Discussion
A single-phase full bridge inverter controls the power in a resistive load. The nominal value of the input DC voltage is 400 V and a uniform pulse width modulation with five pulses per half cycle is used. For the required control, the width of each pulse is \(32^\circ\). What is the RMS voltage of the load?
Answer: C
For uniform pulse width modulation with \(p\) pulses per half cycle, each of width \(d\) (in radians), the RMS output of a full bridge inverter is
\[ V_{rms} = V_{dc}\sqrt{\frac{p\,d}{\pi}} \]
Here \(p = 5\), \(d = 32^\circ = 0.5585\) rad and \(V_{dc} = 400\) V.
\[ \begin{aligned} V_{rms} = 400\sqrt{\frac{5 \times 0.5585}{\pi}} \\ &= 400\sqrt{0.8889} = 400 \times 0.9428 \\ &= 377.1\ \text{V} \end{aligned} \]
The output must be smaller than the 400 V input but not far below it, since 5 pulses of 32 degrees cover 160 of the 180 degrees of the half cycle. That excludes the values of 37.71 V, 3.771 V and 0.3771 V. The answer is 377.1 V, option C.
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