Question

A single-phase semi-converter, connected to 230 V, 50 Hz is feeding a load of resistance R = 10 $\Omega$ in series with a large inductance that makes the load current ripple free. For a firing angle of ${45}^o$, the value of percentage rectification efficiency will be

• A. 80. 58 %
• B. 88.72 %
• C. 93.17 %
• D. 76.83 %

Answer 1

The correct option is A 80. 58 %

For semiconverter,
$V_0=\frac{V_m}{\pi }(1+cos\alpha )=\frac{230\sqrt{2}}{\pi }(1+cos{45}^o)=176.72V$

$I_0=\frac{V_0}{R}=\frac{176.72}{10}=17.672A$

$V_{or}=\frac{V_m}{\sqrt{2\pi }}{\left[\right(\pi -\alpha )+\frac{sin2\alpha }{2}]}^{1/2}$

$=\frac{230\sqrt{2}}{\sqrt{2\pi }}{\left[(\pi -\frac{\pi }{4}+\frac{sin90}{2})\right]}^{1/2}=219.30V$

$I_{or}=I_0=17.672A$
$P_{dc}=V_0{I}_0$
$P_{ac}=V_{or}{I}_{or}$
Rectification efficiency $=\frac{P_{dc}}{P_{ac}}=\frac{V_0{I}_0}{V_{or}{I}_{or}}=\frac{V_o}{V_{or}}=\frac{176.72}{219.30}\times 100=80.58\%$~