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schematic

simulate this circuit – Schematic created using CircuitLab

In the above image, let's assume for namesake that the op-amp is powered and there are inputs at both the terminals.

My question is , in what scenarios do we connect the resistor R1 across the inverting and non-inverting terminals of an op-amp? What is its actual purpose and why do we do it?

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    \$\begingroup\$ It's related to stability but I think we should see the rest of the circuit anyways. \$\endgroup\$ May 25 at 6:35
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    \$\begingroup\$ Depends on context; is R1 line termination perhaps? Was this part of an LVDS receiver perhaps? \$\endgroup\$
    – MarkU
    May 25 at 6:49
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    \$\begingroup\$ It's called "forcing the noise gain". That resistor effectively reduces the feedback fraction (beta) which reduces the loop gain (beta * Aol), improving stability margins. The closed loop gain is largely unaffected but the closed loop bandwidth is reduced. \$\endgroup\$
    – James
    May 25 at 7:06
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    \$\begingroup\$ Does this question relate to your question about why crystal oscillators have a resistor for biasing? \$\endgroup\$
    – Justme
    May 25 at 10:18
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    \$\begingroup\$ But crystal oscillators don't have an op-amp, so there also would not be a resistor between inputs of an op-amp in a crystal oscillator circuit. \$\endgroup\$
    – Justme
    May 25 at 16:36

3 Answers 3

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It's called "forcing the noise gain". That resistor effectively reduces the feedback fraction (beta) which reduces the loop gain (beta * Aol), improving stability margins. The closed loop gain is largely unaffected but the closed loop bandwidth is reduced.

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  • \$\begingroup\$ Can you add more detail on the biasing of the op-amp? Does this resistor actually bias the op-amp? \$\endgroup\$
    – Newbie
    May 25 at 7:26
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    \$\begingroup\$ No, it's not used for biasing purposes. \$\endgroup\$
    – James
    May 25 at 7:29
  • \$\begingroup\$ Ok, but can you explain in simpler terms? \$\endgroup\$
    – Newbie
    May 25 at 7:35
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    \$\begingroup\$ That's not an easy thing to do. It's probably best if you do some research for yourself by googling some of the terms which I've used in my answer such as "forcing the noise gain". \$\endgroup\$
    – James
    May 25 at 7:42
  • \$\begingroup\$ Yup. A very old-fashioned trick from the days when there weren't many types of opamp to choose from. I don't think I've seen it used in 50 years or so. These days, you can almost always do better by selecting the right part. \$\endgroup\$
    – John Doty
    May 25 at 16:10
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Generally, the inputs will have some resistance feeding them, so the resistor shown has the effect of simply reducing the input voltage seen by the amp.

This has the effect of slowing the amp's output response, such as for an open-loop comparator (perhaps presumably, as shown?), or reduces the loop gain when closed.

The term usually used to measure this is noise gain, because the amp's input-referred noise is multiplied by the feedback ratio. When the feedback ratio is lowered (by loading the inputs with extra resistance like this), essentially the amp is less able to control its output, and thus its internal noise makes up a larger fraction of the output.

This is almost always undesirable, so resistors are not often placed in this location.

Two typical examples where it is desirable:

  1. Note that, by employing voltage dividers at the input(s) -- including between the inputs themselves, potentially -- the common mode input voltage VICM can be reduced. By nature of the divider(s), differential voltage is reduced as well, hence the increase in noise gain. Consider the case of a current-sense amp, using a 5V rail-to-rail type op-amp, sensing a 12V supply: we employ a differential amplifier configuration, but VICM will be nearly 12V if we use the diff amp circuit verbatim, greatly exceeding the 5V range of the amp. If instead, we bias the inputs down towards GND, we unfortunately lose some of the input signal we're sensing, but we can bring the inputs within limits, making the circuit function at all. Thus, it can be a functional requirement, while the increased output noise is a tradeoff.
  2. Using a decompensated type amplifier. Unity-gain stable op-amps are abundant, so we don't encounter these too often, but they are handy from time to time. These are amps with excessive gain, so that they will oscillate if wired as a voltage follower, or anything with similarly low noise gain (and thus, usually, overall gain as well). Such amps are primarily chosen when at least a modest overall gain is required, for example whereas the TL081 is good for 5MHz GBW, meaning about 5MHz bandwidth at gain = 1, 500kHz at gain = 10, etc.; a decompensated version might have 50MHz GBW but the same phase shift, so it's wildly unstable at gain = 1, but offers 5MHz BW at gain = 1, a considerable improvement -- for no additional supply current. Anyway, all that said -- suppose we need to use one at lower gain anyway; perhaps we only had a dual or quad device handy, or don't want to add another op-amp to the design and would prefer to keep reusing the one type (BOM item reduction). If we need to use one at a gain of say 3 to 5, we can put it in a circuit with a noise gain of 3 to 2 times higher noise gain, and it will remain stable -- as long as, again, we don't mind the increased noise.
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  • \$\begingroup\$ Do it have anything to do with the biasing of the op-amp? \$\endgroup\$
    – Newbie
    May 25 at 7:34
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As an example of such use of resistor, I compare it with the use of capacitor for reducing "bandwidth". However, this is a function of the specs of the opamp.

NB: sometimes, a capacitor is inserted in series with this resistance, for stability of the system, as suggested by @Rohat Kılıç in comment.

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