0
\$\begingroup\$

I'm building a 1200W portable stereo system powered by repurposed laptop Li-Ion cells. In order to get a minimum 8 hour play time at maximum volume, as well as provide some external power via an inverter, I've designed a 14S4p battery with appropriate BMS, with the idea to put 3 in parallel to meet my 25A discharge requirements and provide even more capacity. I know this is going to be heavy but I'm not trying to build a toss-in-the-backpack kind of bluetooth speaker here.

A 14s pack charging at 0.5c is going to take well over a day to charge, no good if I want to use the system again tomorrow. I find my self wondering if it is possible/feasible/advisable to do the following:

Build 3s1p packs, each with their own BMS, and place 5 packs in series for a 55.5-63V output (to feed the amp through a 48V converter), and then parallel 12 of these for a 24Ah capacity. At 0.5c each 3s1p pack would charge in ~6 hours, as massive win over the 14s' 28 hours.

\$\endgroup\$
  • 3
    \$\begingroup\$ Charging at 0.5C will take 2 hours - regardless of how the cells are arranged. \$\endgroup\$ – Kevin White Aug 31 '19 at 1:59
  • \$\begingroup\$ Actually nearer 3 hours as it'll transition into CV mode about 75% of the charging cycle, but Kevin's right. \$\endgroup\$ – Brian Drummond Aug 31 '19 at 10:59
1
\$\begingroup\$

0.5c is relative to the entire pack, assuming the individual cells capacity is well matched, so for your 14s4p battery, your pack capacity would be 4 times that of your average cell capacity, e.g. 3000mAh cells would be a pack capacity of 12000mAh, this is what the 0.5c is based on, Having batteries in series only increases the voltage you have to charge with,

If you put 2 cells in series, the voltage has increased, but has the same capacity, so if you feed in 0.5c amps, both charge at the same time,

If you put 2 cells in parallel, the voltage is the same, but the capacity has increased, so 0.5c is now twice as much, so you feed in double the current, and both charge at ~0.5c

So it will be roughly 2 hours to charge at 0.5c, however if there are cells with large differences in capacity making up the pack, the balancing while charging will increase this time,

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Thank you for your response. I'd like to ask some clarifying questions: Using 2Ah cells, the 14s4p pack's capacity is 8Ah, meaning 0.5c is 4A and the battery will charge in 2 hours? So the 58.8V charge voltage (14*4.2) does nothing to hurt the 14s? I know over voltage can be terrible for a load but is there no limit to the voltage one can put into an appropriate series of batteries? \$\endgroup\$ – Dustin Kope Aug 31 '19 at 11:15
  • \$\begingroup\$ I think I get it: a nV battery with 2Ah capacity charging at nV and 0.5C will take 2 hours to charge regardless of n value? Seems too easy but I guess it makes sense. Thank you! \$\endgroup\$ – Dustin Kope Aug 31 '19 at 11:19
  • \$\begingroup\$ a lithium charger will have 2 limits, a maximum voltage and a maximum current, you set it to 4A, and however many cells so it knows the max voltage, for all but the last bit of charging, the charger will be current limited with a lower voltage. However, you will need a BMS with balancing connected during charging to make up for any difference in cell capacities, lower capacity cells in a series string will reach 100% charge a little earlier than the rest. \$\endgroup\$ – Reroute Aug 31 '19 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.