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In Learning the Art of Electronics pg.362, the author presents an op-amp follower driving a long cable. I don't understand the explanation given for the cause of oscillation. In particular, the author claims that the inductive output impedance of the op-amp (caused by weakened feedback due to op-amp dominant pole gain roll-off - not actual inductance) combined with the cable's capacitance reduces the phase margin to zero at a frequency where gain is greater than unity, causing oscillation. However, I don't understand how the closed-loop output impedance can cause oscillation.

I was under the impression that stability is determined entirely by the loop-gain, not the closed-loop gain. For instance, to analyze stability, the loop is broken (say, at the inverting terminal of the op-amp), the op-amp output is loaded with the impedance seen looking into the break point, the non-inverting input is grounded, and a signal is applied at the inverting input. The loop gain is then the negative of the gain measured at the break point. But no-where in this analysis does the op-amp output impedance appear inductive because this is not a closed-loop analysis.

enter image description here

enter image description here

How can the inductive output impedance of the closed-loop circuit (which becomes entirely resistive when the loop is opened) affect phase-margin if phase-margin depends solely on the loop-gain and not on the closed-loop gain?

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  • \$\begingroup\$ The author's analogy to an internal inductance only leads to more confusion IMHO. A simpler explanation is that the presence of the external capacitance combined with the output resistance of the op amp produces an extra pole in the frequency response that can bring the phase shift close to 180˚ at high frequencies, what causes oscillation. Notice that the presence of the capacitor does affect the loop gain of the op amp due to the fact that the output impedance is not zero. In other words, the op amp is not an ideal source and the load influences its behavior. \$\endgroup\$
    – joribama
    Nov 6, 2019 at 5:24
  • \$\begingroup\$ Here you have a interesting readout on a similar problem edn.com/electronics-blogs/the-signal/4398048/… \$\endgroup\$
    – G36
    Nov 7, 2019 at 17:04

3 Answers 3

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I reckon you're missing the fact that whatever's connected to the output of an opamp, actually does form part of the feedback loop. It's not obvious from the classic opamp follower diagram, because that diagram does not show such an influence "inserted" into the loop itself:

schematic

simulate this circuit – Schematic created using CircuitLab

However, no opamp has zero output impedance, and all loads on the output are able to modify the opamp's output, from the value it would have had with no load. Anything connected to the output of even a simple follower, that influences the opamp's expected output in some way, is actually influencing the signal that gets fed back, and is therefore surreptitiously introducing some feature of its own behaviour into the "loop gain". Here's a couple of extreme examples that illustrate this:

schematic

simulate this circuit

In the first (left) example, the output is basically just shorted to ground, which may seem stupid (and it is), but clearly that 0Ω is able to alter the loop gain to become 0. There's no way that the signal being fed back can have any other value than 0V.

This is only possible because the opamp is unable to provide infinite output current, due to its finite output impedance. In the second (right) example, I show this inherent output impedance as \$R_O\$, and the entire dotted box represents the opamp. To highlight the problem that \$R_O\$ poses, I loaded an otherwise simple follower with 1µF of capacitance.

That capacitance, together with \$R_O\$, plants a breakpoint firmly at 160Hz, along with the phase shift and attenuation that go with it, and which sits right inside the feedback loop even though the capacitor in the schematic appears visually outside it.

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The rising Rout also contributes a phase shift.

That internal-to-opamp phase shift interacts with the external capacitor to produce a total of 180 degrees.

The unwanted 180 degrees, along with the opamp's negative close-loop gain, becomes 360 degrees.

And Barkhausen tells us ........


Here is a simulation you could be running; the lower left plot shows the virtual inductance.

The upper left plot shows the 30dB peaking, where XL = Xc (the 1uF cap).

enter image description here

Notice, in the upper plot, the 30 db peak frequency is 40KHz..

Then examine the lower plot, and find the reactance is 4 ohms (J*4 ohms).

Now..... what is the reactance of 1uF capacitor at 40,000 Hz?

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  • \$\begingroup\$ Correct me if I'm wrong, but in the expression for the loop-gain, Rout doesn't rise. It's only when you consider the closed-loop output impedance that Rout appears to rise with frequency. But stability depends on loop-gain, not closed-loop gain, so I still don't seem to get it. \$\endgroup\$
    – pr871
    Nov 5, 2019 at 12:55
  • \$\begingroup\$ Who told you that the output resistance would be "entirely resistive, when the loop is open" ? Watch the position of the inductance..is it internal to the opamp yes/no? \$\endgroup\$
    – LvW
    Nov 5, 2019 at 14:48
  • \$\begingroup\$ @LvW Not sure what you mean about the 'position of the inductance'? There is no inductance internal or external to the op-amp. Does the 'virtual' output inductance (open-loop Rout divided by the amount of feedback) combine with the cable capacitance to reduce the phase margin of the loop and cause oscillation? The author says 'yes'. I say that doesn't make sense because in the calculation of the loop-gain (on which stability is based), that virtual inductance doesn't exist. Why am I wrong? \$\endgroup\$
    – pr871
    Nov 5, 2019 at 15:55
  • \$\begingroup\$ have you written the math for a closed-loop unity-gain opamp with non-zero open-loop output resistance? \$\endgroup\$ Nov 6, 2019 at 7:14
  • \$\begingroup\$ @pr871 Look at the first picture you have posted. Can you see an inductance within the opamp? Why do you think it would be effective under closed-loop conditions only? Remark: My answer is based on the given symbol - I know that it is not quite correct...the ouput impedance is not pure inductive, of course. \$\endgroup\$
    – LvW
    Nov 6, 2019 at 10:07
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You are right to think that the closed-loop inductive output impedance is not responsible for the potential instability.

The stability of the loop is determined by the loop gain, which does not contain the inductive element that the opamp looks like after closing the loop. Instead one must look at the open-loop output impedance.

As a final note, the open-loop output impedance of the opamp need not be purely resistive, nor need it be constant with frequency. If \$Z_{OL}\$ is the open loop output impedance, then the closed loop output impedance is given by $$Z_{CL} = \frac{Z_{OL}}{1+A},$$ with \$A\$ the amplifier gain (open-loop, that is). If \$Z_{OL}=R_o\$ is resistive, and \$A(f)\propto 1/f\$ as usual, then \$Z_{CL}\$ will be effectively inductive.

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  • \$\begingroup\$ To make this clear: I am not pretending that resistive OL impedance and inductive CL impedance are independent. But for stability the former, and not the latter, is relevant. \$\endgroup\$
    – polwel
    Sep 20, 2021 at 15:46

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