GATE 2022 IN Q.44: solution and answer

GATE 2022 · INQ.44 · 2 marks · MCQ

In the small signal circuit shown, the enhancement mode n-channel MOSFET is biased in saturation with transconductance $g_{m}$ . If channel length modulation is ignored, the small signal impedance looking into the node P is given by ________

GATE 2022 IN question 44: An enhancement-mode n-channel MOSFET (broken channel line, arrow on the body terminal pointing inward). — Figure from the GATE 2022 IN paper. It is part of the question.

(A)
$R_{S} ∣∣ R_{L} ∣∣ g_{m}^{- 1}$
(B)
$R_{S} ∣∣ g_{m}^{- 1}$
(C)
$(R_{S} + R_{L}) ∣∣ g_{m}^{- 1}$
(D)
$\frac{R _{L} g _{m}}{1 + R _{S} g _{m}} (R_{L} ∣∣ g_{m}^{- 1})$

Answer and solution

ANSWER

The impedance looking into node P is $R_{S} ∣∣ g_{m}^{- 1}$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Small-signal model: the top rail and the gate are both at AC ground. The body is tied to the source, so $v_{b s} = 0$ and there is no body effect (the $g_{mb}$ term drops out). Channel-length modulation is ignored, so $r_{o} \to \infty$ and the MOSFET is just a dependent current source $g_{m} v_{g s}$ from drain to source.
Apply a test voltage $v_{x}$ at P and let $i_{x}$ be the current it delivers into node P.
The gate is at ground and the source is at $v_{x}$ , so $v_{g s} = 0 - v_{x} = - v_{x}$ .
Drain-to-source channel current: $i_{d} = g_{m} v_{g s} = - g_{m} v_{x}$ . A negative drain-to-source current means a current $g_{m} v_{x}$ flows from node P into the source terminal (and up through the channel and $R_{L}$ to AC ground).
Current from P down through $R_{S}$ to ground: $v_{x} / R_{S}$ .
KCL at P: $i_{x} = \frac{v _{x}}{R _{S}} + g_{m} v_{x} \Rightarrow Z_{P} = \frac{v _{x}}{i _{x}} = \frac{1}{\frac{1}{R _{S}} + g _{m}} = R_{S} ∣∣ \frac{1}{g _{m}}$
$R_{L}$ does not appear: it is in series with an ideal current source ( $r_{o} = \infty$ ), so it cannot change the current that source draws from P. Hence $Z_{P} = R_{S} ∣∣ g_{m}^{- 1}$ , which is option (B).

KEY IDEA

Looking into the source of a common-gate stage with $r_{o} \to \infty$ and no body effect (body tied to source), you see $1/ g_{m}$ , whatever resistor sits in the drain. That $1/ g_{m}$ is in parallel with $R_{S}$ , because both go from P to AC ground.

WATCH OUT FOR

Putting $R_{L}$ in parallel (option A) or in series with $R_{S}$ (option C). With $r_{o}$ infinite, $R_{L}$ is in series with an ideal current source and has no effect on the source-side impedance. $R_{L}$ would only matter through $r_{o}$ : the general result is $\frac{r _{o} + R _{L}}{1 + g _{m} r _{o}}$ , which tends to $1/ g_{m}$ as $r_{o} \to \infty$ .

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 20 Amplifiers questions, with notes

GATE 2022 IN Q.44: solution and answer

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