GATE 2026 EC Q.50: solution and answer

GATE 2026 · ECQ.50 · 2 marks · MCQ

A small signal source, $V_{i} (t) = A cos (1 0^{5} t) + B sin (1 0^{7} t)$ is applied to a BJT circuit as shown in the Figure.

Assume zero source resistance, $V_{B E} = 0.7$ V, $β_{d c} = 99$ , Early voltage $= 100$ V and Thermal voltage $= 25$ mV. Effect of internal parasitic capacitances of the BJT may be neglected.

Which expression is the best approximation of the output voltage $V_{o} (t)$ ?

GATE 2026 EC question 50: Single-stage common-emitter amplifier with an NPN BJT. — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
$- 9.1 [A cos (1 0^{5} t) + B sin (1 0^{7} t)]$
(B)
$9.1 [A cos (1 0^{5} t) - B sin (1 0^{7} t)]$
(C)
$- 190.4 [A cos (1 0^{5} t) + B sin (1 0^{7} t)]$
(D)
$190.4 [A cos (1 0^{5} t) - B sin (1 0^{7} t)]$

Answer and solution

ANSWER

Both tones fall in the mid-band. The coupling and bypass reactances are negligible and the BJT's parasitic capacitances are neglected, so both tones see the same inverting gain of a CE stage degenerated by the unbypassed 500 Ω, $A_{v} \approx - g_{m} (R_{C} ∥ r_{o}) / (1 + g_{m} R_{E}) \approx - 9.1$ . That gives $V_{o} (t) \approx - 9.1 [A cos (1 0^{5} t) + B sin (1 0^{7} t)]$ , option (A).

Official answer: ASource: Official answer key, GATE 2026 EC (IIT Guwahati)

DC bias (all capacitors open). Replace the base divider with its Thevenin equivalent: $V_{T H} = 12 \cdot \frac{25}{100 + 25} = 2.4 V$ and $R_{T H} = 100 k Ω ∥ 25 k Ω = 20 k Ω$ . With the 10 µF capacitor open, the DC current flows through both emitter resistors, so $R_{E, d c} = 500 + 1000 = 1.5 k Ω$ .
KVL around the base-emitter loop: $I_{B} = \frac{2.4 - 0.7}{20 k + ( 99 + 1 ) ( 1.5 k )} = \frac{1.7}{170 k} = 10 μ A$ so $I_{C} = β_{d c} I_{B} = 0.99 mA \approx 1 mA$ (the Early effect is ignored for the bias point) and $I_{E} = (β_{d c} + 1) I_{B} = 1 mA$ . Check the region: $V_{C} = 12 - 0.99 \times 5 = 7.05 V$ and $V_{E} = 1 \times 1.5 = 1.5 V$ , so $V_{C E} \approx 5.55 V$ . The transistor is in the active region.
Small-signal parameters (hybrid- $π$ model, taking $I_{C} \approx 1 mA$ ): $g_{m} = I_{C} / V_{T} = 1 mA /25 mV = 40 mS$ , $r_{π} = β / g_{m} \approx 99/0.04 \approx 2.5 k Ω$ , $r_{o} = V_{A} / I_{C} \approx 100 k Ω$ . In the model, $r_{π}$ sits between base and emitter, and the source $g_{m} v_{π}$ and $r_{o}$ both run from collector to emitter. The BJT's parasitic capacitances are left out, as the question says.
Capacitor reactances at the lower tone, $ω = 1 0^{5} rad/s$ : $X_{100 nF} = \frac{1}{1 0 ^{5} \cdot 1 0 ^{- 7}} = 100 Ω$ and $X_{10 μ F} = \frac{1}{1 0 ^{5} \cdot 1 0 ^{- 5}} = 1 Ω$ . At $ω = 1 0^{7} rad/s$ both are 100 times smaller: $1 Ω$ and $0.01 Ω$ .
Both tones are in the mid-band. Input: $R_{in} = R_{T H} ∥ [r_{π} + (β + 1) (500)] = 20 k ∥ 52.5 k \approx 14.5 k Ω$ . The source resistance is zero, so the input capacitor and $R_{in}$ form a divider: $∣ V_{b} / V_{i} ∣ = 1/ 1 + (100/14 500)^{2} > 0.9999$ , with a phase lead of $tan^{- 1} (100/14 500) \approx 0. 4^{\circ}$ at $1 0^{5} rad/s$ and less at $1 0^{7} rad/s$ . Output: there is no load, so no signal current flows through the output capacitor and $V_{o} = V_{c}$ at both frequencies. Emitter: the 1 kΩ in parallel with 10 µF has an impedance of about $1 Ω$ at $1 0^{5} rad/s$ ( $0.01 Ω$ at $1 0^{7}$ ). That is negligible next to the 500 Ω in series with it, so the AC emitter resistance is $R_{E} = 500 Ω$ . High frequencies: the parasitic capacitances are neglected, so the gain does not roll off. Both tones therefore see the same real, inverting gain.
Mid-band gain of a CE stage with an unbypassed $R_{E} = 500 Ω$ . The standard approximation takes $I_{E} \approx I_{C}$ and treats $r_{o}$ as if it were in parallel with $R_{C}$ : $A_{v} \approx - \frac{g _{m} ( R _{C} ∥ r _{o} )}{1 + g _{m} R _{E}} = - \frac{0.04 \times ( 5 k ∥ 100 k )}{1 + 0.04 \times 500} = - \frac{190.5}{21} \approx - 9.07 \approx - 9.1$ Cross-check without those approximations. With $r_{o}$ neglected, the model gives exactly $A_{v} = - \frac{β R _{C}}{r _{π} + ( β + 1 ) R _{E}} = - \frac{99 \times 5000}{2500 + 100 \times 500} \approx - 9.43$ . Putting $r_{o}$ back where it really is (collector to emitter) moves this only to about $- 9.40$ : the emitter degeneration raises the resistance seen looking into the collector to about $r_{o} [1 + g_{m} (R_{E} ∥ r_{π})] \approx 1.8 M Ω$ , far above $R_{C}$ . Every model gives a gain of about $- 9$ . None comes near 190.
Output: the same real gain multiplies both components, so $V_{o} (t) \approx - 9.1 [A cos (1 0^{5} t) + B sin (1 0^{7} t)]$ Option (A) is the only option with a magnitude near 9 and the same inverting sign on both terms.
Why not the others: $g_{m} (R_{C} ∥ r_{o}) = 0.04 \times 4.76 k \approx 190.5$ (printed as 190.4 in (C) and (D)) would be the gain only if the whole emitter resistance were bypassed. Here the 500 Ω is not bypassed, so (C) is out. Options (B) and (D) make the $1 0^{5}$ term non-inverting and flip the sign of only the $1 0^{7}$ term. A CE stage inverts, and a $18 0^{\circ}$ phase difference between two tones that both sit in the flat mid-band is impossible.

KEY IDEA

Bias with the full 1.5 kΩ in the emitter ( $I_{C} \approx 1$ mA, so $g_{m} = 40$ mS). Then check that both tones are in the mid-band: at $1 0^{5}$ rad/s the 100 nF capacitors are 100 Ω, which is tiny next to a 14.5 kΩ input and an open output, and the 10 µF is 1 Ω, which leaves only the 500 Ω unbypassed. With the parasitics neglected, both tones see one real gain, $- g_{m} R_{C}^{'} / (1 + g_{m} R_{E}) \approx - 9$ .

WATCH OUT FOR

Treating the emitter as fully bypassed and taking the gain as $- g_{m} (R_{C} ∥ r_{o}) \approx - 190$ (option C). The 10 µF bypasses only the 1 kΩ, so the 500 Ω still degenerates the gain by a factor of $1 + g_{m} R_{E} = 21$ . Another trap is assuming the $1 0^{7}$ rad/s tone picks up a different phase and flips sign (options B and D). With the parasitics neglected and the coupling capacitors already short at $1 0^{5}$ rad/s, both tones sit in the flat mid-band. Also watch the bias: the DC emitter resistance is the full 1.5 kΩ, not 500 Ω.

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GATE 2026 EC Q.50: solution and answer

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