Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Need help charging my cellphone battery

Status
Not open for further replies.
Audio guru,

A discharge to 3.0 vdc will not destroy LiIon battery.

Most cellphone battery packs have an internal disconnect between 2.2 and 2.4 vdc on battery.
 
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.

Got it! That makes total sense. Thank you very much for taking the time to write that reply.

So when you made your comment that the charging voltage should never pass 4.2V you were meaning to say that for safety sake. That makes sense to me and with that in mind I will just set it below 4.1V as getting it 100% charged is not important to me. I see your point on the `once it hits 4.2V it is 70% charged`, but again -- not an issue for me. I would rather be safe.

That being said, I also see how RCinFLA was getting at that earlier, too.

Thank you to both of you! Now I just need to find a decent way to attach wires to this stupid battery... :)
 
Last edited:
That's fine.

A real LiIon charger will hold it at 4.2v until current drops off to about 10% of current limited stage 1 charge current.

This requires other safety triggers like a timer timeout if current doesn't drop in an hour or two and terminate if battery gets above temp safety limit.

Actually not fully charging a LiIon will make it last longer (aging wise).
 
Last edited:
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 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?

That always happens. You don't get them as accurately as they show.

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.

You won't know. It'll heat up and explode (??? Never happened to me though) and the read voltage is a deceiver. You'll just have to monitor the heat from time to time. In my experience again.
 
The motor in my electric model RC airplane gets extremely hot and my hacked Lithium-ion battery cells get very hot then something throttles down the motor when it is flying. So I land the plane. After resting (cooling?) for a few minutes it flies with a lot of power again for many minutes. Then again after resting (cooling?) it flies again and again for fewer minutes for each flight.
 
Okay everyone, I set up the circuit as follows

+4.1VDC -> 20 ohm resistor -> Pin 1

And then Pin 4 -> Gnd

Measuring the V1-4 while connected showed 3.75V (same as I was measuring V1-4 just prior to doing this) over the entire 20mins I let it run. I was surprised to not see any change so I disconnected it.

Was I just being impatient or did I get something wrong? Normally I'm `Well lets just see what happens!` but I want to be safe.
 
Did the battery get hot? or even warm?

Check the voltage across the resistor to see if there is a voltage drop occurring. (I would think) That would be a more reliable indication of current flowing, at least at the start.
 
I didn't notice it getting any more warm than usual but i was a bit nervous so it may have just been that my fingers were hot! Haha :)

I checked current I1-4 during this time and it came in at a little less than 200mA. I was just surprised that V1-4 was 3.75V during this time, too.

If I put the meter on the bottom leg of the 20ohm resistor and the gnd it would show 3.75V when the batter was connected, 4.1V when it was not. I am was thinking on this but I am not sure what conclusion to draw.

I will have to set it up again tomorrow and report back, but actually I think that it may have been working just more slowly than perhaps I was expecting. I just turned on my phone and made a call and instead of alternating between 1bar and 2bars but predominately staying at 1 bar, it was now at 2bars stable. I know I should be able to tell better than the phones indicator, though!
 
Last edited:
You are seeing the voltage drop across the resistor. 4.1V-3.75=350 millivolts.

The 3.75 is the drop across the internal resistance of the battery.
I think it was working fine.

If you are getting interested in electronics, this is a good lesson in Ohm's Law and Kirchoff's Voltage law.
 
As an experiment you should prove this to yourself by hooking the battery back up and reading the voltage directly across the resistor. If you were measuring 3.75 across the battery (in circuit), I'd wager you'll read 0.35 across the resistor.

If a voltage drop is occurring it is certain proof in a DC circuit that a current flow has been established, as you have demonstrated with your ammeter. I think you'll be happy with the result here as long as you monitor the voltage and temp of the battery. Glad you got some good suggestions, I don't like offering suggestions like that when I'm not 100% sure.
 
You are seeing the voltage drop across the resistor. 4.1V-3.75=350 millivolts.

The 3.75 is the drop across the internal resistance of the battery.
I think it was working fine.

If you are getting interested in electronics, this is a good lesson in Ohm's Law and Kirchoff's Voltage law.
You know it might not seem it but I am pretty familiar with those. The problem that I have is that in my undergrad the courses specifically dealing with these types of problems were not practical at all and thus were pretty ineffective. I have what I consider a `working knowledge` with hands-on circuits and do my day-to-day work as a wireless sensors guy using premade Si chips instead of discrete components.

These types of little projects are exactly what I need to sort of fill in some of the holes I have so that I have the same good sense in looking at something and figured it out with these as I do with other things.

The people on this forum are a big help!

One thing though. I was considering that it may be the voltage drop however when I turned up the source voltage slightly I did not notice a change. If it was the voltage drop only, wouldn't you expect this to affect the voltage I measured? I suppose that it is possible that I am recalling that incorrectly, but I believe that is accurate to what I saw.

Either way, I will go back tomorrow and check it out again.

Edit: I just remade it again and what I said above is correct. I will illustrate it here:

4.1V (prototyping board) -> top of resistor -> bottom of resistor -> (RED METER LEAD) -> Pad 4

And

Pad 1 -> (BLACK METER LEAD) -> ground (prototyping board)

With that setup the meter goes:
Condition 1) Measure as shown -> 3.75V
Condition 2) Increase 4.1V to 4.5V -> 3.75V
Condition 3) Turn OFF 4.1V source -> 3.75V
Condition 4) Pull batter, measure from bottom of resistor to ground -> 4.1V

Does this make sense?
 
Last edited:
Wait a minute, I typed before I thought it through. The battery in this case is more like a charging capacitor. You are reading 200 mA of current with a 4.1 volt supply over a series resistance of 20 ohms that would be more what I would expect from a near fully discharged battery. I would still be interested in knowing what voltage you read across the resistor.

A little more complicated but the formula can be looked at here:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html#c2
 
Last edited:
4.1v -3.75 vdc = 0.350 vdc / 20 ohms = 17.5 mA of charging current.

It's going to take a long time to see any effect.

Assuming your power supply is not current limited already, then put a 1.5 ohm in series with power supply.

Attached is a chart indicating SOC versus open circuit, rested battery voltage for typical cellphone LiIon battery. Rested means not loaded or charged for last hour or two.
 

Attachments

  • Battery Voltage LiIon.pdf
    173.5 KB · Views: 173
Last edited:
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top