Solar simulators are used to study the performance of solar cells in the lab. In the figure below (left picture), the spectral power density of a solar simulator is shown with the blue line. The spectral power density of this solar simulator is given by:
P(λ)=7.5∗1015λ−2.25∗109 [Wm−2m−1] for 300nm<λ<500nm
P(λ)=2.25∗109−1.5∗1015λ [Wm−2m−1] for 500nm<λ<1500nm
Where the wavelength λ is expressed in meters.
a) calculate the irradiation I of the solar simulator (in Wm−2 )
incorrect
b) What is the photon flux of the solar simulator (in 1021m−2s−1)?
incorrect
EQE OF A TANDEM CELL II
The EQE of a tandem cell with junction A and junction B under short circuited (V = 0 V) condition is also presented in the previous figure (right picture).
c) Which junction acts like the top cell in the tandem divide?
A or B
EQE OF A TANDEM CELL III
d) What is the band gap (in eV) of the absorber layer of the junction A?
e) Calculate the short circuit current density Jsc of junction A (in mA/cm2) if the solar cell is measured under the solar simulator.
f) Junction B has a different absorber layer than junction A. Above its band gap, the solar cell B has an EQE = 0.60 that remains constant. Calculate the short circuit current density Jsc of the junction B (in mA/cm2) if the solar cell is measured under the solar simulator.
e) 14.5
f) 18.2
B: 3.47
d) 1.774
Question a)
Alternatively (and much easier), we can just calculate the area under the blue curve:
I=200∗10−9∗1.5∗1092+1000∗10−9∗1.5∗1092=900W/m2
a) Calculate the irradiation of the solar simulator : 900
b) What is the photon flux of the solar simulator : 3.47
A. 937.5
B. 3.47
C. A (Junction A has the highest band gap of the two junctions. Since the more energetic photons have smaller penetration depth in materials, the high band gap junction should always act as the top cell.
Similarly, less energetic photons (near infrared light) have larger penetration depths. Therefore the bottom cell should be the cell with the lowest band gap. The bottom cell will then collect the photons that have not been collected by the top cell since their energy was not larger than the band gap of cell A.)
D. 1.774
E. 14.5 (you have to use the highest wavelength of junction A since this will give the lowest energy needed to excite electrons (the band gap energy).)
F. 18.2
I=200∗10−9∗1.5∗1092+1000∗10−9∗1.5∗1092=900W/m2
C: A
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Can any one help out?
I'm stuck with this question