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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?
Best Answer
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.