I got it... but this is not a constructive answer I finally understood the confusion I made.
Always the XY problem!
This is interesting but, according to me,only focuses on solving a problem and not on learning from his mistakes to solve a problem.
What are these batteries exactly, and what is your expectation for monitoring them? Are you actually using separate 2 V lead-acid deep discharge cells as in a telephone exchange?
The point is, if there is some anomaly, what do you imagine doing about it? Will you actually be able to swap batteries? From where will you swap - given that the batteries swapped in will also need their charge to be maintained whilst in storage?
I will send someone to buy and swap batteries as soon as possible.
And if there is someone available to swap batteries, will they not be able to perform regular monthly checks on the installation in any case, and measure the cells manually?
This is what is currently done but not every month. Measuring the cells manually takes sometime and you can face difficulties to reach remote places... We are not taking about one plant but dozen of plants. (I know XY Problem )
It seems you want to measure individual lead/acid cells in an 48volt stack.
"Flying capacitors" is an easy way to do this.
See post#49 and up on this page.
This is interesting but, according to me,only focuses on solving a problem and not on learning from his mistakes to solve a problem.
What you should be saying is that I want to do this and I am thinking of doing it like this.
That way we get all your misconceptions over on the first post and we can tell you if you are on the wrong tack completely. That fulfills the criteria of you learning by mistakes without you giving us the run around.
One is to build Wawa's "flying capacitor" system which he has successfully used in the past for precisely this purpose. This is a non-trivial approach but he may be prepared to provide you with the PCB files and parts details to reproduce his devices. You need three such boards for each 24 cells.
A second approach is to construct a PCB with two of your 16-way multiplexers and simply measure the voltage at each "tap" using a voltage divider, subtracting one from the next to determine that individual cell voltage. A 12 bit ADC as in the Arduino Nano will give a nominal resolution of 56/1024 or 55 mV which should be adequate for the purpose.
The voltage dividers will impose a constant discharge on the battery - and there are of course, twenty-two of them but using a 110k:10k divider for the full 56 V will make that less than 500µA so this is obviously, no more than 12 mA in total and for what are apparently 220 Amp-hour cells charged daily, this should be absolutely trivial even noting that this load is unevenly distributed across the battery - 12 mA at the top end, 500µA on the fourth cell (and zero on the bottom two cells).
This second approach is clearly much simpler. Of course, with both approaches you require a ½ Amp fuse at every tap point on the battery for your substantial harness. So again, there is a substantial amount of construction involved.
In case you have not realised it in my description, the voltage dividers need not be all the same. The top 12 or so would be a 110k:10k divider, the next six might be a 47k:10k divider, the next two 22k:10k the next two 10k:10k and the "bottom" two are just a 10k in series as they see no more than 5 V. It is a waste of time trying to find dividers to give "neat" divisions because - you have a computer here - you calibrate the software itself to determine the actual voltages from the calibration of the Nano and the known ratios and if you can obtain a resolution of 0.1 V for each cell, this is going to tell you whether it is failing or not. If you do not feel this is sufficient, just get a better ADC.
I don’t think there is any need for a PCB with a flying capacitor circuit, nor a multiplexer circuit. I would just use strip board, at least for the first working prototype.
Well, I think I will go for the simple solution first... Is there any drawback of Solution 2 vs Solution 1? I remember that when I individually measured the voltage of let's say 5 batteries (with a normal voltmeter), made the sum and compared it to the value I got by measuring directly the voltage of the 5 batteries, the result was pretty different... But the relative voltage variation over the time is more important for my purpose (I guess).
Paul__B:
56/1024 or 55 mV
I guess your 56 comes from 24 cells x 2.33 V (which is the maximal voltage of the battery). Right?
I would have said that the resolution is 5 V (input max for the Arduino) / 1024. The voltage divider allows to measure higher voltages but should not influence the resolution. I guess, once again, that I have a false reasoning....
The voltage dividers will impose a constant discharge on the battery - and there are of course, twenty-two of them but using a 110k:10k divider for the full 56 V will make that less than 500µA so this is obviously, no more than 12 mA in total and for what are apparently 220 Amp-hour cells charged daily, this should be absolutely trivial even noting that this load is unevenly distributed across the battery - 12 mA at the top end, 500µA on the fourth cell (and zero on the bottom two cells).
Ok that's clear.
This second approach is clearly much simpler. Of course, with both approaches you require a ½ Amp fuse at every tap point on the battery for your substantial harness. So again, there is a substantial amount of construction involved.
How do you find the value of 1/2 amp?
In case you have not realised it in my description, the voltage dividers need not be all the same. The top 12 or so would be a 110k:10k divider, the next six might be a 47k:10k divider, the next two 22k:10k the next two 10k:10k and the "bottom" two are just a 10k in series as they see no more than 5 V.
This is also clear. But then the discharge would be higher than what you said before (I guess, you did 56/110k * 24) but the precision better (?)
Again, how do you choose 10k (with the experience ?)
I will share the scheme and the software from what I understood of your solution
I finally understood the confusion I made.
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