# Voltage drop

Discussion in 'General Electronics Discussion' started by Zellarman, Feb 13, 2012.

1. ### Zellarman

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Nov 9, 2011
Is voltage drop purely a loss in power because of the resistance in long runs of wire, or is there something else to it? I'm designing LED landscape lighting to operate at 12 volts, and each fixture draws about only 50 ma, but I'm planning on a large number of fixtures and one long run. I connected all of the cords I plan on using into a single run and tested it, I only lost .1 volt on a 120 foot run, but that was without any draw at each junction (without any LEDs actually hooked up) is this right? Can I expect to get 11.9 v at the end of a 120 ft run (I think it's 16 ga wire) that has a .05 Amp draw every 3 feet (40 total)?

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,500
2,840
Jan 21, 2010
Presumably each fixture is designed to operate from 12V (or alternatively you've fitted a series resistor to a LED for that purpose.)

The voltage drop along a length of cable is determined by the resistance and current.

In your case the current is not constant (it drops at every junction where a fitting is connected)

The best thing to do is to measure the resistance of the cable, double it (because there are two conductors) then assume that exactly half the load appears at the end of the cable. So that's 20 fixtures = 1A. The voltage drop = I * R, which is 1 x whatever the *doubled* measures resistance was. If it was 1 ohm, you will drop 2 volts over the run.

Assuming the fixtures are evenly spaced (which I also have above), the drop will vary evenly from the first bulb (essentially no voltage drop) to the last (max voltage drop). It is possible you won't notice the difference, but you might...

You can estimate the effect by getting a resistor with 40 times the measured voltage of the cable and connecting it in series with a fixture. If you use a piece of wire to short out the resistor (the lamp will get brighter) then you can see the amount of difference between the first bulb and the last.

3. ### Zellarman

8
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Nov 9, 2011
Steve, thanks for your reply. I think I understand, in general at least. Though now I have a couple of new questions. First you had said to "use a resistor with 40 times the measured voltage of the cable" I assume you meant 40 times the resistance? Then, maybe the factor of 40 is directly related to the number of fixtures, is it really that simple, or is this just a coincidence?