ISRO EC 2023 Q.15: solution and answer

ISRO 2023 · ECQ.15 · 1 mark · MCQ

The relation between I_D and V_gs of a MOSFET in the saturation region is

(A)
Exponential
(B)
Quadratic
(C)
Logarithmic
(D)
Hyperbolic

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Answer and solution

ANSWER

The relation between $I_{D}$ and $V_{GS}$ of a MOSFET in saturation is quadratic, option (b).

Official answer: BSource: ISRO's revised and final answer key of 16 February 2024

In saturation the channel is pinched off near the drain, so the drain current is set by the gate overdrive and is almost independent of $V_{D S}$ (to first order).
The square-law model for the saturation region gives $I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{t h})^{2} .$
Hence $I_{D}$ is proportional to the square of the overdrive $V_{GS} - V_{t h}$ , so the relation between $I_{D}$ and $V_{GS}$ is quadratic.
An exponential relation belongs to the BJT (and to a MOSFET in weak inversion or subthreshold), not to strong-inversion saturation.
A logarithmic or hyperbolic relation does not arise from the square-law model, so the correct option is (b), quadratic.

KEY IDEA

In strong-inversion saturation $I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{t h})^{2}$ , so $I_{D}$ varies as the square of $V_{GS} - V_{t h}$ .

WATCH OUT FOR

Choosing exponential because the BJT's $I_{C}$ rises exponentially with $V_{B E}$ ; the MOSFET follows a square law above threshold, and is exponential only in weak inversion.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO MOS devices questions, with notes