# Output voltage offset in an op-amp

I am trying to solve section d in the following question by substituting a current source of -Ios/2 at the '+' and +Ios/2 at the '-', as the resistor(s) found in section c would nullify the output for all the other offsets, namely IB and Vos. However, I obtain a result which is different than that asserted in the formal solution and wished to confirm my method here.

For a 150 kΩ resistor connected to '+' and IB flowing away from the ground the output would be zero, except that now Ios is added. Hence, using superposition:

For the -Ios/2 source connected to '+' and flowing away from ground: (Ios/2) × 150 kΩ = Vo

For the +Ios/2 source connected to '-' and flowing away from ground:

Vo/100 kΩ = (Ios/2)

This yields 1.25 mV, whereas the official answer is -1 mV.

Where's my mistake?

• @JohnBirckhead And what about Vos? Where was that taken into account in your analysis? Furthermore, as I pointed out in my question, the resistor on the positive input should be either 50k or 150k, based on the direction of the current $I_B$. Why couldn't I simply add -Ios/2 to the positive input and +Ios/2 to the negative? Jun 30, 2017 at 20:56