TOPIC: Resistor help

I've been looking into a DIY lightsaber for months and decided to do one with parts from each company. After a ton of reading the last thing i need to figure out is which resisitor to use. Ive tried to comprehend figuring it out myself through the formulas posted on different forums but math is not my strong suit. The parts i have are as follows:

Saberforge 12w+ LED Green
Saberforge 18650 Battery
Saberforge 4w 4Ohm Bass Speaker
Sabercore 2.0 Soundboard
Saberforge Lit Momentary Switch

I also am converting a Darth Vader Black Series and am using all the same parts as above but a red LED and need a resistor foor that as well.

Any help is greatly appreciated!

I'm curious on this topic as well, except I'm wiring a Spark2 with 6w green led and 6w red led. I don't think I need a resistor for the green led, but I believe I need one for the red. And from the manual, not clear if I need a resistor for the lit switches or if the Spark2 drives it.

Kouri, hopefully you can provide some guidance

If you get the complete SF LED unit, heatsink and all, they come pre-resistored for a 3.7v battery.

Funny enough, the speaker might need a resistor. As far as I can tell, the Sabercore soundboard expects an 8ohm speaker, so SF wires up the 4ohm speaker in series with a 4ohm resistor to compensate.

As for LEDs, last I remember, specs are similar to XP-E2s, so resistors will be (going off memory - may not be 100% accurate)

Photo Red, Red, Red Orange, Amber: 1-1.2ohm

Green: Can attempt without (or use two 0.5ohm in parallel for 0.25ohm resistance)

PC Amber, Blue, Royal Blue, White: 0.5ohm

Not sure whether the new 12W+ Amber are Standard of Phosphor Converted.

Thanks Kouri any chance you could comment on the Spark2 / provide a diagram? I bought a DIY LED, so have to do the wiring myself and I don't think I need the parallel wiring for the channels the board has. Plan to power with a 3.7v battery with 2 lit switches, a green/red LED and the 2w 8ohm speaker that SF sells. Also going to add a recharge port, but since the switch won't have the extra hole, I plan to stick it in the pommel

From one of your previous posts you advised for an LED with 2 -ve leads to add 0.5 ohms to each, but not sure what to do for the +ve LED.

Not hard rules, but a couple basic guidelines

1) Resistors can go on either the positive or negative line, but shouldn't go on both.
2) If an LED has shared contacts, put the resistor on the individually wired end (since SF LEDs have shared positives, put your resistors on the negative wires).

Following the above diagram (assuming blue wire goes to Green LEDs, orange to Red LEDs), you're probably fine putting the two greens without a resistor and the doubled-up red with a 0.5ohm.

If, however, you've got an X4 with individually addressable negative contacts, you might want to consider putting one Green on TruDrive1, the second Green on TruDrive 2, the first Red with a 1ohm resistor on TruDrive 3, and another Red with 1ohm resistor on TruDrive 4. Then connect TruDrive 1&2 to BoardChannel1, and TruDrive 3&4 to BoardChannel2. This way, the mosfets can dissipate heat a bit better, and your Red LEDs are safe in the event a solder connection comes loose.

Both builds are going into Black Series Conversions so i purchased the SF LEDs to mount on the heatsink that comes with the conversions. I did want to get the prebuilt LED module to open it and see what resistor comes standard but i figured i should probably save the $20 and figure it out myself lol. Definitely harder than expected

What would the wattage need to be with the 4ohm resistor? Or how would i be able to figure that out?

Green should be fine without.

Red, use a 1-1.2ohm 1-2W on each individual diode, or double them up and pop a 0.5ohm 2W+ on each pair.

Not sure on the speaker resistor, since I've never used a mis-matched speaker before. I'll make a wild guess at 2watts being plenty. 1watt might be passable.

I'm honestly still pretty new to audio circuits. My last bit of research suggests a 4ohm speaker won't break anything with an 8ohm amp, but it probably won't sound very good or might produce excess heat on the board's amp without the resistor.

Kouri, you rock

I assume I have an X4 from SF, as there are individual -ve connectors: https://www.shopsaberparts.com/collections/do-it-yourself-electronics/products/12wstar

Last question, what would the resistor be for the lit switch in your opinion? I used one those dynamic ones in my last ASP but perhaps I should have used something different.

Thanks again.

how about lit AV switches? would these need resistors? would it matter which soundboard (sabercore/NBv3/spark2) i use vs the resistor?

I... don't remember voltages on accent LEDs. I want to guess 3v for blue and green, 2v for red and amber?

20ma = 0.02A

Batteries cap at about 4v...

Green calcs being

X ohm = (4v - 3v) / .02A = 1V / 0.2A = 50 ohm
X W = (4v - 3v) * .02A = 1V * 0.02A = 0.02 W+

Red calcs at

X ohm = (4v - 2v) / 0.02A = 2v / 0.2A = 100ohm
X W = (4v -2v) * 0.02A = 2V * 0.02A ~ 0.04 W+

So 50ohm on 3V accents, 100ohm on 2V accents. 1/4W either way? As long as the ohms match up, you're always fine going with higher wattage resistors. Only problem is the higher you go with wattage, the bigger the resistors get.

the words are english but my brain thinks i'm hearing jabba the hutt...

does it matter how "far" the resistor is to the switch? meaning it's a bit tight in the switch section, so i'm thinking if it's possible to move the QCs farther out with about an inch or so of wire so the resistor and QC of the switch come out in the body area. this is for an ASP saber.

Bit confused about these resistors that's from a SF Blood Orange led... I know it's 1 ohm but all I got from the "J" is 5%... But of what? Any help is appreciated!

That is tolerance. So it is saying that the resistor is between .95 and 1.05 Ohms.

Hmmm ok... So there's no indication for wattage?

Gnost-Dural wrote:
It does not matter. The length of wire does increase the resistance, but the increase is pretty much negligible.

For example, if using a 24 gauge wire, you would have to add approximately 42 feet of wire to increase the resistance by 1 ohm (1 inch of wire gives you about 0.002 ohms). So a couple of inches has minimal affect.

I think SF uses 28 gauge wire, therefore, about 17 feet of wire gives you 1 ohm (1 inch of wire gives you about 0.005 ohms).

I digress. To answer your question, it does not matter.

Many of my problems have been figured out and the rest should fall into place, I greatly appreciate the help!

MrBond wrote:
I know I kinda hijacked this thread but does anybody have a clue as to the wattage of these resistors?

MrBond wrote:
Based on what tolerances the resistors would need, I believe these are 2W resistors. I'm basing this on the following since I don't design audio:

We know that we will be running the LEDs at 3.7V and 1A optimally, and that there will be two LEDs paired up per resistor. I'll be assuming these are from a red module since I've seen these on mine and the red takes 1 ohm resistors.

Since a red LED uses ~2.65V at 1A, that means the resistor will be dealing with just over 1V, and since each resistor only deals with two die that means they should only be dealing with 2A ideally. Since P=IV, that means 2W resistors should be roughly enough to deal with the power. Any larger is a safety margin. Since the resistors seem to heat up pretty quickly I'd say they didn't go too much higher than the required tolerances, though.

As for the speaker resistor, we know these from the shopsaberparts website:

I'll assume the speaker is treated as a resistor. We also know that P=(I^2)*R. In that case, that means the 8 ohm and 4 ohm speakers can tolerate 0.5A and 1A respectively. From V=IR, then, we then know these speakers are both intended to be run at 4V. So I think I can assume that the output of the speaker pads runs at the voltage of the battery (~4V), but is limited to 0.5A for the speaker. Doubling adding the 4 ohm resistor to the 4 ohm speaker would limit the current to acceptable levels. Since you have two equivalent resistances in series, they should each deal with half of the voltage (meaning the speaker and resistor would be handling ~2V at 0.5A, or 1W).

With all this in mind, that means the speaker resistor would only have to deal with 1W by my thinking. If Kouri has further input that would be welcome.

Edit: More simply, two equivalent loads divide the power equally. So two 4 ohm loads driven by 2W should only deal with 1W apiece.