coil vary? Y — yes, N — no.

1.12 Calculate the rotor diameter and length of a motor that develops 100 W shaft power at 1500 rpmwith a torque per unit rotor volume of 12 kN-m/m3 and a rotor length/diameter ratio of 1.4.

1.13 An AC motor has an electric loading of 12 A/mm and an average airgap flux-density of 0.6 T.What is the torque per unit rotor volume and the electromagnetic shear stress in the airgap?Assume a fundamental winding factor of 0.9, and assume that the rotor flux is orthogonal to thestator ampere-conductor distribution.

1.14 In the motor of problem 1.13, the slot fill factor is 0.42, the slot depth is 18 mm, and thetooth-width to tooth-pitch ratio is 0.56. Estimate the current density in the winding.

1.15 A brushless servo motor has a peak rated torque of 14 Nm and a torque/inertia ratio of 35,000rad/s2. It is found to be capable of accelerating an inertial load at 4,600 rad/s2 through a 2:1speed-reduction gearbox. Estimate the inertia of the load in kg-m2. What value of gear ratiowould result in the maximum acceleration of this motor and load combination, and what wouldbe the value of the maximum acceleration?

1.16 (a) An AC induction motor has 24 slots, each with a slot area of 62.2 mm2. The stator borediameter is 52.2 mm. The factory can manufacture windings with a slot-fill factor of47%. If the maximum current density allowable in the copper is 4 A/mm2, calculate themaximum electric loading.

(b) The fundamental winding factor of the induction motor in part (a) is 0.911. The motorcan be operated with a peak airgap flux-density of 0.85 T, sine-distributed. Calculate thetorque per unit rotor volume in kNm/m3 and the airgap shear stress in kN/m2, using theelectric loading from part (a).

(c) If the stator OD is 106 mm and the stack length is 57 mm, calculate the power per unitstator volume when the motor delivers its rated torque at 1745 rev/min.

1.17 (a) If the efficiency of the motor in Q1.16 is 85%, calculate the average power loss per unitof stator volume.

(b) Assume that 50% of the power loss is removed by conduction through the motormountings, and 50% at the curved cylindrical surface area of the stator. Calculate theheat transfer rate required at the stator surface in kW/m2.

(c) Estimate the temperature rise if the heat removal at the curved cylindrical surface ofthe stator is by natural convection. Comment on the result.

(d) Estimate the temperature rise if the heat removal at the stator surface is by forcedconvection with an air velocity of 5.5 m/s.

1.18 (a) At rated power, the motor of Q1.16 has a winding temperature rise of 100EC when theambient temperature is 35EC. Calculate its effective thermal resistance with forcedconvection as in Q1.17(d).

(b) If the thermal capacity of the motor is 1.4 kJ/EC, calculate the thermal time-constantcorresponding to the conditions in (a).

(c) Assuming constant efficiency, what is the maximum power output that can be sustainedfor 25 min without exceeding the rated temperature rise?

(d) What is the maximum time for which the motor can operate at 125% of rated powerwithout exceeding the rated temperature rise? (Assume cold start.)

Fig. 1.8

Fig. 1.9

Fig. 1.10

1.19 Fig. 1.8 shows a magnet with pole-pieces arranged to focus flux into a circular airgap of length10mm and diameter 100mm. The magnet dimensions are 100 × 150 × 100 mm. Determine thepermeance coefficient, the magnetic flux-density in the magnet and in the airgap, and themagnet energy-product. Neglect fringing and assume that the pole-pieces are infinitelypermeable. The magnet remanent flux-density is 0.4 T and its relative recoil permeability is 1.0.

1.20 Fig. 1.9 shows a rotary device excited by a permanent magnetwhose demagnetization curve is straight throughout thesecond quadrant of the hysteresis loop. The remanent fluxdensity is 1.0 T and the relative recoil permeability is 1.0.Calculate the airgap flux-density when the rotor is in the'aligned' position as shown. Also determine the permeancecoefficient, the magnet flux-density and the magnet energy-product.

1.21 Fig. 1.10 shows a loudspeaker magnet assembly. If themagnet has a straight demagnetization characteristic witha remanent flux-density of 1.0 T and a relative recoilpermeability of 1.0, calculate the magnetic flux-density in theairgap. Also determine the permeance coefficient and themagnet flux-density. Neglect all fringing and leakage effects,and assume that the steel parts are infinitely permeable.What is the magnet’s energy product when it is installed inthe assembly?

Fig. 1.11

1.22 Fig. 1.11 shows the demagnetization characteristics of a permanent magnet at twotemperatures, T1 = 20EC and T2 = 100EC. The magnet operates initially at point X. Thehorizontal axis is scaled by µ0, i.e., µ0Hm is plotted in [T ]instead of Hm in [A/m].

(a) What is the permeance coefficient at point X ?

(b) The magnet operates initially at point X. Then the temperature increases to 100EC, afterwhich there is a change in the configuration of the magnetic circuit which causes thepermeance coefficient to increase to 3@0. Determine graphically the values of Bm and Hmat the new operating point.

(c) If the magnet temperature is maintained at 100EC while the magnet is keepered (i.e., Hmis reduced to zero), what is the loss of remanence, expressed as a percentage of theoriginal remanence at 20EC?

(d) What fraction of the loss of remanence in (c) is irreversible?


Solutions to Problems

1. Foundation

W '12

B H '12

B 2

µ0

x vol.

'12

1.02

4B x 10!7x 1000 x 10!6

' 398 Joules

S1.9

T '100

1500 x 2 B

60

' 0.637 N&m ' 5.628 in&lb

Rotor vol. Vr 'T

TRV'

0.63712,000

m 3' 53,051.6 mm3

But Rotor vol. Vr ' B r 2 Lstk ' B2r 3Lstk

2r' 2B x 1.4 r 3

ˆ D ' 2r ' 2 x3 53,051.6

2B x 1.4' 36.40 mm ' 1.434 in

Lstk ' 1.4 x D ' 51 mm ' 2.01 in .

S1.12

TRV 'B

2x 0.9 x 12k x 0.6 x 1

' 14.4 kNm/m3

F '14.4

2' 7.7 kNm/m 2

Also F '14.413.8

' 1.04 ps i

S1.13

Use eqn. (1.4):

S1.14

Use eqn. (1.8):

J 'A8

Fslot Aslot

'A

Fslot d( 1 ! J )'

120.42 x 18 x ( 1 ! 0.56)

' 3.6 A/mm 2

motor : TJm

' 35,000 rad/s2 ; T ' 14 N&m

ˆ Jm '14

35,000' 4.0 x 10& 4 kg&m2

n ' 2

" 'Tmp

n Jm %JL

n 2

4,600 '14

2 4 x 10!4%

JL

4

6 JL ' 4.49 x 10!3 kg&m2

Optimum ratio n 'JL

Jm

'4.49 x 10!3

4 x 10!4' 3.35

"max '12

x 14

4 x 10!4x 1

3.35' 5,225 rad/s2

S1.15

Use eqn. (1.14):

Use eqn. (1.16):

S1.16

(a) Copper cross-section area per slot = 62@2 x 0@47 = 29@23 mm2.

Max. amp-conductors per slot = 29@23 x 4 = 116@9 A.

Total amp-conductors around stator periphery = 24 x 116@9 = 2806@5 A.

Stator periphery = π x 52@2 = 164 mm.

Electric loading = 2806@5 / 164 = 17@1 A/mm.

(b) Magnetic loading = Bavg = 2/π x Bpeak = 2/π x 0@85 = 0@541 T.

TRV = (π/%2) kw1 BA = π/%2 x 0@911 x 0@541 x (17@1 x 103) = 18@7 kNm/m3.

Airgap shear stress F = TRV/2 = 9@35 kN/m2.

[Note: in Imperial units F = 9360 /4@45 /1550 = 1@36 lbf/in2 since 1 lbf = 4@45 N and 1 m2 = 1550 in2.]

(c) The rated torque is the torque produced when the TRV = 18@7 kNm/m3, since this correspondsto the maximum permitted electric and magnetic loadings.

Rotor volume Vr = π/4 x Dr2 x Lstk = π/4 x (52 x 10!3)2 x 57 x 10!3 = 1@211 x 10!4 m3.

(note: we’ve assumed the airgap is small so that Dr . 52 mm)

Torque = Vr x TRV = 1@211 x 10!4 x 18,700 = 2@27 Nm.

Output power = 2@27 x (1745 x 2π/60) = 414 W

Stator volume = π/4 x OD2 x Lstk = π/4 x (106 x 10-3)2 x 57 x 10!3 = 0@503 x 10!3 m3.

Power/stator volume = 414 / 0@503 x 10!3 = 823@2 kW/m3. (= 13@5 W/in3).

S1.17

(a) Power loss = Output ! Input = Output x (1/0 ! 1) = 414 x (1/0@85 ! 1) = 73 W.

Power loss/stator volume = 73 / 0@503 x 10!3 = 145@2 kW/m3 (2.38 W/in3)

(b) Cylindrical surface area = π x OD x Lstk = π x 106 x 57 x 10!6 = 19@0 x 10!3 m2.

Required convection rate = (0@5 x 73) / 19@0 x 10!3 = 1@92 kW/m2 (= 1@24 W/in2).

(c) Use eqn. (1.28). The actual heat transfer rate is h)T, and this has to be equal to 1@92 kW/m2.From equation (1.28),

h)T = 7@5 x ()T)1/4 x )T / (OD)1/4 = 7@5 x ()T)5/4 / (OD)1/4

so that )T = [ 1@92 x 103 / 7@5 x 1061/4 ]4/5 = 215 EC.

This is clearly much too hot, indicating that natural convection is not adequate for cooling.

(d) Use equation (1.29). The actual heat transfer rate is h)T, and this has to be equal to 1@92kW/m2. From equation (1.29),

h = 125 x % (5@5 / 57 ) = 38@8 W/m2 /EC

so that with h)T = 1,920 W/m2, )T = 1,920/38@8 = 49@5 EC – much better.

When radiation is taken into account, the temperature rise will be even lower.

S1.18

(a) Effective thermal resistance R = )T/Q where )T is the winding temperature rise (100EC) andQ is the total heat transfer rate (73 W).

ˆ R = 100 / 73 = 1@37 EC/W.

R represents the combined effect of conduction and convection.

(b) T = RC = 1@37 x 1400 = 1,918 s = 32 min.

(c) Use equation (1.49):

k2 = 1 / (1 ! e!ton/J) = 1 / (1 ! e!25/32) = 1@844.

The power loss (dissipation) can increase by the factor k2 to 73 x 1@844 = 134@6 W for 25 min.If the efficiency is constant at 85%, this means that

134@6 = Output x (1/0@85 ! 1) = Output x 0@176

so the Output power = 134@6 /0@176 = 765 W.

(d) Use equation (1.53):

With k2 = 1@25 and J = 32 min,

ton = 32 ln [1@25 / (1@25 ! 1)] = 51@5 min.

" 'Am

Ag

'200 x 150B

4x 1002

' 3.8197

8 'Lm

g'

100.010.0

' 10.0

PC '8

"'

10.03.8197

' 2.618

i.e.Bm

Hm

' ! µ0 x PC

But Bm ' Br % µrec ( µ0 Hm)

' Br % µrec xBm

!PC

'BBBBrrrr xxxx PCPCPCPC

µµµµrecrecrecrec %%%% PCPCPCPC'

BBBBrrrr

µµµµrecrecrecrec %%%%""""

8888

Energy product ' Bm Hm 'Bm

2

µ0 x PC

'0.28942

4B x 10!7 x 2.618' 25.5 kJ/m 3 (3.2 MGOe)

S1.19

Using eqn. (1.63)...

the area ratio is

the length ratio is

the permeance coefficient is

ˆ Bm = Br x 2.618/(1.0+2.618)

= 0.7236 Br

= 0.2894 T

Bg = (Am/Ag) x Bm = "Bm

= 3.8197 x 0.2894

= 1.1056 T

" 'Am

Ag

'

50 x B

180x 11.5 x 10

10 x 10' 1.0036

8 'Lm

2g'

3.82 x 1.0

' 1.9

PC '8

"'

1.91.0036

' 1.8933

i.e.Bm

Hm

' ! µ0 x PC


' Br % µrec xBm

!PC



BBBBrrrr


8888


2

µ0 x PC

'0.65442

4B x 10!7 x 1.8933' 180 kJ/m 3 (22.6 MGOe)

S1.20


the area ratio is

the length ratio is


ˆ Bm = Br x 1.8933/(1.0+1.8933)

= 0.6544 Br

= 0.6544 T


= 1.0036 x 0.6544

= 0.6567 T

" 'Am

Ag

'

B

4x 142

B x 16 x 12' 0.2552

8 'Lm

g'

2.01.0

' 1.0

PC '8

"'

1.00.2552

' 3.9184

i.e.Bm

Hm

' ! µ0 x PC


' Br % µrec xBm

!PC



BBBBrrrr


8888


2

µ0 x PC

'0.79672

4B x 10!7 x 3.9184' 130 kJ/m 3 (16.2 MGOe)

S1.21


the area ratio is

the length ratio is


ˆ Bm = Br x 3.9184/(1.0+3.9184)

= 0.7967 Br

= 0.7967 T


= 0.2552 x 0.7967

= 0.2033 T

S1.22

(a) PC at X = Bm/µ0Hm = 0@27/0@49 = 0@55.

(b) Draw YCYCYCYC parallel to the recoil line through BBBB. Draw load-line with a slope of 3. From theintersection at ZZZZ, read off the scales: Bm = 0@4 T, µ0Hm = 0@14 T, so Hm = 111@4 kA/m.

(c) The magnet ends up keepered at CCCC, with Bm = Br = 0@5 T. Original remanence at AAAA was 0@65 T.The loss is 0@65 ! 0@5 = 0@15 T, i.e., 0@15/0@65 x 100% = 23% of the original remanence at AAAA.

(d) The fraction that is irreversible is BCBCBCBC/ACACACAC = (0@54 ! 0@5)/0@15 = 0@27.


Problems


Answers

2.2 (i) 0.291 T; (ii) 0.07257 Nm/A; (iii) 3,045 rpm; (iv) 231 A; (v) 16.75 Nm

2.3 (a) 2,050 rpm; (b) 0.604; (c) 2.585 A, 3.34 W

2.4 (a) 2.0 ms, 36.3 ms; (b) 5.55 rad/s/V; (c) 40%; (d) 4.4 Hz

2.5 (c) 1.06 mWb; (d) 0.50 T; (e) 3,064 A, !383 A/mm; (f) 31.8 mV-s; (h) 12.7 V; (i) 0.19 mH; (j) 159 mJ

2.6 (a) 3,810 rpm; (b) 153.3 A, 18.4 Nm; (c) (i) 1.029 Nm; (ii) 8.575 A; (iii) 22.06 W; (iv) 85%; (v) 17.15 W; (vi) 81.4%

2.7 191 Nm

2.8 173 V

2.9 (a) 10.8 ohm; (b) 6.612 Nm; (c) 182.33 V, 3.45E (lagging)

2.10 8,260 rpm

2.11 0.921 Nm

2.12 20.29E, 0.313 Nm

2.13 3,668 rpm

2.14 Xd = 8.8347 ohm, Xq = 27.405 ohm


2.1 (a) Draw the drive circuit for a delta-connected brushless DC squarewave motor.

(b) The attached blank circuit diagrams show a wye-connected brushless DC motor anddrive. Highlight the main conduction loops for the following conditions:

(i) normal conduction in lines A and C, with two transistors conducting;

(ii) normal conduction in lines A and C, with only one transistor conducting;

(iii) during the commutation of current from line C to line B; and

(iv) when the machine is generating at high speed, with current in two lines.

2.2 A 3-phase, 4-pole, wye-connected brushless DC motor has a stator bore diameter D = 52 mm, andstack length Lstk = 50mm. The number of turns in series per phase is 48, and the resistancemeasured between terminals is 0.052 ohm. When it is driven at 1,000 rev/min by an auxiliarymotor, the line-line EMF has a peak value of 7.6 V. Calculate

(i) the peak value of the airgap flux-density due to the magnets.

(ii) the torque constant k T

(iii) the no-load speed if the supply voltage is Vs = 24 V

(iv) the locked-rotor current

(v) the locked-rotor torque

2.3 A wye-connected brushless DC squarewave motor has a torque constant of 0.177 Nm/A and itsphase resistance is 0.5 ohm/phase. It operates with a DC supply voltage of 48 V, and the on-stateresistance of each MOSFET in the controller is rDS(on) = 0.5 ohm. Calculate

(a) the base speed in rev/min if the set-point current is 5 A;

(b) the switching duty-cycle of the controller transistors at half the base speed;

(c) the RMS current in each transistor, assuming that all transistors perform the same dutyin sequence during each cycle of 360E elec. Also calculate the conduction loss in eachMOSFET transistor at half speed.

2.4 A wye-connected brushless DC motor has a back-EMF constant of 18.85 V/1000 rpm. Its line-lineinductance is 4.7 mH and its phase resistance is 1.175 ohm. The motor inertia is JM = 0.5 × 10!3

kg-m2 and it is coupled to a load whose inertia is 3JM, referred to the motor shaft.

(a) Calculate the electrical and mechanical time-constants.

(b) The motor is required to move a polishing head on a machine for polishing specialaluminium mirrors. At the beginning of the process the polishing head cycles back andforth at a frequency of 0.1 Hz. As the polishing process continues, the cycle timedecreases until the frequency is 10 Hz. Calculate the value of the gain G(jT) =Tm(jT)/V(jT) at the low initial frequency.

(c) Calculate the amplitude of the polishing stroke at 10 Hz, as a percentage of its value at0.1 Hz.

(d) Estimate the frequency at which the amplitude of the polishing stroke is reduced to 70%of its value at the start of the process.

Fig. 2.5

2.5 A brushless DC motor has the cross-section shown in Fig. 2.5 with D = 60 mm (stator borediameter), Lm = 8 mm (magnet length), g = 0.4 mm (airgap), and Lstk = 35 mm (axial length).Slotting is neglected and a single full-pitch stator coil is shown with 30 turns. The ceramicmagnet has Br = 1.25 T and µrec = 1.05. The current in the stator coil is zero.

(a) Sketch the magnetic field set up by the magnet, by drawing 10 flux lines.

(b) Sketch the variation of the radial component of airgap flux-density B around the insideof the stator, from 0 to 360E.

(c) Draw an equivalent magnetic circuit and use it to calculate the flux crossing the airgapunder each magnet pole, assuming that 10% of the magnet flux fails to cross the airgap.Assume that the permeability of the steel in the rotor and stator is infinite.

(d) Deduce the value of Bg in the airgap at the centre of the magnet arc, i.e. on the directaxis.

(e) Determine the MMF across the magnet and the internal magnetizing force Hm.

(f) Calculate the flux-linkage of the stator coil in the position shown.

(g) Sketch the waveform of the coil flux-linkage as the rotor rotates through 360E.

(h) Determine the waveform and the peak value of the EMF induced in the stator coil if therotor rotates at 4,000 rev/min.

(i) Estimate the inductance of the stator coil.

(j) If the current in the stator coil is maintained constant at 5 A, determine the mechanicalwork that is done in rotating the rotor 180E from the position shown. Is this workpositive or negative? Explain what you mean by “positive” or “negative”.

Fig. 2.7

2.6 A wye-connected PM brushless DC motor operated in squarewave mode has a torque constantof 0.12N-m/A referred to the DC supply. The supply voltage is 48 V DC.

(a) Estimate its no-load speed in rev/min.

(b) If the armature resistance is 0.15 ohm/phase and the total voltage drop in the controllertransistors is 2V, determine the stall current and the stall torque.

(c) The motor is delivering 330W of mechanical power to a load at 3,400rev/min. The voltagedrop across each transistor is a constant 1 V. The friction torque has been separatelymeasured as 0.046N-m at this speed, and the iron loss can be taken as a constantmechanical loss of 20W. Calculate

(i) the electromagnetic torque Te(ii) the current I(iii) the power loss in the winding resistance PCu(iv) the motor efficiency(v) the total power loss in the transistors, and(vi) the overall efficiency of the drive and motor.

2.7 Fig. 2.7 shows the top view of a permanent-magnet actuator used for guiding cables on to avertical-axis cable-laying drum. The length of the arc-shaped magnets in the direction ofmagnetization is 60 mm and that of the rectangular magnets is 55 mm. The vertical height of theelectromagnetic assembly is 250 mm. The airgap length is 10 mm. If the magnet is Ferrite witha remanent flux-density of 0.4 T and relative recoil permeability of 1.0, calculate the flux-densityin the airgap, assuming infinitely permeable iron components. Neglect fringing and leakage.

The magnetic flux path is completed by soft iron rings. Part of the inner ring is encircled by acoil of 200 turns mounted on the rotating 'armature'. If the current in the coil is 20 A, calculatethe torque.

2.8 A 3-phase, 4-pole brushless PM motor has 36 stator slots. Each phase winding is made up of 3coils per pole with 20 turns per coil, and all the coils are in series. The coil span is 7 slots. If thefundamental component of magnet flux is 1.8 mWb, calculate the open-circuit phase EMF E at3,000 rpm.

2.9 The stator bore diameter of the motor of Problem 2.8 is 102 mm and its axial length is 120 mm.The airgap is 1mm, and Carter’s coefficient for slotting is 1.055. The magnet is mounted on therotor surface and has a radial thickness of 9.5 mm and a relative recoil permeability of 1.06.

(a) Calculate the airgap component of synchronous reactance at 100 Hz.

(b) What is the electromagnetic torque at 3,000 rpm if the phase current is 4.0 A and ( = 0?(i.e. the current is in the q-axis).

(c) The phase resistance is 3.7 ohm and the leakage reactance is 5 mH. With the phasecurrent of 4.0 A at ( = 15E at 3,000 rpm, calculate the terminal voltage and thepower-factor angle.

2.10 A brushless permanent-magnet sinewave motor has an open-circuit voltage of 173 V at itscorner-point speed of 3,000 rpm. It is supplied from a PWM converter whose maximum voltageis 200 V RMS. Neglecting resistance and all other losses, estimate the maximum speed at whichmaximum current can be supplied to the motor.

2.11 Show that with rated phase current I, the angle ( which maximises the torque of the hybridPM/reluctance motor satisfies the relationship sin (/cos 2( = )X.I /Eq, where )X = Xq ! Xd.Hence show that for a two-phase motor with Eq = 35.8 V, Xd = 1.18 ohm, and Xq = 2.47 ohm at 3,000rpm, and rated current of 4.0A, the torque is maximised with ( = 7.97E. What is this maximumtorque? Neglect losses.

2.12 If the high-energy magnets in the motor of Problem 2.11 are replaced by ceramic magnets, theohmic values of the reactances remain the same but the open-circuit phase EMF is decreasedfrom 35.8 V to 11.3V. Determine the optimum value of ( and the maximum torque, assumingthat the rated current remains 4.0A. What percentage of this maximum torque is reluctancetorque?

2.13 The absolute maximum speed of a permanent-magnet brushless motor can be estimatedapproximately by assuming that the current is at its rated value and is entirely in the negatived-axis, i.e. all the current capacity of the drive is used up in flux-weakening. If the maximumcontroller voltage is Vc then, neglecting losses, V = 0 + jVq = jVc = jEq + jXdId where Id isnegative. If Vc = 38 V and Eq = 35.8 V at 3,000 rpm, and if Xd = 1.18 ohm at 3,000 rpm, estimate theabsolute maximum speed with I = 4.0 A, remembering that both Eq and Xd are proportional tofrequency or speed.

1This problem can be solved with PC-BDC using the [Alt+7] standard example and changing Embed = Type1, Tw = 3.5,web = 1.0, Slots = 24, gap = 0.4, Bridge = 0.5, RNSQ = Round, TC = 60, wire = 0.5, Coils/P = 2, Throw = 5, Vs = 150, ISP = 1.0,fChop = 24.0; with a NeIGT 30H magnet.

Fig. 2.14

2.14 Calculate the synchronous reactances Xd and Xq of the interior permanent-magnet motor shownin Fig. 2.14 at 1000 rpm, given the following parameters: Stator bore diameter D = 50.8 mm;airgap length g = 0.4 mm; stack length Lstk = 50 mm; pole arc $M = 120E elec. Carter coefficient= 1.177. Winding factor kw1 = 0.837; No. of turns in series per phase = 480. Magnet length indirection of magnetization Lm = 5.5 mm; magnet width = 22 mm; per-unit rotor leakagepermeance prl = 0.091; magnet relative recoil permeability µrec = 1.1. Leakage reactance XF =3.264 ohm. The width of the web in the q-axis is 1.0 mm.1




S2.1

(a)

(b)

S2.2

(i) kE = kT = 4pTphMg /B = 2 Tph Bg DLstk since Mg = Bg x BDLstk /2p.

ˆ Bg = 0@07257/(2 x 48 x 52 x 50 x 10!6) = 0@291 T

(ii) kT = kE = ELL/Tm = 7@6/(2B /60 x 1,000) = 0@07257 Vs/rad or Nm/A

(iii) T0 = Vs / kE = 24/0@07257 = 318@8 rad/s = 3,045 rev/min

(iv) I0 = Vs/RLL = 24/(2 x 0@052) = 231 A

(v) T0 = kT I0 = 0@07257 x 231 = 16@75 Nm

Tb 'Vs & 2RISP

kE

'48 ! 2 x (0.5 % 0.5) x 5

0.177' 214.7 rad/s ' 2,050 rpm

IQ1(rms) ' 5 x 0.604 x 60360

% 1 x 60360

' 2.585 A

S2.3

(a)

(b) At half- speed,

(c) Transistor conduction waveform is

d '

0.177 x 214.72

% 2 x (0.5 % 0.5) x 5

48' 0.604

Conduction loss ' I 2Q1(rms) x rDS(on) ' 2.5852 x 0.5 ' 3.34 W in each transistor

2 transistors and 2 motor phasesin series at any time. RDS(on) + Rph

Virtually “DC”: imaginary parts negligible.

40% amplitude

neglect this term is dominant

S2.4

(a)

(b)

(c)

(d)

The required frequency is approximately the frequency at which

G(jT ) '1/0.18

1 % jT x 0.002) (1 % jT x 0.0363).

When fpolish ' 0.1 Hz, T ' 2B x 0.1 and

G '1/0.18

(1 % j 0.00126) (1 % j 0.0228)• 1/0.18 ' 5.55 rad/sec/V

When fpolish ' 10 Hz, T ' 2B x 10 and

G '1/0.18

(1 % j 0.126) (1 % j 2.28)• 1/0.18 x 0.4 e !73.5E

' 2.22 e !73.5E rad/sec/V

Te 'LR

'4.7

2 x 1.175' 2.0 ms

Tm 'RJ

k 2E

'2 x 1.175 x 0.5

18.851000

x 30B

2' 36.3 ms

Jmotor % Jload ' Jmotor ( 1 % 3)

' 0.125 x 4 ' 0.5 mkg&m 2

1* (1 % j0.002T) (1 % j0.0363T)*

' 0.7

0.7 .1

2'

1* 1 % j 1*

(Recognizing that 70% is closeto the “!3 dB” point)

0.0363T ' 1, i.e. 12B

x 10.0363

' 4.4Hz.

Am 'B

180x 120 x 21.6 x 35 ' 1583 mm 2

S2.5

(a)

(b)

(c)

Since 10% of Mm fails to cross the gap, Mg/Mm = 0.9: i.e. f LKG = 0.9.

To calculate the magnet pole area Am, take a cylindrical surface 1/3 of the way through the magnet,from the inner surface: the radius is 60/2 ! 0.4 ! 8 = 21.6 mm; the arc is 120E (by scaling from thediagram – it is not given in the problem specification); and the length is 35 mm. Thus

Pm0 'µr µ0 Am

Lm

'4B x 10!7 x 1.05 x 1583 x 10!6

8 x 10!3' 2.61 x 10!7 Wb/At

The permeance of one magnet is

To calculate the airgap area Ag (one pole), take a cylindrical surface at the stator bore, radius 30 mm.The length is Lstk = 35 mm and the arc is approximately 120E, so

The airgap reluctance (neglecting fringing) is

From eqn. (2.18),

(d) Bg = Mg/Ag = 1.06 x 10!3/(2119 x 10!6) = 0.500 T.

(e) The flux through the magnet is Mm = Mg/fLKG = 1.06 x 10!3/0.9 = 1.18 x 10!3 Wb, so Bm = Mm/Am =118 x 10!3/(1583 x 10!6) = 0.744 T.

MMF across magnet is Fm = HmLm = 383 x 8 = 3,064 A.

(f) The flux-linkage of the stator coil is 30 x Mg = 30 x 1.06 x 10!3 = 31.8 mV-s.

(g)

This waveform is obtained by integrating where N is the number of turns in the coil.N m Bg rLstk d2

Ag 'B

180x 120 x 30 x 35 ' 2119 mm 2

Rg 'g

µ0 Ag

'8 x 10!3

4B x 10!7 x 2199 x 10!6' 2.895 x 106 At/Wb

Mg 'Mr

1fLKG

% Pm0 Rg

'1.25 x 1583 x 10!6

10.9

% 2.61 x 10!7 x 2.895 x 10!6' 1.06 x 10!3 Wb

Hm 'Br ! Bm

µ0 µrec

'1.25 ! 0.744

4B x 10!7 x 1.05' !383 A/mm

epk 'dRdt

' TmdRd2

' 4000 x 2B60

x 2 x 31.8 x 10!3

120 x B

180

' 12.7 V

(h) The EMF waveform has the same shape as the Bg waveform because the coil has a pitch of 180E.The stator coil flux-linkage changes from !31.8 mVs to +31.8 mVs in about 120E, so the peak value ofEMF is

(i) The permeance P “seen” by the coil is approximately µ0A/h, where A is a cylindrical surface of about180E arc and 35 mm length at radius 30 ! (0.4 + 8)/2 = 25.8 mm, and h = 2 x (8.4) = 16.8 mm, so

Then the inductance is L = N2P = 302 x 0.169 µ0 = 0.19 mH.

(j) Work done = 1 x )Q = 5 x 31.8 x 10!3 = 159 mJ. Since the rotor is in stable equilibrium, this workmust be done by exerting a torque on the rotor. If we regard the work as mechanical input, as thoughthe machine was a generator, then it is positive. If we regard the work as mechanical output, as thoughthe machine was a motor, then it is negative.

P 'µ0 A

h'

25.8 x B x 35 x 10!6

16.8 x 10!3µ0 ' 0.169 µ0 Wb/At

S2.6

(a)

(b)

(c) Electromagnetic (airgap) power Pg = Pmech + Pfriction + Piron loss. Since Tfriction = 0.046 W,

and Pg = 330 + 16.38 + 20 = 366.38 W. So the electromagnetic torque is

(i)

and therefore the current is

(ii)

(iii) The power loss in the winding resistance is PCu = 8.5752 x 0.15 x 2 = 22.06 W.

Total power input to motor = 366.38 + 22.06 = 388.44 W;

(iv) ˆ Motor efficiency = 330/388.44 = 0.85 or 85%

(v) Power loss in transistors = 2 x 1 x 8.575 = 17.15 W

Power delivered by supply = 388.44 + 17.15 = 405.59 W

(vi) ˆ Overall efficiency of drive + motor = 330/405.59 = 81.4%

I0 '48 ! 20.15 x 2

' 153.3 A

T0 ' 153.3 x 0.12 ' 18.4 Nm

T0 'VkT

'480.12

' 400 rad/sec ' 3810 rpm

Pfriction ' 0.046 x 3400 x 2B60

' 16.38 W

Te '366.38

3400 x 2B60

' 1.029 Nm

I 'Te

kT

'1.0290.12

' 8.575 A

Mg ' 2 xFm1 % Fm2

Rg % Rm1 % Rm2

' 2 x 17507 % 19099

(0.8232 % 1.0298) x 106

' 0.0395 Wb

S2.7

The magnets are magnetized N-S-N-S. Analyze one half pole, which has one-half of an arc magnet inseries with one-half of a rectangular magnet in series with an airgap of arc 95/2 = 47.5E.

The airgap reluctance is

The magnets are in series, so we need their “Thevenin” equivalent circuits with parameters Fm1 = Mr1/Pm1and Rm1 = 1/Pm1 and Fm2 = Mr2/Pm2 and Rm2 = 1/Pm2:

We have Am1 = Ag = 47.5 x (B/180) x 280 x 250 x 10!6 = 0.0580 m2, and Lm1 = 60 mm, so

and Mr1 = 0.4 x 0.0580 = 0.0232 Wb so Fm1 = 0.0232 x 8.232 x 105 = 19,099 At.

Also Am2 = 340/2 x 250 x 10!6 = 0.0425 m2, and Lm2 = 55 mm, so

and Mr2 = 0.4 x 0.0425 = 0.0170 Wb so Fm2 = 0.0170 x 1.0298 x 106 = 17,507 At.

The airgap flux (whole pole) is therefore

and the airgap flux-density is Bg = Mg/2Ag = 0.0395/2 x 0.0580 = 0.341 T.

The tangential force on one coilside is F = Bg x L x NI = 0.341 x 250 x 10!3 x 200 x 20 = 341 N, so thetorque is 2 F x r = 2 x 341 x 280 x 10!3 = 191 Nm.

Rg 'g

µ0 Ag

'10 x 10!3

4B x 10!7 x 47.5 x B

180x 280 x 250 x 10!6

' 1.3713 x 105 At/Wb

Rm1 '1

Pm1

'Lm1

µ0 µrec Am1

'60 x 10!3

4B x 10!7 x 1.0 x 0.0580' 8.232 x 105 At/Wb

Rm2 '1

Pm2

'Lm2

µ0 µrec Am2

'55 x 10!3

4B x 10!7 x 1.0 x 0.0425' 1.0298 x 106 At/Wb

E '2B

2x 216.5 x 1.8 x 10!3 x 100 ' 173.1 V rms

S2.8

Total turns = 4 x 3 x 20 = 240 turns.

Frequency f = 3000/60 x 2 = 100 Hz, so

kd1 '

sin q(2

q sin (

2

'sin 30E

3 sin 10E' 0.9598

kp1 ' cos g

2' cos 1

229B ' 0.9397

kw1 ' kd1 kp1 ' 0.9397 x 0.9598 ' 0.9019

kw1 Nph ' 0.9019 x 240 ' 216.5 turns

q = 3 slots/pole/phase( = (360/36) x 2 = 20E(slot-pitch angle. Eelec)

Xsg '6µ0 DLstk f

p 2 gO(kw1 Nph)

2'

6 x 4B x 10!7 x 102 x 120 x 10!6 x 100

2 2 x 10.02 x 10!3x 216.5 2

' 10.8 ohm

S2.9

(a) Effective airgap is

Then the airgap component of synchronous reactance is

(b) The electromagnetic torque is

(c) The total synchronous reactance (including leakage reactance) is

From the phasor diagram,

Power-factor angle N = 108.45 ! (90 + 15) = 3.45E. PF = cos N = 0.998 lag.

Also note the solution in dq components:

VVVV = (0 + jEq ) + (Rph + j Xs)(Id + j Iq) = (RphId ! XsIq) + j (Eq +RphIq + XsId)

In the general case (salient-pole machine) we use Xd on the d-axis with Id, and Xq on the q-axis with Iq;but in this example we have a surface-magnet motor and so Xd = Xq = Xs.

Also Id = ! I sin ( and Iq = I cos (, so

Vd = RphId ! XqIq = 3.7 x (!4 sin 15E) ! 13.94 x 4 cos 15E = !57.6906 V

Vq = Eq + RphIq + XdId = 173.1 + 3.7 x 4 cos 15E+ 13.94 x (!4 sin 15E) = 172.9636 V

ˆ VVVV = Vd + j Vq = !57.6906 + j 172.9636 = 182.33 ej108.45E V

gO ' gN %Lm

µrec

' 1.0 x 1.055 %9.51.06

' 10.02 mm

Xs ' Xsg % XF

' 10.8 % 2B x 5 x 10!3' 13.94 ohm

Te ' 3Eq

Tm

x Iq '3 x 173.1

2B60

x 3000x 4 x sin (0) ' 6.612 Nm

VVVV ' EEEE % (Rph % j Xs )IIII

' j 173.1 % (3.7 % j 13.94) x 4.0 e j(90% 15E)

' 182.33 e j108.45E V

NQ

ND

' u ! 1 ! u2' 0.865 ! 1 ! 0.8652

' 0.3632

S2.10

The per-unit EMF is u = 173/200 = 0.865

The ratio between base speed and maximum speed is

ˆ Max speed = 3,000/0.3632 = 2.753 x 3,000 = 8,260 rpm

)x '(2.47 ! 1.18) x 4

35.8' 0.1441 pu

ˆ s '!1 ± 1 % 8 x 0.14412

4 x 0.1441' 0.1386 or !3.608–––

so ( ' 7.97E .

S2.11

The value of ( that gives maximum torque depends on the current I. To solve for this value of (, let s = sin (; then cos 2( = 1 ! 2 s2 and s/(1 ! 2s2) = )x, which is a quadratic equation in s, with solution

The frequency is 100 Hz and with p = 2, m = 2 the torque is

In this example, with an Xq/Xd ratio of 2.1 and )x = 0.1441, only 2% of the maximum torque is contributedby reluctance torque.

s '!1 ± 1 % 8)x 2

4 )x

T ' mp [Qd Iq ! Qq Id ] (m ' No of phases)

' mp [(Q1Md %Xd IdT

) Iq !Xq IqT

Id ] (Q1Md,Qd and Qq are all RMS)

'mpT

[Eq Iq % (Xd ! Xq ) Id Iq ]

'mpT

[Eq I cos ( % (Xq !Xd).I2 sin ( cos ( ]

'mpT

[Eq I cos ( % )X.I 2 sin ( cos ( ] where )X ' Xq ! Xd > 0

'mpT

[Eq cos ( %)X.I

2sin 2(] I

dTd(

'mpT

[!Eq sin ( % )X. I cos 2( ] I ' 0

ˆEq

)X.I'

cos 2(sin (

or sin (

cos 2('

)X.IEq

' )x (pu) , if the base impedance isEq

I

Tmax '2 x 2

2B x 100[35.8 cos 7.97E %

2.47 ! 1.182

x 4.0 sin (2 x 7.97E ) ] x 4.0

' 0.0255 x [35.4542 % 0.7085]

' 0.921 N&m

)x '(2.47 ! 1.18) x 4.0

11.3' 0.4566

s '!1 ± 1 % 8 x 0.45662

4 x 0.4566' 0.3468 Y ( ' 20.29E

T '2 x 2

2B x 10011.3 cos 20.29E %

(2.47 ! 1.18) x 4.02

sin (2 x 20.29E ) x 4.0

' 0.0255 x [ 10.5987 % 1.6784]

' 0.313 N&m

S2.12

In this example with Xq/Xd = 2.1 and )x = 0.4566 pu, the reluctance torque contributes 14% to themaximum torque.

Vd ' 0

Vq ' Eq % Xd Id

Id ' !4.0 A; Vq ' 38V

38 ' 35.8 x N3000

%N

3000x 1.18 x (!4.0)

' (35.8 ! 1.18 x 4) N3000

ˆ N ' 1.2227 x 3,000 ' 3,668 rpm

S2.13

We’re assuming that all the current is in the d-axis, to make the solution tractable, but this means thetorque will be zero, and therefore the solution is only approximate (to the extent that losses can beneglected). What we’re really estimating is an upper bound for the speed, with the given motor and driveparameters.

Iq = 0, T = 0, R = 0

The phasor diagram shows the flux-weakening effect of Id. Evidently the flux-weakening is morepronounced when XdId is an appreciable fraction of Eq, suggesting that a high reactance extends thespeed range. Though this is generally a valid inference, it says nothing about the variation of torquewith speed. In this motor the speed range extends only 22% above 3,000 rpm, and the torque will fallrapidly between base speed (3,000 rpm) and maximum speed (3,668 rpm).

2Brushless permanent-magnet and reluctance motor drives, Oxford University Press, 19893Design of brushless permanent-magnet motors, Magna Physics Publications, 19944This problem can be solved with PC-BDC using the [Alt+7] standard example and changing Embed = Type1, Tw = 3.5,

web = 1.0, Slots = 24, gap = 0.4, Bridge = 0.5, S-Slot = Round, TC = 60, wire = 0.5, Coils/P = 2, Throw = 5, Vs = 150, ISP = 1.0,fChop = 24.0; with a NeIGT 30H magnet.

Am ' 22 x 50 ' 1100 mm 2

Pm0 'µrec µ0 Am

Lm

'4B x 10!7 x 1.1 x 1100 x 10!6

5.5 x 10!3' 2.7646 10!7 Wb/At

Pm ' ( 1 % prl ) Pm0 ' 1.091 x Pm0 ' 3.0162 10!7 Wb/At

Xd '6µ0 DLstk f

p 2 gdO( kw1 Nph )2

% XF

'6 x 4B x 10!7 x 50.8 x 50 x 10!6 x 33.33

22 x 9.822 x 1.177 x 0.4 x 10!3x (0.837 x 480)2

% 3.264

' 5.5707 % 3.264 ' 8.8347 ohm

Rg 'gN

µ0 Ag

'1.177 x 0.4 10!3

4B x 10!7 x 1330 x 10!6' 2.8169 x 105 At/Wb

Ag '2B3

x 12

x 25.4 x 50 ' 1330 mm 2

1 % Pm Rg ' 1 % 3.0162 x 10!7 x 2.8169 x 105' 1.085

k1 '4B

sin "B

2' 1.1027 (" '

120180

'23

)

k"d '

sin ("B /2)"B /2

'sin 60E23

x B

2

'3B

x sin 60E ' 0.827

k1ad ' " %sin "B

B'

23

%

sin 23B

B' 0.9423

gdO 'gN

k1ad !k1 k"d

1 % Pm Rg

'gN

0.9423 !1.1027 x 0.827

1.085

' 9.822gN

'd 'gN

gdO'

19.822

' 0.102 .

S2.14

Use the formulas in chapter 6 of Miller [1989]2 or chapter 6 of Hendershot and Miller [1994]3 We needto work through quite a few of them... 4

k1aq ' " % S %sin SB ! sin"B

B'

23

% 0.0251 %

sin 0.0251B ! sin 23B

B' 0.4412

gqO 'gNk1aq

'gN

0.4412' 2.2665gN

Xq '6µ0 DLstk f

p 2 gqO( kw1 Nph )2

% XF

'6 x 4B x 10!7 x 50.8 x 50 x 10!6 x 33.33

22 x 2.2665 x 1.177 x 0.4 x 10!3x (0.837 x 480)2

% 3.264

' 24.141 % 3.264 ' 27.405 ohm

'q 'gN

gqO'

12.2665

' 0.441 .


Problems

3. Induction Machines

Answers

3.1 C.

3.2 7920 rpm, 12 Hz

3.3 1200 rpm, 1140 rpm, 3 Hz

3.4 (a) A,B,E; (b) C,D; (c) F.

3.6 1666 W, 0.85 (lagging)

3.7 11.76 kW, 502.6 W, 13.57 kW, 86.7%.

3.8 (a) 1485 rpm; (b) 1469 rpm; (c) 1461 rpm.

3.9 9.0, 0.886 (lagging)

3.11 25.64 kW, 87.7%; 55.8%.

3.12 (a) 0.033; (b) 31.8 kW; (c) 1.06 kW; (d) 336.5 Nm; (e) 30.66 kW; (f) 92.9%.

3.13 (a) 27.0 A; (b) 0.896 lag.; (c) 89.8%; (d) 141.9 Nm; (e) 349 Nm; (f) 159.3 Nm; (g) 156 A.

3.14 1350 rpm; 2250 rpm; freqency and amplitude are the same; phase sequence is reversed.


3.1 A three-phase induction motor is fed from a three-phase supply of fixed voltage and frequency.At what speed is its torque zero?

A StandstillB when s < 0C synchronous speed.

3.2 An induction motor has 2 poles and operates from a 120 Hz supply. If the slip s is 0.1, what is thespeed in rev/min? What is the frequency of the rotor currents?

3.3 What is the speed of the rotating field of a 6-pole, three-phase AC induction motor connected toa 60 Hz supply? Give the answer in rev/min. Calculate the rotor speed if the slip is 5%. Whatis the frequency of the rotor currents?

3.4 (a) In which of the following machines would you expect to find a commutator?

A A permanent-magnet d.c. servomotor;B a series-wound d.c. generator;C A wound-rotor induction motor;D An a.c. steam-turbine generator;E A “universal” motor;F A single-phase induction motor.

(b) In which of the machines in (A) to (F) above would you expect to find slip-rings?

(c) In which of the machines in (A) to (F) above would you expect to find no brushes?

3.5 Discuss two methods for controlling the speed of a cage induction motor. Draw a sketch of thespeed/torque characteristic of an induction motor with

A low rotor resistance; B high rotor resistance;C a double-cage rotor.

Show a typical load characteristic on each sketch.

3.6 A 230-V, wye-connected, 3-phase AC induction motor delivers 8 Nm at 1750 rpm. If the efficiencyis 88% and the line current is 4.92 A, find the input power and power-factor.

3.7 A 4-pole, 50 Hz, 3-phase induction motor develops an electromagnetic airgap torque of 80 Nmwhen running at full load. The frequency of the rotor currents is 2 Hz. Calculate the shaftpower. If the torque absorbed by windage and friction is 2 Nm, and if the stator losses total 1kW, calculate the rotor copper loss, the input power, and the efficiency.

3.8 A 4-pole, 50Hz, high-efficiency induction motor develops full-load torque at 1470 rev/min.

(a) What will be its speed at half rated torque?(b) What will be its speed at half rated torque and 70% voltage? (c) What will be its speed at rated torque and voltage, if the rotor resistance increases by

30% as a result of temperature rise?

3.9 A 7.5-kW, 3-phase, 60 Hz, 460-V, star-connected 4-pole induction motor has a full-load speed of1764 rev/min. In the per-phase equivalent circuit, the stator resistance is 0.25 ohm, the referredrotor resistance is 0.5 Ohm, the total leakage reactance is 2.5 ohm, and the magnetizingreactance is 60 ohm. Calculate the ratio between the standstill current and the full-load current.What is the full-load power factor? Ignore friction and core losses.

3.10 Explain how a rotating magnetic field is set up in the airgap of a 3-phase AC machine.

3.11 A 3-phase, 6-pole, 50 Hz wound-rotor induction motor delivers 22.5 kW at a speed of 950 rev/minwith its slip-rings shorted. Assuming constant friction torque of 1.5 Nm and constant statorlosses of 1.8 kW, find the input power and the efficiency. When the speed is reduced to 600rev/min by increasing the rotor circuit resistance, the load torque remains constant at thefull-load value. Find the efficiency at the reduced speed. State any assumptions used.

3.12 An 8-pole, 3-phase, 60-Hz induction motor is operating at a speed of 870 rev/min. The inputpower is 33 kW and the stator copper loss is 1200 W. Friction and windage loss is 80W. Core lossis negligible. Find

(a) the slip;(b) the airgap power Pgap;(c) the rotor copper loss;(d) the shaft torque in Nm;(e) the shaft power in kW;(f) the efficiency.

3.13 A 6-pole, 3-phase, Y-connected, 460-V, 60 Hz induction motor has the following equivalent-circuitparameters (all in ohms):

Magnetizing reactance = 30.0 Stator resistance 0.80

Total leakage reactance = 1.40 Rotor resistance referred to stator = 0.30

If the rotor speed is 1164 rev/min, calculate

(a) the line current;(b) the power factor;(c) the efficiency;(d) the shaft torque;(e) the breakdown torque;(f) the locked-rotor torque;(g) the locked-rotor current.

Assume rated voltage and frequency, and ignore friction and core losses.

3.14 A 4-pole wound-rotor induction motor is to be used as a frequency-converter. The stator isconnected to a 60 Hz 3-phase supply. The load is connected to the rotor slip-rings via brushes.At what two speeds could the rotor be driven to supply 15 Hz to the load? In what way would the3-phase voltages at the load terminals differ at these two speeds?



3. Induction Machines

N '120 f

P( 1 & s) '

120 x 1202

( 1 ! (! 0.1 ) ) ' 7,920 rpm

fr ' s f ' 0.1 x 120 ' 12 Hz

S3.2

Ns '120 f

P'

120 x 606

' 1200 rpm (synchronous speed)

N ' Ns ( 1 ! s) ' 1200 ( 1 ! 0.05) ' 1140 rpm (shaft speed or rotor speed)

fr ' sf ' 0.05 x 60 ' 3.0 Hz

S3.3

S3.4 : See Answers on front page

S3.5 : See theory manual, chapter 3

Output power ' Pshaft ' T Tm ' 8 x 1750 x 2B60

' 1466 W

Input power 'Output power

Effcy.'

14660.88

' 1666 W

but Pin ' 3 VL IL cos N

i.e. 1660 ' 3 x 230 x 4.92 cos N

ˆ cos N ' 0.85 ( lag. )

S3.6

Tm '2Ts

P( 1 & s) '

2 x 2B x 504

( 1 ! s) ' 150.8 rad/sec

[s ' fr / f ' 2/50 ' 0.040]

Pshaft ' (80 ! 2) x 150.8 ' 11.76 kW

Total mechanical power ' Pmech ' 80 x 150.8 ' 12.06 kW

Airgap power ' Pg 'Pmech

1 & s'

12.061 ! 0.040

' 12.57 kW

Rotor copper loss ' Pr ' Pg ! Pmech ' sPg 's

1 ! sPmech

'0.040

1 ! 0.040x 12.06 kW ' 502.6 W

Input power Pin ' Airgap power % stator loss

' 12.57 % 1 ' 13.57 kW

Effcy. 'Pshaft

Pin '

11.7613.57

' 86.7%

S3.7

s '12

x 1

0.72x 0.02 Y N ' 1500 (1 & 0.02041) ' 1469 rpm

S3.8

It is a high-efficiency motor and the slip is small, so we can assume .T %V 2

s

RR

s

At full load the slip is where ; Ns ! 1470

Ns

Ns '120 x 50

4' 1500 rpm

ˆ s '1500 ! 1470

1500' 0.020

(a) If T is ½, S is ½ x 0.020, so N = 1500 (1 ! 0.01) = 1485 rpm

(b) If T is ½ (i.e. 50% of rated) and Vs is 0.7,

(c) If T is 1 (i.e. 100% of rated) and Vs is 1 and RR = 1.3, s = 0.02 x 1.3 = 0.026

so N = 1500 (1 ! 0.026) = 1461 rpm

Ns '1204

x 60 ' 1800 rpm; s '1800 ! 1764

1800' 0.020 (full load)

IIII M '460/ 3

j 60

' & j 4.426 A

IIII R '460/ 3

0.25 %0.5

0.020% j 2.5

' 10.416 ! j 1.031 A

IIII S ' IIII R % IIII M ' 10.416 ! j 5.458 ' 11.76 e !j27.65E A

IIII R '460/ 3

0.25 % 0.5 % j2.5' 29.24 ! j 97.46 A

IIII s ' IIII R % IIII M ' 29.24 ! j 101.89 A ' 106.0e !j 73.99E A

Ratio Standstill currentfull load current

'106.011.76

' 9.01

Full&load PF ' cos (27.65E ) ' 0.886 lag

S3.9

At full load

At standstill (s = 1)

S3.10 : See theory manual, chapter 3.

Ns '120 x 50

6' 1000 rpm s '

1000 & 9501000

' 0.050

Pmech ' 22.5 k % 1.5 x 950 x 2B60

' 22,649 W

Pgap 'Pmech

1 ! s'

22,6491 ! 0.050

' 23,841 W

Pin ' Pgap % Stator losses ' 23,841 % 1800 ' 25.64 kW

Effcy. 'Pshaft

Pin '

22.525.64

' 87.7%

Pmech ' 14,210 % 1.5 x 600 x 2B60

' 14,305 W

New slip s '1000 ! 600

1000' 0.40

ˆ Pgap '14,305

1 ! 0.40' 23,841 W

Pin ' 23,841 % 1800 ' 25,641 W (Note&same as before)

Effcy. 'Pshaft

Pin '

14,21025,641

' 55.8 %

S3.11

At 600 rpm, Pshaft ' 22500 x 600950

' 14,210 W (at constant torque)

Assume friction torque is constant at 1.5 Nm

Frictionpower

Shaft power delivered toload

Total mech.p o w e r developed

(a) Ns '1208

x 60 ' 900 rpm s '900 ! 870

900' 0.033

(b) Pgap ' Pin ! Pstator copper loss ' 33 k ! 1200 ' 31,800 W

(c) Protor loss ' s Pgap ' 0.033 x 31,800 ' 1,060 W

Pm ' ( 1 ! s) Pgap ' ( 1 ! 0.033) x 31,800 ' 30,740 W

(e) Pshaft ' Pmech & Pfriction ' 30,740 ! 80 ' 30,660 W

(d) T 'Pshaft

Tm

'30,660

2B x 900' 336.5 N&m

(f) Effcy. 'Pshaft

Pin '

30,66033,000

' 92.9 %

S3.12

IIII M '460/ 3

j30

' ! j 8.853 A

Ns '1206

x 60 ' 1200 rpm s '1200 ! 1164

1200' 0.030

IIII R '460/ 3

0.8 %0.3

0.03% j 1.4

' 24.18 ! j 3.135 A ' 24.38 e !j 7.39E A

IIII S ' IIII R % IIII M ' 24.18 ! j (3.135 % 8.853) ' 26.99 e !j26.367E A

(a) IL ' Iph ' 27.0 A

(b) PF ' cos (26.367E ) ' 0.896 lag

Pin ' 3 x 460 x 27.0 x 0.896 ' 19,269 W

Pg ' 3I 2R

RR

sso Pmech ' ( 1 ! s) ' 3 x 24.382 x 0.3

0.030(1 ! 0.030) ' 17

(c) Effcy. 'Pshaft

Pin '

17,29619,269

' 89.8%

(d) T 'Pshaft

Tm

'17,296

2B60

x 1164' 141.9Nm

(e) Tmax '12

x 32B x 60

x 62

x (460/ 3)2

0.82% 1.42

% 0.8' 349 Nm

S3.13

Ignoring friction, Pshaft = Pmech

3 phases

IIII R '460/ 3

(0.8 % 0.3) % j 1.4' 92.157 ! j 117.3 A

IIII s ' IIII R % IIII M % 92.157 ! j 117.3 ! j8.853 ' 156.2 e !j 53.85E A

(g) i.e. Locked&rotor current ' 156.2 A

(f) L.R.torque '3

2B x 60x 6

2x 0.3 (460/ 3)2

(0.8 % 0.3)2% 1.42

' 159.3 Nm

At standstill, s = 1 and

s = 1

Ns '120 x 60

4' 1800 rpm fr ' sf

ˆ N ' Ns ( 1 ± 0.25) ' 1350 rpm or 2250 rpm

S3.14

If fr = 15 Hz then s '1560

' 0.25

But s can be + or ! depending on whether the rotor is rotating slower than, or faster than, synchronousspeed respectively.

The frequency and amplitude of the voltages generated at the slip rings would be the same in bothcases; but the phase sequence would be reversed, because at 1350 rpm the rotating field is overtakingthe rotor, while at 2250 rpm the rotor is overtaking the field.


Problems

4. Switched Reluctance Machines

Answers

4.1 8.33 Nm

4.2 15E, 800 Hz

4.3 18E, 400 Hz

4.4 (a) 0.430 Nm. (b) 0.225 J; (c) 1.608 J, 3.07 Nm

4.6 (b) 1.48 T; (c) 1.09 Nm; (d) 30E

4.7 (a) 8.36 mH; (b) 8.70 A; (c) 2.62 Nm


4.1 A limited-rotation actuator has a rotor winding and a stator winding. The geometry is such thatthe self-inductances are constant but the mutual inductance varies linearly from zero to 109 mHin a rotation of 75E. Calculate the torque when the windings are connected in series carrying acurrent of 10A.

4.2 What is the stroke angle of a 3-phase switched reluctance motor having 12 stator poles and 8rotor poles? What is the commutation frequency in each phase at a speed of 6,000 rpm?

4.3 What is the step angle of a 5-phase switched reluctance motor having 10 stator poles and 4 rotorpoles? What is the commutation frequency in each phase at a speed of 6,000 rpm?

4.4 A switched reluctance motor with 6 stator poles and 4 rotor poles has a stator pole arc $s = 30Eand a rotor pole arc $r = 32E. The unsaturated aligned inductance is Lau = 10.7 mH and theunaligned inductance is Lu = 1.5 mH, and saturation can be neglected.

(a) Calculate the instantaneous electromagnetic torque when the rotor is 15E before thealigned position and the phase current is 7 A. Neglect fringing.

(b) What is the maximum energy conversion in one stroke if the current is limited to 7.0A?Determine the average torque corresponding to this energy conversion.

(c) What is the flux-linkage in the aligned position when phase current is 7.0A? If thisflux-linkage can be maintained constant while the rotor rotates from the unalignedposition to the aligned position at low speed, determine the energy conversion per strokeand the average torque.

4.5 Show that for an unsaturated switched reluctance motor operating with a fixed conductionangle and flat-topped current waveform, the average torque is proportional to V 2/T2 where Vis the supply voltage and T is the angular velocity. Hence show that to maintain constant torqueper ampere it is necessary to maintain the 'volts per Hertz' constant. Deduce that with fixedsupply voltage, a constant-power characteristic can be obtained by making the conduction angleproportional to the speed.

4.6 Fig. 4.6 shows the cross-section of a switched reluctancemotor with the two coils of one phase on opposite statorpoles. The rotor is in such a position that the 'overlapangle' between these stator poles and a pair of rotor polesis 15E. The airgap is g = 0.5 mm, stator bore diameter D =50 mm, and axial length Lstk = 50 mm. There are 98 turnson each stator pole. The stator and rotor pole arcs areboth 30E. Neglect fringing and leakage, and assume thatthe steel parts are infinitely permeable.

(a) A current of 6 A flows through the two coils inseries. Sketch the flux paths on Fig. 4.6 for thiscondition, showing six flux-lines.

(b) Calculate the flux-density in the airgap betweenthe active poles.

(c) Estimate the torque.

(d) Through what angle of rotation is the torque essentially constant, if the current isconstant and there is no fringing?

Fig. 4.6

4.7 The switched reluctance motor in Fig. 4.7A has an airgap g = 0.2 mm, a stator outside diameterof 79.2 mm, a pole arc of 13 mm, and a stack length Lstk = 50 mm. All other dimensions can bescaled from the diagram. Also shown are the two coils of phase 1. Each coil has 32 turns.

(a) Calculate the unsaturated aligned inductance.

(b) Estimate the current is required to bring the airgap flux-density to 1.75 T when the rotoris in the aligned position as shown in Fig. 4.7A.

(c) The maximum permissible flux-density in the stator poles is 2.15 T, and in the alignedposition the current required for this flux-density is 3is. The unaligned inductance is 1/8of the unsaturated aligned inductance. Estimate the maximum average electromagnetictorque that this motor can produce at low speed.

B

A

C

D

E

Fig. 4.7

4.8 The switched reluctance motor in Fig. 4.7A is shown again in Figs. 4.7 B...E. Diagram B is of thenormal “unaligned” position, but diagrams C, D and E all show faults in the winding or itsconnections. In each case, draw a flux-plot with 5 ! 10 lines of B, and comment on the value ofthe inductance of phase 1 (compared with to case A) when the rotor is in each of the positionsshown. The winding conditions are:

A Normal aligned position; coils in series.

B Normal unaligned position: coils in series.

C Aligned position; left-hand coil open-circuited; normal current in coil 1.

D Aligned position; coils in series; left-hand coil connected with wrong polarity.

E Aligned position; series connection; left-hand coil short-circuited; normal current.



4. Switched Reluctance Machines

T ' i1 i2dL12

d2

i1 ' i2 ' i ; T ' i 2dL12

d2

dL12

d2'

109 x 10& 3

75 x B

180

' 0.0833 H/rad

T ' 102 x 0.0833 ' 8.33 Nm

S4.1

g '360mNr

'360

3 x 8' 15E

6000 rev/min ' 100 rev/sec

' 100 x 8 commutations/sec in each phase

' 800 Hz

S4.2

12/8 motor

g '360

5 x 4'

36020

' 18E

6000 rev/min ' 100 rev/sec ' 100 x 4 commutations/sec in each phase

' 400 Hz

S4.3

10/4 motor

T '12

i 2 dLd2

'12

(7)2 x ( 10.7 ! 1.5) x 10!3

B /6

' 0.430 Nm

Ra ' Lau ia ' 10.7 x 7.0 ' 74.9 mV&s

iu '74.91.5

' 49.9 A

W '12

(49.9 ! 7.0) x 0.0749 ' 1.608 J

Te(avg) '12 strokes/rev x W

2B' 3.07 Nm

S4.4

(a)

(b)

(c)

W '12

i 2 ) L

'12

x 7.02 x ( 10.7 ! 1.5) x 10!3 J

' 0.2254 J

V ' iT dLd2

W '12

i @ ) R

'12

iV2D

T

'12

V

TdLd2

@V2D

T

'V 2

T2

S4.5

Neglecting losses, a flat-topped current waveform can be obtained if

As T increases, i decreases in inverse proportion

Let 2D be the dwell angle (i.e., conduction angle of transistors). The energy converted per stroke is

Since the electromagnetic torque is proportional to W, it is proportional to V2/T2.

Bg ' µ0 Hg ' µ02Ni2g

' 4B x 10!7 x 6 x 98

0.5 x 10!3' 1.48 T

T '12

i 2 dLd2

Lmax 'µ0 (2N)2 Apole

2g

'

2 x 4B x 10!7 x 982 x 30 x B

180x 25 x 50 x 10!6

0.5 x 10!3

' 0.0316 H

dLd2

'0.0316B /6

T '12

x 6.02 x 0.0316B /6

' 1.086 N&m

S4.6

(a)

(b)

(c) Assume that the inductance changes from Lmax at maximum overlap to zero when there is nooverlap. Then

(d) 30E.

Lau '2 x 322 x 4B x 10!7 x 650 x 10!6

0.2 x 10!3' 8.36 mH

is 'B x gµ0 Np

'1.75 x 0.2 x 10!3

4B x 10!7 x 32' 8.70 A

Te(avg) ' strokes/rev x W2B

' 12 x 1.3712B

' 2.62 Nm

S4.7

(a) Lau = 2 Np2 Pg where Np is the number of turns on each pole-coil and Pg is the permeance of one

airgap, µ0Ap/g. The pole area is Ap = Pole width x Stack length = 13 x 50 = 650 mm2. So

(b)

(c)

First draw the aligned and unaligned magnetization curves. Then calculate the enclosed area to the leftof the line i = 3 x 8.70 = 26/1 A.

Then W = 89.36 x 26.1 ! ½ (27.27 x 26.1) ! ½ (72.73 x 8.70) ! (89.36 ! 72.73) x (8.70 + 26.1)/2

= 1371 mJ

Maximum available torque (averaged over one revolution) is


Problems

5. DC Machines

Answers

5.1 Yes.

5.2 (d) 0@354T; (e) 225@4A-t; (f) 23@67V-s; (h) 9@47 V; (i) 143@2 µH; (j) 12,800A-t; (k)0@237 J.

5.4 (a) 1,722 rev/min; (b) 1,589 rev/min; (c) 16.0 A.

5.5 (a) 0@0306 Nm/A; (b) 3,121 rev/min; (c) 37@5 A; (d) 3@20 V/1000 rpm; (e) 53@4 mS

5.6 (a) 1,146 rev/min (b) 1,194 rev/min


Fig. 1

5.1 An electric motor contains coils and magnets and the flux is fixed in magnitude.

Can the flux-linkage of any coil vary? Y — yes N — no

5.2 A permanent-magnet DC motor has the cross-section shown inFig. 1 with an armature diameter D = 2r1 = 60 mm; magnetlength Lm = 8 mm; airgap length g = 0.8 mm, and stack lengthLstk = 35 mm. The magnet arc is $m = 120E. A single full-pitchrotor coil is shown with 30 turns. The ceramic magnet has Br= 0.35 T, µrec = 1, and Hc= 278 kA/m. The current in the rotorcoil is zero.

(a) Sketch the magnetic field set up by the magnet, bydrawing 10 flux lines.

(b) Sketch the variation of the radial component of airgapflux-density Bg(2) around the airgap from 0 to 360E.

(c) Draw an equivalent magnetic circuit and use it tocalculate the flux crossing the airgap under each magnet pole. Assume that the leakageflux can be represented by a permeance equal to 0.15 times the magnet internalpermance, and assume that the permeability of the steel in the rotor and stator isinfinite.

(d) Calculate Bg in the airgap at the centre of the magnet arc, i.e. on the direct axis (d-axis).

(e) Determine the MMF across the magnet and the magnetic field strength Hm inside themagnet.

(f) Calculate the flux-linkage R of the rotor coil in the position shown.

(g) Sketch the waveform of the coil flux-linkage R as the rotor rotates through 360E.

(h) Determine the waveform and the peak value of the EMF induced in the stator coil if therotor rotates at 4,000 rev/min.

(i) Estimate the inductance of the stator coil.

(j) If the magnet material requires a magnetizing force of 1600 kA/m to magnetize it fully,estimate the ampere-turns required to magnetize one of the magnets.

(k) If the current in the coil is maintained constant at 5 A, determine the mechanical workthat is done by the rotor in rotating 180E from the position shown.

5.3 Draw the layout diagrams for the following windings in a 4-pole, 15-slot armature:

(a) Progressive lap winding with 1 coilside/layer and span = 3;

(b) Retrogressive lap winding with 1 coilside/layer and span = 3;

(c) Progressive wave winding with 1 coilside/layer and span = 3;

(d) Retrogressive wave winding with 1 coilside/layer and span = 3;

(e) Progressive wave winding with 3 coilsides/layer and span = 3.

Fig. 2

5.4 A permanent-magnet DC motor operates from a supply of 240 V. Its armature resistance is1.2 ohm and the torque constant is k T = 1.31 Nm/A. Friction torque is constant at 1 Nm, and thebrush voltage-drop is 1.4 V per brush. Calculate

(a) the no-load speed;

(b) the speed for a steady load of 20 Nm; and

(c) the armature current for this load.

5.5 Show that if friction and core losses are neglected, the maximum efficiency of a permanent-magnet DC commutator motor is equal to the ratio of the open-circuit voltage E to the supplyvoltage Vs.

A DC commutator motor is to be designed to deliver 300 W at 2,500 rev/min when supplied at12V. The efficiency must not be less than 2/3 (i.e., 66.67 %) when measured on a dynamometerthat eliminates friction torque. The brush material is such that the voltage drop across eachbrush will be 1@0 V, regardless of the current. Calculate

(a) the torque constant k T in Nm/A

(b) the no-load speed

(c) the current

(d) the EMF constant k E expressed in "Volts per 1000 rpm"

(e) the maximum permissible armature winding resistance Ra.

5.6 A 10-hp 230-V DC shunt-wound motor has the equivalent circuit shown in Fig. 2.

The armature resistance is Ra = 0.3 ohm and the field resistance is Rf = 170 ohm. At no-load andrated voltage, the speed is 1200 rev/min and the armature current is Ia = 2.7 A. At full load andrated voltage, the line current is IL = Ia + If = 38.4 A, where If is the field current. Calculate thespeed at full load,

(a) assuming that the flux is constant

(b) assuming that the flux at full-load is 4% less than the no-load value.



5. DC Machines

S5.2

(a)

(b)

(c) Mr = BrAm

Pm = µrecµ0 Am/Rm

PL = > Pm

Rg = Rg/µ0Ag

Bg = Mg/Ag

5.2 (c) cont’d/...

Am = 2π /3 x (60/2 + 0@8 + 8/2) x 35 = 2,551 mm2

Ag = 2π /3 x (60/2 + 0@8/2) x 35 = 2,228 mm2

Pm = 1 x 4π x 10!7 x 2,551 x 10!6 / 8 x 10!3 = 4@007 x 10!7 Wb/A

Rg = 0@8 x 10!3 / (4π x 10!7 x 2,228 x 10!6) = 2@857 x 105 A/Wb

Mg = Mr / (1 + (1 + >)PmRg) where > is the leakage fraction (0@15)

= 0@35 x 2,551 x 10!6 /(1 + 1@15 x 4@007 x 10!7 x 2@857 x 105)

= 0@789 mWb

(d) Bg = Mg /Ag = 0@789 x 10!3 / 2,228 x 10!6 = 0@354 T

(e) Fm = Fg = Mg Rg = 0@789 x 10!3 x 2@857 x 105 = 225@4 A-t

*Hm* = Fm/Rm = 225@4 / 8 = 28@2 A/mm, i.e. 28@2 kA/m

(f) Flux-linkage R = N Mg = 30 x 0@789 = 23@67 V-s, where N = 30 turns and it is assumedthat all the flux Mg links the coil. This is the peak flux-linkage, Rpk.

(g)

(h) Peak EMF is epk = d R/dt = MR/M2 x d2/dt = T x MR/M2 where T = 4,000 x 2B/60 = 418@9rad/s

MR/M2 = 23@67 x 10!3/ (60E x B/180) = 22@60 mVs/rad

ˆ epk = 418@9 x 22@60 x 10!3 = 9@47 V

5.2 cont’d/...

(i) L = 8/i where 8 is the flux-linkage produced by i A

8 = N M = NBAg = N µ0 H Ag = Nµ0 x Ni/2(Rg + Rm) x Ag,

so L = µ0N2Ag/2(Rg + Rm) = 4B x 10!7 x 302 x 2,228 x 10!6/2(8 + 0@8) x 10!3 = 143@2 µH

(j) To get Hm = 1600 kA/m we need Fm = HmRm = 1600 x 103 x 8 x 10!3 = 12,800 At

(k) Work done = 5@0 x (2 x 23@67 x 10!3) = 0@237 J' m e i dt '0

BTMRM2

id2T

' i[)R]0

B

5.3(a) Progressive lap winding in 4-pole, 15-slot armature with 1 coilside/layer; span = 3

S5.3

The windings are shown on the following three pages. Note the last one (which is not part of thequestion, but may be of interest).

5.3(b) Retrogressive lap winding in 4-pole, 15-slot armature with 1 coilside/layer; span = 3

5.3(c) Progressive wave winding in 4-pole, 15-slot armature with 1 coilside/layer; span = 3

5.3(d) Retrogressive wave winding in 4-pole, 15-slot armature with 1 coilside/layer; span = 3

5.3(e) Progressive wave winding in 4-pole, 15-slot armature with 3 coilsides/layer; span = 3

A wave winding is not possible if the number of slots is even.

S5.4

Vs = 240 V Vb = 1.4 V per brush kE = kT = 1@31 V-s/rad or Nm/A

(a) T0 = (Vs ! 2Vb)/kE ! Ra/kE2 x Tf = (240 ! 2 x 1@4)/1@31 ! 1@2/1@312 x 1@0

= 180@4 rad/s = 1,722 rev/min

(b) T = (Vs ! 2Vb)/kE ! Ra/kE2 x (T + Tf) = (240 ! 2 x 1@4)/1@31 ! 1@2/1@312 x (20 + 1@0)

= 166@4 rad/s = 1,589 rev/min

(c) Ia = Te/kT = (T + Tf) /kT = (20 + 1@0)/1@31 = 16@0 A

Check : RaIa = 1@2 x 16@0 = 19@2 V

Ea = Vs ! RaIa ! 2Vb = 240 ! 19@2 ! 2@8 = 218 V

T = Ea/kE = 218/1@31 = 166@4 rad/s .....OK

S5.5

Input power = VsIa

Output power = EaIa = TeT if friction and rotor core loss are neglected

Efficiency = Output/Input = Ea/Vs

(a) At 2,500 rev/min and 300 W we need Effcy $ 0@667, so Ea $ 0@667 x 12 = 8@0 V

kE = Ea/T = 8@0/(2,500 x 2B/60) = 0@0306 V-s/rad = kT

(b) No-load speed T0 = (12 ! 2 x 1@0)/0@0306 = 326@8 rad/s = 3,121 rev/min

(c) Tshaft = 300 W / (2,500 x 2B/60) = 1@146 Nm

Ia = 1@146/0@0306 = 37@45 A

(d) kE = 0@0306 V-s/rad

...= 0@0306 x 2 B/60

...= 0@0032 V/(rev/min)

...= 3@204 V/(1000 rev/min)

(e) The volt-drop RaIa must be limited to 2 V if Ea is to be 8@0 V (with 1@0 V across each brush).

ˆ the winding must be designed so that Ra is no greater than 2/37@45 = 53@4 mohm.

S5.6

At no-load, Ea = Vs ! RaIa = 230 ! 0.3 x 2@7 = 229@2 V.

ˆ kE = kT = Ea/T0 = 229@2/(1,200 x 2B/60) = 1@824 V-s/rad

The field current is If = 230/170 = 1@35 A

(a) At full load, Ia = 38@4 ! 1@35 = 37@05 A

ˆ Ea = 230 ! 37@05 x 0@3 = 218@9 V

and T = Ea/kE = 218@9/1@824 = 120 rad/s = 1,146 rev/min

(c) Ia and If are unchanged but the flux is reduced by 4% and therefore kE and kT are reduced by 4%also; kE = kT = 0@96 x 1@824 = 1@751 V-s/rad. So T is 4% higher at 218@9/1@751 = 125@0 rad/s= 1,194 rev/min.

Documents

Miller, T.J.E. - SPEED's Electric Motors