Resistor Calculation Double Check

[Kyaryo Ysoyav]

So I'm working on two stunt sabers right now. One of them will eventually be upgraded to a saber w/ sound, but just doing it as a stunt to get the hang of soldering. So saber 1 will be running a rebel Cyan LED at 1000mA. I'm still trying to figure out the best battery solution, but I'm thinking either 2AA or a 18500 single Li-ion at 3.7v. So is the correct resistor for that a 1 Ohm 2 W resistor?

Saber 2 is a bit more complicated. It will run off a 14500 pair for 7.4v. It has a red momentary AV ring switch. According to the calculator, I need a 1/4W 270 O resistor for this. I'll be running the two green dice on a Tri-rebel off that same battery. I'm debating direct drive versus a buckpuck. I can't figure out which resistor to use for this, but are there any advantages of the buckpuck? They are quite a bit more expensive.

Thanks for the help!

[Silver Serpent]

The 1 ohm 2 watt resistor is correct for a single li-ion on your first saber. Using 2 AA batteries will not give you enough voltage to properly run that LED. It'll probably still illuminate, but it'll be dim.

Saber 2, if you're using two green rebel dice in series, then you'll need a 1 ohm 2 watt resistor as well with that battery pack. In series, you add the Vf of each die together (3.4v + 3.4v = 6.8v). Your calculations on the switch LED are correct.

Using a buckpuck will give you a more efficient use of your battery. Less energy is wasted as heat, which is how resistors tend to operate. You'll get a longer lasting, more consistently bright blade with the puck. The downside is the price and the extra space it takes up inside the hilt.

[Kyaryo Ysoyav]

Thanks Silver! I went ahead and ordered the resistor. I plan on upgrading to sound before too long, so I'll save the buck puck money for that!