I'm working on a project where I want to mount an LED Light bar to the side of my house. The light bar will be closer to the peak of the roof, and the power supply ideally would be in the basement so all I'd have to do is flick a switch.

I'd like to run a wire down my wall, into the basement. I have an electrician friend that can help me with that part, but I'd like to know is if I can run a DC line that far (approx 30-40 feet)? And can I use Romex wire to do it, or would it have to be something else?

Here is my house with a very crude representation of what I'm trying to accomplish:

enter image description here

Here is the light bar I'd like to use:

Light Bar

And here is the power supply that would be mounted to the wall in the basement:

Power Supply

  • 1
    Please copy the spec of the light and supply into the question. How much current are you trying to draw? how thick are the cables you are planning to use? What is romex wire?
    – Tom Carpenter
    Sep 6, 2016 at 18:58
  • 1
    Tom Carpenter doesn't show up under "@" ... but if you're reading, would it kill you to google "romex" ? Sep 6, 2016 at 19:04
  • 3
    Is there any chance that you could mount the power supply next to the light bar and run mains for the long wires? Sep 6, 2016 at 19:50
  • Your best wire for this is likely "Landscape Lightng Cable". Using vista pro's online calculator recommends 10 gauge for that length. Landscape lighting cable should be in stock at your favorite big box store.
    – Tyson
    Sep 6, 2016 at 20:28
  • @Randell McGlynn I have added an efficient answer for you. Sep 16, 2016 at 4:40

5 Answers 5


At 25A (300W @ 12V), the wiring is going to have to be fairly hefty.

Since the power supply does not have remote sensing, you'll just have to manually crank it up to compensate for the voltage drop in the wiring.

According to the datasheet, it can be set up to 13.2 V, which gives you 1.2 V of "headroom" for the wiring.

1.2V/25A = 48 mΩ total resistance.

2 strands of 40' = 80' total wire, so the wire can have a maximum of 48 mΩ/80' = 0.6 mΩ/ft of resistance.

Using copper wire, you'd have to use AWG8 to get to this level and deliver 12V to the fixture.


Unless you are using the power supply for something else, you can cut your losses and costs by using high voltage transmission technique.

You can dedicate this power supply to the LED fixture and move the power supply up the attic. You can then use commonly available, cheap, rugged Romex NMB 14/2 from your 1st floor to your attic to supply the power to your power supply.


This will save you from using expensive 8 gauge conductors. Also the losses in the transmission line (Romex here) will be lower as compared to the DC on 8AWG. Also you don't have to use a sensing power supply as your power supply will be close to the load.

This technique is used by electric distribution companies to reduce losses in transmission lines. They transmit power at high tension voltages and reduce it at the distribution point using a local transformer.

Here is an excerpt from wikipedia article: "Transmitting electricity at high voltage reduces the fraction of energy lost to resistance, which varies depending on the specific conductors, the current flowing, and the length of the transmission line. For example, a 100-mile (160 km) 765 kV line carrying 1000 MW of power can have losses of 1.1% to 0.5%."


With a 12-30 Volt input range, this led bar likely has a step-up constant current regulator inside, boosting the voltage to a constant current for the leds. This is better than a constant voltage supply, and the led setup doesn't use 12V internally.

In that case, go with a 30 Volt supply for (288 / 30) = 9.6 Amps, which would only require a 14 AWG cable for a 80 foot round trip cable run and a 2 Volt drop . Much cheaper than 8AWG as @Dave suggests. If you put it in the attic as @Chetan wisely suggests, it would only need a 16 AWG wire for a 0.5 Volt drop, marginally cheaper than 14 AWG. The supply will be much cheaper than a 12V and 25 or 30 Amp supply.

Keep in mind, a 300 Watt supply is only 2.5 Amps at mains voltage (not including efficiency loss), and can be plugged in to a standard outlet or a light bulb socket in the attic easier than running a conduit from the basement would be. A single hole in the attic wall is easier than using a ladder to install a conduit.

  • This is the right step, it is not hard to find a "28" volt version of the power supply similar to the 12v version the OP shows. This will lower the current output needed and still meet light voltage requirements. In addition it allows use of 12 or 14 gauge LANDSCAPE CABLE for the run. Which in my opinion is much cleaner then running romex or conduit up the wall. Sep 16, 2016 at 18:09

The calculation is relatively simple. First, you want to be sure your wiring is rated for the total current draw at the voltage you'll be driving. This value (amps) should be provided in the literature or labels that came with your LED bar.
Next, you want to make sure that there won't be excessive voltage drop over the length of the wiring. It's highly unlikely that this will be a problem in such a short run, but grab any rating sheet like this one and calculate the voltage loss. Unless you end up with more than roughly 0.5 Volts less at the LED bar than required, you will be good to go.

But if you have a friend who's a licensed electrician, he can tell you what's needed both for performance and for local code.

  • It's highly unlikely that this will be a problem in such a short run 80 feet round trip at 24 amps is NOT a short run or problem free.
    – cde
    Sep 16, 2016 at 4:25

1) I use 12V LEDS all over the back yard and drive it with 150W adjustable Universal Laptop charger in the basement and allow >2V drop on AWG16 wire with ground wire strapped to Return wire. and get to the end stripLEDs with 14.3V You would need AWG 8 as Dave computed or mount the PSU above the LEDs and route AWG 16 3 wire extension cord.

2) I also use Surplus scrap boards used in Ambulance/Police cars, but I cut the 16x1 arrays down to 5x1 for running at 14.5 to 15V at slightly less than full power for garden lights.

3) Judging by my 10yrs with similar LED experience at this density of power LEDs per sq.in. of aluminum clad board space, I would expect for these LEDs at 25A to burn fingers and age LEDs faster, unless there is a good wind. So I would recommend reducing current to 19A as follows;

4) Meanwell also makes 13.5V 19A CC power supplies. $63 usd


5) Your backyard neighbours may complain about the light glare. A 90 deg downward angle may be what you had in mind.

6) Choosing a supply with remote control or dimmer might be useful.

My former house garden lights all in parallel from CV used 4S Led Alumclad board on 12.5-13V with no noticeable difference in dropped voltage and 20% less power at end did occur. (It is hard to notice unless side by side.)
enter image description here

7) Your backyard will look like stadium lighting. with 288 W.
but I suspect at this price, they will be bluish 6000'K rejects.

Engineering Calculations

If you start with 13.5V and want 12V at Lights. Consider this.

  • 288W for 96LEDs @ 12V translates to an array of 4S24P or 24 parallel ccts. of 4 series LEDs (~3V/LED)

  • Each 3W LED will have an effective series resistance (ESR) of <1/3 Ohm thus *4/24 = 56 milliOhms for the light bar.

  • Thus for a 1.5V drop @25A, the wire dissipates 37.5W which is 13% of the load power and requires a resistance of 1.5V/25A=60 milliOhms (or <=AWG 8 for your length).

  • If you want max brightness but calibrated for a load at 12.0V, you might have to go AWG 6 or shorter cable so you have some margin for error. Crimp contacts have to be done perfectly to achieve < 10 mOhm.

    A PSU inside near the Lights is the best choice for you.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.