From the datasheet, max ratings: Ib = 60pA Ios = 30pA Vos = 30mV Although the solutions list the datasheet offset as the answer (30mV), it considers the bias current quickly to see how much of an effect it has, and I am confused by the method. Essentially the calculation is Ios_(450k||50k) = Voff Which is equal to 2.7uV (negligible) I do not see how that circuit equation was obtained. It also says Ios+ is "absorbed" by Vin, which I don't understand either. What I got was: Vout/450k + Ios_ = 0, solving for Vout = -27uV I treated the node between the 50k and 450k as virtual ground, thus rendering the 50k irrelevant, which I'm guessing is incorrect, but I don't know how to proceed. I simply did superposition, considering the effect of each non-ideality separately, thus turning the rest off.
Start with the definition of an ideal voltage source. It has zero impedance. 60 pa is going somewhere from the non-inverting input. Then consider the inverting input as 60 pa +/- 30 pa going somewhere. The +/- means it will add to the Vos in the worst case, which you were asked for. Then do the Vos between the inputs. You can consider the ideal voltage source as a virtual ground, but you can not consider the inverting input as a virtual ground. Gotcha started?
Ah...I see now how it's "absorbed". Thanks for that! The only reason I considered it as a virtual ground was because of the superposition principal. Yes, it would not be a ground if we considered the circuit fully as it is, but aren't we meant to turn off everything else except the non-ideality (Ios_) we are investigating to consider its individual effect?
The question says, "total input offset voltage". Apparently the point of this exercise is to make you aware that the TOTAL input offset voltage consists of Vos + Ios times Zin. This is important in real world design because both contribute to your error budget. Disclaimer: I never took a class about op amps and have little idea what the teachers want from you, but I know op amps.