Here is what I believe is going on.

When the current is switched from zero to 16A, the temperature of the zeranin sheet rises above the constant-temperature substrate, changing the resistance for reasons as yet unexplained. This small 0.25K rise in zeranin temperature won’t produce any measurable change in resistance by way of the measured R-T curve, because I’m operating on the flat part of the curve where dR/dT=0. As it is only the temperature *difference* between zeranin and substrate that produces a resistance change, I can equally well consider a drop in temperature of the substrate of 0.25K, an approach that simplifies the analysis. Therefore I will analyse what happens to the zeranin resistance when the substrate temperature *falls* by 0.25K, realizing that the result will be the same as if the zeranin temperature was to *rise* by 0.25K.

The aluminium substrate is much thicker than the zeranin, and they are bonded together, so any dimensional change in the substrate in the X-Y direction will be forced to also occur to the Zeranin. The expansion coefficient of aluminium is +22 ppm/K. If the substrate temperature falls, then it will contract by 22 ppm/K equally in all directions. The substrate contraction in the Z-direction will do nothing, because nothing is constrained in the Z-direction, as discussed in previous postings. However, the zeranin will be forced to follow the substrate in X and Y, with the result that the zeranin dimensions will contract by 22ppm/K in X and Y, but this will result in no change in resistance, because these two dimensional changes cancel, one being a Length term, and the other a Width term.

I am confident that the explanation so far is correct, predicting that an increase in the zeranin temperature above the substrate (or fall in substrate temperature relative to zeranin, same thing) will produce no change in resistance, darn it. That is not the answer we wanted, because we observe that the resistance damned well DOES change, that’s the whole problem.

Clearly there is *some other dimensional effect going on in addition*, that has not yet been considered, that causes the observed decrease in zeranin resistance, when the zeranin temperature rises relative to the substrate.

The explanation is found in the ‘Poisson effect’. Imagine elastically stretching a length of wire. Of course, the resistance increases because the length is increased. However, what also happens is that the diameter elastically *decreases*, thus increasing the resistance even further. This is called the Poisson effect, and Poisson’s Ratio is the ratio between the ppm change in length, and the ppm change in diameter, with a value of 0.5 corresponding to an overall conservation of volume. Now apply this to our case where the zeranin has been compressed by 22 ppm/K in X and Y, resulting in an* increase * in thickness in the Z-direction. In effect, when the zeranin is compressed in X and Y, it responds by ‘popping out’ in the Z-direction, in an attempt to maintain the original volume.

The maximum possible extent of this effect would be an increase in zeranin thickness of 44 ppm/K, being a 22 ppm/K contribution from X and Y. However, for typical values of Poisson’s ratio, the actual increase in thickness will be less than that, say around half, leading to an increase in zeranin thickness of 22 ppm/K. The observed rise in temperature is 0.25K, so this would lead to a *decrease* in resistance of 0.25 x 22 = 5.5 ppm, which is remarkably close to the ~5ppm decrease that is observed.

This analysis explains why the zeranin resistance does not change when the zeranin and substrate are heated in unison ( dR/dT=0 on R-T curve), but the resistance **does** change when the zeranin self-heats above the substrate, and both the direction and magnitude of the resistance change are correctly predicted. I feel confident that this explanation is correct.

If we translate the above analysis into a formula, we get :-

**dR = P x Rth x B x EC (equation 1)**

where

dR is the decrease in resistance, in ppm, as a result of self-heating

P is the power dissipation in Watts, self-heating the zeranin

Rth is the thermal resistance from zeranin to substrate, in K/W

B is a constant related to Poisson’s Ratio, ~1.0, but <2.0

EC is the thermal expansion coefficient of the substrate, in ppm/K

For my zeranin shunt example :-

dR = 25 x 0.01 x 1.0 x 22 = 5.5ppm

One can also rearrange equation 1, to give an expression for the Power Coeffcient of Resistance (PCR), in units of (ppm/K) per watt of dissipation.

**dR/P = PCR = Rth x B x EC (equation 2)**

Knowledge is power. Now we can clearly see exactly what will help in reducing the self-heating-induced resistance drift, and what will not.

From equation1, we can reduce the resistance drift, dR, by reducing the dissipated power, P, or the thermal resistance from foil to substrate, Rth. No surprises there, we knew that already.

Some people suspected that the resistance drift was caused by difference in expansion coefficient, EC, of the zeranin and substrate, but not so. The EC of the zeranin foil doesn’t show up in the analysis or equations at all, and therefore there is nothing to be gained from choosing a substrate material that matches the EC of the resistive foil, at least as far as minimizing PCR is concerned. The formula clearly shows the only thing that matters for PCR is the EC of the substrate.

The self-heating-induced resistance drift scales directly with the EC of the substrate. Thus, an aluminium substrate (22ppm/K ) is a poor choice. Copper would be better, and steel significantly better, though the thermal conductivity of steel is less than ideal. Invar would be best, except that the thermal conductivity is so low as to be useless. The resistor manufacturer has further options with ceramics.

Note that whatever substrate material is chosen, it will always be necessary to arrange for dR/dT to be zero or small at the foil operating temperature. If the ECs are not matched, this will modify the ‘naked’ R-T curve which in my case I can account for (within reason) by operating at whatever temperature the sweet spot (dR/dT=0) happens to be at. The resistor manufacturer can account for this by tweaking the resistive alloy.

Please tear this explanation apart, and/or offer an alternative explanation that fits the measurements. Comments, please. If the explanation withstands scrutiny, then the mystery is solved and understood.