GATE 2025 EE Q.16: solution and answer

GATE 2025 · EEQ.16 · 1 mark · MCQ

Which one of the following statements is true about the small signal voltage gain of a MOSFET based single stage amplifier?

(A)
Common source and common gate amplifiers are both inverting amplifiers
(B)
Common source and common gate amplifiers are both non-inverting amplifiers
(C)
Common source amplifier is inverting and common gate amplifier is non-inverting amplifier
(D)
Common source amplifier is non-inverting and common gate amplifier is inverting amplifier

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Answer and solution

ANSWER

(C) The common source amplifier is inverting, with $A_{v} = - g_{m} R_{D}$ . The common gate amplifier is non-inverting, with $A_{v} = + g_{m} R_{D}$ .

Official answer: CSource: Official answer key, GATE 2025 EE (IIT Roorkee)

Model: the MOSFET is biased in saturation. Its small-signal drain current is $i_{d} = g_{m} v_{g s}$ , with $r_{o}$ and the body effect neglected. The drain load is $R_{D}$ and the output is taken at the drain in both configurations. Because the drain voltage is $V_{D D} - i_{D} R_{D}$ , the small-signal output is $v_{o} = - i_{d} R_{D}$ .
Common source: the input drives the gate and the source is at AC ground, so $v_{g s} = v_{i}$ . Then $v_{o} = - g_{m} v_{i} R_{D} \Rightarrow A_{v} = \frac{v _{o}}{v _{i}} = - g_{m} R_{D}$ The negative sign means the output is $18 0^{\circ}$ out of phase with the input, so the amplifier is inverting.
Common gate: the gate is at AC ground and the input drives the source, so $v_{g s} = 0 - v_{i} = - v_{i}$ . Then $v_{o} = - g_{m} (- v_{i}) R_{D} \Rightarrow A_{v} = \frac{v _{o}}{v _{i}} = + g_{m} R_{D}$ The output is in phase with the input, so the amplifier is non-inverting. (With the body effect included the gain becomes $+ (g_{m} + g_{mb}) R_{D}$ , and the sign is still positive.)
Physical check for CG: when the source voltage rises, $v_{g s}$ falls, so the drain current falls. The drop across $R_{D}$ then falls and the drain voltage rises. Input and output move together.
So CS is inverting and CG is non-inverting, which is option (C).

KEY IDEA

The sign of the gain depends on which terminal takes the input. At the gate (CS), $v_{g s} = + v_{i}$ . At the source (CG), $v_{g s} = - v_{i}$ . The output at the drain is always $- g_{m} v_{g s} R_{D}$ , so the extra minus sign in CG cancels the inversion and gives a gain of the same size, $g_{m} R_{D}$ , but positive.

WATCH OUT FOR

Students often assume every transistor amplifier with the output at the collector or drain is inverting, like the CE/CS stage, and choose (A). Driving the source makes $v_{g s} = - v_{i}$ , and that second sign flip makes the common gate stage non-inverting. Another error is to confuse CG with the source follower (common drain). Both have positive gain, but the follower's gain is below 1 and its output is at the source, not the drain.

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GATE 2025 EE Q.16: solution and answer

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