GATE 2022 EC Q.40: solution and answer

GATE 2022 · ECQ.40 · 2 marks · MCQ

Consider an ideal long channel nMOSFET (enhancement-mode) with gate length 10 μm and width 100 μm. The product of electron mobility (μₙ) and oxide capacitance per unit area (C_OX) is μₙC_OX = 1 mA/V². The threshold voltage of the transistor is 1 V. For a gate-to-source voltage V_GS = [2 − sin(2t)] V and drain-to-source voltage V_DS = 1 V (substrate connected to the source), the maximum value of the drain-to-source current is ______.

(A)
40 mA
(B)
20 mA
(C)
15 mA
(D)
5 mA

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Answer and solution

ANSWER

The maximum drain-to-source current is 15 mA (Option C).

Official answer: CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Aspect ratio. $\frac{W}{L} = \frac{100 μ m}{10 μ m} = 10$ .
Overdrive voltage. $V_{o v} (t) = V_{GS} - V_{t n} = (2 - sin (2 t)) - 1 = 1 - sin (2 t)$ . Since $sin (2 t) \in [- 1, 1]$ , the overdrive swings over $V_{o v} \in [0, 2] V$ .
Region check. $V_{D S} = 1 V$ is fixed, so the transistor is in saturation only while $V_{D S} \geq V_{o v}$ , i.e. while $V_{o v} \leq 1 V$ . When $V_{o v} > 1 V$ , the device operates in the triode (linear) region instead, because $V_{D S} < V_{o v}$ there.
Saturation branch ( $V_{o v} \leq 1 V$ ). $I_{D} = \frac{1}{2} μ_{n} C_{o x} (\frac{W}{L}) V_{o v}^{2} = \frac{1}{2} (1) (10) V_{o v}^{2} = 5 V_{o v}^{2} mA$ which is increasing on this range, reaching $5 (1)^{2} = 5 mA$ at $V_{o v} = 1 V$ .
Triode branch ( $V_{o v} > 1 V$ ). $I_{D} = μ_{n} C_{o x} (\frac{W}{L}) [V_{o v} V_{D S} - \frac{V _{D S}^{2}}{2}] = 10 (V_{o v} - 0.5) mA$ using $V_{D S} = 1 V$ .
Continuity check at the boundary $V_{o v} = 1 V$ . Saturation gives $5 (1)^{2} = 5 mA$ ; triode gives $10 (1 - 0.5) = 5 mA$ . The two branches agree, confirming the region split and formulas are applied correctly.
Locating the true maximum. On the triode branch, $I_{D} = 10 (V_{o v} - 0.5)$ increases monotonically with $V_{o v}$ , so the overall maximum current occurs at the largest achievable overdrive, $V_{o v} = 2 V$ (reached when $sin (2 t) = - 1$ ), which lies in the triode region since $2 > V_{D S} = 1$ .
Final value. $I_{D, m a x} = 10 (2 - 0.5) = 10 \times 1.5 = 15 mA$ which matches option (C).

KEY IDEA

A time-varying gate voltage makes the overdrive $V_{o v} (t)$ swing over a range; with $V_{D S}$ fixed, the device moves between saturation and triode as $V_{o v}$ crosses $V_{D S}$ , and since $I_{D}$ keeps rising with $V_{o v}$ in triode too, the true maximum must be checked at the extreme of $V_{o v}$ , not assumed to sit in saturation.

WATCH OUT FOR

Assuming the device stays in saturation for the whole cycle and blindly maximizing $I_{D} = \frac{1}{2} μ_{n} C_{o x} (W / L) V_{o v}^{2}$ at $V_{o v, ma x} = 2 V$ , which gives $20 mA$ (option B) — this ignores that for $V_{o v} > V_{D S} = 1 V$ the transistor has already left saturation and entered the triode region, where the saturation formula no longer applies.

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GATE 2022 EC Q.40: solution and answer

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