The 'all about circuits' page on 'Op-Amp Practical Considerations' [http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/op-amp-practical-considerations/], under the subtopic 'Bias Current' says:

In either case, the compensating resistor value is determined by calculating the parallel resistance value of R1 and R2. Why is the value equal to the parallel equivalent of R1 and R2? When using the Superposition Theorem to figure how much voltage drop will be produced by the inverting (-) input’s bias current, we treat the bias current as though it were coming from a current source inside the op-amp and short-circuit all voltage sources (Vin and Vout). This gives two parallel paths for bias current (through R1 and through R2, both to ground). We want to duplicate the bias current’s effect on the noninverting (+) input, so the resistor value we choose to insert in series with that input needs to be equal to R1 in parallel with R2.

But why does the author consider the current flowing through R2 to be flowing to ground directly? Won't the current go through the load resistance (and maybe even through the op-amp when the op-amp acts as a current sink) before reaching the ground?

 

Link doesn't work. Every opamp product has a bias current spec on its data sheet. On some it flows in, on some it flows out.

To prevent the amplifier seeing an offset voltage between its inverting and non-inverting inputs, the Thevenin equivalent for the bias current source on both the inv and non-inv inputs must be the same. You determine the bias current source impedance based on the feedback resistors, and match the other side so that it has the same Thev. equiv.

 

The input bias current is modeled as a current source in series with each inputs. They go to great lengths to make the two bias currents as equal as they can (a different spec.). So, to make the bias currents cancel each other out, the resistance each input sees must be the same. The current/resistor from the non-inverting input to ground is a no brainer. R1 and R2 must be considered in parallel because the current flowing through each is identical.

 

...R1 and R2 must be considered in parallel because the current flowing through each is identical.

Nope, the current in R1 and R2 is different by the bias current

 

Nope, the current in R1 and R2 is different by the bias current

For equal bias currents and direction at each input and a parallel equivalent resistor in series with the plus input, the bias current divides and goes through R1 and R2 in inverse proportion to their ratio.
Or is that what you are saying?

 

Why is I(R1) ≠ I(R2)?

 

OK OK, I generalized too much. Although, it is the rational for making the non-inverting resistor to ground, equal to, R1 in parallel with R2.

Also, I don't know what is wrong with your sim.

 

Why is I(R1) ≠ I(R2)?

View attachment 87507

Note that the bias currents of the two inputs are apparently equal and in the same direction (out of the inputs) and, in this case, are 12nA each.
Thus the sum of the currents through R1 and R2 is 12nA.
It would appear the offset current is very small for that op amp.