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long distance information...

Discussion in 'EthConGateway by Joshua 1 Systems' started by bobschm, Feb 27, 2013.

  1. bobschm

    bobschm New Elf

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    I'm planning for 8 strings of RGB next year. Looks like an ECGPIXAD8 will do the trick as far as channels/universes/etc. But I'd like to mount it and the power supply(s) in the garage beside my LOR panels...out of the weather. The nearest RGB string will be about 8 feet away. The furthest about 40 feet. WIll power or communications be a problem? What do you advise?
     
  2. fasteddy

    fasteddy I have C.L.A.P Global Moderator Generous Elf

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    Running Pixels from a distance away from the controller - potential issues

    The data can be easily managed by using a dummy pixel in the line if you find that the distance is too far, the 40 feet may be pushing and i would expect that you would need a dummy pixel or 2 in the line.

    A dummy pixel is a pixel you use that is not sequenced but each pixel regenerates the signal so as far as managing and signal degredation then using a dummy pixel is the cheapest and easiest method to use.

    Now the other issue that you must consider is voltage drop, I think this could be a bigger issue for you and you may need a few power injections or a very thick guage cable.

    at 40 feet the voltage drop just in the cable alone will be for a 5vdc of 50 pixels using 0.3 wattls per LED

    22 guage cable resistance = 0.0168 ohms/foot
    0.0168 x (40feet x 2) - the distance is both the +V and ground disance added together
    1.344 Ohms for a 80 foot run in cable resistance for 22 guage
    So now a single pixel draws 0.3 watts @ 5vdc this equates to 0.06 amps per pixel
    0.06 amps x 50 pixels = 3 amps
    voltage drop = current x resistance
    = 3 amps x 1.344 Ohms = 4.032 volts.
    So your voltage at the first pixel will be 0.96 volts using 22 Guage at 5vdc

    18 guage cable resistance = 0.0068 ohms/foot
    0.0068 x (40feet x 2) - the distance is both the +V and ground disance added together
    0.544 Ohms for a 80 foot run in cable resistance for 18 guage
    So now a single pixel draws 0.3 watts @ 5vdc this equates to 0.06 amps per pixel
    0.06 amps x 50 pixels = 3 amps
    voltage drop = current x resistance
    = 3 amps x 0.544 Ohms = 1.63 volts.
    So your voltage at the first pixel will be 3.37 volts using 18 Guage at 5vdc

    16 guage cable resistance = 0.004 ohms/foot
    0.004 x (40feet x 2) - the distance is both the +V and ground disance added together
    0.32 Ohms for a 80 foot run in cable resistance for 16 guage
    So now a single pixel draws 0.3 watts @ 5vdc this equates to 0.06 amps per pixel
    0.06 amps x 50 pixels = 3 amps
    voltage drop = current x resistance
    = 3 amps x 0.32 Ohms = 0.96 volts.
    So your voltage at the first pixel will be 4.04 volts using 16 Guage at 5vdc


    Now we will take a look at using 12vdc light strings and how this effects voltage drop
    (refer to this thread for the differences in 12vdc and 5vdc strings)


    22 guage cable resistance = 0.0168 ohms/foot
    0.0168 x (40feet x 2) - the distance is both the +V and ground disance added together
    1.344 Ohms for a 80 foot run in cable resistance for 22 guage
    So now a single pixel draws 0.3 watts @ 5vdc this equates to 0.025 amps per pixel
    0.025 amps x 50 pixels = 1.25 amps
    voltage drop = current x resistance
    = 1.25 amps x 1.344 Ohms = 1.68 volts.
    So your voltage at the first pixel will be 10.32 volts using 22 Guage at 12vdc

    18 guage cable resistance = 0.0068 ohms/foot
    0.0068 x (40feet x 2) - the distance is both the +V and ground disance added together
    0.544 Ohms for a 80 foot run in cable resistance for 18 guage
    So now a single pixel draws 0.3 watts @ 12vdc this equates to 0.025 amps per pixel
    0.025 amps x 50 pixels = 1.25 amps
    voltage drop = current x resistance
    = 1.25 amps x 0.544 Ohms = 0.68 volts.
    So your voltage at the first pixel will be 11.32 volts using 18 Guage at 12vdc

    16 guage cable resistance = 0.004 ohms/foot
    0.004 x (40feet x 2) - the distance is both the +V and ground disance added together
    0.32 Ohms for a 80 foot run in cable resistance for 16 guage
    So now a single pixel draws 0.3 watts @ 5vdc this equates to 0.025 amps per pixel
    0.025 amps x 50 pixels = 1.25 amps
    voltage drop = current x resistance
    = 1.25 amps x 0.32 Ohms = 0.5 volts.
    So your voltage at the first pixel will be 11.6 volts using 16 Guage at 12vdc

    So in this situation you would be better using 12vdc strings with a 16 guage cable to stop voltage drop or if using 5 vdc strings you would then need a few injection point using 16 guage cable.
     
  3. OP
    OP
    bobschm

    bobschm New Elf

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    As Barbie once said, "math is hard! ". Thank you for doing all that research. Now I'm thinking if I can place a waterproof power supply in the middle, four strings left and four strings right, I might be good. So if my bank account holds up...
     

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