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### Topic: Power a cell phone via arduino (Read 2004 times)previous topic - next topic

#### mmichielin

##### Feb 21, 2011, 10:55 pm
Hello, I'm poking around with arduino since a couple of week and my first game is to power up on old nokia phone (8210) via arduino: I would like to replace the phone battery with an arduino.

So I took a BLB-2 battery, I empty it from the cells leaving only the small circuit that let the phone know that the battery is a real BLB-2 battery (BSI - battery size indication - as a 68kohm resistor and BTEMP - battery temperature indication - as a 47kohm pulldown resistor) and soldering two small wires to power the phone via arduino.
If I connect these two wires to another blb-2 battery the phone switch on.

So I build a simple circuit that reduce 5v arduino to 3.6v-3.7v that this phone tipically accept but I'm not able to switch it on.
I've also tried to connect a 1000uF capacitor to "pump" moe current to the phone but I wasn't lucky: attached you will find the simple circuit.

Clearly there is something that I'm missing: does anybody was successful in this kind of task or does anybody understand what I'm doing wrong?

Thanks in advance to everybody

Ciao

#### Magician

#1
##### Feb 22, 2011, 12:03 am
Do you mean you reduce voltage with potentiometer?
It wouldn't work, as phone drain a lot of current, especially on
start up.
Add up transistor, better darlington, or drop voltage
inserting two diods in series : 5 - ( 0.7 + 0.7 ) = 3.6 V

#### mmichielin

#2
##### Feb 22, 2011, 09:02 am

Do you mean you reduce voltage with potentiometer?
It wouldn't work, as phone drain a lot of current, especially on
start up.
Add up transistor, better darlington, or drop voltage
inserting two diods in series : 5 - ( 0.7 + 0.7 ) = 3.6 V

Thank you very much magician, I also had a suspect of high current requested by the phone.
But, your answer suggest me two additional questions (simple for you):
1) why do you assert that a diode reduce the voltage by 0.7v? (ah, ok, I found it: http://www.satcure-focus.com/tutor/page3.htm)
2) can you give me some hint on how to build a circuit with a transistor (or refer me to some web page)?

Thanks again

#### mmichielin

#3
##### Feb 22, 2011, 09:16 amLast Edit: Feb 22, 2011, 03:27 pm by mmichielin Reason: 1

It is not clear what role the Arduino has in any of this?  It seems rather risky to even have it in the circuit, IMHO.

It would appear that your circuit to produce 3.6-3.7V does not work. You could confirm that with your meter.
If you would like assistance with your voltage regulating circuit, you can start by showing us the schematic diagram.

This small task is a fraction of a bigger project based on arduino that aim to build a water level sensor that can send me SMS through my old nokia 8210 when the water reaches a warning level.
And, not only, it should be able to answer me with an SMS containing the water level when I call him through my cell phone.
To do this, the first step is trying to power my cell phone via arduino board (i could use a charger but, if possible, I would like to avoid it).
My meter confirm that the voltage is correct, but as magician suggested, it is probably a problem of high current needed by the phone.

Thanks

#### Magician

#4
##### Feb 22, 2011, 03:48 pm

Yes, sure. There are two pics, one is what I mean in previous post.
You can use it when 'd like to try something "right now", and you local
shop already closed.
Second is from wikipedia, better option, you will need zener diod
with voltage V = V (out) + V (base emitter) = 3.7 + 0.7 = 4.4

And for long run project try to find IC voltage regulater.
Something like LM317, but I never "solder" it myself and data spec. says
V (out) - V (in) = 5 V, that is pretty high.
Nevertheless, think it will works.

#### balajitenetchat

#5
##### Mar 05, 2011, 10:13 am
You are charging the cell phone using the power supply from Arduino board and an external circuit to regulate. I thought it different. Does it have storage and what does Arduino do in this?

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