GATE 2020 EE Q.51: solution and answer

GATE 2020 · EEQ.51 · 2 marks · NAT

A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is ______.

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Answer and solution

ANSWER

The commutation overlap angle is $μ \approx 4.8 0^{\circ}$ .

Official answer: 4.51 to 5.10Source: Official answer key, GATE 2020 EE (IIT Delhi)

Given: $V_{s} = 230 V$ rms, so $V_{m} = 2302 = 325.27 V$ ; $R = 10 Ω$ ; $ω L_{s} = 2 π (50) (2.28 \times 1 0^{- 3}) = 0.7163 Ω$ ; firing angle $α = 4 5^{\circ}$ . The load inductance is very large, so the load current $I_{d c}$ is constant (ripple-free).
With source inductance, the average output voltage of a single-phase fully controlled bridge is $V_{d c} = \frac{2 V _{m}}{π} cos α - \frac{2 ω L _{s}}{π} I_{d c} .$ The second term is the drop caused by commutation overlap.
Ideal term: $\frac{2 V _{m}}{π} cos 4 5^{\circ} = 207.07 \times 0.7071 = 146.42 V$ . Drop coefficient: $\frac{2 ω L _{s}}{π} = 0.4560 Ω$ .
The load is $R$ in series with a very large inductance, so $I_{d c} = V_{d c} / R = V_{d c} /10$ . Then $V_{d c} = 146.42 - 0.0456 V_{d c}$ , so $V_{d c} = \frac{146.42}{1.0456} = 140.04 V, I_{d c} = 14.004 A .$
During overlap the output voltage is the average of the two source voltages. Over a full half-cycle this gives $V_{d c} = \frac{V _{m}}{π} [cos α + cos (α + μ)]$ . Equating this with the expression in step 2 gives $cos α - cos (α + μ) = \frac{2 ω L _{s} I _{d c}}{V _{m}} .$
Right side: $\frac{2 ( 0.7163 ) ( 14.004 )}{325.27} = 0.06168$ .
$cos (α + μ) = 0.70711 - 0.06168 = 0.64543$ , so $α + μ = cos^{- 1} (0.64543) = 49.8 0^{\circ}$ .
Overlap angle: $μ = 49.8 0^{\circ} - 4 5^{\circ} \approx 4.8 0^{\circ}$ .

KEY IDEA

With source inductance and a constant load current, the overlap angle follows from $cos α - cos (α + μ) = 2 ω L_{s} I_{d c} / V_{m}$ . Here $I_{d c}$ is not given, so first find it from the resistive load together with the voltage-drop equation $V_{d c} = \frac{2 V _{m}}{π} cos α - \frac{2 ω L _{s}}{π} I_{d c}$ .

WATCH OUT FOR

Using the ideal output $V_{d c} = \frac{2 V _{m}}{π} cos α$ to get $I_{d c}$ , which ignores the overlap drop. That gives $I_{d c} = 14.64$ A and an overlap of about 5.0 degrees, which is slightly wrong. Another slip is writing the drop as $\frac{ω L _{s}}{π} I_{d c}$ , which holds for a half-wave circuit, not the full bridge with its factor of 2.

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