## Troll physics: Solar cell and LED in series

So today my brother and I were playing with a mini solar cell that I just got, watching the voltage change with a multimeter, and seeing how even an LED would produce about .5 V in the cell, when he posed a question:

If you were to take a solar cell (i.e. photodiode) capable of producing something like 2 V when hit by an LED, and focus it around an LED that it is connected in series with, put the whole thing in a completely dark box, and "prime" the cell by shining a different LED at it, would you end up with an LED powered perpetually by its own light?

My guess would be no, but I thought about this for a while and couldn't come up with a good reason why not. I know there's something I'm missing since I just don't know enough about diodes, and I'm curious as to what. I know a little about pn junctions and electron/hole travel, but not much more than that.

I've been learning about electronics and building my own circuits for a few years now, and I'm slowly gaining a better understanding of electricity and circuitry, but I'm still a novice.

Can anyone give me an explanation of the interactions going on in this scenario?

Thanks,
Jesse
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 Recognitions: Gold Member Nope. None of your components are 100% efficient.
 i'm not entirely sure about this but consider this scenario suppose you have a box whose inner walls are full of perfect mirrors, you shine a light beam in the box and quickly close it before the light can escape. the photon hence remains trapped &bouncing within the box. in a sense there is light in the box, always & forever, but the total energy is finite. however if you want to see the light you will have to open the box and let the light escape, then it's no longer perpetual since energy is lost. i believe it is the same case with your led problem, as long as energy is lost there is no perpetual lighting

## Troll physics: Solar cell and LED in series

Expanding on what Drakkith said; the conversion of light energy into electrical energy in the solar cell isn't 100-% efficient, similarly the conversion of electrical energy into light energy in the LED isn't 100-% efficient. For these reasons, your "self-powered" system would quickly die out.

A good analogy is the so-called Newton's cradle. The cradle is "primed" by you lifting one of the end balls and the balls transfer energy and momentum in a periodic fashion. However, the system is not 100-% efficient due to air resistance, loss of energy into sound and heat, etc. -- and so the cradle eventually comes to a stop.

Sure, both systems above would run in perpetuity if the processes were 100-% efficient; however, the science of thermodynamics tells us that no real physical processes can be 100-% efficient.
 Even if you could make it absolutely 100% efficient, such that in principle this worked. Then if you peeked into your "dark box" to see if it worked, the light that escaped to allow you to see it would then drain off the light needed to keep it working. Of course, as others have said, this kind of efficiency is not even in principle possible.
 Thanks for the responses. I figured there would be a lot of efficiency issues, I just wondered if I was missing anything basic about how the electricity flowed through the system that would somehow diminish the effect. The reason I brought up the dark box is because I didn't want any extra light powering the photodiode. In practice, an LED shown at the cell reads .5 V even though clearly not all of the LED's light is hitting it (though, I know other light is getting in too). So I thought if some fraction of the light output by the LED was enough to power it through the cell, there would still be light to be seen. I guess I see that I was thinking too much about voltage and current and not enough about energy. If I understand correctly now, all energy absorbed by the cell is transmitted to the LED, which in turn puts that energy right back into the cell, so with no losses, this would be an endless cycling of the same energy (in a sense). Of course, there are always losses, but the thought experiment is an interesting one anyway, and not one I've heard before. I've also heard that solar cells break down rather quickly, so something like this wouldn't last long anyway.
 Yes, when I made the point about peeking at photons I also neglected the idea of testing with a voltmeter. Yet the same principle applies. Attaching a voltmeter would consume power, draining the energy the LED needs to produce light.

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 Quote by tragicmuffin I've also heard that solar cells break down rather quickly, so something like this wouldn't last long anyway.
Not sure where you heard that one from. Modern solar cells last several decades if I remember correctly.
 Recognitions: Science Advisor Real components aren't actually that efficient, but as an idealized case you can picture LED working by having a single electron go down 1V barrier and emit a single photon. While the photovoltaic cell works backwards. It absorbs that one photon and raises electron 1V across the potential difference. If you have 2V from your solar battery, it is actually at least two cells in series, meaning it will take two photons to raise an electron 2V. So for each electron that passes through your circuit, you consume 2 photons and produce 1. So the approach breaks down here already, and we haven't even factored in the losses yet.