Lesson Video: Semiconductor Diodes Physics

Video Transcript

In this video, our topic is semiconductor diodes. These diodes are one of the building blocks of modern electronics, and in this lesson, we’re going to learn how they work down at an atomic and even subatomic level.

We can start thinking about diodes by considering the two main classes or types of semiconductors. There’s the p type, which has a majority of electron holes or vacancies, and then the n type that has a majority of free electrons. So basically, over here, there are a lot of free electrons moving about within the semiconductor. And over here, there are lots of effectively positively charged holes, which would be a good fit for accepting any free electrons they encounter.

To create a semiconductor diode, we take the p type and the n type and we attach them together. When we join them like this, we create what’s called a junction. So sometimes these p and n semiconductors together are called a p-n junction or a p-n diode. Now let’s think about what will happen when we join this p-type semiconductor, which has lots of positively charged holes, we’ll model them with these dots, with an n-type semiconductor that has lots of free electrons moving about all within it. Even though these p- and n-type semiconductors individually are electrically neutral, that is, the net charge of this whole semiconductor is zero as is the net charge of this one, still there will be an electrical attraction between the negatively charged free electrons on the n type and the positively charged holes on the p type.

And so, for the electrons that are near the interface between these semiconductors, there’s a high likelihood that they will be attracted to and then fill the holes nearby this junction on the other side. So let’s say that happens, say that these electrons go to fill these holes. When that happens, we can erase these holes and free electrons, because now that the holes are filled, they no longer exist, and the free electrons now that they’re bound to an atom are no longer free. In this process, though, something interesting goes on. The atoms in the p-type semiconductor that used to have holes but that are now filled by free electrons from the n-side now take on an overall or net negative charge.

We can recall that before the holes in these atoms were filled by free electrons, the atoms themselves were neutral. So if we add a free electron to them, it gives the atoms an overall negative charge. And then at the same time, over on the n-type semiconductor side, because this portion of that semiconductor has lost free electrons as they were transferred over to the holes on the p-type side, the atoms near to this junction acquire an overall positive charge. This happens because they’ve lost, we could say, the free electrons which where they attach to these atoms would give them an overall neutral charge.

So then, this region here right around the junction of these two semiconductors doesn’t have any positively charged holes; those have been filled by free electrons. And it also doesn’t have any free electrons, which have gone to fill those holes. Because of this, a specific name is given to this region. It’s called a depletion region or a depletion layer, and again, that’s because it doesn’t have any holes or free electrons. Notice that in the depletion region, we have positive charges on one side and negative on the other. And that means an overall electric field will be created, always pointing from positive to negative.

Now, this field is important because it actually serves to oppose the movement of free electrons from the rest of the n-type semiconductor toward the positively charged holes on the p-side. Even though the remaining free electrons would be attracted to these positive charges here, they would be repelled by this wall of negative charges at the junction. That would mean they don’t cross over this boundary between the p and the n types. And therefore, we have an unchanging situation as far as our p-n diode goes. And now we get to the heart of how a p-n or semiconductor diode works and how it fits in as a component in an electrical circuit.

Say that we take our p-n diode and we connect it up with a closed circuit. So we could imagine, for example, that our diode is here, and in the rest of the circuit, it’s a part of looks like this. In this setup, we’re going to consider two scenarios, first, with our voltage supply oriented like this, with the positive terminal towards the right. Arranged this way, we know that the direction of conventional current, that is, positive charge flow in the circuit, would be counterclockwise. That would mean then that we effectively have positive charge coming out our diode from this direction and negative charge from the other direction.

Now, when it comes to a p-n diode, that’s part of a closed electrical circuit like this, the key condition for current to exist across the diode, and therefore across the circuit, is that electrons on the n-type side must be able to cross this interface between the two semiconductor types. If that doesn’t happen, charge doesn’t flow through the diode and therefore charge doesn’t flow in the circuit. So let’s consider with positive charge effectively coming in from the right and negative coming in from the left, would that allow the free electrons in the n-type side to cross over this junction?

Well, we know that opposite charges attract, which means that our negatively charged free electrons would actually be drawn to the right towards these incoming positive charges. And moreover, all these incoming negative charges would start to fill holes in the p-type side of our diode, which will lead to the creation of yet more negative ions. That would mean that the barrier to free electron travel across our p-n junction is growing even stronger. And therefore, it simply won’t happen that free electrons will cross this interface flowing through the diode and then allowing charge to flow all throughout our circuit.

So what we’re seeing is when our voltage supply is arranged like this, effectively trying to send positive charges into the n-type side of our p-n diode, it means that no electrons are able to flow across the diode, and therefore current cannot exist in the circuit at all. So that’s one way that things could unfold with our p-n diode. But now let’s take our voltage supply and we’ll flip it around. What we have now is a circuit where conventional current would point in a clockwise direction. So then, looking at our larger sketch, we now have positive charge effectively coming in from the left and negative charge approaching from the right.

So now, thinking about our free electrons over here, these electrons will be repelled from the negative charges approaching from the right, and that will tend to make them move from right to left. And then all these positive charges effectively coming in here will create many more holes on the p-type side of our junction, creating a strong electrical attraction for the negative free electrons on the n-type side. And so the combined effect of these influences will be that the free electrons are able to cross this junction.

Indeed, they’ll do so readily as they’re being pushed or repelled from the right and pulled or attracted from the left. With charge flowing through our diode then, that means charge will flow through our circuit overall. And so, with our diode and our voltage supply set up this way, current exists in this circuit. What we’re seeing is that our p-n diode, our semiconductor diode, can work as a switch in the circuit. Arranged one way relative to the voltage supply, this way, we say that our diode is forward biased, which means it allows charge to flow. And recall that the other arrangement of our voltage supply relative to our diode, this one here, did not let current exist in the circuit because the diode resisted the flow of charge. When a diode resists charge flow, it’s said to be reverse biased.

So then, depending on the way we arrange our circuit, the diode can either allow charge to flow or block it, we can say that a diode functions like a valve in a circuit, blocking flow in one direction while allowing it in the other. Knowing this about how diodes work, let’s get some practice with these ideas through an example.

The diagram shows a lattice of silicon atoms in a semiconductor. The left side of the lattice has been doped with donor atoms. This is called the n-side. The right side of the lattice has been doped with acceptor atoms. This is called the p-side. The regions on either side of the dividing line are of equal size, and the ion concentration is the same on both sides. The semiconductor is at thermal equilibrium.

Alright, before we get to our question, let’s understand this information we’re given here. We’re told that this diagram shows us a lattice of silicon atoms. And we can see that most of the atoms here are indeed silicon, represented by the symbol Si. So we have the silicon lattice, and we’re told that the left side, over here, has been doped with donor atoms. This means that some of the silicon atoms in the lattice have been replaced by what are called impurities. These impurities are called donor atoms because they have one more valence electron in their natural state than silicon does. And therefore, when they’re added into the lattice of silicon, they effectively donate that excess electron to the lattice.

We’re told that this left side of the lattice is called the n-side, so we can label it that way up top. Then, if we look at the right side of our silicon lattice, we’re told that this too has been doped, but this time with acceptor atoms. These are atoms which by themselves have one fewer valence electron than silicon. And therefore, when they’re added into the lattice structure, they create a vacancy or an electron hole. The reason they’re called acceptor atoms is because they tend to accept free electrons into these holes. This right side is the p-side of the semiconductor. Knowing that the p-side and the n-side are of equal size and that they have the same ion concentration and that our semiconductor overall is at thermal equilibrium, let’s now move on to our question.

The first part of our question asks, toward which side of the lattice will free electrons tend to move by diffusion? And then part two says, toward which side of the lattice will holes tend to move by diffusion? Alright, so considering this first part about the movement of free electrons in our semiconductor, we learned earlier that it’s the left side of the semiconductor, the n-side, that features what are called donor atoms. We see these atoms here and here. They’re atoms of phosphorus.

The reason these phosphors atoms are called donors is that while silicon naturally has four electrons in its valence shell, an atom of phosphorus has one more, five, which means that when these phosphorus atoms are inserted into our lattice, each one contributes an extra, we could call it, electron, one more than the silicon atoms they replaced. These extra electrons are separated from the phosphorus atoms and begin to move about freely within the n-side of our semiconductor.

So then, when it comes to which way free electrons will move, we now know what side they’ll begin on. They’ll start on the n-side, contributed by these donor phosphorus atoms. Negatively charged electrons are drawn to positive charges, and it turns out that we can find such positive charges over on the p-side. Here, instead of doping our silicon lattice with phosphorus, we’ve done it with boron. Boron is different from silicon in that it has only three valence electrons, one fewer than silicon. And so, when these boron atoms replace silicon in the lattice, there’s one missing electron, we could say, in their valence shell.

These vacancies are known as holes, and they have an effective positive charge. And that is just what free electrons are drawn to. So over on the n-side, we see this free electron right here and this one here, one each contributed by the phosphorus atoms. And we know that these negatively charged objects will be drawn to the effectively positively charged holes over here on the p-side. So that’s our answer for this first part of our question.

And now let’s think about the second part which asks, toward which side of the lattice will holes tend to move by diffusion? Well, just as the negative free electrons are drawn to the positive holes, so the positive holes will be drawn to the negative free electrons. So the holes then starting out on the p-side, the right side, will be drawn towards the left, and that, as we’ve seen, is the n-side. By diffusion then, holes will be drawn towards the n-side or the left side of our lattice.

Let’s summarize now what we’ve learned about semiconductor diodes. In this lesson, we saw that when p-type and n-type semiconductors are joined together, they create a p-n diode. At the diode junction, the place where the p- and the n-type semiconductors meet, a depletion region is created, which has no holes and no free electrons. And lastly, we saw that when a diode is placed in a circuit, it either allows charge flow, called being forward biased, or it prevents it, called being reverse biased. In this way, a diode functions as a valve. This is a summary of semiconductor diodes.