As an autotransformer, how can a Buck-Boost transformer supply loads significantly higher than its nameplate rating as a low voltage lighting isolation transformer?
With an autotransformer, only a portion of the current acts as a load on the transformer. This portion is roughly proportional to the voltage change. If you increase the voltage from 100VAC to 120VAC, you are roughly adding 20% (20/100). As a result, only about 20% of the current acts as a load on the transformer. This would mean that a 500VA low voltage lighting transformer used in an autotransformer (Buck-Boost) application like this could provide a 2500VA (2.5kVA) load even though the nameplate is only rated for 500VA (.5kVA)
This is a function of changing the voltage by a small amount. For example, if the transformer is connected in such a way that 22 volts is added to a 208 volt primary, a 230-volt output will result. Only a portion of the current goes through a buck-boost autotransformer roughly equivalent to the voltage change. As a result, if a buck boost transformer changes the voltage by 10%, only 10% of the current (kVA) go through the unit. Therefore a transformer rated for 1 kVA when used as an isolation transformer could handle a 10 kVA load if it adjusted voltage by 10% because only 10% of the total load would go through the unit.
Using this example, the calculation for autotransformer kVA is as follows:
KVA = (Output Volts x Secondary Amps)/1000
KVA = (230V x 41.67 Amps)/1000 = 9.58 KVA