Information overload-help complete panel system

Re: Information overload-help complete panel system

So, you have a 12 volt battery bank (how many Amp*Hours, type of Battery?).

Your 145 watt "12 volt panels" are really around 17.5 volt Vmp probably (that is a good thing). Basically, as solar PV panels get hot in the sun (and in hot weather), the Vmp voltage falls... You still need Vmp-array-hot>Vbatt-charging+controller voltage drop+wiring voltage drop.

So, still would be a good idea for you to confirm the panel brand/model/Voc/Vmp/Imp/Isc numbers... But assuming Vmp is at least 17.5 volts and Voc is around 22 volts:

• The Outback FM family (last I read) is most efficient with Vmp-array ~ 2xVbatt-charging. So that would be two 145 watt panels in series.
• You have the option of running 1/2/3/4 panels in series with your battery/controller configuration (5 panels in series, and during cold/freezing weather, you run the risk of Voc-cold over voltaging your charge controller's 140-150 VDC max input).
• Sending power a fair distance (say over ~10 feet), you will need a lot of expensive copper to run your array with all panels in parallel (plus, you should have 1 fuse/breaker per string when you have 3 or more panel strings in parallel).
• You did not say the distance, but you are looking at 2 panels in series with 3 parallel strings (need fused/breakered combiner box), or you can run 3 panels in series/2 strings in parallel which will allow you to send the power farther on smaller gauge wire.
• So, need to know the distance for the wire run from the array to the battery shed/controller/battery bank.
• With a 12 volt battery bank and a 60 amp charge controller, you can build the array out to ~8 panels "cost effectively" before the FM 60 will start current limiting "a lot" during the middle of the day (note, if you want to add panels, they need to meet some sort of valid series/parallel configuration for your controller).

Your Xantrex 1,800 watt inverter should have information about wiring/fusing... But in general, this is how the math would look without reading the manual:

First, the maximum DC input current for the inverter:

• 1,800 watts * 1/10.5 volt batt cutoff * 1/0.85 inverter eff = 202 Amps max continuous
• 1,800 watts * 1/10.5 volt batt cutoff * 1/0.85 inverter eff * 1.25 NEC derating = 252 Amps minimum branch circuit

So, you need to meet two requirements... One is that the wire does not overheat. NEC has a table which is fairly conservative:

Wire Current Ampacities NEC Table 310-16

You can use 4/O cable with the "better" insulation (this table assumes in conduit).

The ABYC (boating) uses this table:

ABYC Wiring Standards

which can allow down to 1/O cable (less than 1/2 the cross sectional area).

The other design factor is voltage drop. What is the maximum power you will draw from the inverter (continuous and peak)... Assuming 1,800 watt continuous and 3,600 peak with 12 volt system (11.5 volt low battery voltage and 10.5 inverter cutoff)--I will suggest 1,800 watts (~202 amps) with a maximum of 0.5 volt drop (allows for surge capacity).

Using a generic voltage drop calculator, 202 amps and 0.5 volt max drop with 1/O and 4/O wiring, the maximum one way distance would be:

• 202 amps, 1/O wire -> 10' one way run and ~0.5 volt drop
• 202 amps, 4/O wire -> 20' one way run and ~0.5 volt drop

And you should use ~250 amp fuse/breaker.

Note, lead acid storage batteries have limitations on how much power they can "reliably" deliver... Assuming 4x 6 volt 225 AH golf cart batteries configured as a 12 volt 450 AH battery bank:

• 12 volts * 450 AH * 1/8 max continuous load * 0.85 inverter losses = 574 watt max continuous (you can exceed this load, but battery bank will discharge pretty fast and may get warm)
• 12 volts * 450 AH * 1/2.5 max surge load * 0.85 inverter losses = 1,836 watt max surge load (again, recommended for normal operation)

And working backwards from the battery bank size, the recommended amount of solar panel for charging would be (based on 5% to 13% rate of charge):

• 450 AH * 14.5 volts charging * 1/0.77 panel+charger derate * 0.05 rate of charge = 424 watt minimum array
• 450 AH * 14.5 volts charging * 1/0.77 panel+charger derate * 0.10 rate of charge = 847 watt "nominal" array
• 450 AH * 14.5 volts charging * 1/0.77 panel+charger derate * 0.13 rate of charge = 1,059 watt "cost effective maximum" array

Again, the above are "conservative" rules of thumb that seem to give a working system for most people. Looking at your system, your inverter is a bit on the "large" size--But everything seems to be sized pretty well otherwise.

So, once we know the distance from the array to the battery/controller location, we can help with the rest of the calcuations.

-Bill