In this explainer, we will learn how to describe the effect that doping a semiconductor has on its electrical properties.
To begin, let us recall some basic information about semiconductors. Semiconductors have conductive properties that fall somewhere between insulators and conductors. Semiconductors’ specialized properties allow us to control their conductivity in some unique and useful ways. Recall that a pure semiconductor consists of a single type of element, most commonly silicon, and contains no other impurities.
A neutral silicon atom has four electrons in its outermost electron shell, and when grouped together, the atoms form a lattice by sharing covalent bonds with adjacent atoms. This lattice configuration is shown in the diagram below. Note that only the outermost electrons for each atom are shown.
Any internal, or central, atom has an adjacent atom to its top, bottom, left, and right, so it has four pairs of electrons surrounding it. This total of eight electrons fills the atom’s outermost shell. Atoms with more complete outermost electron shells are more insulating, so we rely on certain techniques to make a semiconductor more conductive.
One way to increase the conductivity of a pure semiconducting material is to increase its temperature. In this case, if enough thermal energy is transferred to an electron, it can overcome its atomic bond and move around the lattice as a free electron. When an electron becomes freed, it leaves behind a hole, or vacancy, that has an effectively positive charge. When another free electron from the lattice comes near the vacancy, the electron will fill the hole by losing energy and becoming bound to the atom. This process constitutes a cycle of energy being transferred between free and bound electrons and holes. Such free movement of charge is how a pure semiconductor is able to conduct electricity.
It should be specially noted that, for a pure semiconductor in thermal equilibrium, any electron hole corresponds to a free electron that left the hole in its place. Thus, the number of free electrons, represented by , is equal to the number of electron holes, , so the atomic lattice remains electrically neutral.
If the temperature of a semiconductor increases, its conductance also increases because more electrons are able to gain the energy they need to become free. Another, more effective, way to increase the conductance of a semiconductor is to “dope” it, which is what we will explore in the remainder of this explainer.
To dope a semiconductor is to introduce another element to the atomic lattice. Thus, we add “impurities” to the material so that it is no longer “pure,” or composed of one single element. We have seen that a neutral silicon atom has four outermost electrons, which allows for a lattice formation in which each interior atom has a full outermost electron shell. In this explainer, we will explore what happens when an interior silicon atom is replaced with another element that has a different number of outermost electrons—specifically, either one more or one fewer. Thus, we will use an element that (when neutral) has either three or five outermost electrons.
For simplicity, we will focus on one atom of each type; we will use phosphorus () as the atom with five outermost electrons. The diagram below shows a neutral phosphorus atom.
And we will use boron () as the atom with three outermost electrons. The diagram below shows a neutral boron atom.
It should be noted that the number of electrons in the outermost shell determines the way it affects conductivity, which we will explore below. We will examine the effects of doping with phosphorus and boron individually.
To begin, let us focus on the effects of replacing a central silicon atom in a lattice with a phosphorus atom, as shown in the diagram in the example below.
Rather than having four outermost electrons like silicon, a neutral phosphorus atom comes into the lattice with five outermost electrons. Up to eight electrons can occupy the outermost shell, so four of the phosphorus atom’s electrons are shared in covalent bonds, like we have seen before. However, the phosphorus atom came with five electrons, so there is one extra electron that does not fit in the shell. Thus, the extra electron is not bound, or involved in a covalent bond, so instead it moves around the atomic lattice as a free electron.
Four out of the five electrons that the phosphorus atom came with are bound to the nucleus, and one electron is now free and simply not bound to any atoms. Thus, each phosphorus atom that is introduced to the lattice contributes one free electron to the material. Having more free charges increases the conductivity of a material, so we know that doping a semiconductor with atoms increases the conductivity of the material as a whole.
Let us explore some ways to quantitatively express this concept.
Example 1: Understanding Electrons in a Doped Semiconductor
The diagram shows a lattice of silicon atoms that an atom of phosphorus is added to. An atom of phosphorus has five valence electrons. An atom of silicon has four valence electrons.
- How many covalent bonds does the phosphorus atom form with the silicon atoms adjacent to it?
- How many free electrons are added to the lattice when the phosphorus atom is added?
- What is the net relative atomic charge of the phosphorus
after it is added to the lattice?
The diagram shows that an internal atom is immediately surrounded by four other atoms: one to its top, bottom, left, and right. Thus, an internal atom in the lattice, such as the phosphorus atom shown above, can share four sets of covalent bonds with neighboring atoms. Remember that each covalent bond is comprised of two electrons—this makes for a total of eight electrons surrounding the atom, which completes the outermost electron shell.
Thus, the phosphorus atom forms 4 covalent bonds with adjacent silicon atoms.
A neutral phosphorus atom has five electrons in its outermost shell. Thus, when a atom is introduced to the lattice, it has four neighboring atoms to form covalent bonds with. Thus, four out of the five electrons that the atom came with become bonded with adjacent atoms.
The outermost shell of the atom can be occupied by up to eight electrons. Since there are four covalent bonds (and thus eight electrons) immediately surrounding the atom, the outermost shell is full and cannot fit the one last electron that the phosphorus atom came with. Since the one extra electron cannot be bound to the atom, it is now a free electron.
Therefore, the phosphorus atom adds 1 free electron to the lattice.
The phosphorus atom lost one electron that it came with. Since the phosphorus atom has lost one electron, which has a relative charge of , we can say that a charge of was subtracted from the atom, resulting in a net positive charge.
Thus, the net relative charge of the phosphorus atom is +1, and answer choice B is correct.
Before we continue, let us quickly establish a few key vocabulary terms: we will use the terms trivalent, tetravalent, and pentavalent to describe an atom by the number of electrons in its outermost shell. For instance, recall that silicon has four electrons in its outermost shell, so it is considered tetravalent, as “tetra-” means “four.” Further, a boron atom has three outermost electrons, so it is trivalent, and phosphorus is pentavalent because it has five outermost electrons. Finally, any atom used to dope a sample is called a “dopant.”
The example above explains why a neutral phosphorus atom added to a neutral silicon lattice has a net relative charge of +1. Because of this net positive charge, we can refer to this phosphorus atom (or any pentavalent atom used for doping) as a “positive donor ion.” We express the concentration of positive donor ions with the symbol , where gives a quantity, stands for “donor,” (since the atom “donated” an electron to the material), and the + sign refers to the overall positive charge of the phosphorus ion in the lattice. Thus, the concentration of phosphorus in the material is given by .
Remember that for a pure, undoped semiconductor, each free electron creates an electron hole in the lattice. By contrast, when a donor ion adds a free electron to a lattice, it does not create a hole. This is significant, since when we see a free electron in a pentavalent-doped material, we know that it could have come from one of two places: either a positive donor ion or a broken covalent bond in a tetravalent atom such as silicon.
As we dope the sample with positive donor ions, we increase the value of , and so we can tell that the free electron density will be greater than the electron hole density . For this reason, we can call it an “-type” semiconductor, which simply refers to the fact that there are more free electrons than electron holes. This defines one of the two main types of doped semiconductors.
A semiconducting material that has been doped with positive donor ions is known as an -type semiconductor. This is because the free electron density is greater than the electron hole density , following the equation where represents the density of positive donor ions.
Example 2: Free Electron Concentration for an 𝑛-Type Semiconductor
A doped semiconductor that contains donor ions and is at thermal equilibrium is modeled using three variables. The density of free electrons in the semiconductor is represented by . The density of donor ions in the semiconductor is represented by . The density of vacancies in the semiconductor is represented by . Which of the following formulas correctly represents the relationship between these variables in the semiconductor?
Recall that, for a pure semiconductor, each free electron creates and leaves behind a hole in the lattice. Thus, before the donor ions are added, the density of free electrons equals the density of holes, or .
However, when we dope the sample, donor ions contribute free electrons without creating any electron holes, and thus is not equal to .
When we add more free electrons via the donor ions, the density of free electrons is (the amount before doping) plus (the amount of the new electrons added). This is represented in the formula
Therefore, answer choice C is correct.
Thus, we have seen how introducing pentavalent atoms, or atoms with one more outermost electron than silicon, to a lattice increases the overall conductivity by creating an -type semiconductor.
Let us now consider the other type of doped semiconductor, which can be achieved by introducing atoms with one fewer outermost electron than silicon to a lattice. Once again, recall that a neutral silicon atom in a lattice forms covalent bonds with adjacent atoms, which fills its outermost electron shell. We will now explore the effects of replacing an interior silicon atom with a neutral boron atom, which has three outermost electrons.
Example 3: Understanding Electrons in a Doped Semiconductor
The diagram shows a lattice of silicon atoms that contains one atom of boron.
- What is the number of electrons that covalently bond the boron atom with the silicon atoms that surround it?
- What is the number of electrons that would covalently bond with a silicon atom that occupies the position of the boron atom?
A neutral boron atom has three out of eight possible electrons in its outermost shell. Since an interior atom forms covalent bonds with four adjacent atoms, an interior boron atom is surrounded by these four shared electrons, plus the three electrons that it came with.
Thus, the boron atom has 7 electrons covalently bonding it with the neighboring silicon atoms.
Recall that a neutral silicon atom has four out of eight possible electrons in its outermost shell. There would be four atoms surrounding an interior atom, so each pair of atoms would share a pair of electrons in a covalent bond.
Since the atom would have four sets of covalent bonds, there would be 8 electrons covalently bonding it to the lattice.
The example above illustrates how a boron atom in a silicon lattice has seven out of eight possible electrons surrounding it, via covalent bonds. Since it does not contain the full eight electrons that would make the shell complete, we can say that there is one hole in the shell. The presence of this hole means that the boron atom is very likely to “accept” a nearby free electron into its outermost shell. Thus, we can refer to the boron atom (or any trivalent atom used for doping) as a “negative acceptor ion.” We associate this atom with a negative charge because it came into the lattice neural, and with three electrons, but once it accepts a free electron into the hole in its outermost shell, it becomes negatively charged. The symbol often used to represent the concentration of these negative acceptor ions is , where gives a quantity, stands for “acceptor,” and the “” sign refers to the negative charge of the atom after it accepts a free electron.
As we dope a material with negative acceptor ions, we know that the electron hole density will be greater than the free electron density . For this reason, we can say that we have a “-type” semiconductor, which simply refers to the fact that there are more electron holes than free electrons. This defines the second main type of doped semiconductor.
A semiconducting material that has been doped with negative acceptor ions is known as a -type semiconductor. This is because the electron hole density is greater than the free electron density , following the equation where represents the density of negative acceptor ions.
Example 4: Electron Hole Concentration for a 𝑝-Type Semiconductor
A doped semiconductor that contains acceptor ions and is at thermal equilibrium is modeled using three variables. The density of free electrons in the semiconductor is represented by . The density of acceptor ions in the semiconductor is represented by . The density of vacancies in the semiconductor is represented by . Which of the following formulas correctly represents the relationship between these variables in the semiconductor?
Recall that, for a pure, undoped semiconductor, each electron hole was created and left behind by a free electron. Thus, before the acceptor ions are added, the density of holes equals the density of free electrons, or .
However, when we dope the sample, each acceptor ion contributes one hole because there are seven out of eight possible electrons surrounding the dopant. Therefore, when we see a hole in an acceptor ion-doped semiconductor, it could have come from either the dopant itself, or from a broken covalent bond among the original pure semiconductor atoms.
When we add more holes via the acceptor ions, the total density of holes is (the amount before doping) plus (the amount of new holes added). This is represented in the formula and therefore answer choice A is correct.
Thus, we have seen how to create and describe both -type and -type semiconductors.
We can make a generalization that doping a semiconductor is a more efficient way to allow charge flow than simply raising the temperature of a pure sample. Because of this, we can apply some approximations to a couple of the equations we established previously.
First, because the addition of positive donor ions far outweighs the creation of free electrons and holes created by breaking covalent bonds (like we see in a pure sample), we can say that for an -type semiconductor,
This is stating that the great majority of free electrons in our semiconductor are due to doping, so the quantity is much greater than .
Likewise, we can apply this concept to the case of -type semiconductors:
Simply put, the great majority of electron holes in our semiconductor are due to doping, so is much greater than .
Let us now establish one more way to mathematically describe semiconductors, using a formula that nicely summarizes our discussion of both pure and doped semiconductors. As a quick review, recall that, for a doped semiconductor, does not have to be equal to , but for a pure semiconductor, the two quantities are equal.
Pure semiconductors are also called intrinsic semiconductors, and we can use the term (where stands for “intrinsic”) when describing them. For a pure semiconductor, is often used to represent the concentration of free electrons and/or the concentration of holes at the same time, since these two quantities are equal. That is, for a pure semiconductor, the magnitudes of , , and are equal. We can devise an interesting result from these values.
We have seen before that and change as we dope a sample. The following equations will help us determine new, updated values for these quantities.
To begin, we will square the term . Notice that if we square , we can say that also effectively represents the product of the density of free electrons, , and the density of holes, , which can be written as . Thus, we can devise a relationship between and using :
An interesting result of this equation is that it holds true for all semiconductors in thermal equilibrium, not just pure semiconductors.
We have already established that for any doped semiconductor, . The equation above describes the inverse proportionality between these two quantities. If increases, must decrease, like we saw in our discussion of -type semiconductors. Likewise, if increases, must decrease, like we saw in the case of -type materials.
Ultimately, it does not matter whether a semiconductor is -type, -type, or totally pure; the quantity is always constant for a given material at a given temperature. This fact uniquely allows us to explore certain properties of doped semiconductors.
For -type semiconductors, we can relate to the density of electron holes, . Recall that adding positive donor ions to a lattice increases the number of free electrons. With more free electrons present, holes are filled at a greater rate, which decreases the overall concentration of holes. We can find the resulting concentration of holes by using the equation introduced above,
Remember that we are working under the assumption that , so we can make a substitution in the formula
Now we can solve for by dividing the equation by :
This equation relates the quantity to the density of electron holes, , for an -type semiconductor.
Concentration of Electron Holes for an 𝑛-Type Semiconductor
The concentration of electron holes, , for an -type semiconductor is given by where is the concentration of positive donor ions and is the free electron and hole concentration for an undoped sample.
We will use a similar method to relate to the density of free electrons, , in a -type semiconductor. Recall that adding negative acceptor ions to a lattice increases the number of holes. With more holes present, electrons can fill holes at a greater rate, which decreases the overall concentration of free electrons. We can find the resulting concentration of free electrons beginning with the equation
In the case of doping with negative acceptor ions, we use the approximation that . We will use this to substitute the -term out of this equation so that we have
Solving for , the formula becomes
This equation describes the density of free electrons, , for a -type semiconductor.
Concentration of Free Electrons for a 𝑝-Type Semiconductor
The concentration of free electrons, , for a -type semiconductor is given by where is the concentration of negative acceptor ions and is the free electron and hole concentration for an undoped sample.
Let us finish our discussion of doped semiconductors by summarizing some important concepts.
- We can increase the conductivity of a pure semiconductor by raising its temperature or by doping it.
- Doping involves adding “impurities” to a lattice by adding atoms with either three (trivalent) or five (pentavalent) outermost electrons.
- -type semiconductors are doped with pentavalent atoms, or positive donor ions, whose concentration is represented by . The concentration of free electrons is given by and is approximately equal to .
- -type semiconductors are doped with trivalent atoms, or negative acceptor ions, whose concentration is represented by . The concentration of holes is given by and is approximately equal to .
- For both pure and doped semiconductors, , where is the free electron and hole concentration for an undoped sample. As a result, the concentration of electron holes for an -type semiconductor is given by . Likewise, the concentration of free electrons for a -type semiconductor is given by .