September 13, 2022 at 7:54 amliguangrui33Subscriber
I find if the incident angle of the plane wave source is large, i.e. 80 degree, the intensity before the source is in order of e-2 instead of 0 or e-7.
Why is this happening and what should I do to correct it?
September 13, 2022 at 4:57 pmGuilin SunAnsys Employee
It depends on what PML you used so it is PML reflection.
Are you using plane wave plus Bloch BC for a single wavelength, or BFAST for broadband? in either case, since the injection angle is very large, conventional PML settings will not behave well as with small incident angle.
So please try to use Steep Angle PML with more layers, make sure the PML at least half to one wavelength thick, and have larger distance betwen device interface to the PML.
September 14, 2022 at 5:25 amliguangrui33Subscriber
Thank you for your suggestions.
I am using plane wave plus Bloch BC for a single wavelength. Now I have my PML steep angle mode with 16 layers (400nm compared with 266nm wavelength). The distance is 2um away.
It looks like the result is convergence on PML already, since it would not change much if I increase the layer number to 26. But the reflectance is still as much as 0.018 on the monitor before source.
Is there a further step to improve the simulation? Thank you
September 14, 2022 at 2:55 pmGuilin SunAnsys Employee
Since the incident angle is 80 deg, you may need to check if there is any higher order diffraction that creates close to 90 deg orders. In addition, not only PML, but also the source injection can have errors. I would suggest
1: use 48 or 64 layers of PML
2: use smaller autothustoff min, eg, 1e-6 or smaller
3: use longer simulation time to let the autoshutoff min to terminate the simulation.
4: set the source in time domain with pulse length 50fs and off 100fs, or 100fs/200fs. This is because the injected pulse is still broadband, and other wavelength may have resonance. Please use a time monitor and check if there is resonance at other wavelength.
Large incident angle simulation is challenge.
September 16, 2022 at 2:29 amliguangrui33Subscriber
Thank you for your suggestions. I tried with a source setted in time domain with your suggestions in 4. 50fs and off 100fs will still lead to a R=0.011 on the monitor before the source, while 100fs/200fs leads to a R = 0. There are 5 frequencies components shown in the source.
I just realized that when I setted a source not in time domain with a single frequency, the spectrum of the source has actually a significant span (239-299nm). Are the plots on the right of Frequency/Wavelength tab true in case of a single wavelength is setted? Why is it like this?
September 16, 2022 at 4:19 pmGuilin SunAnsys Employee
FDTD is a time-domain method. Therefore the excitation in most cases is always a pulse. A pulse will create a braodband spectrum, thus you see the significant spectrum bandwidth.
The pulse is the most efficient source for FDTD. If you want to have a quasi-monochromatic light, you will need a transient part with half the pulse, then a sinusoidal signal, which must extends much longer time in order to reach the steady state, and use time domain monitor to get the result. Frequency-domain monitors are designed for pulse signal, which uses Fourier transform to get the steady-state result from the time signal. It is rarely used though. Here is an example https://optics.ansys.com/hc/en-us/articles/360042192813-1D-cavity-laser-using-4-level-2-electron-material
Ansys Insight: FDTD 为什么采用脉冲光源激励而不采用正弦波光源 （目前暂时不在论坛上，国内网站上有）
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