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Electric Motors and Drives - Third Edition
Solutions to Review Questions - Chapter 4
1)
Under no-load conditions the speed of a d.c. motor is almost exactly proportional to the
armature voltage, so when the speed reference is increased from 50% to 100% the armature
voltage will double. Assuming the control scheme is good, the actual speed should precisely
track the reference, so the new tacho voltage will be exactly twice what it was.
To compare the armature currents we would need to know how the friction torque varied with
speed. As we have no information all we can say is that if the friction torque was independent
of speed (a reasonable assumption for a separately-ventilated motor of the type usually
employed), the no-load current will be independent of speed. Ideally the no-load current should
be zero, and in practice it is seldom more that a percent or two of full-load current, except
perhaps in very small machines.
2)
With a PI speed controller there will be no steady-state error in the speed, so after the transient
has settled the on-load speed will be exactly 50%: the tacho voltage will therefore be the same
as before the load was applied.
The torque of a d.c. motor is proportional to the armature current, so a load torque of 100%
means that the armature current must be at its rated value.
If we denote the induced e.m.f. in the motor before the load was applied by E1, the
corresponding armature voltage is given by V1  E1  I1R , where I1 is the no-load current.
When the motor is loaded, the control system will adjust the armature voltage to achieve the
same speed, so the induced e.m.f will be exactly the same and the on-load voltage will be given
by V2  E1  I 2 R , where I2 is the full-load current. The increase in armature voltage is thus
V2  V1  ( I 2  I1 ) R. In practice, because the armature resistance is small, the increase in
voltage from no-load to full-load will also be only a few percent of rated voltage.
3)
At first there will be a 100% speed error, so the speed controller output will saturate and
demand full (rated) current to provide maximum acceleration. The current will be maintained
at 100% during acceleration, so the motor torque will be constant and with negligible friction
torque the acceleration will be constant so the speed will increase at a uniform rate.
The output from the speed error amplifier will remain constant, and the current will therefore
remain at full value, until the speed rises to within a few percent of the target. At his stage, the
speed controller comes out of saturation and enters its linear regime, the demanded current (and
the torque) reducing as the final speed is approached smoothly.
While the acceleration is constant the speed increases linearly, and so therefore does the
induced e.m.f, E. The armature voltage is given by V  E  IR , so since I is held constant
during most of the run-up phase, the armature voltage also increases linearly with time, as
shown in Figure 4A.
Fig 4 A here.
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4)
a) The drive will react to the drop in voltage (which will lead to a drop in armature voltage,
torque, and speed) by reducing the firing angle a little so that the armature voltage is returned to
its correct level to return the speed returns to target.
A well-engineered drive will include provision for variation in the supply voltage of at least
10%, by arranging that under normal conditions the full-load armature voltage is obtained with
a converter firing angle of say 15 to 20. In this way, if the mains voltage should fall, the firing
angle can be reduced towards zero in order to maintain the output voltage.
b) If the tacho feedback disappears, the drive will act as if the speed was zero, so with a target
speed of 50% the large speed error would cause the speed controller output to saturate and
demand full current. The motor will then accelerate at full current (see answer to Q3 above)
until the firing angle of the converter has been reduced to zero and the armature voltage is at its
maximum possible value. The motor will then run at somewhat above base speed, drawing a
small no-load current.
If the drive includes tacho loss detection circuitry it may shut-down automatically, or switchover to armature voltage feedback.
c) If the motor was stopped by some mechanical means, there would be a large speed error and
the speed controller output would saturate and demand full current. the current controller
would reduce the output voltage of the converter to a very low level because there would be no
back e.m.f. with the motor stalled. A sophisticated drive would recognise that full current and
no motion indicates trouble, and time-out after a few seconds at most.
d) This would be much the same as (c), except that the output voltage from the converter
would be even lower, depending on the resistance of the ‘short-circuit’. (A large stationary d.c.
motor is almost a short-circuit anyway!)
e) This is very serious, since a principal function of the inner current-feedback loop is to
protect the thyristors from the danger of excess current. With no current feedback, the current
controller will sense a large error, and immediately increase the output voltage from the
converter in an attempt to raise the current. Given the very delicate balance that has to be
maintained between V and E to avoid excessive currents, the current will inevitably shoot up
sufficiently rapidly to blow the expensive fuses that are the last line of defence, but in all
probability some of the thyristors will be lost.
5)
When the armature current is discontinuous, the torque-speed curve for a given converter firing
angle is very poor: a modest increase in torque causes a very large drop in speed. This occurs
because when the current is discontinuous, the output voltage of the converter falls substantially
as the current (i.e. the load) is increased. When the load increases and the current becomes
continuous, the output voltage from the converter is almost independent of the load, and the
speed therefore remains almost constant over a wide rang of load.
Although the undesirable effects of discontinuous current can a largely be masked by the
operation of the closed-loop speed control system, it is more difficult to optimise the control
system (especially the transient response) when the inherent behaviour of the motor itself varies
so markedly according to the load.
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6)
In dynamic braking mechanical energy is converted to electrical form and then dissipated in
resistors. Regenerative braking involves converting mechanical energy into electrical energy
and returning it to the supply system.
7)
The output voltage waveform from a thyristor converter consist of rectified chunks of the
incoming a.c. mains supply, as shown for example in Figure 4B(a). Ideally, since this is to be
the armature voltage for a d.c. motor, we would like pure d.c.(shown by the dotted lines) but as
we can see the actual waveform contains a lot of ‘a.c’ as well.
Fig 4B here.
The corresponding armature current waveforms are shown in Figure 4B(b). The effect of the
armature inductance is to make the current waveform a great deal smoother than the voltage
waveforms, which is very desirable because the torque is proportional to the current, and we
want to minimise torque pulsations. The higher the inductance, the smoother the current, as
shown in Figure 4B. (Those who are familiar with a.c. circuit theory will explain the
smoothing effect of the armature inductance in terms of inductive reactance L. The ‘a.c’
component of the voltage contains a series of harmonic terms (the lowest of which is 100 Hz
when the supply is 50 Hz). The reactance is proportional to frequency, so higher-frequency
voltage harmonics produce very little harmonic current, so the current waveform looks much
smoother than the voltage waveform.)
The disadvantage of a high armature inductance is that to cause the current in an inductance L
to change by an amount I requires a voltage V for a time t such that Vt = LI.
In drives, we usually want the transient response of the inner (current-control) loop to be as fast
as possible, which means that we apply the highest available voltage in order to maximise the
rate of change of current and minimise the time taken to achieve a given change in current.
From the equation above we can see that the higher the inductance, the longer we have to apply
the voltage, so in terms of transient response, the lower the inductance the better.
8)
The axes are as shown in Figure 4C, which is the same as Figure Q8 but rotated through 90°.
(In Figure Q8 the speed axis is vertical, which seems to be the preference of mechanical
engineers.)
Fig 4C here.
The good and bad sections are labelled in Figure 4C: a good characteristic has only a small
change in speed with load, and a bad characteristic is one in which the speed falls significantly
when load is increased.
The abrupt change in character occurs at the point where the armature current changes from
discontinuous to continuous. When current is discontinuous the converter output voltage
depends on the current: the higher the current (i.e. the higher the torque) the lower the voltage,
and hence the lower the speed. When the average current has become large enough that the
current is continuous (i.e. it never falls to zero), the output voltage of the converter no longer
depends on the current, and the speed is therefore almost constant regardless of the load.
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We are told that the converter is fully-controlled, so the mean d.c. voltage with  = 5° is given
by Vdc  Vdo cos 5  1. There is no scale on Figure Q8, but we note that in the continuous
current region, curve B corresponds to roughly half of the speed of curve A. The mean d.c.
voltage for curve B must therefore be half of that for A, so if the firing angle is  B , then
cos  B  0.5, hence  B = 60°.
Adding additional armature circuit inductance makes the armature current smoother, (see Figure
4B), and therefore reduces the likelihood of discontinuous current. The effect on the torquespeed characteristic is to extend the straight portion as shown by the dotted lines.
9)
a) Since both motors are coupled to the same shaft, their speeds will always be the same, so to
share the mechanical work we need to arrange that the torque provided by each motor is
proportional to its power rating: when the original motor is at half rated torque, we want the new
one to be at half rated torque, and so on.
The 150 kW drive has been chosen as the master, i.e. it has its outer (speed control) loop
operational, as well as its inner current-control loop. The 100 kW drive only has its inner loop
operational: its current reference is derived from that of the master.
b) The current reference signal is fed to both drives so that when, for example, 50% of rated
current is demanded in the master, 50% of the (different) rated current is also demanded from the
slave. This can usually be arranged quite easily by fitting appropriate connecting links between the
control boards of the two drives.
c) It would not be a good idea to ask both drives to operate in the speed control mode, as unless
they were precisely matched, there would be a tendency for them to end up fighting one another.
d) It could be argued that by making the slave machine current track the actual current in the
master, rather than the current reference of the master, the slave current would more faithfully
follow the master current. On the other hand, this would mean that if the master current control
went astray, so would the slave. So on balance, it’s probably not a good idea.
- End of Solutions for Chapter 4 -