Power injection, done properly

When a run needs it

When the estimated drop from a single feed passes the point where colour shifts — around 10 % of the rail — and certainly before it passes the point where the pixels stop working, around 25 %. On a typical 10 mm 5 V strip at 60 LEDs/m at full white that happens within the first metre or so; a 5 m WS2812B run cannot be fed from one end at all. A 12 V WS2815 strip reaches several times further, because its current is a third and its rail is 2.4× as high.

There is a second reason that has nothing to do with brightness: the strip’s own power trace is roughly 0.07 mm² of copper — about the ampacity of 28 AWG, good for a few amps. Pushing 18 A into one end of a strip overloads the trace and the connector feeding it long before the supply notices. Injection spreads that current over several entry points.

How many, and where

Feeds cut the run into spans. A span fed from both sides has its worst point in the middle and drops r·i·ℓ² / 8, where i is the current per metre and ℓ the span length; a span fed from one side only — the tail past the last feed — has its worst point at the tip and drops r·i·ℓ² / 2, four times as much for the same length. So the first rule is: never leave a long tail. Feed both ends, or feed the last point close to the end.

Evenly spaced feeds including both ends give p − 1 interior spans of L / (p − 1), and the drop falls with the square of that count: two feeds give a quarter of the single-end drop, three a ninth. Evenly spaced is not optimal, though — two feeds at a quarter and three quarters of the run beat two feeds at the ends by a factor of four, because they turn the long interior span into two short ones plus two short tails.

PixelPower estimates the fewest evenly spaced feeds that keep your run within the threshold you set, then lets you drag them to where the installation allows and shows the voltage at every point for that layout.

What each feed buys

Estimated drop at the worst point of a 5 m WS2812B run at 60 LEDs/m, standard copper, as feeds are added evenly including both ends. A dash means the estimate is past the point where the model describes a working strip.

What each feed carries

Each feed is responsible for half of every interior span beside it and all of any tail beside it. Its cable carries that territory times the current per metre. In an evenly spaced layout the two end feeds therefore carry half what an interior feed does — which the simple I ÷ feeds arithmetic does not capture, and which decides the fuse.

Each feed cable is a branch from the supply, and it drops voltage too: the plain 2 × ρ × length / cross-section × I, nothing halved, because it carries the full current of its territory the whole way. Keep the feeds similar in length and gauge; the model assumes they sit at the same potential, and where they do not, current shifts toward the stiffer one.

Fuse every feed

A supply’s overcurrent protection is set to protect the supply, at a current far above what a strip trace or a JST connector survives. A 5 V 60 A supply will deliver 60 A into a fault on a trace good for four. So each feed needs its own fuse, sized above the current it delivers at worst case with the usual margin and below what its cable and connector can carry. PixelPower recommends a value for the busiest feed and counts one fuse holder per feed in the parts list.

These are planning estimates. Connectors, solder joints and hot copper are not in the model, and real ampacity depends on insulation, bundling and routing. Verify against the parts’ own datasheets and your local regulations before you build.