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schematic

simulate this circuit – Schematic created using CircuitLab

There are two buffers made with the TL084CN, which is part of the conditioning of an LTS25-P transformer. Currently, I have put a 100ohm resistor in the non-inverting input, and the feedback is without the resistor.

Is it advisable to add a resistor in feedback? Will the reading be influenced if I do not?

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  • \$\begingroup\$ What circuit? You can edit your question and add a diagram using the built in circuit editor. \$\endgroup\$ – JRE Nov 15 '18 at 18:40
  • \$\begingroup\$ How bad is the layout? Could there be appreciable capacitance to GND at the - pin \$\endgroup\$ – JonRB Nov 16 '18 at 1:28
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Summary: For your case and your sensor is not important. your 100ohm resistance is negligible.

when your input resistance (\$100\:\Omega\$) * Input Offset Current_max (\$20\:\text{nA}\$) in the worst case generate \$2\:\mu\text{V}\$ offset voltage, your opamp typical input offset is in range \$3\:\text{mV}\$. so it is not important!

Why we compensate the opamp:

opamp input is a symmetric architecture and if you change the symmetry this generates offset and gain temperature drift. the opamp input impedance is base on the substrate, for example for your case is \$10^{12}\:\Omega\$. when you put a resistance in the input, the quiescent current of the input pin generate offset voltage in input and difference of current in two sides generate different behavior in temperature.

When R compensation is important:

  • when input_resistance* Input_Offset_Current_max > Voffset or comparable.
  • When the source has a high built-in impedance
  • If you need a high gain system.
  • If drift in temperature is very important even \$1\:\mu\text{V}\$!
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