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I'm designing a circuit to monitor the performance of lithium polymer battery cells and am stuck on the best way to power it. The battery packs could have between 2 and 8 cells and are supplied pre-wired in series with a primary power connector (pack +ve and -ve) and secondary balance connector (pack -ve plus a tap on the +ve from each cell). My circuit will hook up to to the balance connector so that I can monitor individual cell voltages and I'd also like to power it from this connector both for convenience and because the primary power connector will be in use. So I have a 9-pin header where anywhere between 3 and 9 pins will be connected depending on how many cells are in the battery. I've built similar circuits in the past and used a couple of different approaches, but don't think either is ideal and would like to get opinions on a better design. Here's what I've done in the past...

The simplest solution is to power the board from pins 1 and 3 of the header (pack -ve and cell #2 +ve). This will always provide 6v to 8.4v depending on the state of charge of the pack, plenty to run a 5v LDO. The problem with this approach is that it is actively unbalancing the pack, by drawing power from only the first two cells. This doesn't seem like a very sensible thing to do with a lithium polymer pack as they're notoriously volatile.

Another solution I've used in the past is to connect all the cell +ves together via a diode and use this common cathode and pack -ve for power. This will always provide the full pack voltage to the board, but I'm concerned about what the leakage current through the reverse biased diodes will do to the pack.

Is anyone aware of a better solution to this?

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  • \$\begingroup\$ How much current does your "circuit" draw? Please provide a link to the datasheet for your 5V regulator. \$\endgroup\$ – Elliot Alderson Feb 5 at 21:13
  • \$\begingroup\$ Circuit will draw up to 1A@5V. Currently using a TSR 1-2450 for prototypes because I have some spare, final design may differ. Data sheet here: google.com/url?q=https://www.tracopower.com/products/… \$\endgroup\$ – Owen Feb 5 at 21:34
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It sounds like your best choice might be to replace the LDO with a buck (step down) dc/dc converter. Choose one that operates over a range that includes the minimum voltage you get from 2 cells in series (6V?) and the maximum voltage you get from 8 cells in series (33.6V?). Run the converter with the entire string of cells in series.

Be aware that the current drawn from the cells will be inversely proportional to number of cells in series.

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You could diode OR the cell outputs together. You would then lose 0.7 volts and may need a boost switching regulator. Why don't you use an external supply?

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