1
\$\begingroup\$

I'm trying to figure out the output impedance of this op Amp circuit and I keep getting infinity. I'm not sure if the steps I'm doing are wrong. The reason I ignore everything besides the voltage divider is because those are the only currents I need to find ix. The problem is that I'm finishing ix to be zero. Any ideas of where I'm going wrong?

\$\endgroup\$
  • \$\begingroup\$ Why do you think that 2Vx will magically appear at the output of the op-amp? What is Z4? \$\endgroup\$ – Spehro Pefhany Sep 2 '16 at 21:48
  • 1
    \$\begingroup\$ Wow, what a mistake on my part. I'm supposed to look at the entire thing backwards starting from Vx aren't i? What a newbie mIstake. Thanks guys \$\endgroup\$ – Rayaarito Sep 2 '16 at 21:56
0
\$\begingroup\$

FWIW,an ideal Zout of an Op Amp is 0 and gain=∞

A practical Op Amp is around 300 Ω if open loop and when closed loop Zout is reduced by dividing the open loop gain then multiply by closed loop gain. So it is very low until it saturates internally.

then Z6 is just R3//R4 since Vout is same impedance as ground.

\$\endgroup\$
1
\$\begingroup\$

The output impedance at the opamps output node is Zout=Zo/(1+LG) with Zo=open-loop impedance and loop gain LG=(k x Aol). The feedback factor is k=(Rs||R1)/[(Rs||R1)+Rf] and Aol=open-loop gain.

Therefore, the output impedance at the output node is Z6=R4||(R3+Zout).

Note that Zout in most cases can be neglected.

\$\endgroup\$

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.