Complete Electronics Self-Teaching Guide with Projects
Page 16
Questions
A. Which formula should you use to calculate the resonant frequency of a parallel circuit if the Q of the coil is 20? _____
B. If the Q is 8? _____
Answers
A.
B.
Note Here is another version of the resonance frequency formula that is helpful when Q is known:
11 You can calculate the total opposition (impedance) of an inductor and capacitor connected in parallel to the flow of current by using the following formulas for a circuit at resonance:
At resonance, the impedance of an inductor and capacitor in parallel is at its maximum.
You can use an inductor and capacitor in parallel in a voltage divider circuit, as shown in Figure 7.15.
Figure 7.15
If Vout is plotted against frequency on a graph for the circuit shown in Figure 7.15, the curve looks like that shown in Figure 7.16.
Figure 7.16
Questions
A. What would be the total impedance formula for the voltage divider circuit at resonance? _____
B. What is the frequency called at the point where the curve is at its lowest point? _____
C. Why is the output voltage at a minimum value at resonance? _____
Answers
A. ZT = Zp + R
Note The relationship shown by this formula is true only at resonance. At all other frequencies, ZT is a complicated formula or calculation found by considering a series r, L circuit in parallel with a capacitor.
B. The parallel resonant frequency.
C. The output voltage is at its lowest value at the resonant frequency. This is because the impedance of the parallel resonant circuit is at its highest value at this frequency.
12 For the circuit shown in Figure 7.17, the output voltage equals the voltage drop across the inductor and capacitor.
Figure 7.17
If Vout is plotted on a graph against frequency for the circuit shown in Figure 7.17, the curve looks like that shown in Figure 7.18. At the resonance frequency, the impedance of the parallel inductor and capacitor is at its maximum value. Therefore, the voltage drop across the parallel inductor and capacitor (which is also the output voltage) is at its maximum value.
Figure 7.18
Question
What formula would you use to calculate the resonant frequency? _____
Answer
13 Find the resonant frequency in these two examples, where the capacitor and the inductor are in parallel. (Q is greater than 10.)
Questions
A. L = 5 mH, C = 5 μF _____ fr = _____
B. L = 1 mH, C = 10 μF _____
fr = _____
Answers
A. fr = 1 kHz (approximately)
B. fr = 1600 Hz (approximately)
The Output Curve
14 Now it's time to look at the output curve in a little more detail. Take a look at the curve shown in Figure 7.19 for an example.
Figure 7.19
An input signal at the resonant frequency, fr, passes through a circuit with minimum attenuation, and with its output voltage equal to the peak output voltage, Vp, shown on this curve.
The two frequencies f1 and f2 are “passed” almost as well as fr is passed. That is, signals at those frequencies have a high output voltage, almost as high as the output of a signal at fr. The graph shows this voltage as Vx.
Signals at frequencies f3 and f4 have a low output voltage.
These two frequencies are not passed but are said to be blocked or rejected by the circuit. This output voltage is shown on the graph as Vz.
The output or frequency response curve for a resonant circuit (series or parallel) has a symmetrical shape for a high value of Q. You can make the assumption that the output curve is symmetrical when Q is greater than 10.
Questions
A. What is meant by a frequency that is passed? _____
B. Why are f1 and f2 passed almost as well as fr? _____
C. What is meant by a frequency that is blocked? _____
D. Which frequencies shown on the previous output curve are blocked? _____
E. Does the output curve shown appear to be symmetrical? What does this mean for the circuit? _____
Answers
A. It appears at the output with minimum attenuation.
B. Because their frequencies are close to fr.
C. It has a low output voltage.
D. f3 and f4 (as well as all frequencies below f3 and above f4).
E. It does appear to be symmetrical. This means that the coil has a Q greater than 10.
15 Somewhere between fr and f3, and between fr and f4, there is a point at which frequencies are said to be either passed or reduced to such a level that they are effectively blocked. The dividing line is at the level at which the power output of the circuit is half as much as the power output at peak value. This happens to occur at a level that is 0.707, or 70.7 percent of the peak value.
For the output curve shown in problem 14, this occurs at a voltage level of 0.707 Vp. The two corresponding frequencies taken from the graph are called the half power frequencies or half power points. These are common expressions used in the design of resonant circuits and frequency response graphs.
If a certain frequency results in an output voltage that is equal to or greater than the half power point, it is said to be passed or accepted by the circuit. If it is lower than the half power point, it is said to be blocked or rejected by the circuit.
Question
Suppose VP = 10 volts. What is the minimum voltage level of all frequencies that are passed by the circuit? _____
Answer
V = 10 volts × 0.707 = 7.07 volts
(If a frequency has an output voltage above 7.07 volts, you would say it is passed by the circuit.)
16 Assume the output voltage at the resonant frequency in a circuit is 5 volts. Another frequency has an output of 3.3 volts.
Question
Is this second frequency passed or blocked by the circuit? _____
Answers
V = Vp × 0.707 = 5 × 0.707 = 3.535 volts
3.3 volts is less than 3.535 volts, so this frequency is blocked.
17 In these examples, find the voltage level at the half power points.
Questions
A. Vp = 20 volts _____
B. Vp = 100 volts _____
C. Vp = 3.2 volts _____
Answers
A. 14.14 volts
B. 70.70 volts
C. 2.262 volts
18 Although this discussion started off by talking about the resonance frequency, a few other frequencies have been introduced. At this point, the discussion is dealing with a band or a range of frequencies.
Two frequencies correspond to the half power points on the curve. Assume these frequencies are f1 and f2. The difference you find when you subtract f1 from f2 is important because this gives the range of frequencies that are passed by the circuit. This range is called the bandwidth of the circuit and can be calculated using the following equation:
All frequencies within the bandwidth are passed by the circuit, whereas all frequencies outside the bandwidth are blocked. A circuit with this type of output (such as the circuit shown in Figure 7.17) is referred to as a bandpass filter.
Question
Indicate which of the following pairs of values represent a wider range of frequencies, or, in other words, the wider bandwidth.
A. f2 = 200 Hz, f1 = 100 Hz
B. f2 = 20 Hz, f1 = 10 Hz
Answer
The bandwidth is wider for the frequencies given in A.
When playing a radio, you listen to one station at a time, not to the adjacent stations on the dial. Thus, your radio tuner must have a narrow bandwidth so that it can select only the frequency of that one station.
The amplifiers in a television set, however, must pass frequencies from 30 Hz up to approximately 4.5 MHz, which requires a wider bandwidth. The application or use to which you'll put a circuit determines the band
width that you should design the circuit to provide.
19 The output curve for a circuit that passes a band of frequencies around the resonance frequency (such as the curve shown in Figure 7.20) was discussed in the last few problems.
Figure 7.20
The same principles and equations apply to the output curve for a circuit that blocks a band of frequencies around the resonance frequency, as is the case with the curve shown in Figure 7.21.
Figure 7.21
Questions
A. What points on the curve shown in Figure 7.21 would you use to determine the circuit's bandwidth? _____
B. Would the output voltage at the resonant frequency be above or below these points? _____
Answers
A. The half power points (0.707 Vout(max)).
B. The output voltage at the resonant frequency is the minimum point on the curve, which is below the level for the half power points.
Project 7.2: The Band Pass Filter
Objective
The objective of this project is to determine how Vout changes as the frequency of the input signal changes for a bandpass filter.
General Instructions
When the circuit is set up, you measure Vout for each frequency. You 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.
One function generator.
One oscilloscope.
One breadboard.
Step-by-Step Instructions
Set up the circuit shown in Figure 7.22. 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 7.22
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 until you have recorded Vout for the last row 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 7.23, plot Vout versus 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 7.20.
Figure 7.23
Expected Results
Figure 7.24 shows the breadboarded circuit for this project.
Figure 7.24
Figure 7.25 shows a function generator and oscilloscope attached to the circuit.
Figure 7.25
The input signal is represented by the upper sine wave shown in Figure 7.26, 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 7.26
As you set fin to a new 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 7.27 are adjusted to measure Vout when fin = 500 kHz.
Figure 7.27
Your values should be close to those shown in the following table, and the curve should be similar to that shown Figure 7.28.
fin (kHz) Vout (volts)
100 0.3
150 0.5
200 0.7
250 1.0
300 1.4
350 1.8
400 2.6
450 3.6
500 4.6
550 4.2
600 3.4
650 2.7
700 2.2
750 1.8
800 1.6
850 1.4
900 1.3
Because Q = 3.2 (well below 10), the curve for this circuit is not perfectly symmetrical.
Figure 7.28
20 You can find the bandwidth of a circuit by measuring the frequencies (f1 and f2) at which the half power points occur and then using the following formula:
Or you can calculate the bandwidth of a circuit using this formula:
where:
The formula used to calculate bandwidth indicates that, for two circuits with the same resonant frequency, the circuit with the larger Q will have the smaller bandwidth.
When you calculate Q for a circuit containing a capacitor and inductor in series (such as that shown in Figure 7.29), use the total DC resistance—the sum of the DC resistance (r) of the inductor and the value of the resistor (R)—to calculate Q.
Figure 7.29
When you calculate Q for a circuit containing an inductor and capacitor in parallel, as with the circuit shown in Figure 7.30, you do not include the value of the resistor (R) in the calculation. The only resistance you use in the calculation is the DC resistance (r) of the inductor.
Figure 7.30
When you calculate Q for a circuit containing an inductor, a capacitor, and a resistor in parallel (as with the two circuits shown in Figure 7.31), include the value of the resistor (R) in the calculation.
Figure 7.31
Questions
For the circuit shown in Figure 7.32, all the component values are provided in the diagram. Find fr, Q, and BW.
Figure 7.32
A. fr = _____
B. Q = _____
C. BW = _____
Answers
A.
B.
C.
21 Use the circuit and component values shown in Figure 7.33 to answer the following questions.
Figure 7.33
Questions
Find fr, Q, and BW. Then, on a separate sheet of paper, draw an output curve showing the range of frequencies that are passed and blocked.
fr = _____
Q = _____
BW = _____
Answers
fr = 1590 Hz; Q = 1; BW = 1590 Hz
The output curve is shown in Figure 7.34.
Figure 7.34
22 Use the circuit and component values shown in Figure 7.35 to answer the following questions.
Figure 7.35
Questions
Find fr, Q, and BW for this circuit. Then draw the output curve on a separate sheet of paper.
fr = _____
Q = _____
BW = _____
Answers
fr = 500 Hz; Q = 31.4; BW = 16 Hz
The output curve is shown in Figure 7.36.
Figure 7.36
23 Use the circuit shown in Figure 7.37 for this problem
. In this case, the resistor value is 10 ohms. However, the inductor and capacitor values are not given.
Figure 7.37
Questions
Find BW and the values of L and C required to give the circuit a resonant frequency of 1200 Hz and a Q of 80.
A. BW = _____
B. L = _____
C. C = _____
Answers
A. BW = 15 Hz
B. L = 106 mH
C. C = 0.166 μF
You can check these values by using the values of L and C to find fr.
24 Use the circuit shown in Figure 7.37 for this problem. In this case, the resistor value is given as 10 ohms. However, the inductor and capacitor values are not given.
Questions
Calculate the values of Q, L, and C required to give the circuit a resonant frequency of 300 kHz with a bandwidth of 80 kQ.
A. Q = _____
B. L = _____
C. C = _____
Answers
A. Q = 3.75
B. L = 20 μH
C. C = 0.014 μF
25 A circuit that passes (or blocks) only a narrow range of frequencies is called a high Q circuit. Figure 7.38 shows the output curve for a high Q circuit.
Figure 7.38
Because of the narrow range of frequencies it passes, a high Q circuit is said to be selective in the frequencies it passes.
A circuit that passes (or blocks) a wide range of frequencies is called a low Q circuit. Figure 7.39 shows the output curve for a low Q circuit.