powering the arduino with less than 5 volts

Hi,

I tried to power the arduino with less than 5 volts (3.7 to be exact) via the VIN pin, and everything seems to work properly.

Obviously if I had to interface the board with external logic (expecting 5 volts) it would require 5 volts power source.

But considering that the micro itself can be powered at 3.7 volts (even less), and that my circuit (a simple light trigger) can work at 3.7, can you foresee any issue?

On the "hardware" page it says it could be unstable...but why if the micro is fine at 3.7?

You may ask why I like 3.7V so much...well...that's because I'd like to recicle a nice LiIon battery;)

If the stuff you have connected to the arduino can tolerate the lower voltage it should work fine on 3.7 volts. In my experiences it works perfectly OK down to about 3 volts but I wouldn't try to program it at that voltage.

Your biggest problem is the circuit you'll need to stop the battery becoming over discharged. If the voltage is allowed to drop too low with a Lithium Ion battery, it can destroy the cell and it will never recover.

The Vin pin connects through a diode to the regulator, 3.7 is too low for this input. Connect your 3.7 volt battery to the 5V input – make sure that you have the polarity correct.

Ahh - missed the Vin bit, as mem says, hook it to the 5 volt line and watch the polarity. If its running on 3.7 through a diode it does lend credence to it running on 3 volts though. :)

On the "hardware" page it says it could be unstable...but why if the micro is fine at 3.7?

It's a matter of luck whether your particular chip will run at 16MHz at 3.7V: the official spec says you ought to be supplying more like 4V (or so: it's hard to be exact because there's no grid on the "Safe Operating Area" graph).

Most chip makers try to consistently exceed the minimum specs, and sometimes turn out a batch that's much better than "meets min". But the next chip you buy may not be from such a batch, and may fail in unpredictable ways.

Plus, the battery will drop below than nominal 3.7V as it discharges, pushing you even further outside the safe range.

So, as Dirty Harry said: ya have to ask yourself if you feel lucky. Do ya, punk? ;D

Ran

The Vin pin connects through a diode to the regulator,

Actually the Vin connects between the polarity protection diode and the on-board +5vdc regulator. That is, the external power input jack will experience one diode forward voltage drop before hitting the regulator's input. The Vin will not experience the same voltage drop.

Vin voltage needs to be above the minimum drop out voltage input spec of the regulator.

PS: the problem of wiring a direct DC voltage into the Arduino's +5vdc pin is the possible problems one might have when then plugging into the USB link, as then you have two DC voltages tied together which can lead to problems. Lefty

@retrolefty: that's exactly what I thought, the external plug goes through the diode+regulator, while the Vin pin goes directly to the VCC pin of the micro.

obviously I have to unplug the 3.7 battery when I plug the usb, or when I need the serial (in that case I'll use the FTDI interface which also provides Vcc).

But reading also Ran Talbott's comment I understood that the Vcc needs to be at least 4V (so higher than the "standard" operating voltage) if there's an external resonator (which is my case). Where do I find this? In the micro's datasheet?

[EDIT] at page 30 of the datasheet I read "Note that the Full Swing Crystal Oscillator will only operate for VCC = 2.7 - 5.5 volts."

Tnx!

Vin does not go directly to Vcc, it goes through the regulator (but not the diode) so the Vin voltage should be more than the regulator dropout voltage plus 5 volts. The 5v pin goes directly to Vcc.