Question

The single phase full bridge VSI has an output frequency of 50 Hz is connected to R-load, $R=5\Omega$ through LC filter with value L = 10 mH and $C=63.5\mu F$. If unipolar PWM switching is used with switching frequency of 20 kHz and modulation index of 0.7 and input voltage, $V_{in}=160$ dc, then amplitude of fundamental component in the output voltage in steady state will be ____V.

Answer 1

The rms value of the fundamental component of output voltage of VSI

$V_0=\frac{160}{\sqrt{2}}\times 0.7=79.195V$

Amplitude of peak value of VSI output,

$V_{OP}=\sqrt{2}\times 79.195=112V$

$V_{Rpeak}$ is amplitude of fundamental voltage at the resistor.

$Z=\frac{R(-j{X}_C)}{R-j{X}_C}$

Where, $X_C=\frac{1}{\omega C}=\frac{1}{2\pi \times 50\times 63.5\times {10}^6}=50.127\Omega$

$X_L=\omega L=2\pi \times 50\times 10\times {10}^{-3}=3.141\Omega$

$Z=\frac{5\times (-j50.127)}{5-j50.127}=4.975\angle -5.696\Omega$

$V_{R-peak}=V_{0peak}\times \frac{4.975\angle -{5.696}^{\circ }}{4.975\angle -{5.696}^{\circ }+j3.141}=112\times 0.8862\angle -{34.10}^{\circ }$

$=99.2544\angle -{34.10}^{\circ }V$

The amplitude of fundamental component in output voltage,

$V_0=99.25V$