Throwing in some capacitors to smooth the voltage out when changing the strips rapidly
Taking each supply being able to send 13A each (130 W/ 5 V) and a nominal current of each LED is 20 mA (60 mA at max brightness) - the total draw of all 26m would be about 31.2 A, just short of the total 26A that both power supplies are able to provide. In software, you can tone down the brightness a bit but that also doesn’t leave too much headroom (since no motors are involved it isnt quite as essential but still definitely desired).
Just a heads up regarding the power supplies, make sure a licensed electrician is installing them. Mains power isn’t just lethal, it’s lethal!
When talking about different voltages its better to convert everything to Watts then divide through to get current. A total of 260 W between both power supplies and an efficiency of 75% we’d be looking at about 350 W, RMS of 240V thats about 1.5 A. I’d check out this excellent tutorial on Ohms law: Ohm's Law - Practical Examples for Makers - Tutorial Australia
Just to clarify each LED will draw around 20 Milliamps, with each meter drawing about 1.2 A. So each power supply is able to provide power (and still at 20mA each not the maximum 60mA per pixel) for about 10.5m. Divvying up the strips amongst 3 power supplies would probably be your best bet.
Just to add to Liam’s points, the traces on the flexible PCB that the strip consists of can only take so much current, so it’s usually a good idea to split the supplies across the strip not only for the sake of using small supplies, but also to keep the current along the strip low.
Good luck with the project, and let us know if anything else comes to mind!
-James
2 Likes
And you can get our latest projects and tips straight away by following us on: