Look at the red selection again.The buck/boost will only carry the va associated with the approximately 32 volts added or dropped. The isolation will carry the va of the entire load.
240 volts - 208 volts = 32 volts
32 volts / 240 volts = 13.333 percent of the load
208 to 208 will carry the whole load, thus, must be rated 100%.
Yeah, what he said! :grin:Let me see if I can help.
A buck-boost transformer works in boost by adding its secondary voltage to the primary by being wird in series with it in an additive way.
But, the added voltage doesn't supply free energy. It comes from the primary being energized because it's paralleled with the incoming voltage.
So, whatever load demand is made by the load is manifest by adding 13.3% more current at the lower voltage, which is the line side of the BB.
The current at the lower (incoming) voltage is 13.3% higher than it is at the higher (outgoing) voltage. That power has to come from somewhere.
So, the BB only has to carry the power its secondary adds to the load. This is unlike a standard transformer, which must be sized to carry the entire load.
Let's start with an isolation scenario...I still believe, but I need to wrap the concept.
Technical side note: If I'm not mistaken, the secondary is actually lower, because the 32v secondary is based on the primary receiving 240v, and not the actual 208v.Your volatges are H1-H2 = 208V, X1-X2 = 32V, X1-H2 = 208+32 = 240V.