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I studied electronics but became a programmer instead, and have forgotten a lot.

Now I need to run Arduinos and other 5 V boards in 12, 24 and sometimes 48 V marine and off grid environments, where the battery banks are always 6 V lead acid cells connected in series.

My circuits have to run as efficiently as possible when there is no charge on the batteries.

I would presume I will be using buck(/push?) converters to get a good regulated supply, and these all seem to get more efficient the less voltage they have to drop.

Is it possible or practical to take 6 V from the ground side cell directly?

Am I right to think this would be more efficient than bucking 12 V?

I need to be able to monitor the 12 V rail voltage as well, I hope using an ADC input, and I also need to control 12 V addressable LEDs, which are powered from the 12 V rail.

I cannot seem to find any discussion on this technique online, I presume there is a reason it will not work?

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    \$\begingroup\$ If you look at the actual charts for efficiency in the datasheets of the converter, and calculate lost power given the relatively small load of Arduinos, does it actually become a significant enough concern to outweigh imbalanced discharging using just one cell? Unlike linear regulators which suffer badly when Vin is much larger than Vout, the efficiency of switching buck regulators may drop as Vin increases away from Vout, but remains quite high. \$\endgroup\$
    – nanofarad
    Sep 10 at 17:31
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    \$\begingroup\$ It's not a good idea to individually discharge a single cell (battery) that's connected and charged/discharged with other cells in series. Your 6V cell will be discharged earlier than the other cells in the stack and will probably not be fully charged when other cells in the stack are. This will cause imbalance in the stack and eventually damage the cell. Go for an efficient buck converter. \$\endgroup\$
    – StarCat
    Sep 10 at 17:46
  • \$\begingroup\$ not sure how significant my circuit draw will be, just want the most efficient practical solution. if the alarms in my system \$\endgroup\$ Sep 10 at 18:13
  • \$\begingroup\$ (edit timed out, not used to this interface) not sure how significant my circuit draw will be, just want the most efficient solution, not waste power. my intention is to run other 5v boards, isolated, on the remaining positive cells, and since all my boards are mesh radio connected, they are intended to act as cell imbalance detection and load balancing devices \$\endgroup\$ Sep 10 at 18:48
  • \$\begingroup\$ Taking nominal-6V directly, the voltage will at some point dip too low for the headroom a linear regulator. Better to get a DC/DC module using the full range. If you're pulling like 10 or even 20 mA for some logic, with a car-battery type system, the efficiency should be acceptable. \$\endgroup\$
    – Pete W
    Sep 10 at 18:48
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Use a SEPIC converter, connect at the highest voltage point below its maximum rated input voltage. Be sure to take into consideration the maximum voltage during charging. My preference would be at the 12 to 24V point. Reason it appears there are engines involved in the systems and when the starter(s) kick in battery voltage can drop quite a lot especially if they are not at full charge. I would expect the 6V to drop to maybe 4V during start which would probably cause the Arduino to crash. The efficiency of these SEPIC converters can reach 98%, which is considerably better then any linear solution. The batteries will eventually discharge to the point where stuff will no longer work you might want to look at the recharging schedule. This load should not have much effect on it but I have no idea of the capacity of your batteries.

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  • \$\begingroup\$ brown out crashes on starter and other loads are acceptable, mission critical circuits will be LiPo (sp?) UPS backed, or will run on 3v3, and the whole system is mesh connected and coded to shut itself down before it can hurt the batteries. \$\endgroup\$ Sep 10 at 18:58
  • \$\begingroup\$ thank you very much for the SEPIC acronym, this seems exactly what i need to study, and if i read you correctly, you appear to be confirming it is more efficient to buck/boost a closer supply voltage, than to continously buck a higher supply. i will carry on in this direction, and will report back what i find, thank you all for being there \$\endgroup\$ Sep 10 at 19:32
  • \$\begingroup\$ Thanks for the update, I do not think the efficiency will change to much with the voltage, but the higher voltage generally gives you a higher efficiency because of the diode losses and the additional room keeps stuff happy during a start etc. \$\endgroup\$
    – Gil
    Sep 10 at 19:38

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