Hi was reading this article and I read:

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There is written: Here, the feedback resistor Rf provides a discharge path for capacitor Cf, while the series resistor at the non-inverting input Rn, when of the correct value, alleviates input bias current and common-mode problems. That value is the parallel resistance of Ri and Rf.

Question: What is the reason for this statement, why does the resistor \$\text{R}_\text{n}\$ has to equal: \$\frac{1}{\frac{1}{\text{R}_\text{i}}+\frac{1}{\text{R}_\text{f}}}\$


Input bias currents produced by op-amps (no matter how small) can cause error voltages when there are resistors attached to those inputs. So if the bias current is the same in or out of both inputs and the resistances feeding those inputs are the same then those errors are cancelled.

So, the equivalent resistance of the inverting input is the parallel sum of feedback and input resistor i.e. the formula in your question. If this parallel sum equals the resistance in the non-inverting input then the effects of bias currents can cancel out. I say can and this is usually but not always the case.

  • \$\begingroup\$ Oké thanks, but what is than the reason for \$\text{R}_\text{L}\$? \$\endgroup\$ – klopiu Feb 7 '18 at 20:33
  • \$\begingroup\$ RL represents whatever load the opamp is looking into. i.e. the next stage of signal processing...a meter... headphones... whatever... \$\endgroup\$ – AlmostDone Feb 7 '18 at 20:38
  • \$\begingroup\$ @AlmostDone So it is a load? \$\endgroup\$ – klopiu Feb 7 '18 at 20:39
  • \$\begingroup\$ R_L for load must > Vol/Iol max \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 7 '18 at 20:45
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    \$\begingroup\$ Note that many modern op-amps are internally compensated for input bias current, so the input offset current will result in the most error. As always, check the datasheet. \$\endgroup\$ – Peter Feb 7 '18 at 21:42

On datasheet there are 3 contributions to input offset voltage;

Ii ( input bias current ) ... times source resistance
Vio ( input offset voltage )
Iio ( input offset bias current)... times difference in source resistance

Voffset = gain * Δ(Rin+ - Rin-) * Ii + Iio * (max(Rin+/-)
and Rin-= Ri//Rf whule Rin+= Rn here

Rin- means Norton equivalent R at the inverting input


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