Author Topic: 0.001 ohm 3W resistor at 30A gets too hot; impact on accuracy and alternatives?  (Read 8047 times)

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Offline ShubhamTopic starter

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I'm using a Bourns CRE2512-FZ-R001E-3 resistor (0.001 ohm, 3W, +-1%) for current measurement in a circuit with a maximum current of 45A. However, at 30A, the resistor gets excessively hot, despite calculations indicating a dissipation of only 0.9W. I've also tested with a generic 0.001 ohm, 3W resistor, observing the same issue. Is it normal for a 3W resistor to get very hot at 0.9W? Could the temperature increase impact resistance and affect current measurement accuracy? Any recommendations or alternative for high current measurement?

Datasheet:
https://4donline.ihs.com/images/VipMasterIC/IC/BOUR/BOUR-S-A0013301828/BOUR-S-A0013301828-1.pdf?hkey=6D3A4C79FDBF58556ACFDE234799DDF0

The PCB is made of double-sided FR4 with a board thickness of 1.6mm, standard operating temperature of 140°C, and a copper finish of 35.00µ/1oz. I left polygons on both sides. A screw connector is mounted on the top, and the resistor is on the bottom. While soldering the screw connector and resistor, I applied a dab of solder. However, with a 30A current, it gets extremely hot to the point where I can't touch it for even a second.
« Last Edit: December 14, 2023, 06:04:39 pm by Shubham »
 

Online Ian.M

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That's 0.9W in slightly over 20 mm2 of board area, about 40% of the dissipation of the same area of a 1KW electric bar fire element.  At 35A, without very good heatsinking that resistor (or any other of a similar package size) would glow red hot (and fail) as the power density would be comparable to a bar fire element!   Your problem is getting the heat out of it.  If you aren't on an aluminium core PCB (or over a heavy copper power or ground plane*) with very wide heavy copper tracks to the resistor for the current connections, you'll probably have to derate it. 

On regular double sided FR4, you could try four of the resistors in series/parallel with big 'fans' of copper either side of the middle node to keep that cool, or four 0.004 ohm resistors in series-parallel, spread out some to spread the dissipation.  'Mirror' the power tracks on the back of the board, and heavily via stitch them near the resistor pads to get as much heat as possible out through the copper traces.

* See Siwastaja's comments below
« Last Edit: December 14, 2023, 05:20:05 pm by Ian.M »
 

Online Siwastaja

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Numbers, not adjectives, what do you mean by "excessively hot"? I regularly use 2512 SMD current sense resistors with very little derating, i.e. 1.5W for a 2W rated part, or 2-2.5W for a 3W rated part, and never had any problems, but OTOH my power layouts are 4-layer and the mid layer GND is good at sucking heat away, I also pay attention to large copper fills for the pads, which further couple to the mid layer through the thin prepreg.

Post a picture of your layout + stackup specification, what is your intended maximum ambient temperature, how the PCB is cooled, and what you mean by "excessively hot", and maybe we can help further.
 

Offline tszaboo

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You need to use larger pads. Copper on both sides. 4 layer if possible and connect it together as many vias as practical. Then add even more coper.
 

Online Siwastaja

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That's 0.9W in slightly over 20 mm2 of board area, about 40% of the dissipation of the same area of a 1KW electric bar fire element.  At 35A, without very good heatsinking that resistor (or any other of a similar package size) would glow red hot (and fail) as the power density would be comparable to a bar fire element!   Your problem is getting the heat out of it.  If you aren't on an aluminium core PCB (or over a heavy copper power or ground plane) with very wide heavy copper tracks to the resistor for the current connections, you'll probably have to derate it. 

On regular FR4, you could try four of the resistors in series/parallel with big 'fans' of copper either side of the middle node to keep that cool, or four 0.004 ohm resistors in series-parallel, spread out some to spread the dissipation.  'Mirror' the power tracks on the back of the board, and heavily via stitch them near the resistor pads to get as much heat as possible out through the copper traces.

Whole post is colossally bad advice. Series-parallel configuration is hard to Kelvin sense, and for mere 0.9W, extra ESL and board real estate wasted for no reason whatsoever. For mere 0.9W, even a 1210 sense resistor can be used, in a tight spot, for minimized ESL and switch node ringing caused by that ESL (assuming this is a power converter). Single 2512 should be easy-peasy.
 

Online Ian.M

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Yes series/parallel is far from ideal and near-impossible to kelvin sense, so the resistance (and temperature coefficient) of the solder joints and copper in that intermediate node is a big problem.   If you cant get the heat out of a single 2512 resistor because you cant justify heavier copper or more layers for the entire board, the sane option is probably a physically much larger, higher power rating  resistor, derated for the lack of adequate copper area and thickness to cool it.
« Last Edit: December 14, 2023, 05:49:58 pm by Ian.M »
 

Online Kleinstein

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It is not just the shunt to produce heat, but chances are the traces leading to the shunt will also produce heat.
One may consider an additional copper wire on top of the traces leading to the shunt. This may also help removing some heat from the shunt.

The SMD parts need large pads and quite some copper to get there nominal power rating.
With such high current it may be worth to use more than the normal 1OZ copper, not just for the shunt, but for other parts.

For precision use 0.9 W from a nominal 3 W shunt may already be on the high side, especially if the layout is not including the needed copper area.


 

Online magic

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Just came here to post the same. Too thin PCB traces may easily have more than 1mΩ resistance.
 

Offline ShubhamTopic starter

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1953300-0
The PCB is made of double-sided FR4 with a board thickness of 1.6mm, standard operating temperature of 140°C, and a copper finish of 35.00µ/1oz. I left polygons on both sides. A screw connector is mounted on the top, and the resistor is on the bottom. While soldering the screw connector and resistor, I applied a dab of solder. However, with a 30A current, it gets extremely hot to the point where I can't touch it for even a second.
 

Offline tszaboo

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(Attachment Link)
The PCB is made of double-sided FR4 with a board thickness of 1.6mm, standard operating temperature of 140°C, and a copper finish of 35.00µ/1oz. I left polygons on both sides. A screw connector is mounted on the top, and the resistor is on the bottom. While soldering the screw connector and resistor, I applied a dab of solder. However, with a 30A current, it gets extremely hot to the point where I can't touch it for even a second.
That's not extremely hot, thats somewhere above 80C. Measure it with a thermal camera or thermocouple.
And place more copper.
 

Offline ShubhamTopic starter

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I do not have a thermal camera or thermocouple and I think the temperature should be around 150C and even if it's 80C isn't it too much to affect the measurement? How can I place the more copper on already manufactured PCB's? Is the solder bulk not sufficient?
 

Online Ian.M

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You are out of luck and will probably have to respin the PCB, but you could try soldering heavy copper sheet fins to the tracks right up against the resistor pads, or better  round three edges of the pad.  You are aiming to increase the surface area dissipating heat by at least an order of magnitude.  The fin surfaces should be sanded matte then painted matte black after soldering, with as thin a layer of paint as possible.  This is a horrible bodge and Siwastaja will probably tell you I'm crazy!
« Last Edit: December 14, 2023, 06:21:12 pm by Ian.M »
 

Online coppice

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I do not have a thermal camera or thermocouple and I think the temperature should be around 150C and even if it's 80C isn't it too much to affect the measurement? How can I place the more copper on already manufactured PCB's? Is the solder bulk not sufficient?
What accuracy do you need for your measurement? If you are making some kind of power meter you probably need a shunt with excellent power dissipation to keep it cool enough to maintain the accuracy you require. Look at things like utility energy meters. They use a shunt made of low temp. coeff. material, like manganin, and its a big chunky thing so its temperature stays low enough with maybe a 60A load for its accuracy to remain within a tight spec. If the current measurement didn't need to be so precise, as in things like motor control applications, they could use quite a small shunt.
 

Online Siwastaja

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Layout as shown would heatsink pretty well into the connector, but there is a flipside to this: if the connector itself is used very close to ratings then it will heat up, too. It seems you have some excess room to make the polygons a bit larger, too: use all the easily available board area for heatsinking. Also remove the thermal relief patterns from those connector through holes.

Even without these modifications, I'm pretty certain the resistor does fine, as you would be running it below 1/3 rated power, but you can only be sure by taking a measurement. I know it's hard to measure the resistor's internal temperature, so destructive testing is a good alternative: double the power; it if survives +100% overload for more than a few hours, that's a pretty good sign.

Internal heating is impossible to exactly calculate given the limited data on datasheet, but "Derated to zero power at 170degC" hints that this would be the maximum internal temperature, near full rated power at rated ambient temperature, with a (very) good layout. Use the temperature coefficient number to calculate the error. The result is vastly different depending on your conditions: for example, if your current varies only slightly, and you don't need to take measurements immediately after powerup, then you can calibrate out most of the error. On the other hand, if the thing has to read current a millisecond after power-up while the internal resistor temperature is still +20degC, and it then later heats up to +170degC, error between these readings would be up to 150degC * 50ppm/degC = 0.75%.

Though, note that this type of resistor isn't in practice very accurate because you can never Kelvin sense it perfectly. For 1% tolerance, 1% extra error for TC, and 1-2% variance on mounting, I use parts like this, uncalibrated, for +/- 5% accuracy; with calibration, you can do much better.

You can also buy a better resistor which has a "true" 4-pad layout, smaller TC, and better tolerance, but that's going to cost more.
« Last Edit: December 14, 2023, 06:54:35 pm by Siwastaja »
 

Offline nctnico

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Layout as shown would heatsink pretty well into the connector, but there is a flipside to this: if the connector itself is used very close to ratings then it will heat up, too.
I think you are right on the money here. From the pictures it looks like the connector isn't suitable for 30A in real world applications. Personally I derate such pluggable terminal blocks to half the rated current for reliable operation.
There are small lies, big lies and then there is what is on the screen of your oscilloscope.
 

Online Siwastaja

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Yeah. It is possible that the difference between a good brand part and a cheap part is not as how hot they operate, but rather how well they survive in high temperatures they create. High power resistors are designed to run hot, and high current connectors use plastics that survive high temperature. There are other problems though than the part itself, for example you don't want to melt the PVC insulation of the wires that are screwed into that connector. You also want to mitigate against small user errors like using wrong wire gauge, wrong screwing torque, wire not inserted as far in the connector as it should, etc. Derating gives more margin for all sorts of errors.

So: what is the part number of the connector?
 

Offline ShubhamTopic starter

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I think you are right, the culprit is the screw connector. I am using ZB950 rated for 30A as per UL and 26A as per IEC. But I wonder for 20A also the trace and the resistor are getting warm but not the screw connector body.

here is the datasheet for the screw connector: https://drive.google.com/file/d/1kCU42vpFh89YNWzN2WQoyhoutJVTlmAc/view
 

Offline nctnico

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That is not going to work. Such specifications are determined by loading only 1 pin with 30A in an ideal situation.

There are some pointers here: https://www.electronicdesign.com/content/article/21186362/demystify-current-ratings-for-connector-selection
There are small lies, big lies and then there is what is on the screen of your oscilloscope.
 

Online Siwastaja

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Yeah, that's just a drawing file, usually connectors also have more data in a separate datasheet, might be a part of zillion-page catalog. There it would probably read that pin current rating is for one pin being loaded only, and derating factors depending on number of pins used. This is because in many applications, some of the pins carry less current.

Screw connector body doesn't feel warm because it's plastic, and plastic insulates heat, so heat has nowhere to go except the wire screwed to the connector (making it super important that users follow the cross-sectional area advice, and you should specify thick enough wire, not only based of the heating of wire, but also for heatsinking the connection), and the pin.
 

Offline Berni

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At these sort of currents EVERYTHYNG is a resistor.

The resistor gets hot, the PCB gets hot, vias get hot, connectors get hot, cable terminals get hot, cables themselves get hot.

So it is all about over rating things for more current than you need. At these currents this means rather beefy connections all over. Things like the terminals and cables can even function as heatsinks to pull the heat away from the power dissipated by the contact resistances, but the cables have to be thick enough to not become hot themselves.

One alternative you can also use for these currents is going to hall effect current sensors. Those can have lower internal resistance, since the part the current passes trough is just a piece of copper.
 

Online Siwastaja

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At these currents this means rather beefy connections all over.

Yes. Usual beginner mistake is to only think about making everything thick/wide. Next step is realize that making everything short is more effective. Aspect ratio is what matters! If a 100mm long, 20mm wide trace works, then 50mm long, 10mm wide trace works too, except that it works actually better because it can heatsink to something from its ends. Key to success in high power design is quite often minimization of distances, even if it makes the power density look scary - this is also why I called Ian's comment "bad advice". Extremely small objects also have extremely good ratio between surface area and volume, this is why shrews need to eat all the time in cold conditions, they cool down so easily by just being small. 5x5mm square is excellent in connecting two power components together and can easily carry 30A, even though a "5mm wide track" obviously cannot.

And from this perspective, OP's design is fine, it clearly does not have unnecessary power traces (at least on the part shown here; don't know about the design overall); you can't make "tracks to the resistor" any wider because there are no "tracks to the resistor" to begin with, which is kinda optimal.

OTOH it leads to a situation where, if you connect components together all of which heat up similarly, they won't help heatsink each other. If you placed the two components much further away, they could heatsink separately, but then you'd need really good conductors between the two, the PCB real estate needed grows quadratic per distance unless you stuff like solder thick copper busbars on it, utilizing the third dimension.

I bet the biggest problem is with choosing undersized screw connector; you can mitigate by specifying much larger wire gauge than strictly needed from the wire perspective itself, just make sure people actually follow your recommendation. In other words, make sure the wire doesn't become the third "component that heats up so much it can't act as a heatsink for the others"; given that the resistor heats up, the connector can't push heat into the pad as much as it could otherwise; so heat needs to go to the wire.
« Last Edit: December 15, 2023, 10:41:43 am by Siwastaja »
 
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Offline jonpaul

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for current sense >10A a real current shunt , hall effect, or CT (AC only) is usual.

The the very low R eg  0.1 Ohm, 0.01 Ohm are special types with 4 term, see Isabelhutte, etc.

https://www.isabellenhuetteusa.com/precision-power-resistors/

I have not seen any 1 mOhm R in use as the voltages sensed are overwhelmed with noise in an SMPS.

It may be wiser for the OP to  reconsider the design concept


Jon

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Online Siwastaja

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for current sense >10A a real current shunt , hall effect, or CT (AC only) is usual.

OP has "real current shunt".

Quote
I have not seen any 1 mOhm R in use as the voltages sensed are overwhelmed with noise in an SMPS.

You have not looked hard enough. I have used 1mOhm current sense resistors in 2512 and 1210 packages numerous times, in SMPS designs, without problems with noise. These parts are sold for this purpose, so I guess others use them, too. The trick is to place the sense amplifier close and, as always, average slow readings, and blank small period of time near switching event from decision making if possible. The sense wires are low impedance, can be made short, and can be routed differential, to the differential-input amplifier, and are usually shielded by a ground layer, so noise doesn't need to be a problem even if full range signal level is just some millivolts.
« Last Edit: December 15, 2023, 12:09:56 pm by Siwastaja »
 

Offline Alex Wolf

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As coppice correctly noted above, for accurate/precision current measurements you do not need a resistor, but a full-size current shunt made of a special alloy, such as constantan or manganin. If you are satisfied with the accuracy of measurements with a resistor, then you can try to compensate for heating and temperature drift of the resistance (it is predictable, depending on the current). There is no workaround.
 

Online Siwastaja

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As coppice correctly noted above, for accurate/precision current measurements you do not need a resistor, but a full-size current shunt made of a special alloy, such as constantan or manganin. If you are satisfied with the accuracy of measurements with a resistor

 |O

"Shunt" is just short-hand for "shunt RESISTOR". It is a resistor, not a special component. The resistor as shown is a shunt resistor and obviously made of "special alloy" with moderately low TC, which would be obvious if you looked at the datasheet, the title of which says "current sense resistor", which is the same thing as "shunt resistor". Cheap-ass resistors made of less "special" alloys have worse TC than 50ppm/degC.

How much size is "full size"?

If you want to say "you need 4 separate terminals for better Kelvin sensing", then say it. If you want to say "50ppm/degC TC is not good enough", then tell us what is. 20? 10?
 


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