GATE 2018 EE Q.3: solution and answer

GATE 2018 · EEQ.3 · 1 mark · MCQ

A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?

GATE 2018 EE question 3: A single-phase thyristor bridge fed by an AC source (sine symbol) connected between the two mid-points of the two bridge… — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
v_o ≥ 0 & i_o < 0
(B)
v_o < 0 & i_o < 0
(C)
v_o ≥ 0 & i_o ≥ 0
(D)
v_o < 0 & i_o ≥ 0

Answer and solution

ANSWER

$v_{o} \geq 0$ and $i_{o} \geq 0$ , which is option (C).

Official answer: CSource: Official answer key, GATE 2018 EE (IIT Guwahati)

The circuit is a single-phase fully controlled bridge of four thyristors, with a diode $D_{f}$ across the DC output. $D_{f}$ has its cathode on the top (positive) rail and its anode on the bottom rail, so it is a freewheeling diode. The firing angle $α$ may take any value. The duty ratio of $D_{f}$ depends on $α$ and the load, and the argument below does not need it.
Current direction: thyristors and diodes conduct only from anode to cathode. The upper devices have their cathodes on the top rail and the lower devices have their anodes on the bottom rail. So the bridge can only push current into the top rail and out of the bottom rail, and $D_{f}$ only carries current from the bottom rail to the top rail, in parallel with the load. The load current $i_{o}$ is drawn downward through the load, so it is the sum of the bridge current and the $D_{f}$ current, both non-negative. Hence $i_{o} \geq 0$ at every instant. Option (A) and option (B) are therefore ruled out.
Voltage limit: $v_{o}$ is the voltage from the top rail to the bottom rail. If $v_{o}$ tried to become negative, the top rail would be below the bottom rail. $D_{f}$ would then be forward biased, because its anode is on the bottom rail and its cathode is on the top rail. An ideal diode in conduction holds $v_{o} = 0$ . So $v_{o}$ can never be negative.
While a diagonal pair of thyristors conducts, the output is the rectified source voltage, $v_{o} = v_{s} = V_{m} sin ω t$ for the pair that conducts in the positive half cycle, and $v_{o} = - v_{s} = ∣ V_{m} sin ω t ∣$ for the other pair in the negative half cycle. Both are non-negative. When the thyristors stop conducting, the load current freewheels through $D_{f}$ and $v_{o} = 0$ .
So at every instant $v_{o} \geq 0$ and $i_{o} \geq 0$ . This is option (C). The bridge is effectively a half-controlled rectifier, because the freewheeling diode removes the negative-voltage (inverting) interval.

KEY IDEA

A bridge of thyristors can only carry current in one direction, so $i_{o} \geq 0$ . An ideal freewheeling diode across the output clamps $v_{o}$ at $0$ whenever it would go negative, so $v_{o} \geq 0$ . Together these give $v_{o} \geq 0$ and $i_{o} \geq 0$ .

WATCH OUT FOR

Treating this as an ordinary fully controlled bridge, where $v_{o}$ goes negative after $π$ for $α > 0$ with an inductive load (inverter-like operation). The anti-parallel diode across the load removes that interval. Another slip is to read the diode as reversing the load current. It cannot, because every device in the circuit conducts in one direction only.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 12 Rectifiers questions, with notes

GATE 2018 EE Q.3: solution and answer

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