I'm trying to power 5 WS2812B LED strips (60cm each, 180 LEDs total). The arduino is powered via USB because I need it to communicate with a computer.
Each LED consumes 0.3W so I calculated the whole thing draws 180*0.3W/5V = 10.8A.
I'm wondering which of these two methods is the preferred way to power the strips:
1
My guess would be the second because then the current for each strip is only 10.8A/5 instead of the whole 10.8A flowing through the entire thing.
I have three questions:
is my reasoning here correct?
where should the capacitor be located, in the diagram as well as physically; close to the power source or close to the LED? I read somewhere online that it should be close to the LED so that makes me wonder, in the second setup, do I need a capacitor in front of every strip?
does it matter whether I connect the ground of the arduino in front of or behind the capacitor?
Omit the usb and power the arduino ftom the same +5 as the LED's. Feed it in on the +5V pin.
When programming via usb, put a switch in the RED USB wire to prevent any possibility of power backfeed. I KNOW it shouldn't happen, but it pays to be safe.
follow up question: in my second diagram, for the 2nd and 4th strip, the power goes in the "wrong" end, ie power goes in where data goes out. Is this okay?
It is not like the data that is one way. It does not matter which end or both ends to apply power. The point is to not have the power traces on the strips carrying large currents.
