by Earl Boysen
Questions
A. Which JFET would use electrons as the primary charge carrier for the drain current? _____
B. What effect does changing the voltage on the gate have on the operation of the JFET? _____
Figure 3.28
Answers
A. N-channel because N material uses electrons as the majority carrier.
B. It changes the current in the drain. The channel width is controlled electrically by the gate potential.
30 To operate the N-channel JFET, apply a positive voltage to the drain with respect to the source. This allows a current to flow through the channel. If the gate is at 0 volts, the drain current is at its largest value for safe operation, and the JFET is in the ON condition.
When a negative voltage is applied to the gate, the drain current is reduced. As the gate voltage becomes more negative, the current lessens until cutoff, which occurs when the JFET is in the OFF condition.
Figure 3.29 shows a typical biasing circuit for the N-channel JFET. For a P-channel JFET, you must reverse the polarity of the bias supplies.
Figure 3.29
Question
How does the ON-OFF operation of a JFET compare to that of a BJT? _____
Answer
The JFET is ON when there are 0 volts on the gate, whereas you turn the BJT ON by applying a voltage to the base. You turn the JFET OFF by applying a voltage to the gate, and the BJT is OFF when there are 0 volts on the base. The JFET is a “normally ON” device, but the BJT is considered a “normally OFF” device. Therefore, you can use the JFET (like the BJT) as a switching device.
31 When the gate to source voltage is at 0 volts (VGS = 0) for the JFET (refer to Figure 3.29), the drain current is at its maximum (or saturation) value. This means that the N-channel resistance is at its lowest possible value, in the range of 5 to 200 ohms. If RD is significantly greater than this, the N-channel resistance, rDS, is assumed to be negligible.
Questions
A. What switch condition would this represent, and what is the drain to source voltage (VDS)? _____
B. As the gate becomes more negative with respect to the source, the resistance of the N-channel increases until the cutoff point is reached. At this point, the resistance of the channel is assumed to be infinite. What condition does this represent, and what is the drain to source voltage? _____
C. What does the JFET act like when it is operated between the two extremes of current saturation and current cutoff? _____
Answers
A. Closed switch, VDS = 0 volts, or low value
B. Open switch, VDS = VDD
C. A variable resistance
Inside the JFET
Now take a closer look at the inside of an N-channel JFET. With 0 volts applied to the gate, the channel is at its widest, and the maximum amount of current can flow between the drain and the source. If you apply a negative voltage to the gate, electrons in the channel are repelled from the negative voltage, forming depletion regions on each side of the channel, which narrows the channel, as shown in the following figure.
Further increasing the negative voltage on the gate repels additional electrons, increasing the width of the depletion region and decreasing the width of the channel. The narrower the channel, the higher its resistance. When you apply high enough negative voltage, the depletion regions completely block the channel, as shown in the following figure, cutting off the flow of current between the drain and source. This voltage is called the cutoff voltage.
Before you connect any JFET to other components in a circuit, you must identify the drain, gate, and source leads (referred to as the JFET's pinout) and determine whether the component is an N-channel or a P-channel JFET.
Transistors are marked with a part number. For example, 2N3819, 2N5951, and 2N5460 are all part numbers of JFETs. However, the part numbers don't tell you much about the JFET. For these three transistors, the 2N3819 and 2N5951 are N channel JFETs, whereas the 2N5460 transistor is a P channel JFET. This is not obvious from the part numbers.
Also, the JFET pinout is not identified on the part number. For example, one N-channel JFET, the 2N3819, uses different leads for the gate, drain, and source than the 2N5951 N-channel JFET, as shown in the following figure.
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 JFETs used in this book on the website at www.buildinggadgets.com/index_datasheets.htm.
Summary
At this point, it's useful to compare the properties of a mechanical switch with the properties of both types of transistors, as summarized in the following table.
Switch BJT JFET
OFF (or open)
No current. No collector current. No drain current.
Full voltage across terminals. Full supply voltage between collector and emitter. Full supply voltage between drain and source.
ON (or closed)
Full current. Full circuit current. Full circuit current.
No voltage across terminals. Collector to emitter voltage is 0 volts. Drain to source voltage is 0 volts.
The terms ON and OFF are used in digital electronics to describe the two transistor conditions you just encountered. Their similarity to a mechanical switch is useful in many electronic circuits.
In Chapter 4 you learn about the transistor switch in more detail. This is the first step toward an understanding of digital electronics. In Chapter 8 you examine the operation of the transistor when it is biased at a point falling between the two conditions, ON and OFF. In this mode, the transistor can be viewed as a variable resistance and used as an amplifier.
Self-Test
The following questions test your understanding of the concepts presented in this chapter. Use a separate sheet of paper for your drawings or calculations. Compare your answers with the answers provided following the test.
1. Draw the symbols for an NPN and a PNP bipolar transistor. Label the terminals of each.
2. Draw the paths taken by the base and collector currents, as shown in Figure 3.30.
Figure 3.30
3. What causes the collector current to flow? _____
4. What is meant by the term current gain? What symbol is used for this? What is its algebraic formula? _____
Use the circuit in Figure 3.30 to answer questions 5 through 10.
5. Assume that the transistor is made of silicon. Set RB = 27 kΩ and VS = 3 volts. Find IB. _____
6. If RB = 220 kΩ and VS = 10 volts. Find IB. _____
7. Find VO when RB = 100 kΩ, VS = 10 volts, RC = 1 kΩ, and β = 50. _____
8. Find VO when RB = 200 kΩ, VS = 10 volts, RC = 1 kΩ, and β = 50. _____
9. Now use these values to find VO: RB = 47 kΩ, VS = 10 volts, RC = 500 ohms, and β = 65. _____
10. Use these values to find VO: RB = 68 kΩ, VS = 10 volts, RC = 820 ohms, and β = 75. _____
11. Draw the symbols for the two types of JFETs and identify the terminals. _____
12. What controls the flow of current in both a JFET and a BJT? _____
13. In the JFET common source circuit shown in Figure 3.31, add the correct polarities of the power supplies, and draw the current path taken by the drain current.
Figure 3.31
14. When a base current is required to turn a BJT ON, why is there no gate current for the JFET in the ON state. _____
15. Answer the following questions for the circuit shown in Figure 3.32.
Figure 3.32
A. If the switch is at position A, what will the drain current be, and why? _____
B. If the switch is at position B, and the gate supply voltage is of sufficient value to cause cutoff, what will the drain current be, and why? _____
C. What is the voltage from the drain to the source for the two switch positions? _____
Answers to Self-Test
If your answers do not agree with those that follow, review the problems indicated in
parentheses before you go to Chapter 4.
1. See Figure 3.33.
Figure 3.33
(problems 4 and 5)
2. See Figure 3.34.
Figure 3.34
(problems 13 and 15)
3. Base current. (problem 15)
4. Current gain is the ratio of collector current to base current. It is represented by the symbol β.β = IC/IB. (problems 16 and 17)
5. (problem 7)
6. (problem 7)
7. 5 volts (problems 20–23)
8. 7.5 volts (problems 20–23)
9. 3.1 volts (problems 20–23)
10. 1 volt (problems 20–23)
11. See Figure 3.35.
Figure 3.35
(problem 29)
12. The voltage on the gate controls the flow of drain current, which is similar to the base voltage controlling the collector current in a BJT. (problem 29)
13. See Figure 3.36.
Figure 3.36
(problem 30)
14. The JFET is a high-impedance device and does not draw current from the gate circuit. The BJT is a relatively low-impedance device and does, therefore, require some base current to operate. (problem 28)
15A. The drain current will be at its maximum value. In this case, it equals VDD/RD because you can ignore the drop across the JFET. The gate to source voltage is 0 volts, which reduces the channel resistance to a small value close to 0 ohms. (problem 31)
15B. The drain current now goes to 0 ampere because the channel resistance is at infinity (very large), which does not allow electrons to flow through the channel.
15C. At position A, VDS is approximately 0 volts. At position B, VDS = VDD.
Chapter 4
The Transistor Switch
Transistors are everywhere. You can't avoid them as you move through your daily tasks. For example, almost all industrial controls, and even your MP3 player, stereo, and television may use transistors as switches.
In Chapter 3, “Introduction to the Transistor,” you saw how a transistor can be turned ON and OFF, similar to a mechanical switch. Computers work with Boolean algebra, which uses only two logic states—TRUE and FALSE. These two states are easily represented electronically by a transistor that is ON or OFF. Therefore, the transistor switch is used extensively in computers. In fact, the logic portions of microprocessors (the brains of computers) consist entirely of transistor switches.
This chapter introduces the transistor's simple and widespread application—switching, with emphasis on the bipolar junction transistor (BJT).
When you complete this chapter, you will be able to do the following:
Calculate the base resistance, which turns a transistor ON and OFF.
Explain how one transistor turns another ON and OFF.
Calculate various currents and resistances in simple transistor switching circuits.
Calculate various resistances and currents in switching circuits, which contain two transistors.
Compare the switching action of a junction field effect transistor (JFET) to a BJT.
Turning the Transistor On
1 Start by examining how to turn a transistor ON by using the simple circuit shown in Figure 4.1. In Chapter 3, RB was given, and you had to find the value of collector current and voltages. Now, do the reverse. Start with the current through RC, and find the value of RB that turns the transistor ON and permits the collector current to flow.
Figure 4.1
Question
What current values do you need to know to find RB? _____
Answer
The base and collector currents
2 In this problem circuit, a lamp can be substituted for the collector resistor. In this case, RC (the resistance of the lamp) is referred to as the load, and IC (the current through the lamp) is called the load current.
Questions
A. Is load current equivalent to base or collector current? _____
B. What is the path taken by the collector current discussed in problem 1? Draw this path on the circuit.
Answers
A. Collector current
B. See Figure 4.2. In this figure, note that the resistor symbol has been replaced by the symbol for an incandescent lamp.
Figure 4.2
3 For the transistor switch to perform effectively as a CLOSED switch, its collector voltage must be at the same voltage as its emitter voltage. In this condition, the transistor is said to be turned ON.
Questions
A. What is the collector voltage when the transistor is turned ON? _____
B. What other component does an ON transistor resemble? _____
Answers
A. The same as the emitter voltage, which, in this circuit, is 0 volts
B. A closed mechanical switch
Note In actual practice, there is a small voltage drop across the transistor from the collector to the emitter. This is actually a saturation voltage and is the smallest voltage drop that can occur across a transistor when it is ON as “hard” as possible. The discussions in this chapter consider this voltage drop to be a negligible value; therefore, the collector voltage is said to be 0 volts. For a quality switching transistor, this is a safe assumption.
4 The circuit in Figure 4.3 shows a lamp with a resistance of 240 ohms in place of RC.
Figure 4.3
This figure shows the supply voltage and the collector resistance. Given these two values, using Ohm's law, you can calculate the load current (also called the collector current) as follows:
Thus, 100 mA of collector current must flow through the transistor to fully illuminate the lamp. As you learned in Chapter 3 collector current does not flow unless base current flows.
Questions
A. Why do you need base current? _____
B. How can you make base current flow? _____
Answers
A. To enable collector current to flow so that the lamp lights up
B. By closing the mechanical switch in the base circuit
5 You can calculate the amount of base current flowing. Assume that β = 100.
Question
A. What is the value of the base current IB? _____
Answer
6 The base current flows in the direction shown in Figure 4.4. Base current flows through the base-emitter junction of the transistor as it does in a forward-biased diode.
Figure 4.4
Questions
A. What is the voltage drop across the base-emitter diode? _____
B. What is the voltage drop across RB? _____
Answers
A. 0.7 volt because it is a silicon transistor
B. 24 volts if the 0.7 is ignored; 23.3 volts if it is not
7 The next step is to calculate RB. The current flowing through RB is the base current IB, and you determined the voltage across it in problem 6.
Question
1. Calculate RB. _____
Answer
Figure 4.5 shows the final circuit, including the calculated current and resistance values.
Figure 4.5
8 Use the following steps to calculate the values of IB and RB needed to turn a transistor ON:
1. Determine the required collector current.
2. Determine the value of β.
3. Calculate the required value of IB from the results of steps 1 and 2.
4. Calculate the required value of RB.
5. Draw the final circuit.
Now, assume that VS = 28 volts, that you are using a lamp requiring 50 mA of current, and that β = 75.
Questions
A. Calculate IB. _____
B. Determine RB. _____
Answers
A. The collector current and β were given. Thus:
B.
This calculation ignores VBE.
9 Now, assume that VS = 9 volts, that you are using a lamp requiring 20 mA of current, and that β = 75.
Question
Calculate RB. _____
Answer
In this
calculation, VBE is included.
10 In practice, if the supply voltage is much larger than the 0.7-volt drop across the base-emitter junction, you can simplify your calculations by ignoring the 0.7-volt drop, and assume that all the supply voltage appears across the base resistor RB. (Most resistors are only accurate to within +/- 5 percent of their stated value anyway.) If the supply voltage is less than 10 volts, however, you shouldn't ignore the 0.7-volt drop across the base-emitter junction.
Questions
Calculate RB for the following problems, ignoring the voltage drop across the base-emitter junction, if appropriate.
A. A 10-volt lamp that draws 10 mA. β = 100. _____
B. A 5-volt lamp that draws 100 mA. β = 50. _____
Answers
A.
B.
Turning Off the Transistor
11 Up to now, you have concentrated on turning the transistor ON, thus making it act like a closed mechanical switch. Now you focus on turning it OFF, thus making it act like an open mechanical switch. If the transistor is OFF, no current flows through the load (that is, no collector current flows).
Questions
A. When a switch is open, are the two terminals at different voltages or at the same voltage? _____