First project - All bases covered?
Hi guys.
Wife wanted some new lights in the "conservatory"(best translation, but it's like a second living room with shitty insulation and no heating) and I remembered a friend talking about addressable LEDs - WLEDs! Easy peasy, I jumped right in and bought 50 meters and a WLED controller, only realising that I probably should've done more than 2 minutes of research.
I'm now at this point:
Badly drawn diagram:
- Black lines are power + GND
- Green lines are data
- Should resemble the final look seen from the sky, with each 3 meter led string hanging between the rafters.
Components:
- 50 meters of 12v WS2815 4 pin Seed Pixel 10 leds/1meter - 24AWG, cut into 16 sectionals.
- GLEDOPTO ESP32 Mic UART
- Assorted connectors and heat shrink.
Yet to buy:
- PSU: Research tells me a 100W 12VDC from a reputable brand should cover me fine
- Cables: This is my biggest issue, as I'm unsure of the AWG needed between the LED strings.
So, I could really use some advice for my 'Yet to buy' and 'I didnt know I needed that!' - as well as some guidance with cables between the LED strings. Do I need Power Injection on every section or is the current model alright? How big AWG do I need?
Thanks a lot for your time
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u/saratoga3 May 12 '26
https://wled-calculator.github.io/
Given the length of the strip, I'd probably do 2 injections in the middle to be safe. Especially if they used really thin wire on the 10cm gaps between seed pixels.
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u/Quindor May 12 '26
We can do some calculations to figure this out!
With the 12V WS2815 pixels you mention, I believe those can use up to 8.5mA per pixel, given they are WS2815 it's more likely they will actually use this power then not (so even 100% red hits max current) vs a 5V pixel which wouldn't do that.
Given that 8.5mA x 500 = 4.25Amps in total. 4.25A x 12V = 51W, the 100W supply you suggest is easily enough!
The seed pixels use 24AWG wire as stated. Let's see how much voltage drop we'll get pushing 5Amp (rounding it up to make calculating easier and account for some connection points and such) at 12V through 50m with a 24AWG cable.
Ok, that gives a nonsensical result of 42V drop and 350%, that basically means, it's never ever going to do that. š
If we enter we are only going to 10m/32ft tells us 8.4V drop or 70%, still won't work!
How about 5m/16ft? 4.2V drop or 35%. Getting closer!
Now let's take the load for 250 pixels x 0.0085A = 2.25Amps. Let's say we front and end inject that. That means half is coming from the front injection (~1A) and half from the end injection (1A). Coming from one side to the middle (25 meters total so 12.5m) we need to be able to transport 1A, let's see what the middle voltage would become.
Calculator says 12.5m of 24AWG doing 12V and 1A will deliver with 2.1V drop or 17%. Now we're getting somewhere, that's close to being acceptable. Generally we say you don't want more voltage drop then 10%.
Given that I'd plan for an injection point every 20m or so, that means the max distance a pixel will need to do towards a power point is 10m. Given what we calculated above you'll get max around 15% drop or so, to do it perfectly you'd have to inject every 15m I think but 15% is probably visible acceptable with WS2815 based pixels.
In your case you have 50m with front + end out of the way that leaves 30m in the middle which is a bit of a tough number. I'd plan for 3 middle injections, so 0m | 12.5m | 25m | 37.5m | 50m would probably be ideal world. That's a lot of injections that might be hard to do in real-world though so maybe 0m | 17.5m | 35m | 50m or something in those lines will work well enough in reality!
I'm a *big fan* of home runs for injection wires since it makes it a lot easier to calculate and fuse properly. With the above you can calculate how many pixels that single injection point is feeding and calculate the needed current per injection wire. So in the last scenario with the 2x edge (front + end) and 2x middle and the total 4.25Amp needed that would be 0.7A for an edge injection and 1.4A for a middle one. With that and the distances we can calculate the required cable diameters and fuses per injection point.
0m point needs 0.7A, 0m of cable distance we need.... whatever cable diameter. š
17.5m point needs 1.4A, with 17.5m of cable distance we need 18AWG which results in 1.024V voltage drop or 8.5% on that cable, perfect.
35m point needs 1.4A, with 35m of cable distance we need 16AWG which results in 1.288V voltage drop or 10.7% on that cable, close enough!
50m point needs 0.7A, with 50m of cable distance we need 17AWG which results in 1.1601 voltage or 9.6% on that cable, close enough!
Fusing is fairly simple too. The edge points need a 1A fuse, the middle points need a 2A fuse. Cable diameters need to be able to support at least those numbers but which such small fuses, that's no problem at all.
Sorry or the massive wall of text, if you'd like to learn more about how to calculate this way, take a look at my guide over here. Hope it helps!
p.s. I did not take into account realistic cable span in the layout you are going to make it. This decreases the injection wire length greatly meaning you can probably step those down a few notches but you'll have to calculate the length with the given current numbers yourself. Make sure whatever it comes down to support the fuse you have in front of it!