June 20, 2021 at 4:26 amAlexanderphenomSubscriber
We are using a dipole as the source to calculate the LED light emmision process. Owing to the structure of square layers, we use symmetry and anti-symmetry condition as the FDTD boundary conditions. And one single dipole is placed at the middle point of the FDTD simulation rigion, which is the cross point of the symmetry boundaries. Here's the question, how does the "theta" property of the dipole influence the mode? How can I know whether TE or TM mode is emmitted from the dipole source? And how does the symmetry and anti-symmetry infuence the mode that dipole source emmits. How to decide whether using symmetry boundary condition or anti-symmetry conditon in this process?
Hoping for your reply. Thank you very much.June 21, 2021 at 8:54 pmTaylor RobertsonAnsys EmployeeHello Alexanderphenom It sounds like you are on the right track , and I should say that symmetry conditions can be a subtle topic; especially with dipoles, since they emit spherical waves. They will not tell if you have set something wrong, so it is often helpful to compare with no symmetry conditions when in doubt. If you have set the symmetry conditions correctly then the results of importance will agree between the two cases, and you can proceed to use symmetry for faster more accurate results.
First using theta will break the symmetry of the fields unless theta is 0, or 90 degrees. For a dipole at the origin the symmetry would now be tilted at an angle theta, but FDTD symmetry needs to be along one of the principle axes. In short to use maximum symmetry you would need the dipole axis to be aligned with X, Y or Z, and so I would avoid theta unless there is a very compelling reason to so. It would probably be preferable to use a dipole cloud all set in a single plane, which would still allow you to use some symmetry.
How can I know whether TE or TM mode is emmitted from the dipole source?
This really depends on the structure. I would use a magnetic dipole to try and excite TM modes, and to excite TE modes use an electric dipole. This assumes you know how TE and TM will be polarized in your device and that you have aligned the dipole axis in this orientation. For these dipoles the symmetry should be set tangential to the dipole axis should be the same color, and opposite color would be normal. This is outlined in this article.
If you have done everything correctly then the symmetry conditions should support the modes you are interested in, but this is not guaranteed. Again it may be helpful not enforce any symmetry in the initial exploratory phase. Once you understand the modes you can use symmetry to suppress degenerate modes, and reduce sim time. I would suggest looking at the following example, as it covers lots of these issues through a practical demonstration.
Best RegardsJune 22, 2021 at 1:30 pmAlexanderphenomSubscriberI really appreciate your reply, it helps a lot.
But I still have some questions:
1.Do you mean that if I wanna exite TE mode, I'm supposed to use the electric dipole source, and that TM mode using megnetic dipole source ?
We use dipoles as the light source to calculate the LED lighting process.Owing to the structure of square layers, we use symmetry and anti-symmetry condition as the FDTD boundary conditions. And one single dipole is placed at the middle point of the FDTD simulation rigion.
If we wanna reproduce the situation in the paper "Investigation of Light Extraction Efficiency in AlGaN Deep-Ultraviolet Light-Emitting Diodes". Figures are shown below. How do we set theta and phi or other parameters to get the TE or TM source that matches the circumstance in the paper.
2.Do symmetric and antisymmetric settings affect the mode of dipole emission? Does symmetry cause dipole interference?
June 22, 2021 at 9:33 pmTaylor RobertsonAnsys EmployeeHello
I am not sure about that paper, or the approach the approach they take. I would recommend the MicroLED example as an alternative.
As incoherent devices you need to add the field intensity of 3 orthogonal polarizations, and statistically average over the emission region to replicate the farfield emission profile of the device. See example for more information.
I am not sure how helpful the concept of TE and TM modes, are since these are not really meant to be light guiding structures? You are looking at the non-radiative modes of the stack?
Yes they will. If you use symmetry then you will have an image dipole, so yes they can. Again compare with and without to validate your set-up.
July 2, 2021 at 3:26 amAlexanderphenomSubscriberThank you for your recommondation.
I have some extra confusions,
What is the polarize direction of the dipole source? I noticed that there's a gray cloud region around the dipole source, does it represents the propagation direction? And, there's also two reverse arrows, are they represent the polarization direction?
The light in our structure propagates to any direction, so it is not like a fiber or a waveguide. We assume? that if the light propagates vertically, it is the TE mode. While horizentally for the TM mode. And the reference coordinate system is shown in the below figure. How to recogenize the polariationze situation her? e
July 2, 2021 at 3:28 amJuly 2, 2021 at 3:30 amJuly 7, 2021 at 5:26 pmTaylor RobertsonAnsys EmployeeHello Alexanderphenom The electric dipole will have the electric field polarized along the dipole arrow axis. The grey rings around the dipole indicate the direction of maximum emission. There are is also a small box, this indicates the actual mesh cells used for the excitation.
The dipole emits spherical waves which may or may not couple to guided modes of the structure. These modes may be trapped or easily extracted. In some cases it may be helpful to understand these modes, but ultimately it seems that you are interested in how much light is extracted vs trapped in the substrate; so focusing on the modes may be distracting. A vertically polarized dipole (Z) will emit most of the most of the radiation in the plane of the device and will couple to guided modes of the slab layer which will be lost. A horizontally polarized dipole (X), (Y) will emit more of it's radiation vertically which can be more easily extracted.
The paper notes that light from the InGaN region will be preferentially polarized, so you can ignore the contribution from Z polarized dipole. I am not sure if there is physical reason for this or if it can be ignored because of 2. Might be interesting to measure the contribution from Z?
Your set-up looks correct. In this coordinate system you can add the contribution from an X and Y polarized dipole for each position (x,y).
I might suggest looking at a 1D approximation using STACK as well. This example might be quite helpful.
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