GATE 2025 EC Q.61: solution and answer

GATE 2025 · ECQ.61 · 2 marks · NAT

The diode in the circuit shown below is ideal. The input voltage (in Volts) is given by $V_{I} = 10 sin 100 π t$ , where time $t$ is in seconds.

The time duration (in ms, rounded off to two decimal places) for which the diode is forward biased during one period of the input is ________.

GATE 2025 EC question 61: A single series loop. — Figure from the GATE 2025 EC paper. It is part of the question.

Answer and solution

ANSWER

The diode is forward biased for 40/3 ms, which rounds to 13.33 ms in each 20 ms period.

Official answer: 13.32 to 13.34Source: Official answer key, GATE 2025 EC (IIT Roorkee)

Battery polarity. In the figure the short plate (negative terminal) is on top and connects to the diode cathode. The long plate (positive terminal) is at the bottom and connects to the bottom rail, which is the $-$ terminal of $V_{I}$ . Taking that rail as reference, the cathode sits at $V_{K} = - 5 V$ .
Condition for the ideal diode to conduct. The diode points downward, with its anode towards R and its cathode towards the battery. KVL around the loop gives $V_{I} = i R + v_{D} - 5$ . If the diode is OFF, $i = 0$ , so there is no drop across R and the anode is at $V_{I}$ . The diode is forward biased when its anode is above its cathode, i.e. when $V_{I} > - 5 V$ . This is consistent with the ON state: with $v_{D} = 0$ the current $i = \frac{V _{I} + 5}{R}$ is positive only when $V_{I} > - 5 V$ .
Condition in terms of phase. Let $θ = 100 π t$ . Then $10 sin θ > - 5 \Rightarrow sin θ > - \frac{1}{2}$ .
Portion of the cycle. Over one cycle $0 \leq θ < 2 π$ , $sin θ < - \frac{1}{2}$ only for $\frac{7 π}{6} < θ < \frac{11 π}{6}$ . That interval is $\frac{2 π}{3}$ rad long. So the diode is forward biased for $2 π - \frac{2 π}{3} = \frac{4 π}{3}$ rad, which is $\frac{2}{3}$ of each cycle.
Period. $ω = 100 π rad/s \Rightarrow T = \frac{2 π}{100 π} = 0.02 s = 20 ms$ .
Conduction time. $t_{O N} = \frac{2}{3} \times 20 ms = \frac{40}{3} ms \approx 13.33 ms$ As a check, the diode is OFF from $t = \frac{7}{600} s = 11.67 ms$ to $t = \frac{11}{600} s = 18.33 ms$ . That is $6.67 ms$ OFF, which leaves $20 - 6.67 = 13.33 ms$ ON.

KEY IDEA

An ideal diode conducts whenever its anode is above its cathode. Read the battery polarity from the figure: the long plate is +, and here it is grounded. That puts the cathode at −5 V, so the diode conducts whenever V_I > −5 V. Turn this into a phase condition, sin θ > −1/2, which holds for 4π/3 of every 2π rad. The diode therefore conducts for 2/3 of the 20 ms period.

WATCH OUT FOR

The usual slip is misreading the battery polarity and taking the cathode as +5 V. That leads to the condition sin θ > 1/2, conduction for only 1/3 of the cycle, and an answer of 6.67 ms. The other slip is dropping the battery altogether and treating the circuit as a plain half-wave rectifier, which gives 10 ms.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2025 EC Q.61: solution and answer

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