Upconverting Sunlight Boosts Solar Power

Had a quick look there. I don't think people will be too inclined to add much to jack_of_all_trades well-thought-out comments.

These lanterns are useful if you are away from a source of electricity, and you have to rely in what is available (i.e. the sun) for recharging.

If you have one of these and there isn't a mains-driven battery-charger that is safe to use with the lantern, it might indeed be convenient to use an LED or other electrically-driven light source to recharge it; but even the best LEDs convert more than 50% of input power to heat, and the conversion of mains power to drive the LED is also lossy; plus the solar cell system will have losses and cannot be made very well-matched for battery-recharging. So it will be much more efficient to use a conventional charger if one is available. Even without a conventional battery charger, you 'd still probably do better placing the lantern outside or near a suitable window (unless you find yourself with no other source of portable illumination in the Arctic winter - in which case I would recommend buying a conventional rechargeable lantern rather than LED lamps to recharge this one)

Hi! Physicist, your post contained sound common sense. It is true that LEDs do lose a lot of energy as heat, it is also true that photovoltaic cells are not very efficient. But if you think about how photovoltaic cells work, using LEDs to irradiate them begins to make sense. All PV cells work on the photoelectric principle, that is they depend on the frequency of light they are irradiated with in order to develop a current. In the external photoelectric effect, the higher the frequency or more energetic the photon, the greater the photoelectric effect. So ultraviolet and above works very well, while red has no effect at all ( of course this depends on the material being irradiated). What this means is that even if you have a 1000W red light it would produce no current at all! While irradiation by even a 10V ultraviolet lamp would result in a photoelectric current being present. In the external photoelectric current, the electrons gain enough energy from the incoming photons to leave the material altogether.

PV cells don't work on the same principle, they work on something called the internal photoelectric effect, wherein the electrons have to be irradiated by photons that impart enough energy to the electrons to leave the parent atom and enter the conduction band.(i.e., not leave the material altogether.) It is generally accepted that any photon possessing more than the band gap energy of the photovoltaic material, will be enough to set up a photoelectric current. My experiments show that this is not true. When a silicon solar cell (band gap 1.2eV) is irradiated by red LEDS of about 650nm wavelength, both the voltage and the current produced is much less(negligible in fact)than when the same solar cell is irradiated by blue light. So in fact to work efficiently a silicon photovoltaic cell has to be irradiated with photons possessing at least 2.2eV, that is light in the blue wavelength of about 500nm. My theory is that when a red photon is absorbed by a valence electron, it does not impart enough energy to it to enter the conduction band so it is immediately recaptured. However this action is sufficient to provide enough heat (rather like a micro-wave) for a current to flow.

Silicon is a semi-conductor and mere irradiation with blue light is not enough for current to flow. While irradiation with blue light produces satisfactory charge separation, (i.e voltage) it does not result in a current flow. Current flow in silicon depends on the intrinsic property of silicon to conduct current when it is heated. Therefore in addition to irradiation by blue light the cell also has to be irradiated by infrared LEDs. When this happens both current and voltage in the solar cell are at satisfactory levels and it might even be possible, given the right circumstances to actually produce a slightly higher power from the solar cell than is originally put in to power the LEDs to irradiate it. This is the theory behind the CASER effect solar cell. (Current amplification through stimulated emission of radiation.)

I'm not suggesting you cannot charge your battery in this way - just that it cannot be as efficient as a well-designed electrical battery charger. The basic reason is that the requirements for driving the LED are either very similar to or more onerous than those for recharging the battery. So you would need more than 100% efficiency in the chain from the LED to the battery to gain any advantage; that is tantamount to this chain being a perpetual motion machine.

LEDs may be easier to match to the P-V cell than other light sources, but the fundamental limitations of LEDs and of photocells remain. In fact, 100% power conversion efficiency is not even theoretically possible, either for the LED or for the photocell. I'll only consider the basic limitations photocell here, however. The problem is that approaching the 100% limit would require all the following conditions (some of which are ultimately incompatible) to be met:

a) The photosensor is a direct gap semiconductor (that excludes all present silicon-based photocells - I think silicon becomes direct-gap when the thickness is somewhere in the sub-micron region, but then you start so see other issues).
. Reason: you cannot excite photo-electron pairs in an indirect semiconductor without also exciting phonons (=> phonon energy eventually becomes heat)
b) The photon energy of the light source exactly matches the band-gap
. Reason: any excess photon energy will be lost - normally as kinetic energy of the hole-electron pair which is then converted to heat via collisions with the lattice
c) The semiconductor junction is thick enough for all the light to be absorbed.
. Note that if b) was exactly met, the required thickness would would be infinite.
d) The photo-carrier density is sufficient for the output voltage to become equal to the bandgap.
e) The hole-electron pairs do not recombine before reaching the terminals.
. In the limit, d) and e) are incompatible for a number of reasons; the simplest to understand is that d) requires every possible hole-electron pair to have been excited into the conduction band - so the spacing will be in the order of one atomic spacing (i.e. in the Å region), giving huge numbers of recombination opportunities before the carriers reach the terminals of the photocell
f) There must be minimal reflection from the surface
g) The output resistance of the photocell and its wiring is zero.

BTW, many silicon photovoltaic cells are sensitive to light in the 0.65-um region.
You say that yours has negligible response here, but (at least as I interpret them) the reasons you give are not plausible. In my (grossly out-of-date) view, the most likely reasons for the response at 650-nm (1.9-eV) to be negligible could be:
a) The cell could be made from amorphous silicon* (not quite as efficient as crystalline, but slightly higher output voltage and cheaper for larger areas). Amorphous silicon has a wider bandgap than crystalline silicon, so the response in the red is quite small; typically there are also more problems with recombination than with crystalline silicon, which encourages the use of a thin junction - further reducing the response in the red.
. If crystalline silicon is used, the limitation would be the design of the cell. This will partly be matched to the conditions where power generation is expected (sunlight), and in particular the input power density that is expected. That will mitigate towards a shallow junction (the region in which the light is usefully absorbed), which will result in a reduced absorption in the red. Then the silicon needs to be anti-reflection coated to reduce losses due to reflection from the front surface - and the coating will be optimised for the spectral regions where the device is already most efficient; indeed, it is quite possible that there will be so little advantage in allowing the red light through that the manufacturer would not be too bothered if the most economic coating process happened to absorb in the red.
*You can readily tell this if you can get at the output of a single cell when it is well illuminated but not loaded - crystalline silicon will normally give less than 0.58-Volt, amorphous silicon something above 0.6-Volt.

I was hoping for an expert opinion and I am not disappointed. I have been thinking about the problem in this way:

Suppose you have a 6" x 6" solar cell ( amorphous as you had deduced, manufactured by Sanyo Part No: AM-7008) capable of generating 190ma at 7V in bright sunlight. Suppose you now irradiated this solar cell with a 100W incandescent bulb placed about 2.5 cms – 3.00 cms from the surface of the solar cell and you were able to get an out put of say 6.5V and 150ma, or 0.975W.

Now replace the 100W incandescent with an array of 18 white LEDs connected in parallel and placed in an extremely reflective housing ( of the type one finds in LED torches) in such a manner that all light incident on the solar cell is reflected back to it, no light escapes. Suppose that this array consumes about 10ma per LED or 180ma at 3V (i.e., 0.54W). The solar cell is also irradiated by another array of 14 long infra-red LEDs ( of the type found in infrared massage lamps), similarly placed in the reflective housing and consuming 12ma each at 2.5V or 0.42W, so the total cost of illuminating the solar panel by the two LED arrays is 0.96W. Suppose (there are a lot of suppositions but anyway just suppose) that the output from the solar cell is the same as that when irradiated by the 100W incandescent bulb or 0.975W . That would mean in effect that we have an overall gain in power of 0.975W - 0.96W = 0.015W. Fantastic.

Think about it, is it possible. Maybe! A 5mW green LED diode laser can project a beam of light that is visible at a distance of 2 miles . It works by reflecting the light between two mirrors so that more and more electrons are excited at the same frequency before releasing them.(one of the mirrors is semi-transparent). The system I have described might work in a similar way, allowing no light to escape and reflecting it back onto the surface of the solar cell, meaning more and more light of the correct frequency is available for the cekll to work with. Just a thought.

To make this system work even better, there should be no anti-reflective coating on the solar cell.

Unfortunately, the essential suppositions actually break the only law of physics that causes the US patent office to automatically reject an application. (Maybe there's a genuine reason for this...)