ANOTHER transformer question!!

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I have a pair of transformers marked

440 volt in

66 volt out

265 VA

Using UK mains (mine averages 239VAC), I assume that the voltage out will be in the region of 35-36VAC

My question is will the VA value remain constant at around 265VA, or will it be a different value? IF so, how is the new VA value calculated??

Thank you for your help.
 
I understand the core calculations mentioned. This would apply to the core of the xfmr itself regarding heating, losses, etc... But when you cut the incoming voltage in half and maintain the VA, the secondary current will double. I would be concerned that the secondary wiring is not large enough for the current as it would have been sized for 50% of what the OP is trying to get out it.
 
yes, that is true, but can be checked considering the dc resistance of the secondary winding and the intended load current draw, if your intended load does not cause too much drop in secondary voltage then it's okay....

the plus side is that the primary magnetizing current will be very low so the traffo will run cooler...
 
If the primary volts are 50% of nominal and the OP operates it at full nameplate VA, the secondary current will be double. There is a real limitation of secondary current due to secondary copper diameter. At some point the copper temp will destroy/decay the secondary winding. Any manufacture sizes the copper for intended currents in order to reduce cost.

In electrical distribution applications I would derate the VA due to this. The OP will probably have no problems due to audio use, if it was in a full load condition continuously it would probably fail early.
 
I understand the core calculations mentioned. This would apply to the core of the xfmr itself regarding heating, losses, etc... But when you cut the incoming voltage in half and maintain the VA, the secondary current will double. I would be concerned that the secondary wiring is not large enough for the current as it would have been sized for 50% of what the OP is trying to get out it.

This seems a bit unclear to me. I am trying to get out less than 40VAC to drive a small class A amp. I intend to use dual mono power supplies - [using two of these transformers] - for which as a single PS, 300 VA is recommended.
 
you can parallel connect primaries and secondaries taking care that you observe the proper polarity of secondary windings so you can use two transformers for your amp.

making that a dual power supply is even better.....remember that at idle, your classA amp will be pulling 4 times the output power, so that if your amp is a 25 watter, then it will constantly draw 100watts...

if you can post pictures of your traffo or give dimensions we can tell more....i think you have more than enough.....
 
VA rating is mainly a matter of heat. Two main heat sources: eddy current losses in the core, copper resistance losses in the windings. I believe that most transformer designs aim to get these roughly equal at rated load. Dropping the input voltage will reduce core losses - roughly as the square of the reduction? Raising the output current will increase copper losses, as the square of the current. There is only a limited extent to which you can balance smaller core losses against bigger copper losses.

Consider halving the input voltage. Core losses will be roughly a quarter of the old value. First assume we can take advantage of all of this. So core heat has gone from 50% of VA rating to 12.5%. Copper heat can thus go up from 50% to 87.5%. This means current drawn can increase by a factor of sqrt(87.5/50)=1.323. At half voltage this means that VA is now 0.661 of what it was. That is the best case assumption.

Now assume worst case: that we can't take advantage of reduced core heat at all. Current draw thus has to stay the same, so VA rating is halved.

So we find that halving input voltage means a reduction in VA rating to somewhere between 0.5 and 0.661 of the old rating. I think I would use 60% as a design figure but not approach it too closely.
 
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265VA and 66Vout = 4A out
If you reduce the input voltage from 440V to 240V the output voltage will be reduced to 36V.
Off course the iron losses will be considerable smaller at the to 240V reduced input.
If we presume that the original full load losses are (B= 1.4T) 4W iron and at 4A full load 16W copper = 20W total losses and we now reduce the input voltage to 240V we now have reduced B to about 1T and the iron losses to 2.2W wich means we can increase the copper losses by 1.8W from 16W to 17.8W. The new output load current can therefore be increased from 4A to 4.22A and the new VA rating will be roughly 152VA. The former calculation is based on M6 iron and the assumption that we do not change the operating temperature of the transformer (advisable)
 
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Joined 2006
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VA rating is mainly a matter of heat. Two main heat sources: eddy current losses in the core, copper resistance losses in the windings. I believe that most transformer designs aim to get these roughly equal at rated load. Dropping the input voltage will reduce core losses - roughly as the square of the reduction? Raising the output current will increase copper losses, as the square of the current. There is only a limited extent to which you can balance smaller core losses against bigger copper losses.

Consider halving the input voltage. Core losses will be roughly a quarter of the old value. First assume we can take advantage of all of this. So core heat has gone from 50% of VA rating to 12.5%. Copper heat can thus go up from 50% to 87.5%. This means current drawn can increase by a factor of sqrt(87.5/50)=1.323. At half voltage this means that VA is now 0.661 of what it was. That is the best case assumption.

Now assume worst case: that we can't take advantage of reduced core heat at all. Current draw thus has to stay the same, so VA rating is halved.

So we find that halving input voltage means a reduction in VA rating to somewhere between 0.5 and 0.661 of the old rating. I think I would use 60% as a design figure but not approach it too closely.

I agree with most of what you say but your assumption that most transformer designers aim for losses of 50% iron and 50% copper is, as you said, only a assumption. That is quite far from the real world of smallish transformers. 15-30% iron and 70-85% copper comes more close to now days small transformers with high current densities and quality iron.
 
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