Electronic – Reason for small-valued feedback resistors in low noise Op Amp

About the gain being stated as \$A_V=3\$, the complete relevant text is

R3 and R4 form an output voltage gain stage whose gain, \$A_V=3\$, is reduced to unity at high frequencies by C1 to maintain stability.

What this is saying is that R3 and R4 form a voltage divider so that

$$v_o = \frac{v_{out}}{3}$$

where \$v_{o}\$ is the voltage at the output of the op-amp IC.

Or, turned around,

$$v_{out} = 3 v_o.$$

This works because the negative feedback around the op-amp will cause it to push or pull current from its output pin to make it work.

The overall gain of the circuit is 33, as you calculated.

, it is stated that the gain of the whole circuit is 33 while the gain of the output is 3 . So you can have 2 gains in an op-amp ?

No, the "stage" formed by R3 and R4, with gain 3 doesn't really involve the op-amp.

But even within the op-amp integrated circuit itself, of course every stage in the design can have a different gain value.

So one is the gain of the input to the Vout? and the other is the gain of the op-amp output to the actual Vout?

33 is the gain from \$v_{in}\$ to \$v_{out}\$

3 is the gain from \$v_o\$ to \$v_{out}\$. I think Tim's comment does a better job explaining it than I could:

If the voltage on the op-amp output is not equal to 1/3 of the output voltage (ignoring C1), then current will flow in it's positive or negative power pin. That forms a negative feedback loop composed of the four external capacitors and the op-amps (internal) output stage.

This is a common characteristic of rail-to-rail input op-amps. There are really two front ends and there is a transition between them at some common-mode voltage. R-R output is irrelevant.

If you care about Vos to that extent, you can choose another type of op-amp. Maybe a chopper type, but they have other subtle (and not so subtle) imperfections.