by Earl Boysen
Self-Test
These questions test your understanding of this chapter. Use a separate sheet of paper for your calculations. Compare your answers with the answers provided following the test.
For questions 1–3, calculate the following parameters for the circuit shown in each question.
A. XC
B. Z
C. Vout
D. I
E. tan θ and θ
1. Use the circuit shown in Figure 6.53.
Figure 6.53
A. _____
B. _____
C. _____
D. _____
E. _____
2. Use the circuit shown in Figure 6.54.
Figure 6.54
A. _____
B. _____
C. _____
D. _____
E. _____
3. Use the circuit shown in Figure 6.55.
Figure 6.55
A. _____
B. _____
C. _____
D. _____
E. _____
For questions 4–6, calculate the following parameters for the circuit shown in each question.
A. XC
B. AC Vout
C. DC Vout
4. Use the circuit shown in Figure 6.56.
Figure 6.56
A. _____
B. _____
C. _____
5. Use the circuit shown in Figure 6.57.
Figure 6.57
A. _____
B. _____
C. _____
6. Use the circuit shown in Figure 6.58.
Figure 6.58
A. _____
B. _____
C. _____
For questions 7–9, calculate the following parameters for the circuit shown in each question.
A. DC Vout
B. XL
C. Z
D. AC Vout
E. tan θ and θ
7. Use the circuit shown in Figure 6.59.
Figure 6.59
A. _____
B. _____
C. _____
D. _____
E. _____
8. Use the circuit shown in Figure 6.60.
Figure 6.60
A. _____
B. _____
C. _____
D. _____
E. _____
9. Use the circuit shown in Figure 6.61.
Figure 6.61
A. _____
B. _____
C. _____
D. _____
E. _____
Answers to Self-Test
If your answers do not agree with those provided here, review the applicable problems in this chapter before you go to Chapter 7.
1A. 3 kΩ problems 8, 9, 10, 23
1B. 5 kΩ
1C. 8 volts
1D. 2 amperes
1E. 36.87 degrees
2A. 40 ohms problems 8, 9, 23
2B. 50 ohms
2C. 60 volts
2D. 2 amperes
2.E 53.13 degrees
3A. 5 ohms problems 8, 9, 23
3B. 13 ohms
3C. 10 volts
3D. 2 amperes
3E. 22.63 degrees
4A. 10 ohms problems 26 and 27
4B. 1 volt
4C. 10 volts
5A. 4 ohms problems 26 and 27
5B. 0.4 volt
5C. 10 volts
6A. 64 ohms problems 26 and 27
6B. 5.4 volts
6C. 9.1 volts
7A. 9 volts problems 28–30, 35
7B. 3 ohms
7C. 10.4 ohms
7D. 2.7 volts
7E. 16.7 degrees
8A. 10 volts problems 28–30, 35
8B. 904 ohms
8C. 910 ohms
8D. 1 volt
8E. 83.69 degrees
9A. 0 volts problems 28–30, 35
9B. 1 kΩ
9C. 1.414 kΩ
9D. 7 volts
9E. 45 degrees
Chapter 7
Resonant Circuits
You have seen how the inductor and the capacitor each present an opposition to the flow of an AC current, and how the magnitude of this reactance depends upon the frequency of the applied signal.
When inductors and capacitors are used together in a circuit (referred to as an LC circuit), a useful phenomenon called resonance occurs. Resonance is the frequency at which the reactance of the capacitor and the inductor is equal.
In this chapter, you learn about some of the properties of resonant circuits, and concentrate on those properties that lead to the study of oscillators (which is touched upon in the last few problems in this chapter and covered in more depth in Chapter 9, “Oscillators”).
After completing this chapter, you will be able to do the following:
Find the impedance of a series LC circuit.
Calculate the series LC circuit's resonant frequency.
Sketch a graph of the series LC circuit's output voltage.
Find the impedance of a parallel LC circuit.
Calculate the parallel LC circuit's resonant frequency.
Sketch a graph of the parallel LC circuit's output voltage.
Calculate the bandwidth and the quality factor (Q) of simple series and parallel LC circuits.
Calculate the frequency of an oscillator.
The Capacitor and Inductor in Series
1 Many electronic circuits contain a capacitor and an inductor placed in series, as shown in Figure 7.1.
Figure 7.1
You can combine a capacitor and an inductor in series with a resistor to form voltage divider circuits, such as the two circuits shown in Figure 7.2. A circuit that contains resistance (R), inductance (L), and capacitance (C) is referred to as an RLC circuit. Although the order of the capacitor and inductor differs in the two circuits shown in Figure 7.2, they have the same effect on electrical signals.
Figure 7.2
To simplify your calculations in the next few problems, you can assume that the small DC resistance of the inductor is much less than the resistance of the resistor R, and you can, therefore, ignore DC resistance in your calculations.
When you apply an AC signal to the circuits in Figure 7.2, both the inductor's and the capacitor's reactance value depends on the frequency.
Questions
A. What formula would you use to calculate the inductor's reactance? _____
B. What formula would you use to calculate the capacitor's reactance? _____
Answers
A. XL = 2πfL
B.
2 You can calculate the net reactance (X) of a capacitor and inductor in series by using the following formula:
You can calculate the impedance of the RLC circuits shown in Figure 7.2 by using the following formula:
In the formula, keep in mind that X2 is (XL – XC)2.
Calculate the net reactance and impedance for an RLC series circuit, such as those shown in Figure 7.2, with the following values:
1. f = 1 kHz
2. L = 100 mH
3. C = 1 μF
4. R = 500 ohms
Questions
Follow these steps to calculate the following:
A. Find XL: _____
B. Find XC: _____
C. Use X = XL – XC to find the net reactance: _____
D. Use to find the impedance: _____
Answers
A. XL = 628 ohms
B. XC = 160 ohms
C. X = 468 ohms (inductive)
D. Z = 685 ohms
3 Calculate the net reactance and impedance for an RLC series circuit, such as those shown in Figure 7.2, using the following values:
1. f = 100 Hz
2. L = 0.5 H
3. C = 5 μF
4. R = 8 ohms
Questions
Follow the steps outlined in problem 2 to calculate the following parameters:
A. XL = _____
B. XC = _____
C. X = _____
D. Z = _____
<
br /> Answers
A. XL = 314 ohms
B. XC = 318 ohms
C. X = −4 ohms (capacitive)
D. Z = 9 ohms
By convention, the net reactance is negative when it is capacitive.
4 Calculate the net reactance and impedance for an RLC series circuit, such as those shown in Figure 7.2, using the values in the following questions.
Questions
f = 10 kHz, L = 15 mH, C = 0.01 μF, R = 494 ohms
X = _____
Z = _____
f = 2 MHz, L = 8 μH, C = 0.001 μF, R = 15 ohms
X = _____
Z = _____
Answers
A. X = −650 ohms (capacitive), Z = 816 ohms
B. X = 21 ohms (inductive), Z = 25.8 ohms
5 For the circuit shown in Figure 7.3, the output voltage is the voltage drop across the resistor.
Figure 7.3
In problems 1 through 4, the net reactance of the series inductor and capacitor changes as the frequency changes. Therefore, as the frequency changes, the voltage drop across the resistor changes and so does the amplitude of the output voltage Vout.
If you plot Vout against frequency on a graph for the circuit shown in Figure 7.3, the curve looks like the one shown in Figure 7.4.
Figure 7.4
The maximum output voltage (or peak voltage) shown in this curve, Vp, is slightly less than Vin. This slight attenuation of the peak voltage from the input voltage is because of the DC resistance of the inductor.
The output voltage peaks at a frequency, fr, where the net reactance of the inductor and capacitor in series is at its lowest value. At this frequency, there is little voltage drop across the inductor and capacitor. Therefore, most of the input voltage is applied across the resistor, and the output voltage is at its highest value.
Question
Under ideal conditions, if XC were 10.6 ohms, what value of XL results in a net reactance (X) of 0 for the circuit shown in Figure 7.3? _____
Answer
X = XL − XC = 0, therefore:
XL = XC + X = 10.6 Ω + 0 = 10.6 Ω
6 You can find the frequency at which XL − XC = 0 by setting the formula for XL equal to the formula for XC and solving for f:
Therefore,
where fr is the resonant frequency of the circuit.
Question
What effect does the value of the resistance have on the resonant frequency? _____
Answer
It has no effect at all.
7 Calculate the resonant frequency for the circuit shown in Figure 7.3 using the capacitor and inductor values given in the following questions.
Questions
A. C = 1 μF, L = 1 mH _____
fr = _____
B. C = 16 μF, L = 1.6 mH _____
fr = _____
Answers
A.
B.
8 Calculate the resonant frequency for the circuit shown in Figure 7.3 using the capacitor and inductor values given in the following questions.
Questions
A. C = 0.1 μF, L = 1 mH _____
fr = _____
B. C = 1 μF, L = 2 mH _____
fr = _____
Answers
A. fr = 16 kHz
B. fr = 3.6 kHz
9 For the RLC circuit shown in Figure 7.5, the output voltage is the voltage drop across the capacitor and inductor.
Figure 7.5
If Vout is plotted on a graph against the frequency for the circuit shown in Figure 7.5, the curve looks like that shown in Figure 7.6.
Figure 7.6
The output voltage drops to its minimum value at the resonant frequency for the circuit, which you can calculate with the formula provided in problem 6. At the resonant frequency, the net reactance of the inductor and capacitor in series is at a minimum. Therefore, there is little voltage drop across the inductor and capacitor, and the output voltage is at its minimum value. A circuit with this type of output (such as the circuit shown in Figure 7.5) is called a notch filter, or band-reject filter.
Question
What would you expect the minimum output voltage to be? _____
Answer
0 volts, or close to it
Project 7.1: The Notch Filter
Objective
The objective of this project is to determine how Vout changes as the frequency of the input signal changes for a notch filter.
General Instructions
After the circuit is set up, you measure Vout for each frequency. You can also generate a graph to show the relationship between the output voltage and the input frequency.
Parts List
You need the following equipment and supplies:
One 100 Ω, 0.25-watt resistor.
One 1000 pF capacitor. (1000 pF is also sometimes stated by suppliers as 0.001 μF.)
One 100 μH inductor. You'll often find inductors that use a numerical code to indicate the value of the inductor. The first two numbers in this code are the first and second significant digits of the inductance value. The third number is the multiplier, and the units are μH. Therefore, an inductor marked with 101 has a value of 100 μH.)
One function generator.
One oscilloscope.
One breadboard.
Step-by-Step Instructions
Set up the circuit shown in Figure 7.7. 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 7.7
Carefully check your circuit against the diagram.
After you check your circuit, follow these steps, and record your measurements in the blank table following the steps.
1. Connect the oscilloscope probe for channel 2 to a jumper wire connected to Vin, and connect the ground clip to a jumper wire attached to the ground bus.
2. Connect the oscilloscope probe for channel 1 to a jumper wire connected to Vout, and connect the ground clip to a jumper wire attached to the ground bus.
3. Set the function generator to generate a 5 Vpp, 100 kHz sine wave.
4. Measure and record Vout.
5. Adjust the function generator to the frequency shown in the next row of the table (labeled 150 kHz in this instance). Each time you change the frequency, check Vin and adjust the amplitude knob on the function generator to maintain Vin at 5 Vpp if needed. (If you leave the amplitude knob in one position, the voltage of the signal provided by the function generator will change as the net reactance of the circuit changes.)
6. Measure and record Vout.
7. Repeat steps 5 and 6 for the remaining values until you have recorded Vout in all rows of the table.
fin (kHz) Vout (volts)
100
150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
8. In the blank graph shown in Figure 7.8, plot Vout vs fin with the voltage on the vertical axis and the frequency on the X axis. The curve should have the same shape as the curve shown in Figure 7.6.
Figure 7.8
Expected Results
Figure 7.9 shows the breadboarded circuit for this project.
Figure 7.9
Figure 7.10 shows a function generator and oscilloscope attached to the circuit.
Figure 7.10
The input signal is represented by the upper sine wave shown in Figure 7.11, and the output signal is represented by the lower sine wave. Read the number of divisions for the peak-to-peak output sine wave, and multiply it by the corresponding VOLTS/DIV setting to determine Vout.
Figure 7.11
As you set fin to a ne
w value on the function generator, you may also need to adjust the TIME/DIV control, the VOLTS/DIV control, and vertical POSITION controls on the oscilloscope. The controls shown in Figure 7.12 are adjusted to measure Vout when fin = 500 kHz.
Figure 7.12
Your values should be close to those shown in the following table, and the curve should be similar to that shown in Figure 7.13.
fin (kHz) Vout (volts)
100 4.9
150 4.9
200 4.9
250 4.8
300 4.7
350 4.4
400 4.0
450 2.8
500 0.5
550 2.3
600 3.6
650 4.2
700 4.4
750 4.7
800 4.7
850 4.8
900 4.8
Figure 7.13
Notice the extra data points shown in the graph near the minimum Vout. These extra data points help you to determine the frequency at which the minimum Vout occurs. In this graph, the minimum Vout occurs at a frequency of 505 kHz, which is close to the calculated resonance frequency of 503 kHz for this circuit.
10 You can connect the capacitor and inductor in parallel, as shown in Figure 7.14.
Figure 7.14
You can calculate the resonance frequency of this circuit using the following formula:
In this formula, r is the DC resistance of the inductor. However, if the reactance of the inductor is equal to, or more than, 10 times the DC resistance of the inductor, you can use the following simpler formula. This is the same formula that you used in problems 7 and 8 for the series circuit.
Q, the quality factor of the circuit, is equal to XL/r. Therefore, you can use this simple equation to calculate fr if Q is equal to, or greater than, 10.