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Lithium ion battery inquiry

I have a 3.7, three connectors Lithium battery in my cellular phone that discharges faster than it is charged when I used the phone to browse Internet, so I connected three wires to the contacts in the phone, plugged off the battery and tried to run the phone feeding it with an appropiate power source instead the battery. The problem is that it doesn't work, I guess the middle connector in the battery detects the battery presence, so the phone doesn't work with a regulated power source (1 A, 3.7 VDC) and without a battery. So the question is how to simulate a battery, so the phone works with the power source?

Thanks in advance, Ben

Reply to
Ben
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What did you measure on the third pin of the original battery ?

Rene

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Reply to
Rene Tschaggelar

That third pin will either be a thermistor *or* a serial IO line. If you simply intend to measure the current, then insert an ammeter in series on the positive or negative power line and connect the other two lines directly.

It might be possible to run the unit by providing power to the power and ground pins and connecting the ground and centre pins from the battery, but there's no guarantee of that.

Cheers

PeteS

Reply to
PeteS

Rene,

Thank you for answering.

The third pin is the middle one.

Without any charge the voltage is:

Between the positive and negative terminals 4.08 V Between the positive and middle terminal 3.93 V Between the negative and middle terminal 0 V

Ben

Reply to
berpop

Run the thing with the battery in there, and build a source that provides enough current so that the battery doesn't discharge.

Ed

Reply to
ehsjr

Older phone batteries have a thermistor connected between (-) and the middle contact. It's only relevant during charging, otherwise a 100R resistor will do.

More modern batteries use the middle contact to communicate with a controller in the battery pack. Officially, it's for protecting the customer. Of course, it also protects revenue, as you can't buy cheap third party batteries.

Wether or not the phone will work without the thermistor or the controller depends entirely on the embedded software and is different from phone to phone.

But you have another problem.

The transmitter part of the GPRS modem needs much higher currents than 1A during transmission. It's pulsed, so the average will be below 1A.

You need a very large capacitor across the (+) and (-) battery contacts to cover the high current pulses. Around 4700uF worked for me.

Kind regards,

Iwo

Reply to
Iwo Mergler

The only "appropriate power source" is the mains charger that came with the phone!!

Work out what that does - like what voltage & current it provides. Then emulate that. That will even work.

The middle might be a serial link to the battery monitor or a temparature sensor inside that package that holds the battery.

No, the question is:

Why do you go to great lengts to *not* power your phone in the *obvious* way and add the risk of blowing up the electronics?

Reply to
Frithiof Andreas Jensen

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