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I'm given the following amplifier:

enter image description here

(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:

enter image description here

Or, equivalently,

enter image description here

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?

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  • \$\begingroup\$ 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. \$\endgroup\$ – Dave Tweed Dec 9 '18 at 23:26

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