GATE 2023 IN Q.56: solution and answer

GATE 2023 · INQ.56 · 2 marks · NAT

The figure below shows a feedback amplifier constructed using an nMOS transistor. Assume that $μ_{n} C_{o x} = 1 mA/V^{2}$ , threshold voltage $V_{T} = 1 V$ and $W / L = 2$ . The bias voltage at the drain terminal is 4 V. The capacitors $C_{\infty}$ offer zero impedance at the signal frequency. The ratio $V_{o u t} / V_{in}$ is ________ (rounded off to two decimal places).

GATE 2023 IN question 56: A single enhancement nMOS stage with supply V_DD = 5 V at the top rail. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{o u t} / V_{in} = \frac{g _{m} R _{S}}{1 + g _{m} R _{S}} = \frac{2}{3} \approx 0.67$ .

Official answer: 0.64 to 0.70Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Gate bias (voltage divider, no gate current): $V_{G} = V_{D D} \frac{R _{2}}{R _{1} + R _{2}} = 5 \times \frac{300}{200 + 300} = 3 V$ .
Drain current from the stated drain bias (Ohm's law across $R_{D}$ ): $I_{D} = \frac{V _{D D} - V _{D}}{R _{D}} = \frac{5 - 4}{1 k Ω} = 1 mA$ .
Source voltage and $V_{GS}$ : $V_{S} = I_{D} R_{S} = 1 mA \times 1 k Ω = 1 V$ , so $V_{GS} = 3 - 1 = 2 V$ and the overdrive is $V_{GS} - V_{T} = 1 V$ .
Consistency check (square law): $I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{T})^{2} = \frac{1}{2} (1 mA/V^{2}) (2) (1 V)^{2} = 1 mA$ , which matches the current found from $R_{D}$ . Saturation holds because $V_{D S} = 4 - 1 = 3 V > V_{GS} - V_{T} = 1 V$ .
Transconductance: $g_{m} = μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{T}) = 1 \times 2 \times 1 = 2 mA/V$ (equivalently $g_{m} = \frac{2 I _{D}}{V _{GS} - V _{T}} = \frac{2 \times 1}{1} = 2 mA/V$ ).
Small-signal model: both $C_{\infty}$ are short circuits and $V_{D D}$ is AC ground. $V_{in}$ drives the gate directly (the divider $R_{1} ∥ R_{2}$ only loads the ideal source), and the output is taken at the source with no load, so this is a common-drain (source follower) stage with series-series feedback through $R_{S}$ . Assumptions: $r_{o} \to \infty$ (no channel-length modulation is given), and no body effect because the body is tied to the source. $R_{D}$ sits in the drain and does not affect the source-node gain.
Source-follower gain: with $i_{d} = g_{m} (v_{g} - v_{s})$ and $v_{s} = i_{d} R_{S}$ , $v_{s} = g_{m} R_{S} (v_{in} - v_{s}) \Rightarrow \frac{V _{o u t}}{V _{in}} = \frac{g _{m} R _{S}}{1 + g _{m} R _{S}} = \frac{2 \times 1}{1 + 2 \times 1} = \frac{2}{3} \approx 0.67$

KEY IDEA

The drain bias of 4 V fixes $I_{D} = (5 - 4) /1 k Ω = 1 mA$ straight away. That gives $V_{S} = 1 V$ , $V_{GS} = 2 V$ and $g_{m} = 2 mA/V$ . Because the output is taken at the source, the stage is a source follower with gain $g_{m} R_{S} / (1 + g_{m} R_{S})$ . $R_{D}$ only sets the bias and plays no part in the signal gain.

WATCH OUT FOR

Taking the output as if it came from the drain and using the degenerated common-source gain $- g_{m} R_{D} / (1 + g_{m} R_{S}) = - 0.67$ , which has the right magnitude but the wrong sign and the wrong physical meaning. Other slips: dropping the factor $\frac{1}{2}$ in the square law so that $g_{m}$ comes out wrong, or adding a body-effect term $g_{mb}$ even though the body is tied to the source.

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GATE 2023 IN Q.56: solution and answer

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