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My home builder pre-pulled a length of 18-8 solid copper wire (18 AWG, 8 insulated conductors inside single jacket) into the ceiling of a room where I want to use some LED lighting with a remote power supply (that is, the power supply is at one end of the 18-8 and the LED lighting is at the other end). I can easily calculate the needed current / wattage for the lighting, but I'm having a hard time determining the various limits of the wire (current, voltage drop) to be safe and effective.

If I'm reading the charts correctly, this wire should be about 6.4 milliohms per foot and have an ampacity of 10 amps at 20degreeC insulation temperature and 60degreeC conductor temperature (not sure how that's possible or exactly how they measure that).

My LED lighting to be connected would draw no more than 52 watts, or about 4.3 amps at 12 volts or 2.2 amps at 24 volts.

My proposed power supply will happily provide 150 watts at 12 volts before its own overcurrent protection gets triggered. Therefore, I believe it would be advisable for me to fuse the output connections before the long wire runs to the ceiling at some lower value to prevent the wire from becoming an ignition source in the case where the LED strip itself shorts out.

So my questions are:

  1. How long a run of the 18-8 wire can I used before I run into problems? Since the wire is already there and I'm loathe to tear out my ceiling, what are alternative solutions if my voltage drop is too high? Can I pair up conductors in the 18-8 to have instead of 4 pairs, have 2 pairs of paired conductors?

  2. What current value fuse would make sense? I'm thinking 80% of the 10 amp ampicity, or 8 amps might be a safe value.

  3. Have I read the charts and gotten correct values for resistance and ampacity? Do my numbers sound right?

I'm an amateur at this -- my college EE courses are ancient memories. I probably know only enough to be dangerous, as the saying goes. Feel free to set me straight or advise me in anyway appropriate. I'm not math averse, so if pointing me at the proper equations or references is appropriate, do that. I can do the calculations, but as with anything engineered, there's rules of thumb, acquired wisdom and outside factors to consider beyond something as simple as E=IR.

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Lets simplify this a WHOLE LOT.

As long as you are within 50' of the source, 18 gauge wire can safely carry 10 amps.

That's 1200 watts you can supply with a 50' run. As long as you don't plan on exceeding that power, you have nothing to worry about.

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  • \$\begingroup\$ By the way, even if you wanted to run it 100' or more, you could still pull 10A. You might lose a little voltage, but safety is not an issue. \$\endgroup\$ – Sara Heart May 18 '18 at 22:16
  • \$\begingroup\$ I'm not sure how long the run is, but it might be 100'. The length is not safety issue, but rather a voltage drop problem. 100 feet at 6ohms and 4.3 amps is a ~5.5 volt drop, according to one calculator. 12 volt LEDs are not going to be very happy at 6-7 volts, are they? Should I have asked two separate questions? Safety current fusing, and voltage drop calculations? \$\endgroup\$ – CXJ May 18 '18 at 23:14
  • \$\begingroup\$ P.S. I think you meant 120 watts, there. \$\endgroup\$ – CXJ May 18 '18 at 23:14
  • \$\begingroup\$ You are talking 5.5 Vac if 100'. Again, big deal, the voltage will have to be rectified and dropped to DC, and by that time the DC Voltage will have hardly suffered at all. If it is a switching regulator, it will compensate, and make NO difference. Also 120Vac X 10 Amp = 1200 Watts, So yes, I meant what I said in both instances. You are worrying too much, really. : ) \$\endgroup\$ – Sara Heart May 19 '18 at 1:10
  • \$\begingroup\$ Oh I missed your 2nd question. It sounds like you are not going to be using but a fraction of the 18Ga capability, If so just use a 10A fuse. Be aware that it is common code wiring practice to use a 15A breaker if wiring a room with a 18Ga wire. Just for reference check this out: groverelectric.com/assets/downloads/howto/… \$\endgroup\$ – Sara Heart May 19 '18 at 1:45
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For indoor home lighting I would use Bridgelux EB-Series Gen2 or Samsung F-Series Gen3 strips.

enter image description here

The typical strips that are powered with 12V or 24V DC power supply are horrible solutions for home lighting. These LEDs are cheesy, unreliable, with very low lumen output per watt. They have unnecessary current limiting resistors that waste electricity.

I would power them with a constant current Mean Well HLG driver. No fuse would be needed. I buy my HLG's from arrow.com because they have free shipping on 2 or more.

18 gauge wire is more than sufficient, meaning the wire voltage drop will be insignificant. When using a constant current driver any wire resistance has no affect on the LED current.

This is a 48V low current professional yet economical solution.

Cost of the Bridgelux is less than $4 per foot but provides 1,200 lumens per foot. You cannot come close to this cost per lumen with the 12V strips. I have a 100W fixture using the Bridgelux strips that provide a lot more light than the sunshine coming through my home's front window.

Here I am measuring the irradiance from seven Bridgelux 22" $7 strips ea. (P/N BXEB-L0560Z-35E2000-C-B3) for an horticulture application. Each strip outputs 2400 lumens.

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Below is an experiment with two $4 Bridgelux strips powered by a $28 Mean Well driver mounted to a shelf. This shows tomato plants like warm 3000K lights a lot more than 5700K cool white. All plants were germinated at the same time.

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Below is Samsung F-Series strips in a Klus Aluminum Extrusion.

enter image description here

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    \$\begingroup\$ unless I misunderstood, your solution would bring forth the exact same problem as he had to begin with. \$\endgroup\$ – Sara Heart May 21 '18 at 20:43
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    \$\begingroup\$ @SaraHeart He has no problem. 18 gauge is more than sufficient for low voltage wiring within a single structure. I was just suggesting some strips with very high efficacy vs. the the typical Amazon/Ebay crap. The recommended driver has very high efficiency (≈95%), 7 year warranty, simple and flexible dimming, very reasonable cost per watt, and would not require a fuse. My solution is low current (≈500 mA), low cost, high efficiency (180 lm/W), and high reliability. When using a constant current driver the wire resistance has not affect on the amount of current flowing through the LEDs. \$\endgroup\$ – Misunderstood May 22 '18 at 0:50
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    \$\begingroup\$ This is not AC line voltage he is running, it is 12Vdc, on a long run, which would under power the LED's. Wait if it is constant current, and has the overhead, about 6V, this could work. But he has bought his equipment, is the only problem. \$\endgroup\$ – Sara Heart May 22 '18 at 0:56
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    \$\begingroup\$ @SaraHeart "But he has bought his equipment, is the only problem". I saw the word "proposed" and "Feel free to set me straight or advise me in anyway appropriate". I understand many are using the horrific 12V DC crap, that does not mean it is the way to go. It's not some hobbyist toy he wants. The guy had the builder put in the wiring I would think he seriously wants to do home lighting and wants it done correctly. 12V, 10 Amps is ludicrous. Try 48V, 500 mA. \$\endgroup\$ – Misunderstood May 22 '18 at 1:10
  • \$\begingroup\$ I was under the impression the equipment was purchased, and he could not run AC much closer to the P.S. for the LED's. If I was was wrong, I apologize to you and him. \$\endgroup\$ – Sara Heart May 22 '18 at 1:19

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