GATE 2024 EE Q.35: solution and answer

GATE 2024 · EEQ.35 · 1 mark · NAT

A forced commutated thyristorized step-down chopper is shown in the figure. Neglect the ON-state drop across the power devices. Assume that the capacitor is initially charged to 50 V with the polarity shown in the figure. The load current (I_L) can be assumed to be constant at 10 A. Initially, Th_M is ON and Th_A is OFF. The turn-off time available to Th_M in microseconds, when Th_A is triggered, is ______ (rounded off to the nearest integer).

GATE 2024 EE question 35: A 50 V source (+ on top) feeds the top rail. — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

ANSWER

The turn-off time available to $T h_{M}$ is $t_{c} = 50 μ s$ .

Official answer: 50Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Before $T h_{A}$ is triggered, $T h_{M}$ conducts the load current $I_{L} = 10 A$ . The capacitor holds $V_{C} = 50 V$ with its $+$ plate at the load node, which is at $50 V$ (the source voltage, because $T h_{M}$ has no drop). So the $-$ plate is at $0 V$ .
Triggering $T h_{A}$ joins the capacitor's $-$ plate to the source $+$ rail, which is at $50 V$ . The $+$ plate therefore moves to $50 + 50 = 100 V$ . The cathode of $T h_{M}$ (the load node) is at $100 V$ and its anode is at $50 V$ , so $T h_{M}$ is reverse biased by the capacitor voltage and turns off. The freewheeling diode is also reverse biased, so the load current has to come through the capacitor.
The load current now flows from the source $+$ rail through $T h_{A}$ , up through the capacitor from its $-$ plate to its $+$ plate, into the load node, and through the load back to the source $-$ . Because $I_{L}$ is constant, it removes charge from the $+$ plate at a constant rate:
$I_{L} = - C \frac{d v _{C}}{d t} \Rightarrow v_{C} (t) = 50 - \frac{I _{L}}{C} t$
The voltage across $T h_{M}$ is $- v_{C} (t)$ (cathode higher than anode). The reverse bias lasts as long as $v_{C} > 0$ . The load-node voltage is $50 + v_{C}$ , so $T h_{M}$ would become forward biased once $v_{C}$ reaches $0$ . The turn-off time available to $T h_{M}$ is therefore the time for $v_{C}$ to fall from $50 V$ to $0$ .
$t_{c} = \frac{C V _{C}}{I _{L}} = \frac{10 \times 1 0 ^{- 6} \times 50}{10} s = 50 \times 1 0^{- 6} s = 50 μ s$
The $L$ -diode branch lies between the same two nodes as the conducting $T h_{A}$ (the source $+$ rail and the capacitor's $-$ plate). While $T h_{A}$ conducts it has no voltage across it and carries no current, so it does not affect $t_{c}$ . The turn-off time available to $T h_{M}$ is $50 μ s$ .

KEY IDEA

When the auxiliary thyristor fires, the pre-charged capacitor is put across the main thyristor in reverse. The constant load current then discharges the capacitor linearly, so the turn-off time is $t_{c} = C V_{C} / I_{L}$ .

WATCH OUT FOR

Treating the discharge as an exponential or a resonant half-cycle with $L$ , or using $2 V_{C}$ ( $100 V$ ) in $C V / I_{L}$ and getting $100 μ s$ . The load current is constant, so the capacitor falls linearly from $50 V$ to $0$ . Reverse bias on $T h_{M}$ ends at $v_{C} = 0$ .

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 9 Switches questions, with notes

GATE 2024 EE Q.35: solution and answer

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