Hey Derek,
30 A can not be enough, that’s why I still insist sometimes. Can calculate it for you but not today, Just think of that you have a single phase AC at input, that gets rectified and stored in capacitors, and then a microcontroller switches six IGBTs on and off to form three phases on three different wires at the same time of the max. current that the VFD is rated at output. If you look at the overlying waveforms of the three phases on three-phase electricity, it is clear that at the input, you will at least draw the double current at the input as what the VFD 3-phase rated output current is. And then you add the efficiency of the VFD itself and it gets even more. If you take the strongest motor that is rated for a 2.2 kW VFD which is usually 10 A on 200 V class and 20 A on 100 V class, and consider this with the induction motor formula you find here, it is clear that for a 2.2 kW motor you have more than 4.2 kVA input which at 120 V is already 34 A, and you also see the usual capacity ratings on 2.2 kW VFDs of 4.5 kVA in the manual, and then add the efficiency loss of the VFD itself, it is clear that you would recommend to connect a device of about 40 A max. input rating to a 120 V circuit, knowing that the maximum domestic 120 V circuits are only 30 A strong. An electrician can not do this! Or I need to revise my view about electricians.
I could also try to make it comprehensible in another way, for all who fear the formulae etc.: Here is a single-phase input VFD of the 200 V class for a 2.2 kW motor, it is able to deliver 12/11 A (depending on duty cycle) and at its input, it is rated 24/22 A. In the manual, the fuse recommendation is 30 A for this model.
The same model, if it existed, for 100 V class (they don’t make those), would be for max. 24/22 A spindles, and could be assumed to be rated the double then as 200 V class model, thus 48/44 A input current. And the recommended fuse even bigger.