Thermal EMF for soldered components - Page 1
Forum section: electronics / general

author: justanothername
author_rank: Regular Contributor
timestamp: May 22, 2021, 04:19:23 pm
content: 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-pI 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!

author: dietert1
author_rank: Super Contributor
timestamp: May 22, 2021, 06:00:51 pm
content: 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

author: justanothername
author_rank: Regular Contributor
timestamp: May 22, 2021, 06:30:26 pm
content: 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.htmlAttached is the performance degradation of the circuit in a thermal chamber. Its a zoom in the lower current regions. There must be a thermal gradientacrossRIN, 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 thewidthor theheightof resistor soldered to a PCB, there is no thermal voltage seen. Only when there is a gradient over thelengthof 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.htmlwould this suffice to eliminate all solder joint related thermal voltages?

author: dietert1
author_rank: Super Contributor
timestamp: May 22, 2021, 08:05:18 pm
content: 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

author: justanothername
author_rank: Regular Contributor
timestamp: May 22, 2021, 08:24:14 pm
content: Quote from: dietert1 on May 22, 2021, 08:05:18 pmI 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 overThe datasheet says most of the units are in the 5nV range. Maybe there is a way to preselect them?

author: justanothername
author_rank: Regular Contributor
timestamp: May 22, 2021, 08:41:46 pm
content: Quote from: dietert1 on May 22, 2021, 08:05:18 pmShouldn't that be 0.8µV?

author: dietert1
author_rank: Super Contributor
timestamp: May 22, 2021, 09:44:16 pm
content: 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, DieterPS: Vishay has this:https://www.vishay.com/docs/30175/thermal.pdfand 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.

author: justanothername
author_rank: Regular Contributor
timestamp: May 22, 2021, 10:02:10 pm
content: This was the shunt I was using before:https://www.farnell.com/datasheets/2838152.pdfnow I ordered some new ones, 200m instead of 180m:http://www.farnell.com/datasheets/2339607.pdfhttps://www.yageo.com/upload/media/product/productsearch/datasheet/rchip/PYu-PE_521_RoHS_L_8.pdfIn 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.

author: David Hess
author_rank: Super Contributor
timestamp: May 23, 2021, 04:46:09 am
content: 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.

author: dietert1
author_rank: Super Contributor
timestamp: May 23, 2021, 06:04:35 am
content: 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

author: justanothername
author_rank: Regular Contributor
timestamp: May 23, 2021, 08:20:45 am
content: Quote from: dietert1 on May 23, 2021, 06:04:35 amThats 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?

author: Anders Petersson
author_rank: Regular Contributor
timestamp: May 25, 2021, 06:11:58 am
content: 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?

author: KT88
author_rank: Frequent Contributor
timestamp: May 25, 2021, 07:29:29 am
content: 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...CheersAndreas

author: dietert1
author_rank: Super Contributor
timestamp: May 25, 2021, 08:05:39 am
content: 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

author: 2N3055
author_rank: Super Contributor
timestamp: May 25, 2021, 09:40:01 am
content: Quote from: justanothername on May 23, 2021, 08:20:45 amThere 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.

author: KT88
author_rank: Frequent Contributor
timestamp: May 25, 2021, 10:02:08 am
content: Claiming that the bulk metal resistors show no thermal EMF is utter BS!

author: dietert1
author_rank: Super Contributor
timestamp: May 25, 2021, 10:32:20 am
content: 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

author: justanothername
author_rank: Regular Contributor
timestamp: May 25, 2021, 05:06:54 pm
content: Quote from: dietert1 on May 25, 2021, 10:32:20 amQuote from: 2N3055 on May 25, 2021, 09:40:01 amHow 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?

author: KT88
author_rank: Frequent Contributor
timestamp: May 25, 2021, 06:36:53 pm
content: 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.

author: 2N3055
author_rank: Super Contributor
timestamp: May 25, 2021, 08:52:26 pm
content: Quote from: justanothername on May 25, 2021, 05:06:54 pmLook 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 metalconnections. 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,

author: David Hess
author_rank: Super Contributor
timestamp: May 26, 2021, 12:35:23 am
content: Quote from: justanothername on May 25, 2021, 05:06:54 pmThe 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.

author: justanothername
author_rank: Regular Contributor
timestamp: May 26, 2021, 06:48:21 am
content: Quote from: 2N3055 on May 25, 2021, 08:52:26 pmThank 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 awayQuote from: David Hess on May 26, 2021, 12:35:23 amI will not give up yet.. Let me just wildly fabricate a hypothetical scenario. Just one last timeThere 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)  - 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] - connectorCould 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?

author: Kleinstein
author_rank: Super Contributor
timestamp: May 26, 2021, 04:02:05 pm
content: 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.

author: JohnG
author_rank: Frequent Contributor
timestamp: May 27, 2021, 12:42:43 pm
content: 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

author: dietert1
author_rank: Super Contributor
timestamp: May 27, 2021, 02:22:20 pm
content: 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, DieterPS: 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).

author: justanothername
author_rank: Regular Contributor
timestamp: May 27, 2021, 08:19:04 pm
content: Quote from: dietert1 on May 27, 2021, 02:22:20 pmIn my case, for the application I'm designing this, there will be blind vias on the uppermost layer anyways. So I simply will make them Type VII (filled and capped). The next layer is a maximum of 100µm below, so I have high hopes for very good thermal homogenity.You can stitch gnd planes with type VII vias as much as you want, there will never be solder paste drain. Even when the stitching covers a thermal pad.

author: justanothername
author_rank: Regular Contributor
timestamp: June 20, 2021, 04:44:06 pm
content: Hi!Here is a follow up.I tried to replicate the setup from the Vishay video at home. I recently got a HP419a and restored it, see here:https://www.eevblog.com/forum/other-blog-specific/great-meter!-the-hp-419a-dc-null-voltmeter-(restoration)/msg3588691/#msg3588691My setup consists of a 419A, a resistor (DUT) on a stripe of prototype PCB, and a 100W face warming lamp shining on the DUT from a 50° angle. The bulb is 8cm air distance to the DUT.I used several random SMD current sense resistors I found on my desk.The following steps were done for each measurement:* mount DUT on the stand with double sided sticky tape* let settle for 5 minutes* zero scale* turn on the lamp for exactly 30 seconds* record maximum needle position, watch 30 more secondsFirstly the used wires and pcb material:I cut several stripes from hasl tinned prototype material and put a solder blob as short.For the solder wire, I used ISO-Core Clear Sn100Ni+, as much tin as possible. I tried the following shorts:1) Tinned wire - hasl tinned pcb - tin solder blob as short:+0.2µV2) Tinned wire - copper pcb - tin solder blob+0.5µV3) Tinned wire - manually tinned pcb - tin blob+1.6µVThis is somehow unexpected3) solder wire as connection wire - manually tinned pcb - tin blob+1.2µVTotally unexpected, since there is only one material along the conduction path.So for the tests with the resistors, I used Setup 1).Results:I can only compare the results. There is no value like µV/°C, since I have no clue of real temperature gradient across the resistor. But I tried my very best to make the setup as consistent as possible. I marked all the positions, used the same length wire and so on... Results are best performance first.Vishay WSL0805R2000FEA, 0.2 Ohm 0805+0.5µVThis part is marketed as low thermal EMF. I can confirm.Welwyn LRCS0603-0R5FT5, 0.5 Ohm 0603+1.7µVPanasonic ERJ3RSFR20V, 0.2 Ohm 0603+4µVThick film resistor, but pretty goodTE RLP73N2BR082FTDF, 0.082 Ohm 1206+5.8µVYageo PE0805FRF070R2L, 0.2Ohm 0805+6.3µVStackpole CSR1206FTR500, 0.5 Ohm 1206+7µVYageo PE0402FRF7W0R2L, 0.2Ohm 0402+7.2µVOhmite KDV06DR180ET, 0.18 Ohm 0603+8.2µVThis is disappointing for a metal film 200mW current sense resistor!If you want any resistors checked like this or if you have improvement ideas, contact me.M.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 20, 2021, 05:05:31 pm
content: The tined copper trace is a mixed material. Much of the conductivity is still he copper trace and the composiotion in the solde can vary, as some copper transfers to the solder and some tin can go into the copper. The fresh solder wire can be different from the material of a solder joint. To some degree also the cooling speed of the solder or cold working of copper wire can make a difference to the Sebecke coefficients.In the configuration for the shorts it is also a lot about the symmertry. Even if the materials are a poor choice, with symmetry there would be no resulting thermal EMF. The same applies to the situation with the shunt: if symmetric and thus no temperature difference between the ends of the shunt, there would be no thermal EMF. So for testing the shunt quality one should use an intentionally asymmetric setup. With a nearly symmetric setup like in a real application it is hard to judge the actual shunt quality.

author: justanothername
author_rank: Regular Contributor
timestamp: June 20, 2021, 05:17:18 pm
content: Quote from: Kleinstein on June 20, 2021, 05:05:31 pmI heated the shunts as well as the shorts from the side. See attached picture.

author: dietert1
author_rank: Super Contributor
timestamp: June 20, 2021, 05:31:11 pm
content: The name is Seebeck, not Sebecke. A german physicist who lived from 1770 to 1831. Amazing how he discovered those small voltages. And the link with the voltmeter restauration is:https://www.eevblog.com/forum/other-blog-specific/great-meter!-the-hp-419a-dc-null-voltmeter-(restoration)/msg3588691/#msg3588691.Nice measurements! Did you check the effect of the lamp on the nearby meter?Regards, Dieter

author: justanothername
author_rank: Regular Contributor
timestamp: June 20, 2021, 05:40:46 pm
content: Quote from: dietert1 on June 20, 2021, 05:31:11 pmThank you for reminding me.1) for the actual measurements, the meter was 20cm to the left.2) See attached picture. I shorted the inputs, set the range to 3µV, zeroed, turned the lamp on and took a picture 30 seconds afterwards. Looks good.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 20, 2021, 05:48:09 pm
content: Quote from: dietert1 on June 20, 2021, 05:31:11 pmThe inital measurements of the thermoelectric effect where not measuring the voltage, but measuring the current in the very low resistance circuit. The current was enough to turn a compass needle.

author: dietert1
author_rank: Super Contributor
timestamp: June 20, 2021, 06:31:04 pm
content: Seebecks report is here:https://archive.org/details/abhandlungenderk182233deut/page/n311/mode/2up?view=theater(German language). Nearly 200 years ago!Regards, Dieter