Here is the following assertion :
Question: Why is it commonly stated that in a flyback transformer, the "air gap carries most of the stored magnetic energy"?
Answer: We can intuitively accept the fact that the energy stored is proportional to the volume of the magnetic material. And because of that, we also tend to think the ferrite must be carrying most of the energy, since it occupies the maximum volume — the amount of air enclosed between the ends of the ferrite being very small. However, the stored energy is also proportional to \$B\times H\$, and since the \$H\$-field in the gap is so much larger, it ends up storing typically two-thirds of the total energy, despite its much smaller volume.
I actually do not understand the assertion. The energy into a transformer is proportionnal to \$B \times H\$.
The magnetizing force \$H\$ is given in a transformer by the Ampere's law, \$\oint H\,\mathrm dl = I \text{(Amperes)}\$.
So \$H\$ is the same into the core or into the air gap.
However the flux density \$B\$ depends on the core. \$B= \mu H\$ with \$\mu\$ the permeability. The permeability of the air is lower than the permeability of a core. So, obviously the flux density is higher in the core than in the air gap.
As the magnetizing force is constant in the air or in the core, it means that the energy is higher in the core than in the air gap. Where is my error?
Actually, I'm not certain of what I said. What I know is that the reluctance across the air gap is higher than the reluctance across the core. So if the flux density is constant into the air gap or into the core, we have indeed more energy contained into the air gap because \$H\$ is higher. But why my first assertion is false ? and Why the flux density is constant ? Actually if \$H\$ is higher into the air gap it means that the flux density is constant.
Best Answer
Yes it can be confusing.
For a given un-gapped core, there will be a flux density (B) associated with the applied H field. The ratio of B to H is "permeability" and, if an air-gap is introduced, B becomes much smaller for the same H field because, the effective permeability is also reduced.
With small air-gaps, it's pretty reasonable to assume that the flux density in the gap is the same as the flux density in the core. However, as the gap gets bigger, magnetic fringing reduces the maximum flux density in the gap because field lines become more spread out.
So, if we were to raise the current of the moderately gapped inductor to bring about the same flux density as the un-gapped inductor, the H field would need to be a lot bigger. Basically however much the permeability is reduced due to the gap, the H field has to increase by the same amount to keep B the same.
Noting that the stored magnetic energy per unit volume is \$\dfrac{1}{2}\dfrac{B^2}{\mu}\$, it should be reasonably easy to see that due to the much smaller permeability of air compared to (say) ferrite, the energy per volume is much greater in air for the same flux density.
So then it becomes a case of working out the volume energy of the gap and relating it to the volume energy of the core and comparing the two.
Taking the example of a toroidal core with mean length 0.1 m and cross sectional area of 2 sq cm, it has a volume of about 2 milli cubic metres. The small air-gap might be (say) 1mm long and have an effective volume of 0.02 milli cubic metres.
That's a volme ratio of 100:1 (not surprisingly) but the core might have a relative permeability that is 1000 times that of air hence, 10 times more energy is stored in the air gap.