What exactly goes into a transformer's VA rating? At a fundamental level it seems that there has to be a limit to what the VA rating can represent.

For example if I draw 2 Amps-RMS from the secondary of a 30VA rated transformer at 15 Vrms this makes sense as the maximum Complex Power I can draw. Still, at a power factor of 1 my maximum current in the transformer would be sqrt(2) * 2 Amps.

But what if my power factor is abysmally low (ie in a DC rectifier) due to harmonic distortion and while I still draw 2A-RMS I have brief current spikes of 15 Amps? Surely the transformer's magnetics couldn't handle these spikes as I'd expect the core would saturate.

My question is what is the practical limit for VA ratings of transformers? Would brief current spikes that reduce to only 2A-RMS in this case be fine? Obviously the secondary/primary copper losses play a part in the rating but how do the magnetics factor in?

I ask this because I am designing a simple bench power supply using a 125VA transformer with a 24Vrms secondary and am trying to figure out how much power I can draw before the current spikes into the inductor are too much for the transformer. I also don't want to have to resort to active PFC as I don't want to risk messing up any HVAC PCB design.

Thank you


2 Answers 2


Surely the transformer's magnetics couldn't handle these spikes as I'd expect the core would saturate.

Core saturation has nothing to do with load VA rating. It has everything to do with the magnetization current flowing in the primary. This current is largely constant irrespective of secondary load current.

In short, the ampere turns on the secondary winding (caused by the load) are exactly equal (but opposite in sign) to the ampere turns on the primary due to that secondary load current. Neither of these currents are the magnetization current that can saturate the core.

Imagine a simplified core with a single turn primary: -

enter image description here

At the moment it's just a single turn inductor. With V applied, Im flows and inductance, frequency and voltage all determine how much current (Im) flows. OK so far?

Now imagine that single turn were replaced by 2 closely coupled parallel turns like this: -

enter image description here

You would find that Im/2 flows in each or, in other word,s the same overall current flows. A nice side effect of this is that each individual coil must have twice the inductance of the single coil and, if you happened to make a two turn inductor this way (by wiring them in series) it would have 4x the inductance. Just think about it for a while.

Next scenario: -

enter image description here

So, you drive one of those closely coupled coils and look at the voltage on the other coil. The driving voltage and the secondary voltage are in phase and of equal amplitude (1:1 turns ratio). Do you see why? If not, consider what would have happened in the 2nd scenario if (say) the voltages were out of phase - you'd get a fire and you wouldn't get the inductance rising with turns squared - you'd get zero inductance. This doesn't happen.

Final scenario: -

enter image description here

You've applied a load to that 2nd winding and because in the 3rd scenario you (hopefully) recognized that the voltages were in phase, you have to admit that the currents are COMPLETELY antiphase.

From here, it's a minor leap of faith to recognize that the ampere.turns on the primary (due to the secondary load) are equal and opposite to the ampere.turns on the secondary. As I said earlier, neither of these currents are the magnetization current that can saturate the core - this is due to Im.

It's magnetic field strength that drives the magnetism. It's called "H" and H is measured in ampere.turns per metre. The "per metre" part is irrelevant because it's a core physical dimension and applies equally to primary and secondary.

Basically H never alters one bit due to loading effect. In fact that's not quite true; it gets lower with more load because the copper losses lower the actual terminal voltage and reduce the magnetization current a little bit.

  • \$\begingroup\$ Thank you for the answer but it would do you well in the future to not condescend on people who don't have as firm a grasp on concepts that you do. \$\endgroup\$ Commented Dec 23, 2015 at 22:36
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    \$\begingroup\$ I'm not just talking about you. I see one of the top contributers to this site making the same error plus plenty more. Don't take it personally, it's a wider message than just to you! Happy xmas. \$\endgroup\$
    – Andy aka
    Commented Dec 23, 2015 at 22:42
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    \$\begingroup\$ @AndrewCragg, give Andy the benefit of the doubt here. I've read hundreds of his posts and he is most helpful. I think the comment is meant as a general one and, despite a few decades in electrical engineering, I would include myself in the list of those who 'should really start grasping ...'. I was fine on circuit theory but always felt that magnetics was taught as though we weren't going to understand it. I didn't! (Edit: You got your chat in before my comment.) \$\endgroup\$
    – Transistor
    Commented Dec 23, 2015 at 22:46
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    \$\begingroup\$ @Andyaka, have you got the load connected to the right wires on the last drawing? I think you meant to put it on the 'secondary' but you have it drawn as a series connection between the two coils. \$\endgroup\$
    – Transistor
    Commented Dec 24, 2015 at 0:04
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    \$\begingroup\$ I think that the real question was obscured by the assumption by the asker that the answer had to do with saturation. There is still the question of what does happen when a transformer current has a high harmonic content. The answer is: increased eddy-current losses and skin-effect contribution to copper losses. \$\endgroup\$
    – user80875
    Commented Dec 24, 2015 at 0:20

While the accepted answer provides useful information, I'm not sure it answered the question.

The only thing that goes into a transformer's VA rating is the withstand current of the secondary windings, or equivalently, the acceptable temperature rise.

Because the secondary must carry all the current of the load, regardless of its power factor, the limiting factor in delivering power to the load is the RMS current. The transformer designer decides on an acceptable temperature rise, calculates the current that will produce that rise, multiplies it by the specified secondary voltage (which is considered fixed), and viola, finds the VA rating.

So in your case, provided the RMS of the current drawn is less than the VA rating, you'll meet the VA rating and won't be in danger of cooking the transformer due to excess secondary current (okay, that's not entirely true - I'm ignoring the fact that high frequency currents can generate more heat due to skin effect). However, your hesitation is warranted, because you must also meet all the other requirements of the transformer. Indeed, reaching saturation is a very real concern since if you do manage to saturate the core, the inductance plummets and may lead to smoke.

Some transformers have a "K-rating", which is an indication of their ability to handle non-linear loads. It's usually a better way to accommodate non-linear loads than just de-rating based on VA, but generally only applies to larger transformers. In your case, you might be forced to do the saturation calcs yourself.

  • \$\begingroup\$ My answer answered the main issue that of saturation. The op said "Surely the transformer's magnetics couldn't handle these spikes as I'd expect the core would saturate" and I answered that. The op also said "Obviously the secondary/primary copper losses play a part in the rating but how do the magnetics factor in?" and again I point to my answer as dealing with saturation. You covered copper losses which I deemed unnecessary as the op appeared to understand those re the 2nd quote in this comment. You also appear to ignore the temperature rise due to primary copper loss (para 3). \$\endgroup\$
    – Andy aka
    Commented Dec 24, 2015 at 8:26

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