How to get the secondary current rating of a power transformer?

[abhishekkumar1902]

I have a 220v/20v power transformer which I salvaged from an old appliance. The specification sticker on the transformer is faded now and I wanted to know the secondary (20V) current rating of the transformer (in short the power rating of the transformer). Is there a way to calculate it?

[Zero999]

A good way to guess is to weigh it and compare it to some transformers in a catalogue, of a similar mass and construction i.e. toroidal, E-core etc.

[bktemp]

The power rating of a transformer is defined solely by the maximum temperature it can tolerate and its power losses.
So if you know the resistance of both windings, the voltage at both windings (to calculate the turns ratio) and the surface area, you can calculate the power rating fairly precisely.

If you measure all those values, I can do the calculations and give you an estimate of the output rating.
I need to know: Resistance of primary and secondary, input and output voltage and the outer dimensions of the solid parts (of the iron core and windings, but without terminals or similar parts) and type of the transformer (is it potted, has it a toroidal core etc.).

[ebclr]

The secondary wire gauge and the magnetic core size will give 2 nice gotchas

If you need more accuracy you need to make short circuit and no load characterization

https://en.wikipedia.org/wiki/Transformer

[jeroen79]

Once you made an estimate based on size, weight and wire thickness you can simply test it.
Just apply an increasing load and observe how the transformer maintains voltage and heats up.

[Zero999]

The power rating of a transformer is defined solely by the maximum temperature it can tolerate and its power losses.
So if you know the resistance of both windings, the voltage at both windings (to calculate the turns ratio) and the surface area, you can calculate the power rating fairly precisely.

If you measure all those values, I can do the calculations and give you an estimate of the output rating.
I need to know: Resistance of primary and secondary, input and output voltage and the outer dimensions of the solid parts (of the iron core and windings, but without terminals or similar parts) and type of the transformer (is it potted, has it a toroidal core etc.).

The iron core is also pretty important for determine the maximum power, at the minimum design frequency. If the transformer is designed for 50Hz, you should expect it to be a little heavier than one made for 60Hz, everything else being equal.

[Mechatrommer]

Just apply an increasing load and observe how the transformer maintains voltage and heats up.

agree on real life load testing. but usually big transformer you need to wait sometime until you can feel the heat. if the power rating is based on heat, then simply adding cooling fan will enable us to increase the power rating? i did this on several transformers unit i have working in my lab, who cares its factory spec?. maybe by load testing and observing the 2ndary voltage sag may indicate its power rating, too much sag, we should backup reduce the current to acceptable voltage sag and define that as its power rating. ymmv.

[bktemp]

Just apply an increasing load and observe how the transformer maintains voltage and heats up.

agree on real life load testing. but usually big transformer you need to wait sometime until you can feel the heat.

Yes, it can take hours until you have reached a steady state with an acceptable temperature.
But since the losses are I²R, even small changes in current will result in rather large temperature changes, therefore it is possible to determine the output current quite precisely.

if the power rating is based on heat, then simply adding cooling fan will enable us to increase the power rating? i did this on several transformers unit i have working in my lab, who cares its factory spec?.

Yes, that works well, but the efficiency will go down very quickly, but who cares for short time runs?

maybe by load testing and observing the 2ndary voltage sag may indicate its power rating, too much sag, we should backup reduce the current to acceptable voltage sag and define that as its power rating.

Since the voltage drop is linear (unlike the square law of the power dissipation) it will be much less accurate. And often you don't know the nominal output voltage if it is a custom design.

[Kleinstein]

Usually the maximum load current is thermally limited and the most important part is the ohmic loss in the wires. So measuring the resistance can be a first step. From the size one could get an approximation on the power dissipation possible.

The maximum temperature can also be different - some can tolerate higher temperatures.

One way to measure the temperature rise is by the resistance of the windings just after heating.

[ebclr]

In a transformer wires is in no way the unique limit, The magnetic circuit is very relevant, You can have any wire you want if the magnetic core is saturated no more power will come that's the magnetic limit, that is extremely important. One well-made transformer need to have a balance on electrical limits ( wires ) and magnetic limits ( core)

[Seekonk]

I would measure the secondary voltage and figure out 10% of that. With the known resistance of the secondary, figure out what current gave that voltage.

Some meters go crazy when measuring ohms. Build a 100ma current source and measure the voltage.

[bktemp]

In a transformer wires is in no way the unique limit, The magnetic circuit is very relevant, You can have any wire you want if the magnetic core is saturated no more power will come that's the magnetic limit, that is extremely important. One well-made transformer need to have a balance on electrical limits ( wires ) and magnetic limits ( core)

I have never heard of a transformer saturating due to high output current. Could you please explain how this works? I couldn't find anything about saturation due to overcurrent in a transformer.
The core normally satures at too high voltages or too low frequency: Bmax = V / (4.44 f N A)

[ebclr]

Why do you believe transformer is bigger when power increase

this PDF will help you understand what is going on

http://www.eilor.co.il/Media/Uploads/Eilor_Magnetic_Cores_Metric_Catalog(1).pdf

https://www.ieee.li/pdf/introduction_to_power_electronics/chapter_12.pdf

https://www.google.com.br/url?sa=t&rct=j&q=&esrc=s&source=web&cd=8&cad=rja&uact=8&ved=0ahUKEwjOnOf_nZ3UAhWMgpAKHY5iB1AQFghHMAc&url=http%3A%2F%2Fwww.ymcaust.ac.in%2Felectrical%2Fimages%2Ftransformer_design.pdf&usg=AFQjCNEhtlozfsrep67-s4ioGDZRE4e7bw

[bktemp]

I could not find anything in your links that proved your statement. Instead, is says exactly the opposite, see page 30 of chapter_12.pdf:

Saturation occurs when core flux density B(t) exceeds saturation
flux density Bsat.
• When core saturates, the magnetizing current becomes large, the
impedance of the magnetizing inductance becomes small, and the
windings are effectively shorted out.
• Large winding currents i1(t) and i2(t) do not necessarily lead to
saturation. If
then the magnetizing current is zero, and there is no net
magnetization of the core.
• Saturation is caused by excessive applied volt-seconds

So, unless your transformer has excessive leakage inductance, an increasing load current will not cause saturation.

[ebclr]

I guess you picked up on the word, not the concept.

My point is that the magnetic circuit is part of the problem, not only the wire,

The core will have a loss due to hysteresis, saturation. eddy current etc, It's impossible to specify a transformer spec without taking in count the magnetic core

"Iron losses, also known as hysteresis is the lagging of the magnetic molecules within the core, in response to the alternating magnetic flux. This lagging (or out-of-phase) condition is due to the fact that it requires power to reverse magnetic molecules; they do not reverse until the flux has attained sufficient force to reverse them.

Their reversal results in friction, and friction produces heat in the core which is a form of power loss. Hysteresis within the transformer can be reduced by making the core from special steel alloys.

The intensity of power loss in a transformer determines its efficiency. The efficiency of a transformer is reflected in power (wattage) loss between the primary (input) and secondary (output) windings. Then the resulting efficiency of a transformer is equal to the ratio of the power output of the secondary winding, PS to the power input of the primary winding, PP and is therefore high."

If you believe in anything else good luck but you are wrong

[Zero999]

I would measure the secondary voltage and figure out 10% of that. With the known resistance of the secondary, figure out what current gave that voltage.

Some meters go crazy when measuring ohms. Build a 100ma current source and measure the voltage.

Are you talking about the current drawn, at which the voltage drops below 10% of the no load voltage?

If so, then that will only work if the transformer has regulation factor of 10%. Large transformers have a regulation factor of under 10% and small transformers have a regulation factor of over 10%, so I don't think it's good advice.

Look at a data sheet for a range of transformers of the same type but different sizes:
https://www.rapidonline.com/pdf/82719.pdf

15VA regulation 16%
1000VA regulation 4%

[Seekonk]

So like you said, I'm right on the money.

[Mechatrommer]

I would measure the secondary voltage and figure out 10% of that. With the known resistance of the secondary, figure out what current gave that voltage.

Some meters go crazy when measuring ohms. Build a 100ma current source and measure the voltage.

Are you talking about the current drawn, at which the voltage drops below 10% of the no load voltage?

If so, then that will only work if the transformer has regulation factor of 10%. Large transformers have a regulation factor of under 10% and small transformers have a regulation factor of over 10%, so I don't think it's good advice.

Look at a data sheet for a range of transformers of the same type but different sizes:
https://www.rapidonline.com/pdf/82719.pdf

15VA regulation 16%
1000VA regulation 4%

not quite expert and not quite understand this, but i kinda agree with this. i made a small diy step up transformer, as soon as i put 1Kohm load on the secondary, the voltage drop to like 90% of the initial no load voltage, not a good idea as an indicator, it should be the combination of voltage and current through it, i guess thats why transformers are rated in VA.

[Zero999]

So like you said, I'm right on the money.

What do you mean? You said to measure the transformer's DC resistance and use the current required for the voltage to drop by 10% from the unloaded value, as the maximum rating or did I misunderstand?

If you tried that with the 1000VA transformer on the data sheet I linked to, you would be exceeding its maximum current rating by a factor of 2.5 and the power dissipated, due to the copper losses, would be 6.25 higher than the maximum design value!

To summarise: do not try to load a large transformer, until its secondary drops by 10% of the unloaded value, at least not for a long period of time, because you'll overload it and cause it to overheat.

The correct way is to estimate the transformer's power rating from its mass and the data sheets for similar transformers. Do not attempt to load the transformer, until you're already pretty sure it will be able to take it.

Another thing to note is that the transformer mentioned in the original post has a secondary voltage of 20V, which presumably when it's unloaded. The secondary voltage is always specified when fully loaded, so the voltage rating is likely to be less than 20V.