Hi everyone!
I was reading the Proffie v2 info page and came across this line:
Battery- will carry the combined power of all your LEDs, which is a fair amount of power. It is recommended to use thicker wires, for these wires.
And this got me worried. I am building a double-ended saber using two second-hand SF LEDs that are RRBB. These are wired to LED pads 1, 2, 4, 5. I am now wondering just how thick the wire from BATT- to battery should be, after reading that line, and if mine is too thin. Can I get away with using the same wire as for each of the LED negatives or do I need to step it up?
Now, I’m no expert in electronic wiring, but I understand the basics, and as I see it, both same-colored dies are essentially in parallel on each LED, and the LEDs are in parallel to each other. Let’s assume that the voltage drop is 3V for the blue and 2.6V for the red, and that the LED can be run at 1A. That means that if running both sides blue, I’m using 3V and 4A with a battery that can supply 3.7V and 10A. Am I on track so far?
Since P = I * V, then the power carried by the negative wire to the battery is (3.7-3)*4=2.8W, right? But how does this help me choose the correct wire gauge? Power isn’t reported on any wire that I’ve seen, usually it’s just voltage, and my wire is good for 30V. Unless the power mentioned in the quoted bit means something different; I feel like I’m getting a tad confused at this point and would appreciate some help or clarification from those who know better. Thank you!
Wire thickness is basically entirely based on the number of amperes the wire needs to carry. In your case, that should be somewhere around 8A, assuming each die carries one ampere. There are no hard-and-fast rule, but this handy-dandy chart can help you approximate what wire you will need:
https://www.powerstream.com/Wire_Size.htm
So according to this, 22 gauge wire should be good.
22-gauge wire can be hard to work with in a saber, so you could use 2 24-gauge wires instead. Also, since the wires are fairly short, a single 24-gauge wire will work fine in a pinch.
Current when using pixel strips can run up to ~15A, in which case 22AWG wire is recommended. For your High Power LEDs, you should be fine with smaller gauge, but 22 really isn’t SO big you can’t use it, is it?
Don’t need to overthink this one too much.
Hmmm, I see; yes, it would be 8A if I had all dies running at once, is that what you mean? What I have in reality is one wire for blue and one for red running to each LED. But since I don’t know if the dies are actually connected in series or parallel on the star, I can’t tell if each wire is carrying 2 or 1 ampere.
To make matters even more “fun”, I have a wire bundle that I bought years ago from JQ Sabers, where the wires don’t seem to have gauge markings on them. What I’m currently using feels like it’s at least 24, but probably actually 22, gauge wire, and it’s a bit hard to stuff into the saber, given my interesting setup, so I was considering using a smaller wire, but after this, I suppose I will leave it as is.
Overthinking is what I do. 
Yes, I am aware that 22AWG is recommended for pixels, but I didn’t think I’d need it for a couple of stars. Essentially, I am looking for understanding, rather than a direct solution, so that I can use that understanding in the future to solve similar problems.
By the way, I’ve also seen a nifty chart by Shtok, showing that the max draw of 2 all-white strips at max brightness was around 9.8A. Kind of good to use as a guide, though I also have purchased 15A batteries for future pixel builds, rather than relying on 10A ones.
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If you wish for a deeper understanding, all you have to do is to use the resistance values listed in the table I posted earlier, then apply Ohm’s and Watt’s law.
For example, let’s try using AWG 30 wire:
Table says we have 338.496 ohms per km. Let’s say we have 20cm (back and forth between the battery and the board.) That gives us 338.498 / 1000 * 0.20 = 0.0677 \Omega
Now we applies Ohm’s law: V = R * I = 0.0677 \Omega * 8 A = 0.54 V
That means that we will lose more than half a volt in the cables. So instead of a 3.7 volt battery, we now have a 3.2 volt battery. Since sabers use relatively low voltages, dropping some voltage in the cables is not good.
Time to apply Watt’s law: P = I * V = 8 A * 0.54 V = 4.3 W
So our wires will put out 4.3 watts of heat. That means that it will become about as hot as a 4-watt lightbulb. (Common in nightlights and the like.) The heat dissipation might be a bit better/worse, but it’s not an unreasonable comparison.
So, in conclusion: Don’t use 30 AWG wires for 8A.
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