GATE 2022 EE Q.59: solution and answer

GATE 2022 · EEQ.59 · 2 marks · NAT

For the ideal AC-DC rectifier circuit shown in the figure below, the load current magnitude is $I_{d c} = 15$ A and is ripple free. The thyristors are fired with a delay angle of $4 5^{\circ}$ . The amplitude of the fundamental component of the source current, in amperes, is ______. (round off to two decimal places)

GATE 2022 EE question 59: Single-phase bridge fed from an AC source. — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

ANSWER

The amplitude of the fundamental component of the source current is about 17.64 A.

Official answer: 17.3 to 18Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

The circuit has two thyristors in the upper arms and two diodes in the lower arms, with the source across the two leg midpoints: a single-phase half-controlled bridge. The firing angle is $α = 4 5^{\circ}$ . The load current is ripple free at $I_{d c} = 15 A$ .
In each half-cycle the source current is $\pm I_{d c}$ for a conduction angle $β = π - α = 13 5^{\circ}$ . It is zero for the remaining $α$ , when the load current freewheels through one thyristor and the diode of the same leg. The duty ratio of each half-cycle is $D = \frac{π - α}{π} = \frac{13 5 ^{\circ}}{18 0 ^{\circ}} = 0.75$ . Over one cycle:
$i_{s} (θ) = ⎩ ⎨ ⎧ + I_{d c}, 0, - I_{d c}, 0, α < θ < π π < θ < π + α π + α < θ < 2 π 2 π < θ < 2 π + α θ = ω t$
The average value is $I_{s, avg} = 0$ , because the two half-cycles are equal and opposite.
Fundamental Fourier coefficients, with $i_{s 1} = a_{1} cos θ + b_{1} sin θ$ . The wave has half-wave symmetry, so only odd harmonics appear:
$a_{1} = \frac{2}{π} \int_{α}^{π} I_{d c} cos θ d θ = - \frac{2 I _{d c}}{π} sin α, b_{1} = \frac{2}{π} \int_{α}^{π} I_{d c} sin θ d θ = \frac{2 I _{d c}}{π} (1 + cos α)$
Amplitude of the fundamental:
$I_{s 1, peak} = a_{1}^{2} + b_{1}^{2} = \frac{2 I _{d c}}{π} sin^{2} α + (1 + cos α)^{2} = \frac{2 I _{d c}}{π} 2 + 2 cos α = \frac{4 I _{d c}}{π} cos \frac{α}{2}$
This is the same as $\frac{4 I _{d c}}{π} sin \frac{β}{2}$ , since $β /2 = 9 0^{\circ} - α /2$ .
Numbers for $α = 4 5^{\circ}$ :
$a_{1} = - \frac{2 \times 15}{π} sin 4 5^{\circ} \approx - 6.752 A, b_{1} = \frac{2 \times 15}{π} (1 + cos 4 5^{\circ}) \approx 16.302 A$
$I_{s 1, peak} = 6.75 2^{2} + 16.30 2^{2} \approx 17.64 A$
Check with the closed form:
$\frac{4 \times 15}{π} cos 22. 5^{\circ} = 19.099 \times 0.92388 \approx 17.64 A$

KEY IDEA

In a half-controlled bridge with a ripple-free load current, the source current is a quasi-square wave of height $I_{d c}$ and conduction width $π - α$ per half-cycle. Its fundamental has amplitude $\frac{4 I _{d c}}{π} cos \frac{α}{2}$ .

WATCH OUT FOR

Using the full-converter result $\frac{4 I _{d c}}{π}$ , which gives 19.10 A. That wave conducts for the full $18 0^{\circ}$ , so it ignores the freewheeling interval of $α$ in the semi-converter. Another slip is writing the amplitude as $\frac{4 I _{d c}}{π} cos α$ , which is the full-converter formula for a fully controlled bridge, whose waveform is a square wave shifted by $α$ .

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GATE 2022 EE Q.59: solution and answer

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