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Purpose of potentiometer before non-inverting pin on Op-Amp

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jrjradler:
A circuit I'm replicating ties the output of one op-amp (OPA277) to the input of a potentiometer - that output then acts as the input to the non-inverting input of another op-amp (INA149). I believe it's trying to create a modified howland pump - I'm just a bit confused about what the potentiometer does and if I can replace it with a smaller 0603 resistor.

I attached a photo of the circuit in question.
Here is a link to the OPA277 datasheet -> https://www.ti.com/lit/ds/symlink/opa277.pdf?HQS=dis-mous-null-mousermode-dsf-pf-null-wwe&ts=1669593970133&ref_url=https%253A%252F%252Fwww.mouser.ca%252F
Here is a link to the INA149 datasheet -> https://www.ti.com/lit/ds/symlink/ina149-ep.pdf?HQS=dis-mous-null-mousermode-dsf-pf-null-wwe&ts=1669583195718&ref_url=https%253A%252F%252Fwww.mouser.ca%252F

Doctorandus_P:
Start by drawing a schematic.
Not many people are interested in figuring out a PCB layout and interpreting the results.

And with a bit of luck, if you have drawn the schematic, the reason for the pot may be obvious.

EPAIII:
Some more reading:

https://www.allaboutcircuits.com/technical-articles/the-howland-current-pump/

https://www.ti.com/lit/an/snoa474a/snoa474a.pdf

What does the original circuit do and why are you trying to copy it but with changes?

mikerj:

--- Quote from: jrjradler on November 28, 2022, 12:06:19 am ---A circuit I'm replicating ties the output of one op-amp (OPA277) to the input of a potentiometer - that output then acts as the input to the non-inverting input of another op-amp (INA149).

--- End quote ---

The INA149 is not just an op-amp, it has integrated gain setting resistors so the trim pot is actually in series with a 380k resistor within the INA149.  This means the trimmer will be adjusting gain on the +ve side, presumably to get the best CMRR.  Drawing the schematic including all circuits immediately connected to the INA149 will likely yield clues as to why they found this gain trimming was required.

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