# How to avoid melting a 3D Printed Chandelier with LEDS and DC Voltage Drop Across 7m Cable Run of 100A 12v DC

I am trying to power led strips located inside a custom designed 3D printed chandelier.

I want to find the best solution to power the project and to work out the voltage drop along the cable from the power to the chandelier. Here are my workings and plan:

My led strips require: $$78.75~A \cdot 12~V = 945~W$$

They have a cable run of 7 meters back to two switching power supplies:

100 A @ 12 V provided by two 12 V output at 50 A drawing 600W from mains. These draw 1200 W of mains voltage (240 V).

There is a design problem related to the 3D printed chandelier - these supplies need to be 7 m away from the chandelier because the 3D print has no ventilation and may melt with the build up of heat from the two 50 A, 12 V switching power supplies.

How can I work out the voltage drop along the cable from the power supply to the chandelier and make a choice for the correct thickness of cable?

How would I work out how much heat:

• the leds would produce
• the power supplies would produce if the full 78.75 A was being drawn by the LEDs for say 10 min, 30 min, 1 hour, 24 hours etc.

Could I instead deliver 24 V of DC from switching power supplies, ignore the voltage drop across the 7 m of cable and then transform what ever gets to the chandelier to 12 V (and keep the amperage required, a minimum of 78.75 A).

NB. I couldn't have the AC to DC conversion take place close to or inside the sculpture:

1. the heat generated will be too high, over time the 3D sculpture would melt
2. there is no place to locate the large transformers on or near the sculpture.

• A switching converter wouldn't require large transformers in the chandelier. – JRE Sep 26 '18 at 16:08
• That said, 79A is a LOT of current for continuous use. The cables will need to be thick (like, "car jumper cables" thick.) 7m with that amount of current will have high losses and get warm. Dude, you scare me with that thing. – JRE Sep 26 '18 at 16:13
• Going to 24V only halves the current, and you have to have a buck converter (or a bunchof little ones) to get to 12V. – JRE Sep 26 '18 at 16:15
• 3D print it with GE metal printer, the one with electron cannon, for jet turbines. ge.com/reports/post/94658699280/… – Gregory Kornblum Sep 26 '18 at 17:01
• Car jumper cables are roughly 2AWG (as shown here: youtube.com/watch?v=RNd2NK8-8wQ ) This would give the following: Voltage drop: 0.72 Voltage drop percentage: 5.98% Voltage at the end: 11.28 calculator.net/… i might be able to compensate for this with potentiometer on the power supplies. How can I work out how hot this 7m 2AWG cable would get? – Benjamin Freeth Sep 26 '18 at 17:07

The main power loss is in the LEDs, they are at best 30 % efficient, so 70 % of the 1 kW will end up as heat in the chandelier.

At that wattage I suspect a good PSU to be 90%+ efficient which would be only around 100 W as heat compared to 700 W by the LEDs which are directly mounted on the chandelier.

So I'd say your gut feeling is wrong (the PSU being the main heat generators).

I'm not sure for what you want to use this, but 1 kW LEDs will produce around 100.000 lumens (100 lm/W assumed). That is roughly the equivalent of a 5000 W halogen bulb. This is very bright...

Calculating or guessing the temperature this contraption reaches is quite difficult and for my limited time right now not feasible. Sorry for that, maybe someone else can answer that part.

• +1 for the 100.000 lumens comment. Why is this 3D printed stuff even in there? It's impossible to look at! – Janka Sep 26 '18 at 21:11

The first problem you have is deciding how to dissipate 945W of power in your chandelier. The efficiency of the LEDs at converting this power to light is inconsequential since light itself will heat any target it strikes, and I assume the light from individual LEDs will hit other parts of the chandelier structure. The problem of adding switching regulators will only make the equation worse by 20% (most switching supplies are above 80% efficient). The total power dissipated in and around your device is therefore likely about 1.2kW.

The best solution would appear to be to raise the DC supply to the maximum you can handle at your point of connection to the AC mains. Perhaps direct rectification of the mains (120V??) would be viable. This would provide about 160V DC and require only 7.5A.

Then you can distribute 160V - 12V DC-DC convertors within your structure so the 12V lines are as short as possible.

Note: you don't say how many LEDs there are or give technical details on them. Are you using hundreds of LEDs or only a few large LED preconfigured in strips(say 50W)?

To provide any real advice you'd need to supply much more detail such as:

1. Why 12V to drive the LEDs? Why are you not connecting multiple LEDs (or even LED strips) in series to reduce current flow in the structure? For example: You show in your diagram LED strips, so assuming these are just plain LED strips at 45W they will draw about 3.73A per strip. You could connect 7 strips in series so requiring about 84V @3.75A. You might be required to match the strips against light output and current, but that process is simple. Now you have only 3 * 3.75A distribution points. A much more manageable current configuration.

2. Do you need to modulate the lighting? This includes brightness control and generation of patterns. Do you need PWM control, or data streams to individual LEDs?

3. What is the maximum temperature your plastic can tolerate without losing significant strength?

4. What airflow restrictions exist in the structure? Why can you not provide air into the structure to dissipate heat.

Update_1: Since based on your comments each strip is composed of 220 LEDs, I'd suggest you cut these strips into 10 sections of 22 LEDs each and wire these smaller sections in series. That will raise the voltage requirement to 120V @45W, or about 375mA per strip total.

Since you have 21 strips in all this means that the total current requirement is 21 * 0.04 --> about 7.875A. That can be done with a quite reasonable wire size.

I assume you will want your installation to meet NEC code as much as possible, so you could do no better than using the NEC 310-15(B) to select your wire size. To keep the wire temperature well below 80degC you could select a wire capacity say double your requirement. If we look for wire from the chart then you could use 14G as a very conservative selection. This would support more than twice what you require and this would keep the cable temperature as low as possible.

The voltage drop for these wires is usually calculated at about 2-3% over the run length (you can look up details), so you'd expect to have a voltage drop @8A of only 1% with this underrating and wire temperatures < 50degC.

Raising the delivery voltage to 120V does of course mean a more specialized DC-DC convertor both at the mains entry point and within the chandelier, which might raise difficulties for you if you just want to select existing product.

• Thanks for your response Jack. Here's more information: 1) I'm using 21* 5 metre strips of leds @ 45w per strip (each strip has 220 leds), the strips are rated at 12v- this is where the calculation of 945W originates (78.75A * 12 V = 945W). 2) yes i will modulate the lighting using arduino and IRF520 mosfet - 13 of these to control 13 separate sections of the chandelier – Benjamin Freeth Sep 26 '18 at 17:19
• 3) the plastic is PET en.wikipedia.org/wiki/Polyethylene_terephthalate Heat capacity (C) 1.0 kJ/(kg·K) THERMAL CONDUCTIVITY 0.15[4] to 0.24 W m−1 K−1 Melting point > 250 °C to 260 °C Boiling point > 350 °C (decomposes) 4. The structure has tight narrow sections - air flow will be restricted through these – Benjamin Freeth Sep 26 '18 at 17:22