edit: This has been revised so that it doesn't confuse anyone.
(At the first post)
You can draw as many amps from a battery as you want, just know that the ESR of the battery will cause it to heat up. I got from wikipedia that the ESR of a AA is about 0.9 Ohms. That means that if you draw 2 amps from a AA (P=I squared R) that the energy dissipated in watts would be about 4 watts. That's a lot of heat, and I am SURE the ESR was higher than that. Anyway, if you put batteries in parallel, you cut the ESR just like if you put resistors in parallel. That's how you should calculate it. (Kerm mentions that this is in an IDEAL world, so please leave room for error. also there are physical limits to this as well, involving saturation of the battery.)
Also, keep in mind the discharge curve of a battery. The more you current you draw, it dies about twice as fast. (actually, superlinearly as Kerm would put it. just know that it dies REALLY fast the more you push the envelope. (the one you aren't pushing anyway cuz ur current is so low))
(at the last post)
What I would do is use the 7805 voltage regulator, seeing as how he's a beginner. A zener diode offers no protection, a 7805 can take being wired backwards, short circuits, AND cuts off if the heat is too high.
(omit this paragraph if you are easily confused by my ranting) In terms of the dropoff voltage, who cares? If it drops below about 5.7 volts, the voltage out will drop linearly with the battery. If the battery is at 5.6, the voltage out of the 7805 typically (in my experience) will be 4.9. If the battery is at 5v, the output of the 7805 usually is around 4.3 volts. He'll just have to change his battery a little more often because the dropout wastes 0.7 volts of the usable battery life.
If it was me, I would use a 9v battery because they are a lot smaller anyway, and you won't have to worry about dropout voltages at all because the battery is dead long before it gets to 5.7 volts. Just don't draw much more than an Amp and you'll be fine. He said he was powering a USB device, and that's NEVER more than an amp (and I'm being generous here Kerm
) NEVER more than an amp. You'll blow a fuse in your computer if you draw more than that. Go with the 9v MrMcMan.
If it was myself, I would use a switching regulator with the proper inductor, capacitors, and resistors. Those will give the most efficient regulation possible (more efficient than linear regulators OR zener diodes). Just make sure the capacitor is big enough to smooth out the ripple for your load
. (Kerm likes to use zeners though.