Wire Sizing for LEDS: Long Distance to Switch

[MCI-89]

Hi,

I've got a 40' coach bus, and am working on getting all the electric ready before the spray foam guy comes next week.

The bus will not travel much, and will run on 30A shore power. I have an AC system setup, and in my utility closet, have an outlet with an AC to DC power converter and a fuse block.

I want a light switch at the front of the bus to turn on/off 14 small LED puck lights (.2amps each).

The issue I'm having is the switch is about 30' away from the DC fuse block, and after getting power to the 14 lights, I will have around 80' of wire in one direction.

Using a voltage drop calculator, it is suggesting 5 gauge wire!

Any ideas on what I can do? I've never seen anybody run such thick wire for LED's. Am I missing something?

Is there a way to have a switch in the middle or end of the circuit in order to cut back on that 30' trip to the switch?

Any info would be appreciated. I had 12/2 wire ready to go thinking that was overkill, now I'm not sure what to do!

[PorchDog]

14 lamps X 0.2Amps = 2.8 Amps total load. 80 feet of 12 gauge solid wire has 0.127 ohms of resistance. (Stranded might be a little higher...) V=IR so 2.8 Amps X 0.127 ohms = 0.3556 of voltage drop over the 80 feet. That is negligible.

I think you will be fine with 12/2.

But double check me...

[DeMac]

Originally Posted by PorchDog

14 lamps X 0.2Amps = 2.8 Amps total load. 80 feet of 12 gauge solid wire has 0.127 ohms of resistance. (Stranded might be a little higher...) V=IR so 2.8 Amps X 0.127 ohms = 0.3556 of voltage drop over the 80 feet. That is negligible.

I think you will be fine with 12/2.

But double check me...

-----------

Imho, no worries, using the 12/2.
Double check the ratings on the switch.
Both max wire size and max amps.

Might two switches & two circuits be an option? A second homerun. (Not req'd)

Hopefully DesertDog will offer up more details. 🙏

[ewo1]

Originally Posted by PorchDog

14 lamps X 0.2Amps = 2.8 Amps total load. 80 feet of 12 gauge solid wire has 0.127 ohms of resistance. (Stranded might be a little higher...) V=IR so 2.8 Amps X 0.127 ohms = 0.3556 of voltage drop over the 80 feet. That is negligible.

I think you will be fine with 12/2.

But double check me...

Is that 2.8 amps, lamps wired in series?

What your calculation if wired in parallel?

If you wire the lamps in series, if one goes out they all will go out behind the lamp that failed.

In parralel, if one goes out the rest will stii turn on.

[MCI-89]

Originally Posted by ewo1

Is that 2.8 amps, lamps wired in series?

What your calculation if wired in parallel?

If you wire the lamps in series, if one goes out they all will go out behind the lamp that failed.

In parralel, if one goes out the rest will stii turn on.

I plan on wiring in parallel, but just added up the amps for each light I'm using. Is that the right way to calculate total amps for parallel?

[MCI-89]

Originally Posted by DeMac

-----------

Imho, no worries, using the 12/2.
Double check the ratings on the switch.
Both max wire size and max amps.

Might two switches & two circuits be an option? A second homerun. (Not req'd)

Hopefully DesertDog will offer up more details. 🙏

Thanks. I do have multiple circuits and switches, this is just for one! The lights are tiny 2.5" puck lights.

[ewo1]

Originally Posted by PorchDog

14 lamps X 0.2Amps = 2.8 Amps total load. 80 feet of 12 gauge solid wire has 0.127 ohms of resistance. (Stranded might be a little higher...) V=IR so 2.8 Amps X 0.127 ohms = 0.3556 of voltage drop over the 80 feet. That is negligible.

I think you will be fine with 12/2.

But double check me...

I'll double check ya but only so I can refresh my memory on Ohm Law... it's been a while...

In a series circuit "I" total is equal to the current thru any individual load.

It=I1=I2=I3

In a Parallel circuit then It=I1+I2+I3

So then the formula you gave is for a circuit wired in parallel which is best for the bus because as I stated before, in a series circuit if your first few lights go out so will the rest of them which will be lots of fun troubleshooting which one went out.


I got an As in Electronics engineering and man, have I forgot alot!
I'm glad you made me think today!

[desrtdog]

I agree with DeMac - run multiple switches...multiple zones so you don't have all the lights on at once.

...Smaller bus than yours, but I used 12/2 stranded marine wire for my lights - 12 total lights which are the same as what you have (2.5" LED pucks) - but have 3 separate circuits for different zones (left, right and rear). I also ran the main power to a terminal block like this one (used jumpers to get multiple (+) and (-) terminals as needed)., From that, individual wires to each light, so I think that technically, the only portion of the wiring that sees the full load current is that from the fuse block to the terminal block...which is maybe 10-30 feet total wire run (pos + neg).

Below is a screen capture from my favorite conversion electrical resource faroutride.com's wire size calculator...even at 80ft total run, max wire size you'd need appears to be 10 AWG.

[Jono14]

For puck lights you can be relaxed about the % voltage drop and use 10% in the calculators.
That shifts the recommended wire to 16 AWG which will have ~7% voltage drop over 80 feet at 2.8A.

However, if instead you have say 3-5 switches and run 3-5 pairs of wires to 3-5 different "zones" (as others have suggested) then each zone will only take about 1A (assumes ≤ 5 lights per zone) and then using 18/2 wire will also be fine and cheaper.

Running 3-5 pairs of 12 gauge is overkill for the amps and the costs will add up.