# Why is my calculated wire size a lot bigger than the traces on my LED strip?

I've 3D printed a nanoleaf assembly which will have WS2812B LED strips in them. I'm calculating the correct wire size but my result seems off.

I have 14 hexagonal parts which have sides of 10 cm. I use WS2812B LED strips with 30 LEDs/m. Each led uses a little less than 60 mA (20 mA for each color).

So calculating the needed amperage:

14 x 6 x 0.1 = 8.4 m
30 x 8.4 = 252 LEDs
252 x 0.06 = 15.12 A

I ordered a 5V/20A PSU and now I just need wires and maybe a DC connector. The distance between the PSU and the strips is about 2 m (6.6 feet). The last two evenings I've read all related Q&A on this site, looked at charts I could find and used online wire size calculators. Every source I consulted says I need wires between 10 and 7 AWG, so between 2.6 and 3.7 mm.

The specs of the WS2812B strips says I need to add power every 5 m. So calculating the needed amperage for 5 m:

30 x 5 x 0.06 = 9A

Now when I look up what wire size would be needed to conduct 9 A over 5 m it seems I'm doing something very wrong. The size I find is nowhere near the size of the traces on the LED strips.

What confuses me even more, when I look at e.g. my USB C PD charger for my laptop which supplies 100W (at 20V, so 5A), the 2.5m cable seems super thin.

• In addition to the answers others provided below, you won't have nine amps for the entire length of the thing--each LED you pass takes 60 mA, and the power bus after that LED is carrying 60 mA less. Jan 7, 2022 at 1:33
• You should take in consideration not only wires capability but also wires resistance. If voltage drop lower minimum for IC, chip may not work correctly. Jan 7, 2022 at 1:38

I think this might be a case of your manufacturers taking a shall we say forgiving interpretation of current handling rules. The guides you've probably seen are usually very conservative. If you have a good idea of the capabilities of the specific wire you're using and the environment it's going to be used in, you can usually find that you can get away with thinner wire than what a generic table is going to recommend, which is why your USB-C charger probably isn't using 14 AWG wire.

In the case of your LED strips, the IPC-2221 trace width for 9A on 1oz copper is ~3.5mm for a 25C temp rise; I'm guessing your traces are narrower than this. If you really want to roll the dice, 1.83mm width will give you a 75C rise which gets you uncomfortably close to the typical FR4 PCB transition temperature of 130C. I'd play it safe and supply power to the strips more frequently than every 5m, especially since the LEDs are going to be heating the board too.

Wire (and trace) selection is concerned chiefly about temperature rise. If you can tolerate a 10 or 15°C temperature increase, then you can use a thinner conductor than if you could only tolerate a 5 to 10°C temperature rise. Different calculators may have assumptions about tolerable temperature increases, or may allow you to specify. (And trace width calculators can be based on different standards, such as IPC-2221 and IPC-2152.)

The trace width on LED strips tends to be poor. If you supply 5V at only one end, you will measure an appreciable voltage drop at the opposite end. The heavier gauge wire supplying the LED strips at regular distance intervals will help avoid this problem by providing a less resistive pathway.

Note that unless you are planning on running the LEDs in constant white, your current usage will be less. But planning for the maximum current usage with reasonable overhead is a good strategy for a system that will perform well and not overheat.

Here is a random wire gauge chart:

(Source)

For 9 amps, you will note that 21 gauge could be used for chassis wiring, or 12 gauge for power transmission. Well what do they mean by these terms?

The Maximum Amps for Power Transmission uses the 700 circular mils per amp rule, which is very very conservative. The Maximum Amps for Chassis Wiring is also a conservative rating, but is meant for wiring in air, and not in a bundle.

The main difference here (again) is about acceptable temperature increase. Note that the "chassis wiring" column is meant for situations where the wires are able to dissipate heat easily.

If in doubt, use a heavier gauge wire, but of course it will add expense.

To address your confusion about your USB charger, wire gauge is not about power, it's about current. The insulation on a wire is what provides the voltage rating* (better insulation = higher voltage) while wire gauge is what provides the current rating (heavier gauge = higher current).

*And also its temperature rating. An insulator that melts at some temperature could create a hazard when the conductor reaches its temperature, potentially causing a fire even though the conductor isn't yet at risk of failing.

The simple answer is that you do not need wire that heavy. As you use thinner and thinner wire, it will have higher resistance and heat up more. At very high temperatures the wire will eventually melt the insulation. Those charts are extremely conservative.

Here's a answer I wrote a while ago for what is essentially the same question: https://electronics.stackexchange.com/q/580866

Designing for high current draw led strips is a practice of balancing nominal vs actual values.

The led strips themselves use comparablely high resistance FPC. This prevents the strip from actually drawing the full currents due to the way the cumulative resistance adds up compared to the current draw. On the other hand, each section of the led strip carries less and less current than the previous. It's all linear.

You already know that you should inject power every 5m but in practice you should probably do it at every 2.5m and at both ends, to best reduce the FPCs resistance issues.

As for the total wiring, you seem to be looking at the power transmission chart instead of the chassis wiring chart. The chassis wiring is what you are doing. A relatively short length, inside a case, likely unbundled. So your not running 100 conductors inside a single cable where the heat really really adds up. You are likely using individual wires or zip cable. And depending on your design, the cumulative current draw will be different in each module.

But let's consider your actual use case, 14 independent modules. Each one only 0.6m of led strip or 18 leds or about 1 Amp. Each module can be run independently, can run power in parallel and if arranged wisely will have minimal need for high capacity wires. The input wires will need to carry the whole current, but if you use a tree topology, or branch out in a star topology only one or two modules will see all that current.

If you connect 3 other modules to the primary module, the primary module will see 1 Amp for itself and 14 Amps for the others, with each output possibly carrying between 1 and 12 Amps. If each of those 3 have an equal number of modules connected to them, the average output from the first is 4.33 Amps each. Etc. If you connect two downstream modules together then the power gets divided too. It gets complicated quickly in that. But if you only connect one module to each other in a large series setup then you do have to worry about the current adding up.

Practically speaking there's two answers here. One is that you use a generous size cables inside each. For your target of 15A and never expanding, at 2 m to the module and another 2 meters of just power wiring, then 14 AWG should be sufficient.

The second solution is localized power regulation. The problem with low voltage at high current over long runs is that the conductors needed get bigger and bigger. If you reduce the current needs, then the problem goes away. This is why 120/240V is used instead of 12V or 5V for house wiring. The power requirement stays the same, but the current does not. If you switched your target from 5V 15A or 75W to 12V, for 75W you'd only need 6 Amps. For 24V that's 3 Amps. Now you can use 18 or 20 AWG wire. But the tradeoff is that your module is more complex. You will regulate the 12V down to 5V inside the module for your leds. A switching regulator in each module would do it. You have to adjust for the regulator penalty as well, say 20% of the local power need (5V 1A or 5W out would need about 6W in).