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I am trying to understand power transformers practically.

I have an inverter that outputs 12V AC. I want to step this up to 120V AC.

My load will just be a resistive load that takes no more than 200W (like 2 light bulbs).

Does this mean my transformer has to be rated at least for 200VA?

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    \$\begingroup\$ Is your inverter able to provide 12VAC at 17A RMS at either 50 or 60 Hz? \$\endgroup\$
    – Andy aka
    Jun 3 '14 at 7:31
  • \$\begingroup\$ yes, the inverter mosfets can handle up to 100A and after filtering it is a 60Hz sine wave \$\endgroup\$ Jun 3 '14 at 22:13
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The power rating of a power transformer depends to some amount on how much temperature rise in that transformer you are willing to accept. If the transformer is designed and built to meet a specific UL or ISO (or other standards organisation) standard then its maximum temperature rise is constrained by the limits of the standard.

If the load is intermittent then you may get away with a lower-rated transformer because heat generation within the transformer due to resistive heating of the windings will be less than with a continuous load.

Just be sure your transformer supplier has rated the device on the assumption of continuous load. If in doubt, measure the transformer's temperature over a period of several hours with your load of choice (and your hand on the "Off" switch).

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No, it could be rated for 180VA.

It might have 36 turns on the primary and 324 turns on the secondary, for example, with a primary voltage rating of 12VAC and 180VA rating.

Edit2:

As this is not homework, I've provided a schematic below showing what I had in mind, but for a real application other things would come into play (such as the waveform at the inverter output).

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit the transformer boosts the 12VAC to 108VAC, and that is connected in series with the 12VAC input to give 120VAC, so the transformer only needs to be rated for 90% of the output VA, or 180VA in this case.

Note that it does not give isolation.. this is an autotransformer buck/boost configuration.

The input still has to provide 200W, of course, so 16.8A plus transformer losses.

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    \$\begingroup\$ Could you substantiate your answer a bit? At the moment, it looks like a comment rather than an answer. \$\endgroup\$ Jun 3 '14 at 3:54
  • \$\begingroup\$ Is that appropriate? This looks like homework and I did answer the question numerically. \$\endgroup\$ Jun 3 '14 at 4:00
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    \$\begingroup\$ If it is homework, rather than giving the numerical solution it's better to give pointers, else it's not useful to anyone. I can't (without thinking too much) understand what you mean in your answer. \$\endgroup\$
    – clabacchio
    Jun 3 '14 at 11:31
  • \$\begingroup\$ @DanielGrillo how does this not answer the question? No clarification or critique is necessary, I understand the question and exactly what I think their instructor is going for. \$\endgroup\$ Jun 3 '14 at 11:44
  • \$\begingroup\$ I've flagged your answer when it was only one sentence. I think Nick and clabacchio have given enough clarification better that I would. \$\endgroup\$ Jun 3 '14 at 12:42

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