GATE 2024 EC Q.64: solution and answer

GATE 2024 · ECQ.64 · 2 marks · NAT

An NMOS transistor operating in the linear region has $I_{D S}$ of 5 μA at $V_{D S}$ of 0.1 V. Keeping $V_{GS}$ constant, the $V_{D S}$ is increased to 1.5 V. Given that $μ_{n} C_{o x} \frac{W}{L} = 50 μ A/V^{2}$ , the transconductance at the new operating point (in μA/V, rounded off to two decimal places) is ____.

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Answer and solution

ANSWER

The transconductance at the new operating point is $g_{m} = 52.50 μ A/V$ .

Official answer: 52.40 to 52.60Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

In the triode (linear) region: $I_{D S} = μ_{n} C_{o x} \frac{W}{L} [(V_{GS} - V_{T}) V_{D S} - \frac{V _{D S}^{2}}{2}] = μ_{n} C_{o x} \frac{W}{L} [V_{O V} V_{D S} - \frac{V _{D S}^{2}}{2}]$ , where $V_{O V} = V_{GS} - V_{T}$ .
At the first operating point, $V_{D S} = 0.1 V$ , $I_{D S} = 5 μ A$ : $5 = 50 [(0.1) V_{O V} - \frac{( 0.1 ) ^{2}}{2}] = 50 [0.1 V_{O V} - 0.005]$ .
Solving: $0.1 = 0.1 V_{O V} - 0.005 \Rightarrow V_{O V} = 1.05 V$ . Since $V_{GS}$ is held fixed for the rest of the problem, this $V_{O V}$ stays the same at the new bias point.
Check the region of operation at the new bias point: the triode/saturation boundary is $V_{D S} = V_{O V} = 1.05 V$ . The new $V_{D S} = 1.5 V > V_{O V}$ , so the transistor now operates in saturation, not triode.
In saturation, $I_{D S} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} V_{O V}^{2}$ . Differentiating with respect to $V_{GS}$ (with $\frac{\partial V _{O V}}{\partial V _{GS}} = 1$ ): $g_{m} = \frac{\partial I _{D S}}{\partial V _{GS}} = μ_{n} C_{o x} \frac{W}{L} V_{O V}$ — independent of $V_{D S}$ once in saturation.
Substituting values: $g_{m} = 50 μ A/V^{2} \times 1.05 V = 52.5 μ A/V$ .
Rounded to two decimal places: $g_{m} = 52.50 μ A/V$ .

KEY IDEA

Even though the transistor starts in the triode (linear) region, raising $V_{D S}$ at fixed $V_{GS}$ can push it into saturation once $V_{D S}$ exceeds the overdrive voltage $V_{O V} = V_{GS} - V_{T}$ ; $g_{m}$ must then be computed from the saturation current equation, not the triode one.

WATCH OUT FOR

Assuming the device stays in the linear/triode region for the whole problem and differentiating the triode $I_{D S}$ expression (or scaling the triode current) at $V_{D S} = 1.5$ V without first checking the saturation boundary $V_{D S} \geq V_{O V}$ .

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GATE 2024 EC Q.64: solution and answer

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