by Earl Boysen
Answers
A. TR = 1.83
B. VL = 0.55 Vpp
18 In this problem you use a transformer to match a generator to a 2 kΩ load.
Questions
A. What turns ratio is required to match a 2 kΩ load with a source that has an output impedance of = kΩ? _____
B. If the load requires a power of 20 mW, what should the source be? (First, find the voltage across the load.) _____
C. What are the primary and secondary currents and the power supplied by the source to the primary side of the transformer? _____
Answers
A. TR = 1.58
B.
C. IL = 3.16 mArms, Ip = 2 mArms, Pin = 20 mW
Summary and Applications
In this chapter, you learned about the following topics related to transformers:
The principles that allow an AC signal to be induced in a secondary coil
How the AC voltage across the secondary coil can be stepped up or down depending upon the turns ratio of the transformer
The use of a center tap to produce various voltages from a transformer
The use of transformers to match impedances between a generator and a load
That transformers can cause the output signal to be inverted (out of phase) from the input signal
Self-Test
These questions test your understanding of the material in this chapter. Use a separate sheet of paper for your diagrams or calculations. Compare your answers with the answers provided following the test.
1. How is a transformer constructed? _____
2. What type of signal is used as an input to a transformer? _____
3. If a sine wave is fed into a transformer shown in Figure 10.13, what does the output waveform look like? _____
Figure 10.13
4. What is meant by the term turns ratio? _____
5. If Vin = 1 Vpp and TR = 2, what is the output voltage?
Vout _____
6. Vin = 10 Vpp and Vout = 7 Vpp, what is the turns ratio?
TR = _____
7. In the center-tapped secondary winding shown in Figure 10.14, the voltage between points A and B may be expressed as VA-B = 28 Vpp. What is the voltage between C and A? _____
Figure 10.14
8. In the center-tapped secondary winding shown in Figure 10.14, the voltage between points B and C is VB-C = = Vrms. What is the peak-to-peak voltage between A and B? _____
9. If Iin = 0.5 Arms and Iout = 2.0 Arms, what is the turns ratio? _____
10. Is the transformer in problem 9 a step-up or a step-down transformer? _____
11. If ZL = 600 ohms and ZG = 6 kΩ, find the turns ratio.
TR = _____
12. If ZL = 1 kΩ and the turns ratio is 10:1, what is the generator impedance?
ZG = _____
Answers to Self-Test
If your answers do not agree with those given here, review the problems indicated in parentheses before you go to Chapter 11.
1. Two coils of wire wound around a magnetic core (such as iron or ferrite). (problem 1)
2. An AC voltage—DC does not work. (problem 2)
3. An inverted sine wave. (problem 3)
4. The ratio of the turns in the primary winding to the number of turns in the secondary coil. (problem 6)
5. Vout = 0.5 volts. (problem 7)
6. TR = 1.43:1. (problem 7)
7. VC-A = 14 Vpp. (problem 12)
8. VA-B = 14.14 Vpp. (problem 12)
9. TR = 4:1. (problem 13)
10. It is a step-down transformer. The voltage is lower (stepped down) in the secondary coil than in the primary coil if the current in the secondary coil is higher than the current in the primary coil. This maintains the same power on either side of the transformer. (problem 13)
11. TR = 3.2:1. (problem 16)
12. ZG = 100 kΩ. (problem 16)
Chapter 11
Power Supply Circuits
A power supply is incorporated into many electronic devices. Power supplies convert the 120 volts AC from a wall plug to a DC voltage, providing power for all types of electronic circuits.
Power supply circuits are simple in principle, and those shown in this chapter have been around for many years. Because power supplies incorporate many of the features covered in this book, they make an excellent conclusion to your study of basic electronics.
Diodes are a major component in power supplies. Learning how AC signals are affected by diodes is fundamental to your understanding of how power supplies work. Therefore, this chapter begins with a brief discussion of diodes in AC circuits.
Throughout this chapter, diagrams show how AC signals are affected by diodes and other components in power supply circuits. If you have an oscilloscope, you can breadboard the circuits and observe these waveforms.
When you complete this chapter, you will be able to do the following:
Describe the function of diodes in AC circuits.
Identify at least two ways to rectify an AC signal.
Draw the output waveforms from rectifier and smoothing circuits.
Calculate the output voltage from a power supply circuit.
Determine the appropriate component values for a power supply circuit.
Diodes in AC Circuits Produce Pulsating DC
1 You can use diodes for several purposes in AC circuits, where their characteristic of conducting in only one direction is useful.
Questions
Assume that you apply + 20 volts DC at point A of the circuit, as shown in Figure 11.1.
A. What is the output voltage at point B? _____
B. Suppose that you now apply + 10 volts DC at point B. What is the voltage at point A? _____
Figure 11.1
Answers
A. 20 volts DC (Ignore, for now, the voltage drop of 0.7 volt across the diode.)
B. 0 volts (The diode is reverse-biased.)
2 Figure 11.2 shows the circuit in Figure 11.1 with a 20 Vpp AC input signal centered at + 20 volts DC.
Figure 11.2
Questions
A. What are the positive and negative peak voltages of the input signal? _____
B. What is the output waveform of this circuit? _____
Answers
A. Positive peak voltage is 20 volts + 10 volts = 30 volts. Negative peak voltage is 20 volts − 10 volts = 10 volts.
B. The diode is always forward-biased, so it always conducts. Thus, the output waveform is exactly the same as the input waveform.
3 Figure 11.3 shows a circuit with 20 Vpp AC input signal centered at 0 volts DC.
Figure 11.3
Questions
A. What are the positive and negative peak voltages of the input signal? _____
B. For the positive half wave of the input, draw the output waveform on the blank graph provided in Figure 11.4.
Figure 11.4
Answers
A. Positive peak voltage is +10 volts. Negative peak voltage is − 10 volts.
B. See Figure 11.5.
Figure 11.5
4 When the input is negative, the diode in the circuit shown in Figure 11.3 is reverse-biased. Therefore, the output voltage remains at 0 volts.
Question
Figure 11.6 shows the input waveform for the circuit shown in Figure 11.3. Draw the output waveform on the blank graph provided in Figure 11.6.
Figure 11.6
Answer
See Figure 11.7.
Figure 11.7
5 Figure 11.7 shows the output waveform of the circuit shown in Figure 11.3, for one complete cycle of the input waveform.
Question
Now, draw the output waveform for three complete cycles of the input waveform shown in Figure 11.6. Use a separate sheet of paper for your drawing.
Answer
See Figure 11.8.
Figure 11.8
6 When the diode is connected in the opposite direction, it is forward-biased and, therefore, conducts current when the input signal is negative. In this case,
the diode is reverse-biased when the input signal is positive. Therefore, the output waveform is inverted from the output waveform shown in Figure 11.8.
Question
On a separate sheet of paper, draw the output waveform for three input cycles, assuming that the diode is connected in the opposite direction from the diode shown in Figure 11.3.
Answer
See Figure 11.9.
Figure 11.9
7 Figure 11.10 shows a circuit with a 20 Vpp AC input signal centered at −20 volts DC.
Figure 11.10
Questions
A. When is the diode forward-biased? _____
B. What is the output voltage? _____
Answers
A. Never because the voltage that results from adding the AC and DC signals ranges from − 10 volts to − 30 volts. Therefore, the diode is always reverse-biased.
B. A constant 0 volts.
8 As you have seen, a diode passes either the positive or negative portion of an AC voltage waveform, depending on how you connect it in a circuit. Therefore, the AC input signal is converted to a pulsed DC output signal, a process called rectification. A circuit that converts either the positive or negative portion of an AC voltage waveform to a pulsed DC output signal is called a half-wave rectifier.
Question
Refer to the output waveforms shown in Figures 11.8 and 11.9. Do these waveforms represent positive DC voltage pulses or negative DC voltage pulses? _____
Answer
The waveform in Figure 11.8 represents positive pulses of DC voltage. The waveform in Figure 11.9 represents negative pulses of DC voltage.
9 The circuit shown in Figure 11.11 shows a diode connected to the secondary coil of a transformer.
Figure 11.11
Questions
A. How does the diode affect the AC signal? _____
B. Draw the waveform of the voltage across the load for the circuit shown in Figure 11.11 if the secondary coil of the transformer has a 30 Vpp AC output signal centered at 0 volts DC. Use a separate sheet of paper for your drawing.
Answers
A. The AC signal is rectified.
B. See Figure 11.12. This type of circuit (called a half-wave rectifier) produces an output waveform containing either the positive or negative portion of the input waveform.
Figure 11.12
10 Figure 11.13 shows the waveforms at each end of a center-tap transformer secondary coil. Diode D1 rectifies the waveform shown at point A, and diode D2 rectifies the waveform shown at point B.
Figure 11.13
Questions
A. Which diode conducts during the first half of the cycle? _____
B. Which diode conducts during the second half of the cycle? _____
C. Draw the input waveforms (points A and B), and underneath draw each output waveform (points C and D). Use a separate sheet of paper for your drawing.
Answers
A. During the first half of the cycle, D1 is forward-biased and conducts current. D2 is reverse-biased and does not conduct current.
B. During the second half of the cycle, D2 is forward-biased and conducts current. D1 is reversed-biased and does not conduct current.
C. See Figure 11.14.
Figure 11.14
11 Figure 11.15 shows a circuit in which diodes connected to the ends of a center-tap transformer are connected to ground through a single resistor. The output voltage waveforms from both diodes are therefore applied across one load resistor. This type of circuit is called a full-wave rectifier.
Figure 11.15
Question
On a separate sheet of paper, draw the waveform representing the voltage at point E in the circuit, as shown in Figure 11.15. (This waveform is a combination of the waveforms at points C and D shown in Figure 11.14.)
Answer
See Figure 11.16.
Figure 11.16
12 Full-wave rectification of AC allows a much “smoother” conversion of AC to DC than half-wave rectification.
Figure 11.17 shows a full-wave rectifier circuit that uses a transformer with a two-terminal secondary coil, rather than a center-tapped secondary coil. This type of circuit is called a bridge rectifier.
Figure 11.17
Question
How does this circuit differ from the circuit shown in Figure 11.15? _____
Answer
This circuit has no center tap on the secondary coil, and it uses four diodes.
13 Figure 11.18 shows the direction of current flow when the voltage at point A is positive.
Figure 11.18
Figure 11.19 shows the direction of current flow when the voltage at point B is positive.
Figure 11.19
Notice that the direction of current through the load resistor is the same in both cases.
Questions
A. Through how many diodes does the current travel in each conduction path? _____
B. Draw the voltage waveform at point C. Use a separate sheet of paper for your drawing.
Answers
A. Two diodes in each case.
B. See Figure 11.20.
Figure 11.20
Project 11.1: The Full-Wave Rectifier
Objective
The objective of this project is to compare the outputs of the two types of full-wave rectifiers.
General Instructions
You set up two circuits. One of the circuits is a full-wave rectifier containing a center-tapped transformer and two diodes. The other circuit is a bridge rectifier containing a transformer and four diodes. After each circuit is set up, you apply a 20 Vpp signal to the primary side of the transformer. Then you use your oscilloscope to look at the waveform across the load resistor, and measure the peak voltage of each waveform.
Parts List
You need the following equipment and supplies:
Six 1N4001 diodes.
Two 10 k, 0.25-watt resistors.
One audio transformer with impedance of both the primary and secondary coil rated at 600 (with equal impedance for the primary and secondary coils, the turns ratio will be 1:1), and a center-tapped secondary coil. You can use a transformer with a center tap on both the primary and secondary coil; just don't connect a center tap that's not called for in the schematic.
Two breadboards.
One function generator.
One oscilloscope.
Step-by-Step Instructions
Set up Circuit # 1, the full-wave rectifier circuit shown in Figure 11.21. 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 11.21
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 1 to a jumper wire connected to the end of the resistor nearest the diodes. Connect the ground clip to a jumper wire attached to the ground bus.
2. Connect the oscilloscope probe for Channel 2 to a jumper wire that is connected to one end of the primary coil. This should be the same end to which you've connected the red lead from the function generator. Connect the ground clip for Channel 2 to a jumper wire that is connected to the other end of the primary coil. This should be the end to which you've connected the black lead from the function generator.
3. Set the function generator to generate a 1-kHz sine wave with 20 Vpp.
4. Measure and record Vp for the signal across the resistor.
5. Set up Circuit # 2; the bridge rectifier circuit shown in Figure 11.22. Use the same transformer you used in Circuit #1. You do not connect the center tap on the secondary coil in this circuit.
6. Connect the oscilloscope probe for Channel 1 to a jumper wire connected to one end of the resistor. Connect the g
round clip for Channel 1 to a jumper wire connected to the other end of the resistor.
7. Repeat steps 2 through 4.
Circuit # Vpp (Primary Coil) Vp (Load Resistor)
1
2
Figure 11.22
Expected Results
Figure 11.23 shows the breadboarded full-wave rectifier (Circuit # 1).
Figure 11.23
Figure 11.24 shows a function generator and oscilloscope attached to Circuit #1.
Figure 11.24
In Figure 11.25, the signal across the primary coil is represented by the upper waveform, and the signal across the resistor is represented by the lower waveform. Read the number of divisions for the peak-to-peak voltage of the upper waveform, and multiply it by the corresponding VOLTS/DIV setting to determine Vpp for the signal across the primary coil. Read the number of divisions for the peak voltage of the lower waveform, and multiply it by the corresponding VOLTS/DIV setting to determine Vp for the signal across the resistor.
Figure 11.25
As you measure Vpp and Vp, you may need to adjust the TIME/DIV, VOLTS/DIV, and vertical POSITION controls on the oscilloscope. The controls shown in Figure 11.26 are adjusted to measure Vpp in the primary coil and Vp across the load resistor for Circuit # 1.
Figure 11.26
Figure 11.27 shows the breadboarded bridge rectifier (Circuit # 2).
Figure 11.27