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I have a battery pack comprised of 2 6S 8000mAh LiPo batteries (link) with each one connected to an individual BMS module (link). I'm trying to charge the pack using a 25.5V dc power source as specified by the BMS manufacturer, using a 500W power supply I had which is fairly similar to this one. The BMS are connected parallel to each other.

Technically, the system works. However, the charge rate is very, very slow - to the tune of plugging it in at 22.2V, and 12 hours later it is only at 22.8V. What I am wondering is if there is something I am doing wrong, or perhaps something I need to add to my system to make it charge properly.

A few things that may be relevant:

  • I have added a few 220uf caps parallel to the output voltage of the power supply to reduce ripple
  • The power supply's output voltage sags down to the voltage of the pack while charging it.
  • While testing, I added a small-value resistor in series with the BMS in order to measure the current across it (my meter only goes up to 10A) - the pack was drawing about 16A, however this also caused the voltage supplied to the BMS to no longer sag (is this a good thing? Should I be charging with such a resistor installed?)

Any assistance getting this pack charging at more reasonable rates is much appreciated.

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  • \$\begingroup\$ IF 16A is going into the actual battery cells then they should charge to most of full in 8Ah/16A = 0./5 hours. If they are charging very slowly summat is very very very very wrong. IF you are charging at 16A and the current is going into the BMS and being dissipated then summat is very very wrong AND the about 400 Watts dissipated shoukd be making things melt and/or glow in the dark or get REALLY hot. You need to determine which of these bad/sstrange things is happening.' \$\endgroup\$ – Russell McMahon Nov 2 '19 at 16:16
  • \$\begingroup\$ Update: while waiting for additional components to arrive in the mail, I disconnected one of the batteries' BMS from the power source. While charging, it was drawing around 20A at 24-ish volts. I'm thinking that the power supply is simply running into its current limit and going open-circuit to prevent that. Is this a reasonable guess? \$\endgroup\$ – Jordicious Nov 3 '19 at 0:51
  • \$\begingroup\$ If the pack is charging at 16A and the BMS is drawing 20A you have a perpetuum mobile or are discharging other cells into the BMS as well as the charge current. || The best conclusion is that something is very damaged and that some intelligent troubleshooting will get you further quicker than Edison style exhaustively trying everything. What are the individual cell voltages. What does the battery pack do with no BMS connection. What do YOU mean by disconnected the BMS. Try not to say "it" etc when the meaning is potentially* ambiguous. ... \$\endgroup\$ – Russell McMahon Nov 3 '19 at 2:40
  • \$\begingroup\$ ... You could charge at a lower rate so your meter is useful. You could .... . Plan of attack is in order. || Theory -> model -> test aiming to falsify -> conclusions -> adjust 'model" -> repeat until working. || AT present you have obviously really really really wrong things happening and do not seem to be phased enough to track them down. eg 16A in 20A out to waste slow slow charge ... . That is as sick as si k. Why> || The BMS wires (or BMS PCBA connection points) will give you access to cell voltages without dismantling the pack. \$\endgroup\$ – Russell McMahon Nov 3 '19 at 2:42
  • \$\begingroup\$ Apologies for the ambiguity - with one BMS/pack combo disconnected, the remaining BMS/pack was drawing the ~20A. This current quickly started to decrease as the pack charged up, though - I only left it on for a moment or two as these packs aren't meant to charge that fast. \$\endgroup\$ – Jordicious Nov 4 '19 at 7:38
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The power supply's output voltage sags down to the voltage of the pack while charging it.

If the power supply and battery are directly connected then their voltages must be the same.

Assuming you had the power supply set to 25.2V (6x4.2V), imagine what would happen if it forced that voltage into the battery. Battery voltage 22.2V, battery internal resistance 7.8mΩ, current = (25.2V-22.2V)/0.0078Ω = 384 Amps! The power supply obviously can't put out that much current, so it 'fold-back' limits to a very low current until the 'short' is removed.

Charging through a resistor reduces the current to an amount the supply can put out, so it doesn't go into fold-back limiting. The voltage difference between the power supply and battery is dropped across the resistor. As the battery charges the voltage difference gets less so the charging current reduces. It will work, but the tapering charge current will cause it to take a long time to fully charge.

Be aware that a battery this size is potentially extremely dangerous. Your BMS should prevent it being overcharged and limit current in a short circuit, but doesn't balance the cells. If one or more cells have anomalous voltages then they or other cells could get overcharged while the total voltage is still within spec. If you don't monitor the cell voltages the first you will know about it is when the pack explodes.

The proper way to charge this battery is with a suitably rated charger that has balancing and cell monitoring. They are expensive, but worth the money for battery health, safety, and convenience.

You can continue charging your battery with your power supply and resistor, but you should regularly meaure the individual cell voltages with a meter or cell monitor/balancer to ensure that they are equal (within 0.03V). Don't charge the pack unattended, keep it away from anything flammable, and have a fire extinguisher or bucket of water (to put the battery in) handy in case the unthinkable happens. A fire retardant Lipo bag is also recommended.

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  • \$\begingroup\$ The BMS I currently have is balancing the cells. I've checked them several times during the charge to make sure they're all equal. Out of curiosity, I added an ammeter to the charger and curiously, the current is mostly zero, but increases to an amp or so every now and then. The resistor I previously added has been removed. To me, this sounds like the BMS it's limiting the current - is that a valid guess? Should I replace the cheap BMS with something more high quality? \$\endgroup\$ – Jordicious Oct 23 '19 at 19:28
  • \$\begingroup\$ If I read it correctly the BMS in your link does not do balancing. If it actually does then that's great. The BMS will probably disconnect the battery if current goes too high, then reset and reconnect it after a short period if external voltage is applied. This may be what you are seeing. Either way you need a resistor (or proper charging circuit) to limit the current. The 'cheap' BMS is working fine, the problem is the unlimited charging current! \$\endgroup\$ – Bruce Abbott Oct 23 '19 at 20:23
  • \$\begingroup\$ That would explain a lot of what I'm seeing here. What value of resistor would you suggest to limit the current enough for it not to do that? \$\endgroup\$ – Jordicious Oct 24 '19 at 14:12
  • \$\begingroup\$ BMS is rated for 15A, discharged battery will be ~18V (3.0V per cell), full charge is 25.2V so maximum voltage difference is 7.2V. 7.2V/15A = 0.48 Ohms per pack. However the power supply is only rated for 20A total (500W/25.2V) and 1C charge rate for a 8000mAh battery is 8A, so 7.2/8 = 0.9 Ohms per pack. Nearest standard value is 1 Ohm, producing 7.2A. Maximum resistor power dissipation is 7.2V*7.2A = 52W so you should use a 100W resistor. \$\endgroup\$ – Bruce Abbott Oct 24 '19 at 19:26

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