MrAl
Flashlight Enthusiast
Hi there,
For people still waiting for a charger, here's a procedure
you can use to charge an Li-ion cell if you have some resistors
laying around and a dc wall wart. If you plan to charge at a
high rate you have to make sure the resistors are capable of
handling the power.
The basic idea is to connect the cell in series with a resistor
and dc wall wart and monitor the voltage across the cell. Anytime
the voltage increases above 4.15 volts you increase the value of the
series resistor to twice it's previous value. It takes time
and patience but it works! I charged my first Li-ion cell this
way.
Let's say you have a 6v power supply and a depleted AAA size Li-ion cell
and you wish to charge at a max current of 0.5C .
For finding the initial resistor value for a given type
cell, make sure the cell is depleted to at least as low as
3 volts.
Starting with a series resistor of 20 ohms, 0.5 watt, in series with the cell
apply power and connect the cell. Check current (by measuring voltage across
the 20 ohm resistor and using Ohm's Law, or using a separate meter). Make
sure the current is about 150ma.
With a voltmeter connected directly across the cell's terminals, check
to see that the voltage is less then 4.15 volts.
If at any point you see current over 150ma or voltage over 4.15 volts then
simply INCREASE the value of the resistor. The current through the cell
AND the voltage across the cell will go down when you do this, and this
is normal.
Once you get the initial current set (with the 20 ohm resistor) you'll be
contantly monitoring the voltage and increasing the resistance as the voltage
gets up to 4.16 volts. You'll never allow the voltage to reach 4.19v, that's
the rule. As soon as it reaches 4.16 you'll disconnect and increase the
resistance to either 1.5 times it's current amount or 2.0 times (if you're
not in a hurry).
Here's how a typical run might go:
Connect a 10 ohm resistor across the cell to make sure it goes down to
3.0 volts. Once it does, proceed. (Skip this step once you've determined
the correct value resistor to start with for this kind of cell).
Let's say we got 20 ohms to start with (as probably with a AAA cell).
Now we connect the voltmeter across the cell and the current meter
in series with the cell, turn on the power supply and connect the
cell.
Now let's say we see 160ma flowing and the voltage is 3.2 volts.
The current is just about right and the voltage is under 4.15 volts
so we allow it to charge for some time.
A little later (maybe 1 hour) we see the voltage rose to 3.95 volts and
the current is 100ma (typical). The current is still ok (unless we are
in a hurry) and the voltage is still under 4.15v so we dont do anything.
A little more later (maybe another hour) we see the voltage rose to
4.15v and the current is less than before. Now the voltage is on the
threshold of 4.15v so we get ready to disconnect. A few minutes later
the voltage rises to 4.16v so we disconnect the cell.
Adding another 20 ohms in series with the 20 ohms that is already
there, we then reconnect the cell. Monitoring the voltage, we
see it's less than 4.15v again so we let it charge a little more.
The current probably be around 50ma so it's ok.
Now sometime later the voltage again goes up to 4.16v so we swap
out the resistor again, this time for a total resistance of 80
ohms, then reconnect the cell and monitor voltage again.
We keep repeating this procedure until the current reaches about
25ma and the cell reaches 4.16v again and we're done.
Disconnect everything and you're ready to use the cell in your application.
Have fun,
Al
For people still waiting for a charger, here's a procedure
you can use to charge an Li-ion cell if you have some resistors
laying around and a dc wall wart. If you plan to charge at a
high rate you have to make sure the resistors are capable of
handling the power.
The basic idea is to connect the cell in series with a resistor
and dc wall wart and monitor the voltage across the cell. Anytime
the voltage increases above 4.15 volts you increase the value of the
series resistor to twice it's previous value. It takes time
and patience but it works! I charged my first Li-ion cell this
way.
Let's say you have a 6v power supply and a depleted AAA size Li-ion cell
and you wish to charge at a max current of 0.5C .
For finding the initial resistor value for a given type
cell, make sure the cell is depleted to at least as low as
3 volts.
Starting with a series resistor of 20 ohms, 0.5 watt, in series with the cell
apply power and connect the cell. Check current (by measuring voltage across
the 20 ohm resistor and using Ohm's Law, or using a separate meter). Make
sure the current is about 150ma.
With a voltmeter connected directly across the cell's terminals, check
to see that the voltage is less then 4.15 volts.
If at any point you see current over 150ma or voltage over 4.15 volts then
simply INCREASE the value of the resistor. The current through the cell
AND the voltage across the cell will go down when you do this, and this
is normal.
Once you get the initial current set (with the 20 ohm resistor) you'll be
contantly monitoring the voltage and increasing the resistance as the voltage
gets up to 4.16 volts. You'll never allow the voltage to reach 4.19v, that's
the rule. As soon as it reaches 4.16 you'll disconnect and increase the
resistance to either 1.5 times it's current amount or 2.0 times (if you're
not in a hurry).
Here's how a typical run might go:
Connect a 10 ohm resistor across the cell to make sure it goes down to
3.0 volts. Once it does, proceed. (Skip this step once you've determined
the correct value resistor to start with for this kind of cell).
Let's say we got 20 ohms to start with (as probably with a AAA cell).
Now we connect the voltmeter across the cell and the current meter
in series with the cell, turn on the power supply and connect the
cell.
Now let's say we see 160ma flowing and the voltage is 3.2 volts.
The current is just about right and the voltage is under 4.15 volts
so we allow it to charge for some time.
A little later (maybe 1 hour) we see the voltage rose to 3.95 volts and
the current is 100ma (typical). The current is still ok (unless we are
in a hurry) and the voltage is still under 4.15v so we dont do anything.
A little more later (maybe another hour) we see the voltage rose to
4.15v and the current is less than before. Now the voltage is on the
threshold of 4.15v so we get ready to disconnect. A few minutes later
the voltage rises to 4.16v so we disconnect the cell.
Adding another 20 ohms in series with the 20 ohms that is already
there, we then reconnect the cell. Monitoring the voltage, we
see it's less than 4.15v again so we let it charge a little more.
The current probably be around 50ma so it's ok.
Now sometime later the voltage again goes up to 4.16v so we swap
out the resistor again, this time for a total resistance of 80
ohms, then reconnect the cell and monitor voltage again.
We keep repeating this procedure until the current reaches about
25ma and the cell reaches 4.16v again and we're done.
Disconnect everything and you're ready to use the cell in your application.
Have fun,
Al
