Silicon photonics
Why do optical circuits using ring resonators often place a heater near the ring to control its temperature?
- AKeeping the ring hot makes even indirect-gap silicon emit telecom-wavelength light efficiently
- BRaising the temperature greatly lowers silicon's refractive index, nearly eliminating propagation loss
- CThe resonant wavelength easily shifts with temperature and fabrication variation, so it must be tuned to the wavelength in use
Show answer and explanation
Answer: C. The resonant wavelength easily shifts with temperature and fabrication variation, so it must be tuned to the wavelength in use
Silicon's refractive index changes strongly with temperature (thermo-optic coefficient about 1.8×10⁻⁴ /K), so the resonant wavelength of a C-band ring resonator shifts toward longer wavelengths by about 0.07–0.08 nm per 1 °C. The resonance width is often narrow, about 0.1–0.3 nm, so even a change of a few °C moves it off the wavelength in use. A difference of just a few nm in waveguide width or thickness can shift the resonant wavelength by a few nm. Therefore, a heater raises the temperature to align the resonant wavelength with the laser wavelength, with control applied while monitoring the state. The added heater power is one of the challenges of using rings.