Current sense shunt resistor on high current pcb

[cmorgan]

Hello!

I have a high current pcb, 2oz copper, 7.5mm wide traces each side, designed to comfortably carry 30A @ 240VAC.

I'd like to add a shunt resistor to measure current on the pcb. Resistances from 0.0003 to 0.0001 ohms keep total power below 4W but all of these are surface mount like https://www.digikey.com/product-detail/en/stackpole-electronics-inc/CSS2725FTL250/CSS2725FTL250CT-ND/1923235

Given that I have the 2oz traces on the top and bottom of the pcb how should I handle adding that shunt resistor? Vias to pull the trace to one side of the board, through shunt resistor and then vias on the other side to go back to traces on both sides?

Chris

[001]

Hello!

I have a high current pcb, 2oz copper, 7.5mm wide traces each side, designed to comfortably carry 30A @ 240VAC.

I'd like to add a shunt resistor to measure current on the pcb. Resistances from 0.0003 to 0.0001 ohms keep total power below 4W but all of these are surface mount like https://www.digikey.com/product-detail/en/stackpole-electronics-inc/CSS2725FTL250/CSS2725FTL250CT-ND/1923235

Given that I have the 2oz traces on the top and bottom of the pcb how should I handle adding that shunt resistor? Vias to pull the trace to one side of the board, through shunt resistor and then vias on the other side to go back to traces on both sides?

Chris

use standard 4 wire reading

[coppice]

How accurately do you need to measure the current? It matters a lot in these designs. If you want a rough estimate (say 5%) lots of things will work. If you want decent accuracy (<1%) you will need to take great care to achieve a good 4 wire connection (whether you use an actual 4 wire shunt device or not), and you'll either need to avoid the shunt heating too much, or pay a lot for a shunt with an extremely good temperature coefficient.

[cmorgan]

use standard 4 wire reading

Hi 001.

Would you explain what you mean by a 4 wire reading?

If it helps this is a single phase two wire, so I have either L / N or L1 / L2 through the device. I didn't want to put two shunt resistors, one on the top and one on the bottom as then I'd have to sum up the current through the two paths. I'm looking at using ATM90E26 for the measurements.

Chris

[Marco]

More like 10 vias, or you could use two shunts and then just resistively combine the outputs.

PS. actually on second though, can't easily combine the outputs resistively since you don't have a good common ground.

[cmorgan]

Hi coppice.

5% would be great, this isn't for metering purposes. I was going to try to achieve a kelvin attachment as best as I could as well.

I was originally looking at using CTs but the shunt resistor approach seemed lower cost, less space and less manual complexity to assemble.

At this point I'm most concerned that the smd shunt causes me to redirect the current across one layer of the PCB and could impact the current carrying capability. The rest of the components on the high current path are through hole and connect to top and bottom layers. I found some THT shunts but their resistance is lower and at peak current I was worried about dissipation. It's 30A @ 240VAC, maybe my calculations are broken. I was using a peak voltage of something like 400V, so 400V * 30A, to calculate power through these shunt resistors but its 240VAC RMS so maybe that calculation is indicating far more power than will actually be dissipated and could let me use one of those THT shunt resistors at 0.001 ohms.

Chris

[coppice]

This https://www.analog.com/en/analog-dialogue/articles/optimize-high-current-sensing-accuracy.html is a good source of information about getting good accuracy from SMD current sensors.

[coppice]

The important thing to remember when even trying to get 5% accuracy is copper has a temperature coefficient of 0.4%/C. Even small amounts of it in the measured path can seriously screw up accuracy.

[001]

use standard 4 wire reading

Hi 001.

Would you explain what you mean by a 4 wire reading?

If it helps this is a single phase two wire, so I have either L / N or L1 / L2 through the device. I didn't want to put two shunt resistors, one on the top and one on the bottom as then I'd have to sum up the current through the two paths. I'm looking at using ATM90E26 for the measurements.

Chris

This is trick to eliminate wire error https://en.wikipedia.org/wiki/Four-terminal_sensing

[jbb]

I think LEM make a surface mount IC which senses the magnetic field of current flowing through a PCB trace. It won’t be very accurate but maybe good enough for your purposes.

[Marco]

let me use one of those THT shunt resistors at 0.001 ohms.

Very expensive though, for that cost you could put a SMD shunt on both sides and use some method to measure both (DPDT or two current amplifiers).

[t1d]

TE Connectivity/TE.com may have a resistor to consider. https://www.te.com/usa-en/plp/hs/X262v.html?q=&n=42563&type=products&samples=N&instock=N

[tszaboo]

Hi coppice.

5% would be great, this isn't for metering purposes. I was going to try to achieve a kelvin attachment as best as I could as well.

I was originally looking at using CTs but the shunt resistor approach seemed lower cost, less space and less manual complexity to assemble.

At this point I'm most concerned that the smd shunt causes me to redirect the current across one layer of the PCB and could impact the current carrying capability. The rest of the components on the high current path are through hole and connect to top and bottom layers. I found some THT shunts but their resistance is lower and at peak current I was worried about dissipation. It's 30A @ 240VAC, maybe my calculations are broken. I was using a peak voltage of something like 400V, so 400V * 30A, to calculate power through these shunt resistors but its 240VAC RMS so maybe that calculation is indicating far more power than will actually be dissipated and could let me use one of those THT shunt resistors at 0.001 ohms.

Chris

You do realize that you are not calculating the power the right way?
You never have 240V on a shunt, and the power is not P=U*I*R

You have a shunt with voltage drop of like 1 volt.
And P= I*I*R.
So for a 1 mOhm resistor it is 30A*30A*0.001 = 0.9W

[Zero999]

Yes, the power calculation is wrong. To dissipate <4W at 30A, you need a resistance of <4.444mOhm.

R = P/I² = 4/30² = 4/900 = 0.004444

30A across a 4.444mOhm resistor would be a voltage drop of:
V = I*R = 0.004444*30 = 0.13333V.

Aiming for exceptionally low voltage drops, increases errors due to noise and the offset voltage of the amplifier. For example, if the amplifier has an input offset error of 100µV and you're using a 0.004mOhm shunt, the current offset error will be:
 I = V/R = 0.0001/0.004 = 0.025A = 25mA.

[coppice]

Yes, the power calculation is wrong. To dissipate <4W at 30A, you need a resistance of <4.444mOhm.

R = P/I² = 4/30² = 4/900 = 0.004444

30A across a 4.444mOhm resistor would be a voltage drop of:
V = I*R = 0.004444*30 = 0.13333V.

Aiming for exceptionally low voltage drops, increases errors due to noise and the offset voltage of the amplifier. For example, if the amplifier has an input offset error of 100µV and you're using a 0.004mOhm shunt, the current offset error will be:
 I = V/R = 0.0001/0.004 = 0.025A = 25mA.

Why would amplifier offsets matter when measuring AC? You estimate and remove an DC. The real issue you need to worry about is the CMRR of the input to the electronics. In the OP's case this is not an amp. Its a differential ADC input with a really good CMRR. Its designed for shunts around 500 micro-ohms when the maximum current is 30A.

[nctnico]

The important thing to remember when even trying to get 5% accuracy is copper has a temperature coefficient of 0.4%/C. Even small amounts of it in the measured path can seriously screw up accuracy.

You can use that. In a design which didn't need very accurate current sensing I had a temperature sensor (cheap NTC) near the trace and measured the voltage drop across a piece of PCB trace. Together with the temperature you can get current readings and compensate for the temperature.

[r0d3z1]

The real issue you need to worry about is the CMRR of the input to the electronics. In the OP's case this is not an amp. Its a differential ADC input with a really good CMRR. Its designed for shunts around 500 micro-ohms when the maximum current is 30A.

I confirm this, usually the problem is the CMRR, could you be more precise about the circuit, is it a low side sense ? phase sense ? high side sense ?
For a low side current sense you can use this kind of shunt with a difference amplifier. You must be careful with the resistance due to the soldering which is comparable to the shunt value. For this reason, you have to use a 4 wire sensing as already suggested.

[Zero999]

Yes, the power calculation is wrong. To dissipate <4W at 30A, you need a resistance of <4.444mOhm.

R = P/I² = 4/30² = 4/900 = 0.004444

30A across a 4.444mOhm resistor would be a voltage drop of:
V = I*R = 0.004444*30 = 0.13333V.

Aiming for exceptionally low voltage drops, increases errors due to noise and the offset voltage of the amplifier. For example, if the amplifier has an input offset error of 100µV and you're using a 0.004mOhm shunt, the current offset error will be:
 I = V/R = 0.0001/0.004 = 0.025A = 25mA.

Why would amplifier offsets matter when measuring AC? You estimate and remove an DC. The real issue you need to worry about is the CMRR of the input to the electronics. In the OP's case this is not an amp. Its a differential ADC input with a really good CMRR. Its designed for shunts around 500 micro-ohms when the maximum current is 30A.

Yes, you're right, I missed it was AC, the offset voltage is no problem.

[perieanuo]

Acs723?


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