# Voltage and current gain in two stage BJT amplifier

In the two stage amplifier bellow, by doing the DC analysis you will find the bias currents $$\I_{C,T1}=3.71mA\$$ and $$\I_{C,T2}=5.48mA\$$, if I am correct.

In order to find the voltage gain from Base T1 to Collector T2 $$\\left( A_v=\frac{V_{C2}}{V_{B1}}\right)\$$, I think we should follow the following steps, correct me if I am wrong:

1. We calculate $$\\frac{v_{C1}}{v_{B1}}=A_{v1}=-g_m \left( R_C \parallel R_L \right)\$$, since we have Common Emitter Configuration and $$\R_e=0\$$, since in the AC-analysis, the capacitor C2 becomes a shortcut.
2. We follow similar process to find $$\A_{v2}\$$
3. We have that $$\v_{C1}=v_{B2}\$$
4. Now we calculate $$\A_v=A_{v1}\cdot A_{v2}\$$

Now, what about the current gain?

The image bellow shows the AC-analysis I have done for the circuit by short-circuiting the capacitors and replacing the BJTs with the π model.

I find $$\\frac{i_{C1}}{i_{B1}}=g_{m1}r_\pi\$$ and $$\\frac{i_{C2}}{i_{B2}}=g_{m2}r_\pi\$$.

However how am I supposed to find $$\\frac{i_{B2}}{i_{C1}}\$$?

I thought about a current divisor, but did not know the resistance of the path with the current source.

• Does R1 really connect T1 base to 10V? Perhaps it should connect to T1 collector. Commented Jun 7 at 17:55