Hi, I'm driving a 12V load for ~1s second very infrequently, at a draw of about 30A. The load isn't particularly voltage-sensitive. Total mAh isn't that important.

I have a supply of recycled laptop 18650 cells so was looking for a solution using those (I am aware they will be unreliable). I will charge them individually outside of my circuit, so charge balancing is not a requirement. However I don't think they have over-drain protection built-in, so that needs to be considered. Their positive terminals are flat.

After some searching I think I need a 4 cell holder, wired in series ("4S"), stepped-down to 12V somehow.

Searching eBay reveals hits for "4S 30A 12V balance board". Is this the sort of thing I am after?

  • Continuous discharge current: 30A (MAX)
  • Instantaneous discharge current: 70A
  • Power balance detection voltage: 3.60±0.025V
  • Overcurrent detection voltage: 150mV

I looked into using 3 cells with buck-boost converter, but it seems more expensive.


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    \$\begingroup\$ What makes you think you can get 30A out of these cells, even for a short period of time? Do you have a datasheet for them and, if so, can you share it with us? \$\endgroup\$ Commented Oct 7, 2019 at 17:15
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    \$\begingroup\$ Yes, you can scale up current by putting cells in parallel but then you must add circuitry to balance the load. And I have no idea what a "generic low-quality 18650" looks like, if there is such a thing. \$\endgroup\$ Commented Oct 7, 2019 at 17:22
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    \$\begingroup\$ Unprotected laptop cells may not be trustworthy for this application as it's important that lithium ion cells be matched and balanced, but you can order protected cells online for ~$2-$3 per cell. If you use protected cells with an appropriate charger board($3-$50) in the arrangement Sunnyskyguy suggests, 30A should not be a problem. \$\endgroup\$
    – K H
    Commented Oct 7, 2019 at 18:48
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    \$\begingroup\$ There are 18650's capable of delivering 30A for less than a second without too much voltage sag. But they are not utilized for laptops. Power tools would be a better option. Laptop cells are more optimized for higher capacity rather than higher power delivery. I don't think you should try to cobble this together with leftover parts from a junk bin. For 30A, I am pretty sure you absolutely don't want to use a DC-DC converter. At that power level it will be much better to use a battery pack that can drive the load directly. \$\endgroup\$
    – user57037
    Commented Oct 8, 2019 at 3:00
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    \$\begingroup\$ "Searching eBay reveals hits for "4S 30A 12V balance board". Is this the sort of thing I am after?" - for charging perhaps, but unnecessary (and perhaps even counterproductive) for powering your load. The 30A rating is just the maximum current it can handle. \$\endgroup\$ Commented Oct 8, 2019 at 17:26

2 Answers 2


Li-Ion cells designed for use in power tools are capable of that kind of current - easily. Laptop cells usually are NOT designed for that use. Doesn't mean that they can't supply that kind of current, just that the cells might get hot and will most likely have a short lifetime.

However, put several cells in parallel, then put the groups of parallel cells in series and now you have something usable.

Nominal cell voltage is about 3.6V or 3.7V. Three cells in series is about 10.8V, 4 cells in series is about 14.4V. Personally, I'd lean in the direction of using the higher voltage (4 groups in series rather than 3).

If you use 16 cells as four series groups of four cells each, you should have no problem with either getting that amount of current OR short lifetime.

If you really have a lot of cells available, make the parallel groups more than four cells each.

One final note: putting multiple cells in parallel helps if you have some cells that have higher internal resistance than the rest. The load is spread amongst all the cells.


One second at 30 amps, with one volt sag, requires 30 Farads capacitance, if you don't store energy at higher voltages.

You want to store 12v and 30amp, or 360 watt-seconds.

Energy in a capacitor is 0.5 * C * V^2. What if we use a much higher voltage, and use a buck-converter to provide the 12 volts?

Similarly, at 100 volts and 1 Farad, we store 0.5 * 1F * 100 * 100 = 5,000 watt-seconds.

Thus 0.1Farad at 100 volts will suffice. Now you need the 100v-to-12v buck converter.


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