8
\$\begingroup\$

What is the most effective way of measuring the input/output impedance of an amplifier in a simulation?

When I know the impedance I want to measure is purely resistive, I usually set up an input signal Vin and a test resistor as a resistive divider with the desired impedance. Then I compare the voltage values of the input/output and work my math to get a number for the impedance. Is there a better way?

I am using Orcad Capture with PSpice.

\$\endgroup\$
1

1 Answer 1

10
\$\begingroup\$

Create a current source whose AC magnitude is 1A. Connect one terminal to GND, and connect the other terminal to the appropriate pin of the device under test. Do an AC simulation, and plot the voltage on that pin. Remembering that \$Z(j\omega)=V(j\omega)/I(j\omega)\$, and since \$I(j\omega)=1\$ (because you've set it to 1), then the impedance is simply equal to the voltage: \$Z(j\omega)=V(j\omega)\$.

You might be concerned that 1A is a lot of current for your circuit to handle. But it doesn't matter: AC simulations are, by nature, small-signal analyses. Namely, when doing an AC simulation, the DC operating point is computed first and the AC relationships are found without consideration to AC amplitudes -- i.e., assuming small-signal conditions. So even with a 1A AC current source, the AC analysis is still a small-signal analysis, even if your circuit couldn't handle that level of current in real life (or even in a transient simulation).

\$\endgroup\$
1
  • 1
    \$\begingroup\$ This is a great answer! I should add that the source used should be the one that doesn't affect the bias point. I had to use a VAC source in the input in my simulation. The IAC source was opening some wires it shouldn't be opening in DC. \$\endgroup\$
    – Costagero
    May 9, 2015 at 17:56

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.