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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?

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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