# Determining the output resistance of a BJT amplifier

I'm given the following amplifier: (all the transistors have the same characteristics; Early effect is neglected)

In order to calculate $$\R_{out}\$$, we replace the circuit with its small-signal equivalent: Or, equivalently, In essence, $$\R_L\$$ is the load and $$\R_{out}=\frac{v_{OC}}{i_{SC}}\$$ where $$\v_{OC}\$$ the open-circuit voltage (when $$\R_L\$$ is removed) and $$\i_{SC}\$$ the short-circuit current (when $$\R_L\$$ is replaced with a wire). I managed to find the OC voltage by applying KVL around the loop with $$\R_C, R_2,r_{\pi_2}\$$ and the VCCS:

$$v_{OC} = g_m V_{\pi_2} (R_2 + r_{\pi_2} + R_C) - g_m V_{\pi_1} R_C$$

However finding $$\i_{SC}\$$ seems more complicated. Regardless, every expression that I obtained for $$\i_{SC}\$$ didn't lead to the correct result which is:

$$R_{out}=\frac{r_{\pi_2}+R_2+R_C}{\beta + 1}$$

Judging by this answer, it seems like the term $$\- g_m V_{\pi_1} R_C\$$ in my expression for $$\v_{OC}\$$ is redundant (no expression for $$\i_{SC}\$$ is able to kill this term which appears in the numerator). However ignoring the current flowing into the collector of $$\Q_1\$$ seems wrong to me.

Any suggestions?

• Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached need to be edited back into the question or into an answer. – Dave Tweed Dec 9 '18 at 23:26