A) Explain why the emitter follower is getting cut off when driving the follower

Question

A) Explain why the emitter follower is getting cut off when driving the follower with a sine wave of about 1 V and 1 kHz that is symmetrical about zero volts.

B) At higher amplitude, explain the bumps below ground (hint: see VBE breakdown specification for the 2N440 transistor" the '440 is very similar to the '3904, and in this characteristic is identical)

C) For higher amplitude with V(EE) = -15V, explain why the emiter follower is now giving the full wave. Also explain the voltage drop.

Explanation / Answer

A)ANSWER :FOR THIS 1V SINE WAVE THE EMITTER JUNCTION IS ALWAYS REVERSE BIAS DUE TO HIGHER POTENTIAL(15V) AT N-TERMINAL. AS SINE WAVE IS VARYING ITS AMPLITUDE BETWEN 1V TO -1V, COLLECTOR JUNCTION GETS REVERSE BIAS. SO AS BOTH JUNCTIONS ARE REVERSE BIAS, THE EMITTER FOLLOWER IS GETTING CUTOFF.

B)ANSWER :AT HIGHER AMPLITUDE i.e, AT 1V PEAK THE EMITTER JUNCTION IS FORWARD BIAS AND COLLECTOR JUNCTION IS REVERE BIAS, IN THIS CASE THE EMITTER FOLLOWER ACTS LIKE A ATTENUATOR, SO THE OUTPUT IS ATTENUATED VERSION OF INPUT..

C)ANSWER : FOR HIGHER AMPLITUDE WITH VEE= -15V, THE EMITTER JUNCTION IS FORWARD BIAS AND THE COLLECTOR JUNCTION IS REVERSE BIAS, IN THIS CASE THE TRANSISTOR IS IN ACTIVE REGION OF OPERATION AND ACTS LIKE A AMPLIFIER AND GIVES FULL OUTPUT FOR GIVEN INPUT IF THE Q-POINT IS LOCATED AT MIDDLE OF THE DC LOAD LINE.

BY APPLYING KVL TO I/P LOOP:

1V-IB*270-0.7-IE*3.3K+15=0

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