# Voltage divisor with op amp

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. • 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. – John D Apr 24 '18 at 4:28
• 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 – MCG Apr 24 '18 at 7:49

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: 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 