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Blackbody and absorption problem

  1. Feb 5, 2013 #1
    Well, this question may seem elementary to you but I simply do not have a background for it. Now the question:

    Sun radiation resembles a blackbody at over 5000°K. A solar cell's temperature is around 300°K. If we neglect loss in the cell, its emission and absorption coefficient are equal. Therefore, the cell's absorption is limited by the temperature of the cell and it should be very low because there is a huge gap between the radiation of a blackbody at 5000°K and 300°K. How can we understand the high efficiency of solar cells then?

    By the way, I have a second question: What are the main approximations in derivation of Planck's law of radiation?
  2. jcsd
  3. Feb 5, 2013 #2


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    Why is the cell's absorption limited by the lower temperature?
  4. Feb 5, 2013 #3
    Because it cannot emit/absorb more than a blackbody at 300°K (temperature of the cell).
  5. Feb 6, 2013 #4

    Andy Resnick

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    How about this- a 1m^2 solar cell (if it is a perfect absorber) will absorb (given ideal weather conditions) 1kW of power. Does this energy dissipate, heating up the solar cell, or it is instead converted into some other form?

    Say you had a 1m^2 piece of black paper- a perfect absorber. Will it remain at 300K if you leave it in sunlight? What is the final equilibrium temperature of the paper? The paper, at 300K, radiates about 450 W of power. Using conservation of energy, the final radiated power must be ~1500W, corresponding to a final equilibrium temperature of 400K.

    The solar cell can also be treated using equilibrium concepts; the solar cell is a heat engine (it extracts work from heat) and so the maximum efficiency is given by the usual formula 1- Tc/Th, where Tc is the cold side (the temperature of the solar cell) and Th the hot side (temperature of the sun). The maximum efficiency is when all sunlight is converted into electrical power, or 94% efficiency. Alternatively, we could specify a 'noise temperature' that takes into account all the various inefficiencies of the electronics and use that to modify Tc.
  6. Feb 7, 2013 #5
    The average temperature of the sky with the sun on the sky, let's say it's 5000°K, sun is black, other parts of the sky are white.

    The temperature of the surface of the Earth is maybe 300°K, the color is quite black.

    One side of the solar cell is facing the sky, that has temperature 5000°K, while being quite white on the average.

    Other side of the solar cell is facing the Earth, that has temperature 300°K, while being quite black.

    Well doesn't that sound like a reasonable description about the thermodynamical situation of the solar cell?
  7. Feb 7, 2013 #6

    Andy Resnick

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    This makes no sense whatsoever.
  8. Feb 8, 2013 #7

    Well, let's cover the sun facing side of the solar cell with a lampshade, that kind of lampshade that diffuses the sunlight.

    Now the solar cell will report that: "There's a large and quite white 5000°K object on one side of me, and a large and quite black 300°K object on the other side of me"

    That's what I meant.
  9. Mar 18, 2013 #8
    First of all, thanks for you replies.

    As Andy suggested, maximum efficiency can be found by Carnot's theorem: 1-Tc/Th where Tc is the cell temperature and Th is sun temperature. Shouldn't Tc be temperature of earth? This makes little change in the efficiency but if someone wants the efficiency limit imposed by Carnot's theorem, Tc should be the temperature of a thermal reservoir which can be earth for example but not the cell because cell temperature can easily change. Do you agree with me?

    I have also another question: To convert light into electric energy in a solar cell, one can couple external light to the guided modes of the cell. Do the corresponding coupling coefficients remain unchanged, assuming that material properties and the cell geometry remain constant? I guess yes, but I ask to be sure.

    Thanks in advance for responding :)
  10. Mar 18, 2013 #9
    I'm with Andy, I'm a little lost on that. Here's how I see it:

    A solar cell absorbs light from the sun (which is at about 5800 K). The cell, working like a heat engine, rejects its heat to the surroundings around 300 K. So, Carnot efficiency in this situation, η=1-Tc/Th

    However, there are some things that change this. Out in space, you could potentially argue that the cell is rejecting to space, at 2.7 K (even higher efficiency possible). Also out in space, the earth's atmosphere doesn't attenuate the 1.3 kW/m2 of solar flux coming in.

    However, as Andy was maybe alluding to, the hotter the solar cell, the less conductive the material (typically), and the lower the conversion efficiency. So, depending on how we draw the system, realistically, the heat rejection is taking place at the temperature of the back side of the solar cell (at least that's the way I think of it).

    In addition, once atmosphere is added to the situation, things get weirder. The solar flux is attenuated, and you'll see fluxes significantly reduced depending on how much of the atmosphere the light has to pass through. Some of the highest energy photons (blue, violet, etc) are scattered as they enter the atmosphere, but can still possibly make it to the solar cell, and the very high energy photons (gamma rays, x-rays) are basically absorbed in the atmosphere (becoming cosmic rays), further reducing the possible efficiency of the solar cell on the ground. So, the spectra on the ground is not the same as what the sun emits, and we can't truly assume a source at 5800 K if we account for those losses.

    As for the coupling, I'm not familiar with that. I'll look it up and if I understand what you're talking about, I'll post again.
  11. Mar 19, 2013 #10
    Yes, depending on where the cell is, i.e. earth or space, Tc can change.

    About the role of the atmosphere, there is some loss because of absorption of light in it but solar spectral irradiance data suggests that solar illumination represents more or less a black body at a temperature close to 5000°K. The spectral data is here: http://rredc.nrel.gov/solar/spectra/am1.5/

    The coupling efficient problem can be simplified this way: in everyday optics, we solve Maxwell's equations by considering only geometry of the problem (shape of objects to be simulated for example) and the corresponding material data e.g. refractive indices, etc. There, the effect of temperature seems absent maybe because it is negligible. What if we want to simulate the optical system e.g. a solar cell at hot temperatures? I think then we should not change Maxwell equations but we only need to modify the intensity of incident light according to temperature because the absorption efficiency seems to be relative based on Carnot's theorem. Then the question is: How to scale the incident light intensity based on temperature?

    Another related question: Do materials emit light at the same wavelength that they absorb it? I guess the answer is no. In this case, maybe we need to take into account the spectra of both absorption and emission in the object to be able to scale the incident intensity correctly...
  12. Mar 19, 2013 #11
    photovoltaics are not heat engines. They do not work by converting thermal energy into electrical energy. They turn optical energy into electrical energy. As to how the optical energy is generated, that is irrelevant for what a photovoltaic is.

    Also note that not all EM emissions are thermal. Blackbodies emitting most efficiently applies to THERMAL EM emissions. There are numerous examples of non-thermal spectra. A laser, for example, displays non-thermal spectra. You can have emission efficiencies far superior to blackbodies in non-thermal sources.

    Why are you trying to use a theorem for heat engines in a situation that is clearly not a heat engine?
  13. Mar 19, 2013 #12
    Actually, thermoelectric generators and photovoltaic systems are frequently treated as heat engines. All a heat engine requires is heat transfer from a high temperature reservoir, and heat transfer to a low temperature reservoir. There are three modes of heat transfer:


    Heat transfer, by definition, is just the transport of energy. Whether the source is a "thermal" one (which I would argue the sun is anyway), or a laser (spectrally very selective), in the end, both are energy being transported via EM radiation (and therefore both are heat transfer in the strictest sense).

    The only difference between an "optical" EM source, as you put it, and a "thermal" EM source, as you put it, are the wavelengths the radiation occupies. BOTH are the transport of energy via electromagnetic waves. And, in the case of the sun, the sun is hot enough to have a Plank distribution that occupies what you refer to as "optical" EM wavelengths.

    Below is a reference to a work that treats PV's in just that manner.

    Markvart, Tom (2008) Solar cell as a heat engine: energy-entropy analysis of photovoltaic conversion.Physica Status Solidi (A), 205, (12), 2752-2756.
    Last edited: Mar 19, 2013
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