Sizing wire to allow for Voltage Drop on a Commercial lighting circuit.

W

Will Wire

Senior Member
Location
California: NEC 2020
Occupation
Electrical Contractor
Hello,

I am confused on how to calculate the Circular Mils to determine the size of the wires for a commercial lighting circuit on the outside of a Storage Unit building.

I have seven 120 Volt LED lights with Drivers that draw 1.2 amps each. The total amperage draw is 8.4 amps. It is a continuous load and must be increased 125% for 10.5 amps. I will install a 20 amp circuit.

The lights are equally spaced. The furthest light is 370 feet away. I want no more than three percent voltage drop on the branch circuit.

I am aware of the CM formula, just not sure which amperage value to use.



Question: Do I calculate the size (CM) of the wire based on 1.2 amps, or 8.4 amps, or 10.5 amps at 370 feet?



Thank you for your time.
 
You always use the actual load and not the load at 125%. The fictitious 25% does not create any voltage drop.

If the lights are equally spaced, often you make the calculation at 1/2 the distance using the full load of all of the lights.
 
If they are multi-volt drivers might be worth considering a 208, 240 or 277 volt circuit which would allow smaller conductor and lesser voltage drop.

Just a thought.
 
don_resqcapt19 said:
If the lights are equally spaced, often you make the calculation at 1/2 the distance using the full load of all of the lights.
Click to expand...
To expand on that, if you are running a single size wire for the entire branch circuit, then for the value of "current * distance" in the voltage drop formula, you can just sum up the contribution from each of the loads.

So if all the loads are the same, you can use the total load times the average distance from the source to a load. Equivalently, you can use the current of a single load and the sum of the distances from the source to each load.

Cheers, Wayne
 
Why are we allowed to use half the distance for equally spaced lights? Is it because the different sections of the wire are subject to different current flows?

I think i will suggest wiring at 240 Volts.
 
Do modern high-tech LED lighting systems much care about voltage drop?
If the line voltage drops, they draw more current for the same output.
If the line voltage increases, they draw less current for the same output.
 
Speedskater said:
Do modern high-tech LED lighting systems much care about voltage drop?
Click to expand...
LED lights I typically spec have universal voltage drivers 120 though 277V. So if I use a 277V circuit and the combination of load and circuit distance dictates that the driver at any given light on the circuit receives only 226V (pick any random voltage in the range) there is no issue. It does not care because it falls within the 120-277V range of the universal voltage driver. Technically you can run 16A on a #12 wire a long way if the source is 277V and the voltage at the load (the driver) only needs 120V to properly operate. I am old school and still upsize my wire for 3% drop but in theory there is not much to worry about if all lights on the circuit have the universal voltage driver.
LED's have made voltage drop much less of a problem now since the loads are like 25% or less for the same amount of lumens. Not like when we had to feed 400W Metal Halide site lights (plus ballast load) 700' or more from the building.
 
Will Wire said:
Why are we allowed to use half the distance for equally spaced lights? Is it because the different sections of the wire are subject to different current flows?
Click to expand...
Yes, precisely.

And half the distance only works if the first light is at the source itself; in general it is the average of the distances. I.e. one light at 0 ft from the source, one light at 200 ft, the average is 100 ft. Versus one light at 200 ft from the source, one light at 400 ft, the average is 300 ft.

[And then since the light at the source causes no voltage drop, for the case of equally spaced lights not including one at the source you can use half the distance but add one more light when calculating the load. As seen in the formula for the nth triangular number (sum of the numbers from 1 to n), which is n*(n+1)/2.]

Cheers, Wayne
 
Will Wire said:
Why are we allowed to use half the distance for equally spaced lights? Is it because the different sections of the wire are subject to different current flows?
Click to expand...
It's not about allowed, it's just a handy rough-guessing method. Half of the run is closer than the halfway point, and the other half is farther.

Will Wire said:
I think i will suggest wiring at 240 Volts.
Click to expand...
All other things being equal, the higher the voltage, the lower the current and the resultant voltage drop.
 
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