Phase space considerations for light path lengths in planar, isotropic absorbers

Ian Marius Peters

doi:10.1364/OE.22.00A908

Optics Express
Vol. 22,
Issue S3,
pp. A908-A920
(2014)
•https://doi.org/10.1364/OE.22.00A908

Phase space considerations for light path lengths in planar, isotropic absorbers

Ian Marius Peters

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Ian Marius Peters, "Phase space considerations for light path lengths in planar, isotropic absorbers," Opt. Express 22, A908-A920 (2014)

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- Light Trapping for Photovoltaics
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About this Article
History
- Original Manuscript: February 7, 2014
- Revised Manuscript: March 19, 2014
- Manuscript Accepted: March 20, 2014
- Published: April 14, 2014

Abstract

Fundamental limits for path lengths of light in isotropic absorbers are calculated. The method of calculation is based on accounting for occupied states in optical phase space. Light trapping techniques, such as scattering or diffraction, are represented by the way how the available states are occupied. One finding of the presented investigation is that the path length limit is independent of the light trapping mechanism and only depends on the conditions for light incidence to, and escape from the absorber. A further finding is that the maximum path length is obtained for every light trapping mechanisms which results in a complete filling of the available states in phase space. For stationary solar cells, the Yablonovitch limit of 4dn², with n the refractive index of the absorber, is a very good approximation of this limit.

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Fig. 1 Schematic sketch of the three situations for light incidence and escape investigate in this work.

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Fig. 2 Sketch of Lambertian scattering. The incident light cone (green) is scattered at the scattering surface. Lambertian scattering implies that the available phase space (blue) is completely occupied after a single scattering event. Light that is scattered into the escape cone leaves the system after reaching the front surface again. Note that the hemisphere is only a graphical representation of phase space and does not represent the density of states which is given in Eq. (2).

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Fig. 3 Path length for light in an absorber scattered with Phong characteristics for different Phong exponents p and different refractive indices of the absorber n.

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Fig. 4 Sketch of the 90 degree conical tilt. The incident cone (green) is tilted by 90 degree by the scattering surface and, for geometrical reasons, halved. The available phase space (blue) is not completely filled as the occupied phase space (red) is conserved. b) shows why scattering on a geometrical scatterer results in a “V” shaped path. The red path is the one observed here; the grey paths are its geometrical counterparts. Symmetry requires that red and grey paths are interchangeable; reciprocity requires that the direction of light paths can be reversed. Considering the light intensity in each path it follows, that the solid line is the effective path for the calculation of the path length.

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Fig. 5 Sketch of the 90 degree spherical tilt. The incident light cone (green) is scattered at the scattering surface. Phase space is conserved so that only a fraction of the available phase space (blue) is occupied (red). The spherical tilt results in a ring shaped occupation of available states in phase space.

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Fig. 6 Sketch of the successive filling of phase space concept. a) shows the way how successive filling is treated in the calculation; the path length for non-overlapping rings with equal area on the available phase space is calculated and summed up. When the escape cone is reached, light is exiting the system and the summation stops. The sequence of the summation is of no consequence and the approach can therefore be generalised to all techniques that result in a complete filling of phase space. b) shows a different sketch of how the successive filling works. Blue dots mark scattering events. For light with different angles of incidence α_i, also the scattering angle α_s is different. As phase space is conserved, a unique relation exists between α_i and α_s. As an example the path for one specific angle of incidence is shown.

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Equations (20)

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Ω = \frac{1}{2 π} \int d t Ω_{s u n} (t)

D O S (θ) = \cos θ \sin θ

\frac{N_{a i r}}{N_{a b s}} = \frac{\int_{0}^{arc \sin \frac{1}{n}} \cos θ \sin θ d θ}{\int_{0}^{\frac{π}{2}} \cos θ s i n θ d θ} = \frac{\frac{1}{2} \sin {(arc \sin \frac{1}{n})}^{2}}{\frac{1}{2}} = \frac{1}{n^{2}}

L = 2 d \cdot (\int_{- \frac{π}{2}}^{\frac{π}{2}} d θ \frac{1}{\sin θ} \cos θ \sin θ) \cdot \sum_{k = 0}^{\infty} {(1 - \frac{Ω}{n^{2}})}^{k} = 4 \frac{n^{2}}{Ω} d

L = 4 n^{2} d

L = 4 d \cdot \sum_{k = 0}^{\infty} {(1 - \frac{\sin θ_{s}^{2}}{n^{2}})}^{k} = 4 \frac{n^{2}}{\sin θ_{s}^{2}} d \approx 1.8 10^{5} n^{2} d

ζ = \frac{\int_{0}^{arc \sin \frac{1}{n}} \cos θ^{p} \sin θ d θ}{\int_{0}^{\frac{π}{2}} \cos θ^{p} \sin θ d θ} = 1 - {(1 - \frac{1}{n^{2}})}^{\frac{1 + p}{2}}

L = 4 d \cdot \sum_{k = 0}^{\infty} {({(1 - \frac{1}{n^{2}})}^{\frac{1 + p}{2}})}^{k} = 4 \frac{1}{1 - {(1 - \frac{1}{n^{2}})}^{\frac{1 + p}{2}}} d

S = \frac{Ω_{a i r}}{n^{2}} = \frac{\int_{Ω_{a b s}}^{} \cos θ \sin θ d θ}{\int_{0}^{\frac{π}{2}} \cos θ \sin θ d θ}

L = 2 d \cdot \frac{\int_{Ω_{a b s}}^{} \sin θ d θ}{\int_{0}^{\frac{π}{2}} \cos θ s i n θ d θ}

S = \frac{Ω}{n^{2}} = \frac{\int_{0}^{θ_{m}} \sin θ arc \cos \frac{θ}{θ_{m}} d θ}{\int_{0}^{\frac{π}{2}} \cos θ s i n θ d θ}

L = 2 d \cdot \frac{\int_{0}^{θ_{m}} s i n θ arc \cos \frac{θ}{θ_{m}} d θ}{\int_{0}^{θ_{m}} \cos θ \sin θ d θ} = \frac{4 d θ_{m}}{\sin θ_{m}^{2}}

L = 4 d n \cdot \sqrt{\frac{1}{Ω}} (\sqrt{\frac{1}{π}} + O^{3} [\sqrt{\frac{n^{2}}{Ω}}]) \approx 2.26 \sqrt{\frac{n^{2}}{Ω}}

S = \frac{Ω}{n^{2}} = \frac{\int_{\frac{π}{2} - θ_{d}}^{\frac{π}{2}} \cos θ s i n θ d θ}{\int_{0}^{\frac{π}{2}} \cos θ \sin θ d θ} = \sin θ_{d}^{2}

L = 2 d \cdot \frac{\int_{\frac{π}{2} - θ_{d}}^{\frac{π}{2}} \frac{1}{\cos θ} \cos θ \sin θ d θ}{\int_{\frac{π}{2} - θ_{d}}^{\frac{π}{2}} \cos θ \sin θ d θ} = \frac{2 d \sin θ_{d}}{\frac{1}{2} \sin θ_{d}^{2}} = 4 d \sqrt{\frac{n^{2}}{Ω}}

L = 4 d n = {14 d |}_{n = 3.5}

L = 4 d \frac{n^{2}}{\sin θ_{s}} \approx {851 d n = 2980 d |}_{n = 3.5}

\begin{array}{l} \frac{Ω}{n^{2}} = \frac{\int_{\frac{π}{2} - θ_{d n}}^{\frac{π}{2} - θ_{d n + 1}} \cos θ s i n θ d θ}{\int_{0}^{\frac{π}{2}} \cos θ s i n θ d θ} \\ \to θ_{d n + 1} = \frac{1}{2} arc \cos [\cos 2 θ_{d n} - 2 \frac{Ω}{n^{2}}] \end{array}

\begin{array}{l} L = 2 d \cdot [\frac{\int_{\frac{π}{2} - θ_{d 1}}^{\frac{π}{2}} \sin θ d θ}{\int_{\frac{π}{2} - θ_{d 1}}^{\frac{π}{2}} \cos θ \sin θ d θ} + ... + \frac{\int_{\frac{π}{2} - θ_{d n + 1}}^{\frac{π}{2} - θ_{d n}} \sin θ d θ}{\int_{\frac{π}{2} - θ_{d n + 1}}^{\frac{π}{2} - θ_{d n}} \cos θ \sin θ d θ} + ...] \\ = 8 d \cdot [\frac{\sin θ_{d 1}}{- \cos 2 θ_{d 1}} + ... + \frac{\sin θ_{d n + 1} - \sin θ_{d n}}{- \cos 2 θ_{d n + 1} \cos 2 θ_{d n}} + ...] \\ = 4 d n^{2} Ω [\sin θ_{d 1} + ... + \sin θ_{d n + 1} - \sin θ_{d n} + ...] \\ ​ = 4 d \frac{n^{2}}{Ω} \end{array}

\begin{array}{l} L = 4 d \frac{n^{2}}{Ω} [\sin θ_{d 1} + ... + \sin θ_{d n + 1} - \sin θ_{d n} + ... \sin θ_{c}] \\ = 4 d \frac{n^{2}}{Ω} [\sin π - \sin θ_{c}] \\ ​ = 4 d \frac{n^{2}}{Ω} [1 - \frac{1}{n}] = 4 d \frac{n (n - 1)}{Ω} \end{array}

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Equations (20)

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