Stunt Saber Battery and Driver Options

I need to check if my thinking is correct about the best options for a double bladed stunt saber that I'm trying wire. I have questions around running both blades, separately switched, off the same battery pack. Also, I've been trying read about buckpucks and want to make sure I understand the pros/cons of buckpuck versus direct drive with resistors. I've not wired a stunt saber before, but I have wired sabers off of boards before (but none with constant current control).

The other thing to share is that there is limited space to work with; I'm doing Grand Inquisitor saber, which only needs red so a stunt saber works great. I have some of the wiring options diagrammed below to make sure my descriptions are clear.

So, first is the question about running two blades, each with a separate switch, off of one power source. I'm trying to run 2 red diodes off a Tri-Cree for each blade. So, basically 4A when all diodes are on. This is within the maximum discharge of 2C of my Lion batteries (4.8A max). I could reasonably do a single red if that was pushing it. Is there anything else that I should be concerned about other than the battery discharge rate? I know the battery won't last long so maybe going to one LED per blade is better. I'm not sure how much room I'll have for more batteries. Or does each blade need it's own independent battery, safety wise?

Second, is about making sure I understand buckpuck versus direct drive/resistor. My original design, seen below, was to put the appropriate resistor on each blade (everything wired in parallel), so I needed a 1 ohm, 5W resistor on each blade to lower from 3.7v to the LED forward voltage of 2.65. The chief benefit of the resistor is that I can run off of one 3.7v battery, assuming my first question is good. The downside is that the voltage of the battery is not constant over the life of its charge and this will affect the output of the LEDs. As I understand it, the buckpuck maintains a constant current at 1000mA regardless of the voltage drop in the battery over the life of the charge (which is why the initial voltage requirement is higher than the LED). The tradeoff here is that I would need a buckpuck for each blade, because the 1000mA LEDs attached to a 1000mA buckpuck have to be wired in series to keep the constant current. That wouldn't be possible with two switches off the same buckpuck. So, I'm looking at 2 buckpucks AND my battery source would have to be 7.4v (2x2.65v from LEDs + 2V buffer = 7.3v), again assuming I can run 2 buckpucks from the same battery. That might tax my room inside. Going to one LED per blade does not help with the battery because I still need a minimum 5v (so realistically it has to be a 7.4V).

Is there any meaningful benefit to the buckpuck beyond the consistency of the current over the life of the charge?

Thanks for any comments on this. Wiring diagrams for resistor and buckpuck below, both based on common battery.

Attachment 18033

Ok, a few things.

You should be able to run both LEDs off of the two switches and the one battery, though if you have room for a 7.4V pack, I don’t understand why you wouldn’t have room for 2 18650’s ( they’re the same thing, essentially).

Buckpucks are just for a more constant current. I don’t think you can run two buckpucks off of a 7.4V battery, and it would waste a lot of the power anyway. This sounds like you’re overthinking it. I would go with the two individual 18650’s and appropriate resistors, and go from there.

I'm not actually sure that I will have room for a 7.4v battery. The issue is a combination of a narrow main body in the middle of 1" ID means no side by side and the position of the switch in the middle would cause problems with a long battery. But, separate 18650's should work. I'll do a little math and maybe see how far I can squeeze the switch down the body and cram it all in. Thanks for pointing that out; I don't know why I didn't think of that (tunnel vision, I guess). And thanks for the comment on buckpucks versus resistor.