SPEED's Electric Motors
An outline of some of the theory
in the SPEED software for electric machine design
with problems and solutions
© TJE Miller, University of Glasgow, 2002
SPEED's Electric Motors
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
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 vital
interface between electrical and mechanical engineering. This interface is found wherever mechanical
motion is controlled by electronics, and pervades a vast range of products. A little consideration
reveals a large and important area of technology, in which motor drives are fundamental. In Japan the
term 'mechatronics' is applied to this technology, usually with the connotation of small drives. In the
west the term 'motion control system' is often used for small controlled drives such as position or
velocity 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 not
confined to the control electronics. Two important reasons for the development activity and the
increasing technical variety are:
(1)
Increasing use of computers and electronics for motion control. Automation demands drives
with 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 and
switched 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 into
electricity, and about two-thirds of that is re-converted in electric motors and drives, mostly
integral-kW induction motors running essentially at fixed speed. If a constant-speed motor is used to
drive a flow process (such as a fan or pump), the only ways to control the flow rate are by throttling or
by recirculation, Fig. 1. Since the motor runs at full speed regardless of the flow requirement, there
can be excessive energy losses in the recirculation valve. Similar considerations apply to the control
of airflow by adjustable baffles in air-moving plant.
Fig. 1.1
Flow process controlled by recirculation can produce energy losses in the flow control valve.
1. Sizing, gearing, cooling, materials and design
Page 1.3
In such applications it is often possible to reduce average energy costs by 50 per cent or more by using
adjustable-speed drives, which eliminate the throttling or recirculation loss; see Fig. 1.2. The additional
losses 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 may
be more expensive, but its capital cost can be offset against energy savings and the reduction of
maintenance requirements on mechanical components.
Recirculation in a flow-control process is analogous to the control of a DC motor by means of an
adjustable series resistance. The technique is inherently wasteful, and although it is cheap to
implement, it is increasingly hard to justify in the face of concerns about energy efficiency and
pollution, even at low power levels.
Fig. 1.2
Flow control efficiency is improved by adjustable-speed drive
(b) Velocity or position control. Obvious examples of speed control are the electric train, portable
hand tools, and domestic washing-machine drives. In buildings, elevators or lifts are interesting
examples 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 varying
degrees of precision. Particularly with the trend towards automation, the technical and commercial
growth in drives below about 20 kW is very vigorous. Many system-level products incorporate an
adjustable-speed drive as a component. A robot, for example, may contain between 3 and 6 independent
drives, 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 motor
starts can be eliminated by adjustable-speed drives with controlled acceleration.
A full
adjustable-speed drive is used in this situation only with very large motors or where the start-stop
cycles are so frequent that the motor is effectively operating as a variable speed drive. Most
soft-starting applications are less onerous than this, and usually it is sufficient (with AC motors) to
employ series SCR's (or triacs with smaller motors) which ‘throttle' the starting current to a controlled
value, and are bypassed by a mechanical switch when the motor reaches full speed.1 The soft-starter
is less expensive than a full adjustable-speed drive, which helps to make it economical for short-time
duty during starting.
1
Series control of induction motors is inefficient; produces excessive line harmonics; and is not very stable. These
factors can usually be tolerated for a few seconds during starting, but they render the soft-starter unsuitable for continuous
speed control.
Page 1.4
1.3
SPEED’s Electric Motors
LARGE VERSUS SMALL DRIVES
There are marked design differences between large and small drives. Large motors are almost always
chosen from one of the classical types: DC commutator (with wound field); AC induction; and
synchronous. 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 need
for a wider range of control characteristics. Efficiency and materials utilization are still important, but
so are control characteristics such as torque/inertia ratio, dynamic braking, speed range, acoustic
noise and torque ripple.
There are also several breakpoints in the technology of power semiconductors. At the highest power
levels (up to 20MW) SCRs (thyristors) and GTOs (gate turn-off thyristors) are the only devices with
sufficient voltage and current capability, but IGBTs (insulated-gate bipolar transistors) also now have
voltage ratings measured in kV and current ratings of hundreds of amps. Naturally-commutated or
load-commutated converters are preferred, because of the saving in commutation components and for
reliability and efficiency reasons. In the medium power range (up to a few hundred kW) forced
commutation 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, the
electronic drive, and the control.
Fig. 1.3
Drive system structure
The range of modern motion-control applications is virtually unlimited. Any random list illustrates
the variety—aerospace actuators; washing machines; computer disk and tape drives; printer plattens
and printheads; inertial guidance systems; adjustable-speed pumps, blowers, fans and compressors;
locomotive and subway traction; automatic machine tools and machining centers; servo drives and
spindle drives; robots; automotive auxiliaries; refrigeration and air-conditioning drives; and many
others.
Loads have widely differing requirements. The commonest requirement is for speed control, with
varying degrees of precision and accuracy. Position control is of increasing importance, particularly
in automated plants and processes, and in office machinery and computer peripherals. In some cases
it is the steady-state operation that is most important, for example in air-conditioning and pump drives.
1. Sizing, gearing, cooling, materials and design
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Page 1.5
Compliance with national, EC, USA and industry standards
Maximum continuous power or torque requirements
Forward/reverse operation
Motoring/braking operation
Dynamic or regenerative braking
Overload rating and duration
Supply voltage (AC or DC) and frequency
Type of control: speed, position, etc.
Precision required in controlled speed or position
Programmability: speed and/or position profiles, start/stop ramps etc.
Interface with plant control and communications
Dynamic requirements: torque/inertia ratio, acceleration/deceleration
Gearbox or direct drive; gear ratio
Reliability and redundancy of components
Protection arrangements, both mechanical and electrical
Maximum level of acoustic noise; noise spectrum
Compliance with EMC regulations
Limitation of harmonics in the supply system
Maintenance; spare parts; provision for expansion or reconfiguration
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 because
of the need to minimize the time taken to perform operations or effect control changes. In these cases
the torque/inertia ratio of the motor is an important parameter. In automotive applications the prime
requirements 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 is
increasingly important in servo motors and applications such as automotive power steering, where less
than 1% ripple is typically required.
1.5
DRIVE SYSTEM REQUIREMENTS
Table 1.1 provides a checklist of application requirements. The speed/torque capability diagram is
especially important, Fig. 4. Typically at low speed the drive operates under current limit, and since
torque is generally proportional to current (or nearly so), the torque is controlled in this mode of
operation, such that any value can be obtained up to the value corresponding to maximum current. At
high 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 correct
phase angle or waveform of the current. Base speed is the maximum speed at which rated torque can
be developed. Above this speed, drives are often evaluated according to the speed range over which
they can maintain constant power, the torque decreasing as the speed increases. Some drives (such as
triac-controlled AC universal motors) have almost no constant-torque region and their characteristics
are said to be “highly inverse”. On the other hand, brushless PM motors tend to have a predominantly
constant-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 motion
control. At the 'heavy' end of the spectrum are the multiple drives found in steel rolling mills, paper
mills, and other heavy process plants, where it is normal to coordinate the motion of all the shafts by
means of a computer or a network of computers, some of which may be quite large. At the light end
of the power range are the drives found in office machinery, small computers, and portable goods such
as 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.
Page 1.6
SPEED’s Electric Motors
Fig. 1.4
Speed/torque characteristic
The first functions implemented with microprocessors were low-speed functions such as monitoring
and diagnostics, but digital control has penetrated from outer position loops through the intermediate
velocity loop and is now routinely used even in high-bandwidth current regulators. The development
of field-oriented control for AC induction and synchronous motor drives would not have been practical
without the microprocessor. Field-oriented control is based on reference-frame transformation which
may require rapid computation of trigonometric functions of rotor position. It permits the outer
control loops of AC and DC drives to be the same, both in hardware and software, and improves the
dynamic performance of the AC drive. Its development since the mid 1970's was a key factor enabling
the AC induction motor to compete in precision speed control with the DC motor, which had been
preferred for speed control for at least 50 years before then.
Such is the sophistication and speed of modern microelectronics that the PWM schemes employed to
regulate 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-programmable
gate arrays for the high-speed functions, often combined with microcontrollers. Digital signal
processors (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, preferably
with 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 been
exploited together with new forms of control based on neural networks or fuzzy logic. At the other end
of the drive, communication with computers and other controllers is another area of continuous
development.
Power semiconductor devices. The IGBT is the most popular device in integral-kW drives, and the
power MOSFET in low-power drives, especially at low voltage. GTO thyristors are still used in large
drives especially above 1 MW.
New magnetic materials. The permanent-magnet industry has continued to improve the
characteristics of the main families of magnet used in electric machines: neodymium-iron boron, which
was pioneered by Sumitomo and General Motors; rare-earth/cobalt; and ferrite. At room temperature
NdFeB has the highest energy product of all the common magnet materials. In its early days the
performance was rather sensitive to temperature, but it is now widely used even in automotive and
industrial servo applications where high operating temperatures up to and exceeding 100EC are
common, even in the presence of strong demagnetizing fields.
CAD and numerical analysis in design. Motor design has been computerized since the early days
of computers, initially with the coding of well established design procedures. The last 20 years has seen
a steady development of commercially available finite-element software capable of analysing fields in
machines in two and three dimensions. The most advanced programs can determine eddy-currents and
transients, but the coupling of finite-element solvers to circuit-analysis and simulation algorithms is
1. Sizing, gearing, cooling, materials and design
Page 1.7
still in the research laboratory. In spite of the technical advances it remains the case that finiteelement analysis requires the application of highly skilled personnel, and although machine designers
use it fairly widely, they either follow well-established procedures or rely on specialists to practise
computations 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 the
routine processes. It is fundamentally an analysis tool rather than a design tool: and it suffers from
certain limitations when applied to motor design. It requires detailed input data and the results need
skilled interpretation. It is accurate only in idealized situations where parasitic effects have been
removed. It is too slow to be cost-effective as part of a synthesizing CAD package, and is likely to remain
so for some time. It is most useful in helping to understand a theoretical problem that is too difficult
for conventional analysis, and in this role it has undoubtedly helped to refine many existing motor
designs and improve some new ones.
At the same time the development of special design software for electric machines has produced a
number of products which are much faster and are specificall intended for motor and generator
design. 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 motor
products. One example is the SPEED software from the University of Glasgow, which is used in places
to illustrate this book.2
The primary problems in motor and generator design are not simply electromagnetic, but require an
integrated approach to materials utilization and design-for-manufacture. The philosophy needs to be
a synthesizing one rather than an analytical one. This multidisciplinary problem includes heat
transfer and mechanical design as well as electromagnetics. The situation is more complicated in
adjustable-speed drives where the supply waveforms are 'switchmode' chopped waveforms rather than
pure sinewaves or DC. In these cases time-stepping simulation may be necessary to determine the
expected 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 current
waveform, it could be argued that preoccupation with this class of software tool could hinder the
development of drive systems which escape from the classical DC and sinusoidal waveforms in order
to explore the possibilities of a wider class of drive current waveforms, coupled with new concepts in
motor geometry. The key lies in using the finite-element tool in the correct way to determine
parameters 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 the
true sense. It is much more common to find “optimization” techniques which rely on the automatic
generation of large numbers of “feasible” designs, and then rank them according to a more or less
complex criterion which includes constraints on particular dimensions and other parameters. The
imaginative element in the design process remains in the human mind, and the computer appears to
be far from taking it over.
For simulation of complete systems, such as an automotive power steering system or an aircraft flight
control 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 permit
the simulation of detailed motor models and their drives and controls. They may be used for the
development of control algorithms that are subsequently programmed in a microprocessor or gate
array.
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, but
moulded slot insulation and encapsulation of rotors are also widely used. In brushless motors designed
for high peripheral speeds, the magnets are often restrained against centrifugal force by banding or
tubes made of Kevlar™, glass-reinforced plastics, or carbon fiber.
Motor drives generally require transducers for control and protection, and there has been progress
in current-sensor and shaft position sensor technology. In particular the linearity and
2
Others include design programs from Trimerics in Stuttgart, Germany and Yeadon Engineering Services, Wisconsin.
Page 1.8
SPEED’s Electric Motors
temperature-independence of Hall-effect current sensors has improved greatly, and it is common to
mount these devices in the same package, or on the same printed-circuit card, as the driver stage of the
power electronics in small drives. For larger drives flux-nulling current sensors can be used with
bandwidths 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 by
the rotor magnet or by a separate magnet ring. Alternatively, optical interrupters may be used with
a shaft-mounted slotted disk. At high speeds the commutation sensor can be used to generate a speed
signal via a frequency-to-voltage conversion. For motion control systems and servo-quality drives
separate velocity and position transducers usually have to be used. For such systems the resolver is
attractive because of its ruggedness, resolution, and its ability to provide accurate absolute position
and velocity signals from one sensor.
1.7
WHICH MOTOR ?
The proliferation of new ideas, materials and components obviously generates many opportunities but
also complicates the question, what is the best drive for a particular job? We can perhaps address this
by attempting to trace the evolution of the different motor types in such a way as to bring out their
most important advantages and disadvantages. It is the motor that determines the basic characteristics
of the drive system, and it alsodetermines the requirements on the power semiconductors, the
converter 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 electronic
control has taken place. Together with the PM DC commutator motor in row 2 and the series-wound
AC commutator motor or 'universal' motor, the row 1 motors account for the lion's share of all motor
markets, both fixed-speed and adjustable-speed, even though they represent only a minority of the
many different principles of electromechanical energy conversion on which motor designs may be
based. By contrast, the nonclassical motors are essentially confined to specialist markets and until
recently, few of them except the brushless DC motor have been manufactured in large numbers. Table
1.2 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 permanent
magnets. The synchronous motor immediately becomes brushless, but the DC motor must go through
an additional transformation, from row 2 to row 3 with the inversion of the stator and rotor, before the
brushless 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 three
terminals, together with the switched reluctance motor, which cannot be derived from any of the other
motors. Its awkward placement in Fig. 5 reflects the fact that it has various properties in common with
all the brushless motors. Obviously the emphasis in this book is on the brushless motors, with only
a relatively superficial treatment of the DC motor and the induction motor. Stepper motors are also
excluded.
The DC commutator motor. The traditional DC commutator motor with electronically adjustable
voltage has always been prominent in motion control. It is easy to control, stable, and requires
relatively few semiconductor devices in the drive. For many years the wound-field DC motor held its
own against the challenge of AC drives—arguably for at least fifty years from the mid-1930's until the
mid-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.
1. Sizing, gearing, cooling, materials and design
DC wound field (brush)
AC/DC universal (brush)
AC synchronous
wound-field
DC PM (brush)
[Brushless PM Exterior Rotor]
Brushless PM AC
Page 1.9
AC induction
Switched
reluctance
Brushless PM
Squarewave or sinewave
Fig. 1.5
Synchronous
reluctance
Evolution of brushless motors from classical AC and DC motors
Page 1.10
SPEED’s Electric Motors
The main objections to the commutator motor are brush and commutator wear, and the fact that the
losses arise mostly on the rotor, making cooling more difficult than in AC motors where the losses arise
mostly 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 the
world, 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 and
pole 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. Armature
reaction is reduced and commutation is improved, owing to the low permeability of the magnet. The
loss of field control is not as important as it would be in a larger drive, because it can be overcome by
the controller. In small drives the need for field weakening is less common anyway. The PM DC motor
is 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 of
the 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, brushless
motor 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 and
therefore the size and cost of MOSFETs required in the drive.
AC induction motor drives. AC induction or synchronous motors are often preferred because of the
limitations of commutation and rotor speed in DC motors. Slip is essential for torque production in
the induction motor, and it is impossible, even in theory, to achieve zero rotor losses. This is one of the
limitations 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.
Fig. 1.6
Integrated motor/inverter and hand-held controller
(courtesy of Grundfos A/S, Denmark)
The efficiency and power factor of induction motors falls off in small sizes because of the natural laws
of scaling, particularly at part load. As a motor of given geometry is scaled down, if all dimensions are
scaled at the same rate the MMF required to produce a given flux-density decreases in proportion to the
linear dimension. But the cross-section available for conductors decreases with the square of the linear
dimension, as does the area available for heat transfer. This continues down to the size at which the
mechanical airgap reaches a lower limit determined by manufacturing tolerances. Further
scaling-down results in a more-or-less constant MMF requirement while the areas continue to decrease
with the square of the linear dimension. There is thus an “excitation penalty” or “magnetization
penalty” which becomes rapidly more severe as the scale is reduced. It is for this reason that
permanent magnets are so necessary in small motors. By providing flux without copper losses, they
directly alleviate the excitation penalty.
1. Sizing, gearing, cooling, materials and design
Page 1.11
The induction motor is “brushless” and can operate with simple controls without a shaft position
transducer. The simplest type of inverter is the six-step inverter. With no shaft position feedback, the
motor remains stable only as long as the load torque does not exceed the breakdown torque, and this
must be maintained at an adequate level by adjusting the voltage in proportion to the frequency as the
speed changes. At low speeds, oscillatory instabilities may appear. To overcome these limitations
there 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 between
stator MMF and rotor flux. Field orientation usually requires a shaft position encoder and may include
an in-built control model whose parameters are specific to the motor, and which must be compensated
for changes caused by changing load and temperature. Such controls are complex and generally cannot
be 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 design
requirements. These requirements cannot be met adequately with series- or triac-controlled induction
motors, which are therefore restricted to applications where low cost is the only criterion. Together
these 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 permanent
magnets for excitation. There is no single 'breakpoint' below which PM brushless motors outperform
induction motors, but it is usually in the1!10 kW range. Above this size the induction motor improves
rapidly, while the cost of magnets works against the PM motor. Below it, the PM motor has better
efficiency, torque per ampere, and effective power factor. Moreover, the power winding is on the
stator where its heat can be removed more easily, while the rotor losses are extremely small. These
factors combine to keep the torque/inertia ratio high in small motors. The brushless DC motor is also
easier to control, especially in its ‘squarewave' configuration (Chapter 4). Although the inverter is
similar to that required for induction motors, usually with six transistors for a 3-phase system, the
control algorithms are simpler and readily implemented in 'smartpower' or special-purpose ICs.
Fig. 1.7
The Minas brushless PM motor produced by Matsushita with its stator fabricated from segments. Courtesy of
Matsushita Ltd., Japan
The brushless PM AC synchronous motor. In Row 2 of Fig. 5 the brushless synchronous machine
has permanent magnets instead of a field winding. Field control is again sacrificed for the elimination
of brushes, sliprings, and field copper losses. This motor is a classical salient-pole synchronous AC
motor with approximately sine-distributed windings, and it can therefore run from a sinewave supply
without 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. The
interior construction simplifies the assembly and relieves the problem of retaining the magnets against
centrifugal force. It also permits the use of rectangular instead of arc-shaped magnets, and usually
there is an appreciable reluctance torque which leads to a wide speed range at constant power.
Page 1.12
SPEED’s Electric Motors
The PM synchronous motor operates as a synchronous reluctance motor if the magnets are left out or
demagnetized. This provides a measure of fault-tolerance in the event of partial or total
demagnetization through abnormal operating conditions. It may indeed be built as a magnet-free
reluctance motor, with or without a cage winding for starting 'across-the-line'. Although the power
factor and efficiency are not as good as in the PM motor, synchronous reluctance motors can be
designed 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 the
same shaft, feeding a rotating rectifier which passes DC to a field winding on the main rotor. This
motor has full field control. It is capable of a high specific torque and high speeds. As a generator, this
configuration is popular in high-speed aircraft generators (at 24,000 and 12,000 rpm, 400 Hz) and in a
wide variety of small industrial applications.
All the motors on the diagonal of Fig. 5 operate with inverters that share the same power circuit
topology (three 'totem-pole' phaselegs with the motor windings connected in star or delta to the
midpoints). This gives rise to the concept of a family of motor drives providing a choice of motors and
motor characteristics, but with a high degree of commonality in the control and power electronics and
all the associated transducers. The trend towards integrated phaselegs, or indeed complete three-phase
bridges, with in-built control and protection circuitry makes this concept more attractive. This family
of 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 speed
range); the synchronous reluctance motor (free from magnets and capable of high speeds or
high-temperature operation); and the induction motor. It should be noted that all these drives are
essentially “smooth-torque” systems with low torque ripple.
Stepper motors represent a major class of motors not included in Fig. 5. Steppers are always
brushless and usually operate without shaft position sensing. Although they have many properties in
common with synchronous and brushless DC motors, they cannot naturally be evolved from the motors
in Fig. 5. By definition they are pulsed-torque machines incapable of achieving ripple-free torque by
normal means. Variable-reluctance (VR) and hybrid steppers can achieve an internal torque
multiplication through the use of multiple teeth per stator pole and through the ‘vernier' effect of
having different numbers of rotor and stator poles. Both these effects work by increasing the number
of torque impulses per revolution, and the price paid is an increase in commutation frequency and iron
losses. Steppers therefore have high torque-to-weight and high torque-to-inertia ratios, but are limited
in top speed and power-to-weight ratio. The fine tooth structure requires a small airgap, which adds
to 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 the
high magnetic frequency and the effect of MMF drop in the iron, such motors require expensive
lamination steels to get the best out of them.
Switched reluctance motors are derived from the single-stack VR stepper, in which the current
pulses are phased relative to the rotor position to optimize operation in the 'slewing' (continuous
rotation) mode. This usually requires a shaft position transducer similar to that which is required for
the 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 produce
continuous 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 control
of 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 PM
motor in small sizes.
When the classical motors are interfaced to switchmode converters (such as rectifiers, choppers, and
inverters) they continue to respond to the average voltage (in the case of DC motors) or the
fundamental voltage (in the case of AC motors). The harmonics associated with the switching operation
of the converter cause parasitic losses, torque ripple, and other undesirable effects in the motors, so
that de-rating may be necessary. The nonclassical motors are completely dependent on the switchmode
operation of power electronic converters. In steppers it is acceptable for the torque to be pulsed, but
most brushless drives are designed for smooth torque even though the power is switched.
1. Sizing, gearing, cooling, materials and design
Motor
Page 1.13
Drive or supply
Typical application
Constant DC or variable DC from DC
generator, phase-controlled rectifier,
or chopper
Integral-kW industrial drives; railway
traction
DC commutator motors
(a) Wound-field
(b) Permanent-magnet
DC homopolar motors with slip-rings
or segmented slip-rings
Automotive and aircraft auxiliaries;
small servo and variable-speed drives
Ship-propulsion specials
Brushless PM motors
(a) Squarewave
Electronically commutated rectangular Disk drives, automotive auxiliaries,
current waveform from inverter
small portable goods, servo motors,
spindles
(b) Sinewave
Sinewave current waveform from
inverter
Servo motors, spindles
Universal AC/DC series commutator
motors
Fixed AC; triac-controlled AC; also
variable or constant DC
Domestic appliances (washing
machines, food processors); power
tools
(a) Cage-type, 3-phase
Constant voltage AC or inverter-fed
Pumps, fans, compressors, industrial
machinery 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 starting
conditions; sometimes used with sliprecovery energy systems
Constant voltage AC or inverter-fed
Very large drives
Induction motors
Synchronous motors
(a) Wound rotor with DC winding and
slip-rings
(b) Wound-rotor with brushless exciter Constant voltage AC or inverter-fed
(motoring)
(c) Permanent magnet
Most often used as a generator
Same as Sinewave brushless PM motor
Reluctance motors
(a) Synchronous reluctance (line-start, Inverter-fed AC
with rotor cage)
Multi-machine variable-speed drives
with several motors fed from one
inverter
(b) Synchronous reluctance (cageless) Inverter-fed AC
Servo/spindle motors
(c) Switched reluctance
Electronically regulated and
commutated DC
Washing machine drives, mining
machinery, door openers, automotive
auxiliaries, aircraft actuators, highspeed compressors
(d) Single-phase reluctance
Switched DC or AC
Small actuators, clocks
Switched DC, usually current-limited
Motion-contol applications dependent
on open-loop stepping for position
control (indexing); office equipment,
industrial machinery
(a) Cylindrical rotor
Inverter-fed or constant AC voltage
Simple synchronous motor with good
starting characteristics
(b) Can-type rotor
Constant-voltage AC
Motorized valves and actuators
Stepper motors
(a) VR, single-stack
(b) VR, multiple-stack
(c) Permanent-magnet
(d) Hybrid
Hysteresis motors
TABLE 1.2
A SELECTION OF MOTORS WITH TYPICAL APPLICATIONS
Page 1.14
1.8
SPEED’s Electric Motors
SIZING
When a new electric machine is to be designed from scratch, the requirements usually includes a set
of 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 a
machine can be determined, starting from the performance specifications and working within the
limits of material properties and temperature rise.
In many cases, new machine designs are evolved from existing ones, by modifying existing laminations
and components to minimize the cost of changes in tooling and components. Even so, the same
principles determine how much power and performance can be achieved from a machine of given size
and temperature rise.
The output equation. The classical output equation applies to (and unifies) all electrical machines
from the tiniest micromotors (a few µW) up to the largest AC motors used in process plants or ship
propulsion (up to 20MW). Intuitively it comes from the fundamental law of electromagnetic force which
is often loosely stated as “force = flux × current”, according to the left-hand rule. For engineering
purposes we need to derive a more precise statement of this law. Except in linear motors, we are more
interested in torque than force. It is convenient to work with flux-density and current-density, because
these parameters have values which do not change greatly from one machine to another. Further, the
flux and current densities are closely related to the power loss density which determines the cooling
requirements and temperature distribution throughout the machine.
Specifically, the output equation relates the torque per rotor volume (TRV) to the electric loading A and
the magnetic loading B. We will define A and B first before deriving a precise form of the output
equation. 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 multiplying
constants and interpretation.
The electric loading A is defined as the linear current density around the airgap circumference, that
is, the number of ampere-conductors per metre around the stator surface that faces the airgap.
A '
Total ampere&conductors
'
Airgap circumference
2 m Nph I
BD
A/m
(1.1)
where I is the RMS phase current, m is the number of phases, Nph is the number of turns in series per
phase, and D is the diameter of the airgap. The airgap is assumed to be small compared to the rotor
diameter, so that no distinction is made between the rotor diameter and the stator diameter. The RMS
2
current is used because it determines the I R 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 the
flux-density is distributed sinusoidally so that the fundamental flux/pole is
M1 ' B ×
B D L stk
2p
Wb
(1.2)
where p is the number of pole-pairs and Lstk is the stack length, i.e., the axial length of the active part
of the machine. In slotted stators and rotors, the peak
flux density in the teeth Bt(pk) must be limited to
about 1.6T, otherwise the magnetizing current and/or
the iron losses may become excessive. The peak fluxdensity 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-density
is maximum; see Fig. 1.8. Typically J is of the order of
0.5. Thus B = 2Bg(pk)/B = 2JBt(pk)/B, so B is normally
limited to around 2 × 0.5 × 1.6/B . 0.5T.
Fig. 1.8
Definition of tooth pitch and J
1. Sizing, gearing, cooling, materials and design
Page 1.15
The generated EMF per phase is given by the standard equation3
2B
E '
2
kw1N ph M1 f '
B2 kw1N ph B D Lstk f
p
2
V.
(1.3)
where f is the fundamental frequency, kw1 is the fundamental harmonic winding factor, and the product
kw1Nph is the effective number of turns in series per phase. The maximum available electromagnetic
power at the airgap is mEI. We consider this as being converted into mechanical power TT/p, where
T/p = 2B f is the speed in rad/sec. (Note also that T = 2B/60 × rpm). We can obtain the TRV as
2
T/(BD Lstk/4), and substituting from equations. 1.1, 1.2 and 1.3 we get
TRV
'
T
'
Vr
B
2
kw1 AB
Nm/m 3 .
(1.4)
This equation reflects the “flux-current product” in the form AB. The
multiplying factor is simply a constant multiplied by the winding
factor kw1, which incidentally casts kw1 in the role of a utilization
factor—the higher the winding factor, the greater is the utilization of
flux and current in producing torque. Since kw1 is usually about
0.85!0.95, TRV . 2AB.
The TRV is also related to the airgap shear stess F, which is the
tangential (torque-producing) force per unit of swept rotor surface
area; see Figs. 1.9 and 1.10. For every unit of rotor surface area, the
electromagnetic torque is r F = D F/2, so the total torque is T = BD ×
2
D F/2 = BD F/2, from which it follows that
'
TRV
T
' 2 F.
Vr
Fig. 1.9
Airgap shear stress
(1.5)
The airgap shear stress F is measured in kN/m2. Typical values are
given in Table 1.3.4 The winding factor kw1 is generally between 0.8
and 0.95, so that TRV . 2 BA and F . BA. For example, if the electric
loading 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 F
and TRV would apply with natural convection, while the higher values
would apply with forced-air cooling supplied by an external or shaft- Fig. 1.10 Airgap shear stress
mounted fan.5
Class of machine
kNm/m3
TRV
lbf/in
F
Small totally-enclosed motors (Ferrite magnets)
7 ! 14
0.5 ! 1
Totally-enclosed motors (sintered Rare Earth or NdFeB magnets)
14 ! 42
1!3
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
100 ! 250
7 ! 18
Totally-enclosed motors (Bonded NdFeB magnets)
Large liquid-cooled machines (e.g. turbine-generators)
TABLE 1.3
TYPICAL VALUES FOR TRV AND F (CONTINUOUS OPERATION)
3
See 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 L), so K = TRV × B/4.
2
2
Page 1.16
SPEED’s Electric Motors
The coefficient Bkw1//2 in eqn. (1.11) is peculiar to AC machines where the ampere-conductor
distribution and the flux-density are sinusoidally distributed in space around the airgap; this can be
written
A(2) ' A(pk) sin 2
and
B(2) ' B(pk) sin 2.
(1.6)
2
The product A(2)B(2) is the force per unit of rotor surface in N/m , and therefore the torque can be
2
obtained by integrating rA(2)B(2)d2 = rA(pk)B(pk) sin 2 over the entire rotor surface, where r = D/2,
2
and dividing by the rotor volume BD Lstk/4. This gives TRV = A(pk)B(pk). But the RMS value of A is
A(pk)//2, while the average value of B is BB(pk)/2, giving TRV = BAB//2. The winding factor appears
because 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 winding
in producing the fundamental component is represented by kw1.
2
In DC machines the integral rA(2)B(2)d2 has no sin 2 term, and the result is TRV = 2AB, where B is the
average value of flux-density over the whole rotor periphery. The RMS value of A is equal to the peak
value, since the current is uniformly distributed around the rotor, and
A '
Z I a /a
(1.7)
BD
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 also
related to the current density J in the conductors. Suppose the area of one slot is Aslot. Let d = slotdepth, t = tooth width, w = slot width, and 8 = slot pitch = BD/Ns, where Ns is the number of slots. Also
let J = t/8. Then t + w = 8 and Aslot = wd = (1 ! t)8d. Now if the slot-fill factor Fslot is defined as the ratio
of actual copper cross-section area to the total area of each slot, we can write
J '
A8
'
Fslot Aslot
A
.
Fslot d( 1 ! J )
(1.8)
For example, if the slot depth is d = 15mm, the slot-fill factor is Fslot = 0.4, the tooth-width/slot-pitch
ratio is J = 0.5, and the electric loading is A = 20 A/mm, the current density is
J '
20
0.4 × 15 × ( 1 ! 0.5 )
' 6.7 A/mm 2 .
(1.9)
Typical values of current density for use in AC or brushless machines for different applications are
given in Table 1.4. Note that in machines operated from electronic drives there are usually timeharmonics in the current which increase the current-density without increasing the torque-producing
value of A, and it may be necessary to allow for this by multiplying J by a form factor kf. In AC
machines 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)
1. Sizing, gearing, cooling, materials and design
Page 1.17
These current-density values assume that the windings are varnished for good heat transfer. In aircooled machines, the fan is mounted on the rear of the motor outside the frame with a shroud which
focuses the air over the outside of the motor. Liquid cooled motors may have a passageway around the
outside of the stator with a cooling fluid circulating to remove the heat. The highest values are obtained
with 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 rather
than the peak or RMS value, but the idea behind this is to indicate how well the entire cylinder of steel
is being utilized.6 Its value is limited by the available MMF of the excitation source, and by core losses
which 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 the
torque capability or 'specific output'. As the diameter is increased, both the current and the flux
increase if the electric and magnetic loadings are kept the same. Hence the diameter (or radius) appears
squared in any expression for specific output. On the other hand, if the length is increased, only the flux
increases, not the current. Therefore the length appears linearly in the specific output. Thus the
specific output is proportional to D2Lstk, or rotor volume. In practice as the diameter is increased, the
electric loading can be increased also, because more intense fan-cooling or liquid cooling can be used
without reducing the efficiency. Consequently the specific output (TRV) increases faster than the rotor
volume.
Although it is theoretically possible to write one general equation from which the torque of any electric
motor can be calculated, in practice a different torque equation is used for every different type of
motor. Only in certain cases is it possible to discern in this equation an explicit product of flux and
current, or even of quantities directly related to them. For example, in the DC commutator motor the
electromagnetic torque is given by
T ' k N Ia
(1.10)
where N is the flux and k is a constant. Here the flux-current product is obvious. In rotating-field AC
machines the classical torque equations do not contain this product explicitly. However, the recent
development of 'field-oriented' or 'vector' controls has necessitated the transformation of the classical
equations 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 maximize
the torque, but this is not automatically the case except in field-oriented drives. By contrast, in DC
machines the commutator automatically maintains the optimum relative angle of orientation between
the flux and the ampere-conductor distribution. In the case of doubly-salient motors such as the
switched reluctance motor and stepper motors, the torque cannot be expressed as the explicit product
of a flux and a current. However, the TRV can still be used for initial sizing provided that A and B can
be meaningfully defined (Miller, [1993]).
So far we have restricted attention to the torque per unit rotor volume, a natural consequence of the
fact that the torque appears at the rotor surface. For a very rough estimation of overall size including
the stator, we can use a typical value of 'split ratio' S (i.e., rotor/stator diameter ratio): thus
Stator volume '
Rotor volume
S2
(1.11)
A typical value of split ratio for an AC motor is in the range 0.55!0.65. For switched reluctance motors
rather 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 is
familiar with a particular design of motor will have built and tested several, and the test data provides
values of TRV correlated with temperature rise, electric and magnetic loadings etc. The values quoted
in 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.
6
Switched reluctance machines have very high local flux-densities but a comparatively low magnetic loading, because
the high flux-density is limited to a small fraction of the stator periphery.
Page 1.18
SPEED’s Electric Motors
The TRV determines the volume of the rotor but not its shape. To estimate the rotor diameter and length
separately, a length/diameter ratio should be specified. A value around 1 is common; however, it is also
common to design motors of different ratings using the same laminations but with different stack
lengths. The length/diameter ratio may then vary over a range of 3:1 or more. Very large
length/diameter ratios are undesirable because of inadequate lateral stiffness, but may be used where
a 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 the
electromechanical energy conversion process at the airgap. The actual torque available at the shaft
coupling is less than this in motors, or greater in generators, by the amount of the mechanical losses
which include friction, windage, and certain electromagnetic losses appearing on the rotor. Allowance
should 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) in
electric motors is miserably low in comparison with what engineers would really like to achieve.7 It
always has been low, and it always will be low until someone discovers or invents a material that can
carry ten times as much flux as steel for the same magnetizing force; or a material that has a fraction
of the resistivity of copper. Such inventions would not by themselves be enough to increase the fleapower of the electric motor by an order of magnitude, unless they were manufacturable in reasonable
quantity 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 way
to step up the torque. If the gear ratio is n, and Tm is the motor torque, the torque applied to the load
is nTm. The motor speed Tm is increased over the load speed TL by the same ratio. Thus
TL ' n Tm
and
Tm ' n TL .
(1.12)
In most cases the increased motor speed falls in a standard speed range for 'high-speed' motors, which
may 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 the
load. The choice of gear ratio depends on how the drive operates. If the speed is constant it is usually
a simple matter of matching the load torque TL to the rated continuous motor torque Tmc :
n '
TL
(1.13)
T mc
If, however, the load has a 'dynamic' requirement which specifies a profile of speed or position as a
function 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 rated
torque Tmp, the acceleration of the load is given by
" '
T mp
n Jm %
7
JL
(1.14)
n2
This assertion is valid only in the normal range of sizes. In large intensively-cooled machines such as power-station
generators, the electric machine clearly outperforms the steam turbine — it is typical for the generator to be dwarfed by the
turbines.
1. Sizing, gearing, cooling, materials and design
Page 1.19
Fig. 1.11 Gear ratio
where the term in brackets is the inertia of the motor combined with the load inertia, referred to the
motor shaft. If n is large the gearing makes the load inertia insignificant, but it reduces the load speed
and acceleration relative to those of the motor. If n is small the referred load inertia is large, and this
limits the acceleration. Between the extremes of large and small n, there is a value that gives maximum
acceleration for fixed values of Tmp and the separate inertias. This 'optimum' value can be determined
by equating the differential coefficient d"/dn to zero, giving
JL
n '
(1.15)
JM
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
"max '
1
2
T mp
Jm
1
n
(1.16)
The corresponding acceleration of the motor is n times this value. In this analysis, the inertias of the
pinions (gearwheels) have been ignored. For a very precise evaluation, in the case of a single-stage
gearbox, 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 the
load has a fixed torque TL in addition to its inertia.
" '
T mp & T L /n
n Jm % JL /n 2
.
Again there is one value of gear ratio n that produces maximum acceleration, and by the same
differentiation process it is found to be
n '
TL
T mp
1 %
1 %
JL T mp2
.
.
JM T 2
L
If the inertias are unchanged from the previous case, the gear ratio is increased. The expression for the
optimum ratio can be substituted back in the formula for acceleration to find the maximum load
acceleration. The result is the same as eqn. (1.16); the difference is that with a larger ratio n the load
acceleration will be smaller. It is interesting to note that the maximum acceleration of the motor is
unchanged, and is equal to one-half the torque/inertia ratio of the motor.
Page 1.20
SPEED’s Electric Motors
Peak/continuous torque ratio of motor. In the constant-speed case, the choice of n maximizes the
utilization of the continuous torque rating of the motor, Tmc. In the acceleration case, the choice of n
maximizes the utilization of the motor's peak acceleration capability as expressed by its peak
torque/inertia ratio Tmp /Jm. Consider a load that requires both short periods of acceleration and long
periods at constant speed. Then there is a question, can the two values of n be the same? If so, the
utilization 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 an
example of the kind of analysis that is needed to get a highly optimized system design. Assume that the
load torque is constant at all times, but that short bursts of acceleration (or deceleration) are required
from time to time. The peak rated torque of the motor will be used for acceleration, and the continuous
rated torque for constant speed. If we equate the two separate values of n from the appropriate
formulas given above, and if we write
T mp ' kT mc
(1.19)
where k is the ratio of peak motor torque to continuous motor torque, then the following relationship
can be derived:
n 2J m
JL
'
k2
(k & 1)2 & 1
.
(1.20)
The left-hand expression is the ratio of the referred motor inertia to the load inertia, and we can refer
to it as the 'referred inertia ratio' or just 'inertia ratio'. For a range of values of the inertia ratio, the
equation can be solved to find the values of k that simultaneously optimize n for both the
constant-speed and acceleration periods. The most interesting result of this is that a large range of
inertia ratio is encompassed by only a small range of values of k : as the inertia ratio changes from
infinity down to 2, k changes only from 2 to 4. But values of k in this range are extremely common: so
common, in fact, as to appear to be a natural characteristic of electric motors. This implies that for
most inertia ratios where the referred motor inertia is more than twice the load inertia, the gear ratio
can be chosen to make good utilization of both the continuous torque and the peak acceleration of the
motor, provided k $ 2. If k < 2, the gear ratio must be chosen for constant speed or for acceleration, and
cannot be optimal for both. The property of electric motors to provide short bursts of peak torque for
acceleration 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 restrictive
assumptions that may be too simple in a complex motion-control system. For detailed work it is
desirable 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.
2.
heat removal; and
temperature distribution within the motor.
The main reasons for limiting the temperature rise of the windings and frame of a motor are:
1.
2.
3.
to preserve the life of the insulation and bearings;
to prevent excessive heating of the surroundings; and
to prevent injury caused by touching hot surfaces.
Insulation life. The "life" of electrical insulation is inversely related to the temperature. A sustained
10EC increase in temperature reduces the insulation life by approximately 50%. The extent to which
excessive 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 motor
manufactured by Invensys Brook Hansen, used to extract fumes via the ventilation systems of public
1. Sizing, gearing, cooling, materials and design
Page 1.21
buildings in the event of fire; these motors can operate in an ambient temperature of 300EC for a limited
period of 30 minutes. Similar considerations apply to bearings. Grease-lubricated bearings may be filled
with high-temperature grease for hot-running applications, but in aerospace machines the bearings
are usually lubricated by separately-cooled oil or oil mist.
Heating of the surroundings is obviously undesirable especially if the motor is heating the equipment
it is driving. For this reason it is important to minimize rotor losses conducted along the shaft. PM
motors have cooler rotors than DC or induction motors. In some applications such as hermetic
compressors used in air-conditioning, refrigeration, etc., the motor losses are removed by the working
fluid, reducing the thermodynamic efficiency of the system.
To prevent injury or harm from touching, exposed surfaces must be kept below 50EC. In certain
applications (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 temperature
is generally less than 50EC, and NEMA ratings for electrical insulation assume an ambient temperature
of 40EC. In aerospace applications motors and generators may be directly cooled by oil or fuel and
coolant temperatures can be as high as 100EC.
The increase in winding temperature increases the resistivity of the windings: a 50EC rise by 20%, and
a 135EC rise by 53%, increasing the I 2R losses by the same amount if the current remains the same. The
resistance 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 any
temperature T EC the resistivity of copper can be calculated as
D ' D20 [1 % " ( T ! 20 ) ]
ohm&m
(1.21)
where " = 0.00393 /EC , is the temperature coefficient of resistivity and D20 is the resistivity at 20EC, that
!8
is, 1.728 ×10 ohm-m.
Heat Removal
In most industrial and commercial motors, heat is removed by a combination of
1.
2.
3.
conduction to the frame mountings;
air convection, which may be natural or forced; and
radiation.
In highly-rated machines direct cooling methods are used:
1.
2.
3.
oil mist, especially in aerospace machines;
immersion in refrigerant, in "hermetic" motors used in refrigerator compressors;
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
Q ' k A
dT
dx
. k A
)T
t
W
(1.22)
where )T is the temperature difference through the thickness t. The coefficient k is the thermal
conductivity, with units (W/m2) per (EC/m), i.e. W/EC-m. The thermal conductivity is a material
property, 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 insulating
materials and most fluids have low thermal conductivities.
Page 1.22
SPEED’s Electric Motors
As an example, consider the flow of heat along a conductor whose cross-section area is A = 64 mm2 and
length 50 mm, when the RMS current-density is 7 A/mm2. The electrical resistivity of copper is 1.7 ×
3
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/cm .
In one conductor the I 2R loss is therefore 833,000 × 64 × 10!6 × 50 × 10!3 = 2.7 W. To take the most
pessimistic estimate, assume that all of this heat is generated at the mid-point of the coil-side, half-way
along the motor. The thermal conductivity of copper is 387 (W/m2) per (EC/m). So the temperature
gradient along the coil-side is given by eqn. (1.22) as
dT
'
dx
Q
'
kA
2.7
' 108 EC/m.
387 × 64 × 10!6
(1.23)
Since the heat can flow in both directions, the temperature-gradient is only half this value, and the
temperature 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 along
the length of the coil-side, but this quick calculation reveals that such sophistication is not needed in
the example considered.
Thermal resistance and contact resistance. Eqn. (1.24) can be used to define thermal resistance as
the ratio of temperature difference )T to heat flow rate Q : the symbol used for thermal resistance is
R, with units EC/W. Thus
R '
)T
Q
'
t
kA
EC/W.
(1.24)
The thermal resistance is a "lumped parameter" that can be used to model the conduction through a
region or interface where the individual values of k, A, and t may be difficult to determine. The contact
resistance between two surfaces is usually treated in this way, as, for example, between the frame and
the 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, then
with 40W flowing though it the temperature difference across the interface would be 40EC.
Radiation. Radiation is described by the Stefan-Boltzmann equation
Q
' e F ( T14 ! T24)
A
W/m 2
(1.25)
where F is the Stefan-Boltzmann constant, 5.67 × 10!8 W/m2/K4 for a black body, T1 is the absolute
surface temperature of the radiating body in degrees Kelvin, and T2 is the absolute temperature of the
surroundings.8 A black body is a perfect radiator, that is, one which reflects no radiated heat but
absorbs all the heat radiated towards it. Real surfaces are imperfect radiators, and their radiative
effectiveness relative to that of a black body is called the emissivity e. A black lacquered surface can
achieve an emissivity as high as 0.98, but a more practical rule of thumb is to take 0.9 for black-painted
or lacquered surfaces. For example, a surface with an emissivity of 0.9 that is 50EC above the
surroundings at 50EC, has a net heat transfer rate of
0.9 × 5.67 × 10!8 × ( (50%50%273)4 & (50%273)4 )
(1.26)
2
which is 432 W/m2 or 0.28 W/in . 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:
Q
' h )T
A
W/m 2
(1.27)
where )T is the temperature difference between the cooling medium and the surface being cooled, and
h is the heat-transfer coefficient. The units of h are W/m2/EC. The value of h depends on the viscosity,
8
The absolute temperature in degrees Kelvin (K) is the temperature in EC plus 273.
1. Sizing, gearing, cooling, materials and design
Page 1.23
thermal conductivity, specific heat, and other properties of the coolant, and also on its velocity. In
natural convection the flow of coolant is not assisted by fans, blowers, pumps etc. In forced convection
the flow is assisted by one of these external means.
The heat transfer coefficient for natural convection around a horizontally-mounted unfinned
cylindrical motor can be roughly estimated as
h . 7.5
)T
D
1/4
W/m 2/EC
(1.28)
where D is in mm. For example, for an unfinned cylinder of diameter D = 100 mm and a temperature
rise of 50EC, the natural-convection heat-transfer coefficient is calculated as 6.3 W/m2/EC. For a )T of
50EC, the heat transfer rate is then given by eqn. (1.28) as 6.3 × 50 = 315 W/m2. As a first approximation
this value can be applied to the whole surface including the ends, but if the motor is flange-mounted
then only one end is available for convective cooling.
Forced convection (with a shaft-mounted fan or an external blower) increases the heat-transfer
coefficient by as much as 5!6 times, depending on the air velocity. The increase in heat-transfer
coefficient is approximately proportional to the square-root of the air velocity. An approximate formula
for the forced-convection heat-transfer coefficient is
h . 125
V
W/m 2/EC
L
(1.29)
where V is the actual air velocity [m/s] and L is the frame length [mm] (assumed parallel to the
direction of airflow). For a motor of length 100 mm, if the air velocity is 4 m/s, this formula predicts
h = 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 power
loss of 22 W (0.022 W per mm length) is sufficient to boil the water at the wire surface. The wire surface
temperature is 114EC and the heat transfer coefficient (see below) is 5000 W/m2/EC. The heat flow at the
wire surface is 0.07 W/mm2 and the current-density in the wire is approximately 35 A/mm2. In normal
motors, the rate at which heat can be abstracted is far less than this, and current-densities over 30
2
A/mm2 are achievable only for short bursts. 35 A/mm 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 about
800 W/m2. With forced air convection the rate increases to about 3000 W/m2, and with direct liquid
cooling about 6000 W/m2. A motor that generates more heat than can be removed at these rates must
absorb 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, 0.3
W/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 by
a simple general equation, because the heat transfer rate depends on the shape of the conductors. As
an 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 area
of 4 cm2 while the long wire has a surface area of 40 cm2. The loss density in the conductor is J 2D where
J is the current density and D is the resistivity. With ten times the surface area the long wire can
dissipate ten times the heat, assuming the same heat transfer coefficient in both cases. This suggests
that 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 flow
to keep the motor cool. DC motors often have separate AC-driven fans, because they have to work for
prolonged periods at low speed with high torque. Since most of the heat in a DC motor is generated on
the rotor, good internal airflow is essential. In DC motors the external fan is usually mounted to one
side of the motor, where it is easily accessible, and does not increase the overall length. With vectorcontrolled induction motors a common practice is to mount the fan in line with the motor at the nondrive end, and arrange it to blow air over the outside of the finned frame. The fan may increase the
overall length by up to 60%. Brushless motors have less severe problems because most of the heat at
low speed is generated in the stator windings, and very little on the rotor.
Page 1.24
SPEED’s Electric Motors
Internal temperature distribution
The steady-state temperature distribution within the motor is essentially a diffusion problem. The most
important aspect of the problem is finding the hottest temperature in the motor, given a certain
distribution of losses and a known rate of heat removal. It is difficult to solve precisely, because of
three-dimensional effects and because some thermal resistances (such as the resistance between slot
conductors and slot liner) are hard to calculate.
The differential equation for three-dimensional conduction of heat is the so-called diffusion equation:
L 2T %
where
L 2T '
1 Mq
'
k Mt
M2T
Mx 2
%
1 MT
" Mt
M2T
My 2
%
M2T
Mz 2
(1.30)
(1.31)
and
" '
k
Dc
m 2/s
(1.32)
is the diffusivity in SI units. In SI units, k is the thermal conductivity in W/mEC; c is the specific heat
in kJ/kgEC, and D is the density in kg/m3. In a structure as complex as an electric motor the heat
conduction equation is a complex boundary-value problem that is best solved by computer-based
numerical methods such as the finite-element method.
In electric motors internal convection and radiation may be as important as conduction, and when the
differential equation is extended to include them, matters become very complicated, even for steadystate calculations. During transients the temperature distribution can be very different from the
steady-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 of
the 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 source
is 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 are
represented 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 represented
together as one total source, i.e. the individual sources are taken as being in parallel. The thermal
equivalent circuit is really a lumped-parameter model of all the heat-flow processes within the motor
as well as the heat removal processes discussed earlier.
The thermal equivalent circuit should ideally take into account the anisotropy effects: for example, the
effective thermal conductivity through a lamination stack is lower in the axial than the radial
direction. A more complex thermal equivalent circuit may include provision for direct cooling of the
winding conductors, or for direct cooling of the rotor shaft. If it also includes the thermal masses or
capacities of the winding, the rotor and stator laminations, the frame, the shaft, and other massive
components, then it can be solved for transient as well as steady-state heat transfer. The heat removal
routes by conduction, radiation, and convection are represented by thermal resistances. For convection
the appropriate resistance Rv is given by
Rv '
1
hA
EC/W
(1.33)
where A is the appropriate surface area for convective heat-transfer and the subscript "v" stands for
convection. If h is a function of the temperature-difference, the equivalent circuit becomes non-linear
and requires an iterative solution. For radiation the equivalent thermal resistance Rv is the ratio of
the temperature difference T1 ! T2 to the radiation heat exchange rate Q in eqn. (1.25). Clearly this is
non-linear. However, the non-linearity is often neglected and a fixed value of Rv is calculated assuming
that the case temperature is known.
1. Sizing, gearing, cooling, materials and design
Page 1.25
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 thermal
resistances, e.g., CT = thermal resistance from the conductors to the stator
teeth.
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 1.5
TEMPERATURE 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
TABLE 1.6
0.5
Page 1.26
SPEED’s Electric Motors
SELECTED EMISSIVITIES
Material
(20EC)
k
Sp. Heat
Density
ohm-m × 10
(W/m K)
kJ/kg/EC
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
D
!8
Silicon steel
30!50
20!30
0.49
7700
Cast iron
66
45
0.5
7900
Cobalt-iron
40
30
0.42
8000
4.5
0.8
4900
4
Ceramic magnet
10
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
0.0037
1.88
888
Engine oil
TABLE 1.7 SELECTED MATERIAL PROPERTIES
*Dielectric strength
1. Sizing, gearing, cooling, materials and design
1.11
Page 1.27
INTERMITTENT OPERATION
Fig. 1.13 Intermittent operation
Intermittent operation is normal for brushless PM motors, because most of the applications that use
them 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
tcy ' tON % tOFF .
(1.34)
The duty-cycle d is defined as
d '
tON
tcy
'
tON
tON % tOFF
.
(1.35)
The most efficient use of the thermal capability of the motor will be made if the maximum winding
temperature Tmax just reaches the rated value Tr at the end of each on-time. Because the power
dissipation is interrupted with cool-down intervals tOFF, the power Pd that can be dissipated during the
on-times may exceed the steady-state continuous dissipation rating of the motor Pr, and therefore the
motor may be permitted to exceed its steady-state output power rating during the on-times. The
simplified thermal equivalent circuit model in Fig. 1.14 makes it possible to calculate the permissible
overload factor as a function of the on-time tON and duty-cycle d for a given motor.
Fig. 1.14 Simple thermal equivalent circuit for transient calculations
The thermal equivalent circuit is a parallel combination of thermal resistance R and thermal
capacitance C. R represents the steady-state thermal resistance between the winding and the
surroundings in EC/W. C represents the thermal capacity of the entire motor in J/EC. The thermal timeconstant J is given [in seconds] by eqn. (1.36):
J ' RC
(1.36)
The analysis proceeds by equating the temperature rise during the on-time with the temperature fall
Page 1.28
SPEED’s Electric Motors
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
T ! T0 ' R Pd ( 1 ! e ! t /J ) % ( Tc ! T0 ) e ! t /J .
(1.37)
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,
Tmax ! T0 ' R Pd ( 1 ! e
!tON /J
) % ( Tc ! T0 ) e
!tON /J
(1.38)
.
By definition, the steady-state rated temperature-rise (Tr ! T0) is given by
Tr ! T0 ' R Pr ,
(1.39)
where Pr is the rated steady-state power dissipation in the motor, i.e., the continuous power dissipation
that produces rated temperature rise. We can use this to "calibrate" Pd in eqns. (1.37) and (1.38), by
defining the dissipation overload factor k2, where
k2 '
Pd
Pr
.
(1.40)
The reason for using k2 instead of k is that in most types of brushless servomotor the losses are
dominated by I2R losses while the load torque is proportional to the current I. If the load is increased
by a factor k, it means that the current and torque are increased by the factor k while the losses
increase 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
Tmax ' Tr ,
(1.41)
we obtain the following equation relating the temperature rise to the overload factor and the on-time:
( Tr ! T0 ) [1 ! k 2 ( 1 ! e
! tON /J
) ] ' ( Tc ! T0 ) e
! tON /J
(1.42)
Temperature fall during OFF-time. When the motor is switched off, the power dissipation falls to
zero and the winding temperature falls according to the equation
T ! T0 ' ( Tr ! T0 ) e ! t /J
(1.43)
where t is measured from the end of the on-time, i.e. the beginning of the off-time. At tOFF,
Tc ! T0 ' ( Tr ! T0 ) e
!tOFF /J
(1.44)
.
Steady-state : equating the temperature rise and fall. First, multiply eqn. (1.44) by e
( T c ! T 0) e
!tON /J
' ( T r ! T 0) e
!( tON % tOFF ) /J
.
!tON /J
:
(1.45)
The left-hand side of eqn. (1.45) is identical to the right-hand side of eqn. (1.42), so the right-hand side
of eqn. (1.45) can be equated to the left-hand side of eqn. (1.42). With suitable rearrangement, the result
can 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 of
the on-time and the duty-cycle: writing tON/d instead of tON + tOFF, i.e., instead of tcy, the expression is
k2 '
1 ! e
!tON /J d
1 ! e
!tON /J
(1.46)
1. Sizing, gearing, cooling, materials and design
Page 1.29
For example, if the duty-cycle is 25% (d = 0.25) and tON = 0.2 × J, the dissipation overload factor is
k2 '
1 ! e !0.2/0.25
' 3.04 ,
1 ! e !0.2
(1.47)
which means that the dissipation can be increased to 304% of its rated steady-state value for a period
of 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 be
raised to 304% for 8 minutes followed by a cool-down period of 24 minutes. Increasing the dissipation
to 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
k2 '
1
.
d
(1.48)
This means that when the on/off cycles are very short compared with the thermal time-constant of the
motor, the mean dissipation will be equal to Pr when the peak dissipation Pd = k2Pr is equal to Pr/d. This
simple result is intuitive.
Maximum overload for a single pulse. Eqn. (1.46) can also be used to calculate the maximum
dissipation overload factor for a single pulse, for which d = 0. In this case
1
k2 '
1 ! e
!tON /J
.
(1.49)
For example, if tON = 8 min and J = 40 min, then the maximum dissipation overload factor k2 is 5.5 or
550%, 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 arises
from equating the temperature rise and temperature fall is an expression for the necessary cool-down
time tOFF as a function of the dissipation factor k2 and the on-time tON. The expression is
tOFF ' ! J ln [k 2 ! ( k 2 ! 1 ) e
tON /J
].
(1.50)
Together with eqn. (1.35), this can be used to determine the maximum duty-cycle d that can be used
with a given dissipation overload factor k2 and a given on-time tON, for a motor of thermal time-constant
J. For example, if the dissipation is 200% of rated, and if tON = 8 min, J = 40 min,
tOFF ' ! 40 × ln [2 ! (2 ! 1) × e 8/40 ] ' 10.0 min .
(1.51)
The minimum cycle time is therefore 18 min and the maximum duty-cycle (with 8 minutes' on-time) is
8/18 = 0.44 or 44%.
Maximum on-time for a given overload factor and cool-down time. A third result obtained by
equating the temperature rise and fall is an expression for the maximum on-time tON as a function of
the dissipation overload factor k2 and the off-time tOFF. The expression is
tON ' J ln
k2 ! e
!tOFF /J
k2 ! 1
.
(1.52)
Maximum duration of single pulse. This expression can be used to calculate the maximum duration
of a single pulse having a given dissipation overload factor k2. For a single pulse, tOFF is infinite and
tON ' J ln
k2
k2 ! 1
.
(1.53)
Page 1.30
SPEED’s Electric Motors
For example, if k2 = 5.5 and J = 40 min, then tON =
8 min.
Graphical transient heating curves. Fig. 1.15
shows the relationship expressed by eqn. (1.53)
graphically in terms of the duty-cycle d, the ontime tON as a fraction of the time-constant J, and
the overload factor k.
This graph can be used in a number of ways. For
example, to determine the maximum permissible
duration of a single pulse with a given overload
factor k, the duty-cycle d should be set to zero.
Thus with k = 1.5 the maximum pulse duration is
0.58J. With a time-constant of 40 min this is 23.2
min.
The graph shows the maximum duty-cycle that
can be used with a given overload factor. For
example, at 200% load the maximum duty-cycle is
0.25 or 25%, but in this limiting case the on-time
must be vanishingly small. With an on-time of
0.1J at 200% load, the maximum duty-cycle is
approximately 0.2, which means that the cooldown period in each cycle must be (1!d)J = 0.9J.
If J is 40 min, this means a maximum operating
time at 200% load of 4 min, followed by a cooldown period of 36 min before the cycle can be
repeated. Operations that need a short on-time
with a high duty-cycle must use a lower overload
factor.
1.12
Fig. 1.15 Intermittent heating curves
PERMANENT MAGNET MATERIALS AND CIRCUITS
The permanent-magnet industry has continually improved the properties of PM materials in the past
20!30 years, mainly by painstaking development of the metallurgy of existing materials. Samples of the
main families of PM materials used in electric machines are shown in Table 1.8.
Property
Units
Alnico 5-7
Ceramic
Sm2Co17
NdFeB
1.35
0.41
1.06
1.2
60
325
850
1000
Remanence
Br
T
Coercivity
Hc
kA/m
kJ/m
Energy product (BH)max
Relative recoil permeability
3
µrec
Specific gravity
Resistivity
µS-cm
!6
60
30
210
250
1.9
1.1
1.03
1.1
7.3
4.8
8.2
7.4
4
47
>10
86
150
Thermal expansion coefficient
10 /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 1.8
TYPICAL MAGNET PROPERTIES
1. Sizing, gearing, cooling, materials and design
Page 1.31
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 room
temperature NdFeB has the highest energy product of all commercially available magnets. The high
remanence and coercivity permit marked reductions in motor size, compared with motors using
Ferrite (ceramic) magnets. However, ceramic magnets are considerably cheaper than Rare Earth or
NdFeB.
Both ceramic and NdFeB magnets are sensitive to temperature and special care must be taken if the
working temperature is above 100EC. For very high temperature applications Alnico or Rare
Earth/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 process
similar to that used for amorphous alloys (Magnequench). NdFeB magnets are often made in rings
which 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 magnet
characteristics 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 subjected
to a magnetic field, as for example in a magnetizing fixture, Bm and Hm in the magnet will follow the
initial magnetization curve as the external ampere-turns are increased. If the external ampere-turns
are 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 the
magnet is surrounded by a highly permeable magnetic circuit, that is, if it is 'keepered', then its poles
are 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 after
being magnetized to saturation.
External ampere-turns applied in the opposite direction (i.e., Hm < 0) cause the magnet's operating point
to follow the curve through the second and third quadrants until the magnet is saturated in the
opposite direction. Again, if the current is switched off the operating point returns towards the point
Page 1.32
SPEED’s Electric Motors
(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-density
it can retain when 'keepered' is !Br.
To bring the flux-density to zero from the original positive remanence point (0,Br), the external
ampere-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 +Hc
must be applied. The entire loop is usually symmetrical and can be measured using instruments
designed specially for magnet testing.
If negative external ampere-turns are applied, starting from the positive remanence point (0,Br), and
switched off at R, the operating point of the magnet 'recoils' and will operate along the lower curve of
a 'minor loop'. For practical purposes the minor loop of high-coercivity magnets is very narrow and
can be taken as a straight line, the recoil line, whose slope is equal to the recoil permeability, µrec. This
is usually quoted as a relative permeability, so that the actual slope is µrecµ0 H/m. Operation along the
recoil 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 second
quadrant, which is where the magnet normally operates in service. In very hard magnets that are fully
magnetized, the recoil line is coincident with the second-quadrant section of the hysteresis loop. This
is characteristic of ceramic, Rare Earth/Cobalt, and NdFeB magnets, which usually have µrec between
1.0 and 1.1. 'Soft' PM materials have a 'knee' in the second quadrant, such as Alnico. While Alnico
magnets have very high remanence and excellent mechanical and thermal properties, they have low
coercivity 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' : in
other words, the hysteresis loop of a typical nonoriented electrical steel is very narrow and has a low
coercivity and a high permeability; see Fig. 1.16. The high permeability is desirable in order to
minimize the magnetizing MMF (which is supplied by the magnets in PM motors, or by the magnetizing
current in induction motors). The narrow loop is desirable because the loop area represents an energy
loss 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 second
quadrant, Fig. 1.17. This is called the demagnetization
curve. In the absence of externally applied ampere-turns,
the magnet
operates at the intersection of the
demagnetization curve and the 'load line', whose slope is
the product of µ0 and the 'permeance coefficient' (PC) of the
external circuit: i.e., at (Hm,Bm), with Hm < 0.
Since Bm and Hm in the magnet both vary according to the
external circuit permeance, it is natural to ask what it is
about the magnet that is 'permanent'. The relationship
between Bm and Hm can be written
Bm ' µ0 Hm % J.
(1.54)
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 n
characteristic showing intrinsic curve
The first term is the flux-density that would exist if the
magnet were removed and the magnetizing force remained
at the value Hm. Therefore the second term can be regarded as the contribution of the magnet to the
flux-density within its own volume; accordingly, J is called the magnetization and it is measured in
tesla.9 If the demagnetization curve is straight, and if its relative slope µrec = 1, then J is constant. This
is shown in Fig. 1.17 for negative values of Hm up to the coercivity !Hc. In most hard magnets µrec is
slightly 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
9
Sometimes eqn. (1.54) is written Bm = µ0(Hm + M) and then the magnetization is measured in kA/m instead of T.
1. Sizing, gearing, cooling, materials and design
Page 1.33
quadrant, depending on the material and its grade). Evidently the magnet can recover or recoil back
to its original flux-density as long as the magnetization is constant. The coercive force required to
demagnetize 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
Bm ' µ0 µrec Hm % Br
(1.55)
which can be related to eqn. (1.54) by expanding it as follows:
which indicates that
Bm ' µ0 Hm % µ0 ( 1 ! µrec ) Hm % Br
(1.56)
J ' µ0 ( 1 ! µrec ) Hm % Br ' µ0 PHm % Br
(1.57)
where P is the susceptibility, 1 ! µrec.
Another parameter often calculated is the magnet energy product, BmHm. This is not the actual stored
magnet energy but simply indicates how hard the magnet is working against the demagnetizing
influence 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 the
demagnetization characteristic is tangent to the hyperbola of its (BH)max value. If the relative recoil
permability is unity, this occurs for a permeance coefficient of unity, with Bm = Br/2, provided that
there are no externally applied ampere-turns from windings or other magnets.
Fig. 1.18 Closed and gapped magnetic circuits
Calculation of Magnet operating point. Fig. 1.18 shows a simple magnetic circuit in which the
magnet is 'keepered' by a material or core of relative permeability µr. The core and magnet together
form a closed magnetic circuit. Applying Ampere's law, and assuming uniform magnetizing force in
both the magnet and the core,
Hmlm % HFelFe ' 0 .
(1.58)
Page 1.34
SPEED’s Electric Motors
where Hm is the magnetic field in the magnet, HFe is the magnetic field in the iron core (assumed to be
uniform 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 are
no externally applied ampere-turns. Hence
Hm ' !
lFe
lm
HFe
(1.59)
which establishes that the magnet works in the second quadrant of the B-H loop. Now consider the
gapped magnetic circuit in Fig. 1.18, in which there is an airgap in series with the magnet and the two
sections of iron core. Now
Hmlm % HFelFe % Hglg ' 0 .
(1.60)
where Hg is the magnetic field in the airgap and lg is the airgap length. The permeability of the
electrical steel used in motors is usually several thousand times higher than µ0, so that the term HFe lFe
can be neglected as a first approximation, even though lFe may be much bigger than lg. Then
lg
Hm . !
Hg
(1.61)
lm
Now by Gauss' law, the flux-densities in the magnet and the airgap are related by
BmAm ' BgAg
(1.62)
so that if we take the ratio of Bm/µ0Hm, recognizing that in the airgap Bg = µ0Hg, we get
Bm
Hm
' ! µ0
Ag lm
' ! µ0 × PC
Am lg
(1.63)
where PC is the permeance coefficient. The ratio of magnet pole area to airgap area is sometimes called
the flux-concentration factor or flux-focussing factor:
CM '
Am
Ag
.
(1.64)
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 use
a high value of the flux-concentration factor Am/Ag. But this reduces the permeance coefficient and
eqn. (1.63) shows that this reduces the ratio Bm/Hm, which increases the risk of demagnetization
because it moves the operating point further down the recoil line away from Br towards the knee of the
B-H curve.
To achieve a high permeance coefficient with a high flux-concentration factor we must increase the
ratio lm/lg to compensate for the demagnetizing effect of the airgap: in other words, use a magnet that
is long in the direction of magnetization and also long relative to the airgap length. It does not mean
long in relation to the lateral dimensions of the magnet, and indeed most modern magnets except
Alnico have such high coercivity that the length in the direction of magnetization is the
smallest dimension and is intuitively referred to as the 'thickness'!
The energy product is given by
Bm Hm '
B g H g Ag lg
Am lm
'
2 Wg
Vm
(1.65)
where Wg is the magnetic energy stored in the airgap volume and Vm is the volume of the magnet. This
shows that the minimum magnet volume required to magnetize a given working volume of airspace
1. Sizing, gearing, cooling, materials and design
Page 1.35
is inversely proportional to the working energy product BmHm . Therefore, in these cases it pays to
design the magnet length and pole area in such proportions relative to the length and area of the
airspace, as to cause the magnet to work at (BH)max, which is a property of the particular material at
a given temperature. In motors this principle cannot be applied so simply, because the armature
current produces demagnetizing ampere-turns that may be very great under fault conditions. To
eliminate the risk of demagnetization, motors are designed so that on open-circuit or no-load, the
magnet operates at a high permeance coefficient with an adequate margin of coercive force to resist
the 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 working
conditions, in all cases with no externally applied ampere-turns. Note that B is continuous throughout
the magnetic circuit (because it obeys Gauss' law), but H is not. The discontinuities of H are associated
with the appearance of magnetic poles at the interfaces between different sections of the magnetic
circuit, notably at the 'poles' of the magnet and the working airspace. The polarization of surfaces gives
rise to a magnetic potential difference across the airspace which is useful for calculating flux
distribution in motors. In Fig. 1.18 this potential difference is
u ' H g lg
A&t
(1.66)
If the magnetic potential drop in the steel is neglected, the corresponding magnetic potential difference
across the magnet is
!u . Hm lm
A&t .
(1.67)
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 factors
are as follows:
1 inch
25.4 mm
1T
10 kG
1 kA/m
1 kJ/m
3
4B Oe
B/25 MGOe
TABLE 1.9
CONVERSION FACTORS
Temperature effects; reversible and irreversible losses
High-temperature effects. Exposure to high temperatures for long periods can produce metallurgical
changes which may impair the ability of the material to be magnetized and may even render it
nonmagnetic. There is also a temperature, called the Curie temperature, at which all magnetization
is reduced to zero. After a magnet has been raised above the Curie temperature it can be remagnetized
to its prior condition provided that no metallurgical changes have taken place. The temperature at
which significant metallurgical changes begin is lower than the Curie temperature in the case of the
Rare Earth/Cobalt magnets, NdFeB, and Alnico; but in ceramic ferrite magnets it is the other way
round. Therefore ceramic magnets can be safely demagnetized by heating them just above the Curie
point for a short time. This is useful if it is required to demagnetize them for handling or finishing
purposes. 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
310
TABLE 1.10
METALLURGICAL CHANGE AND CURIE TEMPERATURE
Page 1.36
SPEED’s Electric Motors
Material
Temp. coefft. of Br %/EC
Alnico 5-7
!0.02
Ceramic
!0.19
Sm2 Co17
!0.02
NdFeB
!0.11
TABLE 1.11
REVERSIBLE TEMPERATURE COEFFICIENTS OF BR
Reversible losses. The B-H loop changes shape with temperature. Over a limited range the changes
are reversible and approximately linear, so that temperature-coefficients for the remanence and
coercivity can be used. Table 1.11 gives some typical data. Ceramic magnets have a positive coefficient
of Hc, whereas the high-energy magnets lose coercivity as temperature increases. In ceramic magnets
the knee in the demagnetization curve moves down towards the third quadrant, and the permeance
coefficient at the knee decreases. Thus ceramic magnets become better able to resist demagnetization
as the temperature increases up to about 120EC. The greatest risk of demagnetization is at low
temperatures when the remanent flux-density is high and the coercivity is low; in a motor, this results
in the highest short-circuit current when the magnet is least able to resist the demagnetizing
ampere-turns. In high-energy magnets the knee moves the other way, often starting in the third
quadrant at room temperature and making its way well into the second quadrant at 150 EC. Grades with
a high resistance to temperature are more expensive, yet these are often the ones that should be used
in 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 above
an ambient of 20EC, for instance, a ceramic magnet will have lost about 10%. This is spontaneously
recovered as the temperature falls back to ambient.
Irreversible losses recoverable by remagnetization. (a) Domain relaxation. Immediately after
magnetization there is a very slow relaxation, starting with the least stable domains returning to a
state 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 temperatures
this process is usually negligible, particularly if the magnets have been stabilized (by temperature
cycling and/or AC flux reduction) immediately after magnetization. Elevated temperatures during
subsequent operation may, however, cause an increased relaxation rate. This can be prevented by
temperature-cycling in the final assembly over a temperature range slightly wider than the worst-case
operating range. Subsequent relaxation is reduced to negligible levels by this means. Table 1.12 shows
the stability of different magnet materials at 24 EC.
Material
% loss after 10 years (typ.)
Ceramic
< 0.01
Rare earth/Cobalt
0.2
Alnico
0.5
TABLE 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 remanence
corresponding to point a is at point A. When the temperature is raised from T1 to T2 the operating
point moves from a to b, and the corresponding remanence moves from A to B. Note that because the
knee of the curve has risen above point b, the effective remanence at B' is less than that at B, which
is what it would have been if the magnet had been working at a high permeance coefficient.
1. Sizing, gearing, cooling, materials and design
Page 1.37
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 the
recoil line through A'. The recovery from b to a' is reversible, but there has been an irreversible loss
of flux-density )Bm in the magnet, relative to point a. The remanence at T1 has fallen from A to A'. The
loss can be recovered only by remagnetization at the lower temperature.
Manufacturers' data for irreversible loss should be interpreted carefully to distinguish between the
long-term stability and the effects just described. Irreversible loss is usually quoted at a fixed
permeance coefficient. If the magnet is used at a lower permeance coefficient, the irreversible loss over
the same temperature range will be higher.
Mechanical properties, handling, and magnetization
Magnets are often brittle and prone to chipping, but proper handling procedures are straightforward
enough as long as the rules are followed. Modern high-energy magnets are usually shipped in the
magnetized condition, and care must be taken in handling to avoid injury that may be caused by
trapping fingers. A further hazard is that when two or more magnets are brought close together they
may flip and jump, with consequent risk to eyes. Table 1.13 summarizes some of the important safety
precautions.
The best way to 'tame' magnetized magnets is to keeper them. Fixtures for inserting magnets can be
designed so that the magnets slide along between steel guides which are magnetically short-circuited
together. There still remains the problem of entering the magnets between the guides, but usually there
is enough space to provide for this to be done gently. Obviously it is important to keep magnets clear
of 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 image
on computer terminals and monitors.
Magnets are usually held in place by bonding or compression clips. In motors with magnets on the
rotor, adhesive bonding is adequate for low peripheral speeds and moderate temperatures, but for high
speeds a kevlar banding or stainless steel retaining shell can be used. In motors it is not advisable to
make 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 the
magnet should not move and that it should be protected from excessive temperatures.
Page 1.38
SPEED’s Electric Motors
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 assembling
magnets into a motor. When a large pole magnet is being assembled from smaller magnets, the magnets
have a tendency to flip and jump unexpectedly and may fly a considerable distance.
Work behind a plexiglass screen when experimenting or assembling magnet assemblies. Watch
out 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.
nitrogen dry chemical extinguishers – never use water or halogens.
In case of fire, use LP argon or
Use suitable warning labels, especially on large machines. PM motors generate voltage when the shaft
is rotated, even when disconnected from all power supplies. This may be obvious to an engineer, but is
a potential safety hazard for electricians and maintenance personnel.
Never leave magnetized members open or unprotected. When assembling a rotor to a stator, with
either one magnetized, the rotor must be firmly guided and the stator firmly located.
TABLE 1.13
MAGNET SAFETY
A wide range of shapes is available, but in motors the most common are arcs and rectangles. Close
tolerances of +/!0.1mm can be held in the magnetized direction even for standard magnets. But if the
design permits a relaxation of the required tolerance, particularly in the dimensions perpendicular to
the 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 the
magnetic axis. The coefficients in Table 1.8 are along the direction of magnetization. Most magnets
have 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 and
power supplies are essential, and this is one reason why high-energy magnets are usually magnetized
before shipping. The magnetizing force Hm must be raised at least to the saturation level shown in
Table 1.8, and this normally requires ampere-turns beyond the steady-state thermal capability of
copper coils. Therefore pulse techniques are used, or in some cases superconducting coils. Ceramic
and Alnico magnets can sometimes be magnetized in situ in the final assembly, but this is impractical
with 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, which
is buried in the cost of the motor. It ranges from a few pennies for small ferrite motors, to several
pounds for rare-earth motors. Even so, the cost of magnets is typically only a small fraction of the total
cost of the motor. Broadly speaking, the primary determinants of magnet cost are the torque per unit
volume 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 and
magnetic loadings must be as high as possible. The electric loading is limited not only by thermal
factors, but also by the demagnetizing effect on the magnet. A high electric loading necessitates a long
magnet length in the direction of magnetization, to prevent demagnetization. It also requires a high
coercivity, 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 directly
proportional to the remanence, and is nearly proportional to the pole face area of the magnet. A high
power 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 of
the rotor itself, and the highest power densities cannot be obtained with these magnets. With rare-earth
or other high-energy magnets, the cost of the magnet may be the limiting factor.
1. Sizing, gearing, cooling, materials and design
Page 1.39
The airgap flux-density of AC motors is limited by saturation of the stator teeth. Excessive saturation
absorbs too much excitation MMF (requiring a disproportionate increase in magnet volume); or causes
excessive heating due to core losses. For this reason there is an upper limit to the useable energy of a
permanent magnet. With a straight demagnetization characteristic throughout the second quadrant
and a recoil permeability of unity, the maximum energy-product (BH)max is given by
(BH)max '
B r2
4 µ0
J/m 3 .
(1.68)
Assuming that the stator teeth saturate at 1.6T and that the tooth width is half the tooth pitch, the
maximum airgap flux-density cannot be much above 0.8T and is usually lower than this. Therefore
there will be little to gain from a magnet with a remanent flux-density above about 1 or 1.2 T, implying
3
that the highest useable energy product is about 300 kJ/m . At 100 EC, such characteristics are just
within the range of available high-energy magnets. Evidently it is just as important to develop magnet
materials with 'moderate' properties and low cost, rather than to develop 'super magnets'
Operating temperature range. Because of the degradation in the remanence and coercivity with
temperature, the choice of material and the magnet volume must usually be determined with reference
to 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 own
thermal mass and that of the rest of the motor. The short-time thermal overload capability of the
electronic controller would normally be less than that of the motor, providing a further margin of
protection against magnet overtemperature.
Severity of operational duty. Magnets can be demagnetized by fault currents such as short-circuit
currents produced by inverter faults. In brushless motors with electronic control the problem is
generally limited by the protective measures taken in the inverter and the control. With an
over-running load, or where two motors are coupled to a single load, shorted turns or windings can be
troublesome because of drag torque and potential overheating of the stator. But by the same token, the
dynamic braking is usually excellent with a short-circuit applied to the motor terminals, and motors
may well be designed to take advantage of this. As is often the case, characteristics that are desirable
for one application are undesirable for another. The design must accommodate all the factors that
stress the magnet: electromagnetic, thermal, and mechanical.
1.13
PROPERTIES OF ELECTRICAL STEELS
Fig. 1.20 shows the DC B-H curve in the first quadrant
for two steels. The lower curve is a typical electrical
motor steel having 1.5% Silicon to increase the
resistivity to limit eddy-current losses. The saturation
flux-density of such steels (i.e. the flux-density at which
the incremental permeability becomes equal to µ0) is
typically about 2.1T. The upper curve is for a cobaltiron alloy with a saturation flux-density of about 2.3T.
This material is much more expensive than normal
electrical steel, and is only used in special applications
such as highly rated aircraft generators, where light
weight and high power density are at a premium.
The maximum permeability of electrical steels is of the
order of 5,000 µ0, and usually occurs between 1 and 1.5 T.
In Fig. 1.20, the total permeability of the electrical steel
at 2.0T is about 2.0/3,000 which is approximately 530 µ0.
Fig. 1.20 DC B-H curve for electrical steels
Page 1.40
SPEED’s Electric Motors
Losses. Under AC conditions, a power loss
arises in electrical steel as shown in Fig. 1.21,
which indicates increasing loss as the frequency
and flux-density increase. The loss is attributed
to
(a)
(2)
(3)
hysteresis;
eddy-currents; and
“anomalous loss”.
The hysteresis component is associated with the
changing magnitude and direction of the
magnetization of the domains, while the eddycurrent loss is generated by induced currents.
Eddy-currents can be inhibited by laminating the
steel, so that the eddy-currents become resistance
limited and the loss is then inversely
proportional to the resistivity. If the eddyFig. 1.21 Typical form of variation of losses in electrical steel,
currents are resistance-limited the loss is also
versus frequency and flux-density
2
proportional to 1/t , where t is the lamination
thickness. At higher frequencies the resistance limited condition is lost, and the losses increase rapidly
with frequency. For this reason, very thin laminations, as thin as 0.1 mm, may be used at very high
frequencies (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 empirical
expressions of the iron loss.
Characterization of core loss. Core-loss data from steel suppliers is almost always obtained from
measurements in which a sinusoidal flux waveform is applied to a sample of laminations in the form
of a stack of rings or an “Epstein square” made up from strips interleaved at the corners. The loss may
be characterized by the so-called Steinmetz equation with separate terms for hysteresis and eddycurrent loss:
P ' C h f Bpkn
%C
e
f 2 Bpk2 .
(1.69)
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 often
assumed to be 1.6!1.8, but it varies to a certain extent with Bpk. To a first approximation we can write
n = a + bBpk. With this modification,
P ' C h f Bpk
a % b B pk
%C
e
f 2 Bpk2 .
(1.70)
The flux-density in motor laminations may be far from sinusoidal, and one approximate way to deal
with this is to modify the Steinmetz equation in the following way, recognizing that the eddy-current
loss 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
P ' C h f Bpk
a
% bB
pk
%C
dB
e1
dt
2
.
(1.71)
The hysteresis loss component is unchanged, but the eddy-current component is taken to be
proportional 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 relevant
flux-density waveforms.
1. Sizing, gearing, cooling, materials and design
Page 1.41
The eddy-current loss coefficient Ce1 in the modified form can be derived from the sinewave coefficient
Ce 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
2
2
2
2 2
2
2
2 2
(dB/dt) = 4B f Bpk cos (2B ft), the mean value of which is [dB/dt] = 2B f Bpk . For sinewave fluxdensity, equations 1.70 and 1.71 give the same result if
C e1 '
Ce
2 B2
.
(1.72)
Extracting the core loss coefficients from test data. Two procedures are used for extracting the
coefficients Ch, Ce1, a and b from sinewave loss data. The more elaborate of these requires a complete
set of curves of core loss vs. frequency at different flux-densities. When this data is not available, a
simpler 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 at
50 Hz, measured with Bpk = 1.5 T. There is not enough data to determine the four loss coefficients
uniquely, so we use an estimate for n in eqn. (1.69); for example, n = 1.7. It is further necessary to
estimate the split between hysteresis and eddy-current loss. If h is the fraction of the total loss
attributable to hysteresis, then it can be shown that
Ce '
P(1 ! h)
2
f Bpk
2
and
Ch '
hP
Bpk n f
.
(1.73)
2
Then a = n; b = 0, and Ce1 = Ce/2B .
Procedure used with complete set of core-loss data—The core loss data is usually in the form of graphs
of P vs. f at different flux-densities, or P vs. Bpk at different frequencies. The procedure is to try to
separate the hysteresis and eddy-current components of P. First we divide eqn. (1.70) by f :
P
a % b B pk
' C h Bpk
% C e f Bpk2
f
(1.74)
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 highest
frequency. The graphs should be straight lines and can be represented by
P
' D % E f.
f
(1.75)
The intercept D on the vertical (P/f) axis must be equal to
D ' C h Bpk
a % b B pk
.
(1.76)
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
log D1 ' log C h % ( a % bBpk 1 ) log Bpk 1 .
(1.77)
Page 1.42
SPEED’s Electric Motors
These are solved for log Ch, a and b; Ch is then obtained from log Ch. Next, three values of Ce are
obtained from the gradients of the three graphs of P/f vs. f , eqn. (1.74). The average or the highest value
2
can be taken for Ce. Finally Ce1 = Ce/2B . The loss curves may be re-plotted from the formula as a
check. 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. This
can 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 and
current. This simple equation is the cornerstone of electrical machine design. It leads to the more
advanced work of
machine design
which is concerned with producing the torque with a minimum of material and
power loss; and
drive design
which is concerned with the control of torque and speed, subject to constraints
on the electric and magnetic loadings.
In relation to the machine design, it is always important to minimize power losses and temperature
2
rise caused by I R heating of the conductors, core losses caused by hysteresis and eddy-currents in the
magnetic steel, and other losses. But there are many other aspects, such as the need to minimize torque
pulsations and acoustic noise, and to use materials economically. There is a huge variety of different
types 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-voltage
DC power supply. But in industry, AC induction motors are used primarily because of the availability
of polyphase AC power (which has a natural rotation between the phases), and because the induction
motor 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 supplies
were available, the control equipment was less expensive for DC drives. Since the 1980's, modern power
electronics has become so cost-effective that AC drives have steadily taken over from DC drives even
in the most demanding traction applications.
At the same time the variety in types and designs of electrical machines has greatly increased in many
other fields of application because of the advances made in power electronics and microelectronic
control. This is clearly evident in such products as tape drives, computers, office machinery, and so
on; but there are many others less well known—for example, the use of very high-speed brushless
permanent-magnet motors in machine tools. These motors can run at several tens of kW and several
tens of thousands of revolutions per minute.
In relation to the drive design, one of the fundamental aspects of electrical machines is the orientation
of the flux and the ampere-conductor distribution in relation to one another. The flux and MMF10 must
be 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 angle
is zero, there is no torque and the power factor is zero. In DC commutator machines the orientation
between 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 is
not concerned with orientation and need do no more than regulate the current. By contrast, in AC
induction 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 orient
the MMF and the flux orthogonally. This is quite complex and typically requires the use of
microcontrollers or DSP’s (digital signal processors). The most modern embodiments of field-oriented
control are sophisticated enough to include estimators for important parameters such as the flux, the
direction of the flux, the rotor temperature, and the electromagnetic torque itself.
10
The ampere-conductor distribution is often loosely termed the MMF (magneto-motive force), or MMF distribution.
1. Sizing, gearing, cooling, materials and design
1.15
Page 1.43
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 in
a 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 for
prototypes, which are expensive and time-consuming.
Modern computer methods are rapidly reaching the stage where a new prototype can be designed with
such confidence that it will be "right first time", without the need for reiteration of design and test that
would otherwise be necessary. Computer-aided design goes hand-in-hand with the modern design
engineering environment. Custom designs are often required within a very short space of time, while
cost 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 ask
to see) particular parameters that traditionally were part of the "black art" of the motor builder. Often
these parameters are required for system simulation purposes long before the motor is actually
manufactured. Regulatory pressures on matters such as energy efficiency, acoustic noise, and EMC also
tighten the constraints on the motor designer.
No matter how effective the computer software available, it is always important to check the overall
parameters of a motor design using common sense and fundamental engineering principles. For this
reason it will always be necessary to be able to perform a single set of design calculations on the
computer and check the results against manual calculations. The next stage is to repeat design
calculations, 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 in
this way.
The synthesis of a design by an optimization process is a much more complex undertaking beyond the
scope of this book. However, the development of scripting languages which can run programs such as
SPEED motor design programs automatically opens up new opportunities for user-defined design
automation procedures.
Fig. 1.22 Design loop
Page 1.44
SPEED’s Electric Motors
Index
Acceleration 3, 5, 18-20
Adjustable speed 2, 8
Airgap shear stress 15
Base speed 5
Brushless DC 2, 8, 11, 12
Brushless PM 5, 11, 13, 27, 32, 39
B-H loop 31
CAD 6, 7
Commutator 4, 8, 10, 13, 17, 42
Computer-aided design 43
Conduction 21-25
Convection 15, 21-25
Cooling 1, 2, 10, 14-17, 20-24
Copper losses 10, 11, 24
Core loss 40, 41
Current density 14, 16, 23
DC commutator motor 8, 10, 17, 42
Demagnetization 12, 31-36, 38, 39
Design 1, 2, 4, 6, 7, 11, 17, 18, 20, 35, 38, 39, 42, 43
Digital electronics 5
Drive system requirements 5
Duty-cycle 27-30
eddy-current loss 40
Efficiency 3-6, 10-12, 17, 18, 21, 43
Electric loading 14-17, 38
Electrical steels 39
Energy product 6, 30, 31, 33-35, 39
Energy saving 2
Evolution of motors 8
Fan 2, 15-17, 23
Finite-element 6, 7, 24
Flow process 2
Gearing 1, 2, 18, 19
Heat removal 20-24
Hysteresis 13, 32, 40-42
hysteresis loss 40
Induction motor 6, 8, 10-12, 42
Inertia 4, 5, 11, 12, 18-20
Insulation life 20
Intermittent operation 27
Irreversible losses 35, 36
Magnet operating point 33
Magnet safety 38
Magnetic loading 14, 15, 17, 38
Magnetic materials 6
Magnetization 31-38, 40
Microelectronics 2, 5, 6
Minas 11
Motion control systems 2, 5, 6, 8, 20
New technology 2, 5
Numerical analysis 6
Output equation 14
Overload 5, 12, 27-30, 39
Permanent magnet 13, 30, 39
Permeance coefficient 33-37
Position control 2-4, 13
Power 2-8, 10-15, 18, 23, 27, 28, 38-40, 42
Power density 12, 38, 39
Power factor 10-12, 42
Power semiconductors 2, 4, 8
Radiation 21-25
Reluctance motors 2, 12, 13, 17
Reversible losses 36
Sizing 1, 2, 14, 17
Slip 8, 10, 11, 13
Steinmetz 40
Steinmetz equation 40
Stepper motors 8, 12, 13, 17
Structure of drive systems 4
Switched reluctance 2, 8, 12, 13, 17
Synchronous motors 10, 13
Synchronous reluctance 12, 13
Temperature rise 14, 17, 20-23, 25, 27-29, 42
Thermal equivalent circuit 24, 25, 27
Torque 4-6, 8, 10-12, 14-20, 23, 28, 29, 38, 39, 42
Transients 2, 3, 6, 24
TRV 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
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 are
eliminated. These components include brushes, commutators, slip-rings, etc. used to supply current
to the rotor. Although brushed motors are still widely used, their disadvantages have encouraged the
development 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 high
efficiency.
The development of brushless machines would not have been possible without certain enabling
technologies : in particular, power transistors, microelectronic controls and sensors, and permanentmagnet materials. These components are now so highly developed that they have opened up many new
opportunities 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 we
can recognize four main categories:
(a)
Brushless DC Machines are derived directly from the classical DC machine by replacing the
commutator and brushes with an electronic power supply. The motor is often designed to have
a trapezoidal back-EMF waveform and the current waveforms are rectangular, with alternating
polarity. The current polarity is switched in synchronism with the rotor position, by means of
power semiconductors which are also used to regulate the current. Fig. 2.1.1 shows a typical
example. 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 is
designed to be sinusoidal and the current waveform is also controlled to be sinusoidal. They are
used in servosystems where smooth torque control is required. Resolvers are often used for
shaft 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 wide
speed range—as, for example, in drives for electric vehicles. It is even possible to dispense with
the 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 fed
from a rotating rectifier, which in turn is fed from a rotating AC exciter mounted on the same
shaft. This category includes some large machines (several MW) and high-speed AC machines
such 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 category
covers 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 DC
machine, Fig. 2.1.1. The design principles of machines in category (c) are more in common with those
of classical synchronous machines. For machines in category (b), the design principles are in common
with both categories (a) and (c).
Other important types of "brushless" electric machine include the induction motor, the switched
reluctance motor, and the stepper motor.
1
“Salient” means “sticking out” and refers to the projecting poles on wound-rotor synchronous machines. The term “salientpole” 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 the
synchronous inductances Ld and Lq, and gives rise to reluctance torque.
Page 2.2
SPEED’s Electric Motors
Fig. 2.1
Types of brushless motor.
Brushless permanent-magnet machines
2.2
Page 2.3
BASIC OPERATION OF THE BRUSHLESS DC MOTOR
Fig. 2.2 shows a 2-pole motor with a magnet rotating
counter-clockwise at the instant when the flux-linkage
with coil 1 is at a negative maximum. The magnet is
shown with an arc $M = 180Eelec., and the coil-pitch is also
180Eelec, so that the flux-linkage R1 of each coil varies with
a triangular waveform as the rotor rotates: see Fig. 2.4.
By Faraday’s Law the EMF induced in coil 1 is
ea1 '
dR1
dt
' Tm
dR1
(2.1)
d2
where Tm is the angular velocity in mechanical rad/s and
2 is the rotor position in mechanical radians. As long as
the flux-linkage is varying linearly with rotor position, the
induced EMF is constant. When the flux-linkage reaches a
maximum, it starts to decrease at the same absolute rate,
and the EMF changes polarity. The result is a squarewave
generated EMF, ea1.
Fig. 2.2
Generation of back-EMF
A second coil a2A2 is displaced by a certain angle ( from the first coil. It has the same number of turns
as 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 with
twice the peak value of the individual coil EMF. If each phase is made up of just two coils connected in
series, this is the phase EMF. Its waveform is stepped because of the phase displacement between coils
1 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 . This
is 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 waveform
is just over 120E wide. Then each phase is supplied with a current waveform consisting of blocks of
constant 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 flat
during 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 one
pole-pitch, the airgap flux Mg is given by
Mg ' m
0
B/p
B(2) r d2 L stk ' B g ×
BD L stk
2p
(2.2)
where r is the stator bore radius, D is the stator bore diameter (= 2r), Lstk is the stack length, and p is
the number of pole-pairs. The peak flux-linkage of coil 1 is R1 max = Tc Mg, and if the linear variation
of R1 with 2 is
R1(2) '
p2
B/2
R1 max
(2.3)
× T c Mg
(2.4)
then by eqns. (2.1!2.3) the peak coil EMF is
ea1 ' Tm ×
2p
B
For a machine with Tph turns in series per phase, the peak EMF/phase can be written
E a ' Tm ×
2p
B
× T ph Mg
(2.5)
SPEED’s Electric Motors
Page 2.4
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 wyeconnected they carry the same current I in series, and the line-line EMF during each 60E interval is
E LL ' 2 E a ' k E Tm
(2.6)
where kE is the back-EMF constant in Vs/rad:
kE '
4 p T ph Mg
B
.
(2.7)
The electromagnetic power is ELLI and the electromagnetic torque is Te = ELL I/Tm. This can be written
T e ' kT I
(2.8)
where k T = kE is the torque constant in Nm/A. When driven this way with "two phases on", the motor
behaves very much like a permanent-magnet DC commutator motor. The torque is produced in blocks
60E 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 they
are easy to control, and they can be switched at high frequency. This makes it possible to regulate the
current by chopping with low acoustic noise. MOSFETS are ideal for low-voltage drives because of their
low on-state voltage drop. At higher powers and voltages, IGBTS are used.
Fig. 2.3
Drive circuit for wye-connected brushless DC motor
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 waveforms
are 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 wyeand delta-connected motors. In the wye-connected motor, the magnet arc is 180E. With two phases
conducting, 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 of
ampere-conductors nearest to it. This ensures constant torque over a 60E angle. At the end of each 60E
period, the current is commutated from one phase into the next. In Fig. 2.6, phase 2 will be the next to
turn 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 ampereconductors. To ensure that the overlap is constant for 60E, the magnet arc is reduced to 120E.
Brushless permanent-magnet machines
Fig. 2.4
Flux-linkage, EMF and current waveforms for the motor in Fig. 2.2.
Page 2.5
SPEED’s Electric Motors
Page 2.6
Fig. 2.5
Fig. 2.6
Drive circuit for delta-connected brushless DC motor.
Interaction of arc magnets and ampere-conductor phasebelts in wye- and delta-connected
brushless DC motors.
Brushless permanent-magnet machines
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.
Page 2.7
SPEED’s Electric Motors
Page 2.8
Fig. 2.8
Ideal waveforms of line currents ia, ib, ic; phase current i1; and phase EMF e1 in deltaconnected brushless DC motor and drive. For more detail see Hendershot &
Miller[1994].
Line
Phaseleg A
Phaseleg B
Phaseleg C
Rotor position [Eelec]
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
0
TABLE 2.1
COMMUTATION TABLE FOR BRUSHLESS DC MOTOR DRIVE: 120E SQUAREWAVE LINE CURRENTS
Brushless permanent-magnet machines
2.3
Page 2.9
TORQUE/SPEED CHARACTERISTIC
Fig. 2.9
Torque/speed diagram
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
Vs ' E % R I
(2.9)
where I is the DC current. Using eqns. (2.6) and (2.8), and neglecting losses, the torque/speed
characteristic can be derived in the form
T
T0
'
T
1 !
T0
'
1 !
I
I0
(2.10)
where the no-load speed is
T0 '
Vs
rad/s
kE
(2.11)
and the stall torque is
T0 ' k T I0
(2.12)
and the stall current or locked-rotor current is
I0 '
Vs
R
.
(2.13)
The stall current given by this expression may be large enough to demagnetize the magnets, but usually
this current is far beyond the capability of the power transistors in the drive. Therefore, the current
is 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, the
speed is controlled by the voltage and Fig. 2.9 shows the effect of reduced voltage on the torque/speed
characteristic. 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 is
automatically 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 it
may be permissible to operate at currents higher than the normal limit; accordingly, Fig. 2.9 is divided
into continuous and intermittent operating regions by the current-limit line.
SPEED’s Electric Motors
Page 2.10
2.4
MAGNETIC CIRCUIT ANALYSIS
The most basic magnetic calculation in brushless permanent-magnet motors is to determine the flux
produced by the magnets. An important result of this calculation is the “operating point” of the
magnets and the general saturation level of the iron. It is also important to determine the distribution
of the flux around the airgap, because this in turn determines the waveform of the generated EMF. The
EMF waveform is also affected by the winding distribution, but in this section we will concentrate only
on the calculation of magnet flux, that is, the “open-circuit” condition.
Magnetic circuit calculations in brushless permanent-magnet machines can be classified into three
main classes:
(1)
(2)
(3)
very simple methods which can be executed using a calculator;
more complex analytical methods which may require the use of a computer; and
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 one
used for serious design calculations because it can be executed repetitively and very fast when suitably
programmed on a computer. Finite-element methods are the most accurate, but also the slowest. The
three classes also give a rich variety of physical interpretations, but they are not equally appropriate
for all types of motor. For surface-magnet motors it is often sufficient to work with the first two, but
for embedded-magnet motors the finite-element method is essential because of the complexity of the
rotor geometry, with intense local saturation.
This section describes two methods for the magnetic calculation of surface-magnet motors, the first
being from class 1 and the second from class 2.
Simple magnetic circuit analysis
Fig. 2.10 Simple magnetic circuit
The magnetic circuit calculation begins with a simple estimate of the airgap flux Mg based on the
magnetic equivalent circuit in Fig. 2.10. This is used as the starting-point for a nonlinear calculation
that uses the magnetic equivalent circuit in Fig. 2.11. The permeances in both models are associated
with one half-pole. The magnet permeance is
Pm0 ' µrec µ0
AM
LM
' µrec µ0
$M r M L stk
2
LM
Wb/At
(2.14)
where $M is the pole-arc of the magnet in mechanical radians, AM is the magnet half-pole area, Lstk is
the stack length, and LM is the magnet length in the direction of magnetization. The radius rM is the
effective radius of the magnet.
Brushless permanent-magnet machines
Page 2.11
Similarly the airgap reluctance is
g
Rg '
A&t/Wb
µ0Ag
(2.15)
where Ag is the airgap area. The flux Mg is the flux crossing the airgap, Mm is the magnet flux, and ML
is the leakage flux, i.e., Mm ! ML.
Leakage factor : We can identify two separate (but related) ways of dealing with leakage. For surface-
fLKG '
Mg
Mm
'
Mg
< 1.
Mg % ML
(2.16)
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 it
as a fraction prl of the magnet permeance Pm0: thus PL = prl Pm0. Any saturating bridges (as for example
in 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 is
expressed a fraction of the half-pole remanent flux Mr:
Pg
Mg '
Pg % PL % Pm0
Mr .
(2.17)
For surface-magnet motors Mg is expressed in terms of f LKG as
Mg '
1
1
fLKG
Mr
(2.18)
% Pm0 R g
but for spoke-type and IPM motors it is expressed in terms of prl as
Mg '
1
1 % (1 % p rl ) Pm0 R g
Mr
(2.19)
where Rg = 1/Pg. Eqns. (2.18) and (2.19) are equivalent, with
1 % p rl Pm0 R g '
1
fLKG
.
(2.20)
Note that having f LKG < 1 means that Mg is reduced by the effect of the leakage. A typical value for f LKG
is 0.9 !0.95 for a wide range of surface-magnet motors. If this value is specified, prl can be calculated
from 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
Bm
PC '
.
µ0*H m*
(2.21)
The values of Bm and Hm are obtained from the nonlinear magnetic circuit calculation (below). Having
f LKG < 1 means that Mm is greater than it would be with no leakage, and therefore PC is increased by
leakage.
SPEED’s Electric Motors
Page 2.12
Fig. 2.11 Nonlinear magnetic circuit
Nonlinear calculation : The magnetic circuit in Fig. 2.11 is drawn for half of one Ampère's Law
contour, representing the MMF drops associated with one airgap. The magnet is now represented by a
Thévenin equivalent circuit comprising the "open-circuit" MMF Fma and the reluctance Rm0 = 1/Pm0. The
flux densities in the yoke and teeth sections are calculated from their permeance areas, and the
associated MMF drops F are obtained using the nonlinear BH curve of the steel. The total magnetising
force in the magnet is calculated as
Hm '
( Fg % FSY % FRY % FST )
LM
(2.22)
The circuit is solved iteratively. The basic result of the nonlinear magnetic circuit calculation is the
airgap flux Mg and from this the average flux density over the magnet pole arc can be calculated.
Fig. 2.12 Magnet fringing
Apart from a simple allowance for fringing, the magnetic analysis has so far been described as though
the magnet flux had the ideal rectangular distribution around the airgap shown by the dotted line in
Fig. 2.12. In reality the distribution has rounded corners and there is a fringing zone near the edges
of the magnet. An approximate functional representation of this is the exponential relationship
b '
1
2
1 ! e
!( 2 ! 2a ) / a
(2.23)
in the fringing zone to the right of 2a, where b is the normalized value of the flux-density and a is an
empirical coefficient given by
Brushless permanent-magnet machines
a '
1
2
g [ g % L m / µrec ] .
Page 2.13
(2.24)
To the left of 2a the fringing function is simply
b '
1
2
e
( 2 ! 2a ) / a
(2.25)
and similar functions are used on the right half of the distribution, symmetrical about 2b. This fringing
function 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 the
fundamental space-harmonic component B1(2) and from this the fundamental magnet flux/pole MM1
(oc)
and the peak fundamental open-circuit airgap flux density, B1 .
Demagnetizing effect at locked-rotor: The magnetic circuit model can be used to estimate the
demagnetizing field in the magnet under various conditions, for example, if the rotor is locked or in
any other condition where high current is liable to flow in the windings. The armature ampere-turns
Fa per half-pole are calculated according to the winding type and are assumed to be concentrated at a
single point in the airgap. An example of this calculation is given in Chapter 5 in connection with DC
commutator 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 direct
solution of Maxwell’s equations, which reduce to the Laplace equation in the air region and the Poisson
equation 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 smooth
iron cylinders, for an arbitrary distribution of current-carrying conductors in the airgap or on the
surfaces of the cylinders. This work was applied by Boules [1984], who replaced the magnet by an
equivalent 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: generally
where there is no irregular iron shape in the rotor, the stator has a smooth bore, and the magnet has
a simple geometric shape and a certain direction of magnetization. Boules developed solutions for
certain 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 the
magnet surface. Since M and n are both always in the x,y plane, transverse to the axis of rotation, K
is always in the z direction along the axis of rotation, i.e. K = (0,0,K). M is the actual magnetization
of the magnet, which includes an induced component due to the demagnetizing field of the external
magnetic circuit. Unfortunately this is not known a priori. However, if the recoil permeability is near
1, the susceptibility P m of the magnet is nearly zero, and the induced magnetization is small. Boules
points out that on open-circuit the magnets are normally worked between B r/2 and B r, and he uses the
average magnetization over this range, i.e., M = k mM 0 = k m B r /µ 0 where k m = (1 + 0.75 P m)/(1 + P m).
Note that M is equivalent to the “apparent coercivity” H ca, i.e. the coercivity that the magnet would
have if its recoil line was straight throughout the second quadrant with relative permeability µr . The
value of the susceptibility P m and the constant k m can be seen in Table 2.2 for typical values of µ r. For
most magnets µ r does not exceed 1.1, so the maximum error from this approximation is less than 2.5%.
2
Hague’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 were
published by Zhu et al [1993] and Rasmussen et al [1999]. These later solutions relied on a harmonic series representation of
the magnetization vector and therefore considerably extended the scope of the analysis.
3
Several others have been developed for the PC-BDC program.
SPEED’s Electric Motors
Page 2.14
µr
Pm
km
1
0
1
1.05
0.05
0.988
1.1
0.1
0.977
0.2
0.958
1.2
TABLE 2.2
Boules derived the ampere-conductor distributions for arc magnets whose edges xy lie along radii. For
radially-magnetized magnets, K = 0 on the curved surfaces; and on the edges, K = M. For parallelmagnetized magnets, on the curved surfaces K = M sin 2 and on the edges K = M cos $M/2, where $M is
the magnet pole arc expressed in mechanical degrees. The equations for magnets with parallel edges
are 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 = M
sin 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 between
two concentric cylinders. Fig. 2.13 shows a basic 4-pole distribution of single-turn coils having a radius
c and span 2 >, equivalent to the magnet arrangement in Fig. 2.14. The field produced by the coilset of
Fig. 2.13 at the point (r, 2) is given by
Br ' 2 p
µ0 i
Br
4
j
n
an
cn
@
c 2n % b 2n
a 2n ! b 2n
@
rn
an
%
an
rn
@ kF n sin n > cos n 2
(2.26)
where i is the coil current, p is the number of pole-pairs, and the sum is taken over all odd electrical
harmonics, i.e., n = (2j ! 1), j = 1,2,3.... The factor k Fn is the n’th harmonic skew factor, which is equal
to sin (nF/2) /(nF/2), where F is the skew in electrical radians. When the contributions of all the
filamentary coils are summed, a similar factor arises if the magnetization tapers off from a peak value
to zero over an angle F electrical radians, proving that tapered magnetization and skew are equivalent
in terms of their effect on the airgap field.
Fig. 2.13 Airgap model of surface magnet
Fig. 2.14 Filamentary coil model
The methods developed by Rasmussen [1999] and by Zhu et al [1993] go beyond the Hague-Boules method
just described, by using a direct scalar potential solution that relies on a harmonic series
representation of the magnetization vector. The airgap field is given by expressions of the form
Br ' j
n
q (M nH n % N nK n)
µr (q 2 ! 1)
[r q!1 % a 2qr !q!1]c !q%1H n cos q2
(2.27)
where q = np, p is the number of pole-pairs, Mn and Nn are the n’th harmonic components of the radial
and tangential components of magnetization, and Hn and Kn are functions of q, µr and the various radii
given 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 magnetization
is assumed to be invariant with r, i.e., it does not vary through the thickness of the magnet.
Brushless permanent-magnet machines
Page 2.15
Figs. 2.15!19 show examples of different permanent-magnet brushless motors for which the above
methods are appropriate.
Fig. 2.15 Example with typical proportions. Good results
can be obtained even with the simple magnetic
circuit analysis model.
Fig. 2.16 With very thin magnet and a short airgap, good
results can be obtained with all methods.
Fig. 2.17 With a small rotor diameter to accommodate a
thick non-magnetic retaining ring, the simple
magnetic circuit model is inadequate and the
Hague/Boules or Rasmussen methods give
much better results.
Fig. 2.18 “FullRing” magnet type, a solid 2-pole magnet.
In this case the Hague/Boules method is the
best analytical method although Rasmussen’s
method is also appropriate.
The simple
magnetic circuit model is inadequate in this
case.
Page 2.16
SPEED’s Electric Motors
Finite-element analysis
Because the Hague-Boules and Rasmussen methods are analytical, they are not well adapted to deal
with saturation effects, but the method is nevertheless still useful, either for machines where
saturation is not important, or where simplified allowances can be made for it. The simple magnetic
circuit model can make crude allowances for saturation, but for thorough analysis of the magnetic field
the finite-element method is by far the most powerful. It is particularly effective in computing the
details of local geometric features and the effects of arbitrary distributions of ampere-conductors and
magnetization patterns. These details continually increase in importance, partly because of
competitive pressure to improve performance and cost-effectiveness, but also because of the need to
reduce torque ripple and acoustic noise.
Fig. 2.19 Simple finite-element mesh for brushless PM Fig. 2.20 Magnetic field solution for Fig. 2.19 (open
motor
circuit)
Figs. 2.19 and 2.20 show typical examples of finite-element computations for a simple brushless
permanent-magnet motor on open-circuit, and Fig. 2.21 shows the comparison of the airgap flux-density
distribution obtained by the finite-element and magnetic-circuit methods. The finite-element solution
includes the effect of the slot-openings, which is absent from the analytical solution.
Fig. 2.21 Comparison of finite-element and magnetic circuit model
Brushless permanent-magnet machines
2.5
Page 2.17
WINDING INDUCTANCE OF SURFACE-MAGNET MOTORS
Fig. 2.23 Distribution of armature reaction MMF and flux
Fig. 2.22 Calculation of inductance
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 is
assumed to be infinitely permeable, then the MMF is concentrated across two effective airgaps. Each
effective airgap includes the thickness of the magnet, L M, which is assumed to have a relative recoil
permeability µrec close to 1. Therefore the MMF drop across each airgap is Ni/2. If the flux is assumed
to cross the gap in the radial direction, the magnetizing force in each gap is
H '
Ni
(2.28)
2 gN
where gN is given by
gN ' k c g %
LM
(2.29)
µrec
in which g is the actual airgap modified for slotting by Carter’s coefficient k c. An approximation for
Carter’s coefficient suitable for surface-magnet motors is
kc '
5 % s
5 % s ! s 2g/8
(2.30)
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
B ga '
µ0 N i
2 gN
.
(2.31)
The ideal flux distribution around the airgap is plotted in Fig. 2.23. The flux linkage of the coil is
R ' N × B ga ×
BD
2p
L stk
(2.32)
where p is the number of pole-pairs. The self-inductance is given by Lg = R/i, that is,
Lg '
2
B µ0 T ph L stk D
4
p 2 gN
(2.33)
where Tph is the number of turns in series per phase (= Np/a, where a is the number of parallel paths
in the winding). The subscript g denotes the fact that this is only the "airgap" component of the
inductance. The total self inductance is given by
SPEED’s Electric Motors
Page 2.18
L ph
'
Lg
%
'
LF
Lg
%
[ L slot
%
L end ]
(2.34)
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-linkages
of a second coil placed in the field of the first one, Fig. 2.23. The second coil (phase 2) is located in slots
at the angles !90 + 120 = 30E and +90 + 120 = 210E. Its flux-linkage due to current in coil 1 is
R21 ' N × B ga ×
D
2p
L stk × !
7B
6
!
B
2
%
B
2
!
B
6
(2.35)
from which it can be deduced that the airgap component of mutual inductance is !L g /3. If Mslot and
Mend are the mutual inductance components in the slot and end-turns respectively, then
M ' !
Lg
3
% M slot % M end .
(2.36)
in which Mslot and Mend are separate components of mutual inductance associated with "leakage" fluxes
other than the airgap flux.
Hague's method [5] can be used to calculate the airgap components of the phase inductance and the
mutual inductance between phases, if the ampere-conductors in the slots are represented by current
filaments at the centres of the slot openings, and the stator bore is assumed to be smooth. The airgap
length is augmented by Carter’s coefficient and the magnet length, according to eqn. (2.29). It is
possible to program such a calculation for any arbitrary distribution of winding conductors, but the
computation may require a large number of harmonics (of the order of 100) to produce a satisfactory
result. As in the analytical examples above, the slot and end-winding components of inductance must
be added separately.
The slot-leakage components Lslot and Mslot can be calculated as in induction motors, using the slot
permeance coefficient calculated from the slot geometry. The calculation of Mslot must take account
of the coincidence of conductors from different phases in each slot. The end-turn inductance Lend can
be 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 winding
distribution, it is possible to isolate the contribution of the fundamental, and then the sum of all higher
harmonic terms is called the "differential leakage inductance" Ldiff.4 The fundamental term is the one
that gives rise to the "magnetizing" or "airgap" component of the synchronous inductance, which is
treated 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 winding
inductances vary as the rotor rotates; this property is known as saliency and such motors are classified
as salient-pole motors. It is typical in such motors for the inductance to vary also as a function of
current. As a result, it is difficult to perform a time-stepping simulation. The dq-axis transformation
has long been used for the analysis of salient-pole motors, [9].
4
Classical AC machine theory gives the differential leakage reactance via the differential leakage factor
kd '
1
k w12
j
n>1
k wn2
n2
Brushless permanent-magnet machines
Page 2.19
Fig. 2.25 Reference axes for dq analysis
Fig. 2.24 Cross-section of PM motor showing d-axis
The dq-axis transformation renders the machine inductances independent of rotor position by casting
the equations in a frame of reference that rotates in synchronism with the rotor. Under steady-state
AC conditions the currents and voltages in this frame of reference are constant, as though they were
DC quantities. Although the inductances still vary with current, the functional variation is simpler and
is not time-dependent.
The dq-axis transformation is valid only if the motor is sinewound: i.e., the windings are sinedistributed (or nearly so). In this case, irrespective of the rotor shape, there is only a 2nd harmonic
variation for the phase self-inductances (Laa, Lbb, Lcc) and mutual inductances (Lab, Lbc, Lca) as the rotor
rotates: thus if 2 is the rotor position we can write
L aa ( 2 ) ' ( LF % L g0 ) % L g2 cos 2 2
L bb ( 2 ) ' ( LF % L g0 ) % L g2 cos ( 2 2 % 120E )
L cc ( 2 ) ' ( LF % L g0 ) % L g2 cos ( 2 2 ! 120E )
(2.37)
The angle 2 and the relative orientation of the phase winding axes are shown in Figs. 2.24 and 2.25. The
constant 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 the
conductor distribution. The leakage component includes slot-leakage, end-turn leakage, and
"differential" leakage —i.e., inductance associated with higher-order space harmonics of the winding
distribution. The self-inductance attains extreme values when the d- and q- axes of the rotor are aligned
with the phase axis: thus, for phase a:
Laa[d] ' ( LF % L g0 ) % L g2
Laa[q] ' ( LF % L g0 ) ! L g2
or ( LF % L g0 ) '
Laa[d] % Laa[q]
2
; L g2 '
(2.38)
Laa[d] ! Laa[q]
2
.
(2.39)
The mutual inductance between two phases is given by
L ab ( 2 ) ' MF !
1
2
L g0 % L g2 cos ( 2 2 ! 120E )
(2.40)
where MF represents mutual coupling mainly in the slots. The line-line inductance is
L LL ' L aa % L bb ! 2 L ab
(2.41)
SPEED’s Electric Motors
Page 2.20
When eqns. (2.37) and (2.40) are substituted in eqn. (2.41), LLL simplifies to
L LL ' LLL0 ! 3 L g2 sin ( 2 2 ! 30E )
with
L LL0 ' 2 ( LF ! MF ) % 3 L g0
(2.42)
Like the phase inductance, LLL also has a constant and a second-harmonic term, and it attains extreme
values when the d- and q-axes are aligned such that
LLL[d] ' 2 ( LF ! MF ) % 3 [ L g0 % L g2 ] ' LLL0 % 3 L g2
LLL[q] ' 2 ( LF ! MF ) % 3 [ L g0 ! L g2 ] ' LLL0 ! 3 L g2
(2.43)
For delta-connected sinewound machines it can be shown that
LLL [d,q] ' LLL 0 '
2
3
( LF ! MF ) % L g0 ± L g2 .
(2.44)
The extreme values of LLL could be measured, and then eqns. (2.43) could be solved for LLL0 and Lg2 :
LLL0 '
LLL[d] % LLL[q]
2
; L g2 '
LLL[d] ! LLL[q]
6
(2.45)
Using the d,q-axis transformation [9], we obtain the synchronous inductances
L d ' LF ! MF % L md ;
L q ' LF ! MF % L mq
(2.46)
where
L md '
3
2
( L g0 % L g2 ) ; L mq '
3
2
( L g0 ! L g2 ) .
(2.47)
From eqns. (2.43) with (2.46) and (2.47),
LLL[q] ' 2 L q
and
LLL[d] ' 2 L d
(2.48)
suggesting that static measurement of LLL[d] and LLL[q] can be used to measure Ld and Lq, provided that
the windings are sine-distributed or nearly so.
In most brushless PM motors Lmd< Lmq and Ld < Lq, so Lg2 < 0. The synchronous inductances Ld and
Lq and the reactances Xd and Xq can be calculated from the following:
X d ' 2 B f L d ' X md % XF
X q ' 2 B f L q ' X mq % XF
(2.49)
X md ' 2 B f L md ' 'd X m0
X mq ' 2 B f L mq ' 'q X m0
(2.50)
where
Xm0 is the airgap component of the synchronous reactance of a nonsalient pole machine having an
airgap length equal to gN = k c g, where g is the actual airgap and kc is the Carter coefficient: thus
X m0 ' T L m0 '
3 T µ0 D l
Bp 2g )
( k w1 T ph )2 .
(2.51)
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
'd '
gN
gdNN
;
'q '
gN
gqNN
(2.52)
where gdNN is the effective airgap in the d-axis including the effects of the magnet and the saliency; and
gqNN 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.
Brushless permanent-magnet machines
Page 2.21
Once Lmd and Lmq are calculated from eqn. (2.50), Lg0 and Lg2 can be calculated from eqn. (2.47): thus
L g0 '
1
3
[L md % L mq ] ; L g2 '
1
3
[L md ! L mq ] .
(2.53)
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 winding
harmonics), the respective components of differential leakage inductance can be estimated as
Ldiff ' L g
! Lg0 ;
Mdiff ' M g
! (!Lg0 / 2 ) .
(2.54)
Criterion for sinewound motors: To help decide whether a motor is sinewound, a parameter
2
k sg '
3
L m0
L gg
'
2 X m0
3 T L gg
(2.55)
can be calculated. ksg is unity for sinewound motors. Lm0 is the synchronous inductance and Lgg is the
actual airgap inductance that would be obtained if the winding was perfectly sine-distributed and the
rotor was a smooth iron cylinder with the same airgap as in the actual machine. Both Lm0 and Lgg are
free of end-effects and slot-leakage. For 1- or 2-phase motors the 2/3 factor is omitted. If ksg is close to
1, the winding is nearly sine-distributed, and calculations with d,q equations should agree with those
performed with direct phase variables. On the other hand, if ksg is far from 1, the differential leakage
will not be negligible and calculations with d,q equations will not agree with those performed with
direct 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
L ph ' L g % L slot % L end
;
M ph ' M g % M slot .
(2.56)
The airgap components Lg and Mg can be calculated by summing the self- and mutual flux-linkages of
each coil due to current flowing in every coil, preferably using a method based on Hague’s analysis
discussed earlier, which makes it possible to account for any winding distribution. This procedure
includes the differential leakage reactance in Lg, that is, the reactance associated with all spaceharmonics 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 the
variation is assumed to follow eqns. (2.37) and (2.40). The leakage inductances LF and MF must include
the 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 associated
reactances 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 the
relative permeability of the magnet is slightly higher than 1 and the magnet is not totally rotationally
symmetrical. In salient-pole motors Xd and Xq differ, usually with Xq > Xd, and in embedded-magnet
motors they can vary significantly as a function of current. The component inductances Lg0 and Lg2 can
be recovered using eqn. (2.53). The maximum and minimum values of the phase inductance can be
obtained with eqn. (2.38), and the maximum and minimum values of the line-line inductance with eqn.
(2.48).
SPEED’s Electric Motors
Page 2.22
Classification of motors
Table 2.3 shows a suggested classification of different types of permanent-magnet rotor into “salientpole” and “nonsalient pole” types.
Rotor type
Inductance variation with
2
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 or
parallel
Constant
nonsalient pole
IPM
Strongly variable
salient-pole
Inset magnet
Variable
salient-pole
TABLE 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 of
the 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 phase
inductances. To clarify these differences, the inductance parameters can be classified into two groups
labelled “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]
Ld
Lq
Laa[q]
Xd
Xq
LF
'd
'q
MF
Xm0
XF
Synchronous
kw1
TABLE 2.4
CLASSIFICATION OF INDUCTANCEPARAMETERS
With salient-pole motors the inductances Lph, Mph, Lg and Mg are variable, but Lgg and Mgg are the
approximate 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 circuit
equations can be expressed in direct phase variables. Salient-pole motors, on the other hand, should
be sinewound if at all possible, to facilitate the solution of their circuit equations either by the phasor
diagram, or by time-stepping simulation in dq axes.
Brushless permanent-magnet machines
2.7
Page 2.23
BASICS OF SINEWAVE OPERATION
Fig. 2.26 Phasor diagram on open-circuit
The most fundamental parameters of a brushless permanent-magnet motor drive are the number of
phases and the method of controlling the current. The drive circuit configuration depends primarily
on the number of phases. Broadly speaking there is a division into two groups:
(1)
(2)
Low-cost brushless DC drives with one or two phases, widely used in blowers and fans; and
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 drives
use bridge circuits for high efficiency and maximum controllability. In some instances, full H-bridge
circuits are used with 1-phase or 2-phase motors to achieve high efficiency. With 1 or 2 phases, it is
generally not possible to achieve reliable starting in both directions from any position, unless auxiliary
methods 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 motor
operating from such a supply can be represented by its phasor diagram, provided that it is sinewound
or approximately so. Even before the motor is connected to the supply, the open-circuit phasor
diagram can be drawn as shown in Fig. 2.26. Since the motor is on open-circuit there is no current, and
the phasor of the generated EMF E leads the phasor of the fundamental flux-linkage Q1Md by 90E. The
fundamental flux-linkage Q1Md is the product of the effective number of series turns/phase kw1Tph and
the fundamental component of magnet flux in the airgap M1Md, where kw1 is the fundamental harmonic
winding factor and Tph is the number of series turns per phase.5 Thus if B1Md is the peak value of the
fundamental flux-density produced by the magnet on open-circuit,
Q1Md ' k w1 T ph M1Md '
k w1 T ph B1Md D L stk
2
p
[V&s r.m.s.]
(2.57)
and is centred on the d-axis. The subscript 1 refers to the fundamental space-harmonic component. The
phasor relationship between E and Q1Md is
E ' j E q1 ' j T Q 1Md
(2.58)
where the subscript 1 emphasizes the fundamental, q the q-axis, d the d-axis, and M the magnet. The
90E phase lead comes from Faraday’s law, when the phasors are expressed as complex numbers, and
it is normal to consider the magnet flux-linkage phasor to be aligned with the d-axis while the generated
6
EMF is aligned with the q-axis.
5
Because of the distribution of conductors the winding links only a fraction kw1 of the fundamental flux. It acts as a filter. To
get a sinusoidal EMF requires low winding factors kwn for the harmonic fluxes (n > 1) and a value of kw1 close to 1.
6
The phase lead results from the use of the “sink” convention in which e = dR/dt rather than !dR/dt. This is appropriate for
motors, but less intuitive for generators. However, permanent magnet motors are vastly more common than generators.
Page 2.24
SPEED’s Electric Motors
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 in
Fig. 2.27. If the motor is nonsalient pole it can be represented electrically
by its generated EMF E, its phase resistance R and its synchronous
reactance Xs (which is equivalent to Xd in a salient-pole machine). The
AC voltage source is assumed to have a constant voltage V and no
internal impedance. The phasor diagram for the circuit in Fig. 2.27 is
shown in Fig. 2.28. It shows that the supply voltage V comprises the
series combination of the volt-drops RI, jXsI, and E, which must be added
“vectorially” because they are not in phase with one another. The voltdrop RI is in phase with the current I and its phasor is parallel to the
current phasor, but the volt-drop jXsI is at right-angles to the current and
leads it in phase because Xs is an inductive impedance.
The angles shown in Fig. 2.28 are important. The angle N between V and
I 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 in
drives which control the phase and magnitude of the current relative to
the shaft position. The angle * between V and E is called the “load angle”
and is also sometimes called the “torque angle” especially when the
motor is operating from an AC voltage source (i.e., without current
control).
Fig. 2.28 Phasor diagram
Operation from an AC voltage source is useful for understanding the basic concepts of the phasor
diagram, but in practice it is very rare. Brushless PM motors operated in this way are inherently
unstable and tend to “hunt” or oscillate about the synchronous speed unless they are fitted with
damper windings on the rotor.8 Moreover, without the damper winding the motor has no means of
starting. With damper windings, such motors are referred to as “line-start” motors. They are used in
certain applications where several motors operate in synchronism from a single inverter, or direct
from the mains supply where, for some reason, exact synchronous speed is required. Several attempts
have been made to develop such machines to compete with induction motors on the basis of high
efficiency, even with single-phase capacitor motors, and although some very good technical results
have been reported, there has been little commercial development because of the cost and certain
application problems.
7
This definition of power factor is valid only if the voltage and current are sinusoidal.
8
Also known as “amortisseur” windings, these are usually in the form of a cast cage similar to the cage of an induction motor
rotor. See, for example, Jordan [1983], Miller [1984].
Brushless permanent-magnet machines
2.8
Page 2.25
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 field
system as defined by the field winding or excitation winding. In the DC machine the excitation winding
is on the stator9 and therefore the d.q axes are fixed to the stator. In the AC synchronous machine the
excitation winding is on the rotor and therefore the d.q axes are fixed to the rotor, Fig. 2.29. The d-axis
is 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 two
axes of symmetry, the rotor is said to have two-axis symmetry. In electrical terms the d- and q-axes are
orthogonal, i.e., separated by 90 “electrical degrees”. Indeed the definition of “electrical degrees” is
such that there are 180 electrical degrees between consecutive d-axes.
Fig. 2.29 Wound-field synchronous machine
Fig. 2.30 Surface-magnet motor
Fig. 2.31 Interior-magnet motor
In the permanent-magnet machine the wound poles are replaced by permanent magnets, Figs. 2.30 and
2.31, but the meaning of the d and q axes is unchanged. The interior-magnet motor in Fig. 2.31 is a
salient-pole machine with different inductive properties along the d- and q-axes, but the surface-magnet
motor in Fig. 2.30 is nonsalient-pole, being rotationally symmetric apart from the magnetization of the
magnets 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 is
slightly more complex than the one in Fig. 2.28 in that it is drawn for a salient-pole machine with Xd
different from Xq. The voltage drop XsI is replaced by two separate voltage drops XdId and XqIq.
9
AC induction motors do not have obvious d,q axes because (i) both the stator and the rotor are cylindrical, i.e., rotationally
symmetric; 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, the
labels d,q are applied extensively, but several different frames of reference are used and there is probably no single definition
of d,q axes that would work equally well for all variants. In dealing with PM brushless machines it is natural to fix the dq axes
to the rotor, as in the wound-field synchronous machine of Fig. 2.29, but this is not the only convention that can be adopted.
SPEED’s Electric Motors
Page 2.26
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 simple
equation V = TQ where T is the frequency in electrical rad/sec. We have seen that the phase angle of
this 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 value
j(* + B/2)
, the angle (* + B/2) being the phase angle of V relative to the
of V is the complex number V = V e
reference axis (the d-axis). Both V and Q are sinusoidal quantities in time, and both are represented
in the diagram by their r.m.s. values, while the continual advance of their phase angles is represented
by the rotation of all the phasors at the angular velocity T rad/sec. During this rotation the phase
displacement 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 into
two parts makes the diagram clearer. The flux-linkage part of the diagram is usually omitted, but it can
be considered to have a physical reality, in that the flux actually rotates in space, at an angular velocity
of T elec. rad/s. The flux-linkages can therefore be interpreted as space vectors (if considered as
rotating 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 there
is a flux due to the magnets which links all the windings in turn, and gives rise to the flux-linkage Q1Md
in each phase, even when there is no current flowing. Corresponding to this flux-linkage is the “opencircuit” 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 phasor
diagram 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 (of
flux-linkages) both rotate synchronously in their respective coordinate systems, we tend to blur the
distinction 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 the
distinction between them.
Brushless permanent-magnet machines
Page 2.27
When current flows in the stator windings it creates an additional flux which is easier to analyse if we
first “resolve” the current into two components in the phasor diagram: Id along the d axis and Iq along
the q axis. In terms of the complex phasor I, this is written
I ' Id % j Iq
(2.59)
The flux-linkage produced by Id is LdId, where Ld is the d-axis synchronous inductance. It is in phase
with Id and induces a voltage XdId which is 90E ahead of Id (i.e., parallel to the q axis); then Xd = TLd
is the d-axis synchronous reactance. Likewise the current Iq produces a flux-linkage LqIq and a voltage
XqIq which is parallel to the negative d axis, with Xq = TLq the q-axis synchronous reactance. The total
voltage 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. Similarly
the 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 current
available from the drive. Neglecting losses, the electromagnetic torque is given by
T e ' m p ( Qd I q ! Qq I d )
(2.60)
where m is the number of phases, p is the number of pole-pairs, and Qd and Qq are the d- and q- axis
components 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 no
magnet flux but only the armature-reaction component LqIq. Thus
Qd ' Q1Md % L d I d ;
Qq ' L q I q .
(2.61)
If we substitute the expressions for Qd and Qq in eqn. (2.60), we get
T e ' m p [ Q1Md I q % I d I q ( L d ! L q ) ]
(2.62)
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 the
magnet 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 torque
and reluctance torque can be adjusted by changing the phase angle of the current (() as well as its
magnitude. Noting that ( is measured from the q-axis in the positive (CCW) direction, we can write
I d ' ! I sin ( ;
I q ' I cos (
(2.63)
and if we substitute these expressions in equation (2.60) we get
T e ' m p [ Q1Md I cos ( ! I 2 sin ( cos ( ( L d ! L q ) ]
(2.64)
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
( ' sin
!1
1
4
!
Q1Md
)Q
%
Q1Md
)Q
2
% 8
(2.65)
where )Q = (Ld ! Lq)I.
10
The above theory is called the two-axis theory, two-reaction theory, or dq-axis theory of the synchronous machine. Historically
it dates back to A. Blondel in the 1890's, but its modern exposition is generally attributed to R.H. Park along with others around
1920!1930, notably Doherty, Nickle, and W.V. Lyon. See Park RH, Two-reaction theory of synchronous machinery; I ! Generalized
method of analysis, Transactions IEEE, Vol. 48, pp. 716!730, July 1929. One of the most famous accounts of it is given by Charles
Concordia in Synchronous Machines, John Wiley, 1951.
SPEED’s Electric Motors
Page 2.28
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 seen
in 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 has
a certain value. If this is the rated r.m.s. current Im, the tip of the current phasor lies on a circle, centre
0, radius Im, Fig. 2.33. In practical terms this corresponds to the case where the drive is operating
under current limit and the current regulators have complete control of the current waveform.
Fig. 2.33 Current-limit circle and voltage-limit circle at the change-over speed; non-salient-pole motor with Xd = Xq.
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 must
therefore examine what happens when the drive is voltage-limited rather than current-limited. Imagine
that the drive is supplying maximum voltage Vm to each phase of the motor, such that the phase angle
between Vm and E is * (Fig. 2.32).11 From the phasor diagram
Id '
and
so that
11
Vq ! E
Xd
'
V m cos * ! E
Xd
;
Iq '
!V d
Xq
'
V m sin *
Xq
.
(2.66)
V m2 ' V d2 % V q2 ,
(2.67)
( X q I q )2 % ( E % X d I d )2 ' V m2 .
(2.68)
Vm is the fundamental harmonic component of the actual phase voltage.
Brushless permanent-magnet machines
Page 2.29
According to these equations, the tip of the voltage phasor lies on a circle of radius Vm, but the tip of
the 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 by
substituting eqns. (2.66) in eqn. (2.62) to eliminate Id and Iq. The result is
Te '
mp
E Vm
T
Xd
sin * %
V m2
1
2
Xd
!
1
Xq
sin 2 * .
(2.69)
Again we can differentiate this expression to find the phase angle * which maximises the torque. After
some simplification the result is
* ' cos!1
J2 % 8
! J ±
4
(2.70)
where
E / Vm
J '
1 ! Xd / Xq
.
(2.71)
If there is no saliency, the angle which gives maximum torque is * = 90E. The phase angle of the current
will vary in a more complex manner as * varies.
For machines with no saliency (such as surface-magnet motors), Xd = Xq and we can summarise the
guidelines for maximum torque as follows:
Low speed
Control current with ( = 0
High speed
Control voltage with * = 90E
TABLE 2.5
The change-over between “low speed” and “high speed” is the maximum speed at which the rated
current Im can be driven into the motor with ( = 0. For a surface-magnet (non-salient-pole) motor with
Xd = Xq the voltage-limit ellipse becomes a circle, and the change-over frequency can be determined
from its intersection Q with the current-limit circle, Fig. 2.33. At this speed the voltage-limit circle is
just large enough to intersect the current-limit circle at point Q, where Iq = Im and Id = 0. Therefore
( V m / X d )2 ' ( E / X d )2 % I m2
(2.72)
In this equation Xd and E are both proportional to speed or frequency T, but Im and Vm are fixed. If we
substitute E = TQ1Md and Xd = TLd we can re-arrange eqn. (2.72) to give the change-over frequency:
TQ '
Vm
Q1Md
2
% ( Ld Im )
rad/sec .
2
(2.73)
The subscript Q identifies this value of frequency as the change-over value, sometimes known as the
corner-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 the
optimum angle for maximum torque (( = 0 in a non-salient-pole motor). As the frequency increases the
radius of the voltage-limit circle decreases and the intersection with the rated-current circle moves
along the arc QD. To illustrate this, Fig. 2.34 shows the conditions at four different speeds, N1 < N2 (=
NQ) < N3 < N4.
SPEED’s Electric Motors
Page 2.30
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, which
means that this current can be driven into the motor with any phase angle. In fact the current could
be 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 be
driven 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 (. Eventually
a speed is reached at which the current Im can be driven only along the negative d axis, circle 4. All
the current is now used to suppress the flux (flux weakening), and none of it is available to produce
torque. The intersection is at point D and the torque is zero. To achieve this point the current
regulator 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
NQ
ND
' u !
1 ! u2
(2.74)
where u = TQ Q1Md/Vm = EQ/Vm, EQ being the value of E at the change-over speed. For a solution to
exist at a positive speed, we must have
1
< u < 1
(2.75)
2
For example, if u = 0.8, ND = 5 NQ, but if u = 0.9, ND = 2.155 NQ. Alternatively suppose that the motor
must 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 the
motor. Define the per-unit synchronous inductance as
Brushless permanent-magnet machines
x '
Ld Im
Q1Md
Page 2.31
.
(2.76)
The per-unit synchronous reactance is identical to x, and by equating CD to OP in Fig. 2.34 we get
ND
NQ
'
x2 % 1
x ! 1
.
(2.77)
Even with x = 0.3 (a high value for a surface-magnet motor), ND/NQ = 1.491, which is quite a narrow
speed range above the corner-point speed. For a motor which must maintain constant torque up to
3,000 rpm and just be able to reach 6,000 rpm ( ND/NQ = 2), eqn. (2.77) prescribes that x must be at least
2.347, which is unrealistically high. This helps to explain why in some instances, additional inductance
has 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 total
flux, 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, and
apply only to non-salient-pole (surface-magnet) motors, they help to resolve the question as to whether
phase advance is needed “to overcome the rising EMF” or “to compensate for inductance” as the speed
increases. The answer is both.
Fig. 2.35 Constant-torque loci (salient-pole motor)
A constant-torque locus can be superimposed on the circle diagram, as in Fig. 2.35, by writing the
torque equation as
mp
Te '
I dN I q ) X
(2.78)
T
where
) X ' Xd ! Xq
and
I dN ' I d %
E
)X
.
(2.79)
This is a rectangular hyperbola asymptotic to the negative d-axis and to a q-axis which is shifted to the
right 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.
SPEED’s Electric Motors
Page 2.32
The constant-torque loci in Fig. 2.35 are drawn for three torques T1 < T2 < T3. The middle one goes
through 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 phase
advance between Q and P, with the current maintained at the rated value Im. The additional torque
is reluctance torque, the second term in eqn. (2.64), which comes from the saliency. As ( increases, the
sin ( 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 the
per-unit synchronous reactance x for a non-salient-pole motor, it is a simple matter to extend this to
xd 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 as
defined earlier, eqn. (2.65) gives
( ' sin!1
1
4
!
u
)x
2
u
%
)x
% 8 .
(2.80)
For example, suppose u = 0.9 and xd = 0.5 and xq = 1.5. Then )x = !1.0 and
( ' sin!1
1
4
!
0.9
1.0
%
0.9
1.0
2
% 8
' 31.13E .
(2.81)
The phase advance can be used to achieve the same torque at a lower current, or the same torque at
a 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 magnitudes
and phase angles, I and ( or V and *; in other words, we have represented the voltage and current in
polar coordinates. It is equally straightforward to describe the controlled variation of current in terms
of 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,
I ' I d % j I q ' I e j $ ' I e j (( % 90E) ;
V ' V d % j V q ' V e j (* % 90E) .
2.9
(2.82)
HISTORY OF BRUSHLESS PM MOTOR DRIVES
The permanent-magnet synchronous machine was certainly known in the early 1950's.12 Although most
permanent-magnet machines at that time were generators, motors were also manufactured,13 and the
classical two-axis theory was used to analyze them. The motors were “line-start” motors supplied
directly from the AC mains, without electronics. In the early 1970's the discovery of high energy cobaltsamarium magnets gave new impetus to the development of permanent-magnet AC motors. Lowerenergy ferrite magnets were already used in DC brush-type motors and improvements in these magnets
also 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 “linestart” motors without shaft position sensing: therefore they were not self-synchronous in the sense
that 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 the
machine is physically similar to the AC brushless permanent-magnet machine, and in many cases
12
See 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.
13
See DD Hershberger, Design considerations of fractional horsepower size permanent-magnet motors and generators, AIEE
Transactions, pp. 581!584, June 1953.
Brushless permanent-magnet machines
Page 2.33
identical, the method of driving it is fundamentally different. The brushless DC or “electronically
commutated” motor, sometimes also known as the “squarewave” or “trapezoidal” motor, does not have
a rotating ampere-conductor distribution and since it does not have sine-distributed windings, phasor
analysis and dq-axis theory are not applicable to it. To a purist, the squarewave drive is strictly
appropriate only with surface-magnet motors which have no “saliency” (i.e. Ld = Lq) and no reluctance
torque. 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 engineers
envisaged the possibility of removing the rotor cage from the “line-start” AC motor and of feeding this
motor with sinewave currents phase-shifted to maximize the torque per ampere. An account of such
an 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 his
controller is reproduced in Fig. 2.36, and it is evident that it is an “I-( controller” in the sense that it
provides for the adjustment or control of the magnitude and phase of the current, as discussed in the
previous 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 the
current are controlled independently, one of them being substantially oriented to control the flux and
the other one being oriented to control the torque. In general this gives rise to a control block diagram
in which the separate d- and q-axis components are identifiable, as well as the means of controlling
them. A controller of this type is described by Leonhard [1983].15
Fig. 2.36
Lajoie-Mazenc’s I!( controller (1983)
14
Lajoie-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 earlier
description of this system.
15
Since about 1985 a good many more papers have been published specifically on field-oriented dq-axis control, with
developments in sensorless operation (i.e., the elimination of the encoder or resolver), and more sophisticated methods of
determining Id and Iq and of controlling the current regulators. See Jahns and Van Nocker [1990], Morimoto et al [1990!1994].
Page 2.34
SPEED’s Electric Motors
Fig. 2.37 dq controller described by Jahns et al.
Another example of a true dq controller is shown in Fig. 2.37.16 In the simplest case the d-axis
component Id can be held at zero ( fd = 0) while the torque would be varied by controlling Iq via fq. At
high 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 aligns
the d-axis with the magnet flux, and in non-salient-pole machines the q-axis is the natural axis for the
torque-producing component of current. This reference frame is the only one which separates the fluxcontrolling 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 of
reference: 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 a
torque/speed characteristic with no phase advance. The torque starts to fall off at about 1400 rpm, and
has fallen to zero when the speed reaches 2500 rpm. Also shown is the torque/speed characteristic with
a phase advance angle ( which increases linearly from 1200 rpm to 2400 rpm. The torque does not reach
zero. 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 no
phase advance, at 1800 rpm the torque is 0.7 Nm and the power is about 132 W and is less than the
power 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.
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 motor
is of the type shown in Figs. 2.24 and 2.31.
16
See, for example, Jahns TM, Kliman GB and Neumann TA, Interior permanent magnet motors for adjustable speed drives, IEEE
Transactions, Vol. IA-22, No. 4, pp. 738!747, July/August 1986.
Brushless permanent-magnet machines
Page 2.35
The principle of phase advance is equally applicable to the squarewave drive. But since squarewave
operation 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 current
can be expressed in terms of voltage vectors, as shown in
Fig. 2.39.17 This diagram can be explained with the help of
the connection diagrams in Fig. 2.40, which show the
polarities of the motor line terminals corresponding to the
states of the six transistors in the bridge circuit, Fig. 2.3.
Note that the numbering of the transistors in Fig. 2.3
reflects the order in which they switch on, as in the
squarewave drive; but in the sinewave drive it is normal to
have three transistors conducting instead of only two, and
the conduction period for each transistor is 180E instead of
120E. Such switching schemes are called “3-phase-on” and
“2-phase-on” respectively; or, for short, “3Q” and “2Q”.
Fig. 2.39 Voltage vectors
Fig. 2.40 Six-step connections
Each of the six vectors in Fig. 2.39 corresponds to one of the connections in Fig. 2.40. For example, Q612
means that transistors 6,1 and 2 are on, so that line A is connected to the positive terminal and lines
B 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 lines
A and B are connected to the positive terminal and line C to the negative terminal. In the motor, the
orientation of the stator ampere-conductor distribution (or “MMF vector”) advances 60E, corresponding
to the transition from the 1,0,0 to the 1,1,0 state. Over a full 360 electrical degrees of rotation the six
states 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 controller
can switch between states at a much higher frequency than 6 times per cycle in such a way as to make
the current track a sinusoidal reference signal whose amplitude and phase are determined by the
torque demand and the speed. The coding example on the following page describes a simple, intuitive
current regulator algorithm to achieve this. Only the four highlighted vectors are used during the base
segment.18 In the regulator algorithm, the currents (ik1 and ik2) in two lines are sensed at each step
of the time-stepping solution, and the decision is made as to which transistors to switch on or off. The
coding is as follows:
17
The term “switch control strategy” means much the same thing as “pulse-width modulation strategy”. There is a large
number of different PWM strategies and the subject has been extensively researched, especially in relation to the control of
field-oriented AC motor drives. The particular strategy described here is a simple intuitive one.
18
The “base segment” is a 60E interval during which PC-BDC computes the current waveforms recursively, reconstructing the
entire period from this segment by a substitution process once the solution has converged.
Page 2.36
SPEED’s Electric Motors
{Current regulation in Sixstep}
begin
iCR1 := ISP * Sin(ThRe);
iCR2 := ISP * Sin(ThRe - 2*pi/3);
If (k*j) mod kTs = 0 then
If ik1 > iCR1 then
if ik2 > iCR2 then
VoltageVector := Q456
else
VoltageVector := Q234
else
if ik2 > iCR2 then
VoltageVector := Q612
else
VoltageVector := Q123;
end;
{Set reference in line 1}
{Set reference in line 2}
{Check switching freq. }
The transistors do not always begin to conduct when they are switched on, because there may be a
freewheeling current in the antiparallel diode. At high speed the diode currents may be flowing all the
time so that although the transistors receive turn-on signals they may never conduct. Note that it is
not immediately obvious from the current waveforms whether, say, a positive current in phase 1 is
flowing in Q1 or D4, etc.
Although this description is written for a wye connection, the same control logic applies with a delta
connection. Also note that the 3Q strategy uses complementary switching of the upper and lower power
transistors 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 a
squarewave 2Q drive, normally two transistors are conducting (PERIOD B) and two terminal voltages and one current are
determinate, instead of the three terminal voltages. Therefore vd and vq are not determinate and the d,q equations cannot be
used. During commutation or freewheeling (PERIOD A), however, a diode connects the freewheeling line terminal to the
positive or negative rail and all three terminal voltages are known. Therefore PC-BDC switches from dq-axes in PERIOD A to
direct phase variables in PERIOD B. With two phases conducting, the inductance is the line-line inductance of two phases in
series. It varies sinusoidally with rotor position (assuming that the machine is sinewound), and it and its derivative with respect
to 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 used
throughout. Note that the actual phase self- and mutual inductances Lph and Mph are used when he dynamic simulation is
working in actual phase variables, while the synchronous inductances Ld and Lq. are used when the dynamic simulation is in
dq axes.
Fig. 2.41 summarizes these considerations. The solution is no faster in dq axes than in direct phase variables, because under
transient conditions the reference-frame transformations to vd, vq from va,vb,vc and to id,iq from ia,ib,ic (and their inverses) must
be performed every time-step. This neutralizes any saving in computation time which might be expected with the dq solution.
Fig. 2.41 Usage of dq equations and direct phase variables
Brushless permanent-magnet machines
Page 2.37
We have already seen that permanent-magnet brushless motors are often driven with squarewave
currents, and in the next section we will return to the subject of squarewave control and current
regulation. 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 this
mode 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 synchronism
with the rotation of the rotor. For 3-phase sinewave drives it is often the practice to have three
transistors conducting at any time, but with squarewave drives normally two transistors are
conducting: 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 in
series, during the interval from 30E to 90E in Fig. 2.7. The line currents and the states of the transistors
are identical for wye- and delta-connected motors, but the way in which the current divides between
the phases is different in the two cases. (A detailed analysis is given in Hendershot and Miller, [1994]).
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.
With both transistors Q1 and Q6 conducting, the voltage across the series combination of phases 1 and
2 is v12 = Vs. When Q1 switches off, v12 = 0, and the current freewheels through diode D4, as shown in
Fig. 2.42(b). Instead of switching off Q1, we could equally well switch off Q6; then v12 = 0 and the current
would freewheel through D3. If both transistors Q1 and Q6 are turned off, the current freewheels
through both D4 and D3, but now v12 = !Vs. This is summarized in Table 2.6.
19
The theory of space vectors relies only on the assumption of sinusoidally distributed windings, so that the voltages and
currents need not be sinusoidal. The theory of space vectors could be applied to the squarewave drive, but only if the motor
had sine-distributed windings — in which case it would be a sinewave motor fed with squarewave currents, not a squarewave
motor in the normal sense of the electronically commutated brushless DC motor.
20
The “dwell” or conduction period of the transistors in a 2Q drive can be shortened below 120E. It can also be extended beyond
120E, in which case there are periods with 3 transistors conducting. If the dwell is extended to 180E the 2Q scheme becomes a
3Q scheme all the time. All these possibilities are included in the PC-BDC program.
SPEED’s Electric Motors
Page 2.38
Q1
D4
Q6
D3
v12
1
0
1
0
Vs
0
1
1
0
0
1
0
0
1
0
0
1
0
1
!Vs
TABLE 2.6
TRUTH TABLE FOR SWITCHES IN FIG. 2.42
Fig. 2.43 Switching duty-cycle
If the transistors are switched at sufficiently high frequency, the motor inductance keeps the current
waveform smooth and the motor responds to the average applied voltage V12. Suppose the circuit is
switched between the states in Fig. 2.42(a) and 2.42(b) at high frequency, by chopping Q1 with a dutycycle d, where d is the ratio ton /ts with which Q1 is switched: see Fig. 2.43. The switching frequency is
fs = 1/ts. Then the average voltage V12 is given by
V12 '
V s × t ON % 0 × (t s ! t ON)
ts
'
t ON V s
ts
' d V s.
(2.83)
The effect of the chopping on the motor current can be determined by applying this voltage to the
equivalent circuit of phases 1 and 2 in series. The simplified circuit is shown in Fig. 2.44.
Fig. 2.44 Simplified circuit for chopping
Fig. 2.45 Average di/dt
We have
v1 ' e1 % R i1 % L p1 % M p2
v2 ' e2 % R i2 % L p2 % M p1
(2.84)
where p1 = di1/dt and p2 = di2/dt. R is the phase resistance, L is the phase self-inductance, and M is the
mutual inductance between phases. Since i3 = 0 during the 60E interval of interest, we have i2 = !i1 and
therefore p2 = !p1. Then if we write e12 = e1 ! e 2,
v12 ' v1 ! v2 ' e12 % 2 R i1 % 2 ( L ! M ) p1 .
(2.85)
Now if the circuit is switched at high frequency between Fig. 2.42(a) and Fig. 2.42(b), we can replace v12
by 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 constant
and equal to a set-point value Isp, as shown in Fig. 2.46.
Brushless permanent-magnet machines
Page 2.39
In this case P1 = 0 and from eqns. (2.83) and (2.85), the
required duty-cycle d is given by
d '
e12 % 2 R I sp
Vs
.
(2.86)
The maximum value of d is 1. When d is 1, the chopping
transistor (Q1) is on all through the 60E interval. Since
e12 (the line-line voltage eLL) is proportional to speed,
this condition occurs at a particular speed, called the
base speed Tb. Evidently
Tb '
V s ! 2 R I sp
kE
Fig. 2.46 Chopping to maintain constant current
.
(2.87)
The base speed is the maximum speed at which the setpoint current Isp can be forced into the motor. If Isp is
equal to the rated current, Tb is therefore also the
maximum 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 the
motor, and the torque decreases quickly to zero as the
speed rises. This is shown in Fig. 2.47, which also shows
the hyperbola of constant power Pmax passing through the
base-speed point. Sometimes this is called the "corner
point".
Fig. 2.47 Torque/speed curve
Commutation
At the end of each 60Einterval the current must switch from one pair of lines to another pair. In the
preceding 60E interval the current would have been flowing in lines BC, and at the end of that interval
it is “commutated” into lines AB.
Ideally, the current in line B remains constant while the current in line C switches to line A. In
practice, however, iC does not fall to zero instantaneously, neither does iA rise from zero to Isp
instantaneously. Rather, there is a brief interval when iC is falling and iA is rising, and there is current
in all three lines at once. This interval is called the freewheeling or commutation interval: Fig. 2.48.
Fig. 2.48 Commutation interval between line C and line A
The circuit can be analyzed during the commutation interval by writing the voltage equations for two
meshes: 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 Q5
has reached the end of its conduction period and Q1 is switched on. The current in line C freewheels
through 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 applies
to this condition, i.e., after the extinction of iC.
SPEED’s Electric Motors
Page 2.40
The voltage Vbld controlling the build-up of current in phases 1 and 2 (lines A and B) is the line-line
voltage across these two phases in series, which can be written
V bld ' v12 ' v1 & v2 ' v AB .
(2.88)
Similarly the voltage Vfwh ("fwh" = "freewheel") controlling the decay of current in phases 3 and 2 (lines
C and B) is the line-line voltage across these two phases in series, which can be written
V fwh ' v23 ' v2 & v3 ' v BC .
(2.89)
If Q1 is the controlling (chopping) transistor, the value of Vbld is determined by the state of Q1, which
is labelled sQ1 :
* V & 2V & 2R i
s
q
q 1
V bld ' *
&V q & R qi1 & V d
*
(s Q1'1)
(s Q1'0)
(2.90)
Note that each transistor is modelled by a resistance Rq and a voltage-drop Vq ; and each diode by a
voltage-drop Vd. These equations can be combined:
V bld ' s Q1 (V s & 2V q & 2R qi1) % (1 & s Q1) (&V q & R qi1 & V d ) .
(2.91)
The value of Vfwh is not affected by the state of Q1:
V fwh ' V d % V q % R q (&i2) .
(2.92)
If Q1 is chopped at a sufficiently high frequency then sQ1 can be set equal to the duty-cycle d, and Vbld can
be interpreted as the average voltage across the terminals AB, or V12. When we make sQ1 equal to d we
are using the principle of "state space averaging", to replace a binary value sQ1 by a real number d, as
suggested by eqn. (2.83). d is the effective value of sQ1 under the stated conditions, where the chopping
is 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-state
currents, the very same formulas for Vbld and Vfwh can be used in digital simulation, where sQ1 is the
actual 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 by
chopping Q1. The current i1 is essentially a DC value. This is possible only if d < 1. Evidently
i1 '
V bld & ( e1 & e2 )
2R
.
(2.93)
If we set i1 = Isp then d can be calculated as
d '
e1 & e2 % 2R I sp % V d % V q % R q I sp
V s & V q % V d & R q I sp
.
(2.94)
A simplified version of this arises if the voltage drops across the transistors and diodes are negligible
compared 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,
d '
e12 % 2 R I sp
Vs
(2.95)
which is exactly the same as eqn. (2.86). In 12V systems this approximation is often not admissible
because the volt-drop across power transistors and diodes can be appreciable in comparison with Vs.
Brushless permanent-magnet machines
Page 2.41
Conduction mode diagrams
Fig. 2.50 (case 1) shows the circuit diagram. For purposes of analysis one cycle is divided into six
segments and the interval from 30E to 90E is the “base segment” (see Fig. 2.7). The base segment starts
when Q5 switches off and Q1 switches on. Fig. 2.50 (case 2) shows the "build" loop (solid line) via lines
A 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 C
is still flowing: i.e., during the commutation interval shown in Fig. 2.48. Case (4) is where Q1 is
switched 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 diodes
are 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 these
cases the base segment is calculated twice, with a different switch control in the two segments. When
2
both passes are complete, the integrals of current and current in the lines, transistors, and diodes are
constructed as indicated by the shaded areas in Fig. 2.49.
Fig. 2.49 Accumulation of current and current
2
Accumulation of Ii dt and Ii 2 dt for mean and r.m.s. currents
During any 60E commutation interval three accumulations are made forIi dt and Ii2 dt:
AQn
ADn
AQo
ADo
incoming transistor (e.g. Q1)
diode that freewheels the current for the incoming transistor (e.g. D4 for Q1)
outgoing transistor
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 only
during the commutation interval. The currents in lines A and B are accumulated as ALA and ALB
respectively. Note that ALA = AQn + ADn, and ALB = AQo + ADo. The label “C60 Q1" in Fig. 2.49
refers to one of the standard switching strategies in PC-BDC, in which each transistor has control of
the chopping for 60E, and during the base segment it is Q1.
Page 2.42
SPEED’s Electric Motors
Fig. 2.50 (1) circuit diagram and (2) forward conduction mode in 3-phase squarewave drive
Fig. 2.51 Soft chopping with Q1 or Q6
Fig. 2.52 Hard chopping and over-running (rectifying)
Brushless permanent-magnet machines
2.12
Page 2.43
BACK-EMF SENSING WITH SQUAREWAVE DRIVE
Motors with 120E square-wave currents are often
operated with “back-EMF sensing” to eliminate the
need for a shaft position sensor. The schematic in
Fig. 2.53 produces a waveform of the detection
voltage e det, whose zero-crossings are used for
commutation. This voltage is obtained by switching
the detection circuit to the idle line every 60E.
The detection EMF e det is multiplexed from the
voltages e AM, e BM, and e CM according to the states of
the power transistors. The waveforms of the
current and EMF repeat every 60E (commutated into
different phases), so it is necessary to analyse only
one 60E period—the base segment described earlier.
At the beginning of this period Q5 has just switched
off, and i C continues to freewheel through D2,
forcing e det = e CM = !Vs/2. During the freewheeling
period (“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 the
Fig. 2.53 Back-EMF sensing
remainder of the conduction segment, “PERIOD B”,
starts. During PERIOD B, only two lines are conducting, A and B. There is only one mesh current, that
is, the one flowing through phase 1 and negatively through phase 2. The detection voltage is given by
edet '
where
Vs
2
! v1 % e3
v1 ' e1 % R ph i1 %
dR1
dt
(2.96)
.
(2.97)
The flux-linkage R1 includes a self- and a mutual term, and since i2 = !i1 this is
R1 ' ( L1 ! L12 ) i1 .
(2.98)
In a nonsalient pole motor, L1 = Lph and L12 = Mph, and both these inductances are constant so that
v1 ' e1 % R ph i1 % ( L ph ! M ph )
di1
dt
.
(2.99)
In a salient-pole motor, if it is a sinewound motor then
L1 ' LF % L g0 % L g2 cos 2 2
L12 ' MF !
1
2
(2.100)
L g0 % L g2 cos ( 2 2 ! 2 B/3 )
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 phase
1 during PERIOD B, then
dR1
dt
' L1B
di1
dt
% i1
dL1B
dt
' L1B
di1
dt
% i1 Te
dL1B
(2.101)
d2
where
L1B ' ( LF ! MF %
dL1B
d2
3
2
L g0 ) % L g2 3 cos ( 2 2 % B/6 )
' ! L g2 2 3 sin ( 2 2 % B/6 ) .
and
(2.102)
SPEED’s Electric Motors
Page 2.44
By substituting all these equations back into eqn. (2.96), we arrive at an equation for the detection
voltage e det during PERIOD B. However, we still need the differential voltage equation for the whole
circuit so that the current waveform i1 can be calculated. This is simple enough if we combine all the
inductances in the single conducting loop AB into one line-line inductance:
L LL ' L1 % L2 ! 2 L12
' 2 ( LF ! MF ) % 3 L g0 ! 3 L g2 sin ( 2 2 ! B/6 )
(2.103)
The derivative of LLL with respect to 2 is
dL LL
d2
' ! 6 L g2 cos ( 2 2 ! B/6 )
(2.104)
and the required voltage equation is
V bld ' e1 ! e2 % 2 R ph i1 % L LL
di1
dt
% Tei1
dL LL
d2
(2.105)
which can be rearranged for integration by Euler's method. The "build" voltage V bld is the voltage
applied 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 the
supply voltage Vs minus the voltage-drops in the power transistors. Eqn. (2.96) is valid only for
chopping strategies in which Q1 remains on during PERIOD B, which requires that any chopping must
be performed by Q6 during the base segment. It is straightforward to program the solution for i1 in
PERIOD 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 any
second-harmonic variation in the self- and mutual inductances of the phases. Other sources of
variation in the zero-crossings include even-harmonic distortion of the EMF, causing the waveform over
one half-pole to differ from the waveform over the second half-pole. This would happen if the magnets
were not centred, or if their magnetization varied over their width, or if the magnetization of magnets
was not uniform under all the poles. These effects can be checked by running the motor open-circuit
and 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 in
motors with one or two phases. In any phase winding the EMF e1 is an
alternating quantity in which the positive and negative half-cycles are
similar in shape.21 This being the case, equal impulses of torque in the
positive and negative half-cycles of EMF require current pulses of
opposite polarity and similar waveshape, as shown by the symmetrical
current squarewave i1. The torque produced by the interaction of e1 and
i1 is shown as T1. The EMF and the current both require a finite time to
change from positive to negative, and this results in a dip in the torque
waveform every half-cycle, i.e., at twice the fundamental frequency.
The most common means of driving a symmetrical alternating current
waveform such as i1 is a bridge circuit such as the one labelled “H- Fig. 2.54 Bifilar motor windings
bridge” in Fig. 2.55. This circuit has four transistors, two of which are
referenced to the negative supply rail and the other two to the positive rail. The upper transistors are
called “high-side” transistors and the lower ones “low-side” transistors. The gating and protection
circuits for the high-side devices are more complicated than for the low-side ones, and for this reason
the bifilar circuit is sometimes used. In this case the phase winding is split into two equal parallel
paths which are connected within the motor with opposite polarity, as shown in Fig. 2.54. The half21
i.e., there are no even harmonics in the EMF waveform unless they are deliberately introduced by means of geometric or other
imbalances or asymmetry in the magnetic circuit. Generally such imbalances are acceptable only in very small motors.
Brushless permanent-magnet machines
Page 2.45
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 current
i1b flows in the other path. In Fig. 2.54 the paths are shown as separate phases 1 and 2, but this is a
matter of convention; the bifilar motor in Fig. 2.54 is often termed “1-phase bifilar”. Complementary
switching is used with the transistors controlling the two paths or phases: normally each conducts for
half a cycle in any full cycle. When one switches off, the current ideally transfers immediately to the
other, but in practice the mutual coupling between the two paths is imperfect and the resulting leakage
inductance retains a fraction of the inductive energy, which must be dissipated. There is also an
overvoltage 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-powered
fans or blowers, provided that the motor can start satisfactorily. Sometimes a tapered airgap is used
to “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, with
windings displaced 90Eelec from those of the first phase, and currents 90E out of phase with those of the
first phase. The current, EMF and torque contribution of a second phase are shown in Fig. 2.55. The
drive 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 with
two orthogonal sets of windings is often termed “2-phase bifilar” but it can equally well be regarded
as 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 considerable
ripple because of the particular waveforms of the individual phase torques. This ripple can be reduced
by reducing the conduction angle from 180E to 90E in each half-cycle in each phase, which is easily
accomplished with the H-bridge full-wave circuit. In the bifilar circuit, however, with less than 180E
conduction the stored inductive energy in the winding is not transferred to the complementary
winding. Unipolar circuits usually have no means of returning it to the supply, being designed for low
component 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 wye
or delta connection of the windings), and in this case the 6-transistor bridge can be used with no
restriction on the current control, in the sense that positive and negative current and voltage can be
applied to the motor terminals by appropriate switching controls. The three-phase motor has better
SPEED’s Electric Motors
Page 2.46
overlap for covering the natural torque dips, as we can see in Fig. 2.7 where each line need conduct for
only 120E in each half-cycle, leaving 60E for the EMF waveform to change polarity. Because of the
control 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 Single-ended circuit with no
separate freewheeling path
Fig. 2.57 Single-ended circuit with Zener
diode and damping resistor
Fig. 2.56 shows the simplest form of unipolar circuit,22 in which the winding W is controlled by a single
transistor 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 supply
Vs when Q is switched on. When Q switches off, a reverse voltage v < 0 must be developed across the
winding in order to reduce the current to zero. The only source of the necessary reverse voltage is the
avalanche voltage of the transistor or diode, Vz.
It is possible to protect the transistor and diode by connecting a separate voltage-limiting circuit in
parallel with them, with a Zener diode Vz and a damping resistor Rd as shown in Fig. 2.57. To ensure
that v < 0 it is obviously necessary to have Vz + Rd × i > Vs ! Rs × i where e is the generated EMF and
Rs is the supply resistance. The peak forward voltage appearing across the transistor when it turns
off is Vz + Rd × i.
As we have already seen in Fig. 2.55, the EMF,
current and torque waveforms for a single-ended
circuit such as the one in Fig. 2.56 are of the form
shown in Fig. 2.58, which shows the additional
possibility of spurious conduction during the
negative half-cycles of EMF, if the sum of Vs and the
peak EMF epk exceeds the avalanche voltage Vz. This
uncontrolled current produces a pulse of negative
torque.
The circuit equations for the circuit in Fig. 2.57 are:
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)
where Vq is the forward voltage drop in the transistor during conduction and Rq is the transistor
resistance. 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 with
the winding so that the supply voltage does not appear in the freewheeling path. The suppression
voltage is developed across the damping resistor Rd together with a small additional forward voltage
drop 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.
22
Unipolar or “half-wave” circuits are also sometimes called “single-ended”.
Brushless permanent-magnet machines
Fig. 2.59 Unipolar circuit with
plain freewheeling diode
Fig. 2.60 Unipolar circuit with
Zener suppression diode
Page 2.47
Fig. 2.61 Current in bifilar circuit with
negative spike at commutation
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-cycles
of 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
OFF :
v ' !V d ! V z ! R d i
(2.108)
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 displaced
by 90E, or 4-phase motors with phases uniformly displaced by 45E, because it cannot deliver the
alternating (full-wave) currents required in these cases.
It is important to account for mutual coupling between phases during the freewheeling period after
each phase transistor turns off. The circuit equations for the bifilar motor can be taken in pairs, since
there is no coupling between the first pair (phases 1&3) and the second pair (phases 2&4) in a 4-phase
motor. With no loss of generality we can consider the 2-phase motor, labelling the phases 1&2.23 Thus
v t1 ' e1 % R ph i1 % L ph p1 % M ph p2
v t2 ' e2 % R ph i2 % M ph p1 % L ph p2
(2.109)
where p = di/dt and vt1 and vt2 are the terminal voltages of the phase windings, which depend on the
states of the transistors and diodes. These equations are rearranged in the canonical Euler form for
stepwise integration of p1 and p2 to give new currents i1 and i2, and in a simulation program the states
of 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 transistor
is 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
p1 '
( 1 ! k ) Vs ! Vz
L(1 ! k 2)
;
p2 '
( 1 ! k ) Vs % k Vz
L(1 ! k 2)
.
(2.110)
If k is nearly !1, these equations show that if Vz # 2Vs, p1 = p2 $ 0, so that commutation requires Vz > 2Vs.
2
For example if Vz = 3Vs, p1 = p2 = !Vs/L (1 ! k ), and since both of these are negative it is evident that
when one transistor switches off, a negative current is induced in the complementary phase. This
current appears as a spike which disappears and is followed by the positive current pulse after a few
degrees of rotation, Fig. 2.61. If chopping is employed, the negative spike current recurs every time a
transistor is switched off, and the inductive energy in the leakage inductance is dissipated. To avoid
chopping, phase advance can be used to control the torque, (also at the expense of efficiency). High
voltage stress on the transistor (>Vz) is inevitable with these unipolar circuits during turn-off.
23
This is the one that is sometimes termed “single-phase bifilar”!
Page 2.48
2.14
SPEED’s Electric Motors
WINDINGS
Fig. 2.62 Definition of some terms used in motor windings
Fig. 2.62 shows some of the terms used in connection with motor windings. Although the subject of
windings is vast, brushless permanent-magnet motors basically use two types of winding: the
concentrated winding for squarewave drive and the distributed winding for sinewave drive. Because
the number of slots/pole/phase is usually quite small, the practical differences between these windings
are 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 are
said to be “concentric” because of the way in which the outer coil embraces the inner one. The outer
coil in each pole-group in Fig. 2.62 has a span of 5 slots, and the inner one a span of 3 slots.24
24
The terms “coil span” and “coil pitch” are synonymous. They are usually measured in slot-pitches. The term “throw” is also
often used with the same meaning. However, in some factories “throw” is one less than “span” : a coil that is said to be wound
in slots “1 and 6” has a “throw” of 6 and a span of 5. In SPEED software, however, “throw” means the same as “span”.
Brushless permanent-magnet machines
Page 2.49
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 coilsides constitute what is known as a phase belt. In this winding the phase belt spans 4 slot-pitches, and
since there are 24 slots covering 4 poles, this is 4/6 or 2/3 of a pole-pitch or 120 electrical degrees. In a
perfectly concentrated winding the phase belt would have a span of zero: in other words, all the
coilsides 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 span
of 6 slot-pitches (“full-pitch” coils). In that case only 4 slots would be occupied by any phase, and half
the slots in the machine would be empty. As it is, every slot holds two coil-sides, one from each of two
different 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 a
significant influence on the winding inductance and the mutual inductance between phases. A fullpitch concentrated winding produces an EMF time-waveform on open-circuit that is a replica of the
spatial 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 equivalent
distribution of full-pitch coils. The phase-shifts between the coils can be cleverly combined to eliminate
selected harmonics from the resulting phase EMF waveform. For example, windings in which all the
coils 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 shows
the reference axes of phases a, b, and c together with the rotor reference axes d and q. These axes are
consistent with the equations for variation of phase inductances with rotor position in §2.6. The phasor
diagram is reproduced in Fig. 2.63 for a normal motoring condition.
Fig. 2.63 Phasor diagram for salient-pole brushless PM machine
25
Such windings are suitable for delta connection since there will be no zero sequence EMF to drive current around the delta.
SPEED’s Electric Motors
Page 2.50
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. fundamental
flux-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 is
measured positive if I leads E. In Fig. 2.63, ( < 0. The angle $ is the angle between the current phasor
and the d-axis, such that $ = B/2 + (. Thus
I d ' I cos $ ' ! I sin (
and
I q ' I sin $ ' I cos ( .
(2.111)
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 magnetizing
direction, and Id = I with Iq = 0. When $ = 180E, Id = !I with Iq = 0, and the stator MMF is aligned with
the negative d-axis in the demagnetizing direction. When $ = 90E and ( = 0, the current is "in the qaxis": Iq = I with Id = 0. In surface-magnet sinewave motors at low speed, the current is normally
oriented in the q-axis with ( = 0.
The phasor voltage V = Vd + jVq is the airgap voltage produced by the r.m.s. fundamental airgap fluxQ1 = Q1d + jQ1q. The phasor diagram is completed by the equation
linkage Q1 under load: thus V = jTQ
V ' E % j X d I d % j X q I q % R ph I
(2.112)
where Id is written as the phasor Id = Id + j0 and Iq is written as the phasor Iq = 0 + jIq. Ra is the phase
resistance. In this equation, both Xd and Xq should incorporate the slot, end-turn, and differential
leakage components (and any skew-leakage component) in addition to the main airgap component
associated 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 fundamental
flux-linkage per phase Q1 and the phase current I using the equation
T ' m p [ Q1d I q ! Q1q I d ]
(2.113)
where p is the number of pole-pairs and m is the number of phases, and Q1d and Q1q are the components
of Q1 resolved along the d- and q-axes. This is the average torque produced by the fundamental
harmonic components of the airgap flux and the stator MMF distribution. It does not include cogging
torque 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-designed
motor 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 components
contributed by the magnet and the d-axis armature current; neither does it even require a knowledge
of 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 a
current-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 of
j(B/2 + ()
, from ( = !90E to +90E (i.e., $ ranges from 0 to 180E). For
values of I and (, where I = Id + jIq = I e
each solution the airgap flux-density waveform is Fourier-analyzed to give B1d and B1q, where B1d is
the peak value of the d-axis component and B1q is the peak value of the q-axis component. Then Qd1 and
Qq1 are obtained using the same form as eqn. (2.57), i.e.
Q1d '
k w1 T ph B1d D L stk
2
p
;
Q1q '
k w1 T ph B1q D L stk
2
p
(2.114)
The leakage components IdLF and IqLF should be added to Q1d and Q1q respectively. If the leakage
reactance is saturable, the flux-MMF diagram technique should be used instead (see below). These
values are substituted in eqn. (2.113) to produce a set of torque curves as illustrated in Fig. 2.64.
Brushless permanent-magnet machines
Page 2.51
The values of currents in the individual slots
are obtained from the conductor location
vector (CLV), that is, an array C which has m
columns, one for each phase, and n rows,
where n is the number of slots. The value of
th
element is the number of
the (j,k)
th
th
conductors belonging to the j phase in the k
slot. C[j,k] is signed according to the direction
th
of the conductors. Then for the k slot the
total current is
I k ' C [a,k] ia % C [b,k] ib % C [c,k] ic . (2.115)
Instantaneous phase currents ia, ib and ic are
obtained from the phasor values by Park’s
transformation:
Fig. 2.64 Torque vs. torque angle (, for various currents
ia ' I d cos 2 ! I q sin 2
ib ' I d cos (2 ! 2 B / 3) ! I q sin (2 ! 2 B / 3)
ic ' I d cos (2 % 2 B / 3) ! I q sin (2 % 2 B / 3)
(2.116)
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
ia ' I d
ib ' I d cos (!2 B / 3) ! I q sin (!2 B / 3) ' (!I d %
3 I q) / 2
(2.117)
ic ' I d cos (2 B / 3) ! I q sin (2 B / 3) ' (!I d ! 3 I q) / 2 .
Substituting from eqn. (2.111), with $ = ( + 90E the angle between the current and the d-axis,
ia ' I sp cos $
ib ' I sp cos ($ ! 2 B / 3)
ic ' I sp cos ($ % 2 B / 3)
(2.118)
where Isp = /2* I * is the setpoint current of the current-regulator. In PC-BDC the sign conventions are
such that the main magnet pole analyzed in the finite-element model is a south pole. Fig. 2.65 shows
the directions of the current for three different values of (: ( = 0 is the normal "q-axis" control, ( = !90E
is "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 be
sinusoidal, both I and ( are controlled and only one finite-element computation is needed at that point.
Fig. 2.65
Direction of currents and stator MMF around a S magnet for three values of (.
Page 2.52
(b)
SPEED’s Electric Motors
Flux-MMF diagram technique
Method (a) is fast because it uses a fixed mesh and
simple post-processing of the finite-element field
solution, but it fails to give adequate results if the
machine is not sinewound, or if the airgap fluxdistribution is not sinusoidal, or if there is significant
saturation.
Moreover. under saturated conditions the values of E
and X d become indeterminate, because the
superposition 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, one
Fig. 2.66 i-psi loop
attributed 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 of
sinewound machines tacitly assumes that there is no Ed produced by "magnet flux" in the q-axis. But
under 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 a
sequence of finite-element field solutions at intervals of rotation throughout an electrical cycle. The
current (in all phases) set to the correct value for each rotor position, and the i-R loop is plotted, that
is, 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 "energyconversion loop" is elliptical, its major axis being oriented at an angle that depends on the phase angle
between the current and the d-axis. The average electromagnetic torque is proportional to the area W
of the i-R loop: thus
Te '
m
2B
× W.
(2.119)
The phase flux-linkage R is extracted from the finite-element solution. The flux through a single coil
is N = Lstk (Ar ! Ag) [Wb], where Ar is the vector potential at the "return" coilside and Ag is the vector
potential at the "go" coilside, and Lstk is the stack length. The flux-linkage is R = Tc N [V-s], where Tc
is the number of turns in the coil.
If required, the d! and q-axis components of flux-linkage can be approximated using the
transformation, Rd = 2 Ra/3 ! (Rb + Rc)/3 and Rq = (Rb ! Rc)//3. However, this transformation is not
of interest unless the winding is sinewound.
The flux-MMF diagram technique works for any
machine with any drive. It is not restricted to
sinewound machines or sinewave drives, and works
equally well with machines having non-sinusoidal EMF
waveforms and squarewave drive, regardless of the
level of saturation.
The main limitation of the method is that the phase
currents must be known at every step of rotation,
because the finite-element computation is "currentdriven".
Fig. 2.67 shows an example of the i-R loop computed for
a "squarewave" brushless DC motor, with the current
regulated to follow a 120E squarewave by chopping.
Fig. 2.67 i-psi loop
Brushless permanent-magnet machines
2.16
Page 2.53
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 phase
currents i1, i2,... and the EMF’S e1, e2,... generated by the rotation of the magnets. Ideally the energy
conversion is described by the equation
T e Tm ' e1 i1 % e2 i2 % e3 i3 % ... % e m im
(2.120)
where Tm is the speed in rad/sec and m is the number of phases. It is important to recognize Te as the
instantaneous torque and not the average torque. The average torque (averaged over one or more
revolutions) is the main output parameter,
T avg '
2B
0
T e ( 2 ) d2
(2.121)
and in most cases Tavg is the parameter that is quoted, with never any mention of the instantaneous
torque. 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 gave
a second thought to the existence of torque ripple, and the whole subject of instantaneous torque and
the 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 torque
ripple, 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 smooth
torque, 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 though
this equation includes only the “alignment” torque of the permanent magnets and does not include any
reluctance torque.26 In AC machines the most classical instance of “constant torque” is the threephase machine wherein the EMF’s and currents are balanced polyphase quantities:
e1 ' e pk cos T t
e2 ' e pk cos ( T t ! 2 B/3 )
e3 ' e pk cos ( T t % 2 B/3 )
i1 ' ipk cos ( T t ! ( )
i1 ' ipk cos ( T t ! 2 B/3 ! ( )
i1 ' ipk cos ( T t % 2 B/3 ! ( )
(2.122)
Substituting in eqn. (2.120), and recognizing the r.m.s. values E = epk//2 and I = ipk//2, we get
Te ' 3
EI
Tm
cos (
(2.123)
which is constant. The contributions of the individual phases, however, are not constant: for example,
e1 i1 ' e pk ipk cos T t cos ( T t ! ( )
'
26
e pk ipk
2
[ cos ( % cos (2 T t ! () ]
Eqn. (2.120) therefore applies to surface-magnet motors but is incomplete for embedded-magnet motors.
(2.124)
SPEED’s Electric Motors
Page 2.54
which has an average value (EI/2) cos ( with a double-frequency oscillatory component whose
amplitude is greater than the average value unless ( = 0 (i.e., unless e1 and i1 are in phase). This
enormous torque ripple is the reason why pure single-phase AC motors are used in only a limited
number of applications. It also explains the use of auxiliary capacitors with split-phase induction
motors operating from a single-phase supply. With two phases the phase displacement between phases
is ideally B/2 or 90E instead of the 2B/3 or 120E used in three-phase machines, and if
e1 ' e pk cos T t
e2 ' e pk cos ( T t ! B/2 )
i1 ' ipk cos ( T t ! ( )
i1 ' ipk cos ( T t ! B/2 ! ( )
(2.125)
cos ( .
(2.126)
then the sum e1i1 + e2i2 yields
Te ' 2
EI
Tm
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, which
is 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 is
accomplished in the 3-phase brushless motor. The angle of rotation over which one of the product
terms (such as T1) can be kept constant is limited by the magnet arc and the winding distribution, and
although the current can be held constant for very nearly 180E of rotation, it is not possible to maintain
a flat-topped phase EMF waveform wider than about 175E or so. For this reason it is impossible to
achieve constant torque in a two-phase brushless motor, and it is better to settle for three phases and
120E 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 current
waveform may depart from the ideal sinusoid or 120E squarewave because of chopping (PWM) and
commutation in the drive. Moreover, at high speeds the ideal sine or 120E squarewave current cannot
be achieved because of the combination of series inductance and the growth in the EMF relative to the
available supply voltage.
In sinewound motors with sinusoidal supply currents the torque equation has already been expressed
in various forms such as eqns. (2.60), (2.62), (2.69), (2.78), and (2.113): all of these equations refer to the
average torque. Another variant of these equations is
Te '
m
Tm
[ E q I q % (X d ! X q) I d I q]
(2.127)
The reluctance torque is identifiable as the second term in this equation, and this is the average value
since Eq, Iq and Id are all r.m.s. quantities. However, the instantaneous reluctance torque Trel can be
determined if the instantaneous currents are used instead of the r.m.s. values:
T rel '
m
Tm
× 2 B f [ L d ! L q ] id iq '
mp
2
[ (L d ! L q) i d i q]
(2.128)
where Tm = 2B f /p and id and iq are the instantaneous d- and q-axis currents, obtained from the
instantaneous phase currents by Park's transformation:
id '
2
3
iq ' !
[ ia cos 2 % ib cos ( 2 ! 2 B / 3 ) % ic cos ( 2 % 2 B / 3 )] ;
2
3
[ ia sin 2 % ib sin ( 2 ! 2 B / 3 ) % ic sin ( 2 % 2 B / 3 )] .
(2.129)
and 2 is the angle between the d-axis and the axis of phase a. The inverse Park transformation is given
by eqn. (2.116). This is an interesting illustration of the fact that Park’s transformation is valid under
transient conditions, or with nonsinusoidal currents, provided only that the windings are sinewound.
Brushless permanent-magnet machines
2.17
Page 2.55
COGGING TORQUE CALCULATIONS USING FINITE-ELEMENT ANALYSIS
Fig. 2.68 Representation of magnet by equivalent air-cored coil
A magnet (or any element thereof) can be represented by an equivalent distribution of linear surface
current density K = M × n, together with a volume distribution of current density J = curl M, where
M 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 of
the 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
Wf '
1
2
m
BH dV
(2.130)
in all regions of the motor, and adding the resulting integrals together. In the air regions, B = µ0H. In
the iron regions, B and H are related by the B/H characteristic of the iron. But within the volume
occupied by the magnet,
B ' B r % µrec µ0 H
(2.131)
where Br is the remanence and µrec is the relative recoil
permeability. It is not obvious that we can substitute
this equation in eqn. (2.130) and get the correct value for
the energy in the volume occupied by the magnet. An
energy analysis is required, and this must use the
principle of virtual work, because a change in stored
energy can be brought about only by an exchange of
mechanical work when the motor is on open-circuit.
Consider a small displacement )2 such that the
operating point of the magnet moves from X to Y, as a
result of the change in the permeance of the magnetic
circuit as the magnet moves past the slot openings. The
magnet flux changes by )Mm, and if the change takes
place in a time )t, an induced voltage e = )Mm/)t
appears at the terminals of the equivalent coil in Fig.
2.68, if we assume that the coil has one turn. The power
at these terminals is just p = e i and the energy
exchange during the time interval )t is e i)t = i )Mm.
But i = Fc, the coercive MMF of the magnet. Therefore
Fig. 2.69 Result of a small perturbation in magnet
the energy exchange with the fictitious excitation coil
operating point from X to Y.
is Fc )Mm. Because the coil and current source are
fictitious, there is no external electrical exchange of
energy 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 space
occupied by the magnet, and a further change )Wgap in the remainder of the magnetic circuit, (i.e., the
airgap and the iron). Let Wmag = ½FmagMmag and Wgap = ½FgapMgap, where Fmag and Fgap are the
respective MMF drops across these sections of the magnetic circuit, and Mmag = Mgap = Mm, the total flux.
SPEED’s Electric Motors
Page 2.56
Then
) W mag '
) W gap '
The sum of these is
) W mag % ) W gap '
1
2
1
2
1
2
( Fmag @ ) Mmag % )Fmag @ Mmag )
( Fgap @ ) Mgap % )Fgap @ Mgap )
(Fmag % Fgap) ) Mm %
1
2
( ) Fmag % ) Fgap ) Mm
(2.132)
(2.133)
But )Fmag + )Fgap = )Fc = 0, since Fc is by definition constant and Fmag + Fgap = Fc. Therefore
) W mag % ) W gap '
1
2
F c ) Mm
(2.134)
We can now complete the energy audit using the law of energy conservation. If the movement XY is
from a position of higher permeance to one of lower permeance, then mechanical work is supplied
to the motor via the shaft. The stored magnetic energies increase while the flux Mm decreases.
Therefore
) W m ' ) W mag % ) W gap ! Fc ) Mm ' !
1
2
F c ) Mm
(2.135)
In order to move the rotor and cause the flux change )Mm, mechanical work ) Wm must have been
exerted via a torque at the shaft. This torque is the cogging torque Tcog, and if we regard torque exerted
by the motor as positive, then Tcog )2 = ! )Wm and in the limit as )2 tends towards zero,
T cog '
1
2
Fc
d Mm
d2
(2.136)
The shaded triangle )Wm in Fig. 2.69 is equal to
½Fc )Mm, and therefore represents the work of cogging
torque during the rotation from X to Y. Fig. 2.70 shows
that the area )Wm is also equal to the triangle OXY.27
The change of energy )Wmag in the volume Vm
occupied by the magnet is suggested by classical theory
as
) W mag '
mmmV
m
1
2
m
H dB d V m
(2.137)
This is represented by the shaded area in Fig. 2.71. The
difference between )Wm and )Wmag is the energy
exchanged 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, even
though the expression ½Fc )Mm contains only magnet
parameters. Indeed B and H do not need to be known in
Fig. 2.70
the airgap and iron regions in order to calculate )Wm
(and from it the cogging torque). All the necessary
information about the mechanical energy exchange is included in the expression ½Fc )Mm. This
expression also applies to magnetically linear variable-reluctance devices that are supplied from a
constant-current source. The factor ½ reflects a partition of energy between mechanical energy
conversion and field storage. Evidently permanent magnets have additional storage capability beyond
the integral in eqn. (2.137). It is associated with the internal current source in Fig. 2.68 and it is, of
course, what distinguishes them as “permanent”.
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.
Brushless permanent-magnet machines
Fig. 2.71
Page 2.57
Fig. 2.72
The equivalent-coil model helps remove doubt about the absolute value of the stored energy in the
magnet. Consider a change in the external circuit permeance that takes the operating point up to the
point R, where B = Br, the remanence, and H = 0 in the magnet. The mechanical energy exchange along
XR is shown as Wmag in Fig. 2.72. For the purposes of calculating cogging torque, with the motor
windings open-circuited, the magnet energy at R can be taken to be zero, because any displacement
from R requires mechanical work to be provided. Of course, in a motor with a finite airgap, the magnet
never 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 energy
exchange with the air and iron regions is Wm ! Wmag , which is shown as Wgap in Fig. 2.72. Fig. 2.72 also
shows the separate components of the coercive MMF Fc : that is, Fgap across the air + iron part of the
magnetic circuit and Fmag across the magnet itself. When the magnet is working near the remanence
point 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 total
stored energy is large: an external torque must have been applied to the shaft to make this change, and
the energy stored in the magnet is larger than that stored in the gap. The gap energy is low because
the flux-density is reduced to a low level, as a result of the reduction in the circuit permeance. The
potential energy given to the magnet will be recovered if the shaft is allowed to turn back to its original
position. During continuous rotation the cogging torque is generated as the operating point cycles
between points such as X and Y. Point X is a point of maximum permeance and point Y is a point of
minimum permeance. When the rotation is from X to Y, the stored energy is increasing and the cogging
torque is a retarding torque in the direction of rotation. At Y the rotor continues in the same physical
direction but the operating point moves towards X, which corresponds to a new maximum permeance
position. The angle of rotation between two X’s depends on the number of slots and poles; in an
integral- 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 an
increment in the total stored field energy, or as an equal and opposite increment !)Wm in the
coenergy. The total coenergy is represented by triangle OCX and its components WNgap and WNmag are
shown in Fig. 2.72. Since there is no external electrical energy exchange, it does not matter which one
is used, and the cogging torque is the total derivative dW m/d2. The PM motor on open-circuit differs
from the electrically excited actuator or motor, where it is necessary to use partial derivatives
MWN(i, 2)/M2 or !MWf(M , 2 )/M2 and keep to the rule of “constant MMF” or “constant flux” respectively. In
the PM motor on open-circuit there is no distinction between these partial derivatives, except as to
their sign.
Page 2.58
SPEED’s Electric Motors
Even when the magnetic circuit is linear, the coenergy and energy are in general unequal, i.e. OXR is
not generally equal to OCX. This is a further difference between the PM motor on open-circuit and the
normal electrically-excited device.
The sign of the cogging torque is not particularly important, because it is purely oscillatory when the
rotor rotates continuously, and since it cannot sustain any continuous energy conversion, its average
value must be zero. Suffice it to say that the cogging torque will be such that mechanical work must
be supplied to the motor when the magnetic stored energy is increasing, and mechanical work will be
supplied 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 position
calculate the change in the total stored energy. By evaluating the triangle OCX for each magnet
element, we are automatically including the coenergy stored in the “airgap” (i.e., in the rest of the
magnet circuit), and therefore it is not necessary to evaluate Wgap separately by integrating BH/2 in
the air and iron regions. The same is true if we use triangle OXR to evaluate the total stored field
energy due to the current magnet element. The required process is as follows: for each magnet
element, per metre of stack length,
Coenergy
Triangle OCX
½Hc C B × element area
Stored field energy
Triangle OXR
½ Br C H × element area
TABLE 2.7
Fig. 2.73
The coenergy and energy calculations are shown in Fig. 2.73. Since Br and Hc are constants, the areas
require the integration of a linear function over each element. The integral on the right of Fig. 2.73 is
2
2
the coenergy evaluation that would be done in air regions using ½B H = ½B /µ0 = ½µ0H , and it is
quadratic. 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 coenergy
integral of Fig. 2.73 or Table 2.7 has been summed over the entire magnet region, it represents the
coenergy of the entire machine in the same way that OCX represents this coenergy for a lumped magnet
in Fig. 2.70. When the rotor moves from position X to position Y, the change in this coenergy sum will
be of the form OCX ! OCY, and this will give the average torque through the interval )2 = XY. On the
other hand, if we use the alternative magnetic energy integral in Fig. 2.73, the total energy change from
X 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 and
fitted with a cubic spline; then the differentiation to get the cogging torque will be smoother. This
method helps avoid the discretization noise which troubles the Maxwell stress method.
Brushless permanent-magnet machines
Deodhar’s method: A simple method for evaluating the
triangles OXY treats the entire magnet as a single element.
The magnet flux Mm is evaluated the from the difference of
vector potential at the magnet edges (multiplied by the stack
length), and used together with the magnet MMF obtained from
the relation F = Lm/µrecµ0 × (Mm /Am ! Br) to plot the triangle
OXY directly, moving from position to position and calculating
the torque at each step. The vector potentials can be evaluated
at points shown in Fig. 2.74 on the edges of the magnets, using
the expression Mm = (A1 ! A2) × Lstk, where Lstk is the stack
length or magnet length in the axial direction. If the centreline
of the magnet is a line of symmetry, generally A = 0 there and
A2 = ! A1, so that Mm = 2A1 Lstk. An alternative is shown in Fig.
2.75, where Mm is approximately equal to 2(AQ ! AZ) × Lstk. In
many cases AZ will be zero.
This method could be made more efficient by recognizing that
OXY = ½Fc )Mm where )Mm = MY!MX. Alternatively, OXY =
½Mr (FY ! FX) where FY = HY × Lm and FX = HX × Lm, and HX and
HY are the values of H in the magnet at the positions X and Y
respectively, evaluated by H = (B ! Br)/µrec µ0. Together with
the integrals in Table 2.7, evidently there are several simple
processes for determining the cogging torque once the FE
solution is available.
Fig. 2.76
Page 2.59
Fig. 2.74 Flux calculation
Fig. 2.75
Variation of airgap flux-density
PC-BDC’s approximate method: The rotor is stepped round in intervals of 1Eelec. At each position
the airgap flux-density distribution is calculated including the effect of slotting. In the interests of rapid
calculation the slot modulation is very crude. It is estimated using an effective airgap that varies as
shown in Fig. 2.76 as the point P sweeps across the slot opening. Mm is evaluated at each rotor position
by means of a ratio function Lm/(Lm + gN), where gN is the effective airgap at each position of the point
P. Then the total airgap flux is evaluated by integrating the airgap flux over the pole pitch, and the
rotor is stepped to the next position.
Fractional-slot motors can be accommodated by evaluating the cogging torque for the N and S pole
magnets independently, and then adding them together. Both methods allow for skew by summing the
effects taken at several axial positions.
SPEED’s Electric Motors
Page 2.60
2.18
GETTING INDUCTANCE FROM FINITE-ELEMENT CALCULATIONS
Plain inductance (phase self inductance or mutual inductance): Most finite-element programs
claim to be able to calculate stored field energy W, (and coenergy), suggesting the extraction of
inductance from the formula W = ½LI 2, if the current I is known. However, this method is ambiguous
if there is any saturation, because the formula W = ½LI 2 relies on the assumption that L (the ratio
between flux-linkage and current) is constant. Moreover, as we have seen in §2.17, the stored energy
in permanent magnets introduces another ambiguity.
A more rigorous approach is to use the vector potential A directly,
with the equation
N '
m
A @ dl
(2.138)
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) is
given by N = Ac1 ! Ac2, where Ac1 and Ac2 are the vector potential values
at the coilside positions, Fig. 2.77. If there is a complete winding with
coilsides in different locations, the method can be extended by
summing the fluxes with appropriate polarities according to the
direction of the conductors. If the coils are in series carrying current Fig. 2.77 Calculation of flux-linkage
from vector potential
I, and all have Nc turns, the inductance is L = Nc EN /I. This
inductance 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 important
to know the incremental inductance dR/dI when the machine is fully fluxed with full current, because
this is the inductance presented to the current regulator. Using the A method described above, the
incremental inductance is given by )R/(I1 ! I2), where )R is the difference in computed flux-linkage at
two slightly different current levels I1, I2.
Synchronous inductance: The extraction of the synchronous inductances Ld and Lq has been
described earlier. The vector-potential method calculates total flux linkage, and therefore it is not
possible 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-element
calculation with two phases connected in series line-line, and determine the total line-line inductance
LLL for the two rotor positions that give the maximum and minimum values LLL(max) and LLL(min). Then
if Lq > Ld, Lq = LLL(max)/2 and Ld = LLL(min)/2. End-winding inductances (not calculated in 2-D FE) must
be added to these values of Ld and Lq. Since these are generally small, approximate methods of
estimation 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-phase
sinewave motor is
T/I '
3
2
2
B r1 L stk B1Md k w1 T ph
2
Nm/A
(2.139)
where r1 is the stator bore radius and B1Md is the peak fundamental flux-density produced by the
magnet 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 peak
ampere of phase current is
T/i ' 4 r1 B Mg L stk T ph Nm/A
where BMg is the peak flux-density produced by the magnet in the airgap.
(2.140)
Brushless permanent-magnet machines
Page 2.61
The r.m.s. phase current I is derived from the 120E squarewave as:
I ' i
2
(2.141)
3
where i is the peak or flat-top value of the phase current. Therefore the torque per r.m.s. ampere of
phase current is
3
T/I ' 4
2
r1 B Mg L stk T ph Nm/A
(2.142)
With these expressions the ratio of torque per ampere in the two motors can be compared. With equal
r.m.s. phase currents, the torque of the squarewave motor exceeds that of the sinewave motor by the
factor
4
3
2
3
2
× 2 ×
B
' 1.47
(2.143)
2
With equal peak currents, the factor is 1.27. The squarewave motor appears to have a significantly
higher torque per ampere. However, the comparison assumes equal peak magnet flux-densities in the
airgap, which is likely to require significantly more magnet and a thicker stator yoke in the
squarewave motor. To rectify this imbalance it is perhaps better to compare the motors on the basis
of the same flux per pole. For the same peak flux-density the flux per pole of the square-wave motor
exceeds that of the sinewave motor by the factor B/2. Now with equal r.m.s. phase currents the torque
ratio 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, and
magnet, 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 the
drive 'rating' can be made in terms of the total kVA rating of its main switches, per kW of power fed to
the motor. The relevant parameters are defined as follows. With respect to the r.m.s. current in each
switch, if q is the phase number,
r.m.s. kVA/kW ' 2q × I s × V s
(2.144)
where Is is the r.m.s. current in each switch and Vs is the peak voltage across each switch. For the
drives normally used with brushless DC motors the peak device voltage is nominally equal to the DC
supply voltage, because each switch must block this voltage while the other one in the same phaseleg
is conducting. Obviously there must be a margin for voltage spikes caused by stray inductance and
reverse-recovery of diodes. While these effects are not small, they are parasitic and not fundamental
to 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,
peak kVA/kW ' 2q × iˆs × V s
(2.145)
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 half
sinewave 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 is
star-connected). The peak device current is equal to the peak phase current. The peak line-line voltage
of the motor is approximately equal to V.28 Thus
v̂LL ' V '
28
The ratio actually depends on the depth of modulation.
3 v̂ ph '
6V ph
(2.146)
Page 2.62
SPEED’s Electric Motors
We can now write
r.m.s. switch VA ' 6V
I
2
;
peak switch VA ' 6Viˆ ;
(2.147)
drive power output ' 3V phI ' 3
V
6
I
From the appropriate ratios between these quantities, the r.m.s. switch kVA/kW is 3.5 and the peak
kVA/kW is 6.9.
For the squarewave motor the corresponding equations are:
r.m.s. switch VA ' 6Vi
2
3
;
peak switch VA ' 6Vi ;
(2.148)
drive power output ' Vi
From the appropriate ratios between these quantities, the r.m.s. switch kVA/kW is 4.9 and the peak
kVA/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 drive
switches. Although the sinewave motor appears from this analysis to have a much better utilization
of their r.m.s. current capability, this advantage may be offset by the greater duty cycle in the sinewave
motor, where three drive switches are conducting at any instant, not two. In the analysis it has been
assumed that the ideal waveforms are achieved by pulse-width modulation at a sufficiently high
frequency, and of course this incurs switching losses which tend to lessen the significance of the lower
r.m.s. switch currents required by the sinewave motor. More detailed comparison requires the use of
computer simulation.
2.20
PERMANENT MAGNETS VERSUS ELECTROMAGNETIC EXCITATION
Because of the laws of electromagnetic scaling there is an 'excitation penalty' associated with small
motors. As the size is decreased, the cross-section area available for conductors decreases with the
square of the linear dimension, but the need for MMF decreases only with the linear dimension, being
primarily determined by the length of the airgap. As the motor size is further decreased, the airgap
length reaches a minimum manufacturable value (typically of the order of 0.15!0.3 mm). Past this point
the MMF requirement decreases only slowly while the copper area continues to decrease with the
squared linear dimension. The per-unit copper losses increase even faster, and the efficiency decreases
rapidly. The loss-free excitation provided by permanent magnets therefore increases in relative value
as the motor size is decreased.
In larger motors magnets improve efficiency by eliminating the losses associated with electromagnetic
field windings. But in larger motors the relative excitation penalty is small. At the same time the
required volume of magnets with adequate properties increases with motor size to the point where PM
excitation 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 railway
or transit-car traction, a PM motor with a short-circuit winding fault would have a large braking
torque, possibly with a high ripple component and a definite risk of overheating which could
demagnetize the magnets or even cause a fire. There is no simple way to protect against this type of
fault. A conventional synchronous or DC machine can be de-excited under fault conditions, and the
induction motor is self-protecting.
Brushless permanent-magnet machines
Page 2.63
There is no hard-and-fast power level below which permanent-magnet excitation becomes
advantageous, but it is possible to examine the excitation penalty in ways which indicate roughly
where the breakpoint lies, and why. For a given level of excitation the choice can be made between
magnets 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 not
be 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 peak
fundamental flux-density produced by the magnet in the airgap. If rotor leakage is neglected this flux
can be taken to be approximately equal to the flux through the magnet in a surface-magnet motor, that
is, Bm Am. If the magnet is operated at a fraction ( of its remanent flux-density, Bm = (Br and hence
B1DL stk
Am '
(2.149)
p(B r
From Ampere’s law we have
H mlm ' H g g ' 0
(2.150)
from which
gB g µrec
lm '
(2.151)
(1 & () B r
where Bg is the average airgap flux-density. The required volume of magnet is then
V m ' 2pAmlm
(2.152)
If B1 = k1 Bg, and if we assume that l = D, then
2
Vm '
2 B1 D 2 g µrec
(2.153)
2
k1 B r ((1 & ()
Now we can calculate the amount of copper needed to magnetize the airgap to the same level. Assuming
a sinusoidal distribution of conductors,
Fg ' g Hg ' g
Bg
µ0
'
g
µ0
B1 cos p2
(2.154)
But with N conductors in series per phase,
Fg '
2
0
i
N
2
sin p2 d2 '
iN
2p
cos p2
(2.155)
Hence
g B1
µ0
'
iN
2p
(2.156)
and the total ampere-conductors required are
i N × 2p ' 2piN
(2.157)
If Ac is the cross-section of the copper winding in the whole motor cross-section and J is the current
density,
AcJ ' 2piN
and hence
(2.158)
SPEED’s Electric Motors
Page 2.64
Ac '
4p 2gB1
µ0J
(2.159)
2
The copper cross-section required is proportional to p , unlike the magnet volume which is independent
of the number of poles. The volume of copper can be estimated by assuming a mean length of conductor
equal to twice the stack length (to allow for end-turns); thus
Vc '
8p 2gDB1
µ0J
(2.160)
The relative volumes of magnet and copper required can be compared by taking the ratio
Vm
Vc
'
µ0 µrec J B1 D
2
4 k1 B r p 2 ( (1 & ()
(2.161)
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 with
2
Br = 0.8T and µrec = 1.05. The electrically-excited motor has J=4 A/mm , giving
Vm
Vc
'
D
488
(2.162)
where D is measured in mm. This means that for motors less than about 500 mm in rotor diameter the
magnet 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 times
that of copper. For the magnet cost to be less than the cost of the copper in a separate field winding, the
rotor diameter must therefore be less than 500/25, i.e. only 20 mm, giving a stator diameter of about 40
mm. The technical potential of high-energy magnets is thus offset by their high cost in all but the
smallest motors. In very small motors a smaller value should be used for the current density J; with
2
J = 2.5 A/mm the stator diameter for equal cost would be increased from 40 mm to 64 mm (2.5 in). In
general, high-energy magnets can only be justified where there is a special premium on efficiency or
compactness. Of course this argument is simplistic, ignoring factors such as process and manufacturing
costs and many others, but it provides a basic physical understanding of the application potential of
magnets, and the effects of scale. Motors magnetized with ceramic magnets must settle for a lower
2
airgap flux-density. Using values of J = 4 A/mm ; B1 = 0.3 T; µrec = 1; Br = 0.35 T, the result is
Vm
Vc
'
D
230
(2.163)
For motors of less than 460 mm stator diameter the magnet volume indicated is less than the volume
of copper in a separate field winding. Ceramic magnets are much less expensive than high-energy
magnets, the cost per unit volume being of the order of 0.6 times that of copper, so that the magnet cost
will be less than the cost of field copper in motors of diameter less than 460 × 0.6, i.e. 275 mm. In practice
PM motors as large as this are relatively uncommon. With ferrite, the flux-density is too low; with
rare-earth magnets the cost is too high.
If running costs are taken into account, the comparison between PM and electrically excited motors
changes significantly. With the present cost of raw materials and present kWh tariffs, the kWh cost of
electrical excitation would outstrip the raw-material cost of the copper in just a few months, assuming
the motor runs at full excitation 24 h/day. Even when all the manufacturing costs are added up, the PM
motor should eventually pay for itself in this way.
Brushless permanent-magnet machines
2.21
Page 2.65
SLOTLESS MOTORS
The availability of very high energy rare-earth and Neodymium-Iron-Boron magnets has re-awakened
interest in the slotless motor, in which the stator teeth are removed and the resulting space is partially
filled with additional copper. At least one such motor is manufactured commercially. The slotless
construction permits an increase in rotor diameter within the same frame size, or alternatively an
increase in electric loading without a corresponding increase in current density. The magnetic
flux-density at the stator winding is inevitably lessened, but the effect is not so drastic as might be
expected.
For a motor with an iron stator yoke and an iron rotor body the magnetic field and its harmonic
components can be calculated by Hague’s method described in §2.4, or the methods described by Hughes
and Miller [1977]. Considering the fundamental radial component of B, the value is greatest at the rotor
surface (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 is
given by:
b '
2(r/R)p%1
[1 % (r/R)2p]
(2.164)
Consider a rotor body of 40 mm diameter with a high-energy magnet of remanent flux-density 1.2 T and
thickness 5 mm. If the radial thickness of the stator winding is 5 mm (including the airgap), then for
a 4-pole magnet b = 0.78. The magnet flux-density will be about half the remanent flux-density with
these proportions, so that the radial flux-density in the stator winding varies from about 0.6 T near the
bore to 0.47 T just inside the stator yoke, giving a mean value of fundamental flux-density of about 0.53
T. The electric loading may be increased relative to that of a slotted stator, because of the additional
space available for copper; but the increase may not be much because the close thermal contact with
the teeth is lost, and the cooling of the slotless winding by conduction to the stator steel may not be as
effective. Taking these factors into account, the power density should be roughly the same as that of
the 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 of
slotting and the relatively low flux-density in the stator yoke. The reactance is also lessened by the
elimination 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 conventional
slotted motor, because the there are no teeth to limit the flux by saturating; indeed the concept would
not 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 the
elimination of cogging effects against the stator teeth). A further possibility is a completely helical
winding such as that proposed for superconducting AC generators [Ross, 1971], or as used in very small
PM commutator motors. Because the helical winding has no end-turns its utilization of copper is higher
than 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 the
rotor.
Sign conventions follow the mathematical ones: in the radial direction, outwards is positive; in the
circumferential direction, counter-clockwise (CCW) is positive; and in the axial direction, out of the
paper 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 reaction
force on the magnet is upwards and to the left, producing positive (CCW) torque T.
Page 2.66
SPEED’s Electric Motors
Fig. 2.78 i > 0, T > 0 producing positive torque.
Fig. 2.79 T > 0, e < 0
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 direction
of 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 a
generated EMF and i is in the direction of positive e. In Figs. 2.78 and 2.79, pelec[out] is negative. The
instantaneous electrical input power is pelec[in] = ! pelec[out], and this is positive. Therefore the
conditions 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 the
same direction. If there were no losses, pelec[in] = pmech[out] and T = ei/T. Note that if the EMF is
proportional to speed, e = kE T, then T = kE i, where kE is the EMF constant or torque constant kT = kE.
Fig. 2.80 Armature reaction field of a single coil
Fig. 2.81 Torque-producing armature current with magnet
Fig. 2.80 shows the direction of the magnetic field produced by a single armature coil with a plain
airgap. In Fig.2.81 two magnets are introduced into the airgap with polarities as shown. Using the
result of Fig. 2.78, the torque is positive. We can deduce the direction of the torque from the polarity
of the armature reaction flux, which tends to make the stator surface a south pole, as indicated by the
lower-case s. Since like poles sS repel, and opposite poles sN attract, the rotor is pushed in the CCW
direction.
The arrows show that the armature reaction field tends to strengthen the magnet flux at one edge of
each 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).
Brushless permanent-magnet machines
Page 2.67
With perfect symmetry the net flux through the
magnet would be unchanged. But perfect symmetry
is not achieved in general, for the following reasons:
(1) The stator slots may not be symmetrically
arranged about the magnet centre-line. With
squarewave drive, the current pattern remains
fixed for typically 60E intervals of rotation. In
sinewave motor drives, the fundamental
component rotates in synchronism with the
rotor but space-harmonics in the winding
distribution may distort it.
(2) The stator teeth are liable to saturate more on
the side with higher flux-density, that is, where
the magnet flux and armature reaction flux are
in the same direction.
(3) If the demagnetization characteristic of the
magnet is not linear, the magnet may suffer
irreversible loss of magnetization on the side
which has the lowest flux-density.
Fig. 2.82 Armature reaction with the current advanced 90E
in phase relative to the generated EMF.
Fig. 2.82 shows the armature reaction with the current advanced 90E in phase relative to the generated
EMF. The conductors are in the same position, with the same currents as in Figs. 2.80 and 2.81, and the
phase advance is represented by a 90E rotation of the rotor in the negative (CW) direction. The
armature-reaction field is demagnetizing across the entire width of each magnet. We say that the
armature reaction is “in the negative d-axis”. In a sinewave motor drive in the steady-state, this
relationship is fixed. The MMF axis is aligned with the magnet centre-line (the d-axis), and the armature
MMF 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 the
magnet flux is radially inwards and it is symmetrically aligned with the MMF axis of the coil. As the
rotor moves forwards (CCW) the flux-linkage R changes in the positive direction unti l it reaches zero
at 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 in
relation 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 is
at a positive maximum in Fig. 2.81. The back-EMF e = +dR/dt is convenient in motor theory, because it
is 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 current
relative to the back-EMF. This is exploited in the technique known as flux-weakening, which permits
brushless 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 phase
advance angle ( is less than 90E because otherwise there would be no magnet-alignment torque. The
current phasor is resolved into d- and q-axis components I = Id + jIq. The demagnetizing effect of Id is
clearly seen in that the voltage generated by its associated flux-linkage is jXdId which is in phase
opposition 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 the
bottom 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 to
the right than the trailing edge, since the x-axis is the azimuth or circumferential coordinate. The top
trace 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.
Page 2.68
SPEED’s Electric Motors
In the sinewave motor in the steady-state, the flux pattern remains more or less fixed and rotates at
synchronous speed, but in the squarewave motor the mmf remains fixed intervals typically of 60E, and
as the magnet sweeps past the airgap flux-density distribution changes shape. If no phase advance is
applied, the top trace in Fig. 2.84 represents the conditions at the middle of the commutation interval.
Fig. 2.83 Phasor diagram of sinewave brushless PM
motor drive, in which ( > 0 and the
current I has a negative d-axis component
Id 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 axis
at 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 advance
of the current relative to the back-EMF. The armature reaction is demagnetizing, with the south-pole d-axis at !90E
or 270E.
Brushless permanent-magnet machines
2.22
Page 2.69
PM GENERATORS
A PM brushless machine can operate as a motor or a generator. When it is motoring, the
electromagnetic torque is in the same direction as the rotation. When it is generating, the
electromagnetic torque opposes the rotation. In either case, there is a generated EMF in each phase
proportional to speed. This EMF is an AC quantity. Its fundamental frequency is given by
f ' n × p
Hz
(2.165)
where n is the speed in rev/sec, i.e. rpm/60, and p is the number of pole-pairs. Alternatively,
rpm
f '
60
×
Poles
2
rpm × Poles
'
120
.
(2.166)
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
T ' 2B f
rad/sec
(2.167)
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” is
also known as 2B “electrical radians” or 360 “electrical degrees”. This is equal to 2B/p mechanical or
actual radians, or 360/p mechanical degrees.
To begin, we will assume that the EMF is sinusoidal, because this is normally the case for AC
generators, 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 the
flux produced by the magnet, and it obeys the classical equation for AC machines:
E q1 '
2B
2
k w1T ph f M1 ' 4@44 k w1T ph f M1
V.
(2.168)
Here Tph is the number of “turns in series per phase”. If each phase has a total of T turns and they
are connected in a parallel paths, then Tph = T/a. The factor kw1 is the fundamental harmonic winding
factor. 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 compared
to that which would be obtained if all the coils were “fully pitched” and concentrated together. kw1
expresses this reduction. Usually it has a value between 0.8 and 1, so the slight loss of EMF is not a high
price to pay for the elimination of unwanted harmonics which would distort the waveform. Sometimes
the 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 is
running on open-circuit. Only the fundamental harmonic component of airgap flux contributes to the
fundamental component of EMF. M1Md can be obtained from the total open-circuit airgap flux by Fourier
[oc]
by the equation
analysis. It is related to the peak open-circuit airgap flux-density B1
M1Md '
B1[oc] D L stk
p
where D is the stator bore diameter and Lstk is the stack length.
(2.169)
SPEED’s Electric Motors
Page 2.70
The subscript q in Eq1 tells us that the EMF lies along the q-axis in the phasor diagram; the flux M1 lies
along the d-axis. In phasor terms, E is written 0 + jEq1, or simply jEq or even just jE. Also
E ' j T Q 1Md
(2.170)
where Q1Md = Q1Md + j0 is the fundamental flux-linkage per phase, and Q1Md = kw1TphM1//2 represents
the product of the magnet flux and the effective series turns/phase. The units of Qd1 are volt-seconds
and the /2 converts the peak value into the r.m.s. value, since Q1 represents a flux-linkage that is
varying sinusoidally in time. All the quantities in the phasor diagram are r.m.s. values. Loosely
speaking, eqn. (2.170) means that the generated EMF is proportional to speed times flux, or frequency
times flux. Strictly speaking, it states that the EMF is proportional to speed times flux-linkage, because
it 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 Q1 by 90E. The phasor diagram on
open-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 load
connect to it. Four important kinds of load can be identified as being important in understanding the
operation 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 fixed
voltage and frequency. Physically it is approximated by a very
large network such as the U.K. National Grid, which is so large
(60,000 MW) that neither the voltage nor the frequency can be
perceptibly altered by connecting one small additional generator Fig. 2.85 Generator connected to infinite
bus
(or load) to it. Fig. 2.85 shows the connection of a generator to an
infinite bus. This diagram is only a schematic diagram because
it does not show the internal impedance of the generator; (it also
omits the circuit-breaker!)
A fixed-impedance load is represented electrically as a passive
electrical circuit containing resistance R and inductance L in
each phase. This is represented in Fig. 2.86.
Fig. 2.87 shows a generator connected to another synchronous
machine, and Fig. 2.88 shows a generator connected to a rectifier.
The rectifier itself is loaded with an impedance comprising a
resistance R and inductance L. The complex representation R +
jTL is not used on the DC side, since it applies only when the
voltages and currents are sinusoidal AC.
Fig. 2.86 Generator connected to R!L load
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 to
calculate the current. The DC current id is also not a pure DC current, because it contains harmonics.
29
Note that a spinning PM generator is electrically "hot" (i.e. "live") whenever it is spinning, and it is dangerous to assume that
it is safe just because it is disconnected or isolated from its load. PM generators should ideally have warning labels on the
terminal 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/constantfrequency) aircraft power generating systems; but this is beyond the scope of these notes.
Brushless permanent-magnet machines
Fig. 2.87 Generator
machine
connected
to another synchronous
Page 2.71
Fig. 2.88 Generator connected to a rectifier with a DC load
having resistance R and inductance L.
The computation of rectifier loads requires computer simulation. To get round this difficulty we design
the PM generator in two stages. The first stage involves designing as though the generator was going
to be connected to a linear AC load. The generator can be "rated" to operate with this linear load. We
call 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, leading
to 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 have
resistance R, and they also have self-inductance and mutual inductance between phases. The simplest
type of generator is a nonsalient pole generator, in which the phase inductances are unaffected by the
rotor position. In this case, in the steady state with AC sinusoidal current and EMF, the generator has
a very simple equivalent circuit for each phase, Fig. 2.89.
Fig. 2.89 Equivalent circuit of one phase of nonsalient-pole generator
The total internal impedance of the generator is Z = R + jXd. The reactance Xd is the synchronous
reactance and it is equal to TLd where Ld is the synchronous inductance. The synchronous inductance
is not simply the self-inductance per phase, but includes the effect of armature-reaction flux in
generating an internal voltage jXdI which is in series with E and the resistance voltage-drop RI. The
internal voltage jXdI is a sinusoidal voltage represented as a phasor, and therefore Xd incorporates the
filtering effect of the winding distribution (and any skew).
When current flows, the terminal voltage V deviates from the open-circuit value E, and according to
Ohm’s law applied to Fig. 2.89, the relationship between V and E is
V ' E ! ( R % j Xd ) I .
(2.171)
SPEED’s Electric Motors
Page 2.72
Fig. 2.90 Phasor diagram of nonsalient-pole generator with lagging power factor angle N.
This is represented in the phasor diagram, Fig. 2.90, which is drawn to emphasize the relationships
between the phasors, so the resistance R and reactance Xd have been exaggerated relative to their
normal 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 voltage
and 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 voltage
E. The “armature reaction” (i.e., the internal voltage-drop across Xd) has a demagnetizing effect
and the machine is said to be “overexcited” (i.e., with E > V). If the load is disconnected, the
terminal voltage will jump up to the higher value E. This condition must be considered in
practice 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 (such
as a phase-controlled rectifier) on order to maintain constant output voltage as speed varies.
(d)
The output power per phase is given by
P ' V I cos N
(2.172)
jN
in watts, where N is the power-factor angle. If the load is an impedance ZL = ZLe = RL + jXL,
2
2
then cos N = RL/ZL = RL//(RL + XL )
(e)
If R is much smaller than Xd, eqn. (2.172) can be expressed in terms of E and V in watts/phase:
P '
EV
Xd
sin *
(2.173)
where * is the load angle. This equation is commonly used in power systems engineering to
express the fact that there is a maximum power Pmax which can be generated stably. In power
systems 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 power
per phase can be written as
P ' E I cos ( ! R I 2
(2.174)
2
in which RI represents the resistive power loss per phase. The term EI cos ( represents the
mechanical power per phase, which is written Pm/m, where m is the number of phases (usually
3). Now I cos ( = Iq, the q-axis component of the current, and E = Eq = TQ1Md. If Te is the
electromagnetic torque and Tm is the angular velocity, then Tm = T/p and TmTe = Pm, so
T e ' m p Q1Md I q
Nm.
(2.175)
Brushless permanent-magnet machines
Page 2.73
This equation is commonly used in connection with PM motors. It shows that the torque is
maximized if the current phasor is oriented along the q-axis such that it is “in quadrature with
the 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)
2
The phasor diagram includes resistive losses (RI ) but it does not include mechanical losses or
iron losses. The mechanical losses can easily be allowed for by a friction torque Tf which must
be added to Te to give the shaft torque T. Thus
T ' Te % Tf .
(2.176)
The simplest way to deal with the iron losses is to treat them as a mechanical loss and include
the corresponding torque along with Tf. Thus the power loss in the core is WFe = TmTFe and
W Fe
T ' Te % Tf %
Tm
.
(2.177)
It is also possible to represent the core loss as an electrical loss, but this modifies the phasor
diagram 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 qaxis or interpolar axis. In a salient-pole machine the phase inductances vary with rotor position, but
if the machine has sinusoidally distributed windings the analysis can be simplified by Park’s
transformation 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 drop
into 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
V d ' V sin * ;
I d ' I sin ( ;
V q ' V cos *
I q ' I cos (
(2.178)
then from the phasor diagram
V d ' !R I d % X q I q ;
Vq ' E ! R I q ! X d I d .
(2.179)
*
The electrical power output per phase is P = Re{VI } = VdId + VqIq, giving
P ' E I q % ( X q ! X d ) I d I q ! R ( I d 2 % I q 2)
(2.180)
and if R is negligible then
P '
E Vd
Xd
1
! Vd Vq
Xd
1
!
Xq
(2.181)
in watts per phase. This is often written
EV
P '
Xd
sin * !
V2
1
2
Xd
!
1
Xq
sin 2 * .
(2.182)
The electromagnetic torque is
T e ' m p ( Qd1 I q ! Qq1 I d )
Nm
(2.183)
where Qd1 is the total fundamental d-axis flux-linkage and Qq1 is the fundamental q-axis flux-linkage.
SPEED’s Electric Motors
Page 2.74
Fig. 2.91 Phasor diagram of salient-pole generator with a load having a lagging power-factor angle N
Rectifier loads and “rectifier rating”
Rectifiers draw nonsinusoidal current waveforms, and a rigorous analysis would involve complex timestepping simulation which is a difficult and time-consuming process, especially if there is significant
saturation in the machine. To continue to use the simpler theory of the phasor diagram it is desirable
to 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 diode
rectifier we can equate the DC power to the AC input power:
3VLL I L cos N ' VDC IDC
(2.184)
where VDC is the mean DC voltage and IDC is the mean DC current, VLL is the r.m.s. AC line-line
voltage at the terminals of the generator, and IL is the AC line current. The input power factor
of the rectifier is
3
cos N '
cos "
(2.185)
B
where " is the phase-delay angle. In a diode rectifier " = 0 and cos N = 3/B = 0.955. The AC line
current is therefore
IL '
(2)
VD ID
A [rms]
3 V LL cos N
(2.186)
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
Q '
2/3
( 6 / B ) cos "
'
1
cos N
.
(2.187)
The generator is designed for a sinewave current of IL and power factor cos N, but the calculated I2R
loss in the generator is increased by the factor Q2 to allow for the additional losses caused by harmonics
in the current waveform.
The process may need to be repeated several times to determine a complete operating chart covering
a range of currents and power factors, especially if the generator is to operate at different load levels.
Brushless permanent-magnet machines
Page 2.75
Control and protection
If the generator is running at a certain speed, the generated EMF is fixed and the power output depends
on the load impedance, which determines the current and the power factor. If the load draws more
power, then the prime mover must respond by providing more torque, otherwise the generator will
slow down. For isolated generators it is normally necessary to provide a governor to make sure that
the 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 whatever
torque 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 most
likely to occur if the generator loses its load, which can happen quite normally if the load is switched
off, or because of a malfunction that causes the load to disappear or to be disconnected from the
generator. A sudden loss of load causes all the prime mover power to be applied in accelerating the
generator/prime mover, and unless the prime mover is shut down quickly, a dangerous overspeed can
result very quickly. For this reason the governor must be designed with a sufficiently fast response
to 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, the
voltage can be varied by changing the field current. The power factor, however, is still determined by
the load impedance. If a wound-field generator is connected to an infinite bus, the voltage cannot be
changed by varying the field current. Instead, the power factor changes. Increasing the field current
tends to make the power factor more leading (i.e. "overexcited"), whereas decreasing it makes the
power factor more lagging ("underexcited").31 When a generator is connected to an infinite bus, the
power 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 the
generator terminals cannot be varied without changing the load impedance or the speed. Changing
the speed also changes the frequency. If a PM generator is connected to an infinite bus, the frequency
and 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 the
internal 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 the
generator remains synchronized with the frequency and phase of the infinite bus. With isolated
generators the issue of maintaining synchronism does not arise.
Another hazard with all generators is the over-voltage which occurs when the load is suddenly
disconnected. If the load current was at its maximum value just before disconnection, there may be
considerable energy stored in the internal reactance of the generator. When the disconnection takes
place, 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 the
terminals. 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 the
other hand PM generators are also used at extremely high speeds, so the potential overvoltage must
be allowed for in the insulation system. There is no possibility of switching it off, since the magnets
are permanently excited. If there is a filter capacitor on the DC side, overvoltage protection is
essential. This can be provided by a fast-acting “crowbar” circuit which detects the overvoltage and
connects 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 by
the mechanical system to which it is connected. In this case there is also a possibility of destroying
the freewheel diodes which rectify the generated current and feed it to the DC link capacitor. This is
one reason for caution in the use of “embedded magnet” motors, which have higher per-unit inductance
than surface-magnet motors, so that if the load is lost at high speed the generator terminal voltage
tends to rise more than that of a surface-magnet machine.
31
By convention, an overexcited generator is said to be “generating VArs” (reactive power), while an underexcited generator
is “absorbing VArs”. For example, if the load is inductive, the VArs generated by the generator are absorbed by the load.
Page 2.76
SPEED’s Electric Motors
With all generators the load current must be monitored, and if the current exceeds the generator
rating, appropriate protective measures must be activated. If the current is only a few percent above
the rated value, it may be sufficient to do no more than display a warning light. More sophisticated
protection would use inverse-time overcurrent relays and automatically disconnect the generator by
opening a circuit-breaker. Generator protection may also include differential current relays (to detect
internal generator faults); over-temperature relays; and negative-sequence relays to protect against
excessive unbalance between phase currents. In addition, the overall system will generally require a
ground fault protection scheme. One of the things to bear in mind with electrical generators is that
there 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 uncontrolled
release.
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, 1-881855-031)
3. Mohan N, Undeland T.M. and Robbins W.P.: Power electronics: converters, applications, and design, John
Wiley & 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 on
Industry 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, IEEE
Transactions, 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 motors
and 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 airgap
magnetic fields in brushless permanent-magnet motors, to be presented at IEEE Industry Applications Society
Annual Meeting, Phoenix, Az, October 1999.
12. Reliance Motion Control Inc.: DC Motors, SPEED Controls, Servo Systems: The Electro-craft Engineering
Handbook, 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, IEEE
IEMDC 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 Publishing
Company, Tokyo, 1984.
Brushless permanent-magnet machines
Page 2.77
Index
2Q 36
AC brushless machines 1
AC Vector control 25
Airgap flux distribution 3
Alignment torque 27, 67
Armature reaction 17, 61, 66-68
Back-emf
trapezoidal 3
Back-EMF 1, 3, 4, 40, 43, 67, 68
Back-EMF constant 4
Back-EMF sensing 43
Basic operation 3
Bifilar 23, 44, 45, 47
Blowers 1, 23, 45
Boules 13-16, 76
Brushed motors 1
Brushless DC machines 1
Chopping 4, 9, 37-42, 44, 47, 52, 54
Classification of inductances 22
Cogging torque 50, 53, 55-59
Commutation 4, 8, 32, 36, 39-41, 43, 47, 54, 68
Computer disk drives 1, 53
Constant-torque locus 31
Core loss 73
Current regulation 36, 37
Delta 4, 6, 8, 20, 36, 37, 45, 49, 62, 69
Demagnetization 65, 67
Differential leakage inductance 18, 21
dq-axis transformation 18, 19
Drive 4, 6-9, 23, 27-29, 33, 35-37, 42-46, 48-50, 52, 54, 61, 62,
67, 68
Duty-cycle 9, 38-40
Effective airgap 17, 20, 59
Efficiency 1, 23, 24, 45, 47, 62, 64
Electromagnetic excitation 62
Electromagnetic torque 3, 4, 7, 27, 52, 53, 69, 72, 73
Embedded-magnet 10, 11, 21, 53
Exciter 1
Fans 1, 23, 45
Finite-element 10, 11, 16, 49-52, 55, 58, 60
Finite-element analysis 11, 16, 55, 58
Flux-MMF diagram 50, 52, 56
Fringing 3, 12, 13
Generated EMF 3, 10, 23, 24, 41, 46, 66, 67, 69, 70, 75
Generators 1, 23, 32, 65, 69, 70, 75, 76
Hague 13-16, 76
Inductance 17-22, 27, 30, 31, 36, 38, 40, 43-45, 47, 49, 54,
60, 61, 70, 71, 75
Infinite bus 70, 72, 75
IPM 11, 22
Laplace 13
Leakage factor 11, 18
Line-start 32
Locked-rotor 9, 13
Magnet 1, 1, 3, 4, 9-18, 20-23, 25-29, 31-34, 37, 48-61, 63-70,
75, 76
Magnetic circuit analysis 10, 15
Magnetization 10, 13, 14, 16, 25, 44, 55, 67, 73
Magnets 1, 6, 9, 10, 13, 14, 25, 26, 31, 32, 44, 53, 55, 56, 59,
60, 62-66, 75
Nonlinear magnetic circuit 11, 12
Nonsalient pole 20, 22, 24, 31, 32, 34, 43, 71
No-load speed 9
Period A 36
Permanent magnet 23, 34, 63, 76
Permanent magnets 1, 25, 53, 56, 60, 62
Permeance coefficient 11, 18
Phase advance 28, 30-32, 34, 35, 47, 67, 68
Phase inductances 22, 49, 71, 73
Phasor diagram 21-28, 37, 49-51, 67-70, 72-74
Phasors 23, 26, 27, 35, 70, 72
Power 1, 3, 4, 9, 24, 34, 36, 37, 39, 40, 43, 44, 53, 55, 60, 61,
63, 65, 66, 70, 72-76
Power factor 24, 72, 74, 75
Rasmussen 13-16, 76
Rectifier 1, 33, 70-72, 74
Reluctance torque 1, 27, 32, 33, 53, 54
Saliency 1, 18, 20, 27-29, 32, 33
Salient-pole 18, 21, 22, 24, 25, 28, 29, 31, 32, 34, 43, 49, 73,
74
Salient-pole motors 18, 21, 22
Self-synchronous 1, 32
Servo-motors 32
Shaft position feedback 1
Sign conventions 51, 65
Sinewave drive 35, 48
Sinewave operation 23
Sinewound 19-23, 36, 43, 49, 50, 52, 54, 60
Single-ended drive 23
Sliding contacts 1
Slotless motors 65
Slots 18, 19, 48, 49, 51, 57, 67
Squarewave drive 33, 35, 37, 42, 43, 48, 52, 67
Stall torque 9
Stepper motor 1
Surface-magnet 10, 11, 17, 21, 25, 28, 29, 31, 33, 50, 53, 75
Switching strategy 35
Synchronous inductance 18, 21, 27, 30, 60, 71
Synchronous reactance 20, 24, 27, 31, 32, 71
Time-stepping 18, 22, 35, 36, 74
Torque 1, 3, 4, 7, 9, 16, 23, 27-35, 39, 44-47, 49-62, 65-67, 69,
72, 73, 75, 76
Torque constant 4, 27, 28, 66
Torque control 1
Torque per ampere 33, 60, 61
Torque ripple 16, 53, 54, 65
Torque/speed characteristic 9, 34
Traction 1, 62
Trapezoidal 1, 3, 37, 54
trapezoidal back-emf 3
Turns in series per phase 3, 17, 20
Unipolar drive 44
Vector control 25
Winding
inductance 17
Windings 13, 18-20, 23, 24, 26, 27, 33, 37, 44, 45, 47-49, 54,
57, 62, 63, 71, 73
Wye 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
3.
INDUCTION MACHINES
3. 1
BASIC THEORY
Page 3.1
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.
Fig. 3.1
Fig. 3.2
Sine-distribution of ampere-conductors
2-pole sine-distributed field
Rotating magnetic field
The flux distribution of a single sine-distributed winding is fixed in space: in particular, its orientation
is fixed along the winding axis. In Figs. 3.1 and 3.2 the winding axis is the x-axis, 2 = 0. The flux-density
at 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-sheet
around the stator bore, K1 = K sin p2 [A/m], such that K d2 is the number of ampere-conductors in an
arc d2 at the stator bore. p is the number of pole-pairs. In Figs. 3.1 and 3.2, p = 1. K1 can be interpreted
as the product of the phase current i1 and the conductor distribution C sin p2. If the current varies
sinusoidally in time, i1 = Imax cos Tt, then the current sheet remains fixed in space but its amplitude at
any point varies sinusoidally: thus K1(2, Tt) = K sin p2 cos Tt.
2-phase winding: If a second "phase" is added, with its axis rotated 90E from that of the first phase, and
if it has the same number of turns as the first phase, it will produce an ampere-conductor distribution
orthogonal to the first one: thus K2 = K cos p2. The axis of phase 2 is the angle at which cos p2 = 0, and
since cos p2 = sin (p2 + 90E), this is at !90E/p. If phase 2 carries a sinewave current which leads i1 by
90E 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:
K (2, T t) ' K (sin p 2 cos T t ! cos p 2 sin T t) ' K sin (p 2 ! T t)
(3.1)
which is interpreted as a rotating ampere-conductor distribution. The rotation speed is T/p rad/s: or
Ns '
60
2B
×
T
p
'
120 × f
poles
[rpm]
(3.2)
since T = 2B f, where f is the supply frequency in [Hz]. This is the synchronous speed. If the polarity
of 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.
Page 3.2
SPEED’s Electric Motors
3-phase windings: Three-phase windings are much more
common in practice than 2-phase windings. Fig. 3 shows the
arrangement of three sine-distributed phase windings
arranged symmetrically so that their axes are displaced by
120E from each other. Balanced three-phase currents supply
these windings, so that the total ampere-conductor
distribution is given by
K(2, T t) ' K sin p 2 cos Tt
% K sin (p 2 ! 2B /3) cos (Tt & 2B /3)
% K sin (p 2 % 2B /3) cos (Tt % 2B /3)
3
'
K sin (p 2 & Tt) .
2
(3.3)
Fig. 3.3
3-phase winding
This represents a rotating ampere-conductor distribution whose peak value is 3/2 times the peak value
produced 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 distributions
because there is only one North and one South. Windings can have any number of pole-pairs. For
example, 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 think
of a pole-pitch as spanning B electrical radians, where one electrical radian is equal to 2/P mechanical
radians or 1/p mechanical radians. The synchronous speed is T electrical rad/s, or T/p mechanical
rad/sec. In rpm it is given by eqn. (3.2). Common examples are tabulated below:
Poles
Frequency
2
4
6
8
Speed [rpm]
Frequency
3,000
50
1,500
1,000
Speed [rpm]
3,600
60
750
1,800
1,200
900
Table 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 to
be 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 links
the flux produced by the stator windings, but is not connected to the supply. The currents flowing in
the rotor are induced, and in order to obtain a driving EMF the rotor must rotate at some speed different
from synchronous speed. For if the rotor were rotating at synchronous speed, the flux linkage of the
rotor 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 is
Tr elec rad/sec, the slip is defined as
T ! Tr
s '
[p.u.] .
(3.4)
T
In general, the induction motor rotates slightly slower than synchronous speed and s > 0. The induction
generator 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 also
expressed as Tr = (1 ! s) T or Nr = 120 f /P × (1 ! s) rpm. The relative speed between the rotating flux
and the rotor is T ! Tr = s T elec rad/s. Therefore the frequency of EMF's and currents induced in the
rotor is equal to s f ; i.e., the slip is the ratio of the rotor frequency to the stator frequency.
Induction machines
Page 3.3
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 at
2.5% slip from a 60-Hz supply is rotating at (1 ! 0.025) × 1,800 = 1,755 rpm. The slip speed is 45 rpm and
the 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 the
stator ampere-conductor distribution, i.e. at synchronous speed. The rotor ampere-conductor
distribution therefore rotates forwards relative to the rotor, at the relative speed s T electrical rad/s.
Because both ampere-conductor distributions rotate at synchronous speed relative to the stator, any
voltages induced in the stator windings by will be at the line frequency f. There is only one resultant
magnetic flux, produced by the combined effect of the rotor and stator. Provided that the windings are
sine-distributed, it rotates at synchronous speed.
Fig. 3.4
Equivalent circuit of one phase of polyphase induction motor, with different frequencies in the primary and
secondary. 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 magnetizing
current.
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 leakage
inductances are higher than in a typical power transformer. Consequently the magnetizing
reactance 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 one
phase. 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 the
transformation ratios for the voltage, current, and impedance. By Faraday's Law, the sinusoidal
induced voltage in each winding is proportional to the turns and the frequency. In a transformer the
voltage ratio is n1 f/n2 f = n1 /n2, the turns ratio, where n1 and n2 are the effective numbers of turns in
the stator and rotor windings, respectively: the frequency cancels because it is the same in both
windings. 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 effect
controlled 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/sn22.
Page 3.4
Fig. 3.5
SPEED’s Electric Motors
Equivalent circuit of one phase of polyphase induction motor. The frequency is the line frequency in both primary
and secondary, and the turns ratio is eliminated. The EMF Erb is used to represent leakage flux in saturated rotor slotbridges, and does not contribute to torque production. Boxed variables appear in the phasor diagram, Fig. 3.9.
It is now possible to refer all voltages, currents, and impedances in Fig. 3.4 to the stator winding, as in
Fig. 3.5. All voltages and currents in Fig. 3.5 are at line frequency f, and all reactances have their linefrequency values. While the secondary EMF and reactance lose their dependence on slip, the rotor
resistance becomes R2 /s, and this is further divided into two components in Fig. 3.5, using
R2
s
1
' R2 % R2
s
! 1.
(3.5)
2
The first term represents I R loss, while the second term represents electromechanical power
conversion. The referred equivalent circuit (Fig. 3.5 ) shows that any increase in rotor current appears
as an increase in the stator current I1 drawn from the supply. It can be used to calculate all the
important 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
Pgap ' 3
I22 R2
(3.6)
s
where the 3 accounts for all three phases. This is the electromechanical power or airgap power, i.e. the
power transferred across the airgap from the stator to the rotor. The actual copper loss in the rotor,
however, is only 3I22R2, and since s < 1 (in motoring operation), the airgap power exceeds the rotor
2
2
copper loss by 3I2 R2/s ! 3I2 R2 = Pgap (1 ! s). The electromechanical power is therefore
Pmech ' Pgap ( 1 & s ) .
(3.7)
Since the actual rotor speed is (1 ! s) T/p rad/s, electromagnetic torque must be
T gap '
Pmech
(1 ! s) T
'
p
T
Pgap
[Nm] .
(3.8)
The airgap power Pgap divides into two. A fraction s goes as rotor copper loss (heating the rotor). The
remaining fraction (1 ! s) is converted into mechanical power at a speed that is equal to (1 ! s) times
the synchronous speed, producing the electromagnetic torque Tgap. The shaft torque Tshaft is less than
the electromagnetic torque because of friction torque Tfric :
T shaft ' T & T fric .
(3.9)
Induction machines
Page 3.5
The efficiency is now calculated as
Tm × T shaft
0 '
3 V1 I1 cos N1
× 100% ,
(3.10)
where cos N1 is the power factor and Tm = Tr/p is the speed in mechanical radians/sec. Note that V1 is
the 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 the
line-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. This
increases the fraction of airgap power that is converted to mechanical power, and decreases the
fraction that is dissipated as rotor I 2R loss, helping to maximise the efficiency. A small value of s
increases 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 only
slightly 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,
V1
I2 '
(3.11)
( R1 % R2 / s ) % j X L
If this is substituted into eqn. (3.6) for Pgap, then from eqn. (3.8) the airgap torque is given by
T gap '
3p
T
×
R2
s
×
V12
( R1 % R2 / s )2 % X L2
[Nm] .
(3.12)
This equation relates the electromagnetic torque to the slip, when the supply voltage and frequency
are fixed. It can therefore be used to plot a graph of torque vs. slip. Since the speed in rev/min is given
by (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
T
6
3p
T
×
V12
R2
× s,
(3.13)
i.e., the torque is proportional to slip. Since the slip is generally small for torques which are less than
or equal to the rated load torque, the speed remains near synchronous speed as the torque varies. The
actual torque is determined by the load, and the operating point is at the intersection of the motor's
torque/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. This
is 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
T
6
3p
T
×
V12 R2
X L2
×
1
s
,
(3.14)
i.e., the torque is inversely proportional to slip at low speed. The locked-rotor torque is the value at
standstill, with s =1. This is the torque when the motor is first switched on.
Page 3.6
SPEED’s Electric Motors
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 the
linear section of the torque/slip curve. This value of slip can be estimated by differentiating the torque
equation with respect to slip, and setting the derivative to zero. The result is
s Tmax '
R2
(3.15)
R12 % X L2
The corresponding maximum torque is called the breakdown torque:
Tmax '
1
2
×
3p
T
V12
×
R1 %
R12 % X L2
[Nm] .
(3.16)
If the load is gradually increased above the rated load, the slip increases and the motor produces more
torque. The operating point moves up the torque/slip curve until the slip reaches sTmax. Any increase
of load torque then causes the motor to slow down still more, increasing the slip. Beyond sTmax the
motor 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 torque
is roughly half the breakdown torque. This provides a margin of safety against stalling due to transient
changes of load torque, or undervoltage or underfrequency conditions. Since Tmax is proportional to
2
V1 , 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 constant
supply voltage and balanced conditions. Iterative solution of the phasor diagram makes it possible to
allow for nonlinearities such as saturation of the magnetizing reactance and leakage reactance, the
effect 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 spaceharmonic MMF and is limited to sinusoidal waveforms of voltage and current.
Induction machines
Page 3.7
Speed control
Fixed supply: If the supply voltage and frequency are fixed, the speed can be controlled by varying
the rotor resistance by means of an external resistor connected into the rotor circuit by slip-rings and
brushes. This technique is used in large wound-rotor induction motors, especially for controlling the
rate 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 resistance
is shorted out and the characteristic changes to a "low-slip, high-efficiency" characteristic. Wound-rotor
machines are expensive and are relatively less common than cage-rotor machines.
Fig. 3.7
Varying R2 to achieve high starting torque
Fig. 3.8
Varying the supply frequency at constant V/Hz
Pole amplitude modulation: With a fixed supply, the speed of a cage-rotor machine can be changed
by 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 there
is 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. The
range of speed variation is small, unless the motor is designed with a high rotor resistance, but this
makes 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 supply
frequency. This causes the torque/slip curve to be translated along the speed axis, Fig. 3.8. If the
voltage/frequency ratio is kept constant ("constant volts/Hz"), the breakdown torque remains constant
over most of the speed range; at lower speeds it tends to fall as the stator resistance begins to become
significant compared with the leakage reactance. With this type of drive the slip for a given torque can
be 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 operate
by orienting the stator MMF distribution at an optimal angle relative to the flux trapped by the rotor
currents, and under transient conditions they are not limited to sinusoidal current. However, the
equivalent 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 inner
cage has a high leakage reactance and a low resistance, and the outer cage has a low reactance and a
high resistance. The resulting torque/speed characteristic is similar to the sum of the high-resistance
and low-resistance curves in Fig. 3.7, providing high starting torque and low operating slip (therefore
high efficiency) in the one motor. As an alternative to the double cage, skin-effect is used in the deep-bar
rotor to increase the rotor R/X ratio as the slip (and therefore the rotor frequency) increases.
Page 3.8
3.3
SPEED’s Electric Motors
EQUIVALENT-CIRCUIT PARAMETERS
Although the equivalent circuit (Fig. 3.5) is simple and elegant, the calculation of its parameters is a
formidable and complex task. They are normally calculated from the motor geometry, the winding
parameters, and the appropriate material properties. The phasor diagram for one phase of a balanced
polyphase 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 saturation
of 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-vector
analysis (as used in the analysis of field-oriented drives), provided that the windings are sufficiently
sine-distributed. If the voltage and current waveforms are non-sinusoidal in time, then of course the
losses, saturation levels, and frequency-dependent impedances will be less accurate.
Several alternative methods are known for
calculating leakage reactance components,
including belt-leakage, zig-zag, skew and
differential components.1 Saturation of
magnetizing reactance is calculated using
the "60E" method (Alger [1970]; Say [1948])
or a specially developed method with more
detail, (Ionel et al [1998]). Saturation of
leakage reactance is calculated by means of
Erb, or by a modification of Norman’s [1934]
method. (See also Boldea et al, [2000]).
Deep-bar effect is calculated using a special
method that works with slots of any shape,
(Boldea [1995]). Alternatively, the classical
rectangular-bar theory can be used,
(Kostenko and Piotrovsky [1974]; Schuisky
[1960]). For double-cage rotors the rotor
branch containing Erb, X2 and R2 is
duplicated in parallel with the one in the
circuit of Fig. 3.5.
V1
VZ1
E1
VX1
Erb
φ
VR1
VZ2
VX2
ER2
VR2
I2
Irc
Inl
Imag
I1
Fig. 3.9 Phasor diagram for one phase of a balanced polyphase
Closed rotor slots and Erb: For closed
induction motor. The phasor Erb represents rotor slot
rotor slots Erb can be used to represent the
bridge leakage.
flux through the bridges that close the
rotor slots. These bridges cannot be
modelled by a fixed component of the rotor slot permeance, because they tend to saturate. It is assumed
that there will be a sine-distributed component of the airgap flux that enters the rotor radially and then
travels circumferentially around the rotor, via the bridges. At a position 90Eelec from the point of peak
airgap flux-density, the bridge flux density will be a maximum, and this density is arbitrarily assumed
to be 2.1T. When this is multiplied by the cross-section area of the bridge, the resulting flux is a
component 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 extended
equivalent circuit such as the one in Fig. 3.10, which includes two sets of phase-belt harmonics (5th and
7th); the forward and backward slot permeance harmonics of order S1/p±1, and the forward and
backward slot-MMF harmonics of order S1/p±1, where S1 = stator slots and p = pole-pairs.
1
The classical references are Alger, Veinott, Richter and others. To understand these calculations it is virtually essential to
study the original references, but some idea of the procedures can be seen in §3.6 which describes the calculation of slot
permeance. 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.
Induction machines
I1
R1
Page 3.9
I2
jX 1
Second cage
V1
jX m
Rc
Im
b
jX pm
jX 2
jX 2m
R2m
1 + m(1!s)
Slot permeance
R2
harmonics
f
jX pn
jX 2n
R2n
1 ! n(1!s)
b
jX m5
jX 25
R25
6 ! 5s
Phasebelt
harmonics
f
jX m7
jX 27
R27
7s ! 6
b
jX 2m
jX mm
R2m
1 + m(1!s)
Slot MMF
(zigzag)
harmonics
f
jX 2n
jX mn
R2n
1 ! n(1!s)
Fig. 3.10 Harmonic equivalent circuit for polyphase machines given by Alger [1970]
In a 3-phase motor the MMF wave produced by the 5th harmonic of the winding distribution rotates in
the backward direction at 1/5 synchronous speed, and the MMF wave produced by the 7th harmonic
rotates 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 rotates
in the forward direction at 1/5 synchronous speed, and the MMF wave produced by the 7th harmonic
rotates in the backward direction at 1/7 synchronous speed. These directions are opposite to those in
the three-phase motor.
Page 3.10
3.4
SPEED’s Electric Motors
WINDING FACTORS AND OTHER WINDING-RELATED MATTERS
For standard concentric and lap windings these are defined as in the classical literature.
Pitch factor
nB"
k pn ' sin
2
(3.17)
where " is the per-unit coil pitch (i.e., the span in electrical radians divided by B), and n is the order of
the 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
nq(
sin
2
k dn '
n(
q sin
(3.18)
2
where ( is the slot pitch in electrical radians and q is the number of slots per pole per phase. This is used
only with lap windings.
Skew factor
nF
sin
k sn '
2
nF
(3.19)
2
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 distribution
of 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 in
series. The resultant magnitude of the n'th harmonic MMF coefficient is cn = %(an2 + bn2). The winding
factor, is the ratio of cn and the n'th harmonic winding factor of a "base" winding with the same number
of series turns distributed equally in full-pitch coils among the 2p poles. The "base" winding is assumed
to start in slot 0, so that it has only sine coefficients Bn. Thus kwn = cn/Bn. Note that the phase
information in an and bn is lost in this process, so kwn is always positive, even with a winding for which
negative values of kwn are possible. For example, in a 24-slot 2-pole motor we could wind two coils each
with a span of 8 slots (i.e., 2/3 pitch) diametrically opposite to each other. (Coil 1 in slots 1!9; coil 2 in
slots 21!13). This is a concentric winding for which " = 2/3 and so k5 = sin (5 × 2/3 × B/2) = !0.866. The
above procedure gives a5/B5 = !0.836 and b5/B5 = !0.224, so that c5/B5 = +0.866. When the winding is
skewed, 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 Al
absorbs iron and other impurities that reduce the conductivity to 58!59%. A figure as low as 50% is
often 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 totally
2
enclosed fan-ventilated motor, like the general purpose 3-phase induction motor, a value of 6-7 A/mm
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 of
2
3-3.5A/mm can be used for initial sizing, but the influence of the total surface area is important, as is
the airflow and the emissivity of the paint (for cooling by radiation).
Induction machines
3.5
Page 3.11
ROTOR AND STATOR SLOT NUMBERS
For a given stack length, rotor diameter and electromagnetic loading (air-gap flux density and stator
current density) the number of stator slots has little influence on the stator resistance or the copper
weight. The number of stator slots can then be used to control the value of the leakage reactance: it
controls the differential component directly, and the slot component indirectly, through the ratio of slot
depth 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 use
an integral number of stator slots per pole per phase (with a minimum value of 2), the choice being
made, 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 effects
such 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 of
stator slots in order to maximise the secondary zig-zag reactance for the stator slot MMF harmonics and
to 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 slots
is 1,2, or 4 per pole; while standstill locking will occur if the difference is 3 per pole". (Ns = synchronous
speed). The choice of stator and rotor slot numbers is discussed in Heller and Hamata [1977] and
Kopilov et al [1980].
Two tables from these references with recommended numbers for stator and rotor slots are provided
below. These values were derived from theory, mainly developed for line start machines, and practical
experience. It should be noted that on this subject there are divergences between various authors as
can be seen for example by inspecting the two tables. The tabulated slot combinations should not be
necessarily regarded as “extremely safe” and also it should be kept in mind that there are known cases
of successful motor designs that do not fulfill these recommended combinations.
Poles
2
4
6
8
Stator Slots
Rotor Slots (Bars)
24
(16), [20], ([22]), (28), [30]
30
(16), [20], (22), [26], [34], [36]
36
[24], 26, [28], 30, ([32]), 42, (44), [46]
48
(32), 34, [36], 38, [40], ([44]), (56), 58, [60]
36
24, [26], [46]
54
38, 40, [44], [64], 66, [68]
48
34, [62]
72
50, 52, 54, [56], 58, 86, 88, [90]
TABLE 3.2
SUITABLE 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 large
synchronous parasitic torques in the braking region). The number of rotor slots in square brackets
maybe 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 small
as well as large machines".
Page 3.12
Poles
SPEED’s Electric Motors
Stator Slots
Rotor Slots (Bars)
Unskewed
2
4
6
8
10
12
14
16
1
Skewed
1
12
9 , 15
18
1
11 , 12 , 15 , 21 , 22
14 , 18 , 19 , 22 , 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
—
1
1
1
1
1
1
2
1
1
—
37, 39, 41, 55, 57, 59
12
9
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
2
36
26, 46, 48
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
2
48
34 , 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
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
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
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
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 3.3
RECOMMENDED COMBINATIONS OF STATOR AND ROTOR SLOT NUMBERS
1
2
used especially for fractional horse power machines.
might cause increased motor vibrations.
Induction machines
3.6
Page 3.13
SLOT PERMEANCE CALCULATIONS
The slot-leakage inductance of a single coil is given by an equation of the form
Lc_slot ' µ0 L stk N 2 ( P1 % P2 )
(3.20)
where N is the number of turns, Lstk is the stack length, and P1 and P2 are the permeance coefficients
for the two slots in which lie the two coilsides. The permeance coefficient for each slot permits the
inductance 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 in
calculating the permeance coefficient the distribution of flux within the slot must be taken into account.
Fig. 3.11
Slot sections
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 the
x-direction (that is, B = (B x,0,0). The contribution )P of any section depends on the variation B x(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 driving
this 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 flux
dN is linked by the fraction A(y)/Aw of the total turns N, so the contribution dR to the total flux-linkage
2
2
is equal to dR = dN.N A(y)/Aw = [µ0LstkN I/Aw ] 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 integrate
dP = (1/Aw2) [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 shapes
may 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/Aw2) [A(y)]2dy/x. It is convenient to
integrate 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 these
substitutions and performing the integration with respect to 2 from 0 to $, we get
)P '
$ [ 4$2 / 3 % 1/2 % 2 cos 2$ ] & (5/4) sin 2$
2 [ 2 $ & sin 2 $ ]2
.
(3.21)
Page 3.14
SPEED’s Electric Motors
When $ = B/2, the slot-bottom is semicircular and )P = 0.1424. With $ = B we get the "classical" value
for 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 from
x = w0 to x = w1, we get the following expression (with B = U – kw02):
)P '
2k
Aw2
B 2 ln
w1
w0
%
h
2
( w1%w0 ) { B %
k
4
2
2
( w1 % w0 ) }
(3.22)
When the trapezium has parallel sides w1 = w0 = w, so k 6 4 and if a = hw, )P simplifies to
)P '
1
Aw2
U(U % a) %
a2
h
3
w
.
(3.23)
For a rectangular section at the bottom of the slot, U = 0; and if this is the only section there are no
conductors above it, so Aw = a and
1 h
)P '
(3.24)
3 w
which is the well known formula for a rectangular slot. Another special case arises at the bottom of
a slot if w0 = 0; then the section is triangular and
3
1 h w1
)P '
(3.25)
Aw2 16
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
w1
h
)P '
ln
(3.26)
w0
w1 ! w0
and if w1 and w0 are nearly equal this becomes
)P '
2h
w0 % w1
(3.27)
which is commonly quoted in textbooks. Veinott in his VICA-31 program for slot constants uses a
modified form
4h
)P '
; w1 < w0
(3.28)
3 w1 % w0
in which w1 is equal to the slot opening and w0 is the width at the bottom of the slot wedge. By giving
three times more weight to w1 than to w0, he increases the value of )P and makes an allowance for
fringing 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 the
slot opening, and h equal to the depth of the tooth-tip: i.e., )P = h/w if there is no conductor in the slot
opening. 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 this
theory assumes infinitely permeable iron. Closed slots, and even slots with significant saturation of
the tooth tips, require a different treatment and their effective permeance depends on the slot current.
Induction machines
Page 3.15
Comparison with finite-element calculations
The analysis assumes that the flux crosses the slot in the x-direction with no fringing. In practice
fringing increases the permeance, and finite-element studies of all the standard example slots in PC-IMD
indicate 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 expression
2
E/µ0I , 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
Open
custom
2.42
2.16
Rectangular
slot
2.166
2.167
Fig. 3.12 Typical finite-element flux plots.
The permeance coefficients are a few percent
higher than those calculated on the assumption
that the flux crosses the slot in parallel tubes.
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
2.85
FE
2.87
PC-IMD
PC-IMD
Classical formula
Slot opening region empty
Slot opening full
Kostenko & Piotrovsky
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 the
classical method for a rectangular slot, and the other is an integration of the complex diffusion equation
throughout the slot, using a layered model developed for SPEED by Prof. I. Boldea. This is similar to
the 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.
Page 3.16
3.7
SPEED’s Electric Motors
LOCATION OF AMPERE-CONDUCTORS
It is sometimes useful to be able to visualize how the ampereconductors in the rotor and stator are distributed, and how they are
oriented with respect to each other, especially in finite-element
calculations.
Fig. 3.13 shows the basic model of a sine-distributed amp-conductor
distribution. When a 3-phase winding is fed with balanced 3-phase
currents, the fundamental ampere-conductor distribution rotates at
synchronous speed. The axis of the airgap MMF lags 90E behind the axis
of the amp-conductor distribution.2
The current Im in the stator sets up a rotating ampere-conductor
distribution ACm whose MMF axis is along the x-axis at the instant
shown in Fig. 3.13. This distribution rotates in the positive direction
(i.e., anticlockwise). The subscript ‘m’ means ‘magnetizing’.
Fig. 3.13 Sine-distributed ampereconductor distribution
A rotor conductor experiences a magnetic flux-density B (produced by Im) and rotates clockwise relative
to B at slip speed, so its velocity in the magnetic field is the vector v. The EMF v × B in this conductor
is in the positive (dot) direction. It causes current to flow in the same direction. Because of the sinedistribution of B, the resulting induced current in the rotor also follows a sine-distribution, and it is
labelled AC2. AC2 lags behind the rotor EMF distribution by a small angle because of the rotor leakage
reactance in series with the rotor resistance. Note that the force F = i × B on the rotor conductor is in
the right direction to produce positive torque (counterclockwise).
The principle of ampere-turn balance (i.e., Ampère's law) requires that an equal and opposite ampconductor 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 current
I2N : thus I1 = Im + I2N. The resulting ampere-conductor distribution in the stator is AC1 = ACm + AC2N
as shown in Fig. 3.14.
Only two of t he ampere- conductor
distributions physically exist : AC1 and AC2.
The other two, ACm and AC2N, are abstract
components of the stator current distribution
which are used to explain the "mechanism" of
induction. The ACm or Im component is the
"magnetizing" component since it alone is
responsible for the mutual flux.
Fig. 3.14
shows that the rotor ampereconductor distribution is largely in opposition
to the stator ampere-conductor distribution, as
though the rotor tries to cancel the flux
produced by the stator. This must be so if the
magnetizing current is to be small compared
with the load current. Since the positive
ampere-conductors of AC1 and the negative
ampere-conductors of AC2 are nearly opposite
each other across the airgap, they tend to force
flux in the peripheral direction around the
airgap, i.e. into the leakage paths.
Fig. 3.14
2
Distribution of ampere-conductors.
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).
Induction machines
3.8
Page 3.17
TRANSIENTS AND EIGENVALUE ANALYSIS
PC-IMD can calculate starting transients. To model the capability of soft-starters with point-on-wave
switching, 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 close
simultaneously, 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 at
zero slip over a range of frequencies from the normal line frequency down to zero. The resulting rootlocus diagram is indicative of the relative stability of the motor when fed from a variable-frequency
inverter 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 changes
of notation: see Fig. 3.15.4 Although this approach is rather old-fashioned it is practical and clear. PCIMD does not include zero-sequence voltages or currents, so the theory works only for three-wire
connections and the transformation matrices are not square.
Fig. 3.15 Orientation of axes
The transformation from (a,b,c) variables (i.e., direct phase variables) to dq axes is eqn. (3.29), where
2 is the angle between the d-axis and the axis of phase a in elec. rad. The same transformation (and its
inverse, eqn. (3.30)) is used for voltages, currents, and flux-linkages.
vd '
2
3
vq ' !
4
[ v a cos 2 % v b cos (2 ! 2 B/3) % v c cos (2 % 2 B/3) ] ;
2
3
[ v a sin 2 % v b sin (2 ! 2 B/3) % v c sin (2 % 2 B/3) ] .
(3.29)
The main ones are: 2r instead of 22 (rotor angle); subscripts d,q instead of 1d,1q; and subscripts D,Q instead of 2d,2q. The
phases are labelled a,b,c here, but in the SPEED software they are generally labelled 1,2,3 (lines being labelled a,b,c).
Page 3.18
SPEED’s Electric Motors
Inverse:
v a ' v d cos 2 ! v q sin 2 ;
v b ' v d cos (2 ! 2 B/3) ! v q sin (2 ! 2 B/3) ;
v c ' v d cos (2 % 2 B/3) ! v q sin (2 % 2 B/3) .
(3.30)
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 reference
is to constrain 2 appropriately.
The total electrical instantaneous power at the terminals is
v a ia % v b ib % v c ic '
3
2
[ v d id % v q iq ]
(3.31)
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,
Ra ' L aa ia % L ab ib % L ac ic % L aA iA % L aB iB % L aC iC
Rb ' L ba ia % L bb ib % L bc ic % L bA iA % L bB iB % L bC iC
Rc ' L ca ic % L cb ib % L cc ic % L cA iA % L cB iB % L cC iC
(3.32)
The self-inductances are constant and symmetrical: Laa = Lbb = Lcc; LAA = LBB = LCC. The stator-stator
mutual inductances are constant: Lab = Lbc = Lca = Lba = Lcb = Lac. The rotor-rotor mutual inductances
are constant: LAB = LBC = LCA = LBA = LCB = LAC.
The stator-rotor mutual inductances vary sinusoidally with rotor position 2r: for example,
L aA ( 2 ) ' L aA cos 2r
L aB ( 2 ) ' L aA cos ( 2r % 2 B/3 )
L aC ( 2 ) ' L aA cos ( 2r ! 2 B/3 )
(3.33)
A similar set of equations applies for LbA(2), LbB(2), LbC(2),LcA(2), LcB(2), and LcC(2), with appropriate
angles taken from Fig. 3.15. Also an equation similar to eqn. (3.32) is written for the rotor flux-linkages
RA,RB, and RC.
The three stator phases will be replaced by two fictitious, orthogonal coils d and q fixed to (and rotating
with) the d- and q- axes respectively. The flux-linkage of the d-coil is obtained by applying the voltage
transformation to Ra, Rb and Rc :
Rd '
2
3
[ Ra cos 2 % Rb cos ( 2 ! 2 B / 3 ) % Rc cos ( 2 % 2 B / 3 ) ] ' L11 id % L12 iD
(3.34)
where L11 = Laa ! Lab and L12 = 3/2 LaA and much algebra has been omitted. Similarly in the q-axis,
Rq ' L11 iq % L12 iQ .
(3.35)
For the rotor D and Q coils, the transformation is the same but with 2r instead of 2; thus
RD ' L22 iD % L12 id ;
RQ ' L22 iQ % L12 iq .
where L22 = LAA ! LAB, and again L12 = 3/2 LaA.
(3.36)
Induction machines
Page 3.19
Physical attributes of the fictitious d and q coils
Consider the instant when the d-axis is aligned with the axis
of phase a, and ia = I, ib = ic =!½I in the 3-phase winding,
Fig. 3.16. According to the reference frame transformation
id = I. If the flux-linkage in phase a is Q, the flux-linkages of
phases b and c must be !½Q each, so that by the reference
eframe 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 acoil. If we consider I to be a DC current then the voltage
Fig. 3.16
d-axis aligned with phase a
drop in phase a is va = RaI and the voltage-drops in phases
b 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 with
having the same number of turns, provided that the total cross-section of copper in the conductors of
the 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 of
phases b and c, for a total of (Laa ! Lab)I = L11I. The flux-linkage of phase b is Laa × (!½I) due to self
flux-linkage, with Lab × (!½I) mutually coupled from phase c, and LabI mutually coupled from phase
The flux-linkage of phase c is the same. So Rd = 2/3 × L11I ×
a, for a total of !½L11I.
[1+(!½)×(!½)+(!½)×(!½)] = L11I. According to the transformation equation the d-coil current is I, so
the 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 coil
A 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 × LaA I. The fluxlinkage 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 !¾LaA I. The flux-linkage of coil c is the same, so that according to the
reference 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 that
the apparent mutual inductance is 3/2 × LaA.
Voltage equations
For the stator
v a ' R a ia % p Ra ; v b ' R a ib % p Rb ; v c ' R a ic % p Rc
(3.37)
where p is the operator d/dt, and we have assumed that Ra = Rb = Rc. For the stator d- and q-coils the
voltage equations are derived by incorporating eqns. (3.37) into the reference-frame transformation
equations for vd and vq. When p operates on the trigonometric functions of 2, we get the speed voltages.
Missing out a lot of algebra,
v d ' R d id % p Rd ! p 2 . Rq ;
v q ' R q iq % p Rq % p 2 . Rd .
(3.38)
In synchronously rotating axes p2 = T = 2Bf. In d,q axes fixed to the rotor, p2 = Tr, i.e., the rotor angular
velocity 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 rotating
at angular velocity Tr, the physical angular velocity of the rotor in elec rad/s. The result is
SPEED’s Electric Motors
Page 3.20
v D ' 0 ' R D iD % p RD ! p 2s . RQ
v Q ' 0 ' R Q iQ % p RQ % p 2s . RD .
(3.39)
where p2s is the slip velocity in elec rad/s, i.e. the angular velocity of the d,q axes relative to a point
fixed on the rotor. Thus in synchronously rotating axes p2s = T!Tr, with T = 2B f, but in d,q axes fixed
to the rotor, p2s = 0. Note that RD = RQ = RA.
The electromagnetic torque is determined from the power associated with the stator current and the
speed voltages, divided by the angular velocity in mechanical rad/s. If P is the number of poles,
Te '
3 P
2 2
[ Rd iq ! Rq id ]
[ Nm ] .
(3.40)
The dynamical equation of motion is
Te ' J
d Tr
dt
×
2
P
% D Tm
(3.41)
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.
p Rd
p Rq
p RD
p RQ
v d ! R d id
v q ! R q iq
! R D iD %
! R Q iQ !
P
p Tr '
[ Te !
2J
'
'
'
'
% T Rq
! T Rd
( T ! Tr ) RQ
( T ! Tr ) RD
(3.42)
D Tm ]
where T is the velocity of the rotating reference frame in elec rad/s. At the end of each timestep, new
values 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 be
0.002 s, but it depends on the time-constants of the particular motor. The system of equations is nonlinear because of products like Rd iq in the torque equation and T Rd 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. The
integration 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
v a ' v pk cos ( T t % * ) ;
v b ' v pk cos ( T t % * ! 2 B / 3 ) ;
v c ' v pk cos ( T t % * % 2 B / 3 ) .
(3.43)
where T = 2B f and * is an arbitrary phase angle. According to the transformation equations
v d ' v pk cos ( T t ! 2 % * ) ;
v q ' v pk sin ( T t ! 2 % * ) .
(3.44)
If 2 = 2r the d,q axes are fixed to the rotor. If 2 = Tt they rotate at synchronous speed, with T = 2B f.
v d ' v pk cos * ;
v q ' v pk sin * .
Then vd and vq are constant. In particular, if * = 0, vd = vpk and vq = 0.
(3.45)
Induction machines
Page 3.21
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 be
already closed, with no current flowing. Those in line b close at the "point on wave" Tt = 21, and line
c closes at the angle Tt = 21 + 22 ; that is, after a delay 22/T measured from the closure of line b.
Fig. 3.17
Non-simultaneous switching of the supply phases
With line-neutral voltages given by eqns. (3.43), then with * = 0 the line-line voltages are
v ab ' v pk 3 cos ( T t % B / 6 ) ;
v bc ' v pk 3 cos ( T t ! B / 2) ;
(3.46)
v ca ' v pk 3 cos ( T t % 5 B / 6 ) .
During the interval 21 # Tt # 22 vc is undefined, but ic = 0 and ia = !ib. Current flows in the loop ab. No
torque is produced, and the rotor remains stationary. A particular case of interest is to close line b at
a 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 resistance
is RLL = 2Ra and the line-line inductance LLL is
L LL ' L aa % L bb ! 2 L ab ' 2 L11 .
(3.47)
Obviously current is induced in the rotor. The rotor circuit can be reduced to a single circuit by fixing
the d-axis to the rotor and aligning it with the axis of the effective stator coil ab that results from the
loop current flowing in lines a and b. Then the whole circuit is reduced to that of a single-phase
transformer. With ic = 0 we have
id '
2
i [ cos 2 ! cos ( 2 ! 2 B / 3 ) ] ' 2 ia / 3 cos ( 2 % B / 6 ) ;
3 a
2
iq ' ! ia [ sin 2 ! sin ( 2 ! 2 B / 3 ) ] ' ! 2 ia / 3 sin ( 2 % B / 6 )
(3.48)
3
and the required alignment is achieved by setting 2 = 2r = !B/6, with id =2 ia/%3 and iq = 0: that is, all
the 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, while
eqn. (3.42) holds for RD with T = Tr = 0:
p RD ' ! R D iD .
(3.49)
The mutual inductance between the rotor D-coil and the stator d-coil is L12 = 3/2 × LaA. Since the d- and
a- coils have the same number of turns, the mutual inductance between the D-coil and phase a is also
3/2 × LaA = L12 when their axes are aligned. With 2 = 2r = !B/6 this mutual inductance is decreased to
Page 3.22
SPEED’s Electric Motors
Fig. 3.18 Alignment of dq axes with the conducting loop in phases a and b.
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 with
current in the stator loop ab and ib = !ia phase b makes an equal contribution to Rd. Therefore the total
mutual inductance MD-LL is 2 × L12 × %3/2 = %3 L12, and
RLL ' L LL ia % 3 L12 iD
RD ' L12 id % L D iD
(3.50)
where RLL is the flux-linkage of the series connection of phases ab. Using the inverse reference-frame
transformation it can be shown that
RLL ' Rab ' Ra ! Rb '
3 Rd
(3.51)
and since ia = /3/2 id and LLL = 2 L11, we can write eqns. (3.50) as
Rd ' L11 id % L12 iD ;
RD ' L12 id % L D iD .
(3.52)
This agrees with eqns. (3.34) and (3.36), since LD = L22 . The voltage equation for the stator loop is
p RLL ' v LL !
3
2
R LL id .
(3.53)
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 line
c closes, all three motor terminal voltages are known: eqn. (3.45) can be used for vd and vq, and the
solution can proceed in d,q axes as before, provided that the final currents at the end of the delay are
transformed into initial values for id and iq.
Induction machines
Page 3.23
Steady-state operation
The theory so far assumes that we know the inductances L11, L12 etc., but we need to relate these to the
familiar "equivalent circuit" parameters of the induction motor: R1, R2, X1, X2, and Xm. In the steady
state the terms in eqns. (3.42) containing the p operator are zero so that
v d ' R d id ! T Rq ' R d id ! T ( L11 iq % L12 iQ ) ;
v q ' R q iq % T Rd ' R q 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)
We can combine these equations by writing V = Vd + jVq where Vd = vd/%2 and Vq = vq/%2 and similarly
with the currents. This can be seen as follows: if 2 = Tt then
From eqn. (3.54) it follows that
V d % j V q ' [ R1 % j T L11 ] ( I d % j I q ) % j T L12 ( I D % j I Q )
(3.56)
where R1 = Ra and L12 = 3/2 × LaA. We can write I2 = ID + j IQ directly for the "rotor current", while
for the "stator current" I1 = Id + jIq. The construction of the equivalent circuit requires also the
equivalent of eqn. (3.56) for the rotor, i.e.
0 ' [ R2 % j s T L22 ] ( I D % j I Q ) % j s T L12 ( I d % j I q ) .
(3.57)
where R2 = RA = RD. Eqns. (3.56) and (3.57) represent the conventional equivalent circuit and so we
have 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, the
rotor circuit is "referred" to the stator with the same frequency T and the familiar R2/s appears as the
only manifestation of rotation.
Small-signal analysis
For this we assume small perturbations (denoted by )) about a steady operating point (denoted by
subscript 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 missing
because it is considered fixed and equal to 2Bf, but Tr can vary. If we substitute these into the voltage
eqn. (3.42) and the torque eqn. (3.40) and simplify by subtracting the "steady-state" components, while
ignoring products of the form )x)y, then after some grinding we get
p)Rd
!1/JsN
T
k r/JsN
0
0
)Rd
)vd
p)Rq
!Tr0
!1/JsN
0
k r/JsN
0
)Rq
)vq
k s/JrN
0
!1/JrN
T ! Tr0
!Rq0
)RD
p)RQ
0
k s/JrN
0
!1/JrN
Rd0
)RQ
0
p)Tr
a51
a52
a53
a54
!D/J
)Tr
0
p)RD
=
+
0
(73)
where
a51 '
a52 ' !
3 ( P / 2 )2
2
J
[ iq0 !
3 ( P / 2 )2
2
J
[ id0 !
Rq0
L sN
]
Rd0
L sN
(3.59)
]
(3.60)
SPEED’s Electric Motors
Page 3.24
a53 '
3 ( P / 2 )2 k r
J
2
a54 ' !
L sN
Rq0
3 ( P / 2 )2 k r
2
J
L sN
Rd0
(3.61)
(3.62)
and JsN = LsN/R1, JrN = LrN/R2, LsN = (LsLr ! M2)/Lr, LrN = (LsLr ! M2)/Ls, ks = M/Ls, kr = M/Lr, and P is
the number of poles. The notation has been changed using Ls = L11; Lr = L22, and M = L12 so that the
matrix eqn. (3.58) can be more easily compared with Vas' eqn. 2.10-36 [9] which is derived using spacevector theory.
In the calculation of the coefficients a51..a54, the determination of Rd0 and Rq0 generally requires the
inversion of a 4 × 4 matrix but this can be avoided if the eigenvalues are calculated for the zero-slip
condition. This makes iD0 = iQ0 = 0 and T!Tr0 = 0. From eqn. (3.57) the steady-state conditions are
v d0 ' R d id0 ! T Rq0 ; Rq0 ' L11 iq0
v q0 ' R q iq0 % T Rd0 ; Rd0 ' L11 id0
(3.63)
for which the solution is
id0 '
3.9
T L11 v q0 % R a v d0
R a2 % T2 L112
; iq0 ' !
T L11 v d0 ! R a v q0
R a2 % T2 L112
(3.64)
SPLIT-PHASE MOTORS
Fig. 3.19 shows the equivalent circuit for a split-phase induction motor and Fig. 3.20 shows a simplified
form of the phasor diagram for balanced operation. Split-phase motors are calculated by the method
of forward and backward rotating fields (Veinott [1959], Morrill [1929]); or by the method of symmetrical
components (Fitzgerald and Kingsley [1961]); or by the cross-field method. The harmonics of the MMF
distribution are ignored.
Fig. 3.19 Equivalent circuit of split-phase motor. During start-up, the cut-out switch switches from "start" to "run" at a certain
percentage of the synchronous speed. The OC (open-circuit) position of the cut-out switch simply represents here the
electrical connections of a single-phase motor.
Induction machines
Page 3.25
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 speed
is below the cut-out speed, this auxiliary impedance is usually that of a start capacitor Cstart. Above the
cut-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 following
equation in which "m" and "a" refer to the main and auxiliary windings, respectively; . is the angle
between the winding axes, and " is the phase difference between the currents ia and im :
F ' im0N mcos 2 cos Tt %
% ia0N acos (2%.) cos (Tt%")
'
%
1
2
1
2
(3.65)
im0N m[cos (2%Tt)%cos (2!Tt)]
ia0N a[cos (2%Tt%.%")%cos (2!Tt%.!")]
Let im0Nm = ia0Na = Fm : then
1
F '
%
2
1
2
Fm[cos (2%Tt) % cos(2%Tt%.%")]...[b]
Fm[cos (2!Tt) % cos(2!Tt%.!")]...[f]
(3.66)
For the backward component b to be zero,
" % . ' B
(3.67)
Then if . = B/2, " = B/2, i.e., the auxiliary current must lead the main current by 90E. In this case the
MMF reduces to F ' F m cos (2!Tt) .
Let
a '
Na
Nm
'
Im
Ia
'
Va
Vm
(3.68)
SPEED’s Electric Motors
Page 3.26
then
Ia '
Im
a
e j" ; V a ' a V m e j"
(3.69)
and if . = B/2 then
Ia ' j
Im
a
; V a ' ja V m
(3.70)
These relationships are shown in the phasor diagram, Fig. 3.20.
Capacitor reactance and turns ratio: The circuit connection (Fig. 3.19) makes
Vm ' Vs ' Va % Vc
(3.71)
If we write V m ' Z m I m ' Z m I m e j N and V c ' !j X c I a , eqn. (3.71) reduces to
e !j " ! a ' !j
Xc
a Zm
e !j N
(3.72)
Taking real and imaginary parts,
cos " ! a ' !
Xc
a Zm
sin N ;
sin " '
Xc
a Zm
cos N
(3.73)
Taking the squares of these equations. and adding them, and substituting " = B ! ., we get
Xc '
aZ m sin .
cos N
(3.74)
and if the first of eqns. (3.73) is divided by the second we get
a ' sin . . tan N ! cos .
If . = 90E then
Xc '
a Zm
cos N
and
a ' tan N
(3.75)
(3.76)
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
V c ' V m ! V a ' V m [1 ! a e j"]
(3.77)
V c ' V m [1 ! j a]
(3.78)
Vc ' Vm 1 % a 2
(3.79)
and
Induction machines
Page 3.27
Capacitor reactive power (with . = B/2)
Im
Qc ' Vc I a ' Vm 1 % a 2 ×
a
' I m2 Z m
1 % a2
a
(3.80)
In the balanced condition the total input power to the motor is
Pin ' 2 I m2 Z m cos N
(3.81)
But a ' tan N so cos N ' 1/ 1 % a 2 and
Qc ' Pin .
1 % a2
' Pin .
2 a cos N
1 % a2
2a
(3.82)
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
I m2 Z m
Qc '
(3.83)
sin N
Supply current (with . = B/2)
1 % a2
Is ' Im
'
a
Im
sin N
(3.84)
The supply current does not have a minimum value. For all practical cases (N < B/2), the supply current
exceeds the main phase current.
Supply power factor (with . = B/2) :The apparent impedance at the primary terminals is
Z s ' Zs e
'
'
j Ns
'
Vs
Is
Vm
'
I m(1 % j/a)
Z me j N
(3.85)
(1 % j/a)
a Zm
2
a % 1
!1
e j[N ! tan
(1/a)]
In general a = tan N , so in the special case when a = 1, N = 45E, then Ns = 0 and the supply power-factor
is 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.
Page 3.28
SPEED’s Electric Motors
Forward and backward revolving-field theory
Fig. 3.21 Single-phase equivalent circuit, modified with core-loss resistance
The calculation methods available in PC-IMD include the forward- and backward-rotating field method
of Morrill, [1929]; the symmetrical-component method; and the cross-field method. All of these are
described by Veinott [1959] in their basic form, but PC-IMD implements them with a core-loss
calculation in which the resistance Rc is computed iteratively until the circuit equations converge. The
core losses are calculated independently from the flux-density waveforms and are represented as an
electrical 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 easily
incorporated in all three methods, and PC-IMD’s calculations are all independently based on original
analysis with several ad hoc modifications and improvements.6
In PC-IMD the forward- and backward-rotating field model and the cross-field model are further
extended to cover tapped-winding motors of various configurations.
Fig. 3.21 shows the circuit representation of the pure single-phase motor using forward and backward
rotating fields. The impedance Zf is the parallel combination (jXm/2 ) 2 (R2/s + jX2)/2, and Zb is the
parallel combination (jXm/2 ) 2 (R2/(2 ! s) + jX2)/2, in accordance with the fact that the actual pulsating
airgap flux in a single-phase motor is resolved into two equal counter-revolving fields, such that half
the voltage induced by the airgap flux is due to the forward field and half to the backward field. As
Veinott 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 method
mixes direct phase variables (in the main and auxiliary windings) with the forward and backward
rotating 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 the
positive and negative sequence currents by Im = Ip + In and Ia = j (Ip ! In) /a, where a is the effective
auxiliary/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 important
case is the cross-field theory of the tapped-winding motor, which is described in the next section.
5
6
The desirability of this approach was recognized 60 years ago, but it was impractical on desktop computers until recently.
In principle the models could be derived from one another by reference-frame transformations of generalized-machine theory.
Induction machines
z
Im
1m
Page 3.29
z
I2
Ia
1a
Ic
Z
E fm
Zf
a2Z f
c
E fa
Va
Vm
!j
E fa
jaE fm
a
Rc
E 2m
E bm
j
Zb
a2Z b
E ba
E ba
!jaE bm
a
SplitPh.wpg
Fig. 3.22 Rotating-field model of capacitor motor, including core-loss resistance
z
Ip
1m
If
Zp
I cf
Vm
R cf
Vp
Ef
Zd
I p ! In
Zc
I cb
Vn
R cb
z
In
Zf
Zb
Eb
1m
Zn
Ib
Fig. 3.23 Symmetrical-component model of capacitor motor
Page 3.30
3.10
SPEED’s Electric Motors
CROSS-FIELD THEORY OF TAPPED-WINDING CAPACITOR MOTOR
Fig. 3.24
Tapped-winding capacitor motor circuits. (a) Base circuit analyzed; (b) G-tap circuit, [1].
Circuits
Fig. 3.24(a) shows the generic circuit of a tapped-winding capacitor motor, in which the auxiliary
winding 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 that
the aux winding current and voltage normally lead the main winding voltage and current in phase, and
the rotor rotates from the aux towards the main. The directions of positive current (closed arrows) and
positive MMF and flux (open arrows) are shown in Fig. 3.24(a), and the polarities of the winding sections
are 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 tapping
is represented by the tap ratio t. When t = 0 the tap is at full voltage, but when 0 < t < 1 the tap is at a
lower voltage.
Fig. 3.25(a) shows a more flexible circuit that can represent several different configurations of tappedwinding capacitor motor connections. The intention is to have a basic “core” analysis for the circuit
of 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). This
circuit can be transformed into the base circuit by exchanging the “aux” and “main” labels and
reflecting it about a vertical line. Mathematically this reflection is the same as reversing the direction
of rotation.7 The exchange of labels means that the impedances must be exchanged before the analysis
starts, and the appropriate currents must be exchanged after it finishes. Note that the “main” capacitor
will 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-unit
speed from positive to negative, so that the voltages and currents of the "aux" winding always lead those
of the main winding.
7
This makes the per-unit speed S equal to (s ! 1) instead of (1 ! s), where s is the slip.
Induction machines
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.
Page 3.31
Page 3.32
SPEED’s Electric Motors
Construction of voltage equations
The tapped winding is decomposed into a T-equivalent as shown in Fig. 3.27. The magnetizing reactance
is 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 with
the rotor circuits. The resulting circuit is suitable for analysis by either the cross-field method or the
double revolving-field method. The dot convention in Fig. 3.27 means that the two sections of the tapped
winding produce MMF’s in the same direction when they both conduct positive current. The direction
of 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 checking
the signs of the speed voltages.10 The main winding 1m has a self-impedance volt-drop and a mutual voltdrop due to current in the “rotor main” winding 2m. Likewise the auxiliary winding 1a has a selfimpedance volt-drop and a mutual volt-drop due to current in the “rotor aux” winding 2a. There are
no speed voltages in the stator windings.
The rotor main circuit 2m has a self-impedance volt-drop and a mutual impedance volt-drop due to
current in the “stator main” 1a. In addition it has speed voltages of the form SxI induced by rotation
at the per-unit speed S through flux-linkages xI established by the orthogonal windings 1a and 2a. For
winding 1a, reactance x excludes the stator leakage; but the rotor leakage is included in x for rotor
winding 2a: in other words all the flux produced by current in rotor aux 2a generates a speed voltage
in rotor main 2m. The direction of the speed voltage in 2a induced by Im1 is shown in Fig. 3.26(c). The
speed voltage in 2m induced by Ia1 is shown in Fig. 3.26(d). Fig. 3.26(e) summarizes the directions of the
speed voltages the rotor circuits in each axis. With this we have enough to construct the voltage
equations 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 with
the polarity indicated by the cross and the dot. The arrow representing the flux defines the axis of the coil. The axes
of all the four windings are shown in (b). Note that the auxiliary axis is ahead of the main axis, since positive rotation
is counter-clockwise. Also, the axes of the stator and rotor coils 1m and 2m are in the same direction, as are those of
1a and 2a. This means that positive current in coil 1m produces flux in the same direction as positive current in coil
2m. In (c), positive current in coil 1m induces a speed voltage in coil 2a in the negative direction. Likewise, positive
current in coil 2m will also induce a negative speed voltage in coil 2a. In (d), positive current in coil 1a induces a speed
voltage in coil 2m in the positive direction. Likewise. positive current in coil 2a will induce a positive speed voltage
in coil 2m. No speed voltage is induced in coils 1m or 1a.
9
Flux associated with the reactance !t(1 !t)Xm does not link the auxiliary winding or any rotor circuits, and therefore does not
generate any speed voltages. This reactance is merely an artefact of the T-equivalent circuit and accounts for the coupling
between the tapped and untapped sections of the main winding.
10
The sign convention for the rotor currents is opposite to that used in [3]. All coils on the same axis produce MMF and flux in
the same direction, if their currents are in the same direction.
Induction machines
I 1m
Page 3.33
main
t(r1m + jx1m)
V 1m
I
j t x MI2m
tapped section
ct
E
t
jtxM
I1a
!jt(1 ! t)xM
untapped section
aux
Ica
j (1 ! t) x M
E
j (1 ! t) x MI2m
Icu
E
u
a
jxA
V1a
(1 ! t) ( r 1 m + j x 1 m )
Z cm
stator
j x AI2a
r 1a + j x 1 a
Z ca
I2a
I2m
r 2m + jx 2 m
S a t x M (I 1 m ! I ct )
r 2a + j x 2 a
S a x M (I 1 a ! Ic a)
jxM
Sa(1 ! t)xM(I1m ! I1a ! I cu )
Sa(x2m+xM)I2a
j t x M(I1m ! Ict)
S a (xM + x2m)I2m
j (1 ! t) x M(I 1 m ! I 1 a ! I cu )
rotor
Fig. 3.27
jxA
j x A(I1a ! I ca )
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 fluxgenerating elements in each branch of the circuit. Consider a winding wound on a core in which the
2
core losses are W. If the EMF in the winding is E, the expression W = GE reflects the assumption that
the core loss is proportional to the square of the flux, which itself is proportional to the EMF. The scaling
factor G is a fictitious conductance, which draws current from the circuit in phase with the EMF E.
SPEED’s Electric Motors
Page 3.34
2
2
If the winding is re-wound with a times as many turns, then W = (G/a )E . By this logic the conductance
2
associated with the auxiliary winding is G/a , and it represents that fraction of the core loss that is
assumed 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 winding
circuit, but where the main winding is tapped it is assumed that a fraction t of the core loss Wm
attributable to the component of flux in the main axis is associated with the tapped part of the winding.
2
2
2
The remaining fraction (1 ! t) is associated with the untapped part. Thus tWm = t(G/t )Et = (G/t)Et , so
that the conductance associated with the tapped part of the winding is G/t. Similarly the conductance
associated with the untapped part is G/(1 ! t). Note that the relationship between induced EMF and the
corresponding component of core loss is assumed to be the same for all three winding sections.
The total core loss satisfies the equation
W Fe ' G
E t2
t
%
E u2
1 ! t
%
E a2
a2
(86)
.
The total core losses WFe are calculated separately from an estimate of the flux-density waveforms in
the 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 practice
a 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
Pgap ' Re [ E1 I1m( % E2 ( I1m ! I1a )( % E3 I1a( ] ' Re [ (E1 % E2) I1m( % (E3 ! E2) I1a( ]
(87)
where the core-loss currents have been omitted for clarity, and
E1 ' v1 % j txM I1m ' j txM ( I1m % I2m ) ,
E2 ' v2 % j ( 1 ! t ) xM ( I1m ! I1a ) ' j ( 1 ! t ) xM ( I1m % I2m ! I1a ) ,
(88)
and
E3 ' v3 % j xA I1a ' j xA ( I1a % I2a ) .
(89)
(90)
Removing products of voltages and currents which are obviously in quadrature, we get
Pgap ' Re [ v1 I1m( % v2 ( I1m ! I1a )( % v3 I1a( ] .
(91)
Torque
In a balanced polyphase motor the electromagnetic torque is equal to Pgap in “synchronous watts”, but
in unbalanced motors this is not so, and it is necessary to differentiate between the separate torques
contributed by the forward and backward revolving fields.10 The torque can also be expressed in terms
of forward and backward fluxes Mf and Mb and MMF’s Ff and Fb, related to the main-axis flux Mm and MMF
Fm and the auxiliary-axis flux Ma and MMF Fa by
M f ' Mm ! jM a;
Mb ' Mm % jMa
Ff ' Fm ! jFa;
Fb ' Fm % jFa.
(92)
10
An alternative is to use dq-axis theory. Puchstein & Lloyd [1941] quote a torque formula that is effectively based on dq-axis
theory, and Veinott [1959] provides a physical explanation of this method; but in many early works the formal mathematical
development 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 speed
in rad/s, i.e. Te = TSW × p/T. In balanced polyphase machines TSW = Pgap.
Induction machines
The inverses are
Mm '
1
Fm '
1
2
2
(M f % M b );
Ma ' j
1
(F f % F b );
Fa ' j
1
(Mf ! Mb)
2
2
Page 3.35
(Ff ! Fb).
(3.93)
The average electromagnetic torque is then given in synchronous watts by 11
Te '
T
2
(
Re ( j M f Ff
! j M b F(b ) .
(3.94)
where T = 2Bf. The amplitude of the double-frequency pulsating torque is given in Nm by
p
2 /
T pls '
j M f Fb
! j M b Ff / ,
(3.95)
and the airgap power is given by
Pgap '
T
2
(
Re ( j M f Ff
(
% j M b Fb ) .
(3.96)
With Te in synchronous watts, the power conversion is (1 ! s)Te, so the rotor copper loss is
WCu R ' Pgap
!
! s ) Te .
(1
(3.97)
The fluxes Mm and Ma are obtained as
Mm '
Et
j T t Nm
and
Ma '
Ea
j T a Nm
(3.98)
where
E t ' j t xM ( I1m ! I ct % I2m ) ! j t ( 1 ! t) xM ( I1a ! Ict % Icu )
(3.99)
and
E a ' j t xA ( I1a ! Ica % I2a )
(3.100)
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
Eu ' Et ×
1 ! t
t
.
(3.101)
Finally the MMF’s Fm and Fa are given by
F m ' t N m ( I1m ! Ict ) % ( 1 ! t ) N m ( I1m ! I1a ! Icu )
(3.102)
F a ' a N m ( I1a ! Ica ) .
(3.103)
and
3.11
INTERBAR CURRENTS
See [31].
3.12
SATURATION OF LEAKAGE REACTANCE
See [32].
11
Eqns. (3.94)—(3.97) apply to all AC induction machines provided that Mm, Ma, Fm and Fa are defined as the actual fluxes and
Equivalent equations are obtained with symmetrical component analysis except that the factor ½ is replaced by a factor
that depends on the form of the symmetrical components transformation; for capacitor motors the factor is usually 2.
MMF’s.
Page 3.36
SPEED’s Electric Motors
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, Second
Edition, 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 the
title 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, 3rd
edition, 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 to
switching 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, New
York, 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 Physics
Publications/Oxford University Press, 1994 ISBN0-19-859389-9
16.
Veinott CG and Martin JE, Fractional and subfractional horsepower electric motors, Fourth
edition, 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 the
computation of airgap reactances in 3-phase induction motors, IEEE Industry Applications
Society, 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 capacitor
motor, 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: Proektirovanie
elektriceskih 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.
Induction machines
Page 3.37
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, IEEE
Industry Applications Society, IAS 2001, Chicago, USA, 30 Sep – 5 Oct. 2001.
32.
Miller TJE, Boldea I, Dorrell DG, Rasmussen CB [2000] Leakage Reactance Saturation in
Induction Motors, ICEM 2000, Helsinki, Finland, 28-30 August 2000 pp. 203-207.
Page 3.38
SPEED’s Electric Motors
Index
2-phase winding 1
3-phase windings 2
Airgap power 4, 5, 34, 35
Airgap torque 5
Alger 8, 9, 11, 36
Balancing theory 25
Breakdown torque 6, 7
Cage-rotor 7
Capacitor motor 28-31, 33, 36, 37
balance theory 25
Capacitor reactance 26
Capacitor voltage 26
Closed rotor slots 8
Closed slots 14
Conductivity 10
Constant Volts/Hz 7
Core loss 3, 6, 33, 34
Core losses 28, 33, 34
Cross-field theory 28, 30, 36, 37
Current density 10, 11, 15
Deep-bar effect 15
Deep-bar 7, 8, 15, 28
Differential 8, 11, 20
Distribution factor 10
Double-cage 7, 8
Efficiency 3-5, 7
Eigenvalue analysis 17
Electromagnetic torque 4, 5, 20, 34, 35
Equivalent circuit 3-9, 23, 24, 28, 32, 33
Finite-element 15, 16
Forward and backward components 25
Forward and backward revolving-field theory 28
Fractional-slot 10
Harmonics 8, 10, 11, 24
Interbar currents 35, 37
Leakage reactance 6-8, 11, 16, 35, 37
Location of ampere-conductors 16
Magnetizing reactance 3, 6, 8, 28, 32
Non-simultaneous switching 21
Park 17
Phase-belt harmonics 8
Phasor diagram 4, 6, 8, 16, 24-26
Pitch factor 10
Pole amplitude modulation 7
Power 3-5, 12, 18, 20, 26, 27, 34, 35
Power factor 5, 26, 27
Rotating magnetic field 1
Saturation 6, 8, 14, 28, 35, 37
Shaft torque 4
Sine-distributed windings 1
Skew 8, 10
Skew factor 10
Slip 2-7, 16, 17, 20, 23, 24, 30
Slot designs 25
Slot numbers 11, 12
ratio 11
Slot permeance 8, 13, 14
Slot-MMF harmonics 8
Speed control 7
Split-phase 24, 25
Split-phase motors 24
Starting torque 7
Stray load loss 6, 34
Synchronous speed 1-5, 7, 9, 11, 16, 20, 24, 25, 34
Synchronous watts 35
Tapped-winding 28, 30, 31, 33, 36
Torque 2, 4-7, 16, 17, 20-23, 34, 35
Torque/slip curve 6, 7
Torque/speed characteristic 5-7
Transients 17, 20
Winding factors 10
Wound-rotor 7
Zig-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
4.
SWITCHED RELUCTANCE MACHINES
4.1
DEFINITIONS
A reluctance machine is an electric machine in which torque is
produced by the tendency of its moveable part to move to a position
where the inductance of the excited winding is maximized. As we
have seen in chapter 1, this definition covers both switched and
synchronous reluctance machines. The switched reluctance motor
has salient poles on both the rotor and the stator, and operates like
a 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 two
coils wound on opposite stator poles are excited simultaneously,
and generate magnetic flux as shown. There is only one phase. In
the position shown, the resulting torque tends to rotate the rotor in
the counterclockwise direction towards the aligned position, Fig.
4.2. This machine can produce torque only over a limited arc of
rotation, roughly corresponding to the stator pole arc $s. However,
it is the basic model on which the theory of torque production is
based, so we will analyze it first, and then consider methods of
starting and the extension to multiple poles and phases.
Fig. 4.2
Aligned position
Fig. 4.1
Fig. 4.3
Simple reluctance machine
with one phase and two poles
on both the stator and rotor.
Unaligned position
Aligned and unaligned positions
For the primitive reluctance machine in Figs. 4.1!4.3, the aligned and unaligned positions are
characterized 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 equilibrium
TABLE 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 shown
in Fig. 4.4. Positive rotation is in the counterclockwise direction. Assume that the coil carries a
constant current. Positive motoring torque is produced only while the inductance is increasing as the
rotor approaches the aligned position; that is, between positions J and A. At J, the leading edge of the
rotor 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
Fig. 4.4
Page 4.3
Variation of inductance and torque with rotor position; coil current is constant. The
small icons show the relative positions of the rotor and stator poles, with the rotor
moving to the right. A = aligned position; U = unaligned position; J = start of
overlap; K = end of overlap.
The torque changes direction at the aligned position. If the rotor continues past A, the attractive force
between the poles produces a retarding (braking) torque. If the machine rotates with constant current
in the coil, the negative and positive torque impulses cancel, and therefore the average torque over a
complete cycle is zero. To eliminate the negative torque impulses, the current must be switched off
while the poles are separating, i.e., during the intervals AK, as in Fig. 4.5.
Fig. 4.5
Variation of inductance, current, flux-linkage, torque, and EMF with rotor position, with ideal pulsed unidirectional
current
Page 4.4
SPEED’s Electric Motors
The ideal current waveform is therefore a series of pulses synchronized with the rising inductance
intervals. The ideal torque waveform has the same waveform as the current. The cycle of torque
production associated with one current pulse is called a stroke. Evidently the production of continuous
unidirectional torque requires more than one phase, such that the gaps in the torque waveform are
filled 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 is
generally flowing for only a fraction of the rotor pole-pitch. Note that the current and inductance
waveforms imply a sawtooth waveform of flux-linkage R = Li. Such a waveform is not practical
because 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 beginning
of the stroke (J) is zero. In practice the inductance along UJ is very small so the leading-edge di/dt can
be very large, presenting a possible problem for the power semiconductors. The rectangular current
waveform in Fig. 4.5 can be approximated at low speed by chopping the current along JA, which has
the effect of reducing the average forward applied voltage along JA to a value Va much lower than the
supply voltage Vs. If there is no chopping after commutation at the end of the stroke, the reverse
voltage !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 magnetic
saturation. For simplicity we also ignore the effect of fringing flux around the pole corners, and
assume that all the flux crosses the airgap in the radial direction. Mutual coupling between phases is
normally zero or very small, and is ignored. The voltage equation for one phase is
v ' Ri %
dR
' R i % Tm
dt
' R i % Tm
d (L i)
dR
d2
' Ri % L
di
% Tm i
(4.1)
dL
d2
dt
d2
where v is the terminal voltage, i is the current, R is the flux-linkage in volt-seconds, R is the phase
resistance, 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:
e ' Tm i
dL
d2
.
(4.2)
It is helpful to visualize the supply voltage as being dropped across the three terms in (4.1): namely, the
resistance voltage drop, the L di/dt term, and the back-EMF e. The instantaneous electrical power vi is
vi ' R i 2 % L i
di
dt
% Tm i 2
dL
d2
.
(4.3)
The rate of change of magnetic stored energy at any instant is given by
d
1
dt 2
Li2
'
1
2
i2
dL
dt
% Li
di
dt
1
'
2
i 2 Tm
dL
d2
% Li
di
dt
(4.4)
According to the law of conservation of energy, the mechanical power conversion p = TmTe is what is
2
left after the resistive loss Ri and the rate of change of magnetic stored energy are subtracted from the
power input vi , Te being the instantaneous electromagnetic torque. Thus from eqns. (4.2) and (4.3),
2
2
writing Te = p/Tm = vi ! Ri ! d(½Li )/dt, we get
Te '
1
2
i2
dL
d2
.
Note that dL/d2 is the slope of the inductance graph in Fig. 4.5.
(4.5)
Switched reluctance machines
Page 4.5
Drive circuit — unidirectional current, bidirectional voltage
Eqn. (4.5) says that the torque does not depend on the direction
2
of the current, since i is always positive. However, the voltage
must be reversed at the end of each stroke, to return the fluxlinkage to zero. By Faraday’s law, this requires a negative
voltage applied to the coil, to ensure that dR/dt < 0. Fig. 4.6
shows the half-bridge phaseleg circuit that accomplishes this.
When Q1 and Q2 are both on, the voltage across the motor
windings is v = Vs, the supply voltage. When Q1 and Q2 are both
off, 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, the
reverse voltage must be much larger than the forward voltage,
otherwise the flux-linkage will persist beyond the aligned
position, producing an unwanted negative pulse of torque. At low
speeds this is achievable by chopping the forward voltage,
reducing its effective value compared to the reverse voltage.
Fig. 4.6
Single phaseleg circuit
The circuit of Fig. 4.6 can operate the machine as a motor or as a
generator, since the electrical power vi can be positive or
negative. If the average power is negative (i.e., generating), the energy supplied during transistor
conduction in one stroke must be less than the energy recovered during freewheeling. The transistor
conduction period (with positive applied voltage) is still necessary to establish the flux, which is built
up from zero and returned to zero each stroke. The voltage-time integrals during transistor conduction
and freewheeling must be approximately equal (apart from resistive volt-drop), regardless of whether
the machine is motoring or generating. As regards control, the main difference between motoring and
generating is the phasing of the conduction pulse relative to the rotor position. From Fig. 4.4 it appears
that generating current pulses must coincide with AK, just as motoring pulses coincide with JA.
Fig. 4.7
Alternative single-phaseleg circuits
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 two
parts of the winding could be bifilar-wound or they could have completely different numbers of turns
and wire sizes. Unfortunately the leakage inductances of the two parts of the winding are usually quite
large, 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.
SPEED’s Electric Motors
Page 4.6
Additional phases simply use additional drive circuits of the same form as the first phase, usually with
a common voltage source. With four-phase motors it is possible to use a common chopping transistor
between two phases, so that only six power transistors are required; but the control range of
conduction angles must be limited limited to allow de-fluxing of the “complementary” phase in each
pair.
Limitations of the ideal linear model
At the aligned position A in Fig. 4.5, the current
must be switched off quickly to avoid the
production of negative torque after the poles have
passed the aligned position. The magnetic stored
2
energy ½Lai must be returned to the supply. In a
nonsaturating reluctance machine of this type, the
magnetic energy stored at A is large, because both
L and i are at their maximum values. We can get
further insight from an energy audit taken over one
stroke as the rotor moves from J to A. The process
is shown in the energy conversion diagram, Fig. 4.8,
which plots flux-linkage against current. The slope
of OU is the inductance at the unaligned position,
Lu, and the slope of OA is the inductance at the
aligned position, La. At intermediate positions the
Fig. 4.8
Linear energy conversion diagram
inductance is represented by a line of intermediate
slope between Lu and La. At the J position (start of overlap), if fringing is neglected (as in the idealized
inductance variation in Fig. 4.4), LJ = Lu. The complete stroke is represented by the locus OJAO. In
motoring operation it is traversed in the counterclockwise direction, and in generating operation in
the 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. 4.8
2
this step is along OJ and energy OJC = ½Luim must be supplied to the magnetic circuit as the current
increases from 0 to im. The step cannot be accomplished in zero time, since that would require infinite
voltage, but if the angular velocity is low, the angle of rotation along OJ is small. Along JA the
electromechanical energy conversion is W or OJA, given by
W ' OJA ' T e )2 '
1
i 2
2 m
dL
d2
)2 '
1
i 2
2 m
)L .
(4.6)
Along JA there is a back-EMF e which absorbs energy equal to the area ABCJ:
e i )t ' ABCJ ' im2 Tm
dL
d2
×
)2
' im2 )L
Tm
(4.7)
where )t = )2/Tm is the time taken to rotate through the interval )2 = JA, and )L = La ! Lu is the
change in L. The total energy supplied is the sum of areas OCJ and ABCJ, i.e., area S = W + R = OJAB,
and
S ' im2 )L %
1
2
im2 L u .
(4.8)
We can now define the energy ratio as
Q '
energy converted
energy supplied
W
'
S
%
W
W % R
(4.9)
and if we write 8 = La/Lu we can substitute for W and S from (4.6) and (4.8) to give
Q '
8 ! 1
28 ! 1
.
(4.10)
Switched reluctance machines
Page 4.7
In this type of nonsaturating motor, less than half the energy S supplied by the drive is converted into
mechanical 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 from
the 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 makes
Q < 0.5. The inverse of the energy ratio is the converter volt-ampere-seconds per joule of energy
conversion 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 end
of the stroke by commutating the current into the diodes, so that the voltage reverses and forces the
flux-linkage to fall to zero. In the ideal locus this fall is along AO. However this is not possible with
finite supply voltage. If the current is chopped with a duty-cycle d along JA, the average forward
applied voltage along JA is d × Vs. In a circuit of the form of Fig. 4.6, the reverse voltage after
commutation is !Vs. By integrating Faraday’s law the rise and fall periods of flux-linkage can be shown
to be in the ratio
tf
' d.
(4.11)
tr
This shows that “instant” suppression of the flux is effectively achieved at very low speed when d is
small. But at higher speed there is no chopping: the forward and reverse voltages are equal in
magnitude, and d =, so tf = tr and the flux-linkage waveform is triangular. The time taken along OA is
the same as the time taken along OJA, The waveforms of flux-linkage and current corresponding to
Fig. 4.8 are shown in Fig. 4.9, and show a tail in the current waveform extending past the aligned
position 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 to
maintain a flat-topped current waveform. With full voltage applied, the speed at which this occurs is
called the base speed. The current im could be called the base current. (It is not necessarily the rated
current, because that depends on the cooling arrangements).
Fig. 4.9
Current and flux-linkage waveforms corresponding to Fig. 4.8
Page 4.8
SPEED’s Electric Motors
In practical terms the nonsaturating switched reluctance motor makes poor use of the power
semiconductors, because they have to supply more than 2J of energy in order to get 1 J of mechanical
work, 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 the
energy conversion W: in fact
R ' W ×
1 ! Q
Q
.
(4.12)
For example, if 8 = 8, Q = 0.467, C = 2.14, and R = 1.14W. This implies that for zero ripple voltage at the
DC link, the filter capacitance must be large enough to absorb R joules with negligible change of
voltage. This requirement may be reduced by overlapping charge/discharge requirements of adjacent
phases, but still it is a serious consideration.
Eqns. (4.1) and (4.2) imply the existence of an equivalent
circuit of the form shown in Fig. 4.10, in which there is a
back-EMF e = Tm idL/d2. Unfortunately e is not an
independent parameter, but depends on the current. In
a normal equivalent circuit we interpret the product e i as
the electromechanical power conversion TmTe, implying
2
that Te = i dL/d2. However, eqn. (4.5) states that the
2
torque is only ½i dL/d2. Of the power ei, only half is
converted into mechanical power during the “working”
part of the stroke JA. The other half is being stored as
Fig. 4.10 Equivalent circuit
magnetic field energy in the increasing inductance. With
L 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 reluctance
machines and their drives requires the direct solution of eqns. (4.1) and (4.5), even when saturation is
ignored. An elegant and thorough solution for the nonsaturating motor was presented by Ray and
Davis [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 has
a low energy ratio and makes poor utilization of the
drive. Practical switched reluctance machines are
more effective but they are far from being
magnet ically linear. To underst and t he
electromechanical energy conversion properly, we
need a nonlinear analysis that takes account of the
saturation of the magnetic circuit. One such analysis
is based on the magnetization curves. A magnetization
curve is a curve of flux-linkage R versus current i at
a particular rotor position, Fig. 4.11.
We also need to define the stored magnetic energy Wf Fig. 4.11
and the coenergy Wc graphically, as in Fig. 4.12.
A magnetization curve at one rotor position
Mathematically,
Wf '
Wc '
m
m
i dR ;
R di .
(4.13)
In a magnetically linear device with no saturation, the magnetization curve is straight, and Wf = Wc.
Switched reluctance machines
Page 4.9
The effect of saturation is to make Wf < Wc. In
machines with magnetic circuits similar to the one in
Fig. 4.1, saturation of a typical magnetization curve
occurs in two stages. When the overlap between rotor
and stator pole corners is quite small, the
concentration of flux saturates the pole corners, even
at low current. When the overlapping poles are closer
to the aligned position, the yokes saturate at high
current, tending to limit the maximum flux-linkage.
Magnetization curves near the aligned position may
appear “double-jointed” as in Fig. 4.15, if the airgap is
small 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
)W e '
m
ei dt '
m
i
dR
dt
dt '
m
i dR ' ABCD (4.14)
Fig. 4.12
Definition of energy Wf and coenergy Wc
Fig. 4.13
Determination of electromagnetic torque
and the change in magnetic stored energy is
)W f ' OBC ! OAD .
(4.15)
The mechanical work done must be
)W m '
'
'
'
) We ! ) Wf
ABCD ! ( OBC ! OAD )
OABCD ! OBC
OAB
(4.16)
and this is equated to Te )2, so that when )2 6 0,
Te '
MW c
M2
i'const
(4.17)
Ideal cases: In a motor with no saturation the magnetization
curves would be straight lines, Fig. 4.14a. At any position, Wf =
2
2
Wc = ½L(2)i , and in this case (4.17) reduces to Te = ½i dL/d2,
which we saw earlier.
In a motor with a very small airgap, and especially if the steel
has a "square" B/H curve, the magnetization curves
approximate 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 the
permanent-magnet motor which has an EMF that is independent
of the current. Because of the small airgap, only a small fraction
of the current is needed to raise the flux-density to the
saturation level in the overlap region, and the flux-linkage
varies linearly with the overlap angle. The energy ratio is 1, and
the utilization of the power semiconductors in the drive is high.
Practical reluctance motors, especially highly-rated ones, are
often designed to try to approximate this ideal condition.
Fig. 4.14
Ideal magnetization curves
SPEED’s Electric Motors
Page 4.10
Average torque: So far we have seen that the torque
is produced in impulses as the rotor rotates, and we
have determined the instantaneous torque Te in terms
of the rate of change of coenergy at constant current.
The average torque could be calculated by integrating
Te over one cycle, (i.e., one rotor pole-pitch J = 2B/Nr,
where Nr is the number of rotor poles), and dividing by
J. However, it is difficult in practice to calculate Te
accurately, and it is better to calculate the average
torque from the enclosed area W in the energyconversion diagram, Fig. 4.15.
In one cycle of operation the maximum possible energy
conversion at a current I is the area W enclosed
between the unaligned magnetization curve U, the
aligned magnetization curve A, and the vertical line UA
at the current I. One cycle of operation, i.e., one
execution of this loop, is called a stroke. If S is the
number of strokes per revolution, the average
electromagnetic torque in Nm is
Te[avg] '
SW
2B
Fig. 4.15
Calculation of average torque
.
(4.18)
The drive must switch the current on and off at the correct rotor angles to cause the operating point
to follow this loop as closely as possible. Along UA, the current can be regulated by chopping, but at
high speed this may not be achievable and the loop may be smaller than the maximum loop, Fig. 4.30.
Fig. 4.16
Current, torque, and flux-linkage waveforms with a naturally-determined flat-topped
current waveform. i = phase current; T = phase electromagnetic torque; o = per-unit
overlap between active stator and rotor poles; R = phase flux-linkage; W = energy
conversion loop area; U = unaligned; A = aligned.
Fig. 4.16 shows an example of a motor operating with a naturally flat-topped current waveform, which
is obtained when the back-EMF is approximately equal to the DC supply voltage. This motor is analyzed
in more detail later. The speed in this example is 1015 rpm.
Switched reluctance machines
4.4
Page 4.11
CONTINUOUS TORQUE PRODUCTION
The motor in Fig. 4.1 is useful for analyzing torque
production, and although it can maintain a nonzero
average torque when rotating in either direction, this
torque is discontinuous, which means that continuous
rotation depends on the momentum or flywheel effect.
Moreover it cannot self-start from every rotor position. For
example, at the unaligned and aligned positions the torque
is zero. Unidirectional torque can be produced only over a
limited angle where the overlap angle 8 between the rotor
and stator poles is varying. To provide continuous
unidirectional torque, with starting capability from any
position, 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 the
number of rotor poles Nr and the number of phases m, and
in general
S ' m Nr .
Fig. 4.17
3-phase 6/4 switched reluctance motor
(4.19)
This can be substituted in eqn. (4.18) to give the average torque including all m phases, provided that
S 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 is
g = 360/12 = 30E. The three phases are labelled AA', BB ' and CC ,' and the ideal current/torque pulses
are 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 computation
requires a numerical simulation of the transient electromagnetic behaviour throughout one stroke.
Fig. 4.18
Waveforms in 3-phase 6/4 switched reluctance machine
Magnetic frequency: The fundamental frequency f1 of the current in each phase is evidently equal
to the rotor pole passing frequency, i.e.,
rpm
f1 '
Hz .
× Nr
(4.20)
60
The number of strokes per second is given by
f ' m f1
Hz .
(4.21)
This frequency and its harmonics appear in the flux waveforms in various parts of the magnetic circuit.
Page 4.12
SPEED’s Electric Motors
Stator/rotor pole numbers
To provide a structure for ordering the numbers of stator and rotor poles, we can start by defining a
regular switched reluctance motor as one in which the rotor and stator poles are symmetrical about
their centre-lines and equally spaced around the rotor and stator respectively. An irregular motor is
one which is not regular. This chapter, indeed the whole book, is more concerned with regular
machines, since they usually have the most sophisticated power electronic control requirements; and
therefore it focuses on machines with m = 3 or 4 phases. Machines with m = 1 or 2 are usually irregular
and 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-zero
torque 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 torque
comparable to the rated torque. The effective torque zone is comparable to the lesser pole-arc of two
overlapping 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
2B
2B
g '
,
'
S
m Nr
(4.22)
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 of
producing torque at all rotor positions. In practice a value of 1 is not sufficient, because one phase can
never provide rated torque throughout the absolute torque zone in both directions. The effective
overlap ratio De is defined as the ratio of the effective torque zone to the stroke angle, De = Je/g . For
regular motors with $s < $r this is approximately equal to $s/g. For example, in Fig. 4.17 the effective
overlap ratio is 30E/30E = 1. Note that De < Da. A value of De of at least 1 is necessary to achieve good
starting torque from all rotor positions with only one phase conducting, and it is also a necessary (but
not 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 3phase motors can be made for 4-quadrant operation. In the 6/4 motor in Fig. 4.17 , forward rotation
corresponds to negative phase sequence. This is characteristic of vernier motors, in which the rotor
pole-pitch is less than B/m. The 3-phase 6/4 motor has S = mNr = 12 strokes/rev, with a stroke angle
g = 30E, giving De = $s/g = 30/30 = 1.
With regular vernier motors there is always the
choice of having either Nr = Ns ! 2, as in the 6/4; or
Nr = Ns + 2, which gives the 6/8 motor shown in
Fig. 4.19; it has S = 24 strokes/rev and g = 15E. The
advantage of the larger Nr is a smaller stroke
angle, leading possibly to a lower torque ripple; but
inevitably the price paid is a lower inductance
ratio which may increase the controller voltamperes and decrease the specific output. The
stator pole arc has to be reduced below that of the
6/4 motor and this decreases the aligned
inductance, the inductance ratio, and the
maximum flux-linkage (although it increases the
slot area). The consequent reduction in available
conversion energy tends to offset the increase in
the number of strokes/rev, and the core losses may
be higher than those of the 6/4 motor because of
the higher switching frequency.
Fig. 4.19
3-phase 6/8 motor. Each phase has two coils on
opposite poles.
Switched reluctance machines
The 12/8 three-phase motor is effectively a 6/4 with
a "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 motor
discussed earlier. A high inductance ratio can be
maintained and the end-windings are short: this
minimizes the copper losses, shortens the frame,
and decreases the unaligned inductance.
Moreover, the magnetic field in this machine has
short flux-paths because of its four-pole magnetic
field configuration, unlike the two-pole
configuration in the 6/4 (or the 8/6; see below), and
the four-pole magnetic circuit helps to minimize
acoustic noise (see chapter 4). Although the MMF
per pole is reduced along with the slot area, the
effects of long flux-paths through the stator yoke
are alleviated. The 12/8 is possibly the most
popular configuration for three-phase machines.
Fig. 4.20
Page 4.13
3-phase 12/8 motor. Each phase has four coils, and
the magnetic flux-pattern is 4-pole.
Four-phase regular motors: The 4-phase regular
8/6 motor shown in Fig. 4.21 has 24 strokes/rev and
a stroke angle of 15E, giving Da = 2. With $s = 21E, De
= 1.33, which is sufficient to ensure starting torque
from any rotor position, and it implies that there
will be no problem with torque dips. However, it is
generally impossible to achieve the same fluxdensity waveform in every section of the stator
yoke, because of the polarities of the stator poles
( NNNNSSSS , NNSSNNSS , or NSNSNSN ). This
configuration was one of the first to be produced
commercially.1
With Ns = Nr + 2 = 10, S = 32 strokes/rev and g =
11.25E. The inductance ratio is inevitably lower
than in the 8/6, and the poles are narrower, while
the clearance between pole-corners in the
unaligned position is smaller, increasing the
unaligned inductance. This motor is probably on
Fig. 4.21 4-phase 8/6 motor
the borderline where these effects cancel each
other out; with higher pole-numbers, the loss of
inductance ratio and energy-conversion area tends to dominate the gain in strokes/rev. For this
reason, 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 = 7
phases. The parameter NwkPP is the number of working pole-pairs: that is, the number of pole-pairs in
the basic magnetic circuit. For example, the 4-phase 8/6 has NwkPP = 1 (a 2-pole flux pattern), while the
3-phase 12/8 has NwkPP = 2 (a 4-pole flux pattern). The unshaded boxes in Table 4.2 are probable the best
choices, the others having too many poles to achieve a satisfactory inductance ratio or too high a
magnetic frequency.
1
i.e., the well-known OULTON motor introduced in 1983 by Tasc Drives Ltd., Lowestoft, England.
SPEED’s Electric Motors
Page 4.14
m
Ns
Nr
NwkPP
gE
S
m
Ns
Nr
NwkPP
gE
S
2
4
2
1
90.00
4
4
8
6
1
15.00
24
2
8
4
2
45.00
8
4
16
12
2
7.50
48
2
4
6
1
30.00
12
4
24
18
3
5.00
72
2
8
12
2
15.00
24
4
32
24
4
3.75
96
2
12
18
3
10.00
36
4
8
10
1
9.00
40
2
16
24
4
7.50
48
m
Ns
Nr
NwkPP
gE
S
m
Ns
Nr
NwkPP
gE
S
5
10
4
1
18.00
20
3
6
2
1
60.00
6
5
10
6
1
12.00
30
3
6
4
1
30.00
12
5
10
8
1
9.00
40
3
6
8
1
15.00
24
5
10
12
1
6.00
60
3
12
8
2
15.00
24
3
18
12
3
10.00
36
m
16
4
7.50
48
Nr
NwkPP
S
24
Ns
gE
3
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 4.2
EXAMPLES 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 energy
conversion process in which the unaligned
magnetization curve is assumed to be
straight, while the aligned curve is
composed of two sections, a straight line OS
and a parabola SA. For a given peak
current
im the energy conversion
capability W is completely defined by the
three points U,S,A. It is shown in Miller
and McGilp [1990] that the parabola section
SA is represented by
( R ! Rs0 )2 ' 4 a ( i ! is0)
(4.23)
2
with is0 = is ! a/Lau , Rs0 = Rs ! 2a/Lau, a =
2
Rms /4(ims ! Rms/Lau), Rms = Rm ! Rs, and
ims = im ! is. These relationships ensure
that the first derivatives of the segments OS
and SA are equal at S. The area R can be
calculated by direct integration:
R ' Rms
Rm 2 % Rs ( Rm % Rs )
12 a
Fig. 4.22
% Rs0
Nonlinear energy conversion analysis, with aligned
magnetization curve represented by a straight line OS and
a parabola SA.
Rs0 ! (Rm % Rs )
4a
% is0
%
1
2
L au is2
(4.24)
Switched reluctance machines
and then the energy-conversion area W can be obtained as
1
W ' im Rm ! R ! L u im2 .
2
Page 4.15
(4.25)
The energy ratio Q is defined as W/(W + R), with inverse C = 1 + R/W. For example, in Fig. 4.22 the
unsaturated 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), and
the 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 conversion
process in which the unaligned magnetization
curve is again straight, while the aligned curve
is fitted by straight lines OS and SA. For a given
peak current the energy conversion capability W
is again defined by the three points U,S,A. This
model is like the one which J.V. Byrne [1970]
used to describe controlled saturation. The
saturation effect is characterized by the ratio
Rm / im
F '
(4.26)
Rs / is
which is effectively the ratio of the saturated to
the unsaturated inductance in the aligned
position. At the base speed the current
waveform is naturally flat-topped with peak
value im, because the back-EMF e of the motor Fig. 4.23
equals the supply voltage (resistance is
neglected).
If the angular rotation JA is
assumed equal to the stator pole-arc $s, then it
can be shown that e = Vs when
V s $s
' im L u ( 8 F ! 1 ) '
Tm
Nonlinear energy-conversion analysis, with aligned
magnetization curve represented by two straight
lines.
) Rm
(4.27)
Commutation is at point C such that the change of flux-linkage between J and C is c)Rm, where c # 1.
After commutation the current continues to flow throughout the angle (1 ! c)$s and for an
undetermined interval k$s thereafter. Since the peak flux-linkage Rc is c)Rm + Lui, we can write
V s $s
Rc '
( 1 ! c % k ) ' c ) Rm % L u im
(4.28)
Tm
which describes the de-fluxing interval with !Vs applied via the diodes. Eqns. (4.27) and (4.28) can be
used to solve for c:
(1 % k)(8F ! 1) ! 1
c '
.
(4.29)
2(8F ! 1)
When the rotor reaches the aligned position the flux-linkage is
k
RE '
Rc
1 ! c % k
and it follows that the current at the aligned position is
RE
iE '
.
L au
The example in Fig. 4.22 has F = 0.350 and 8 = 8. We can consider two extreme cases:
(4.30)
(4.31)
SPEED’s Electric Motors
Page 4.16
(i)
Commutation at the aligned position — In this case c = 1 and from eqn. (4.29) k = 1.556, so the
current continues to flow for 1.556 times the angle $s after the aligned position. The flux-linkage
at 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 case
k = 0 and from eqn. (4.29), c = 0.222, and RE = iE = 0. Although all negative torque is completely
eliminated, 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 between
these extremes, with k > 0 and some negative torque. In the example shown in Fig. 4.16, the negative
torque is very small and c = 2/3, giving k = 0.89 and RE = 0.727Rc. Although resistance accelerates the
flux suppression, it can be seen that the current tail in the negative torque zone has little impact on the
overall energy conversion.
Basic torque/speed characteristic
An interesting simplified analysis of the speed range at constant power was given by Byrne and
McMullin [1982], in which they derived a formula for the speed range at constant power. If Tp is the
maximum 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,
n '
Tp
'
Tb
1 2p
2
2 2b
(8 F)2
8F ! 1
(4.32)
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 probably
optimistic but it shows the importance of phase advance and the saturated inductance ratio 8F. A
typical 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 average
torque. At low speeds it is possible to work with c = 1, so that the entire available energy conversion
area 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
T a ' im2 L u ( 8 F ! 1 )
m Nr
4B
c (2 ! c/s)
(4.33)
where s = (8 ! 1)/(8F ! 1). At the base speed c has a value cb < 1 and from (4.33) the ratio of the torque
at zero speed to the torque at base speed is derived as
T0
Tb
'
2s ! 1
cb ( 2 s ! cb )
.
(4.34)
In the example motor cb = 2/3 and s = (8 ! 1)/(8 × 0.35 ! 1) = 3.89, and T0/Tb = 1.43. The motor can
evidently 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, the
peak torque is the same in both cases, implying that the torque waveform must be peakier at the base
speed than at standstill. Miller [1985b] goes on to compare the volt-ampere requirements of the
switched reluctance motor with those of a comparable high-efficiency induction motor, and concludes
that 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.
Switched reluctance machines
Page 4.17
Analysis of the energy-conversion loop
During a typical motoring stroke the locus of the
operating point (i,R) follows a curve similar to the one
in Fig. 4.24, which is drawn with the aligned and
unaligned magnetization curves and another
magnetization curve at the commutation angle C. At
this point the supply voltage is reversed and the
current freewheels through the diode. At C the
accumulated energy from the supply is equal to the
total area U = Wmt + WfC. The stored magnetic energy is
equal to WfC. Therefore the mechanical work done
between turn-on and commutation is Wmt, during the
period of transistor conduction. In Fig. 4.24 this is
roughly comparable to WfC, meaning that only half of
the energy supplied has been converted to mechanical
work. The other half is stored in the field.
Fig. 4.24
A nalysis of energy-conversion loop:
transistor conduction
Fig. 4.25
Analysis of energy-conversion loop: diode
freewheeling
Fig. 4.26
Analysis of energy-conversion loop: the
combined loop
After commutation, Fig. 4.25, the supply voltage is
reversed and the energy Wd is returned to the supply
via the diodes. The mechanical work done during the
freewheel interval is Wmd = WfC ! Wd. In Fig. 4.25 this
is less than half of WfC. The energy balance can be
deduced from the areas in Figs. 4.24 and 4.25. Suppose
that the energy supplied from the controller during the
"fluxing" interval (transistor conduction) is U = Wmt +
WfC = 10 Joules. At C, 5J has been converted to
mechanical work and 5J are stored in the field. During
the "de-fluxing" period (diode freewheeling), Wd = 3.5J
is returned to the supply and Wmd = 1.5J is converted to
work. The total work is therefore W = Wmt + Wmd = 5 +
1.5 = 6.5J or 65% of the energy supplied by the
controller. 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 combines
the two previous diagrams. The energy conversion is
now shown as the area W, while the energy returned to
the supply is R = Wd. The original energy supplied by
the 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 by
lumped-parameter magnetic circuit analysis, with an
allowance for the stator slot-leakage which increases at
high flux levels. The unaligned curve is more difficult
to calculate because of the complexity of the magnetic
flux paths in this position, but practical results are
reported by Miller and McGilp [1990] using a dualenergy method based on a quadrilateral discretization
of the slotted region. Earlier work by Corda and
Stephenson [1979] also produces adequate results for
many practical or preliminary design calculations.
Page 4.18
SPEED’s Electric Motors
The computation should include the “partial linkage” effect, meaning that at certain rotor positions
the turns of each coil do not all link the same flux. In a finite-element calculation this implies that the
flux-linkage should be calculated as the integral of A@dl along the actual conductors, with the coilsides
in the finite-element mesh in their correct positions in the slot.
End effects are important in switched reluctance motors. (See Michaelides and Pollock [1994]; Reece
and 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 the
unaligned 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 analysis
cannot help with this problem, and 3-dimensional finite-element calculations tend to be expensive and
slow. 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 or
near the aligned position at high flux levels, the stator and rotor poles can be highly saturated and the
external 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 simple
end-effect factors for Lu. For example, the PC-SRD computer program [Miller, 1999] splits the end-effect
calculation into two parts. For Lu,
L u ' Lu0 × (e u % fu)
(4.35)
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 the
inductance 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 before
being normalized to Lu0. fu is a factor that accounts for the axial fringing in the end-region. It is
calculated by the approximation
R1 ! R0 % L stk
fu '
(4.36)
L stk
which is derived by analogy with the fringing formula for two opposite teeth or poles. For the aligned
position the procedure is similar, with a factor fa = (g + Lstk)/Lstk.
At positions intermediate between the
aligned and unaligned positions, the
calculation of individual magnetization
curves is not practical by analytical methods
and the finite-element method should be
used. Fig. 4.1 shows a simple example of a
finite-element flux-plot at a position of
partial overlap, and Fig. 4.27 a more complex
example. Considerable success has been
achieved with interpolating procedures
especially in computer programs for rapid
design, e.g., [Miller and McGilp, 1990],
[Miller et al 1998].
Measurement of the magnetization curves is
described in [Cossar 1992].
Fig. 4.27
Finite-element flux-plot in a partial-overlap position
Switched reluctance machines
4.7
Page 4.19
SOLUTION OF THE MACHINE EQUATIONS
The PC-SRD computer program has been widely used for design and analysis of switched reluctance
machines for many years. Each phase is treated as a variable inductance in which the flux-linkage R
is a nonlinear function of both the current i and the rotor position 2: thus R = R(i, 2). PC-SRD computes
the function R(i, 2) from the geometry, the winding details, and the B/H curve of the steel, and it
represents the result graphically as a set of static magnetization curves in which R is plotted vs. i at
several rotor positions between the unaligned and the aligned positions. For a given machine, the
curves are a fixed property. It is not necessary to recalculate them unless changes are made to the
geometry, the windings, or the steel. For precise work PC-SRD can import external magnetiztion
curves 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’s
method. Each integration step produces a new value of flux-linkage R, and PC-SRD computes a new
value of current i from the function R(i,2), i.e., from the magnetization curves. The method of solving
for each new current value i depends on whether the curves are internal or external. With internal
magnetization curves PC-SRD uses a fast algebraic interpolation method based on so-called gauge
curves [Miller and McGilp, 1990]. With external curves, PC-SRD fits the curves with a set of cubic
splines, 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 internal
magnetization curves, the derivative is evaluated using approximate algebraic expressions derived
from the gauge curve model. With external mag curves it is evaluated from a precalculated set of spline
functions that represent the coenergy Wc as a function of current and rotor position, Wc (i,2). The
average electromagnetic torque is computed from the loop area W in Fig. 4.15, and a typical example
is given in Fig. 4.16. Several computed examples are given in chapter 5. The average electromagnetic
torque is given by (4.18). Since W is evidently an integral quantity, errors in the distribution of the
magnetization 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 the
intermediate 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-shifting
the current waveform of phase 1, which is calculated as though it were the only current flowing. The
fluxes and torques of the other phases are added to those of phase 1 without taking account of
interactions in shared magnetic circuits. This is one of the main limitations of PC-SRD, but it also
explains the extraordinary speed of computation. A full magnetic model of a polyphase switched
reluctance motor, including all magnetic interactions between phases, requires a multi-dimensional
set of magnetization curves and is a formidably complex thing to contemplate. PC-SRD’s simple model
is successful when the yokes are sufficient to avoid mutual interaction between phases. Under
conditions of extreme loading or (e.g., faults), the PC-SRD model cannot be expected to give accurate
results, 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 by
a more direct method based on a coenergy map. The voltage equation for one phase is
v ' e % Ri
(4.37)
where the back-EMF is given by
e '
MR
Mt
' Tm
MR
M2
(4.38)
and this can be obtained from the R(i,2) curves by differentiating with respect to 2 at constant current.
Page 4.20
SPEED’s Electric Motors
The solution proceeds at each timestep by calculating
i '
v ! e
(4.39)
R
using the current values of v and e. Once the “new” current is calculated from (4.39), e is re-evaluated
using (4.38) for the next integration step. The torque is calculated using
T '
MW c
(4.40)
M2
which is also evaluated at constant current. It is interesting to note that the flux-linkage can also be
evaluated from the coenergy using
MW c
R '
(4.41)
Mi
evaluated at constant 2. This suggests that the back-EMF can be evaluated using
M2W c
e ' Tm
M2 Mi
(4.42)
This suggests that the machine can be represented by a surface Wc (i,2) whose derivatives can be used
at any position 2 and any current i to determine the back-EMF e (4. 38) the flux-linkage R (4.41) and the
torque T (4.40). The finite-element solution of the magnetization curves can therefore be expressed in
terms of the surface Wc (i,2) without even computing the flux-linkage R, given that the finite-element
method can compute Wc directly by means of a global integration. A difficulty with this approach, as
with the previous one, is the representation of the coenergy function by a sufficiently smooth
interpolating function that is differentiable both 2 and i. Also, it does not naturally provide data which
can be compared with measurements.
4.8
CONTROL PRINCIPLES
Torque in the switched reluctance machine is produced by pulses of phase current synchronized with
rotor position. The timing and regulation of these current pulses are controlled by the drive circuit
and the torque control scheme. Usually there are also outer feedback loops for controlling speed or
shaft position, as shown in Fig. 4.1. The outer loops are generally similar to those used in other types
of motor drive, but the inner torque loop is specific to the switched reluctance machine.
T*
Fig. 4.28
Nested control loops. T * = torque demand; Tm* = speed demand, Tm = speed; 2* = position demand; 2 = shaft
position. Tacho = tachometer or speed transducer; Enc = encoder or position transducer.
The torque demand signal generated by the outer control loops is translated into individual current
reference 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/position
relationships must be compensated in the feed-forward torque control algorithm. This implies that the
torque control algorithm must incorporate some kind of “motor model”.
Switched reluctance machines
Page 4.21
Unlike the DC or brushless DC motor drive, the switched reluctance motor drive cannot be
characterized by a simple torque contstant kT. (torque/ampere). The drive must be specifically
programmed for a particular motor, and even for particular applications. One cannot take a switched
reluctance motor from one source and connect it to a drive from another source, even when the voltage
and current ratings are matched. On the contrary, the motor and drive control must be designed
together, 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. These
circuits can supply current in only one direction, but they can supply positive, negative, or zero voltage
at the phase terminals. Each phase in the machine may be connected to a phaseleg of this type, and the
phases together with their phaseleg drive circuits are essentially independent. The circuits in Fig. 4.7
can be adapted to operate the phases with separate DC supplies of different voltages, although the most
usual case is to connect them all to a common DC supply. Figs. 4.7a and 4.7b also show the possibility
of “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 PWM
or current regulation. As the speed increases the back-EMF increases to a level at which there is
insufficient voltage available to regulate the current; the torque can then be controlled only by the
timing 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 require
a combination of the high-speed and low-speed control modes. Even at lower speeds with voltage PWM
or 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) are
assumed to remain constant during each stroke. This corresponds to the operation of a “variable-speed
drive”, 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 classical
theory of torque control in electric drives is based on the DC machine, in which torque is proportional
to flux × current. The flux and current are controlled independently, and the “orientation” of the flux
and the ampere-conductor distribution, both in space and in time, is fixed by the commutator. In AC
field-oriented control, mathematical transformations are used, in effect, to achieve independent control
of flux and current, and the commutator is replaced by a shaft-position sensor which is used by the
control processor to adjust the magnitude and phase of the currents to the correct relationship with
respect 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 current
is varied in response to the torque demand. In both cases the torque control theory is characterized
by 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 meaning
which depends on the particular form or model of the system.
In switched reluctance machines, unfortunately there is no equivalent of field-oriented control. Torque
is produced in impulses and the flux in each phase must usually be built up from zero and returned to
zero each stroke. The “orthogonality” of the flux and current is difficult to contemplate, because the
machine is “singly excited” and therefore the “armature current” and “field current” are
indistinguishable from the actual phase current. Although this appears to be the case also with
induction machines, the induction machine has sine-distributed windings and a smooth airgap, so that
the 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 can
be aligned with the flux or orthogonal to it. The switched reluctance machine does not have sinedistributed windings or a smooth airgap, and there is virtually no hope of “field-oriented” control. To
achieve continuous control of the instantaneous torque, the current waveform must be modulated
according to a complex mathematical model of the machine.
Page 4.22
4.9
SPEED’s Electric Motors
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 is
shown in Figs. 4.12 and 4.13. The objective of “average torque control” is a simple current pulse
waveform which produces the required value of W corresponding to the torque demand. Even in
simple cases, this is more complex than simply determining the required “value of current”, since the
torque/ampere varies with both position and current. The following sections describe the general
properties of the current waveform at different points in the torque/speed diagram, Fig. 4.35.
Fig. 4.29
Low-speed motoring waveforms. i = phase current, R = phase flux-linkage, T = phase torque, and o = overlap
between stator and rotor poles. Horizontal axis is rotor angle (degrees). Unaligned position U = 45E; aligned position
A = 90E. The position J is the start of overlap between the active rotor poles and the stator poles of this phase.
Low-speed motoring — At low speed the motor EMF e is low compared to the available supply voltage
Vs, and the current can be regulated by chopping. If voltage-drops in the semiconductor devices are
neglected, the drive can apply three voltage levels +Vs, !Vs or 0 to the winding terminals to raise or
lower 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 typical
low-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 10E
before the aligned position (at 90E). At first no torque is produced because the inductance is low and
unchanging, but when the corners of the stator and rotor poles are within a few degrees of conjunction
J, torque suddenly appears. It is controlled by the regulating the current. When the transistors are
switched 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 transistors
are first switched on, R grows linearly at first because the full supply voltage is applied across the
winding terminals. When the current regulator starts to operate, R is also regulated to a constant
value at first because the constant current is being forced into an inductance that is still almost
constant at the low value around the unaligned position, before the poles begin to overlap. As soon as
the pole corners approach conjunction J, the inductance starts to increase, so the flux-linkage R also
increases as constant current is now being forced into a rising inductance. The flux-linkage continues
to 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 the
resistive voltage-drop is small, and therefore the rate of fall of flux-linkage is almost linear. At low
speed the dwell is made approximately equal to $s, since this is “width” of the “torque zone”, and this
angle might typically be a little less than 30E in a typical 6/4 motor. De-fluxing is completed over only
a small angle of rotation since the speed is low, so the entire conduction stroke occupies only about 30E.
Switched reluctance machines
Page 4.23
The process is summarized in the energy-conversion loop, which fits neatly between the aligned and
unaligned magnetization curves as a result of the selection of the firing angles. It appears that the
energy conversion W could be increased slightly, by retarding the commutation angle to extend the
loop up to the aligned magnetization curve. This would not require any increase in peak current, but
it would increase the average and r.m.s. values. It is also possible that delayed commutation could
incur a period of negative torque just after the aligned position, which would appear as a re-entrant
distortion of the energy-conversion loop, limiting the available gain in torque. Operation is at point
M1 in the torque/speed characteristic, Fig. 4.35. It is possible to maintain torque constant with
essentially the same current waveform as the speed increases up to a much higher value, since the
motor 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 be
insufficient for chopping, so that the torque can be controlled only by varying the firing angles of a
single pulse of current. Fig. 4.32 shows a typical example, in which the speed is 1300 rev/min.
Fig. 4.30
High-speed motoring waveforms.
The driving transistors are switched on at 50E and off at 80E, the same as in Fig. 4.29. At first the
overlap 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 backEMF suddenly appears, with a value that quickly exceeds the supply voltage and forces di/dt to become
negative, 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 the
supply 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 which
the transistors must be kept on throughout the stroke in order to sustain the rated current. Any
chopping 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 during
the stroke is given by Vs )2/T, where )2 is the “dwell” or conduction angle of the transistors. If the peak
flux is to be maintained at higher speeds, the “dwell” must be increased linearly with speed above the
base speed. At high speed the turn-on angle can be advanced at least to the point where the sum of the
fluxing and de-fluxing intervals is equal to the rotor pole-pitch, at which point conduction becomes
continuous (i.e. the current never falls to zero). This corresponds to a dwell of 45E and a total
conduction stroke of 90E, neglecting the effect of resistance (which tends to shorten the de-fluxing
interval).
Page 4.24
SPEED’s Electric Motors
Fig. 4.31
Thus it appears that the dwell or “fluxbuilding angle” can increase from 30E at low
speed to 45E at high speed, an increase of
50% or 1.5:1. Over a speed range of 3:1, the
peak flux-linkage might therefore fall to
1.5/3 = 0.5, or one-half its low-speed value.
This is illustrated in Fig. 4.31 for a speed of
3900 rev/min.
The peak current is
approximately unchanged but the loop area
W is only about one-third of its low-speed
value. The comparison between the loop
areas at 1300 and 3900 rev/min is shown
more clearly in Fig. 4.32. The average
torque is therefore only about one-third of
its low-speed value, but the power remains
almost unchanged. Operation is at point M3
in Fig. 4.35.
Very high speed motoring
Fig. 4.32
Energy-conversion loops at low and high speed, 1300 and
3900 rev/min.
Low-speed generating — Low-speed generating is similar to low-speed motoring except that the firing
angles are retarded so that the current pulse coincides with a period of falling inductance. Fig. 4.33
shows a typical example. The average torque is negative and the energy-conversion loop is traversed
in the clockwise direction. At the start of the stroke, there is a slight positive torque because the
current is switched on shortly before the aligned position, while the inductance is still rising. In this
example the torque falls to zero before the current is commutated, indicating that the commutation
angle could be advanced slightly without reducing the average torque. The reduction in copper loss
would increase the efficiency. During that “tail” period when there is current but no torque, the
current is maintained by the drive which is simply exchanging reactive energy with the DC link filter
capacitor. Operation is at G1 in Fig. 4.35.
Switched reluctance machines
Fig. 4.33
Page 4.25
Low-speed generating waveforms
High-speed generating — High-speed generating is similar to high-speed motoring, except that the
firing 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 traversed
in the clockwise direction. At the start of the stroke, there is a slight positive torque because the
current is switched on a few degrees before the aligned position, while the inductance is still rising.
Operation is at G2 in Fig. 4.35.
Fig. 4.34
Energy-conversion loop: high-speed generating
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 and
rated torque can be achieved at rated voltage. In this region the torque is controlled by regulating the
current, with relatively minor adjustments in the firing angles as necessary to alleviate noise or
improve the current or torque waveform, or to improve efficiency.
Page 4.26
SPEED’s Electric Motors
Fig. 4.35
Torque-speed characteristics
Constant power region—As the speed and back-EMF increase, the dwell is increased to maintain the
peak flux-linkage at the highest possible level. If the dwell is equal to half the rotor pole-pitch and the
de-fluxing angle is negligible at the base speed, then in principle the dwell can be doubled before the
onset of continuous conduction. Therefore if the dwell is increased in proportion to speed, the peak fluxlinkage can be maintained up to about twice the base speed. However, constant power can be
maintained to a higher speed than this, because the loss of loop area dW/dT is compensated by the
increase in speed. If power is taken as TT and T % W, then P % TW and for constant power we require
that )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 rate
of loss of loop area is balanced by the rate of increase of speed. The rate of increase in back-EMF is less
than proportional to the speed, because the current decreases with speed and MR/M2 is reduced. (In the
linear analysis e = iT dL/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 very
high speeds can be attained against a light load. The maximum phase advance depends on the drive
controller. If the turn-on angle is advanced beyond the point where the dwell becomes equal to about
half the rotor pole-pitch, continuous conduction will begin: the phase current never falls to zero and
the energy-conversion loop “floats” away from the origin. As it does so, it moves to a region where the
separation between the aligned and unaligned curves is increased, and the torque per ampere actually
increases. For this reason, operation with continuous conduction is a possible means of increasing the
power density, not only at high speeds but even at low speeds. The increase in copper loss is acceptable
if there is a greater gain in converted power and the machine can withstand the temperature rise. A
similar 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, corresponding
to 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 can
provide the correct reference position and direction sense. The firing angles for motoring in one
direction become generating angles in the reverse direction, at least at low speed. The machine is thus
reversible 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 must
be ‘covered’ by segments of rising inductance from different phases, as shown in Fig. 4.16, and the phase
currents must be sequenced to coincide with the appropriate segments. The total torque averaged over
one revolution is usually assumed to be the sum of the torque contributions from each phase. Although
the calculation and control of torque are both referred to one phase, some degree of overlap is required
in practice to minimise notches in the instantaneous torque waveform when the phases are
commutated, and to produce adequate starting torque at all rotor positions.
Switched reluctance machines
4.10
Page 4.27
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 are
switched on/off, usually at a high frequency compared with the fundamental frequency of the phase
current waveform. The voltage applied to the winding terminals is +Vs if both transistors are on, 0 if
one is on and the other is off, and !Vs if both transistors are off and the phase current is freewheeling
through both diodes. In the zero-volt state the phase current freewheels through one transistor and
one diode. These three conduction modes are shown in Fig. 4.36, and Table 4.3 shows the states of the
power transistors and diodes in the three conduction modes.
Fig. 4.36
Conduction modes
Soft chopping is when only one transistor is chopping. The other transistor remains on, and it is
called the "commutating" transistor because its only function is to steer or commutate the current into
its associated phase winding at the beginning and end of the conduction period. The voltage applied
to the winding switches between +Vs and 0. During the zero-volt period the rate of change of fluxlinkage is very small (in fact it is equal to !Ri), and therefore the current falls slowly. This means that
the chopping frequency and DC link capacitor current can both be greatly reduced for a given current
ripple or hysteresis band (see below).
State
A
B
C
D
Q1
1
1
0
0
Q2
1
0
1
0
D1
0
0
1
1
D2
0
1
0
1
TABLE 4.3
TRUTH TABLE FOR THE STATES OF THE TRANSISTORS AND DIODES
V
Vs
0
0
-Vs
Hard chopping is when both transistors are switched on/off together. It generally produces more
acoustic and electrical noise, and increases the current ripple and DC link capacitor current for a given
current ripple or hysteresis band. It is necessary in certain conditions particularly during
regeneration, to prevent loss of control of the current waveform, and of course the final “chop” at 2c
at the end of the conduction period is a hard chop.
SPEED’s Electric Motors
Page 4.28
Single-pulse control at high speed
The flux must be established from zero every stroke. Its build-up is controlled by switching both power
transistors on at the turn-on angle 20 and switching them off at the commutation angle 2c. In motoring
operation the dwell )2 = 2c ! 20 is timed to coincide with a period of rising inductance, and in
generating operation with a period of falling inductance. At a sufficiently high speed, the waveforms
of 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 with
infinitely permeable iron has a waveform similar to that of the pole-overlap waveform, and provides
a 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:
2c
1
Rc '
( V s ! R i) d2 % R0
(4.43)
T 20
where R0 is the flux-linkage pre-existing at 20 (ordinarily zero). Vs is the supply voltage, R is the phase
resistance, and i is the instantaneous current. All impedances and volt-drops in the controller and the
supply are ignored at this stage. Eqn. (4.43) can be written as
TRc ' V s (1 & u1 ) . 2D
(4.44)
where 2D = (2c ! 20)is the dwell and v1 = u1Vs is the mean volt-drop in the resistance and transistors
during 2D. If u1 << 1 the flux-linkage rises linearly. In motoring operation the flux should ideally be
reduced to zero before the poles are separating, otherwise the torque changes sign and becomes a
braking torque. To accomplish this the terminal voltage must be reversed at 2c, and this is usually done
by the action of the freewheeling diodes when the transistors turn off. The angle taken for the negative
voltage to drive the flux back to zero at the "extinction angle" 2q is again governed by Faraday's Law:
2q
1
0 ' Rc %
( ! V s & R i) d 2
(4.45)
T 2c
and this can be written as
T Rc ' V s (1 % u2 ) ( 2q & 2c ).
(4.46)
where v2 = u2Vs is the mean volt-drop in the resistance and diodes in the de-fluxing period (2q - 2c). If
u2 << 1 the flux-linkage falls linearly, and at constant speed the angle traversed is nearly equal to the
dwell angle, both being equal to Rc/Vs. The peak flux-linkage Rc occurs at the commutation angle 2c. The
total angle of phase current conduction covers the fluxing and de-fluxing intervals and is equal to
TRc
2 ! u1 % u2
.
2q ! 20 •
.
(4.47)
V s ( 1 % u2 )( 1 ! u1 )
If u1 = u2 = 0 this reduces to 2TRc/Vs. The entire conduction period must be completed within one rotor
pole-pitch "r = 2B/Nr, otherwise there will be a ratcheting or pumping effect in which R0 has a series
of non-zero values increasing from stroke to stroke.2 This condition is also called "continuous
conduction". That is, 2q ! 20 # "r. Eqns. (4.44) and (4.47) combine to give the maximum permissible dwell
angle,
1 % u2
2D max ' "r .
.
(4.48)
2 ! u1 % u2
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
(1 % u)
.
2D max ' "r .
(4.49)
2
For example, in a symmetrical 6/4 motor the pole-pitch is "r = 90E (360 elecE) and if u = 0 the maximum
2
Note that "r is the angle of rotation between two successive aligned positions.
Switched reluctance machines
Page 4.29
dwell angle is 2D = 45E, giving a total angle of conduction in the phase winding of 90E. But if u = 0.2 the
maximum dwell angle is 54E. In a regular switched reluctance motor the angle of rising inductance is
only "r/2. Ideally the flux should be zero throughout the period of falling inductance, because current
flowing in that period produces a negative or braking torque. To avoid this completely, the conduction
angle must be restricted to "r/2 and the maximum dwell angle is then
"r 1 % u
.
.
2D #
(4.50)
2
2
In the 6/4 motor, with u = 0.2 this indicates a maximum dwell angle of 27E (108 elecE) and a conduction
angle of 54E. In practice, larger dwell angles than this are used because the gain in torque-impulse
during the rising-inductance period exceeds the small braking-torque impulse, which generally occurs
in a region when the torque/ampere is low (i.e., near the aligned and/or unaligned positions). This
condition is shown in Figs. 4.30 and 4.31, where the current has a "tail" extending beyond the aligned
position. 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 until
the poles begin to overlap. The rising inductance generates a back-EMF which consumes an increasing
proportion 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 the
speed is constant. What was an excess of applied forward voltage now becomes a deficit, and the current
begins to decrease. At the point of commutation the applied terminal voltage reverses, and there is a
sharp increase in the rate of change of current. At the aligned position the back-EMF reverses, so that
instead of augmenting the negative applied terminal voltage, it diminishes it, and the rate of fall of
current decreases. In this period there is a danger that the back-EMF may exceed the supply voltage and
cause 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 be
advanced as the speed increases.
Figs. 4.30 and 4.31 also show the importance of switching the supply voltage on before the poles begin
to overlap. This permits the current to grow to an adequate level while the inductance is still low. For
as long as the inductance remains nearly constant, there is no back-EMF and the full supply voltage is
available to force the increase in current. The turn-on angle may be advanced well ahead of the
unaligned 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 speaking
there are two main methods: current-hysteresis control and voltage-PWM control, but many
variations 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 the
current 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 peak
value. This interval is called the “dwell” or “transistor conduction angle”. At 2c both transistors are
switched off, and the freewheeling diodes connect the reverse of the supply voltage to the phase
terminals, causing the flux to decay to zero. The “de-fluxing interval” lasts from 2c to 2q, and in general
is shorter then the fluxing interval.
Voltage PWM (Fig. 4.37): In voltage-PWM, at least one of the two transistors in a phaseleg is switched
on 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. In
voltage-PWM there is no closed-loop control of the instantaneous current. The current waveform has
its "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 current
reaches a predetermined level iHi, the current will be limited by switching off at least one of the
phaseleg transistors.
SPEED’s Electric Motors
Page 4.30
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 is
connected to the phase winding. When it is off, the winding is short-circuited through Q2 and D2. Q1
is called the "chopping transistor" and D2 the "chopping diode". Q2 is called the "commutating
transistor" and D1 the "commutating diode", because they change state only at the commutation angles
20 and 2c. During the dwell angle the average voltage applied to the phase winding is D × Vs. Again
using 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:
TRc ' 2D ( D & u ) V s ' ( 2q & 2c ) ( 1 % u ) V s .
(4.51)
This can be rearranged to show that the total conduction angle is
1 % D
2q & 20 ' 2D
.
1 % u
To prevent continuous conduction, 2D must be restricted to
1 % u
2D < "r .
.
1 % D
(4.52)
(4.53)
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 prevent
any braking torque, 2D must be restricted to
"r
1 % u
.
2D <
,
(4.54)
2
1 % D
i.e. one-half of the absolute maximum, or 36E in the example. The dwell can be increased as the dutycycle 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 switched
together at high frequency. In both soft and hard chopping, the flux-linkage waveform increases in
regular steps with a more-or-less constant average slope. Before the start of overlap, the average slope
of the current waveform is also nearly constant as the linearly-increasing flux is forced into a constant
inductance. Thereafter, the inductance increases more or less linearly while the flux-linkage continues
to rise linearly. Consequently the current tends to become constant or flat-topped. Voltage-PWM tends
to 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 control
strategy, the whole of the absolute torque zone can be used. As is evident from eqn.(4.51), the ratio of
the slopes of the rising and falling parts of the flux-linkage waveform is approximately equal to D, so
that with a low duty-cycle (needed to "throttle" the voltage at low speed), the de-fluxing is accomplished
in a very few degrees, permitting late commutation.
Although the pole arcs do not appear in any of the equations constraining the limiting values of the
firing angles, they are important in determining their optimum values.
Switched reluctance machines
Fig. 4.38
Page 4.31
Architecture of voltage-PWM controller
The duty-cycle is typically set by the outer speed and position control loops, while the firing angles can
be scheduled with speed to optimise efficiency. Fig. 4.38 shows the concept of average torque control
with voltage PWM, with typical voltage and current waveforms as shown in Fig. 4.37. The torque
demand is represented by the duty-cycle command signal D*, which may vary as the torque demand
varies, with consequent variation in the current. Because no attempt is made to control the currents
instantaneously, there is no need for current sensors in individual phases. Voltage-PWM schemes may
therefore 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 a
phaseleg is switched off when the current exceeds a specified set-point value iHi. It is switched on again
when 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 naturally
determined by the changing inductance and back-EMF as the rotor rotates (this is sometimes called
single-pulse mode). Fig. 4.39 shows the waveforms obtained with a hysteresis-type current regulator
and 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 fed
back 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 shown
in Fig. 4.39. The waveforms for hard chopping are similar. As in the case of voltage-PWM, soft chopping
decreases the current ripple and the filter requirements, but it may be necessary in braking or
generating modes of operation.
Fig. 4.39
Device current waveforms
SPEED’s Electric Motors
Page 4.32
Fig. 4.40
Architecture of current-hysteresis controller
Delta modulation: A variant of the current hysteresis controller is delta modulation in which the
current 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 frequency
is not fixed but is limited by the sampling rate.
Soft braking: Soft braking is used for regenerative braking and uses the same zero-voltage loop
principle as soft chopping. Initially both transistors are on and full supply voltage is applied until the
phase current in the winding exceeds a predetermined limit, i > iHi. Both transistors are then switched
off and !Vs is applied until the current falls below iLo. Thereafter one transistor is switched on and off
to regulate the current until the end of the conduction period. The state of the switches alternates
between B and D, or between C and D, in Table 4.4. During the freewheeling state B or C, the fluxlinkage remains approximately constant and at low speed the rate of rise of current is low, so that this
strategy 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 become
necessary, in which the supply voltage Vs is used to suppress the rate of rise of current and the switch
states 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, zero
voltage 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 torque
without current chopping and its associated losses. The peak flux is limited to a lower value. The
energy conversion loop still makes good utilization of the available energy. This technique has been
used also for noise reduction. [Horst, 1995].
Fig. 4.41
Zero-volt loop used to achieve an interval of approximately constant flux-linkage.
Switched reluctance machines
4.11
Page 4.33
MATHEMATICAL DESCRIPTION OF CHOPPING
Fig. 4.42
Definition of firing angles
For analysis or computer simulation the firing angles must be defined with respect to a reference value
of rotor position, Fig. 4.42. The rotor position reference is the graph of “per-unit overlap” between the
stator 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 convenient
to start at the “previous aligned position” and end at the “current aligned position”, A. When 2 is
outside the principal range it can be “reduced” to the principal range by adding or subtracting integral
multiples of "r. In a symmetrical machine the magnetization curves are generally defined only between
the unaligned and aligned positions, and it is convenient to divide the principal range into two sections
AU and UA. If 2 is in AU, it is replaced by A!2 so that the reduced rotor angle is always in the range
UA. A flag ST is set to !1 to represent the sign of the torque when 2 is in the AU range, and +1 when
it is in the UA range. Otherwise all calculations can proceed as if the rotor was in the range UA. If the
machine is not symmetrical, the principal range cannot be divided and must extend over a complete
rotor 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 the
unaligned position and one electrical cycle is equal to the rotor pole-pitch. Therefore the conversion
from electrical degrees to mechanical degrees is
2elec ' ( 2mech ! U ) × N r .
(4.55)
The interval between unaligned and aligned positions is always 180E(elec). In a 3-phase 6/4 motor with
a 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. In
current hysteresis control, the applied voltage during the conduction interval 2c ! 20 is Vs. In a
computer 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 chopping
transistor is switched on and off at the frequency fchop, with a duty-cycle D. At 2c, Q1 and Q2 are
switched off, and the reverse ("de-fluxing") voltage is !Vs. The voltage equation for one phase is:
SPEED’s Electric Motors
Page 4.34
T
dR
d2
' d1[V s ! R ph i ! 2R q i ! 2V q]
% d2 [ !R ph i ! R q i ! V q ! V d]
% d3 [ !V s ! R ph i ! 2 V d]
(4.56)
' (d1 ! d3)V s ! R ph i ! 2 (d1 % d2)(R q i % V q) ! (d2 % 2 d3)V d
where d1 is the duty-cycle with Q1 and Q2 on, d2 with Q1 off, and d3 with Q1 and Q2 off. This equation
caters for all combinations of the states of Q1 and Q2. At each integration step d1, d2 and d3 are assigned
the correct values, and d1+d2+d3 = 1 within each integration step. The compact voltage equation
expressed in this form with d1, d2 and d3 embodies all the switching states and logic required for both
current hysteresis control and voltage- PWM control. In current hysteresis control, d1, d2 and d3 are
scalar values that multiply the voltage terms in the equation. This is the principle of "state space
averaging", and is based on the notion of an upstream chopper controlling the DC source voltage, with
infinite 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 by
the combined states of the transistors and diodes. Only one of the three states d1,d2 and d3 can be
non-zero during one integration step.
The transistor and diode currents and their squares are accumulated in each integration step using
iQ1 = d1 × i; iQ2 = (d1+d2) × i; iD1 = d3 × i; iD2 = (d2+d3) × i; and iDC = (d1!d3) × i {= iQ1 ! iD1 = iQ2 ! iD2}. When
the integration is finished, the mean and mean-squared values are calculated from the accumulations
by dividing by the number of steps. This process is the same for both current hysteresis control and
voltage-PWM. An exception to this calculation is the DC link ripple current. This has to be constructed
from the phase-shifted sums of the phaseleg currents, which flow in both directions in the DC link. It
can 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 based
on the simple linear relationship between torque and current, and the torque demand produced by the
velocity loop is translated directly into a current command by a simple constant of proportionality, the
torque constant kT. A similar principle is implemented in field-oriented AC motor drives. The
translation 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 therefore
no torque feedback. In the switched reluctance drive the relationship between torque and current is
not linear. An example is shown in Fig. 5.16.
Fig. 4.43
Typical variation of torque with current
Switched reluctance machines
Page 4.35
The torque also depends on the firing angles 20 and 2c. To complicate matters, 20 and 2c may be required
to vary for reasons other than torque control — for example, to minimize noise or to compensate for
back-EMF at high speed. The result is that the feedforward torque control must usually be implemented
as a mapping from T* to i* with, with additional links to the firing angles and possibly also the supply
voltage. 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 a
particular motor and drive, and usually also specific to a particular application.
Fig. 4.44
Graphical representation of look-up table for turn-on angle
Fig. 4.44 shows a graphical representation of a look-up table for the turn-on angle 20 as a function of
speed and torque demand. With single-pulse control the variation of torque with 20 and 2c is equally
complex. 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 to
eliminate the error between the reference and actual currents, and to give a smooth variation of phase
voltage with reduced torque ripple and electromagnetic noise. The main disadvantage stems from the
variation of inductance with rotor position, which can cause the electrical time constant to vary by
as much as 10:1, making it difficult to tune for satisfactory transient performance.
Fig. 4.45
Architecture of single-pulse controller
SPEED’s Electric Motors
Page 4.36
Generator control
The switched reluctance machine will regenerate power to the DC supply if the current pulse is timed
to 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, and
generated power is returned to the DC source when they are both switched off. The power circuits of
Figs. 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 the
DC source disconnected, but the DC link capacitor must be retained to provide excitation power during
the “fluxing interval” during the first part of each stroke. Generally the load will be connected in
parallel with the DC link capacitor, and in general its impedance will be variable and not under the
control of the switched reluctance generator controller. Inevitably the DC voltage will decrease during
the fluxing interval, and increase during the de-fluxing interval when power is being returned through
the diodes. The variation or ripple in the DC link (capacitor) voltage depends on the energy conversion
per stroke, the energy ratio, and the capacitance. The controller must maintain the average DC voltage
constant in much the same way as it must maintain constant average torque in motoring mode. This
is equivalent to the maintenance of constant torque, averaged over at least one stroke. Therefore the
architecture of a generator controller can be similar to that of a motor controller. However, the DC
link capacitance has an integrating effect such that it requires lower bandwidth to control the DC
voltage than to control the DC current (or torque). At low speed, therefore, the DC link voltage is
controlled by varying the set-point current i* or the duty-cycle D*, and at high speed by varying the
control angles 20 and 2c. See Fig. 4.46. As in motoring operation, a control map is required to determine
how the control variables must vary in response to the voltage error )Vd. Various linearising schemes
have been presented to simplify the control, [Radun, 1993], [Kjaer, Cossar, Gribble, Li and Miller, 1994].
Control mode
Control variables
Current hysteresis
control
iHi, iLo
Delta-modulation
i*
Voltage-PWM control
D
Zero-volt-loop mode
iHi, iLo, 2z
Single-pulse control
20, 2c
TABLE 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 by
varying any one or indeed all of the control variables in a given mode, but the configuration depends
on the performance requirements, the acceptable level of complexity, and the cost. Since there are
many possible combinations of control variables which produce the same torque, a secondary control
objective 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 7506
50737 and Miller TJE, Switched reluctance motors and their control, Oxford University Press/Magna
Physics Publishing, 1993, ISBN 1-881855-02-3 (available from Motorsoft). All references in brackets are
listed in this book, together with many others.
Switched reluctance machines
Page 4.37
Index
Acoustic noise 13, 36
Aligned 2, 3, 5-7, 9-12, 14-19, 21-26, 28, 29, 33
Average torque 3, 10, 11, 16, 21, 24, 31, 36
Back-EMF 4, 6-8, 10, 15, 19-21, 23, 26, 29, 35
Braking 3, 26, 28-32
Brushless DC motor 21
Byrne 15, 16
Capacitor 24, 27, 36
Chopping 4-7, 10, 22, 23, 27, 29-34
Classical machines 21
Coenergy 8-10, 19, 20
Commutation angle 15-17, 23, 24, 28, 29
Constant power 16, 26
Constant torque 25, 36
Control 5, 6, 12, 20, 21, 25-36
Converter volt-ampere requirement 14
Current hysteresis 30-34, 36
Current regulation 21, 27, 29
Definition 2, 9, 33
Drive circuit 5, 20, 21, 29
Electromagnetic torque 4, 9, 10, 16, 19, 22
Energy conversion 6-8, 10, 14-17, 23, 32, 36
Energy ratio 6-9, 14, 15, 17, 36
Energy-conversion diagram 10
Energy-conversion loop 17, 22-26
Equivalent circuit 8
Euler’s method 19
Finite-element 18-20
Firing angles 21, 23-26, 30, 31, 33, 35
Flat-topped current 7, 10
Half-bridge phaseleg 5
Hard chopping 27, 30-33
High-speed generating 25
High-speed motoring 23, 25
Inductance 2-4, 6, 8, 12, 13, 15, 16, 18, 19, 22-26, 28-31, 35,
36
Instantaneous torque 10, 19, 21, 26
Linear analysis 4, 26
Look-up table 35
Low-speed generating 24, 25
Low-speed motoring 22, 24
Magnetization curves 8, 9, 17-20, 23, 33
Motor 2, 5, 7-13, 15, 16, 19-23, 28-30, 33-36
Multiple-phase operation 26
Multiplicity 13
Nonlinear analysis 8
Number of strokes per revolution 10, 11
Overlap ratio 12
PC-SRD 18, 19
Pole arc 2, 12
Pole numbers 12
Ripple 8, 12, 27, 31, 32, 34-36
Saturation 4, 8, 9, 15
Single-pulse 28, 29, 35, 36
Soft chopping 27, 29-32
Stator/rotor pole number 14
Stator/rotor pole numbers 12
Stepper motor 2
Stepper-motor 5
Stored field energy 7, 9
Stroke 4-8, 10-13, 17, 21-25, 28, 32, 36
Stroke angle 11-13
Strokes/rev 12, 13
Suppression voltage 5
Switching frequency 12, 29, 32
Synchronous reluctance 2
Torque 2-13, 16, 19-26, 28-36
Torque zone 7, 12, 16, 22, 30
Torque/speed characteristic 16, 23, 25, 26
Turn-on angle 23, 26, 28, 29, 35
Unaligned 2, 3, 6, 10, 11, 13-19, 22, 23, 26, 29, 33
Vernier 12
Voltage equation 4, 19, 33, 34
Voltage PWM 21, 29, 31
Zero-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
5.
COMMUTATOR MACHINES
5.1
DC Commutator machines
Page 5.1
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-control
applications, where they can be controlled with low-cost electronic drives such as choppers and phasecontrolled converters.1 The theory of the DC machine is also fundamental in the understanding of many
other 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 electric
circuit 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.
Fig. 1
Electrical equivalent circuit model of DC motor
The motor model is represented by a single differential equation,
V ' ea % ( Ra % Rleads ) ia % La
dia
dt
% 2 Vb
(5.1)
in which ia is the instantaneous armature current, ea is the instantaneous back-EMF, Vb is the volt-drop
in each brush, Ra is the armature resistance, Rleads is the resistance of the leads, and La is the armature
inductance. 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 chopper
or 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 the
properties of the steel and magnet materials specified via the materials database. Ea is obtained from
the magnetic circuit model in the classical form
E a ' k E Tm
(5.2)
where the back-EMF constant kE is given by
kE '
p Z Mg
aB
.
(5.3)
The flux/pole Mg is calculated in the magnetic circuit model. Z is the total number of armature
conductors, p is the number of pole-pairs, and a is the number of parallel paths. Tm is the angular
velocity in rad/sec, equal to 2B × RPM/60.
1
Large DC machines generally have field windings. They have traditionally been used in rail traction and high-power motion
control applications such as steel rolling mills, elevators, excavators, etc. Many of these now use AC motor drives.
Page 5.2
5.1.2
SPEED's Electric Motors
TORQUE
In S.I. units the back-EMF constant kE is equal to the torque constant k T, which is used to calculate the
electromagnetic torque Te = kT ia. The torque constant is one of the most important parameters of a DC
motor and is usually quoted in catalogue data. It can be measured by running the machine on a
dynamometer and plotting the torque vs. current, then measuring the slope of the resulting graph. The
EMF constant kE can be measured by driving the machine with an auxiliary motor and measuring the
generated voltage. Generally this has a linear variation with speed. The measured k T is often less than
kE because of friction, and at high current the armature reaction distorts the magnetic flux and reduces
k T still further. In spite of the "saturation" of k T at high current, the linearity of Te vs. ia is an
important 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, which
includes the windage and friction loss. The iron loss is also often lumped in Trot.
T ' kT ia ! Trot
5.1.3
Nm .
(5.4)
EFFICIENCY
The mechanical power output is TmT and the electrical input power is VIa, so the efficiency is
0 '
Tm T
output power
'
input power
V Ia
× 100% .
(5.5)
2
The copper loss is calculated from I Ra, using the r.m.s. value of Ia. The iron loss WFe can be calculated
using the modified Steinmetz equation, taking into account the variation of the flux-density in the teeth
and 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 :
Tm '
Ea
kE
'
or
V ! 2 Vb ! Ra Ia
kE
V ! 2 Vb
'
Tm ' T0 !
kE
Ra
kE2
T
!
Ra
kE2
( T % Trot )
(5.6)
(5.7)
where T0 is the no-load speed:
T0 '
V ! 2 Vb
kE
!
Ra
kE2
Trot .
(5.8)
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 boload speed, preferably in the neighbourhood of maximum efficiency.
The speed can be varied by introducing series resistance, which steepens the negative slope of the
torque/speed characteristic. This is inherently wasteful. If, for example, the series resistance is
sufficient to reduce the voltage at the motor terminals to 50% of the supply voltage, then the power
dissipated in the series resistance must be equal to the entire input power to the motor, so the overall
efficiency 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 T0
and therefore raises or lowers the whole speed curve. DC choppers can be used for this when the supply
is DC; or phase-controlled rectifiers when the supply is AC.
Commutator machines
Fig. 2
5.1.5
Page 5.3
Torque/speed characteristics of PM DC motor
OPERATING CHARACTERISTICS
Normalized (i.e., per-unit) operating characteristics at constant voltage are shown in Fig. 2. There is
a peak efficiency and a peak output power, occurring at different loads. The load torque at which the
peak 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
2
= kT , i.e., the variable loss varies with the square of the torque (effectively, with the square of the
current since torque is proportional to current). The efficiency 0 is given by
0 '
Tm T
Tm T % W1 % k T 2
(5.9)
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
Tm '
V
.
kE
(5.10)
This steady-state equation gives no indication as to how the speed varies dynamically if the supply
voltage V is changed, as it might be by the action of the electronic controller. The dynamic response
can be determined from the transfer function between speed and voltage, which is derived as follows
from the equations that describe the system. These equations are:
V ' Ra Ia % kETm
(5.11)
Te ' kT Ia
(5.12)
Te ' J " ' J
dTm
dt
.
where J = Jm + JL, is the total inertia of motor + load, and " is the angular acceleration, dTm/dt.
(5.13)
Page 5.4
SPEED's Electric Motors
Assuming that the speed is zero at time t = 0, in terms of Laplace transforms,
" ' s Tm
(5.14)
so that
Te
Tm '
kT Ia
'
Js
'
Js
kT (V & kETm)
Ra J s
.
(5.15)
This can be rearranged to give the transfer function between speed and voltage as
Tm (s)
V (s)
1/kE
'
1 % s
'
Ra J
1/kE
1 % s Jm
.
(5.16)
kT kE
The denominator has only one root, so the transfer function has only one pole at s = !1/Jm, and the
system is said to be a first-order system. Jm is called the mechanical time constant:
Jm '
Ra J
kT kE
.
(5.17)
After a disturbance such as a step-change in voltage, the speed will settle exponentially to a final steadystate value, because the transfer function is of the form 1/(s + a), which has an inverse transform that
is a decaying exponential function of time. After an infinite time, in the steady-state all derivatives are
zero 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-state
expression 1/k E, 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. The
electrical equation of the motor is
V ' La
dIa
dt
% Ra Ia % kETm .
(5.18)
where V is the terminal voltage and brush volt-drops are neglected. By Laplace transformation,
V ( s ) ' ( La s % Ra ) Ia ( s ) % kETm( s ).
(5.19)
The mechanical equation of the motor is
Te ' J
dTm
dt
% D Tm % Trot % TL .
(5.20)
where DTm is an additional "viscous" torque sometimes described as a damping torque that is
proportional to speed. Of most interest is the transfer function between speed and voltage, and the
simplest case is at no load, i.e. with TL = 0. If we also neglect Trot, then by Laplace Transformation,
Te( s ) ' ( J s % D ) Tm( s )
(5.21)
From equations (5.14),(5.21) and (5.21),
Te( s ) '
( V ! kETm( s ) ) kT
La s % Ra
' ( J s % D ) Tm ( s ) .
(5.22)
Commutator machines
Page 5.5
From this the transfer function between speed and voltage can be extracted as
Tm(s)
'
V
kE
( La s % Ra ) J s % D % kE kT
.
(5.23)
The poles of the system are the roots of the denominator of eqn. (25), which is quadratic. There are
therefore two roots, and it is common to write eqn. (25) in a form which separates these two roots: thus
Tm(s)
G(s) '
'
V(s)
1/kE
( s Je % 1 ) ( sJm % 1 )
(5.24)
where Je is the electrical time constant and Jm is the mechanical time constant. As before, 1/kE is the
system gain. Assuming no motor damping (D = 0) and low inductances, the time constants are
Je .
5.1.7
La
Ra
,
Jm .
Ra J
kE kT
.
(5.25)
RESISTANCE AND INDUCTANCE
The armature resistance Ra is calculated from the winding details at the operating temperature:
Ra ' DCu ×
Z × LT / 2 a 2
Nsh × B dCu2 / 4
(5.26)
where Nsh is the number of strands-in-hand in each conductor, dCu is the strand diameter, DCu is the
resistivity 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. This
calculation is considerably more complex than the resistance calculation, and the theory is given in full
in Ref. [8]. Basically there are three components of armature inductance,
La ' Lslot % Lgap % Lend ,
(5.27)
where Lslot is the slot-leakage component, Lgap is the airgap component, and Lend is the end-winding
component. The airgap component Lgap is liable to be affected by frame saturation especially in 2-pole
motors, or where the frame is very thin. Therefore, for work requiring very accurate values of La it
might 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 timeconstant Je = La /Ra , which is sometimes important in determining the dynamic response, especially
for servo-type applications. The other is in the calculation of the current waveform when the motor is
driven from a chopper or phase-controlled rectifier. In these cases the current waveform may be
sensitive to the value of La, and if an accurate value is not used, the current and the torque will be
calculated incorrectly.
Page 5.6
5.1.8
SPEED's Electric Motors
MAGNETIC CIRCUIT ANALYSIS
The flux produced by the magnet at no-load has the general form shown in Fig. 3. The magnet flux Mm
can be separated into two components: the larger of these is the airgap flux Mg = f LKG × Mm, which
crosses the airgap and links the armature winding. The fraction f LKG is called the leakage factor, and
a typical value is 0.9. The leakage flux component (1 ! f LKG)Mm does not link the armature winding.
Fig. 3
No-load flux distribution in PM DC motor
Fig. 4 shows a graph of the airgap flux-density under one pole at no-load. The peak value is Bg and the
airgap 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 the
armature length and 2p being the number of poles. Darm is the rotor diameter.
Fig. 4
Flux-density waveform under one pole at no-load
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 model
correspond roughly to the main flux-paths in Fig. 3. They are calculated from the geometry and from
the 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") that
distorts the airgap flux-density waveform. This effect is less severe in PM motors than in wound-field
machines. For accurate analysis of the magnetic field, it is desirable to use a finite-element analysis.
Commutator machines
Fig. 5
Page 5.7
Magnetic equivalent circuit
Magnet operating point
The magnet operates at a point P on the recoil line shown in Fig. 7, where Bm = Mm/Am. The recoil line
is described by the equation
Bm ' Br % µrec µ0 Hm ,
(5.28)
where µrec is the relative recoil permeability and Br is the remanence. The magnetic circuit must be
designed so that the operating point (Hm,Bm) lies on the straight part of this line, to avoid irreversible
loss 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 ×
Fig. 6
PC
.
µrec % PC
(5.29)
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.
Page 5.8
5.1.9
SPEED's Electric Motors
DEMAGNETIZING EFFECT OF ARMATURE REACTION
The armature ampere-conductors are distributed uniformly around the rotor in blocks of alternating
polarity, 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 :
Z Ia / a
$M
.
×
Fd '
(5.30)
180
2p × 2
where p is the number of pole-pairs and $M is the magnet arc in Eelec. The effect of Fd on the magnet
operating point can be calculated by combining eqn. (33) with Ampère’s Law,
Hm LM % Hg g ' F d
(5.31)
where g is the effective airgap length. Assuming that Bm . Bg = µ0 Hg,
±Fd !
Hm '
g
Br
µ0
LM % µrec g
.
(5.32)
The positive value of Fd applies at the leading pole-tip and the negative value at the trailing pole-tip. The
demagnetization current Idemag can be extracted from eqns. (5.30) and (5.32) after setting Hm to the
coercivity, which is usually taken to be the intrinsic coercivity HcJ.
Fig. 7
Effect of armature reaction on magnet operating point
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
TRV
' 2BQ
Nm/m 3
(5.33)
where B is the average flux-density in the airgap in [T] and Q is the electric loading in [A/m], given by
Q '
Z × Ia / a
B Darm
.
(5.34)
If J is the RMS current density in the armature conductors, and each conductor has a cross-section
Acond, then JAcond = kf Ia/a, where kf = Irms/Ia is the current form factor, i.e., the ratio of RMS to average
current. Note that the average magnetic flux-density in the airgap at no-load is called the magnetic
loading.
Commutator machines
Page 5.9
5.1.11 ARMATURE WINDINGS
The conductors are laid in slots on the rotor. The ampere-conductors under each magnet pole must all
be in the same direction, so that all the ampere-conductors produce torque in the same direction. Since
the magnet polarities alternate NSNS..., the polarities of the ampere-conductors must also alternate
with 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 lap
winding the coil ends are terminated on adjacent commutator segments, but in wave windings they are
approximately 360Eelec apart; see Figs. 11, 13 and 14. In small machines the winding is usually wound
in one operation on a winding fixture, but large machines use separate form-wound coils.
Fig. 8
Armature amp-conductor distribution
Fig. 9
Fig. 10
Single coil in lap-wound DC armature
Single coil in a wave-wound DC armature
Page 5.10
SPEED's Electric Motors
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 or
back pitch of the coil, yB. The throw must be chosen so that the coil links all the flux of one magnet
when its axis is aligned with the magnet axis. If there are Ns slots and 2 p poles, the pole-pitch is Ns/2p
slots or 2 B/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).
Fig. 11
Front pitch yC and back pitch yB. (a) Progressive lap; (b) retrogressive lap; (c) progressive wave
The ends of the coils are connected to commutator segments. The front pitch of a coil is the number of
commutator 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 commutator
segments per pole. Table 1 summarizes the permissible values of yC. C is the number of commutator
segments, and x is known as the plex or multiplicity of the windings. If x = 1 the winding is simplex; if
x = 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 retrogressive
TABLE 1
Multiple coilsides per layer
It is common to use multiple coilsides per layer (CSL) in each slot and to connect the ends of the
coilsides in one layer to CSL adjacent commutator segments, Fig. 12. This reduces the leakage
inductance of the coil which is being shorted by the brush as it undergoes commutation, that is, the
reversal of the direction of its current. It is usually at the position where its flux-linkage is maximum
and its induced voltage dN/dt is ideally zero, so that the brush can reverse the polarity of the current
under zero-voltage conditions, which helps to avoid sparking. However, the coil flux-linkage includes
leakage flux components in the end-windings and in the slots. Unlike the main flux, these are directly
proportional to the current in the coil or coils undergoing commutation, and the reversal of current
therefore induces a voltage (sometimes called the reactance voltage) between the commutator segments
that are shorted by the brush. If the reactance voltage exceeds about 2!3 V, sparking will occur. Clearly
this is liable to increase when the armature current increases. Therefore it is always desirable to
design for minimum leakage inductance by using multiple coilsides per slot, and by other means such
as wide slot-openings and shallow slots. In general
CSL '
C
.
Ns
(5.35)
Commutator machines
Fig. 12
Page 5.11
Multiple coilsides/layer; CSL= 3
CSL is also known as the "number of bars per slot". It should not be confused with the multiplicity or
plex 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 windings
the number of parallel paths is equal to the number of poles: a = 2p. The number of brushes b in lap
windings is also equal to the number of poles: thus a = b = 2p. In wave windings only two brushes are
needed 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. These
relationships 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 throw
or 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 retrogressive
wave winding with yC = 7. In both examples CSL = 1.
Page 5.12
SPEED's Electric Motors
Fig. 13
Fig. 14
Progressive lap winding with 2 coilsides/slot in a 15-slot, 4-pole armature. yB = 3, yC = 1
Retrogressive wave winding with 1 coilside/slot in a 15-slot, 4-pole armature. yB = 3, yC = 7
Commutator machines
Page 5.13
REFERENCES
[1]
Hamdi, ES [1994] Design of small electrical machines, John Wiley & Sons, Chichester, ISBN 0 471
95202 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: motorsoft@msn.com]
[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 permanentmagnet 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 permanentmagnet 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, ISBN
0 85296 921 X
[10]
The Electro-craft engineering handbook, Reliance Motion Control, Inc.
Page 5.14
SPEED's Electric Motors
Index
Armature inductance 2, 6
Armature resistance 2, 6
Armature windings 10
Automotive 2
Back pitch 11, 12
Back-EMF 2, 3
Back-EMF constant 2, 3
Battery 2
Brush 2, 5, 11, 12
Brushes 12
Chopper 2, 6
Demagnetizing 9
Duplex 11
Efficiency 3, 4
Electric circuit 2
Electric loading 9
Flux/pole 2
Form factor 9
Front pitch 11
Inertia 4
Iron loss 3
kE 2, 3, 5, 6
kT 3
Lap winding 10, 12, 13
Losses 4
Magnet operating point 8, 9
Magnetic circuit analysis 7
Magnetic loading 9
Maximum efficiency 3, 4
Multiplex 11
No-load speed 3
Operating characteristics 4
Parallel paths 2, 12
Permeance coefficient 8
Phase-controlled rectifier 2, 6
Pitch 10-12
Plex 11, 12
Power 2-4, 15
Progressive 11-13
Reactance voltage 11
Rectifier 2, 6
Retrogressive 11-13
Saturation 3, 6, 8
Series resistance 3
Shaft torque 3
Simplex 11
Slot-leakage 6
Span 11
Throw 11, 12
Time constant 5, 6
Torque 3-6, 9, 10
Torque constant 3
Torque per rotor volume 9
Torque/speed characteristic 3
Triplex 11
Wave winding 10, 12, 13
Windings 2, 10-12
SPEED's ELECT ROMAGNET IC PRIMER
1.
INTRODUCTION
Electromagnetics can be a bit confusing?
Not to worry! It took the philosophers about nineteen centuries to figure out even
the rudiments of the subject. So we shouldn’t feel too discouraged.
After the philosophers, the engineers got to work. Working about 100 times faster
than the philosophers they applied the subject so vigorously that by 1900 many of
the fundamental torque-producing mechanisms had been developed, including the
d.c. commutator machine, and a.c. synchrono u s a n d i nduction machines. The
transformer, too, came at about this time. Even though it produces no torque, it's
quite a useful thing.
Thanks to the early engineers, the form of all these electromagnetic energy
converters is such as to exploit the fundamental electromagnetic principles in
their simplest form. In other words, electric machines is a good place to begin the
study 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
2.
SPEED's Electromagnetic Primer
ENGINEERING EFFECTS OF MAGNETIC FIELDS AND ELECTRIC CURRENTS
There were three natural philosophers who did more than any others to get the
engineers started in their exploitation of electromagnetic phenomena. They were
Oersted, Ampere and Faraday 1 , and their most important work was done between
1820 and 1831. Our modern view of electromagnetics owes much to Clerk Maxwell,
who formulated the basic laws mathematically as Maxwell's Equations. Others
made fundamental contributions in the application of the electromagnetic laws.
One of these is S t e i n m e t z , who is credited with some of the basic early work on
phasors and the properties of electrical steels.
It is perhaps best if we take the th r e e o r i ginal discoverers in the wrong
chronological order.
Faraday discovered that an electromotive force ( E M F ), and from it a current, could
be generated by means of a changing magnetic field. The change in the magnetic
field can arise in two ways, corresponding to the transformer and the "dynamoelectric" machine.
A coil of copper wire which is stationary in a magnetic field of varying intensity
has a voltage ( E M F ) induced in it. This is "transformer action".
A coil which is moving relative to a magnetic field of fixed intensity also has an
induced in it. This is "generator action" (sometimes called "flux-cutting").
EM F
Faraday formulated this discovery precisely: the E M F induced is proportional to
the number of turns, N, on the coil; and is also proportional to the rate of change
of what he called “magnetic flux”, N. Thus
e ' N
dN
dt
(1)
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 talk
about it so freely, that it is easy to come to think that it really exists. In reality
the only evidence fo r i ts existence is the very phenomenon it is supposed to
explain — induced voltage!
1
I n th e Un i te d State s, J ose p h H e n r y m ad e th e d i scov e r y of e l e ctr om ag n e ti c i n d u cti on at
ab ou t th e sam e ti m e as Far ad ay . Far ad ay i s accor d e d p r i or i ty b e cau se h e w as f i r st to p u b l i sh
h i s r e su l ts. H e n r y ' s n am e i s u se d as th e u n i t of i n d u ctan ce .
2
T h e w or d "f l u x " m e an s "f l ow ", an d m ag n e ti c f l u x i s m ath e m ati cal l y an al og ou s to oth e r
q u an ti ti e s w h i ch ob e y L ap l ace ' s e q u ati on an d ar e k n ow n as "str e am l i n e s ". Many examp l e s e x i st
i n ae r od y n am i cs, f l u i d f l ow , e l e ctr ostati cs, an d oth e r f i e l d s.
SPEED's Electromagnetic Primer
Page 3
Flux
i
i
Fi g . 1 Fl u x l i n k i n g a coi l
We usually think of flux in terms of flux lines, or “lines of force” which link the
coil. 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’s
Law itself. (We can measure e, but we have no independent way of measuring N).
Solving eqn. (1) for N, we get
N '
1
N
e d t % constant
(2)
T h e concept of magnetic flux is suggested by another (earlier) experiment of
Faraday’s which is of equal important in applications: that t here is a physical
force acting on a conductor which carries current through a magnetic field. This
is the historica l basis of motor technology. Faraday formulated this discovery
precisely also: the f o r ce i s p r o portional to the current, to the length of the
conductor and to what we now call the magnetic “flux-density”, B:
F ' BLi
(3)
The t h r e e q u a n t ities F, B and i are mutually at right angles in space. The fluxdensity B can be thought of as the number of flux lines passing through an area
of one square metre. In general the flux passing through any area A is given by
the simple integral,
N '
B dA.
(4)
If the flux-density is uniform (i.e. if B does not vary across the area A), then this
relationship simplifies to
N ' B A.
3
(5)
T h e l i n e s ar e d e f i n e d b y an oth e r p r op e r ty of th e f i e l d w h i ch i s th at th e y ar e th e p ath s
w h i ch w ou l d b e f ol l ow e d b y i sol ate d m ag n e ti c p ol e s f r e e to m ov e i n th e f i e l d .
Page 4
3.
SPEED's Electromagnetic Primer
THE ESTABLISHMENT OF MAGNETIC FIELDS BY ELECTRIC CURRENT
Ni
a
H
Fi g . 2 Mag n e ti c f i e l d p r od u ce d b y a cu r r e n t
Although Faraday’s discoveries give us a precise way of quantifying flux and fluxdensity [eqns. (1) a n d ( 3 ) ] , they cannot be used for engineering design and
analysis unless we can calculate N or B by some other, independent, method. For
example, we cannot predict the E M F induced in a coil unless we already know the
rate of change of the flux linking it.
Up to now we’ve probably been vaguely aware that the flux discussed in Section
2 was caused or established by another cu rrent not shown in the diagrams, or
pe r h a p s by a permanent magnet. This is true. The basic law governing the
establishment of magnetic fields by electric current was formulated by Ampere,
following experiments by Oersted which shows that electric currents do indeed
establish magnetic fields.
Ampere’s Law says that there is a magnetizing force, or magnetic field strength
H, encircling any electric current and that the integrated value of H around any
closed contour surrounding the current is equal to the current itself. Thus,
Š H. ds
(6)
' i
where ds is an element or the path. The current is sometimes called the total
"magnetomotive force" ( M M F ) in this context. If there are I strands or turns of
conductor, each carrying the current i, then the M M F i s F = N i. The simplest
e x a m p le is that of a long, straight wire. If we take our closed contour to be a n y
circle o f r a d i us a with the current Ni at the centre, then by symmetry H is the
same all the way around and
Š H. ds
' H × 2 B a ' N i,
so
H '
Ni
.
2Ba
H in this case is in the circumferential direction, as shown in the diagram.
(7)
SPEED's Electromagnetic Primer
4.
Page 5
MATERIAL EFFECTS : PERMEABILITY
B
Iron
µ0
Air
H
Fi g . 3 B /H r e l ati on sh i p f or i r on an d ai r
We now have two apparently independent systems of equations for describing the
magnetic field; one in terms of B (deri v e d f r o m Faraday) and one in terms of H
(derived from Oersted and Ampere). We have to relate these two systems to one
another to get a complete set of equations capable of describing the magnetic field.
It turns out that the relationship b e t w e e n B and H is a property of the medium
(i. e ., material) in which they are measured. This property is called the
permeability, µ, and is the ratio between B and H:
B ' µ H.
(8)
2
Since B has the units Wb/m and H has the units A/m, it turns out that µ has the
units of inductance per unit length, H/m, and it has sometimes be e n ca l l ed the
“specific inductivity” of a material.
In air, µ has the value
µo ' 4 B× 10! 7 H/m
(9)
In iron, µ is not constant and we often represent the relationship between B and
H graphically:
For low flux-densities, µ in iron can b e h u n d r e d s and even thousands of times
greater than µ 0 . We say that iron is "highly permeable" and observe that in iron
only a small value of H is required to achieve a large value of B.
Page 6
SPEED's Electromagnetic Primer
As an example, suppose we enclose the current of a previous diagram by an iron
ring. H inside the iron will have the same value as before (by symmetry), i.e. H =
i/2Ba. But B h a s t he value µH, which is much larger than the value outside the
ring, 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 :
Ni
Flux
φ
Iron ring
g
Fi g . 4 Ef f e ct of an ai r g ap i n a m ag n e ti c ci r cu i t
What is the flux-density in the gap?
If we assume now that the iron is infinitely permeable then, since Ampere’s Law
still holds for H, the integral ŠH.ds must be developed entirely across the gap, and
is equal to H × g. This is equal to i (or Ni if there are N turns or strands). In other
words, all the M M F is expended in forcing flux across the gap and none in the iron.
The flux density in the gap is just µ 0 H g = µ 0 Ni/g.
If, instead of a long straight wire, we have a complete coil wound around the iron,
the topology becom e s e s s e n t ially the same as that of an electric machine or
transformer, except that the moveable parts are still missing.
Another way of looking a t t h e high permeability of iron is in terms of "induced
magnetization". Permanent magnets acquire this induced magnetization when
they are magnetized (by a coil with a high current), and they have the property of
retaining this magnetization when the current is switched off, so they tend t o
sustain the flux. This shows that the permanent m a g n e t h a s an internal
"excitation" similar to that of the original magnetizing current.
SPEED's Electromagnetic Primer
Page 7
It may bother some people as to why we have two parameters (B and H) to describe
magnetic fields. What’s the difference between them anyway?
(a)
B obeys Faraday’s Law (through N = BA):
(b)
H obeys Ampere’s Law:
H does not.
B does not.
Therefore we have no choice but to use both. To solve magnetic problems we must
relate Ampere’s Law with F a r a d a y’s Law and this we do through the
"constitutive" relationship B = µH.
5.
GAUSS’ LAW
Gauss' Law states that magnetic
Mathematically it is stated as
flux
always
flows
B . dS ' 0
in
clos ed
loops.
(10)
where dS is an element of the area over w h i ch B is being integrated. This is a
mathematical way of stating that th e r e a r e no isolated magnetic poles or
"charges". If we draw any closed surface, the flux e n t e r i n g m u s t equal the flux
leaving. A simple example is the polepiece of a laboratory magnet that is tapered
to concentrate the flux, i.e. to increase the flux-density :
B . dS ' B1 A1 & B2 A2 ' 0
B1 A1 ' B2 A2
B1
B2
'
A2
A1
> 1
(11)
B 1 A1
B 2 A2
An alternative to the "integral" formulation of Ga u s s ’ Law is the "differential"
formation: div B = 0, which is one of Maxwell's equations.
U n i t s — In the SI system of unit s , A is measured in square metres and N in
Webers. One Weber equals one volt-second, which is clearly the correct unit for
eqn. (2), if e is measured in volts and t in seconds. Consequently, B is measured
2
in Webers per square metre (Wb/m ) and this unit i s ca lled the Tesla, after the
inventor of the induction m otor. In the USA, B is often measured in lines per
square inch, the flux being measured in lines (also called "maxwells"). One Weber
8
equals 10 lines.
Page 8
6.
SPEED's Electromagnetic Primer
SUMMARY
The basic laws of magnetics are :
7.
(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
EXTRAS
P e r m a n e n t m a g n e t s behave like sources of flux, but with very low permeability
called the recoil permeability. They are described by the equation
B ' µrec µ0 H % Br
(12)
where B r is the remanence and µ rec is the relative recoil permeability (i.e., relative
to µ 0 ).
This equation describes the so-called recoil line,
w h i ch is an
approximation to a very narrow minor hysteresis loop along w h i ch t he magnet
operates.
F l u x - l i n k a g e is often defined as the product of "flux × turns", NN. In magnetic
devices it is always the case t h a t "not all the flux links all the turns", and this
gives 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).
V e c t o r p o t e n t i a l is another magnetic field variable like B or H (and like them
it is a vector quantity). It is defined such that B can be derived from it by
differentiation. 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 finiteelement method; and
(b)
the flux linkage of a coil can be calculated directly from the difference
in vector potential between the two coil sides.
SPEED's Electric Motors
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
1.7
n H@@dl '
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/m ; 7.7 kN/m or 1.04 psi
1.14
3.6 A/mm
1.15
4.49 × 10!3 kg-m ; 3.35; 5,225 rad/s
1.16
(a) 17@1 A/mm; (b) 18@7 kNm/m , 9.35 kN/m ; (c) 823 kW/m .
1.17
(a) 145 kW/m ; (b) 1@92 kW/m ; (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/m (3.2 MGOe)
1.20
PC = 1.8933; Bm = 0.6544 T; Bg = 0.6567 T; Energy product = 180 kJ/m (22.6 MGOe)
1.21
PC = 3.9184; Bm = 0.7967 T; Bg = 0.2033 T; Energy product = 130 kJ/m (16.2 MGOe)
1.22
(a) 0.55; (b) Bm = 0.4 T, Hm = 111.4 kA/m; (c) 23%; (d) 0.27.
1.6
E Ni
3
2
2
2
2
3
3
2
3
2
3
3
3
Full solutions are given at the end
1.1
Sketch the complete B-H curve for a typical ‘hard’ permanent magnet. Indicate the remanent
flux-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
B
C
1.3
Which of the following motors never has permanent magnets?
A
B
1.4
DC commutator motor
induction motor
Which of the following materials is a 'hard' permanent magnet?
A
B
C
1.5
Induction motor
Brushless DC servomotor
Variable-reluctance stepper motor
Samarium Cobalt
Stainless steel
Gallium Arsenide
In the following motors, are the permanent magnets on the rotor or the stator?
A
B
DC commutator motor
Brushless DC motor
1.6
State Ampere's Law.
1.7
Which of the following units are identical to the henry?
A
B
C
Wb/At
At/Wb
V-s/A
(Webers per ampere-turn)
(Ampere-turns per Weber)
(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 bottle
containing 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 wall
is built from bricks. It is necessary to test the polarity of each small magnet before adding it to
the assembly. If the material is high-energy Neodymium-Iron-Boron, which of the following
polarity tests would you recommend? Give reasons for your choice.
1.11
A
floating the small magnet on a cork in water and noting which way it points in the
earth's field;
B
offering the small magnet up to a compass and noting which way the compass needle
swings;
C
offering the small magnet up to another small magnet whose polarity is known, and
noting whether the poles attract or repel.
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.
1.12
Calculate the rotor diameter and length of a motor that develops 100 W shaft power at 1500 rpm
3
with a torque per unit rotor volume of 12 kN-m/m 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 the
stator 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 the
tooth-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 35,000
2
2
rad/s . It is found to be capable of accelerating an inertial load at 4,600 rad/s through a 2:1
2
speed-reduction gearbox. Estimate the inertia of the load in kg-m . What value of gear ratio
would result in the maximum acceleration of this motor and load combination, and what would
be the value of the maximum acceleration?
1.16
(a)
An AC induction motor has 24 slots, each with a slot area of 62.2 mm . The stator bore
diameter is 52.2 mm. The factory can manufacture windings with a slot-fill factor of
2
47%. If the maximum current density allowable in the copper is 4 A/mm , calculate the
maximum electric loading.
(b)
The fundamental winding factor of the induction motor in part (a) is 0.911. The motor
can be operated with a peak airgap flux-density of 0.85 T, sine-distributed. Calculate the
3
2
torque per unit rotor volume in kNm/m and the airgap shear stress in kN/m , using the
electric loading from part (a).
(c)
If the stator OD is 106 mm and the stack length is 57 mm, calculate the power per unit
stator volume when the motor delivers its rated torque at 1745 rev/min.
(a)
If the efficiency of the motor in Q1.16 is 85%, calculate the average power loss per unit
of stator volume.
(b)
Assume that 50% of the power loss is removed by conduction through the motor
mountings, and 50% at the curved cylindrical surface area of the stator. Calculate the
heat transfer rate required at the stator surface in kW/m2.
(c)
Estimate the temperature rise if the heat removal at the curved cylindrical surface of
the stator is by natural convection. Comment on the result.
(d)
Estimate the temperature rise if the heat removal at the stator surface is by forced
convection with an air velocity of 5.5 m/s.
(a)
At rated power, the motor of Q1.16 has a winding temperature rise of 100EC when the
ambient temperature is 35EC. Calculate its effective thermal resistance with forced
convection as in Q1.17(d).
(b)
If the thermal capacity of the motor is 1.4 kJ/EC, calculate the thermal time-constant
corresponding to the conditions in (a).
(c)
Assuming constant efficiency, what is the maximum power output that can be sustained
for 25 min without exceeding the rated temperature rise?
(d)
What is the maximum time for which the motor can operate at 125% of rated power
without exceeding the rated temperature rise? (Assume cold start.)
1.17
1.18
2
Fig. 1.8
1.19
Fig. 1.8 shows a magnet with pole-pieces arranged to focus flux into a circular airgap of length
10mm and diameter 100mm. The magnet dimensions are 100 × 150 × 100 mm. Determine the
permeance coefficient, the magnetic flux-density in the magnet and in the airgap, and the
magnet energy-product. Neglect fringing and assume that the pole-pieces are infinitely
permeable. The magnet remanent flux-density is 0.4 T and its relative recoil permeability is 1.0.
Fig. 1.9
1.20
Fig. 1.9 shows a rotary device excited by a permanent magnet
whose demagnetization curve is straight throughout the
second quadrant of the hysteresis loop. The remanent flux
density 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 permeance
coefficient, the magnet flux-density and the magnet energyproduct.
1.21
Fig. 1.10 shows a loudspeaker magnet assembly. If the
magnet has a straight demagnetization characteristic with
a remanent flux-density of 1.0 T and a relative recoil
permeability of 1.0, calculate the magnetic flux-density in the
airgap. Also determine the permeance coefficient and the
magnet 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 in
the assembly?
Fig. 1.10
Fig. 1.11
1.22
Fig. 1.11 shows the demagnetization characteristics of a permanent magnet at two
temperatures, T1 = 20EC and T2 = 100EC. The magnet operates initially at point X. The
horizontal 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, after
which there is a change in the configuration of the magnetic circuit which causes the
permeance coefficient to increase to 3@0. Determine graphically the values of Bm and Hm
at the new operating point.
(c)
If the magnet temperature is maintained at 100EC while the magnet is keepered (i.e., Hm
is reduced to zero), what is the loss of remanence, expressed as a percentage of the
original remanence at 20EC?
(d)
What fraction of the loss of remanence in (c) is irreversible?
SPEED's Electric Motors
Solutions to Problems
1. Foundation
S1.9
W '
'
1
BH '
2
1 B2
x vol.
2 µ0
1
1.02
x 1000 x 10!6
2 4 B x 10!7
' 398 Joules
S1.12
T '
100
2B
1500 x
60
Rotor vol. Vr '
' 0.637 N&m ' 5.628 in&lb
T
'
TRV
0.637
12,000
m 3 ' 53,051.6 mm3
But Rotor vol. Vr ' B r 2 Lstk ' B 2 r 3
3
ˆ D ' 2r ' 2 x
Lstk
2r
' 2 B x 1.4 r 3
53,051.6
' 36.40 mm ' 1.434 in
2 B x 1.4
Lstk ' 1.4 x D ' 51 mm ' 2.01 in .
S1.13
Use eqn. (1.4):
TRV '
B
x 0.9 x 12k x 0.6 x 1
2
' 14.4 kNm / m3
F '
Also F '
14.4
' 7.7 kNm/m 2
2
14.4
' 1.04 p s i
13.8
S1.14
Use eqn. (1.8):
J '
A8
'
Fslot Aslot
A
'
Fslot d ( 1 ! J )
12
' 3.6 A/mm 2
0.42 x 18 x ( 1 ! 0.56 )
S1.15
Use eqn. (1.14):
motor :
T
' 35,000 rad / s2 ; T ' 14 N&m
Jm
ˆ Jm '
14
' 4.0 x 10& 4 kg&m2
35,000
n ' 2
Tmp
" '
n Jm %
JL
n2
14
4,600 '
2 4 x 10!4 %
JL
6
JL ' 4.49 x 10!3 kg&m2
4
Use eqn. (1.16):
Optimum ratio n '
"max '
JL
Jm
'
4.49 x 10!3
4 x 10!4
' 3.35
1
14
1
x
' 5,225 rad / s2
x
!4
2
3.35
4 x 10
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 corresponds
to 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
(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
(c)
(2.38 W/in3)
(= 1@24 W/in2).
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
so that
=
7@5 x ()T)5/4 / (OD)1/4
)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@92
kW/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) and
Q 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.
S1.19
Using eqn. (1.63)...
Am
" '
Ag
the area ratio is
Lm
8 '
the length ratio is
g
PC '
the permeance coefficient is
i.e.
Bm
Hm
200 x 150
' 3.8197
B
2
x 100
4
'
'
8
'
"
100.0
' 10.0
10.0
10.0
' 2.618
3.8197
' ! µ0 x PC
But Bm ' Br % µrec ( µ0 Hm)
' Br % µrec x
'
B r x PC
µrec % PC
Bm
! PC
'
Br
µrec %
"
8
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
Energy product ' Bm Hm '
'
B m2
µ0 x PC
0.28942
4 B x 10
!7
x 2.618
' 25.5 kJ/m 3 (3.2 MGOe)
S1.20
Using eqn. (1.63)...
" '
the area ratio is
Am
Ag
50 x
'
8 '
the length ratio is
Lm
2g
Bm
i.e.
Hm
'
8
'
"
PC '
the permeance coefficient is
B
x 11.5 x 10
180
' 1.0036
10 x 10
3.8
' 1.9
2 x 1.0
1.9
' 1.8933
1.0036
' ! µ0 x PC
But Bm ' Br % µrec ( µ0 Hm)
' Br % µrec x
'
B r x PC
µrec % PC
Bm
! PC
'
Br
µrec %
"
8
ˆ Bm = Br x 1.8933/(1.0+1.8933)
= 0.6544 Br
= 0.6544 T
Bg
= (Am/Ag) x Bm = "Bm
= 1.0036 x 0.6544
= 0.6567 T
Energy product ' Bm Hm '
'
B m2
µ0 x PC
0.65442
4 B x 10
!7
x 1.8933
' 180 kJ/m 3 (22.6 MGOe)
S1.21
Using eqn. (1.63)...
Am
" '
the area ratio is
Ag
B
x 142
4
' 0.2552
B x 16 x 12
'
Lm
8 '
the length ratio is
g
8
'
"
PC '
the permeance coefficient is
Bm
i.e.
Hm
'
2.0
' 1.0
1.0
1.0
' 3.9184
0.2552
' ! µ0 x PC
But Bm ' Br % µrec ( µ0 Hm)
' Br % µrec x
'
B r x PC
µrec % PC
Bm
! PC
'
Br
µrec %
"
8
ˆ Bm = Br x 3.9184/(1.0+3.9184)
= 0.7967 Br
= 0.7967 T
Bg
= (Am/Ag) x Bm = "Bm
= 0.2552 x 0.7967
= 0.2033 T
Energy product ' Bm Hm '
'
B m2
µ0 x PC
0.79672
4 B x 10
!7
x 3.9184
' 130 kJ/m 3 (16.2 MGOe)
S1.22
(a)
PC at X = Bm/µ0Hm = 0@27/0@49 = 0@55.
(b)
Draw YC parallel to the recoil line through B. Draw load-line with a slope of 3. From the
intersection at Z, 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 C, with Bm = Br = 0@5 T. Original remanence at A 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 A.
(d)
The fraction that is irreversible is BC/AC
BC AC = (0@54 ! 0@5)/0@15 = 0@27.
SPEED's Electric Motors
Problems
2. Brushless Permanent-Magnet Machines
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
Full solutions are given at the end
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 and
drive. 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
2.3
2.4
A 3-phase, 4-pole, wye-connected brushless DC motor has a stator bore diameter D = 52 mm, and
stack length Lstk = 50mm. The number of turns in series per phase is 48, and the resistance
measured between terminals is 0.052 ohm. When it is driven at 1,000 rev/min by an auxiliary
motor, 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
A wye-connected brushless DC squarewave motor has a torque constant of 0.177 Nm/A and its
phase resistance is 0.5 ohm/phase. It operates with a DC supply voltage of 48 V, and the on-state
resistance 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 duty
in sequence during each cycle of 360E elec. Also calculate the conduction loss in each
MOSFET transistor at half speed.
A wye-connected brushless DC motor has a back-EMF constant of 18.85 V/1000 rpm. Its line-line
inductance 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 special
aluminium mirrors. At the beginning of the process the polishing head cycles back and
forth at a frequency of 0.1 Hz. As the polishing process continues, the cycle time
decreases 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 at
0.1 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.
2.5
A brushless DC motor has the cross-section shown in Fig. 2.5 with D = 60 mm (stator bore
diameter), 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 ceramic
magnet has Br = 1.25 T and µrec = 1.05. The current in the stator coil is zero.
Fig. 2.5
(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 inside
of the stator, from 0 to 360E.
(c)
Draw an equivalent magnetic circuit and use it to calculate the flux crossing the airgap
under 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 direct
axis.
(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 the
rotor 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 mechanical
work that is done in rotating the rotor 180E from the position shown. Is this work
positive or negative? Explain what you mean by “positive” or “negative”.
2.6
A wye-connected PM brushless DC motor operated in squarewave mode has a torque constant
of 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 controller
transistors 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 voltage
drop across each transistor is a constant 1 V. The friction torque has been separately
measured as 0.046N-m at this speed, and the iron loss can be taken as a constant
mechanical loss of 20W. Calculate
(i)
(ii)
(iii)
(iv)
(v)
(vi)
the electromagnetic torque Te
the current I
the power loss in the winding resistance PCu
the motor efficiency
the total power loss in the transistors, and
the overall efficiency of the drive and motor.
Fig. 2.7
2.7
Fig. 2.7 shows the top view of a permanent-magnet actuator used for guiding cables on to a
vertical-axis cable-laying drum. The length of the arc-shaped magnets in the direction of
magnetization is 60 mm and that of the rectangular magnets is 55 mm. The vertical height of the
electromagnetic assembly is 250 mm. The airgap length is 10 mm. If the magnet is Ferrite with
a remanent flux-density of 0.4 T and relative recoil permeability of 1.0, calculate the flux-density
in 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 a
coil of 200 turns mounted on the rotating 'armature'. If the current in the coil is 20 A, calculate
the torque.
2.8
A 3-phase, 4-pole brushless PM motor has 36 stator slots. Each phase winding is made up of 3
coils per pole with 20 turns per coil, and all the coils are in series. The coil span is 7 slots. If the
fundamental component of magnet flux is 1.8 mWb, calculate the open-circuit phase EMF E at
3,000 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 the
rotor 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 phase
current of 4.0 A at ( = 15E at 3,000 rpm, calculate the terminal voltage and the
power-factor angle.
2.10
A brushless permanent-magnet sinewave motor has an open-circuit voltage of 173 V at its
corner-point speed of 3,000 rpm. It is supplied from a PWM converter whose maximum voltage
is 200 V RMS. Neglecting resistance and all other losses, estimate the maximum speed at which
maximum current can be supplied to the motor.
2.11
Show that with rated phase current I, the angle ( which maximises the torque of the hybrid
PM/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 3,000
rpm, and rated current of 4.0A, the torque is maximised with ( = 7.97E. What is this maximum
torque? Neglect losses.
2.12
If the high-energy magnets in the motor of Problem 2.11 are replaced by ceramic magnets, the
ohmic values of the reactances remain the same but the open-circuit phase EMF is decreased
from 35.8 V to 11.3V. Determine the optimum value of ( and the maximum torque, assuming
that the rated current remains 4.0A. What percentage of this maximum torque is reluctance
torque?
2.13
The absolute maximum speed of a permanent-magnet brushless motor can be estimated
approximately by assuming that the current is at its rated value and is entirely in the negative
d-axis, i.e. all the current capacity of the drive is used up in flux-weakening. If the maximum
controller voltage is Vc then, neglecting losses, V = 0 + jVq = jVc = jEq + jXdId where Id is
negative. If Vc = 38 V and Eq = 35.8 V at 3,000 rpm, and if Xd = 1.18 ohm at 3,000 rpm, estimate the
absolute maximum speed with I = 4.0 A, remembering that both Eq and Xd are proportional to
frequency or speed.
2.14
Calculate the synchronous reactances Xd and Xq of the interior permanent-magnet motor shown
in 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 in
direction of magnetization Lm = 5.5 mm; magnet width = 22 mm; per-unit rotor leakage
permeance 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
Fig. 2.14
1
This 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.
SPEED's Electric Motors
Solutions to Problems
2. Brushless Permanent-Magnet Machines
S2.1
(a)
(b)
S2.2
(i)
kE = kT = 4pTph Mg /B = 2 Tph Bg D Lstk 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 / k E = 24/0@07257 = 318@8 rad/s = 3,045 rev/min
(iv)
I0 = Vs /RLL = 24/(2 x 0@052) = 231 A
(v)
T0 = k T I0 = 0@07257 x 231 = 16@75 Nm
S2.3
(a)
Tb '
(b)
2 transistors and 2 motor phases
in series at any time.
Vs & 2 R ISP
kE
'
48 ! 2 x (0.5 % 0.5) x 5
' 214.7 rad/s ' 2,050 rpm
0.177
At half- speed,
0.177 x
d '
(c)
RDS(on) + Rph
214.7
% 2 x (0.5 % 0.5) x 5
2
' 0.604
48
Transistor conduction waveform is
IQ1(rms) ' 5 x
0.604 x
60
60
% 1 x
' 2.585 A
360
360
2
x rDS(on) ' 2.5852 x 0.5 ' 3.34 W in each transistor
Conduction loss ' IQ1(rms)
S2.4
(a)
Te '
L
'
R
4.7
' 2.0 ms
2 x 1.175
Jmotor % Jload ' Jmotor ( 1 % 3 )
' 0.125 x 4 ' 0.5 m kg&m 2
Tm '
(b)
G (j T ) '
RJ
kE2
2 x 1.175 x 0.5
' 36.3 ms
2
18.85
30
x
1000
B
1/0.18
.
1 % j T x 0.002) (1 % j T x 0.0363 )
When fpolish ' 0.1 Hz,
G '
'
T ' 2 B x 0.1 and
1/0.18
• 1/0.18 ' 5.55 rad/sec/V
(1 % j 0.00126 ) (1 % j 0.0228 )
Virtually “DC”: imaginary parts negligible.
(c)
When fpolish ' 10 Hz,
G '
T ' 2 B x 10 and
1/0.18
• 1/0.18 x 0.4 e !73.5E ' 2.22 e !73.5E rad/sec/V
(1 % j 0.126 ) (1 % j 2.28 )
40% amplitude
(d)
1
' 0.7
* (1 % j 0.002 T) (1 % j 0.0363 T) *
this term is dominant
neglect
0.7 .
1
2
'
1
*1 % j1*
(Recognizing that 70% is close
to the “!3 dB” point)
The required frequency is approximately the frequency at which
0.0363 T ' 1, i.e.
1
1
x
' 4.4 Hz .
2B
0.0363
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 the
diagram – it is not given in the problem specification); and the length is 35 mm. Thus
Am '
B
x 120 x 21.6 x 35 ' 1583 mm 2
180
The permeance of one magnet is
µr µ0 Am
Pm0 '
4 B x 10!7 x 1.05 x 1583 x 10!6
'
Lm
8 x 10
!3
' 2.61 x 10!7 Wb/At
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
Ag '
B
x 120 x 30 x 35 ' 2119 mm 2
180
The airgap reluctance (neglecting fringing) is
Rg '
g
'
µ0 A g
8 x 10!3
4 B x 10
!7
x 2199 x 10
!6
' 2.895 x 106 At/Wb
From eqn. (2.18),
Mg '
Mr
1
fLKG
% Pm0 Rg
'
1.25 x 1583 x 10!6
1
% 2.61 x 10!7 x 2.895 x 10!6
0.9
' 1.06 x 10!3 Wb
(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.
Hm '
B r ! Bm
µ0 µrec
'
1.25 ! 0.744
4 B x 10!7 x 1.05
' ! 383 A/mm
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 N
m
Bg r Lstk d 2 where N is the number of turns in the coil.
(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 of
EMF is
epk '
dR
dR
2B
2 x 31.8 x 10!3
' Tm
' 4000 x
x
' 12.7 V
dt
d2
60
B
120 x
180
(i) The permeance P “seen” by the coil is approximately µ0A/h, where A is a cylindrical surface of about
180E 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
P '
µ0 A
h
'
25.8 x B x 35 x 10!6
16.8 x 10!3
µ0 ' 0.169 µ0 Wb/At
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 work
must be done by exerting a torque on the rotor. If we regard the work as mechanical input, as though
the machine was a generator, then it is positive. If we regard the work as mechanical output, as though
the machine was a motor, then it is negative.
S2.6
(a)
T0 '
(b)
I0 '
V
'
kT
48
' 400 rad / sec ' 3810 rpm
0.12
48 ! 2
' 153.3 A
0.15 x 2
T0 ' 153.3 x 0.12 ' 18.4 N m
(c) Electromagnetic (airgap) power Pg = Pmech + Pfriction + Piron loss.
Pfriction ' 0.046 x 3400 x
Since Tfriction = 0.046 W,
2B
' 16.38 W
60
and Pg = 330 + 16.38 + 20 = 366.38 W. So the electromagnetic torque is
(i)
Te '
and therefore the current is
(ii)
I '
366.38
2B
3400 x
60
Te
kT
'
' 1.029 Nm
1.029
' 8.575 A
0.12
(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%
S2.7
The magnets are magnetized N-S-N-S. Analyze one half pole, which has one-half of an arc magnet in
series with one-half of a rectangular magnet in series with an airgap of arc 95/2 = 47.5E.
The airgap reluctance is
Rg '
g
'
µ0 A g
4 B x 10!7
10 x 10!3
' 1.3713 x 105 At/Wb
B
x 280 x 250 x 10!6
x 47.5 x
180
The magnets are in series, so we need their “Thevenin” equivalent circuits with parameters Fm1 = Mr1/Pm1
and 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
R m1 '
1
'
Pm1
L m1
µ0 µrec Am1
'
60 x 10!3
4 B x 10
!7
' 8.232 x 105 At/Wb
x 1.0 x 0.0580
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
Rm2 '
1
Pm2
'
Lm2
µ0 µrec Am2
'
55 x 10!3
4 B x 10
!7
' 1.0298 x 106 At/Wb
x 1.0 x 0.0425
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
Mg ' 2 x
' 2 x
Fm1 % Fm2
R g % Rm1 % R m2
17507 % 19099
(0.8232 % 1.0298) x 106
' 0.0395 Wb
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 the
torque is 2 F x r = 2 x 341 x 280 x 10!3 = 191 Nm.
S2.8
Total turns = 4 x 3 x 20 = 240 turns.
q(
2
kd1 '
(
q sin
2
sin
kp1 ' cos
'
sin 30E
' 0.9598
3 sin 10E
g
1
' cos
2
2
2
B
9
' 0.9397
kw1 ' kd1 kp1 ' 0.9397 x 0.9598 ' 0.9019
kw1 Nph ' 0.9019 x 240 ' 216.5 turns
Frequency f = 3000/60 x 2 = 100 Hz, so
E '
2B
2
x 216.5 x 1.8 x 10!3 x 100 ' 173.1 V rms
q = 3 slots/pole/phase
( = (360/36) x 2 = 20E
(slot-pitch angle. Eelec)
S2.9
(a) Effective airgap is
gO ' gN %
Lm
µrec
' 1.0 x 1.055 %
9.5
' 10.02 mm
1.06
Then the airgap component of synchronous reactance is
Xsg '
6 µ0 D L stk f
2
p gO
(kw1 Nph)2 '
6 x 4 B x 10!7 x 102 x 120 x 10!6 x 100
2
2 x 10.02 x 10
!3
x 216.5 2 ' 10.8 ohm
(b) The electromagnetic torque is
Te ' 3
Eq
Tm
x Iq '
3 x 173.1
x 4 x sin ( 0 ) ' 6.612 Nm
2B
x 3000
60
(c) The total synchronous reactance (including leakage reactance) is
Xs ' Xsg % XF ' 10.8 % 2 B x 5 x 10!3 ' 13.94 ohm
From the phasor diagram,
V ' E % ( R ph % j Xs )II
' j 173.1 % ( 3.7 % j 13.94 ) x 4.0 e j (90 % 15E)
' 182.33 e j 108.45E V
Power-factor angle N = 108.45 ! (90 + 15) = 3.45E. PF = cos N = 0.998 lag.
Also note the solution in dq components:
V = (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 V = Vd + j Vq = !57.6906 + j 172.9636 = 182.33 ej108.45E V
S2.10
The per-unit EMF is u = 173/200 = 0.865
The ratio between base speed and maximum speed is
NQ
ND
' u !
1 ! u2 ' 0.865 !
1 ! 0.8652 ' 0.3632
ˆ Max speed = 3,000/0.3632 = 2.753 x 3,000 = 8,260 rpm
S2.11
T ' m p [ Qd I q ! Qq I d ]
( m ' No of phases )
X d Id
Xq I q
) Iq !
I ] ( Q1Md,Qd and Qq are all RMS )
' m p [ (Q1Md %
T
T d
mp
[ E q Iq % ( X d ! Xq ) I d Iq ]
'
T
mp
[Eq I cos ( % (Xq !Xd).I 2 sin ( cos ( ]
'
T
mp
[Eq I cos ( % ) X.I 2 sin ( cos ( ] where ) X ' Xq ! Xd > 0
'
T
mp
) X.I
[Eq cos ( %
sin 2 (] I
T
2
'
ˆ
dT
'
d(
mp
[ !Eq sin ( % ) X. I cos 2 ( ] I ' 0
T
Eq
cos 2 (
or
sin (
'
) X.I
sin (
'
cos 2 (
Eq
) X.I
' ) x (pu) , if the base impedance is
Eq
I
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
s '
!1 ± 1 % 8 ) x 2
4 )x
)x '
(2.47 ! 1.18) x 4
' 0.1441 pu
35.8
!1 ± 1 % 8 x 0.14412
' 0.1386 or !3.608
–––
ˆ s '
4 x 0.1441
so ( ' 7.97E .
The frequency is 100 Hz and with p = 2, m = 2 the torque is
Tmax '
2 x 2
2.47 ! 1.18
[ 35.8 cos 7.97E %
x 4.0 sin ( 2 x 7.97E ) ] x 4.0
2 B x 100
2
' 0.0255 x [ 35.4542 % 0.7085 ]
' 0.921 N&m
In this example, with an Xq/Xd ratio of 2.1 and )x = 0.1441, only 2% of the maximum torque is contributed
by reluctance torque.
S2.12
)x '
s '
T '
( 2.47 ! 1.18 ) x 4.0
' 0.4566
11.3
!1 ±
1 % 8 x 0.45662
' 0.3468 Y
4 x 0.4566
( ' 20.29E
2 x 2
( 2.47 ! 1.18 ) x 4.0
sin ( 2 x 20.29E ) x 4.0
11.3 cos 20.29E %
2 B x 100
2
' 0.0255 x [ 10.5987 % 1.6784 ]
' 0.313 N&m
In this example with Xq/Xd = 2.1 and )x = 0.4566 pu, the reluctance torque contributes 14% to the
maximum torque.
S2.13
We’re assuming that all the current is in the d-axis, to make the solution tractable, but this means the
torque will be zero, and therefore the solution is only approximate (to the extent that losses can be
neglected). What we’re really estimating is an upper bound for the speed, with the given motor and drive
parameters.
Iq = 0, T = 0, R = 0
Vd ' 0
Vq ' E q % Xd I d
Id ' !4.0 A ;
38 ' 35.8 x
Vq ' 38 V
N
N
%
x 1.18 x ( !4.0 )
3000
3000
' ( 35.8 ! 1.18 x 4 )
N
3000
ˆ N ' 1.2227 x 3,000 ' 3,668 rpm
The phasor diagram shows the flux-weakening effect of Id. Evidently the flux-weakening is more
pronounced when XdId is an appreciable fraction of Eq, suggesting that a high reactance extends the
speed range. Though this is generally a valid inference, it says nothing about the variation of torque
with speed. In this motor the speed range extends only 22% above 3,000 rpm, and the torque will fall
rapidly between base speed (3,000 rpm) and maximum speed (3,668 rpm).
S2.14
Use the formulas in chapter 6 of Miller [1989]2 or chapter 6 of Hendershot and Miller [1994]3 We need
to work through quite a few of them... 4
Am ' 22 x 50 ' 1100 mm 2
µrec µ0 Am
Pm0 '
Lm
'
4 B 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
Ag '
2B
1
x
x 25.4 x 50 ' 1330 mm 2
3
2
Rg '
gN
1.177 x 0.4 10!3
'
' 2.8169 x 105 At/Wb
!7
!6
µ0 Ag
4 B x 10 x 1330 x 10
1 % Pm Rg ' 1 % 3.0162 x 10!7 x 2.8169 x 105 ' 1.085
k1 '
4
"B
' 1.1027
sin
2
B
k" d '
sin " B
2
'
%
B
3
gN
g dO '
k1 k" d
k1ad !
'
120
2
' )
180
3
sin ( " B / 2 )
sin 60E
3
'
x sin 60E ' 0.827
'
"B/2
B
2
B
x
3
2
k1ad ' " %
Xd '
(" '
1 % Pm R g
6 µ0 D Lstk f
p g dO
2
'
sin
2
B
3
' 0.9423
B
gN
' 9.822 gN
1.1027 x 0.827
0.9423 !
1.085
gN
1
' 0.102 .
'd '
'
g dO
9.822
( k w1 Nph )2 % XF
6 x 4 B x 10!7 x 50.8 x 50 x 10!6 x 33.33
2
2 x 9.822 x 1.177 x 0.4 x 10
!3
x ( 0.837 x 480 )2 % 3.264
' 5.5707 % 3.264 ' 8.8347 ohm
2
Brushless permanent-magnet and reluctance motor drives, Oxford University Press, 1989
3
Design of brushless permanent-magnet motors, Magna Physics Publications, 1994
4
This 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.
k1aq ' " % S %
g qO '
Xq '
'
sin S B ! sin" B
'
B
2
% 0.0251 %
3
gN
gN
' 2.2665 gN
'
k1aq
0.4412
6 µ0 D Lstk f
p gqO
2
6 x 4 B x 10!7 x 50.8 x 50 x 10!6 x 33.33
2 x 2.2665 x 1.177 x 0.4 x 10
' 24.141 % 3.264 ' 27.405 ohm
!3
2
B
3
B
'q '
( k w1 Nph )2 % XF
2
sin 0.0251 B ! sin
' 0.4412
gN
1
' 0.441 .
'
g qO
2.2665
x ( 0.837 x 480 )2 % 3.264
SPEED's Electric Motors
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.
Full solutions are given at the end
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
B
C
Standstill
when s < 0
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 the
speed 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 to
a 60 Hz supply? Give the answer in rev/min. Calculate the rotor speed if the slip is 5%. What
is the frequency of the rotor currents?
3.4
(a) In which of the following machines would you expect to find a commutator?
A
B
C
D
E
F
A permanent-magnet d.c. servomotor;
a series-wound d.c. generator;
A wound-rotor induction motor;
An a.c. steam-turbine generator;
A “universal” motor;
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 the
speed/torque characteristic of an induction motor with
A
B
C
low rotor resistance;
high rotor resistance;
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 efficiency
is 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 Nm
when running at full load. The frequency of the rotor currents is 2 Hz. Calculate the shaft
power. If the torque absorbed by windage and friction is 2 Nm, and if the stator losses total 1
kW, 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)
(b)
(c)
What will be its speed at half rated torque?
What will be its speed at half rated torque and 70% voltage?
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 of
1764 rev/min. In the per-phase equivalent circuit, the stator resistance is 0.25 ohm, the referred
rotor resistance is 0.5 Ohm, the total leakage reactance is 2.5 ohm, and the magnetizing
reactance 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/min
with its slip-rings shorted. Assuming constant friction torque of 1.5 Nm and constant stator
losses of 1.8 kW, find the input power and the efficiency. When the speed is reduced to 600
rev/min by increasing the rotor circuit resistance, the load torque remains constant at the
full-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 input
power is 33 kW and the stator copper loss is 1200 W. Friction and windage loss is 80W. Core loss
is negligible. Find
(a)
(b)
(c)
(d)
(e)
(f)
3.13
the slip;
the airgap power Pgap;
the rotor copper loss;
the shaft torque in Nm;
the shaft power in kW;
the efficiency.
A 6-pole, 3-phase, Y-connected, 460-V, 60 Hz induction motor has the following equivalent-circuit
parameters (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)
(b)
(c)
(d)
(e)
(f)
(g)
the line current;
the power factor;
the efficiency;
the shaft torque;
the breakdown torque;
the locked-rotor torque;
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 is
connected 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 the
3-phase voltages at the load terminals differ at these two speeds?
SPEED's Electric Motors
Solutions to Problems
3. Induction Machines
S3.2
N '
120 f
(1 & s) '
P
120 x 120
( 1 ! (! 0.1 ) ) ' 7,920 rpm
2
fr ' s f ' 0.1 x 120 ' 12 Hz
S3.3
Ns '
120 f
'
P
120 x 60
' 1200 rpm
6
(synchronous speed)
N ' Ns ( 1 ! s ) ' 1200 ( 1 ! 0.05 ) ' 1140 rpm
fr ' s f ' 0.05 x 60 ' 3.0 Hz
S3.4 : See Answers on front page
S3.5 : See theory manual, chapter 3
(shaft speed or rotor speed)
S3.6
Output power ' Pshaft ' T Tm ' 8 x
Output power
'
Effcy.
'
but
Pin
'
3 VL IL cos N
i.e.
1660
'
3 x 230 x 4.92 cos N
' 0.85 ( lag. )
2B
60
1466
' 1666 W
0.88
Input power
ˆ cos N
1750 x
' 1466 W
S3.7
Tm '
2 Ts
P
(1 & s) '
2 x 2 B x 50
( 1 ! s ) ' 150.8 rad / sec
4
[ 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
'
12.06
' 12.57 kW
1 ! 0.040
Rotor copper loss ' Pr ' Pg ! Pmech ' s Pg '
'
Input power Pin
0.040
x 12.06 kW ' 502.6 W
1 ! 0.040
' Airgap power % stator loss
' 12.57 % 1 ' 13.57 kW
Effcy. '
Pshaft
Pin
'
11.76
' 86.7%
13.57
s
P
1 ! s mech
S3.8
It is a high-efficiency motor and the slip is small, so we can assume T
At full load the slip is
ˆ s '
Ns ! 1470
Ns
where
Ns '
RR
s.
120 x 50
' 1500 rpm ;
4
1500 ! 1470
' 0.020
1500
(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,
s '
(c)
%
Vs2
1
1
x 0.02 Y N ' 1500 ( 1 & 0.02041 ) ' 1469 rpm
x
2
0.72
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
S3.9
Ns '
120
x 60 ' 1800 rpm;
4
s '
1800 ! 1764
' 0.020
1800
IM '
(full load)
460 / 3
j 60
' & j 4.426 A
At full load
IR '
460 / 3
' 10.416 ! j 1.031 A
0.5
0.25 %
% j 2.5
0.020
I S ' I R % I M ' 10.416 ! j 5.458 ' 11.76 e !j 27.65E A
At standstill (s = 1)
IR '
460 / 3
' 29.24 ! j 97.46 A
0.25 % 0.5 % j 2.5
I s ' I R % I M ' 29.24 ! j 101.89 A ' 106.0 e !j 73.99E A
Ratio
Standstill current
'
full load current
106.0
' 9.01
11.76
Full&load PF ' cos ( 27.65E ) ' 0.886 lag
S3.10 : See theory manual, chapter 3.
S3.11
Ns '
120 x 50
' 1000 rpm
6
Pmech ' 22.5 k % 1.5 x
Total mech.
power
developed
Pgap '
Pin
Pmech
1 ! s
Shaft power
delivered to
load
'
950 x
2B
60
s '
1000 & 950
' 0.050
1000
' 22,649 W
Friction
power
22,649
' 23,841 W
1 ! 0.050
' Pgap % Stator losses ' 23,841 % 1800 ' 25.64 kW
Effcy. '
Pshaft
Pin
'
22.5
' 87.7%
25.64
At 600 rpm, Pshaft ' 22500 x
600
' 14,210 W
950
(at constant torque)
Assume friction torque is constant at 1.5 Nm
Pmech ' 14,210 % 1.5 x
600 x
2B
60
s '
1000 ! 600
' 0.40
1000
ˆ Pgap '
14,305
' 23,841 W
1 ! 0.40
New slip
Pin
' 23,841 % 1800 ' 25,641 W
Effcy. '
Pshaft
Pin
'
14,210
' 55.8 %
25,641
' 14,305 W
(Note&same as before)
S3.12
120
x 60 ' 900 rpm
8
Ns '
(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 '
(f)
Effcy. '
Pshaft
Tm
'
Pshaft
Pin
30,660
2 B x 900
'
30,660
33,000
s '
900 ! 870
' 0.033
900
(a)
' 336.5 N&m
' 92.9 %
S3.13
IM '
460 / 3
j 30
' ! j 8.853 A
Ns
'
IR
'
120
x 60 ' 1200 rpm
6
460 / 3
0.8 %
s '
1200 ! 1164
' 0.030
1200
' 24.18 ! j 3.135 A ' 24.38 e !j 7.39E A
0.3
0.03
% j 1.4
IS
' I R % I M ' 24.18 ! j ( 3.135 % 8.853 ) ' 26.99 e !j 26.367E 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
' 3 IR2
RR
s
so
Pmech ' ( 1 ! s ) ' 3 x 24.382 x
0.3
( 1 ! 0.030 ) ' 17
0.030
3 phases
Ignoring friction, Pshaft = Pmech
(c) Effcy. '
(d) T
'
(e) Tmax
'
Pshaft
Pin
Pshaft
Tm
'
17,296
' 89.8%
19,269
'
17,296
' 141.9Nm
2B
x 1164
60
1
3
6
x
x
x
2
2 B x 60
2
( 460 / 3 )2
2
2
0.8 % 1.4 % 0.8
' 349 Nm
At standstill, s = 1 and
IR '
460 / 3
' 92.157 ! j 117.3 A
( 0.8 % 0.3 ) % j 1.4
I s ' I R % I M % 92.157 ! j 117.3 ! j 8.853 ' 156.2 e !j 53.85E A
(g) i.e. Locked&rotor current ' 156.2 A
(f) L.R.torque '
s=1
3
6
( 460 / 3 )2
x
x 0.3
2 B x 60
2
( 0.8 % 0.3 )2 % 1.42
' 159.3 Nm
S3.14
Ns '
120 x 60
' 1800 rpm
4
If fr = 15 Hz then s '
fr ' s f
15
' 0.25
60
But s can be + or ! depending on whether the rotor is rotating slower than, or faster than, synchronous
speed respectively.
ˆ
N ' Ns ( 1 ± 0.25 ) ' 1350 rpm
or
2250 rpm
The frequency and amplitude of the voltages generated at the slip rings would be the same in both
cases; but the phase sequence would be reversed, because at 1350 rpm the rotating field is overtaking
the rotor, while at 2250 rpm the rotor is overtaking the field.
SPEED's Electric Motors
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
Full solutions are given at the end
4.1
A limited-rotation actuator has a rotor winding and a stator winding. The geometry is such that
the self-inductances are constant but the mutual inductance varies linearly from zero to 109 mH
in a rotation of 75E. Calculate the torque when the windings are connected in series carrying a
current of 10A.
4.2
What is the stroke angle of a 3-phase switched reluctance motor having 12 stator poles and 8
rotor 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 rotor
poles? 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 = 30E
and a rotor pole arc $r = 32E. The unsaturated aligned inductance is Lau = 10.7 mH and the
unaligned inductance is Lu = 1.5 mH, and saturation can be neglected.
(a)
Calculate the instantaneous electromagnetic torque when the rotor is 15E before the
aligned 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 this
flux-linkage can be maintained constant while the rotor rotates from the unaligned
position to the aligned position at low speed, determine the energy conversion per stroke
and the average torque.
4.5
Show that for an unsaturated switched reluctance motor operating with a fixed conduction
2
2
angle and flat-topped current waveform, the average torque is proportional to V /T where V
is the supply voltage and T is the angular velocity. Hence show that to maintain constant torque
per ampere it is necessary to maintain the 'volts per Hertz' constant. Deduce that with fixed
supply voltage, a constant-power characteristic can be obtained by making the conduction angle
proportional to the speed.
4.6
Fig. 4.6 shows the cross-section of a switched reluctance
motor with the two coils of one phase on opposite stator
poles. The rotor is in such a position that the 'overlap
angle' between these stator poles and a pair of rotor poles
is 15E. The airgap is g = 0.5 mm, stator bore diameter D =
50 mm, and axial length Lstk = 50 mm. There are 98 turns
on each stator pole. The stator and rotor pole arcs are
both 30E. Neglect fringing and leakage, and assume that
the steel parts are infinitely permeable.
(a)
A current of 6 A flows through the two coils in
series. Sketch the flux paths on Fig. 4.6 for this
condition, showing six flux-lines.
(b)
Calculate the flux-density in the airgap between
the active poles.
Fig. 4.6
(c)
Estimate the torque.
(d)
Through what angle of rotation is the torque essentially constant, if the current is
constant and there is no fringing?
4.7
The switched reluctance motor in Fig. 4.7A has an airgap g = 0.2 mm, a stator outside diameter
of 79.2 mm, a pole arc of 13 mm, and a stack length Lstk = 50 mm. All other dimensions can be
scaled 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 rotor
is 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 aligned
position the current required for this flux-density is 3is. The unaligned inductance is 1/8
of the unsaturated aligned inductance. Estimate the maximum average electromagnetic
torque that this motor can produce at low speed.
A
B
D
C
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 the
normal “unaligned” position, but diagrams C, D and E all show faults in the winding or its
connections. In each case, draw a flux-plot with 5 ! 10 lines of B, and comment on the value of
the inductance of phase 1 (compared with to case A) when the rotor is in each of the positions
shown. 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.
SPEED's Electric Motors
Solutions to Problems
4. Switched Reluctance Machines
S4.1
T ' i1 i2
d L12
d2
i1 ' i2 ' i ;
d L12
d2
'
T ' i2
d L12
d2
109 x 10& 3
' 0.0833 H / rad
B
75 x
180
T ' 102 x 0.0833 ' 8.33 Nm
S4.2
12/8 motor
g '
360
'
m Nr
360
' 15E
3 x 8
6000 rev / min ' 100 rev / sec
' 100 x 8 commutations / sec in each phase
' 800 Hz
S4.3
10/4 motor
g '
360
'
5 x 4
360
' 18E
20
6000 rev/min ' 100 rev / sec ' 100 x 4 commutations / sec in each phase
' 400 Hz
(a)
T '
1 2 dL
i
'
2
d2
1
( 10.7 ! 1.5 ) x 10
( 7 )2 x
2
B/6
' 0.430 N m
(b)
W '
'
1 2
i )L
2
1
x 7.02 x ( 10.7 ! 1.5 ) x 10!3 J
2
' 0.2254 J
(c)
Ra ' L au ia ' 10.7 x 7.0 ' 74.9 mV&s
iu '
74.9
' 49.9 A
1.5
W '
1
(49.9 ! 7.0) x 0.0749 ' 1.608 J
2
Te(avg) '
12 strokes/rev x W
' 3.07 N m
2B
!3
S4.4
S4.5
Neglecting losses, a flat-topped current waveform can be obtained if
V ' iT
dL
d2
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
W '
1
i@)R
2
'
1 V 2D
i
2
T
'
1
2
'
V 2D
V
@
dL
T
T
d2
V2
T2
Since the electromagnetic torque is proportional to W, it is proportional to V2/T2.
S4.6
(a)
(b)
Bg ' µ0 Hg ' µ0
2Ni
6 x 98
' 1.48 T
' 4 B x 10!7 x
2g
0.5 x 10!3
(c) Assume that the inductance changes from Lmax at maximum overlap to zero when there is no
overlap. Then
T '
Lmax '
1 2 dL
i
2
d2
µ0 (2 N)2 Apole
2g
2 x 4 B x 10!7 x 982 x 30 x
'
B
x 25 x 50 x 10!6
180
0.5 x 10!3
' 0.0316 H
dL
'
d2
T '
(d) 30E.
0.0316
B/6
1
0.0316
x 6.02 x
' 1.086 N&m
2
B/6
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
L au '
(b)
2 x 322 x 4 B x 10!7 x 650 x 10!6
is '
0.2 x 10!3
B x g
'
µ0 N p
1.75 x 0.2 x 10!3
4 B x 10!7 x 32
' 8.36 mH
' 8.70 A
(c)
First draw the aligned and unaligned magnetization curves. Then calculate the enclosed area to the left
of 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
Te(avg) ' strokes/rev x
W
1.371
' 12 x
' 2.62 N m
2B
2B
SPEED's Electric Motors
Problems
5. DC Machines
Answers
5.1
Yes.
5.2
(d) 0@354 T; (e) 225@4 A-t; (f) 23@67 V-s; (h) 9@47 V; (i) 143@2 µH; (j) 12,800 A-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
Full solutions are given at the end
5.1
An electric motor contains coils and magnets and the flux is fixed in magnitude.
Can the flux-linkage of any coil vary?
5.2
5.3
Y — yes
N — no
A permanent-magnet DC motor has the cross-section shown in
Fig. 1 with an armature diameter D = 2r1 = 60 mm; magnet
length Lm = 8 mm; airgap length g = 0.8 mm, and stack length
Lstk = 35 mm. The magnet arc is $m = 120E. A single full-pitch
rotor 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 rotor
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 Bg(2) around the airgap from 0 to 360E.
Fig. 1
(c)
Draw an equivalent magnetic circuit and use it to
calculate the flux crossing the airgap under each magnet pole. Assume that the leakage
flux can be represented by a permeance equal to 0.15 times the magnet internal
permance, and assume that the permeability of the steel in the rotor and stator is
infinite.
(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 the
magnet.
(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 the
rotor 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 work
that is done by the rotor in rotating 180E from the position shown.
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.
5.4
5.5
A permanent-magnet DC motor operates from a supply of 240 V. Its armature resistance is
1.2 ohm and the torque constant is k T = 1.31 Nm/A. Friction torque is constant at 1 Nm, and the
brush 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.
Show that if friction and core losses are neglected, the maximum efficiency of a permanentmagnet DC commutator motor is equal to the ratio of the open-circuit voltage E to the supply
voltage Vs.
A DC commutator motor is to be designed to deliver 300 W at 2,500 rev/min when supplied at
12V. The efficiency must not be less than 2/3 (i.e., 66.67 %) when measured on a dynamometer
that eliminates friction torque. The brush material is such that the voltage drop across each
brush will be 1@0 V, regardless of the current. Calculate
5.6
(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.
A 10-hp 230-V DC shunt-wound motor has the equivalent circuit shown in Fig. 2.
Fig. 2
The armature resistance is Ra = 0.3 ohm and the field resistance is Rf = 170 ohm. At no-load and
rated voltage, the speed is 1200 rev/min and the armature current is Ia = 2.7 A. At full load and
rated voltage, the line current is IL = Ia + If = 38.4 A, where If is the field current. Calculate the
speed 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.
SPEED's Electric Motors
Solutions to Problems
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 assumed
that 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@9
rad/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 = N B Ag = N µ0 H Ag = N µ0 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 ' m e i dt '
B
T
0
MR d2
B
i
' i[)R] 0 = 5@0 x (2 x 23@67 x 10!3) = 0@237 J
M2 T
S5.3
The windings are shown on the following three pages. Note the last one (which is not part of the
question, but may be of interest).
5.3(a) Progressive lap winding in 4-pole, 15-slot armature with 1 coilside/layer; span = 3
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
(a)
Vb = 1.4 V per brush kE = kT = 1@31 V-s/rad or Nm/A
T0 = (Vs ! 2 Vb)/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 ! 2 Vb)/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.