Discussion
A single-phase full bridge inverter controls the power in a resistive load. The nominal value of the input DC voltage is 600 V and a uniform pulse width modulation technique is used to control the output voltage, with 4 pulses per half cycle and duration of each pulse is \(25^\circ\). If the DC supply is decreased by 20%, determine the pulse width to be maintained for the same load power.
Answer: A
The load power for a resistive load depends on the square of the RMS voltage, and for uniform pulse width modulation \(V_{rms}^2 = V_{dc}^2\,p\,d/\pi\). To keep the same power with the same number of pulses, \(V_{dc}^2 d\) must be constant.
\[ V_{dc}' = 0.8 \times 600 = 480\ \text{V} \]
\[ \begin{aligned} 600^2 \times 25^\circ \\ &= 480^2 \times d' \Rightarrow d' \\ &= 25^\circ \times \left(\frac{600}{480}\right)^2 \\ &= 25 \times 1.5625 = 39.06^\circ \end{aligned} \]
So the pulse width must be increased to about \(39^\circ\) to make up for the lower supply. The smaller angles of 35, 32 and 30 degrees would reduce the output further. The answer is option A.
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