I'm trying to make a display out of 2811 LEDs from China (15 meters long x 7 LED strips). Each strip consumes 18W@5V(3.6A) per meter and I plan to use 7 strips, each 15m long. My calculations show we'll need a 5V power supply capable of producing ~460A...

(15 meters * 18 watts per meter * 7 strands of LEDs / 0.8 power supply efficiency) ~= 2300W @5V ~>460A.

I was planning on purchasing a 600A 5V power supply, running a power bus behind the LEDs and tack it to the strips every 2 meters to ensure even power distribution, but when I tried to calculate necessary wire gauge online, all the calculators say values "out of Bounds" suggesting that the cable needed would be too large. Would I really need to have a separate power bus for each strip?

How do we make this work? :-/ We're installing the LEDs between two elevators so it's not like we can hide power supplies in the middle. It would have to be at either end of the 15m runs...

Any crazy suggestions would be appreciated.

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    \$\begingroup\$ 3.6A is a lot of current. You're going to need multiple power buses. \$\endgroup\$ Commented Feb 21, 2014 at 4:27
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    \$\begingroup\$ It'd be easier to find a strand with all the LEDs in series and drive that with a higher voltage constant current. \$\endgroup\$
    – Theran
    Commented Feb 21, 2014 at 5:45
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    \$\begingroup\$ Or use multiple series strands strung end to end to work with even a bit higher voltage. \$\endgroup\$ Commented Feb 21, 2014 at 6:00
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    \$\begingroup\$ Look at the cables on a car battery. A car starter draws 100 - 200 A temporarily. You want more than 2x that continually. You would need to have 2 cables thicker than a thumb to carry that. Plus well-connected terminals. Anything over 32A is a bad idea in general. \$\endgroup\$
    – posipiet
    Commented Feb 21, 2014 at 8:01
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    \$\begingroup\$ These are smart rgb strips, with a controller (ws2811) in each led. This is why they are in parallel instead of series, and why they are 5v instead of a higher voltage. @posipiet \$\endgroup\$
    – Passerby
    Commented Mar 23, 2014 at 15:42

3 Answers 3


The typical rule of thumb for these strips is injecting power every 5m or better. Power should be injected at both ends, for the best color matching. FPC, flexible printed copper has relatively high resistance and it quickly adds up.

You need 1890W ( (15m * 7) * 3.6A ) * 5v. You mentioned the supply's power efficiency at 80%, but that should already be accounted for in it's stated output capacity. On the other hand, a 20% margin of safety, as to not drive the supplies at their maximum output, is a good idea. 2300W it is.

You have two options. Get a few 5v high current supplies, and run some cables in parallel to each 4~5m section. Due to the high current, this can get pretty expensive, as each 5m section takes 18 Amps. A quick calculation with a AWG calculator, At a 32 foot run (5 Meters to and from, you have to double the cable run) with 18A at 6V, you get a 0.6V drop... USING 10 AWG. That's 10% wasted in pretty thick and expensive cable. You could go with a 12v supply instead, at 18A, 10M, 20AWG, would result in 50% voltage drop, giving you 6v at the target end. Still not efficient.

The other option, is to get some high voltage, low current supplies (Say 48v), and some small dc-dc buck switchers able to handle 18A output. You only need 90W at the target. At 48V, 90 Watts is a little under 2 amps. 2.5 Amps, to adjust for the cable drop and switcher efficiency. This way you only need some 22AWG cable, resulting in a 2.5% voltage drop, which doesn't effect the switcher at all. You could even run it all in series, as the resistance drop at 22 AWG and 48v can be very negligible with the right switchers. And you only need a 48v 50A (2400W) supply to power it all. Way better than 5v 480A and high gauge cables.


You could consider something like copper bar or tubing. The resistivity of Cu \$\rho\$ is \$1.7\times 10^{-8} \Omega \cdot m\$

Recall R = \$\rho \cdot L \over A\$ , so to get a voltage drop of 300mV over 15m you'd need cross-sectional area A= \$ \rho \cdot 600A \cdot 15m\over 0.3V\$ or about \$500mm^2\$.

So, a copper bar 6mm x 80mm (about 1/4" x 3") would about do it. Total drop will be double (>0.6V), so you'll still lose more than 10% of your power.

It would be nice to find a higher voltage solution. You could consider distributing power at (say) 48V and having local buck regulators along the 15m supplying the low voltage. Or try to find higher voltage LED strips.


You haven't selected an answer yet so I'll offer some additional input. I would seriously consider rethinking your architecture. 460A is a tremendous amount of current. Don't do it this way if you want to transmit over more than a few cm otherwise you will need thick bus bars to distribute the electricity or have massive losses in the wire.

A standard architecture from the LED industry is to use higher voltage strips e.g 24, 48 or 60V that are driven from a constant current. Then each stip is supplied using a constant current LED supply. Then you can distribute the power to each strip using the mains. This way is more efficient than distributing the power to each strip on the low voltage dc side.

If you are adamant about using the 5V LED strips and are willing to tolerate more losses, then I would recommend using a separate 5V supply for each strip. By the way, using a separate power supply for each strip will also improve your system redundancy. If one supply goes down, the rest will still be alive.


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