I have been investigating recapturing energy from house lighting using solar panels.

Are there solar panels specifically designed to capture indoor lighting with a higher efficiency?

If so, what are the differences between the outdoor and indoor solar panels designed for this purpose?


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    $\begingroup$ It may also depend on the type of lighting you have -- incandescent, fluorescent, halogen, and LED all have different frequency profiles. $\endgroup$ Commented Jan 21, 2015 at 1:30
  • $\begingroup$ @MatthewRead +1. See also spectra here. $\endgroup$ Commented Jan 21, 2015 at 3:31
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    $\begingroup$ The energy density of solar is already extremely small (compared to other energy resources). The collectable energy from your 60W bulbs is going to be orders of magnitude smaller. $\endgroup$
    – Rick
    Commented Jan 21, 2015 at 16:40
  • $\begingroup$ Is this an energy harvesting application? $\endgroup$ Commented Jun 8, 2022 at 16:25

4 Answers 4


I don't think they existed, and it has its reason.

First: a solar panel can be characterized mainly by its efficiency spectrum: on which wavelength, which ratio of light energy can it convert to electric power.

This needs to have its maximum around the visible light, because the Sun gives most of its energy in this wavelength interval.

This is because our eyes can see best in this spectrum. We simply evolved to the sunlight.

And this is why home light is also in this wavelength: this is what we, human, like the most.

There was no need for different solar panels.

But the power of the sunlight is around some hundred $\frac{W}{m^2}$, although it varies very heavily:

enter image description here

The light power of a house bulb is around sometimes 10W - not for a $m^2$, but for a whole room! Maybe, we see in a well lighted room just as good as in sunlight, but it is only because our eye is very adaptive. The actual light power density is a tenth, or even hundredth smaller, compared to sunlight.

And the efficiency of most solar panels is around 10-20%. There are experimental, very costly versions reaching 40%. A solar panel in a room couldn't produce valuable energy, at most some watts - at the cost of the price of a solar panel on the roof. And the cost is their main problem even with the many times bigger solar power.

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    $\begingroup$ Information only: A 100W equivalent LED bulb may have about 20W input and actual optical light output about 5 Watt. 10Watt light energy out would be a VERY powerful LED light. $\endgroup$ Commented Feb 9, 2015 at 19:14
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    $\begingroup$ In addition, lighting in most inhabited areas is designed counting on light bouncing off of mostly light-colored surfaces. If you replaced those surfaces with light-absorbing solar cells, your room would have harsher shadows and potentially require more or brighter light sources. There are a bunch of other strategies to reduce energy consumption for lighting more efficiently. $\endgroup$
    – Ethan48
    Commented Feb 9, 2015 at 19:46
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    $\begingroup$ Your factor 100 is optimistic. Direct sunlight is 100.000 lux, indoor light can be 100 lux and rarely runs as high as 500 lux. That's 200-1000 times less. $\endgroup$
    – MSalters
    Commented Feb 24, 2015 at 0:34

My gut feel is that you would be able to save more energy than you could re-capture by eliminating light wastage and lowering the energy usage of the light source accordingly.

Let me use a thought experiment to explain:

If you were in a white cube-shaped room with a light in the roof and a large window in one wall with black curtains. At night when there's no light from the window if your curtains are open, you lose all that light, if you draw the curtains it's no better as they are black so they absorb it and the room is no lighter. However, if you make the curtains white the room will be a bit brighter and you can lower the power consumption of your light to get the same brightness in the room.

So, if you put a solar panel all over one of the white walls of your room, you now have something "re-capturing" the energy, but the room is darker because that light is no longer adding to the brightness of the room by being reflected, so you need to increase the energy of your light to get the same brightness.

Due to inefficiencies of energy conversion from electricity to light and in re-capturing that and converting it back to electricity, you will always lose more than the extra you have to put in in order to maintain the same room-brightness. Therefore it's more efficient to lower the light usage and optimise the room for reflecting more of the light back in (white surfaces).

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    $\begingroup$ This is of course true -- otherwise you would be well on the way to a perpetual motion machine. $\endgroup$
    – Chris H
    Commented Feb 12, 2015 at 17:00
  • $\begingroup$ @ChrisH for a simpler attempt at the same perpetual motion approach, see this question: sustainability.stackexchange.com/questions/2658/… :-) $\endgroup$
    – Flyto
    Commented Feb 24, 2015 at 14:10

Some materials are somewhat better than others in lower light levels BUT in a typical indoor situation, even an extremely optimised material simply lacks the availability of input energy to convert. eg Mono-crystalline silicon is somewhat better at very low light levels than Poly-crystalline but neither can overcome the fundamental lack of energy.

Full sunlight ~= 100,000 lux (lumens/metre^2).
Room-light is say 250 lux (quite bright).
A modern LCD monitor displaying all white and set to "nice and bright" will have a brightness at the screen surface of about 250 lux. The difference between full sunlight & this is 400:1.

A good PV (photovoltaic) panel in bright sunlight will provide 150W per square metre delivered. (Over 200 W at the cell in the best cases). At 250 lux the BEST you can expect is around 500 milliWatts per square metre.

To get the 250 lux = 250 Watts/m^2 I mentioned above you'd need 250 Watts of actual light energy shining on exactly one square metre. The very best LEDs available convert about 50% of their DC input to light. A good commercial product is very approximately 25% efficient. So you'd need to illuminate a one square metre area with 1,000 Watts DC input** of good commercial LED lighting to get 500 milliWatts output. An about 2000:1 ratio of power in to power out.

** eg a top "150 Watt replacement" bulb (Philips, Cree, ...) may have about 25 Watts of input. You'd need 1000 / 25 = 40 of these "150 Watt replacement" bulbs shining on a good 1 square metre PV panel to produce about 500 milliWatts output.


I have this 150 Watt panel, 12 volt(said to be). In testing it with two 100 Watt (equal too regular 100 watt old kind) bulbs used for lighting the room. This one panel can light up two other LED bulbs from the two on the ceiling. WOW, I thought that was impossible. That you only get 10% from the panel. Not so...

enter image description here

enter image description here

Two of these to power solar panel. The LEDs look a little dimmer then normal. But the panel is only getting 11.61 Volts (about).

Plugged LEDs in a 12 volt power supply and there's a big difference in lighting strength.

  • $\begingroup$ Welcome to Engineering.SE! This does not provide an answer to the question, which is about spectral response, not efficiency. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. $\endgroup$
    – Dave Tweed
    Commented Sep 13, 2015 at 11:17
  • $\begingroup$ These panel normally max to 21.5 volts outside sun light. Putting a skylight in the roof would do better then panels and then light bulbs, then more indoor panels to power more lights. with Skylights(roof window) you get heat and light, two good sources of energy. With little lose. Light comes in skylight, you use it to see, and it heats the walls, and you use it again to warm your house. Panel can't do that. $\endgroup$ Commented Oct 6, 2015 at 12:51

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