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Leakage inductance computation of a toroidal transformer
Posted Oct 18, 2011, 8:08 p.m. EDT Low-Frequency Electromagnetics, Geometry Version 4.2 5 Replies
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Hello friends,
I'm trying to simulate a 2D top view of a toroidal transformer and compute the leakage inductance from the total energy. The equation used is W = 1/2*L(I^2). Energy (W) is computed from the total magnetic energy from comsol and multiplied by the height of transformer. In this case, the primary and secondary coils are excited using multi turn coil domains with a current of 5A through each coil (Short circuit test is modeled). The appropriate direction for current is taken into consideration. In the model primary and secondary winding are wound on opposite sides of the core as required.
The energy obtained from Comsol = 2.6456, using total magnetic energy from global evaluation. But the expected value for the same dimension transformer using experiment is almost three times higher. Could you check the model and suggest what could be the reason behind this reduction in the energy ?
The model is attached here.
Thanks a lot for any suggestions !!!
Noel
I'm trying to simulate a 2D top view of a toroidal transformer and compute the leakage inductance from the total energy. The equation used is W = 1/2*L(I^2). Energy (W) is computed from the total magnetic energy from comsol and multiplied by the height of transformer. In this case, the primary and secondary coils are excited using multi turn coil domains with a current of 5A through each coil (Short circuit test is modeled). The appropriate direction for current is taken into consideration. In the model primary and secondary winding are wound on opposite sides of the core as required.
The energy obtained from Comsol = 2.6456, using total magnetic energy from global evaluation. But the expected value for the same dimension transformer using experiment is almost three times higher. Could you check the model and suggest what could be the reason behind this reduction in the energy ?
The model is attached here.
Thanks a lot for any suggestions !!!
Noel
Attachments:
5 Replies Last Post Nov 1, 2011, 4:18 a.m. EDT
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Posted:
1 decade ago
Hi
A few comments: for the geometry, you do not need to take all those differences, its enough to verlay all circles, and the default "union" finish will generate all the different domains, its quicker to set up ;)
your current conducting Cu layer is very thin you could perhaps consider a "thin" surface layer too, but if you have enough RAM, this way is OK ;)
your B and Az display looks strange (the symmetry pattern), have you checked how your current is flowing ? And its sign ? plot out mf.Jx, mf.Jy and mf.Jz
In ACDC I always check my currents before solving, the easiest is to set up your model, and solver, get COMSOL to generate the default solver sequence, then right click on the "Dependent Variables" node and "compute to here", this loads all initial conditions and set's up your model ready to solve, then you can also see the postprocessing plots with initial values
I also often use "arrows" on my plots to visualise the current flow, but for your case here, these are along Z so they do not show up as "arrows" in the plane ;)
--
Good luck
Ivar
A few comments: for the geometry, you do not need to take all those differences, its enough to verlay all circles, and the default "union" finish will generate all the different domains, its quicker to set up ;)
your current conducting Cu layer is very thin you could perhaps consider a "thin" surface layer too, but if you have enough RAM, this way is OK ;)
your B and Az display looks strange (the symmetry pattern), have you checked how your current is flowing ? And its sign ? plot out mf.Jx, mf.Jy and mf.Jz
In ACDC I always check my currents before solving, the easiest is to set up your model, and solver, get COMSOL to generate the default solver sequence, then right click on the "Dependent Variables" node and "compute to here", this loads all initial conditions and set's up your model ready to solve, then you can also see the postprocessing plots with initial values
I also often use "arrows" on my plots to visualise the current flow, but for your case here, these are along Z so they do not show up as "arrows" in the plane ;)
--
Good luck
Ivar
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Posted:
1 decade ago
Thank you Ivar for your valuable comments.
The current excitation given is in the z direction, since the model is the top-view of a toroidal transformer. The primary and secondary windings are modeled with +z and -z direction currents. Since it is a short circuit test in a 1:1 transformer, both windings are excited with same current. Also primary and secondary windings are modeled without overlapping, i.e one half side of the core is primary and the other half side is the secondary. So I hope the excitation given is correct as it is a toroid. Please correct me if you think this is not the right way.
Actually I tried the surface current density method by modeling arcs instead of Copper winding bands by defining the surface currents to it. Even when I simulate that, the magnetic energy value is very small than expected and thus the leakage inductance too. And I uses our big server which is 96 GB RAM, so it solves real quick.
As you said, I cannot see the arrow plots since current is in the Z-direction.
Could you predict any other reasons for the lesser energy value than expected. I appreciate your help.
Thanks a lot,
Noel
The current excitation given is in the z direction, since the model is the top-view of a toroidal transformer. The primary and secondary windings are modeled with +z and -z direction currents. Since it is a short circuit test in a 1:1 transformer, both windings are excited with same current. Also primary and secondary windings are modeled without overlapping, i.e one half side of the core is primary and the other half side is the secondary. So I hope the excitation given is correct as it is a toroid. Please correct me if you think this is not the right way.
Actually I tried the surface current density method by modeling arcs instead of Copper winding bands by defining the surface currents to it. Even when I simulate that, the magnetic energy value is very small than expected and thus the leakage inductance too. And I uses our big server which is 96 GB RAM, so it solves real quick.
As you said, I cannot see the arrow plots since current is in the Z-direction.
Could you predict any other reasons for the lesser energy value than expected. I appreciate your help.
Thanks a lot,
Noel
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Posted:
1 decade ago
Hi
but what I see is that the current is positive in the left part of the circle, and negative in the other and then inverse int he inner cylinder, I'm not sure you have got the current density signs correct for the 4 quadrants
--
Good luck
Ivar
but what I see is that the current is positive in the left part of the circle, and negative in the other and then inverse int he inner cylinder, I'm not sure you have got the current density signs correct for the 4 quadrants
--
Good luck
Ivar
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Posted:
1 decade ago
Hi Ivar,
Thank you for the comments.
I rechecked my model and the current density-z component plot is attached here. The left two plots corresponds to winding 1(primary) current density, and the right two plots for the winding 2 current density.
In this model I'm trying to model windings covering only 180 degrees of the entire core. Also both the windings are on opposite sides. The core is modeled as two circles making a ring while viewing from top.
From the Jz plot, its clear that the current density varies such that current remains constant. So to my knowledge, I understand that the direction of current seems ok. I appreciate any thoughts from you.
Thanks,
Noel
Thank you for the comments.
I rechecked my model and the current density-z component plot is attached here. The left two plots corresponds to winding 1(primary) current density, and the right two plots for the winding 2 current density.
In this model I'm trying to model windings covering only 180 degrees of the entire core. Also both the windings are on opposite sides. The core is modeled as two circles making a ring while viewing from top.
From the Jz plot, its clear that the current density varies such that current remains constant. So to my knowledge, I understand that the direction of current seems ok. I appreciate any thoughts from you.
Thanks,
Noel
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Posted:
1 decade ago
Hi
well I probbaly do not understand your model, what puzzles me is the graph below I just do not understand hpw you get the Jz current to flow like that with positive on the left side and negative on the right half ring side
My understanding was that it looped around radially with different signes for inneror outer, or that it went in on the inner and "out" on the outer ring.
--
Good luck
Ivar
well I probbaly do not understand your model, what puzzles me is the graph below I just do not understand hpw you get the Jz current to flow like that with positive on the left side and negative on the right half ring side
My understanding was that it looped around radially with different signes for inneror outer, or that it went in on the inner and "out" on the outer ring.
--
Good luck
Ivar