Page 184 - Power Electronics Handbook

176 Static switches
The r.m.s. current rating of the thyristors depends on the conduction point
a and is given by equation (8.12) for TH3, TH, and by equation (8.13) for
THl, TH2, where R is the load resistance.

(8.12)

[:
4fr
-
z=(,,,,~) - - - {sin 2a - 2a) + 2x3 (8.13)
2R
The voltage rating of devices need only be (1-A)V since with one or
other arm always conducting the transformer tap voltage appears across
them.
This control method produces a lower distortion than traditional
phase-control techniques, the Fourier series of the output being given by
equation (8.14), the values Of Un and bn being given by equations (8.15) and
(8.16) respectively, for n equal to 3, 5, 7, and so on.

a0
VO = y + (uncosnot + b,sinnot) (8.14)
L
2V(1 - A)
an = [cos acosna + n sin a sinnar - 11 (8.15)
II (1 - n2)

2V(l - A)
b, = [cos a sin Ita - n sin a cos na] (8.16)
II (1 - n”)
From this equation the magnitude of any harmonic for a fixed value of a
and A can be found. Harmonics may be reduced by increasing the number
of tapping points at the expense of increased circuit complexity.
Fundamental coefficients are given by equations (8.17) and (8.18), the
amplitude of the fundamental being given by (8.19).

2V(1 + A) sin’a
a1 = (8.17)
2n
2V(1 - A)[(asin2a - cos’a) - sinacosa]
61 = (8.18)
2.n

[a: + b:l% (8.19)

The operation of the circuit shown in Figure 8.20 becomes slightly more
complicated on inductive load. For example, at r2 thyristor TH1 is
conducting. Even though the line voltage now reverses, this thyristor will
be kept on by the stored energy in the inductive load. To turn it off and so
limit the output voltage to AVsin0, thyristor TH3 is turned on. This means,
however, that provided TH3 is still on, thyristor TH2 cannot be fired at r3,
as would normally be done, or it would short-circuit through TH3.
Therefore although the control circuit for the thyristors is very similar to