Question

A 3-phase fully controlled bridge converter with free wheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 60°. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be

• A.
  IDF = 0.5; IPF = 0.478
• B.
  IDF = 0.867; IPF = 0.828
• C.
  IDF = 0.867; IPF = 0.552
• D.
  IDF = 0.5; IPF = 0.318

Answer 1

The correct option is A

IDF = 0.5; IPF = 0.478
Load current is constant
$I_0$ = 10 A

In $3-\varphi$ full converter with free wheeling diode,
Input displacement factor (IDF)

$=\cos \alpha =\cos 60{}^{\circ }=0.5$
RMS value of source current

$I_s=\sqrt{I_0^2\times \frac{2\pi }{3}\times \frac{1}{\pi }}=I_0\sqrt{\frac{2}{3}}$
RMS value of nth harmonic

$I_{sh}=\frac{4I_0}{\sqrt{2}n\pi }\sin \frac{n\pi }{3}$
RMS value of fundamental current

$I_{s1}=\frac{2\sqrt{2}}{\pi }{I}_0\sin 60{}^{\circ }=\frac{\sqrt{6}}{\pi }{I}_0$

Current distortion factor (CDF)

$=\frac{I_{s1}}{Is}=\frac{\sqrt{6}}{\pi }{I}_0\times \left(\frac{1}{I_o}\sqrt{\frac{3}{2}}\right)$

$=\frac{3}{\pi }=0.955$
Input power factor (IPF)
= CDF × IDF
$=0.955\times 0.5\approx 0.478$