GATE 2022 IN Q.64: solution and answer

GATE 2022 · INQ.64 · 2 marks · NAT

In the circuit shown, the switch is initially closed. It is opened at $t = 0$ s and remains open thereafter. The time (in milliseconds) at which the output voltage $V_{o u t}$ becomes LOW is ________ (round off to three decimal places)

GATE 2022 IN question 64: A +5 V supply rail runs along the top. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

The output voltage goes LOW at $t = 0.5 ln (25/12) ms \approx 0.367 ms$ .

Official answer: 0.365 to 0.369Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Reference at the inverting input. An ideal comparator draws no input current, so the 600 kΩ / 400 kΩ divider is unloaded: $V_{-} = 5 \times \frac{400}{600 + 400} = 2 V$ . This does not change with time.
Steady state before $t = 0$ (switch closed, capacitor acts as an open circuit): the 1 kΩ and 5 kΩ resistors form a divider, so $v_{C} (0^{-}) = 5 \times \frac{5}{1 + 5} = \frac{25}{6} \approx 4.167 V$ . Since $V_{+} = 4.167 V > V_{-} = 2 V$ , $V_{o u t}$ is HIGH (+5 V).
Capacitor voltage cannot change instantly: $v_{C} (0^{+}) = v_{C} (0^{-}) = \frac{25}{6} V$ .
For $t > 0$ the switch is open, so the source is disconnected from node X. The comparator input draws no current, so the capacitor discharges through the 5 kΩ resistor alone: $τ = R C = 5 k Ω \times 0.1 μ F = 0.5 ms$ , and $v_{C} (\infty) = 0$ .
Source-free RC response: $v_{C} (t) = \frac{25}{6} e^{- t / τ} V$ , with $τ = 0.5 ms$ .
The output stays HIGH while $V_{+} > V_{-}$ and switches LOW when $V_{+}$ falls to $V_{-} = 2 V$ : $\frac{25}{6} e^{- t / τ} = 2 \Rightarrow t = τ ln \frac{25}{12} = 0.5 \times 0.73397 ms$
So $t \approx 0.36698 ms$ , which is $0.367 ms$ to three decimal places.

KEY IDEA

The comparator switches when the capacitor voltage crosses the fixed 2 V reference. Before the switch opens, the 1 kΩ / 5 kΩ divider charges the capacitor to 25/6 V. After it opens, the capacitor discharges through 5 kΩ alone with $τ = 0.5 ms$ . Setting $\frac{25}{6} e^{- t / τ} = 2$ gives the crossing time $t = τ ln \frac{25}{12}$ .

WATCH OUT FOR

The usual slips are: taking the initial capacitor voltage as the full 5 V and forgetting the 1 kΩ / 5 kΩ divider (that gives $0.5 ln 2.5 \approx 0.458$ ms); using $(1 + 5) k Ω$ or $1 k Ω ∥ 5 k Ω$ as the discharge resistance, when the open switch leaves only 5 kΩ; and taking the threshold as 3 V by reading the divider upside down (600 kΩ is on top, so the reference is 2 V).

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2022 IN Q.64: solution and answer

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