GATE 2026 EC Q.60: solution and answer

GATE 2026 · ECQ.60 · 2 marks · NAT

An $n$ -channel MOSFET is connected as shown in the Figure.

Assume $V_{T H} = 1 V$ , $V_{D D} = 5 V$ , and $μ C_{o x} (\frac{W}{L}) = 2 mA V^{- 2}$ and neglect channel length modulation effects.

The gate voltage $(V_{G})$ of the $n$ -channel MOSFET (in Volt) is ____. (rounded off to two decimal places)

GATE 2026 EC question 60: One n-channel MOSFET, drawn inside a circle, with its drain at the top and its source at the bottom. — Figure from the GATE 2026 EC paper. It is part of the question.

Answer and solution

ANSWER

$V_{G} = 1 + \frac{- 1 + 17}{2} \approx 2.56 V$ .

Official answer: 2.50 to 2.60Source: Official answer key, GATE 2026 EC (IIT Guwahati)

Circuit reading. The gate is wired to the drain and the source is grounded, so $V_{G} = V_{D} = V_{GS} = V_{D S}$ . This is a diode-connected MOSFET fed from $V_{D D}$ through $R = 1 k Ω$ . No current flows into the gate, so all of the resistor current is the drain current $I_{D}$ .
Region of operation. The device conducts only when $V_{GS} > V_{T H}$ . Because $V_{T H} > 0$ , it always holds that $V_{D S} = V_{GS} > V_{GS} - V_{T H}$ , so the triode condition $V_{D S} < V_{GS} - V_{T H}$ can never be met. Once the device is on, it is in saturation.
Square law in saturation ( $λ = 0$ , channel length modulation neglected): $I_{D} = \frac{1}{2} μ C_{o x} \frac{W}{L} (V_{GS} - V_{T H})^{2} = \frac{1}{2} (2 mA/V^{2}) (V_{G} - 1)^{2} = (V_{G} - 1)^{2} mA$
Ohm's law on the resistor: $I_{D} = \frac{V _{D D} - V _{G}}{R} = \frac{5 - V _{G}}{1 k Ω} = (5 - V_{G}) mA$
Set the two equal. Let $x = V_{G} - 1$ , the overdrive voltage. Then $5 - V_{G} = 4 - x$ , so $x^{2} = 4 - x \Rightarrow x^{2} + x - 4 = 0 \Rightarrow x = \frac{- 1 \pm 17}{2}$
Choose the physical root. $x = \frac{- 1 + 17}{2} = \frac{- 1 + 4.1231}{2} = 1.5616 V$ . The other root, $x = - 2.56 V$ , would make $V_{GS} < V_{T H}$ , which means the device is off. That contradicts the square-law equation it came from, so it is rejected. (If the device were off, $I_{D} = 0$ would give $V_{G} = 5 V > V_{T H}$ , so the device must be on.)
Gate voltage: $V_{G} = V_{T H} + x = 1 + 1.5616 = 2.5616 V \approx 2.56 V$ .
Check. $I_{D} = (1.5616)^{2} mA = 2.438 mA$ . The drop across $1 k Ω$ is $2.438 V$ , so $V_{G} = 5 - 2.438 = 2.562 V$ , which agrees. Also $V_{D S} = 2.56 V > V_{GS} - V_{T H} = 1.56 V$ , so the device is in saturation as assumed.

KEY IDEA

When the gate is tied to the drain, $V_{D S} = V_{GS}$ , so an n-channel MOSFET that is on must be in saturation. Set the square-law current $\frac{1}{2} μ C_{o x} \frac{W}{L} (V_{G} - V_{T H})^{2}$ equal to the resistor current $(V_{D D} - V_{G}) / R$ . Solve the quadratic and keep the root with $V_{GS} > V_{T H}$ .

WATCH OUT FOR

Writing $I_{D} = μ C_{o x} \frac{W}{L} (V_{GS} - V_{T H})^{2}$ without the factor $\frac{1}{2}$ . That gives $2 x^{2} + x - 4 = 0$ , so $x = 1.19 V$ and $V_{G} \approx 2.19 V$ , which is outside the key. The key uses the standard $\frac{1}{2} μ C_{o x} \frac{W}{L}$ form. Other mistakes: keeping the negative root (it makes $V_{G} < V_{T H}$ , so the device would be off), and checking for the triode region, which a diode-connected device can never enter.

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GATE 2026 EC Q.60: solution and answer

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