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March 15, 2024 at 10:42 am
sicli21
Subscriber
Thank you for your response. In https://optics.ansys.com/hc/en-us/articles/360042326694-Ferroelectric-modulator , it says
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Due to the same reason as the one mentioned above, the Si layer in the waveguide also requires a voltage boundary condition. Instead of providing the boundary condition directly from a contact (gnd) we provided the boundary condition through another semiconductor region
"""
If I make my SiO2 insulator, my semiconductor waveguide won't have any BC and according to the example that may cause inaccurate results.
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