# 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