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.
-
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?
-
Does only voltage drop(not current)? do you subtract the voltage drop from the original voltage, and assume the original current?
-
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?
Best Answer
Your answers:
Yes, the current flows through both lengths of cable so you have to account for both.
Only voltage drops. Current is drawn through.
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.