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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 __________.
Correct answer is between '74.5,75.5'. Can you explain this answer?
Most Upvoted Answer
Quantum efficiency of a photodiode (ratio between the number of libera...
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

Calculation:

I = 75.18 μA
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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 __________.Correct answer is between '74.5,75.5'. Can you explain this answer?
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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 __________.Correct answer is between '74.5,75.5'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about 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 __________.Correct answer is between '74.5,75.5'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 __________.Correct answer is between '74.5,75.5'. Can you explain this answer?.
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