GATE 2020 EE Q.14: solution and answer

GATE 2020 · EEQ.14 · 1 mark · MCQ

A common-source amplifier with a drain resistance, $R_{D} = 4.7 k Ω$ , is powered using a 10 V power supply. Assuming that the transconductance, $g_{m}$ , is $520 μ A/V$ , the voltage gain of the amplifier is closest to:

(A)
$- 2.44$
(B)
$- 1.22$
(C)
$1.22$
(D)
$2.44$

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Answer and solution

ANSWER

The voltage gain is $A_{v} = - g_{m} R_{D} = - (520 μ A/V) (4.7 k Ω) \approx - 2.44$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2020 EE (IIT Delhi)

Assumptions: the MOSFET is biased in saturation, the output resistance $r_{o}$ is neglected ( $r_{o} \to \infty$ ), there is no load other than $R_{D}$ , and the source is at AC ground (no unbypassed source resistance). The question gives no $r_{o}$ , load or source resistor, so this is the intended model.
Small-signal model: replace the MOSFET with a voltage-controlled current source $g_{m} v_{g s}$ flowing from drain to source. The source is grounded, so $v_{g s} = v_{in}$ .
The 10 V supply is a DC source, so it is AC ground and the top of $R_{D}$ sits at $0$ in the small-signal circuit. The current $g_{m} v_{g s}$ drawn into the drain has to come from that node through $R_{D}$ . KCL at the drain gives $\frac{v _{o u t} - 0}{R _{D}} + g_{m} v_{g s} = 0$ , so $v_{o u t} = - g_{m} R_{D} v_{g s}$ .
Voltage gain: $A_{v} = \frac{v _{o u t}}{v _{in}} = - g_{m} R_{D}$ The minus sign is the $18 0^{\circ}$ phase inversion of a common-source stage.
Substitute the values: $A_{v} = - (520 \times 1 0^{- 6} A/V) (4.7 \times 1 0^{3} Ω) = - 2.444$
The supply voltage (10 V) only sets the DC operating point, and its effect is already inside the given $g_{m}$ . It does not appear in the small-signal gain. $A_{v} \approx - 2.44$ , which is option (A).

KEY IDEA

With $r_{o}$ neglected and the source at AC ground, a common-source stage has small-signal gain $A_{v} = - g_{m} R_{D}$ . The DC supply is AC ground, so it sets the bias but never appears in the gain.

WATCH OUT FOR

Dropping the minus sign and choosing +2.44 (D). A common-source amplifier inverts the signal. Another mistake is trying to use the 10 V supply in the gain, or halving the result to about 1.22 (options B and C). The supply only fixes the DC bias, and nothing in the circuit halves $g_{m} R_{D}$ .

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GATE 2020 EE Q.14: solution and answer

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