Question

Q2
Consider a solar cell made of semiconducting nanocrystals with a bandgap energy of Wg = 0.67 eV. Determine the theoretical efficiency when the solar cell is exposed to the radiation of a 6000-K black body? Assume that photons with less than 3.3 Wg create each one single electron/hole pair and that those with more than 3.3 Wg create 2 electron/hole pairs each owing to impact ionization. [50 marks]

Answer 1

solution:

here Photons with energy < 0.67 eV do not interact with the solar cell. Photons with energy > 0.67 eV create, each, one exciton. Let the flux of these photons be . The electric power density derived from these photons is

.....Equ(1)

In addition, photos with energy > 3.3Wg eV, create an additional exciton through impact ionization. Let the flux of these photons be . The electric power density derived from these photons is

...Equ(2)

The total electric power density will be

.....Equ(3)

then We have to calculate and

....Equ(4)

where ,,

is the energy of the a photon at the low frequency edge of the band considered. For the band starting at 0.67 eV,at

,

and for the band starting at

,

..Equ (5)

and ,....Equ(6)

The values are 1.873 for the lower limit of 1.295, and 0.409 for the lower limit of 4.274.(calculating by your book appendixe), Thus

..Equ(7)

.........Equ(8)

and,....Equ(9)

...Equ(10)

.....Equ(11)

The incident light power is,

....Equ(12)

then the efficiency is,

So the efficiency is 45.6%