1. Introduction
Industrial brushless servomotors can be divided into two main types. One operates
in a similar way to the threephase synchronous motor and the other is a relatively
simple development of the brushed DC motor. Both types of brushless motor have
the same sort of construction and have an identical physical appearance. Both
have many characteristics similar to those of a permanent magnet brushed DC
motor, and both are operated from a source of direct current. A review of the
features of the permanent magnet brushed motor is therefore a convenient first
step in the approach to the brushless type. In this first section, the relationships
between the supply voltage, current, speed and torque of the brushed motor
are developed from fundamental electromagnetic principles. Attention is also
given to the factors controlling the steadystate speed of the unloaded motor.
The later part of the section is devoted to the question of DC motor rating.
Only the basic ideas are covered at this stage, in preparation for the more
detailed treatment in Section 5. The power losses which lead to motor temperature
rise are identified, and the main factors affecting the final steadystate
temperature are explained for both continuous and intermittent operations of
the motor. The scope of this section is confined to cases where the losses
during periods of speed change are insignificant in Comparison to those generated
during the periods of constant motor speed.
2. Operational principles
Motor construction
FIG. 1 shows the essential parts of a rudimentary permanent magnet DC motor.
Two conductors are connected in series to form a winding with one turn. The
winding has a depth ! and width 2r meters and is mounted between the poles
of a permanent magnet. The winding is free to rotate about the dotted axis
and its ends are connected to a DC source through sliding contacts to form
a circuit carrying current I A. The main diagram is drawn for the moment when
the conductors are passing the center of the poles.
The contacts allow the direction of current in the winding to reverse as it
moves through the vertical position, ensuring that the direction of flow through
the conductors is always the same relative to the direction of the magnetic
field. In other words, it does not matter in the diagram which side of the
winding is to the left or right when we look at how torque is produced.
Torque production
The torque produced by the motor in FIG. 1 is the result of the interaction
between the magnetic field and the current carrying conductors. The force
acting on each conductor is shown as F. Some simple magnetic principles are
involved in the evaluation of the torque.
FIG. 1 Principle of the permanent magnet brushed DC motor
Magnetic flux Φ
The amount of magnetic flux in a magnetic field tells us how much magnetism
is present. By itself, it does not give the strength of the field. The flux
may be represented by lines drawn between the poles of the magnet and in the
old British system the unit of flux was, in fact, the line. In the SI system
the unit is the weber, denoted by Wb, where one weber is equivalent to 10 lines
in the old system.
Magnetic flux density β
As its name suggests, the term magnetic flux density describes the concentration
of the magnetic field. The SI unit of magnetic flux density is the tesla, denoted
by T, where a tesla is equal to one weber per square meter.
The force on a conductor
When a conductor of length l, carrying a current/, is placed in a magnetic
field of uniform flux density B, it is found that the conductor is acted on
by a force which is at right angles to both the field and the conductor. The
force is greatest when the conductor and field are also at right angles, as
in FIG. 1.
In this case, the force is given by
f = B/I (N)
The unit of force is the newton, denoted as N. The direction of F can be found
by the 'lefthand motor rule'. This states that the thumb of the left hand
points in the direction of the force, if the first finger of the hand is pointed
in the direction of the field and the second finger in the direction of the
current.
Torque
Force F acts on each conductor of the winding shown in FIG. 1. The torque
produced at each conductor is T= Fr (Nm)
The unit of torque is the newton meter, denoted as Nm. The radius of action
of F around the axis falls as the winding moves away from the horizontal position,
reducing the torque. In the figure, the winding lies in a plane between the
centers of the fiat poles of the magnet, where B is greatest.
With such a pole shape the flux will be less dense at other winding positions,
reducing the torque still further.
FIG. 2 shows three practical DC motors with the circular type of pole faces
shown in FIG. 3. These give a substantially radial and uniform pattern to the
flux so that B and T remain constant in the ideal case. The winding has a number
of turns, with the conductors distributed in slots (not shown in crosssection)
around a cylindrical iron carrier, or rotor. For simplicity, the crosssection
shows only seven turns, each with two conductors arranged diametrically. The
current directions are shown by the use of a cross and a dot for current flowing
into and out of the paper respectively.
The turns of the rotor winding are connected to the segments of a commutator
which rotates between springloaded brushes. The current in each turn of the
winding reverses each time the turn passes the brush axis, and the pattern
of crosses and dots in FIG. 3 will be the same for any rotor position. The
reversals give a rectangular AC waveform to the current in the individual turns
of the motor winding.
Only the brushes carry a unidirectional current.
FIG. 2 Permanent magnet DC motors
FIG. 3 Crosssection and rotor of a twopole, permanent magnet DC motor
For a winding with N turns, there are 2N conductors. The finish of each turn
is joined to the start of its neighbor at a segment of the commutator. Two
circuits of N/2 turns appear in parallel between a pair of brushes which touch
segments at opposite sides of the commutator, and so each of the 2N conductors
carries a current of 1/2. The combined torque is
T = NBlIr
Assuming that the poles of the motor in FIG. 3 are the same length l (into
the paper) as the conductors, we can write the flux density around the face
of each pole in terms of webers per square meter as Φ / π * r * l. The torque
expression for the two pole motor with one winding of N turns becomes
T = N Φ I / π
The torque constant
For any given motor, the only variable in the last expression is the current
I. The torque can be expressed as
T=K_T l
lit is the torque constant, expressed in Nm/A. It is one of the most important
constants in the motor specification.
Motor speed
When the voltage is switched on to an unloaded DC motor, the rotor speed rises
from zero and quickly reaches a 'noload' terminal value. The normal losses
associated with the DC motor itself would not be enough to prevent the speed
from rising to a point very much higher than the noload value, and the question
arises of how the limit in speed occurs. To answer, we must look at a second
aspect of the behavior of a moving conductor in a magnetic field.
Voltage generation
FIG. 4 shows a conductor of length l which is being moved with velocity v
meters per second (m/s) across and at fight angles to a uniform magnetic field
of density B. As the conductor moves across the field, a voltage known as the
electromotive force or emf will be generated along its length equal to
= Blv (V)
FIG. 4 Conductor moving across a magnetic field
In the 'righthand generator rule', the second finger points in the direction
of E if the forefinger is pointed in the direction of the field and the thumb
in the direction of movement. The rotor of a twopole motor with a winding
of N turns has 2N conductors, and there are always two parallel paths of N
conductors connected in series between the brushes. The conductors travel at
a speed of ~or, where v is the angular velocity expressed in radians per second
or rad/s. The total voltage induced between the start and finish of the winding
is therefore
E = NBlv
Substituting for B as before gives
E= N Φ ω / π
The voltage constant
In the last equation above, all quantities except ~o are constant for any
given motor and so the induced voltage is
E= K_E ω
where KE is the voltage constant expressed in volts/radian per second or V/rad
s^1.
KT and KE
Comparing the expressions above for T and E shows that for a twopole motor
with a single winding,
KT=KE = N Φ / π
The equality is maintained when the number of pole pairs and a number of parallel
windings are taken into account. Note that the constants have the same numerical
value in SI units, but not in other systems of units.
Back emf and the terminal speed of the unloaded motor
FIG. 5 shows a motor connected to a voltage source VDC. E is generated in
the direction which opposes the cause of its generation, namely the movement
of the rotor. Accordingly, E acts against the applied voltage VDC and is normally
referred to as the back emf. Note that AC emfs are generated across the individual
turns of the rotor winding. The emfs are commutated in the same way as the
AC currents in the turns, so that the total back emf E appears as a direct
voltage at the motor brushes.
If the mechanical losses due to friction and windage are ignored, steadystate
conditions would be reached at a speed sufficient to make the induced voltage
KEa; equal the supply voltage Voc, that is when the motor speed w = VDc/KE.
In practice, the terminal speed and the induced voltage will be slightly lower
to allow a small current to flow to supply the losses.
FIG. 5 Unloaded motor at steadystate
3. The loaded motor at steady state
The power required to supply a torque of T Nm at a speed of ω
rad/s is
P = T ω (W)
The unit of power is the watt, denoted by W. FIG. 6 shows a DC motor connected
to a load. Current flows to the motor following the application of the constant
voltage VDC, and the motor accelerates to a constant speed. The final steady
current and speed occur when the motor output torque equals the opposing torque
at the load, at which point the power output from the motor is equal to the
power supplied to the load.
Steadystate characteristics
In FIG. 7 the motor is represented by the resistance R of the rotor winding
conductors, and the back emf E. The supply voltage is
V= RI+E
FIG. 6 The loaded motor
FIG. 7 Simple equivalent circuit of a DC motor
At steady state,
V= RI + K_E ω
The motor torque at steady state is
T=K_TI
ω = (V /KE) – TR/ KT KE
from which we see that the speed of the permanent magnet brushed motor varies
linearly with torque. The speedtorque characteristic shown in FIG. 8 is plotted
by drawing a straight line between two reference points. At the first point,
when T is zero, the noload speed is given by
&ω;_NL = V / K_E
FIG. 8 DC motor speedtorque diagram
The second reference is taken by imagining the load to increase to the point
where the motor is forced to stall, making w zero. T would then be at a theoretical
maximum of:
T'= VK_T/R
The equation for the speedtorque curve can now be written as
ω = ω_NL – R_RC T
where R_RC is the speed regulation constant of the motor, equal to the slope
R/KT KE of the speedtorque characteristic in FIG. 8. The current carried by
the rotor conductors rises with the motor torque. The last expression above
does not take account of motor losses due to, for example, brush contact and
rotor bearing friction, which in practice would cause a reduction in ω_NL.
FIG. 8 has been drawn for a fixed value of supply voltage. For any particular
motor, a family of linear speed torque characteristics can be drawn for a
range of operating voltages. The smallest of the motors shown in FIG. 2 is
a twopole, 24 V motor with the following constants"
KT  0.07 Nm/A
KE = 0.07 V/rad
R  0.70 ohm
Using these constants, the noload speed and the torque developed at the point
of stall can be found at several supply voltages up to 24 V. FIG. 9 shows the
resulting characteristics. These must be applied with caution, as damage to
the motor may result from the flow of high current at the low speed, high torque
end.
FIG. 9 Speed versus torque at various supply voltages
Small permanent magnet DC motors have a wide range of applications such as
door operators, tape drives, floor scrubbers, conveyors, as well as in small
battery powered vehicles. As an example, we will take the case of an automatic
sliding door which is to be driven by the small 24 V motor described above.
FIG. 10 shows the profile of door velocity. The door opens rapidly, and then
crawls in readiness for its stop at the fully open position. The same action
occurs in the reverse direction. For safety reasons the door closes relatively
slowly, and finally crawls to its closed position. The most obvious feature
of the diagram is that the motor is not required to work continuously at a
constant speed, which raises the question of its rating. We should now look
at DC motor ratings in general, before returning to the example.
FIG. 10 Velocity profile for an automatic sliding door
4. Motor rating
This section deals with ratings for continuous or intermittent motor operation,
the work being broadly relevant to both the brushed and brushless motor. The
intermittent operations are limited to duty cycles in which the electrical
energy supplied to the motor during acceleration and deceleration may be ignored
in comparison to the amount supplied over the complete cycle.
Section 5 covers the rating of the brushless motor in more detail, and includes
cases where the duty cycle demands a relatively high input of energy during
periods of speed change.
Power losses FIG. 11 shows how the electrical power input is distributed as
the DC motor performs its normal task of converting electrical energy into
mechanical energy. The output power is lower than the input power by the amount
of the losses, which appear mainly in the form of heat within the motor.
FIG. 11 Power distribution in the DC motor
The I^2R winding loss
The flow of current I through the rotor winding resistance R results in a
power loss of I2R. Note the dependence of this loss on the motor torque K_TI.
Friction, windage and iron losses
As well as friction and windage, there are other effects of the physical rotation
of the rotor. For example, as the rotor position changes with respect to the
permanent magnetic field, flux reversals take place inside the iron core which
encourage the flow of eddy currents. The consequent losses and rotor heating
increase with rotor speed.
Torque loss and power loss at constant speed
The iron, friction and windage losses result in a reduction in the available
output torque. The loss at constant speed is
T_loss = Tf + D ω
where Tf is the torque due to constant friction forces, such as those produced
at the rotor bearings, and D is a constant of proportionality for speeddependent
torque losses due to viscous effects such as iron losses. The constant D is
known as the damping constant expressed as Nm/rad s^1. The product of the
torque loss and the motor speed is known as the speed sensitive loss. Adding
the I^{2}R loss gives the total power loss in SI units as:
P_loss = ω( Tf + D ω) + I^{2}R)
P_loss is the difference between the electrical power at the motor input and
the mechanical power at the output shaft. Over a period of time, more energy
is supplied to the motor than reaches the load. Most of the difference results
in motor heating and a rise in temperature, which continues until as much heat
is passed from the motor to the surrounding air as is produced internally.
As there is always a designed maximum limit to the motor temperature, limits
must also be set on the performance demands which lead to temperature rise.
The last equation above shows that the power loss depends on motor speed and
the square of the current. The current is directly related to the motor torque
and we can conclude that motor speed and the square of the torque are the factors
which control the temperature rise.
Continuous operation
The limits of continuous speed and torque which give rise to the maximum permissible
temperature at any part of the motor are determined experimentally and plotted
as a boundary on a speedtorque plane. The region to the left of the boundary
is the Safe Operating Area for Continuous operation, the boundary being known
as the Soac curve. FIG. 12 shows two areas of safety, one with and one without
forced air cooling. The curve takes account of the i2R and speed sensitive
loss at all speeds and can always be used down to the stall point, unlike the
basic speedtorque characteristics of FIG. 9.
FIG 12 Safe operating areas on the speedtorque plane
Intermittent operation
While the area to the right of the Soac boundary may not be used for continuous
running, the higher torques may still be intermittently available if the overall
losses do not raise the temperature of any part of the motor above the safe
limit, normally 150 degr. C. For the brushed motor, the speedsensitive loss
is usually low in comparison to the I^2R loss. The motor losses and heating
therefore depend largely on the square of the current, or effectively on the
square of the motor torque. It is clearly wrong therefore to base the rating
for intermittent operation on the average torque requirement. The rating on
the righthand side of the Soac boundary should be based on the rootmeansquare
(rms) value of the torque supplied over a complete duty cycle. Note that this
applies automatically on the lefthand side, where rms and continuous torques
have the same values.
At this point we may return for a moment to the example of the automatic door
with the velocity profile shown in FIG. 10.
Maximum demand on the motor occurs when the door is required to open and close
continuously, with the fully open periods at a minimum. The ideal motor current
waveform is shown in FIG. 13. If the current is supplied from an electronic
drive, a tipple may be present on the waveform. As the same method applies
for any waveform, assume for simplicity that the motor current follows the
pattern shown in the figure.
The average current over the 16 second period of the cycle is
I_av – I_M(1.0 x 2 + 0.5 x 3 + 0.7 x 3 + 0.5 x 3) / 16 = 0.44I_M
The rms current over the same period is
I_rms  ___/ [/I^2M(1.02 • 2 + 0.52 • 3 + 0.72 • 3 + 0.52 • 3)]/16 = 0.56I_M
It is now clear that extra fiR losses will be produced as a result of the
intermittent nature of the load. The motor must be able to accept an rms current
which is greater than the duty cycle average by the factor 0.56/0.44, or 1.3.
FIG. 13 Maximum demand on door operator
The form factor
Although the above example is for one particular current waveform, the same
arguments for motor rating would apply for any other waveforms. As much as
possible, ratings should take into account the waveform shape defined by the
term:
form factor = I_rms / I_av
In practice, the form factor depends on a number of variables and is not always
a simple, constant value.
Motor temperature
When the motor runs continuously at a fixed speed, its temperature gradually
rises towards a steadystate value.
When the operation is intermittent, a ripple occurs in the plot of temperature
against time. The evaluation of the temperatures relies on the use of two important
motor constants.
Thermal resistance and thermal time constant
FIG. 14 shows a rise in motor temperature for continuous operation at a constant
load, from the ambient value of Θ0 to the final steadystate value Θss. The
final temperature rise in degrees centigrade (above ambient) is
(Θss
Θ0) = Rth P_loss (degr. C)
where P_loss is the constant power loss at temperature Θss and Rth is
the thermal resistance in degr. C/W. Rth is usually quoted as the value of
thermal resistance from the hottest part, normally the rotor winding, to the
air surrounding the motor case. In FIG. 14, Θss is therefore the final
temperature of the winding.
FIG. 14 Motor temperature rise at a constant load
If the curve is assumed to rise exponentially towards Oss, the temperature
at time t is
Θ = Θ0 + (Θ_ss Θ_0(1  e^{t/τ th}))
where tau_th is the thermal time constant of the motor, normally given
in minutes on the motor specification. The magnitude of the time constant is
a measure of how slowly the temperature rises to the steadystate value. The
value of τth is normally quoted for the main mass, which for brushed motors
is taken to be the rotor as a whole. Note particularly that the temperature
curve has the overall rate of rise of the rotor temperature, but terminates
at the final value of the winding temperature.
Winding temperature ripple
When the motor runs on a duty cycle with an intermittent torque demand, the
losses are also generated intermittently.
In FIG. 15, the torque pulses and the losses are assumed to follow the same
waveform. The figure shows the limits of the steadystate, aboveambient temperature
of the winding as Θmin and Θpk.
If the shapes of the curves of winding temperature rise and fall over the
pulse times tp and ts are assumed to be exponential, and to have the same time
constant τw, we may write
Θpk  Θmin = (RthPloss(pk)  Θmin)(1 e ^{tp/τw})
and...
Θpk  Θmin = Θpk (1  e^{ t_s/τw})
Combining the last two equations and writing tp + ts as t' gives the peak
rise above ambient of the winding temperature as
Θpk = RthP_{loss(pk) }[(1  e ^{t_p/rw}) / (1  e ^{ts/rw})]
The i2R loss arises in the winding, which has a relatively low thermal capacity.
The winding temperature rises faster than the rotor iron temperature, and also
falls faster during the time ts. The thermal time constant for the winding
is therefore
FIG. 15 Steadystate temperature of the rotor winding
lower than for the rotor as a whole. The above expression can be used to predict
the limiting conditions for the ripple at an assumed value of ~'w. If the average
loss is kept at a constant level, the ripple on the winding temperature becomes
more pronounced as t' is increased and/or as tp is reduced. As a rule of thumb,
the ripple in the steadystate temperature can normally be assumed to be within
a band of
+/ 10 degr. C
when τth > 50t'
where τth >= 25 minutes. The thermal time constant of the motor used
in the example of the sliding doors is given as 25 minutes, or 1500 seconds,
and the period of the duty cycle in FIG. 13 is 16 seconds. We can conclude
that the winding temperature may be designed to reach 140 degr. simply as a
result of the average I^2R loss, assuming there is no significant speedsensitive
loss.
Servomotor ratings
For both brushed and brushless servomotors, extra losses are generated if
an application demands rapid changes in the speed of the motor and load. When
rating the motor it is important to add the extra losses to those for the periods
of steady speed, especially if the transient periods form a significant part
of the duty cycle. We will look at such cases for the brushless servomotor
in Section 5.
5. The brushed servomotor
So far we have studied the permanent magnet, brushed DC motor, mostly without
reference to its role as a servomotor.
In the example of the sliding door operator, the speed of the doors was determined
by the balance between the motor output and the frictional forces developed
in the door slide mechanism. No other control of the speed, or rate of change
of speed, was required and the system can be described as open loop in the
sense that speed control is achieved without the need for information feedback
from the load to the motor.
For servo applications, precise control of load speed may be required at various
stages of an operation and the servomotor must be capable of responding to
calls for high transient torques. Two typical brushed servomotors are shown
in FIG. 16. The most striking difference between these and the normal DC motor
is in the long and narrow shape, which gives the rotor a relatively low moment
of inertia, increasing the output torque available for acceleration of the
load itself.
The stators of the motors illustrated carry four permanent magnets made from
a highly coercive ferrite material designed to withstand high demagnetizing
fields. Also on the stator are four brushes which form the main point of motor
maintenance. Depending on the motor duty, inspection is recommended up to eight
times during the life of the brushes.
The speed of a servomotor must be controllable at all times.
The speed is measured using the signal from a tachometer mounted on the motor
shaft in the rear housing. The tacho has its own permanent magnetic field and
brushes, and is a precision instrument which must be maintained in the same
way as the motor itself.
FIG. 16 Permanent magnet, brushed servomotors
The thermal characteristics of a typical DC servomotor are shown in FIG. 17.
The motor speed axis is marked in krpm, or revolutions per minute x 10^3.
The curves are drawn for a winding temperature rise of 110 degr. C. There are
two continuous duty characteristics, one with and one without forced cooling.
Both assume that the motor has a pure DC, unity form factor supply and derating
may be needed if this is not the case.
FIG. 17 Thermal characteristics for a brushed servomotor
Example 1
Continuous operation is required at a speed of 1000 rpm. What is the maximum,
average torque indicated by the Soac curve if cooling is unforced and the form
factor of the current supplied by the electronic drive is 1.1? Kr = 0.43 Nm/A.
FIG. 18 shows the type of current waveform provided by the electronic drive.
The ripple is produced by the action of electronic switches as they operate
to control the average value of current.
The maximum torque at 1000 rpm is found from FIG. 17 to be 2 Nm. The maximum
rms current which may be supplied to the motor at 1000 rpm is therefore
I_rms  T/K_T = 2/0.43 = 4.65 A
The usable average current is
I_av= I_rms/Form factor = 4.65/1.1 = 4.23 A
and so the maximum average torque is
τmax = 0.43 • 4.23 = 1.82 Nm
FIG. 18 Current supply for Example 1
Example 2
A torque of 8 Nm at 500 rpm is required once every 9 seconds. What pure DC
current would be required and what would be the maximum intermittent operation
time? The maximum ambient temperature is 30 degr. C and τth = 50 minutes.
FIG. 19 shows the current in this case is assumed to be ripple free, and
to consist of a series of rectangular pulses of length tp. The maximum, continuous
DC at 500 rpm is
I=T/K_T = 2.05/0.43 = 4.77
At the torque of 8 Nm, the motor current is:
Ip = 8/0.43  18.6 A
For an average winding temperature rise of less than 110 degr. C the rms value
of the intermittent pure DC should be no more than the maximum continuous value,
and so giving:
(τpmax • 18.62/9) 0.5 4.77 tpmax  0.6 s
FIG. 19 Current supply for Example 2
The 9 second period of the duty cycle is less than τth/50 and we can
assume that any steadystate temperature ripple will be less than +/ 10 degr.
C. The peak winding temperature is therefore less than 30 + (110 + 10), or
150 degr. C. Note that τpmax would be less than 0.6 s if the current of 18.6
A is supplied as an average value by a source of impure DC, e.g. the electronic
drive in example (1).
