Theoretical efficiency of a solar cell

Electricity and Magnetism Level 4

A solar cell consists of a semiconductor, which has no free conductor carriers without external excitations. However, the incident light excites electrons from the conduction band into the valence band, so that these are now available as conduction carriers and allows electric currents. For such excitation to occur, the photons must have an energy E = h ν E = h \nu , that is at least as large as the bandgap E g E_g of the semiconductor. However, the electrons immediately lose a part of their excitation energy through collisions, so that each excited electron stores only the energy E g E_g .

The energy yield of a solar cell depends strongly on the band gap E g E_g of the semiconductor used. If the band gap is too large, only a small part of the radiation is absorbed. If the band gap is too small, the electrons can only store a small amount of energy in each case. Therefore, there is an optimal band gap for the sunlight, for which the energy turnover is greatest.

What is the maximum theoretical efficiency η \eta of a solar cell for the optimal semiconductor? The efficiency is the ratio η = P electrons P radiation \eta = \frac{P_ \text {electrons} }{ P_ \text {radiation}} of usable power of the electrons P electrons P_ \text {electrons} and the power of the incident solar radiation P radiation P_ \text {radiation} .

Assumptions: The spectral power density of solar radiation can be approximated by Wien's radiation law I ( ν ) d ν = h ν f ( ν ) d ν ≈ 2 h ν 3 c 2 e − h ν / k T d ν I (\nu) d \nu = h \nu f (\nu) d \nu \approx \frac {2 h \nu ^ 3} {c ^ 2} e ^ {- h \nu / k T} d \nu with the surface temperature of the sun, T T , the Planck constant, h h , the Boltzmann constant, k k , and the speed of light c c . The density function f ( ν ) d ν f (\nu) d \nu is the relative number of photons within the frequency interval [ ν , ν + d ν ] [\nu, \nu + d \nu] . The power of the incident sunlight on the solar cell with surface area A A then gives P radiation = A ⋅ ∫ 0 ∞ I ( ν ) d ν = A ⋅ ∫ 0 ∞ h ν f ( ν ) d ν P_ \text {radiation} = A \cdot \int_0^\infty I (\nu) d \nu = A \cdot \int_0^\infty h \nu f (\nu) d \nu

η max ≈ 25 % \eta_\text{max} \approx 25\,\% η max ≈ 46 % \eta_\text{max} \approx 46\,\% η max ≈ 32 % \eta_\text{max} \approx 32\,\% η max ≈ 85 % \eta_\text{max} \approx 85\,\%

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
May 15, 2018

The total solar radiation corresponds to a power P radiation = A ⋅ ∫ 0 ∞ h ν f ( ν ) d ν P_\text{radiation} = A \cdot \int_0^\infty h \nu f(\nu) d\nu with the area A A of the solar cell. All photons with an enery h ν ≥ E g h \nu \geq E_g provide an useable electrical energy E g E_g . Therefore, the electrical power yields P electrons = A ⋅ ∫ E g / h ∞ E g f ( ν ) d ν P_\text{electrons} = A \cdot \int_{E_g/h}^\infty E_g f(\nu) d\nu The energy efficiency results to η = P electrons P radiation = ∫ E g / h ∞ E g f ( ν ) d ν ∫ 0 ∞ h ν f ( ν ) d ν = ν g ∫ ν g ∞ ν 2 e − α ν d ν ∫ 0 ∞ ν 3 e − α ν d ν = ν g ∫ ν g ∞ ( ∂ 2 ∂ α 2 e − α ν ) d ν ∫ 0 ∞ ( − ∂ 3 ∂ α 3 e − α ν ) d ν = ν g ∂ 2 ∂ α 2 ( ∫ ν g ∞ e − α ν d ν ) − ∂ 3 ∂ α 3 ( ∫ 0 ∞ e − α ν d ν ) = ν g ∂ 2 ∂ α 2 ( 1 α e − α ν g ) − ∂ 3 ∂ α 3 ( 1 α ) = ν g ( 2 α 3 + 2 ν g α 2 + ν g 2 α ) e − α ν g 6 α 4 = 1 6 ( 2 x + 2 x 2 + x 3 ) e − x \begin{aligned} \eta &= \frac{P_\text{electrons}}{P_\text{radiation}} \\ &= \frac{\int_{E_g/h}^\infty E_g f(\nu) d\nu} {\int_0^\infty h \nu f(\nu) d\nu}\\ &= \frac{\nu_g \int_{\nu_g}^\infty \nu^2 e^{- \alpha \nu} d\nu}{ \int_0^\infty \nu^3 e^{- \alpha \nu} d\nu} \\ &= \frac{\nu_g \int_{\nu_g}^\infty \left(\frac{\partial^2}{\partial \alpha^2} e^{- \alpha \nu} \right)d\nu}{ \int_0^\infty \left(- \frac{\partial^3}{\partial \alpha^3} e^{- \alpha \nu} \right) d\nu} \\ &= \frac{\nu_g \frac{\partial^2}{\partial \alpha^2} \left( \int_{\nu_g}^\infty e^{- \alpha \nu} d\nu \right)}{ - \frac{\partial^3}{\partial \alpha^3} \left(\int_0^\infty e^{- \alpha \nu} d\nu \right)} \\ &= \frac{\nu_g \frac{\partial^2}{\partial \alpha^2} \left(\frac{1}{\alpha} e^{-\alpha \nu_g}\right) }{ - \frac{\partial^3}{\partial \alpha^3} \left(\frac{1}{\alpha}\right) } \\ &= \frac{\nu_g \left(\frac{2}{\alpha^3} + 2 \frac{\nu_g}{\alpha^2} + \frac{\nu_g^2}{\alpha} \right) e^{-\alpha \nu_g}}{\frac{6}{\alpha^4}} \\ &= \frac{1}{6} (2 x + 2 x^2 + x^3 ) e^{-x} \end{aligned} Here, we used the abbreviationen α = h / k T \alpha = h/k T , ν g = E g / h \nu_g = E_g/h and x = α ν g = E g / k T x = \alpha \nu_g = E_g/k T . The maximum of the function η ( x ) \eta(x) can be solved graphically and results to ( x max , η max ) ≈ ( 2.27 , 0.457 ) (x_\text{max}, \eta_\text{max}) \approx (2.27, 0.457) . Therefore, almost 46 % 46\% of the incident sunlight can be converted into electrical power, if there are no further losses.

However, the real solar spectrum differs from the blackbody radiation, as the atmosphere absorbs part of the solar radiation. Performing the same calculation with the real solar spectrum yields a theoretical efficiency of 41%. By combining several semiconductors with different band gaps, however, even higher efficiency ratios can be achieved

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