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If you consider a black body as energy source and photoelectric conversion at room temperature as the energy receiver.

Wouldn't it be possible to:

  • use an ideal prism/diffraction grating to separate the incident light into its spectrum and
  • use photoelectric conversion with 100% quantum efficiency at each particular wavelength ?

At which point does the carnot efficiency limit the conversion efficiency, if at all ?

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Prism or diffraction grating does not separate the frequencies into parallel beams of rays falling on the same area as before. It increases beam divergence, which increases the area needed to capture the light. This new area could have been used to capture different light before, so it actually hurts energy capture efficiency (useful energy captured per unit area).

It is better to work with stacked cells where each layer captures different part of light and let the other part to pass through.

Quantum efficiency is the number of electrons released per radiation energy $hf$, but these electrons need not have the energy $hf$ (and they usually don't). So quantum efficiency can be even higher than 100% without great energy efficiency.

What you meant is probably that the cell has energy conversion efficiency, or thermodynamic efficiency of 100%. But Carnot efficiency for Sun-Earth pair is around 95%. If the cell or cell stack efficiency was higher than the Carnot limit, this would violate 2nd law of thermodynamics and lots of "impossible things" would suddenly be possible.

The Carnot efficiency is the maximum possible efficiency of a periodically working engine between two reservoirs that does not violate 2nd law of thermodynamics.

Photovoltaic conversion isn't exactly a periodically working engine, because nothing moves macroscopically there, but it is similar in that it does not contribute its own energy to output in the long run, just as periodically working engines don't. So the Carnot limit should apply.

It would be very surprising if photovoltaic cell or cell stack could beat the Carnot efficiency, because this would then mean that 2nd law is violated, and more work can be extracted from radiation source than a reversible heat engine is able to.

This would open a can of worms, as it would be possible to use this energy to power additional Carnot engine in reverse mode to extract heat from Earth and heat up the Sun. This pair of engine would thus remove heat from the Earth, add it to the Sun, without any external power source. It would be the forbidden "heat flows from a colder body to a hotter body" process.

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  • $\begingroup$ The cell stacking is indeed a better implementation but I went for for the simpler principle of splitting the rays spatially in the question. So does it mean that photoelectric efficiency cannot be 100 % at any wavelength ? I do agree btw that Thermodynamics hold, but I am looking for where exactly the limit manifests in the above scenario. $\endgroup$
    – tobalt
    May 7 at 16:29
  • $\begingroup$ Yes. 100% energy efficiency does not happen. Theoretically it could happen if the light was perfectly predictable like a sine wave, but real light isn't. $\endgroup$ May 7 at 16:53
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    $\begingroup$ But monochromatic light is a sine wave right ? For the sake of the thought experiment, one could make the precision of the grating arbitrarily high $\endgroup$
    – tobalt
    May 7 at 17:10
  • $\begingroup$ Diffraction grating does not produce monochromatic light in the mathematical sense. Real light has finite temporal coherency time, so even the light resolved by the grating will manifest chaotic behaviour. $\endgroup$ May 7 at 18:39
  • $\begingroup$ "This new area could have been used to capture different light before". This isn't necessarily the case. In some cases it can increase depth as opposed to frontal area. That being said, cell stacking in practice tends to be better. $\endgroup$
    – TLW
    May 7 at 19:41
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100% quantum efficiency does not translate to 100% efficiency. It simply means 'one electron out per photon in'.

A standard single-junction cell has a maximum thermodynamic efficiency of ~33% at 100% quantum efficiency and a bandgap of ~1.34eV. (The maximum for single-junction silicon cells is slightly worse, at ~32% and 1.1eV.)

Why the difference? Solar radiation is a spectrum. If you have a solar cell with a bandgap of 1.34eV, an incoming photon with less energy will be 'wasted' (won't produce useful output energy), and an incoming photon with more energy... will still only produce 1.34eV of useful energy. So there's a tradeoff - higher bandgaps mean each absorbed photon produces more useful energy, but more photons are 'wasted' and don't produce any useful energy. 1.34eV is the optimum here.


Well, what about something like your design? Something that splits into an (near)infinite series of different cells optimized for specific wavelengths.

Let's simplify for a moment. Instead of a giant splitter, let's just imagine sunlight aimed through a material that passes a narrow window at 500nm and reflects anything else, at a 2.5eV bandgap solar cell. (Assume that the solar cell is illuminated like this from all directions.)

Well... as it turns out that's still not a free lunch. See, we've been talking about solar cells as an irreversible process. Photons come in, useful energy comes out. But this is actually a reversible process. It's actually more like photons kick electrons, forming electron-hole pairs. These pairs then wander around, potentially dissipating energy (in the form of heat) as they cross the PN junction and either recombine (forming more photons), or escape the solar cell (producing useful energy).

You can view this system as having a sort of 'effective' temperature. A high temperature corresponding to many electrons and holes wandering about, and a low temperature corresponding to few. Short-circuit operation corresponds to ambient temperature, and open-circuit operation corresponds to the temperature of the sun (or whichever input source).

Interestingly, when you work out the overall efficiency, comparing energy flows and useful output work, you get the same as a Carnot engine working between said temperature and a cold side consisting of the solar cell.

...which in turn means that your maximum possible efficiency is at 'nearly' open-circuit operation, and at (near) Carnot efficiency between the sun and solar cell ambient temperature.

And of course, splitting the input into an series of engines each operating at no more than Carnot efficiency, itself operates at no more than Carnot efficiency.

(Note that this is not the maximum output power point. Near-open-circuit operation results in near-zero output power. The maximum power point for said stack of monochromatic cells is at ~87% efficiency for 'standard' conditions.)

For more details and the backing calculations, see "Solar cell as a quantum converter" in Theoretical Calculation of the Efficiency Limit for Solar Cells here.

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  • $\begingroup$ Thanks for this answer and the paper. So it means, fundamentally, the losses due to premature exciton recombination are temperature dependent and cannot be engineered away?! Interesting. I don't quite understand yet how the source temperature comes to matter though, as each receiver only receives monochromatic radiation and cannot know the source temperature 🤔 ( I haven't yet read the paper) $\endgroup$
    – tobalt
    May 8 at 5:42
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    $\begingroup$ @tobalt Even monochromatic radiation has an effective temperature, equal to the temperature that would produce its intensity given that it's a slice of a Planck spectrum. When we transmit power using monochromatic radiation (like the 50/60 Hz guided from generator to you by a transmission line), the intensity we use is so enormous that we may ignore its tiny entropy. The electric power to your house has an intensity of >10^24 times the intensity of the thermal energy in the line. $\endgroup$
    – John Doty
    May 9 at 11:59
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    $\begingroup$ @tobalt But you can't do that. The mathematical idea that captures this is Liouville's Theorem. See crossfield.ku.edu/8901_2019A/readings/… (and this reminds me that I should have used "specific intensity" above). It is unfortunately rather abstract. The way I learned to appreciate it as physics was photography in the days of manual exposure selection. While the intensity of light from a source varies as 1/r^2, the specific intensity doesn't change. Thus, you don't adjust your camera's exposure setting for distance. $\endgroup$
    – John Doty
    May 11 at 12:33
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    $\begingroup$ @tobalt - also see conservation of etendue. Focusing radiation doesn't quite do what you seem to think it does. There's a decent discussion here: physics.stackexchange.com/questions/338865/… $\endgroup$
    – TLW
    May 12 at 0:19
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    $\begingroup$ @JohnDoty and TLW thanks for these further links. Indeed, these concepts are rather abstract and I must do more reading to fully understand them. But it is interesting, in how various ways thermodynamics makes its subtle appearence in physics :) $\endgroup$
    – tobalt
    May 12 at 6:30
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The carnot limit only applies to heat engines which use thermal energy flows to produce mechanical work. So it does not apply to electrical power generation using moving fields to drive the flow of current, nor to things like water turbines that extract mechanical work from falling water. Other limits will apply in those cases.

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  • $\begingroup$ Erm.. Ok so suppose I have a big black sphere in vacuum and I burn wood inside of it. All of the thermal energy will be radiated away and eventually one comes back to my original question, no ? $\endgroup$
    – tobalt
    May 7 at 16:26
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    $\begingroup$ That radiation would transfer both energy and entropy (all heat transfer does). But (1) work doesn’t carry entropy, and (2) entropy can’t be destroyed. This puts you in a jam when trying to get work out of this scenario, and this is the essence of the Carnot limit. Lenses and prisms don’t change that. In contrast, a waterwheel that runs by turning kinetic energy (rather than heat) into work doesn’t have this limitation. $\endgroup$ May 7 at 17:47
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    $\begingroup$ But the question is not about moving fields / water driving current/turbine, it's about efficiency of photovoltaic energy generation from blackbody radiation, which is a heat engine - in the sense that it takes energy from high temperature reservoir and rejects lower amount of energy into the lower temperature reservoir. If some engine was better than Carnot with this pair of reservoirs, it would be possible to set up a violation of 2nd law. $\endgroup$ May 9 at 12:29
  • $\begingroup$ @JánLalinský Well, you can analyze the water turbine as a heat engine, but it's not a particularly illuminating point of view. The bulk flow of the water represents a very large energy attached to a very small number of degrees of freedom, so the effective temperature is enormous. $\endgroup$
    – John Doty
    May 11 at 17:04
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In theory, the hottest you can make an object by focusing solar energy on it is the photospheric temperature of the sun, ~6000K. Any hotter, and it would heat the Sun! The coldest theoretical radiator on Earth would reflect radiation at all wavelengths except those where the atmosphere is so transparent that the effective temperature is that of the cosmic background, ~3K, thus cooling to that temperature. So, the theoretical thermodynamic efficiency limit for a solar engine is (6000-3)/6000, or 99.95%.

Real solar cells are, of course, nowhere near that efficient. We don't usually think of them as heat engines, even though their temperature is an important factor in their practical efficiency.

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  • $\begingroup$ solideas.com/papers/GPS_ThermCon_SEMSC.pdf $\endgroup$
    – UVphoton
    May 7 at 22:48
  • $\begingroup$ That is all good but I don't see the connection to this question. $\endgroup$
    – tobalt
    May 8 at 4:54
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    $\begingroup$ @tobalt Well, the point is that the Carnot efficiency limit applies, theoretically, but the limit is so close to 100% that it has no impact on the actual efficiency of a solar cell. $\endgroup$
    – John Doty
    May 8 at 12:27

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