hi 123.
Surely you can answer this question.

Would you say the amplifier is designed for AC or DC amplification.?


The square wave is due to over driving the input.

In either case, at AC you have an unstable second stage amplifier. It is going to vary all over the place based on changes in transistor gain and it will change with temperature as well. Distortion is guaranteed because the second stage gain is also determined by the bias voltage from base to emitter with will be all over the place for different amplitude AC signals. In fact the circuit might go into oscillation. Depending on cap values. It is not in any way a well designed amplifier.

 

Thank you

Could you provide some information on what would be the purpose and working of such circuit?

I think the professor is looking to see if you can identify a badly designed AC amplifier. Pay careful attention to the second stage. At high frequencies what is the second stage emitter AC voltage and what effects does this have?

 

Honestly? I can't answer to any of your questions. I haven't studied electronics for a few semesters so I've forgotten many things.

Here's the simulation I've done. result was unexpected. output is cut off, I wonder why. It became like a rectangular as if there are diodes in the circuit!

The reason for the cutoff and asymmetric output has to do with the second stage amplifier having an unstable gain. See my other posts.

 

Ok, in theory this can be done. Though in real life we never do it. But it will still depend upon the transistor and the temperature. But taking the average transistor at say room temperature you first have to calculate 'transconductance' this is nothing more than the inversion of the 'dynamic' resistance from base to emitter. That is deltaV / delta I. This only works for really small signals. If the signal is to large it will fail. The common gm calculation is -Ic/Vt where Vt at room temperature is typically 26mV. We calculated Ic for Q2 at 890 microamps so transconductance is -0.00089/0.026 or -0.0342 mhos. The gain is then gm * Rc. In this case 2.5K * -0.0342 (notice mohms cancels with ohms giving your unitless gain) and the voltage gain is thus -85.
So final stage amplification is -85 with output impedance of 2.5K with the 4K load you end up with 4K/(4K+2.5K) = 0.615 reduction in gain. So the entire circuit is a inverting AC voltage amplifier with a 'gain' of around 52.
But, this is only true for extremely small input signals, you will have distortion due to changes in Ic current using larger signals as the Ic is not constant meaning that -gm is NOT constant. And I am not talking about saturation or cutoff, I am literally talking about real AC distortion because -gm depends upon Ic current!