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gashmore

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The project: Install 12VDC 3 Watt Low voltage LED pathway lights, one every 20', along a 420' boardwalk through a mangrove forest. There are 24 lights and to minimize taps in the main line they will be installed in groups of 3 with one light at the tap, one light 20 feet behind and one 20 feet ahead. Taps will be 60 feet apart.

I am assuming that the full 72 watt load in the first 60' will cause significantly more voltage drop that the 9 watts in the last 60' so I treated the setup as 7 separate segments and calculated the voltage drop for each segment with the input voltage being the reduced voltage from the previous segment and the wattage being what remains to be powered. Assuming this logic is correct, an input voltage of 13.8V and #8 AWG copper will result in voltage at the end of right at 12V. A total of 13% drop.

First question: Is my methodology reasonable?

Second, will the LED lights at the first tap be able to handle the 13.8V or do I need to add some resistance to bring the voltage down?
 
A way of doing this to get even voltage/current to each lamp is to use a 3 core cable, using one core to run to the far end, unbroken, and connect it to one of the other cores there.
Feed the power into the first core and the third core, and tap the lamps off of the second and third cores.
So in effect you are feeding from both ends. I can see how the increasing length and decreasing wattage on the supply side would be compensated by the reducing length and increasing wattage on the return would even out the voltage but what that does is add 420 of wire carrying the full 72 watts.

If I understand it, core 1 carries the whole 72 watts to the the far end. 420*72/18960=1.6 volt drop before I get to the first tap. Then I get 420*9W/18960= .199V more voltage drop on the return. That totals to 1.79V. Exactly what I had calculated for the far end on my model.
BUT then I have to deal with another 1.6V drop to get back home.
This is gonna take more thinking.
 
A way of doing this to get even voltage/current to each lamp is to use a 3 core cable, using one core to run to the far end, unbroken, and connect it to one of the other cores there.
Feed the power into the first core and the third core, and tap the lamps off of the second and third cores.
I’ve read and re read. Don’t understand.

the bit “run to the far end, unbroken, and connect it to one of the other cores there. “

what do you mean by “ one of the other cores“ .

if your talking about one of the other 3 cores, why not say take two cores to the farthest point.

sorry maybe a picture would help.
 
I am assuming that the full 72 watt load in the first 60' will cause significantly more voltage drop that the 9 watts in the last 60' so I treated the setup as 7 separate segments and calculated the voltage drop for each segment with the input voltage being the reduced voltage from the previous segment and the wattage being what remains to be powered. Assuming this logic is correct, an input voltage of 13.8V and #8 AWG copper will result in voltage at the end of right at 12V. A total of 13% drop.

First question: Is my methodology reasonable?
Yes, you can do each section's computation and get to the final voltage that way. A spread sheet or similar allows you to define a cable resistance per length, each length, and the drops, etc and makes it easier to try a change of cable size (resistance) as needed.

Though often the approximation for a lot of similar loads uniformly spread is to compute the volt drop from the total load and half the cable length.
Second, will the LED lights at the first tap be able to handle the 13.8V or do I need to add some resistance to bring the voltage down?
You would need to check the specifications for the specific choice of lights.

If you need to drop a more-or-less fixed voltage for DC you can make use of a silicon rectifier diode as approx 0.7V drop over a wide range of current, so a couple in series can give you about 0.7/1.4/2.1V drop as needed. Obviously you need to get it the right way round!
 
1. @Paignton pete, see pic for method suggested by @brianmoooore which does significantly even out the voltage, although not completely. As @gashmore says, it adds a full length run carrying the full current incurring significant loss.

2. For LED lights fed with a constant voltage supply, that use internal resistors to set the current, neither the brightness nor the current are proportional to voltage. The current is proportional to the voltage across the internal resistor, which is only part of the supply voltage. Therefore a 20% change in supply voltage could cause a 50% or more change in current.

3. I would avoid sending 12V over that distance. I guess you don't want to send 120V AC along the boardwalk but even so, I would prefer to cover as much of the distance as possible at 24, 36 or 48VDC and use a step-down DC-DC converter or buck regulator at one or more remote points, such as your taps at 60 foot intervals. This has two advantges. a) the higher launch voltage reduces the current in the main run of cable, so a smaller cable can be used for the same drop. b) the regulator corrects for the voltage drop up to that point, so that only the drop from the regulator to the LED has any effect on brightness.

Example.
Launch at 36V. Place one buck-regulator the size of a sugar cube within each 3-light tap-off point every 60 feet, to step the 36V down to 12V. Let's aim for maximum 10% voltage drop in the main 36V cable run. This won't have any effect on the brightness, only on energy losses. Even if you don't care about energy (which is trivial since the lights are small) the voltage drop must be kept reasonable for correct regulator operation.

LEDs 3W @ 12V. Regulator efficiency 95%.
Current input to regulator 3 * 3 / (36 * 0.95) = 0.26A
7 loads of 0.26A at 60 foot intervals equivalent to single load of 7 * 0.26 = 1.8A at 200 feet distance (=400 feet round trip of wire)
Resistance of #16AWG = 4.016mΩ / foot
7 * 0.26 * 400 * 0.004016 = 2.92V drop at far end of 36V circuit.
2.92/36 = 8.1% drop.
Total power loss 8.1% (cable) + 5% (regulator) x 63W (LEDs) = 8.25W

The voltage drop from regulator to LED is going to be trivial.
Resistance of #20AWG = 10.15mΩ / foot
Each LED uses 3 / 12 = 0.25A at 20 feet from regulator (=40 feet round trip of wire)
0.25 * 40 * 0.01015 = 0.1V
0.1/12 = 0.8% drop

So you can use #16 cable for the main 36V run, place one regulator at each tap point, run 20 feet of #20 either side of the regulator and the voltage of all LEDs will be within 1%. The total power loss will be less than with your #8 main run, and the cost saving will be the difference in cost of #16 wire vs. #8, less approximately $70 for the regulators.

If you fit a 500mA fuse at the input to each regulator, any LED or regulator failure will be effectively contained within that 3-LED cluster. Here's a suggestion for the regulator (Traco is a world leading brand, there are many US stockists too)

Traco 2A 12V buck regulator

By using a 2A rated part at only 0.75A, it is running at less than 50% capacity and will have a long, reliable life and will be able to withstand higher temperatures of the junction box in which it is located. You should check the temperature derating curve of the regulator and consider whether the box is going to be in direct sunlight. You can solder leads onto the regulator and connect them to lever nuts or whatever for the outgoing circuits.
 
If you use these Traco PCB-mounted regulators (or any other similar type), I would recommend soldering them to a little piece of stripboard or square pad board and then soldering the wires to that, to avoid breaking the pins from the inside the regulator if the wires are stressed during installation. You can also mount the fuseholder on the board. Note that the fuse only prevents the whole circuit shutting down if a regulator fails short-circuit. It should not blow if an LED becomes short-circuited because the electronic short-circuit protection in the regulator will act faster. I never worry about making the fuse easy to replace, because it would only need replacing if the regulator was already toast, which I have never seen happen.

Another failure mode worth mentioning is that in theory a buck regulator can go short-circuit from input to output and feed 36V to the 12V circuit. In practice I have never seen it happen, but here what would hopefully occur is that the current will be high enough to rapidly blow the 500mA fuse before the LEDs overheat. For expensive equipment downstream it would be worth adding a protective crowbar circuit or using a sophisticated regulator with one built in. For three LED lights I would say it's overkill. Make sure the common negative connection is solid. Losing that could potentially send 36V to the LEDs. I would solder two wires to the common pin, one for the input and one for the output, to avoid the LEDs remaining connected if the regulator is detached.
 
What happened to KISS? The solution I suggested just requires the PSU to be bumped up a volt or two to compensate for the extra voltage drop, and no other components, other than some IP rated joint boxes.
 
If you use these Traco PCB-mounted regulators (or any other similar type), I would recommend soldering them to a little piece of stripboard or square pad board and then soldering the wires to that, to avoid breaking the pins from the inside the regulator if the wires are stressed during installation. You can also mount the fuseholder on the board. Note that the fuse only prevents the whole circuit shutting down if a regulator fails short-circuit. It should not blow if an LED becomes short-circuited because the electronic short-circuit protection in the regulator will act faster. I never worry about making the fuse easy to replace, because it would only need replacing if the regulator was already toast, which I have never seen happen.

Another failure mode worth mentioning is that in theory a buck regulator can go short-circuit from input to output and feed 36V to the 12V circuit. In practice I have never seen it happen, but here what would hopefully occur is that the current will be high enough to rapidly blow the 500mA fuse before the LEDs overheat. For expensive equipment downstream it would be worth adding a protective crowbar circuit or using a sophisticated regulator with one built in. For three LED lights I would say it's overkill. Make sure the common negative connection is solid. Losing that could potentially send 36V to the LEDs. I would solder two wires to the common pin, one for the input and one for the output, to avoid the LEDs remaining connected if the regulator is detached.
I believe I will go with sealed 36 to 12V buck converters used on golf carts. They come with breakers. This whole installation being only about 24 inches above a salt marsh so everything has to be very well sealed. Even with the buck converters and gel filled connectors the price came out less than a spool of #8.
 
What happened to KISS?

Nothing, it's a good policy. But sometimes banishing all technology from a solution results in a brute-force approach. If that happens to be costly, it can divert resources and attention away from the more critical aspects of the project. The achilles' heel of an outdoor lighting installation is usually ingress protection. The MTBF of the regulators is overshadowed by that of the glands and gaskets, so if the regulators save copper and cost, more can be spent on sourcing the very best products and techniques for keeping the moisture out.

The feed-from-both-ends method is useful but it does not completely even out the voltage drop. Therefore it is not possible to use a very small cable, ignore the difference in drop and simply launch at whatever voltage is needed to get 12V at the LEDs. I haven't done the maths for that. I will do presently, as I would like to think it's still a viable solution. I wanted to throw the voltage step-down method in as it had not been mentioned, and the advantage that a higher voltage offers is rarely appreciated. VD percentage for a given load power and cable size decreases as the inverse square of the voltage. 36V is nine times 'better' than 12V at transporting energy, while 120V is 100 times 'better'.

I believe I will go with sealed 36 to 12V buck converters used on golf carts.

I would not like to guess at the MTBF of those units, mainly because they contain electrolytic capacitors of unknown quality. Could be great, could be useless, unless you can get proper engineering documentation. The Traco modules I linked have a (calculated) MTBF > 13,000,000 hours, permit operation with a case temperature of 105°C and contain no electrolytic capacitors. They are also small enough to fit inside any reasonable tap-off connection box therefore do not add ingress vulnerabilities. YMMV.

They come with breakers.

I would tend to avoid using breakers in this environment because they will introduce a risk of poor contact due to oxidation that need not be there. They should never trip, therefore using fuses instead would be no inconvenience but avoids the poor contact concerns.

E2A, there's yet another approach. If the LED lights are passive i.e. contain nothing but LEDs and resistors, how about connecting the three LEDs of each group in series and powering them directly from the 36V?
 
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The project: Install 12VDC 3 Watt Low voltage LED pathway lights, one every 20', along a 420' boardwalk through a mangrove forest. There are 24 lights and to minimize taps in the main line they will be installed in groups of 3 with one light at the tap, one light 20 feet behind and one 20 feet ahead. Taps will be 60 feet apart.

I am assuming that the full 72 watt load in the first 60' will cause significantly more voltage drop that the 9 watts in the last 60' so I treated the setup as 7 separate segments and calculated the voltage drop for each segment with the input voltage being the reduced voltage from the previous segment and the wattage being what remains to be powered. Assuming this logic is correct, an input voltage of 13.8V and #8 AWG copper will result in voltage at the end of right at 12V. A total of 13% drop.

First question: Is my methodology reasonable?

Second, will the LED lights at the first tap be able to handle the 13.8V or do I need to add some resistance to bring the voltage down?
I did a job in another state changing out VFD’s and the head engineer paralleled #10 x 3 using 12dc controls and it worked like a champ with no voltage drop. Per the NEC your not supposed to parallel less than 1/0 but that doesn’t apply to plants. This one was good year tires
 
Per the NEC your not supposed to parallel less than 1/0 but that doesn’t apply to plants.
That is interesting as the UK regs has various rules about paralleling but no specific limit on size. Your 1/0 AWG is 53.5mm CSA which (to me at least) is fairly large.

Ours seem to revolve around equal-sharing of current (so you should be paralleling identical conductors, and no branch or switch on just one of them, etc) and the arrangements for over-current protection to make sure individual conductors are fault-protected, which might be:
  • Single OCPD, so long as a mid-cable fault is able to safely disconnect (probably for long runs when volt drop is the limit and reason for paralleling, not current carrying capacity)
  • Each run has OCPD at both ends to clear both feed and back-feed of the faulted section (so one conductor would need to be able to clear both end's OCPD more or less in parallel), but such as fault might not be detected immediately if other parallel section(s) limp along powering the load.
  • Use of linked breakers at feed end only, so long as a fault anywhere along the length of one conductor is able to safely trip one (or more) of them and disconnect the lot.
Our 18th regs has Appendix 10 going over the details of these protection requirements as it is a lot more complicated than for a single conductor.
 
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Right, series groups of 3 lights run direct from 36V supply. All three LEDs in a group must be same wattage, ideally same model. Passive lights only (just LEDs and resistors, optionally a rectifier), AC/DC types are OK but multi-voltage no good as they contain active electronics.

Stick with #16 but 3-core this time, send one leg to the far end and back. 7 taps at 60 foot intervals starting at 0 feet, each taking 0.25A. Resistance 4mΩ/foot. Voltage drop per conductor 1mV per foot per tap supplied. Ignore temperature coefficient of resistance.

Cumulative drops along one distributor leg (volts):
0 / 0.36 / 0.66 / 0.90 / 1.08 / 1.20 / 1.26
Drop at each tap due to both distributor legs (volts)
1.26 / 1.56 / 1.74 / 1.80 / 1.74 / 1.56 / 1.26

Voltage difference between highest and lowest taps 1.8 - 1.26 = 0.54V
Voltage difference at LEDs = 0.54 / 3 = 0.18V
0.18 / 12 = 1.5%.
Voltage drop along untapped return leg + 20 foot 2-wire feeder to first tap
= 440 * 7 /1000 = 3V
Power loss due to return leg = 0.25 * 7 * 0.25 = 5.4W (offset by saving in converter loss)

Launch at: 36 + 3 + (1.26+1.8)/2 = 40.5V
LED voltages will range from 11.9 to 12.1V

This is my current favourite method. No converters needed. Slightly higher risk of electrolytic corrosion if capillary leaks exist, due to higher voltages at LEDs. Most important limitation is that all LEDs in each group should be identical. Credit to @brianmoooore for recommending feed-opposite-ends method
 
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Here is a diagram of (what I think is) Lucien's solution:
 

Attachments

  • Visio-LED-lights-example.pdf
    4.7 KB · Views: 11
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Thanks @pc1966.

Obviously if there is a long run of feeder between the power brick and the start of the boardwalk, the 40.5V can be adjusted up to compensate. Actually I omitted to add in the drop in the local wiring of each LED group. If we still assume this to be #20, carrying 0.25A over 80 feet of conductor (there and back 20 feet either side of the tap point) it adds 10 * 0.25A * 80 = 200mV extra needed at launch, or 66mV lower at each LED. TBH that kind of difference will be lost in the noise, not least of which is the temperature coefficient of resistance of the main run of cable.

40.7V is not actually a very convenient launch voltage because it is slightly beyond the range of a 36V PSU with 10% adjustment range. Some will adjust over a wider range though and even launching at 39.6V only robs the LEDs of 0.36V. If the lights will accept AC, I would recommend it as electrolytic corrosion is less severe. In that case it would be a question of finding the right primary and secondary tappings, allowing for regulation. Optionally, decrease the voltage drop by upgrading to #14AWG.
 
Having restored a 70 year old Hammerlund receiver, I understand the problem of electrolytics. Traco is hard to find this side of the pond but found a couple of similar converters. Marine grade wire and potting compound should make a solid package. Running at less than 20% of capacity heat should not be a problem.
 
But what do you think about the series groups without converters?
 
But what do you think about the series groups without converters?
That's what I am looking at now. It seems that most of these 12V nominal LED landscape lamps have internal power regulation that can accept 12 to 24V DC or AC which eliminates wiring in series but does open up some possibilities for the 3 wire solution. What about running some 12/3 for the supply line then parallel 3 watt 12W to 24V lamps every 20 feet coming back.
 
If they will accept 24V AC then they can handle a peak voltage of 24 * sqrt(2) = 34V and would probably be OK on 34V DC, but we should not assume this and it would probably invalidate the warranty.

Assuming the light output is independent of voltage, which it should be, it is better to distribute positive and negative from the supply end as uniformity of voltage drop is no longer important. If using 12/3, parallel two of the conductors subject to code. I would think it's acceptable since either conductor will carry the very low short-circuit current alone, and by being paralleled at each tap point the total short-circuit loop resistance from just one side of one severed conductor anywhere will barely exceed the resistance at the end point.

Launch at 24V. Resistance of #12 = 1.6mΩ / foot, one leg single conductor, other leg two in parallel. Aim for similar loss percentage as before, so assume voltage at lights >22V and all lights at midpoint. Total current 63/22=2.9A
Drop = 420 * 1.6 * 3/4 * 2.9 = 1.46V
Cable power loss 1.46 * 2.9 = 4.2W

Those numbers are based on DC. For optimum corrosion resistance, it is better to use AC. However, depending on the exact configuration of the integral LED driver circuits, the power factor might be significantly below unity due to limited conduction angle of the input rectifier. Therefore, the peak current and hence RMS drop in the cable might be higher than we have allowed for.

Anyhow, 24V and 12/3 is probably the way to go, thanks to the inbuilt voltage regulation in the lights. You could maybe even use 14/3.
 
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