Phone company left me 500' of 4 strand telephone wire with a heavy insulation that is rated tough enough where all you need to do is role it out on the ground ( no burying or hanging or conduits needed). I want to draw 12v dc at 0.2 amps (one fifth of a amp) over 200' and want to calculate or know the voltage drop . I am guessing 30 AWG?

The correct terminology is multi conductor "cable." With four single strand wires.

Phone company uses lots of different wire - but 30 awg sounds too small, try 22-20 gauge. Or here's a novel idea - measure the diameter and look it up.

Make sure it is copper also - they do use other materials, especially when they want an indestructible cable.

Or if you can get at the ends, measure the resistance and calculate your voltage drop.

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You really do not need to know the guage of the wire. Logic would suggest that you utilize the 500' roll, run the 12 volts through it, load it with your 200ma load and measure the voltage drop.... then you can interpolate the approximate voltage drop for any length including the

200 ft length you are planning to use. The 500 ft length will not permit the intended load to reach the full 200ma because of the additional voltage drop of the extra length but this method will get you close....... you could use a variable resistor to create exactly a 200ma load and then your interpolation would be dead on. If you use a resitance chart for the guage (size) of the wire then you will have to multiply by TWO because your circuit will use two wires. Best Regards, Daniel Sofie \- - - - - -

----- Original Message ----- From: "davidlaska" Newsgroups: sci.electronics.repair Sent: Thursday, August 31, 2006 12:33 AM Subject: What AWG is "Drop on ground" phone line wire?

"davidlaska" wrote in news:1157009586.378668.258860 @p79g2000cwp.googlegroups.com:

If it really is copper wire, then the following numbers might be helpful.

20 AWG, .032 in dia, 10.15 ohms per 1000 ft, 2 ohms per 200 ft., at 0.2 amp the voltage drop will be 0.4 volts each wire, total 0.8 volts for two-wire setup. (If you parallel two wires out and two wires back, making use of all 4 wires in the cable, total voltage drop is 0.4 volts.)

22 AWG .025 in dia, 16 ohms per 1000 ft. 3.2 ohms per 200 ft., total voltage drop 1.2 volts (2 wire) or 0.6 volts (4 wire).

24 AWG .020 in dia, 25.7 ohms per 1000 ft, 5 ohms per 200 ft., total drop
2 volts (2-wire) or 1 volt (4-wire).

26 AWG .016 in dia, 41 ohms per 1000 ft, 8 ohms per 200 ft., total drop

3.2 volts (2-wire) or 1.6 volts (4-wire).

As you can see, you might get by with 20 or 22 AWG, but 26 AWG or smaller is totally hopeless.

The unit I am power is a sony bullet cam that uses a switching power source, I suppose that means it would be more susceptible to voltage drops. I wonder if on a hot sunny day if voltage was border line in the cool morning, it would drop further during the afternoon when things get the hottest. From all the help here, the easiest less thinking approach is to take a voltage reading at the end of the line with a 20 watt 12v light and see the drop and compensate any number of ways to get close enough for the the camera to be safely powered up.

The cable's 4 wire and you only want 2 (12v + and -), correct?

If you parallel up the cores to make 2 wires of 2 cores, that'll reduce the resistance of the cable and thus the voltage drop.

I did this with a long run to some garden lights. With 2 wires, the voltage drop made the lamps dim. Fortunately, I'd run 4-core 1.5mm^2 cable out to the lights (it was in my spares pile) and so was able to reduce the voltage loss by doubling up the cores. This made the lamps run at about normal brightness.

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