Home › Previous year questions › GATE 2017 EE (EE2) › Question 49GATE 2017 EE (EE2) – Question 49 Power Electronics · Single-phase and three-phase voltage and current source inverters, sinusoidal pulse width modulation · 2 marks · Multiple choice
A half-bridge voltage-source inverter feeds an RL load with R=40 Ω and $L=0.3/\pi$ H. Sinusoidal PWM has modulation index 0.6 and fundamental frequency 50 Hz. Fundamental active load power is 1.44 kW. Each split DC source is labeled $V_{DC}$ as shown. Its voltage is
$300\sqrt2$ V 500 V $500\sqrt2$ V $1000\sqrt2$ V Practise this question in The GATE Grind →
Show answer and explanation Correct answer: (C) $500\sqrt2$ V
Explanation Fundamental current rms is $I_1=\sqrt{P/R}=\sqrt{1440/40}=6$ A. At 50 Hz, XL=2π50(0.3/π)=30 Ω, so |Z|=50 Ω and fundamental output rms is 6×50=300 V.
The fundamental peak is 300√2 V. In this diagram the rails are ±VDC, so sinusoidal PWM gives fundamental peak M VDC, not M VDC/2.
Thus $V_{DC}=300\sqrt2/0.6=$ **500√2 V, C**. The total rail-to-rail voltage is twice the labeled individual source.
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