# Quantum efficiency of a photodiode (ratio between the number of liberated electrons and the number of incident photons) is 0.75 at 830 nm. Given Planc

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Quantum efficiency of a photodiode (ratio between the number of liberated electrons and the number of incident photons) is 0.75 at 830 nm. Given Planck’s constant h = 6.624 × 10-34 J-sec, the charge of an electron e = 1.6 × 10-19 C and the velocity of light in the photodiode Cm = 2 × 108 m/s. For an incident optical power of 100 μW at 830 nm, the photocurrent in μA is __________.

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Given:

η = 0.75

λ = 830 nm

h = 6.626 x 10-34 J-sec

e = 1.6 x 10-19 C

cm = 2 x 108 m/s

Pin = 100 μW

Concept:

The ratio of output current to input power is given by Responsivity

$R = \frac{I}{{{P_{in}}}} = \frac{{\eta e\lambda }}{{hc}}$

$I = \frac{{\eta e\lambda }}{{hc}}{P_{in}}$

Calculation:

$I = \frac{{0.75 \times 1.6 \times {{10}^{ - 19}} \times 830 \times {{10}^{ - 9}} \times 100 \times {{10}^{ - 6}}\;}}{{6.626 \times {{10}^{ - 34}} \times 2 \times {{10}^8}}}$

I = 75.18 μA