Black silicon photodetector breaks the 100% efficiency limit | Hacker News Reader

Black silicon photodetector breaks the 100% efficiency limit

phys.org

84 pointsgmays28 comments

The most relevant single sentence from the paper seems to be:

> Special emphasis is put on the UV range (λ=200-350nm) where we show that extremely high response, more than 130%, can be achieved at zero bias.

The material they constructed turns out to react well to UV light, so well that it proves their technique is worth further investment.

Can someone decode/ELI5 please? (or did i just miss the joke...)

Normal p-n junctions (including solar panels and ordinary semiconductors) are light sensitive because one photon of light can strike and transfer its energy into creating one electron-hole pair (promoting one electron from the valence band to conduction band). This is called "optical generation", in contrast to "thermal generation". These electron-hole pairs are generated in the depletion region of the junction where they are quickly swept across by strong electric fields that arise from the junction itself. But in general, one photon = one electron-hole pair -- at most.

Avalanche photodiodes already exist and convert a single incident photon into multiple charge carriers. They do this using an externally-applied electric field. This externally-applied electric field enables "impact ionization" which means that one mobile charge carrier seems to be able to knock another one loose.

(As a thought experiment: imagine a bed of pebbles. Toss in another pebble, and it's likely to kick one of the existing pebbles free. Now, pretend the bed of pebbles is at a 45-degree angle to gravity. Toss a pebble in and it's more likely that the ejected pebble will itself kick another pebble loose when it lands. Make the angle steep enough and you get an "avalanche", hence the name.)

As far as I can tell, the novelty here is that the high electric field is created by the geometry and materials at the surface of the nanoscale needle structure, so some version of the avalanche effect happens without needing to apply an external electric field.

In the headline, efficiency refers to quantum efficiency (electrons per photon); conservation of energy still applies.

Going to try this explanation on my 5yo and see if this is really ELI5.

But in all seriousness, thanks for such a detailed explanation! It'll be a while before I understand it, but it's pretty great.

Great explanation — although only a week or so ago I wouldn't have understood a lot of it. Knowing how transistors work helps a lot: https://www.youtube.com/watch?v=DXvAlwMAxiA

Pebbles are hard for me to imagine, instead I think of those coin games at the carnivals where you'd put in one going and a pusher device would then push things off the edge (if you were lucky).

Not quite ELI5 -- buttttt nice analogy - and I see. Particles not energy > 100% Thanks!

how does the avalanche photodiode provide more electrons that required to generate the external field?

It doesn't. The role of this kind of diode is not to generate energy but to detect faint amounts of light, to achieve that you spend additional energy on the external field.

Thanks, that makes way more sense.

They made a detector that can convert one photon into more than one electron (about 1.3 electrons per photon on average). Note that this is not energy efficiency as in the 2nd Law of Thermodynamics; they haven't created a system that generates more energy out than it receives. The electrons have lower energy than the original photons. This "efficiency" is purely about the number of particles involved.

More electrons per photon means more accurate detection of low amounts of light and also lower power requirements for light sensors in general. That would include, for example, the cells that make up a digital camera's sensor grid. Economics permitting, this could lead to better low-light performance or more efficient optical equipment (e.g. fiber optic transceivers).

Is it also a relative improvement of the energy conversion into electricity vs. into heat? Intuitively the photons energy will be partially converted into both (conservation of energy) and this efficiency improvement seems like it would skew that ratio into favor of electricity generation.

> Is it also a relative improvement of the energy conversion into electricity vs. into heat?

I would think so, since the reason the efficiency was less than 100% before was that some of the electrons were being recaptured rather than emitted from the detector. On the other hand, the article didn't say anything about the amount of energy captured. It could be that more of the photon's energy is converted into heat, with the remainder being spread across more electrons.

Ah, that's a good point. I wasn't thinking about the relative energies of the electrons themselves.

THANKS! This one stuck/made sense :)

Would this also help with solar panels?

If you throw an apple at an orange tree, you can knock down an orange. Someone did even better and figured out how to get two oranges to fall about 3/10 of the time.

I feel like this is the only actual ELI5 answer so far.

It is not >100% efficient in the energy sense. But it does create more than one electron of current* per incoming photon.

* Since current is charge per time, technically it creates more than one electron per second of current per incoming photon per second.

"The external quantum efficiency of a device is 100% when one incoming photon generates one electron to the external circuit. 130% efficiency means that one incoming photon generates approximately 1.3 electrons."

Another way of saying they can generate more energy from a single photon

No, same (or less) energy, more electrons. Splitting the same energy into more electrons is useful for detectors, since it increases their sensitivity.

They say more electrons, not more energy.

"Aalto University researchers have developed a black silicon photodetector that has reached above 130% efficiency. Thus, for the first time, a photovoltaic device has exceeded the 100% limit..."

Well, looks like the apparently correct physics which, up until now, defined the limit on photovoltaic efficiency -- was wrong... <g>

Physics is such a fickle subject... <g>

"You cannot change the laws of physics!"

-Star Trek's Scotty (James Doohan)

This is super cool.

Excited to see the way this breakthrough will be absorbed into different industries.

Can anyone comment on the impact of this with regards to LIDAR tech?

It is probably not very useful for lidar, because the effect only works for UV wavelengths. At the normal eye-safe infrared wavelengths used for lidar the quantum efficiency of this type of detector is basically 0.

It does work well for general low light imaging and nightvision applications, you can get a black silicon sensor quite high QE (~80%-ish depending on what wavelength ranges you are interested in) commercially from https://www.sionyx.com/

So if "100%" efficiency is defined as one photon generating one electron...

...then what would total energy conversion be according to this metric? Where the energy of the electric current equals the energy of the photons? No energy lost? If this is above 130%, what's the ceiling? (Even if no materials exist that could ever achieve it.)

Total energy efficiency would entail extracting electron-volt from the photon. There isn’t any real cap here except how much energy you can stuff into a vaguely stable photon and how much energy your PV cell can withstand.

PV cells usually excite at one one energy level; any photon with more energy just gives the electron extra speed, which is wasted. This paper is claiming to be able to extract the excess kinetic energy from the electron through collision by colliding with another electron and knocking it loose.

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