# Finding AC equivalent circuit of a BJT amplifier

Given this circuit,

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 .

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.

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.

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
• Is Ro always in parallel with the transconductance and calculated using beta/gm? Second, if this was PNP, then the current source direction would just be flipped right?
– user91567
Mar 13, 2016 at 20:14
• Yes, Ro always is in parallel with the transconductance. For a PNP the model would be same. The base-emitter voltage is reversed and the current direction as well, therefore the same model results. If the current source was flipped the direction of Vpi would have to be flipped as well. Mar 13, 2016 at 20:26

(1) C1 is not necessary because the circuit is an "equivalent diagram" that is valid for small signals only (no dc values within the circuit).

(2) The product (gm * Vpi) is identical to the product (beta * Ib) because of beta/gm=Vpi/Ib=r,pi

(3) In the transconductance model the controlling voltage is the base-emitter voltage Vbe=Vpi.

(4) The emitter resistor is given and fulfills Ohms law: Re=Ve/Ie (DC values!).