Conductivity of a photodiode starts changing | vedclass

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Question

(A) The energy of the incident photon corresponds to the band gap energy $E_g$ of the photodiode.$E_g = \frac{hc}{\lambda}$Given $\lambda = 660 \, nm

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) The energy of the incident photon corresponds to the band gap energy $E_g$ of the photodiode.$E_g = \frac{hc}{\lambda}$Given $\lambda = 660 \, nm = 660 	imes 10^{-9} \, m$, $h = 6.6 	imes 10^{-34} \, Js$, and $c = 3 	imes 10^8 \, m/s$.$E_g = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{660 	imes 10^{-9}} \, J$To convert this energy into $eV$, divide by the elementary charge $e = 1.6 	imes 10^{-19} \, C$:$E_g (in \, eV) = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{660 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \, eV$$E_g = \frac{19.8 	imes 10^{-26}}{1056 	imes 10^{-28}} \, eV = \frac{19.8 	imes 100}{1056} \, eV = \frac{1980}{1056} \, eV$Simplifying the fraction: $\frac{1980}{1056} = \frac{15}{8} \, eV$.Comparing this with $\left(\frac{x}{8}ight) eV$, we get $x = 15$.

Conductivity of a photodiode starts changing only if the wavelength of incident light is < 660 \, nm. The band gap of the photodiode is found to be $\left(\frac{x}{8}\right) eV$. The value of $x$ is: (Given, $h=6.6 \times 10^{-34} \, Js, c=3 \times 10^8 \, m/s, e=1.6 \times 10^{-19} \, C$)

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