Due to the ferromagnetic materials from which they’re made, many magnetic cores are electrically conductive. When exposed to an AC magnetic field, these cores produce small current loops known as eddy currents. In transformers, eddy currents can be a significant source of power loss.
Particularly at low frequencies, a common method of reducing eddy current loss is to use cores made of thin sheets of magnetic material. The sheets, known as laminations, are usually a fraction of a millimeter thick and are insulated from each other by a thin layer of oxide. In this article, we’ll learn how different parameters affect the eddy current loss in laminated cores. In the interest of keeping things simple, we’ll limit our discussion to low-frequency operation.
What Is Eddy Current Loss?
According to Faraday’s law, a changing magnetic field induces an electromotive force—and, therefore, a current—in a conductive wire. Similarly, a changing magnetic field can also produce circulating currents in a bulk piece of conductive material that’s placed within it (Figure 1).
Figure 1. A changing magnetic field produces eddy currents in an electrically conductive core.
As eddy currents flow through the material, they inevitably encounter some electrical resistance along their paths. This causes I2R energy loss, where I is the current and R is the resistance of the path.
To calculate the eddy current loss in a structure, we need to know the current and resistance of each path in the material. Once we know the losses for all of the different paths, we add them up to get the total eddy current loss. We’ll shortly use this method to derive the formula for the eddy current loss in a laminated core.
Before we move on, it’s worth mentioning that eddy currents can be beneficial in certain applications. For example, electromagnetic braking systems use eddy currents to slow down or stop moving objects such as trains and roller coasters. Another application is induction heating, where eddy currents are used to intentionally heat up metal objects. Eddy currents are undesirable in inductors and transformers, however, because they dissipate energy as heat and degrade the efficiency and performance of the device.
What Are Laminated Cores?
For low-frequency, high-power applications, laminated cores are an economical way of reducing eddy current loss. Figure 2 shows how eddy currents flow in a laminated core.
Figure 2. Eddy currents in a laminated core.
Note that the laminations are in parallel with the magnetic flux, while the eddy currents are perpendicular. This arrangement restricts the eddy currents to the width of the laminations. As Figure 3 illustrates, a laminated core with the sheets perpendicular to the flux does little to reduce eddy current loss.
Figure 3. Laminations normal to the flux are not very effective in reducing eddy current loss.
To better understand eddy current loss in laminated cores, let’s analyze it mathematically. For those who are less interested in the details of the math, we’ll also go through a simpler explanation of mathematical cores after we examine the key takeaways from our initial analysis.
Understanding Eddy Current Loss in Low-Frequency, Laminated Cores
Consider a lamination with thickness t, height h, and depth l, as shown in Figure 4. The final core is formed by stacking many of these laminations.
Figure 4. A cross-sectional view of a single lamination.
The externally applied magnetic field is uniform and perpendicular to the front face of the lamination. For a sufficiently low frequency of operation, we can assume that the skin depth is much larger than the thickness of the lamination (δ ≫ t). The flux density distribution will therefore be uniform across the lamination’s cross-section.
The gray area in the figure shows a possible path for eddy currents. Let’s examine the current flowing through this path, which is at a distance of x from the centerline of the front face and has a width of dx. The area of this path is:
$$A ~=~ 2x ~times~ h$$
Equation 1.
Assume that the magnetic flux density in the material is sinusoidal:
$$B ~=~ B_m sin (omega t)$$
Equation 2.
Where:
Bm is the peak value of the flux density
ω is the angular frequency
t is the time (not the thickness!).
Therefore, the magnetic flux through the gray path is:
$$Phi ~=~A ~times~ B ~=~ 2xhB_m sin (omega t)$$
Equation 3.
When the magnetic flux in the enclosed area varies with time, it induces an electromotive force (EMF) that can be calculated using Faraday’s law:
$$v~=~Nfrac{d Phi}{dt}~=~ 2xhB_m omega ; cos(omega t)$$
Equation 4.
In our case, the number of turns in Faraday’s formula is N = 1. Therefore, the peak value of the induced voltage is:
$$V_m~=~ 2xhB_m omega$$
Equation 5.
As expected, the induced voltage is proportional to x. This is because a path further away from the centerline corresponds to a larger enclosed area and consequently experiences a larger change in the magnetic flux.
We’re now ready to calculate the resistance of the current path. Because the height of the lamination is much greater than its thickness, the total resistance is approximately equal to the sum of the resistances of the AB and CD paths in the figure. The familiar equation for calculating the resistance of a conductor is:
$$R ~=~ rho frac{l}{A}$$
Equation 6.
where:
ρ is the resistivity of the conductor in Ω·m
l is the length of the conductor in meters
A is its cross-sectional area in m2.
If we refer back to Figure 4, the AB and CD paths have the same parameters. Because of this, the resistance of each is equal to:
$$R_{AB}~=~R_{CD}~=~ rho frac{h}{l ; dx}$$
Equation 7.
Therefore, the eddy current power dissipated in the gray current path is:
$$begin{eqnarray}dP_e~ &=& ~frac{{V_m}^2}{2 R_{total}} ~=~ frac{(2xhB_m omega)^2l}{4 rho h} dx \&=& ~frac{(B_m omega)^2lh}{ rho }x^2 dx end{eqnarray}$$
Equation 8.
The total eddy current power dissipated in the lamination is found by integrating the above equation over x = 0 to x = t/2, where t is the lamination thickness:
$$P_e ~=~ int_{0}^{t/2} dP_e ~=~ frac{(B_m omega)^2lh}{ rho } int_{0}^{t/2} x^2 dx ~=~ frac{(B_m)^2 omega^2lh t^3}{ 24rho }$$
Equation 9.
Finally, noting that (l × h × t) is actually the volume of the lamination, we can determine the density of the power loss:
$$p_e ~=~ frac{(B_m)^2 omega^2 t^2}{ 24rho }$$
Equation 10.
Important Observations
Equation 10 shows how different parameters affect the eddy current loss in a laminated core:
- Frequency (ω): Eddy current loss is proportional to the square of frequency. This is in contrast to the hysteresis loss, which increases linearly with frequency.
- Lamination thickness (t): The eddy current loss is proportional to the square of the lamination thickness. To minimize the core losses, the laminations should be as thin as possible.
- Resistivity (⍴): To reduce eddy current loss, materials with high resistivity, such as silicon steel, are preferred.
At high frequencies, we reduce eddy current loss by using ferrite cores rather than laminated iron cores. Ferrites are ceramic-type materials made from iron oxides and other metal oxides. They have a very low electric conductivity—about 1 S/m, which is a millionth that of iron. For that reason, eddy current losses in ferrites are usually easily manageable.
An Intuitive Explanation of Laminated Cores
Eddy currents produce ohmic power loss. The loss depends on the amount of EMF induced in the core and the resistance of the core. For a resistance (R) connected to a sinusoidal voltage with a peak voltage of Vm, the dissipated power is:
$$P ~=~ frac{{V_m}^2}{2R}$$
Equation 11.
When we replace a solid core with a laminated one having n laminations, the cross-sectional area of each lamination is (frac{1}{n} ) that of the solid core. This means that the induced EMF in each lamination (VL) is (frac{1}{n} ) of what it is in the solid core (VS):
$$frac{V_L}{V_S} ~=~ frac{1}{n}$$
Equation 12.
Also, the resistance of each lamination (RL) is n times the resistance provided by the solid core (RS):
$$R_L ~=~ n R_S$$
Equation 13.
Let PS and PL be, respectively, the power dissipated in the solid core and the power dissipated in a single lamination. From Equation 11, we have:
$$frac{P_L}{P_S} ~=~ (frac{V_L}{V_S})^2 ~times~ frac{R_S}{R_L}~=~frac{1}{n^3}$$
Equation 14.
Since there are a total of n laminations, the total power dissipated in the laminated core is (frac{1}{n^2}) that of the solid core. Note that this result is consistent with Equation 10. They both show that when we reduce the lamination thickness by a factor of n (or, equivalently, increase the number of laminations by a factor of n), the eddy current loss reduces by a factor of (frac{1}{n^2}).
Let’s finish the article with an illustrative example.
Example Problem: Finding the New Eddy Current Loss
A magnetic core with lamination thickness of t1 = 0.5 mm exhibits eddy current loss of p1 = 50 W/m3 when operated at a frequency of f1 = 50 Hz. If we increase the frequency of operation to f2 = 250 Hz and use t2 = 1 mm thick laminations, what would the new eddy current loss be? Assume that the maximum flux density and resistivity remain unchanged.
From Equation 10, we have:
$$frac{p_2}{p_1} ~=~ (frac{f_2}{f_1})^2 ~times~ (frac{t_2}{t_1})^2$$
Equation 15.
Plugging in the values given at the start of the example, we obtain:
$$frac{p_2}{50} ~=~ (frac{250}{50})^2 ~times~ (frac{1}{0.5})^2 ~Rightarrow~ p_2~=~5000 ; ~text{W/m}^3$$
Equation 16.
Due to the increase in frequency and lamination thickness, the eddy current loss has gone from 50 W/m3 to 5,000 W/m3.
Wrapping Up
Analyzing the eddy current loss in its general form can be challenging. In the above discussion, the assumption of having a uniform field distribution across the cross-section of the core significantly simplified the analysis. At high frequencies, this assumption may not be valid. Such simplifying assumptions, however, help us gain better insight into the factors that determine eddy current loss.
All images used courtesy of Steve Arar
