Electronics > Projects, Designs, and Technical Stuff
Charging a Lifepo4 with an Lm2596 module
Tj80:
Yes, it's 3.6v for the LiFePO4 cell (I'm using one of these instead of a Li-ion battery because I want to run the Wemos D1 Mini without a power-sapping voltage converter). Agree on reducing the charge voltage, I was going to go for 3.5v but used 3.6 to get a higher charge rate while testing - interesting you suggest 3.45v, I may give that a try.
My next option is to use a simple relay to completely disconnect the battery from the LM2596 module unless the module itself is powered, but that feels like a horribly clunky solution! It will also consume a bit more power, but at least it won't discharge the battery.
Many thanks,
Tim
magic:
It doesn't work because it takes some voltage drop across the diode to force current through it. See "typical forward characteristic" in the datasheet, it's gonna be 0.6~0.8V for reasonable currents. And no, you can't increase LM2596 output by 0.6V because it will also be 0.4V for slightly lower current, which my still end up slowly overcharging the battery by 0.2V.
In theory you could take feedback to the LM2596 from behind the diode, maybe, but then the feedback divider would be discharging the battery so it's pointless.
Some kind of "ideal diode" circuit would help, if you can find one that has negligible leakage when off. I'm not familiar with those things.
A relay doesn't even look that bad at this point.
Tj80:
Interesting. Many thanks - relay it is then!
Cheers,
Tim
Doctorandus_P:
The voltage drop over the diode is the most likely cause of your troubles.
The voltage to current curve is actually exponential, but for practical reasons it's usually taken as a constant 600mV voltage drop in forward direction (for silicon diodes) and some 300mV for Scottky diodes.
You can make an "ideal" diode with a mosFET. MosFet's do not have an inherent voltage drop, they have a (relatively constant) resistance when "on". The "on" resistance of a Mosfet can easily be as low as 10mOhm, even for small mosfets (SOT023 housing).
But in your case a simple mechanical relay seems a good solution.
The relay is then directly powered from your external power source, so the current drain won't be a big problem.
Also, think about any electronic component connected to the battery in your circuit.
Current leakage through an electrolytic capacitor can be 100uA, and can be bigger than the current of your uC in deep sleep.
bin_liu:
I'm wrong.
mistake it is Li-ion. ;D
Lifepo4 OCV is 3.65V.after stopping charging.voltage will drop.is ~3.4V/CELL.
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