GATE 2014 EC (EC4) – Question 49
The common-collector amplifier shown has high β, negligible V_CE(sat), and V_BE = 0.7 V. Find maximum undistorted peak-to-peak output v₀ (V).
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Correct answer: 9.39 to 9.41
Explanation
Highβmakes the bias-divider voltage approximately12×10/(5+10)=8V, so the emitter rests at7.3V. With negligible VCE,sat, the upper limit is12V and lower cutoff limit is0V. The smaller headroom is12−7.3=4.7V, permitting a symmetric peak-to-peak swing of **2×4.7=9.4V**.