Thermal EMF for soldered components

[justanothername]

Hi Everybody!
I'm recently investigating measuring very low currents with even lower shunts. Please don't aks my why i do this, it is purely "academical". So far I'm sucessful measuring 10µA across a 180mOhm shunt with a zero drift amplifier.
Lately I was conducting some tests with a thermal chamber and I was noticing a degradation in performance when cooling down, nothing unexpected, since i completely forgot the EMF matching resistor in my amplifier, shame on me.
The new PCB boards are on the way. Meanwhile:
The connections to the operational amplifier, I'll be able to properly compensate thermal EMF-wise.
But what about the Shunt itself?
Look at this question here:
https://electronics.stackexchange.com/questions/348382/how-to-calculate-estimate-thermocouple-junction-of-a-solder-junction-between-a-p
I think it is not properly answered. Lets say that resistor is my shunt and there is a thermal gradient from PCB towards the upper surface of the resistor. How is the thermal EMF voltage actually calculated? Does it superpose on the shunt voltage? And if yes: how can I mitigate this effect?
Lets say that the shunt is pulsed by high currents and gets heated up shortly. How can I mitigate a thermal EMF that propably follows up (besides of changing to a shunt with bigger thermal mass)?
Thanks for reading.
M.

Edit: actually, thinking about the upmost answer to the linked topic, what is actually meant is that if the copper traces and the pads are strictly symmetrical then the PCB can get as hot as it wants in respect to the Shunt surface, the thermal voltages WILL cancel out completeley, as long as there is no thermal gradient ACROSS the resistor. So symmetry is the key!

[dietert1]

Yes symmetry helps, but as you know in engineering this kind of compensation is always limited to 1 % or so. So it's both: symmetry and low temperature (for low temperature gradients). If it's academic, you could design a board with options for an SMD shunt and through hole shunt to see which one performs better. Or you could glue the SMD part to the board for a test. Depending on the type of SMD part, even mounting it upside down can help to reduce temperature gradients. Eager to see your results!

Regards, Dieter

[justanothername]

So this is a current to frequency converter. The troubling stage is the current controlled current source that loads a capacitor. The following stage is a comparator with treshold discharge and this works fine, I evaluated it separetely.
With the shunt amplifier stage, I totally missed out to place a match resistor. It's a structure similar to this, where the match resistor would be R1:
https://www.analog.com/en/design-notes/dn-precision-ultralow-power-high-side-current-sense.html
Attached is the performance degradation of the circuit in a thermal chamber. Its a zoom in the lower current regions. There must be a thermal gradient across RIN, maybe because of inhomogenous thermal coupling in the thermal chamber.
Nothing I can fix with existing PCBs where I cut traces with a knife and solder a resistor in place. So I'm waiting for a new set of PCBs.

So am I right to assume when there is a thermal gradient over the width or the height of resistor soldered to a PCB, there is no thermal voltage seen. Only when there is a gradient over the length of that resistor?

Edit: just a thought experiment: I can choose chemical tinning as surface finish for the PCB. The resistor terminals (example: shunt resistor - http://www.farnell.com/datasheets/2339607.pdf) and the ICs terminals are tinned, and there is pure tin solder wire available:
https://www.alibaba.com/product-detail/pure-tin-wire-sn99-95-lead_60674746728.html
would this suffice to eliminate all solder joint related thermal voltages?

[dietert1]

In the AN you linked in the paragraph "Zero Point" they explain that the zero point inaccuracies start below 50 uA. Your shunt is 1.8 times more so in your case the limit would be about 30 uV. Your measurements seem to confirm that.
One reason for the observed temperature dependence may be the temperature drift of the zero drift amplifier: 20 nV/K times 40 °C = 8 uV and 8 uV / 0.18 Ohm = 44 uV again. As the spec is an upper limit, the real effect is a fraction of that.
Unless you study thermal EMF of the shunt separately, it's difficult to tell how much it contributes.

Regards, Dieter

[justanothername]

One reason for the observed temperature dependence may be the temperature drift of the zero drift amplifier: 20 nV/K times 40 °C = 8 uV and 8 uV / 0.18 Ohm = 44 uV again. As the spec is an upper limit, the real effect is a fraction of that.
Unless you study thermal EMF of the shunt separately, it's difficult to tell how much it contributes.
Regards, Dieter

I was hoping it is the shunt and the layout. Not that I have any possibility to measure that though. If its the amplifier, then its game over
The datasheet says most of the units are in the 5nV range. Maybe there is a way to preselect them?

[justanothername]

20 nV/K times 40 °C = 8 uV

Shouldn't that be 0.8µV?

[dietert1]

Yes, so that would be the equivalent of 4.4 uA. From your data the effect of a 20 °C change seems to cause a shift of about 20 uA and at most 10 % of that can be offset drift.
Those theories about using similar metals and the like need confirmation by experiment. Web research gave me this: mouser.cn/pdfdocs/PU_whitepaper_20170126.pdf . So the problem is known and there seem to be large differences. What kind of shunt did you select?

Regards, Dieter

PS: Vishay has this: https://www.vishay.com/docs/30175/thermal.pdf and their measurement indicates 3 uV of thermal EMF with an estimated 8 °C temperature difference, probably less than you saw. Mouser stocks the 180 mO WSL2512.

[justanothername]

This was the shunt I was using before:
https://www.farnell.com/datasheets/2838152.pdf

now I ordered some new ones, 200m instead of 180m:
http://www.farnell.com/datasheets/2339607.pdf
https://www.yageo.com/upload/media/product/productsearch/datasheet/rchip/PYu-PE_521_RoHS_L_8.pdf

In the end, the shunt should be as small as possible, 0603 or 0402 even. But for experimenting I'm OK with any standard package. For now, I'm positive that the new layout will vastly improve thermal EMF performance. I made everything prettily symmetric.
This aside, it would be interesting if there is a theoretical possibility to rule out all thermal EMF caused by solder joints on a PCB board if you use only tinned surfaces and solder wire. I may prepare a test PCB or similar if anyone gives me an idea how to measure this. My best Multimeter (in working condition) has 6.5 digits only.

[David Hess]

Low thermal EMF solder will help but the current shunt itself has junctions with dissimilar conductors which will dominate.  Besides laying out the circuit symmetrically to cancel thermal EMFs, keeping the self heating temperature low by using a higher power current shunt helps.

[dietert1]

Generally speaking, for soldered connections with good thermal contact the surface treatment (gold tin or so) makes a small change. More important to solder a copper wire to a copper trace/via. The thermal gradient inside the solder and the surface treatment is small.
Watch this video to see the differences between shunts of different make. The setup looks pretty symmetric. Is he cheating by using different solder?


Regards, Dieter

[justanothername]

Is he cheating by using different solder?

Thats really what I hoped that will NOT happen. The resistor in his setup is uniformely heated from above, the thermal voltages should cancel out. In the other hand, the setup looks quite simple, does anyone here have a nullvoltmeter like this?

[Anders Petersson]

Shouldn't thermals stay more uniform if the components were covered in thermal paste? That doesn't seem to be common practice so what am I missing?

[KT88]

Thermal paste wouldn't most likely help. It is still a bad heat conductor - it just has a lower Rth than air. In thin layers it improves heat transport by displacing air between the heat source- and sink.

The issue is the difference in temperature between the terminals - a thermal gradient is only causing the difference. David Hess describes the viable options to reduce the temperature difference between the terminals.
There is another effect that is unknown to most engineers: Slight differences in the alloy composition of the solder (it dissolves tiny amouns of copper and terminal plating) causes the peltier effect which actvely creates a temperature difference. This can be tens of degrees centigrate for highly loaded shunts.

Keeping the the shunt resistor cool (referred to ambient) and providing a good thermal coupling between the terminals is key.
Also the ambient temperature should stay as uniform as possible - yes, no large gradients...

Cheers

Andreas

[dietert1]

In the context of "Thermal EMF for soldered components" we should not only think about current measurement shunts, but more general about resistors in precision circuitry, for example those LTZ1000 boards. The gurus spent some time to determine the influence of resistor tolerance and TC, but i don't remember a discussion about selecting parts with low thermal EMF. A board with a hot reference does have temperature gradients. There was the idea to place resistors orthogonal to gradient and of using slots to confine the gradients, but a comparison of different parts/makes in a systematic fashion is missing.

Regards, Dieter

[2N3055]

Is he cheating by using different solder?

Thats really what I hoped that will NOT happen. The resistor in his setup is uniformely heated from above, the thermal voltages should cancel out. In the other hand, the setup looks quite simple, does anyone here have a nullvoltmeter like this?

There is nothing uniform in that experiment. For a first resistor, light is shining to the side...  Why didn't he get a cup of warm mineral oil and dunked that resistor in... There wouldn't be voltage then.
If he also moved the lamp around, there would be place where voltage drops to zero...

Bulk metal foil resistors have better thermal conductivity, so they will be less sensitive to small mismatches.
Since there is no perfect design, that helps, but in truly isothermal situations there is no Seebeck voltage.

Also, Seebeck voltage doesn't happen because you connected two different materials, i.e. in a connection.
It happens because same temp gradient creates different voltages in two sides from dissimilar metals, so they don't cancel out. What you see is uncanceled difference.

[KT88]

Claiming that the bulk metal resistors show no thermal EMF is utter BS!

[dietert1]

There are differences. For example with THT resistors i usually try to get them with copper wires if thermal EMF matters. Saturday i tested one of those and got a 1.7 uV effect when "heating" one end with my fingers. That was a Dale RS-02B wirewound resistor. As far as i understand that is a good result, less than the 3 uV they claim in one of the Vishay videos.

But again, one would need a setup that can reproduce temperature gradients with parts of various shapes. Maybe with a heater resistor on one end of a PCB and two temperature sensors near both ends of the DUT.

Regards, Dieter

[justanothername]

Saturday i tested one of those and got a 1.7 uV effect when "heating" one end with my fingers. That was a Dale RS-02B wirewound resistor. As far as i understand that is a good result, less than the 3 uV they claim in one of the Vishay videos.

There is nothing uniform in that experiment. For a first resistor, light is shining to the side...

How do you distinguish the thermal EMF caused by the structure inside the shunt from the thermal EMF caused by the metal joints/junctions between resistor terminals and the conductor? On the other hand, the guy in that video presents his resistor with ZERO thermal emf with the same (lets assume) assymetrical setup (regarding the heating gradient). If hes not faking, he must have a recipe for solder connections with zero thermal emf both for resistor to PCB trace and for PCB trace to test wire. Is it pure tin? Tinned wire, tinned resistor terminals, tinned PCB traces? Could it be that simple?

Edit: Just another thought. I am now wondering why all these reference designs around the LTZ1000 do use gold plated PCBs. Copper to gold has 0.5µV/°C, but gold to most available solder alloys has 3-5µV/°C. This would mean that you really need to take care about the positioning of every pad of the critical components. Why don't they just use chemical tinned PCBs as they are widely available even from pool manufacturers?

[KT88]

The BS part of this video is the fact that the Tc of the resistance and the thermal EMF are not at all correlated.
I even doubt that the measurement was real.
First, they don't show the whole setup during the measurement.
Second, specifically whatching the thick film example -the turn on the light and a second later the needle starts all over sudden to move. If one would think that the time is not well alligned one would see some change in brightness at the scale.
Third, the needle moves quite fast and stops fast - I would expect a more exponential decay of the movement of the needle.

[2N3055]

Saturday i tested one of those and got a 1.7 uV effect when "heating" one end with my fingers. That was a Dale RS-02B wirewound resistor. As far as i understand that is a good result, less than the 3 uV they claim in one of the Vishay videos.

There is nothing uniform in that experiment. For a first resistor, light is shining to the side...

How do you distinguish the thermal EMF caused by the structure inside the shunt from the thermal EMF caused by the metal joints/junctions between resistor terminals and the conductor? On the other hand, the guy in that video presents his resistor with ZERO thermal emf with the same (lets assume) assymetrical setup (regarding the heating gradient). If hes not faking, he must have a recipe for solder connections with zero thermal emf both for resistor to PCB trace and for PCB trace to test wire. Is it pure tin? Tinned wire, tinned resistor terminals, tinned PCB traces? Could it be that simple?

Edit: Just another thought. I am now wondering why all these reference designs around the LTZ1000 do use gold plated PCBs. Copper to gold has 0.5µV/°C, but gold to most available solder alloys has 3-5µV/°C. This would mean that you really need to take care about the positioning of every pad of the critical components. Why don't they just use chemical tinned PCBs as they are widely available even from pool manufacturers?

Look at the video again, you can clearly see bright spot is not in the middle. Halogen light is a spot source and it definitely didn't heat uniformly, INCLUDING connection wires going to microvoltmeter. Light is not in the same spot for other two resistors..
That is why I said that he had to have immersed them in heated Galden or mineral oil and then see what happens.. I bet you results wouldn't show much difference.

Second thing, Seebeck voltage DOESN'T appear across different metal connections. It appears across thermal gradients in material, and if you connect two dissimilar materials, two sides won't be the same and won't null differentially. 
Difference will be in both Seebeck voltage coefficient for the metal, and it thermal resistance, making temperature gradients different. In sensitive circuits soldered joint might make more difference if pads are different surface and have different amount of solder than what solder you used. Any imbalance will show, dissimilar materials only make thing more visible.

So if you have copper trace, solder pad with solder to one side of resistor, resistor, other side soldered  to solder pad going to copper trace, make it all symmetric, make all of that on aluminum PCB, and than, you put that PCB on heated plate larger than PCB and measure that... I bet you you won't see much of thermoelectric voltage, regardless of what resistor you solder in.

Full metal foil resistors, with metallic connection to pads will conduct temperature better than small ceramic resistors made from ceramic NOT designed for thermal conductivity. So any thermal differences will be more visible. So resistor that conducts temperature better will be less sensitive to Seebeck effect. And that applies to ANY resistor made to be current shunt, they all are designed for that. He compared two unspecified resistors with expensive dedicated metal foil shunt resistor. I guess it would be better comparison to compare to other dedicated shunts....

It is shady marketing video. Which is shame, really, because those metal foil Vishay resistors are top quality stuff. No need for misleading marketing.

Best regards,

[David Hess]

I am now wondering why all these reference designs around the LTZ1000 do use gold plated PCBs. Copper to gold has 0.5µV/°C, but gold to most available solder alloys has 3-5µV/°C. This would mean that you really need to take care about the positioning of every pad of the critical components. Why don't they just use chemical tinned PCBs as they are widely available even from pool manufacturers?

The gold plating is suppose to be thin enough to completely dissolve into the tin base solder.

As 2N3055 pointed out, the thermal EMF is not generated at the junction anyway; the junction itself has zero voltage across it because there is no temperature across it.  The thermal EMF is generated along the temperature difference of the wires and if the wires are not the same material, then there is a difference in voltage.

[justanothername]

So if you have copper trace, solder pad with solder to one side of resistor, resistor, other side soldered  to solder pad going to copper trace, make it all symmetric, make all of that on aluminum PCB, and than, you put that PCB on heated plate larger than PCB and measure that... I bet you you won't see much of thermoelectric voltage, regardless of what resistor you solder in.

Thank you for the explanation above, it gets clearer. It is not a subject that I'm intuitively familiar yet on board level. I understand that the thermal gradient across a conductor is the cause, and will only be seen when the conductor is terminated to different metals. For instance in thermocouple sensors, where the cold junction often is meters away

As 2N3055 pointed out, the thermal EMF is not generated at the junction anyway; the junction itself has zero voltage across it because there is no temperature across it.  The thermal EMF is generated along the temperature difference of the wires and if the wires are not the same material, then there is a difference in voltage.

I will not give up yet.. Let me just wildly fabricate a hypothetical scenario. Just one last time There may be is a hot zone (do you call this isothermal zone?) with a trace coming out to a cold zone with a connector. So there is:
Resistor terminal (or IC pin) [tinned] - solder [tin alloy] - solder pad [gold (or copper if the gold is dissolved)] ---trace with gradient--- solder pad [gold (or copper if the gold is dissolved)] - solder [tin alloy] - connector[copper]
Could you find such a situation in practice, maybe in precision reference designs? Is there a thermal EMF related problem to be expected? If yes, how to mitigate? Could PCB tinning be a solution?

[Kleinstein]

One has to look at the full closed voltage path. So there can not be just one section with a temperature gradient - the temperature must come back.

Tinned copper traces are more like a bad idea, at least the HASL type, as the thickness of the tin can vary and one has mixture of copper trace and a variable thickness tin layer in parallel. So better have just copper traces with solder mask for corrosion protection. Usually a temperature difference along a pure copper trace is usually no problem. The thing to avoid is having the temperature gradiens at connectors or solder joints. Relatively thin traces can help to carry not so much heat to the connectors.

The gold plating is usually very thin. Chances are is dissolves quick in the tin - so similar amounts of tin may be a good idea. Usually there is little thermal gradient in the solder joints, unless there are stong heat sources around.

Connectors can be a nasty source of thermal EMF.  Pure copper is too soft for many uses and thus different materials are used. Cables can also carry heat over longer distance.

Current shunts for higher currents usually use low therma EMF (relativ to copper) manganin. So even with a high current there is not that much heat from the peltier effect. The main source is usually the resistive heat and asymmetriy in the thermal conditions around the shunt.

Most voltage reference ciruits don't care much about thermal EMF as this is still a relative small effect - the only exception are heated refrences like LTZ1000 because of the large gradients near the chip and kovar leads that have quite high thermal EMF if paired with copper.

For the circuit with the AZ amplifier there can be an effect of humidity and board stress on the offset. It is not only the temperature.

[JohnG]

If you want isothermal performance in the vicinity of your shunt, there are some things you can do with PCBs to help this.

Use a multilayer PCB and have a copper plane or planes to spread heat. Use a lot of vias to tie the planes together, and use 2 oz copper at least. Keep the seperation between the component layer and the next layer down as thin as possible (3-4 mils is commonly available). You can use a 4 layer PCB to get this if you need. Use as much copper around the components as you can so that the thermal resistance to the first plane layer is low. While PCB material has poor thermal conductivity, you can make up for this because the area of contact can be made large.

If the above is not good enough, consider using an IMS substrate (metal substrate PCB), like this: https://www.pcbway.com/pcb_prototype/General_introduction_of_Aluminum_PCB.html. PCBway even has it in their fast quote option. If you need better than that, there are more options, for example: https://dm.henkel-dam.com/is/content/henkel/Bergquist%20Comprensive%20Selection%20Guide%20-%20Thermal%20Cladpdf.

Hope this is useful.

Cheers,
John

[dietert1]

The pcbway link includes a comment with this drawing:



Interesting how people try to improve thermal conduction in a metal PCB using a pillar. Maybe one can learn something from high power LED designs.

Regards, Dieter

PS: When looking at the Henkel Bergquist PDF, they call it "pedestal" on p. 13. In the diagram on p. 15 they compare thermal conductivity of different materials. It shows how much better copper works than anything else, including alumina (ceramics).