# Building a power rail for a project – help choosing wire gauge please

Hi all,

I'm tackling my first major arduino project and need some help with the electrical safety side of things. Specifically, I need to figure out what gauge wire to use to safely build a power rail. I'm a lawyer – not an engineer – so the electrical side is a bit new to me, though I'm trying to learn.

Right now I'm building a proto board to distribute power. The project uses both 12v and 5v components, so the board takes a 12v input and then uses a 12 -> 5v regulator I bought to split it off into a 5v rail.

The two rails have the following draws at peak load:

5v: up to 30 30ma LEDs (so about 900ma), 4 shift registers @ ~80ma, 6 moisture sensors @ 50ma. Total current of 1.52A.

12v: 1.1A pump, up to 6 500ma solenoids, and (indirectly) everything on the 5v side. Total current, including the 5v side, of 5.62A.

It's very unlikely that all of that would be powered on at once, but I'd like to build for that case.

This website:

seems to indicate that 22ga wire can handle up to 7A for "chassis wiring," which I assume is what I'm doing here. My questions are 1) is that correct? and 2) is that 1.5A over my theoretical peak load enough of a safety margin?

Thanks for any help you can provide. If I haven't been clear, happy to provide any additional info required.

Use a SMPS.

A 12 to 5 regulator @ 1.5 amps means you have to dissipate 7*1.5=10.5watts.
You will have big problems!

LarryD, thanks for your reply. I'm trying to avoid buying components to the greatest extent possible, so would like to find an alternative to a SMPS. Am I correctly understanding that dissipating 10.5W generates an unacceptable amount of heat?

If I power my LEDs with the 12v rail instead and use a larger resistor, I think I can get the load on the voltage regulator down to about 500mA peak and dissipate only 3.5W. Is that a manageable amount?

Then you're dissipating that much heat in the resistor, unless you can also rearrange the LEDs so they're in a few groups in series, so the applied voltage is higher and current is lower. Ofc, if you need to turn them on and off individually, this is a non-starter. And besides, the shift registers probably can't switch 12v, if that's what you've got them doing.

You need to use a DC-DC converter (SMPS typically refers to designs that have a transformer in them, this isn't needed for a DC-DC converter, though the term is sometimes used to encompass all kinds of switching supplies) for this. But they're very, very cheap - just like Arduino pro mini clones, the assembled boards cost less than you or I could get the bare part for: http://www.ebay.com/sch/i.html?_nkw=buck+converter+step+down

20mA thru LEDs, say LEDs have Vf of 2.7V, then resistor has (12 - 2.7) = 9.3V across it.
Power dissipated in the resistor P = IV = .02A x 9.3V = 186mW.
Can you wire the LEDs so 3 or 4 are in series and share the same 20mA? Then less power is dissipated in the resistor.
Will also need a transistor to switch current on & off, 12V will smoke an Arduino output.

DrAzzy, thanks so much. Turns out I am not very clever and used imprecise language – the part I got is a DC-DC converter rated up to 3A. The part is here: http://www.amazon.com/gp/product/B00CXKBJI2?psc=1&redirect=true&ref_=oh_aui_detailpage_o04_s00 The LEDs will be at different locations and pretty dispersed, so I don't think heat will become an issue for them under any conceivable operating conditions. I am using a 74HC595, so I think you're right that it wouldn't care much for 12v switching.

CrossRoads, thanks to you as well. The LEDs need to be individually addressable so I can't run them in series. However, based on what everyone has told me, I am going to test the project running power through my 12V -> 5V/3A converter at a smaller scale and monitor the heat buildup. If it's acceptable, I will scale up accordingly.

I really appreciate all the help! Will keep you posted with my results.

Hi,
One of the governing factors of wire gauge and current is the length of the wire you are going to be driving that current.