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I'm reading a book*, and it derives the output resistance of a non-inverting op-amp amplifier by grounding the input voltage, applying a test current source at the output, and measuring the resulting output voltage:

enter image description here

Here L is the loop gain (i.e. open-loop gain a x feedback gain b).

Here's my take on finding the output resistance, which gives a different answer. We can think of the closed-loop amplifier as a black box, as shown below.

enter image description here

We can measure the output resistance R_o by shorting the output and measuring the current that flows through. The resistance is then:

$$ R_o = \frac{Av_i}{i_{o,\text{short}}} \approx \frac{v_i}{bi_{o,\text{short}}} $$

Where I have used A = 1/b, with b being the feedback gain.

If we short the output of the actual circuit (with v_I included), we have: $$ i_{o,\text{short}}=\frac{av_D}{r_o} = \frac{av_I}{(1+L)r_o} $$

(Note that \$R_1\$ and \$R_2\$ appear in series with each other and in parrallel with the short, and therefore do not affect the output current).

Then:

$$ R_o = \frac{(1+L)r_0}{ab} = \frac{1+L}{L}r_o \approx r_o $$

*"Design with Operational Amplifiers and Analog Integrated Circuits", by Sergio Franco.

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Your analysis will work when the output is forced to 1V instead of 0V. I was able to get the correct answer (output impedance is indeed lowered by loop gain!) by analyzing a 1V voltage source on the output.

The way that I prevent this sort of mistake is by being strict about how I write my equations. Among other rules, I don't introduce new terms like L and b, I don't use any numbers until the end, and I don't make approximations in the algebra. When the algebra becomes tedious, I use the Maxima computer algebra program. I think there is also a Mathematica package somewhere that does this type of analysis correctly.

In this problem, the rule that was violated is that the Vout term was replaced by a number before the equations were solved. In this case the number choice was unfortunate and it ruined the equation.

The more intuitive explanation is that Zout=Vout/Iout, so Vout=0 is a problematic choice.

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  • \$\begingroup\$ Nice analysis, Tom! I was too lazy to dig into the math :) \$\endgroup\$ – John D Dec 1 '16 at 6:30
  • \$\begingroup\$ Excellent! I had in fact realized my mistake before you posted this, and re-worked it to convince myself that the output resistance is indeed reduced by a factor of the feedback gain. Nice to have confirmation. Thank you. \$\endgroup\$ – MGA Dec 1 '16 at 6:52
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If you're going to model the amplifier as a black box you have to model the closed loop output impedance, which is approximately ro/(loop gain) as found in the book you are reading. You can't just transform the topology to a black box then try to find the output impedance of the new topology. It's indisputable that negative voltage feedback reduces the open-loop output impedance by the loop gain (approximately).

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  • \$\begingroup\$ The output impedance is what I want to find, and I include it in the model as R_o (note: uppercase R_o = closed loop output impedance, lowercase r_o = open loop (i.e, op-amp) output impedance). I'm questioning precisely why it's claimed that the closed-loop output impedance is equal to r_o/(loop gain). \$\endgroup\$ – MGA Dec 1 '16 at 5:13
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    \$\begingroup\$ If you think about it intuitively if the gain were infinite (the op-amp ideal except for ro) then you could draw infinite current from the output and the output voltage wouldn't change at all, right? So the negative feedback "compensates" for the finite output impedance, and would reduce the output impedance to zero. So your black box model and/or math is incorrect. \$\endgroup\$ – John D Dec 1 '16 at 5:55
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I guess while saying \$i_{short} = \dfrac{a V_d}{r_o}\$, you are not including the current passing through \$R_2\$ resistor.

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  • \$\begingroup\$ No, this is seen in series with R_1 (assuming no bias current into the op-amp) and in parallel with the short, so the short 'wins'. \$\endgroup\$ – MGA Dec 1 '16 at 5:16
  • \$\begingroup\$ Yeah, the current due to V_d passes directly to GND instead of R_2. But the current from your V_i will pass through R_1 and then R_2 and then mix with your i_short(I mean the current due to V_d) right? \$\endgroup\$ – user3219492 Dec 1 '16 at 5:19
  • \$\begingroup\$ No, if there is a short in parallel with any resistance, that resistance can be taken out of the picture. All the current will pass through the short, and none through the resistance. \$\endgroup\$ – MGA Dec 1 '16 at 5:20
  • \$\begingroup\$ Yes, but what you are saying is valid for the current from V_d. Not for the current from V_i right? So what will be the path for the current from V_i ? \$\endgroup\$ – user3219492 Dec 1 '16 at 5:23
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Now change the gain model, from constant gain, to gain that begins to roll off at some low frequency and continues rolling off for decade after decade of frequency. And include in the analysis a Cload capacitor.

For various parameter ranges, the OpAmp will not longer be able to control the output. You likely will encounter this issue in your lab experiments.

Suppose the open-loop output resistance is 100 ohms (some opamps, running at 1uA, have Rout of 80,000 ohms, but Rout as low as 10 ohms is common), and let UGBW be 1MHz. We can accurately model the closed-loop OpAmp as having INDUCTIVE output impedance, of value 16microHenry at 1MHz because that produces 100 ohms.

At UGBW, the opamp has gain=1, and has no ability to control the Vout behavior. At UGBW/10, the opamp has gain=10, and has SOME ability to control Vout, which in fact looks like 10 ohms. (if configuration is unity-gain-buffer). At UGBW/100, the opamp has gain=100, making Zout look like 1 ohm. This dropoff as frequency drops..............is inductive.

enter image description here Now hang 1uF capacitor on the output. You get a troubling peak; solution is to dampen it with a series resistor, value sqrt(L/C). enter image description here

Using Rdamp = sqrt(L/C), we need 4 ohms. Here is that comparison: enter image description here

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