I'm trying to measure the DC resistance of an inductor I made out of AWG 11 wire.

From my calculations its impedance should be around 0.0012 ohm, and the ohmmeter measured precisely 0.0012 ohm.

But that at ambient temperature (which was 18ºC). But I had to measure its impedance at different temperatures too.

Then I tried heating the inductor with an hot air gun from a distance, at the lowest power. It heated to mere 30 ºC and stabilized at that temperature, but the resistance went up to 0.0041 ohm. That sure was impossible, it's too much of an increase due to such a small heating.

When that happened it occurred to me it should be some thermoelectric effect. Without removing the hot air I inverted the polarity of both pairs of leads and read again. Now a negative resistance appeared while it was being heated, but after cooling down the resistance went back to positive 0.0012 ohm (as it should be, cause inverting both pairs should not make a difference).

The 4 leads are all very close together in inductor (all like 2-3 cm apart at max) and being roughly uniformly heated. There's no reason for a difference in temperature to arise in there, but there's a difference in temperature from all of them at the inductor and the other ends in the ohmmeter.

I can't heat the ohmmeter together, what can I do to compensate for the seebeck effect?

I imagine that, If the seebeck effect voltage is being added in one way and subtracted in the other, adding the resistances measured in both ways (the negative one and the large positive one) should give me the correct value.

Am I correct?

  • \$\begingroup\$ what kind of ohmmeter are you using? 0.0012 \$\Omega\$ is awful small. How are you handling the contact resistance between your probes and the wire? \$\endgroup\$ Commented Dec 3, 2020 at 23:04
  • \$\begingroup\$ @MathKeepsMeBusy the contact resistance shouldn't be a problem while using 4-lead mode, but anyway I'm making sure they're the best possible \$\endgroup\$ Commented Dec 3, 2020 at 23:06
  • \$\begingroup\$ @MathKeepsMeBusy It's an Agilent 34461A multimeter, and the measure itself an average of a great number of measures. The contacts has not been an Issue, I tested moving them to see if any variation arised but no, they're quite firm in there. The problem seems to be solely from seebeck effect. \$\endgroup\$ Commented Dec 3, 2020 at 23:07
  • \$\begingroup\$ The lowest range on your meter is 100\$\Omega\$. You are measuring a value approximately 5 orders of magnitude smaller than that. I personally would maintain a certain degree of skepticism regarding the values I obtained. I am not saying they are wrong, but I would maintain some skepticism. \$\endgroup\$ Commented Dec 3, 2020 at 23:24
  • \$\begingroup\$ I agree with Math. The specified accuracy of the 100 ohm range is 0.003% of reading + 0.003% of range. In this case, the reading error is negligible but the range error is 0.0030 ohms which is almost 3 times your results. \$\endgroup\$
    – Barry
    Commented Dec 3, 2020 at 23:34

1 Answer 1


If your generating a negative voltage on the ohmmeter, then yes it's most likely thermoelelectric. My reasoning is I see this all the time when I'm measuring peltiers that have a thermal gradient across them and are generating a voltage with my meter.

The effect is usually measured in mV but usually uV (here for solders on copper) and results from a thermal gradient across a metal. The easiest thing to do would be to put the meter on mV and see how much it changes.

If you really do think it is the thermoelectric effect or have a problem with seebeck voltages, then raising the voltage while measuring resistance would get you well out of the mV's that could be generated. So do a 4 wire measurement and use 5V or 10V (if the device under test can handle the current). make sure the leads are placed as close as they can to the DUT.

Another thing that would help is use the same metal for the contacts because dissimilar metals usually make the best thermoelectric generators. (like use copper on copper.


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