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I'm trying to review some old concepts, and thus far, there was one that I couldn't really understand: impedance of circuits involving op-amps. Assuming an ideal op-amp, it has infinite input impedance and no output impedance, but that's only for the op-amp itself. I'm having trouble understanding how to find the input/output impedance of some op-amp circuits as a whole.

Take the standard non-inverting amplifier for instance:

schematic

simulate this circuit – Schematic created using CircuitLab

Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals. However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case. Replacing the load resistor with a current source, you just see an internal op-amp 'output' resistance of 0 ohms in parallel with Z2 to ground, so is that set of parallel impedances the cause of the 0 ohm output impedance? Is this logic correct?

It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance. Is this always the case, or are there some exceptions? I'm trying to develop some methodologies for measuring such impedances in circuit problems as it's hard for me to wrap my head around.

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The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.

Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals.

That's OK.

However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case.

I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Rout is the output impedance of the op-amp.

With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = \$ \frac {Z_3}{Z_2+Z_3} V_{IN} \$. That means that the output of the op-amp actually has to go some distance beyond Vout to compensate for the voltage drop across the output resistance.

It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance.

Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.

Does that help?

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  • \$\begingroup\$ One comment: Yes - the ouput impedance of an opamp with feedback is very low and can be neglected - as long as the loop gain large enough !! With rising frequencies the loop gain drops and the ouput impedance gets larger (and gets an inductive component). Perhaps one thinks that this is not a problem as long as the whole circuit is operated at lower frequencies - however, the described effect has an influence on the stabiliy margin of the circuit, in particular if a capacitive load is connected. \$\endgroup\$ – LvW Dec 14 '18 at 8:51
  • \$\begingroup\$ Thanks for that, @LvW. My op-amp experience is all on audio and only for hobby. I had, however, struggled briefly many years ago with the same question the OP had - although only in the simple resistive feedback configuration so I hope that the answer may help. \$\endgroup\$ – Transistor Dec 14 '18 at 12:04
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Just to add a little more to Transistor's answer (a mathematical approach).

You can look at it this way. This is the circuit model of the op amp (with intrinsic output impedance) plus the external resistors:

schematic

simulate this circuit – Schematic created using CircuitLab

Obviously from the circuit, \$V^+=0\$. You can try to find and expression for \$\dfrac{V_{\text{test}}}{I_{\text{test}}}\$.

Also \$V^-=V_{\text{test}}\dfrac{R_1}{R_1+R_2}\$, \$I_o=\dfrac{V_{\text{test}}+AV^-}{R_o}\$, and \$I_f=\dfrac{V_{\text{test}}}{R_F+R_1}\$.

You combine those, and obtain:

$$ \dfrac{V_{\text{test}}}{I_{\text{test}}}=\dfrac{R_o(R_1+R_F)}{R_1+R_F+AR_1+R_o}$$

Given the fact that \$A\$ is a really large number (the op amp open-loop gain), the \$AR_1\$ term dominates in the denominator.

So,

$$ \dfrac{V_{\text{test}}}{I_{\text{test}}} \approx \dfrac{R_o(R_1+R_F)}{AR_1}$$

Remember that \$G=\dfrac{R_1+R_F}{R_1}=1+\dfrac{R_F}{R_1}\$ is the closed loop gain. Which allows to further re-write this as:

$$ \dfrac{V_{\text{test}}}{I_{\text{test}}} \approx \dfrac{R_oG}{A}\to 0$$

And that approaches zero or a very low value. As you can see the output impedance under negative feedback is even smaller than the intrinsic output impedance of the op amp given the fact that \$A\$ is large and \$G\$ is no doubt much smaller. If you were to just ignore \$R_o\$ this would be ideally zero.

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The 90 degree phase_shift of the OPAMP's open loop gain has the result of producing an Inductive Zout.

This Zout, when loaded with a capacitor, results in ringing that may need dampening.

A good value, to place between the OpAMP output pin and the Capacitor, is R = sqrt(L / C).

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  • \$\begingroup\$ For my opinion, this is somewhat like an oversimplification. I think, the output impedance of the opamp without feedback (at least at low and midrange frequencies) is small (some tenth of ohms) and is primarily ohmic. For higher frequencies it gets an inductive component but it is not "inductive". The problem wirh a capacitive load is caused by a lowpass effect created by the (ohmic) output resistance and the load capacitance. Hence, the loop gain has an increased phase shift - and the phase margin is reduced. \$\endgroup\$ – LvW Dec 14 '18 at 8:46
  • \$\begingroup\$ Without feedback, a bipolar output classB circuit running at 0.5mA will have 2 'reac' (1/gm) in parallel, from the pullup and the pulldown bipolars, each of reac 52 ohms, thus total in-parallel of 26 ohms. I've seen FET opamps have 80,000 ohms Rout at/near DC (opamp ran on 1uA), and I've seen some FET opamps with less than 100 ohms at/near DC. All these numbers are WITHOUT feedback. \$\endgroup\$ – analogsystemsrf Dec 14 '18 at 12:50

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