INA219 breakout - remove shunt resistor?

[allanhh]

I wish to use the INA219 breakout board to measure currents up to 50A, in a 12V battery monitoring system.
I have an external shunt of .001 ohm, which should get up to 50mV across it.
So, I need to remove the onboard 0.1 ohm surface mount shunt resistor without destroying the board.
Can anyone help? I only have a standard soldering iron.
adafruit - is there any chance of producing a future breakout board with a removable link to disable the onboard resistor?

Thanks in advance, Allan

[adafruit_support_mike]

Is your 1mohm shunt a through-hole or SMT device?

If it's a through-hole, check to see if its leads will fit through the vias on either side of the 0.1ohm shunt already on the board. If it fits, just solder it in. Putting a 0.1 ohm resistor in parallel with a 1mohm resistor produces a parallel resistance of about 0.99mohms.. a change small enough that you can ignore it or compensate for it in software.

[allanhh]

Hi, thanks for the suggestion.
My 1mohm shunt is a large external one (50A) and I need to mount it about
6 feet away from the measuring circuit. I tried connecting it in parallel to
the onboard resistor using wires, but the resistance of the connecting
wires was too much ,relative to the onboard shunt, and it was very inaccurate.

Allan

[adafruit_support_mike]

Hmm.. that arrangement would put the 50A shunt and the INA219's sense resistor in parallel. That's generally a good thing for situations like this, but yes, you'd have series resistance from the wire too.

To keep the wire resistance from introducing too much error, you'd want it to be less than about 5% of the INA219's sense resistor, or about 5mohm. Taking the wire coming in and going back into account, you have a total wire length of about 4m, so you'd need wire with a resistance less than 1.25mohm/m. From the tables I have, you'd need to go to about 6 gauge.

It would be much more effective to put the INA219 right up next to the 50A shunt, and run the signal from that the 6' to the rest of your circuit.

[john444]

Hi Allan HH,

Another possibility is to replace the INA219s 0.1-ohm shunt resistor with a much
larger one, say, 10 or 100-ohm. Then the lead-wire resistance will have much
less impact on the 50-mV produced by your 50-A shunt.

I wouldn't recommend simply eleminating the on-board shunt. It is desireable to
have some significant resistance always connected to the INA219. It becomes
important if (when) the connection to the external shunt comes off or
becomes loose.

I have two 100-Amp / 100-mV shunts on my battery bank. I replaced the 0.1-ohm
on-board shunt (on each INA169) with a 100-ohm resistor. I also used 18-AWG
stranded wire from the 100-A shunts to the INA169's terminal-block.
It seems to work well.

Good Luck with your project.
John

[allanhh]

Thanks for the advice,
If I locate the INA219 beside the shunt, will the I2C protocol work ok over the 2m distance? It's the inside of a small motorhome and could be electrically noisy.
Allan

[adafruit_support_mike]

You should be okay if you send VCC and GND along with SDA and SCK.

If you supply those four wires, the only electrical connection between the INA219 and the noisy system will be through the current shunt. That connection is a closed loop though, so there's no way current can flow from the shunt to any other part of the INA219/Arduino circuit.

You might see some problems from inductive noise.. 50A emits a lot of EMF.. but you can limit that with shielding. A length of CAT5 phone cable is a good candidate for your connection since it's inexpensive, easy to find, and specifically designed to avoid picking up noise.

Connect the Arduino's VCC and GND lines to one pair in the cable, SDA to one wire from another pair, and SCK to one wire from a third pair. Leave the ends of the other four wires unconnected at the Arduino end. Connect VCC, GND, SDA, and SCK normally at the INA219, then connect all four unused wires to the GND wire at the INA219 end. That arrangement turns the unused wires into a shield that blocks EMF noise from entering the signal lines.