Chapter 8: Choppers
I
CHOPPER CONTROL OF D.C.
SERIES MOTOR
The development of choppers has made it possible to apply in practice an old
idea concerning the speed control of d.c. motors operated from a d.c. source.
The choppers are extensively used for speed control of d.c. motors in industrial
and traction drives. When the source voltage is d.c. since the series motors have
very high starting torques, they are best suited for traction drives. Figure 8.1
shows the basic chopper circuit for the control of a d.c. series motor. Here also,
one can employ fixed frequency TRC or variable frequency TRC. In the following
discussion, fixed frequency TRC is assumed. Figure 8.2 shows the voltage and
current waveforms for chopper control of a d.c. series motor.
Fig. 8.1 Chopper control of d.c. series motor
Fig. 8.2
Voltage and current waveforms
During the on-time of the chopper CH, the supply voltage Edc is applied to the
motor. During the off-period of the chopper, no voltage is applied to the motor.
Thus, the voltage is applied in pulses. As shown in Fig. 8.2, the current waveform
is not of rectangular type because of the presence of the inductor L. During the
off-period of the chopper, the freewheeling diode Df provides the conducting
path for the load current and hence, protect the motor from the voltage peaks,
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which may present due to inductive energy during the on-period. Here, chopper
frequency must be kept sufficiently high to avoid the discontinuous operation.
1
Circuit Analysis
The main problem in the analysis of a chopper-controlled series motor arises
due to the nonlinear relationship between the armature induced voltage and
armature current because of the saturation in the magnetization characteristic.
The analysis of the series motor for TRC is now carried out by making the
following assumptions:
(i) Thyristor T and diode Df are assumed to be ideal one.
(ii) It is also assumed that the motor’s armature and field inductance, and
armature and field resistance are constant.
The field inductance does vary considerably due to the saturation of the
magnetic circuit and eddy currents. However, the variation of the inductance affects the steady-state performance curves of the motor only by
a small amount.
(iii) The motor speed is assumed to be constant during one complete chopper cycle.
(a) During the on-period of the chopper, we can write
Edc = (La + L f)
dia
+ ia(R a + R f) + eb
dt
(8.1)
where L f and L a are the field and armature inductances, respectively.
R f and R a are the resistance of the field and armature, respectively,
eb is the back emf of the d.c. motor.
(b) During the off-period of the chopper, we can write
dia
+ ia(R a + R f) + eb
dt
Integrating Eq. (8.1) over one complete period,
0 = (L a + L F)
Eav =
1
T
Ton
Ú
0
(8.2)
dia
È
˘
Í( La + L f ) dt + ia ( Ra + R f ) + eb ˙ dt
Î
˚
where T = T on + T off, and Eav = average value of the armature voltage.
or
Eav =
1
T
Ton
Ú
0
dia
È
˘
Í( La + L f ) dt + ia ( Ra + R f ) + KNia ˙ dt
Î
˚
where K = motor constant, and N = speed of motor
or
Eav = Iav [R a + R f ] + Ebav
(8.3)
Now, the steady state torque,
Tav µ f ◊ Iav
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Choppers
where f = flux in the motor field and is proportional to the armature current in
series motor.
(8.5)
or
Tav = K Iav2 µ Iav2
Also, we have the relation, Eav = Edc
Ton
(Ton + Toff )
(8.6)
From Eqs (8.5) and (8.6) it becomes clear that the Eav and hence speed and
torque can be controlled by varying on-period of the chopper (T on).
Speed–torque characteristics of a series motor with constant frequency TRC
is shown in Fig. 8.3, for two different values of T on.
Fig. 8.3
Speed–torque characteristics
SOLVED EXAMPLES
Example 8.1 A d.c. series motor used for a rapid transit system is fed through
a d.c. chopper. The series motor has total circuit resistance of 3 W and inductance of 1.5
mH. What external inductance should be inserted in series with the armature circuit in
order to limit the per unit ripple in armature current to 10% for a duty-cycle ratio of 0.4.
The chopping period is 1000 ms.
Solution: From Eq. (8.37) of the book, per unit ripple in current is given by
=
(
)(
1 - e - a T / t 1 - e - (1 - a )T / t
I 0 max - I 0 min
=
Edc / R
1 - e- T / t
(
)
)
Given per unit ripple = 10% = 0.1
Let us define, T/t = x.
0.1 =
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(1 - e
)(1 - e ) (1 + e
(1 - e )(1 + e )
- 0.4 x
- 0.6 x
-x
-x
)
-x
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0.1 – 0.1 e–2x = 1 – e–0.6x , e–0.4x + 2e–x – e–1.6x – e–1.4x + 2e–x
– 2.303 + 0.2 x = 0.6 x + 0.4 x – 2 x + 1.6 x + 1.4 x – 2 x
2 x = 2.303, x = 1.152
1000 ¥ 10- 6
= 868.06 ¥ 10–6 s
1.152
Now
t = L/R a , \ L = t × R a = 868.06 ¥ 10–6 ¥ 3 L = 2.6 mH.
Therefore, external inductance that should be inserted for keeping the per unit
ripple within limits = 2.6 – 1.5 = 1.1 mH.
Now
x = T/t = 1.152 \t = T/x =
Example 8.2 The speed of the separately excited d.c. motor is controlled by a
chopper. The d.c. supply voltage is 100 V, armature circuit resistance is R a = 0.4 W,
armature circuit inductance is La = 10 mH and the motor constant is K a . f = 0.05 V/rpm.
The motor drives a constant torque load requiring an average armature current of 30 A.
Assume that motor current is continuous. Compute
(a) the range of speed control.
(b) range of duty cycle a.
Solution: The average value of armature voltage is given by
E a = R a Ia + Eb
Minimum speed is zero at which Eb = 0
\
E a = R a Ia = 30 ¥ 0.4 = 12 V
Also,
E 0 = a Edc , \ 12 = 100 a
\ a = 0.12
Maximum speed corresponds to a = 1, at which Ea = Edc = 100 V.
E b = Ea – Ia Ra = 100 – (30 ¥ 0.4) = 88 V
\
Now, average back (emf ) voltage,
Eb = Ka f N
\N=
Eb
88
=
= 1760 rpm.
k a f 0.05
\ The range of speed is 0 < N < 1760 rpm, and the range of duty cycle is 0.12 < a < 1.
Example 8.3 A constant frequency TRC system is used for the control of a d.c.
series traction motor from 230 V d.c. supply. The motor is having armature and field
resistance of 0.03 W and 0.02 W, respectively. The average current in the circuit is 80 A
and chopping period is 1000 ms. Calculate the pulse width if the average value of back
emf is 80 V.
Solution: We know the relation, Eqav = Iav (Ra + Rf) + Ebav. = 80 (0.03 + 0.02) + 80 = 84 V.
Ton
Edc = T on . Edc . f
T
1
84 = T on ¥ 230 ¥
1000 ¥ 10- 6
E0 =
Also,
II
\ T on = 0.365 ms.
CHOPPER FIRING CIRCUIT
From the previous sections, it has been observed that most of the d.c. choppers
consist of at least two thyristors, one main power carrying thyristor and one
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Choppers
auxiliary thyristor used to turn-off the main SCR. The time interval between the
triggering of the two SCRs determine the duty cycle and hence the output voltage.
A control voltage is used to control the duty cycle of the chopper.
A typical chopper firing circuit is shown in Fig. 8.5. This circuit is mainly
divided into following four parts:
(i) Triangular wave generator
(ii) Voltage comparator
(iii) Edge detector, and
(iv) Pulse amplifiers.
Figure 8.6 shows the waveforms observed at various parts of the circuit.
(i) Triangular Wave Generator As shown in Fig. 8.4, the three operational
amplifiers OP1, OP2, and OP3 together form a triangular wave generator that
generates the triangular wave eT shown in Fig. 8.5. The inverting pin of the
operational-amplifier OP2 is connected to the anode of diode D2, whose forward
bias voltage is 0.6 V. Now, as the voltage eT decreases below 0.6 V, the output of
operational amplifier OP2 changes from 13.5 V to –13.5 V, which, in turn triggers
OP3 to change state. The output of OP3 which is now –13.5 V (negative) makes
diode D1 forward-biased, and the 2.2 K path takes control of the integrator input
summing junction. The output of OP1 quickly rises to 13.5 V which, in turn
triggers OP2 and OP3 and changes their outputs to positive voltages.
Now, diode D1 is reverse biased and therefore, the feedback loop through D1
is also reverse biased and hence open. Control of integrator OP1 reverts to the
200 K path due to the reverse-biased diode D1 and therefore, the output voltage
eT has a constant slope that depends on the values of the capacitor C, the input
resistor R i, and the input voltage Ei. In fact, this oscillator can be used as a
voltage-controlled oscillator (VCO). Here, the time delay is introduced by the
OP2 so that there is enough time to charge up the capacitor C and therefore, eT
rises to 13.5 V. Diode D2 is used for the offset adjustment so that eT is always
above zero-voltage.
(ii) Voltage Comparator Here, the operational amplifier OP4 is used as the voltage
comparator. Ec is the control voltage. Now, if the control voltage Ec exceeds the
voltage eT , the output of OP4 changes, as shown by the waveform eP in Fig. 8.5.
As shown in Fig. 8.4, the two monostable multivibra-tors
are used (M V1 and M V2). These multivibrators are connected in such a way that
one of them is triggered by the rising edge and the other by the falling edge of the
signal ep. On receiving the rising or falling edge, the monostable multivibrator
produces output signals eQ and eR whose pulse widths can be adjusted. A pulse
width in the range of 20 to 200 ms is sufficient for triggering the SCRs. The pulse
eQ is used to trigger the main SCR, whereas pulse eR is used to trigger the
auxiliary SCR of the chopper circuit.
With reference to the waveforms of Fig. 8.29 of the book of current
commutated chopper circuit of Fig. 8.28 of the book, the pulse width of eR for
(iii) Edge Detection
the auxiliary SCR T 2 should be less than (t4 – t1). The pulse width of p LC =
t2 – t1 is satisfactory.
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&&
Fig. 8.4
Chopper firing circuit
Power Electronics
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Choppers
Fig. 8.5
Observed waveforms
(iv) Pulse Amplifier Stage In Fig. 8.4, the pulse amplifier circuit consists of
Darlington transistors Q1, Q2, Q3, Q4 and pulse transformers T 1 and T2. Now,
the firing pulses eQ and eR obtained from the edge detection circuit are given to
the pulse amplifier circuit. As shown in Fig. 8.4, the pulse eQ for the main SCR
is “ANDed” with a signal from an overcurrent protection logic circuit so that
whenever overcurrent condition occurs, the pulse eQ is blocked.
1
A.C. Chopper Firing Circuit
Figure 8.6 shows the firing circuit of an a.c. chopper. This firing circuit consists
of the following four main sections:
(a) Voltage comparator
(b) RC differentiating network
(c) Waveform generator (IC 555), and
(d) Pulse amplifier.
For achieving the symmetric output, the rectified sine-wave of the input is
given to the two comparators at point marked M. This signal obtained is compared
with a d.c. level which can be obtained by using a variable potentiometer.
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Fig. 8.6
Triggering-circuit for a.c. chopper
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Choppers
The output voltage of comparator is differentiated by R–C network. The
differentiated signal is now ‘ANDed’ with the signal produced by the waveform
generator IC 555. Now, the firing pulse obtained from the AND-gate circuit is
given to the pulse amplifier circuit.
As shown, the pulse amplifier circuit consists of a pulse transformer
connected directly to the collector of an output drive transistor. When the
transistor is switched into saturation, the voltage +12 V is applied across the
primary of the pulse transformer. This causes a corresponding voltage pulse to
appear across the secondary of the transformer which is connected to the gate
of the SCR to be fired.
When the transistor is switched OFF, the magnetizing current flowing in the
transformer primary is diverted into diode D, as a result, the voltage across the
winding reverses, and the flux in the transformer core is reset by this
“backswing” voltage, which corresponds to the forward voltage drop of
the diode.
Here, the main SCR is triggered first and then the auxiliary SCR. The turning
ON of the auxiliary SCR commutates the main SCR.
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