In an attempt to develop an automated potentiometer measuring device I have implemented the following op-amp based voltage controlled current source, controlled by the output of a DAC. The voltage across two terminals of the potentiometer is then measured using a differential ADC (not pictured) from which one of the resistances is found.


simulate this circuit – Schematic created using CircuitLab

I'm having a problem with this circuit in that the DC offset introduced in the op-amp is causing problems when the potentiometer resistance is low. The measured voltage goes negative instead of positive, or remaining at 0V.

To correct this I am thinking of using a chopper-stabilized op amp to minimise the DC offset. This would also need to be rail-to-rail as the supply it is driven from is a single supply +15V. Do these exist?

My question is: would this suitably reduce the DC offset? Or is there a compensation technique I could use in this circuit to reduce the offset?

  • \$\begingroup\$ The measured voltage goes negative instead of positive, or remaining at 0V. Is this the voltage of the DAC? \$\endgroup\$ – Huisman Apr 17 at 10:42
  • \$\begingroup\$ Where is the potentiometer? Where are you measuring? Show us the whole circuit. \$\endgroup\$ – Finbarr Apr 17 at 10:54
  • \$\begingroup\$ The potentiometer is represented by the 10k resistor above the FET. And yes the DAC voltage goes low, with it usually sitting at maybe 20mV when V1 is 0 \$\endgroup\$ – loudnoises Apr 17 at 11:06
  • \$\begingroup\$ DAC = 20mV when V1 = 0V ... I thought V1 was the DAC output? (op-amp ... controlled by the output of a DAC). I'd recommend to update the question/schematic indication R2 is the potentiometer (also refered to as "10k load" ??). V1 being the DAC? And the DAC being controlled by the ADC? \$\endgroup\$ – Huisman Apr 17 at 20:45
  • \$\begingroup\$ @Huisman apologies for not making the question clear enough. I'm going to try a chopper-stabilized op-amp and try and see if I can generally improve the circuit myself since this question has gone poorly. Next time I ask about it I'll have a better schematic and some measurements to clarify. \$\endgroup\$ – loudnoises Apr 18 at 7:11

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