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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?.

Solutions 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? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE). Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.

Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of 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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an ample number of questions to practice 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? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.