0
\$\begingroup\$

I cannot seem to find a solid conclusive answer anywhere on this, but if I build a pack of 6 18650 batteries in parallel which are 3500 mAh each, that would make the pack 21 Ah in total. Then I take those 6 and make 13 more packs and wire those in series so I have a total of 78 batteries making 48v (3.7v * 13). This would make a 48v 21Ah battery.

If each battery has a maximum discharge of 10A. Does that mean the pack has a higher discharge rate? Or does the maximum discharge of the pack stay at 10A?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Under ideal conditions, parallel packs designed for the same voltage will increase allowed discharge rate by the number of packs. In series the discharge rate stays the same but the voltage doubles. This is ignoring any kind of thermal effects of creating larger packs \$\endgroup\$
    – Stonie
    Commented Apr 15, 2019 at 11:30
  • 2
    \$\begingroup\$ In addition to the thermal effects, each cell in a parallel pack will have a slightly different internal resistance. The cell with the lowest IR will be taxed with the most current, especially pulse currents. This means that you should de-rate the current capability of a parallel pack. For a 6P pack of 10A cells, you theoretically get 60A combined, but you should subtract a significant percentage away from that to stay on the safe side. If in doubt, cut it in half. This is provided you use new, identical cells. If you recycle cells, you should already have a big safety factor on top of that. \$\endgroup\$
    – Dampmaskin
    Commented Apr 15, 2019 at 11:38

1 Answer 1

1
\$\begingroup\$

All strings MUST* have the same number of cells.
All cells MUST* be identical (model, date of manufacture, usage)

Recycled cells may be allowed to violate some aspects of MUST after careful capacity testing and with all cells at the same state of charge at time of pack manufacture. Mixing cells of different models and remaining capacity will often enough attract Murphy's attention. This is not usually wise.

*MUST === That thy days may be long on the face of the land and that thou shalt not experience the magic smoke and vent with flame.

______________________________________

N Cells in a series string:

  • N cells in series in a string increase voltage to Vstring = N x Vcell

  • Current calculations in a string are the same as for a single cell.
    eg I_cell_max = I_string_max etc

  • Ah of N cells in a string = Cell_Ah **

  • Wh of N cells in a string = N x cell Wh
    Wh is equivalent to energy so increases with number of cells

K Parallel strings in a battery

  • Battery V = string V

  • Battery currents = K x string current.
    eg I_battery_max = K x I_string_max

  • Ah in a battery - Ah_Battery = Ah_string x K

  • Wh in a battery - Whbattery = Wh_string x K

_____________________________

Back derivable from the above:

Battery of Q cells,
N cells per string (Ncps)
K Strings (Ks)
Q = N x K
Vx = some Vparameter (Vmax, Vmin, Vmean, ...)
Ix = some I parameter

  • Ah_battery = Ah_cell x Kstrings
  • Wh_battery = Wh_cell x Qcells
  • Vx_battery = Vcell x N_cps
  • Ix_battery = Ix_cell x Kstrings

SPECIFIC QUESTIONS:

If I build a pack of 6 18650 batteries in parallel which are 3500 mAh each, that would make the pack 21 Ah in total.

Yes.

Then I take those 6 and make 13 more packs and wire those in series so I have a total of 78 batteries making 48v (3.7v * 13). This would make a 48v 21Ah battery.

Yes.
But, you meant "12 more" and so "13 total".

If each battery has a maximum discharge of 10A. Does that mean the pack has a higher discharge rate? Or does the maximum discharge of the pack stay at 10A?

See above. The discharge rate PER SERIES STRING is the same as per cell, but increases with the number of strings. The fact that the cells are cross connected physically makes the string count a little less obvious, but you functionally have 6 strings, so total allowed current = 10A x 6 = 60A.


E&OE


** Edited - prior statement should have related to Wh.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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