Semiconductor Photodiodes For a particular photodiode a puls

Semiconductor Photodiodes: For a particular photodiode, a pulse of light containing 5 x 1012 incident photons at wavelength of 550 nm gives rise to, on average, 1.5 x 1012 electrons collected at the terminals of the device.

a) What is the energy incident to the photodiode in Joules? What is the quantum efficiency of the photodiode?

b) The diffusion length of a charge carrier is the distance that it will travel, on average, before it recombines with another oppositely charged carrier and stops carrying any current. The diffusion length of charge carriers is 0.5 µm in this detector. The detector is roughly a circle with a 0.5 µm radius. (Making the detector any larger would mean that you lose a lot of carriers before they reach electrodes). If the electron diffusion velocity is 8 x 106 cm/s, estimate the response time of the detector (do not take drift into account).

Solution

The energy in each photon = 1240 eV*nm / 550 nm = 2.25 eV

The incident energy of the pulse = 2.25 eV/photon * 5 * 10¹² photons

= 1.125 * 10¹³ eV = 1.80 * 10^-6 J

The quantum efficiency is the fraction of photons that contribute to the current

= 1.5 * 10¹² / 5 * 10¹² = 0.3

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A charge carriermust travel 0.5 mum at the diffusion velocity to reach an electrode

= 0.5 * 10^-6 m / 8 * 10⁴ m/s

= 6.25 * 10^-12 s = 6.25 ps