Output of opamp circuits often relies on ratio of resistors (or of impedances when using capacitors and inductors) used in them e.g inverting amplifier.

At the same time resistors may be used e.g to create a vref for single supply opamp inverting amplifier so the noninverting terminal is at vcc/2 rather than at gnd potential.

My question is, what constraint applies to absolute values of resistors (impedance in general to be more inclusive) used in opamp circuits? I guess it has something to do with something called noise gain but I do not understand what that really is and I could be wrong.

  • \$\begingroup\$ Noise, both thermal and radiated/coupled, errors due to offset and leakage currents, even PCB contamination when R values get very large. \$\endgroup\$
    – John D
    Commented Jul 28, 2016 at 0:08
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/102508/… \$\endgroup\$ Commented Jul 28, 2016 at 11:48

2 Answers 2


There are several factors at play here -- the main ones that serve as upper bounds, though, are:

  • Input bias currents -- trying to use large value resistors with an opamp with significant bias currents will lead to large voltage offset errors at the output
  • Noise -- large resistors produce more thermal noise
  • EMI pickup -- high impedances are more vulnerable to EMI pickup than low impedances are (takes less induced loop current to produce a significant RF voltage that can then be rectified by internal junctions in the opamp).
  • Sensitivity to environmental conditions -- extremely large resistors increase leakage current errors

All of these conspire to limit resistor values to the 100s of kΩs for bipolar op amps and the 10s of MΩs for JFET (incl. BiFET) or CMOS op amps.

The primary lower bound comes from the current drive of your components -- excessively low resistor values will overload the output stage of both the opamp and whatever's driving your opamp stage. A practical lower bound is 1kΩ, at least in the voltage-coupled baseband world. RF, OTOH, is run interstage at 50ohms due to stripline characteristics -- this might be a factor with some very fast CFB amps used in RF apps.

  • \$\begingroup\$ What about lower bounds? \$\endgroup\$ Commented Jul 28, 2016 at 0:44
  • \$\begingroup\$ Those I'm not so sure about -- someone else here should be able to fill us in on that \$\endgroup\$ Commented Jul 28, 2016 at 0:59
  • \$\begingroup\$ One limit to how low you can go is the output capabilities of the active components. A feedback resistor that is too low might pass more current than the op-amp can supply. The input impedance of the circuit needs to be high enough that the input source is capable of driving it. The lower the resistor values, the more current the circuit will draw. \$\endgroup\$
    – Mark
    Commented Jul 28, 2016 at 1:22
  • \$\begingroup\$ @user2943160 Lower bound are determined by the load driving ability of the op-amp, which is typically in hundreds of Ohms, or a few K. \$\endgroup\$
    – user207421
    Commented Jul 28, 2016 at 1:37
  • \$\begingroup\$ What about this enigmatic property called noise margin which is different from signal margin? \$\endgroup\$
    – quantum231
    Commented Jul 28, 2016 at 7:52

There is actually a special case that limits the value of resistors at the input of an operational amplifier that has a certain type of rail to rail input:

LT1638 with inputs highlighted

From V- to about V+ (-1.5V), the PNP pair is conducting; above this the NPN pair is conducting, but the input bias current of the two pairs is very different.

Consider this part in a buffer application for a DC voltage:


simulate this circuit – Schematic created using CircuitLab

During power up, the voltage at the non-inverting input will lag behind the amplifier power a bit due to the filter; while the voltage is rising on this filter toward the set point of 3.75V, the PNP input is conducting and therefore current is flowing out of the non-inverting pin, adding to the total current in R2 .

As the voltage rises above the crossover threshold (which is highly thermally dependent), the NPN pair will start to conduct, and now current will flow into the non-inverting input.

I noted above that the currents in each pair are different:


So the current flowing into the non-inverting input can be as high as 60nA which can easily cause a voltage shift over a large resistor.

This subtracts current from R2, which will make the voltage across R2 decrease.

If this voltage is at the threshold of the two input pairs, this effect will continue indefinitely, causing a low level oscillation at the input of the device and at just what voltage this will occur is thermally dependent. (I have actually observed this behaviour).

So if you are using a rail to rail amplifier for this type of application (which is very common), make sure it will be stable for DC buffer applications by analysing the circuit (if possible) or by keeping the divider pair at a low enough resistance such that the oscillation cannot occur.

  • \$\begingroup\$ Or spend extra money on a rail-to-rail op-amp that only has one input pair and some other feature (e.g. charge pump) to bias the input stage. I can't remember which one I looked at recently that does this. \$\endgroup\$ Commented Jul 31, 2016 at 3:11

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