The figure below presents the EQE of a triple junction solar cell with junctions A, B and C under short circuited (V = 0 V) condition.
EQE table is:
A: EQE = 0.7 wavelength= 300 to 650 nm
B: EQE = 0.9 Wavelength= 650 to 850 nm
C: EQE = 0.8 Wavelength= 850 to 1250 nm
a) What is the bandgap (in eV) of the absorber layer of the junction A?
b) What is the bandgap (in eV) of the absorber layer of the junction B?
c) What is the bandgap (in eV) of the absorber layer of the junction C?
d) Which of the following statements is TRUE?
Junction C acts as the top cell, Junction B as the middle cell, and Junction A as the bottom cell.
Junction B acts as the top cell, Junction C as the middle cell, and Junction A as the bottom cell.
Junction A acts as the top cell, Junction B as the middle cell, and Junction C as the bottom cell
TRIPLE JUNCTION SOLAR CELL - III
Each junction is illuminated under standard test conditions. Given the photon fluxes below, calculate the short-circuit current density (in mA/cm2) of each (separate) junction (A, B and C):
ϕ=9.3∗1020m−2s−1 for 300nm<λ<650nm
ϕ=8.4∗1020m−2s−1 for 650nm<λ<850nm
ϕ=1.4∗1021m−2s−1 for 850nm<λ<1250nm
e) Jsc of Junction A:
f) Jsc of Junction B:
g) Jsc of Junction C:
TRIPLE JUNCTION SOLAR CELL - IV
h) The Voc of each junction in V can be roughly estimated by the equation
Voc=Egap(J)2q=Egap(eV)2
where q is the elementary charge, Egap(J) is the bandgap energy expressed in Joules, and Egap(eV) is the bandgap energy expressed in eV. Assume a fill factor of FF=0.75. What is then the efficiency (in %) of the triple junction solar cell?
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Voc=Egap(J)2q=Egap(eV)2
where q is the elementary charge, Egap(J) is the bandgap energy expressed in Joules, and Egap(eV) is the bandgap energy expressed in eV. Assume a fill factor of FF=0.75. What is then the efficiency (in %) of the triple junction solar cell?
What is the meaning solar cell absorbs photons
shows the mass of the solar spectrum streamlined air AM1.5 radiation at 1000 watts per square meter. Spectrum is divided into three spectral areas:
A for wavelengths [0nm <λ <620nm]
B for wavelengths [620nm <λ <1240nm]
C for wavelengths [1240nm <λ <1860nm]
a) To find the bandgap of the absorber layer of junction A, we need to find the corresponding wavelength where the EQE is maximum. Looking at the given table, the EQE for junction A is highest in the wavelength range of 300 to 650 nm. So, the bandgap energy can be found using the equation:
E = (1240 / λ)
where E is the energy in eV and λ is the wavelength in nm. Since the range is given as 300 to 650 nm, we can use the midpoint of the range.
Taking the midpoint wavelength, λ = (300 + 650) / 2 = 475 nm
Substituting this value in the equation, we get:
E = 1240 / 475 ≈ 2.61 eV
Therefore, the bandgap of the absorber layer of junction A is approximately 2.61 eV.
b) Similarly, for junction B, the EQE is highest in the wavelength range of 650 to 850 nm. Taking the midpoint wavelength, λ = (650 + 850) / 2 = 750 nm
Substituting this value in the equation, we get:
E = 1240 / 750 ≈ 1.65 eV
Therefore, the bandgap of the absorber layer of junction B is approximately 1.65 eV.
c) For junction C, the EQE is highest in the wavelength range of 850 to 1250 nm. Taking the midpoint wavelength, λ = (850 + 1250) / 2 = 1050 nm
Substituting this value in the equation, we get:
E = 1240 / 1050 ≈ 1.18 eV
Therefore, the bandgap of the absorber layer of junction C is approximately 1.18 eV.
d) Looking at the EQE values, we can determine the order of the junctions from highest to lowest EQE. In this case, Junction C has the highest EQE, followed by Junction B, and Junction A has the lowest EQE. Therefore, the correct statement is:
Junction C acts as the top cell, Junction B as the middle cell, and Junction A as the bottom cell.
e) To calculate the short-circuit current density (Jsc) of Junction A, we need to use the photon flux and the wavelength range for Junction A. From the given data, the photon flux for Junction A is ϕ = 9.3 * 10^20 m^-2 s^-1 for 300 nm < λ < 650 nm.
The short-circuit current density (Jsc) can be calculated using the equation:
Jsc = (q * ϕ * EQE) / E
where q is the elementary charge, ϕ is the photon flux, EQE is the external quantum efficiency, and E is the energy of the photon.
Substituting the values into the equation:
Jsc = (1.6 * 10^-19 C * 9.3 * 10^20 m^-2 s^-1 * 0.7) / (2.61 eV * 1.6 * 10^-19 C) ≈ 1.601 mA/cm^2
Therefore, the Jsc of Junction A is approximately 1.601 mA/cm^2.
f) Following a similar calculation process, for Junction B, the photon flux is ϕ = 8.4 * 10^20 m^-2 s^-1 for 650 nm < λ < 850 nm.
Jsc = (1.6 * 10^-19 C * 8.4 * 10^20 m^-2 s^-1 * 0.9) / (1.65 eV * 1.6 * 10^-19 C) ≈ 3.787 mA/cm^2
Therefore, the Jsc of Junction B is approximately 3.787 mA/cm^2.
g) For Junction C, the photon flux is ϕ = 1.4 * 10^21 m^-2 s^-1 for 850 nm < λ < 1250 nm.
Jsc = (1.6 * 10^-19 C * 1.4 * 10^21 m^-2 s^-1 * 0.8) / (1.18 eV * 1.6 * 10^-19 C) ≈ 7.800 mA/cm^2
Therefore, the Jsc of Junction C is approximately 7.800 mA/cm^2.
h) To estimate the Voc of each junction, we can use the equation:
Voc = Egap(J) / (2q) = Egap(eV) / 2
where Egap(J) is the bandgap energy in Joules, Egap(eV) is the bandgap energy in eV, and q is the elementary charge.
Since we already calculated the bandgap energy for each junction in part a), b), and c), we can use those values to calculate the Voc.
For Junction A, Voc = 2.61 eV / 2 ≈ 1.305 V
For Junction B, Voc = 1.65 eV / 2 ≈ 0.825 V
For Junction C, Voc = 1.18 eV / 2 ≈ 0.590 V
Now, to calculate the efficiency of the triple junction solar cell, we can use the formula:
η = (Voc * Jsc) / (FF * Pin)
where η is the efficiency, Voc is the Voc of the triple junction solar cell, Jsc is the Jsc of the triple junction solar cell, FF is the fill factor, and Pin is the incident power.
Assuming a fill factor of FF = 0.75, we need the incident power to calculate the efficiency. However, the incident power is not given in the question.