While studying amplifiers, I came across the AC equivalent circuit of an amplifier:

enter image description here

To calculate the output resistance of the amplifier, the input signal was set to zero and the load resistance was removed as shown in the figure:


What is the reason for doing so?

Link to the PDF I am referring to is here.

  • \$\begingroup\$ By driving the output with another signal of known impedance, with 0 input, one can compute Ro by the attenuation. \$\endgroup\$ May 16, 2021 at 10:58
  • \$\begingroup\$ Welcome to Electrical Engineering. I guess you should also show us the derivation of \$R_O\$. \$\endgroup\$ May 16, 2021 at 10:59
  • \$\begingroup\$ @user2233709 I have linked the PDF. It shows the calculation of output resistance. \$\endgroup\$
    – Essar
    May 16, 2021 at 11:10
  • \$\begingroup\$ @TonyStewartEE75 I don't get it. Could you please explain? \$\endgroup\$
    – Essar
    May 16, 2021 at 11:12
  • \$\begingroup\$ It tells you esentially to use Fig 6 b) to test for Ro just like Ri \$\endgroup\$ May 16, 2021 at 11:34

1 Answer 1


To calculate the Thévenin equivalent circuit, you set voltage sources to zero (make voltage sources shorts) and calculate the impedance between the nodes with no load. That's all that's happening here. It's just explicitly showing that the dependent voltage source is also zero (a short). That means the output impedance is whatever R0 is.

In the PDF they also describe how you can use that to measure the output impedance as Tony Stewart points out in the comments.

  • \$\begingroup\$ So we have to consider the thevenin equivalent circuit while calculating the output impedance ? \$\endgroup\$
    – Essar
    May 16, 2021 at 13:53
  • \$\begingroup\$ It's a way of doing it (assuming the model is appropriate). Thevenin equivalence basically reduces things to voltage sources + output resistance. It works if your model is linearized (in other words approximated) around some operating point (that is, q point). \$\endgroup\$
    – KD9PDP
    May 16, 2021 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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