Wiring check

[Kael-Duvar Hawkens]

Ok well I planned to test a series of resistors on the red LED to find the color that I want. So if I have the two blues in series and the red in parallel wouldn't all 3 get 700mA without a buck puck?

[FenixFire]

Do all three have the same fv? If not I dont know if I would trust wiring them to the same puck until someone confirms they had it wired successful for without burning up one or more LEDs. I have not seen anyone post a diagram like that yet?

For any type of color mixing most people just use resisters.

Without a puck no. They will get the full voltage and amperage the batteries are allowed to dump through the protection circuit. The resistors restrict the voltage by converting a certain amount into heat. If you wire the blues in series you will need a 7.4v batter and a very large resistor for the red, which will run very inefficiently.

3.7v battery, The two blues run parelell with a resitor each, and the red in parelell with its resistor is the most common I have seen for color mixing. Though the 7.4v with large resitor a comes up too, just waists more power into heat on the red.

I think you are confusing mA and mAH. mAH is the amount of storage capacity vs time, XXXX milliamperes per hour. So if you have a 1400mAH battery it run a 1000mA LED for about 1.4 hours, or 2 1000mA LEDs for about 40 minutes, or 3 1000mA LEDs for 25 minutes. It's not the amount of mA it constantly outputs.

[Kael-Duvar Hawkens]

Ok, well I already have the 7.4v battery and the store doesn't have any 3.7v in stock so I'll just have to run the three in parallel with either a single or multiple buck pucks then, is that about right?

[Silver Serpent]

Well...you *can* do it with multiple Buckpucks instead of resistors. It's gonna be expensive and take up a ton of room. You'd need a Buckpuck for the two blues in series, and another Buckpuck for the Red. Consider that each Buckpuck is $14-15 apiece, versus less than $1 apiece for resistors, and they take up WAY more room in your hilt than a resistor does.

Long story short, it sounds like you're going a very expensive route just to avoid calculating your resistor value.

[Kael-Duvar Hawkens]

I did calculate the resistor's needed but had the wrong current, I've calculated the new ones as they are "supposed" to be at 1000mA, I know the red one is gonna be different than the calculation because I'm gonna be adding more to find the purple that I want. Otherwise please tell me if I calculated it right

-Math below-
Blue (2 in series)
R=[7.4V-(3.41V+3.41V)/1A]
R=0.6 ohm

P=0.6ohms*1A^2
P=0.6 watts

Red
R=[7.4V-(2.59V)/1A]
R=4.81 ohms

P=4.81ohms*1A^2
P=4.81 watts

so with rounding for the blue I need
1 ohm
1 watt

and red I need at least
5 ohm
5 watt

[Silver Serpent]

That math is correct. Depending on the specific shade of purple you want, you'll probably end up with a value of around 7-10 ohms or so on your red.