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Anditechnovire

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Please I have some couple of 3.6v 3Ah cylindrical li-ion battery, but since I bought them, charging it became a huge problem. I first connected a single one to a 2A infinix phone charger (made in china. According to the ratings, it capable of adjusting output current and voltage to load)thinking that if it's capable of charging a 4Ah infinix phone battery then a 3Ah shouldn't be a problem.
Until I connected the battery directly to the charger output, after some time, i measured the output voltage of both the battery and the charger, the battery was not charging and the charger decreased it output voltage from 5v to 2v and was also giving me an unusual jolt in my finger. I concluded that the charger has damage. I tried it again with a similar charger, but this time a TECNO phone charger, but the same result.
I then found another SMPS charger 12v, 1.2A. In my desperation, I ignorantly implemented a '7805' regulator ic to drop the voltage to 5v(since using a smps to regulate a smps sounds absurd). When I connected the battery, I measured the charging current, and I was so surpries to see 2A.From a 1.2A(short circuit current) SMPS.
So please where did I went wrong and how can I charge my batteries before it damage.
 
Phone "chargers" with USB sockets or cables are simply 5V power units.
The actual charge controller is in the phone (or other device) alongside the battery.

A lithium cell needs an accurately regulated voltage, usually 4.2V, plus a current limit circuit to prevent it drawing too much current when flat.
Without voltage regulation, they can be destroyed by over-charge - often bursting or even catching fire.

Also, if they are over-discharged (much below around 3V per cell) and left for any time like that, the internal structure can break down in another way and cause them to short out.
Again, one that happens, the cell is useless.

Your cells may well already be wrecked if they are only showing 2V on charge...

More information:
**broken link removed**

 
As already mentioned, charging Li-Ion is fairly critical, although simple to do - and letting batteries go as low as 2V has probably already ruined them - and incorrect attempts at charging will also ruin them.
 
Phone "chargers" with USB sockets or cables are simply 5V power units.
The actual charge controller is in the phone (or other device) alongside the battery.

A lithium cell needs an accurately regulated voltage, usually 4.2V, plus a current limit circuit to prevent it drawing too much current when flat.
Without voltage regulation, they can be destroyed by over-charge - often bursting or even catching fire.

Also, if they are over-discharged (much below around 3V per cell) and left for any time like that, the internal structure can break down in another way and cause them to short out.
Again, one that happens, the cell is useless.

Your cells may well already be wrecked if they are only showing 2V on charge...

More information:
**broken link removed**

So please what circuit do you suggest for me, I have some little experience with SMPS.
Or can I as well build a similar charge controller used internally by the phone? Will that be possible.
 
I really appreciate your help, but I don't think am in the right situation to actually purchase any online shipping product now. Not just now. The location I am right now, doing that will be a long protocol or I'll just have to wait after the pandamic which will go a long way in completely discharging my battery to death.
I needed something more urgent.Any kind of a hack.
 
I saw this circuit in one site. It looks so simple to be able to charge a 3.7v 3A li-ion battery. What's your opinion?
 

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That circuit makes no sense and would not work, unfortunately.

You have mentioned you have a 7805; connect two normal rectifier diodes in series with the output; that will reduce it by roughly 1.2V (so around 3.8).

That will give roughly a 40% charge. The cell voltage should reach that and then the current drops off to near zero after a while.

If that works and the cells charge to that, then the voltage can be increased slightly to get nearer full charge - but try that first to be sure the cells are OK, then we can consider adjusting the output voltage some more.

If they won't charge to over 3V and hold that voltage, then they are scrap.
 
Find or make any 4.2V fixed regulated or maximum voltage supply and apply to cells, in parallel or one at a time. Any supply that can go above 4.2V will be a danger. Never exceed 4.20V.

Or try several 7805 ICs, perhaps you will find a not very precise one that will yield 4.9V and add a silicone diode like 1N4001 to bring voltage down by 0.7V. A 20 to 50 ohm resistor in series will help to keep all components cool if cell is overdischarged.

(+)-------------in7805out-------------+4.9V--------------|>|--------------+4.2V-------------/\/\/\/\---------> to cell (+)
 
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That circuit makes no sense and would not work, unfortunately.

You have mentioned you have a 7805; connect two normal rectifier diodes in series with the output; that will reduce it by roughly 1.2V (so around 3.8).

That will give roughly a 40% charge. The cell voltage should reach that and then the current drops off to near zero after a while.

If that works and the cells charge to that, then the voltage can be increased slightly to get nearer full charge - but try that first to be sure the cells are OK, then we can consider adjusting the output voltage some more.

If they won't charge to over 3V and hold that voltage, then they are scrap.
also told you my experience while using 7805.
From a power supply of 12v 1.2A(with heat sink, the 7805 can handle 1.2A), the 7805 regulated it to 5v, when I connected just one of the cells, I measured the charging current(connected a multimeter in series with the battery), and I saw a whooping 2A, and the ic was really heating up. I was very suppries, the battery draws the current up to that level. That shows that even the 7805 could not regulate the current and I don't know why.

So please will connecting a diode in series to drop the voltage to 3.8v change the story?
 
So please will connecting a diode in series to drop the voltage to 3.8v change the story?

Two diodes will reduce the voltage to around 3.8V, which is a safe level for the Li-Ion cells.

Current limiting is a separate thing; the simplest method is a filament lamp in series with the 12V supply to the 7805; eg. a 12V 6W lamp would limit it to half an amp or a bit less, so a safe charging current.

A power resistor with a suitable value would do the same, 10 Ohm should also give a limit of roughly half an amp with a 12V supply feeding the 7805.

The 7805 should be on a heatsink & have caps at the input and output, no matter what circuit you use.
 
Two diodes will reduce the voltage to around 3.8V, which is a safe level for the Li-Ion cells.

Current limiting is a separate thing; the simplest method is a filament lamp in series with the 12V supply to the 7805; eg. a 12V 6W lamp would limit it to half an amp or a bit less, so a safe charging current.

A power resistor with a suitable value would do the same, 10 Ohm should also give a limit of roughly half an amp with a 12V supply feeding the 7805.

The 7805 should be on a heatsink & have caps at the input and output, no matter what circuit you use.
Ok, but am trying to find a way to stop the li-ion battery from drawing too much current, and damaging either the power supply or the 7805 or even both.
That was my main problem. I needed to make the battery accept the amount of current it being given by the power supply. I don't know if you get my point?
 
Find or make any 4.2V fixed regulated or maximum voltage supply and apply to cells, in parallel or one at a time. Any supply that can go above 4.2V will be a danger. Never exceed 4.20V.

Or try several 7805 ICs, perhaps you will find a not very precise one that will yield 4.9V and add a silicone diode like 1N4001 to bring voltage down by 0.7V. A 20 to 50 ohm resistor in series will help to keep all components cool if cell is overdischarged.

(+)-------------in7805out-------------+4.9V--------------|>|--------------+4.2V-------------/\/\/\/\---------> to cell (+)
Ok, with that ,will it be able to make the li-ion battery accept the given current without drawing excess current and damaging the 7805 or the power supply?
 
You need a PROPER Li-Ion charging circuit, Li-Ion charging is fairly critical, simple enough - but you MUST do it properly, or you will damage the cells, set them on fire, or explode them!.

There are three stages:

1) Constant current - this charges the batteries with a constant current (520mA is a decent choice for standard cells), during this time the voltage across the cells. Keep doing this until you reach the terminal voltage for the cells (4.2V per cell).

2) Constant voltage - once you reach the terminal voltage you maintain that voltage, and the current gradually falls as the cells charge further, keep doing so until the cells are fully charged.

3) Fully charged - once the charging current drops low enough (about 130mA) the cells are fully charged, and all charging should be discontinued.

You can do the first two stages quite easily with a bench PSU, set the current to 520mA, and the voltage to 4.2V per cell (so 16.8V on my case using four in series) - then keep a regular eye on the charging, as you have to manually notice that it's dropped down to 130mA and is fully charged.

Your crappy 7805 circuits are down right dangerous, and completely unsuitable for Li-Ion. I would also seriously suggest you add protection boards to your batteries, as that will at least give a degree of protection - but you still need to charge them properly.
 
Ok, but am trying to find a way to stop the li-ion battery from drawing too much current, and damaging either the power supply or the 7805 or even both.
That was my main problem.

I gave you the solution:

Current limiting is a separate thing; the simplest method is a filament lamp in series with the 12V supply to the 7805; eg. a 12V 6W lamp would limit it to half an amp or a bit less, so a safe charging current.

A power resistor with a suitable value would do the same, 10 Ohm should also give a limit of roughly half an amp with a 12V supply feeding the 7805.

The 7805 should be on a heatsink & have caps at the input and output, no matter what circuit you use.

Note that the 7805 with the two diodes (& resistor or lamp) are purely to allow you to part-charge the cells with safe limits, and see if they are still usable - as that seems very unlikely, from some of the info you have given previously.

IF it turns out that they will reach around 3.8V and the current drop right down after some hours, then it would be worth looking in to precision charging circuits, as Nigel suggests.
 
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