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.


3 Answers 3


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 magnetic permeability is 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 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 volume 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.

  • \$\begingroup\$ Nice explanation ! :D \$\endgroup\$
    – Jess
    Commented Oct 7, 2019 at 11:54

It is the other way around, the magnetic current is constant (Flux and B = Flux/Area) in the core. And the magnetic voltage (magnetic field strength H) will drop across the core length (\$I * N = H_c*l_c+H_g*l_g\$).

And we will have the larges "magnetic voltage drop" across the larges magnetic resistance (reluctance). The ferrite core will have lower magnetic reluctance compared to the air gap magnetic reluctance.

$$ R = \frac{l}{\mu_o\mu_r A}$$

So if the core length is \$l_e = 37mm\$, the Area of a core is \$A_e = 20mm^2\$

And the air gap is length \$l_g = 0.1mm\$

We have the air gap magnetic reluctance will be equal to about if we ignore the Flux fringing.

$$R_g \approx \frac{0.1mm}{4\pi 10^{-7} 20mm^2} \approx 398000 \:H^{-1}$$

The ferrite core magnetic reluctance will be equal to if the relative permeability of a ferrite core is \$\mu_r = 2000\$

$$R_c \approx \frac{37mm}{4\pi 10^{-7} *2000* 20mm^2} \approx 73609\:H^{-1}$$

Therefore the magnetic voltage drop across the air gap will be 398000/73609 = 5.4 times large in the air gap compared to the core magnetic voltage drop.

Or we can use the fact that the flux (the magnetic current) is the same in the caore is is inr the air gap we will have:

$$ B = \mu_o \mu_r H_c = \mu_o H_g $$


$$H_g = H_c \mu_r$$

This means that the magnetic field strength in the air gap will be larger than the \$Hc\$ in the core.

In our example, we have \$\mu_r = 2000\$ and the air gap length is \$370\$ times smaller than the core length. The magnetic field strength in the air gap will be \$ \frac{2000}{370} \approx 5.4 \$ large than the H in the core.


According to Faraday's law, the amount of flux swing is given by dividing the flux rate of change (Volts/turn) by the core cross-section area and multiplying by the switch ON time. The amount of air gap introduced has no effect on the amount of flux swing Bac. The air gap changes the slope.

Observe the difference in the area of the BH curve for each polarity due to the addition of a small air gap.

enter image description here

the B-H curve so that the gapped core can support a much larger value of H without saturation. A DC current component in the windings gives rise to a DC magnetizing force HDC on the H-axis of the B-H loop. This DC component, in turn, results in a mean flux density Bdc. Therefore, for the gapped core, a much larger DC current is required to produce the same mean flux density as the non-gapped core.

In continuous mode, the ripple current is small enough that AC loss (in the core) is not significant, but in discontinuous mode AC losses may dominate. Sufficient turns and core area are provided to support the applied pulse conditions, and sufficient air gap is provided in the core to prevent saturation and support the DC components.

The flyback transformer is a misnomer and ought to be considered as a switched inductor with coupling, as it does store energy unlike an ideal transformer. However the addition of a small air gap allows more current with greater H fields now occupied in the air gap Not all the energy is in the gap but optimally it can be 2x as much as in the core.

This may be a useful 0 formula for air gap
REF: https://www.edn.com/design/components-and-packaging/4333799/Designing-high-current-chokes-is-easy


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