# Currents around transformer

If I use a transformer to step the voltage down from 220V to around ~30V and I expect the 30V end to draw up to ~30A, what kind of current magnitude will I have to plan for or expect to see on the 220V side?

I am very noob. Please feel free to help make this question better/clearer or give suggestions.

• You are a beginner and plan to use a transformer of, give or take 1kW? I dare say you should not be doing whatever you plan on doing. Jun 8 '16 at 21:31
• Since it's 900VA of AC, you may also need to consider power factor correction. Be sure to read up on uniform electrical code esp. saftey, before attempting to deal with high power circuits. Jun 8 '16 at 22:05

For a ideal transformer, current scales inversely proportional to voltage. Put another way, current x voltage is constant. That's because current x voltage is power, and for a ideal transformer, power in and power out are the same.

Of course no transformer is ideal. There will always be some loss, so power in will be a bit higher than power out, with the remainder ending up heating the transformer.

The easiest approach is to start with assuming ideal transformer, then derate according to the efficiency you think you will actually attain. 30 A at 30 V out is 900 VA. 900 VA / 220 V = 4.1 A. At 80% efficiency, for example, that would be (4.1 A)/80% = 5.1 A. 5 A would be cutting it close, but possible with a good transformer. Unless it's a hardship, I'd plan for 6 A input capability at least.

Another factor is inrush current. This can be large when the core is not magnetized, or even magnetized in the opposite direction when power is turned on at the peak of the waveform. This is a real issue, especially in toroidal transformers, which are common for these kinds of applications. The inrush only lasts up to 1/2 cycle, but can be several times larger than the normal operating current. I once blew a 15 A breaker with a toroidal transformer that didn't even have a load connected to it.

Power in has to equal power out. V1*I1 = V2*I2

Solving the equation (30*30)/220 = 4 amps.

This is theoretical. In reality there will be losses. Provision for 5 to 6 amps to have enough overhead at the output.