Only 23 mA


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Masterno

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May 5, 2022
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Hello,
I just discovered this page, and i'am a newbi.

I want to charge an 1850 cell (used).
I use a bulk converter with a diode set to 4.2V.

Without a battery, i could reach 1 A, but when plugin the cell, i only have 23 mA.

Someone had an idea ?

Thanks !
 

Korishan

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Welcome to the forum

It would help if we had more details about your setup. But I'm guessing the cell has high IR or a busted CID. Or there's something else wrong
 

Masterno

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May 5, 2022
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Thanks for you answer.

My bulk converter is a LM2596, its just that. It was set to 4.2v.
The cell's voltage is at 3.66 v
I measure my IR with a 1 ohm ressistor and i think i's 200 miliohm
 

ajw22

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bulk converter is a LM2596
Such _buck_ converter modules usually consume 10~30mA even when nothing is attached to the output. So there's a good chance the cell isn't getting charged at all.

with a diode
Every diode has a so called "forward voltage" that lowers the voltage a bit. Depending on the diode type, it's typically ~0.7v for silicon diodes (~0.2v for Schottky type). So even if the buck converter is set to 4.2V, it could be dropping down to 3.5V if behind a silicon diode... and that would not charge a 3.66V cell.
Or the diode could be connected the wrong way :)

Having said that, connecting a buck converter directly to a lithium cell is generally speaking a very very bad idea, because a LOT of current could flow when feeding 4.2V into an empty (3.0V) cell, damaging the cell. So make sure to monitor the current very closely, or better yet, get dedicated charging modules - they're about as cheap as buck converter modules.
 

Redpacket

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Like ajw22 says, it would be better to change your converter design/type to a current limiting type or type designed as a charger to deliver a controlled current to the cell. Eg max 1 amp. Cells will be happier with slower charging, eg 100-200mA.
Many charger type units will also taper/ramp down current & some will also switch off when the cell is full.

For longer cell life, consider charging your cells to about 4.1V vs 4.2V.
 

Masterno

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May 5, 2022
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Thanks for all your answer. The idea is for educational purpose. This is not intended to not be monitored.

The 4,2 v was set with the diode to cover the voltage drop. Then i connected a 4,2v to the cells
 

Redpacket

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So did you have:
converter (4.9V) > diode (-0.7V) > 18650 cell (4.2V)
or
converter (4.2V) > diode (-0.7V) > 18650 cell (3.5V)
 

ajw22

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Nov 16, 2018
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With the numbers you cited, the theoretical charge rate would be around 3A - Pushing the LM2596 very hard, perhaps tripping the overcurrent/overheat protection. Or perhaps tripping another protection device somewhere (power supply?).
I'd try with a lower voltage, perhaps 3.8V, then ramp up.

A picture of the setup might be helpful in spotting issues.
 

ajw22

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Supply voltage is 4.2V. Subtracting the existing battery voltage of 3.66V results in voltage differential of 0.54V. You mentioned the cell has around 0.2 Ohm resistance. Then it's just basic calculation using the last 2 values, I like using this helper:

It's bit of a ballpark figure, due to weird properties of batteries, buck module operating at the limits, variable nature of forwardV-vs-current of the diode etc.
 

Redpacket

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Like ajw22 says, most LM2596 converter circuits do not limit current except at high max 3A.
This can push high current into the cell which is not good.
Can you current limit the source (4.9V) supply?
Or at least add a series resistor, eg 1 ohm.

Maybe you should change to using a converter/charger using a TP5100 chip instead. It does all this much better.
You don't need a diode with this one.
 

Masterno

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Thanks all of you.
@ajw22 Thanks, knowing the ohm laws is one thing, figure how to apply in a field i begin to learn is anotherthing. Thanks again
@Redpacket i tried to limit the current with a potentiometer or a resistor and i have the same result.

I have a tp4056 on the way. I will check if i could get more current. thanks again !!
 

Korishan

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Some are fake but good luck.
Not fake, clones. The function mostly as the original. But they may not be quite as accurate. Instead of 1%, they may be closer to 2% variance on results. Or get much hotter during charge.
I kind of wonder if the "clones" are actually the original ones that didn't meet quality specs during random batch tests.
 

Redpacket

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clones. The function mostly as the original. But they may not be quite as accurate. Instead of 1%, they may be closer to 2% variance on results. Or get much hotter during charge.
Exactly, that's why I suggested the TP5100 instead, there's less clones of that & it doesn't get as hot (switching vs linear circuit).
 
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