I am a beginner in electronics and I want to ask you a simple question. Can I power my old SE w200i directly from arduino, using resistive divider like this one: https://en.wikipedia.org/wiki/Voltage_divider

r1=100Ω r2=3*100Ω v2= 3.75V. Orginal battery from this phone is 3.7V. Is there any possibility that it'll work?

// also i was thinking about something like this: http://i.minus.com/jb2rCe3kCTwwXs.jpg


1 Answer 1


Is there any possibility that it'll work?

I think not.

Your 400 ohm divider with no load (i.e. no phone connected) draws 0.012 Amps. If you increase the load even slightly, your voltage divider will produce a much lower voltage.

A cellphone can draw up to 1.6A from it's battery. You can't supply that from a 5V supply through a 100 ohm resistor (consider V=IR)

The maximum current that 5V can drive through 100 ohms is 0.05 Amps.

Even with a high efficiency regulator instead of a voltage divider - this doesn't look to me like it will work.

If you power the Arduino (say a Duemilanove or similar) from 12V, then I think you have 11.3V available on the Vin pin of the headers. I'm not sure what the current limit is on this. You'd need to check that it can cope with the max/peak draw by the phone.

According to this answer Vin can not supply more than 1A. So you need to be sure your phone has untypically low peak-current requirements.

An LM317 supplying 3.7V 1A from a 11.3V supply will waste a lot of power. It will get hot if that current is drawn for long (which I would guess it isn't). I'd size the LM317 heatsink for an average current of 300 mA in talk mode since that seems to be a typical average current draw. You should probably measure current draw peak and average before prototyping this.

Someone with more experience will hopefully comment if I've made too many blunders in this answer.


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