Electric motors are so much a part of everyday life that we seldom give
them a second thought. When we switch on an electric drill, for example,
we confidently expect it to run rapidly up to the correct speed, and we
don't question how it knows what speed to run at, or how it's that once
enough energy has been drawn from the supply to bring it up to speed, the
power drawn falls to a very low level. When we put the drill to work it
draws more power, and when we finish the power drawn from the mains reduces
automatically, without intervention on our part.

The humble motor, consisting of nothing more than an arrangement of copper
coils and steel laminations, is clearly rather a clever energy converter,
which warrants serious consideration. By gaining a basic understanding
of how the motor works, we will be able to appreciate its potential and its limitations, and (in later sections) see how its already remarkable
performance can be further enhanced by the addition of external electronic
controls.

This section deals with the basic mechanisms of motor operation, so readers
who are already familiar with such matters as magnetic flux, magnetic and
electric circuits, torque, and motional e.m.f can probably afford to skim over
much of it. In the course of the discussion, however, several very important
general principles and guidelines emerge. These apply to all types of motors
and are summarized in Section 1.8. Experience shows that anyone who has a good
grasp of these basic principles will be well equipped to weigh the pros and
cons of the different types of motor, so all readers are urged to absorb them
before tackling other parts of the guide.

Wiki Series: Laboratory Manual for Electronics

Three-Phase Circuits

Objectives:

Connect a wye connected, three-phase load.

Calculate and measure voltage and current values for a wye connected
load.

Connect a delta connected load.

Calculate and measure voltage and current values for a delta connected
load.

Before beginning the study of three-phase transformers, it's appropriate
to discuss three phase power connections and basic circuit calculations.
This unit may be review for some students and new ground for others. Whichever
is the case, a working knowledge of three phase circuits is essential before
beginning the study of three-phase transformers.

Most of the power generated in the world today is three-phase. Three-phase
power was first conceived by a man named Nikola Tesla. There are several
reasons why three-phase power is superior to single-phase power.

1. The kVA rating of three-phase transformers is about 150% greater than
for a single phase transformer with a similar core size.

2. The power delivered by a single-phase system pulsates. The
power falls to zero three times during each cycle. The power delivered
by a three-phase circuit pulsates also, but the power never falls to zero
(ill. 2). In a three-phase system, the power delivered to the load is the
same at any instant.

ill. 1 Single-phase power falls to zero three times each cycle.

ill. 2 Three-phase power never falls to zero.

3. In a balanced three-phase system, the conductors need be only about
75% the size of conductors for a single-phase two-wire system of the same
kVA rating. This helps off set the cost of supplying the third conductor
required by three-phase systems.

A single-phase alternating voltage can be produced by rotating a magnetic
field through the conductors of a stationary coil as shown in ill. 3. Since
alternate polarities of the magnetic field cut through the conductors of
the stationary coil, the induced voltage will change polarity at the same
speed as the rotation of the magnetic field. The alternator shown in ill.
3 is single-phase because it produces only one AC voltage.

If three separate coils are spaced 120° apart as shown in ill. 4, three
voltages 120° out of phase with each other will be produced when the magnetic
field cuts through the coils.

This is the manner in which a three-phase voltage is produced. There are
two basic three phase connections, the wye or star, and the delta.

ill. 3 Producing a single-phase voltage.

Wye Connection

The wye or star connection is made by connecting one end of each of the
three-phase windings together as shown is ill. 5. The voltage measured
across a single winding or phase is known as the phase voltage as shown
in ill. 6. The voltage measured between the lines is known as the line-to-line
voltage or simply as the line voltage.

ill. 4 The voltages of a three-phase system are 120° out of phase with
each other.

ill. 5 A wye connection is formed by joining one end of each winding together.

ill. 6 Line and phase voltages are different in a wye connection.

In ill. 7 ammeters have been placed in the phase winding of a wye connected
load and in the line supplying power to the load. Voltmeters have been
connected across the input to the load and across the phase. A line voltage
of 208 volts has been applied to the load. Notice that the voltmeter connected
across the lines indicates a value of 208 volts, but the voltmeter connected
across the phase indicates a value of 120 volts.

In a wye connected system, the line voltage is higher than the phase voltage
by a factor of __ 3 (1.732). Two formulas used to compute the voltage in
a wye connected system are:

E_PHASE = E_LINE/ sqr_rt (3)

E_LINE = E_PHASE x sqr_rt (3)

ill. 7 Line current and phase current are the same in a wye connection.

Notice in ill. 7 that there is 10 amps of current flow in both the phase and the line.

In a wye connected system, phase current and line current are the same.

Helpful Hint: In a wye connected system, the line voltage is
higher than the phase voltage by a factor of sqr_rt 3 (1.732).

Helpful Hint: In a wye connected system, phase current and line
current are the same.

Voltage

Relationships in a Wye Connection ill. 8 Single-phase transformer
with grounded center tap.

Many students of electricity have difficulty at first understanding why
the line voltage of the wye connection used in this illustration is 208
volts instead of 240 volts. Since line voltage is measured across two phases
that have a voltage of 120 volts each, it would appear that the sum of
the two voltages should be 240 volts. One cause of this misconception is
that many students are familiar with the 240/120 volt connection supplied
to most homes. If voltage is measured across the two incoming lines, a
voltage of 240 volts will be seen. If voltage is measured from either of
the two lines to the neutral, a voltage of 120 volts will be seen. The
reason for this is that this connection is derived from the center tap
of an isolation transformer, as shown in ill. 8. If the center tap is used
as a common point, the two line voltages on either side of it will be in
phase with each other. Since the two voltages are in phase, they add similar
to a boost connected transformer, as shown in ill. 9. The vector sum of
these two voltages would be 240 volts.

Three-phase voltages are 120° apart, not in phase. If the three voltages
are drawn 120° apart, it will be seen that the vector sum of these voltages
is 208 volts, as shown in ill. 10.

Another illustration of vector addition is shown in ill. 11. In this illustration
two-phase voltage vectors are added and the resultant is drawn from the
starting point of one vector to the end point of the other. The parallelogram
method of vector addition for the voltages in a wye connected three-phase
system is shown in ill. 12.

ill. 9 The two voltages are in phase with each other.

ill. 10 Vector sum of the voltages in a three-phase wye connection.

ill. 11 Adding voltage vectors of two-phase voltage values.

ill. 12 The parallelogram method of adding three-phase vectors.

Delta Connection

In ill. 13 three separate inductive loads have been connected to form
a delta connection. This connection receives its name from the fact that
a schematic diagram of this connection resembles the Greek letter delta
(?). In ill. 14, voltmeters have been connected across the lines and across
the phase. Ammeters have been connected in the line and in the phase. In
the delta connection, line voltage and phase voltage are the same. Notice
that both voltmeters indicated a value of 480 volts.

LINE, PHASE

ill. 13 Three-phase delta connection.

Helpful Hint: In the delta connection, line voltage and phase voltage
are the same.

ill. 14 Voltage and current relationships in a delta connection.

ill. 15 Division of currents in a delta connection.

Notice that the line current and phase current are different, however.
The line current of a delta connection is higher than the phase current
by a factor of __ 3 (1.732). In the example shown, it's assumed that each
of the phase windings has a current flow of 10 amperes. The current in
each of the lines, however, is 17.32 amperes. The reason for this difference
in current is that current flows through different windings at different
times in a three-phase circuit. During some periods of time, current will
flow between two lines only.

At other times, current will flow from two lines to the third, seen in
ill. 15. The delta connection is similar to a parallel connection because
there is always more than one path for current flow. Since these currents
are 120° out of phase with each other, vector addition must be used when
finding the sum of the currents (ill. 16). Formulas for determining the
current in a delta connection are:

Three-Phase Power

Students sometimes become confused when computing values of power in
three-phase circuits. One reason for this confusion is because there are
actually two formulas that can be used. If LINE values of voltage and current
are known, the apparent power of the circuit can be computed using the
formula:

If the PHASE values of voltage and current are known, the apparent power
can be computed using the formula:

VA = 3 x E_PHASE × I_PHASE

Notice that in the first formula, the line values of voltage and current
are multiplied by the square root of 3. In the second formula, the phase
values of voltage and current are multiplied by 3. The first formula is
the most used because it's generally more convenient to obtain line values
of voltage and current since they can be measured with a voltmeter and clamp-on ammeter.

Watts and VARs

Watts and VARs can be computed in a similar manner. Watts can be computed
by multiplying the apparent power by the power factor:

3 PHASE × PHASE

Note: When computing the power of a pure resistive load, the voltage and current are in phase with each other and the power factor is 1.

VARs can be computed in a similar manner, except that voltage and current
values of a pure reactive load are used. E.g., a pure capacitive
load is shown in ill. 17. In this example, it's assumed that the line
voltage is 480 volts and the line current is 30 amperes.

Capacitive VARs can be computed using the formula:

VAR c = 29,097.6

ill. 16 Vector addition is used to compute the sum of the currents in
a delta connection.

ill. 17 Pure capacitive three-phase load.

Three-Phase Circuit Calculations

In the following examples, values of
line and phase voltage, line and phase current, and power will be computed
for different types of three-phase connections.

Example #1. A wye connected, three-phase alternator supplies power to
a delta connected resistive load, as shown in ill. 18. The alternator has
a line voltage of 480 volts.

Each resistor of the delta load has 8 ohm of resistance. Find the following
values:

E_L(LOAD) - Line voltage of the load

E_P(LOAD) - Phase voltage of the load

ill. 18 Computing three-phase values: Example Circuit #1.

I_P(LOAD) - Phase current of the load

I_L(LOAD) - Line current to the load

I_L(ALT) - Line current delivered by the alternator

E_P(ALT) - Phase voltage of the alternator P - True power

Solution: The
load is connected directly to the alternator. Therefore, the line voltage
sup plied by the alternator is the line voltage of the load.

The three resistors of the load are connected in a delta connection. In
a delta connection, the phase voltage is the same as the line voltage.

Each of the three resistors in the load comprises one phase of the load.
Now that the phase voltage is known (480 volts), the amount of phase current
can be computed using Ohm's law.

L (LOAD) 480 volts =

P LOAD () I LOAD ()

P LOAD () 480 volts =

In this example the three load resistors are connected as a delta with
60 amperes of current flow in each phase. The line current supplying a
delta connection must be 1.732 times greater than the phase current.

The alternator must supply the line current to the load or loads to which
it's connected. In this example, there is only one load connected to the
alternator. Therefore, the line current of the load will be the same as
the line current of the alternator.

The phase windings of the alternator are connected in a wye connection.
In a wye connection, the phase current and line current are equal. The
phase current of the alternator will, therefore, be the same as the alternator
line current.

L LOAD () P LOAD () 1.732 × =

L LOAD () 60 1.732 × =

L LOAD () 103.92 amps =

LALT () 103.92 amps =

P_ALT () 103.92 amps =

The phase voltage of a wye connection is less than the line voltage by
a factor of the square root of 3 (__ 3 ). The phase voltage of the alternator
will be:

In this circuit the load is pure resistive. The voltage and current are
in phase with each other, which produces a unity power factor of 1. The
true power in this circuit will be computed using the formula: Example
#2. In the next example, a delta connected alternator is connected to a
wye connected resistive load, as shown in ill. 19. The alternator produces
a line voltage of 240 volts and the resistors have a value of 6 ohm each.
The following values will be found:

E_L(LOAD) - Line voltage of the load EP(LOAD) - Phase voltage of the load
IP(LOAD) - Phase current of the load I_L(LOAD) - Line current to the load
IL(ALT) - Line current delivered by the alternator EP(ALT) - Phase voltage
of the alternator P - True power As in the first example, the load is connected
directly to the output of the alternator. The line voltage of the load
must, therefore, be the same as the line voltage of the alternator.

P_ALT () 277.13 volts =

1.732 LALT ()

× LALT ()

× × =

1.732 480 × 103.92 × 1 × =

86,394.93 watts =

L LOAD () 240 volts =

ill. 19

Example #2.

The phase voltage of a wye connection is less than the line voltage by
a factor of 1.732.

Each of the three 6 ohm resistors comprises one phase of the wye connected
load. Since the phase voltage is 138.57 volts, this voltage is applied
to each of the three resistors. The amount of phase current can now be
determined using Ohm's law.

The amount of line current needed to supply a wye connected load is the
same as the phase current of the load.

In this example there is only one load connected to the alternator. The
line current supplied to the load is the same as the line current of the
alternator.

The phase windings of the alternator are connected in delta. In a delta
connection the phase current is less than the line current by a factor
of 1.732.

L LOAD () 23.1 amps =

LALT () 23.1 amps =

P LOAD () 23.1 amps =

The phase voltage of a delta is the same as the line voltage.

Since the load in this example is pure resistive, the power factor has
a value of unity or 1.

Power will be computed by using the line values of voltage and current.

PALT () 240 volts =

1.732 L × L × × =

1.732 240 × 23.1 × 1 × =

9,602.21 watts =

LABORATORY EXERCISE

Name _

Date _ Materials Required

2 AC voltmeters

AC ammeter, in-line or clamp-on. (If a clamp-on type is used, it's recommended
to use a 10:1 scale divider.) 6 100-watt lamps

In this experiment six 100 watt lamps will be connected to form different
three-phase loads.

Two lamps will be connected in series to form three separate loads. These
loads will be connected to form wye or delta connections.

1. Connect the two 100 watt lamps in series to form three separated load
banks. Connect the load banks in wye by connecting one end of each bank
together to form a center point, as shown in ill. 20. It is assumed that
this load is to be connected to a 208 VAC three-phase line. Connect an
AC ammeter in series with the line supplying power to the load.

2. Turn on the power and measure the line voltage supplied to the load.

E(LINE) volts

3. Calculate the value of phase voltage for a wye connected load.

E(PHASE) __ volts

4. Measure the phase voltage and compare this value with the computed
value.

E(PHASE) volts

5. Measure the line current.

I(LINE) amp(s)

6. Turn off the power supply.

7. In a wye connected system, the line current and phase current are the
same. Reconnect the circuit as shown in ill. 21.

8. Turn on the power and measure the phase current.

I(PHASE) amp(s)

9. Turn off the power supply.

10. Reconnect the three banks of lamps to form a delta connected load,
as shown in ill. 22.

11. Turn on the power and measure the line voltage supplied to the load.

E(LINE) volts

ill. 20 Measuring the line current in a wye connected load.

ill. 21 Measuring the phase current in a wye connected load.

12. Measure the phase value of voltage.

E(PHASE) volts 13. Are the line and phase voltage values the same or different?

14. Measure the line current.

I(LINE) amp(s) 15. Turn off the power supply.

16. In a delta connected system, the phase current will be less than the
line current by a factor of 1.732. Calculate the phase current value for
this connection.

ill. 22 Measuring the voltage and line current values of a delta connected
load.

I(PHASE) amp(s) 17. Reconnect the circuit as shown in ill. 23.

18. Turn on the power supply and measure the phase current. Compare this
value with the computed value.

I(PHASE) amp(s) 19. Turn off the power supply.

20. Disconnect the circuit and return the components to their proper place.

QUIZ

1. How many degrees out of phase with each other are the voltages of a
three-phase system?

2. What are the two main types of three-phase connections?

3. A wye connected load has a voltage of 480 volts applied to it. What
is the voltage dropped across each phase?

4. A wye connected load has a phase current of 25 amps. How much current
is flowing through the lines supplying the load?

5. A delta connection has a voltage of 560 volts connected to it. How
much voltage is dropped across each phase?

6. A delta connection has 30 amps of current flowing through each phase
winding. How much current is flowing through each of the lines supplying
power to the load?

7. A three-phase load has a phase voltage of 240 volts and a phase current
of 18 amperes.

What is the apparent power of this load?

ill. 23 Measuring the voltage and phase current values of a delta connected
load.

8. If the load in question 7 is connected in a wye, what would be the
line voltage and line current supplying the load?

9. An alternator with a line voltage of 2,400 volts supplies a delta connected
load. The line current supplied to the load is 40 amperes. Assuming the
load is a balanced three-phase load, what is the impedance of each phase?

10. What is the apparent power of the circuit in question 9?

Nearly all motors exploit the force which is exerted on a current-carrying
conductor placed in a magnetic field . The force can be demonstrated by
placing a bar magnet near a wire carrying current, but anyone trying
the experiment will probably be disappointed to discover how feeble the
force is, and will doubtless be left wondering how such an unpromising
effect can be used to make effective motors.

We will see that in order to make the most of the mechanism, we need to
arrange a very strong magnetic field , and make it interact with many conductors,
each carrying as much current as possible. We will also see later that
although the magnetic field (or working of the motor, it acts only as a
catalyst, and all of the mechanical output power comes from the electrical
supply to the conductors on which the force is developed. It will emerge
later that in some motors the parts of the machine responsible for the
excitation and for the energy converting functions are distinct and self-evident.
In the DC motor, for example, the excitation is provided either by permanent
magnets or by field coils wrapped around clearly defined projecting field poles on the stationary part, while the conductors on which force is developed
are on the rotor and supplied with current via sliding brushes. In many
motors, however, there is no such clear-cut physical distinction between
the ‘excitation’ single stationary winding serves both purposes. Nevertheless,
we will find that identifying and separating the excitation and energy-converting
functions is always helpful in understanding how motors of all types operate.

ill. 1 Mechanical force produced on a current-carrying wire in
a magnetic field.

Returning to the matter of force on a single conductor, we will first
look at what determines the magnitude and direction of the force, before
turning to ways in which the mechanism is exploited to produce rotation.
The concept of the magnetic circuit will have to be explored, since this
is central to understanding why motors have the shapes they do. A brief
introduction to magnetic field , magnetic flux, and flux density is included
before that for those who are not familiar with the ideas involved.

Magnetic field and magnetic flux

When a current-carrying conductor is placed in a magnetic field , it experiences
a force. Experiment shows that the magnitude of the force depends directly
on the current in the wire, and the strength of the magnetic field , and that the force is greatest when the magnetic field is perpendicular to the
conductor.

In the set-up shown in ill. 1, the source of the magnetic field is
a bar magnet, which produces a magnetic field.

The notion of a ‘magnetic field’ idea that helps us to come to grips with
the mysterious phenomenon of magnetism: it not only provides us with a
convenient pictorial way of picturing the directional effects, but it also
allows us to quantify the ‘strength ’of the magnetism and hence permits
us to predict the various effects produced by it.

The dotted lines are referred to as magnetic flux lines,
or simply flux lines. They indicate the direction along which iron filings
(or small steel pins) would align themselves when placed in the field of
the bar magnet. Steel pins have no initial magnetic field of their own,
so there is no reason why one end or the other of the pins should point
to a particular pole of the bar magnet.

However, when we put a compass needle (which is itself a permanent magnet)in
the field we find that it aligns itself. In the
upper half of the figure, the S end of the diamond-shaped compass settles
closest to the N pole of the magnet, while in the lower half of the figure,
the N end of the compass seeks the S of the magnet. This immediately suggests
that there is a direction associated with the lines of flux, as shown by
the arrows on the flux lines, which conventionally are taken as positively
directed from the N to the S pole of the bar magnet.

The sketch in ill. 2 might suggest that there is a ‘source ’near the
top of the bar magnet, from which flux lines emanate before making their
way to a corresponding ‘sink ’at the bottom. However, if we were to look
at the flux lines inside the magnet, we would find that they were continuous,
with no ‘start ’or ‘finish ’.(In ill. 2 the internal flux lines have
been omitted for the sake of clarity, but a very similar field pattern
is produced by a circular coil of wire carrying a DC. See ill. 6 where
the continuity of the flux lines is clear.).Magnetic flux lines always
form closed paths, as we will see when we look at the ‘magnetic circuit
’,and draw a parallel with the electric circuit, in which the current is
also a continuous quantity.(There must be a ‘cause ’of the magnetic flux,
of course, and in a permanent magnet this is usually pictured in terms
of atomic-level circulating currents within the magnet material.

Fortunately, discussion at this physical level isn't necessary for our
purpose.)

Magnetic flux density

Along with showing direction, the flux plots also convey information about
the intensity of the magnetic field. To achieve this, we introduce the
idea that between every pair of flux lines (and for a given depth into
the paper) there is a same ‘quantity ’of magnetic flux. Some people have
no difficulty with such a concept, while others find that the notion of
quantifying something so abstract represents a serious intellectual challenge.
But whether the approach seems obvious or not, there is no denying of the
practical utility of quantifying the mysterious stuff we call magnetic
flux, and it leads us next to the very important idea of magnetic flux
density (B ).

When the flux lines are close together, the ‘tube ’of flux is squashed
into a smaller space, whereas when the lines are further apart the same
tube of flux has more breathing space. The flux density (B) is
simply the flux in the ‘tube ’(phi )divided by the cross sectional area
(A )of the tube, i.e. B = Phi / A [1.1]

The flux density is a vector quantity, and is therefore often written
in bold type: its magnitude is given by equation, and its direction
is that of the prevailing flux lines at each point. Near the top of the
magnet in ill. 2, for example, the flux density will be large (because
the flux is squashed into a small area),and pointing upwards, whereas on
the equator and far out from the body of the magnet the flux density
will be small and directed downwards.

It will be seen later that in order to create high flux densities in motors,
the flux spends most of its life inside well-de fined ‘magnetic circuits
’made of iron or steel, within which the flux lines spread out uniformly
to take full advantage of the available area. In the case shown in ill. 3, for example, the cross-sectional area at bb ’is twice that at aa ’,but
the flux is constant so the flux density at bb ’is half that at aa ’.

ill. 3 Magnetic flux lines inside part of an iron magnetic circuit.

It remains to specify units for quantity of flux, and flux density. In
the SI system, the unit of magnetic flux is the weber (Wb).If one weber
of flux is distributed uniformly across an area of 1m^2 perpendicular to
the flux, the flux density is clearly one weber per square meter (Wb=m2).
This was the unit of magnetic flux density until about 40 years ago, when
it was decided that one weber per square meter would henceforth be known
as one tesla (T), in honor of Nikola Tesla who is generally credited with
inventing the induction motor. The widespread use of B (measured in tesla)
in the design stage of all types of electromagnetic apparatus means that
we are constantly reminded of the importance of tesla; but at the same
time one has to acknowledge that the outdated unit did have the advantage
of conveying directly what flux density is, i.e. flux divided by area.

In the motor world we are unlikely to encounter more than a few milliwebers
of flux, and a small bar magnet would probably only produce a few microwebers.
On the other hand, values of flux density are typically around 1T in most
motors, which is a reflection of the fact that although the quantity of
flux is small, it's also spread over a small area.

Force on a conductor

We now return to the production of force on a current-carrying wire placed
in a magnetic field , as revealed by the setup.

The direction of the force is shown in ill. 1: it's at right angles
to both the current and the magnetic flux density. With the flux density
horizontal and to the right, and the current flowing out of the paper,
the force is vertically upward. If either the field or the current is reversed,
the force acts downwards, and if both are reversed, the force will remain
upward.

We find by experiment that if we double either the current or the flux
density, we double the force, while doubling both causes the force to increase
by a factor of four. But how about quantifying the force? We need to express
the force in terms of the product of the current and the magnetic flux
density, and this turns out to be very straightforward when we work in
SI units.

The force on a wire of length l, carrying a current I and exposed to a uniform magnetic flux density B throughout its length
is given by the simple expression

F = BIl [1.2]

… where F is in newtons when B is in tesla, I in amperes, and l in meters.

This is a simple formula, and it may come as a surprise to some readers
that there are no constants of proportionality involved in equation 2.
The simplicity isn't a coincidence, but stems from the fact that the unit
of current (the ampere) is actually defined in terms of force.

Strictly, equation 2 only applies when the current is perpendicular
to the field . If this condition isn't met, the force on the conductor
will be less; and in the extreme case where the current was in the same direction
as the field , the force would fall to zero. However, every sensible motor
designer knows that to get the best out of the magnetic field it has to
be perpendicular to the conductors, and so it's safe to assume in the
subsequent discussion that B and I are always perpendicular. In the remainder
of this book, it will be assumed that the flux density and current are
mutually perpendicular, and this is why, although B is a vector quantity
(and would usually be denoted by bold type), we can drop the bold notation
because the direction is implicit and we are only interested in the magnitude.

The reason for the very low force detected in the experiment with the
bar magnet is revealed by equation 2. To obtain a high force, we must
have a high flux density, and a lot of current. The flux density at the
ends of a bar magnet is low, perhaps 0.1 tesla, so a wire carrying 1 amp
will experience a force of only 0.1 N/m (approximately 100 gm wt). Since
the flux density will be confined to perhaps 1 cm across the end face of
the magnet, the total force on the wire will be only 1 gm. This would be
barely detectable, and is too low to be of any use in a decent motor. So
how is more force obtained?

The first step is to obtain the highest possible flux density. This is
achieved by designing a Secondly, as many conductors as possible must be
packed in the space where the magnetic field exists, and each conductor
must carry as much current as it can without heating up to a dangerous
temperature. In this way, impressive forces can be obtained from modestly
sized devices, as anyone who has tried to stop an electric drill by grasping
the chuck will testify.

MAGNETIC CIRCUITS

So far we have assumed that the source of the magnetic field is a permanent
magnet. This is a convenient starting point as all of us are familiar with
magnets, even if only of the fridge-door variety. But in the majority of
motors, the working magnetic field is produced by coils of wire carrying
current, so it's appropriate that we spend some time looking at how we
arrange the coils and their associated iron ‘magnetic circuit’ Welds which
then interact with so as other current-carrying conductors to produce force, and hence rotation.

ill. 4 Magnetic flux lines produced by a straight,
current-carrying wire.

First, we look at the simplest possible case of the magnetic field surrounding
an isolated long straight wire carrying a steady current.
(In the figure, the + sign indicates that current is flowing into the paper,
while a dot is used to signify current out of the paper: these symbols
can perhaps be remembered by picturing an arrow or dart, with the cross
being the rear view of the fletch, and the dot being the approaching
point.) The flux lines form circles concentric with the wire, the field
strength being greatest close to the wire. As might be expected, the field
strength at any point is directly proportional to the current. The convention
for determining the direction of the field is that the positive direction
is taken to be the direction that a right-handed corkscrew must be rotated
to move in the direction of the current.

ill. 4 is somewhat artificial as current can only flow in a complete
circuit, so there must always be a return path. If we imagine a parallel
‘go’ field can be obtained by and superimposing the field produced by the
positive current in the go side with the field produced by the negative
current in the return side, as shown in ill. 5.

ill. 5 Magnetic flux lines produced by current in a parallel go and return circuit.

ill. 5 Magnetic flux lines produced by current in a parallel go and return circuit

ill. 6 Multi-turn cylindrical coil and pattern of magnetic flux
produced by current in the coil. (For the sake of clarity, only
the outline of the coil is shown on the right.)

We note how the field is increased in the region between the conductors, and reduced
in the regions outside. Although ill. 5 strictly only applies to an infinitely
long pair of straight conductors, it will probably not come as a surprise
to learn that the field produced by a single turn of wire of rectangular,
square or round form is very much the same as that shown in ill. 5. This
enables us to build up a picture of the field that would be produced in
air, by the sort of coils used in motors, which typically have many turns.

The coil itself is shown on the left in ill. 6 while the flux pattern
produced is shown on the right. Each turn in the coil produces a field
pattern, and when all the individual field components are superimposed
we see that the field inside the coil is substantially increased and that
the closed flux paths closely resemble those of the bar magnet we looked
at earlier. The air surrounding the sources of the field offers a homogeneous
path for the flux, so once the tubes of flux escape from the concentrating
influence of the source, they are free to spread out into the whole of
the surrounding space. Recalling that between each pair of flux lines there
is an equal amount of flux, e see that because the flux lines spread out
as they leave the con fines of the coil, the flux density is much lower
outside than inside: for example, if the distance ‘b ’is say four times
‘a ’the flux density B b is a quarter of B a.

Although the flux density inside the coil is higher than outside, we would
find that the flux densities which we could achieve are still too low to
be of use in a motor. What is needed firstly is a way of increasing the
flux density, and secondly a means for concentrating the flux and preventing
it from spreading out into the surrounding space.

Magnetomotive force (MMF)

One obvious way to increase the flux density is to increase the current
in the coil, or to add more turns. We find that if we double the current,
or the number of turns, we double the total flux, thereby doubling the
flux density everywhere.

We quantify the ability of the coil to produce flux in terms of its magnetomotive
force (MMF).The MMF of the coil is simply the product of the number of
turns (N )and the current (I ),and is thus expressed in ampere-turns. A
given MMF can be obtained with a large number of turns of thin wire carrying
a low current, or a few turns of thick wire carrying a high current: as
long as the product NI is constant, the MMF is the same.

Electric circuit analogy

We have seen that the magnetic flux which is set up is proportional to
the MMF driving it. This points to a parallel with the electric circuit,
where the current (amps)that flows is proportional to the EMF (volts) driving
it.

In the electric circuit, current and EMF are related by Ohm's Law, which
is

Current = EMF / Resistance i.e., :I = V/R

For a given source EMF (volts), the current depends on the resistance
of the circuit, so to obtain more current, the resistance of the circuit
has to be reduced.

We can make use of an equivalent 'magnetic Ohm's law 'by introducing the
idea of reluctance (R).The reluctance gives a measure of how difficult
it's for the magnetic flux to complete its circuit, in the same way that
resistance indicates how much opposition the current encounters in the
electric circuit. The magnetic Ohm 's law is then

Flux = MMF / Reluctance i.e., Φ = NI/R

We see from equation 1.4 that to increase the flux for a given MMF, we
need to reduce the reluctance of the magnetic circuit. In the case of the
example (ill. 6), this means we must replace as much as possible of
the air path (which is a 'poor 'magnetic material, and therefore constitutes
a high reluctance) with a 'good 'magnetic material, thereby reducing the
reluctance and resulting in a higher flux for a given MMF.

The material which we choose is good quality magnetic steel, which for
historical reasons is usually referred to as 'iron '.This brings several
very dramatic and desirable benefits, as shown.

ill. 7 Flux lines inside low-reluctance magnetic circuit with air-gap
Firstly, the reluctance of the iron paths is very much less than the air
paths which they have replaced, so the total flux produced for a given
MMF is very much greater. (Strictly speaking therefore, if the MMFs and cross-sections of the coils in ills 1.6 and 1.7 are the same, many more
flux lines should be shown in ill. 7 than in ill. 6,but for the
sake of clarity similar numbers are indicated.) Secondly, almost all the
flux is confined within the iron, rather than spreading out into the surrounding
air. We can therefore shape the iron parts of the magnetic circuit as shown
in ill. 7 in order to guide the flux to wherever it's needed. and finally, we see that inside the iron, the flux density remains uniform
over the whole cross-section, there being so little reluctance that there
is no noticeable tendency for the flux to crowd to one side or another.

Before moving on to the matter of the air-gap, we should note that a question
which is often asked is whether it's important for the coils to be wound
tightly onto the magnetic circuit, and whether, if there is a multi-layer
winding, the outer turns are as effective as the inner ones.

The answer, happily, is that the total MMF is determined solely by the
number of turns and the current, and therefore every complete turn makes
the same contribution to the total MMF, regardless of whether it happens
to be tightly or loosely wound. Of course it does make sense for the coils
to be wound as tightly as is practicable, since thisn't only minimizes
the resistance of the coil (and thereby reduces the heat loss) but also
makes it easier for the heat generated to be conducted away to the frame
of the machine.

The air-gap

In motors, we intend to use the high flux density to develop force on
current-carrying conductors. We have now seen how to create a high flux
density inside the iron parts of a magnetic circuit, but, of course, it's physically impossible to put current-carrying conductors inside the
iron.

We therefore arrange for an air-gap in the magnetic circuit. We will
see shortly that the conductors on which the force is to be produced will
be placed in this air-gap region.

If the air-gap is relatively small, as in motors, we find that the flux
jumps across the air-gap as shown in ill. 7, with very little tendency
to balloon out into the surrounding air. With most of the flux lines going
straight across the air-gap, the flux density in the gap region has the
same high value as it does inside the iron.

In the majority of magnetic circuits consisting of iron parts and one
or more air-gaps, the reluctance of the iron parts is very much less than
the reluctance of the gaps. At first sight this can seem surprising, since
the distance across the gap is so much less than the rest of the path through
the iron. The fact that the air-gap dominates the reluctance is simply
a reflection of how poor air is as a magnetic medium, compared to iron.

To put the comparison in perspective, if we calculate the reluctances
of two paths of equal length and cross-sectional area, one being in iron and the
other in air, the reluctance of the air path will typically be 1000 times
greater than the reluctance of the iron path.(The calculation of reluctance
will be discussed in later.) Returning to the analogy with the electric
circuit, the role of the iron parts of the magnetic circuit can be likened
to that of the copper wires in the electric circuit. Both offer little
opposition to flow (so that a negligible fraction of the driving force
(MMF or EMF) is wasted in conveying the flow to where it's usefully exploited) and both
can be shaped to guide the flow to its destination. There is one important
difference, however. In the electric circuit, no current will flow until
the circuit's completed, after which all the current is confined inside
the wires. With an iron magnetic circuit, some flux can flow (in the surrounding
air) even before the iron is installed. and although most of the flux
will subsequently take the easy route through the iron, some will still
leak into the air.

We won't pursue leakage flux here, though it's sometimes important,
as will be seen later.

ill. 8 Air-gap region, with MMF acting across opposing pole-faces

Reluctance and air-gap flux densities

If we neglect the reluctance of the iron parts of a magnetic circuit,
it's easy to estimate the flux density in the air-gap. Since the iron
parts are then in effect 'perfect conductors 'of flux, none of the source
MMF (NI ) is used in driving the flux through the iron parts, and all of
it's available to push the flux across the air-gap. The situation depicted
in ill. 7 therefore reduces to that shown in ill. 8, where an MMF
of NI is applied directly across an air-gap of length g .

To determine how much flux will cross the gap, we need to know its reluctance.
As might be expected, the reluctance of any part of the magnetic circuit
depends on its dimensions, and on its magnetic properties, and the reluctance
of a rectangular 'prism 'of air, of cross-sectional area A and length g
as in ill. 8 is given by

Rg = g / A μ0 [1.5]

…where μ0 is the so-called 'primary magnetic constant 'or 'permeability
of free space '.Strictly, as its name implies, m 0 quantifies the magnetic
properties of a vacuum, but for all engineering purposes the permeability
of air is also μ0 .The value of the primary magnetic constant (μ0)
in the SI system is 4 10 7 H/m; rather surprisingly, there is no name for
the unit of reluctance.

In passing, we should note that if we want to include the reluctance of
the iron part of the magnetic circuit in our calculation, its reluctance
would be given by…

Rfe = lfe / A μfe

and we would have to add this to the reluctance of the air-gap to obtain
the total reluctance. However, because the permeability of iron (μfe
)is so much higher than 0,the iron reluctance will be very much less than
the gap reluctance, despite the path length l being considerably longer
than the path length (g )in the air.

Equation 1.5 reveals the expected result that doubling the air-gap would
double the reluctance (because the flux has twice as far to go), while
doubling the area would halve the reluctance (because the flux has two
equally appealing paths in parallel).To calculate the flux, F ,we use the
magnetic Ohm's law (equation 1.4),which gives

Phi = MMF/R = NI A μ0 / g [1.6]

We are usually interested in the flux density in the gap, rather than
the total flux, so we use equation 1.1 to yield

B = phi / A = μ0 NI / g [1.7]

Equation 1.7 is delightfully simple, and from it we can calculate the
air gap flux density once we know the MMF of the coil (NI )and the length
of the gap (g ).We don't need to know the details of the coil-winding
as long as we know the product of the turns and the current, nor do we
need to know the cross-sectional area of the magnetic circuit in order
to obtain the flux density (though we do if we want to know the total flux,
see equation 1.6).

E.g., suppose the magnetizing coil has 250 turns, the current is
2A, and the gap is 1 mm. The flux density is then given by

B = (4 pi x 10^-7 x 250 x 2) / (1 x 10^-3) = 0.63 tesla

(We could of course obtain the same result using an exciting coil of 50
turns carrying a current of 10A, or any other combination of turns and
current giving an MMF of 500 ampere-turns.) If the cross-sectional area
of the iron was constant at all points, the flux density would be 0.63
T everywhere. Sometimes ,as has already been mentioned, the cross-section
of the iron reduces at points away from the air-gap, as shown for example
in ill. 3. Because the flux is com pressed in the narrower sections,
the flux density is higher, and in ill. 3 if the flux density at the
air-gap and in the adjacent pole-faces is once again taken to be 0.63 T,
then at the section aa'(where the area is only half that at the
air-gap)the flux density will be 2 x 0.63 = 1.26T.

ill. 9 Sketch showing how the effective reluctance of iron increases
rapidly as the flux density approaches saturation

Saturation

It would be reasonable to ask whether there is any limit to the flux density
at which the iron can be operated. We can anticipate that there must be
a limit, or else it would be possible to squash the flux into a vanishingly
small cross-section, which we know from experience isn't the case. In
fact there is a limit, though not a very sharply defined one.

Earlier we noted that the iron has almost no reluctance, at least not
in comparison with air. Unfortunately this happy state of a V airs is only
true as long as the flux density remains below about 1.6 -1.8 T, depending
on the particular steel in question. If we try to work the iron at higher
flux densities, it begins to exhibit significant reluctance, and no longer
behaves like an ideal conductor of flux. At these higher flux densities,
a significant proportion of the source MMF is used in driving the flux
through the iron. This situation is obviously undesirable, since less MMF
remains to drive the flux across the air-gap. So just as we wouldn't recommend
the use of high-resistance supply leads to the load in an electric circuit,
we must avoid overloading the iron parts of the magnetic circuit.

The emergence of significant reluctance as the flux density is raised
is illustrated qualitatively in ill. 9.

When the reluctance begins to be appreciable, the iron is said to be beginning
to 'saturate '.The term is apt, because if we continue increasing the MMF,
or reducing the area of the iron, we will eventually reach an almost constant
flux density, typically around 2T. To avoid the undesirable effects of
saturation, the size of the iron parts of the magnetic circuit are usually
chosen so that the flux density does not exceed about 1.5T. At this level
of flux density, the reluctance of the iron parts will be small in comparison
with the air-gap.

ill. 10 Evolution of d.c. motor magnetic circuit from gapped C-core

Magnetic circuits in motors

The reader may be wondering why so much attention has been focused on
the gapped C-core magnetic circuit, when it appears to bear little resemblance
to the magnetic circuits found in motors. We will now see that it's actually
a short step from the C-core to a magnetic motor circuit, and that no fundamentally
new ideas are involved.

The evolution from C-core to motor geometry is shown in ill. 10, which
should be largely self-explanatory, and relates to the field system of
a d.c. motor.

We note that the first stage of evolution (ill. 10,left)results in
the original single gap of length g being split into two gaps of length
g /2, reflecting the requirement for the rotor to be able to turn. At the
same time the single magnetizing coil is split into two to preserve symmetry.

(Relocating the magnetizing coil at a different position around the magnetic
circuit's of course in order, just as a battery can be placed anywhere
in an electric circuit.)Next,(ill. 10,centre)the single magnetic path
is split into two parallel paths of half the original cross section, each
of which carries half of the flux; and finally (ill. 10, right),the
flux paths and pole-faces are curved to match the rotor. The coil now has
several layers in order to fit the available space, but as discussed earlier
this has no adverse effect on the MMF. The air-gap is still small, so the
flux crosses radially to the rotor.

TORQUE PRODUCTION

Having designed the magnetic circuit to give a high flux density under
the poles, we must obtain maximum benefit from it. We therefore need to
arrange a set of conductors, fixed to the rotor, as shown in ill. 11, and to ensure that conductors under a N-pole (at the top of ill. 11)
carry positive current (into the paper),while those under the S-pole carry
negative current. The tangential electromagnetic ('BIl ') force
(see equation 1.2)on all the positive conductors will be to the left, while
the force on the negative ones will be to the right. A nett couple, or
torque will therefore be exerted on the rotor, which will be caused to
rotate.

(The observant reader spotting that some of the conductors appear to have
no current in them will find the explanation later, in section 3.)

ill. 11 Current-carrying conductors on rotor, positioned to maximize
torque.(The source of the magnetic flux lines (arrowed) isn't shown.)

At this point we should pause and address three questions that often crop
up when these ideas are being developed. The first is to ask why we have
made no reference to the magnetic field produced by the current carrying
conductors on the rotor. Surely they too will produce a magnetic field,
which will presumably interfere with the original field in the air-gap,
in which case perhaps the expression used to calculate the force on the
conductor will no longer be valid.

The answer to this very perceptive question is that the field produced
by the current-carrying conductors on the rotor certainly will modify the
original field (i.e. the field that was present when there was no current
in the rotor conductors.) But in the majority of motors, the force on the
conductor can be calculated correctly from the product of the current and the
'original 'field. This is very fortunate from the point of view of calculating
the force, but also has a logical feel to it. E.g. in ill. 1, we wouldn't expect any force on the current-carrying conductor if there was no
externally applied field, even though the current in the conductor will
produce its own field (upwards on one side of the conductor and downwards
on the other).So it seems right that since we only obtain a force when
there is an external field, all of the force must be due to that field
alone.

The second question arises when we think about the action and reaction
principle. When there is a torque on the rotor, there is presumably an
equal and opposite torque on the stator; and therefore we might wonder
if the mechanism of torque production could be pictured using the same
ideas as we used for obtaining the rotor torque. The answer is yes; there
is always an equal and opposite torque on the stator, which is why it's
usually important to bolt a motor down securely. In some machines (e.g.
the induction motor)it's easy to see that torque is produced on the stator
by the interaction of the air-gap flux density and the stator currents,
in exactly the same way that the flux density interacts with the rotor
currents to produce torque on the rotor. In other motors, (e.g. the d.c.
motor we have been looking at), there is no simple physical argument which
can be advanced to derive the torque on the stator, but nevertheless it's equal and opposite to the torque on the rotor.

The final question relates to the similarity between the set-up shown
in ill. 10 and the field patterns produced for example by the electro
magnets used to lift car bodies in a scrap yard. From what we know of the
large force of attraction that lifting magnets can produce, might not we
expect a large radial force between the stator pole and the iron body of
the rotor? and if there is, what is to prevent the rotor from being pulled
across to the stator? Again the affirmative answer is that there is indeed
a radial force due to magnetic attraction, exactly as in a lifting magnet
or relay, although the mechanism whereby the magnetic field exerts a pull
as it enters iron or steel is entirely different from the 'BIl 'force
we have been looking at so far.

It turns out that the force of attraction per unit area of pole-face is
proportional to the square of the radial flux density, and with typical
air gap flux densities of up to 1 T in motors, the force per unit area
of rotor surface works out to be about 40N/cm^2

.This indicates that the total radial force can be very large: for example
the force of attraction on a small pole-face of only 5 x10 cm is 2000N,
or about 200 Kg. This force contributes nothing to the torque of the motor, and is merely an unwelcome by-product of the 'BIl 'mechanism we
employ to produce tangential force on the rotor conductors.

In most machines the radial magnetic force under each pole is actually
a good deal bigger than the tangential electromagnetic force on the rotor
conductors, and as the question implies, it tends to pull the rotor onto
the pole. However, the majority of motors are constructed with an even
number of poles equally spaced around the rotor, and the flux density in
each pole is the same, so that in theory at least the resultant force on
the complete rotor is zero. In practice, even a small eccentricity will
cause the field to be stronger under the poles where the air-gap is smaller, and this will give rise to an unbalanced pull, resulting in noisy running and rapid bearing wear.

Magnitude of torque

Returning to our original discussion, the force on each conductor is given
by equation 1.2, and it follows that the total tangential force F depends
on the flux density produced by the field winding, the number of conductors
on the rotor, the current in each, and the length of the rotor. The resultant
torque or couple (T) depends on the radius of the rotor (r), and is given
by:

T = Fr [1.8]

We will develop this further in Section 1.5, after we examine the remark
able benefits gained by putting the conductors into slots.

The advantages of slotting

If the conductors were mounted on the surface of the rotor iron, as in
ill. 11, the air-gap would have to be at least equal to the wire diameter, and the conductors would have to be secured to the rotor in order to transmit
their turning force to it. The earliest motors were made like this, with
string or tape to bind the conductors to the rotor.

Unfortunately, a large air-gap results in an unwelcome high-reluctance
in the magnetic circuit, and the field winding therefore needs many turns and a high current to produce the desired flux density in the air-gap.
This means that the field winding becomes very bulky and consumes a lot
of power. The early (Nineteenth-century) pioneers soon hit upon the idea
of partially sinking the conductors on the rotor into grooves machined
parallel to the shaft, the intention being to allow the air-gap to be reduced
so that the exciting windings could be smaller. This worked extremely well
as it also provided a more positive location for the rotor conductors, and thus allowed the force on them to be transmitted to the body of the
rotor.

Before long the conductors began to be recessed into ever deeper slots
until finally (see ill. 12) they no longer stood proud of the rotor
surface and the air-gap could be made as small as was consistent with the
need for mechanical clearances between the rotor and the stator. The new
'slotted 'machines worked very well, and their pragmatic makers were unconcerned
by rumblings of discontent from skeptical theorists.

(1. Older readers will probably have learned the terms Couple and Moment
(of a force) long before realizing that they mean the same as torque.)

ill. 12 Influence on flux paths when the rotor is slotted to accommodate
conductors

The theorists of the time accepted that sinking conductors into slots
allowed the air-gap to be made small, but argued that, as can be seen from
ill. 12, almost all the flux would now pass down the attractive low-reluctance
path through the teeth, leaving the conductors exposed to the very low
leakage flux density in the slots. Surely, they argued, little or no 'BIl'
force would be developed on the conductors, since they would only be exposed
to a very low flux density.

The skeptics were right in that the flux does indeed flow down the teeth;
but there was no denying that motors with slotted rotors produced the same
torque as those with the conductors in the air-gap, provided that the average
flux densities at the rotor surface were the same. So what could explain
this seemingly too good to be true situation? The search for an explanation
preoccupied some of the leading thinkers long after slotting became the
norm, but finally it became possible to verify theoretically that the total
force remains the same as it would have been if the conductors were actually
in the flux, but almost all of the tangential force now acts on the rotor
teeth, rather than on the conductors themselves.

This is remarkably good news. By putting the conductors in slots, we simultaneously
enable the reluctance of the magnetic circuit to be reduced, and transfer
the force from the conductors themselves to the sides of the iron teeth,
which are robust and well able to transmit the resulting torque to the
shaft. A further benefit's that the insulation around the conductors no
longer has to transmit the tangential forces to the rotor, and its mechanical
properties are thus less critical. Seldom can tentative experiments with
one aim have yielded rewarding outcomes in almost every other relevant
direction.

There are some snags, however. To maximize the torque, we will want as
much current as possible in the rotor conductors. Naturally we will work
the copper at the highest practicable current density (typically between
2 and 8A =mm^2), but we will also want to maximize the cross sectional
area of the slots to accommodate as much copper as possible.

This will push us in the direction of wide slots, and hence narrow teeth.

But we recall that the flux has to pass radially down the teeth, so if
we make the teeth too narrow, the iron in the teeth will saturate, and lead to a poor magnetic circuit. There is also the possibility of increasing
the depth of the slots, but this cannot be taken too far or the centre
region of the rotor iron which has to carry the flux from one pole to another
will become so depleted that it too will saturate.

In the next section we look at what determines the torque that can be
obtained from a rotor of a given size, and see how speed plays a key role
in determining the power output.