How to get the effective index of the grating coupler


#1

Hi all
Now I have a structure of the grating coupler.How can I get the effecive index neff of the grating coupler?
Can the lumerical software solve the question and how to do it? I am hope to get the answer of the question.
Wish to get your feedback.Thanks in advance.


#2

Dear @jbwei

You can use MODE solutions and FDE solver to find the effective index of the desired mode. The webinar linked below is also quite useful to learn more about grating couplers and might answer some of your potential questions:

https://www.lumerical.com/solutions/communications/video/silicon_photonics_grating_coupler_design_video.html

Please let me know if you had further questions.

Thanks


#3

Dear bkhanaliloo
Thanks for your reply.Now I have built a GC by using MODE solution. I have upload the model file but I dont usually use the mode solution.and can you show me how to get the effective index from the file.Thanks a lot.GC_neff_extract.lms (1.6 MB)


#4

Hi all
Can anyone help me solve the question.Thanks a lot.


#5

Dear @jbwei

Thank you for providing simulation file and sorry for late reply.

As is explained in the webinar (link that I provided earlier, watch after ~ 11min 20sec), effective index of grating coupler can be estimated by averaging the effective index of thin and thick section of slab:

This means that you need to use FDE solver to calculate the effective index in these two sections and then find the average. Of course this is an approximation (it is explained in the webinar).

If you want to calculate the exact value of effective index, you can use FDTD to find the far field emission angle (θ). Then you can use the equation below to find the neff:

These are two methods that comes to my mind. In the best of my knowledge, I do not think there is a method to directly calculate neff of the grating coupler.

Please let me know of your thoughts and I will be glad to be of a help.

Thanks


#6

Dear bkhanaliloo
Thanks very much for your reply.Now I will have a try to replicate the results in the video.If I have more questions,I will ask you for a help. Thanks again for your feedback.:grinning:


#7

You are more than welcome @jbwei, and let me know if you had any further questions.

Thanks