I'm trying to know the capacity of each cell of a bank power, the finality is to replace them:

  • 6x3.7V cells, 12.35V fully charged. (tested with multimeter)
  • step-up circuit 12v -> 19v (efficiency: let's say 85%)
  • 3S2P configuration

The battery can work 30 min @ 0.7A and the 19V output.

Is the capacity of each cell can be estimate, or must I know the amperage at 12V ?

  • \$\begingroup\$ Why don't you just read the cell markings to determine the capacity? \$\endgroup\$ – Andy aka Apr 23 '15 at 8:34
  • \$\begingroup\$ There's nothing written. It's Lipo slim cells, 6x60x90mm. I just know this! \$\endgroup\$ – Guillaume Apr 23 '15 at 8:36
  • \$\begingroup\$ How are the cells wired with each other? \$\endgroup\$ – Andy aka Apr 23 '15 at 8:39
  • \$\begingroup\$ 3S2P, means 3xserial(2xparallel) \$\endgroup\$ – Guillaume Apr 23 '15 at 8:41
  • \$\begingroup\$ BTW, it's not "amperage". It's called "current". A lot of people call it amperage, but only because it's a slang term that has been spread. This basic knowledge should help you understand literature on it. \$\endgroup\$ – CL22 Apr 23 '15 at 11:17

Yes, you should be able to get a reasonable estimate of the battery's capacity as it is now - not what it was brand new.

It runs at 19 volts at 0.7 amps for 1/2 hour. So: 19 V x 0.7 amps x 0.5 hour = 6.65 Wh

6.65 Wh at 19 volts after it has been boosted from its 12 volt nominal state. You assume an 85% efficiency which is reasonable which means that you lost 15% in the boost, so 6.65/.85 = 7.82 Wh from the battery pack.

7.82 Wh for a 12 volt lithium 3s battery is really at 11.1 or 10.8 volts nominal. Since we don't know which one, and it doesn't make a big difference either way, I'll use 11.1 volts as the nominal voltage so: 7.82 Wh ÷ 11.1 v = 702 mAh.

Remember that it is a 3S2P so 702 ÷ 2 = 351 mAh per battery.

This seems a little low to me, so perhaps I have made a mistake somewhere, but I don't see where.

A way to confirm is to use the powerbank to charge a battery that you know the capacity of. If it charges that battery to 50% for example, you can then add the 15% loss for the boost to calculate the powerbank's capacity.

  • \$\begingroup\$ It means that the worst cell (or pair of cells in parallel) at the current state of aging is 351mAh/cell. Capacity may once have been higher than this (unless the battery is new!) \$\endgroup\$ – Brian Drummond Apr 23 '15 at 9:56
  • \$\begingroup\$ Yes, this result seems low, but I founded the same result. The battery is pretty new, I would like to change cells with bigger capacity. (to run 5-6h, not 1/2h). Thanks \$\endgroup\$ – Guillaume Apr 23 '15 at 11:43
  • \$\begingroup\$ Now that I think about it, those cells are probably only 351 mAh. They only ran for 1/2 hour at a relatively low rate. Was the power bank relatively inexpensive? Also, because it runs 3 series, as Brian Drummond point out, there may be only 1 parallel pair that small and the other two are larger, but when that pair is empty, the whole series string will drop voltage and read empty even if the other two parallel groups still have capacity. \$\endgroup\$ – Filek Apr 23 '15 at 21:09
  • \$\begingroup\$ By the way, I have taken several of these cheap powerbanks and replaced the batteries with higher capacity cells with no problems. just make sure you solder them back in the correct polarity. \$\endgroup\$ – Filek Apr 23 '15 at 21:12

You need to know the amperage at 12 volts.

If the circuit can output 19 volts at 0.7 amperes, that's 13.3 watts.

If the converter is operating at 85% efficiency, then the battery must supply 15.7 watts and, at 12 volts, that's 1.3 amperes into the converter.

Then, since the battery only has to supply that current for 1/2 hour, its capacity must be at least 0.65 ampere-hours.

Finally, since you have two series strings in parallel, each string has to supply half the current, so the capacity of each cell in the battery must be at least 325 milliampere-hours, assuming its rate of discharge is C/1.

  • \$\begingroup\$ Thanks, I will try to make this, but this means to cut an internal wire to plug the multimeter. The precision should be better I think, because 85% is a 'supposed' value. \$\endgroup\$ – Guillaume Apr 23 '15 at 11:39

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