What is the purpose of a resistor in the feedback path of a unity gain buffer?

I often see unity-gain followers with a resistor in the feedback path. For an ideal op-amp, of course, there is no current into the input, and this resistor does nothing. What is its effect with a real op-amp, and how do I choose its value?

What does R1 do in this circuit?

• The designer owns stock in a resistor company. Jan 31, 2013 at 22:54
• They want to increase the noise of the stage? Nov 9, 2015 at 17:18

You will rarely see a circuit with just one resistor as you show it; usually there will be another resistor (or equivalent source resistance) of the same value on the noninverting input, too.

Most (nonideal) opamps have a finite input resistance, and this means that a tiny current flows into or out of the input terminals. This current is called "input bias current", and it varies with the voltage at the inputs. Since most opamp circuits use negative feedback to keep the two inputs at the same voltage, this means that for any given voltage, the current through both inputs will be the same.

The current through each input flows through whatever resistance is connected to that input, and this introduces a voltage shift at the input. If the resistance at the two inputs is different, this voltage shift will be different, too, and the difference between those two shifts will appear as an additional input offset error in the operation of the circuit.

For this reason, an effort is made in all opamp circuits to make sure that the resistances connected to the two inputs are the same, eliminating this additional source of error. Even in a unity-gain buffer, if the source resistance is 100Ω, a 100-Ω resistor will be used in the feedback path.

• This is definitely one reason to include the resistor, but it seems there are others. Feb 4, 2013 at 16:28

Here's an excerpt from the OP27 data sheet, showing that the answer is more involved than equalizing the impedances seen by the two inputs:

And another example, from the AD797 data sheet:

• Though the diode protection on the inputs and low Rbb are unusual features of these ultralow noise op-amps, right? If this resistor is necessary it will probably be specified in the datasheet? Nov 9, 2015 at 17:26

One reason the feedback resistor may be used is to match the output impedance of Vin. Real Op-amps have input current bias and input current offset.

Take for example this representative circuit:

Here, I've create a more realistic model of an op-amp by adding current sources which simulate the current flowing into a real op-amp's terminals. The difference between the two input currents is the offset input current.

The input voltage at the positive input terminal actually is:

$$Vin_{actual} = Vin - I_1 \cdot R_1$$

Through ideal op-amp action, the negative input terminal voltage is the same. We can then calculate the resultant output voltage:

$$Vout = Vin_{actual} + I_2 \cdot R_2\\ Vout = Vin - I_1 \cdot R_1 + I_2 \cdot R_2\\$$

By closely matching R1 and R2 the effect of input bias current is effectively nulled. Note that this doesn't solve input offset current, though. To solve both problems ensure that the resistance of R1 and R2 are both small. This will solve both of the issues of input offset current and input bias current. With a small enough R1 there may not be any need for an actual discrete matched R2, though you will of course get better results if there is one.

I found another reason! In JFET op-amps, the input capacitance changes with the voltage, which creates distortion in the non-inverting configuration (where the voltage at the input changes with the signal).

It is possible to cancel this distortion by placing a resistance equal to the source impedance in the op amp’s feed-back loop. This produces an identical distortion signal at the op amp’s inverting input. Because the distortion is now common to both inputs, it is removed by the op amp’s common-mode rejection. Unfortunately, the resistance in the feedback path introduces additional noise and also can cause stability issues if it is very large.

TI: Distortion and source impedance in JFET-input op amps

• Very interesting... Another example of the passive compensation technique... Feb 12, 2021 at 2:06
• Interesting to see you add a positive technical answer 5+ years after your initial comment on the question. I wonder what Olin would have to say now if asked. (Note: thats a genuine comment - not meant to be snide.) Feb 12, 2021 at 5:27
• @RussellMcMahon My initial comment on the question was just a joke :) Feb 12, 2021 at 16:12

There are 2 kinds of op amps: voltage feedback and current feedback. In current feedback, as one can guess from their name, the current driven from the output to the input through the Rf "feedback resistor" determines 1) the bandwith 2) the gain when associated in a current divider with Rg "ground resistor". In current feedback op amps, if there is very low feedback resistor this application note http://www.ti.com/lit/an/slva051/slva051.pdf says that the op amp will oscillate. Choosing a high value will decrease bandwith.

In fact there are so many different types of op amps nowadays that the simple model of infinite impedance inputs implicating Vin+ = Vin- is far from true in many cases. Many op amps have low input impedance values, others have very high impedance on +In and very low on -In. VHF op amps are all current feedback and very touchy to implement, like the LMH6703

This feedback entirely on OA's Bandwidth with its unity gain internal compensation. There is no single solution for all OA's

http://www.ti.com/lit/ds/sbos267d/sbos267d.pdf

G = +1, RF = 25Ω

Conclusion; Each OA is unique. Use recommended spec or trial & error.