synchro
Senior Member
- Location
- Chicago, IL
- Occupation
- EE
Larry's suggestion in Post #21, along with using H2-H3 as the "neutral" would create the 3 new voltage points A', B', and N' shown in the diagram below. I just eyeballed the length of the new black line (which represents the autotransformer supplied by A and B).
I didn't write in all the voltages (given by the distance between two points in the diagram), but (A,B,C,N) are as usual for 208Y/120V. The new points have voltages to ground (to N) of A' = 134V, N' = 60V, and B' = 134V. But A'-N' and B'-N' are 120V as desired.
Cheers, Wayne
View attachment 2570412
Bingo! And as was said above, the system is grounded. Nothing says the load must use that conductor.
Yes, in that configuration N' would efffectively just be a tap on the primary of the buck-boost transformer.
I don't think it would be a good idea to put 120V loads across A'-N' or across B'-N', even aside from the restrictions of 210.9 and 215.11 on autotransformers.
If you put a load current IL across A'-N', then a current IL would flow through the 16V secondary winding A-A', 0.42 x IL through winding N'-A, and 0.58 x IL through winding N'-B, as a consequence of balancing the ampere-turns of those three windings. But the primary windings on A-N' and B-N' would only need to be capable of 16/104 = 0.154 times the rated current through each 16V secondary winding (which can fully support the rated L-L load crrents). Therefore, with a single 120V load from A' or B' to N', the allowable 120V load current would only be 0.154/0.58 = 0.265 times the rated current for L-L loads (and obviously would be further reduced by any 240V L-L loads).