by Earl Boysen
Figure 7.39
Recall the discussion in problem 18 (comparing the bandwidths of radio tuners and television amplifiers) to help you answer the following questions.
Questions
A. Which is the more selective, the radio tuner or the television amplifier? _____
B. Which would require a lower Q circuit, the radio tuner or the television amplifier? _____
Answers
A. The radio tuner
B. The television amplifier
26 The inductor and capacitor shown in Figure 7.40 are connected in parallel, rather than in series. However, you can use the same formulas you used for the series circuit in problem 20 to calculate fr, Q, and BW for parallel LC circuits.
Figure 7.40
Questions
Find fr, Q, and BW for the circuit shown in Figure 7.40.
A. fr = _____
B. Q = _____
C. BW = _____
Answers
A. fr = 1.6 MHz
B. XL = 10 ohms, so Q = 10/0.1 = 100 (The only resistance here is the small DC resistance of the inductor.)
C. BW = 16 kHz (This is a fairly high Q circuit.)
27 In the last few problems, you learned how to calculate fr, BW, and Q, for a given circuit, or conversely, to calculate the component values that would produce a circuit with specified fr, BW, and Q values.
When you know the resonant frequency and bandwidth for a circuit, you can sketch an approximate output curve. With the simple calculations listed in this problem, you can draw a curve that is accurate to within 1 percent of its true value.
The curve that results from the calculations used in this problem is sometimes called the general resonance curve.
You can determine the output voltage at several frequencies by following these steps:
1. Assume the peak output voltage Vp at the resonant frequency fr to be 100 percent. This is point A on the curve shown in Figure 7.41.
2. The output voltage at f1 and f2 is 0.707 of 100 percent. On the graph, these are the two points labeled B in Figure 7.41. Note that f2 – f1 = BW. Therefore, at half a bandwidth above and below fr, the output is 70.7 percent of Vp.
3. At f3 and f4 (the two points labeled C in Figure 7.41), the output voltage is 44.7 percent of Vp. Note that f4 – f3 = 2 BW. Therefore, at 1 bandwidths above and below fr, the output is 44.7 percent of maximum.
4. At f5 and f6 (the two points labeled D in Figure 7.41), the output voltage is 32 percent of Vp. Note that f6 – f5 = 3 BW. Therefore, at 1.5 bandwidths above and below fr, the output is 32 percent of maximum.
5. At f7 and f8 (the two points labeled E in Figure 7.41), the output voltage is 24 percent of Vp. Note that f8 – f7 = 4 BW. Therefore, at 2 bandwidths above and below fr, the output is 24 percent of maximum.
6. At f10 and f9 (the two points labeled F in Figure 7.41), the output is 13 percent of Vp. Note that f10 – f9 = 8 BW. Therefore, at 4 bandwidths above and below fr,, the output is 13 percent of maximum.
Figure 7.41
Questions
Calculate fr, XL, Q, and BW for the circuit shown in Figure 7.42.
Figure 7.42
A. fr = _____
B. XL = _____
C. Q = _____
D. BW = _____
Answers
A. fr = 1 MHz
B. XL = 1607 ohms
C. Q = 100
D. BW = 10 kHz
28 Now, calculate the frequencies that correspond with each percentage of the peak output voltage listed in steps 1 through 6 of problem 27. (Refer to the graph in Figure 7.41 as needed.)
Questions
A. At what frequency will the output level be maximum? _____
B. At what frequencies will the output level be 70 percent of Vp? _____
C. At what frequencies will the output level be 45 percent of Vp? _____
D. At what frequencies will the output level be 32 percent of Vp? _____
E. At what frequencies will the output level be 24 percent of Vp? _____
F. At what frequencies will the output level be 13 percent of Vp? _____
Answers
A. 1 MHz
B. 995 kHz and 1005 kHz (1 MHz – 5 kHz and + 5 kHz)
C. 990 kHz and 1010 kHz
D. 985 kHz and 1015 kHz
E. 980 kHz and 1020 kHz
F. 960 kHz and 1040 kHz
29 You can calculate the output voltage at each frequency in the answers to problem 28 by multiplying the peak voltage by the related percentage for each frequency.
Questions
Calculate the output voltage for the frequencies given here, assuming that the peak output voltage is 5 volts.
A. What is the output voltage level at 995 kHz? _____
B. What is the output voltage level at 980 kHz? _____
Answers
A. V = 5 volts × 0.70 = 3.5 volts
B. V = 5 volts × 0.24 = 1.2 volts
Figure 7.43 shows the output curve generated by plotting the frequencies calculated in problem 28 and the corresponding output voltages calculated in this problem.
Figure 7.43
Introduction to Oscillators
In addition to their use in circuits used to filter input signals, capacitors and inductors are used in circuits called oscillators.
Oscillators are circuits that generate waveforms at particular frequencies. Many oscillators use a tuned parallel LC circuit to produce a sine wave output. This section is an introduction to the use of parallel capacitors and inductors in oscillators.
30 When the switch in the circuit shown in drawing (1) of Figure 7.44 is closed, current flows through both sides of the parallel LC circuit in the direction shown.
Figure 7.44
It is difficult for the current to flow through the inductor initially because the inductor opposes any changes in current flow. Conversely, it is easy for the current to flow into the capacitor initially because with no charge on the plates of the capacitor there is no opposition to the flow.
As the charge on the capacitor increases, the current flow in the capacitor side of the circuit decreases. However, more current flows through the inductor. Eventually, the capacitor is fully charged, so current stops flowing in the capacitor side of the circuit, and a steady current flows through the inductor.
Question
When you open the switch, what happens to the charge on the capacitor? _____
Answer
It discharges through the inductor. (Note the current direction, shown in drawing [2] of Figure 7.44.)
31 With the switch open, current continues to flow until the capacitor is fully discharged.
Question
When the capacitor is fully discharged, how much current is flowing through the inductor? _____
Answer
None.
32 Because there is no current in the inductor, its magnetic field collapses. The collapsing of the magnetic field induces a current to flow in the inductor, and this current flows in the same direction as the original current through the inductor (remember that an inductor resists any change in current flow), which is shown in drawing (2) of Figure 7.44. This current now charges the capacitor to a polarity that is opposite from the polarity that the battery induced.
Question
When the magnetic field of the inductor has fully collapsed, how much current will be flowing? _____
Answer
None.
33 Next, the capacitor discharges through the inductor again, but this time the current flows in the opposite direction, as shown in drawing (3) of Figure 7.44. The change in current direction builds a magnetic field of the opposite polarity. The magnetic field stops growing when the capacitor is fully discharged.
Because there is no current flowing through the inductor, its magnetic field collapses and induces current to flow in the direction shown in drawing (3) of Figure 7.44.
Question
What do you think the current generated by the magnetic field in the inductor will do to the
capacitor? _____
Answer
It charges it to the original polarity.
34 When the field has fully collapsed, the capacitor stops charging. It now begins to discharge again, causing current to flow through the inductor in the direction shown in drawing (2) of Figure 7.44. This “seesaw” action of current will continue indefinitely.
As the current flows through the inductor, a voltage drop occurs across the inductor. The magnitude of this voltage drop will increase and decrease as the magnitude of the current changes.
Question
What would you expect the voltage across the inductor to look like when you view it on an oscilloscope? _____
Answer
A sine wave
35 In a perfect circuit, this oscillation continues and produces a continuous sine wave. In practice, a small amount of power is lost in the DC resistance of the inductor and the other wiring. As a result, the sine wave gradually decreases in amplitude and dies out to nothing after a few cycles, as shown in Figure 7.45.
Figure 7.45
Question
How might you prevent this fade-out? _____
Answer
By replacing a small amount of energy in each cycle.
This lost energy can be injected into the circuit by momentarily closing and opening the switch at the correct time. (See drawing [1] of Figure 7.44.) This would sustain the oscillations indefinitely.
An electronic switch (such as a transistor) could be connected to the inductor as shown in Figure 7.46. Changes in the voltage drop across the inductor would turn the electronic switch on or off, thereby opening or closing the switch.
Figure 7.46
The small voltage drop across the few turns of the inductor (also referred to as a coil), between point B at the end of the coil, and point A about halfway along the coil, is used to operate the electronic switch. These points are shown in Figure 7.30.
Using a small part of an output voltage in this way is called feedback because the voltage is “fed back” to an earlier part of the circuit to make it operate correctly.
When you properly set up such a circuit, it produces a continuous sine wave output of constant amplitude and constant frequency. This circuit is called an oscillator. You can calculate the frequency of the sine waves generated by an oscillator with the following formula for determining resonant frequency:
The principles you learned in the last few problems are used in practical oscillator circuits, such as those presented in Chapter 9.
Summary
In this chapter, you learned about the following topics related to resonant circuits:
How the impedance of a series LC circuit and a parallel LC circuit changes with changes in frequency.
At resonant frequency for a parallel LC circuit, the impedance is at its highest; whereas for a series LC circuit, impedance is at its lowest.
The concept of bandwidth enables you to easily calculate the output voltage at various frequencies and draw an accurate output curve.
The principles of bandpass filters and notch (or band-reject) filters.
The fundamental concepts integral to understanding how an oscillator functions.
Self-Test
These questions test your understanding of the concepts covered 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. What is the formula for the impedance of a series LC circuit? _____
2. What is the formula for the impedance of a series RLC circuit (a circuit containing resistance, inductance, and capacitance)? _____
3. What is the relationship between XC and XL at the resonant frequency? _____
4. What is the voltage across the resistor in a series RLC circuit at the resonant frequency? _____
5. What is the voltage across a resistor in series with a parallel LC circuit at the resonant frequency? _____
6. What is the impedance of a series circuit at resonance? _____
7. What is the formula for the impedance of a parallel circuit at resonance? _____
8. What is the formula for the resonant frequency of a circuit? _____
9. What is the formula for the bandwidth of a circuit? _____
10. What is the formula for the Q of a circuit? _____
Questions 11–13 use a series LC circuit. In each case, the values of the L, C, and R are given. Find fr, XL, XC, Z, Q, and BW. Draw an output curve for each answer.
11. L = 0.1 mH, C = 0.01 mF, R = 10 ohms _____
12. L = 4 mH, C = 6.4 mF, R = 0.25 ohms _____
13. L = 16 mH, C = 10 mF, R = 20 ohms _____
Questions 14 and 15 use a parallel LC circuit. No R is used; r is given. Find fr, XL, XC, Z, Q, and BW.
14. L = 6.4 mH, C = 10 mF, r = 8 ohms _____
15. L = 0.7 mH, C = 0.04 mF, r = 1.3 ohms _____
16. Use the output curve shown in Figure 7.47 to answer the following questions.
Figure 7.47
A. What is the peak value of the output curve? _____
B. What is the resonant frequency? _____
C. What is the voltage level at the half power points? _____
D. What are the half power frequencies? _____
E. What is the bandwidth? _____
F. What is the Q of the circuit? _____
Answers to Self-Test
If your answers do not agree with those given here, review the problems indicated in parentheses before you go on to Chapter 8, “Transistor Amplifiers.”
1. Z = XL − XC (problem 2)
2. (problem 2)
3. XL = XC (problem 5)
4. Maximum output (problem 5)
5. Minimum output (problem 11)
6. Z = minimum. Ideally, it is equal to the resistance. (problem 5)
7. (problem 10)
In this formula, r is the resistance of the coil.
8. (problem 6)
9. (problem 20)
10. (problem 20)
or
To draw the output curves for Questions 11–13, use the graph in Figure 7.41 as a guide and insert the appropriate bandwidth and frequency values. (problems 21–29)
11. fr = 160 kHz, XL = XC = 100 ohms, Q = 10, BW = 16 kHz, Z = 10 ohms (problems 21–29)
12. fr = 1 kHz, XL = XC = 25 ohms, Q = 100, BW = 10 Hz, Z = 0.25 ohms (problems 21–29)
13. fr = 400 Hz, XL = XC = 40 ohms, Q = 2, BW = 200 Hz, Z = 20 ohms (problems 21–29)
14. fr = 600 Hz, XL = 24 ohms, XC = 26.5, Q = 3, BW = 200 Hz, Z = 80 ohms (problems 21–29)
Because Q is not given, you should use the more complicated of the two formulas shown in problem 10 to calculate the resonant frequency.
15. fr = 30 kHz, XL = 132 ohms, XC = 132, Q = 101.5, BW = 300 Hz, Z = 13.4 ohms (problems 21–29)
16A. 10.1 volts (problems 27 and 28)
16B. 148 kHz (problems 27 and 28)
16C. 10.1 × 0.707 = 7.14 volts (problems 27 and 28)
16D. Approximately 135 kHz and 160 kHz (not quite symmetrical) (problems 27 and 28)
16E. BW = 25 kHz (problems 27 and 28)
16F. about 5.9 (problems 27 and 28)
Chapter 8
Transistor Amplifiers
Many of the AC signals you'll work with in electronics are small. For example, the signal that an optical detector reads from a DVD disk cannot drive a speaker, and the signal from a microphone's output is too weak to send out as a radio signal. In cases such as these, you must use an amplifier to boost the signal.
The best way to demonstrate the basics of amplifying a weak signal to a usable level is by starting with a one-transistor amplifier. When you understand a one-transistor amplifier, you can grasp the building block that makes up amplifier circuits used in electronic devices such as cellphones, MP3 players, and home entertainment centers.
Many amplifier circuit configurations are possible. The simplest and most basic of amplifying circuits are used in this chapter to demonstrate how a transistor amplif
ies a signal. You can also see the steps to design an amplifier.
The emphasis in this chapter is on the bipolar junction transistor (BJT), just as it was in Chapter 3, “Introduction to the Transistor,” and Chapter 4, “The Transistor Switch,” which dealt primarily with the application of transistors in switching circuits. Two other types of devices used as amplifiers are also examined: the junction field effect transistor (JFET) (introduced in Chapters 3 and 4), and an integrated circuit called the operational amplifier (op-amp).
When you complete this chapter, you will be able to do the following:
Calculate the voltage gain for an amplifier.
Calculate the DC output voltage for an amplifier circuit.
Select the appropriate resistor values to provide the required gain to an amplifier circuit.
Identify several ways to increase the gain of a one-transistor amplifier.
Distinguish between the effects of a standard one-transistor amplifier and an emitter follower circuit.
Design a simple emitter follower circuit.
Analyze a simple circuit to find the DC level out and the AC gain.
Design a simple common source (JFET) amplifier.
Analyze a JFET amplifier to find the AC gain.
Recognize an op-amp and its connections.
Working with Transistor Amplifiers
1 In Chapter 3 you learned how to turn transistors ON and OFF. You also learned how to calculate the value of resistors in amplifier circuits to set the collector DC voltage to half the power supply voltage. To review this concept, examine the circuit shown in Figure 8.1.
Figure 8.1
Use the following steps to find the value of RB that will set the collector DC voltage (VC) to half the supply voltage (VS):