some question about the ring resonator


#1

dear all
Recently I have learned about the ring resonator example on your website.But I have some question whci I dont understand and want to ask.


In fact I dont understand why a channel spacing of 200GHz means 1.6nm at 1550nm,the how to get the FSR from 3200GHz and how to get the Q value from it.

Secondly, I have download the resonator file including VarFDTD and FDTD file,I change the parameter value of both the two file, but I get different results,which is shown below.

I dont know what make the two results different.Can you help me check it. Thanks a lot.
1.lms (532.0 KB)
2.fsp (488.1 KB)


#2

Hi @jbwei,

I had a look at the files, it seems you increased the radius of the ring, but the other elements were not modified. You would need to move the source and the monitors and also to increase the size of the FDTD and varFDTD regions in a same way.

In the current files, the sources and the monitors could possibly truncate the fields and therefore, not get the proper result. Once you make sure both simulation are set up in the same way, then you can compare the results.

I hope this helps.


#3

dear gbaethge
Thanks for your reply.firstly I have two question in the post above and you just answer the second question. Can you help me solve the first question or provide the derivation of formulas.

Secondly now I have increased the area of FDTD region and move the monitor to the proper position.I get the result like this below.
ring_resonator.lms (537.5 KB)

furthermore I want to ask how to judge whether the sources and monitors truncate the fields.
Hope to get your feedback. Thanks a lot.


#4

Hi @jbwei,

Sorry I missed the first question!

To get the channel spacing in nm, we start from λ=c/f , so

dλ = c * df / f^2, and
dλ= λ^2 * df / c

For the FSR:

We want to design the system to drop every 16th channel

So FSR = 16 * df = 3200 GHz

Regarding the Q factor:

We would like the FWHM of the drop to be 100GHz, corresponding to a Q of approximately 1550nm/0.8nm ~ 2000.

and we have

We know λ and δλ (FWHM is 100GHz that is 0.8nm). That said, I believe Q should be λ/δλ . I’d have to check.

Finally, regarding the size of the source and monitors, you want it to be large enough so the fields decay enough before reaching the limits. For instance, if you look at the mode in from the file you attached in the first post:

In the last file, it is much better:


#5

dear gbaethge
Thanks so much for your reply.Now I have understood most of the relationship of the relative parameters. But there exist a small question I want to ask.From the previous section,FSR is given as the formula below


but in the calculation of ring length,you use the formula

It means lambda^2=c ?? I am confused about it. Can you explain it to me?

Secondly,thanks to find out the difference of the two varFDTD files for me. Now I understand that why I set wrong in the first varFDTD file. Does it mean that the second varFDTD file is set right?But the result is still not as same as the result got from the 3D FDTD file.Can you help me check it.Thanks a lot

3D_ring_resonator.fsp (770.5 KB)
Hope to get your feedback.Thanks in advance.


#6

Hi @jbwei,

Sorry for the delay, we had a long weekend. Regarding your questions:

It’s a bit confusing, but in the first definition, FSR is expressed in wavelength unit. In the second, we use FSR expressed in frequency unit.

Regarding the second question, it is a bit more tricky, actually: the simulations are run with a coarse mesh, so you would need first to make sure all the parameters are set correctly (source, monitor, material fit, etc), then you would need to reduce the mesh size (convergence testing) then you could really compare both results.
The main issue will be with the FDTD simulation, at it may take some time to run, since it is a full 3D calculation.