Lesson Explainer: Transistors Physics • 9th Grade

In this explainer, we will learn how to describe how transistors can be used as electrical switches in circuits.

The most important property of a transistor is that it can act like a switch. More precisely, a transistor can make a small change in current cause a much larger change in current to occur.

A transistor contains three doped semiconducting regions.

An n-type doped semiconductor consists of an atomic lattice that contains more free electrons than it contains atoms with vacancies in their outer shells.

A p-type doped semiconductor consists of an atomic lattice that contains more atoms with vacancies in their outer shells than it contains free electrons.

A transistor can be formed by placing a p-type semiconductor between two n-type semiconductors. A transistor can also be formed by placing an n-type semiconductor between two p-type semiconductors.

These types of transistors are shown in the following figure.

Both NPN and PNP transistors consist of three regions.

When connected to a circuit, a transistor has a connection to the circuit from each of its regions.

The three regions of a transistor are called

  • the emitter,
  • the collector,
  • the base.

The base is much less strongly doped than the emitter.

A circuit that connects these three regions in this way is called a common emitter configuration circuit. This is shown in the following figure.

We see that the transistor in the circuit is an NPN transistor.

For there to be current in this circuit, there must be potential difference across some part of the circuit. In a transistor circuit, there is in fact a potential difference source in each loop of the circuit. A resistor is also included in each loop of the circuit.

The circuit including all its components is shown in the following figure.

A transistor circuit can also be represented using a transistor circuit symbol. This is shown in the following figure.

For a PNP transistor, the symbol is slightly different, as the following figure shows.

We recall that applying a potential difference across a boundary of p-type and n-type semiconducting materials applies either a forward bias or a reverse bias across the boundary of the materials.

Let us look at an example involving a transistor circuit.

Example 1: Identifying the Regions of a Transistor

An NPN transistor is connected to two direct current sources, as shown in the diagram. The two n-regions are identical.

  1. Which of the regions of the transistor is the collector region?
  2. Which of the regions of the transistor is the emitter region?

Answer

Part 1

The positive terminal of the source that connects to both N and N is connected to N.

For a common emitter NPN transistor, the positive terminal of the source that connects to both N and N connects to the collector.

Part 2

The negative terminal of the source that connects to both N and N is connected to N.

For a common emitter NPN transistor, the negative terminal of the source that connects to both N and N connects to the emitter.

The current in a transistor circuit is affected by the biases at the boundaries of the base and the regions adjacent to it.

The following figure shows how the p-type and n-type materials in an NPN transistor respond to the potential difference sources in the circuit. Free electrons are shown by blue circles. Vacancies are shown by red rings.

The diagram shows four important things:

  • The base region is thinner than the collector and emitter regions. For a real transistor, the base region is extremely thin compared to the other regions. The difference in thickness is much greater than is shown in the diagram.
  • The concentration of vacancies in the base region is much lower than the concentration of free electrons in the emitter and collector regions.
  • The emitter is forward biased, and the collector is reverse biased.
  • The negative terminals of both potential difference sources are at the same potential.

The currents in different parts of this circuit depend on the semiconducting properties and dimensions of the emitter, base, and collector regions.

In the circuit, the current directions for each connection of the transistor are as follows:

  • There is a current out of connection to the emitter region. This can be denoted 𝐼E.
  • There is a current into the connection of the collector region. This can be denoted 𝐼C.
  • There is a current into the connection of the base region. This can be denoted 𝐼B.

These currents are shown in the following figure. The flow of free electrons is also shown.

We see that 𝐼E is due to free electrons moving from the emitter into the base region.

Free electrons that travel from the emitter to the base are accelerated by the forward bias at the emitter, toward the collector. Most of these electrons have sufficient energy to overcome the effect of the reverse bias across the collector and pass into the collector region.

A small portion of the electrons from the emitter region recombine with vacancies in the base. The base current consists of these electrons.

The magnitude of 𝐼B compared to 𝐼C depends on the thickness of the base region and on the difference in the doping concentration of the emitter and base regions.

There is a formula relating the currents in a transistor circuit.

Formula: The Relation between the Emitter, Base, and the Collector Currents

The values of the collector current, 𝐼C, the emitter current, 𝐼E, and the base current, 𝐼B, are related as follows: 𝐼=𝐼+𝐼.ECB

The ratio of 𝐼C to 𝐼B is an important value for a transistor circuit. For a common emitter transistor circuit, the value of 𝐼B is usually much less than that of 𝐼C. This is because the base region has a low doping concentration and thickness compared to the collector region.

The ratio of 𝐼C to 𝐼B can be determined by expressing 𝐼C as a fraction of 𝐼E. The constant of proportionality between 𝐼C and 𝐼E is called 𝛼. This means that 𝐼=𝐼𝛼.CE

It must be therefore that 𝐼=𝐼(1𝛼).BE

The ratio of 𝐼C to 𝐼B is therefore given by 𝐼𝐼=𝐼𝛼𝐼(1𝛼)𝐼𝐼=𝛼1𝛼=𝛽,CBEECB where 𝛽 is called the current gain of the circuit.

Formula: Current Gain in Common Emitter Connection

The current gain of a transistor circuit, 𝛽, is given by 𝛽=𝐼𝐼,CB where 𝐼C is the collector current and 𝐼B is the base current.

The magnitude of 𝐼B compared to 𝐼C depends on the thickness of the base region and on the difference in the doping concentration of the emitter and base regions.

For a circuit where 𝐼𝐼,BC it must be the case that 𝛼1 and hence that 𝛽 is a very large value.

The following figure shows a common emitter transistor circuit, with various circuit values labeled.

The currents 𝐼C, 𝐼E, and 𝐼B are shown, as are

  • 𝑉CC, the potential difference supplied across the collector and the emitter,
  • 𝑉CE, the potential difference across the collector and the emitter,
  • 𝑉BE, the potential difference supplied across the base and the emitter,
  • 𝑅C, the resistance to the collector current,
  • 𝑅B, the resistance to the base current.

The emitter region contact of the transistor is at zero potential compared to both 𝑉CC and 𝑉BE.

𝑉BE is called the input potential, and 𝑉CE is called the output potential.

Let us look at an example involving the currents in a transistor circuit.

Example 2: Determining Currents in a Transistor Circuit

An NPN transistor is connected to a power supply with voltage 𝑉CC. A power supply with voltage 𝑉EB is connected across the transistor’s emitter and base terminals, as shown in the diagram. There is a current 𝐼=99.5CmA between 𝑉CC and the collector terminal, a current 𝐼E between 𝑉EB and the emitter terminal, and a current 𝐼=0.5BmA between 𝑉EB and the base terminal.

  1. Calculate 𝐼E.
  2. Find the rate at which free electrons diffusing through the base region recombine with holes. Use 1.6×10 C for the charge of an electron. Answer in scientific notation to one decimal place.

Answer

Part 1

The currents in the circuit are related by the equation 𝐼=𝐼+𝐼.ECB Substituting the values stated in the question, we see that 𝐼=99.5+0.5=100.EmAmAmA

Part 2

The base current is assumed here to entirely consist of free electrons that recombine with holes in the base. The current in the base region is 0.5 mA, which is 5×10 A. One ampere is equal to one coulomb per second.

The number of electrons, 𝑛, recombining per second to produce this current is given by 𝑛=5×10/1.6×10.CsC

In scientific notation, to one decimal place, 𝑛 is 3.1×10 s−1.

Let us look at an example involving the current gain in a transistor circuit.

Example 3: Determining the Current Gain for a Transistor Circuit

An NPN transistor is connected to a power supply with voltage 𝑉CC. A power supply with voltage 𝑉EB is connected across the transistor’s emitter and base terminals, as shown in the diagram. There is a current 𝐼=99.5CmA between 𝑉CC and the collector terminal, a current 𝐼=100.0EmA between 𝑉EB and the emitter terminal, and a current 𝐼B between 𝑉EB and the base terminal.

  1. Calculate 𝐼B.
  2. The dc current gain of the transistor is equal to the ratio of 𝐼C to 𝐼B. Calculate the dc current gain of the transistor.

Answer

Part 1

The currents in the circuit are related by the equation 𝐼=𝐼+𝐼.ECB

We can make 𝐼B the subject of this equation, giving us 𝐼𝐼=𝐼.ECB

Substituting the values stated in the question, we see that 𝐼=10099.5=0.5.BmAmAmA

Part 2

The base current gain, 𝛽, is given by the equation 𝛽=𝐼𝐼.CB

Substituting the values stated in the question, we see that 99.50.5=199.mAmA

We can see from Kirchhoff’s second law that, in a transistor circuit, 𝑉=𝑉𝐼𝑅.CECCCC

We know that, in such a circuit, the collector current and base current are related by the current gain according to 𝐼=𝛽𝐼.CB

This means that 𝐼C can be changed by increasing the input potential, as increasing 𝑉BE increases 𝐼B.

We can call 𝐼B the input current and call 𝐼C the output current.

For a constant 𝛽, the ratio of 𝐼B to 𝐼C is a constant for a transistor. We see then that increasing the input current increases the output current.

Let us look at an example involving the current changes in a transistor circuit.

Example 4: Relating Current Changes in a Transistor Circuit

An NPN transistor is connected to a power supply with voltage 𝑉CC. A power supply with voltage 𝑉EB is connected across the transistor’s emitter and base terminals, as shown in the diagram. There is a current 𝐼C between 𝑉CC and the collector terminal, a current 𝐼E between 𝑉EB and the emitter terminal, and a current 𝐼B between 𝑉EB and the base terminal. An external resistance 𝑅C is placed between 𝑉CC and the collector terminal, and an external resistance 𝑅B is placed between 𝑉EB and the base terminal. The potential difference across the collector and emitter terminals is 𝑉CE.

  1. If the value of 𝑅B is reduced, which of the following most correctly describes the effect on the value of 𝐼C?
    1. 𝐼C increases.
    2. 𝐼C decreases.
    3. 𝐼C is constant.
  2. If the value of 𝑅B is increased, which of the following most correctly describes the effect on the value of 𝐼C?
    1. 𝐼C is constant.
    2. 𝐼C increases.
    3. 𝐼C decreases.

Answer

Part 1

Decreasing 𝑅B increases 𝐼B.

From the equation 𝐼=𝛽𝐼,CB we see that, for a constant 𝛽, increasing 𝐼B will increase 𝐼C.

Part 2

Increasing 𝑅B decreases 𝐼B.

From the equation 𝐼=𝛽𝐼,CB we see that, for a constant 𝛽, decreasing 𝐼B will decrease 𝐼C.

The relationship between the input and output values of current are not directly proportional.

This means that the value of the current gain is not actually constant but is approximately constant for some values of 𝐼B and 𝐼C.

To show how the change in the value of 𝐼B corresponds to a large change of 𝐼C, let us show the effect of a small change in a small number that is used to divide a much larger number.

For example, consider the equation 𝑛=𝑎𝑏.

Let 𝑎=1 and let 𝑏=0.002.

We have then 𝑛=10.002=500.

Now, let us suppose we have a value Δ𝑚=Δ𝑎=Δ𝑏.

Let Δ𝑚 be 0.001.

This means that 𝑎 increases by 0.001 and 𝑏 decreases by 0.001.

We have then 𝑛=1.0010.001=1001.

We see that a change of 𝑚 of 0.001 has increased 𝑛 by 501.

Now, suppose that we let Δ𝑚 be 0.0015.

We have then 𝑛=1.00150.0005=2003.

We see that a change of 𝑚 of 0.0015 has increased 𝑛 by 1‎ ‎503.

A graph of 𝑛 vs 𝑚 shows how much greater the change of 𝑛 can be than the change in 𝑚.

We can see that this graph mostly consists of two regions. In one region, the value of 𝑛 is approximately constant as the value of 𝑚 changes, and in the other region, the value of 𝑚 is approximately constant as 𝑛 changes. These regions correspond to approximately constant values of current gain for a transistor when it is acting as a closed switch and as an open switch.

We can also consider the input and output potentials instead of the input and output currents.

From the equation 𝑉=𝑉𝐼𝑅,CECCCC we see that, for the maximum value of 𝐼C, the minimum value of output potential is obtained.

If the input potential is decreased, both the input current and the output current decrease. The output current is zero for zero input current.

For zero output current, the maximum value of output potential is obtained.

A graph of the change of output potential of a transistor with its input potential is shown in the following figure.

Let us now summarize what has been learned in this explainer.

Key Points

  • A transistor consists of either two n-type semiconductors on either side of a p-type semiconductor (NPN) or two p-type semiconductors on either side of an n-type semiconductor (PNP).
  • A transistor has collector, emitter, and base terminals. There is one terminal for each semiconductor region.
  • A transistor is used in a circuit that contains two potential difference sources. The potential difference sources forward bias the emitter and reverse bias the collector.
  • The currents in different parts of a transistor circuit depend on the semiconducting properties and dimensions of the emitter, base, and collector regions.
  • The currents at the emitter, 𝐼E, collector, 𝐼C, and base, 𝐼B, terminals are related by the formula 𝐼=𝐼+𝐼.ECB
  • The currents at the emitter, 𝐼E, and collector, 𝐼C, terminals are related by the formula 𝐼=𝐼𝛼,CE where 𝛼 is a constant.
  • The current at the collector, 𝐼C, and base, 𝐼B, terminals are related by the formula 𝐼𝐼=𝛼1𝛼=𝛽,CB where 𝛽 is the current gain of the circuit.
  • A small change in 𝐼B can cause a much greater change in 𝐼C, allowing a circuit with a transistor to effectively switch 𝐼C on or off.

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