For the blackbody Sun ( )

PHYSICS · Multi-part · Fermi University Physics · Earth, Climate & Energy Physics · Energy resources and conversion · Energy resources and conversion

Energy from Physics, Ch The illuminated junction: generation, collection and the cell's characteristic

Problem

For the blackbody Sun (T⊙=5772 KT_\odot = 5772\,\mathrm{K}), the photon flux above a threshold energy EgE_g falls in the Wien tail of the Planck spectrum once xg≡Eg/(kBT⊙)≳2x_g \equiv E_g/(k_BT_\odot) \gtrsim 2, where it is well approximated by Φ(Eg)∝e−xg(xg2+2xg+2)\Phi(E_g)\propto e^{-x_g}(x_g^2+2x_g+2). Using this scaling, a silicon cell's photon ceiling is known to be Jceiling=eΦ(Eg)=53 mA cm−2J_{\rm ceiling}=e\Phi(E_g)=53\,\mathrm{mA\,cm^{-2}} for Eg=1.12 eVE_g=1.12\,\mathrm{eV}. ((a)) Using the Wien-tail scaling above, estimate the photon ceiling eΦ(Eg)e\Phi(E_g) for a GaAs cell, Eg=1.42 eVE_g=1.42\,\mathrm{eV}, by scaling from the silicon value. ((b)) A record single-junction GaAs cell has a measured short-circuit current density of about 29.6 mA cm−229.6\,\mathrm{mA\,cm^{-2}} at one sun under the standard AM1.5G spectrum. What fraction of the blackbody photon ceiling from part (a) does it reach, and how does this compare with silicon's 41/53=77%41/53 = 77\%, found the same way? Both fractions set a ground-level AM1.5G current against a blackbody count outside the atmosphere, so they compare the two cells with each other, not with their own AM1.5G ceilings.

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