Lithium ion battery inquiry

[Ben]

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

 

[Rene Tschaggelar]

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?

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

Rene

 

[PeteS]

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

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

 

[[email protected]]

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

Rene

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

 

[ehsjr]

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

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

Ed

 

[Iwo Mergler]

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?

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

 

[Frithiof Andreas Jensen]

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 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 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.

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

So the question is how to simulate a battery, so the phone works with the
power source?

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?