The RF Signal amplifier

 I was talking to another amateur the other day, about a preamp stage for a xtal filter and a following amplifier stage. The xtal filter needed to see a source impedance of 250R and a load of the same.

Now the basic transistor, common emitter, transistor and equivalent circuit looks like this

Where the collector is a current source/sink. A complete circuit would look like this

Transistor AC gain is B

Where Rs is the input source impedance, RBIAS is the bias resistor(s), re is as above, Rc is the collector load (which determines the output impedance) and RL is the load (for maximum power transfer Rc = RL).

Using this circuit we have

Gain Av = Vout/Vin = Rc * ic / re * ie, 

  and since ic ~ ie, Av = Rc/re

Input impedence Zin = re*ie/ib = re*B (when RBIAS >> B*re and B >> 1)

Zout = Rc

So an amplifier with 50R source impedance and 250R output impedance (for the xtal filter source impedance) would look like this

Amplifier

Assuming

B = 100

Ic = 10mA

re = 26/10 = 2.6R

Transistor Zint = 26/10 * 100 = 260R

Amplifier Zin = 50R with a 1:3 transformer

Amplifier Zout = 250R

The transistor bias can be arranged however you want provided the bias resistors are large compared to Zint.  

The output of the xtal filter also needs to see a load impedance of 250R, but this about the same as our amplifier input impedance (260R), so we could have this simple design

Xtal filter post-amplfier

The output side of the post-filter amplifier can be whatever is needed for the following circuits.

But the input impedance is still critical on the collector/emitter current and on the AC Beta (gain) of the transistor. Chosing Ic = 10mW has determined the input impedance.

Feedback

So the next thing to look into is building a feedback amplifier where it might be possible to determine the input/output impedances using some sort of negative feedback across the transistor. And so it is.

This is a simple amplifier (sorry poor screen grap from youtube...)

You can see

1. A 1k feedback resistor, meaning with 50R input source, the gain is Rf/Rs = 1000/50 = 20.

2. The bias is set with the Ve = 3V, and bias is set by the 470R resistor, which does not affect the source resistance much

3. The amplifier gain has to equal the the feedback gain, i.e. 20. This is determined by the RL/Re. RL is 200R (chosen so that a 1:2 transformer drives a 50R output load). And so Re has to be 200/20 = 10R.