I am looking at adding another array to catch some winter sun.
To accomplish this, I will need to place the array further uphill, meaning a run of at least 200'.
As I am looking at wire costs, I am trying to come up with the most economical wire size.
SO, as I am looking at size, do I calculate for the nominal operating voltage, the lowest operational voltage, or the maximum operational voltage, same on amperage. We do get some mighty cool temps here; it has been known to hold at -40 F for a few days at a time.
If I plan on maximum amperage and nominal voltage, is there any risk of damage when I am in low sun, or is it just that I am losing the low end until there is enough solar to overcome the voltage drop and start producing?
Thanks much,
To get the smaller wire gauge from array to controller, I would have the solar array at the highest possible voltage ( use the string calculator for your particular pv panels) . I think you calculate using the Vmpp and the string calculator should tell you that. The highest voltage will have the lowest current . Then you will have to use a wire loss calculator to find out what you need based on the current.
I like to use this one
https://www.southwire.com/calculator-vdro
Larry
Cool, thank you, I like that calculator.
So am I correct in thinking that by going to the smaller wire (but one that will still handle the amperage), all I am potentially losing is in the lower light area. Since as the wire size decrease voltage drop increases. I will need more light to get the voltage required to actually start charging the batteries.
Example: I may be getting 125V at the panels, but due to voltage drop I may only be getting 112V at the Midnite but still at a set amperage, so the wire is still safe? As long as the wire meets the amperage needs, I am not building a fusible link?
As light increases, voltage drop decreases and the system (wires) gets more efficient.
Quote from: LilMT on October 10, 2023, 03:19:59 PMCool, thank you, I like that calculator.
So am I correct in thinking that by going to the smaller wire (but one that will still handle the amperage), all I am potentially losing is in the lower light area. Since as the wire size decrease voltage drop increases. I will need more light to get the voltage required to actually start charging the batteries.
Example: I may be getting 125V at the panels, but due to voltage drop I may only be getting 112V at the Midnite but still at a set amperage, so the wire is still safe? As long as the wire meets the amperage needs, I am not building a fusible link?
As light increases, voltage drop decreases and the system (wires) gets more efficient.
I doubt you will see too much difference in the output of the pv no matter what voltage you have them wired for. As long as you have the wire sized for the current then you are okay.
My 8 pv panels are all wired in series and that is about 240v ( into Hawkes Bay) . The voltages add in series but the current stays the same so the current is only around 10 amps at 240v. I just looked at the voltage drop calculator and made sure I used a wire size where the drop wasn't too excessive . In my case I went with 10 gauge at 100 feet of wire , though I may have gotten by with 12 gauge. I forget the exact calculations since it was last year I put that new PV up.
Larry
Thank you very much, I thought I was on the right track but always helps when someone else agrees. :)