by Earl Boysen
4 For simplicity, consider all inputs at this time to be pure sine waves. The circuit shown in Figure 6.2 shows a sine wave as the input signal to a capacitor.
Figure 6.2
Question
If the input is a pure sine wave, what is the output? _____
Answer
A pure sine wave
5 The output sine wave has the same frequency as the input sine wave. A capacitor cannot change the frequency of the signal. But remember, with an AC input, the capacitor behaves in a manner similar to a resistor in that the capacitor does have some level of opposition to the flow of alternating current. The level of opposition depends upon the value of the capacitor and the frequency of the signal. Therefore, the output amplitude of a sine wave will be less than the input amplitude.
Question
With an AC input to a simple circuit like the one described here, what does the capacitor appear to behave like? _____
Answer
It appears to have opposition to alternating current similar to the behavior of a resistor.
6 If you connect a capacitor and resistor in series (as shown in Figure 6.3), the circuit functions as a voltage divider.
Figure 6.3
Question
What formula would you use to calculate the output voltage for a voltage divider formed by connecting two resistors in series? _____
Answer
7 You can calculate a total resistance to the flow of electric current for a circuit containing two resistors in series.
Question
What is the formula for this total resistance? _____
Answer
8 You can also calculate the total opposition to the flow of electric current for a circuit containing a capacitor and resistor in series. This parameter is called impedance, and you can calculate it using the following formula:
In this equation:
Z = The impedance of the circuit in ohms
XC = The reactance of the capacitor in ohms
R = The resistance of the resistor in ohms
Questions
Use the following steps to calculate the impedance of the circuit, and the current flowing through the circuit, as shown in Figure 6.4.
Figure 6.4
A. _____
B. _____
C. _____
Answers
A. 400 ohms
B. 500 ohms
C. 20 mApp
9 Now, for the circuit shown in Figure 6.4, calculate the impedance and current using the values provided.
Questions
A. C = 530 μF, R = 12 ohms, Vin = 26 Vpp, f = 60 Hz _____
B. C = 1.77 μF, R = 12 ohms, Vin = 150 Vpp, f = 10 kHz _____
Answers
A. Z = 13 ohms, I = 2 App
B. Z = 15 ohms, I = 10 App
10 You can calculate Vout for the circuit shown in Figure 6.5 with a formula similar to the formula used in Chapter 5 to calculate Vout for a voltage divider composed of two resistors.
Figure 6.5
The formula to calculate the output voltage for this circuit is as follows:
Questions
Calculate the output voltage in this circuit using the component values and input signal voltage and frequency listed on the circuit diagram shown in Figure 6.6.
Figure 6.6
A. Find XC: _____
B. Find Z: _____
C. Use the formula to find Vout: _____
Answers
A. XC = 500 ohms (rounded off)
B. Z = 1120 ohms (rounded off)
C. Vout = 8.9 Vpp
11 Now, find Vout for the circuit in Figure 6.5 using the given component values, signal voltage, and frequency.
Questions
A. C = 0.16 μF, R = 1 kΩ, Vin = 10 Vpp, f = 1 kHz _____
B. C = 0.08 μF, R = 1 kΩ, Vin = 10 Vpp, f = 1 kHz _____
Answers
A. Vout = 7.1 Vpp
B. Vout = 4.5 Vpp
Note Hereafter, you can assume that the answer is a peak-to-peak value if the given value is a peak-to-peak value.
12 The output voltage is said to be attenuated in the voltage divider calculations, as shown in the calculations in problems 10 and 11. Compare the input and output voltages in problems 10 and 11.
Question
What does attenuated mean? _____
Answer
To reduce in amplitude or magnitude (that is, Vout is smaller than Vin.).
13 When you calculated Vout in the examples in problems 10 and 11, you first had to find XC. However, XC changes as the frequency changes, while the resistance remains constant. Therefore, as the frequency changes, the impedance Z changes and also so does the amplitude of the output voltage Vout.
If Vout is plotted against frequency on a graph, the curve looks like that shown in Figure 6.7.
Figure 6.7
The frequencies of f1 (at which the curve starts to rise) and f2 (where it starts to level off) depend on the values of the capacitor and the resistor.
Questions
Calculate the output voltage for the circuit shown in Figure 6.8 for frequencies of 100 Hz, 1 kHz, 10 kHz, and 100 kHz.
Figure 6.8
A. 100 Hz: _____
B. 1 kHz: _____
C. 10 kHz: _____
D. 100 kHz: _____
E. Plot these values for Vout against f, and draw a curve to fit the points. Use a separate sheet of paper to draw your graph.
Answers
A. Vout = 0.1 volt
B. Vout = 1 volt
C. Vout = 7.1 volts
D. Vout = 10 volts
E. The curve is shown in Figure 6.9.
Figure 6.9
Note You can see that Vout is equal to Vin for the highest frequency and at nearly zero for the lowest frequency. You call this type of circuit a high-pass filter because it will pass high-frequency signals with little attenuation and block low-frequency signals.
Project 6.1: The High-Pass Filter
Objective
The objective of this project is to determine how Vout changes as the frequency of the input signal changes for a high-pass filter.
General Instructions
When the circuit is set up, measure Vout for each frequency; you will also calculate XC for each frequency value to show the relationship between the output voltage and the reactance of the capacitor.
Parts List
You need the following equipment and supplies:
One 1 kΩ, 0.25-watt resistor.
One 0.016 μF capacitor. (You'll probably find 0.016 μF capacitors listed as polypropylene film capacitors. A polypropylene film capacitor is made with different material than the more typical ceramic capacitor but performs the same function. If your supplier doesn't carry 0.016 μF capacitors, you can use the closest value the supplier carries. Your results will be changed slightly but will show the same effect.)
One function generator.
One oscilloscope. (You can substitute a multimeter and measure Vout in rms voltage rather than peak-to-peak voltage.)
One breadboard.
Step-by-Step Instructions
Set up the circuit shown in Figure 6.10. 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 6.10
Carefully check your circuit against the diagram.
After you have checked 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 gener
ate a 10 Vpp, 25 Hz sine wave.
4. Measure and record Vout.
5. Adjust the function generator to the frequency shown in the next row of the table.
6. Measure and record Vout.
7. Repeat steps 5 and 6 until you have recorded Vout for the last row of the table.
8. Calculate the values of XC for each row and enter them in the table.
fin XC Vout
25 Hz
50 Hz
100 Hz
250 Hz
500 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
20 kHz
30 kHz
50 kHz
100 kHz
200 kHz
500 kHz
1 MHz
9. In the blank graph shown in Figure 6.11, 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 6.8, but don't worry if your curve is shifted slightly to the right or left.
Figure 6.11
Expected Results
Figure 6.12 shows the breadboarded circuit for this project.
Figure 6.12
Figure 6.13 shows a function generator and oscilloscope attached to the circuit.
Figure 6.13
The input signal is represented by the upper sine wave shown in Figure 6.14, and the output signal is represented by the lower sine wave.
Figure 6.14
As you change fin, you may need to adjust the TIME/DIV, VOLTS/DIV, and vertical POSITION controls. The controls shown in Figure 6.15 are adjusted to measure Vout when fin = 7 kHz.
Figure 6.15
Your values should be close to those shown in the following table, and the curve should be similar to Figure 6.16:
fin XC Vout
25 Hz 400 kΩ 0.025 volts
50 Hz 200 kΩ 0.05 volts
100 Hz 100 kΩ 0.1 volts
250 Hz 40 kΩ 0.25 volts
500 Hz 20 kΩ 0.5 volts
1 kHz 10 kΩ 1 volts
3 kHz 3.3 kΩ 2.9 volts
5 kHz 2 kΩ 4.5 volts
7 kHz 1.4 kΩ 5.6 volts
10 kHz 1 kΩ 7.1 volts
20 kHz 500 Ω 8.9 volts
30 kHz 330 Ω 9.5 volts
50 kHz 200 Ω 9.8 volts
100 kHz 100 Ω 10 volts
200 kHz 50 10 volts
500 kHz 20 10 volts
1 MHz 10 10 volts
Figure 6.16
Notice the relationship between XC and Vout in this circuit. Low values of Vout and the voltage drop across the resistor in this circuit occur at frequencies for which XC is high. When XC is high, more voltage is dropped across the capacitor, and less voltage is dropped across the resistor. (Remember that XC changes with frequency, while the value of the resistor stays constant.) Similarly, when XC is low, less voltage is dropped across the capacitor, and more voltage is dropped across the resistor, resulting in a higher Vout.
14 Refer to the curve you drew in Project 6.1 for the following question.
Question
What would cause your curve to be moved slightly to the right or the left of the curve shown in Figure 6.16? _____
Answer
Slightly different values for the resistor and capacitor that you used versus the resistor and capacitor used to produce the curve in Figure 6.16. Variations in resistor and capacitor values are to be expected, given the tolerance allowed for standard components.
15 The circuit shown in Figure 6.17 is used in many electronic devices.
Figure 6.17
For this circuit, you measure the output voltage across the capacitor instead of across the resistor (between point A and ground).
The impedance of this circuit is the same as that of the circuit used in the last few problems. It still behaves like a voltage divider, and you can calculate the output voltage with an equation similar to the one you used for the high-pass filter circuit discussed in the last few problems. However, by switching the positions of the resistor and capacitor to create the circuit shown in Figure 6.17, you switch which frequencies will be attenuated, and which will not be attenuated, making the new circuit a low-pass filter, whose characteristics you explore in the next few problems.
Questions
A. What is the impedance formula for the circuit? _____
B. What is the formula for the output voltage? _____
Answers
A.
B.
16 Refer to the circuit shown in Figure 6.17 and the following values:
Questions
Find the following:
A. XC: _____
B. Z: _____
C. Vout: _____
Answers
A. 795 ohms
B. 1277 ohms
C. 6.24 volts
17 Again, refer to the circuit shown in Figure 6.17 to answer the following question.
Question
Calculate the voltage across the resistor using the values given in problem 16, along with the calculated impedance value. _____
Answer
18 Use the information from problems 16 and 17 to answer the following question.
Question
What is the formula to calculate Vin using the voltages across the capacitor and the resistor? _____
Answer
The formula is .
19 Vout of the circuit shown in Figure 6.17 changes as the frequency of the input signal changes. Figure 6.18 shows the graph of Vout versus frequency for this circuit.
Figure 6.18
Question
What parameters determine f1 and f2? _____
Answer
The values of the capacitor and the resistor
Note You can see in Figure 6.10 that Vout is large for the lowest frequency and nearly zero for the highest frequency. This type of circuit is called a low-pass filter because it will pass low frequency signals with little attenuation, while blocking high-frequency signals.
Project 6.2: The Low-Pass Filter
Objective
The objective of this project is to determine how Vout changes as the frequency of the input signal changes for a low-pass filter.
General Instructions
After the circuit is set up, measure Vout for each frequency. You also calculate XC for each frequency value to show the relationship between the output voltage and the reactance of the capacitor.
Parts List
You need the following equipment and supplies:
One 1 kΩ, 0.25-watt resistor. (You can use the same resistor that you used in Project 6.1.)
One 0.016 μF capacitor. (You can use the same capacitor that you used in Project 6.1.)
One function generator.
One oscilloscope. (You can substitute a multimeter and measure Vout in rms voltage rather than peak-to-peak voltage.)
One breadboard.
Step-by-Step Instructions
Set up the circuit shown in Figure 6.19. 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 6.19
Carefully check your circuit against the diagram.
After you have checked 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 10 Vpp, 25 Hz sine wave.
4. Measure and record Vout.
5. Adjust the function generator to the frequency shown in the next
row of the table.
6. Measure and record Vout.
7. Repeat steps 5 and 6 until you have recorded Vout for the last row of the table.
8. Enter the values of XC for each row in the table. (Because you used the same capacitor and resistor in Project 6.1, you can take the values XC from the table in Project 6.1.)
fin XC VOUT
25 Hz
50 Hz
100 Hz
250 Hz
500 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
20 kHz
30 kHz
50 kHz
100 kHz
200 kHz
500 kHz
1 MHz
9. In the blank graph shown in Figure 6.20, 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 6.18.
Figure 6.20
Expected Results
Figure 6.21 shows the breadboarded circuit for this project.
Figure 6.21
Figure 6.22 shows a function generator and oscilloscope attached to the circuit.
Figure 6.22
The input signal is represented by the upper sine wave, as shown in Figure 6.23, and the output signal is represented by the lower sine wave. Reading the number of divisions for the peak-to-peak output sine wave and multiplying it by the corresponding VOLTS/DIV setting allows to you measure Vout.
Figure 6.23
As you change fin adjustments in the TIME/DIV control, the VOLTS/DIV and vertical POSITION controls for channel 1 may be needed. The controls shown in Figure 6.24 are adjusted to measure Vout when fin = 20 kHz.