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I would like to use multiple shunt resistors in parallel to reduce power dissiption to the power specified per resistor. However these resistors are 4-terminal resistors WSL40265L000FEB. The reason being I need the accuracy they provide. My CS amp has input range of +/- 50mV . Current passing is from 0 to +/-200A, which yields 10watts.

Idea 1: Use 1 of the 4-terminal resistors in parallel with a normal(2 pin) shunt resistor of the same value. Does this defeat the purpose of the 4 Kelvin connection?

Idea 2: Use 2 of the mentioned resistors in parallel, and connect the Kelvin pins in parallel before going to the amplifier (however it might be difficult in layout.)

Idea 3: Using 2 of the 4-terminal resistors in series and connect the Kelvin pins in series, also difficult in layout.

Please advise what would be the best method.

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  • \$\begingroup\$ I'm having a hard time envisioning your ideas, but keep in mind that these resistors are only rated to 3 W, so you would need at least 4 to meet your power needs. Also, the lowest value they go to is 0.2 mohm which is 40mV, so even just two in series is 80 mV which is above the input range of the CS amp, meaning they all have to be in parallel. I'd also be worried about loss and heating across the PCB trace/plane carrying the current. 200 A is a lot. \$\endgroup\$
    – Ryan
    Commented Jan 27, 2022 at 17:26
  • \$\begingroup\$ So what would you suggest ? there are offboard screw mount shunts that handle the current . but i am worried the sense lines will pick up noise from switching nearby Regarding trace/plane I am using Solder-able copper bus bars that look like hair comb. \$\endgroup\$
    – Eng Sam
    Commented Jan 28, 2022 at 16:18

3 Answers 3

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It would be best to find a resistor that does not require paralleling. If you absolutely must, I suggest arranging for the high current connections to be as symmetrical and close to each other as possible and then adding series resistance to the 'sense' terminals so that they don't conduct significant current due to any mismatch. For example, you might use 100 ohms or 1K 1% resistors (one pair per resistor). So, for 2 shunts you'd need 4 equal-value resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ thank u , the drawn schematic helped alot . I might go this route . however regarding your suggestion of using 1 resistor , there are resistors that handle the current however they are offboard shunts they have screw mounts and they are not for pcb assembly . i am afraid that using off board resistors may increase the noise pickup of sense lines ( mosfets switching inductive load nearby) \$\endgroup\$
    – Eng Sam
    Commented Jan 28, 2022 at 16:11
  • \$\begingroup\$ Can you comment on Idea1 ? \$\endgroup\$
    – Eng Sam
    Commented Jan 28, 2022 at 16:19
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    \$\begingroup\$ Yes. The accuracy will be affected. If you 'had' to do that, it would be better to parallel it with an exactly similar shunt and ignore the sense terminals on one of the shunts. But for the sake of a few pennies worth of resistors I'd rather not. \$\endgroup\$
    – Spehro Pefhany
    Commented Jan 28, 2022 at 16:22
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    \$\begingroup\$ Even 10 ohms would be much, much better than letting the copper decide where the current flows. They're in parallel so 10 || 10 = 5\$\Omega\$. \$\endgroup\$
    – Spehro Pefhany
    Commented Jan 28, 2022 at 17:08
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    \$\begingroup\$ Yes, I think so. You might want to match the sense trace lengths and keep them reasonably wide so they are << 10 ohms and the difference is <<< 10 ohms. \$\endgroup\$
    – Spehro Pefhany
    Commented Jan 28, 2022 at 17:12
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Your sense resistor, if I'm reading it correctly, is 0.005 ohms. At 200 amps, it would dissipate 200W and have a voltage of 1V across it. You want to cut the voltage range by at least a factor of 20 and reduce the power about the same to get it into the ranges you're expecting above.

That series of parts goes down to 0.0002 ohms. Is there a particular reason you don't just switch to that value?

If you're already using the 0.0002 ohm part, and just listed the wrong part number above, you're probably best off with option 1. The 2-terminal resistor doesn't need to be the same value, but it needs the same accuracy or your readings will be off correspondingly.

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  • \$\begingroup\$ Hi , i am using a lower value , but i attached the link to any resistor in that series (sorry for that) , but even the 200uohm resistor is still 3W . and the power developed across it is 200x200x200x10^-6 = 8W . So i have to use at least 3 in parallel. \$\endgroup\$
    – Eng Sam
    Commented Jan 28, 2022 at 16:14
  • \$\begingroup\$ When I said the 2-term didn't have to be the same value, I was alluding to the idea of using e.g. a single 0.0001 ohm part instead of two 0.0002s in parallel. However, even a single 0.0002 in parallel would reduce your dissipation to 2W for each resistor, and wouldn't reduce your usable ADC range as much. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Jan 28, 2022 at 17:04
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As we know, shunts have the problems of: Temperature coefficient for resistance, Resistance tolerance and paralleling has problems of unequal current sharing and uneven temperature rise. These problems become apparent only on testing the PCB in lab and require deeper understanding and mitigation for achieving high accuracy (say 0.05% FSR) for high currents (upto 120A). I have been working on this problem of paralleling shunt resistors for measuring current and have been lead to this question. In my testing, Idea 2 discussed above is a better way to have higher accuracy. Adding 10-15 ohm series resistance (2 per shunt) improves accuracy when checked against a calibrated current measurement. Having no resistor gave poor accuracy (>0.3%) whereas higher resistances like 100-500ohms also gave poor accuracy and a major drop in the expected differential voltage in my testing, suggesting an optimal resistance exists. Further, the layout and ensuring proper soldering was found to be very important for the sharing of currents and temperature rise.

I would like to know if the original poster had any better results or solutions for his problem, perhaps in comments.

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  • \$\begingroup\$ thank you for the input . however i have not tested the results since the chip shortage the required current sense amplifier is out of stock :( \$\endgroup\$
    – Eng Sam
    Commented Jul 15, 2022 at 7:34

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