Assume that you are dealing with perfect blackbody radiation $($T $6000$ K$).$ We want to examine the theoretical limits of a photodiode. Assume no light losses by surface reflection and by parasitic absorption in the diode material. Consider silicon $($bandgap energy, Wg$,$ of $1\mathrm{.}1$ eV$).$ Clearly, photons with energy less than Wg will not interact with the diode because it is transparent to such radiation. a$.$ What percentage of the power of the black$-$body radiation is associated with photons of less than $1\mathrm{.}1.$

ev$?$

b$.$ What would the photodiode efficiency be if all the energy of the remaining photons were converted to electric energy?

e$.$ Would germanium $($Wg $=0\mathrm{.}67$cV$)$ be more or less efficient? d$.$ Using Table $12\mathrm{.}1$ in the text, determine the percentage of the solar energy absorbed by silicon.

e$.$ A photon with $1\mathrm{.}1$ eV will just have enough energy to produce one electron$-$hole pair, and under ideal conditions, the resulting electron would be delivered to the load under $1\mathrm{.}1$ V of potential difference. On the other hand, a photon of$,$ say, $2$ eV$,$ will create pairs with $0\mathrm{.}9$ eV excess energy. This excess will be in the form of kinetic energy and will rapidly be thermalized, and, again, only $1,1$ eV will be available to the load. Thus, all photons with more than Wg will, at best, contribute only Wg units of energy to the load. Calculate what fraction of the black$-$body radiation is available to a load connected to an ideal silicon photodiod