Resistor heat

[Luke-SkyMarcher]

This may seem like a rather stupid question, but is the resistor supposed to get really hot? I am asking this because of the trouble I have had with the saber this wiring goes to. When I had it wired with the freshly charged 4.8v battery pack and the 1.2 ohm 3 watt resistor, it worked fine for a while, then unexplainably stopped. When I took it apart, the resitor was discolored, its leads looked rusty, and the Hasbro board which was right next to it was blackened (and also didn't work). Now today, I rewired my red lux III with a spare 5w 2.2 ohm resistor, (the white box kind), and was carrying the bare wiring around, and the resistor began to be close to painfully hot, and the heatsink was more than just a little warm....
So, is anything wrong with this, or will somthing get fried? I just dont want my saber to fry...

-Luke

[LAN-ED-TUL]

did you refer to wiring guides section for any diagrams that would suit your situation?

i wouldnt think the resistor would get hot, maybe a lil bit warm, and the heatsink will get warm on you.

maybe the resident experts might know more.

[Do-Clo]

You need to recheck your wiring, and make sure what the voltage really is on your battery pack when fully charged. I use a 2.2 ohm 5 wattt resister with a 6 volt battery pack to power a luxeon 3 watt and it works great, no heat isues at all.

[Novastar]

Do-Clo is right... metering your pack will likely reveal to you that even though a "1.5v" battery (or whatever cell) is classified as such... they can often be more like 1.6.

No big deal, right? Well--it depends on:

1. How MANY cells you are using
2. If you are using Li-Ion as well

Examples:

One "3.6v" Li-Ion... is effectively 4.2v at full charge. They also tend to maintain voltage throughout a cycle, so keep that in mind.

When in doubt, use a resistor calculator online.

Another thing you can do is simply give the LED *SUPER CLOSE* (and slightly more) to what its forward voltage is (for example Lux III red is around 2.95). So a 3v setup will work very well, and you will not lose much energy via heat.

Ultra has done this before, and it works very well, I'd imagine.

[Luke-SkyMarcher]

I double checked wiring, polarity is correct, and the resistor is on the + side. Right now it is a simple curcuit, not even a switch. The battery pack is Tim's 4.8v pack, which does meter to around 5.3v when fully charged.

[Novastar]

Right. Batteries tend to "truly" meter well over their rated voltage... especially when combining cells (i.e. 3AA vs. 1AA). But, as you may know, most batteries lose voltage as they are run. 1.5v alks drop to like 1.2 or whatever.

However, Li-Ion cells maintain their voltage over a cycle (for the mostpart), and therefore... if you figure out what you need voltage-wise for your LED, and match it with a Li-Ion setup--you should be good for the entire usage until the battery nears the end of its charge.

Most "3.6v" or "3.7v" Li-Ion cells... meter to about 4.2v. I find that *perfect* for LEDs requiring 3.85v, or even 3.4v. IN these cases, you may not even need a resistor. Please someone correct me if I'm wrong.

You would definitely want a resistor if using a 2.95v (Lux III red for example) LED... and a 3.6v (4.2) Li-Ion.

About 30 minutes into a Li-Ion charge cycle, the voltage will end up around 4v, stay there for a long time... and then have a drastic drop near the end, which will trigger the PCB to prevent the battery from becoming useless...

But at that point--it's time to recharge the cell anyhow!

[Novastar]

Right, Dregan--a driver is always the best solution for an LED.

But a resistor is pennies, and a driver is anywhere from $10 to $130 depending on what it does and all that jazz.

Personally, I say if someone is doing a saber, why not do it right--Alex's board or Erv's... sound + light.