Coupling coefficient of grating assisted coupler


#1

I have previously used the following example to calculate the coupling coefficient of integrated bragg gratings.
https://kb.lumerical.com/en/pic_passive_bragg_initial_design_with_fdtd.html
Now, I need to calculate the three coupling coefficients of a contra-directional grating assisted coupler (k11, k22, k12). I was wondering if I can use the same example and apply it to contra directional grating assisted coupler? You can find an example of contra-directional grating assisted coupler, in the following link:
https://www.osapublishing.org/oe/abstract.cfm?uri=oe-21-3-3633


#2

Hi @mehrnoosh.taghiboroo,

It is possible to use the same technique, based on the band structure calculation; however, since you have two waveguides, it is more difficult to interpret the results from the band structure.

In the case of a single Bragg grating when you look at the spectrum from the band structure calculation you see two peaks that can be directly associated with the two bands that form the band gap of the structure. Using the separation of the two peaks, you can calculate the coupling coefficient as done here.

For the contra-directional coupler you will have a more complicated band structure, with possibly more than two peaks in the spectrum, so you need to identify which peaks are related to extract the required bandgaps. For example, if there is reflection along the same waveguide and along the other waveguide (due to cross coupling), there will be two band gaps that you need to identify. One thing that could help is comparing the band structure of the waveguides together with that of the two separate waveguides.

Hope this helps!


#3

Thank you for your response.
I ran the simulation for a contra directional grating assisted coupler which has back reflection from both waveguides, as well as cross coupling between the two waveguides. But I am only able to see two bangaps instead of three. Can you help me fix this issue?
Thanks


#4

Hi @mehrnoosh.taghiboroo,

Can you share your simulation file so that I can take a look at the setup and the results?


#5

W450_W600_g100_Dw0_DW50.fsp (283.1 KB)

I uploaded the file.

Thanks for your help


#6

band_structure.lsf (2.2 KB)

and this is the file I use for band structure calculation


#7

Hi @mehrnoosh.taghiboroo,

I discussed your simulation with my colleague @xwang and this is what we think about the results.

Your structure is formed by two waveguides: the upper one (WG1) is a regular waveguide, while the lower one (WG2) is a Bragg grating with misaligned corrugations. If we run band structure calculations with each waveguide alone, we see peaks at two different wavelengths without any splitting, which is what we expect since there will be no reflection. When we put the two waveguides together, the two peaks split as shown below:

The two bandgaps correspond to different channels for reflection. The band gap at lower frequency is probably the one associated with the contra-directional coupling, as shown below. The band gap at higher frequency, on the other hand, is probably associated with light coming along WG1 reflected back along the same waveguide, due to the mode perturbation caused by WG2.

We don’t see an additional band gap because we don’t expect WG1 to perturb the mode in WG2 in a way that will create a reflection back along the same waveguide WG2.

Hope this helps!


#8

Thank you for your comment. This is the simulation result of this configuration using EME solver. This structure definately has reflection from both waveguides. because the electric field distribution in waveguide 2 is not symmetrical, the misaligned grating cannot calcel the back reflection in waveguide 2.
I was wondering mayb I must change something in the fdtd simulation to be able to see the three bandgaps?


#9

Hi @mehrnoosh.taghiboroo,

Can you share your EME simulation file? I would like to take a look at it to compare it with the FDTD simulation.


#10

CDGAC_W450_W600_g100_D0_D50.lms (372.4 KB)
EME_wavelength_sweep.lsf (1.5 KB)
I uploaded the files.


#11

I would really appreciate it if you respond this question faster
Thanks


#12

Hi @mehrnoosh.taghiboroo,

I took a look at your EME simulation and the first thing I noticed is that the structure is different from the one in FDTD. In the EME simulation you have two Bragg gratings with misaligned corrugations instead of one. This will change the behavior so I don’t think we can compare the two simulations directly. Which situation is the one you are primarily interested in?


#13

Hi,

I am sorry for the confusion.

would you please compare the results of these two files:
W450_W600_g100_Dw24_DW24.fsp (284.0 KB)


#14

here is the second file
https://drive.google.com/file/d/0BxzP7y9-AsNqUWxiM3RldmljOVk/view?usp=sharing
please compare the results of FDTD and EME


#15

Hi @mehrnoosh.taghiboroo,

Sorry for the wait. I checked your simulation files and I am still not quite sure about what is the reason for the discrepancy between the FDTD and EME simulations. Have you tried looking at a full band structure of the coupler in FDTD? It is possible that the missing band gap is not at the band edge. I think it would be good to look at several Bloch vectors close to the band edge in order to see how the different bands are behaving.