DC lighting transfromers

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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
. . . as you can see this ring is fed to both sides of the ring, and the voltage and resistance will be equal around this ring circuit, . . .
But, wouldn't those lights nearer the transformer's side of the circle still be brighter?


I like Gar's method, which, as a circle, would be just like yours, except that one ring would have a break just clockwise of the transformer, and the other would have a break just counter-clockwise of it.

In other words, the lights are supplied via a typical single parallel pair of wires, except one wire gets fed at the end closest to the supply, and the other, via a third wire, at the far end.

This way, every light receives the same voltage, regardless of overall circuit length.
 

dicklaxt

Senior Member
Now that I have been introduced to this circuitry,I ask myself why this is not used in plain vanilla 120VAC lighting.I guess when VD is not a problem the cost of the 3rd conductor is a total waste but when VD is a problem then the added cost of the 3rd wire would be accepted.I guess at this point a comparison of calculating VD to the load center and upsizing conductors as needed is a wash.It would be interesting to see where the break even point was with all factors stirred in.

dick
 

WinZip

Senior Member
Dick,

What voltage would you need to send through the entire 400 foot loop to make this work , and wire size.

12 - 13 -14 - 15 -16 - 17 volt
 
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dicklaxt

Senior Member
That would take a bit of calcs but doable,I would first divide the wattage of the lamp by a desired voltage to determine the current,then with that known value I could solve a second equation and determine the resistance of the lamp,you can now substitute different voltage levels and solve for current at that voltage,do a summation of current in resistors in parallel to get total load at a given voltage,then using that solve for VD allowed for the threshhold voltage, which would be the acceptable voltage required to give you a satisfactory light output.A lot of basic Ohm's Law math.

I think I said that right,if not someone will correct it LOL'

dick
 

WinZip

Senior Member
No offense but that seems like a lot of thinking an calculating to make that set up work, then you have the person that comes back years later working on it an he is baffled LOL
 
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