# Calculation of common emitter output impedance with output resistance

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I don't understand why $g_{m}$ survives. Since $v_{\pi}$ is equivalent to $\Delta V$ , which is perturbation of signal, which is then set as AC ground. Doesn't that mean $v_{\pi}= \Delta V = 0$? Thus make $g_{m}$ open for calculation? How can I calculate the output impedance like so? With inspection method assuming $g_{m}$ is open, I'm getting $R_{out} = ( r_{\pi} || R_{E} ) + r_{0}$ also could you show how to calculate the input impedance as well?

No. What is happening here is the test voltage Vx will create a current Ix. These are used to calculate the output impedance. What you are not taking into consideration here is that Vx will cause a voltage change at node P. Since the other end of r_pi is connected to ground, this cause a change in delta V. Delta V is not zero.

• What if R_o is not there? Then is g_m open in that case? Where I can define Delta V = v_pi as zero? – Sysnaptic May 30 '17 at 1:43