Can one Li polymer 18V 3100mAH charge another? If so, contact to contact ,or induction? Drawbacks? How long to charge?
Thanks for your help
Can one Li polymer 18V 3100mAH charge another? If so, contact to contact ,or induction? Drawbacks? How long to charge?
Thanks for your help
Hi Randall, These batteries have some strict charging protocols.While connecting 2 batteries together will cause some energy to move from the charged one into the uncharged one this is not good way of doing it and may damage the batteries. If you have 2 batteries why not swap the charged one for the uncharged one and then put the uncharged one on the correct charging circuit? I see in the stores that they have aux battery boosters for cell phones. These work as the power from the extra battery is higher voltage and is applied to the charging circuit of the phone not directly to the battery.
John
John,
Thanks for your reply. Unfortunately switching the batteries out is not an option. If we assume all the charging control circuits are in place, using contact to contact charging only, can it be done? If so, would the charging battery need to be 2X the receiving battery as I have seen in other discussions? Again, thanks for your interest.
Randal
Hi Randal, Yes if you are supplying votage to the charging circuit of a battery you will need a higher voltage than the battery in order to charge it. For example: if we could charge one battery with another battery of equal voltage and one was fully charged and the other is at its minimun the best we could hope for would be 2 batteries with half a charge. I suspect that 2 times the voltage might be a little more than needed depending on the design of the battery and its charger but yes that is the right idea.
John
Thanks John
Thanks John
Randal, what are you working on? what type of battery, what type of chargeing system, ect.
Think about amp hours, i have a Lio batt, with 1.5 Ah capacity.
If it was drained, and i was to connect in parallel, another identical battery that is fully charged, Some of the charge of the charged battery will charge up the dead battery.
But most battery's are not designed to charge this way. Now, assuming that the battery's would not destruct, the best you could hope for would be a 50% charge of the dead battery, and a 50% drain of the charged battery, however.
In reality, that would not be the case, because you will loose some in heat, so it would be closer to 1\3 would be given off in the battery's getting hot., thus, 66% left, divided into 2 battery's, leaves us at around 33% charge at both battery's.
Problem number 2 is battery's are not meant to be charged this way. They use other charging circuits to charge them, because there life would be dramatically reduced.
Problem number 3, is, it would be relay hard to achieve a full charge anyways. Because no matter what the charge of the Dead battery is, the best you could hope for is half the difference of the two battery's.
Example. :
0%+1/2 of 100%=50%...
50%+1/4 of 100%=75%....
75%+1/8 of 100%=87.5%...
87.5%+1/16 of 100%=93.75%...
93.75%+ 1/32 of 100%=96.875%
Now, that is 5 times that you use a fully charged battery to charge a dead battery, just to get over 95% charge.
And that is assuming no loss in heat, or anything else.
So, it is not recommended to charge a battery that way. That is why we are not suppose to mix dead battery's, and new battery's in a flashlight. if you use it for any time, the battery's may leak, explode, short out, and if nothing else, go dead very quickly.
If i have miss understood your question, i am sorry. But i hope this helps.
Hi Lupe, very nice synopsis.I liked that you brought the reality of energy lose due to heating into the discussion too.
John
Hi,
Don't forget that injecting or sinking uncontrolled current in these kind of batteries can be very dangerous, so don't connect them directly together.
If swapping the battery is not an option, why don't you use ideal diode circuits that are used to switch from a supply to an other (for redundant supply for example ?)
Guillaume.
Thanks much for your help Lupe. The information was very helpful.