Transformer Power Rating

What sort of load is it: inductive, resistive, capacitive? You are thinking in the right direction, but just need some detail to tell what the likely power factor actually is.

The output of the transformer goes into a rectifier plus couple of capacitors and used as DC power supply for other systems. So I have no idea how to work out power factor from that. Guessing it's resistive/capacitive?

For Safety reasons, I am thinking of choosing transformer power rating of 220 to 250VA. You reckon that would be over estimated bit too much?

Sounds reasonable. If the transformer isn't fully loaded, it will be slightly less efficient, and cost a bit more initially; but better safe than sorry.

As a rule of thumb, if you take the average dc amps out of the rectifier and multiply that by the secondary volts rms will yield the required VA. Usually that is very close to the power you calculated already, but adding 10 to 20% never hurts. Are there intermittent loads beyond the nominal you have measured that you may need to include?

If you are trying to replace the transformer, the replacement should be about the same physical size/ weight or larger. If it is smaller you need to do some head scratching as to why.

"Guessing it's resistive/capacitive?"

it's inductive (current charging the capacitors lags voltage).

brgds

Snel

Yes I believe that is correct, however your efficiency might be a bit low....You can determine the efficiency by dividing the input va by the output watts....

Do you want the power rating or throughput? Because they are a bit different.

What is REALLY needed is the "VA" rating of the transformer.

Remember, the VA remains the same regardless of the load.....

Your power supply may not be designed to use the maximum power that the transformer could safely deliver...for example....

With the VA, you know just what (pun intended) the transformer can/was designed to deliver....

A reasonably simple explanation can be found here:-

Why_are_transformers_rated_in_VA_instead_of_watts?

For example:-

If you take a drinking glass to the tap in the kitchen and just turn the tap partially on to fill the glass, that is not using the full potential of the tap/water supply.

Turning the tap on fully would be using it fully, (assuming that even fully on the tap does not slow the amount of water passing through it!) the amount of water delivered would now be constricted only by the water pressure and the cross section of the pipe.....(Voltage and thickness of supply cable/wire in the transformer example! )

I hope this helps you to understand better.....

I'm off to get bigger glass. Half an hour should see me dried out again.

I understood the problem before you introduced the water. Still understand it electrically but away you go with your glass of water .......

...but does the OP (TDoan)??

IF you multiply the rated current by the rated voltage for the same transformer side (high voltage side or low voltage side) you will get the '' VA '' which is the apparent power

Now multiply that with whatever power factor you have for the load and you will get the real power