What color is that? 3.3 must be white or blue. 7000mcd is 'fairly bright' for white and blue, but there are brighter out there. Usually the rated voltage is just barely enough to get the gate to open, then there's very little current left to drive the light. Two fresh alkaline will be 3.4ish V, mid life 3.0, end of life 2V. Rechargeable AA wouldn't even get there with just 2.
I'd say add another battery, or the 9V. You definitely should have the resistor in line. The resistor controls the current, without the resistor you can't be sure how much or how little current you're getting though the LED (except for a visual on the light produced).
9V - 3.3V = 5.7V, 5.7V/30mA = 190ohm, but will last on the order of 10hours. You're burning a lot of efficiency on across the resistor.
4.5V (3AA) - 3.3V = 1.2V, 1.2V/30mA= 40ohm. Will last on the order of 50hours
Don't worry about the resistors wattage rating. Even the inefficient method above is only burning .005Watts across the resistor.
Jay W
Jeff Shyu wrote:follow up, though i have kinda changed the plan:
electrical savvy people, here's what i've got to play with right now, all bought from radio shack.
3.3v 7000mcd 25mA LED (i've got 2, but only using 1 to test right now)
battery box that holds (2) AAA batteries in series
when i connect the stuff up, it's not very bright. I didn't put any resisters inline, since i think it's actually a little lower than the required to power the LED.
do I stand to get much better output via 9v + 1/2w resistor? or am i pretty much getting what i'm gonna be able to squeeze out of the little LED?