Hi guys

I am reading a tutorial from ADI here about copper stabilized amp. on page 2, figure 2, I can understand how to null the Vos of the nulling amplifier (A2) when the two switches are at Z(zero) position.

But I don't understand how the nulling amplifier (A2) null out the Vos of the main amplifier(A1) at position S(sample), I mean the the input is always connected to the main amplifier (A1), how does the nulling amplifier (A2) measure the main amplifer(A1)'s Vos and put it in C2?

Thanks guys!

 

The input to A2 measures the differential input voltage of A1. If the loop around A1 is closed (as would normally be the case) then this differential voltage corresponds to the offset of A1, independent of any signal voltage. Thus A2 can correct for the offset voltage of A1.

That make sense?

 

The input to A2 measures the differential input voltage of A1. If the loop around A1 is closed (as would normally be the case) then this differential voltage corresponds to the offset of A1, independent of any signal voltage. Thus A2 can correct for the offset voltage of A1.

That make sense?

Arrhh... yes it all make sense now, why didn't I think of it, thanks a lot!