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Electric Motors and Drives - Third Edition
Solutions to Review Questions - Chapter 9
1)
In one-phase-on operation the equilibrium (rest) positions are where several rotor and stator
teeth (which are usually of the same width) are directly aligned with each other. The alignment
positions are therefore well-defined at the design/manufacture stage, and do not depend on the
magnitude of current in the excited phase.
When two phases are excited simultaneously, the rotor comes to rest at a position intermediate
between the two positions it would occupy if only one of the phases were excited. If the two
exciting windings are identical and the currents are equal the rest position will be mid-way
between the two one-phase-on positions. In practice, however, even if the windings are identical
it is unlikely that the two currents will be exactly equal, so the final equilibrium position will
tend to be slightly closer to the phase that has the higher current. The rest positions are thus less
well-defined than during one-phase-on operation.
2)
Detent torque is the (small) torque that exists in hybrid or permanent-magnet motors when none
of the phase-windings carries any current. It can easily be detected by turning the rotor slowly
by hand with the motor unexcited.
The rotor equilibrium (or rest) positions that result from the detent torque are the same as the
normal step positions: this is an advantage if the motor is left unexcited, because the existence
of the detent torque helps to prevent the rotor from losing registration by being inadvertently
moved to another position.
3)
The holding torque is the maximum static torque that the motor can develop, i.e. the peak of the
torque/angle curve, i.e. Tmax in the typical torque/angle curve shown in Figure 9.8. When a
stepping motor is referred to as, say, ‘a 2 Nm motor’, it is understood to mean that the holding
torque is 2 Nm.
4)
We are not told the step angle so clearly the answer to this question will have to be expressed as
a fraction of a step. However we know that the motor is a 3-phase one, so one step corresponds
to one-third of the angle between successive stable alignment positions for one phase. Hence if
we use the sinusoidal approximation suggested, the static torque can be written as T  0.8 sin 
Nm, where   360 corresponds to three steps and one step is represented by 120 .
With a steady load torque of 0.25 Nm the equilibrium position will therefore be given by
0.25  0.8 sin  , i.e.   18.2. Ideally, of course the equilibrium position should be at   0 :
expressed as a fraction of step, the step position error is therefore 18.2/120 = 0.15 step.
5)
When driven from an ideal constant-current source, the output torque effectively jumps
instantaneously from the static torque curve of one phase onto the static torque curve of the
incoming phase, the lag angle (see section 9.4.2) adjusting such that the average motor torque is
equal and opposite to the load torque. With the assumption that the static torque/angle
relationship is sinusoidal, the maximum torque will be obtained when the rotor lag angle is such
that the instantaneous torque follows the solid line in Figure 9A.
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Fig 9A here
We can find the average of the solid line by integration, the mean torque being given by
 /3
Tav 
 0.8 cos d
0
 /3
 0.66 Nm,
The pull-out torque is therefore 0.66 Nm, and if the load torque exceeds this value, the motor
will stall.
6)
If the torque/angle curve is assumed to be linear about the step position, the rotor will oscillate
in much the same way as a simple pendulum as it comes to rest.
Mathematically, if the restoring torque is given by T   k  , where  is the angle of
displacement from the equilibrium position, the equation of motion is  k   J
d
d 2
J 2 ,
dt
dt
k
.
J
the solution of which is   A sin n t , where n 
We can therefore calculate the angular frequency of oscillation provided we know k, the
gradient of the torque/angle curve, and the total inertia, J. We must use SI units, so we first
convert 2 Nm/degree to 2  180/π = 114.6 Nm/rad. Substituting in the expression above yields
114.6
n2 
, n  252.3 rad/ sec, or just over 40 Hz.
1.8  10 3
7)
The step angle is given by:-
Step Angle 
360
.
Rotor teeth  Stator phases
a) Applying this formula the step angle is given by
360
 15.
83
b) Each stack of a multi-stack motor is a separate phase, so this is a 3-phase motor and the step
360
 7.5.
angle is given by
16  3
c) Using the same formula, the step angle is given by
8)
360
 1.8.
50  4
Probably the easiest thing to do is to spin the rotor with the stator windings open circuited. If
there is negligible mechanical resistance apart from friction, it is almost certainly a VR type. If a
3
cyclic detent torque is evident, it must be a hybrid or a permanent-magnet type. To be absolutely
sure, the windings can be short circuited and if when the rotor is turned the mechanical resistance
has increased we can deduce that the machine is acting as a generator, the currents in the windings
being due to motional e.m.f. This can only happen if the motor includes a permanent-magnet to
provide the excitation: the VR type has no excitation unless it is connected to the drive circuit, so it
cannot generate in isolation.
9)
When operating in the stepping mode, a hybrid motor with two phases would normally run from a
bipolar supply, the rectangular one-phase-on pattern of excitation typically being  A,  B ,  A,
 B ,  A, etc , i.e. one complete cycle of excitation of each phase corresponds to four steps.
When supplied form a sinusoidal source rather than a rectangular one, the torque will be smoother
and rather than an incremental motion the rotor velocity is effectively smooth. In one cycle of the
supply, the rotor will therefore move four steps, so at 60 Hz the effective stepping rate is 240
steps/sec. The step angle is 1.8°, so the motor speed will be 240  1.8° = 432°/sec or 72 rev/min.
(It is interesting to note that the 1.8° hybrid motor was originally conceived to produce a lowspeed mains-fed direct drive motor, offering 72 rev/min from 60 Hz or 60 rev/min from 50 Hz.
The mains-fed version has more turns of thinner wire to suit the higher voltage, but is otherwise
the same as the stepping version.)
10)
In its original form the drive applies 4.5 V to the winding resistance of 1.5 , yielding a steadystate current of 3 A. If the inductance of the winding is L, the time-constant will be L/1.5,
which means that it takes a time of about 4L/1.5 for the current to rise to 3 A. This rise-time is
of little consequence at very low stepping rates.
However, for the motor in question, at 50 steps/sec the time-constant is undoubtedly
comparable with the ‘on’ time, so that the current never actually reaches its full value and the
pull-out torque is therefore less than at very low speeds.
If we wish to maintain the same pull-out torque at a speed of 250 steps/sec, where the ‘on’ time
is only a fifth of what it is at 50 steps/sec, we need to ensure that the effective time-constant is
reduced by a factor of 5, i.e. to L/7.5, by ensuring that the total circuit resistance is 7.5 . We
therefore need to add a further 6  of ‘forcing’ resistance in series with each motor phasewinding, and at the same time the supply voltage must be increased by a factor of five, to 22.5
V in order to maintain the steady-state current at 3 A.
The price to be paid for the improved performance is the extra input power, most of which is
wasted as heat in the forcing resistance.
11)
Perhaps it should have been pointed out to the scientist that the claim that stepping motors
‘typically complete each single step in a few milliseconds’ does presuppose that the motor was
being used in a ‘typical’ application, which normally means that the load inertia is of similar
order to the motor inertia.
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To estimate the inertia of the pointer (which we will assume is fixed to the rotor at one end) we
Ml 2
use the formula for the inertia of a uniformly-distributed linear mass, i.e., i.e.
= 18,000
2
gmcm2.
We will have to guess that the rotor diameter is at most 1.5 cm, and its length perhaps 6 cm.
Assuming it is made of steel with a relative density of 7.8 its mass is approximately 85 gm. The
85  (0.75) 2
 24 gm cm 2 .
rotor inertia is therefore given by
2
We see that the load inertia is 750 times the rotor inertia, so in no sense is this a typical
application.
The acceleration at the beginning of the step will be only 1/750 of what it is for the unloaded
motor, and if this massive reduction applied throughout the whole step trajectory, the transit
time would therefore be increased by a factor of 750 , or approximately 27 times.
In practice however, the frictional damping in VR motor is low and therefore because of the
high inertia the damping factor will be very low and the pointer must be expected to oscillate
several times before coming to rest.
A viscous-coupled inertia damper will make matters much better, though the load inertia is so
high that it may be difficult to find an ‘off the shelf’ damper that is optimum.
- End of Solutions Chapter 9 -