Electrical – Finding AC equivalent circuit of a BJT amplifier

biasbjtcircuit analysis

Given this circuit,
enter image description here

I want to find the AC equivalent circuit. Now, the rules I apply are

  1. Short the DC voltages
  2. Short-circuit the capacitors
  3. Open-circuit the inductors
  4. Simplify
  5. Find amplifier parameters.

The solution given was enter image description here.

My questions are the following.

  1. Why is C1 still left in the circuit?
  2. I assumed the current would be beta * Ib. But they are using gm * Vpi. I know gm = Ic/Vt, with Vt = 26mV. However, I am confused on which convention to use. Or are they both the same thing?
  3. Why is gm multiplied by Vpi? If it's voltage controlled current source, then shouldn't it be gm * Vo?
  4. Lastly, would Re be calculated by doing Vo/Ie or Vt / Ie?

Different sources are giving me different ways of answering my questions and I want to narrow down to the most accurate method.

Best Answer

In order to find the operating point for your circuit you have to perform the steps 1-5 given above. The operating point is equivalent to the DC analysis of the circuit. Inductors and capacitors are therefore removed by a short or open circuit, respectively.

For this DC solution the circuit is linearized. Your transistor is replaced by a simple transconductance (voltage-controlled current source). The AC components are of course required for the AC model of the circuit and need to be included again.

To answer your questions:

  1. C1 is left in the circuit because we need the AC model of the circuit
  2. The transistor is now a voltage-controlled current source, with a transconductance gm
  3. Vpi is simple the controlling voltage (the base-emitter voltage) of the transistor
  4. Re is the emitter resistor which is given as 1kOhm
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