# Transformer Magnetization Flux

As far as I understand, an ideal transformer wouldn't show any net flux inside its core, as any such net flux would induce electric fields and in turn currents in the windings such that it will disappear. So an ideal transformer would show no (net) flux at all, correct? So it would kind of act like a control loop trying to keep the flux at 0.

A real transformer requires energy in order to magnetize its core. This energy is provided to the transformer in the form of the magnetizing current $$\ I_\mu \$$ and a corresponding magnetizing flux $$\ \phi_m \$$.

It was introduced to me as $$\phi_m = \phi_1 - \phi_2$$

with $$\\phi_1 \$$ being the flux put into the system by the primary winding, and $$\\phi_2 \$$ being the induced flux coming back from the secondary winding. So what's left after taking this difference is the magnetization flux.

We were shown this visualization in class:

It suggests this flux flows in the whole ferrite core. My internal model was that the energy provided by the additional current $$\ I_\mu \$$ would be "absorbed" by the core in order to magnetize itself and the flux wouldn't make it to the secondary winding. If it does flow throughout the whole core, and also through the secondary winding, wouldn't it play into the "control loop" described above and vanish as soon as it arises?

What am I missing? Thank you for your time in advance!

• First of all get rid of this term "net" flux and make it More clearer i.e flux at no load and flux at loaded condition , half of your confusion go away from this only Dec 31, 2020 at 15:06
• There's nothing wrong with the term net flux @user215805 Dec 31, 2020 at 16:01
• @Andy aka yes you are right but if I just edit this question and replace every "net flux" by flux at no load and flux at loaded condition suitably and it turn out to be very easy question.Ambitious choice of words makes this question a more complicated at least for beginners Dec 31, 2020 at 16:44
• My advice: don't edit it. Dec 31, 2020 at 16:54

## 2 Answers

My internal model was that the energy provided by the additional current Iμ would be "absorbed" by the core in order to magnetize itself and the flux wouldn't make it to the secondary winding.

Magnetization flux couples both primary and secondary. It induces voltage in the secondary as per the turns ratio. If the secondary is open circuit then that should be easy to see.

The problem most engineers have is realizing that when the secondary is attached to a load, the current in the secondary would "appear" to produce another flux that would "appear" to alter the core flux and screw around with the magnetization.

It doesn't because as soon as the secondary current forms, an extra primary current forms in opposition to the secondary current and the two individual fluxes caused by the secondary current cancel.

What is left is (still) the same magnetization flux and we still get a voltage transformation as per the turns ratio.

So an ideal transformer would show no (net) flux at all, correct?

This is called an ideal power converter or impedance transformer and, as much as EEs like to break down things into smaller manageable lumps, I don't think "an ideal transformer" brings anything to the party when trying to understand non-ideal transformers.

A real transformer requires energy in order to magnetize its core.

I think this misses the point a little. A real transformer has a secondary winding and, if this secondary winding isn't connected to a load, it might just as well be not there at all. So, the "real-transformer" is really just an inductor when it comes to what current it draws from an AC supply in order to produce core magnetism. Nothing more complicated than that.

I mean... do we say that an inductor requires energy to magnetize its core? No we don't; we say that current flows due to the applied voltage and the inductive reactance and that current (along with the number of turns) produces a H field that magnetizes the core. We don't need to think in transformer terms when thinking about core flux. And we don't need to think about energy when defining core flux; current and turns is sufficient.

• I will continue to upvote everytime you have to repeat this. Don't give up Andy. The core flux is probably the most misunderstood about transformers. Dec 31, 2020 at 15:56
• @Marla tis sad but true but not sad to receive an upvote of course. Merry xmas/new year despite all the bad news. Dec 31, 2020 at 16:00
• I really was trying to argue with the "ideal transformer" too much. I was thinking about the magnetizing flux as just a byproduct of the ideal world and was starting to think the coupling mechanism lies elsewhere. Thank you very much! Jan 1, 2021 at 11:39

This energy is provided to the transformer in the form of the magnetizing current $$\ I_\mu \$$ and a corresponding magnetizing flux $$\ \phi_m \$$.

It was introduced to me as $$\phi_m = \phi_1 - \phi_2$$

with $$\\phi_1 \$$ being the flux put into the system by the primary winding, and $$\\phi_2 \$$ being the induced flux coming back from the secondary winding. So what's left after taking this difference is the magnetization flux.

Not exactly correct. The flux due to the secondary current is cancelled by the primary current, what remains is the magnetizing flux.

Not exactly. The flux due to the secondary current is cancelled by the primary extra current (load), what remains is the magnetizing flux.

The magnetizing flux remains unchanged (with load or no load condition) and it is 90 degrees out of phase with respect to the load current.

The magnetizing flux remains unchanged (with load or no load condition) and it is 90 degrees out of phase with respect to the voltage.

• Haha not exactly true either: The flux due to the secondary current is cancelled by a flux produced by an extra current in the primary (to be exact)!! Dec 31, 2020 at 15:52
• And doesn't the phase relationship depend on the load impedance (not always 90)? Dec 31, 2020 at 16:07
• @relayman357 The magnetizing flux is always 90 deg. off with respect to the voltage, and it is the only flux in the transformer. Dec 31, 2020 at 16:11
• 90 degrees with respect to the primary voltage but, not necessarily so with the load current if the load is a reactive component. Dec 31, 2020 at 16:16
• I agree friends, @MarkoBuršič answer incorrectly says “90 degrees...with respect to the load current”. Happy new year! Dec 31, 2020 at 17:30