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I am fairly new to battery management systems. I am planing to use two cordless drill 18v battery (5A each) in parallel for an IoT project.

For the time being, I've just coupled the batteries through Schottky diodes in series in order to prevent stronger battery starts charging the weaker one. But, I know this is not enough for even out the load sharing.

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I think I need to use ballasting resistors in series as well. I like to test this addition. But, I am not sure what that resistors are really doing and how to calculate required value and wattage for my two batteries.

Any input greatly appreciated.

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  • \$\begingroup\$ With the Einhell batteries I have, there is a fair amount of electronics in them that might take care of any issues with paralleled packs. I’ve not investigated deep enough to determine if this is the case. It also seems some devices require 36V using two packs in series so i expect the electronics copes with this scenario. \$\endgroup\$
    – Kartman
    Commented May 11, 2022 at 0:25
  • \$\begingroup\$ @Kartman, yes this was another thought in my mind. But, If I am not mistaken, as I have Makita batteries, they might not have BMS inside the batteries but in the tool instead. \$\endgroup\$
    – Sener
    Commented May 11, 2022 at 7:01
  • \$\begingroup\$ The average 18V tool battery has three terminals. This suggests that there is at least balancing circuitry inside. Note that your batteries are not 5A, but 5Ah. They can supply much more than 5Amps. \$\endgroup\$
    – Kartman
    Commented May 11, 2022 at 11:49

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For the time being, I've just coupled the batteries through Schottky diodes in series in order to prevent stronger battery starts charging the weaker one. But, I know this is not enough for even out the load sharing.

This will essentially prevent load sharing. What will happen is one battery will supply until it drains below the other battery and then the other battery will supply. It will alternate back and forth like this. Combined with the fact that batteries reduce in voltage more while supplying more current than less current or no current, things may behave erratically as batteries are rapidly switched in and out.

Balancing resistors, not ballasting resistors. And if you always charge them together you do not need balancing resistors. Balancing resistors just give some slack so minor voltage differences between the batteries have somewhere where it can be eaten up. Without them the lower voltage battery will try to eat up the difference...and that's not good.

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  • \$\begingroup\$ Thank you for explaining some facts. Yes, load sharing is not there for sure. My initial response was to prevent charging effect between batteries. Could you please also elaborate your answer with some possible solution remarks (either passive or active)? \$\endgroup\$
    – Sener
    Commented May 11, 2022 at 7:16
  • \$\begingroup\$ Balancing resistors or just always charge them and use together \$\endgroup\$
    – DKNguyen
    Commented May 11, 2022 at 13:57
  • \$\begingroup\$ For 18V and 5Ah battery and not more than 500mA load (once a while max. 2A will be drawn for a minute), how can I calculate the Balancing resistor value and wattage? For example, if there is %2 voltage difference between the battery outputs and I also proportionally calculate the nominal load current, I calculated this; R=(V1out-V2out)/(Iload1-Iload2), R = 18-17.64 / 490mA-10mA = 0.00075 Ohms. Does this make sense? \$\endgroup\$
    – Sener
    Commented May 11, 2022 at 14:37
  • \$\begingroup\$ I assume you used this (which also may be where you got the term ballast from) techweb.rohm.com/knowledge/dcdc/dcdc-ldo/dcdc-ldo01/9235 Note it is not a concrete calculation. You can choose for greater efficiency vs more even current balancing with less or more resistance \$\endgroup\$
    – DKNguyen
    Commented May 11, 2022 at 14:44
  • \$\begingroup\$ I got it. From what you say, either case, I am choosing efficiency or more even current balancing I will be on the reasonably good side rather than not using a resistor at all as you also mentioned earlier. Thank you. \$\endgroup\$
    – Sener
    Commented May 11, 2022 at 15:36

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