MrsAlaskanNoob again. Understood re higher voltage/lower current = smaller wires. My question, in deciding whether we use AC or DC to run the 500 feet is this:

AC, if we step it up, we can do it with two wires, I think (single phase, live wire and neutral wire). If we are able to step it up high enough, this wire can be a reasonable size. Or so I think.

This option would mean the batteries, inverter, generator and the balance of system would need to be in a heated structure at the site of the panels.

The inverter can only deliver 8000 watts in inverter mode and the generator can only do 7000 max so lets size for 8000 watts as a gesture to surge capacity.

For simple single phase 230VAC@60hz

1

**un-grounded** conductor(live)

1

**grounded** conductor(neutral)

1

**grounding** conductor(ground)

8000 ac watts / 230VAC = 34.782608696 service amps

34.782608696 service amps / .8 fuse headroom = 43.47826087 fault amps

The means the live and neutral have to be copper 1 awg or larger, so better than 2000 feet of 1 awg wire

The ground wire can be smaller but it still has to be larger enough to clear a ground fault. so lets say 2 awg. So 1000 feet of 2 awg.

This free voltage drop calculator estimates the voltage drop of an electrical circuit based on the wire size, distance, and anticipated load current.

www.calculator.net

For split phase 120/240VAC@60hz

2

**un-grounded** conductors(live)

1

**grounded** conductor(neutral)

1

**grounding** conductor(ground)

8000 ac watts / 240VAC = 33.333333333 service amps

33.333333333 service amps / .8 fuse headroom = 41.666666667 fault amps

The means the both lives and neutral have to be copper 1 awg or larger, so better than 3000 feet of 1 awg wire

The ground wire can be smaller but it still has to be larger enough to clear a ground fault. so lets say 2 awg. So 1000 feet of 2 awg.

This free voltage drop calculator estimates the voltage drop of an electrical circuit based on the wire size, distance, and anticipated load current.

www.calculator.net

I think it comes down to is 3000 feet of 1 awg worth more than a auto-transformer and the additional complexity and and another point of failure.

Folks have said DC would be better. But for DC, if we put all the equipment by the cabin, we have to run 500 feet between the panels and the charge controllers. The charge controllers' capacity limit us in terms of voltage, i.e., we can't do one massive string of all the panels - too many volts.

I suggest that you have 2 solar charge controllers for redundancy.

Have the voltage as high as possible which also means the current can be lower.

Lets take 24000 watts.

Each run requires 2 wires, 12000 watts@600 volts = 20 amps service amps

20 amps / .8 fuse headroom = 25 fault amps.

That requires a little over 2000 feet of 6 awg thhn in conduit in a trench.

This free voltage drop calculator estimates the voltage drop of an electrical circuit based on the wire size, distance, and anticipated load current.

www.calculator.net

So doesn't running DC mean running many wires 500 feet, even if they're smaller, compared to the two-wire AC option?

See previous.

The ac is not a 3 or 4 wire option.

My question is - does anyone know of a way to run just two wires for all of the DC power that our array can produce, i.e., is there any charge controller that can take that much voltage? (2829 volts, by my calculation ... or say half that, 1415 volts, given I think PV wire is rated at 2000 volts).

Not that I know of.

Direct bury aluminum is the cheapest option.

Much cheaper than copper even though you need larger conductors.

The direct bury part means no conduit which is another expense and a pita to install.

So you can't just pull another wire.

Once the trench is back-filled its a done deal.

I championed it a couple of times but people seem to not like aluminum although I've not heard any clear objections.