GATE 2022 BM – Question 56
The shown PNP BJTs are biased with VEB=0.7 V. Neglect base currents in DC calculations and assume infinite transistor input and output impedance. Input is 10 mV and thermal voltage 25 mV. Find small-signal output in mV to one decimal place. Supply is 5 V; bias resistors are 3.9 kΩ,1.1 kΩ; emitter resistors 6.4 kΩ and 2.5 kΩ; output load 50 Ω.
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Correct answer: 6.2 to 6.8
Explanation
The divider sets the first base at 1.1 V. Its emitter is 1.8 V, giving current (5−1.8)/6400=.5 mA and incremental emitter resistance rₑ1=25 mV/.5 mA=50 Ω. First follower gain≈6400/(6400+50). The second emitter is 2.5 V, giving current one milliampere and rₑ2=25 Ω. Its AC load is 2500||50=49.0196 Ω, so second gain≈49.0196/(49.0196+25). Overall gain≈.6571, producing about 6.6 mV. Coupling/bias sources are AC shorts and both follower stages preserve sign.