What does the expression R1 = RF||RG mean?

While reading the datasheet for the Op Amp AD8029, I came across the following expression:

$$R_1 = R_\text{F} \parallel R_\text{G}$$

I am confused about what this means.



Does this mean that \$R_1\$ is the equivalent resistance of the parallel-connected resistors \$R_\text{F}\$ and \$R_\text{G}\$? And, accordingly, can it be calculated using the following formula?

$$1/R_1 = 1/R_\text{F} + 1/R_\text{G}$$

Yes, that's exactly what it means.

The reason you want this is that any offsets due to input bias currents automatically cancel out.

That value (parallel combination RF and RG) minimizes the error due to the input bias current.

Yes, it means resistance of R1 must be equal to the parallel combination of resistors RF and RG.

This is to have the inverting and non-inverting input driven with equal source impedance, so that the voltage drops due to bias currents are equal.

Yes your understanding is correct.

I'll keep going since this is a short answer.

R1 is included to balance the input impedance a signal will "see" on the Non-inverting input verses the inverting input. The value of R1 comes from the equivalent impedance "seen" at the inverting input node. To prove that: Remember in small signal analysis we treat the DC power sources as ideal sources unaffected by your AC signal, so as far as your AC signal is concerned, voltage sources become shorts and current sources are open circuit. An op-amps output can be thought of as a voltage source in this case, so the output is shorted to ground. You are left with Rf and Rg in parallel, and from the inverting input the impedance to ground is Rg in parallel with Rf.