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CBSE Class 12 Physics 2019 Outside Delhi Set 2 Paper

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Question : 7 of 9
Marks: +1, -0
What is the reason to operate photodiodes in reverse bias?
A       p\;\;\;p - nn photodiode is fabricated from a semiconductor with a band gap of range of 2.52.5 to 2.8 eV2.8\,\mathrm{eV}. Calculate the range of wavelengths of the radiation which can be detected by the photodiode.
Solution:  
(a) Reason
(b) Formula
(c) Calculation of range of wave length.
Reason:
Under reverse bias change in current with the change in intensity of light is more significant.
Calculation:
  λ=  hcE\;\lambda=\;\frac{h c}{E}
  λ1=  6.63×10−34×3×1082.5×1.6×10−19\;\lambda_1=\;\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}} =497×10−9 m=497 \times 10^{-9} \,\mathrm{m}
  λ2=  6.63×10−34×3×1082.5×1.6×10−19\;\lambda_2=\;\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}} =444×10−9 m=444 \times 10^{-9} \,\mathrm{m}
Range is 444×10−9 m444 \times 10^{-9} \,\mathrm{m} to 497×10−9 m497 \times 10^{-9} \,\mathrm{m}
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