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Copy of Inductor Model using Magnetic Circuit Modeling Method - on Sat, 02/15/2020 - 11:33

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Description

This example shows that you can build an inductor model from magnetic components (i.e. a winding and a core), and with an air-gap if needed.You can run the nominal simulation with n = 23 turns on the winding, with the core model state set to "Linear" and a near-zero air-gap. The current transient in the winding matches the current in an ideal inductor model, both are representative of 1mH inductance. But in that configuration, the flux density "b" in the core is over 1.8 Tesla, well beyond the limits of a ferrite material.You can set the magnetic core to "Non-linear with Saturation", and note that the saturation level is 525mT for the modeled core material. Then you can re-run the simulation to see the actual current profile that would result from this more realistic core behavior.Finally, set the number of winding turns to 96. Then set the air-gap to 0.25mm, instead of the nominal 0.25um. This will restore the current rise profile that is expected for a 1mH inductor, but also keep the flux density in the core below the 525mT level.It is also instructive to observe the energy stored in the ideal inductor, as well as the magnetic core and the air-gap, in each of the cases above.

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Understanding SV magnectics

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transformer model comparison -- perfect coupling

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Description

This circuit compares two different ways to specify a two winding transformer. One way is by specifying the inductance of each winding, with a coupling coefficient between the windings.The other way is by specifying the number of windings and the size and permeability of the core on which the windings are wound. For a perfectly coupled transformer, the flux in the core is completely shared by both windings. In the first model, this corresponds to a coupling coefficient of 1.0. In the second model, this corresponds to a single flux path in the magnetic circuit.L = u0*ur*N^2*A/Lu0 = 1.25663706 × 10-6 [m kg s-2 A-2]L_primaryur = 3000 [no units]N = 10 [turns]A = 100.0E-6 [meter^2]L = 100.0E-3 [meter]L_primary =1.25663706E-6 *3000*10^2*1000E-6/100.0e-3 = 3.76991E-04L_secondaryur = 3000 [no units]N = 100 [turns]A = 100.0E-6 [meter^2]L = 100.0E-3 [meter]L_secondary = 1.25663706E-6 *3000*100^2*1000E-6/100.0e-3 = 3.76991E-02The third circuit shows how a magnetic circuit can be used to represent imperfect winding coupling. In this example, the length of the core is divided between the two horizontal core segments. A third (vertical) core segment is also shown. This magnetic path carries flux that is not common to both windings, thus reducing the coupling between the two windings. When the length of this segment is very long, its effect is negligible, approximating the simpler magnetic circuit above it..

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