Hi fellow makers and engineers,

I'm trying to charge a 6S Li-ion battery pack (6 NCR18650B cells in series) using a rectified DC voltage of 16 Volts and boosting it to about 25.2V to charge the pack using a DC to DC converter stated HERE. Currently I'm getting a very low charging current to the pack (about 20mA) and I would like to understand why and how I could fix it so as to achieve a higher current. I know it is common practice to use some kind of PCM to protect the pack from over charging, etc., but I am trying to follow a simple approach to charging the pack and I'm monitoring my cells to disconnect power manually when max limits of cells are reached. Essentially I'm trying to achieve what is described in this lecture by battery university.

Is my approach correct or is a more sophisticated approach like a CC-CV algorithm crucial? Is the algorithm necessary to achieve a proper charging current for the pack? Any help is appreciated.

  • \$\begingroup\$ A simple approach may indeed be a stupid approach when it comes to lithium cells. Read the battery data sheet and follow the guidelines. On another note, I've never heard of a "battery university". \$\endgroup\$ – Andy aka Jun 14 '16 at 7:49
  • \$\begingroup\$ @Andyaka Battery University \$\endgroup\$ – Russell McMahon Jun 14 '16 at 10:46
  • \$\begingroup\$ Notes: 25.2V is insufficient for a good charging. You need higher voltage than that (even 30V is good). The 25.2V should be the stop-charge disconnect limit measured when not connected to charger. Also, this assumes that all 6 cell are in perfect shape/condition, which is rarely the case. If 1 or more cells is not good, charging in series would cause some of the cells to overcharge and some to be insufficiently charged. That's why recent notebook batteries die so fast - because the charge pairs in series instead of individual pairs. inevitably, at least 1 pair gets to be way undercharged. \$\endgroup\$ – Overmind Jun 14 '16 at 12:17
  • \$\begingroup\$ @Overmind Your comments about balancing are valid and worth noting. | Your comment about needing more than 25.2V is incorrect. While a charger MAY have a higher Voc and be V limited to the proper levl on charge (as you say) this is not essential and risks bad things happening if things go wrong. (Things go wrong regardless :-).). As long as the charger can deliver 25.2V AT the battery terminals at desired may current (here ~= 1650 mA) then it is adequate for LiIon. [[LiFePO4 cells can be charged briefly to above Vmax nominal but that is a different situation]]. . \$\endgroup\$ – Russell McMahon Jun 15 '16 at 3:13
  • \$\begingroup\$ @Joel - Overmind's comments about balancing are worth noting. With 6 cells in series it is possible for some to reach full voltage (4.2V) before others do and for cell imbalance to occur with possibly disaterous consequences. This is much less of an issue if you stop at the end of the CC mode so that no cell is at full charge. This discussion can follow once you have got more current flowing. \$\endgroup\$ – Russell McMahon Jun 15 '16 at 3:16

You are on the right track but it is unclear from what you say what you are actually doing. The BU paper says terminate at 3% of C/1 rate - which is far too low for sensible use. You say you are trying to copy their approach BUT also that you are getting 20 mA etc.

You need to say WHEN you are getting 20 mA - when cells are discharged or fully charged or ...?

It helps to give battery specs AND a link. Panasonic NCR18650B datasheet here.

If the cells are genuine Panasonic then they are nominal 3250 mAh capacity
and are specified to be charged at C/2 max = 1625 mA.

If the cells are less than fully charged (say Vpack < 6 x 4 = 24V) then applying a source able to charge at up to 1625 mA at 6 x 4.2 = 25.2V WILL produce a charge current of 1625 mA except in one case. If any cell has been discharged to a dangerously low level the internal protection circuit may limit current to a trickle up rate - often Cmax/10 = 160 mA here. This will jhopefully not be the case here.

Your cited power supply has the makings of what you need but needs to be set up correctly. If the Iamx pot is set to minimum it may limit current to the 20 mA you are seeing. So ...

  • Measure voltage of all individual cells. If any cell is under 3V report back. If under 2.5V report back carefully :-).

  • Disconnect battery from charger.

  • Adjust PSU current limit pot to some mid range point
  • Power up PSU and adjust Vout to 4.2 x 6 = 25.2v.
  • Short psu into an ammeter (via a low value series resistor optional).
  • Adjust Isc to say 1600 mA.
  • Apply 25.2V, 1600 mA max psu to battery pack and measure current.

Note that some ammeters have quite high resistance on mA ranges (I have seen a 17 Ohm milliameter!). 10A range is usually low Ohms.
A good method is to add a series eg 0.1 Ohm or 0.01 Ohm series resistor and measure the voltage across it.
A 0.1 Ohm will drop V = IR = 1 x 0.1 = 0.1V at 1 Amp and a 0.01 Ohm resistor will drop 0.01 V at 1 A.

As Diego says, stoppingh charge when Vbattery first rises to 4.2V/cell will only provide about 80% of full capacity BUT the cycle life may be doubled if you always stop there and 80% capacity may be fine.

ON NO ACCOUNT should you leave the voltage source connected to the battery indefinitely. A LiIon battery will have its cycle life badly reduced by not removing Vchg when Ichg has dropped to a small fraction of initial value. Imax/10 is usually fairly aggressive. Imax/4 or Imax/2 provide most of max capacity and are much kinder to the battery.

  • \$\begingroup\$ Thanks for the thorough answer Russell. The cells are NCR18650 and are genuine because I've tested their capacity. I'm getting 20mA when charging them. The cells are discharged to about 2.9Volts although some of them may have a bit higher voltage like 3V or 3.1V. I'm kind of confused why such a low current. Also, I'm connecting the boost converter output voltage (25.2V) directly to the battery pack. Should I be charging at a higher voltage or raising the voltage while it charges to satisfy CC mode? Also, I don't need to charge the batteries at 1C and I could settle for 0.2 or 0.1C \$\endgroup\$ – Joel Pou Jun 14 '16 at 19:59
  • \$\begingroup\$ @JoelPou Did you follow my instructions re setting up the power supply current limit? If you do not do this there is no certainty what the supply will do. | The data sheet (link was supplied) says that these are designed for a max charge rate of C/2 or about 1600 mA. Many cells are specified to charge at Imax = C/1 but for what ever reason, these are not. | Also - you can test that the supply will do what it needs to as follows. At 25.2V to get 1600 mA you have an effective load of R = V/I = 25.2/1.6A 1= 16 Ohms. So a 16 Ohm load should have 25.2V across it. Power is ... \$\endgroup\$ – Russell McMahon Jun 15 '16 at 3:01
  • \$\begingroup\$ @JoelPou W = V x I = 25.2 x 1.6 = 40 Watts so you'd need a rather solid resistor. Instead, using 2 x say 15 Watt auto bulbs in series would give 12.6V / bulb so they'd run at maybe slightly under rated wattage (usually designed for 13.xV but who can be sure. 30 W (2 x 15W) is a good enough test to start. The psu MAY have trouble starting these as Roff is very low and rises as they heat. Adding as large a capacitor as you have available across the capacitor, starting supply and then adding load may help. If the supply will not operate at 25.2V at this sort of load it needs addressing. ... \$\endgroup\$ – Russell McMahon Jun 15 '16 at 3:07
  • \$\begingroup\$ If you have 12v LED modules or 12V LED strips, 2 of these in series should make an ~= 24V load. | Turning PSU voltage down, adding bulbs and then ramping V up over 1+ seconds should allow bulbs to be started. | Again - DID YOU SET CURRENT LIMIT AS ABOVE? \$\endgroup\$ – Russell McMahon Jun 15 '16 at 3:09

Regarding the manual disconnect versus the CC-CV algorythm, the problem in interrupting the charge when the voltage reaches 25.2V is that your battery will miss the CV phase of the charging cycle and therefore it will not be fully charged (let's say 80% charge, as a rule of thumb).

To have your battery fully charged, it should reach 25.2V at zero charging current: if you manually interrupt at 25.2V, you are charging at that voltage minus the voltage drop on the internal resistance of your cell (which depends on the charging current value, too).

  • \$\begingroup\$ LiIon will never taper to 0% and should not be allowed to get anywhere near 10% of Imaxchg. Even 10% of Imax gives a very well charged cell at Vmax (say 4.2V) and is hard on cell life. Stopping at 25% or 50% of Imax will get most of the available capacity with useful gains in cycle life. Battery University say 3% of Imax in their paper cited above - this is very low and I do mnot know how they chose that figure. \$\endgroup\$ – Russell McMahon Jun 14 '16 at 10:50
  • \$\begingroup\$ I'm ok with 80% charge for now and I'm planning to add a PCM to properly charge them on the long run. I would just like to raise the charging current entering the batteries from my boost converter \$\endgroup\$ – Joel Pou Jun 14 '16 at 20:08

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