I'm currently using an Arduino Dueminalove with Atmega 328 and I was wondering what would be the appropriate approach on supplying my board with batteries? The only available rechargeable batteries at the store is 6v 1800mAH Ni-MH. The clerk told me I could buy 2 of those batteries and put it in series and connect it to the Arduino. Is that really advisable? Although I feel there's a need for a regulator like the LM7805 but I'm not really sure.
Thanks in advance for those who would help me.
you could also just use the battery as is. 12V is sort of overkill anyway.
NiMH batteries can't supply 6.0V, since cell voltage is about 1.3V. 4 cells give around 5.2--5.5V, 5 cells 6.5--6.8V
A 4-cell NiMH battery should be safe to connect direct to the +5V supply pin (I do this all the time)
For alkaline batteries you have to reduce this to 3 cells.
If you want to benefit from the on-board regulator for a nice stable supply (if using analog inputs this is often desirable) then you need 7V or more on Vin (6 or more NiMH, 5 or more alkaline cells)
I am surprised you can only find "6V" NiMH batteries, individual AA and AAA cells are commonplace I thought - finding a batteryholder is a little harder.
Well based on the schematics, the board is using a NCP1117 regulator (NCP1117) which has a drop-out voltage of 1.2V. That means you need a voltage supply greaterthan (5+1.2) = 6.2V so that the regulator will operate properly. And yes the clerk was right, series 2 batteries then connect them to the DC jack.
And yes the clerk was right, series 2 batteries then connect them to the DC jack.
I doubt the battery voltage is a true 6V. But if it is and you double that up to 12V you'll be right on the top edge of the recommended voltage range. If, as seems likely, what you really have is 5.2V then connecting a single pack to the 5V line would be more sensible.
@MarkT: I'm talking about the battery pack that gives out 6 volts not the individual battery itself, sorry for the confusion.
@josephdzn: thanks for confirming the suggestion.
@PeterH: I've got what you said on your first statement. What did you mean by having 5.2v and connecting it to the 5v line?