That would be fine, if you are doing 3 reds. Resistor properly.
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Thank you guys as I continue to feed upon your knowledge
I tried for one of my sabers to have a 3-Cree driven with an NB. My impression is that the NB will limit the overall current to 2A no matter how many dice you have in parallel. I had a potenti-o-meter :) regulating the series resistance to the dice, when I gradually opened up the 3 LED's, up to a given point the light intensity increased, but then I could not tweak out any more, so the board limited the current which could be sourced by it. It's of course an experimental result, without knowing the exact details how Erv designed NB's safety features.
Thanks for all the info in this thread. This is something that has been confusing the heck out of me for a while.
I'd been reading that you could drive 1A to two dice on an LED...but that would already meet the max rating on the NB. I was wondering how in the world you could safely run 2A to the leds while still running the board and a 2W speaker without damaging the NB. It's good to know that there is a nice factor of safety...though it would be nice to know exactly what the limits are; The user manual just says 2A with the high power LED. I was considring resistoring all the dice to keep every led below 750mA to be safe. Resistors no more!..cept of course for reds & such.
Use this as a rule of thumb. You can use up to 2A for your main LEDs. Use external resistors, to prevent overheating the board, even for the higher voltage colors. Overheating the board causes un-triggered swings and such.
For colors that could take 1A, you would be better off running 2 at 1A. I said you could do 3 reds, because they don't need that much. You could try the 2 reds at 700 mA, and see how you like it.
I quicky looked up the datasheet of a Cree-MC-E, which I built in into my latest addition. There is a chart showing the light intensity scaled to a nominal value (350mA for Cree HP-LEDs). On the chart it can be seen, that the intensity is close to a linear function of the current, slagging down at the last Quadrant only. I guess you have to have a closeer look at the datasheet of the LEDs you bought and make the maths, maybe you find an optimum using all the 3, but due to the linearity and the space the additional wiring of the 3rd LED takes up, the 2 blues each with 1A is probably the best you could do.