Hi all! I've got a bit of a question regarding wiring when upgrading my saber to a rebel trip-star LED. I have a stunt, blazing red, prophecy V3 saber (review here:
https://www.youtube.com/watch?v=BnuqRwe-dD8) which I modified replacing the single blazing red (or bright red) LED with a deep red LED. I did this to get rid of the slightly orange tinge of the blazing red, and it's beautiful, not a bit of orange can be spotted, just red, really looks like Vader's lightsaber.
However, the disadvantage of this is that the deep red LED loses quite a bit of brightness as you can imagine, so I'd like to upgrade it to a tri-LED, namely a rebel tri-star with three deep red diodes (not two deep reds and a white, but three deep reds) in order to increase brightness significantly, just stunt, no sound, no flash on clash, just bright LEDs. Ultrasabers itself does not provide it, but Luxeon (its LED provider) does. I'm really into DIY and wiring, so I know most of the drill. HOWEVER, I've only worked with single LEDs and normal Ohm resistors, not tri-LEDs or buckpucks. I know how to wire the tri-LEDs in series and what lens to use, but I'm not sure about the driver. Each diode is 2.1V to 2.8V, making them in series 6.3V to 8.4V. First off, I use standard battery setup, 4 Ni-mh AAA batteries, a total of 6V, so I can already see a problem there, not enough voltage, right? I could try Lithium ion batteries, each 3.7V. Using two would nearly make it, and three would over do it, wouldn't it? I know I need a 700 mA powerpuck or buckpuck driver (the deep red LEDs must be run at 700mA), but do these drivers maintain the 700mA throughout the circuit regardless to the voltage supplied? For example, let's say I use 4 Li-ion AAA batteries with a total voltage of 14.8V (despite being kind of a waste of energy). Would the current still be kept at 700mA? And would the buckpuck/powerpuck compensate for all that heat dissipation without a resistor?
I hope you guys can help me find out if my logic is correct or not, thank you all!