GATE 2020 EC Q.37: solution and answer

GATE 2020 · ECQ.37 · 2 marks · MCQ

Using the incremental low frequency small-signal model of the MOS device, the Norton equivalent resistance of the following circuit is

GATE 2020 EC question 37: One n-channel MOSFET, drawn as a three-terminal symbol with the arrow on the source pointing outward and labelled with its… — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
$r_{d s} + R + g_{m} r_{d s} R$
(B)
$\frac{r _{d s} + R}{1 + g _{m} r _{d s}}$
(C)
$r_{d s} + \frac{1}{g _{m}} + R$
(D)
$r_{d s} + R$

Answer and solution

ANSWER

The Norton equivalent resistance at the source terminal is $R_{N} = \frac{r _{d s} + R}{1 + g _{m} r _{d s}}$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2020 EC (IIT Delhi)

Small-signal circuit. To find the Norton resistance, set every independent source to zero. $V_{D D}$ becomes AC ground, so $R$ runs from the drain to ground, and the gate is also at ground. Replace the MOSFET with its low-frequency model: a VCCS $g_{m} v_{g s}$ from drain to source in parallel with $r_{d s}$ . No body-effect parameter is given, so the body effect is neglected. The Norton resistance $R_{N}$ is the resistance seen looking into the source terminal.
Test source. Apply a test voltage $v_{x}$ at the source and let $i_{x}$ be the current it drives into the source terminal. Since the gate is at ground, $v_{g s} = 0 - v_{x} = - v_{x}$ .
One path for the current. The only way $i_{x}$ can leave the source node is up through the transistor (VCCS and $r_{d s}$ together) and then down through $R$ to ground. So the drain voltage is $v_{d} = i_{x} R$ , and the current flowing from drain to source inside the device is $i_{d} = - i_{x}$ .
Drain-to-source current. The current from drain to source is the VCCS current plus the $r_{d s}$ current: $i_{d} = g_{m} v_{g s} + \frac{v _{d} - v _{s}}{r _{d s}} \Rightarrow - i_{x} = - g_{m} v_{x} + \frac{i _{x} R - v _{x}}{r _{d s}}$
Solve. Multiply by $r_{d s}$ : $- i_{x} r_{d s} = - g_{m} r_{d s} v_{x} + i_{x} R - v_{x} \Rightarrow v_{x} (1 + g_{m} r_{d s}) = i_{x} (r_{d s} + R)$
Result. $R_{N} = \frac{v _{x}}{i _{x}} = \frac{r _{d s} + R}{1 + g _{m} r _{d s}}$ This is option (B).
Limit checks. As $r_{d s} \to \infty$ , $R_{N} \to \frac{1}{g _{m}}$ , the familiar resistance looking into the source of a common-gate or source-follower stage. With $R = 0$ , $R_{N} = \frac{r _{d s}}{1 + g _{m} r _{d s}} = r_{d s} ∥ \frac{1}{g _{m}}$ , which is correct for a gate-grounded device whose drain is at AC ground. Both limits agree.

KEY IDEA

Looking into the source of a MOSFET whose gate is at AC ground, the drain-side resistance $r_{d s} + R$ is divided by $(1 + g_{m} r_{d s})$ , which is about the device's intrinsic gain. So the source is a low-resistance node, close to $1/ g_{m}$ unless the drain load $R$ is comparable to $g_{m} r_{d s}$ times $1/ g_{m}$ . This is the dual of looking into the drain, where the source-side resistance is multiplied up by the same factor.

WATCH OUT FOR

Picking (A), $r_{d s} + R + g_{m} r_{d s} R$ . That has the form of the resistance looking into the DRAIN of a source-degenerated MOSFET, where resistance is multiplied up. Here we look into the SOURCE, and resistance is divided down. Other slips: taking $v_{g s} = + v_{x}$ instead of $- v_{x}$ , which gives $(r_{d s} + R) / (1 - g_{m} r_{d s})$ , and ignoring the dependent source altogether, which gives (D), $r_{d s} + R$ .

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GATE 2020 EC Q.37: solution and answer

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