Hello I want to power a circuit (that runs at 3.3v) with the balancing cable of a 3s Lipo Battery, my plan is to use only 1 cell through an LDO to power my circuit.

My question is can I power it through the balancing cable?

This circuit only draws about 700 mA (more realistically 400 - 500) with a 1A max burst. I need to minimize weight and complexity that's the reason why I cant just regulate the 12v coming from the battery.

From my research i found out that the balancing connector is rated up to 3A so that shouldn't be an issue, but can i draw this much current from the battery? I also do not care about unbalancing it.

My circuit

Circuit Diagram

Thank you very much! This is my first time using this website and constructive critics are more than welcome!

Edit: I'm using a 3s 450mAh Lipo battery

  • 1
    \$\begingroup\$ 700mAh is a unit of energy capacity, not current. Do you mean that you are drawing 700mA current? \$\endgroup\$ Jun 13, 2023 at 15:50
  • \$\begingroup\$ Yes 700mA @ 3.3v, it was a typo \$\endgroup\$
    – Rafael
    Jun 13, 2023 at 15:52

2 Answers 2


You say that you care about weight an complexity.

If you do like you planned, your 3.3V circuit will draw its energy from a single cell of the Lipo. Whats more, according to your drawing, you accept an input range down to 3.5V (that's already a quite good LDO to have a voltage drop of only 200mV@1A). So as soon as the battery reaches 3.5V on the lowest cell, you have to stop operating.

So you end up waiting a lot of the energy of your Lipo.

So instead of doing exotic stuff, I would recommend using a smaller lipo with a circuit that can use it down to about 3V. I think you can easely compensate the weight of a simple buck converter by the smaller Lipo.

Just in case : if your total current is far bigger than 700mA (ie the current on the 1 cell is negligible), then it might be worse digging further into your idea).

NB : if it's the size of the BOM that matters, then there are plenty of small, all-integrated DC-DC converters you can solder on your PCB

  • \$\begingroup\$ That's a great reply, and in accord with what the other reply said, perhaps i really took the wrong approach to the problem and going with an external module is better. Unfortunately the smallest lipo i could find is a 300 mAh, and it would still leave energy unused. \$\endgroup\$
    – Rafael
    Jun 13, 2023 at 16:26
  • \$\begingroup\$ I'll mark this question as answered \$\endgroup\$
    – Rafael
    Jun 13, 2023 at 16:26
  • \$\begingroup\$ If you need so little energy in total, you might consider going the other way around : use a 1S Lipo, an LDO for your 3.3V circuit, and a boost converter to generate your 12V or whatever voltage you want for your motors (nb : be careful, running motors out of a voltage regulator brings it own potential complications : if you want to go this way, ask a separate question specifying your motor and motor controller) \$\endgroup\$
    – Sandro
    Jun 13, 2023 at 16:40
  • \$\begingroup\$ That's something worth considering! thank you. \$\endgroup\$
    – Rafael
    Jun 13, 2023 at 16:52

This would not be a sensible design choice. You end up loading one of the three cells more than the other two.

Drawing power from just once cell results in that cell discharging faster than the others, causing it to prematurely reach its minimum voltage.

If this were a very small current (a couple of milliamps), it would probably not cause much issue, but 700mA is quite a significant load.

Best case, this effectively reduces your overall battery capacity as if you'd drawn the 700mA from all of the cells. This is because if one cell reaches minimum, it doesn't matter where the other cells are, you have to disconnect the load.

If the individual cells down't have their own built in protection, you can also overdischarge and kill the bottom cell before the overall battery protection knows what has happened - e.g. top two cells being at 3.7V, bottom at 2.7V would appear to the battery protection circuit as all cells being at ~3.4V if it assumes a balanced pack.

You would be much better off using a DC-DC buck converter from the 3S output (9-12.6V) down to 3.3V.

I also do not care about unbalancing it.

You really should.

At some point you end up with a pack which is so unbalanced that, even using a balance charger, you will have permanently degraded performance.

  • \$\begingroup\$ Thank you for the reply, I fully understand your point, but my circuit is for a robot that will only run for about 5 - 8 minutes so i do not think the extra load to a cell would unbalance it too much, designing a buck converter would add too many components to my BOM, my only other option would be to add an external dc-dc buck module, but that's less than ideal, I might need to reconsidere the design. \$\endgroup\$
    – Rafael
    Jun 13, 2023 at 16:15

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