# Alternative to powering an Arduino with a 9V battery; Survival calculation time

I'm not able to find the usual duracell 9V battery to power my Arduino. Is it okay to connect in series 6 AA batteries and power the Arduino? My Arduino has a lot of load on it ,for example, bluetooth shields and a transciever etc. I'm not sure how long these batteries will actually last. Is there a way to calculate the life-time by measuring the current that my Arduino usually draws when powered by a USB?

USB can only supply 500mA. Can’t you use a 9V wall-wart ?

Energizer lists their industrial Alkaline AA’s as having 2779 ma/h or 2.7 amp/hr.

(that’s the same for your case because they are all in series, so 2.5A/hour. If your circuit is drawing 500mA then it will run 4 to 5 hours on batteries.

Instead of wasting nearly half the 9V supply as heat in a regulator, consider running from 3 batteries in series - AA, Cs, Ds, some rechargeable - directly and don't go thru a regulator at all. Or use a high amperage 3.7V LiPo/LiIon, in the thousands of mAH capacity, with a switching stepup, or boost, regulator to get to 5V, or a similar 7.4V battery with a switching stepdown regulator. (vs a Linear regulator, which can only dissipate excess voltage as heat)

http://www.pololu.com/category/84/regulators-and-power-supplies

raschemmel: USB can only supply 500mA. Can't you use a 9V wall-wart ?

Energizer lists their industrial Alkaline AA's as having 2779 ma/h or 2.7 amp/hr.

(that's the same for your case because they are all in series, so 2.5A/hour. If your circuit is drawing 500mA then it will run 4 to 5 hours on batteries.

2.5Ah - ampere-hour, not amps per hour (that would be a current-slew-rate spec!)

Charge (ie capacity) = current x time

2.5Ah - ampere-hour, not amps per hour (that would be a current-slew-rate spec!)

Perhaps I misspoke but I think it is obvious from my statement that if I said it would power a 500mA load for 4 hours from a 2.5Ah battery that I meant 2.5 Amps for a period of one hour. (which may not be 2.5A/per hour). I don't think there is any confusion as to what I meant.

CrossRoads: Instead of wasting nearly half the 9V supply as heat in a regulator, consider running from 3 batteries in series - AA, Cs, Ds, some rechargeable - directly and don't go thru a regulator at all. Or use a high amperage 3.7V LiPo/LiIon, in the thousands of mAH capacity, with a switching stepup, or boost, regulator to get to 5V, or a similar 7.4V battery with a switching stepdown regulator. (vs a Linear regulator, which can only dissipate excess voltage as heat)

http://www.pololu.com/category/84/regulators-and-power-supplies

From your post,you indirectly imply that I bring the voltage up/down to 5V when powering mean the the Arduino with the barrel jack? or did you mean the USB? I am quite confused now. Can I not solder the 9V battery lines to barrel jack and directly power the Arduino? nevermind the wastage of heat from the regulator.

You need >= ~7.5V when powering via the barrel jack to account for voltage drop in the reverse polarity diode and the 5V regulator. You can chop off the A-connector on a USB cable and bring 5V in via the USB connector. Or bring in 4.5-5V via the 5V pin on the power header. Or Connect however you'd like to the barrel connecter with 7.5 to 9V.

You might try directly powering the arduino with four rechargeable AA batteries that will supply 4.8v when fully charged. You might also use six of the rechargeable AA batteries for 7.2v and run that thru a UBEC regulator outputting 5v. The UBECs are inexpensive on ebay and are power efficient.

From your post,you indirectly imply that I bring the voltage up/down to 5V when powering mean the the Arduino with the barrel jack?

I think the part you are missing is that you are not aware that when you use the barrel jack, the onboard regulator needs a 2.5V overhead so 5V+2.5V=7.5V. If you plug in 9V (or 12V) , the difference between the 9V (or 12V) and the 7.5V optimum input voltage is dissipated as heat because the regulator needs to bring it down to 7.5V , hence the comment

Instead of wasting nearly half the 9V supply as heat in a regulator

Other things being equal, the capacity of batteries is well correlated to their weight. All of the above suggestions are good. Any scheme using 3, or 4, or 5, or 6, or 7 AA size batteries, is going to weigh more than one 9V battery. They are probably going to be better, and almost certainly no worse, than the 9V battery.

I can't think of any convenient way to measure the power being drawn through the USB connection. It would be handy if there was such a method ... a suggestion for someone to go and invent.

michinyon:
I can’t think of any convenient way to measure the power being drawn through the USB connection. It would be handy if there was such a method … a suggestion for someone to go and invent.

One could peel open up USB cable and break the power line, taking it out to 2 test points for multimeter probes, I guess. Equip the break with a jumper somehow, so that the cable can still be used when not being used with a meter.

Something like the attached.

One could peel open up USB cable and break the power line, taking it out to 2 test points for multimeter probes, I guess. Equip the break with a jumper somehow, so that the cable can still be used when not being used with a meter

Isn't there a LAW against that ? XD