# What is the reason to operate photodiodes in reverse bias? A p-n photodiode is fabricated from a semiconductor with a bandgap of range of 2·5 to 2·8 eV. Calculate the range of wavelengths of the radiation which can be detected by the photodiode.

The change in current with the change in intensity of light is more significant in reverse bias condition.

Now, calculating the range of wavelength:-

We know,

$\lambda =\frac{hc}{E }\; \; \; \; \; \; \; \; \; \; \left [ \because h=6.63\times 10^{-34},c = 3\times 10^{8}\right ]$

Then,

$\lambda_{1} =\frac{6.63\times 10^{-34} \times 3\times 10^{8}}{2.8\times 1.6\times 10^{-19}}=497\times 10^{-9}m$

$\lambda_{2} =\frac{6.63\times 10^{-34} \times 3\times 10^{8}}{2.5\times 1.6\times 10^{-19}}=444\times 10^{-9}m$

Hence, the range of wavelengths of the radiation which can be detected by the photodiode is $444\times 10^{-9}m$ to $497\times 10^{-9}m$.

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