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Calculate the responsivity of a p-i-n photodiode at [tex]1.3 \mu m[/tex] and [tex]1.55 \mu m[/tex] if the quantum efficiency is 80%. Why is the photodiode more responsive at [tex]1.55 \mu m[/tex]?

Draw a block diagram of a digital optical receiver showing its various components. Explain the function of each component. How is the signal used by the decision circuit related to the incident optical power?

Answer :

Final answer:

The responsivity of a photodiode is calculated using a formula incorporating quantum efficiency and wavelength, with higher wavelength causing higher responsiveness. A digital optical receiver consists of a photodetector, amplifier, filter, and decision circuit which translates the signal into digital form. The decision circuit uses the received signal's amplitude to decode the transmitted bits.

Explanation:

The responsivity (R) of a photodiode can be calculated using the formula: R = (quantum efficiency η * Planck's constant h * speed of light c) / (wavelength λ * electronic charge e). Given that the quantum efficiency is 80%, the responsivity at 1.3 μm and 1.55 μm can be calculated using this formula.

The photodiode is more responsive at 1.55 μm because responsivity is inversely proportional to the wavelength, hence a larger wavelength leads to a larger responsivity.

A digital optical receiver mainly consists of four key components: a photodetector, an amplifier, a filter, and a decision circuit. The photodetector converts the optical signal into an electrical signal, the amplifier strengthens this signal, the filter cleans the signal by removing noise, and the decision circuit decodes the signal into digital form.

The decision circuit uses the amplitude of the received signal to determine whether the transmitted bit was a '0' or a '1', which is related to the incident optical power - a higher optical power will produce a larger amplitude signal.

Learn more about photodetector here:

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