# Relation between transformer core material and application frequency

Regarding the following excerpt from a text:

The text mentions that at RF air core transformers are adequate, but for 50/60Hz low freq. applications iron core needed to obtain a low reluctance.

The text does not explain the reason. What is the relation between the frequency being low and the need for low reluctance medium? Low reluctance intensifies the magnetic field but why particularly at low frequencies it is needed?

• I think you should edit the question to add the title and author and edition number of that text excerpt. Commented Jan 13, 2018 at 17:09

At high frequencies, hysterisis and eddy current losses come into play if a ferromagnetic material like iron is used a core. Hence Air core is preferred at high frequencies, despite its higher reluctance. But at low frequencies, hysterisis and eddy current losses are nullified and hence Iron core is a better choice. Way better flux-linkage and better efficiency than air core ones.

• Very good point. Commented Jan 13, 2018 at 17:22
• Do you also have an idea why using transformer cores as lamination stacks reduces eddy currents? Commented Jan 13, 2018 at 17:24
• You see here by using laminations on the right side the eddy currents loops in narrower loops. But they still exist: upload.wikimedia.org/wikipedia/commons/6/67/… But why this means they are reduced I dont know. Commented Jan 13, 2018 at 17:38
• I think I got why.. The currents are induced by dphi/dt and phi is proportional to area. Laminations reduces the area so much that the eddy currents induced in each lamination becomes so small. Commented Jan 13, 2018 at 17:43
• Good question. If you look at eddy current loss equation, it has a square dependence of thickness of the material. If a core is divided and stacked into thinner plates of thickness t, with insulation in between, the sum of their losses will always be less than the loss of the core as a whole. Mathematically speaking, the sum of squares is lesser. Commented Jan 13, 2018 at 17:47

It is difficult to explain in an intuitive way. But I like to think of it in terms of primary inductance. When you apply a voltage to the primary, even if the secondary is open-circuit, a certain amount of current flows in the primary, because of the primary inductance. In other words, when the secondary is open circuit, the primary just looks like an inductor. You want that current to be small, which means that the inductance needs to be large.

As it turns out, for 50 or 60 Hz, it is not really practical to do an air core, because the inductance will be too low for practical core sizes. But when a low reluctance core is used, the primary inductance is much higher, and the transformer size is more manageable.

The cross section area of the core and the number of turns are usually fine tuned so that the magnetic field strength in the core is within a range that is suitable for the transformer core material.

• Really interesting point. To formulate what you tell when the secondary is open: the current is I=|V|/|[j(2×pi×f×L)+R]|. Since f is too low, to obtain a smaller current L needs to be large. And L increases with the reluctance. Ant iron has lower reluctance than the air. Commented Jan 13, 2018 at 17:01
• Exactly. The other piece of the puzzle is to consider the saturation field strength in the core. Under full load, the field strength must be less than saturation. So that has to be considered, too. In theory, you could get a very large inductance by putting more and more windings on a small core, but in practice, that will saturate the core, so it won't really work. Commented Jan 13, 2018 at 17:07
• Current transformers are not subject to the same constraint, by the way, since the primary does not have to be inductive. Commented Jan 13, 2018 at 17:11

The volts per turn of the transformer are the product of core flux, and operating frequency.

If you have a very high frequency, you don't need much flux, so can get away with an air core. This is just as well, as a material core is expensive and lossy at RF.

If you have a very low frequency, you need the higher B field from the high permeability core. Fortunately, the losses of very high permeability materials are low at low frequency, so they're OK to use.

Inductor impedance is Z(f)=ωL where L is a product of the core permeability μ.

In order for power transformers to be efficient at low frequencies, the no load impedance must be > 10x rated load Z otherwise the excitation losses at no load would be excessive.

As frequency rises the trend to lower relative permeability towards air=1 is due to the non-linear eddy current losses.

Thus the result is a limited useful range $µ_r/ω$ for any give material.

You may look up the various relative mu values from 1 for air to 600 for ferrite to 4000 for transformer steel to 100k for special mu-metal foil used to raise the impedance and bandwidth of submarine cable surrounded by high k=80 water.