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?

  • \$\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

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}\$!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.