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Topic: Feasibility of measuring large DC current with Arduino (Read 4656 times) previous topic - next topic

JohnDeere630

#15
Nov 29, 2016, 06:09 pm Last Edit: Nov 29, 2016, 06:15 pm by JohnDeere630
The 300A hall transducer can be ranged to 100A by simply looping your power line through the torus 3 times.
In effect the sensor then detects the flux produced by 3 x 100A and "thinks" there is 300A flowing.

Or to 75A by looping 4 times

If you set Vref low enough then you get full resolution provided you aren't interested in measuring reverse current flow.

Your original enquiry stated "reasonably accurately"  What do you consider that to be.
Resolution can never be better than 0.1% and component tolerances probably reduce accuracy to around 1%


Right, I have used the multiple loop trick  to boost sensitivity of amp-detect on some past projects. Regarding accuracy, obviously, I'd like it to be as accurate as I can without exotic solutions, but 1% is probably as close as I can get. As this is just a monitor function to see how may amps the solar array is putting into the battery bank, +/- 1/2 ~1 amp resolution is good enough.

JohnDeere630

The reason I asked about location is the shunt needs to be the last thing in the ground lead, right at the Batt neg. terminal if possible and Arduino GND connected to batt neg., but sounds like you already know that. Single supply op amps are cheap and easy to use, would not be too difficult to rig up a 33 to 1 amplifier.
Like this:

@Outsider,
The more I think on it, the more I like the approach you mentioned. The circuit looks easy to build, but I have a few questions: I cannot find a part# for a 1/2 MCP6002, has that part # been discontinued? Can I use a 6001 chip instead? The data sheets list the pinouts, and the 6001 has just the five you are using.

When you say a 33:1 ratio, I notice that there are a 33k and a 1k resistor...coincidence? Or is adjusting the ratio really that easy?

Where do the grounds all connect? The arduino GND rail? In other words, are they all referencing the 5 volt supply from the arduino, or the circuit being measured? It would seem that at least one referencing the arduino would be needed for the 2.56 v analog signal.....

Sorry to be a pain, but this stuff is interesting. Thanks!

Last, (for now) how do you know the values needed for the caps?

JohnDeere630

I drew up this diagram, referencing the MCP spec sheet and Outsider's sketch...

If you can actually see this, am I on the right track?

Wawa

80Amp and an opamp gain of 34 gives ~418 A/D values.
Lowering the 1k resistor to e.g. 470ohm could give you twice that resolution.

Are you sure you can safely ground (to USB/computer/mains) the negative wire on the solar cell side?

High-side measurement (shunt in the positive line) could be a safer way.
https://www.adafruit.com/product/1164
Leo..

JohnDeere630

I just noticed this on the MCP data sheet:

Gain = 1+R1/R2

So I guess it really is that easy. Now if I can figure out how Outsider knew the rating for the caps....

JohnDeere630

80Amp and an opamp gain of 34 gives ~418 A/D values.
Lowering the 1k resistor to e.g. 470ohm could give you twice that resolution.
Right...I imagine it will take some figuring and tinkering to fine tune the circuit to my requirement. Fortunately, the parts are just a few cents each, even the arduino board is just a few bucks.

Are you sure you can safely ground (to USB/computer/mains) the negative wire on the solar cell side?
I am not sure of anything, LOL...this is all new territory for me. What would you suggest? I figured even at 100 amps, the secondary is only .075 volts.....

High-side measurement (shunt in the positive line) could be a safer way.
https://www.adafruit.com/product/1164
Leo..
Thanks, I will look into that.

jackrae

80Amp and an opamp gain of 34 gives ~418 A/D values.
Lowering the 1k resistor to e.g. 470ohm could give you twice that resolution.

Are you sure you can safely ground (to USB/computer/mains) the negative wire on the solar cell side?

High-side measurement (shunt in the positive line) could be a safer way.
https://www.adafruit.com/product/1164
Leo..
But using a hall torus would give 'perfect' isolation and be totally "safe"

JohnDeere630

@Wawa, I looked at that adafruit part you mentioned. It maxes out at 5 amps, how would one use that with the larger shunt?

Please forgive my 1st grade questions. Electronics design is something that has always interested me, but I have little experience and zero formal training; like Arduino programming, it is learn as I go. LOL

JohnDeere630

But using a hall torus would give 'perfect' isolation and be totally "safe"
I ordered a pair of these, and plan to modify one of them to solder each of the chip traces onto a 1/8" copper strap, bolted to a 4 Ga. lug.

None of the three approaches I have been presented with are expensive, so I plan to try all 3 as time allows. Right now, I am on a fact finding mission. I have already greatly increased my small database of electronics design knowledge!

Wawa

@Wawa, I looked at that adafruit part you mentioned. It maxes out at 5 amps, how would one use that with the larger shunt?
Just connect it to the larger shunt.

The big shunt is 0.075volt / 100Amp = 0.00075ohm.
The 0.1ohm shunt on that breakout board will be the fly on the elephant.
100Amp through the main shunt is 0.075volt / 0.1ohm = 0.75Amp through the breakout board shunt.
Leo..

JohnDeere630

#25
Nov 29, 2016, 11:45 pm Last Edit: Nov 29, 2016, 11:55 pm by JohnDeere630
Just connect it to the larger shunt.

The big shunt is 0.075volt / 100Amp = 0.00075ohm.
The 0.1ohm shunt on that breakout board will be the fly on the elephant.
100Amp through the main shunt is 0.075volt / 0.1ohm = 0.75Amp through the breakout board shunt.
Leo..
That's what I suspected...Thank you! Just some simple math on the arduino to convert the analog voltages to amp values, and I can change things around with a simple resistor change.

Wawa

If you decide to use hall sensors...
You could use two 50Amp hall sensors (and two analogue inputs) if you split/divide current over two parallel cables.
Leo..

JohnDeere630

If you decide to use hall sensors...
You could use two 50Amp hall sensors (and two analogue inputs) if you split/divide current over two parallel cables.
Leo..
In my general practice, I have tried to avoid that if possible. Inevitably over time, one connection gets loose/corroded/etc. and throws the load to the other leg, with spectacular results. 30,000 amps dumped onto a single 4/0 cable is not something you quickly forget. It wasn't my doing, as I was just a helper at that time, but was magnificent....put the 4th of July fireworks to shame. Who knew molten lead and copper could spatter so far and stick to so much?

outsider

MCP6002 has 2 amps on the one 8 pin chip, you only need one thus "1/2 MCP6002" it can be had with 1, 2 or 4 amps per chip as MCP6001,2 and4.
BTW since the old timers here seem to know a lot more than me it would be best to listen to them.

MarkT

In my general practice, I have tried to avoid that if possible. Inevitably over time, one connection gets loose/corroded/etc. and throws the load to the other leg, with spectacular results. 30,000 amps dumped onto a single 4/0 cable is not something you quickly forget. It wasn't my doing, as I was just a helper at that time, but was magnificent....put the 4th of July fireworks to shame. Who knew molten lead and copper could spatter so far and stick to so much?
Sometimes paralleling helps prevent premature failure due to deterioration - if one wire
degrades the other helps, postponing failure to when both are deteriorating.  You'll often seen multi-pronged
wipers on switches and potentiometers for this reason, since if one contact point corrodes the other
still has a low resistance and prevents further damage on the corroded one.

Of course if everything is too close to its maximum rating you no longer have diversity/redundancy, and get
cascading failiures, whether one wire or two.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

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