2
\$\begingroup\$

I'm trying to understand why the output impedance of non-inverting and inverting opamp circuit is almost zero. I found the following text on the derivation:

enter image description here enter image description here

β is the closed loop gain above. But I couldn't understand their derivation here. Is there an easier or clearer way to reach this solution? I would be glad to see a more verbal and clearer explanation.

Edit:

If this equivalent circuit below is correct, how can we proceed from this equivalent circuit and conclude Zout of this inverting opamp circuit is almost zero? Or should I say Zout = Rout. (Rout is the opamp's internal output resistance, Zout is the circuit's output impedance/resistance)

enter image description here

\$\endgroup\$
6
  • \$\begingroup\$ Clue me in on what part of the derivation you don't understand. Help me help you. \$\endgroup\$
    – Jason_L_Bens
    Commented Apr 8, 2017 at 14:20
  • \$\begingroup\$ This explanation is very implicit. I need a opamp model to comprehend this derivation. \$\endgroup\$
    – user16307
    Commented Apr 8, 2017 at 14:52
  • \$\begingroup\$ Do you understand that both configurations are using negative feedback, and that this is why the same equation regarding output impedance applies? The opamp model is also the same -- a voltage source in series with an internal impedance \$Z_{out}\$. \$\endgroup\$
    – Dave Tweed
    Commented Apr 8, 2017 at 15:05
  • \$\begingroup\$ I accept an ideal opamp has almost zero output impedance as a premise. But when it becomes a circuit such as in an inverting opamp I need a circuit model like showing what is parallel to what so that I can conclude Zout is in parallel with R2 ect. I need to see this relation on a model \$\endgroup\$
    – user16307
    Commented Apr 8, 2017 at 15:08
  • \$\begingroup\$ In other words, I need to see the derivation of the output impedance of an inverting amplifier by using an equivalent circuit of an inverting amplifier. The explanation I provided is implicit, I cannot picture it. \$\endgroup\$
    – user16307
    Commented Apr 8, 2017 at 15:16

1 Answer 1

Reset to default
1
\$\begingroup\$

I think the images you posted cover the math reasonably. Let me just try to give you a bigger picture view.

Output impedance is the change in output voltage due to a change in the output current. (I don't think your sources do a good job making this point, at least within the bits you cut and pasted).

So if you have negative feedback circuit, and the load starts to draw more current (say it has a switch inside it that gets closed), then what happens?

At first, the output voltage might drop, because of the internal resistance of the op-amp output circuit. But when it does, that will be conveyed by the feedback network back to the op-amp inputs, telling it to increase its output voltage. Which counteracts (mostly) that initial change. If the load current change is slow enough, you wouldn't even see the "initial" change before the feedback responded and eliminated it.

So the output voltage change of the closed-loop circuit is much smaller than it would have been without the feedback. This is the point your sources are trying to convey.

Of course, this all depends on the changes in load current being slow enough for the op-amp and feedback network to keep up with. That means you can only count on this improvement in output impedance for load changes with frequencies in the operating bandwidth of your circuit (op-amp and feedback network).

\$\endgroup\$
8
  • \$\begingroup\$ Regarding the equivalent circuit in my question(the last figure), can we say that this equation is correct: Zout = Ro//(Rf+Rs)? This is also supports your argument Zout<Ro \$\endgroup\$
    – user16307
    Commented Apr 8, 2017 at 16:27
  • \$\begingroup\$ No that doesn't account for the loop gain (\$\beta A_{OL}\$) of the circuit responding to the disturbance. \$\endgroup\$
    – The Photon
    Commented Apr 8, 2017 at 16:29
  • \$\begingroup\$ This is more complicated than I was assuming \$\endgroup\$
    – user16307
    Commented Apr 8, 2017 at 16:33
  • 1
    \$\begingroup\$ Big picture: The negative feedback causes the amplifier to respond to correct a disturbance in the output voltage. This reduces (dramatically) the change in output voltage when load current changes. Which is mathematically indicated by low output impedance. \$\endgroup\$
    – The Photon
    Commented Apr 8, 2017 at 16:35
  • \$\begingroup\$ I see your point. When the current increases since feedback fixes the output Voltage and Zout= dVout/dIout; and dVout=0 in feedback case Zout becomes zero. If there were no feedback dVout wouldnt be zero and Zout would be bigger. Conceptually I see the point. But mathematically not. Maybe I dont need the math behind. \$\endgroup\$
    – user16307
    Commented Apr 8, 2017 at 17:08

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.