by Earl Boysen
Answers
A. Reactance
B. Resistance
13 Just as with resistance, you determine reactance by using an equation.
Questions
A. What is the equation for reactance? _____
B. What does each symbol in the equation stand for? _____
C. How does the reactance of a capacitor change as the frequency of a signal increases? _____
Answers
A.
B. XC = the reactance of the capacitor in ohms.
f = the frequency of the signal in hertz.
C = the value of the capacitor in farads.
C. The reactance of a capacitor decreases as the frequency of the signal increases.
14 Assume the capacitance is 1 μF and the frequency is 1 kHz.
Question
Find the capacitor's reactance. (Note: 1/(2π) = 0.159, approximately.) _____
Answers
f = 1 kHz = 103 Hz
C = 1 μF = 10−6 F
Thus,
15 Now, perform these simple calculations. In each case, find XC1 (the capacitor's reactance at 1 kHz) and XC2 (the capacitor's reactance at the frequency specified in the question).
Questions
Find XC1 and XC2:
A. C = 0.1 μF, f = 100 Hz. _____
B. C = 100 μF, f = 2 kHz. _____
Answers
A. At 1 kHz, XC1 = 1600 ohms; at 100 Hz, XC2 = 16,000 ohms
B. At 1 kHz, XC1 = 1.6 ohms; at 2 kHz, XC2 = 0.8 ohms
A circuit containing a capacitor in series with a resistor (as shown in Figure 5.6) functions as a voltage divider.
Figure 5.6
Although this voltage divider provides a reduced output voltage, just like a voltage divider using two resistors, there's a complication. If you view the output and input voltage waveforms on an oscilloscope, you see that one is shifted away from the other. The two waveforms are said to be “out of phase.” Phase is an important concept in understanding how certain electronic circuits work. In Chapter 6, “Filters,” you learn about phase relationships for some AC circuits. You also encounter this again when you study amplifiers.
Using the Oscilloscope
You use an oscilloscope to measure AC signals generated by a circuit, or to measure the effect that a circuit has on AC signals. The key parameters you measure with an oscilloscope are frequency and peak-to-peak voltage. An oscilloscope can also be used to show the shape of a signal's waveform so that you can ensure that the circuit works properly. When using an oscilloscope to compare a circuit's input signal to its output signal, you can determine the phase shift, as well as the change in Vpp.
The following figure shows an oscilloscope whose probe connects to the output of an oscillator circuit to measure the frequency of the signal generated by the oscillator. (Oscillator circuits are discussed in Chapter 9, “Oscillators.”) This example uses an analog oscilloscope, but you can also use a digital oscilloscope, which automates many of the measurements.
This oscilloscope has two channels, which provide the capability to measure two waveforms at once. Only channel 2 was used for the measurement in the preceding figure. The oscilloscope probe was clipped to a jumper wire connecting to
Vout for the circuit, and the oscilloscope ground clip was clipped to a jumper wire connecting to the ground bus.
The following figure shows the oscilloscope control panel. You use the VOLTS/ DIVcontrol to adjust the vertical scale and the TIME/DIVcontrol to adjust the horizontal scale. Set the vertical position knob and the horizontal position knob to adjust the position of the waveform against the grid to make it easier to measure.
You can determine the period of this waveform by counting the number of horizontal divisions the waveform takes to complete one cycle, and then multiplying the number of divisions by the TIME/DIVsetting. In the following figure, the period of the sine wave generated by this oscillator circuit is approximately 3.3 divisions wide.
Because the TIME/DIVknob is set at 10 μs, the period of this sine wave is 33 μs. The frequency of this sine wave is therefore calculated as follows:
You can also measure the effect of a circuit on a signal of a particular frequency. Supply the signal from a function generator to the input of the circuit. Attach the oscilloscope probe for channel 2 to the input of the circuit. Attach the oscilloscope probe for channel 1 to the output of the circuit. The following figure shows a function generator and oscilloscope attached to the voltage divider circuit shown in Figure 5.6.
In this example, the red lead from the function generator was clipped to a jumper wire connected to the resistor in the voltage divider circuit, and the black lead was clipped to a jumper wire connected to the ground bus. The oscilloscope probe for channel 2 is clipped to a jumper wire connected to the resistor, and the ground clip is clipped to a jumper wire attached to the ground bus. The oscilloscope probe for channel 1 is clipped to a jumper wire connected to the voltage divider circuit Vout, and the ground clip is clipped to a jumper wire connected to the ground bus.
The function generator supplies an input signal at a frequency of 10 kHz and an amplitude of 10 Vpp. The input signal is represented by the upper sine wave on the oscilloscope. Many function generators (such as the one shown here) have an amplitude adjustment knob without a readout. You set the input signal amplitude to 10 Vpp with the amplitude knob on the function generator while monitoring the amplitude on the oscilloscope.
The output signal is represented by the lower sine wave in the following figure. Adjust the VOLT/DIVcontrols and vertical position controls for channels 1 and 2 to fit both sine waves on the screen, as shown here.
You can measure Vpp for each sine wave by multiplying the number of vertical divisions between peaks by the setting on the VOLT/DIVknobs. For the input sine wave in this example, this measurement is two divisions at 5 VOLTS/DIV, for a total of 10 volts. For the output sine wave, this measurement is 3 divisions at 2 VOLTS/DIV, for a total of 6 volts. This indicates that the circuit has decreased the input signal from 10 Vpp to 6 Vpp.
Also note that that the peak of the output waveform shifts from the input waveform, a phenomenon called phase shift. You learn more about how to calculate phase shift in Chapter 6.
The Inductor in an AC Circuit
16 An inductor is a coil of wire, usually wound many times around a piece of soft iron. In some cases, the wire is wound around a nonconducting material.
Questions
A. Is the AC reactance of an inductor high or low? Why? _____
B. Is the DC resistance high or low? _____
C. What is the relationship between the AC reactance and the DC resistance? _____
D. What is the formula for the reactance of an inductor? _____
Answers
A. Its AC reactance (XL), which can be quite high, is a result of the electromagnetic field that surrounds the coil and induces a current in the opposite direction of the original current.
B. Its DC resistance (r), which is usually quite low, is simply the resistance of the wire that makes up the coil.
C. None
D. XL = 2πfL, where L = the value of the inductance in henrys (H). Using this equation, you can expect the reactance of an inductor to increase as the frequency of a signal passing through it increases.
17 Assume the inductance value is 10 henrys (H) and the frequency is 100 Hz.
Question
Find the reactance. _____
Answer
XL = 2πfL = 2π × 100 × 10 = 6280 ohms
18 Now, try these two examples. In each case, find XL1 (the reactance of the inductor at 1 kHz) and XL2 (the reactance at the frequency given in the question).
Questions
A. L = 1 mH (0.001 H), f = 10 kHz _____
B. L = 0.01 mH, f = 5 MHz _____
Answers
A. XL1 = 6.28 × 103 × 0.001 = 6.28 ohms
XL2 = 6.28 × 10 × 103 × 0.001 = 62.8 ohms
B. XL1 = 6.28 × 103 × 0.01 × 10−3 = 0.0628 ohms
XL2 = 6.28 × = × 106 × 0.01 × 10−3 = 314 ohms
A circuit containing an inductor in series with a resistor functions as a voltage divider, just as a circuit containing a capacitor in series with a resistor does. Again, the relationship between the input and output voltages is not as simple as a resistive divider. The circuit is discussed in Chapter 6.
Resonance
19 Calculations in previous problems demonstrate that capacitive reactance decreases as frequency increases, and that inductive reactance increases as frequency increases. If a capacitor and an inductor are connected in series, there is one frequency at which their reactance values are equal.
Questions
A. What is this frequency called? _____
B. What is the formula for calculating this frequency? You can find it by setting XL = XC and solving for frequency. _____
Answers
A. The resonant frequency
B. 2πfL = 1/(2πfC). Rearranging the terms in this equation to solve for f yields the following formula for the resonant frequency (fr):
20 If a capacitor and an inductor are connected in parallel, there is also a resonant frequency. Analysis of a parallel resonant circuit is not as simple as it is for a series resonant circuit. The reason for this is that inductors always have some internal resistance, which complicates some of the equations. However, under certain conditions, the analysis is similar. For example, if the reactance of the inductor in ohms is more than 10 times greater than its own internal resistance (r), the formula for the resonant frequency is the same as if the inductor and capacitor were connected in series. This is an approximation that you use often.
Questions
For the following inductors, determine if the reactance is more or less than 10 times its internal resistance. A resonant frequency is provided.
A. fr = 25 kHz, L = 2 mH, r = 20 ohms _____
B. fr = 1 kHz, L = 33.5 mH, r = 30 ohms _____
Answers
A. XL = 314 ohms, which is more than 10 times greater than r
B. XL = 210 ohms, which is less than 10 times greater than r
Note Chapter 7, “Resonant Circuits,” discusses both series and parallel resonant circuits. At that time, you learn many useful techniques and formulas.
21 Find the resonant frequency (fr) for the following capacitors and inductors when they are connected both in parallel and in series. Assume r is negligible.
Questions
Determine fr for the following:
A. C = 1 μF, L = 1 henry _____
B. C = 0.2 μF, L = 3.3 mH _____
Answers
A.
B.
22 Now, try these two final examples.
Questions
Determine fr:
A. C = 10 μF, L = 1 henry _____
B. C = 0.0033 μF, L = 0.5 mH _____
Answers
A. fr = 50 Hz (approximately)
B. fr = 124 kHz
Understanding resonance is important to understanding certain electronic circuits, such as filters and oscillators.
Filters are electronic circuits that either block a certain band of frequencies, or pass a certain band of frequencies. One common use of filters is in circuits used for radio, television, and other communications applications. Oscillators are electronic circuits that generate a continuous output without an input signal. The type of oscillator that uses a resonant circuit produces pure sine waves. (You learn more about oscillators in Chapter 9.)
Summary
Following are the concepts presented in this chapter:
The sine wave is used extensively in AC circuits.
The most common laboratory generator is the function generator.
f = 1/T
Capacitive reactance is calculated as follows:
Inductive reactance is calculated as follows:
Resonant frequency is calculated as follows:
Self-Test
The following problems test your understanding of the basic concepts presented in this chapter. Use a separate sheet of paper for calculations if necessary. Compare your answers with the answers provided following the test.
1. Convert the following peak or peak-to-peak values to rms values:
A. Vp = 12 V
Vrms = _____
B. Vp = 80 mV
Vrms = _____
C. Vpp = 100 V
Vrms = _____
2. Convert the following rms values to the required values shown:
A. Vrms = 120 V
Vp = _____
B. Vrms = 100 mV
Vp = _____
C. Vrms = 12 V
Vpp = _____
3. For the given value, find the period or frequency:
A. T = 16.7 ms
f = _____
B. f = 15 kHz
T = _____
4. For the circuit shown in Figure 5.7, find the total current flow and the voltage across R2, (Vout). _____
Figure 5.7
5. Find the reactance of the following components:
A. C = 0.16 μF, f = 12 kHz
XC = _____
B. L = 5 mH, f = 30 kHz
XL = _____
6. Find the frequency necessary to cause each reactance shown:
A. C = 1 μF, XC = 200 ohms
f = _____
B. L = 50 mH, XL = 320 ohms
f = _____
7. What would be the resonant frequency for the capacitor and inductor values given in A and B of question = if they were connected in series? _____
8. What would be the resonant frequency for the capacitor and inductor values given in A and B of question 6 if they were connected in parallel? What assumption would you need to make? _____
Answers to Self-Test
If your answers do not agree with those provided here, review the problems indicated in parentheses before you go to Chapter 6, “Filters.”
1A. 8.5 Vrms (problems 4–6)
1B. 56.6 Vrms
1C. 35.4 Vrms
2A. 169.7 Vp (problems 4–6)
2B. 141.4 mVp
2C. 33.9 Vpp
3A. 60 Hz (problem 7)
3B. 66.7 μsec
4. IT = 0.1Arms, Vout = 12Vrms (problems 9–11)
5A. 82.9 ohms (problems 14 and 17)
5B. 942.5 ohms
6A. 795.8 Hz (problems 14 and 17)
6B. 1.02 kHz
7. 5.63 kHz (problem 19)
8. 711.8 Hz. Assume the internal resistance of the inductor is negligible. (problem 20)
Chapter 6
Filters
Certain types of circuits are found in most electronic devices used to process alternating current (AC) signals. One of the most common of these, filter circuits, is covered in this chapter. Filter circuits are formed by resistors and capacitors (RC), or resistors and inductors (RL). These circuits (and their effect on AC signals) play a major part in communications, consumer electronics, and industrial controls.
When you complete this chapter, you will be able to do the following:
Calculate the output voltage of an AC signal after it passes through a high-pass RC filter circuit.
Calculate the output voltage of an AC signal after it passes through a low-pass RC circuit.
Calculate the output voltage of an AC signal after it passes through a high-pass RL circuit.
Calculate the output voltage of an AC signal after it passes through a low-pass RL circuit.
Draw the output waveform of an AC or combined AC-DC signal after it passes through a filter circuit.
Calculate simple phase angles and phase differences.
Capacitors in AC Circuits
1 An AC signal is continually changing, whether it is a pure sine wave or a complex signal made up of many sine waves. If such a signal is applied to one plate of a capacitor, it will be induced on the other plate. To express this another way, a capacitor will “pass” an AC signal, as illustrated in Figure 6.1.
Figure 6.1
Note Unlike an AC signal, a D
C signal is blocked by a capacitor. Equally important is that a capacitor is not a short circuit to an AC signal.
Questions
A. What is the main difference in the effect of a capacitor upon an AC signal versus a DC signal? _____
B. Does a capacitor appear as a short or an open circuit to an AC signal? _____
Answers
A. A capacitor will pass an AC signal, whereas it will not pass a DC voltage level.
B. Neither.
2 In general, a capacitor will oppose the flow of an AC current to some degree. As you saw in Chapter 5, “AC Pre-Test and Review,” this opposition to current flow is called the reactance of the capacitor.
Reactance is similar to resistance, except that the reactance of a capacitor changes when you vary the frequency of a signal. The reactance of a capacitor can be calculated by a formula introduced in Chapter 5.
Question
Write the formula for the reactance of a capacitor. _____
Answer
3 From this formula, you can see that the reactance changes when the frequency of the input signal changes.
Question
If the frequency increases, what happens to the reactance? _____
Answer
It decreases.
If you had difficulty with these first three problems, you should review the examples in Chapter 5.
Capacitors and Resistors in Series