Rockbox mail archiveSubject: Re: Charging FMR via hardware
Re: Charging FMR via hardware
From: [IDC]Dragon <idc-dragon_at_gmx.de>
Date: Sun, 28 Sep 2003 12:18:53 +0200 (MEST)
> to charge the battery ? Now both cells are about 3.3V which seems to be
> insufficent to spin up disk, required (don't know why ...) by rockbox's
> charging process, which definitly doesn't work.
> I suppose the charging circuit would be like this, with U being the power
> sypply's voltage, B1 and B2 both battery cells and R a resistor, chosen to
> control intensity :
> --------|(+) U (-)|--------
> | |
> | |
> | (R)
> | |
> | |
> --------|(+) B1 (-)|-------
> | |
> --------|(+) B2 (-)|-------
> Just tell me if that circuit may work, and which values I need for R and
> how long this would be, and if there are some issues I'd have to know
> LiIon batteries. For information, the values written on the cells are :
> Archos - 2 Lithium-ion rechargeable batteries
> 2200 mAh - 4.2 V
Hello again Fred,
(for the others: there's been some private email discussion beween us before
the circuit should be OK. This won't give you a high performance charger,
nor full capacity load, but should serve your purpose.
The charger in the FM has a cutoff voltage of 4.2V, so your U must under no
circumstances be higher. Go a little below, but not too much, maybe 4.1V.
The difficult point with LiIons is that a good part of the charging is done
right before reaching the cutoff voltage. This is why charging circuits have
to be very accurate with this (the LTC1734 claims 1%), else they waste
capacity (cutoff too early) or damage the cell (cutoff too high).
Your simple circuit is safe in that respect, but the higher the cell voltage
rises to U, the less current you get through the resistor. It is asymptotic,
you'll never really reach it. But for your purpose you're probably happy
with a partial charge.
Now about Ohm's law (the resistor R): since your cells now have 3.3V and a
charge current of max. 700 mA is normal, your resistor can go as low as R= U/I
= (4.1V-3.3V)/0.7A = 1.14 Ohm. Its power dissipation at the beginning will
be P= U*I = (4.1V-3.3V)*0.7A = 0.56W, so it should be a little bit bigger than
the standard 1/4W type (or use some in parallel resulting in minimum 1.14
Ohm). Monitor the cell voltage while charging to find out when to stop (it may
take a long time).
If you have a bench power supply with a current limiter, you can as well set
that to 700mA and need no resistor. In that case the current will stay high
until the end, giving you a rapid charge. In this case, I would be more
careful, voltage a bit lower, current maybe 500 mA. Charging will take maybe 3
hours in this case, since you get 500mAh in for each hour.
And remember to "parallelize" the cells again, if you just take it out it's
like two cells side by side, but not connected to each other.
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