The responsivity will be 0.45 A / W. Then the quantum efficiency will be 80%.
A Si pin photodiode has an active light-receiving area of a diameter of 0.4 mm.
When radiation of wavelength 700 nm (red light and intensity 0.1 mW cm^-2 is incident it generates a photocurrent of 56.6 nA.
Diameter (d) = 0.4 mm = 0.04 cm
Then the area will be
Area (A) = π / 4 x d²
Area (A) = π / 4 x 0.04²
Area (A) = 1.26 x 10⁻³ square cm
Then the incident power will be
P₀ = intensity of light x area
P₀ = 1.26 x 10⁻³ x 0.1 x 10⁻³
P₀ = 0.126 x 10⁻⁶ μW
Then the responsivity will be
R = photocurrent / power
R = 56.6 x 10⁻⁹ / 0.126 x 10⁻⁶
R = 0.4492
R ≅ 0.45 A / W
Then the quantum efficiency will be
η = Rhc / qλ
h = plank constant
c = speed of light
q = charge of an electron
Then we have
η = 0.45 x 6.62 x 10⁻³⁴ x 3 x 10⁸ / 1.6 x 10⁻¹⁹ x 700 x 10⁻⁹
η = 0.7979
η = 0.8
η = 80%
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