# Class B amp

Lunat1c
Would someone be kind enough to explain what happens to the operation of both transistors in both the positive cycle and the negative cycle of the input signal please? I'm having a hard time understanding how this works exactly.

What I know is that with no input signal, both transistors have a quiescent current going through them. Since they're complementary they will have the same voltage drop across them and in between them (i.e. the point where RL is connected) will be at Vcc/2. Now, once the input signal is applied, the positive cycle is supposed to be applied across TR1. TR2 would have a base voltage greater than the collector voltage and therefore it will go into cutoff.

When on the other hand there's the negative cycle, TR1 goes into cutoff which is equivalent to an open circuit. How will TR2 maintain the quiscent current given that now there's an open circuit instead of TR1?

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There is a mistake in that circuit. The bottom supply rail would have to be negative, not earthed, if you have the load connected like that.
If it was a speaker and with a single supply, there would be a large electrolytic capcitor in series with the load.

Each transistor acts as an emitter follower with the load alternately getting power from the two transistors but at opposite polarity.
It doesn't matter that the non operating transistor is cut off while the operating one sends a lot of current into the load resistor.

When there is a signal, neither transistor gets quiescent current. This is just a small current that flows when there is no signal.

Lunat1c
if we were to consider the loadlines of both transistors.. would it be correct to say that the quiescent voltage of the first one is Vcc/2 and that of TR2 to be -Vcc/2 for all values of collector current?

that would mean that the signal is applied to both transistor. however the first one clips the negative cycle while the other one clips the positive cycle and my problem would be solved ^^