Problem 3.11#
Fundamentals of Solar Cells and Photovoltaic Systems Engineering
Solutions Manual - Chapter 3
Problem 3.11
Assume a perfect blackbody at temperature T = 5780 K. We want to examine the theoretical limits of a solar cell. We will assume no losses by reflection at the surface. The solar cell is made of silicon (\(E_g\) = 1.12 eV).
(a) What percentage of the power of the blackbody radiation is associated with photons of less than 1.12 eV?
(b) What would be the efficiency of an imaginary silicon solar cell that could convert all the absorbed energy into electricity?
(c) If the solar cell was made of germanium instead (\(E_g\) = 0.67 eV), would that result in a higher or a lower efficiency?
(d) Discuss the results considering the main loss mechanism in actual solar cells.
We will use the package scipy.integrate to perform a numeric integration and the package matplotlib.pyplot to plot the results
import numpy as np
import matplotlib.pyplot as plt
import scipy.integrate as integrate
(a) What percentage of the power of the blackbody radiation is associated with photons of less than 1.12 eV?
We define the relevant constants.
h=6.63*10**(-34) # [J·s] Planck constant
c =299792458 # [m/s] Light speed
k = 1.380649 *10**(-23) #[J/K] Boltzmann constant
sigma = 5.6703744*10**(-8) # [W/(m²K⁴)] Stefan-Boltzmann constant
The blackbody temperature is 6000 K
T = 5780 # [K]
The total power emitted by a blackbody is obtained from the Stefan-Boltzmann law (Eq. 2.4 in Chapter 2).
\(P=\sigma · T^4\)
P_total = sigma * T**4 # [W/m²]
P_total
63288250.26248946
To calculate the fraction of power P_gap emitted with photons of less than than a given energy, we integrate the equation for the blackbody spectral radiant emittance (Eq. 2.2 in Chapter 2).
\(M_\lambda = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{exp(\frac{hc}{k_B\lambda T})-1}\)
First we convert the wavelength into energy
lambda_gap = 1240/1.12 # [nm]
We can now do the integration
constant = 2*np.pi*h*c**2
P_gap = integrate.quad(lambda x: (constant/(x**5))*(1/(np.exp(h*c/(k*x*T))-1)), lambda_gap*10**(-9), np.inf)
P_gap[0] # [W/m²]
14601459.733338425
The percentage of the power of the black-body radiation associated with photons of less than 1.12. eV is
perc = P_gap[0]/P_total*100
print ('%.1f' % perc + '%')
23.1%
(b) What would be the efficiency of an imaginary silicon solar cell that could convert all the absorbed energy into electricity?
The efficiency would be equal to the percentage of power that is absorbed, which corresponds to the fraction of photons with energy higher than 1.12 eV: 76.9%.
(c) If the solar cell was made of germanium instead (\(E_g = 0.67\) eV), would that result in a higher or a lower efficiency?
It would result in a higher efficiency, as more light would be absorbed.
(d) Discuss the results considering the main loss mechanism in actual solar cells.
The previous efficiency calculation only considered the transmission losses of solar cells, which affect the production of photocurrent. When the thermalization losses, which affect the voltage, are also included, a silicon solar cell has a greater efficieny limit than a germanium solar cell.