Powering my 12v DMX Devices

[Steven]

I measured the current my new DMX devices draw and now I need to figure out how to get power to them. My results were: Rainbow Flood Extremes: 1.09A (x2); Rainbow Floods: 820mA (x3); RGB strips around windows: 1.70A (x3) for a total of about 10A.

I figure my 25A power supply from Taiwan should work nicely. I will use a MultiPort Blender from Seasonal Entertainment to spit the DMX+power signal 3 ways and group my fixtures into 3 groups based on location.

The tricky fixture group is about 30 feet from the power supply, and is 2 RGB strips and 2 Rainbow Floods for a total of about 5A. If I use Cat5 cable, the computed voltage drop will be 30 ft * 2 / 3 * .02567 Ω/ft * 5A = 2.6V. (The factor of 2 is for the power and ground, and the factor of 3 is because I have 3 24AWG in parallel.) This voltage drop seems too high for proper operation, so my idea is to use SPT-1 18-gauge cable in parallel with the 12v supply lines. My calculations say that this will reduce the voltage drop to .4V, which is acceptable.

I need somebody to give me a little sanity check here, to let me know if I am going to completely screw this up. This is the first time I've tried running low voltage power over these distances (although I did run about 30 feet from a DC controller to a (single) Rainbow Flood last year, and it worked fine).

[edvas69]

I would tend to think you are correct here and is one of the reason CAT 5 is not a good choice for power.

you are correct the actual toatal run is 60 feet, 30 feet for +V and 30 feet for return ground.

having 3 cables in parrallel effectively divides the overall resistance by 3

Cat 5 is really only good for 1 amp per core so even using 3 cores you are effectively 2 amps below the specified load you require. people must remember that CAT5/6 cable is designed for signals and is wound differently to standard power cables so the current capacity of the actual copper is derated.

If using the 18 guage cable you can expect at least 7.5 amps which definetly gives you some head room with your load

Its all to do with ohms law and the relationship between voltage, current and resistance. The voltage and the current are set due to the elements in your display you are supplying power to, so its the resistance that we can change by using larger guage cable or more cores.

Volts = current x resistance

So the CAT5 cable you referred to has a resistance of 0.02567 Ω/ft so at 5 amps we could expect

5 amps x 0.02567 = 0.12835 volts dropped per foot

= 0.12835 x 60 feet = 7.701 volts dropped

That same cable with 2 amps load

2 amps x 0.02567 = 0.05134 volts dropped per foot

= 0.05134 x 60 feet = 3.0804 volts dropped

So if using 18 guage cable with a resistance of 0.00762 ohms/foot we could then expect

5 amps x 0.00762 ohms = 0.0381 volts per foot

= 0.0381 x 60 feet = 2.286 volts dropped

So as you can see from the example above the more head room you have as far as the current capacity of the cable and the load you are actually putting down the cable will determine the amount of voltage drop. The higher the voltage the less % of voltage drop we will see over the same distance with lower voltages.

[Ken Benedict]

Online voltage drop calculator: http://calculator.net/voltage-drop-calculator.html?material=copper&wiresize=52.96&voltage=24&phase=dc&noofconductor=1&distance=100&distanceunit=feet&amperes=3&x=96&y=20

[Ponddude]

Steven your math appears to be correct there and the reason we never recommend long distances for the basic Rainbow lights. Another reason we introduced the waterproof cabling system that is on the Blenders.

In any event, whatever method you choose, the math you did above appears correct. (Although math was never a strong point of mine. )