GATE 2026 EC Q.65: solution and answer

GATE 2026 · ECQ.65 · 2 marks · NAT

Consider the ideal diodes D1 and D2 as shown in the Figure with cut-in voltage $V_{γ} = 0$ Volt and $v_{i} (t)$ is in Volt.

The maximum voltage (Volt) of the output $v_{o} (t)$ is ____. (rounded off to two decimal places)

$v_{i} (t) = 6 sin (ω t)$

Answer and solution

ANSWER

The maximum output voltage is $v_{o, m a x} = 4.00 V$ .

Official answer: 3.70 to 4.30Source: Official answer key, GATE 2026 EC (IIT Guwahati)

Read the battery polarities. The 2 V battery has its long (+) plate on top, so D1's cathode is held at $+ 2 V$ . The 4 V battery has its short (−) plate on top, so D2's anode is held at $- 4 V$ .
Find when each ideal diode conducts ( $V_{γ} = 0$ ). D1 (anode towards the output through 10 kΩ, cathode at $+ 2 V$ ) can conduct only when $v_{o} > 2 V$ . D2 (cathode at the output, anode at $- 4 V$ ) conducts only when $v_{o} < - 4 V$ . So on the positive half-cycle only D1 matters, and D2 stays reverse-biased.
Region $- 4 V \leq v_{i} \leq 2 V$ (both diodes off): neither shunt branch carries current, so no current flows in the series 10 kΩ and $v_{o} = v_{i}$ . (For $v_{i} < - 4 V$ , D2 conducts and holds $v_{o}$ at $- 4 V$ ; this affects only the minimum, not the maximum.)
Region $v_{i} > 2 V$ (D1 on, D2 off): one loop current flows from the source, through the series 10 kΩ and the 10 kΩ in D1's branch, into the 2 V battery. By KVL, $i = \frac{v _{i} - 2}{10 k Ω + 10 k Ω} = \frac{v _{i} - 2}{20 k Ω}$
Output voltage: $v_{o} = v_{i} - (10 k Ω) i = v_{i} - \frac{v _{i} - 2}{2} = \frac{v _{i} + 2}{2}$ Above 2 V the output therefore keeps rising, with slope $1/2$ . D1 does not clamp it flat, because the 10 kΩ in D1's branch sits between the output and the battery: the two 10 kΩ resistors form a divider between $v_{i}$ and $2 V$ .
Put in the peak $v_{i} = 6 V$ . Since $v_{o}$ increases with $v_{i}$ in this region, the maximum output occurs at the input peak: $v_{o, m a x} = \frac{6 + 2}{2} = 4.00 V$
Check the assumed diode states at the peak: D1 current is $i = (6 - 2) /20 k Ω = 0.2 mA > 0$ , so D1 is on. D2 has $V_{A K} = - 4 - 4 = - 8 V < 0$ , so D2 is off. Both assumptions hold, and $v_{o, m a x} = 4.00 V$ .

KEY IDEA

A diode clipper with a resistor in series with the diode and its reference battery does not clip flat. Once D1 turns on, the series 10 kΩ and the branch 10 kΩ form a voltage divider between $v_{i}$ and the 2 V battery, so above 2 V the output follows $v_{o} = (v_{i} + 2) /2$ and reaches 4 V at the 6 V input peak. D2, with its anode at −4 V, affects only the negative half-cycle.

WATCH OUT FOR

Treating D1 as a hard clamp and answering 2 V, which ignores the 10 kΩ resistor in D1's own branch. Another mistake is misreading the battery symbols (long plate = +). Taking D1's cathode at −2 V gives the wrong divider result, and taking D2's anode at +4 V wrongly brings D2 into the positive half-cycle.

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