Lesson Explainer: Semiconductor Diodes Physics • 9th Grade

In this explainer, we will learn how to model a semiconductor diode as a junction between n-type and p-type doped semiconductors.

Diodes are electrical circuit components that allow current in one direction but not in the other. In this explainer, we will discuss how they work at the subatomic level.

Diodes are made of semiconductors. The most commonly used semiconductor is based on a lattice of silicon atoms. A silicon atom has four electrons in its outer, or valence, shell that are available to form bonds with adjacent atoms. In a lattice of silicon atoms, every electron in the outer shell is involved in bonding with an adjacent silicon atom. This is illustrated in the following diagram.

When we dope this lattice with another element like phosphorus, which has five electrons in its outer shell, there is a “spare” electron not involved in bonding with an adjacent silicon atom. This electron (indicated by an arrow in the diagram below) is then free to move through the lattice. These free electrons are negatively charged, so this type of doped semiconductor is called an n-type.

It is important to remember that the material as a whole does not have an electrical charge: the numbers of protons and electrons overall are the same. When we refer to it as n-type, we only mean that the charge carriers that are free to move—in this case, the electrons—have negative charge.

We can also dope silicon with an atom that has three electrons in its outer shell, like boron. In this case, there will be a vacancy. A vacancy itself has no charge, but relative to an electron, it is positively charged. Other electrons will be attracted towards this vacancy and will sometimes jump across from another atom to fill it. Although, in practice, it is the electrons that are moving, we think of the vacancy moving around the lattice as different electrons jump in to fill it. This means the charge carrier is positively charged, so we call this kind of doped semiconductor a p-type. In the diagram below, we indicate the vacancy with a dashed circle.

These doped semiconductors have interesting properties when we join them together. In the diagram below, we have a p-type on the left and an n-type on the right. Where they meet, we have a junction we refer to as the PN junction.

For simplicity, we have left out the atomic nuclei and bound electrons and shown only the charge carriers: positively charged vacancies on the left and negatively charged electrons on the right. Remember, though, that both sides are overall electrically neutral.

When arranged like this, we have vacancies on one side that attract electrons and lots of free electrons on the other side. Some of the electrons close to the boundary will therefore diffuse across the junction to fill the vacancies on the p-side, as indicated in the following diagram.

The previously neutral p-side has now acquired an electron and so has a negative charge. Similarly, the n-side has lost an electron, so it has more protons than electrons and is now positively charged. In the next diagram, the filled vacancy is illustrated as a negative charge, and the absence of an electron on the n-side is illustrated as a positive charge.

More electrons close to the boundary cross over to fill vacancies on the p-side, as shown in the diagram below.

We end up with an accumulation of negative charge close to the junction on the p-side and positive charge on the n-side of the boundary. This is shown in the following diagram.

This region where there is negative charge on one side and positive charge on the other is known as the depletion region. The buildup of charge creates an electric field, labeled 𝐸 in the diagram below, across the depletion region.

The negative charge buildup close to the junction on the p-side now acts as a barrier, repelling further electrons that might jump across.

At this point, let’s look at a couple of examples that relate to doped semiconductors and properties of the PN junction.

Example 1: Understanding Doped Semiconductors

The diagram shows a lattice of silicon atoms in a semiconductor. The left side of the lattice has been doped with donor ions. This is called the n-side. The right side of the lattice has been doped with acceptor ions. This side of the lattice 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.

  1. What is the ratio of free electrons on the n-side to vacancies on the p-side?
  2. What is the difference between the net relative electronic charge in the two regions?

Answer

Part 1

In this example, we have a doped semiconductor in which the n-side is on the left and the p-side is on the right. The n-side has been doped with phosphorus, which has five electrons in its outer shell. Each atom therefore provides one “free” electron. The p-side on the right is doped with boron, which has three electrons in the outer shell. Each boron atom therefore contributes one vacancy, or positive charge carrier. The question states that the two regions are of equal size and have equal ion concentration—that is, the density of acceptor or donor ions. This means that there will be equal numbers of free electrons on the n-side and vacancies on the p-side. The ratio of free electrons on the n-side to vacancies on the p-side is therefore 1.

Part 2

We now need to consider the net electronic charge in each region. There is a common misconception that having free electrons makes the n-type region negatively charged. In fact, the region is electrically neutral, as the donor atoms have the same number of protons as electrons. The same is true on the p-side: the acceptor ions have fewer electrons than the silicon atoms, but also fewer protons. Overall, both regions are electrically neutral. The difference between the net electronic charge in the two regions is therefore zero.

Example 2: Understanding PN Junctions

In a PN junction, both free electrons and vacancies can diffuse through the junction, as shown in the diagram.

  1. Toward which side of the junction is the direction of the net diffusion current?
  2. In which of the following regions is the concentration of free electrons greatest?
    1. The p-side
    2. At the middle of the junction
    3. The n-side
  3. In which of the following regions is the concentration of vacancies greatest?
    1. The n-side
    2. The p-side
    3. At the middle of the junction
  4. In which of the following regions is the concentration of both free electrons and vacancies smallest?
    1. At the middle of the junction
    2. The p-side
    3. The n-side

Answer

Part 1

Here, we have a semiconductor with an n-region full of free electrons on the left and a p-region full of vacancies on the right. Free electrons from the n-region will diffuse across the junction to fill vacancies in the p-region. We can equivalently say that vacancies diffuse from the p-region towards the n-region. When discussing electrical currents, the convention is that they are in the direction from positive to negative. The direction of the net diffusion current is therefore toward the n-region.

Part 2

There are free electrons distributed throughout this PN junction. They are at their highest concentration toward the left, in the n-region: we can see this in the diagram or recall that an n-region is so named because it is doped such that it has free electrons. The answer is therefore C, the n-side.

Part 3

We have vacancies throughout the PN junction, but they occur in the highest concentration on the p-side. The answer is therefore B, the p-side.

Part 4

Given a high concentration of free electrons on the n-side and a high concentration of vacancies on the p-side, electrons will tend to diffuse across the junction and combine with vacancies on the p-side. At this point, the electrons are no longer “free.” Similarly, once they have filled the vacancies, they are no longer vacant. The concentration of both free electrons and vacancies will therefore be smallest in the middle where this occurs, so the answer is A, at the middle of the junction.

Let’s now connect the PN junction to a power source in an electrical circuit. An example circuit diagram is shown below.

Recall that conventional current is in the direction from positive to negative, so current will be in the counterclockwise direction in this circuit.

Let’s consider the response of positive and negative charge carriers in a PN junction that has a potential difference applied across it.

Positively charged vacancies on the p-side of the junction are attracted to its negative terminal, and hence away from the depletion region. This allows more free electrons from the n-side to exist in the p-side of the depletion region.

Similarly, negatively charged free electrons on the n-side of the junction are attracted toward its positive terminal, and hence away from the depletion region. This allows more vacancies from the p-side to exist in the n-side of the depletion region.

We see then that the depletion region expands as a result of the applied potential difference. This is illustrated in the diagram below.

The wider depletion region acts as an even stronger barrier, preventing electrons from crossing the junction. There will therefore be no current in this circuit. A PN junction in this configuration is known as “reverse biased.”

Let’s now reverse the polarity of the power source in the circuit, as in the following diagram.

In this case, we have the p-side connected to the positive terminal and the n-side connected to the negative terminal.

As shown in the following diagram, the free electrons in the n-side are repelled by the negative terminal, pushing them toward the middle of the junction. This “push” overcomes the boundary at the PN junction, allowing the electrons to cross to the other side and fill vacancies there.

As the electrons are now able to cross the boundary, there will be current through this circuit. We refer to a PN junction in this orientation as “forward biased. ”

These properties of a PN junction make it useful in electrical circuits where we want to allow current in one direction and not the other. They, therefore, form the basis of a component known as a diode. A diode acts as a valve in a circuit: when forward biased, current can pass through it, but when reverse biased, there can be no current through it.

Example 3: Identifying Forward- and Reverse-Biased PN Junctions

The diagram shows a PN junction in a circuit. The positively charged side of the junction’s depletion region is shown in red, and the negatively charged side is shown in blue. Is the junction forward biased or reverse biased?

Answer

We have a circuit consisting of a power source and a PN junction. The positively charged side of the depletion region is on the left, and the negatively charged side is on the right. Recall that the region of positive charge occurs because free electrons from that side have crossed the junction to fill vacancies on the other side. The left is therefore the n-side, or the side that contains free electrons.

Recall also that conventional current is the flow of positive charge, in the direction from the positive terminal to the negative, which is clockwise in this circuit. This means the flow of electrons is counterclockwise. We therefore have electrons entering the p-side on the right, where they will combine with vacancies and strengthen the negatively charged side of the depletion region. This will act as a barrier, repelling further electrons. The inverse occurs on the p-side, where the vacancies are repelled by the positively charged terminal, strengthening the barrier on that side.

This strengthened barrier impedes current, so the junction is reverse biased.

If we consider a circuit containing a PN junction in which the potential difference can be varied, how will the current change? We can see this in a current–voltage, or 𝐼𝑉, graph.

If the junction acts as a switch, we might expect the graph to look like the one below. There is zero current when the junction is reverse biased (indicated by negative voltage) and a large current when it is positive.

We call this an “ideal diode.” In practice, diodes do not work perfectly. An 𝐼𝑉 graph for a real diode may look more like the one below.

Here, when the voltage is negative (or the PN junction is reverse biased), the current is almost, but not exactly, zero. The small current that exists is negative, to indicate that its direction is opposite to the forward-biased case. When there is a sufficiently large negative voltage, the junction breaks down and stops impeding current. The voltage at which this happens is known as the “breakdown voltage.”

When the junction is forward biased, a real diode does not immediately allow a large current. There is some threshold voltage below which the current will be small, and above this threshold, there will be a large current.

This final example will give some practice interpreting a current–voltage graph for a PN junction.

Example 4: Understanding Current–Voltage Graphs for PN Junctions

The graph shows the variation in the current through a PN junction with the external voltage applied across the junction. The graph contains a region where the current is nearly zero. Does this region correspond to the PN junction being forward biased or reverse biased?

Answer

The graph shows what happens to the current through a PN junction as the voltage increases. When the voltage is negative, there is a small negative current. When the voltage is positive, the current is also positive and increases rapidly with increasing voltage.

The question asks about the region where the current is nearly zero. This refers to the portion where the voltage is negative and the current is also negative but very close to zero. In this region, the PN junction offers very high resistance so that there is almost no current. When the polarity of the external voltage supply is reversed, the PN junction offers almost no resistance and the current is very high. These correspond respectively to the junction being reverse and forward biased. So, when the current is nearly zero, the PN junction is reverse biased.

Key Points

  • A diode consists of a junction between two types of doped semiconductor: p-type and n-type.
  • Both p-type and n-type semiconductors are electrically neutral.
  • The p-type region contains vacancies that transport charge, whereas in the n-type region, charge is carried by free electrons.
  • At the junction, free electrons from the n-side will cross over to the p-side to fill some of the vacancies. This results in a depletion region adjacent to the junction.
  • The depletion region acts as a barrier, preventing further electrons from crossing the junction.
  • When a PN junction is reverse biased, electrons fill vacancies and strengthen the barrier in the depletion region so that no current is permitted.
  • When a PN junction is forward biased, free electrons are able to overcome the barrier in the depletion region so that current is permitted.

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