0
\$\begingroup\$

[CE]

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?

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ 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? \$\endgroup\$ – Sysnaptic May 30 '17 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.