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## Q. 5.9

A three-phase bridge fully controlled rectifier is fed from 415 V 50 Hz supply through line inductors of 0.3 mH per phase that have negligible resistance. The rectifier feeds a load through a smoothing inductor that has negligible resistance. At a particular operating condition the rectifier firing angle is $45^{\circ}$ and the load current is pure dc of 300 A. Assuming that the load voltage is smooth, calculate the following:
a) The rectifier overlap angle.
b) The average output voltage.

## Verified Solution

Solution

a) The overlap angle m can be calculated using again Eq. (5.105) as follows:

$\cos \alpha-\cos (\alpha+\mu)=\frac{2 \omega L _{ S } \overline{ I }_{ o }}{\sqrt{6} \widetilde{ V }_{ i }}$ (5.105)

\begin{aligned}\cos \alpha-\cos (\alpha+\mu) &=\frac{\overline{ I }_{ o }\left(2 \omega L _{ s }\right)}{\sqrt{6} V _{ i }} \\\text { or } \quad \cos (\alpha+\mu) &=\cos \alpha-\frac{\overline{ I }_{ o }\left(2 \omega L _{ s }\right)}{\sqrt{6} \widetilde{ V }_{ i }}=\cos 45^{\circ}-\frac{(300)\left(2 \cdot 2 \pi \cdot 50 \cdot 0.3 \cdot 10^{-3}\right)}{\sqrt{6} \frac{415}{\sqrt{3}}} \\&=0.707-\left(\frac{56.52}{587}\right)=0.61\end{aligned}

$\text { or } \quad \alpha+\mu=\cos ^{-1}(0.61)$

b) Using Eqs. (5.62) and (5.106), the average output voltage is:

$\bar{V}_{o(\text { loss })}=3 \frac{\sqrt{6} \tilde{V}_{i}}{2 \pi}[\cos \alpha-\cos (\alpha+\mu)]$ (5.106)

\begin{aligned}\overline{ V }_{ o } &=\frac{1}{ T } \int_{0}^{ T } v _{ o }( t ) d t =\frac{\text { Area } A }{\pi / 3}=\frac{1}{\pi / 3} \int_{-\frac{\pi}{6}+\alpha}^{\frac{\pi}{6}+\alpha} v _{ ab } d (\omega t ) \\&=\frac{1}{\pi / 3} \int_{-\frac{\pi}{6}+\alpha}^{\frac{\pi}{6}+\alpha} \sqrt{6} \widetilde{ V }_{ i } \cos (\omega t ) d (\omega t )=\frac{3 \sqrt{6}}{\pi} \widetilde{ V }_{ i } \cos \alpha=2.34 \widetilde{ V }_{ i } \cos \alpha\end{aligned} (5.62)

\begin{aligned}\overline{ V }_{ o (\text { net })} &=\text { average output voltage }-\text { voltage loss due to the } 6 \text { overlaps } \\&=\frac{3 \sqrt{6}}{\pi} \widetilde{ V }_{ i } \cos \alpha-\frac{3 \sqrt{6} \tilde{ V }_{ i }}{2 \pi}[\cos \alpha-\cos (\alpha+\mu)] \\&=\left(\frac{3 \sqrt{6}}{\pi}\right)\left(\frac{415}{\sqrt{3}}\right) \cos 45^{\circ}-\left(\frac{3 \sqrt{6}}{2 \pi}\right)\left(\frac{415}{\sqrt{3}}\right)\left[\cos 45^{\circ}-\cos \left(45^{\circ}+7.4^{\circ}\right)\right] \\&=396.5-27.16=369.34 V\end{aligned}