GATE 2025 EE Q.46: solution and answer

GATE 2025 · EEQ.46 · 2 marks · MCQ

The 3-phase modulating waveforms ( $v_{a} (t), v_{b} (t)$ and $v_{c} (t)$ ), used in sinusoidal PWM in a Voltage Source Inverter (VSI) are $v_{a} (t) = 0.8 sin (ω t)$ V, $v_{b} (t) = 0.8 sin (ω t - \frac{2 π}{3})$ V, $v_{c} (t) = 0.8 sin (ω t + \frac{2 π}{3})$ V, where $ω = 2 π \times 40$ rad/s is the fundamental frequency. The modulating waveforms are compared with a 10 kHz triangular carrier whose magnitude varies between +1 and -1. The VSI has a DC link voltage of 600 V and feeds a star connected motor. The per phase fundamental RMS motor voltage, in volts, is closest to

(A)
169.71
(B)
300.00
(C)
424.26
(D)
212.13

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Answer and solution

ANSWER

The per-phase fundamental RMS motor voltage is about 169.71 V, option (A).

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

The carrier swings between $- 1$ and $+ 1$ , so its peak is $1$ . The modulating peak is $0.8$ , so the modulation index is $m_{a} = \frac{0.8}{1} = 0.8$ . Since $m_{a} \leq 1$ , the inverter works in the linear region of sinusoidal PWM, with no over-modulation.
In the linear region, each pole voltage measured from the DC mid-point contains a fundamental whose peak is $m_{a} \frac{V _{d c}}{2}$ . This is because the leg voltage switches between $+ \frac{V _{d c}}{2}$ and $- \frac{V _{d c}}{2}$ , and its local average follows the modulating wave.
$V_{a o, 1 (peak)} = 0.8 \times \frac{600}{2} = 240 V$
The load is a balanced star. The three modulating waves have the same amplitude and are displaced by $12 0^{\circ}$ , so the three pole voltages have the same fundamental amplitude and sum to zero. The star point therefore sits at the mid-point potential for the fundamental, and the fundamental phase voltage equals the fundamental pole voltage. Any common-mode or triplen component appears equally on all three poles and does not drop across the motor phase. So the peak fundamental phase voltage is $240 V$ .
$V_{p h, 1 (rms)} = \frac{240}{2} \approx 169.71 V$ This is option (A).

KEY IDEA

In the linear region of sinusoidal PWM ( $m_{a} \leq 1$ ), the fundamental phase voltage of a balanced star load has peak $m_{a} V_{d c} /2$ . Divide by $2$ for the RMS value.

WATCH OUT FOR

Using $V_{d c}$ in place of $V_{d c} /2$ as the pole-voltage base gives a peak of $480$ V. Another slip is to convert the RMS value of $m_{a} V_{d c} /2$ as if it were already RMS, or to multiply by $3$ and quote the line voltage. Both lead to options (B), (C) or (D) instead of (A).

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 6 Inverters & PWM questions, with notes

GATE 2025 EE Q.46: solution and answer

Answer and solution

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