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battery packs cell shorting?

Hi,

I made a 14.8V (4s) lithium ion battery pack (18650 cells 4s3p) and after gluing the batteries together into the pack, I used a multimeter to check if any of the cases were shorted anywhere (some of the thin plastic covering on the batteries was removed). Before conneting the batteries in series, I noticed that the third and fourth section of the battery (11.1V and 14.8V) had a measureable 0.1V and 0.5V voltage across the terminals of the batteries.

gnd-3.7 (3p 19650 batteries)

3.7-7.4 (3p 19650 batteries)

7.4-11.1(a) (3p 19650 batteries)

(b)11.1-14.8(c) (3p 19650 batteries)

The four 3p packs above are glued together but not electrically connected in series yet, but when measuring the voltage from a to b, the multimeter shows 0.1V and from a to c shows 0.5V.

All four of the 3p battery packs are individually at 3.8V.

Mystery!? :)

I haven't tried to short (through a load) the 0.5V to see if it is a low impedance short, but the main thing I was curious about is where the 0.5V can come from. I am guessing that if I put a 220ohm resistor across it the multimeter would read zero, but still 0.5V seems high!

cheers, Jamie

Reply to
Jamie M
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on 2/01/2015, Jamie M supposed :

Jamie, Please try and use the correct terms. SHORT means put a piece of wire between the terminals.

CONNECT is the way to connect a resistor or other device between the terminals.

Using words so wrong leaves us poor people without a clue and it may leave you without an eye.

SHORTing a battery can easily result in an explosion. :-?

--
John G Sydney.
Reply to
John G

Protected cells or unprotected?

I use cellophane packing tape or shrink tubing.

That should be 18650 cells.

Not really, In order to have a voltage across something that isn't connected to the meter, you have to have a leakage current somewhere. Conductive glue perhaps? Your hand on the volts guesser probe providing a conduction path to the battery? Cells "insulated" with conductive aluminum coated mylar tape? Difficult to tell what you're doing. Photos?

Put about 10K ohms across the volts guesser leads to swamp out any leakage path. If it then reads zero, you probably have a leakage path, somewhere.

Hint: Think about a balance charger for whatever it is you're building. Something like this:

Drivel: Don't assume that your 18650 cells run at the claimed capacity. I tested some 3000 ma-hr Ultrafire cells and measured about

850 ma-hr at 1.3A. The 1.3A is higher than the normal test current of 0.2 * ma-hr rating or 600 ma but that's what the flashlight under test was normal drawing. 
--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558
Reply to
Jeff Liebermann

no idea what's going on in this post, but one cheap 'n good way to make battery packs is by "bottle-shrinking" them with pop or water bottles. Youtube should have videos of this. PET shrinks good and is quite tough, without leaving a sticky mess like tape and glue.

Reply to
Cydrome Leader

Hi,

I put a series 10K resistor with the multimter and still measured 0.5V, I did a current test with the resistor and there is 0.6uA of leakage current shown on the multimeter. I think the glue I used must be slightly conductive. I think it is safe to leave with that small leakage current, eventually the batteries will discharge unevenly but I think the leakage current might be lower than the self discharge rate anyway.

The batteries are CGR18650CF high drain types, I already have that same balance charger you linked to.

I tried the ultrafire batteries before, but I think I got quite a bit less than the 850mAHr you measured.

cheers, Jamie

Reply to
Jamie M

Oops meant to say I did a current test *without* the resistor.

Reply to
Jamie M

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