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What is the Q-point for a fixed-bias transistor with IB = 75 µ A ,βDC = 100, VCC = 20 V, and RC = 1.5 K Ohm?(a) VC = 0 V(b) VC =12.25V(c) VC = 8.75 V(d) VC = 20VThis question was posed to me in an interview for internship.This question is from Input Bias and Offset Currents of the Bipolar Differential Amplifier topic in division Differential and Multistage Amplifiers of Electronic Devices & Circuits

Answer»

Right answer is (C) VC = 8.75 V

Easiest EXPLANATION: VCE=Vcc-IC Rc(VCE = VC , β= IC/IB => IC= β*IB)

Vc = Vcc – β* IB * Rc

=20-100x75x10^-6 x1.5x 10^-3

=8.75V.



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