A layperson's introduction to LEDs
Introduction
I want to give an introduction on several physics topics at a level understandable to laypeople (high school level physics background). Making physics accessible to laypeople is a much discussed topic at universities. It can be very hard to translate the professional terms into a language understandable by people outside the field. So I will take this opportunity to challenge myself to (hopefully) create an understandable introduction to interesting topics in modern physics. To this end, I will take liberties in explaining things, and not always go for full scientific accuracy, while hopefully still getting the core concepts across. If a more in-depth explanation is wanted, please ask in the comments and I will do my best to answer.
Previous topics
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Today's topic
Today's topic will be light emitting diodes, better known as LEDs. As the name suggests, we'll have to discuss light and diodes. We will find out why LEDs can only emit a single colour and why they don't get hot like other sources of light. Let's start by discussing diodes, in case you are already familiar with diodes note that I will limit the discussion to semiconductor (p-n with a direct bandgap) diodes as that's the type that's used in LEDs.
What's a diode?
A diode is an electronic component that, ideally, only lets electric current through in one direction. In other words it's a good resistor when the current flows in one direction and a really good conductor when the current flows in the other direction. Let's look a bit closer at how diodes function.
Semiconductors
Diodes are made out of two different semiconducting materials. In everyday life we tend to classify materials as either conducting (metals being the prime example) or non-conducting (wood, plastics, rubber). Conductance is the flow of electrons through a material, a conducting material has a lot of electrons that can move freely through a material while an insulator has none. Semiconducting materials fall in between these two categories. They do conduct but not a lot, so in other words they have a few electrons that can move freely.
N-type semiconductors
We are able to change a semiconductor's conductivity by adding tiny amounts of other materials, this is called doping. As an example, we can take silicon (the stuff that the device you're reading this on is made out of) which is the most well-known semiconductor. Pure silicon will form a crystal structure where each silicon atom has 4 neighbours, and each atom will share 1 electron with each neighbour. Now we add a little bit of a material that can share 5 electrons with its neighbours (how generous!). What will happen? Four of its shareable electrons are busy being shared with neighbours and won't leave the vicinity of the atom, but the fifth can't be shared and is now free to move around the material! So this means we added more freely flowing electron and that the conductivity of the semiconductor increases. An illustration of this process is provided here, Si is chemistry-talk for silicon and P is chemistry-talk for phosphorus, a material with 5 shareable electrons. This kind of doping is called n-type doping because we added more electrons, which have a negative charge, that can freely move.
P-type semiconductors
We can do the same thing by adding a material that's a bit stingy and is only willing to share 3 electrons, for example boron. Think for a moment what will happen in this case. One of the silicon atoms neighbouring a boron atom will want to share an electron, but the boron atom is already sharing all of its atoms. This attracts other electrons that are nearby, one of them will move in to allow the boron atom to share a fourth electron. However, this will create the same problem elsewhere in our material. Which will also get compensated, but this just creates the same problem once more in yet another location. So what we now have is a hole, a place where an electron should be but isn't, that is moving around the crystal. So in effect we created a freely moving positive charged hole. We call this type of doping p-type. Here's an illustration with B the boron atoms.
Creating a diode
So what would happen if we took a n-type semiconductor and a p-type semiconductor and pushed them against one another? Suddenly the extra free-flowing electrons of the n-type semiconductor have a purpose; to fill the holes in the p-type. So these electrons rush over and fill the holes nearest to the junction between the two semiconductors. However, as they do this a charge imbalance is created. Suddenly the region of p-type semiconductor that is near the junction has an abundance of electrons relative to the positive charges of the atom cores. A net negative charge is created in the p-type semiconductor. Similarly, the swift exit of the electrons from the n-type semiconductor means the charge of the cores there isn't compensated, so the region of the n-type semiconductor near the junction is now positively charged. This creates a barrier, the remaining free electrons of the n-type cannot reach the far-away holes of the p-type because they have to get through the big net negative charge of the p-type near the junction. Illustration here. We have now created a diode!
How diodes work
Think for a moment what will happen if we send current* (which is just a bunch of electrons moving) from the p-type towards the n-type. The incoming electrons will face the negative charge barrier of the p-type and be unable to continue. This means there is no current. In other words the diode has a high resistance. Now let's flip things around and send electrons through the other way. Now they will come across the positive charge barrier of the n-type semiconductor and be attracted to the barrier instead. The electrons' negative charge compensates the net positive charge of the barrier on the n-type and it will vanish. This destroys the equilibrium situation of the barrier. The p-type holes are no longer repelled by the positive barrier of the n-type (as it no longer exists) and move closer to the junction, this means the entire barrier will fade and current can move through. We now have a conductor.
OK, but I don't see what this has to do with light
Now let's find out how we can create light using this method. When current is applied to a diode what happens is that one side of the diode is at a higher energy than the other side. This is what motivates the electrons to move, they want to go from high energy to low energy. If the p-type semiconductor is at a higher energy than the n-type the electron will, upon crossing the junction between the two types, go from a high energy level to a lower one. This difference in energy must be compensated because (as @ducks mentioned in his thermodynamics post) energy cannot be destroyed. So where does the energy go? It gets turned into light!
The energy difference between the p-type and n-type is fixed, meaning a fixed amount of energy is released each time an electron crosses the junction. This means the light is of a single colour (colour is how we perceive the wavelength of light, which is determined by the energy of the light wave). Furthermore, none of the energy is lost so there is no energy being turned into heat, in other words the LED does not get warm.
Conclusion
So now we know why the LED is so power-efficient; it does not turn any energy into heat, it all goes into light. We now also know why they only emit a single colour, because the energy released when an electron crosses the junction is fixed.
Next time
I think next time I will try to tackle the concept of wave functions in quantum mechanics.
Feedback
As usual, please let me know where I missed the mark. Also let me know if things are not clear to you, I will try to explain further in the comments!
Addendum
*) Yes, current flow is defined to be opposite to the flow of the electrons, but I don't want to confuse readers with annoying definitions.
How does the fixed energy difference between the p-type and n-type layer dictate that the energy release must be in the form of light and not heat? In other words, how does the electron movement across this n-p barrier differ from electron movement across normal conductor, such that one must produce light and the other heat?
Very interesting question. To answer it I will first rephrase it. Why does the energy release result in photons, light particles, being released and not phonons? Phonons are """heat particles""", or more technically collective vibrations of the crystal.
When an electron meets a hole and decides how to release the energy it needs to keep some things in mind. As I mentioned in the post, energy needs to be conserved. However, there is a second quantity that needs to be conserved as well. This is the so-called crystal momentum. Photons carry energy, but they cannot carry crystal momentum. Meanwhile phonons carry both energy and crystal momentum. In fact, a phonon must have crystal momentum to exist just like a photon must have energy to exist.
For a LED to work it must then force the electron to lose energy but not crystal momentum; so that phonons cannot be created but photons can be created. This is exactly what happens in so-called direct bandgap materials (in fact we define this class of materials for having this property). If we created a LED out of an indirect bandgap material it would instead create phonons, as now the electron must also dump some crystal momentum when it goes across the junction, and this crystal momentum cannot go into a photon. As such calling such a device a LED would be a serious misnomer, it wouldn't emit any light but instead create phonons which we observe as heat.
In fact, phonon-electron interactions are a source of "friction" for the electrons, reducing their ability to flow freely and as such reducing the material conductivity. This is why conductivity reduces as temperatures increase; there are more phonons hanging around at higher temperatures.
If you have more questions or need more clarification feel free to ask :)
Thank you. So it seems like in all cases, electrons passing a material will create a combination of both photon and phonon. It's just that some materials are "soft", where the "impact" of the electron can "deform" the lattice significantly, draining most of its energy in the form of "shock waves" going through the lattice, while other materials are "hard", where most of the energy can be channeled as light "sparks" (I was thinking of flint and steel as analogy). Then we just find and use those "hard" material to make LED. Am I close?
That's a lot of classical analogies for an inherently quantum mechanical process.
Note that having a junction between two materials is critical for light emission. The recombination of the holes and electrons is what drives the emission.
Electrons just moving through a material will interact with phonons (the crystal vibrations) but they won't necessarily create phonons. The junction is needed to force photon or phonon emission.
If you are nterested in more detail on how the junction is critical, I can try to explain.
So to create the photons we need the electrons to drop in energy. In order to explain this properly I'll first have to introduce a few more concepts.
A semiconductor has 2 energy bands, the conductance band and the valence band. If an electron has a high enough energy it will end up in the conductance band and be able to, well, conduct. If it doesn't have the required energy it ends up in the valence band where it's stuck being unable to move. Crucially, in semiconductor there's a gap between these two bands; the bandgap. What this means is that electrons cannot have an energy that falls within this gap.
So, now back to the junction. The two materials making up the junction do not have their valence and conductance bands at the same energies. The p-type's conductance band will start at a higher energy than that of the n-type. This means that an electron that flows into the barrier will end up in one of two scenarios. If it has enough energy to end up in the n-type's conductance band it will continue flowing. Otherwise, if it has enough energy to be in the p-type's conductance band but not in the n-type's, it is forced into valence band of the n-type. This means it will have to drop in energy - else its energy would end up in the band gap - and while doing this it will emit a photon. This is how the junction forces the electron to emit a photon.
What your previous comment seemed to imply is that you think electrons go bouncing around creating phonons. However, they usually just scatter off of phonons, making them change directions elastically (without a change in energy). Phonon creation can also happen at a junction, as I discussed previously, when the material has an indirect bandgap.
A vibration from sound wave can be modelled as a certain kind of phonon, an acoustic phonon (with infinite wavelength).
Okay so the electrons jump the junction, some energy gets turned into a single, consistent wavelength of light. How do white LEDs happen? Multiple junctions?
That's a good question. For other readers; white light is a mix of different colours of light so as @Diff noted it would be impossible to create with a LED.
There are two ways we can create white light LEDs, one is indeed to use multiple LEDs (RGB) and then mixing the light they emitted. However, this only produces the illusion of white light and not actually white light that has all the colours in it like the light from the sun does. This means that the colours of objects viewed in this light can appear off.
A better way to produce white light is by taking just one blue or UV LED and letting its light fall onto a fluorescent material. This material will absorb the light and re-emit at lower frequencies. The frequencies of visible light are lower than that of UV (and blue has the highest frequency) so this way we produce a lot more colours of light. However, fluorescence is a quantum mechanical process too so it will not emit the full spectrum of visible light either.
So if you need to be able to see all colours, sunlight or good old lightbulb light is still the best.
I'm not really sure how to phrase this but are "slanted" diodes possible? You said
So would it be possible to design a diode so that the left side had a smaller difference than the right side? Or would that just make all the electrons cross on one side?
That's a really cool idea. I hadn't considered it before.
With current techniques we can create blue, green, red etc. LEDs but these are all made out of different semiconductors. So fusing them together into one diode is just impractical and would probably introduce secondary effects on the boundary between the materials.
However, in theory I don't see a problem with this approach - as long as your material can have a wide enough bandgap tunability (i.e. we can change the energy gap to range all the way from blue light to red light being emitted). In fact, I found a paper from 2011 where exactly this is done on a nanowire. The material they used has such a wide bandgap tunability that it can produce light in the entire visible spectrum. To quote:
Emphasis mine. It seems that the limiting factor here is how smoothly they can change the bandgap over the width of the material. Here's a photo they made of the nanowire they produced.
So thank you very much for asking this question, it led me to discover this very interesting paper :)
I don't really understand this bit. What is different about the LED from the diode that doesn't let the energy just pass through?
Quoting myself from elsewhere in the thread:
I think that should answer the question. If not, please let me know!