# How to best run long distance low voltage electric

I am looking into options for running 12 VDC approximately 2000 feet to power some remote equipment. The equipment will have a low current draw (2A max).

I am seeking to keep the cost low. Copper wire is quite expensive, especially as size gets larger.

Here are the options I can think of... Can anyone suggest a better alternative?

1. Run 1AWG wire the entire distance and input about 14VDC (Too expensive!)

2. Run 14AWG or smaller (24AWG?) wire and input 120VAC with a step-down transformer and rectifier/smoothing circuitry at the destination (is this crazy?)

3. Use solar power with deep-cycle battery and charger (unreliable?)

4. Battery-only system: swap out and recharge batteries as they discharge. (too much maintenance)

Any suggestions for a better way to do this?

• It appears that you've already considered the inherent voltage drop in your question. Kudos for prior research. – JYelton May 21 '14 at 21:00
• What's the average current draw, and how long do the 2 A peaks last? – The Photon May 21 '14 at 21:05
• Even if you boost the voltage (or use Mains voltage), I would recommend using at least #18AWG wire. It's standard practice (and often a code requirement) in industrial equipment to use a minimum of #18AWG or larger wire for mechanical strength, even if thinner wire would be electrically acceptable. – bitsmack May 21 '14 at 21:34
• Adapting some power-over-Ethernet gear might be viable: it's boosted to 44V, although the maximum power is supposedly only 25W. – pjc50 May 21 '14 at 22:03
• Run 1/2 inch PVC pipe and put a little turbine/generator at the distant point (die grinder and generator for example). Pump air through it. Might be cheaper than a mile of wire. – C. Towne Springer May 22 '14 at 5:39

14 AWG has 2.525 ohms per 1000 foot so clearly, at 2A, this is not going to work because the volt drop at 2000 foot is twice this and then twice again because 2 wires need to be used for send and return.

You need to push 24 watts to the far end so maybe consider a boost converter. It will step up the dc voltage at the sending end and then, at the receiving end a buck converter can restore it back to 12 volts.

I'm also thinking that to get 24 watts at the far end, it's not unreasonable to push 30 watts into the cable at the sending-end - this means a loss of 6 watts in the cable.

Given 2000 ft of 14 awg cable (doubled) has a resistance of 10.1 ohms, you could argue that to dissipate 6 watts, the current should be: -

Power = I$^2$ R therefore I = $\sqrt{P/ R}$ = 0.77 amps

Given that you need to feed in 30 watts, the boosted voltage at the sending-end will be: -

30/0.77 = 38.9 volts.

This seems perfectly doable using a booster to raise the DC voltage to maybe 45 volts (gives a little bit of extra power for inefficiencies in the switchers).

I'd use a booster because it's fairly simple, can easily have current limiting to protect the cable and uses a safe voltage range. You could repeat the calcs for 24 AWG wire using the above method to determine sending voltages for other gauges. If you are in EU, a good safe DC voltage limit is 60 volts - this is known as SELV (safety extra low voltage) - paying a little bit more than lip-service to safety specs could save you a law suit!!

Gut feeling is that 24 AWG might need more than 60V.

• Thank you Andy i was leaning toward a step-up/step-down solution as probably the simplest way to accomplish this. Now i feel more confident. More research is now on my agenda per your info. – Ryan Griggs May 21 '14 at 21:50