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Registered Member #108
Joined: Thu Feb 09 2006, 11:44PM
Location: Billings, MT
Posts: 61
4hv! This is about as much Computer Science as it is Chemistry perhaps - ha
I have a 3.7v Li-ion battery which I need to charge for a specific electronic device whose charging circuitry is kaput.
There is a whole host of things laying around the house which use 3.7v Li-ion batteries - phones, cameras, etc. However, they all have special circuitry involved in the charging.
I tried charging the battery through the leads on my cell phone but as one might expect it fails responding "Insert genuine battery." Can anyone think of an easy way to force-charge the battery at least part way by bypassing the circuitry in conventional chargers?
Could I just wire up an adapter and feed 3.7v back into the battery without causing a disaster? I know I've found on the internet before that this is not a good method for charging Li-ion batteries.
Thanks
--EDIT-- I finally got a little more google savvy and found information I needed. I have a couple ideas now anyway. Just need to find my multimeter and a 4.2v power supply... will 50ma be too much current? Still open to feedback
Registered Member #27
Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058
Make sure the voltage never goes above 4.2 V and limit the current to about half the capacity of the battery. So if your battery is 700 mAh, try to keep the current below 350 mA. If you limit the voltage a little below 4.2 V there is a small reduction in capacity but you will extend the lifetime of the battery.
If you often charge small lithium batteries, building a charger like this is safer:
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
You don't have to use a 4.2 V power supply (good luck finding one) you can use any voltage greater than 4.2 V and a suitable series (dropper) resistor. e.g. 12 Vdc power supply, 100 mA (0.1 Amp) charging current;
R = V/I = (12 - 4.2) / 0.1 = 78 ohms, use 100 ohms for I = (12 - 4.2) / 100 = 78 mA NOTE : 78 mA x 7.8 V = 600 mW ... use at least 1W resistor.
OR
Use a 12V 60 mA lamp/bulb instead of a resistor. etc.
P.S. If the battery gives problems rip out the electronics and use the cell/battery direct.
Registered Member #108
Joined: Thu Feb 09 2006, 11:44PM
Location: Billings, MT
Posts: 61
Perfect. I'll try throwing something in series with the power supply I am currently using.
Right now I have a variable AC-DC adapter, that varies from 3v - 12v. When I set the adapter to 3v, it puts out 5v.
The adapter is rated at 500ma. That's too much, even at 5v. I suppose I test to see how much current the supply will put out when it's actually under a load? The voltage will drop for sure...
I've never understood why power adapters put out more voltage then that which they read on the case.
Anyways what I have is working (if I set the adapter to 3v, the adapter puts out 5v; then when I connect to the battery, it the voltage stays about 0.1v above the charge read across the cells when disconnected if that make sense) however I am probably charging the cell a little too quickly.
Registered Member #27
Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058
If you don't limit the voltage to 4.2 V it may start a fire if you forget about it (you will sooner or later). So better put it in a fire proof container.
If the protection circuit in the battery still works it will cut out at 4.2 V but it is not a great idea to depend on it.
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3.7v Lithium Ion battery
