Ic of Common Emitter Amplifier

[sspj9]

I am trying to design the second stage of a pre-amplifier, after already designing the power stage (class AB). The pre-amp should consist of two class A stages (common emitter biasing). What do you recommend I assume the value of Ic to be for the second stage? The impedance of the power stage is approx 2800ohms and the required output is of 8vp-p and I am working with a Vcc of 12V.
Thanks.

[Zero999]

This seems like an odd specification, is it coursework by any chance?

A class A amplifier draws the same current, irrespective of the load current so the quiescent current needs to be greater than the maximum possible load current.

By the way, to get 8Vp-p from a 12V output you'll needs bridged output stage.

Silly me, I was thinking you needed 8V peak, not peak to peak, so it should be possible to do this with a single ended amplifier but only just.

[T3sl4co1l]

Eh?  8Vpp is 2V from each rail (+12 and 0)..

What's 2800, the input resistance to the output stage??

How much gain does this preamp need?

Is this purely an open-loop, discrete stage exercise, or do you "know" about negative feedback?

Tim

[orolo]

As Tim hinted, maybe it's better thinking about gain and feedback.

Anyway, as a quick exercise I tried building such an amplifier. Using resistors, to get such swing you need very small resistances, a huge current in the order of 120mA, and the input impedance of the amplifier will be so low that you will have to drive it with a big buffer.

So a current source at the collector is the way to go. In that case, the only AC load will be 2800 Ohm, which with a swing of 4V peak will draw a max of about 4/2800  = 1.4 mA. If you don't want your transistor to saturate when the collector is at 2V from ground, Vce should be about 1V, and that means that the emitter resistor souldn't take more than about 1.2V or so. A good value seems to be a 100 Ohms, since at a quiescent current of 10mA it will take 1V, and at max sink it will take 1V + 1.4*100 = 1.14V, avoiding saturation. So I finally came up with the circuit attached below: the CE amplifier and the current source are complementary, to bias the circuit at 6V. The voltage gain is about 28, and the circuit can swing the full 8V, tough with quite a bit of distortion. I consider this a spice trick, since I had to tweak the emitter resistor of the pnp to get to the correct bias, and a small error causes a big change in the bias point: at this time in the morning I don't come up with a clean way to stabilize it  .

[T3sl4co1l]

You need voltage feedback from the collector.  First try: change base bias divider from +V to collector (and adjust resistor ratio to about half).  Second try: build a differential that compares the output voltage to a half-VCC divider; couple in signal with a capacitor and another resistor to either side.  (At this point, you basically have an op-amp, and using one proper isn't much of a stretch.)

Tim

[orolo]

Thank you for the ideas! I tried to follow this line: the previous circuit consists of a current sink taking current from a current source: the least mismatch will saturate one of them and force a rail hit. So we should take a bit of current from between source and sink, and use it in a feedbacky way, without spoiling impedance or amplification.

So I kept the upper transistor as current source, somewhat above 10mA, and used that current to bias the lower transistor via a voltage divider. Since that lowers the input impedance quite a bit, I bootstrapped the CE amplifier. The final circuit works as the previous one, and seems to be more stable, but I haven't got the time to test it thoroughly.

Thanks for the tips!