 Hello and welcome to the lecture on optical detectors part 3, learning outcome of this session. By the end of this session student will be able to compute the performance parameters of a photo detecting material. You may here pause the video and try to find out the answer to this question. What is Planck's constant? Planck constant is a term used to relate the energy of a photon to its frequency. The value of a Planck constant is denoted by the alphabet H which is equals to 6.626 into 10 raised to minus 34 meter square kilogram per second. Today we will see how to solve a numerical to find out the performance parameters of a photo detecting material. Here we have a problem when 3 into 10 raised to 11 photons each with the wavelength of 0.85 micrometer are incident on a photodiode. On an average 1.2 into 10 raised to 11 electrons are collected at the terminal of the device. Determine the quantum efficiency and the responsivity of a photodiode at 0.85 micrometer. Here we are asked to calculate the quantum efficiency and the responsivity. We know the formula of quantum efficiency which is denoted by the later eta as the number of photons absorbed. It is a ratio of number of photons absorbed to the number of photon incident. So 3 into 10 raised to 11 photon incident. So in the denominator 3 into 10 raised to 11 and 1.2 into 10 raised to 11 number of electrons are collected means generated electron hole pairs. So we get the quantum efficiency as 1.2 into 10 raised to 11 by 3 into 10 raised to 11. So this results to 0.4. If we express the quantum efficiency in percentage we can express as a 40 percent of quantum efficiency. 40 percent of quantum efficiency means out of 100 photon incident 40 are absorbed. So we have a 40 percent of quantum efficiency. The second term we are asked is responsivity. We know the formula for responsivity is eta e times lambda by hc. Here quantum efficiency is 0.2 into 10 raised to 11 by 3 into 10 raised to 11.4. We have calculated over here. E value of charge on electron is 1.6 into 10 raised to minus 19 and lambda which is wavelength. We have wavelength of 0.85 micrometer. So 0.85 into 10 raised to minus 6 plan constant. We know the value of plan constant as 6.8 into 10 raised to minus 34. C is nothing but speed of light which is 3 into 10 raised to 8 meter per second. So if we calculate this we get the value as 0.41. So the responsivity is 0.41. The unit for responsivity is ampere per watt. So the quantum efficiency we calculated as a 40 percent and responsivity is 4.1 ampere per watt. We will solve another example. A photodiode has a quantum efficiency of 65 percent. When photons of energy 1.5 into 10 raised to minus 19 joules are incident upon it. At what wavelength is the photodiode operating? Here we are asked to find out the operating wavelength of a photodiode with quantum efficiency of 65 percent with a photon energy of 1.5 into 10 raised to minus 19 joules. We know the formula for energy of a photon which is E equals to h f. We can rewrite this formula as h c by lambda because f is nothing but c by lambda where c is speed of light and lambda is a wavelength. So we can rearrange this equation to get h c by E equals to lambda. So we know the value of plane constant 6.62 into 10 raised to minus 34. Speed of light is 3 into 10 raised to 8 meter per second and the energy of a single photon is 1.5 into 10 raised to minus 19 joules which is given in the problem study. So we calculate this and we get the lambda as 31300 nanometers. So a photodiode with a quantum efficiency of 65 percent when photons of energy 1.5 into 10 raised to minus 19 joules are incident upon it the operating wavelength will be 1300 nanometers. We will go for another numerical. In a photo detection device having absorption layer of thickness 10 micrometer with a refractive index of 3.5 and absorption coefficient alpha equals to 10 raised to 5 per meter. Then calculate the quantum efficiency if reflectivity is 0.31. So here we are asked to calculate quantum efficiency. In the previous session we have derived the formula for quantum efficiency which is the formula is 1 minus r times 1 minus e to the power minus alpha w where r is Fresnel's reflection coefficient which is 0.31. Alpha is absorption coefficient which is given in the statement is 10 raised to 5 per meter and w is width or thickness of absorption layer which is 10 micrometer. We will put these values in a formula 1 minus 0.31 times 1 minus r times 1 minus r times 1 minus r times 1 minus e to the power 10 raised to 5 which is absorption coefficient alpha into 10 micrometer which is a thickness of absorption layer. So we will calculate this value and we get the value as 0.43. So for a photo detection device we have a thickness with a absorption coefficient of 10 raised to 5 per meter with a thickness of absorption layer with 10 micrometer and Fresnel's reflection coefficient of 0.31 we get the quantum efficiency of 0.43. We also know quantum efficiency linear times is expressed in terms of percentage. So this 0.43 of quantum efficiency in percentage becomes 43 percent which means out of 100 incident photons 43 photons are absorbed to generate the electron hole pair which in turn contribute to the photo current generation. These are the references. Thank you.