GATE 2025 EC Q.21: solution and answer

GATE 2025 · ECQ.21 · 1 mark · MCQ

A simplified small-signal equivalent circuit of a BJT-based amplifier is given below.

The small-signal voltage gain $v_{o} / v_{s}$ (in V/V) is ________.

GATE 2025 EC question 21: Small-signal hybrid-pi circuit. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$\frac{- β R _{L}}{R _{S} + r _{π}}$
(B)
$\frac{+ β R _{L}}{R _{S}}$
(C)
$\frac{- β R _{L}}{R _{S}}$
(D)
$\frac{+ β R _{L}}{R _{S} + r _{π}}$

Answer and solution

ANSWER

$\frac{v _{o}}{v _{s}} = \frac{- β R _{L}}{R _{S} + r _{π}}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Model as drawn: the transistor appears as its hybrid-pi input resistance $r_{π}$ and a dependent current source $β i_{b}$ . No output resistance $r_{o}$ is drawn, so it is taken as infinite (open circuit). The input and output halves are joined only through the common ground.
KVL around the input loop ( $v_{s} \to R_{S} \to r_{π} \to$ ground). The same current $i_{b}$ flows through $R_{S}$ and $r_{π}$ : $v_{s} = i_{b} (R_{S} + r_{π})$ , so $i_{b} = \frac{v _{s}}{R _{S} + r _{π}}$
Direction of the dependent source: its arrow points down, so the current $β i_{b}$ enters the source's top terminal from the $v_{o}$ node and leaves at the bottom terminal into ground. In other words, the source pulls $β i_{b}$ out of the output node.
KCL at the output node, summing currents leaving it: $\frac{v _{o}}{R _{L}} + β i_{b} = 0 \Rightarrow v_{o} = - β i_{b} R_{L}$
Substitute $i_{b}$ from step 2: $\frac{v _{o}}{v _{s}} = \frac{- β R _{L}}{R _{S} + r _{π}} V/V$
The gain is negative, because current pulled out of the load node inverts the phase as in a common-emitter stage. Its magnitude has $R_{S} + r_{π}$ in the denominator. That is option (A).

KEY IDEA

The source resistance $R_{S}$ and $r_{π}$ are in series, so $i_{b} = v_{s} / (R_{S} + r_{π})$ . The dependent source forces $β i_{b}$ through $R_{L}$ , which gives $∣ v_{o} ∣ = β i_{b} R_{L}$ . The source arrow points away from the output node, so the output voltage is negative.

WATCH OUT FOR

One mistake is to drop $r_{π}$ from the input loop and write $i_{b} = v_{s} / R_{S}$ , which leads to option (C). The other is to misread the source's direction and lose the minus sign, which leads to (D). The downward arrow means the source pulls current OUT of the $v_{o}$ node through $R_{L}$ from ground, so $v_{o}$ is negative.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 20 Amplifiers questions, with notes

GATE 2025 EC Q.21: solution and answer

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