480vac transformer used with 120vac? for a chipamp?!

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I have a couple 480VAC input transformers which are rated at 24VAC output. They are each 50VA.

If I were running them @ 120VAC input would each transformer then support 200VA? or does the 50VA hold true, but with the lesser voltage they can support much more current?

It seems to me that if they are running with 1/4 the voltage they're designed to that they should support more power then they are rated at.



[http://www.audiosector.com/lm4780 psu.pdf]
Finally, I'd like to know if I can run one transformer into the AC1 inputs and a second into the AC2 inputs.

will the amp then see:
24/4= 6vac -> 6*1.414= 8.484vdc -> 8.484*2=16.968vdc

TIA
 
zenon said:
If I were running them @ 120VAC input would each transformer then support 200VA?

No

zenon said:
or does the 50VA hold true, but with the lesser voltage they can support much more current?

No, they were rated to give 2 amps, that's all you will get.

As for building a chipamp with them, DC voltage out of one will be roughly 6 * 1.414 - 1.4 = approx 7 volts. Even with 2, I wouldn't bother except maybe a tweeter amp.
 
zenon said:

...
It seems to me that if they are running with 1/4 the voltage they're designed to that they should support more power then they are rated at.
...
TIA

This violates most electronics laws.

Transformers have a volts/second per turn rating (shared by all windings) that should not be exceeded in order to prevent core saturation, and a rms current rating for each winding that should not be exceeded to prevent overheating. The VA rating is something that was invented to allow people not understanding transformers to talk about them.

Since the maximum allowed rms current through each winding is limited, maximum power output is reduced proportionally to input voltage. Thus the higher the input voltage, the more optimum the usage of the transformer is.
 
Hi,
each secondary winding must not exceed it's design current limit.

If you run a 480Vac transformer at 25% of primary voltage you get 25% of output voltage. Multiply V times A and you end up with 25% of VA rating.

The proportion of undervoltage equals the proportion of effective VA rating. i.e. half voltage=half VA and quarter voltage =quarter VA.
 
Another way of looking at it, which supports the advice above; the secondary is wound with a certain gauge wire, with an associated resistance. The copper loss in the secondary will be I (the current) squared times that resistance, generating a certain amount of heat. Irrespective of the voltage, this heat is purely generated by the current. So doubling the output current draw would increase the loss by four times, and things might start smelling funny. And of course doubling the output current would double the input current, with the same problem on the primary winding.
 
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