Hello
Why using a radius with 5 μm in the waveguide at photonic integrated circuits while a radius with 7 μm has minimum loss?
Optimized coupling to a standard SM fiber and its mode field diameter?
If you mean the core radius then as the waveguide radius increases more field will be confined inside the core. however, at some point, you will have more than one mode propagating inside the core.
If you provide more details the answer will be more specific.
Loss may not be the most significant factor for an integrated circuit if the length is short. How much does the total loss increase with this change of radius?
Dear Fatimah,
As colleagues wrote above there are several possible reasons for that ( if you asked about a cross-section of a waveguide) such as:
1) Amount of modes (normally it should be single-mode to avoid beats between modes and optical losses for high order modes due to scattering on walls what is explained lower factor of mode confinement ),
2) Propagated polarizations (it depends on requirements for PIC as well as technology peculiarities),
3) Dispersion engineering (that is the crucial thing for success if there are tasks in microcombs in optical resonators),
4) Stresses in waveguide films that appear during sputtering/oxidation/deposition.
If you wrote about bending radius then the answers of Malcolm White and Mursal Ayub Hamdani described it.
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