Cable Feeding the Inverter

To calculate the size of cable feeding the inverter, we need to know 1) the maximum amount of power being provided by the inverter 2) the efficiency of the inverter at configured values and 3) the voltage we are going to be feeding the inverter with.

The power being supplied by the inverter is equal to the input power derated by the inverter efficiency.

`P_"in"=(P_"out")/(Efficiency)`

We know we plan to provide a maximum of 220 VAC at 100 amps.  

That gives us a wattage of 22,000 watts at the output of the inverter

The manufacturer should publish the efficiency of the inverter in its specs sheet.  A typical efficiency falls between 85% and 95%

I will use 90% efficiency for this design.

I will also use 48 VDC for the input voltage.  There are some inverters that go as high as 384 VDC input voltages, but that comes with another set of issues in the battery array.  48 VDC is a pretty common input voltage.

`P_"in" = (P_"out")/(Efficiency)`

`P_"in" = (22000 W)/.90`

`P_"in" = 24,444.44 W`

`I_"in" = (P_"in")/(V_"in")`

`I_"in" = (24,444,44 W)/(48 VDC)`

`I_"in" = 509.25 A`

The feed wire to the inverter will need to carry over 500 amps.  This could be a single wire of 1000 MCM (about 1.32 inches in diameter) or multiple smaller wires.  It is not uncommon to use the same wire as was used to feed the loadcenter.  In this case you would use as many as would be needed to carry 509 amps or more.  

Do not forget that this wire needs to be derated based on the length of the run.  So keep the runs short to save money and increase efficiency.

As you might suspect, this wire is quite expensive (Over $50/ft. as of this writing) but there are ways to reduce its size.  

Place the inverter as close to the battery array as possible

This will reduce the length needed considerably, but it will reduce efficiency on the AC side due to line losses.

Increase inverter input voltage

Increasing the inverter voltage will reduce the diameter of the wire needed, but may increase the number of batteries needed in the array.  This is because the batteries are placed in series, to increase the voltage.  So, increasing the voltage will increase the number of batteries in each chain.  Of course, you can reduce the number of chains, but the number of batteries may still need to increase slightly.

EXAMPLE:  

Suppose we need 8-12 volt batteries and we currently run a 48-volt array.  We have 2 chains of 4 batteries each.  We can easily switch this to a 96-volt array by placing 8 batteries in series.  We now have a single chain of 8 batteries. No battery count increase is needed.

This has cut our inverter infeed cable's current requirements by 50% (to 254.5 amps) without adding any batteries.  This reduces the cable size to 250 MCM (About $9/ft)

Now, suppose we need to add more battery capacity. We must add 8 new batteries...instead of the 4 the old configuration required.  Batteries must be added in complete chains.  

There are possibly other options, though.  For instance, we could just increase the array voltage to 108 volts and add a single battery.