ISRO EC 2025 Q.72: solution and answer

ISRO 2025 · ECQ.72 · 1 mark · MCQ

Pinch-off voltage in an FET is :

(A)
The gate-to-source voltage that gives zero drain current
(B)
The drain voltage that gives zero drain current
(C)
The gate source voltage that gives unity drain current
(D)
The drain voltage that gives infinite drain current

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Answer and solution

ANSWER

Pinch-off voltage is the gate-to-source voltage that gives zero drain current, option (a).

Official answer: ASource: ISRO's final answer key of 15 May 2026

The pinch-off voltage $V_{P}$ of a JFET (or depletion-type FET) appears in its transfer characteristic in the saturation region: $I_{D} = I_{D S S} (1 - \frac{V _{GS}}{V _{P}})^{2}$ .
Putting $V_{GS} = V_{P}$ gives $I_{D} = 0$ . At this gate bias the depletion regions meet and the channel is fully closed, so no drain current flows. Hence $V_{P}$ is a gate-to-source voltage at which the drain current is zero.
Option (b) is wrong: $V_{P}$ is set by the gate-to-source bias, not by a drain voltage. A zero drain voltage ( $V_{D S} = 0$ ) does not describe pinch-off.
Option (c) is wrong: a drain current of 'unity' has no special meaning for the device. The defining condition is zero drain current.
Option (d) is wrong: an infinite drain current is physically impossible. Beyond pinch-off in the output characteristic the drain current saturates at a finite value, $I_{D S S}$ when $V_{GS} = 0$ .
Answer: (a).

KEY IDEA

$V_{P}$ is the gate-to-source voltage at which the channel is fully depleted and $I_{D} = 0$ , so that $I_{D} = I_{D S S} (1 - V_{GS} / V_{P})^{2}$ vanishes at $V_{GS} = V_{P}$ .

WATCH OUT FOR

Confusing the pinch-off voltage $V_{P}$ (a gate-to-source value that switches the channel off) with the drain-source voltage $V_{D S}$ at which the channel pinches and the current saturates, and so picking an option about drain voltage.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO MOS devices questions, with notes