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current density simulation

时间:03-31 整理:3721RD 点击:
I am wondering if there is a simulator which simulate current distribution in 1D. IE3D can simulate current distribution in 3D. But I need to know the current distribution along a t-line. Thanks.

i can't understand what u meen by 1D ! u mean 1 dimension thing !?
is that a point or something ?
or line stip like T ?
i thing u chose the most powerfull software IE3d, but try out feko too.

For 1D, I mean current distribution along a line in 3D structure (eg. transmission line). What is feko?

Hi, danda821:

IE3D solves the surface current distribution. You can also visualize the current distribution on the surface too. In case you want to find the current distribution at a specific point or along a line, you can display the current distribution on MGRID and select File->Save Current Density Data for the values in ASCII files. You can write a simple parser to post processing the data. For example, you can try to find the current along a line, etc. Regards.

Thanks. Jian, I save the current density data in cdd format. It has many columns:
No. Sel Nv V1 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V2 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V3 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V4 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz

Can you tell me the meaning of each column? Is there any way I can save the current distribution at a certain plane (Z=certain value) only, instead of in the whole range of z?

Hi, danda821:

The exlainatation is in the following:

No. Sel Nv V1 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V2 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V3 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V4 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz

No. is the cell's No. Sel is whether the cell is selected on the screen. Nv is the number of vertices. V1 X Y Z ReJx ImJx ... are for the No.1 vertex's x, y, z and current distriution. It lists all the 3 or 4 vertices of the cell. It should be easy for you to write a parser in any language to post processing the data. Thanks!

Best regards.

So it calculates current density at all vertex of a cell.
See below vertice 1 of cell 1 and cell 5 are at the same position, but current density is different. Is this right?

Do I need to do a current averaging in each cell?

No. Sel Nv V1 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V2 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V3 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz V4 X Y Z ReJx ImJx ReJy ImJy ReJz ImJz
1 0 3 1 3.00E+00 2.30E+01 3.07E+02 5.44E-01 2.38E+00 7.08E+00 1.48E+01 0.00E+00 0.00E+00 2 1.28E+01 2.60E+01 3.07E+02 7.84E-01 3.20E+00 7.15E+00 1.50E+01 0.00E+00 0.00E+00 3 1.28E+01 0.00E+00 3.07E+02 7.84E-01 3.20E+00 6.51E+00 1.28E+01 0.00E+00 0.00E+00
2 0 3 1 1.28E+01 0.00E+00 3.07E+02 4.81E-01 2.88E+00 7.23E+00 1.36E+01 0.00E+00 0.00E+00 2 3.00E+00 0.00E+00 3.07E+02 2.60E-01 2.06E+00 7.23E+00 1.36E+01 0.00E+00 0.00E+00 3 3.00E+00 2.30E+01 3.07E+02 2.60E-01 2.06E+00 7.75E+00 1.55E+01 0.00E+00 0.00E+00
3 0 3 1 2.88E+01 4.60E+01 3.07E+02 4.21E+00 1.37E+01 9.55E-01 3.37E+00 0.00E+00 0.00E+00 2 4.19E+01 2.60E+01 3.07E+02 4.53E+00 1.48E+01 4.64E-01 1.68E+00 0.00E+00 0.00E+00 3 1.28E+01 2.60E+01 3.07E+02 3.81E+00 1.24E+01 4.64E-01 1.68E+00 0.00E+00 0.00E+00
4 0 3 1 3.00E+00 4.60E+01 3.07E+02 4.01E+00 1.12E+01 1.53E+00 2.85E+00 0.00E+00 0.00E+00 2 2.88E+01 4.60E+01 3.07E+02 4.67E+00 1.33E+01 1.53E+00 2.85E+00 0.00E+00 0.00E+00 3 1.28E+01 2.60E+01 3.07E+02 4.26E+00 1.20E+01 1.02E+00 1.18E+00 0.00E+00 0.00E+00
5 0 3 1 3.00E+00 2.30E+01 3.07E+02 9.53E-01 4.11E+00 7.20E+00 1.53E+01 0.00E+00 0.00E+00 2 3.00E+00 4.60E+01 3.07E+02 9.53E-01 4.11E+00 7.81E+00 1.73E+01 0.00E+00 0.00E+00 3 1.28E+01 2.60E+01 3.07E+02 1.21E+00 4.96E+00 7.28E+00 1.56E+01 0.00E+00 0.00E+00

Hi, danda821:

IE3D is using the roof-top basis functions. For roof-top basis functions, they do not guarantee the current to be continuous across a cell's boundary. It only guarantees the normal component of the current is continous across the boundary. Interestingly, when I first implemented an EM simulator using cell-based basis functions in 1989, I tried to use basis functions guarantee all components of current distribution to be continuous. The simulator did not yield accurate results. Eventually, I tried roof-top basis funcitons and they yield good results. I think many people may have tried it and know the fact. That is why roof-top basis functions are the most popular ones now. Thanks!

Jian,

So if I want to calculate average current density at a certain point, can
I just find the cell containing the point and calculate the average current density on all vertices of the cell? Thanks.

Hi, danda821: You can do it in that way or you can use a linear interpolation for it. For better smoothing, I do suggest you to find all the currents belong to different cells at the same vertex and do an average on it. Then, you use the linear interpolation for it. For the .cdd file saved in IE3D, we just provide the raw data to you so that you know what it is originally. Regards.

Considering the example in the following figure, I plan to calculate the current in the following way:

I(1)
I(2)=(I(Cell_1,2)+I(Cell_2,2))/2
I(3)
I(4)=(I(Cell_1,4)+I(Cell_1,4))/2

I(5)=(I(1)+I(2)+I(4))/3
I(6)=(I(2)+I(3)+I(4))/3

Is it right? Or how to do the linear interpolation on I(5) and I(6)

Yes. You are right. For any point 4 inside a triangle (1, 2, 3), you can find

V4 = C1 * V1 + C2 * V2 + C3 * V3

C1 = area_4_2_3 / area_1_2_3
C2 = area_4_3_1 / area_1_2_3
C3 = area_4_1_2 / area_1_2_3

area_1_2_3 = area of the triangle formed by vertices 1, 2 and 3.

apologies for hijacking this thread, but to extend the discussion, is the computation for currents at a point 5 inside a quadrilateral (1_2_3_4) mesh element similar?

(see attd. fig)

V5=C1*V1+C2*V2+C3*V3+C4*V4

where C1=(area_2_3_5+area_3_4_5)/area_1_2_3_4?

While on the topic, it would be really useful to add a J (current density) option in the near-field calculation so that the current distribution can be extracted for a user-generated uniform rectangular grid (just like H-field or A-potential). Assured it's not exactly 'near-field' (because it's right on the surface) but this takes out the guesswork involved in parsing the cdd file (please suggest if there is a better way of extracting the current distribution for uniform grid directly in current version of IE3D).

TIA,
Sree

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