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THUNDERRr:
Based on https://en.wikipedia.org/wiki/Continuity_equation
you take the divergence of current, and then integrate (not differentiate - my mistake) to get charge. You can take divergence of current in the field calculator. I believe you can "integrate" this over time by dividing by the frequency, since everything is of the form e^(jwt).
Yashwanth:
Why do that, when you can do ∫J•dA over the cross section? Also, I believe your equation is wrong because it does not take into account varying electric field with time.
http://en.wikipedia.org/wiki/Maxwell's_equations
Thank you for replying. In HFSS cook book it is said that to find the current you need to use that value. Also I think the displacement current will be small and can be ignored. So to see what I am doing is right or not? I simulated a 1/2-wave dipole antenna and measured current along the length of the dipole using this procedure(draw circular polylines around the dipole(PEC or copper) at regular postions and measured ∫H.dl) and I got the sinusoidal variation as per text book.
The current behavior I am looking for is constant along the length of the via.i.e., as you move from top to bottom of the via the current shouldn't change much.
Then I did this procedure for the sivenpiper structure and simulated it, its not helping either. The other way I did is I drew a straight polyline along the length of the vias and calculated mag_Jvol and complx_jvol. This too didn't help. The way I am doing is correct or wrong? or is there any way to do this. I am also attaching the simulation file to this post. I am simulating for TM polarization 30 degrees. When I measured the phase I got good result with a MTT paper.
Looking at your HFSS file, I do not think your method can work, for mathematical reasons. You draw a loop and calculate ∫H•dl around it. According to Ampere's law, this should equal the integral of current through a surface bounded by the loop. But WHICH surface? Any surface whose edge is the loop should work - not just a flat disc.
However, different surfaces will give different results for the integral. Let's say you draw a balloon-like surface whose "mouth" is your loop, which has the entire sievenpiper structure inside it. Since this surface does not cross any conductive structure, ∫J•dA will be zero. However, if you choose a flat disc bounded by your loop, it crosses the sievenpiper stem, so ∫J•dA will be nonzero.
How is it possible that you get different results to ∫J•dA if ∫H•dl is the same? The answer is that displacement current MUST be significant.
Define a cross-section surface (if you haven't already - I don't remember from the file). Calculate ∫J•dA over it using the field calculator. I'm not sure offhand how to calculate the dot product with the normal vector (I think integrating scalar |J| over area will give inaccurate results), but it can surely be done. In the field calculator you can save the result as a "named expression" and then make a plot ("Fields Report") if you want.
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