My wife’s new Pantech Jest cell phone won’t charge its battery. This has been a recurring problem with new batteries we get for it, too: inevitably it always ends up refusing to charge the battery. If you plug the phone in to charge then the phone will turn on and power up, but it will not give any indication of charging. Then no matter how long it is plugged in, the moment the phone gets unplugged it instantly turns off. The first time this happened our wireless provider replaced the phone, but the new phone exhibited the exact same behavior right out of the box. This meant that the battery was at fault, not the phone, so then our wireless provider replaced the battery which did fix the problem. However, a few weeks later the exact same thing happened. So they gave us another replacement battery (which fixed the problem again). But now this week the same problem has happened a third time. Fortunately there is a solution.
The Real Culprit
It turns out that this problem is caused by the phone not recognizing any battery present if the battery has too little of a charge left on it, so the phone doesn’t attempt to charge the battery (because it doesn’t think that there is a battery there) which means that the battery stays dead and the phone continues not recognizing it / not charging. There is actually nothing wrong with the battery – the problem is a poorly designed charging circuit inside the phone. (I learned that this was the problem when I stumbled across Phoneyguy’s post at http://www.howardforums.com/showthread.php/1655999-Pantech-Jest-Release-Thread?p=14073937#post14073937.)
As you can see from the picture of the battery below, its voltage is supposed to be at about 3.7V, but when I measured our battery voltage is was very low, at about 2.8V:
So although the phone is the faulty piece of equipment, we kept having to replace the battery (not the phone) to get around the problem. But not anymore:
How to Fix the Problem
What all this means is that the "bad" battery is actually a good battery and will work again if you can just get sufficient charge back onto it. Good news – you can do this yourself if you have some spare AAA, AA, C, or D batteries, some wire, a few common circuit components and at least a small level of familiarity with circuits.
WARNING: Proceed with these instructions at your own risk! Manually charging your cell phone’s battery may void its warranty and/or cause damage to the battery and phone.
The simplest way to manually charge the cell phone battery requires three AA batteries, some wire, and a resistor. AA batteries provide 1.5 V each, so putting three of them in series will give 3 * 1.5 V = 4.5 V. You don’t have to use strictly AA batteries, it can be AAA, AA, C or D, as long as putting the three of them together in series yields around 4.5 V total. The reason we need 4.5 V is because the cell phone battery is supposed to be charged to at least 3.7 V, and we need our AA batteries to provide a voltage larger than that.
Some people or web sites suggest to connect the AA batteries directly to the cell phone battery without using a resistor to limit the current or drop down the voltage, but this is not very safe because you probably do not know exactly what voltage and current levels the cell phone battery can handle. For this reason you should use a resistor, and choose a resistor value as described in the section "Calculate Resistor Value to Use" later in this write-up.
I used the circuit shown below to charge my wife’s cell phone battery.
Connect the (+) and (-) terminals of Vout to the (+) and (-) terminals on the cell phone battery, respectively (see picture below). The resistor is an 8 ohm, 1/2 Watt resistor. I calculated 8 ohms and 1/2 Watt for the resistor value (see the "Calculate Resistor Value to Use" section below), but I didn’t have any 8 ohm resistors on hand. The lowest value of resistor I had was 33 ohms, so I first used one of those.
Even using a 33 ohm resistor, the voltage on the cell phone battery started going up pretty quickly, but after about 15 or 20 minutes it slowed to a crawl. At that point I added another two or three 33 ohm resistors in parallel with the first so that the equivalent resistance was about 8 ohms. As the battery voltage got closer to 3.7 V it kept slowing down more so I then added even more resistors in parallel.
Every few minutes I would disconnect the cell phone battery and measure its voltage and then put it in the phone to see if the phone would start charging it yet. However:
Just as the battery reached 3.7 V, I stopped charging it, put it in the phone, and the phone finally detected it and showed it as being very low. It showed up as one "bar" on the battery indicator, and the phone finally was able to charge the battery!
Note that it took about 4 hours or so for this to work for me. It may take less time for you if you start off with an 8 ohm resistor or if your battery isn’t as dead as my wife’s was.
Let me know if you have any questions!
If you want to understand more about why the circuit above works, how you can make the circuit be safer, and other neat tidbits then continue reading below.
Calculate Resistor Value to Use
It is important that you choose the right value of resistor to charge your battery. If you choose too small of a resistance then you might provide too much current to the cell phone battery and might also exceed the power rating for the resistor itself. If you choose too large of a resistance then it will take too long to charge the cell phone battery.
To calculate the best resistor to use, remember from the pictures near the top of this post, that the battery is nominally at 3.7 V, and stores 920 mAh of charge, but that the dead battery was at 2.8 V. Also recollect that the three AA batteries give us 4.5 V. Next we also have to consider how long a dead-to-full charge normally takes, which I decided must be around 3 hours when I reflected on charge times from the past. From these values we can calculate how we should design the circuit:
Max desired current for DIY charger:
Of course, we need to leave some margin in the design, so let’s use 0.2 A instead of 0.3 A. Then the minimum initial resistance choice is:
The power rating for the resistor must meet worst-case scenario for power dissipation:
The nearest standard power ratings for resistors are 1/4 W and 1/2 W. We need the resistor to handle at least 0.36 W, which means we need to go with the 1/2 W rating. Alternatively, you can double the resistance value in order to halve the required power rating (e.g., use a 16 ohm resistor that is rated for 1/4 Watt).
Alternative Circuits to Consider Using
Instead of using the simple manual charging circuit I described above, you could design a more sophisticated circuit that automatically stops charging when it reaches 3.7 V, or one that includes LEDs to indicate whether or not the charge has reached 3.7 V. I will not go into a lot of detail here. These could be accomplished by doing the following:
- Add an N-channel MOSFET in series with the resistor.
- Add a comparator that compares the cell phone battery voltage with a separate reference voltage of 3.7 V.
- The reference 3.7 V can be obtained by using a resistor divider on the 4.5 V from the AA batteries.
- The comparator output can drive the MOSFET mentioned above.
- Add the LEDs that indicate whether the circuit is busy charging or is finished charging. The LEDs would be driven by the comparator previously mentioned.