Hey Jared,
I was only talking of the input of the VFD, the supply circuit for the VFD.
The output (connection to the spindle) of a VFD is always a three-phase current, as induction motors only work with this. The three-phase current is taken from the capacitor that stores the DC and is switched by six IGBTs to form three quasi-sine waves of AC on three output wires. Also according to IEC 60034-1 the power rating on spindles does not mean the electrical power they draw, but the mechanical power they are able to deliver at their shaft (see here for details).
A 200-240 V VFD for 2.2 kW spindle may draw more than 4 kVA. Here you see the nameplate of two typical 2.2 kW VFDs:
Image 1: Michael’s Hitachi WJ200 VFD nameplate.
Image 2: Aiph5u’s Omron MX2 VFD nameplate.
As you can see, on single-phase input, they are rated with 24 A current draw, and in their manual, a 30 A fuse is recommended. The wire size I use to supply this VFD is a 2+PE wire in the strength 4 mm².
At 110 V (which these VFD manufacturers do not offer), the same power would require double the current, double the fuse rating and double the wire cross section area (i.e. 48 A supply circuit, 60 A fuse, and 8 mm² wire cross-section area). Cheap chinese manufacturers are the only ones that offer 110 V VFDs for 2.2 kW (3 hp) spindles, and since there exist no domestic supply circuit in the U.S. to drive such a VFD at 110 V, they simply omit the VFD input current rating both in the manual and on the nameplate of the VFD.
(This has nothing to do with the output of the VFD and the spindle input.)