GATE 2018 EC Q.1: solution and answer

GATE 2018 · ECQ.1 · 1 mark · MCQ

Two identical nMOS transistors $M_{1}$ and $M_{2}$ are connected as shown below. The circuit is used as an amplifier with the input connected between G and S terminals and the output taken between D and S terminals. $V_{bias}$ and $V_{D}$ are so adjusted that both transistors are in saturation. The transconductance of this combination is defined as $g_{m} = \frac{\partial i _{D}}{\partial v _{GS}}$ while the output resistance is $r_{o} = \frac{\partial v _{D S}}{\partial i _{D}}$ , where $i_{D}$ is the current flowing into the drain of $M_{2}$ . Let $g_{m 1}$ , $g_{m 2}$ be the transconductances and $r_{o 1}$ , $r_{o 2}$ be the output resistances of transistors $M_{1}$ and $M_{2}$ , respectively.

Which of the following statements about estimates for $g_{m}$ and $r_{o}$ is correct?

GATE 2018 EC question 1: Two nMOS transistors stacked in series (cascode). — Figure from the GATE 2018 EC paper. It is part of the question.

(A)
$g_{m} \approx g_{m 1} \cdot g_{m 2} \cdot r_{o 2}$ and $r_{o} \approx r_{o 1} + r_{o 2}$ .
(B)
$g_{m} \approx g_{m 1} + g_{m 2}$ and $r_{o} \approx r_{o 1} + r_{o 2}$ .
(C)
$g_{m} \approx g_{m 1}$ and $r_{o} \approx r_{o 1} \cdot g_{m 2} \cdot r_{o 2}$ .
(D)
$g_{m} \approx g_{m 1}$ and $r_{o} \approx r_{o 2}$ .

Answer and solution

ANSWER

Official answer: CSource: Official answer key, GATE 2018 EC (IIT Guwahati)

Identify the circuit. $M_{1}$ is a common-source stage driven at G. $M_{2}$ has its gate at the DC voltage $V_{bias}$ , which is an AC ground, so it is a common-gate stage stacked on top of $M_{1}$ . Together they form an nMOS cascode. Both devices are in saturation, so each one's small-signal model is a controlled current source $g_{m} v_{g s}$ (drain to source) in parallel with $r_{o}$ . Assumptions: body effect is neglected (no body terminals are drawn), and $g_{m} r_{o} ≫ 1$ for each device.
Let $X$ be the node joining the source of $M_{2}$ to the drain of $M_{1}$ , and let $v_{x}$ be its small-signal voltage. Since the gate of $M_{2}$ is at AC ground, $v_{g s 2} = - v_{x}$ . The current $i_{D}$ enters the drain of $M_{2}$ and flows down through $M_{2}$ and then $M_{1}$ to S. The same current flows in both devices.
Transconductance (output shorted, $v_{d s} = 0$ ). The current through $M_{2}$ is $i_{D} = g_{m 2} (- v_{x}) + \frac{0 - v _{x}}{r _{o 2}} = - v_{x} (g_{m 2} + \frac{1}{r _{o 2}})$ so $v_{x} \approx - \frac{i _{D}}{g _{m 2}}$ . This is small, because node $X$ looks into the source of $M_{2}$ and sees only about $1/ g_{m 2}$ .
The same current flows down through $M_{1}$ : $i_{D} = g_{m 1} v_{g s} + \frac{v _{x}}{r _{o 1}}$ . Substituting $v_{x}$ gives $i_{D} (1 + \frac{1}{g _{m 2} r _{o 1}}) \approx g_{m 1} v_{g s} \Rightarrow g_{m} = \frac{i _{D}}{v _{g s}} \approx g_{m 1} (g_{m 2} r_{o 1} ≫ 1)$ Almost none of $M_{1}$ 's signal current is lost in $r_{o 1}$ , so $M_{2}$ passes it through to the output.
Output resistance (input shorted, $v_{g s} = 0$ ). With $v_{g s} = 0$ the controlled source of $M_{1}$ is off, so $M_{1}$ is just $r_{o 1}$ carrying $i_{D}$ : $v_{x} = i_{D} r_{o 1}$ .
Current through $M_{2}$ : $i_{D} = g_{m 2} (- v_{x}) + \frac{v _{d s} - v _{x}}{r _{o 2}}$ , so $v_{d s} = v_{x} + r_{o 2} (i_{D} + g_{m 2} v_{x})$ . Substituting $v_{x} = i_{D} r_{o 1}$ : $r_{o} = \frac{v _{d s}}{i _{D}} = r_{o 1} + r_{o 2} + g_{m 2} r_{o 2} r_{o 1} \approx g_{m 2} r_{o 2} r_{o 1}$ The product term dominates because $g_{m 2} r_{o 2} ≫ 1$ and $g_{m 2} r_{o 1} ≫ 1$ .
Result: $g_{m} \approx g_{m 1}$ and $r_{o} \approx r_{o 1} g_{m 2} r_{o 2}$ . This is option (C). A cascode keeps the transconductance of $M_{1}$ and multiplies its output resistance by the intrinsic gain $g_{m 2} r_{o 2}$ of $M_{2}$ .

KEY IDEA

This is a cascode: $M_{1}$ is common-source and $M_{2}$ , with its gate at AC ground, is common-gate. Looking into the source of $M_{2}$ , node $X$ sees only about $1/ g_{m 2}$ , so all of $M_{1}$ 's signal current $g_{m 1} v_{g s}$ reaches the output and $g_{m} \approx g_{m 1}$ . Looking down from the drain, $M_{2}$ is a transistor with source degeneration $r_{o 1}$ . That multiplies the resistance by $M_{2}$ 's intrinsic gain: $r_{o} = r_{o 1} + r_{o 2} + g_{m 2} r_{o 2} r_{o 1} \approx g_{m 2} r_{o 2} r_{o 1}$ .

WATCH OUT FOR

Adding the two output resistances in series ( $r_{o} \approx r_{o 1} + r_{o 2}$ , as in (A) and (B)). That treats $M_{2}$ as a plain resistor. When $v_{x}$ rises, $v_{g s 2} = - v_{x}$ falls, and $M_{2}$ 's controlled source $g_{m 2} v_{g s 2}$ works against the change in current. That feedback produces the dominant $g_{m 2} r_{o 2} r_{o 1}$ term. Picking (D), $r_{o} \approx r_{o 2}$ , ignores $M_{1}$ altogether. Writing $g_{m} \approx g_{m 1} + g_{m 2}$ treats the devices as if they were in parallel, but they are stacked, carry the same current, and only $M_{1}$ is driven by the input.

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GATE 2018 EC Q.1: solution and answer

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