Using AA batteries to power Arduino and GPS unit

Question

I'm more of a software guy, so I need your help to figure out the power issue with my Arduino.

What I would like to do is using a GSM/GPS unit (Ai-Thinker A7) with Arduino Nano as a car tracking system. I have USB cables in the car, so I can plug it into my Arduino and power both while I'm driving. To power the unit while I'm not driving, I thought using rechargeable batteries would be a good idea. I'm planning to use GPS via SMS activation, so I hope fully charged batteries will be enough to power it for at least a week. But I'm confused about a couple of things.

Rechargable batteries are 1.2V. So technically I can use 5 of them and power my Arduino and the GPS unit (which requires 5V), right? However, for a similar unit at Adafruit they say that a LiPo battery is a must. Well, I think it is because GPS unit draws more power while using GPS and stuff. So, in this case are AA batteries out of the question? Will I fry anything if I build the system as follows?

If AA batteries happen to be fine, do I need another piece in my design, such as this one?

At last, I have no idea if I need any diode, capacitor, resistance or sth else for this system. I would appreciate further comments about my design.

Answer 1

I suggest using 6 or 8 AA's (eg 6 alkalines or 8 NiMH) and a MP1584EN DC ­DC Power Stepdown Buck Converter (about a dollar on ebay, 1.5 to 3 times that on Amazon). Note, the hole pattern does not match up with breadboard spacing:

The batteries would tie in via a couple of diodes (preferably Schottky), as shown below. When the 12 V supply is off, the converter would be powered by 9.6 V from eight rechargeable AA's. When the 12 V supply is on, no current is drawn from the AA's. The converter output should be adjusted to 5V before being connected to the Nano or other devices it powers.

Using a buck regulator has two advantages vs a linear regulator: battery current is lower (allowing the battery pack to be more efficient as it discharges) and less energy is wasted as heat.

Note, it might make sense to remove the power LED from the Nano, to save 3 to 5 mA of current.

^{simulate this circuit – Schematic created using CircuitLab}