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Current sense shunt resistor on high current pcb
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Marco:

--- Quote from: cmorgan on November 12, 2018, 03:24:32 pm ---let me use one of those THT shunt resistors at 0.001 ohms.

--- End quote ---
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:

--- Quote from: cmorgan on November 12, 2018, 03:24:32 pm ---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

--- End quote ---
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:

--- Quote from: Hero999 on November 13, 2018, 11:43:10 am ---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.

--- End quote ---
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.
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