Op amp resistor on output... but not like "normal"

[composition4]

I use a "trick" when I need to reduce the load on an op amp  output which seems too good to be true, so I think I must be missing something.

Instead of putting a series resistor outside of the feedback loop which increases output impedance, I place it inside the feedback loop before the feedback resistor, but take the output from between the series resistor and the feedback resistor. A clearer example is as below, with the load-reducing resistor being R₀.



This seems to work to keep the output impedance to just that of the opamp and ignore the resistor... at least the stage that follows the opamp reacts that way. The feedback loop gain is also unaffected by the series resistor in this configuration. What am I missing, or is this really a free (or cheap) lunch?

[antonis]

Does this remind you anything..??



P.S.
In your scheme:
Considering Ao output zero, R2 "sees" R_o in parallel with R_l..

[Rob Strand]

This seems to work to keep the output impedance to just that of the opamp and ignore the resistor... at least the stage that follows the opamp reacts that way. The feedback loop gain is also unaffected by the series resistor in this configuration. What am I missing, or is this really a free (or cheap) lunch?

What you lose is output swing when driving low impedance loads.

When you place a capacitive load on the output there is a higher risk of instability ie. oscillations
because the added R and capacitive load cause more phase shift inside the feedback loop.
However you can add a small feedback capacitor *before* the output series resistor.

Go here, and about 60% down the page you will find a section "In-loop compensation:"
https://www.analog.com/en/analog-dialogue/articles/ask-the-applications-engineer-25.html

Also notice the added resistor Rx is the same as the resistor you added.

For single opamps you can place resistors in series with the supply.  These are outside of the feedback loop but it needs more care not let the supply get too low.   The idea is seen with and without caps across the opamp supply.

[amptramp]

In remote sensing applications, we always used to use a capacitor directly from op amp output to inverting input and the feedback taken from after the resistor connected to the output.  That allows you to drive capacitive loads while maintaining gain accuracy and gain stability under differing loads.  This was important when you had a number of optical detectors and they were matched to each other, so the gains had to be matched as well or the output image would have horizontal bands when displayed.  You rarely run into this consideration in audio.

You don't have to match the pole frequencies of Rin times input capacitance equals feedback cap times feedback resistor unless you wanted to get the maximum closed-loop bandwidth.  We always designed in more feedback capacitance than that so we would err on the side of stability.  In most cases, you can achieve stability with a feedback capacitor that was more than sufficient for stability but did not limit the bandwidth more than necessary.

You don't need high-fidelity bandwidth from a guitar.  When Phil Collins released "No Jacket Required", it was the first album where there was almost nothing in the whole album that went above 5 KHz.  Nobody complained about the sound or thought it was muffled, so bandwidth can be limited without causing a problem.

[ElectricDruid]

+1 what Amptramp and others said.

This is a pretty common circuit, but it needs the little feedback cap too.

One place I've seen it is for modular synth circuits. If you're outputting a Pitch CV, you want a low output impedance because you don't want the Pitch CV to get reduced at all (the dreaded "pitch droop"), but the typical way to protect an output for modular gear is a 1K series resistor. So what's to be done? If you use the 1K resistor, you lose 1% of your pitch CV into the standard 100K input, but if you don't use it, you risk someone blowing up the module if they plug something else into the output.
This circuit is the solution - it provides the series protection but the feedback keeps the Pitch CV accurate, and it can drive capacitive loads like long patch cables without any trouble.

[antonis]

I think OP didn't have any capacitive loading in mind..

[ElectricDruid]

I think OP didn't have any capacitive loading in mind..

No, maybe not, but it got mentioned because that's one of the useful features of this circuit.

[Rob Strand]

What am I missing, or is this really a free (or cheap) lunch?

I think OP didn't have any capacitive loading in mind..

No, maybe not, but it got mentioned because that's one of the useful features of this circuit.

In the context of the OP's post more of a pitfall.

[PRR]

> I need to reduce the load on an op amp

I don't understand. Is the opamp complaining?

> a small feedback capacitor *before* the output series resistor.

Here is an over-elaborate version of that plan. The MOSFET gate capacitance distresses the opamp up around 1MHz; it oscillated. The R-C network takes NFB from "jack" through R4 over the audio band and more, for good DC-Audio performance; then cross-over to bypass MOSFET loading and take NFB from C5 for higher frequencies.

[Rob Strand]

Let's not forget the most common reason for adding an output resistor is to prevent opamp oscillation with capacitive loads.  Yes it does other things: reduce power dissipation, limit short-current, protect the opamp against various external abuses.

The down side is some attenuation with under load,


If we put the output resistor inside the feedback loop it removes the attenuation but now the resistor no longer prevents oscillations, in fact it makes oscillation more likely.   In that case we need to use the standard trick documented on the analog design page.

One thing you cannot restore in the loss in output swing with low impedance loads - regardless if the output resistor is inside or outside of the feedback path.  (attenuation is not swing)

[Eb7+9]

This seems to work to keep the output impedance to just that of the opamp and ignore the resistor... at least the stage that follows the opamp reacts that way. The feedback loop gain is also unaffected by the series resistor in this configuration. What am I missing, or is this really a free (or cheap) lunch?

what you're describing is predicted by control theory ...

Zout of your servo loop is Ro/Ao
where Ro is combined sum of open-loop output stage impedance and added resistor
and Ao is the open loop gain of the op-amp

if Ro is 1k and Ao is 10^6 then Zout is still well below 1ohm AC, even with 10k added ...
but that's all in terms of small-signal theory

once you try driving an external load, or feedback resistor, with any real current the output node voltage will tank

there are other issues as Rob points out ...

[composition4]

Thanks for the insights, knew there was no such thing as a free lunch!