GATE 2018 EC Q.43: solution and answer

GATE 2018 · ECQ.43 · 2 marks · NAT

In the circuit shown below, the $(W / L)$ value for $M_{2}$ is twice that for $M_{1}$ . The two nMOS transistors are otherwise identical. The threshold voltage $V_{T}$ for both transistors is 1.0V. Note that $V_{GS}$ for $M_{2}$ must be $> 1.0 V$ .

Current through the nMOS transistors can be modeled as

I_{D S} = μ C_{o x} (\frac{W}{L}) ((V_{GS} - V_{T}) V_{D S} - \frac{1}{2} V_{D S}^{2}) for V_{D S} \leq V_{GS} - V_{T}

I_{D S} = μ C_{o x} (\frac{W}{L}) (V_{GS} - V_{T})^{2} /2 for V_{D S} \geq V_{GS} - V_{T}

The voltage (in volts, accurate to two decimal places) at $V_{x}$ is ______.

GATE 2018 EC question 43: Two enhancement nMOS transistors stacked in series between a 3.3 V supply and ground. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

$V_{x} = 1 - \frac{1}{3} \approx 0.42 V$ (M2 saturated, M1 in triode).

Official answer: 0.41 to 0.435Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Both gates are tied to the $2.0 V$ node, so $V_{GS 1} = 2 V$ and $V_{GS 2} = 2 - V_{x}$ . M2 and M1 are in series, so they carry the same drain current: $I_{D 1} = I_{D 2}$ . Let $k = μ C_{o x} (W / L)_{1}$ ; then $μ C_{o x} (W / L)_{2} = 2 k$ . Body effect is ignored, since the given current model has no body term.
Region of M2: $V_{D S 2} = 3.3 - V_{x}$ and $V_{GS 2} - V_{T} = 1 - V_{x}$ . The saturation condition $V_{D S 2} \geq V_{GS 2} - V_{T}$ reduces to $3.3 \geq 1$ , which is always true. So whenever M2 conducts, it is in saturation.
Region of M1: $V_{GS 1} - V_{T} = 2 - 1 = 1 V$ and $V_{D S 1} = V_{x}$ . The question requires $V_{GS 2} > 1 V$ , i.e. $2 - V_{x} > 1$ , so $V_{x} < 1 V$ . That gives $V_{D S 1} < V_{GS 1} - V_{T}$ , so M1 is in the triode (linear) region.
Set the two currents equal using the given models: $\frac{2 k}{2} (2 - V_{x} - 1)^{2} = k (1 \cdot V_{x} - \frac{1}{2} V_{x}^{2})$
Cancel $k$ and simplify: $(1 - V_{x})^{2} = V_{x} - \frac{1}{2} V_{x}^{2} \Rightarrow 1 - 2 V_{x} + V_{x}^{2} = V_{x} - \frac{1}{2} V_{x}^{2} \Rightarrow 1.5 V_{x}^{2} - 3 V_{x} + 1 = 0$ .
Solve the quadratic: $V_{x} = \frac{3 \pm 9 - 6}{3} = \frac{3 \pm 3}{3} = 1.577 V or 0.423 V$
Reject $V_{x} = 1.577 V$ : it gives $V_{GS 2} = 0.423 V < V_{T}$ , so M2 would be off, and $V_{x} > 1 V$ contradicts the triode assumption for M1. Keep $V_{x} = 1 - \frac{1}{3} = 0.4226 V$ . Check: $V_{D S 1} = 0.423 < 1$ (M1 triode, consistent) and $V_{GS 2} = 1.577 > 1$ (M2 on and saturated, consistent).
Therefore $V_{x} \approx 0.42 V$ .

KEY IDEA

Series transistors carry the same current. M2 has its drain on the 3.3 V rail, so $V_{D S 2} - (V_{GS 2} - V_{T}) = 2.3 V > 0$ and it is always saturated when on. M1 has $V_{GS 1} - V_{T} = 1 V$ , and the requirement $V_{GS 2} > V_{T}$ forces $V_{x} < 1 V$ , so M1 is in triode. Setting the saturation current equal to the triode current gives a quadratic in $V_{x}$ . Keep only the root that satisfies both region assumptions.

WATCH OUT FOR

Assuming both transistors are saturated. Equating $k (1 - V_{x})^{2}$ with $\frac{k}{2} (1)^{2}$ gives $V_{x} \approx 0.29 V$ , which breaks M1's saturation condition ( $V_{D S 1} = 0.29 < 1$ ). Two other mistakes: keeping the root $1.577 V$ , which switches M2 off, and forgetting the factor 2 in $(W / L)_{2}$ .

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 13 Biasing questions, with notes

GATE 2018 EC Q.43: solution and answer

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