# Offset and Bias current of Op Amp

Given this circuit : OA3 is not ideal and has a bias current (5μA) and an offset voltage (10mV).The other two are ideal. I have to calculate the maximum deviation voltage on RL (load resistance). . I'm struggling with the understanding of the bias current. Let's consider IB+(OA3), since Vs is zero, and virtual ground concept still holds for 0A1 is it okay to assume there is no current through R8 and as a result V1 is 0V ? I'm not really sure about that. How might the current IB+ behave ? Would it just enter the output terminal of 0A1 or does it have any other path to flow through ?

• JustCurious, The answer I get is:$$\Delta V_{R_L}\le\frac{R_7}{R_1+R_7}\cdot\left(\bigg| I_{_\text{BIAS}}\bigg|\cdot R_5+\bigg| V_{_\text{OFFSET}}\bigg|\cdot\left[1+\frac{R_5}{R_4}\right]\right)\quad\quad\lt 107\:\text{mV}$$If you are still around and respond, I can walk you through the process I applied. Or, you can look it over and see if you can work it out on your own.
– jonk
Mar 7, 2022 at 4:24

since Vs is zero, and virtual ground concept still holds for 0A1 is it okay to assume there is no current through R8 and as a result V1 is 0V ?

Yes.

How might the current IB+ behave ? Would it just enter the output terminal of 0A1 or does it have any other path to flow through ?

OA1's output will take/sink/source all of the bias current produced by OA3's +Vin input. OA1's output voltage remains unaffected.

• JustCurious - if this question has been answered then you should choose the most appropriate answer and formally accept it using this method. Of course if you have residual queries then raise a comment. Jul 26, 2022 at 8:47

The bias current has to be supplied by the whatever circuit is connected to the opamp input. The source impedance of that circuit then determines the voltage.

It is irrelevant if the nominal voltage of that node is 0 V.

• Looking out of the non-inverting input of OA3, we see, in parallel, the output of OA1, and R8 connected to OA1's inverting input. The output of OA1 is low impedance, especially so because of feedback. It is reasonable to assume that this node is close to 0 Ohms. Hence the $$\I_{b+}\$$ bias current of OA3 does not play a role. It is simply provided by OA1, without any other effects.
• At the inverting input, things are slightly more complicated. Again, the impedance we see is R4, in parallel with R5. One might be tempted to conclude that the impedance thus is simply R4 || R5 = 9 kOhm. However, because of feedback, the inverting node must remain at the same voltage, so the extra current must go through R5 exclusively. It will create a voltage error of $$\I_{b^-}\cdot R_5\$$. Another way to see this is that for the $$\I_{b^-}\$$ bias current, OA3 is wired as a transimpedance amplifier.