GATE 2022 EE Q.60: solution and answer

GATE 2022 · EEQ.60 · 2 marks · NAT

A 3-phase grid-connected voltage source converter with DC link voltage of 1000 V is switched using sinusoidal Pulse Width Modulation (PWM) technique. If the grid phase current is 10 A and the 3-phase complex power supplied by the converter is given by (−4000 − j3000) VA, then the modulation index used in sinusoidal PWM is ___________. (round off to two decimal places)

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Answer and solution

ANSWER

The modulation index is $m \approx 0.47$ (exactly $0.4714$ ).

Official answer: 0.46 to 0.48Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

Magnitude of the three-phase complex power: $∣ S ∣ = (- 4000)^{2} + (- 3000)^{2} = 5000 VA$ . The negative signs only say which way the active and reactive power flow, so they do not change $∣ S ∣$ .
An AC current given without a qualifier, such as 'the grid phase current is 10 A', is an rms value. So $I_{p h} = 10 A (rms)$ .
For a balanced three-phase system, $∣ S ∣ = 3 V_{p h} I_{p h}$ , with both values rms. The rms phase voltage at the converter terminals is therefore $V_{p h} = \frac{∣ S ∣}{3 I _{p h}} = \frac{5000}{3 \times 10} = 166.67 V (rms)$
Peak value of the fundamental phase voltage: $V_{p h, m} = 2 \times 166.67 = 235.70 V$ .
In sinusoidal PWM the modulation index is $m = \frac{V _{control, peak}}{V _{carrier, peak}}$ . In the linear range ( $m \leq 1$ ), each pole voltage measured from the DC-link midpoint has a fundamental of peak $m \frac{V _{d c}}{2}$ . For a balanced load this is also the peak of the fundamental phase voltage: $V_{p h, m} = m \frac{V _{d c}}{2}$
Modulation index: $m = \frac{2 V _{p h, m}}{V _{d c}} = \frac{2 \times 235.70}{1000} = 0.4714 \approx 0.47$
Check using line voltage: $V_{L} = 3 \times 166.67 = 288.68 V (rms)$ . In linear SPWM, $V_{L} = \frac{3}{2 2} m V_{d c} = 0.6124 m \times 1000$ , which gives $m = 0.4714$ . This lies inside the key range 0.46 to 0.48.

KEY IDEA

From $∣ S ∣ = 3 V_{p h} I_{p h}$ (rms values), find the rms phase voltage. Convert it to a peak value. Then use the linear-range sinusoidal PWM relation $V_{p h, m} = m V_{d c} /2$ to get $m = 22 V_{p h} / V_{d c}$ .

WATCH OUT FOR

Reading the 10 A as a peak value gives $V_{m} = 333.3 V$ and $m = 0.67$ , which is wrong. The other common slip is putting the rms phase voltage straight into $m = 2 V / V_{d c}$ without multiplying by $2$ , which gives $m = 0.33$ . Grid currents and voltages are rms unless the question says otherwise. The SPWM relation $V_{p h, m} = m V_{d c} /2$ uses the peak value.

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GATE 2022 EE Q.60: solution and answer

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