I am trying to measure the wiper resistance of a potentiometer. This is performed by leaving the potentiometer wiper in one position, and measuring from each terminal to the wiper (\$R_1\$ and \$R_2\$), as well as end terminal to end terminal \$R_\mathrm{T}\$. The wiper resistance is given by

$$ R_\mathrm{w} = \frac{R_1 + R_2 - R_T}{2} $$


The issue lies in taking the three resistance readings. As I measure the resistance any part of the potentiometer, the reading decreases with time. Let's say I measure \$R_\mathrm{T}\$ first, the wiper resistance can come out as negative due to the decrease in resistance when measuring \$R_1\$ and \$R_2\$. This result is non-physical and generally makes no sense.

What is the best method of holding these resistances at fixed values such that the results all work together to produce an accurate value for \$R_\mathrm{w}\$?

Probable issue

The LCR meter I am using to take the measurement is heating up the potentiometer while taking the measurement which decreases the resistance. From this it would seem that a lower measurement voltage may solve the issue, but the LCR meter only offers one setting of amplitude at ~0.5 Vrms.

I could wait until the measurement stabilises at a given value, then switch which resistance I am measuring, hoping to maintain that equilibrium while changing connections. Changing where the voltage is applied across the potentiometer will surely change the heating effect as well though, so the thermal equilibrium will not be maintained. This problem is giving me a headache!

EDIT: The potentiometers of interest have so far been 1 MΩ, and some wiper resistances I have seen have been 252 Ω, but if I measure too long on some measurements I have also found a wiper resistance to be -248 Ω.

Taking measurements while applying the LCR meter for the shortest period possible seems like one option, but I don't like it because I don't know how much the temperature of the resistances have changed

  • \$\begingroup\$ Maybe adding more current so the current (and heating) of the LCR meter is negligible? Or supplying it a fixed known current (which is lower than the LCR to lower the influence of heating) and measure its voltages, hoping the voltage measurement doesn't influence the measurement? \$\endgroup\$ – Huisman Feb 22 at 13:39
  • \$\begingroup\$ It might be helpful to add into your question the range of pot resistances you are interested in and the typical wiper resistances you have been seeing. \$\endgroup\$ – Transistor Feb 22 at 13:51
  • \$\begingroup\$ Mmm, applying a fixed current requires more circuitery, connected to the swiper \$\endgroup\$ – Huisman Feb 22 at 13:51
  • \$\begingroup\$ @Transistor I've added total potentiometer resistances and initial wiper values. \$\endgroup\$ – loudnoises Feb 22 at 14:08
  • \$\begingroup\$ @Huisman I don't think adding a constant current source would interface very well with an LCR meter. If I were to build a dedicated tool for this however it could be good: grab the two voltages across each terminal and currents around two loops, probably yielding enough information to find each resistance. \$\endgroup\$ – loudnoises Feb 22 at 14:22

You're trying to measure to within a couple hundred ppm. Unless it's very expensive, I doubt your LCR meter is really anywhere near that accurate or stable. They're not designed to take precise DC measurements.

I suggest using a decent benchtop multimeter. My old Agilent 34401A uses a 5uA current source on the 1M\$\Omega\$ range. It should be stable in the short term to a few ppm, which would allow a measurement accuracy in the 1% range if the pot resistance stays put.

Temperature changes are a possibly limiting effect, but more likely to be a serious issue if your pot is a carbon type. Typical tempco of a conductive plastic pot might be in the 100ppm/°C range, so you have to make sure it stays relatively constant during the measurement.

The 5uA causes heating of I2R = 12uW and your 0.5V RMS causes heating of 0.5uW. Neither one is very large, but the LCR meter is considerably less. If the tempco of your DUT is really bad you may need to make some kind of custom measurement apparatus. Unfortunately, I don't see any way to keep the dissipation constant and make the necessary measurements so it might involve switching tiny currents very accurately and/or perhaps immersing the element in something like oil.

Interesting problem!


It is difficult to accurately measure high resistance values (> 1 megohm) with common lab meters. This is because the input resistance of the meter is effectively in parallel with the resistance to be measured. If your meter has a \$10\,\mathrm{M}\Omega\$ input resistance then your resistance readings will be too low, by almost 10% when measuring \$1\,\mathrm{M}\Omega\$.

You are trying to observe changes on the order of 0.01% so you need a meter with an input resistance at least \$10\,\mathrm{G}\Omega\$ input resistance.


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.