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CYCLOCONVERTER CONTROL
Cycloconverter allows variable frequency and variable voltage supply to be obtained from a
fixed voltage and frequency ac supply. A half-wave cycloconverter is shown in Fig. 6.42 along
with the nature of its output voltage waveform. Because of low harmonic content when operating
at low frequencies, smooth motion is obtained at low speeds. Harmonic content increases with
frequency, making it necessary to limit the maximum output frequency to 40% of the source
frequency. Thus, maximum speed is restricted to 40% of synchronous speed at the mains trequency.
A motor with large leakage inductance is used in order to minimize derating and torque pulsations
due to harmonics in motor current. The drives has regenerative braking capability. Full four-
quadrant operation is obtained by reversing the phase sequence of motor terminal voltage. Since
cycloconverter employs large number of thyristors, it becomes economically acceptable only in
large power drives.
Induction
motor
cycloconverter-fed induction motor
(a) Half wave
Fundamental
Phase voltage component
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Phase voltage waveform
(b)
motor drive
Fig. 6.42 Cycloconverter controlled induction
Cycloconverter drive has applications in high power drives requiring good dynamic response
The low speed operation is
Dut only low speed operation e.g. in ball millnumbers from plant.
in a cement
a cycloconverter operating at low
Obtained by feeding a motor with large pole
unlike conventional drives, the low
requencies. These drives are called gearless drives because,
SDeed operation of load is obtained without a reduction gear, thus eliminating the associated
maintenance.
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EXAMPLE 6.8
A 2.8 kW, 400 V, 50 Hz, 4 pole, 1370 rpm, delta connected squirrel-cage induction motor has
tollowing parameters referred to the stator: R, = 2 2, R = 5 2. X, = X; = 5 2, Xm =80Se.
Motor speed is controlled by stator voltage control. When driving a fan load it runs at rated peed
at rated and
voltage. Calculate (i) motor terminal voltage, current and torque at 1200 TP
(11) motor speed, current and
torque for the terminal voltage of 300 V.
Solution T vRs
(ms
R,+ (X, +x{)*
Synchronous speed = =x S0 = 1500 rpm = 50 T rad/sec
P 4
At full load
. 1500-1370 0.0867
1500
4002 x 5/0.0867
At full load T= x = 48.13 N-m
50T 2
+(5 5)2
2 0.0867 +
For a fan load torque is proportional to (speed).
Thus TL K(1 - s)2
At full load T= TL
K(1 - 0.0867) = 48.13
K = 57.7
Hence
TL 57.7(1 - s) ()
(i) At 1200 rpm
s = 15001200 = 0.2
1500
At this speed from Eq. (i)
TL 57.7(1 - 0.2)? = 36.9 N-m
Since T= TL. T= 36.9 N-m
350T x
y2x5/0.2 = 36.9
Now
(22 +(10)2
which gives V = 253.2 V
253.2 = 8.246 j3.054
i=7 5
2532
8.246 -j3.054-j3.165
=
10.328 2 -
37
I, 1} + Im
=
=
10.328 17.89 A
Line currept = 3 x
=
i) At 300v
3002 x 5/s 27
x 104s (ii)
3 107t(104s +20s+ 25)
T 507 (+2+(10)
In steady state T = T. Therefore, from Eqs. (i) and (1)
27x 10s = 57.7(1 - s)"
10r(104s+20s + 25)
or 104 188s+ 89s2- 179s+ 25 =0
which gives s = 0.147.
Hence torque produced by the motor
T= 57.7 (1 - 0.147) = 41.94 N-m
Speed = N, (1 - s) = 1500 (1 - 0.147) = 1279 rpm
300 300
5
=
9.75 2-37.3
2+0147+j10
Line current = 3 x 9.75 = 16.88 A
ExAMPLE 6.15
motor has following ratings and parameters:
A Y-connected squirrel-cage induction
3 2, X, X, = 3.5 2, X,, =
5
400 V, 50 Hz, 4-pole, 1370 rpm, R,
= 2 2, R = =
inverter at a constant flux. Calculate
t is controlled by a current source
when operating at 30 Hz and rated slip sDeed
() Motor torque, speed and stator current
rated motor torque and motor speed of 1200 mm
(11) Inverter frequency and stator current for
the region of interest
be parallel straight lines in est,
Assuming motor speed torque curves to
when operating at
calculate motor speed
(iii) 30 Hz and half the rated motor torque.
(iv) 45 Hz and braking torque equal to rated motor torque.
Solution
1500 rpm or 507 rad/sec
Synchronous speed =
Full load slip se = S70= 0.0867
1500
Full load slip speed = 1500 1370 = 130 rpm
From Fig. E.6.15 motor impedance
j55 0.0867 +3.5
Z 2 +j3.5 +3
0.0867
+j(S5 + 3.5)
24.65+ j20.19 = 31.86< 39.3° Q
R jX XR/s
wo00TO00
Fig. E.6.15
Full load stator current
I 400/3 7.2486 A
31.86 =
Full load rotor current
jXm
L(R/s) +j(X; + X)=7.2486
j55 =5.865 A
L o.0867j(58.5)
Full load torque
7 ms R/sl =x(5.865)x a0R722.73
0.0867 22.73 N-m
Nmx
) According to Example 6.14 at rated slip speed, torque and I, will have same values as at
50 Hz operation. Thus
T 22.73, 1, = 7.2486 A
Now at 30 Hz synchronous speed = 30
0 x 1500 =900 rpm
Full load slip speed = 130 rpm
Motor speed = 900 - 130 = 770 rpm
(i) At rated motor torque, slip speed and I, will be same as at 50 Hz operation.
Therefore
I, 7.2486, slip specd = 130 rpm
Synchronous speed = 1200 + 130 = 1330 rpm
1330x
Frequency = 1500 50 = 44.33 Hz
(ii) When speed-torque curves are assumed to be straight lines,
Slip speed at half the rated torque = = 65 rpm
At 30 Hz, synchronous speed = 900 rpm
Motor speed = 900 - 65 = 835 rpm
(iv) At rated braking torque, slip speed = - 130 rpm
Synchronous speed at 45 Hz = x 1500 = 1350 rpm
Motor speed = 1350 + 130 = 1480 rpm
