with 5V? Only its common-mode input voltage range is specified as 0.3V to 2.8V.
What peak value of a superimposed AC signal is allowed for fully differential
mode? There must be an upper limit, I assume.
The input common-mode signal must be contained within the range of 0.3V to
2.8V. For a differential-to-differential application, the common-mode voltage
“to the left” of the gain resistors and the output common mode voltage are
generally DC voltages. You can find the DC common-mode level directly at the
amplifier input by simply taking the voltage division between the output
common-mode voltage and the common-mode voltage “to the left” of the gain
resistors. For example, if the output common-mode voltage is set to 0.9V,
which it most often is with the ADA4930 and the common-mode voltage “to the
left” of the gain resistors is 2.5V, and the gain is one (Rf = Rg), then the
difference between the two common mode voltages is divided in half (it would be
a different fraction for different Rf/Rg ratios) at the amplifier inputs. In
this case, the amplifier’s input common-mode voltage would be half-way between
0.9V and 2.5V, or at 1.7V.
It’s a little more complicated for single-ended-to-differential applications
because the amplifier’s input common-mode voltage has ripple on it. You can
also view the webinars of Jonathan Pearson, he did last January & February,
which cover similar material – these are available on the ADI website. I’ve
pasted in a couple of slides that show single-ended-to-differential application
with bipolar and unipolar input signals and how the input common-mode voltage