Need help charging my cellphone battery

[Atheist]

The battery in my cell phone no longer charges. I took the phone apart and found the female connector for the charger to be broken off the PCB with some of the pads being torn apart in the process. I am sure that this is because of the junky charger I picked up on eBay that never fit quite right. I am planning to get a new phone mid-January anyhow, so this is not a huge problem, but I would like to charge this one a few times before then for emergency calls, etc.

The charger specs are:
5.2 +- .3V
460 +- 70mA

So I set up a circuit to mimic these settings -- ~5.2V through a 10 ohm (nominal) resistor and measured .45A at the terminals which I would like to connect to the battery.

The battery has 4 pads on it and two labels, + and -. I have made an illustration below. I would of course like to have pictures, but I do not have a camera. The numbers I show below are for reference and are not actually marked.

**broken link removed**

Where the circles indicating + and - are above the plastic divider between pads 1 and 2 (+) and pads 3 and 4 (-)

Using a meter, I find
V1-2 = 3.25V
V1-3 = 3.57
V1-3 = 3.78

So the question is, which pads do I apply power to? Also, as an aside, what do the voltage readings tell you? I work on embedded systems and this is not my strong suit!
V2-3 = 0V
V2-4 = 0V
V3-4 = 0V

 

[audioguru]

The battery is a Lithium type which is a very active metal. If you don't use a charging circuit that is made for it (the charging circuit is probably inside the phone) then the battery will probably get over-charged and will catch on fire. You don't want to be near a Lithium fire because it burns with a white-hot flame like Magnesium. Water on it makes it burn hotter. Look at all the videos of Lithium battery fires in Google.
Merry Christmas you athiest.

 

[Atheist]

I only want to charge it enough to make an emergency phone call if needed. I am about to head home for Christmas and the drive is 10 hours each way.

10 minutes on the charger would be more than enough for that! I could monitor the voltage and temperature carefully.

My thinking is that number 1 is and that 2,3,4 are the different cells however that is not even close to what the markings shown.

P.S. I did check out those videos just now. Cool/frightening!

 

[ke5frf]

Christians never have this sort of trouble at Christmas time.

(just pickin')

 

[Atheist]

haha okay okay, I get it.. I need to think of a new user name. I have had this one for years.

Can I get a little help though?

 

[ke5frf]

I don't know how old your phone is, or the model, but if it were me I might call some friends/family and ask if they happen to have the particular model phone (or a similar phone with the same model battery) and use their phone to charge your battery. Another option is that in mid to major cities nowadays there is a big business in old cellphone recycling for parts etc. You might check the internet for a store in your area and see if they have an old phone you can pick up for little to nothing, even a non-working phone might be used as a charger only. Another option might be to go to the vendor you purchase phone airtime from, if they have a retail store, and they MIGHT help you charge up the battery if they have an unpackaged phone (display etc) and you plead your case. Just don't tell them you are atheist, play like the humble Christian on his way to an orphanage to give little kiddies Christmas toys.

These are "non-technical" solutions that might prevent damage to your phone.

 

[Atheist]

Thank you for the reply.

I bought the phone for 30$ used on eBay. It is quite old. I don't know anyone with one and I live in an area too rural to have a phone recycling center. Charging it is not urgent, but I would like to give it a try.

 

[audioguru]

I charge the lithium battery cells from an old laptop pc with the current limited by the max output of a wall-wart AC adapter (1/4 the capacity rating of the cells) and the voltage limited to 4.2V per cell with an LM317 voltage regulator IC. I use two cells in series in my electric model RC airplane.

10% to 15% of the population around me are not Christians. But many stores advertise Merry Holidays, Holiday tree or Holiday Sale instead of saying Christmas.
Are all these stores athiests?

 

[Atheist]

What is going on with all of the username discussion? I don't mind, but its a bit odd. More-so than I have ever seen. Let me just say this:

Merry Christmas to all!

Where should I be applying power to these cells?

 

[RCinFLA]

Of the four battery connections, two are the + and - of battery, probably the outer two, and the two are for a thermistor connection used as safety feature of the charger.

The actual charger section is within the phone. The wall power supply, which you said was 5 vdc is the input to the phones charger circuitry. There is a series pass FET in the phone charger. The 5 vdc is selected to give enough headroom for the charger to achieve 4.2 vdc on the LiIon battery, and not be to much higher which would increase the power dissipation (heat) within the phone's internal series pass FET.

You can charge the battery out of the phone with a current of about 20-30% of the rated mA-Hrs. of the battery (probably battery is around 800 mA-Hrs).

You must terminate the charge when the battery gets to 4.2 vdc. There are several other criteria for charging LiIon batteries but assuming the initial battery discharged voltage is greater then 3.0 vdc you will be okay with a current source charging at about 200 mA that is terminated when battery reaches 4.2 vdc. Do not leave it connected indefinately at 4.2 vdc.

On a GSM phone, the transmit current pulse can be as high as 1.8-2.0 amps. As a LiIon battery is cycled it's internal resistance gradually increases. After 200-400 cycles the internal battery resistance is high enough (about 0.35 ohms) to where the high Tx current pulse causes the low battery voltage cutout to trip within the phone.

 

[audioguru]

I don't know your battery cells.
The ones I hacked from a laptop battery had a few complicated circuit boards for protection and charging that I removed. I used only the bare battery cells that had two terminals.

 

[ke5frf]

I think we are just picking at you. I'm sure I'd grab attention if my username was "Right wing fundamentalist" or "Flaming g**". Kind of a self imposed target for wise cracks. I'm not real big on advertising my personal beliefs and preferences, but that is just me Don't take it personal. Merry Christmas!!!

 

[Atheist]

Of the four battery connections, two are the + and - of battery, probably the outer two, and the two are for a thermistor connection used as safety feature of the charger.

The actual charger section is within the phone. The wall power supply, which you said was 5 vdc is the input to the phones charger circuitry. There is a series pass FET in the phone charger. The 5 vdc is selected to give enough headroom for the charger to achieve 4.2 vdc on the LiIon battery, and not be to much higher which would increase the power dissipation (heat) within the phone's internal series pass FET.

You can charge the battery out of the phone with a current of about 20-30% of the rated mA-Hrs. of the battery (probably battery is around 800 mA-Hrs).

You must terminate the charge when the battery gets to 4.2 vdc. There are several other criteria for charging LiIon batteries but assuming the initial battery discharged voltage is greater then 3.0 vdc you will be okay with a current source charging at about 200 mA that is terminated when battery reaches 4.2 vdc. Do not leave it connected indefinately at 4.2 vdc.

Oh okay, good deal, I appreciate the reply. That makes a lot of sense to me and, along with a PM from another user, I am seeing why they indicated the + and - the way that they did. The only part I am still a bit confused about is how to determine which of them are the correct two to connect to power given the voltage readings I find.

As an embedded systems programmer, I find this stuff just fascinating and really enjoy reading the posts on this forum. As a student, I look forward to becoming better at this!

 

[Atheist]

I think we are just picking at you. I'm sure I'd grab attention if my username was "Right wing fundamentalist" or "Flaming g**". Kind of a self imposed target for wise cracks. I'm not real big on advertising my personal beliefs and preferences, but that is just me Don't take it personal. Merry Christmas!!!

I am not really either. I guess I was more of a cock at 12 when I picked this name than I am now 10 years later.. Just haven't adjusted the name to show the change!

 

[Vizier87]

Yeah that username is gonna be a problem when you're so willing it to be portrayed to the rest of the netizens. Shows how adamant you are at showing your stand. This is the first time I've seen AG use smileys! (No offense, AG )

About the pads, I think it might have internal diodes which block the meter's reading between terminal 2 and the rest.

Try applying voltage at any combination for 10 seconds, and test the output on the respective terminals using a LED. It will light up at the right combination.

And yes, ten seconds is enough to make a difference from off to on. I played with 20 Li-ion batteries. All of them are working good.

 

[RCinFLA]

Check the terminals with a voltmeter. You won't get any voltage on the thermistor connection. Verify polarity with your voltmeter pos,neg leads.

To charge you force current backwards so positive of power supply goes to positive of battery.

 

[audioguru]

You must terminate the charge when the battery gets to 4.2vdc.

No!
The charging voltage must be limited to 4.2V. When the battery voltage reaches 4.2V then it is about 70% fully charged and still needs much more charging time and it draws current. The battery is fully charged and should be disconnected from the charger when its charging current drops to about 1/40th of its rated capacity.

 

[Atheist]

Yeah that username is gonna be a problem when you're so willing it to be portrayed to the rest of the netizens. Shows how adamant you are at showing your stand. This is the first time I've seen AG use smileys! (No offense, AG )

About the pads, I think it might have internal diodes which block the meter's reading between terminal 2 and the rest.

Try applying voltage at any combination for 10 seconds, and test the output on the respective terminals using a LED. It will light up at the right combination.

And yes, ten seconds is enough to make a difference from off to on. I played with 20 Li-ion batteries. All of them are working good.

The phone is not COMPLETELY dead. The LED idea is a GREAT one! I have to say that I didn't think of it because after metering and finding a voltage between pad 1 and all of the other 3 pads I got a bit confused. I will have to think on how I could get a voltage measurement from all of those, but only get usable current to light the LED from one.

I connected an LED just now and it lights up between posts 1 and 4.

I am thinking this:

4.5V source, ~20ohm resistor, should = ~200mA

I will connect the lead from the bottom of the resistor to pad 1, and then connect pad 4 to gnd.

I will do that for maybe 1 minute and re-check voltage to see how it has changed. The battery label says 3.7V, 800mA however V1-4 shows 3.78V. I get the 4.2V for Li-I, but I am a bit confused about the label rating of 3.7V o the back of the battery. What are they getting at?

What is the safest way to know when to say when? I am thinking the phone measures battery life via voltage measurement (the inaccuracy of which I do understand) but, again, the 3.7V label is throwing me off.

Again I want to say how much I appreciate the help here. I am still learning this stuff! I wish I knew it better so I could reason my way through things that I have little experience with.

 

[audioguru]

A lithium battery cell is 4.2V when it is fully charged. It drops to 3.2V when it is almost dead and should be charged. A discharge voltage of 3.0V or less will destroy it. The average voltage is 3.7V.

What about a "1.5V" alkaline cell?
It is 1.6V when brand new.
It drops quickly to 1.2V when it is used.
The manufacturer rates its capacity when its voltage has dropped to only 0.8V which is half the voltage when new.

 

[RCinFLA]

Audio guru, I am trying to keep it simple for him. He says he only wants to get by until he gets a new phone.