GATE 2022 EC Q.59: solution and answer

GATE 2022 · ECQ.59 · 2 marks · NAT

Consider the circuit shown with an ideal long channel $n$ MOSFET (enhancement-mode, substrate is connected to the source). The transistor is appropriately biased in the saturation region with $V_{GG}$ and $V_{D D}$ such that it acts as a linear amplifier. $v_{i}$ is the small-signal ac input voltage. $v_{A}$ and $v_{B}$ represent the small-signal voltages at the nodes A and B, respectively. The value of $\frac{v _{A}}{v _{B}}$ is ________ (rounded off to one decimal place).

GATE 2022 EC question 59: Enhancement-mode n-channel MOSFET, body tied to source. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

ANSWER

$\frac{v _{A}}{v _{B}} = - \frac{R _{D}}{R _{S}} = - \frac{4 k Ω}{2 k Ω} = - 2.0$ .

Official answer: -2.1 to -1.9Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Assumptions: the MOSFET is in saturation. It is an ideal long-channel device, so $r_{o} \to \infty$ (no channel-length modulation). The body is tied to the source, so $v_{b s} = 0$ and there is no body effect ( $g_{mb} v_{b s} = 0$ ). The small-signal model is then a single dependent current source $i_{d} = g_{m} v_{g s}$ from drain to source, with the gate open. $V_{D D}$ and $V_{GG}$ are dc sources, so both are ac ground.
The gate draws no current and there is no $r_{o}$ path, so the same small-signal current $i_{d}$ flows down through the $4 k Ω$ drain resistor, through the channel, and down through the $2 k Ω$ source resistor.
Node A (drain): $i_{d}$ flows from the ac-grounded $V_{D D}$ rail into A through $4 k Ω$ . By Ohm's law, $v_{A} = 0 - i_{d} (4 k Ω) = - i_{d} (4 k Ω)$ .
Node B (source): the same $i_{d}$ flows from B to ground through $2 k Ω$ , so $v_{B} = + i_{d} (2 k Ω)$ .
Take the ratio. $i_{d}$ cancels, and with it $g_{m}$ and $v_{i}$ : $\frac{v _{A}}{v _{B}} = \frac{- i _{d} ( 4 k Ω )}{i _{d} ( 2 k Ω )} = - \frac{R _{D}}{R _{S}} = - 2.0$
Check with the individual gains. With $v_{g s} = v_{i} - v_{B}$ and $v_{B} = g_{m} v_{g s} R_{S}$ , we get $v_{g s} = \frac{v _{i}}{1 + g _{m} R _{S}}$ . So $\frac{v _{A}}{v _{i}} = - \frac{g _{m} R _{D}}{1 + g _{m} R _{S}}$ and $\frac{v _{B}}{v _{i}} = \frac{g _{m} R _{S}}{1 + g _{m} R _{S}}$ . Their ratio is $- R_{D} / R_{S} = - 2$ at any bias point, as long as the device stays in saturation.

KEY IDEA

With $r_{o} \to \infty$ and no body effect, one small-signal current $i_{d}$ flows through both $R_{D}$ and $R_{S}$ . The drain voltage is $- i_{d} R_{D}$ (current flows into the node from ac ground) and the source voltage is $+ i_{d} R_{S}$ (current flows out of the node to ground). So the drain-to-source voltage ratio is simply $- R_{D} / R_{S}$ . You don't need $g_{m}$ , the bias point or $v_{i}$ .

WATCH OUT FOR

Answering $+ 2$ by dropping the sign: the drain (common-source output) is inverted and the source (source-follower output) is not, so the ratio is negative. Another mistake is working out $v_{A} / v_{i}$ or $v_{B} / v_{i}$ separately and getting stuck because $g_{m}$ is not given. It cancels in the ratio. Adding an $r_{o}$ or a body-effect term also breaks the single-current argument, but the question rules both out ('ideal long channel', 'substrate connected to the source').

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

GATE 2022 EC Q.59: solution and answer

Answer and solution

More Amplifiers questions