GATE 2020 EC Q.35: solution and answer

GATE 2020 · ECQ.35 · 2 marks · MCQ

For the BJT in the amplifier shown below, $V_{B E} = 0.7 V$ , $k T / q = 26 mV$ . Assume that BJT output resistance ( $r_{o}$ ) is very high and the base current is negligible. The capacitors are also assumed to be short circuited at signal frequencies. The input $v_{i}$ is direct coupled. The low frequency voltage gain $v_{o} / v_{i}$ of the amplifier is

GATE 2020 EC question 35: Common-emitter amplifier with an NPN BJT and split supplies. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
$- 89.42$
(B)
$- 128.21$
(C)
$- 178.85$
(D)
$- 256.42$

Answer and solution

ANSWER

The low-frequency voltage gain is $v_{o} / v_{i} = - g_{m} (R_{C} ∥ R_{L}) = - 89.42$ , option (A).

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

Assumptions. Hybrid- $π$ model with $r_{o} \to \infty$ (given), base current negligible so $I_{C} \approx I_{E}$ , $V_{T} = k T / q = 26 mV$ , and all capacitors are open at DC and short at signal frequencies (given).
DC bias. The source $v_{i}$ is direct coupled and has no DC component, so the base sits at $V_{B} = 0 V$ . With the capacitors open at DC, the emitter is at $V_{E} = V_{B} - V_{B E} = - 0.7 V$ .
Emitter current (Ohm's law across $R_{E}$ ). $I_{E} = \frac{V _{E} - V _{E E}}{R _{E}} = \frac{- 0.7 - ( - 10 )}{20 k Ω} = \frac{9.3 V}{20 k Ω} = 0.465 mA$ Base current is negligible, so $I_{C} \approx I_{E} = 0.465 mA$ .
Check the operating region. $V_{C} = 10 - (0.465 mA) (10 k Ω) = 5.35 V$ , so $V_{C E} = 5.35 - (- 0.7) = 6.05 V$ , well above saturation. The BJT is in the forward-active region, so the small-signal model applies.
Transconductance. $g_{m} = \frac{I _{C}}{V _{T}} = \frac{0.465 mA}{26 mV} = 17.885 mA/V$
Small-signal circuit. $C_{E}$ shorts the emitter to AC ground, so $R_{E}$ drops out of the signal path. $C_{1}$ shorts the collector to $v_{o}$ , so the collector sees $R_{C} ∥ R_{L} = 10 k Ω ∥ 10 k Ω = 5 k Ω$ . Because $v_{i}$ drives the base directly with the emitter at AC ground, $v_{b e} = v_{i}$ (the value of $r_{π}$ does not affect the gain here).
Common-emitter gain. The collector current $g_{m} v_{b e}$ flows out of the collector node through $R_{C} ∥ R_{L}$ , so $\frac{v _{o}}{v _{i}} = - g_{m} (R_{C} ∥ R_{L}) = - (17.885 mA/V) (5 k Ω) = - 89.42$
This matches option (A).

KEY IDEA

The DC base voltage is 0 V because $v_{i}$ is direct coupled, which fixes $I_{E} = (0 - 0.7 + 10) /20 k Ω = 0.465 mA$ and $g_{m} = I_{C} / V_{T}$ . With $C_{E}$ bypassing $R_{E}$ and $v_{b e} = v_{i}$ , the gain of this common-emitter stage is $- g_{m}$ times the total AC collector load $R_{C} ∥ R_{L}$ .

WATCH OUT FOR

Leaving out $R_{L}$ : $C_{1}$ is a short at signal frequencies, so $R_{L}$ sits in parallel with $R_{C}$ . Using $R_{C}$ alone gives $- g_{m} (10 k Ω) = - 178.85$ , which is option (C). Other slips are ignoring $V_{B E}$ in the bias ( $I_{E} = 10/20 k Ω = 0.5 mA$ ), and keeping $R_{E}$ in the gain even though $C_{E}$ bypasses it.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 20 Amplifiers questions, with notes

GATE 2020 EC Q.35: solution and answer

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