Complete Electronics Self-Teaching Guide with Projects
Page 24
Answers to Self-Test
If your answers do not agree with those provided here, review the problems indicated in parentheses before you go on to Chapter 10.
1. An amplifier, feedback, and a resonant load. (problem 1)
2. Positive feedback is “in phase” with the input, and negative feedback is “out of phase” with the input. (problems 2–3)
3. Positive feedback. (problem 3)
4. (problem 11)
5. See Figure 9.15. (problem 20)
6. See Figure 9.27. (problem 25)
7. See Figure 9.35. (problem 25)
8A. (problems 27–30)
AV cannot be calculated.
C1/C2 = 0.047/0.1 = 0.47
Z cannot be calculated because r is unknown.
fr = 8.8 kHz (approximately).
8B.
AV = 2.2 (approximately).
C1/C2 = 1/3 (approximately).
Z cannot be calculated.
fr = 10 kHz (approximately).
8C.
Av cannot be calculated.
C1/C2 = 1
Z cannot be calculated.
fr = 3 kHz
8D.
Av cannot be calculated.
C1/C2 = 0.3
Z cannot be calculated.
fr = 1 kHz (approximately)
9. C1 = 0.0033 μF; C2 = 0.01 μF; CB = CC = 0.1 μF (problems 26–30)
Chapter 10
The Transformer
Transformers are used to “transform” an AC voltage to a higher or lower level. When you charge your cellphone, you use a transformer to reduce the 120 volts supplied by the wall outlet to the 5 volts or so needed to charge your cellphone's battery. Most electrical devices that you plug into wall outlets use transformers to reduce power coming from an outlet to that required by the electrical components in the device.
You can also use transformers to increase voltage. For example, some of the equipment used to manufacture integrated circuits requires thousands of volts to operate. Transformers are used to increase the 240 volts supplied by the power company to the required voltage.
When you complete this chapter, you will be able to do the following:
Recognize a transformer in a circuit.
Explain and correctly apply the concepts of turns ratio and impedance matching.
Recognize two types of transformer.
Do simple calculations involving transformers.
Transformer Basics
1 Consider two coils placed close to each other, as shown in Figure 10.1. If you apply an AC voltage to the first (or primary) coil, the alternating current flowing through the coil creates a fluctuating magnetic field that surrounds the coil. As the strength and polarity of this magnetic field changes, it induces an alternating current and a corresponding AC voltage in the second (or secondary) coil. The AC signal induced in the secondary coil is at the same frequency as the AC signal applied to the primary coil.
Figure 10.1
Both transformer coils are usually wound around a core made of a magnetic material such as iron or ferrite to increase the strength of the magnetic field.
Questions
A. When the two coils are wound around the same core, are they connected electrically? _____
B. What type of device consists of two wire coils wound around an iron or ferrite core? _____
C. If you apply an AC voltage to the terminals of the primary coil, what occurs in the secondary coil? _____
Answers
A. No.
B. A transformer.
C. An alternating current is induced in the secondary coil, which produces an AC voltage between the terminals of the secondary coil.
2 A transformer is used only with alternating currents. A fluctuating magnetic field (such as that generated by alternating current flowing through a primary coil) is required to induce current in a secondary coil. The stationary magnetic field generated by direct current flowing through a primary coil will not induce any current or voltage in a secondary coil.
When a sine wave signal is applied to a primary coil, you can observe a sine wave of the same frequency across the secondary coil, as shown in Figure 10.2.
Figure 10.2
Questions
A. What will be the difference in frequency between a signal applied to a primary coil and the signal induced in a secondary coil? _____
B. What will be the voltage difference across a secondary coil if 10 volts DC is applied to the primary coil? _____
Answers
A. No difference. The frequencies will be the same.
B. Zero volts. When a DC voltage is applied to the primary coil, there is no voltage or current induced in the secondary coil. You can summarize this by saying that DC does not pass through a transformer.
3 You can compare the output waveform measured between the terminals of the secondary coil to the output waveform measured between the terminals of the primary coil. If the output goes positive when the input goes positive, as shown in Figure 10.3, then they are said to be in phase.
Figure 10.3
The dots on the coils in Figure 10.3 indicate the corresponding end of each coil. If one coil is reversed, then the output will be inverted from the input. The output is said to be out of phase with the input, and a dot is placed at the opposite end of the coil.
Question
In Figure 10.4, the output sine wave is out of phase with the input sine wave. Place a dot in the correct location in the secondary coil to show that it is out of phase.
Figure 10.4
Answer
The dot should be at the lower end of the right coil.
4 The transformer shown in the right side of Figure 10.5 has three terminals. The additional terminal, in the middle of the coil, is called a center tap.
Figure 10.5
Question
What is the difference between the two output waveforms shown for the transformer on the right side of Figure 10.5? _____
Answer
The two waveforms are 180 degrees out of phase. That is, the positive peak of the upper output occurs at the same time as the negative peak of the lower waveform.
5 In a transformer, the output voltage from the secondary coil is directly proportional to the number of turns of wire in the secondary coil. If you increase the number of turns of wire in the secondary coil, a larger output voltage is induced across the secondary coil. If you decrease the number of turns of wire in the secondary coil, a smaller output voltage is induced across the secondary coil.
Question
How does increasing the number of turns of wire in a secondary coil affect the output voltage across the secondary coil? _____
Answer
It increases the output voltage across the secondary coil.
6 Figure 10.6 shows the number of turns in the primary and secondary coils as Np and Ns.
Figure 10.6
Question
The ratio of the input to output voltage is the same as the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. Write a simple formula to express this. _____
Answer
Note The ratio of primary turns to secondary turns is called the turns ratio (TR):
7 Use the formula from problem 6 to answer the following question.
Question
Calculate the output voltage of a transformer with a 2 to 1 (2:1) turns ratio when you apply a 10 Vpp sine wave to the primary coil. _____
Answer
8 Use the input voltage and turns ratio for a transformer to answer the following questions.
Questions
Calculate Vout in the following:
A. Vin = 20 Vpp, turns ratio = 5:1.
Vout = _____
B. Vin = 1 Vpp, turns ratio = 1:10.
Vout = _____
C. Vin = 100 Vrms. Find Vout when the primary and secondary coil have an equal number of turns.
Vout = _____
Answers
A. 4 Vpp (This is a s
tep-down transformer.)
B. 10 Vpp (This is a step-up transformer.)
C. 100 Vrms (This is an isolation transformer, which is used to separate or isolate the voltage source from the load electrically.)
9 Almost all electronic equipment operated from 120 volts AC house current requires a transformer to convert the 120 volts AC to a more suitable, lower voltage. Figure 10.7 shows a transformer that steps down 120 volts AC to 28 volts AC.
Figure 10.7
Question
Calculate the turns ratio for this transformer. _____
Answer
10 Figure 10.8 shows an oscilloscope trace of the output waveform from the 28-volt secondary coil.
Figure 10.8
Questions
A. Is 28 volts a peak-to-peak or an rms value? _____
B. What is the peak-to-peak value of the 28 volts across the secondary coil? _____
Answers
A. rms
B. 2 × 1.414 × 28 = 79.184 volts
11 Like the 28-volt transformer output value, the 120-volt wall plug value is an rms measurement.
Question
What is the peak-to-peak value of the voltage from the wall plug? _____
Answer
Approximately 340 volts
12 The actual voltage measured across the secondary coil of a transformer depends upon where and how you make the measurement. Figure 10.9 illustrates different ways to measure voltage across a 20 Vpp secondary coil that has a center tap.
Figure 10.9
If the center tap is grounded as shown in diagram (1) of Figure 10.9, then there is 10 Vpp AC between each terminal and ground. You can see that the two output waveforms in diagram (1) are out of phase (180 degrees out of phase, in this case) by comparing the two sine waves shown next to the two terminals. If the bottom terminal is grounded as it is in diagram (2) of Figure 10.9 and the center tap is not used, then there is 20 Vpp between the top terminal and ground.
Questions
A. Assume a center-tapped secondary coil is rated at 28 Vrms referenced to the center tap. What is the rms voltage output when the center tap is grounded? _____
B. Assume the 28 Vrms is the total output voltage across the entire secondary winding. What will be the output voltage between each end of the coil and the center tap? _____
C. Assume the output voltage of a center-tapped secondary coil is 15 Vrms between each end of the coil and the center tap. What is the peak-to-peak output voltage when the center tap is not connected? _____
Answers
A. 28 Vrms between each end of the coil and the center tap.
B. 14 Vrms (one half of the total Vout).
C. When the center tap is not connected, the output is 30 Vrms. Therefore, Vpp = 2 × 1.414 × 30 = 84.84 volts.
13 When the magnetic field induces an AC signal on the secondary coil, there is some loss of power. The percentage of power out of the transformer versus the input power is called the efficiency of the transformer. For the sake of this discussion, assume the transformer has an efficiency of 100 percent. Therefore, the output power of the secondary coil equals the power into the primary coil.
Power in = Power out (or Pin = Pout)
However, P = VI. Therefore, the following is true:
You can rearrange this to come up with the following formula:
Questions
A. What would be the input current for a transformer if the input power was 12 watts at a voltage of 120 Vrms? _____
B. What would be the transformer's output voltage if the turns ratio was 5:1? _____
C. What would be the output current? _____
D. What would be the output power? _____
Answers
In AC power calculations, you must use the rms values of current and voltage.
A.
B.
C. Iout = Iin (TR) = 0.1 × = = 0.5 Arms
D. Pout = VinIout = 24 × 0.5 = 12 watts (same as the power in)
Inside the Transformer
In addition to the turns ratio discussed in this chapter, the design of transformers incorporates a few more aspects. The frequency at which a transformer is expected to operate has a big impact on the design and composition of the core. Transformers with iron cores work well at low frequencies, such as 50 or 60 Hz household AC, and even at audio frequencies. Transformers used at these frequencies are often made of laminated sheets of iron, instead of one solid piece of iron, which increases the electrical resistance of the core, which reduces the eddy current. The eddy current is an electrical current induced in the core by the fluctuating magnetic field that reduces the efficiency of transformers.
Reducing the eddy current is especially important at high frequencies. A transformer designed to work in the MHz range requires a core with higher electrical resistance to reduce the eddy current. Therefore, high-frequency transformers have cores made of different materials, such as iron oxides (called ferrites) or powdered iron. Transformers are rated for a particular frequency range, which you can find either in the supplier's catalog or in the manufacturer's data sheet.
The maximum power that can pass through a transformer is stated as a VA rating. VA stands for “volts x amperes” and is dependent upon factors such as the gauge of wire used. The VA rating makes it easy to calculate the maximum amperage when you know the voltage your circuit requires.
Transformers may also be rated by their input and output impedance at a particular frequency stated in the data sheet.
Transformers in Communications Circuits
14 In communications circuits, an input signal is often received via a long interconnecting wire (usually called a line) that normally has an impedance of 600 ohms. A typical example is a telephone line between two cities.
Question
Communications equipment works best when connected to a load that has the same impedance as the output of the equipment. What output impedance should communications equipment have? _____
Answer
600 ohms output impedance, to be connected to a 600-ohm line
15 Because most electronic equipment does not have a 600-ohm output impedance, a transformer is often used to connect such equipment to a line. Often, the transformer is built into the equipment for convenience. The transformer is used to “match” the equipment to the line, as shown in Figure 10.10.
Figure 10.10
To work correctly, the output of the transformer secondary coil should have a 600-ohm impedance to match the line. The output impedance of the transformer (measured at the secondary winding) is governed by two things. One of these is the output impedance of the equipment.
Question
What would you expect the other governing factor to be? _____
Answer
The turns ratio of the transformer. (The DC resistance of each coil has no effect, and you can ignore it.)
16 Figure 10.11 shows a signal generator with an output impedance of ZG connected to the primary coil of a transformer. A load impedance of ZL is connected to the secondary coil.
Figure 10.11
You know that Pin = Pout and that P = V2/Z. Therefore, you can write an equation equating the power of the generator to the power of the load in terms of V and Z, as shown here:
You can rearrange this equation to give the ratio of the voltages, as shown here:
And, because VG = Vin, and VL = Vout, and Vin/Vout = Np/Ns, the following is true:
Therefore, the ratio of the input impedance to the output impedance of a transformer is equal to the square of the turns ratio. As you can see in the following question A, you can determine the turns ratio for a transformer that matches impedances between a generator and a load. In this way, the generator “sees” an impedance equal to its own impedance, and the load also “sees” an impedance equal to its own impedance.
For the following problem, a generator has an output impedance of 10 kΩ and produces a 10 Vpp (3.53 Vrms) signal. It will be connected to a 600-ohm line.
Questions
&n
bsp; A. To properly match the generator to the line, what turns ratio is required? _____
B. Find the output voltage across the load. _____
C. Find the load current and power. _____
Answers
A.
B. , which is 0.866Vrms
C.
Note For the power calculation, you must use the rms value of the voltage.
IL = Iin (TR) = 0.354 × 4.08 = 1.445 mArms, which is 4.08 mApp
which is the same as the input power. This circuit is shown in Figure 10.12.
Figure 10.12
Note The generator now sees 10 kΩ when it looks toward the load, rather than the actual 600-ohm load. By the same token, the load now sees 600 ohms when it looks toward the source. This condition allows the optimum transfer of power to take place between the source and the load. In practice, however, the optimum condition as calculated here rarely exists. Because it may be impossible to obtain a transformer with a turns ratio of 4.08:1, you would have to select the closest available value, which might be a turns ratio of 4:1. The difference in the turns ratio affects the conditions at the load side, but only slightly.
17 In this problem, you use a transformer to match a generator to a load.
Questions
A. What turns ratio is required to match a generator that has a 2 kΩ output to a 600-ohm line? _____
B. If the generator produces 1 Vpp, what is the voltage across the load? _____