Why is it not right to use the classic Vo = Vin*R2/(R1+R2) voltage divider formula in the marked area? I want to obtain the voltage, in terms of Vs, in the non inverting input of the op amp. But if I use a normal voltage divider, I’m not getting the expected result. circuit

  • \$\begingroup\$ Barring unreasonably high bias currents or unreasonably low input impedance it is right to use the normal voltage divider formula to get the voltage at the non-inverting input. If you're not getting the expected result (which is what, where?) then something else is wrong. \$\endgroup\$
    – John D
    Apr 24, 2018 at 4:28
  • 2
    \$\begingroup\$ What value resistors are you using? What is the value of Vs? What is the result you are expecting vs what is the result you are getting? Using a voltage divider should work fine here. Knowing more information may help \$\endgroup\$
    – MCG
    Apr 24, 2018 at 7:49

2 Answers 2


The exact equation would also include the bias current of the non-inverting input of the op-amp, which should be negligible in most cases (by design, you want to keep the resistor values low enough that the bias current is not significant, because the bias current is not known accurately and is not stable with temperature.

However, other than possible erroneous measurements or wiring, wrong parts, or a defective op-amp, there are a couple other possibilities (and with more information, supplied by you, it would not be necessary to guess), but for completeness..

  1. If the divider voltage is outside the supply range then it's possible the input current will be much larger than the stated operational bias current. For example, if the negative rail is 0V, Vs is -10V and 2Rb = 10K then the voltage measured at the non-inverting input will likely be more like -0.7V because the internal protection diodes or isolation diodes will conduct.

  2. If the circuit is such that the op-amp cannot balance, then some op-amps will pass current between inputs if the voltage differential exceeds a diode drop. There are effectively back-to-back (inverse-parallel) diodes across the inputs. There are usually current-limiting resistors in series with the inputs but the current can be quite significant. This is common in op-amps with super-\$\beta\$ bipolar transistor front ends, since breaking down the base-emitter junction of the transistors can degrade the transistors.

OPA202 internal circuit showing the diodes as mentioned above: enter image description here


If the final op-amp is saturated then it's normal that the ideal Vp=Vm doesn't hold. Check that the opamps are working in the linear region enter image description here


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