Complete Electronics Self-Teaching Guide with Projects
Page 7
Note The β introduced here is referred to in manufacturers' specification sheets as hFE. Technically, it is referred to as the static or DC β. For the purposes of this chapter, it is called β. Discussions on transistor parameters in general, which are well covered in many textbooks, will not be covered here.
17 The mathematical formula for current gain is as follows:
In this equation, the following is true:
IB = base current
IC = collector current
The equation for β can be rearranged to give IC = βIB. From this, you can see that if no base current flows, no collector current flows. Also, if more base current flows, more collector current flows. This is why it's said that the base current controls the collector current.
Question
Suppose the base current is 1 mA and the collector current is 150 mA. What is the current gain of the transistor? _____
Answer
150
Inside the Bipolar Transistor
Now take a closer look inside a bipolar transistor. In a bipolar transistor, with no voltages applied, two depletion regions exist. As shown in the following figure, one depletion region exists at the emitter and base junction, and one at the collector and base junction. There are no free electrons or holes in these depletion regions, which prevents any current from flowing.
The following figure shows a bipolar transistor with a positive voltage applied to the base and a negative voltage applied to the emitter (forward-biasing the base-emitter diode), as well as a positive voltage applied to the collector. The forward bias on the base-emitter diode allows electrons to flow from the emitter into the base. A small fraction of these electrons would be captured by the holes in the base region and then flow out of the base terminal and through the base resistor. The base of a transistor is thin, which allows most of the electrons from the emitter to flow through the base and into the collector. The positive voltage on the collector terminal attracts these electrons, which flow out of the collector terminal and through the collector resistor.
For a transistor with β = 100, only one electron flows out to the base terminal for every 100 electrons that flow to the collector terminal. β is controlled by two factors: the thickness of the base region and the relative concentration of the impurities providing holes in the P region to the concentration of the impurities providing electrons in the N regions. A thinner base region plus the lower relative concentration of holes allow more electrons to pass through the base without being captured, resulting in a higher β. (Remember, the direction that conventional electrical current flows in is opposite to the direction in which electrons flow.)
Before you connect any bipolar transistor to other components in a circuit, you must identify the emitter, base, and collector leads (referred to as the transistor's pinout) and determine whether the transistor is NPN or PNP.
Transistors are marked with a part number, such as 2N3904, 2N3906, BC337, and PN2222. However, the part numbers don't tell you much about the transistor. For these transistors, the 2N3904, BC337, and PN2222 are NPN, whereas the 2N3906 is a PNP, which is not obvious from the part number.
Also, the transistor pinout is not identified on the part number. For example, one NPN transistor, the BC337, uses the opposite leads for the emitter and collector than the 2N3904 transistor, as shown in the following figure.
Your best bet is to refer to the manufacturer's data sheet for the transistor pinout and other characteristics. You can easily look up data sheets on the Internet. Also, you can find links to the data sheets for the transistors used in projects in this book on the website at www.buildinggadgets.com/index_datasheets.htm.
18 Current gain is a physical property of transistors. You can find its value in the manufacturers' published data sheets, or you can determine it by experimenting.
In general, β is a different number from one transistor part number to the next, but transistors with the same part number have β values within a narrow range of each other.
One of the most frequently performed calculations in transistor work is to determine the values of either collector or base current, when β and the other current are known.
For example, suppose a transistor has 500 mA of collector current flowing, and you know it has a β value of 100. To find the base current, use the following formula:
Questions
Calculate the following values:
A. IC = 2 ampere, = 20. Find IB. _____
B. IB = 1 mA, β = 100. Find IC. _____
C. IB = 10 μA, β = 250. Find IC. _____
D. IB = 0.1 mA, IC = 7.5 mA. Find β. _____
Answers
A. 0.1 ampere, or 100 mA
B. 100 mA
C. 2500 μA, or 2.5 mA
D. 75
19 This problem serves as a summary of the first part of this chapter. You should be able to answer all these questions. Use a separate sheet of paper for your drawing and calculations.
Questions
A. Draw a transistor circuit utilizing an NPN transistor, a base resistor, a collector resistor, and one battery to supply both base and collector currents. Show the paths of IB and IC.
B. Which current controls the other? _____
C. Which is the larger current, IB or IC? _____
D. IB = 6 μA, β = 250. Find IC. _____
E. IC = 300 mA, β = 50. Find IB. _____
Answers
A. See Figure 3.17 and Figure 3.18.
B. IB (base current) controls IC (collector current).
C. IC
D. 1.5 mA
E. 6 mA
Project 3.1: The Transistor
Objective
The objective of this project is to find β of a particular transistor by setting several values of base current and measuring the corresponding values of collector current. Next, you divide the values of collector current by the values of the base current to determine β. The value of β will be almost the same for all the measured values of current. This demonstrates that β is a constant for a transistor.
General Instructions
While the circuit is set up, measure the collector voltage for each current value. This demonstrates (experimentally) some points that are covered in future problems. As you perform the project, observe how the collector voltage VC drops as the collector current increases.
Parts List
One 9 V battery (or a lab power supply)
One multimeter set to μA
One multimeter set to mA
One multimeter set to measure DC voltage
One 10 kΩ resistor
One 510 ohm resistor
One 2N3904 transistor
One breadboard
One 1 MΩ potentiometer
Step-by-Step Instructions
Set up the circuit shown in Figure 3.19 on a breadboard. If you have some experience in building circuits, this schematic (along with the previous parts list) should provide all the information you need to build the circuit. If you need a bit more help building the circuit, look at the photos of the completed circuit in the “Expected Results” section.
Figure 3.19
Follow these steps, recording your measurements in the blank table following the steps.
1. Set the potentiometer to its highest value; this sets IB to its lowest possible value.
2. Measure and record IB.
3. Measure and record IC.
4. Measure and record VC. This voltage is sometimes referred to as the collector-emitter voltage (VCE), because it is taken across the collector-emitter leads if the emitter is connected to ground or the negative of the power supply.
5. Adjust the potentiometer to give the next targeted value of IB. You do not need to hit these values exactly. For example, for a target of 20 μA, a measured value of 20.4 μA is fine.
6. Measure and record the new values for IB, IC, and VC.
7. Adjust the potentiometer to give the next targeted value of IB.
8. Measure and record the new values fo
r IB, IC, and VC again.
9. Repeat steps 7 and 8 for each of the targeted values of IB.
10. For each value of IB and its corresponding value of IC, calculate the value of β (β = IC/IB). The values will vary slightly but will be close to an average. Did you get a consistent β?
Save this circuit. You use it later in this chapter in Project 3.2.
Expected Results
Figure 3.20 shows the breadboarded circuit for this project.
Figure 3.20
Figure 3.21 shows the test setup for this project.
Compare your measurements with the ones shown in the following table.
Figure 3.21
Don't worry if your results give a different value of β. The manufacturing process that produces transistors can allow variation of the base thickness and doping levels, which causes variation of β in the finished transistors.
20 In Project 3.1, you measured the voltage level at the collector (VC) and recorded your measurements. Now, examine how to determine the voltage at the collector, when it's not possible to measure the voltage level.
Use the values shown in the circuit in Figure 3.22 to complete the following steps:
1. Determine IC.
2. Determine the voltage drop across the collector resistor RC. Call this VR.
3. Subtract VR from the supply voltage to calculate the collector voltage.
Figure 3.22
Here is the first step:
1. To find IC, you must first find IB.
Now, perform the next two steps.
Questions
A. VR = _____
B. VC = _____
Answers
A. To find VR:
VR = RC × IC = 1 kΩ × 5 mA = 5 volts
B. To find VC:
VC = VS – VR = 10 volts – 5 volts = 5 volts
21 Determine parameters for the circuit shown in Figure 3.22 using the value of β = 75.
Questions
Calculate the following:
A. IC = _____
B. VR = _____
C. VC = _____
Answers
A.
B. VR = 1 kΩ × 7.5 mA = 7.5 volts
C. VC = 10 volts – 7.5 volts = 2.5 volts
22 Determine parameters for the same circuit, using the values of RB = 250 kΩ and β = 75.
Questions
Calculate the following:
A. IC = _____
B. VR = _____
C. VC = _____
Answers
A.
B. VR = 1 kΩ × 3 mA = 3 volts
C. VC = 10 volts−3 volts = 7 volts
23 From the preceding problems, you can see that you can set VC to any value by choosing a transistor with an appropriate value of β or by choosing the correct value of RB.
Now, consider the example shown in Figure 3.23. The objective is to find VC. Use the steps outlined in problem 20.
Figure 3.23
Questions
Calculate the following:
A. IB = _____
IC = _____
B. VR = _____
C. VC = _____
Answers
Your results should be as follows:
A.
B. VR = 1 kΩ × 10 mA = 10 volts
C. VC = 10 volts −10 volts = 0 volts.
Here the base current is sufficient to produce a collector voltage of 0 volts and the maximum collector current possible, given the stated values of the collector resistor and supply voltage. This condition is called saturation.
24 Look at the two circuits shown in Figure 3.24 and compare their voltages at the point labeled VC.
Figure 3.24
Consider a transistor that has sufficient base current and collector current to set its collector voltage to 0 volts. Obviously, this can be compared to a closed mechanical switch. Just as the switch is said to be ON, the transistor is also said to be “turned on” (or just ON).
Questions
A. What can you compare a turned on transistor to? _____
B. What is the collector voltage of an ON transistor? _____
Answers
A. A closed mechanical switch
B. 0 volts
Project 3.2: The Saturated Transistor
Objective
Normally, for a transistor, IC = β × IB. However, this relationship does not hold when a transistor is saturated. The objective of this project is to determine the relationship of the collector current to the base current when a transistor is saturated.
General Instructions
Using the same breadboarded circuit you built in Project 3.1, set the base current to several values, starting at 90 μA and increasing the base current. Record measurements of the collector current and collector voltage at each value of the base current.
Step-by-Step Instructions
Follow these steps and record your measurements in the blank table following the steps.
1. Set up the circuit you built in Project 3.1.
2. Adjust the potentiometer to set IB at approximately 90 μA.
3. Measure and record IB.
4. Measure and record IC.
5. Measure and record VC.
6. Adjust the potentiometer slightly to lower its resistance, which sets a larger value of IB.
7. Measure and record the new values for IB, IC, and VC.
8. Repeat steps 6 and 7 until you reach the lower limit of the potentiometer, which is also the highest value of IB.
IB (μA) IC (mA) VC (volts)
Expected Results
Figure 3.25 shows the test setup for this project with the potentiometer set at its lower limit, providing the highest value of IB. The test setup is the same as that used in Project 3.1; however, the value of IB is considerably higher than in Project 3.1.
Compare your measurements to the ones shown in the following table.
IB (μA) IC (mA) VC (volts)
91 14.5 1.53
101 15.9 0.843
126 16.9 0.329
150 17.0 0.256
203 17.1 0.211
264 17.1 0.189
389 17.2 0.163
503 17.2 0.149
614 17.2 0.139
780 17.2 0.127
806 17.2 0.126
Your data will probably have slightly different values but should indicate that IC stays constant for values of VC of 0.2 and below, whereas IB continues to rise. In this region, the transistor is fully ON (saturated) and IC can't increase further. This agrees with the data sheet published by Fairchild Semiconductor for the 2N3904 transistor, which indicates that the transistor saturates at VC = 0.2 volts.
Figure 3.25
25 Now, compare the circuits shown in Figure 3.26.
Figure 3.26
Because the base circuit is broken (that is, it is not complete), there is no base current flowing.
Questions
A. How much collector current is flowing? _____
B. What is the collector voltage? _____
C. What is the voltage at the point VC in the mechanical switch circuit? _____
Answers
A. None.
B. Because there is no current flowing through the 1 kΩ resistor, there is no voltage drop across it, so the collector is at 10 volts.
C. 10 volts because there is no current flowing through the 1 kΩ resistor.
26 From problem 25, it is obvious that a transistor with no collector current is similar to an open mechanical switch. For this reason, a transistor with no collector current and its collector voltage at the supply voltage level is said to be “turned off” (or just OFF).
Question
What are the two main characteristics of an OFF transistor? _____
Answer
It has no collector current, and the collector voltage is equal to the supply voltage.
27 Now, calculate the following parameters for the circuit in Figure 3.27, and compare the results to the examples in problems 25 and 26. Again, the objective here is to find VC.
; Figure 3.27
Questions
A. IB = _____
IC = _____
B. VR = _____
C. VC = _____
Answers
A.
B. VR = 1 kΩ × 5 mA = 5 volts
C. VC = 10 volts−5 volts = 5 volts
Note The output voltage in this problem is half of the supply voltage. This condition is important in AC electronics and is covered in Chapter 8.
The Junction Field Effect Transistor (JFET)
28 Up to now, the only transistor described has been the BJT. Another common transistor type is the JFET. Like the BJT, the JFET is used in many switching and amplification applications. The JFET is preferred when a high input impedance circuit is needed. The BJT has a relatively low input impedance as compared to the JFET. Like the BJT, the JFET is a three-terminal device. The terminals are called the source, drain, and gate. They are similar in function to the emitter, collector, and base, respectively.
Questions
A. How many terminals does a JFET have, and what are these terminals called? _____
B. Which terminal has a function similar to the base of a BJT? _____
Answers
A. Three, called the source, drain, and gate.
B. The gate has a control function similar to that of the base of a BJT.
29 The basic design of a JFET consists of one type of semiconductor material with a channel made of the opposite type of semiconductor material running through it. If the channel is N material, it is called an N-channel JFET; if it is P material, it is called a P-channel.
Figure 3.28 shows the basic layout of N and P materials, along with their circuit symbols. Voltage on the gate controls the current flow through the drain and source by controlling the effective width of the channel, allowing more or less current to flow. Thus, the voltage on the gate acts to control the drain current, just as the voltage on the base of a BJT acts to control the collector current.