Page 156 - Power Electronics Handbook

A.C. contactors 149

non-conducting the load is effectively isolated from the supply lines A and
B, if the leakage of the devices is ignored. When thyristors 1 and 2 are fired
they can be considered as short circuits and the load is connected to the
supply.
Figure 7.l(b) shows a three-phase static contactor. Only one thyristor
and a reverse diode is required in each line, since one of the lines must
conduct in the forward direction before any load current can flow. As in
single-phase contactors, with no thyristor gate drive the load is isolated or
off, and when the thyristors are turned on the load is connected to the
supply.
If it is required to reverse the supply to the load, for instance where the
load is a motor and its direction of rotation is to be changed, the circuit of
Figure 7.l(c) must be used. With thyristors 1 to 6 conducting and thyristors
7 to 10 off, line A feeds E, B feeds F and C feeds G. To reverse the supply
to the load, thyristors 1,2 and 7 to 10 are on and thyristors 3 to 6 are off.
Now A feeds E, B feeds G and C feeds F. It is necessary to use
back-to-back thyristors, or triacs, in all the lines since if, for instance, 10
was a diode then firing thyristor 3 would cause a short circuit across lines B
and C.
The devices in Figure 7.1 must all be rated to withstand the peak line
voltage. For 240 V single-phase and 440 V three-phase supplies this means
peak voltages of 34OV and 63OV respectively. To allow for line transients
it is usual to employ MOV and 8OOV devices. Additionally, surge
suppressors must be used between lines and across devices if the transients
are troublesome. In this context it is perhaps worth noting that the circuit
of Figure 7.l(b) is protected against reverse voltage transients by the
diodes. Overvoltage in the forward direction could cause a thyristor to
break over into conduction for a half cycle, but now it need not be
damaged and the load may be such as not to be appreciably affected by the
half cycle of power.
The current rating of the devices can be calculated as follows, with
reference to the single-phase circuit. Assuming the line current to the load
to be given by Z,,&ne when the thyristors are conducting, then the r.m.s.
line current is given by equation (7.1).

IPk
=-
J2
The thyristors carry current for only half a cycle each, so their mean
rating is given by IT(.v, in equation (7.2), and this can be simplified to
equation (7.3) by using equation (7.1).

1Pk
=- (7.2)
n