Question

A single phase full bridge inverter is supplying power to RLC load with R = 5 $\Omega$, $X_L=10\Omega$. The bridge operates with a periodicity of 0.15 ms, the circuit turn off time required for transistor is 20 $\mu$s. The suitable value of capacitance 'C' that results in load commutation is _____ $\mu$F. (Up to two decimal places)

Answer 1

For proper load commutation
$\theta =\omega t_c={\tan }^{-1}\left(\frac{X_C-X_L}{R}\right)$

or $T=0.15ms$

$\omega =\frac{2\pi }{T}=\frac{2\pi }{0.15\times {10}^{-3}}rad$

$\omega t_c=\frac{2\pi }{0.15\times {10}^{-3}}\times 20\times {10}^{-6}rad=48{}^{\circ }$

${\tan }^{-1}\left(\frac{X_C-10}{5}\right)=48{}^{\circ }$

or $X_C=5\tan 48{}^{\circ }+10=15.55\Omega$

$X_C=\frac{1}{\omega C}$

$C=\frac{1}{\omega X_C}=\frac{0.15\times {10}^{-3}}{15.55\times 2\pi }=1.54\times {10}^{-6}$

$C=1.54\mu F$