This might work if the phase converter used the neutral. Do most do this?
Most phase converters don't use the neutral, but IMHO that is irrelevant to the point I was making.
Ideally a phase converter takes single phase input and synthesizes the third leg. If you plot all the voltages on a phase diagram, then you've been given 2 initial points (the input single phase) and the phase converter synthesizes a third point that lies on the vertex of an equilateral triangle with the other two points. This is exactly equal to finding the midpoint between the two initial points, making a 90 degree angle, and going 0.866x the distance between the two initial points.
The phase converter doesn't know anything about the source 3 phase system, and doesn't care. It just takes 2 input legs and synthesizes the third, and the set of three (ideally) lie on an equilateral triangle drawn on a phase diagram.
If the input legs are referenced to ground, that defines the ground reference of the derived 3 phase system.
If you start with ordinary split 120/240V single phase, the derived leg will 'look' like the high leg of a 3 phase delta. The two initial legs are at 120V to ground and opposite phase, so the derived third leg ends up 208V to ground and 90 degrees out of phase.
If you start with 120/208V single phase split off a 208/120V three phase system, the derived leg will _by coincidence_ 'look' like the original 3rd leg of the three phase system. Not because the phase converter is connected to neutral, but because the initial two legs are 120V to ground/supply neutral and 120 degrees apart in phase; take those two points and derive an equilateral triangle, and the third leg ends up 120V to original supply neutral and 120 degrees out of phase from the other two. (I just realized that there is a _huge_ exception to this!)
If the synthesized rotation created by the phase converter matches the phase rotation of the original system, then my above analysis holds. But the phase converter doesn't know the phase rotation of the input system; it can't. So it could just as easily derive the third leg with the _opposite_ rotation. Draw this on the phase diagram and the derived third leg would _not_ coincide with the original supply third leg. If I did my math right, then in this case the 3rd leg to supply neutral voltage is about 300V.
-Jonathan