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My background: I'm engineer of electronic HW engineering and embedded computer systems. During repair of a vintage digital voltmeter (resulution 1µV) I discovered an effect that probably only a physicist can answer. I isolated the problem and condensed it to a simple experiment:

  1. Take a piece of copper wire (about 20cm) and cut it in two halves.

  2. Join the two halves again by soldering. But don't let the two wires come in direct contact. They should be connected via the solder bead alone.

  3. Measure the voltage across the joint with a 1µV resolving voltmeter.
  4. Apply heat via hot air (about 100 degrees Celsius) to the solder joint.
  5. There is a voltage of about -20µV.
  6. Swap the wire ends at the voltmeter side.
  7. The voltage is about 20µV (to be expected).

My question: I assumed, the thermoelectric voltages from copper to solder and solder to copper would cancel out. But they don't. Why ?

Thanks !

P.S. The solder is regular solder, leaded, out of the pre-lead-free age.

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  • $\begingroup$ The soldering can be asymmetrical. Different geometry, different amount of intermetallic compounds. Is the experiment repeatable with a different soldering? $\endgroup$
    – my2cts
    Commented Nov 20, 2021 at 22:50

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Mario, welcome to physics SE. You are right, the effects should cancel because of the symmetry of the junctions. In fact, a similar technique is commonly used to null out the effects of ambient temperature at the measuring gauge when a long run of thermocouple wire is used between the junction and the gauge. So you have uncovered a subtle and interesting effect! Here are some thoughts.

Zeroth, try the same measurement with a DIFFERENT voltmeter and leads to see if the meter itself is properly functional.

First, it is always good & prudent to check the null-point calibration of the meter, when making measurements of order ~1 mV- especially when the device has been recently repaired.

Second, how are the meter leads attached to the ends of the copper wire? with toothed clamps? If so, are the clamps metallurgically identical?

Third, how are the meter lead plugs connected to the input jacks on the meter housing? Same commentary applies.

Try these things and then write us back here with your results so we can ponder the imponderables some more, as needed.

Good luck!

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    $\begingroup$ Another possibility is that the heating of the junction is not perfectly symmetric. It wouldn't take much of an asymmetry to result in a voltage of 20 $\mu V$. For example, 20 $\mu V$ corresponds to a temperature change of just about 1 ˚C for a K-type thermocouple. So if the heating of the junction is slightly asymmetric so that there is a thermal gradient of just 1 ˚C across the solder junction, that could conceivably result in a voltage of 20 $\mu V$ across the wires. $\endgroup$
    – user93237
    Commented Aug 26, 2018 at 21:20
  • $\begingroup$ to zeroth: The meter is ok. By the way no other meter of this quality is around here. $\endgroup$
    – Mario Blunk
    Commented Aug 27, 2018 at 21:22
  • $\begingroup$ to zeroth: The meter is ok. By the way no other meter of this quality is around here. to second and third: The leads are connected via regular 4mm lab plugs on the meter side and toothed clamps of equal material and finish on the object side. If I connect the leads on the side of the specimen of interest the meter reads stable zero µV. The internal null-point calibration is also ok and stable. The fact that by swapping the two leads of the "unwanted thermocouple" the meter displays the reverse polarity tells me that the the effect is contributed by the object of interest alone. $\endgroup$
    – Mario Blunk
    Commented Aug 27, 2018 at 21:31
  • $\begingroup$ @MarioBlunk - I would try adjusting the direction of the hot air from the heat gun slightly with respect to your wire junction in order to see how sensitive that 20 $\mu V$ reading is to slight heating asymmetries in your setup. As a null test, you might also consider just heating the middle of a solid length of copper wire (with no solder junction) just to see what level of background voltages your voltmeter gives in your setup because 20 $\mu V$ is probably starting to get close to the sensitivity limit of your instrument. $\endgroup$
    – user93237
    Commented Aug 27, 2018 at 23:02

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