Hi,
input voltage noise is given in datasheet.
input current noise is given in datasheet. This has to be multiplied with the input impedance. (To make it more simple you can choose a single frequency (noise) of interest) to give a corresponding voltage noise.
you need to add them with square and root
e(t) = sqrt( e_volt^2 + e_curr^2) = input referred
for two opamps: you can multiply it with 1.414
for output referred you need to multiply the input referred noise with the OPAMP gain. (simplified)
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a simple rule:
your values: en =3.9nV/sqrt(Hz), in = 0.4pA(sqrt(Hz) --> r = U/I --> 3.9nV/0.4pA = 9750 Ohms
if your input impedance is > 9750 Ohms, then current noise dominates
if your input impedance is < 9750 Ohms, then voltage noise dominates
Although this means that this OPAMP has best noise performance with about 10kOhms source impedance but to compare two OPAMPs you allways need to calculate both input noise values.
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for more information i recommend to read:
<Noise Calculations in Op Amp Circuits – Design Note 15 / Alan Rich>
https://www.linear.com/docs/4201
(linear technology,
www.linear.com)
Klaus