Voltage drop on 120v landscape lights over 1000' run.

[jaylectricity]

Looking to install 17 LED 120v bollard post lights along a 1000' driveway.

The posts that are spec'd are 28 watts. Using a voltage drop calculator I found online I did some quick trial and error calculations.

It looks like I can put the four lights furthest from the panel on a #12 thhn at 1 amp and suffer less than 5% voltage drop. Then I can put the next four furthest from the panel and suffer even less voltage drop. Then I can put the closest 9 lights at a little over 2 amps on a #12 and still suffer less than 5%.

I figure three runs, I can give them each their own neutral and fit it in a 3/4" PVC easily. Even at 40% allowed ampacity due to the # of CCC in the conduit they should meet code

Does this sound right? Am I being too cautious? Could I put the two furthest runs on the same circuit just with separate conductors from the breaker?

 

[GoldDigger]

Your solution would work and it would be acceptable to feed both runs from the same breaker. Just don't try to put two wires on a breaker lug that is only listed to accept one. Instead join the two sets of wires together at a box or inside the panel enclosure.

But there is no reason that you cannot instead use different wire gauges at different points in the circuit and just have one run of wire going the whole distance.
Especially given that most LED drivers include current regulation it may not even be necessary to keep the voltage drop low in the first place.

Example: #10 to first group of nine and extend it with #12 from there.
You would have to play around with the actual wire sizes and where to make the changeover, but it may come out to less wire in total and keep the conduit cleaner and easier to pull.

 

[kwired]

A little hard to give specific answers when we don't know distance of each run,

but do consider cost differences between making three separate runs vs larger conductor or even multiwire circuits.

If 10 AWG will allow for acceptable voltage drop it may cost less then two circuits of 12 AWG.

120/240 3 wire MWBC, with load balanced will have less voltage drop and may even be able to use 14 AWG depending on details.

Even starting out with a larger conductor and reducing along the way may be a way to go.

 

[mpoulton]

LED drivers are switching power supplies with active current regulation, so they are not very sensitive to input voltage at all. They may be able to tolerate very large voltage drop without any problems. They might even be officially rated for "universal" input (90-300VAC, for example). The voltage drop limitations in the code are recommendations, not enforceable hard rules. It would probably be worth checking this first since you may be able to just put all of them on one run of #12.

 

[jaylectricity]

A little hard to give specific answers when we don't know distance of each run,

but do consider cost differences between making three separate runs vs larger conductor or even multiwire circuits.

If 10 AWG will allow for acceptable voltage drop it may cost less then two circuits of 12 AWG.

120/240 3 wire MWBC, with load balanced will have less voltage drop and may even be able to use 14 AWG depending on details.

Even starting out with a larger conductor and reducing along the way may be a way to go.

I tried #10 all the way through and it came out to 6%, but I suppose that was applying the full load of the whole run and attributing it to the farthest light.

Can I figure each separate fixture then add them together? Or some other algebraic equation? If I only had one light at 30W at the end of a 1000' of #10 the VD is .5%.

 

[jaylectricity]

LED drivers are switching power supplies with active current regulation, so they are not very sensitive to input voltage at all. They may be able to tolerate very large voltage drop without any problems. They might even be officially rated for "universal" input (90-300VAC, for example). The voltage drop limitations in the code are recommendations, not enforceable hard rules. It would probably be worth checking this first since you may be able to just put all of them on one run of #12.

Or make sure we buy fixtures that have universal input. The customer showed me the lights that were spec'd but were open to suggestions. I think they just want lights that work and look nice.

 

[GoldDigger]

I tried #10 all the way through and it came out to 6%, but I suppose that was applying the full load of the whole run and attributing it to the farthest light.

Can I figure each separate fixture then add them together? Or some other algebraic equation? If I only had one light at 30W at the end of a 1000' of #10 the VD is .5%.

There are calculators for that function out in the wild somewhere. One of our members posted a spreadsheet here to do that, but I can't find it right now.

The clean manual way to do it is to take the actual current for a given wire segment and use it to calculate the VD for that segment only. Add up all of those numbers to get the voltage drop to the last light.
Or you could "cheat" and calculate by taking groups of 3 lights and treating them as if the whole draw was at the middle one. That gives you far fewer segments to add up, but does not sacrifice much accuracy.

 

[jaylectricity]

120/240 3 wire MWBC, with load balanced will have less voltage drop and may even be able to use 14 AWG depending on details.

Even starting out with a larger conductor and reducing along the way may be a way to go.

I'm definitely feeling better about something like either of those suggestions. I've never done this type of landscape lighting so I don't want to look like a fool when the fixtures at the end don't light up properly or worse the inspector looks at the conduit and says, "How do you expect to run four different circuits through 1/2" PVC?" when I thought they'd all fit on a 15A circuit of 14 gauge.

Does anybody know if those types of fixtures typical fit over the conduit stubs or do the conduits go into KO's on the base? I believe we're pouring small concrete pads for each light.

 

[Smart $]

...
120/240 3 wire MWBC, with load balanced will have less voltage drop...
...

Running a 3-wire MWBC compared to a 2-wire circuit all the way almost cuts the voltage drop in half. What makes it more than half is the voltage drop in the neutral between loads that balance each other. It is almost like powering the same load at twice the voltage.

 

[wwhitney]

Suppose n (= 17) lights demanding a fixed current of I (= 28/120) per light are placed along a run of electrical wire of a single size, with total resistance R.

If all the lights were at the end of the run, then the voltage drop would be (n*I)^2 * R.

If the lights are instead spaced evenly, then the resistance of each wire segment is R/n, and the current through each segment starts at n*I and decreases to I for the farthest segment. The sum of the first n squares is n * (n+1) * (2n+1) / 6, so the voltage drop at the end light will be (n+1)*(2n+1) / 6 * I^2 * R. Thus it is a bit more than 1/3 of the voltage drop for the first case.

Cheers, Wayne

 

[GoldDigger]

Why are you squaring the current to get the voltage drop?
Wouldn't that be the way to get the lost power instead?

Sent from my XT1585 using Tapatalk

 

[Dennis Alwon]

Jay, there is no code on the 5% limit to voltage drop. With leds many of them will run with much lower voltages. I am willing to bet that a #10 all the way will work. That is an 8 volt drop and most power supplies are 124v or thereabout. 124-8 still leaves you with 116V. IMO, this is plenty for those led's

 

[wwhitney]

Why are you squaring the current to get the voltage drop?
Wouldn't that be the way to get the lost power instead?

Whoops, good call. That makes the math simpler:

If all the lights were at the end of the run, then the voltage drop would be n * I * R.

If the lights are instead spaced evenly, then the resistance of each wire segment is R/n, and the current through each segment starts at n*I and decreases to I for the farthest segment. The sum of the first n integers is n * (n+1) / 2, so the voltage drop at the end light will be (n+1) / 2 * I * R. Thus it is a bit more than 1/2 of the voltage drop for the first case.

Cheers, Wayne

 

[wwhitney]

I tried #10 all the way through and it came out to 6%, but I suppose that was applying the full load of the whole run and attributing it to the farthest light.

If the above is true, then the end voltage drop with the lights evenly distributed along the whole run would be a bit over 3%.

BTW, I think the discussion so far has assumed the power factor on the led lights is 1, is that a reasonable assumption?

Cheers, Wayne

 

[FionaZuppa]

Running a 3-wire MWBC compared to a 2-wire circuit all the way almost cuts the voltage drop in half. What makes it more than half is the voltage drop in the neutral between loads that balance each other. It is almost like powering the same load at twice the voltage.

and if the neutral lifts or breaks for any reason? if they can run on 240 then using 240v would also reduce waste.

28w LED lights? do they have a massive heat sink on them. ~1500 lumens each ? 25.5k lumens total ? sounds very very bright.

these 28w fixtures sound very not efficient.

in general, ~9w 120v LED gives roughly 80lm/w

at that rate, 28w LED fixture should give roughly 2240 lumens, but i suspect the fixture is not anywhere near that.

i would amp clamp a fixture to see what the actual amps are, then go from there. it could be 28w, but if it is that to me sounds like a very inefficient fixture.

 

[kwired]

If the above is true, then the end voltage drop with the lights evenly distributed along the whole run would be a bit over 3%.

BTW, I think the discussion so far has assumed the power factor on the led lights is 1, is that a reasonable assumption?

Cheers, Wayne

It is something I have been thinking of bringing up, I think PF is going to be less then 1.0 in most applications.

 

[Smart $]

Whoops, good call. That makes the math simpler:

If all the lights were at the end of the run, then the voltage drop would be n * I * R.

If the lights are instead spaced evenly, then the resistance of each wire segment is R/n, and the current through each segment starts at n*I and decreases to I for the farthest segment. The sum of the first n integers is n * (n+1) / 2, so the voltage drop at the end light will be (n+1) / 2 * I * R. Thus it is a bit more than 1/2 of the voltage drop for the first case.

Cheers, Wayne

We want to know the end voltage as volts or percent lost from the supply voltage. Relative voltage drop as a first to last ratio has little to no meaning.

 

[wwhitney]

We want to know the end voltage as volts or percent lost from the supply voltage. Relative voltage drop as a first to last ratio has little to no meaning.

I computed the end voltage drop with a constant wire size for two different cases: the simple case where all the fixtures are at the end (which any voltage drop calculator can do), and the case of equally spaced fixtures. Then since the OP indicated that a voltage drop calculator gave him 6% voltage drop for #10 copper with all the fixtures at the end, I pointed out that correcting for equally spaced fixtures the voltage drop would be a bit over 3%.

With n fixtures, the ratio of end voltage drop in the two cases is (n+1)/2n. A simple way to see this is that with equally spaced fixtures each drawing I amps, the current in the first segment is nI, the current in the last segment is I, and so the average current is (n+1)/2 * I. Versus a constant current of nI if the fixtures were all at the end.

Cheers, Wayne

 

[jaylectricity]

Thank you Wayne for the equations. That's helpful.

28w LED lights? do they have a massive heat sink on them. ~1500 lumens each ? 25.5k lumens total ? sounds very very bright.

these 28w fixtures sound very not efficient.

in general, ~9w 120v LED gives roughly 80lm/w

at that rate, 28w LED fixture should give roughly 2240 lumens, but i suspect the fixture is not anywhere near that.

i would amp clamp a fixture to see what the actual amps are, then go from there. it could be 28w, but if it is that to me sounds like a very inefficient fixture.

I don't know the specifics of the fixture at the moment. I looked at their plans and saw the lights that were spec'd and just looked for wattage or amperage to work with. I could very well purchase a different fixture.

 

[Smart $]

...
With n fixtures, the ratio of end voltage drop in [equally spaced vs. all at the end] is (n+1)/2n. ...

Now that makes more sense.

Another way to do it that works for equally and unequally spaced is to multiply the load at each point in the run by the distance from the source. Sum the results and divide by the total load. Use the resulting distance and total load to calculate voltage drop

Working the problem as equal distance...

1×1,000ft/17segments×28W/120V=13.7ft-A
2×1,000ft/17segments×28W/120V=27.5ft-A
3×1,000ft/17segments×28W/120V=41.2ft-A
...
17×1,000ft/17segments×28W/120V=233.3ft-A

Anyway, the sum is 2,100ft-A (I cheated using n(n+1)/2 times the first line item ft-A value ).
2,100ft-A÷(17×28W/120)=529ft

For proof...

(n+1)/2n=18/34=0.529
0.529×1,000ft=529ft