Page 428 - Handbook of Electrical Engineering

HARMONIC VOLTAGES AND CURRENTS 417

Therefore the commutating reactance = 2.0 × 0.245 × 0.4306 = 0.211 ohms/phase = X c
3X c I d
V d = V do cos α −
π
3 × 0.211 × 425.0
262.3 = 560.45 cos α −
π
= 560.45 cos α − 43.685

Therefore
262.3 + 85.623
cos α = = 0.6208
560.45
α = 51.63 ◦

Also

V do
V d = (cos α + cos(α + u))
2
560.45
262.3 = (0.6208 + cos(51.63 + u))
2
cos(51.63 + u) = 0.3152
51.63 + u = 71.626

u = 20 ◦

The resulting waveform is shown in Figure 15.4.

15.3.4 Simplified Waveform of a 12-pulse Bridge

The six-pulse rectifier bridges can be connected in such a manner as to produce a 12-pulse DC output
voltage. The average value of DC ripple voltage is thereby reduced. From the AC power system point
of view the magnitude of the harmonic components is reduced and some harmonics are eliminated.
Figure 15.6 shows a typical circuit of a 12-pulse bridge.

◦
The upper bridge is fed by a Dyll delta-star transformer T u which has a 30 phase shift
between the primary and secondary line currents. The lower transformer T l has zero phase shift. See
sub-section 6.4 for an explanation of phase shifts in transformer windings.
The primary currents for transformer T u are added as follows,

I 12 = I 1 − I 2
I 23 = I 2 − I 3

I 31 = I 3 − I 1
Where, I 12 etc. can be either the rms values or the instantaneous values, but displaced by their
◦
◦
◦
appropriate phase angles, i.e. 0 , −120 and −240 .