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I am working on an antenna and ATMEGA-based reader. It is powered off of 2 6v deep-cycle marine batteries, and the reader/antenna draws 1 amp at 6 volts. I'm not sure about the amp-hours in the batteries, but they can power the board easily for a week or two. we know this because we need to swap out the batteries with charged ones on the systems that have not been converted to solar. As it is an annoyance to keep hauling out charged batteries, we are trying to convert the rest of the systems to solar. I'm trying to find out the length of cable with which the board will still be getting enough juice from the panel.

Here's a link to the panel and charge controller. This charges the batteries - the charging circuit has already been figured out for me. The max output of the panel is 5 amps at 17.2 volts. Using this calculator, I have been trying to find out the voltage drop.

  1. do I have to account for both + and - leads? For example, if my solar panel is 100' away from my charge controller, do I have to account for 200' of voltage drop?

  2. Does only voltage drop(not current)? do you subtract the voltage drop from the original voltage, and assume the original current?

  3. What is the longest length of wire that you would trust to run the system, considering that the antenna draws 1 amp at 6 volts, the panel can output 5 amps at 17.2 volts maximum, and the cable is 14AWG stranded copper wiring?

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Your answers:

  1. Yes, the current flows through both lengths of cable so you have to account for both.

  2. Only voltage drops. Current is drawn through.

  3. Basically I wouldn't go longer than a drop that would take it below the charge threshold of the batteries / charge circuit.

Basic ohms law is your friend:

For 1 amp of draw:

  • 14AWG has a resistance of 2.525mΩ/ft

  • So 200' has a resistance of 0.002525 * 200 = 0.505Ω

  • V=IR

  • Therefore V=1 × 0.505

  • Which equals 0.505V

For the full 5 amps of draw:

  • V=5 × 0.505

  • = 2.525V

Which is obviously quite a drop.

This table has a nice list of the resistances of different cables. As you can see increasing the cross-sectional area (decreasing the AWG) decreases the resistance. Even just going from 14 to 12 AWG has a big difference (1.588mΩ/ft as opposed to 2.525mΩ/ft), and fatter cables are even better.

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    \$\begingroup\$ The answer to #2 is from Kirchoff's Current Law. \$\endgroup\$ – Mike DeSimone Jun 14 '11 at 13:01
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I see you have already gotten a direct answer to your question, so here is something else to think about.

Eventually you are transferring power from one place to another. In your case, it seems you need 6W (1A at 6V). The same power can be transferred at a higher voltage and therefore a lower current. The voltage drop on the cable is only a function of the current thru it. Using higher voltage and therefore lower current would allow for using a smaller cable and less power wasted in the cable.

For example, even going to only 12V at 500mA makes a big difference. Now the cable can use half the copper to get the same voltage drop, and that drop wastes half the power as before.

The lower cost of the copper cable and the slightly lower power the panel has to produce might offset the cost of the appropriate converter at the device. You can probably get a panel with higher voltage and lower current for about the same price since it would require the same total solar cell area. It would be more smaller cells wired in series. The cost of a converter from a higher voltage to the proper float voltage for the 6V batteries may well less than the cost of a thicker cable.

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    \$\begingroup\$ This is exactly why the power distribution companies use such insanely high voltages to transfer power around the country. \$\endgroup\$ – Majenko Jun 14 '11 at 12:59
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    \$\begingroup\$ Another thing you could do is use switching DC/DC regulators to step up the voltage from the panel and step it back down at the load. Linear Technologies has a lot of parts that are god for this application. \$\endgroup\$ – Mike DeSimone Jun 14 '11 at 13:03
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You have gotten good answers to all your questions, except the answer to question 3 could be a little more direct; so I will chime in on that one.

So, the 5A/17.2V supply is just for your single 1A max load? If it is 1A continuous, then you are going to need maybe 3A during effective insolation hours to average 1A over 24 hours. In that case I would derate the 17.2V to 15V. that gives you 9V of headroom (allowable V drop) over your 6V/3A load.

V = IR ---> 9V = (3A)(R) ---> R = 3 Ohms

The resistance of your 14AWG copper is 0.00505 Ohm/foot. (That is (0.002525 Ohms/foot)(2).)

Allowable run = (3 Ohms)/(0.00505 Ohms/ft) = about 600 feet.

Note that we have all been assuming copper wire. If you go with aluminum, increase R by a factor of 1.6. So the allowable run with 14AWG Aluminum would be about 600/1.6 = 375 feet.

You can go to larger AWG sizes or double up on your 14AWG to get larger distances, or lower IR drops for a given distance.

If you go for a long enough run that your IR drop exceeds, say, 0.5V, you will probably need to run 2 extra voltage sensing wires in parallel with your current carrying wires, so that your controller turns on and off based on the voltage at the terminals of the marine battery you are recharging.

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