PCR versus TCR - Page 1
Forum section: electronics / general

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 01:04:18 am
content: There is much discussion here about TCR (Temperature coefficient of Resistivity), as to be expected, but much less about PCR (Power coefficient of Resistivity).From the point of view of a Metrologist, arguably TCR is more important, as Metrology is only concerned with accurately measuring and comparing resistance values, and to this end, power dissipation and thus self-heating in the resistors is arranged to be as low as possible.In the Real World though, PCR is often at least as important, sometimes more so. In real world applications, resistors commonly dissipate significant power, and the resultant change in temperature from self-heating often exceeds changes in temperature from the environment.Just so, I hear you say, but it’s really all just the same thing, isn’t it? If you know the thermal resistance from the resistive element to the environment, and you know the power dissipation, then you can easily calculate the temperature rise of the resistive element for any given power dissipation, or even simpler, you can directly measure the temperature rise for a given power dissipation. Then it’s just a matter of looking at the R-T (resistance versus temperature) curve (obtained from manufacturer or measurement) and you can confidently predict the change in resistance, as a result of the self-heating-induced change in temperature. In other words, the self-heating-induced change is resistance is easily predicted knowing the self-heating-induced change in temperature and the R-T curve. Right?Well, that’s what I used to think too, but have found that in practice it doesn’t quite work like that, to the point that in some situations, the self-heating-induced change in resistance bears no relationship to the R-T curve. In short, the change in resistance with temperature of the resistive element depends on whether that change in temperature was caused by self-heating, or by changing the temperature of the environment is which the resistor resides, which is how the R-T curve is measured. To put it another way, when the temperature change is as a result of self-heating, there appear to be other mechanisms at work that change the resistance, that have nothing to do with the R-T curve. That is a bold claim, and I will spend some time presenting experimental evidence to justify this claim, and then bounce ideas around with the experienced resistorologists here as to what I believe those other mechanisms are.This is not merely of academic interest. The application is a 0R1 shunt resistor, used to measure up to 16A, in an ultra-high-precision current driver circuit. To reduce noise and thermoelectric potentials to an acceptably low level requires 1.6V across the shunt at full current of 16A, thus the choice of 0R1. Inevitably though, the shunt dissipation is 16x16x0.1 = 22.5W, leading to self-heating of the shunt. If the application called for a constant current of 16A over a long period of time there would be no particular problem, as the shunt temperature and resistance (and therefore the controlled current) would eventually stabilize.Unfortunately, the applications call for the controlled current to be stable within a few ppm, on all timescales after the current is abruptly switched from zero to 16A, meaning that the shunt resistance must be stable within ppm immediately after the current is switched from zero up to 16A. Suddenly the application becomes very challenging indeed, and a very low PCR becomes critically important.The adopted solution is a very large, custom build resistor made from Zeranin sheet, about 0.3mm thick. The Zeranin sheet is first bonded to a 1.6mm thick aluminium plate, using a 0.07mm thick heat-bonding, electrically insulating film. The Zeranin is then etched to the required shape, being a zig-zag pattern, with the conductor being about 25mm wide, and a total conductor length of about 1.4m. Overall dimensions are about 350mm x 100mm x 1.6mm. The thermal resistance from the Zeranin sheet to the aluminium substrate is about 0.01 K/W, and the assembly is firmly bolted/clamped down onto a 12mm thick aluminium heatsinking plate which itself is temperature controlled to within 0.1K, by way of an array of Peltier modules driven by a PID temperature controller. A fairly serious setup, which keeps the Zeranin temperature constant to within 0.3K under all conditions, even when the dissipation is abruptly changed from zero to 22.5W. The R-T curve of the Zeranin bonded to the aluminium substrate has been carefully measured, and the heatsink/resistor are operated at a temperature of 34.0 DegC, where the R-T curve has a minimum slope of about 1ppm/K. Easy-peasy. When the Zeranin temperature changes by 0.3K, then any text book will tell you that the resistance will change by 0.3K x 1.0 ppm/K = 0.3ppm, which I would be more than happy with.Unfortunately, what is actually observed is very different, and cannot be explained by way of the R-T curve. After the current is abruptly switched from zero to 16A, the shunt resistance drifts downward by about 5ppm over 50 seconds, and thereafter remains stable. Furthermore, it makes not one zot of difference whether I operate the Zeranin resistor at 20 DegC, 34 DegC, or 47 DegC, the 5ppm drift is exactly the same, although the slope of the R-T curve is many times greater at temperatures other than 34 degC. In other words, as I stated early in this posting, in this particular case the measured R-T curve has nothing to do with the self-heating-induced change in resistance – some other mechanism must be responsible.I have some ides as to what this mechanism might be, but don’t want to pre-empt ideas that others might have. Comments, gentlemen?

author: zlymex
author_rank: Frequent Contributor
timestamp: April 19, 2016, 02:06:19 am
content: Very interesting topic.Does the downward drifts of 5ppm, although independent of the operate temperatures, depend on the operate current?That is to say, will the drifts become about 2.5ppm when operate at 8A instead of 16A?My guess of the possible cause, it got to be heat or temperature related, not electro-mechanical related such as the length or distance/thickness change by electromagnetic, since that change would be instantaneous.And another related question is, why you design you shunt to operate at 1.6V instead of 0.8V? I'm asking this because it is related to my partially answered topic/question of 'Why output voltages of precision shunts are so high?'

author: chickenHeadKnob
author_rank: Super Contributor
timestamp: April 19, 2016, 02:34:25 am
content: How are you connecting (conductors) to the zeranin? I would look to dis-similar junctions to cause most of the mischief. Mr. Pettis who sometimes posts here and has deep wire wound resister making experience has differentiated his resisters from the others by his proprietary lead welding. Search the forum for his posts and links to his EDN? articles. The other thing is that by gluing the zeranin sheet down you now have to deal mechanical strain and different expansion rates. This is why immersion in a thermal transfer fluid like an oil bath would be better. Let your resistor creep on its own volition.

author: gilbenl
author_rank: Regular Contributor
timestamp: April 19, 2016, 03:11:56 am
content: Nice first post!I think what you're describing (non-linerarity) is the consequence of the Joule Effect. Check this article out:http://www.digikey.com/Web%20Export/Supplier%20Content/VishayPrecisionGroup_804/PDF/vishay-tech-non-linearity-characteristic.pdf

author: zlymex
author_rank: Frequent Contributor
timestamp: April 19, 2016, 03:52:36 am
content: I once tore open an Isabellenhütte RUG-Z shunt resistor, 250W, 0R01, 0.1%, TCR 3ppm/K.They claim the shunt is made of Zeranin too and it seems to me the conducting sheet is about 0.3mm in thickness as well.

author: gilbenl
author_rank: Regular Contributor
timestamp: April 19, 2016, 04:17:02 am
content: Quote from: zlymex on April 19, 2016, 03:52:36 amI apologize for hijacking, but zlymex-How are you able to tear apart all of these high end items? You've posted tens of thousands of dollars worth of beneficent destruction. Are the items out of spec/BER?

author: zlymex
author_rank: Frequent Contributor
timestamp: April 19, 2016, 05:03:47 am
content: Quote from: gilbenl on April 19, 2016, 04:17:02 amI'm known locally as the detructor, someone send me items for tear apart only such as that VHA518-7, sometime we group-buy the item to be teardown as in the case of a VHP202Z. However in this case, the RUG-Z had already broken(by sheer mechanical force I think, not burnt down, no smell of that sort), I tore open for repair

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 08:07:30 am
content: Hi. Yes, the magnitude of the drift scales with the current, and I agree that the effect is thermal, even if the exact explanation is subtle.As to the choice of 1.6V across the shunt, everything in life is a compromise. On the one hand, I want the largest voltage across the shunt, because this means that noise and thermoelectric effects become less significant. In this design, very low broadband noise is required, set by the Johnson noise of the resistors in the preamplifier that amplifies the voltage across the shunt, so the higher the shunt voltage, the better the SNR. Total hum and noise is around 1 part in 2E7 over a bandwidth of 10 kHz, not achievable if I needed to amplify up a much smaller shunt voltage. Simultaneously, I also want the shunt voltage to be as low as possible, to minimise self-heating induced changes in resistance, as described. In my application, the ‘sweet spot’, for the best trade-off between these conflicting requirements turns out to be with about 1.6V across the shunt. Having 0.8V across the shunt would also be OK, and I have that option, as I have an identical 0R05 shunt.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 08:14:05 am
content: Quote from: chickenHeadKnob on April 19, 2016, 02:34:25 amHi, I agree that thermoelectric potentials can be a nasty source of error, but I can prove that this is not the case here. A thermoelectric potential produces a fixed offset error, that would also be present in my ‘zero current’ measurement, but there is no drift in my zero current measurements. What I actually do is produce a 16A ‘square wave’, zero current for 100 seconds followed by 16A for 100 seconds, over a number of cycles. If thermoelectric effects were present, they would show up equally in the zero current measurement, and this does not happen. The zero-current measurement is rock-solid, so thermoelectric effects can be ruled out here.I agree that immersion of naked zeranin in an oil bath would solve this problem, but causes others. The thermal resistance from (temperature controlled) oil bath to the resistive element will be too high unless vigorous stirring or pumped circulation of the oil is employed, and that is complexity and size I want to avoid.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 09:00:01 am
content: Quote from: gilbenl on April 19, 2016, 03:11:56 amI did indeed check it out, and am familiar with all that is written there. However, what I am observing is specificallynotthe textbook stuff described there. The Joule Effect simply means I^2R self heating, and yes, we all agree that the observed drift is as a result of self-heating. However, the article speaks only of effects that can be predicted from the R-T curve, whereas the effect I observe specifically cannotbe predicted from the R-T curve.

author: zlymex
author_rank: Frequent Contributor
timestamp: April 19, 2016, 09:33:44 am
content: Quote from: Zeranin on April 19, 2016, 08:07:30 amIs the noise in peak to peak or in rms?What kind of preamplifier are you using?

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 11:01:49 am
content: The quoted noise is RMS. If you are really interested, I'll look back at my notes to find the exact numbers. Thinking back, although the circuit bandwidth is around 10 kHz, the noise spec is quoted in ~1.5kHz bandwidth, this being the maximum frequency that the Bose Einstein Condensate atoms can 'see'. The current driver drives current through magnetic coils that are part of a Physics experiment producing BECs. The shunt preamp is part of the precision current driver circuit. It uses multiple amplifiers in parallel to get a root(n) noise reduction. To keep resistor noise down, no resistor anywhere in the driver op-amp circuits can be >500 ohms, and the entire circuit is inside a temperature controlled oven. Transformers or fans can't be placed within 1.5m, or the stray AC fields will get into the circuit. As the shunt is inevitably quite large, it forms a loop, so by Faradays Law this induction loop picks up stray AC magnetic fields, that arealwayspresent, so this is cancelled by an 'antiphase cancellation loop' that follows the same path as the zeranin. From memory, the mains hum in the output current is down somewhere in the 1E-8 level. As the load is highly inductive, a thumping great magnetic coil, the driver must be able to produce up to 180V at 16A, to maximize current slew rate through the inductive load coils. I'm lucky. I get paid to design and build this stuff.But I have drifted off topic. The most troublesome part of this project has been the 16A temperature controlled shunt resistor. As far as I'm concerned, if I can measure an imperfection, then it should not be there.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 11:51:26 am
content: Quote from: DiligentMinds.com on April 19, 2016, 03:30:41 amI broadly agree with what you say, but feel that the exact mechanism requires a more detailed explanation. Like others, you refer to thedifferent TCE’s between the different materials causing the bonding strain to be different,but I’m not convinced that this is relevant. Keep in mind that the R-T curve that I measured was with the zeranin sheet already bonded to the aluminium substrate, and mounted in the current driver circuit. It is true that differential expansion between the zeranin and aluminium will modify the zeranin R-T curve, and indeed this is the case, with my measured curve being significantly different to the manufacturer zeranin curve, taken on a ‘naked’ sample of zeranin. Therefore, this different rate of expansion is already accounted for in my measured R-T curve, isn’t it?Let me explain how my R-T curve was measured. As explained, the zeranin sheet is pre-bonded to a 1.6mm thick aluminium substrate and clamped/bolted down to a temperature-controlled 12mm thick heatsinking aluminium plate. This zeranin shunt is the current measuring element in a precision current driver circuit, that for measuring the R-T curve is set to produce a steady 16A. In series with the same current is my massive (100A rated) Leeds&Northrup (L+N) naked master-reference-shunt, temperature controlled to 0.02K. If the resistance of the zeranin shunt in the current driver should change, then the controlled current changes, and this is detected via the voltage across the L+N master shunt. The zeranin shunt is clamped down onto the temperature-controlled aluminium plate with a 12mm thick plate of copper, and a thermistor is embedded deep within this copper plate, essentially touching the zeranin sheet, so the thermistor therefore accurately measures the zeranin temperature, say within 0.1K. The heat flux flows from the zeranin to the temperature controlled aluminium plate, with a thermal resistance of around 0.01K/W, so at full dissipation of 25.6W, the zeranin temperature rises by ~0.25K with respect to the 12mm thick aluminium plate to which it is clamped, not much at all, and completely negligible on the R-T curve. Measuring the R-T curve is real easy. The heatsink temperature (and therefore the Zeranin temperature) is ‘dialled up’ on the Peltier temperature controller setpoint to values between 20 and 50 DegC. At each zeranin temperature, the voltage across the L+N master shunt is recorded, being an accurate measure of the zeranin resistance.As you would thus appreciate, this is exactly the same physical setup as when the 5ppm drifts in the zeranin resistor are observed. In this case, the current driver is commanded to produce a 16A ‘square wave’, zero current for 100 seconds, followed by 16A for 100 seconds, repeated for a number of cycles. If the resistance of the zeranin shunt resistor within the driver circuit should change, then the current produced by the driver circuit changes, and this is detected by logging the voltage across the L+N master shunt. If I log the voltage across the 0R1 zeranin shunt, then of course I see a perfect square wave of 1.600000V amplitude, with no measurable drift, exactly as we should expect, as the driver circuitry is of very high quality. However, immediately after the current is commanded from zero to 16A, the voltage across the L+N master reference resistor is observed to drift upward about 5ppm, over a time of ~50 seconds, and then stabilize. As the actual current is drifting upward, this means that the resistance of the zeranin shunt within the driver circuit must be drifting downward. Sorry for long explanation, but it’s not fair of me to ask questions without describing details of the equipment.So, why is it that my measured R-T curve, taken with the zeranin sheet already bonded to the aluminium substrate and bolted to the temperature controlled heatsinking plate, is not relevant to the situation wherethe very same resistor, mounted in exactly the same way, is self heated? Differential expansion caused by the different coefficients of expansion of zeranin and aluminium is already accounted for in my measurement of the R-T curve, and yet this R-T curve cannot possibly explain the observed 5ppm change in resistance. According to my R-T curve we would need a temperature shift of several degrees to get a 5ppm resistance shift, yet nothing changes temperature by more than a few tenths of a degree.I predict that we would see the same 5ppm drift, even if the zeranin resistive element was bonded to a zeranin substrate, and clamped down onto a zeranin temperature controlled plate, such that everything has the same temperature coefficent of expansion (TCE). Presumably you don’t agree, as you specifically mention the different TCE’s of the zeranin and aluminium but, as I said, this is already accounted for in my R-T measurement, isn’t it? Can you (or anyone) explain your ideas in more detail, in the light of the points raised in this email?I have a possible (detailed) explanation for what is happening, but do not wish to not pre-empt the ideas of others just yet. I do agree that the problem is caused by having the zeranin bonded to a thick metal substrate.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 19, 2016, 11:58:20 am
content: Hi and welcome to the forum!1) Does the current need to be set quickly at "any" value between 0 and 16A? Or is it an "on/off" situation for a particular current which can be set and just steered in/out of the load?2) Right now your amplifier noise is equivalent to about 40 Ohm resistor, isn't it?3) Are you sure that you are not observing a drift in your L+N master shunt?CheersAlex

author: zlymex
author_rank: Frequent Contributor
timestamp: April 19, 2016, 12:19:46 pm
content: Quote from: Zeranin on April 19, 2016, 11:01:49 amIf that is the case, one part in 2E7 times 6, we get about 0.3ppm pp noise, that is not a big deal. There are many precision opamps capable doing the job alone. Take the very old OP27A for example, 0.08uVpp typical noise(0.18uVpp max), if amplifying 0.8V, it will 'add' 0.1ppm pp typical noise, much less than 0.3ppm pp. So it is very safe to use 0.8V shunt voltage instead of 1.6V as far as the noise is concerned.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 19, 2016, 12:35:16 pm
content: The difference in temperature between the resistor element and the aluminum backing can have an influence, but the time constant should be rather short, as it does not take that long for the temperature to equilibrate.The time constant might be a good hint to locate the problem. It could be as well be the other shunt.A thermal effect should be proportional to the square of the current - so a test with a smaller current (e.g. 10 A) might help.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 19, 2016, 12:47:50 pm
content: Quote from: zlymex on April 19, 2016, 12:19:46 pmNot in a 0-10kHz bandwidthCheers,Alex

author: zlymex
author_rank: Frequent Contributor
timestamp: April 19, 2016, 02:16:16 pm
content: Quote from: Alex Nikitin on April 19, 2016, 12:47:50 pmThanks for that. I misread for 10 Hz.Quote from: Zeranin on April 19, 2016, 08:07:30 am

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 19, 2016, 03:43:00 pm
content: Quote from: DiligentMinds.com on April 19, 2016, 02:25:52 pmThis 80 seconds time frame sort of hints to the "reference" shunt problem. An easy way to check would be to precondition the L&N shunt at about 16 A (doesn't need to beveryaccurate) for several minutes and then measure the current from the source.CheersAlex

author: Zeranin
author_rank: Regular Contributor
timestamp: April 19, 2016, 11:23:08 pm
content: Several have rightly pointed out that the measured drift could be the L+N reference shunt, rather than the zeranin shunt in the current driver. However, I had already performed the following experiment, showing that the L+N shunt is not the cause of the problem.The current driver is guaranteed to produce a rock-stable current when producing a steady 16A for a long period of time, because all thermal effects in the zeranin shunt will reach an equilibrium after a few minutes, after which no further drift will occur. For this experiment to determine if the L+N shunt is OK, I set the current driver to produce a steady 16A. I installed an SPST switch, that can switch the driver current output either through a 'dummy' 10 mohm resistor, or through the L+N master shunt 10 mohm resistor. Initially, I send the current through the dummy resistor, for 10 minutes or so, until the current stabilizes to better than 1ppm, which it does. Then, I flick the switch, so the current is now sent through the L+N resistor, and I data log the voltage across the L+N. Ignoring the 1st 100ms or so after the switch is settling down, the current pulse sent through the L+N will be perfect, without any drift at all on the leading edge, or thereafter. This we know, because the zeranin shunt in the current driver sees a steady 16A at all time (except for a few ms when the switch is switching), so there cannot be any thermal drift effects occurring in the zeranin shunt, and I know from many measurement that the current stability produced by the Driver is set entirely by the stability of it's shunt resistor - the rest of the current drive circuitry may be regarded as perfect.The resultant square wave logged across the L+N master shunt, produce by operating the toggle switch, is observed to be perfect, with no drift on the leading edge, after the current is abruptly switched from zero to 16A. I therefore conclude that the L+N master shunt is OK, and that the 5ppm drift issue that I have previously described is caused by resistance drift in the temperature-controlled zeranin shunt in the current driver.I'll talk more about peoples relies, which I appreciate. It's great to have a bunch of sharp people that find this kind of stuff interesting, that I can bounce ideas off. I think we are all in broad agreement that the problem comes about because the zeranin is mechanically bonded to a metal substrate, and I'll attempt to fill in some details as to exactly why this causes the observed problem.

author: Cerebus
author_rank: Super Contributor
timestamp: April 20, 2016, 02:17:55 am
content: I need to read everything again to be sure I precisely understand what's going on. That said, if my understanding of the 'experimental' setup is correct and the drifts are happening when I think they are I have a suspicion that the problem is caused by creep in the bonding layer. That is, you hit it with a 16A square wave, the shunt heats up, the aluminium heats up and sets up a stress on the bonding layer because of differential expansion of the shunt and aluminium. Then the bonding layer slowly relaxes giving the observed shift as the stress in the shunt changes with movement of the binding layer. Very much a first guess but I think it's consistent with the observed behaviour, if I've understood the observed behaviour properly.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 20, 2016, 07:56:19 am
content: Quote from: Cerebus on April 20, 2016, 02:17:55 amA good point, and a possible solution would be to use a heatsink material with the same thermal expansion coefficient as Zeranin (18ppm/C). Silver looks like a perfect match with 18ppm/C and even plain copper (19ppm/C) looks much better than aluminium with 23ppm/C.CheersAlex

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 09:41:59 am
content: Quote from: Alex Nikitin on April 19, 2016, 11:58:20 amI know where you are coming from, but sadly the current needs to be set from any current I1, to any current I2, either in a step, or a linear ramp, or some other form of ramp.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 09:57:18 am
content: Quote from: Kleinstein on April 19, 2016, 12:35:16 pmYou claim that if the drift effect is thermally induced from self-heating, then the effect would scale as the square of the current. Actually, that is not really right. In ABSOLUTE terms, that is correct, but results and plots of this type are always in units of ppm or ppk/K. If you halve the current, then the ABSOLUTE effect in Amps is reduced by a factor of 4, but the current itself is halved, so the effect when expressed in ppm or ppm/K is halved, in proportion to the current, and this has been confirmed experimentally.On the time constant, it's more complicated. When the current is switched from zero to 16A, the temperature of the zeranin sheet starts to rise above the constant-temperature substrate. This will not happen instantaneously, because of the thermal mass of the zeranin. In my particular situation, a 12mm thick copper plate is used to clamp the zeranin onto the 12mm thick, constant-temperature heatsinking plate, so the effective thermal mass of the zeranin is considerable, and we would expect the temperature rise of the zeranin to be relatively slow, a few 10’s of seconds. I have a thermistor embedded in the copper block, almost touching the zeranin, so I can confirm that the temperature of the zeranin and the copper block do indeed rise, by about 0.25K, over a few 10’s of seconds.I explained in another posting the experiment showing that the L+N reference shunt is not at fault.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 10:11:32 am
content: Quote from: DiligentMinds.com on April 19, 2016, 02:25:52 pmYes, your posting makes perfect sense, and I thank you for thinking about it. I think the topic still has some way to run though. I am writing up a long explanation of exactly what I think is happening, and hope to post it soon.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 10:17:05 am
content: Quote from: Cerebus on April 20, 2016, 02:17:55 amI understand what you are saying, and it’s a good suggestion. However, the electrically insulating layer that bonds the zeranin to the aluminium substrate is organic and would soften more at higher temperature, so we would expect any such creep would be faster and further at 47 DegC compared to 20 DegC, and yet no difference is observed. I would also need to be convinced that such a theory matches the observed direction of the drift, and approximately predicts the correct magnitude. Keep in mind also that the zeranin temperature rise is only 0.25K, so the induced strain would be very small, and just as well. I’ll bet creep effects would start to happen for temperature swings of ten’s of degrees. This is certainly not the way to build a metrology style resistor, but most metrology style resistors won’t cut the mustard here because the significant self-heating at 16A would produce an unacceptable temperature rise. Whatever else you can say about this zeranin resistor, you won't find another resistor whose temperature rises by only 0.25K on application of 25.6 Watts, ie, a thermal resistance from zeranin to backing plate of 0.01 K/W.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 10:43:30 am
content: Quote from: Alex Nikitin on April 20, 2016, 07:56:19 amI think you have missed the point here, and my own suspicion is that using a heatsink material with the same thermal expansion coefficient as Zeranin will not help. If the heatsink and zeranin change temperature in unison, as when the R-T curve is measured, then difference in expansion coefficient will indeed modify the manufacturer R-T curve that was taken with naked zeranin. However, the R-T curve that I measured is with the zeranin already bonded to the aluminium substrate, and I use thisactualR-T curve to select an operating temperature (~34 DegC) where the actual dR/dT is almost zero. The effect of the different thermal expansions is simply to modify the R-T curve, and shift the position of the sweet spot where dR/dT=0.It is my suspicion, and some others have said this as well, that what causes the unexpected resistance drift is the temperature rise of the zeranin with respect to the constant-temperature substrate, and the stresses and strains that this creates are not related to the difference in expansion coefficients. This temperature rise is tiny, just 0.25K, so one could be forgiven for thinking it should have a negligible effect. The thread may still have some way to run to fully explain what is going on.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 20, 2016, 10:52:11 am
content: Another suggestion: as the dissipation in the shunt goes from 0 to maximum and the temperature rises by 0.25K, it creates a shift in the temperature distribution in the heatsink and a shift in the temperature regulation loop. Also 0.25K makes for ~5ppm linear extension in Zeranin...CheersAlex

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 11:08:29 am
content: Quote from: Alex Nikitin on April 20, 2016, 10:52:11 amA fair suggestion. However, the x12 Peltier modules sitting on the opposite side of that 12mm thick constant temperature aluminium heatsinking plate are arranged in a grid that nicely and uniformly covers the area of the zeranin shunt, deliberately designed that way to ensure an even temperature across the entire area of the shunt. To further iron out temperature differentials, there is the 12mm thick aluminium plate itself, plus that 12mm thick copper plate on the other side of the zeranin resistor, used to clamp it down to the constant temperature aluminium plate, so all in all we can be confident that the entire zeranin resistor is damned close to being at a single, uniform temperature.You are correct that 0.25K makes for 5ppm linear expansion in zeranin ..... an astute observation, but as yet, only hand waving. You will need to stitch that into a full, detailed explanation of what is going on.

author: amspire
author_rank: Super Contributor
timestamp: April 20, 2016, 12:01:51 pm
content: Quote from: Zeranin on April 20, 2016, 11:08:29 amIs there a layer between the copper layer and the Zeranin to allow for the different expansion rate of copper to aluminium? Copper has a lower expansion rate then aluminium, so it may be compressing the top of the Zeranin which would lower the resistance. Perhaps if you can try different top plates, you can get one to give the best coefficient. Zinc, for example, is a higher expansion rate then aluminium so it should work towards a positive temperature coefficient.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 20, 2016, 12:10:49 pm
content: Quote from: amspire on April 20, 2016, 12:01:51 pmIt is certainly a good question why different materials are used on two sides of the shunt. Why not copper on both sides?!CheersAlex

author: d-smes
author_rank: Regular Contributor
timestamp: April 20, 2016, 04:44:46 pm
content: Quote from: amspire on April 20, 2016, 12:01:51 pmThere must be something there or the Copper would electrically shunt the Zeranin! Regardless, I like this theory. If the Copper is bolted to the Aluminum at the corners, then the assembly would bow as it heats- The Al expands more and faster than the Cu both because of the higher expansion rate and because the Al is a fraction of a degree warmer than the Cu. With the Al now being slightly longer than the Cu between clamping points, the assembly curls/bows (long dimension is no longer straight).Having just written this, resistance shift due to the bending should have showed up in the R-T test. Maybe it did which is why the clamped R-T didn't match the Zeranin in free space R-T. So we're back to the internal heating of the shunt with current as causing some additional stress / strain forces that causes the shift.I like Alex's suggestion of using Copper on both sides; especially since the Cu has a closer TCE to Zeranin. In addition, I'd suggest cooling both sides symmetrically and use the same 0.07mm thick electrically insulating film on the clamping side. If that doesn't help, maybe you need to invest in another L+N shunt and use that in your machine (or attempt to copy how it is made).

author: Zeranin
author_rank: Regular Contributor
timestamp: April 20, 2016, 11:57:01 pm
content: Here is a more precise description of how the beast is built. There are 3 major components.(a) The 12mm thick temperature-controlled aluminium heatsinking plate, with an array of Peltier modules on the underside. This plate is 350 x 230mm, which is wider than the 100mm wide resistor. This plate doubles as the base of the temperature controlled box that houses the electronics for the current driver, which is why it is 230mm wide. This arrangement produces a temperature controlled box for the electronics 'for free'. I don't have easy access to 12mm thick copper plate of this dimension, so was made of aluminium. The current driver will soon need to be delivered to the customer, so there is not time to rebuild this plate in copper, and in any event, I would hesitate to embark on such a major and expensive design change unless firmly convinced it would make a difference.(b) The zeranin resistor assembly. This consists of the etched ziz-zag patern of zeranin sheet, bonded to a 1.6mm thick sheet of aluminium (AKA the substrate), of dimension 250 x 100mm. The 0.07mm thick heat-bonded film provides electrical isolation b/w the zeranin and aluminium.(c) The 350 x 100 x 12mm thick copper clamping plate.The object is to minimize self-heating induced temperature changes of the zeranin. Therefore it was decided that the zeranin (rather than the aluminium substrate) would be in contact with the constant-temperature heat-sinking plate, separated of course by a 0.06mm thick electrically insulating PVC plastic film, with small quantity of thermal grease. Thus, the zeranin sees aluminium on both sides. The copper clamping plate is in contact with the aluminium substrate, and x8 of M4 screws pull the clamping plate down onto the constant-temperature heatsinking plate, with the zeranin resistor assembly sandwiched between. The screws pass through the zeranin resistor assembly, with mating 5mm clearance holes etched in the zeranin sheet, and subsequently drilled through the substrate.At this stage, it is not practical to change the metals that have been used. I originally requested that the substrate be made of copper, but that was not possible, as the etching process would have removed the copper, as well as the zeranin!This project has a long history. It was originally designed to use a Powertron FHR4-80370 shunt resistor, but this was found to be unsatisfactory, with a drift on the leading edge of the 16A square wave of almost 15ppm. We had a custom version of this resistor made at great expense from Zeranin, yet found it performed no better than the cheap bog-standard foil, again reinforcing the point that the R-T curve has nothing to do with the drift that we observe. By this time I was sure in my mind that the problem was due to the temperature rise of the zeranin with respect to the substrate, so designed a 'super-sized' version of the 80370, with almost 3 times the surface area of zeranin. The thermal resistance from zeranin to substrate scales as the surface area, so this new resistor with x3 surface area results in a x3 reduction in the temperature rise of the zeranin. Sure enough, the new resistor does perform x3 better, with a drift of only 5ppm vs the original 15ppm. Further improvement could be made by scaling up even larger, but the existing super-sized resistor is already at the limits of the equipment used to manufacture it, and I would also have to completely redesign and rebuild the heatsinking plate and peltiers.Further improvement could be made by decreasing the thermal resistance of the 0.06mm thick PVC plastic film between the zeranin and the constant temperature heat-sinking plate. The best material to use here is the thermally conductive version of the common Kapton (Polyimide) 'HN' film. The thermally conductive version is known as 'Kapton MT' with a thermal resistance x3 lower than the common HN version. Unfortunately, the MT version is hard to get, and the largest sheets I have been able to obtain (without spending k$ for large quantities) are A4 size, which is just slightly too small, damn it.The customer can live with the 5ppm drift, and typically the largest step change in current is <8A, with drift of 2.5ppm. That said, I live by the philosophy that if I can measure an imperfection, then it should not be there.I'll mention that most precision designers don't use shunts at >10A, precisely because it is so damned difficult to avoid self-heating induced changes in temperature and resistance. The usual approach is to use a magnetically based sensor that does not suffer from any self heating at all, such as the Danfysik (now LEM) 'Ultrastab' range of current sensors. Look inside a million$ MRI machine, or commercial current driver for magnetic coils used in science research, and you will find a Danfysik current sensor. I have used these sensors in the past, but they have issues of their own. I have placed a Danfysik current sensor in series with the 16A current driver output, and was horrified to find that the measured current drifted by 200ppm (not a misprint) over the first 300 ms after a step current change, then overshoots, then converges to an approximately stable value. I contacted the manufacturer, who at first blamed my measurements, but after a month of email communication, were eventually able to independently verify that I was correct. To this day there is no mention of this imperfection in their data sheets though. The technical term that describes this effect is 'settling time'. They need to measure and specify the settling time to 100 ppm, 10ppm, 5ppm and 1ppm, but the trouble is, if they were to do so, then demanding customers may not buy the product. Or maybe they would anyway, because you can't buy a shunt resistor that does not drift immediately after application of 16A or more. The customer is presently using a 16A current driver that I built many years ago, employing a Danfysik Ultrastab current sensor, so will be well satisfied with better than order-of-magnitude improvement. The Ultrastabs are noisy, too, with the new current driver having better than order-of-magnitude lower noise as well.At an academic level, I am never happy with any observed effect that I do not understand, so regardless of customer needs, I would go mad if I did not at least fullyunderstandwhat was going on. I'll get to explaining a possible detailed explanation, really just a refinement of some of the ideas already presented, and see if it withstands scrutiny.Cheers, Colin

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 12:21:00 am
content: Quote from: d-smes on April 20, 2016, 04:44:46 pmA good point, but it is not that easy. The reference shunt is 10 mohms, so the heating is a mere 2.5W at 16A. It is also huge, the naked manganin strip being 100mm x 0.7mm x 1400mm, folded 6 times. I also temperature control the manganin strip to within 0.02K, just to be sure. Self-heating induced changes in temperature and resistance (at 16A) are truly negligible.Unfortunately though, I can't use a 10 mohm shunt in the current driver, because the voltage developed (0.16V at 16A) is too low, and would result in excessive noise. Therefore the zeranin shunt in the current driver is 0R1, giving 1.6V of signal, and improving the SNR by an order of magnitude. Also, thermoelectric effects become very significant with a 10 mohm shunt, and I need to work hard to keep that under control when using the 10 mohm L+N at 16A, an issue that I can do without, and that all but disappears with a 0R1 shunt.Life was not meant to be easy.I do have an identical 0R05 zeranin shunt that would have the initial drift to 2.5ppm, at the expense of slightly higher noise. Re 0R1 or 0R05, its 6 to one, and half a dozen to the other.With unlimited time and budget, I would use a large, naked oil-immersed Zeranin shunt, with pumped circulation of the oil past the zeranin to reduce thermal resistance to an acceptable level. The whole show would be temperature controlled with the sensor attached to the zeranin.

author: amspire
author_rank: Super Contributor
timestamp: April 21, 2016, 12:48:22 am
content: It is pretty difficult to track down temp coefficient issues when the cost means you cannot keep making experimental shunts until you crack the problem.If it is possible to run test with a temp sensor moved to various locations - a few different places on the aluminium, the copper, the Peltier cooling. Then apply a load and plot the different temperatures versus drift, you may find a correlation between one of the temperature drift curves and the shunt drift. It can help narrowing down the problem.The other thing that could be useful is if it is possible to etch out a very thin extra Zeranin resistor - say along one edge of the current shunt resistor. You could then be monitoring the resistance changes in this thin resistor at the same time as load is applied to the main resistor. This would probably make it easier to see the Zeranin resistance changes with temperature, and it would also show of the effect is inherent in the bulk TCR of the Zeranin, or if the effect only shows in the track with the power applied - like a PCR effect.If the monitor resistance track does match the main shunt resistor for TCR, then this monitor resistance track could be used to generate compensation for the TCR in the main shunt.Richard

author: zlymex
author_rank: Frequent Contributor
timestamp: April 21, 2016, 01:19:59 am
content: Quote from: Zeranin on April 20, 2016, 11:57:01 pmHave you tried the blue silicone pad? although may be thick(I'm sure there will be thinner version), it conduct heat very efficiently and also with very high insulting resistance(I measured at >1E12 in anyway)http://www.aliexpress.com/item/FREE-SHIPPING-400-200-2-5mm-Blue-Thermal-Conductive-Silicone-Pad-mat-for-computer-and-laptop/651156177.htmlAnd because it is soft, any gaps will be filled by apply not very large force, also the joint will be seamless if smaller pieces to be used.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 03:13:43 am
content: Quote from: DiligentMinds.com on April 20, 2016, 05:29:05 pmYes, I too have wondered about the effect of the pressure provided by the clamping plate, but we need to think more carefully. The pressure exerted by the plate is at all times provided by the x8 of brass M4 screws, that pull the plate onto the constant-temperature aluminium heat-sinking plate, with the zeranin resistor assembly sandwiched between.The total cross sectional are of these screws is negligible compared to the area of zeranin, so these screw can never produce a large force, relative to the area of zeranin. You appear to be analysing the problem as if the zeranin, the aluminium substrate, and the copper plate, were sandwiched between two totally immovable plates, in which case temperature changes in the zeranin, alum-substrate or copper would indeed result in massive forces, but that is not the case at all. For all practical purposes, the zeranin/substrate/copper are unconstrained and free to expand and contract as they choose in the direction of their thickness – the clamping screws could as well be made of plasticine in this respect, because cross sectional area of the screws is so negligible compared to the area of the zeranin.As further, very convincing evidence of this, I can even loosen off these screws, and observe no measurable effect in the resistance of the zeranin shunt. Another enticing theory bites the dust.If we agree that the zeranin is effectively unconstrained in the Z-direction, and I insist this is the case, then at first glance the explanation for the 5ppm drift is easy. The zeranin is mechanically constrained in the X &Y directions, in the plane of the material, because it is mechanically bonded to an aluminium substrate that is much thicker than it is. Easy peasy. Zeranin expansion coefficient is +18ppm/K. The zeranin heats by 0.25K with respect to the substrate, and as a result, expands by 4.5ppm (18x0.25) in the Z direction, but is prevented from expanding in X and Y, and this will have the result of decreasing the resistance by 4.5ppm, just as observed.Unfortunately there is a serious problem with this explanation. What’s good for the goose is also good for the gander. We would expect the same behaviour when measuring the R-T curve, but not so in practice. When the zeranin temperature is changed in unison with the heatsinking aluminium plate, as per the R-T curve, there is essentially no change in resistance, yet there will indisputably still be the same expansion of the zeranin in the Z-direction, with any increase in temperature. The zeranin will still be constrained in the X & Y directions, though there will now be a small change in length in X and Y due to the different COEs of zeranin and aluminium. However, this won’t have any effect on the zeranin resistance, because an equal change in X and Y (a length and crosss section term) cancel, the increase in length cancelling the increase in width as far as resistance is concerned. Drats! You can try until the cows come home, but this explanation just won’t work. Sure, you can explain as above why the zeranin changes resistance when it self-heats above it’s substrate, but then you won’t be able to explain why the same thing doesn’t happen when both are heated in unison.Curiouser and curiouser said Alice. (a favourite quote of mine from Alice in Wonderland) There must be something else going on.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 10:02:30 am
content: Quote from: zlymex on April 21, 2016, 01:19:59 amI always regarded the rubbery thermal interface materials as inferior, and some of the older products of this type were almost useless, appealing to those that wanted a quick, easy, non-messy solution, whereas my own philosophy is to use the best, meaning lowest thermal resistance available, period. That said, I'm amazed by the high thermal conductivity of the materials used in some of these higher performance rubbery sheets, even if the large thickness offends me. The product you reference has a thermal conductivity of 1.5 W/mK, whereas the plastic film that I used to mount the zeranin shunt is a mere 0.19 W/mK, but MUCH thinner. Even so, when I do the calculations, I find that a 0.5mm thick film of your material would perform as well as the 0.06mm PVC film that I used. I really should make an effort to use something better. I was not able to get hold of a sufficiently large sheet of Kapton MT, that would gain me a x3 advantage, but there are rubbery sheet materials out there with thermal conductivities as high as 3.0 W/mK, and some available in thinner than 0.5mm, so I could do maybe x3 better than now by shopping around for one of the thinner, very-high-conductivity rubbery sheet materials. Thank you for alerting me. Such materials might have creep problems in this application, but never know without trying.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 10:47:20 am
content: Quote from: DiligentMinds.com on April 21, 2016, 04:49:59 amWhen the shunt self-heats, there is a temperature difference between the substrate and the element. When it is heated in an oven, there is no temperature difference.I agree with this 100%, and from a practical point of view, this tells us what options are available for reducing the problem. However, It is quite a stretch to sayMystery solved.Beyond 'handwaving', no one has offered a detailed explanation of what is going on. No one has explained the nature of the stresses and strains that result in the 5ppm drift that I observe, or why the R-T curve does not apply, and so on. I deliberately posted an analysis of the stresses and strains that we should expect, pointing out that the stresses and strains that we predict and expect specifically donotexplain what is happening, which specifically makes this a case of Mysterynotsolved, doesn't it?The best way to fix it is to reduce the root cause of the problem, only then should one consider adding 'corrections', at least that's my philosophy.I am most interested in those CSNG shunts that you mention. They are relatively recent, and I was not aware of them. The PCR claims to be 4 ppm/W, which is only 20 times worse than the 0.2 ppm/W (5ppm/25W) that I achieve, and certainly the best off-the-shelf shunt that I have seen, and perhaps the best off-the-shelf shunts on Planet Earth. Apparently you can get (on special order, at unknown price) x6 of them on a single substrate, so it would be possible to match what I have now with x3 or x4 of those. There are complications paralleling multiple 4-terminal resistors, basically you need a separate differential amplifier for each, but that is quite practical if you only need 3 or 4. However, to do significantly better than I presently achieve by using those VPG shunts instead would be very messy and expensive, and probably not practical.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 21, 2016, 11:10:44 am
content: I still think that your amplifier noise can be improved and the shunt voltage (and dissipation) can be lowered, reducing all power-related effects. The self-noise of the shunt is very low compared to 40 Ohm or so of the amplifier equivalent noise. Thermoelectric effects may be easier to deal with.CheersAlex

author: zlymex
author_rank: Frequent Contributor
timestamp: April 21, 2016, 11:40:51 am
content: I have cut open one of the CSNG shunts six chip version, 0R0500, looks good inside.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 12:34:06 pm
content: Quote from: zlymex on April 21, 2016, 11:40:51 amI am absolutely amazed! Did you just happen to have this CSNG resistor sitting on your bench? How many more have you got, and how and why do you come to have them? Assuming you bought it, do you mind me asking how much it cost?

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 12:50:04 pm
content: Quote from: zlymex on April 21, 2016, 11:40:51 amThe Kelvin sensing appears to be incredibly primitive. It looks as if the sense connection is made to the high-current conductor, just beyond where the wires enter the package. The sense connections should be at the resistive foil, not several inches back along the tinned-copper high-current connecting wire. What am I missing?

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 01:01:19 pm
content: Quote from: Alex Nikitin on April 21, 2016, 11:10:44 amWe appear to be at cross purposes. The Johnson self noise of the 0R1 shunt itself is so low as to be neglected. If we lower the shunt resistance, then the shunt noise is of course even less, but that's irrelevant. It doesn't matter how you cut it though, if you lower the shunt resistance, then you get less signal, and this degrades SNR at the shunt preamp output. Reducing the shunt to 0R05 (which I have) wouldn't degrade the noise performance dramatically, and may provide a better all round balance.

author: zlymex
author_rank: Frequent Contributor
timestamp: April 21, 2016, 01:19:30 pm
content: Quote from: Zeranin on April 21, 2016, 12:34:06 pmI have another CSNG shunt(also six-chips) together with 20+ pieces of similar shunts(in the same datasheet as CSNG): VCS331Z, VCS332Z and VFP4.I cut it open about 2 years ago, someone sent it to me for free just let me cut open because it was new, strange and we have not seen inside photo of it before. I cannot remember the exact price of my other CSNG because I have bought many other shunts in these several years time, price tag roughly around $5 to $80 each

author: zlymex
author_rank: Frequent Contributor
timestamp: April 21, 2016, 01:35:07 pm
content: Quote from: Zeranin on April 21, 2016, 12:50:04 pmOne current conductor is on top side of all the chips, covering with thick tin. partially can be seen.The other current conductor is similar, go thru all the chips at bottom side, torn from the chips.Kevin sensing are two thin wires(one can be seen) soldered to the middle of the two current conductors.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: April 21, 2016, 02:16:03 pm
content: Quote from: Zeranin on April 21, 2016, 01:01:19 pmWhat I've meant that your amplifier appears relatively noisy (~80nV RMS noise in 10kHz BW, or 0.8nV/rtHz, equivalent of ~40 Ohm resistor noise at 300K) . If you use multiple amplifiers in parallel to achieve a lower noise, this figure looks excessive.CheersAlex

author: Kleinstein
author_rank: Super Contributor
timestamp: April 21, 2016, 03:00:37 pm
content: The Kelvin sensing looks a little primitive, but it is not that bad as one might think at first. The copper / solder part is mainly responsible for spreading the current over the parallel connected resistor elements. As the length of the copper path is the same for all elements the current is evenly divided in the resistors, even if the copper part changes resistance. Then there is not much tin /copper in the path sensed by the voltage sense connectors. I agree it could have been done better, but as long as the parts stays stable this should not be such a bad thing. The copper/tin part might add a little to the TC, but not much.The main part I would be afraid of would be aging in the tin part, as this is a relatively low meting point alloy with normally a fine structure, it can change even at room temperature or slightly above. As it's potted there should be no tin whiskers growing.As for the amplifier a value of 0.8 nV / Sqrt(Hz) is not that bad, but a parallel connection of 4 amps could give you about halve the noise and thus allow for the reduced shunt size. It's a trade of in spending money for the shunt or the amplifiers.

author: d-smes
author_rank: Regular Contributor
timestamp: April 21, 2016, 08:19:54 pm
content: Quote from: Zeranin on April 21, 2016, 03:13:43 amZeranin- Thanks for bringing your problem to the forum and for your detailed descriptions and responses. Everyone loves a mystery!I would like to question your assumption that the Zeranin is constrained in the X and Y axis. As-built, the Aluminum X & Y are locked to the Zeranin X & Y through the heat-bonding, electrically insulating film. But that film has mechanical compliance and its own TCE (which is safe to ignore). The mechanical compliance can be thought of as an array of stiff springs connecting each Al X-Y coordinate to the same Zeranin X-Y coordinate at room temperature (as-built). Now when heated uniformly in an oven (no current), the Al and Zeranin want to grow to two different sizes because of the differing TCEs but they can't because they are constrained by all these stiff springs. So, does the Al get compressed to match the Zeranin hot dimensions or does the Zeranin get stretched to match the Al hot dimensions? My guess is that the Al, even though it's softer, is thicker and ultimately has the higher stiffness. So the Zeranin gets stretched when the assembly is heated. But the springs of the insulating film also get stretched such that the hot X-Y coordinates no longer match; the Zeranin is stretched to slightly smaller X-Y dimensions than the Al. In this regard, the rubbery thermal interface may have and advantage because it has more "give" (weaker springs).Also recognize that in the dimension of the zig-zags of the Zeranin, there are gaps which will take up a lot of the differential TCE in that dimension. However, these put additional strain on the corners where the zig bends to become a zag. But I agree, that within each 25mm zig and zag the Zeranin is still getting stretched in both dimensions.Now apply current. The Zeranin gets 0.25K warmer and would have slightly larger dimensions, if it weren't constrained, by an amount I'll call dx and dy. In this case, the Al X-Y coordinates stay the same (assumed constant temperature) but what will the Zeranin's X&Y's become? I assume the spring constant of the insulating film remains constant and stretches the Zeranin by the same amount. It follows then that the differential heated dimensional difference of dx and dy shows up as Zeranin's X&Y coordinates growing by dx and dy (in reality, slightly less than dx & dy due to slightly less spring force).I believe there is also a slight temperature difference between the cooled side/face of the Zeranin and the top clamp side of the Zeranin. This means the clamped face area expands slightly more than the cooled face causing the Zeranin to want to curl or cup (convex on the warm side, concave on the cool side). Given the thinness and good thermal conductivity of the Zeranin, this is probably a second-order effect at best. But it also results in different stress/strain and dimensional changes that may explain the powered vs un-powered difference in R-T.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 09:42:06 pm
content: Quote from: zlymex on April 21, 2016, 01:35:07 pmI feel much happier now we can clearly see that the sensing has been done properly. I could not believe that it was done as badly as first appeared. Thanks so much for providing these pictures.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 10:23:30 pm
content: Quote from: DiligentMinds.com on April 21, 2016, 05:53:07 pmAnd I maintain that you are 'hand-waving', without understanding what is actually going on.It's time for me to stick my neck out. I have always claimed to know what was going on, and that I would get around to explaining my ideas. What I have been doing so far is shamelessly using the excellent brains here to see what ideas others would come up with, which helps crystallize my own thoughts. I now feel sufficiently confident to present what I believe is happening, in full detail. Any detailed explanation will need to correctly predict the observeddirectionof resistance shift, as well as get the magnitude approximately right. I welcome critical scrutiny, that is one of the great benefits of presenting ideas on a forum.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 10:39:06 pm
content: Quote from: d-smes on April 21, 2016, 08:19:54 pmI also love a technical/engineering mystery, and glad others do too. Personally I need to understand everything I observe and measure, otherwise I would go mad, even if it means considerable head-bashing and lost sleep until I see the light.Thank you for thinking in detail about the stresses and strains that we would expect. I take your point that the thin bonding film will be very slightly compliant, so that the zeranin does not exactly follow the movement of the aluminium substrate. It remains my belief though, that for all practical purposes, the two are completely bonded together. The full explanation of what I believe is happening is rather long and subtle, and so far I have been flat out reading and responding to the many replies on this topic. Time has come for me to give my explanation, which does completely explain all observations, and I hope you will be convinced by it. Cheers.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 21, 2016, 11:56:15 pm
content: I have started a sister thread entitledResistivity vs Temperature – flatter is better?,as a precursor to explaining what I believe is going on to produce the 5ppm drift that I observe immediately after the current through my Zeranin 30 shunt is switched from zero to 16A, and why the R-T curve does not apply.If and when we have agreement on the sister thread, I'll resume my discussion here.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 22, 2016, 04:48:02 am
content: Quote from: DiligentMinds.com on April 21, 2016, 05:53:07 pmI agree with all of that. I am not in a position to make major changes at this stage, though it would make sense for me to make an effort to reduce thermal resistance from zeranin to constant-temperature heatsink. I still have time and budget to at least have a go at that, and conceivably could gain a factor of x2 improvemnt. I hear what you are saying re compensation. Can't do it in software, as there is no software in this current driver. As the effect is non-linear because the heating goes as I^2, hardware compensation would be ugly and messy, and I have no intention of re-designig and re-building the current driver PCB.Your suggested shunt design would perform superbly. To properly eliminate PCR effects, we agree that you really need to go naked.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 23, 2016, 01:10:21 am
content: 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 temperaturedifferencebetween 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 temperaturefallsby 0.25K, realizing that the result will be the same as if the zeranin temperature was toriseby 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 issome 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 elasticallydecreases, 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 anincreasein 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 adecreasein 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 resistancedoeschange 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)wheredR is the decrease in resistance, in ppm, as a result of self-heatingP is the power dissipation in Watts, self-heating the zeraninRth is the thermal resistance from zeranin to substrate, in K/WB is a constant related to Poisson’s Ratio, ~1.0, but <2.0EC is the thermal expansion coefficient of the substrate, in ppm/KFor my zeranin shunt example :-dR = 25 x 0.01 x 1.0 x 22 = 5.5ppmOne 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.

author: zlymex
author_rank: Frequent Contributor
timestamp: April 23, 2016, 02:10:40 am
content: I'm wondering how your ‘naked’ R-T curve is measured.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 23, 2016, 02:44:29 am
content: Quote from: zlymex on April 23, 2016, 02:10:40 amI have not personally measured the 'naked' R-T curve for the zeranin, firstly because it came to me already bonded to an aluminium substrate, and secondly because the only R-T curve that matters in this case is the curve measured with the zeranin already bonded to the substrate, as that is how it will be used. When I speak of my measured R-T curve, I mean the curve with the zeranin already bonded to the substrate, and mounted in the current driver, clamped against the constant-temperature heatsinking aluminium plate. I explained how my R-T curve was measured, in my posting #13, the relevant part copied below :-Let me explain how my R-T curve was measured. As explained, the zeranin sheet is pre-bonded to a 1.6mm thick aluminium substrate and clamped/bolted down to a Peltier-temperature-controlled 12mm thick heatsinking aluminium plate. This zeranin shunt is the current measuring element in a precision current driver circuit, that for measuring the R-T curve is set to produce a steady 16A. In series with the same current is my massive (100A rated) Leeds&Northrup (L+N) naked master-reference-shunt, temperature controlled to 0.02K. If the resistance of the zeranin shunt in the current driver should change, then the controlled current changes, and this is detected via the voltage across the L+N master shunt. The zeranin shunt is clamped down onto the temperature-controlled aluminium plate with a 12mm thick plate of copper, and a thermistor is embedded deep within this copper plate, essentially touching the zeranin sheet, so the thermistor therefore accurately measures the zeranin temperature, say within 0.1K. The heat flux flows from the zeranin to the temperature controlled aluminium plate, with a thermal resistance of around 0.01K/W, so at full dissipation of 25.6W, the zeranin temperature rises by ~0.25K with respect to the 12mm thick aluminium plate to which it is clamped, not much at all, and completely negligible on the R-T curve. Measuring the R-T curve is real easy. The heatsink temperature (and therefore the Zeranin temperature) is ‘dialled up’ on the Peltier temperature controller setpoint to values between 20 and 50 DegC. At each zeranin temperature, the voltage across the L+N master shunt is recorded, being an accurate measure of the zeranin resistance.The 'naked' R-T curve can be found on the website of the zeranin manufacturer, Isabellenhutte. In this case, the resistance of a naked sample of zeranin is accurately measured as a function of the zeranin temperature, probably with the zeranin sample in a variable-temperature controlled oven.

author: Zeranin
author_rank: Regular Contributor
timestamp: April 28, 2016, 02:41:17 am
content: All of the discussions here predict that my PCR problems scale with the rise in temperature of the zeranin foil, and that therefore I should be able to reduce the problem by reducing the thermal resistance from foil to constant-temperature heatsinking plate.To this end, I have ordered an 18” x 18” sheet of 0.003” thick Tpcm 580 Series Phase Change Material, part number Tcpm583, manufactured by Laird Technologies, priced at US $49.81 from Digikey, the highest performance thermal interface material that I have been able to find.I have started a thread about on the Technical Stuff forum for a general discussion about the best thermal interface materials available. That thread has the calculation showing that the temperature rise of the zeranin should be reduced from ~0.25K down to 0.037K, a x6 improvement.I will report back on this thread as to what improvement this gives to my 5ppm drift in shunt resistance, after the current is switched from zero to 16A. I’m not expecting a x6 improvement because there are other imperfections such as the constant temperature plate is not atexactlyconstant temperature under transient conditions, and there is still a thermal ‘contact resistance’ on each side of the thermal interface material, but I do expect a measurable improvement in my PCR induced drift of 5ppm. The result will be interesting.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 03, 2016, 01:39:55 am
content: There is a wise old saying, that if something appears too good to be true, then it probably is. As it turns out, the Laird Tcpm-583 thermal interface product that I ordered is utterly useless, being merely an expensive and inferior alternative to thermal grease.Despite being an electrical insulator, this material (as it turns out) is apparently not intended to provide electrical isolation, though nowhere do Laird actually say this. The ‘film’ is supplied between 2 transparent protective plastic sheets. To use the material, you peel off the top plastic sheet, then press the sticky surface of the grey ‘thermal interface material’ onto one surface. Then you peel off the other protective plastic sheet, and finally, press your second surface down onto the tacky grey material. The interface ‘film’ has no mechanical strength or integrity whatsoever. I tried a 25mm square test sample of the material, clamped between 2 metal surfaces and heated to 65 DegC, and just as predicted, a few 10’s of volts was enough to break down the film, with the 2 metal surfaces thereafter shorted together. Oh, and once assembled, you can’t get the surfaces apart again, another great ‘feature’ of the product.Apparently, the purpose of this material is simply to provide good thermal contact between two metal surfaces, but provide no electrical isolation. That being so, it is difficult to comprehend why they make the material electrically insulating in the first place, as better thermal conductivity can be obtained by using thermal grease loaded with microfine copper or silver particles. Furthermore, a metal-loaded grease will provide a thinner interface, as the excess grease is squeezed out under pressure, while with the Laird product you are lumbered with the thickness of the thermal interface material, 0.003” in this case, and much thicker again for most of their Tcpm-580 products. No thanks.In summary, what we have here is an expensive and inferior alternative to a high-quality, metal-loaded thermal grease. I am extremely unimpressed with both the product and the company. As described in detail in a previous posting, the published thermal resistance specifications are nonsense. I sent two emails to Laird, to 2 different enquiry email addresses, seeking clarification, and still have received no reply to either. Their documentation, product, and customer service are all complete rubbish as far as I am concerned, and I won’t be buying anything from Laird again. I am seriously unimpressed.My conclusion from all of this, is that the only advantage of the modern thermal interface materials is ease of use. Sure, some of the modern insulating interface materials have quite good thermal conductivity (W/mK), but in order to get useable mechanical integrity and electrical isolation, the thickness is so great that the resultant thermal resistance is at best as good, and generally inferior, to the old-fashioned approach of using a thin film of Kapton or similar, with thermal grease. My original plan was to use the thermally conductive grade of Kapton, known as ‘Kapton MT’, and it looks like that would still be the best solution. The only reason that I didn’t do it that way in the first place was because I could not obtain a sample sheet of Kapton MT bigger than A4 size, which is too small.Maybe I need to spend $500 or so on to get the minium order quantity of 37 mictron Kapton MT. At least I would have plenty left over for the next job.