by Earl Boysen
Figure 6.24
Your values should be close to those shown in the following table, and the curve should be similar to Figure 6.25.
fin XC Vout
25 Hz 400 kΩ 10 volts
50 Hz 200 kΩ 10 volts
100 Hz 100 kΩ 10 volts
250 Hz 40 kΩ 10 volts
500 Hz 20 kΩ 10 volts
1 kHz 10 kΩ 10 volts
3 kHz 3.3 kΩ 9.4 volts
5 kHz 2 kΩ 9.1 volts
7 kHz 1.4 kΩ 8.2 volts
10 kHz 1 kΩ 7.1 volts
20 kHz 500 Ω 4.5 volts
30 kHz 330 Ω 2.9 volts
50 kHz 200 Ω 2.0 volts
100 kHz 100 Ω 1 volt
200 kHz 50 0.5 volt
500 kHz 20 0.2 volt
1 MHz 10 0.1 volt
Figure 6.25
Notice the relationship between XC and Vout in this circuit. Low values of Vout (The voltage drop across the capacitor in this circuit.) occur at frequencies for which XC is also low. When XC is low, more voltage is dropped across the resistor and less across the capacitor. (Remember that XC changes with frequency, whereas the value of the resistor stays constant.) Similarly, when XC is high, less voltage is dropped across the resistor, and more voltage is dropped across the capacitor, resulting in a higher Vout.
Phase Shift of an RC Circuit
20 In both of the circuits shown in Figure 6.26, the output voltage is different from the input voltage.
Figure 6.26
Question
In what ways do they differ? _____
Answer
The signal is attenuated, or reduced. The amount of attenuation depends upon the frequency of the signal. Circuit 1 will pass high-frequency signals while blocking low-frequency signals. Circuit 2 will pass low-frequency signals while blocking high-frequency signals.
21 The voltage is also changed in another way. The voltage across a capacitor rises and falls at the same frequency as the input signal, but it does not reach its peak at the same time, nor does it pass through zero at the same time. You can see this when you compare the Vout curves to the Vin curves in Figure 6.27.
Note The numbered graphs in Figure 6.27 are produced by the corresponding numbered circuits in Figure 6.26.
Figure 6.27
Questions
A. Examine graph (1). Is the output voltage peak displaced to the right or the left? _____
B. Examine graph (2). Is the output voltage peak displaced to the right or the left? _____
Answers
A. To the left
B. To the right
22 The output voltage waveform in graph (1) of Figure 6.27 is said to lead the input voltage waveform. The output waveform in graph (2) is said to lag the input waveform. The amount that Vout leads or lags Vin is measured in degrees. There are 90 degrees between the peak of a sine wave and a point at which the sine wave crosses zero volts. You can use this information to estimate the number of degrees Vout is leading or lagging Vin. The difference between these two waveforms is called a phase shift or phase difference.
Questions
A. What is the approximate phase shift of the two waveforms shown in the graphs? _____
B. Do you think that the phase shift depends on the value of frequency? _____
C. Will an RC voltage divider with the voltage taken across the capacitor produce a lead or a lag in the phase shift of the output voltage? _____
Answers
A. Approximately 35 degrees.
B. It does depend upon frequency because the values of the reactance and impedance depend upon frequency.
C. A lag as shown in graph (2).
23 The current through a capacitor is out of phase with the voltage across the capacitor. The current leads the voltage by 90 degrees. The current and voltage across a resistor are in phase. (That is, they have no phase difference.)
Figure 6.28 shows the vector diagram for a series RC circuit. θ is the phase angle by which VR leads Vin. ϕ is the phase angle by which VC lags Vin.
Figure 6.28
Note Although the voltage across a resistor is in phase with the current through the resistor, both are out of phase with the applied voltage.
You can calculate the phase angle by using this formula:
As an example, calculate the phase angle when 160 Hz is applied to a 3.9 kΩ resistor in series with a 0.1 μF capacitor.
You can calculate the inverse tangent of 2.564 on your calculator and find that the phase angle is 68.7 degrees, which means that VR leads Vin by 68.7 degrees. This also means that VC lags the input by 21.3 degrees.
In electronics, the diagram shown in Figure 6.28 is called a phasor diagram, but the mathematics involved are the same as for vector diagrams, with which you should be familiar.
Question
Sketch a phasor diagram using the angles θ and ϕ resulting from the calculations in this problem. Use a separate sheet of paper for your diagram.
Answer
See Figure 6.29. Note that the phasor diagram shows that the magnitude of VC is greater than VR.
Figure 6.29
24 Using the component values and input signal shown in Figure 6.30, answer the following questions.
Figure 6.30
Questions
Find the following:
A. XC: _____
B. Z: _____
C. Vout: _____
D. VR: _____
E. The current flowing through the circuit: _____
F. The phase angle: _____
Answers
A.
B.
C.
D.
E.
F.
Therefore, θ = 53.13 degrees.
25 Use the circuit shown in Figure 6.31 to answer the following questions.
Figure 6.31
Questions
Calculate the following parameters:
A. XC: _____
B. Z: _____
C. Vout: _____
D. VR: _____
E. The current flowing through the circuit: _____
F. The phase angle: _____
Answers
A. XC = 265 ohms
B.
C. VC = 125 volts
D. VR = 83 volts
E. I = 0.472 ampere
F.
Therefore, θ = 56.56 degrees.
Resistor and Capacitor in Parallel
26 The circuit shown in Figure 6.32 is a common variation on the low-pass filter circuit introduced in problem 15.
Figure 6.32
Because a DC signal will not pass through the capacitor, this circuit functions like the circuit shown in Figure 6.33 for DC input signals.
Figure 6.33
An AC signal will pass through both the capacitor and R2. You can treat the circuit as if it had a resistor with a value of r (where r is the parallel equivalent of R2 and XC) in place of the parallel capacitor and resistor. This is shown in Figure 6.34.
Figure 6.34
Calculating the exact parallel equivalent (r) is complicated and beyond the scope of this book. However, to demonstrate the usefulness of this circuit, you can make a major simplification. Consider a circuit where XC is only about one-tenth the value of R2 or less. This circuit has many practical applications, because it attenuates the AC and the DC differently.
The following example can help to clarify this. For the following circuit, calculate the AC and DC output voltages separately.
For the circuit shown in Figure 6.35, you can calculate the AC and DC output voltages separately by following the steps outlined in the following questions.
Figure 6.35
Questions
A. Find XC. Check that it is less than one-tenth of R2.
B. For the circuit in Figure 6.35, determine through which circuit components DC signals will flow. Then use the voltage divider formula to find DC Vout.
C. For the circuit in Figure 6.35 determine which circuit components AC signals will flow through. Then use the voltage divider
formula to find AC Vout.
D. Compare the AC and DC input and output voltages. _____
Answers
A. XC = 106 ohms and R2 = 1000 ohms, so XC is close enough to one-tenth of R2.
B. Figure 6.36 shows the portion of the circuit that a DC signal passes through.
Figure 6.36
C. Figure 6.37 shows the portion of the circuit that an AC signal passes through.
Figure 6.37
D. Figure 6.38 shows the input waveform on the left and the output waveform on the right. You can see from the waveforms that the DC voltage has dropped from 20 volts to 10 volts and that the AC voltage has dropped from 10 volts to 1.05 volts.
Figure 6.38
27 Figure 6.39 shows two versions of the circuit discussed in problem 26 with changes to the value of the capacitor or the frequency of the input signal. The DC input voltage is 20 volts, and the AC input voltage is 10 Vpp. Use the same steps shown in problem 26 to find and compare the output voltages with the input voltages for the two circuits shown in Figure 6.39.
Figure 6.39
Questions
1.
A. XC = _____
B. DC Vout = _____
C. AC Vout = _____
D. Attenuation: _____
2.
A. XC = _____
B. DC Vout = _____
C. AC Vout = _____
D. Attenuation: _____
Answers
1.
A. XC = 10.6 ohms.
B. DC Vout = 10 volts.
C. AC Vout = 0.1 volts.
D. Here, the DC attenuation is the same as the example in problem 26, but the AC output voltage is reduced because of the higher frequency.
2.
A. XC = 10.6 ohms.
B. DC Vout = 10 volts.
C. AC Vout = 0.1 volts.
D. The DC attenuation is still the same, but the AC output voltage is reduced because of the larger capacitor.
Inductors in AC Circuits
28 Figure 6.40 shows a voltage divider circuit using an inductor, rather than a capacitor.
Figure 6.40
As with previous problems, consider all the inputs to be pure sine waves. Like the capacitor, the inductor cannot change the frequency of a sine wave, but it can reduce the amplitude of the output voltage.
The simple circuit, as shown in Figure 6.40, opposes current flow.
Questions
A. What is the opposition to current flow called? _____
B. What is the formula for the reactance of the inductor? _____
C. Write out the formula for the opposition to the current flow for this circuit. _____
Answers
A. Impedance
B. XL = 2πfL.
C.
In many cases, the DC resistance of the inductor is low, so assume that it is 0 ohms. For the next two problems, make that assumption in performing your calculations.
29 You can calculate the voltage output for the circuit shown in Figure 6.41 with the voltage divider formula.
Figure 6.41
Question
What is the formula for Vout? _____
Answer
30 Find the output voltage for the circuit shown in Figure 6.42.
Figure 6.42
Use the steps in the following questions to perform the calculation.
Questions
A. Find the DC output voltage. Use the DC voltage divider formula.
B. Find the reactance of the inductor.
C. Find the AC impedance.
D. Find the AC output voltage.
E. Combine the outputs to find the actual output. Draw the output waveform and label the voltage levels of the waveform on the blank graph shown in Figure 6.43.
Figure 6.43
Answers
A.
B. XL = 1 kΩ (approximately).
C.
D.
E. The output waveform is shown in Figure 6.44.
Figure 6.44
31 For the circuit shown in Figure 6.45, the DC resistance of the inductor is large enough that you should include that value in your calculations.
Figure 6.45
Questions
For the circuit shown in Figure 6.45, calculate the DC and AC output voltages, using the steps listed in problem 30.
A. DC Vout = _____
B. XL = _____
C. Z = _____
D. AC Vout = _____
E. Draw the output waveform and label the voltage levels of the waveform on the blank graph in Figure 6.46.
Figure 6.46
Answers
A.
Note The 500 Ω DC resistance of the inductor has been added to the 1 kΩ resistor value in this calculation.
B. XL = 2 kΩ,
C.
Note
The 500 Ω DC resistance of the inductor has been added to the 1 kΩ resistor value in this calculation.
D. AC Vout = 1.6 Vpp,
E. See Figure 6.47.
Figure 6.47
32 To calculate Vout, in problems 30 and 31, you also had to calculate XL. However, because XL changes with the frequency of the input signal, the impedance and the amplitude of Vout also change with the frequency of the input signal. If you plot the output voltage Vout against frequency, you should see the curve shown in Figure 6.48.
Figure 6.48
The values of the inductor and resistor determine the frequency at which Vout starts to drop (f1), and the frequency at which Vout levels off (f2).
The curve in Figure 6.48 shows that using an inductor and resistor in a circuit such as the one shown in Figure 6.42 produces a low-pass filter similar to the one discussed in problems 15 through 19.
Question
What values control f1 and f2? _____
Answer
The values of the inductor and the resistor
33 You can also create a circuit as shown in Figure 6.49, in which the output voltage is equal to the voltage drop across the inductor.
Figure 6.49
Questions
A. What formula would you use to find Vout? _____
B. If you plot the output voltage versus the frequency, what would you expect the curve to be? Use a separate sheet of paper to draw your answer. _____
Answers
A.
B. See Figure 6.50.
Figure 6.50
The curve in Figure 6.50 demonstrates that using an inductor and resistor in a circuit, such as the one shown in Figure 6.49, produces a high-pass filter similar to the one discussed in problems 6 through 13.
Higher-Order Filters
Filter circuits that contain one capacitor or inductor are called first-order filters. Filter order numbers reflect the number of capacitors, inductors, or operational amplifiers (a component discussed in Chapter 8, “Transistor Amplifiers”) in the filter. For example, a filter that contains four capacitors is a fourth-order filter, whereas a filter that contains six capacitors is a sixth-order filter.
If you want a sharper drop-off between frequencies, you can connect first-order filters in series. This effect is demonstrated in the following figure.
This graph shows how Vout changes as fin changes for the first-order low-pass filter used in Project 6.2, and for an eighth-order low-pass filter. Vout for the eighth-order filter drops by 80 percent at approximately 10 kHz, whereas Vout for the first-order filter doesn't drop by 80 percent until the frequency reaches approximately 50 kHz.
Phase Shift for an RL Circuit
34 Filter circuits that use inductors (such as those shown in Figure 6.51) produce a phase shift in the output signal, just as filter circuits containing capacitors do. You can see the shifts for the circuits shown in Figure 6.51 by comparing the input and output waveforms shown below the circuit diagrams.
Figure 6.51
Question
In which circuit does the output voltage lead the input voltage? _____
Answer
In graph (1), the output voltage lags the input voltage, an
d in graph (2), the output voltage leads.
35 Figure 6.52 shows a vector diagram for both the circuits shown in Figure 6.51. The current through the inductor lags the voltage across the inductor by 90 degrees.
Figure 6.52
The phase angle is easily found:
Question
Calculate the phase angle for the circuit discussed in problem 30. _____
Answer
45 degrees
36 Refer to the circuit discussed in problem 31.
Question
Calculate the phase angle. _____
Answer
Therefore, θ = 53.1 degrees.
Summary
This chapter has discussed the uses of capacitors, resistors, and inductors in voltage divider and filter circuits. You learned how to determine the following:
The output voltage of an AC signal after it passes through a high-pass RC filter circuit
The output voltage of an AC signal after it passes through a low-pass RC circuit
The output voltage of an AC signal after it passes through a high-pass RL circuit
The output voltage of an AC signal after it passes through a low-pass RL circuit
The output waveform of an AC or combined AC-DC signal after it passes through a filter circuit
Simple phase angles and phase differences