Complete Electronics Self Teaching Guide with Projects

Complete Electronics Self Teaching Guide with Projects
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Complete Electronics Self Teaching Guide with Projects

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    ffirs.indd 26/14/2012 7:14:41 PM

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    Complete ElectronicsBook AuthorffirsV1June 14, 2012 7:14 PM ffirs.indd 16/14/2012 7:14:41 PM

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    ffirs.indd 26/14/2012 7:14:41 PM

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    Complete Electronics SELFTEA CHING GUIDE WITH PROJEC T SEarl Boysen | Harry KybettBook AuthorffirsV1June 14, 2012 7:14 PM ffirs.indd 36/14/2012 7:14:43 PM

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    Complete Electronics executive editor: carol longproject editor: kevin shafertechnical editor: rex millerproduction editor: kathleen wisorcopy editor: san dee phillipseditorial manager: mary beth wakefieldfreelancer editorial manager: rosemarie grahamassociate director of marketing: david mayhewm...

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    To my science and engineering teachers. I’d particularly like to thank Jim Giovando, my physics and chemistry teacher at Petaluma Senior High School, who, even decades later, I remember as having been an inspiration. I also dedicate this book to the physics and chemistry faculty of Sonoma State...

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    viAbout the AuthorEarl Boysen spent 20 years as an engineer in the semiconductor industry, and currently runs two websites, BuildingGadgets.com(dedicated to electronics) and UnderstandingNano.com (covering nanotechnology topics). Boysen holds a Masters degree in Engineering Physics from the Unive...

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    viiAbout the Technical EditorRex Miller was a Professor of Industrial Technology at The State University of New York, College at Buffalo for more than 35 years. He has taught on the technical school, high school, and college level for more than 40 years. He is the author or co-author of more tha...

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    viiiAcknowledgmentsI want to first thank Harry Kybett for authoring the original version of this book many years ago. It’s an honor to take over such a classic book in the electronics field. Thanks also to Carol Long for bringing me on board with the project, and Kevin Shafer for his able pro...

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    ixContents at a GlanceIntroductionxviiCHAP TE R 1DC Review and Pre-Test1CHAP TE R 2The Diode47CHAP TE R 3Introduction to the Transistor91CHAP TE R 4The Transistor Switch135CHAP TE R 5AC Pre-Test and Review187CHAP TE R 6Filters211CHAP TE R 7Resonant Circuits267CHAP TE R 8Transistor Amplifiers319CH...

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    xAPPE N D IX CPowers of Ten and Engineering Prefixes517APPE N D IX DStandard Composition Resistor Values519APPE N D IX ESupplemental Resources521APPE N D IX FEquation Reference525APPE N D IX GSchematic Symbols Used in This Book529Index533Book AuthorffirsV1June 14, 2012 7:14 PM ffirs.indd 106/14...

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    xiContentsIntroductionxviiCHAP TE R 1DC Review and Pre-Test1Current Flow ............................................... 2Ohm’s Law ................................................. 5Resistors in Series .......................................... 10Resistors in Parallel ............................

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    xiiCHAP TE R 3Introduction to the Transistor91Understanding Transistors.................................. 92The Junction Field Effect Transistor (JFET) ................... 123Summary .................................................. 129Self-Test......................................................

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    xiiiPhase Shift for an RL Circuit ................................. 258Summary .................................................. 260Self-Test.................................................... 260CHAP TE R 7Resonant Circuits267The Capacitor and Inductor in Series ........................ 268The...

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    xivSummary and Applications ................................. 432Self-Test.................................................... 432CHAP TE R 10The Transformer435Transformer Basics ......................................... 436Transformers in Communications Circuits ................... 447Summary an...

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    xvAPPE N D IX FEquation Reference525APPE N D IX GSchematic Symbols Used in This Book529Index533Book AuthorftocV1June 14, 2012 9:55 AM ftoc.indd 156/14/2012 9:55:54 AM

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    Book AuthorflastV1June 14, 2012 7:11 PM flast.indd 166/14/2012 7:11:27 PM

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    xviiIntroductionThe rapid growth of modern electronics is truly a phenomenon. Electronic devices (including cell phones, personal computers, portable MP3 players, and digital cameras) are a big part of many of our daily lives. Many industries have been founded, and older industries have been reva...

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    xviiiHelpful sidebars are interspersed throughout the book to provide more information about how components work, and how to choose the right component. Other sidebars provide discussions of techniques for building and testing circuits. If you want this addi-tional information, be sure to read th...

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    xixCHAP TE R 6Filters—This chapter looks at how resistors, capacitors, and inductors are used in high-pass filters and low-pass filters to pass or block AC signals above or below a certain frequency.CHAP TE R 7Resonant Circuits—This chapter examines the use of capacitors, induc-tors, and re...

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    xxAPPE N D IX DStandard Resistor Values—This appendix provides standard resistance values for the carbon film resistor, the most commonly used type of resistor.APPE N D IX ESupplemental Resources—This appendix provides references to helpful websites, books, and magazines.APPE N D IX FEquatio...

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    xxishould read a book such as Electronics for Dummies (Indianapolis: Wiley, 2009) to get the necessary background for this book. You can also go to the author’s Website actionURI(http://www.BuildingGadgets.com):(www.BuildingGadgets.com) and use the Tutorial links to find useful online lessons ...

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    xxiiYou can work through this book alone, or you can use it with a course. If you use the book alone, it serves as an introduction to electronics but is not a complete course. For that reason, at the end of the book are some suggestions for further reading and online resources. Also, at the back ...

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    1DC Review and Pre-TestElectronics cannot be studied without first under-standing the basics of electricity. This chapter is a review and pre-test on those aspects of direct current (DC) that apply to electronics. By no means does it cover the whole DC theory, but merely those topics that are es...

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    CHAPTER 1 DC REVIEW AND PRETEST2 ■ Power ■ Small currents ■ Resistance graphs ■ Kirchhoff’s Voltage Law ■ Kirchhoff’s Current Law ■ Voltage and current dividers ■ Switches ■ Capacitor charging and discharging ■ Capacitors in series and parallelCURRENT FLOW1  Electrica...

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    CURRENT FLOW3ANSWERThe following is a list of the most common ways to generate current: ■ Magnetically—This includes the induction of electrons in a wire rotating within a magnetic field. An example of this would be generators turned by water, wind, or steam, or the fan belt in a car. ■ C...

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    CHAPTER 1 DC REVIEW AND PRETEST4QUESTIONSA. Draw arrows to show the current flow in Figure 1.1. The symbol for the battery shows its polarity.VR+−FIGURE 1.1B. What indicates that a potential difference is present? C. What does the potential difference cause? D. What will happen if the bat...

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    OHM’S LAW5OHM’S LAW 4  Ohm’s law states the fundamental relationship between voltage, current, and resistance.QUESTIONWhat is the algebraic formula for Ohm’s law? ANSWERV I R5 3 This is the most basic equation in electricity, and you should know it well. Some electronics books state O...

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    CHAPTER 1 DC REVIEW AND PRETEST6ANSWERSA. 10 voltsB. 11.2 voltsC. 10 voltsD. 30 volts6  You can rearrange Ohm’s law to calculate current values.QUESTIONSWhat is the current for each combination of voltage and resistance values? A. V 5 1 volt, R 5 2 ohmsI5B. V5 2 volts, R 5 10 ohmsI5C. V5 1...

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    OHM’S LAW7QUESTIONSWhat is the resistance for each combination of voltage and current values? A. V5 1 volt, I 5 1 ampereR5B. V5 2 volts, I 5 0.5 ampereR5C. V5 10 volts, I 5 3 amperesR5D. V5 50 volts, I 5 20 amperesR5ANSWERSA. 1 ohmB. 4 ohmsC. 3.3 ohmsD. 2.5 ohms8  Work through these examples. ...

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    CHAPTER 1 DC REVIEW AND PRETEST8ANSWERSA. 1.2 amperesB. 3 ohmsC. 375 voltsINSIDE THE RESISTORResistors are used to control the current that flows through a portion of a circuit. You can use Ohm’s law to select the value of a resistor that gives you the cor-rect current in a circuit. For a g...

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    OHM’S LAW9film determines the resistance (the thicker the carbon film, the lower the resis-tance). Carbon film resistors work well in all the projects in this book.On the other hand, precision resistors contain a metal film deposited on an insu-lator. The thickness and width of the metal ...

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    CHAPTER 1 DC REVIEW AND PRETEST10RESISTORS IN SERIES9  You can connect resistors in series. Figure 1.3 shows two resistors in series.10Ω5ΩR1R2FIGURE 1.3QUESTIONWhat is their total resistance? ANSWERRT 5 R1 1 R2 5 10 ohms 1 5 ohms 5 15 ohmsThe total resistance is often called the equivalent ...

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    RESISTORS IN PARALLEL11QUESTIONWhat is the total resistance here? ANSWER11112121th112RRRus RohmTT515 15=;RT is often called the equivalent parallel resistance. 11  The simple formula from problem 10 can be extended to include as many resistors as wanted.QUESTIONWhat is the formula for three r...

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    CHAPTER 1 DC REVIEW AND PRETEST12QUESTIONSWhat is the total or equivalent resistance? A. R1 5 1 ohm, R2 5 1 ohm RT 5 B. R1 5 1,000 ohms, R2 5 500 ohms RT 5 C. R1 5 3,600 ohms, R2 5 1,800 ohms RT 5 ANSWERA. 0.5 ohmsB. 333 ohmsC. 1,200 ohmsRT is always smaller than the smallest of the resisto...

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    POWER1314  The first formula shown in problem 13 allows power to be calculated when only the voltage and current are known.QUESTIONSWhat is the power dissipated by a resistor for the following voltage and current values? A. V5 10 volts, I 5 3 amperesP5B. V5 100 volts, I 5 5 amperesP5C. V5 120 v...

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    CHAPTER 1 DC REVIEW AND PRETEST14ANSWERSA. 5 wattsB. 0.224 wattsC. 0.5 wattsD. 0.4 watts16  Resistors used in electronics generally are manufactured in standard values with regard to resistance and power rating. Appendix D shows a table of standard resistance values for 0.25- and 0.05-watt re...

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    SMALL CURRENTS15SMALL CURRENTS17  Although currents much larger than 1 ampere are used in heavy industrial equip-ment, in most electronic circuits, only fractions of an ampere are required.QUESTIONSA. What is the meaning of the term milliampere?B. What does the term microampere mean? ANSWERSA. A...

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    CHAPTER 1 DC REVIEW AND PRETEST16 19  The following exercise is typical of many performed in transistor circuits. In this example, 6 volts is applied across a resistor, and 5 mA of current is required to flow through the resistor.QUESTIONSWhat value of resistance must be used and what power ...

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    THE GRAPH OF RESISTANCE17Consider a simple circuit in which a battery is connected across a 1 kΩ resistor.QUESTIONSA. Find the current flowing if a 10-volt battery is used. B. Find the current when a 1-volt battery is used. C. Now find the current when a 20-volt battery is used. ANSWERSA. 10...

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    CHAPTER 1 DC REVIEW AND PRETEST18ANSWERYou should have drawn a straight line, as in the graph shown in Figure 1.6.510201Volts201051MilliamperesABFIGURE 1.6Sometimes you need to calculate the slope of the line on a graph. To do this, pick two points and call them A and B. ■ For point A, let V...

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    THE VOLTAGE DIVIDER19+−R1R2VOVSFIGURE 1.7The object of this circuit is to create an output voltage (V0) that you can control based upon the two resistors and the input voltage. V0 is also the voltage drop across R2.QUESTIONWhat is the formula for V0? ANSWERVVRRRoS531212R1 1 R2 5 RT, the tota...

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    CHAPTER 1 DC REVIEW AND PRETEST20ANSWERVVRRRvoltsOS53153153521210646106106 25  Now, try these problems.QUESTIONSWhat is the output voltage for each combination of supply voltage and resistance? A. VS 5 1 volt, R1 5 1 ohm, R2 5 1 ohm V0 5 B. VS 5 6 volts, R1 5 4 ohms, R2 5 2 ohms V0 5 C. VS ...

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    THE VOLTAGE DIVIDER2126  The output voltage from the voltage divider is always less than the applied voltage. Voltage dividers are often used to apply specific voltages to different components in a cir-cuit. Use the voltage divider equation to answer the following questions.QUESTIONSA. What is...

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    CHAPTER 1 DC REVIEW AND PRETEST22Contact group+ V busGroundbusBreadboards contain metal strips arranged in a pattern under the contact holes, which are used to connect groups of contacts together. Each group of five con-tact holes in a vertical line (such as the group circled in the figure) ...

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    THE VOLTAGE DIVIDER23group of contact holes that is also connected by a jumper wire to the ground bus. In this example, a 1.5 kΩ resistor was used for R1, and a 5.1 kΩ resistor was used for R2.Ground busoutR2R1TerminalblockVo outBatterypack wiresA terminal block is used to connect the battery p...

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    CHAPTER 1 DC REVIEW AND PRETEST24THE CURRENT DIVIDER27  In the circuit shown in Figure 1.9, the current splits or divides between the two resistors that are connected in parallel.+−I1R2R1I2ITITVSFIGURE 1.9IT splits into the individual currents I1 and I2, and then these recombine to form IT....

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    THE CURRENT DIVIDER252. Rearrange the ratio to give I2 in terms of I1: IIRR2112533. From the fact that IT 5 I1 1 I2, express IT in terms of I1 only.4. Now, find I1.5. Now, find the remaining current (I2).QUESTIONThe values of two resistors in parallel and the total current flowing through the ...

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    CHAPTER 1 DC REVIEW AND PRETEST26 29  Now, try these problems. In each case, the total current and the two resistors are given. Find I1 and I2.QUESTIONSA. IT 5 30 mA, R1 5 12 kΩ, R2 5 6 kΩ B. IT 5 133 mA, R1 5 1 kΩ, R2 5 3 kΩ C. What current do you get if you add I1 and I2? A...

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    THE CURRENT DIVIDER27QUESTIONWrite the equation for the current I2. Check the answers for the previous problem using these equations.ANSWERIIRRR2T21251()()()The current through one branch of a two-branch circuit is equal to the total current times the resistance of the opposite branch, divided...

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    CHAPTER 1 DC REVIEW AND PRETEST28R1R2NOTE The circuit used in a multimeter to measure voltage places a large-value resistor in parallel with R2 so that the test itself does not cause any measurable drop in the current passing through the circuit.TIP Whenever you perform tests on a circuit, att...

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    THE CURRENT DIVIDER29R1R2NOTE The circuit used in a multimeter to measure current passes the current through a low-value resistor so that the test itself does not cause any measur-able drop in the current.RESISTANCEYou typically use the resistance setting on a multimeter to check the resistance o...

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    CHAPTER 1 DC REVIEW AND PRETEST30SWITCHES31  A mechanical switch is a device that completes or breaks a circuit. The most familiar use is that of applying power to turn a device on or off. A switch can also permit a signal to pass from one place to another, prevent its passage, or route a si...

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    SWITCHES31B. What is the voltage at point A and point B with respect to ground? C. What is the voltage drop across the switch? ANSWERSA. 10101voltsohmsampere5B. VA 5 VB 5 10 voltsC. 0 V (There is no voltage drop because both terminals are at the same voltage.) 32  The circuit shown in Figu...

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    CHAPTER 1 DC REVIEW AND PRETEST32ANSWERSA. VA 5 10 volts; VB 5 0 volts.B. No current is flowing because the switch is open.C. 10 volts. If the switch is open, point A is the same voltage as the positive battery terminal, and point B is the same voltage as the negative battery terminal.33  Th...

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    CAPACITORS IN A DC CIRCUIT33CAPACITORS IN A DC CIRCUIT34  Capacitors are used extensively in electronics. They are used in both alternating cur-rent (AC) and DC circuits. Their main use in DC electronics is to become charged, hold the charge, and, at a specific time, release the charge.The capa...

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    CHAPTER 1 DC REVIEW AND PRETEST34B. What is the time constant for the circuit shown in Figure 1.15? C. How long does it take the capacitor to reach 10 volts? D. To what voltage level does it charge in one time constant? ANSWERSA.τ 5 R 3 C.B.τ 5 10 kΩ 3 10 μF 5 10,000 Ω 3 0.00001 F 5 0.1...

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    CAPACITORS IN A DC CIRCUIT35ANSWERSA. Both will be at 0 volts if the capacitor is totally discharged.B. It will rise toward 10 volts.C. It will stay at 0 volts.D. About 6.3 volts.37  The capacitor charging graph in Figure 1.16 shows how many time constants a voltage must be applied to a capacito...

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    CHAPTER 1 DC REVIEW AND PRETEST36ANSWERSA. It is called an exponential curve.B. It is used to calculate how far a capacitor has charged in a given time.38  In the following examples, a resistor and a capacitor are in series. Calculate the time constant, how long it takes the capacitor to full...

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    CAPACITORS IN A DC CIRCUIT37QUESTIONSA. With the switch in position X, what is the voltage on each plate of the capacitor? B. When the switch is moved to position Y, the capacitor begins to charge. What is its charging time constant? C. How long does it take to fully charge the capacitor? ANSWERS...

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    CHAPTER 1 DC REVIEW AND PRETEST38ANSWERSThe circuit powering a camera flash is an example of a capacitor’s capability to store a charge and then discharge upon demand. While you wait for the flash unit to charge, the camera uses its battery to charge a capacitor. When the capacitor is char...

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    CAPACITORS IN A DC CIRCUIT3941  Capacitors can be connected in parallel, as shown in Figure 1.18.(1)(2)1μF2μFC11μFC12μFC23μFC3C2FIGURE 1.18QUESTIONSA. What is the formula for the total capacitance? B. What is the total capacitance in circuit 1? C. What is the total capacitance in circuit 2?...

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    CHAPTER 1 DC REVIEW AND PRETEST40 42  Capacitors can be placed in series, as shown in Figure 1.19.C1C22 μF1 μFFIGURE 1.19QUESTIONSA. What is the formula for the total capacitance? B. In Figure 1.19, what is the total capacitance? ANSWERSA. 11111123CCCCCTN5111 1B. 111121232;23C1thusCTT...

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    SUMMARY41ANSWERSA. 3.3 μFB. 103.06 μFC. 0.15 μFSUMMARYThe few simple principles reviewed in this chapter are those you need to begin the study of electronics. Following is a summary of these principles: ■ The basic electrical circuit consists of a source (voltage), a load (resistance), and ...

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    CHAPTER 1 DC REVIEW AND PRETEST42 ■ You can find the power delivered by a source by using the following formula: PVI5 ■ You can find the power dissipated by a resistance by using the following formula: PI RorPVR5 225 ■ If you know the total applied voltage, VS, you can find the voltag...

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    DC PRETEST43 ■ At one time constant in an RC circuit, the values for current and voltage have reached 63 percent of their final values. At five time constants, they have reached their final values. ■ Capacitors in parallel are added to find the total capacitance. CCCCT12N11 1 ■ Capac...

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    CHAPTER 1 DC REVIEW AND PRETEST443. R1 5 12 kΩ, R2 5 8 kΩ, VS 5 24 volts V1 5 , V2 5 4. VS 5 36 V, I 5 250 mA, V1 5 6 volts R2 55. Now, go back to problem 1. Find the power dissipated by each resistor and the total power delivered by the source. P1 5 , P2 5 , PT 5 Questions ...

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    DC PRETEST45B. What is the voltage across the capacitor 60 μsec after the switch is closed? VC 5C. At what time will the capacitor be fully charged? T511. In the circuit shown in Figure 1.22, when the switch is at position 1, the time con-stant should be 4.8 ms. R1R215 V12C0.16μF10 kΩVSFIGU...

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    CHAPTER 1 DC REVIEW AND PRETEST46ANSWERS TO DC PRETESTIf your answers do not agree with those provided here, review the problems indicated in parentheses before you go to Chapter 2, “The Diode.” If you still feel uncertain about these concepts, go to a website such as actionURI(http://w...

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    2The DiodeThe main characteristic of a diode is that it conducts electricity in one direction only. Historically, the first vacuum tube was a diode; it was also known as a rectifier. The modern diode is a semiconductor device. It is used in all applications where the older vacuum tube diode was...

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    CHAPTER 2 THE DIODE48of adjacent regions of semiconductor crystals, which allow the manufacture of small diodes, as well as transistors and integrated circuits.When you complete this chapter, you can do the following: ■ Specify the uses of diodes in DC circuits. ■ Determine from a circuit dia...

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    UNDERSTANDING DIODES49PPPNNNNPSiGeSiNPGe(a)(b)(c)(d)(e)(f )FIGURE 2.1ANSWERDiagrams (b) and (e) only2  In a diode, the P material is called the anode. The N material is called the cathode.QUESTIONIdentify which part of the diode shown in Figure 2.2 is P material and which part is N material. Ano...

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    CHAPTER 2 THE DIODE50Figure 2.4 shows the circuit symbol for a diode. The arrowhead points in the direction of current flow. Although the anode and cathode are indicated here, they are not usually indicated in circuit diagrams.AnodeCathodeFIGURE 2.4QUESTIONIn a diode, does current flow from ano...

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    UNDERSTANDING DIODES51ANSWERSA. Yes.B. The anode connects to the positive battery terminal, and the cathode connects to the negative battery terminal. Therefore, the anode is at a higher voltage than the cathode.5  When the diode is connected so that the current flows, it is forward-biased. In...

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    CHAPTER 2 THE DIODE52QUESTIONDraw a reverse-biased diode in the circuit shown in Figure 2.7. +−FIGURE 2.7ANSWERYour drawing should look something like Figure 2.8.+−FIGURE 2.87  In many circuits, the diode is often considered to be a perfect diode to simplify calcula-tions. A perfect diode ha...

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    53UNDERSTANDING DIODES8  A forward-biased perfect diode can thus be compared to a closed switch. It has no voltage drop across its terminals, and current flows through it.A reverse-biased perfect diode can be compared to an open switch. No current flows through it, and the voltage difference ...

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    CHAPTER 2 THE DIODE54If you do not have access to equipment, do not skip this project. Read through the project, and try to picture or imagine the results. This is sometimes called “dry-labbing” the experiment. You can learn a lot from reading about this project, even though it is always bett...

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    55UNDERSTANDING DIODES❑ One 330-ohm, 0.5-watt resistor❑ One 1N4001 diode❑ One breadboard❑ One 1 MΩ potentiometer❑ One terminal blockSTEPBYSTEP INSTRUCTIONSSet up the circuit as shown in Figure 2.11. The circled “A” designates a multimeter set to measure current, and the circl...

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    CHAPTER 2 THE DIODE564. Adjust the potentiometer slightly to give a higher voltage.5. Measure and record the new values of voltage and current.6. Repeat steps 4 and 5 until the lowest resistance of the potentiometer is reached, taking as many readings as possible. This results in the highest volt...

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    57UNDERSTANDING DIODESmAVolts012348121620240.2 0.3 0.4 0.5 0.6 0.7 0.8FIGURE 2.12EXPECTED RESULTSFigure 2.13 shows the breadboarded circuit for this project.330Ω resistor1N4001 diodeBanded endof diodeFIGURE 2.13Book Authorc02V107/04/2012 2:49 PM c02.indd 577/4/2012 2:49:46 PM

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    CHAPTER 2 THE DIODE58Figure 2.14 shows the test setup for this project.1 MΩpotentiometerVoltageTo multimeterset to voltsTo multimeterset to mACurrentFIGURE 2.14Compare your measurements with the ones shown in the following table:V (volts)I (mA)0.440.000.460.010.500.060.520.110.550.230.580.490.60...

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    59UNDERSTANDING DIODESFurther reductions in the resistance below the 330 Ω included in the circuit causes little increase in the voltage but produces large increases in the current.Figure 2.15 shows the V-I curve generated using the measurements shown in the pre-ceding table.mAVolts0123481216202...

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    CHAPTER 2 THE DIODE60VActual diodecurvePerfect or idealdiode curveHigh resistanceregionKneeregionLowresistanceregionIFIGURE 2.169  The knee voltage is also a limiting voltage. That is, it is the highest voltage that can be obtained across the diode in the forward direction.QUESTIONSA. Which has ...

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    61UNDERSTANDING DIODES10  Refer back to the diagram of resistance regions shown in Figure 2.16.QUESTIONWhat happens to the current when the voltage becomes limited at the knee? ANSWERIt increases rapidly.11  For any given diode, the knee voltage is not exactly 0.7 volt or 0.3 volt. Rather, it v...

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    CHAPTER 2 THE DIODE62 12  Calculate the current through the diode in the circuit shown in Figure 2.17 using the steps in the following questions.+−VSVRVD5 VSi1 kΩFIGURE 2.17QUESTIONSA. The voltage drop across the diode is known. It is 0.7 volt for silicon and 0.3 volt for ger-manium. (“Si...

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    63UNDERSTANDING DIODES 13  In practice, when the battery voltage is 10 volt or above, the voltage drop across the diode is often considered to be 0 volt, instead of 0.7 volt.The assumption here is that the diode is a perfect diode, and the knee voltage is at 0 volts, rather than at a threshold v...

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    CHAPTER 2 THE DIODE64B. Calculate the current through the perfect diode, as shown in Figure 2.18. VD 5 VR 5 VS 2 VD 5 IVRR55 ID 5 ANSWERSA. 0.7 volt; 9.3 volt; 9.3 mA; 9.3 mAB. 0 volt; 10 volt; 10 mA; 10 mA 14  The difference in the values of the two currents found in problem 13 is less ...

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    65UNDERSTANDING DIODESANSWERSA. Yes, it can be considered a perfect diode.B. I 5 10 mA 15  When a current flows through a diode, it causes heating and power dissipation, just as with a resistor. The power formula for resistors is P 5 V 3 I. This same formula can be applied to diodes to find t...

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    CHAPTER 2 THE DIODE66Provided the current in the circuit does not exceed this, the diode cannot overheat and burn out.QUESTIONSuppose the maximum power rating of a germanium diode is 3 watts. What is its highest safe current? ANSWERIwattsvoltamperes5530310 17  Answer the following questions f...

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    67UNDERSTANDING DIODES18  The next several examples concentrate on finding the current through the diode. Look at the circuit shown in Figure 2.20.+−VSSiI2R2R1ITIDVDFIGURE 2.20The total current from the battery flows through R1, and then splits into I2 and ID.I2 flows through R2, and ID fl...

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    CHAPTER 2 THE DIODE68To find ID in the circuit shown in Figure 2.21, go through these steps, and then check your answers.+−5 VSiI2R2R1ITVD = 0.7 V43 Ω70 ΩFIGURE 2.21QUESTIONSA. I2 5 B. VR 5 C. IT 5 D. ID 5 ANSWERSA. IVRvoltvoltohmsamperemAD2220707700011055555RB. VR 5 VS − VD 5 5 volts ...

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    69UNDERSTANDING DIODESANSWERP 5 VD 3 ID 5 (0.7 volt)(90 mA) 5 63 milliwatts 21  To find the current in the diode for the circuit shown in Figure 2.22, answer the following questions in order.1.6 VGeI2R2R1ITID440 Ω250 ΩFIGURE 2.22QUESTIONSA. I2 5 B. VR 5 C. IT 5 D. ID 5 ANSWERSA. ImA2032...

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    CHAPTER 2 THE DIODE70DIODE BREAKDOWN22  Earlier, you read that if the circuit in Project 2.1 was not working correctly, then the diode may be in backward. If you place the diode in the circuit backward—as shown on the right in Figure 2.23—then almost no current flows. In fact, the current ...

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    DIODE BREAKDOWN71QUESTIONThe diode in the circuit shown in Figure 2.25 is known to break down at 100 volts, and it can safely pass 1 ampere without overheating. Find the resistance in this circuit that would limit the current to 1 ampere. +−200 V100 VRFIGURE 2.25ANSWERVVV volts volts00 voltsR...

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    CHAPTER 2 THE DIODE72The breakdown voltage is often called the peak inverse voltage (PIV) or the peak reverse voltage (PRV). Following are the PIVs of some common diodes:DiodePIV1N400150 volts1N4002100 volts1N4003200 volts1N4004400 volts1N4005600 volts1N4006800 voltsQUESTIONSA. Which can permanen...

  • Page 97

    DIODE BREAKDOWN73++++++ ++ ++ + ++ + + ++ ++ ++ + ++ + + ++ ++ ++ + ++ + + ++ ++ ++ + ++ + + ++ + + + + +−−−−−− −− −− − −− −− −−−− − − −− −− −−− − − −− − − − −−− − − − − −− − − − − −− − − − ...

  • Page 98

    CHAPTER 2 THE DIODE74Note that an interesting aspect of diodes is that while the N and P regions of the diode have mobile charges, they do not have a net charge. The mobile charges (elec-trons in N regions and holes in P regions) are simply donated by the impurity atoms used to dope the semicondu...

  • Page 99

    THE ZENER DIODE75THE ZENER DIODE24  Diodes can be manufactured so that breakdown occurs at lower and more precise voltages than those just discussed. These types of diodes are called zener diodes, so named because they exhibit the “Zener effect”—a particular form of voltage breakdown. At ...

  • Page 100

    CHAPTER 2 THE DIODE76+−VL = 20 VVRVLRLRGDCgeneratorVS = 50 VI = 1.5 AFIGURE 2.27You need to determine a suitable value for the resistance. Follow these steps to find a suitable resistance value:1. Find RL, the lamp resistance. Use the following formula:RVILL52. Find VR. Use VS 5 VR 1 VL.3. Fin...

  • Page 101

    THE ZENER DIODE77 26  Assume now that the 20-ohm resistor calculated in problem 25 is in place, and the voltage output of the generator drops to 35 volts, as shown in Figure 2.28. This is similar to what happens when a battery gets old. Its voltage level decays and it will no longer have suffic...

  • Page 102

    CHAPTER 2 THE DIODE7827  In many applications, a lowering of voltage across the lamp (or some other compo-nent) may be unacceptable. You can prevent this by using a zener diode, as shown in the circuit in Figure 2.29.+−VZRG35 VgeneratorLampFIGURE 2.29If you choose a 20-volt zener (that is, one...

  • Page 103

    THE ZENER DIODE79design is a common practice in electronics. It is used to ensure that equipment can work under the most adverse conditions.) The worst case here would occur when the generator puts out only 35 volts. Figure 2.30 shows the paths that current would take in this circuit.VZIZILIRR35 ...

  • Page 104

    CHAPTER 2 THE DIODE80 29   Now, take a look at what happens when the generator supplies 50 volts, as shown in Figure 2.31.IL = 1.5 AR = 7.5 Ω50 V20 VFIGURE 2.31Because the lamp still has 20 volts across it, it can still draw only 1.5 amperes. But the total current and the zener current change....

  • Page 105

    THE ZENER DIODE81ANSWERSA. IT has increased by 2 amperes.B. IZ has increased by 2 amperes.The increase in IT flows through the zener diode and not through the lamp.31   The power dissipated by the zener diode changes as the generator voltage changes.QUESTIONSA. Find the power dissipated when th...

  • Page 106

    CHAPTER 2 THE DIODE82QUESTIONFor the circuit shown in Figure 2.32, what power rating should the zener diode have? The current and voltage ratings of the lamp are given. ANSWERAt 24 volts, assuming a zener current of 0.5 ampere:Rohms5 590 57515 7At 60 volts:I amperes therefore I amperesRZ55 4515...

  • Page 107

    83THE ZENER DIODE❑ One 56-ohm, 0.5-watt resistor❑ One 1N4735A zener diode❑ One breadboard❑ One lamp rated for approximately 25 mA at 6 volts. (Part # 272-1140 from Radio Shack is a good fit for this project.)❑ Two terminal blocksSTEPBYSTEP INSTRUCTIONSSet up the circuit shown on ...

  • Page 108

    CHAPTER 2 THE DIODE846. Repeat steps 4 and 5 four times.Time (Minutes)VS (Volts)IL (mA)IZ (mA)03060 (1 hr)90120 (2 hr)EXPECTED RESULTSFigure 2.34 shows the breadboarded circuit for this project.To lamp1N4735A diodeBanded endof diode56Ω resistorFIGURE 2.34Book Authorc02V107/04/2012 2:49 PM c02.in...

  • Page 109

    85THE ZENER DIODEFigure 2.35 shows the test setup for this project.To multimeterset to voltsTo multimeter set to mATo multimeterset to mAILIZVSFIGURE 2.35Compare your measurements with the ones shown in the following table. You should see a similar trend in the measured values, but not exactly th...

  • Page 110

    CHAPTER 2 THE DIODE86Time (Minutes)VS (Volts)IL (mA) IZ (mA) 120 (2 hr)7.9124.97.01507.8224.95.7180 (3 hr)7.7624.94.62107.7024.93.7240 (4 hrs)7.6524.92.9As you can see in this data, even though the supply voltage dropped by approximately 15 percent, the lamp current stayed roughly constant, showi...

  • Page 111

    SELFTEST87 ■ Building high-speed switches ■ Rectifying radio frequency signalsSELFTESTThe following questions test your understanding of this chapter. Use a separate sheet of paper for your diagrams or calculations. Compare your answers with the answers that follow the test.1. Draw the ...

  • Page 112

    CHAPTER 2 THE DIODE888. Calculate question 7 using these values: VS 5 3 volts and R 5 1 kΩ. 9. In the circuit shown in Figure 2.37, find the current through the diode.VS 5 10 voltsR1 5 10 kΩR2 5 1 kΩSiR2R1VSFIGURE 2.3710. In the circuit shown in Figure 2.38, find the current through the zene...

  • Page 113

    SELFTEST8911. If the supply voltage for question 10 increases to 45 volts, what is the current in the zener diode? 12. What is the maximum power dissipated for the diode in questions 10 and 11?ANSWERS TO SELFTESTIf your answers do not agree with those given here, review the problems indicat...

  • Page 114

    CHAPTER 2 THE DIODE906.Si 5 0.7 volt; Ge5 0.3 volt (These are approximate.)(Project 2.1)7.ID 5 100 mA.(problem 12)8.As VS 5 3 volt, do not ignore the voltage drop across the diode. Thus, ID 5 2.7 mA.(problem 12)9.Ignore VD in this case. Thus, ID 5 0.3 mA. If VD is not ignored, ID 5 0.23 mA.(probl...

  • Page 115

    3Introduction to the TransistorThe transistor is undoubtedly the most important modern electronic component because it has enabled great and profound changes in electronics and in your daily lives since its discovery in 1948.This chapter introduces the transistor as an electronic component that a...

  • Page 116

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR92“The Transistor Switch,” and Chapter 8, “Transistor Amplifiers”) illustrates how BJTs and JFETs function and how they are used in electronic circuits. Because JFETs and MOSFETs function in similar fashion, MOSFETs are not covered here.Projects in...

  • Page 117

    UNDERSTANDING TRANSISTORS93CollectorCollectorcase is thecollectorBaseBaseBaseEmitterEmitterEmitterFIGURE 3.1QUESTIONSA. How many leads are there on most transistors? B. Where there are only two leads, what takes the place of the third lead? C. What are the three leads or connections called? D. Wh...

  • Page 118

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR942  You can think of a bipolar junction transistor as functioning like two diodes, connected back-to-back, as illustrated in Figure 3.2.NNPPFIGURE 3.2However, in the construction process, one important modification is made. Instead of two separate P regi...

  • Page 119

    UNDERSTANDING TRANSISTORS95QUESTIONWhy don’t two diodes connected back-to-back function like a transistor? ANSWERThe transistor has one thin P region, whereas the diodes share two thick P regions.4  The three terminals of a transistor (the base, the emitter, and the collector) connect, as show...

  • Page 120

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR96QUESTIONWhich transistor terminal includes an arrowhead? ANSWERThe emitter5  It is also possible to make transistors with a PNP configuration, as shown in Figure 3.6.collectorbaseemitterPNPFIGURE 3.6Both NPN and PNP type transistors can be made from eit...

  • Page 121

    UNDERSTANDING TRANSISTORS97C. Silicon and germanium are not mixed in any commercially available transistors. However, researchers are attempting to develop ultra-fast transistors that contain both silicon and germanium. NPNPNPFIGURE 3.76  Take a look at the simple examples using NPN transistors ...

  • Page 122

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR98 7  In the circuit shown in Figure 3.9, you can calculate the base current using the techniques covered in Chapter 2, “The Diode.”+−3 V3 V1 kΩ0.7 VFIGURE 3.9QUESTIONFind the base current in the circuit shown in Figure 3.9. (Hint: Do not ignore the...

  • Page 123

    UNDERSTANDING TRANSISTORS99QUESTIONCalculate the base current. IB 5 ANSWERIkkAB52V5V5()100110110 m 9  Look at the circuit shown in Figure 3.11.+−Base-collectordiodeBase-emitterdiode+−FIGURE 3.11QUESTIONWill current flow in this circuit? Why or why not? ANSWERIt cannot flow because the ...

  • Page 124

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR100+−+−FIGURE 3.12When you connect batteries to both the base and the collector portions of the circuit, currents flowing through the circuit demonstrate a key characteristic of the transistor. This characteristic is sometimes called transistor action...

  • Page 125

    UNDERSTANDING TRANSISTORS101ANSWERSA. IC (the collector current).B. IB and IC. Both of them flow through the base-emitter diode.C. Base current causes collector current to flow.No current flows along the path shown by the dotted line in Figure 3.14 from the col-lector to the base.+−+−FIGUR...

  • Page 126

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR102QUESTIONSCompare Figure 3.15 with Figure 3.13. How are the circuits different relative to the following? A. Battery connections: B. Current flow: ANSWERSA. The battery is reversed in polarity.B. The currents flow in the opposite direction.12  Figur...

  • Page 127

    UNDERSTANDING TRANSISTORS103As stated earlier, NPN and PNP bipolar transistors work in much the same way: Base current causes collector current to flow in both. The only significant difference in using a PNP versus an NPN bipolar transistor is that the polarity of the supply voltage (for both ...

  • Page 128

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR104ANSWERSA. The battery, the resistor RB, and the transistorB. BaseC. EmitterD. Collector14  Take a moment to recall the key physical characteristic of the transistor.QUESTIONWhen base current flows in the circuit shown in Figure 3.17, what other current...

  • Page 129

    UNDERSTANDING TRANSISTORS105QUESTIONSA. List the components through which the collector current flows. B. What causes the collector current to flow? ANSWERSA. The resistor RC, the transistor, and the battery.B. Base current. (Collector current doesn’t flow unless base current is flowing.)1...

  • Page 130

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR106NOTE The β introduced here is referred to in manufacturers’ specification sheets as hFE. Technically, it is referred to as the static or DC β. For the purposes of this chapter, it is called β. Discussions on transistor parameters in general, which ...

  • Page 131

    UNDERSTANDING TRANSISTORS107DepletionregionsBase terminalCollectorterminalEmitterterminal++++++++ ++++− −−−− −−−−−−−−−−−−−−−−−−−−−−−−−−− −−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−...

  • Page 132

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR108For a transistor with β5 100, only one electron flows out to the base terminal for every 100 electrons that flow to the collector terminal. β is controlled by two fac-tors: the thickness of the base region and the relative concentration of the impuri...

  • Page 133

    UNDERSTANDING TRANSISTORS109For example, suppose a transistor has 500 mA of collector current flowing, and you know it has a β value of 100. To find the base current, use the following formula: b5IICBIImAmABC5b55 5001005QUESTIONSCalculate the following values: A. IC 5 2 ampere, β 5 20. Find I...

  • Page 134

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR110C. Which is the larger current, IB or IC?D. IB 5 6 μA, β 5 250. Find IC.E. IC 5 300 mA, β 5 50. Find IB.ANSWERSA. See Figure 3.17 and Figure 3.18.B. IB (base current) controls IC (collector current).C. ICD. 1.5 mAE. 6 mAPROJECT 3.1: The Transistor OBJ...

  • Page 135

    111UNDERSTANDING TRANSISTORS❑ One multimeter set to mA❑ One multimeter set to measure DC voltage❑ One 10 kΩ resistor❑ One 510 ohm resistor❑ One 2N3904 transistor❑ One breadboard❑ One 1 MΩ potentiometerSTEPBYSTEP INSTRUCTIONSSet up the circuit shown in Figure 3.19 on a brea...

  • Page 136

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR1123. Measure and record IC.4. Measure and record VC. This voltage is sometimes referred to as the collector-emitter voltage (VCE), because it is taken across the collector-emitter leads if the emitter is connected to ground or the negative of the power sup...

  • Page 137

    113UNDERSTANDING TRANSISTORSEmitterBaseCollector 10KΩ resistor510Ω resistor2N3904 transistorFIGURE 3.20Figure 3.21 shows the test setup for this project.Compare your measurements with the ones shown in the following table. Target IB (μA)IB (μA)IC (mA)VC (volts)βLowest possible8.71.58.41172.4...

  • Page 138

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR114To multimeterset to voltsTo multimeterset to mATo multimeterset to μA1 MΩpotentiometerVCIBICFIGURE 3.21Don’t worry if your results give a different value of β. The manufacturing process that produces transistors can allow variation of the base thic...

  • Page 139

    UNDERSTANDING TRANSISTORS115RB =100 kΩRC = 1 kΩVC10 Vβ = 50FIGURE 3.22Here is the first step: 1. To find IC, you must first find IB. IvoltskmAB5V51010001IImAmACB5b3 5 35500 15Now, perform the next two steps. QUESTIONSA. VR 5 B. VC 5 ANSWERSA. To find VR: VR 5 RC 3 IC 5 1 kΩ 3 5 mA 5...

  • Page 140

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR116 21  Determine parameters for the circuit shown in Figure 3.22 using the value of β 5 75. QUESTIONSCalculate the following: A. IC 5 B. VR 5 C. VC 5 ANSWERSA. IvoltskmAB5V51010001ImAmAC535750175B. VR 5 1 kΩ 3 7.5 mA 5 7.5 voltsC. VC 5 10 volts – ...

  • Page 141

    UNDERSTANDING TRANSISTORS117B. VR 5 1 kΩ 3 3 mA 5 3 voltsC. VC 5 10 volts–3 volts 5 7 volts 23  From the preceding problems, you can see that you can set VC to any value by choosing a transistor with an appropriate value of β or by choosing the correct value of RB.Now, consider the example ...

  • Page 142

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR118ANSWERSB. VR 5 1 kΩ 3 10 mA 5 10 voltsC. VC 5 10 volts –10 volts 5 0 volts.Here the base current is sufficient to produce a collector voltage of 0 volts and the maximum collector current possible, given the stated values of the collector resistor and...

  • Page 143

    UNDERSTANDING TRANSISTORS119ANSWERSA. A closed mechanical switchB. 0 voltsPROJECT 3.2: The Saturated TransistorOBJECTIVENormally, for a transistor, IC 5 β× IB. However, this relationship does not hold when a transistor is saturated. The objective of this project is to determine the relationship...

  • Page 144

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR120IB (μA)IC (mA)VC (volts)EXPECTED RESULTSFigure 3.25 shows the test setup for this project with the potentiometer set at its lower limit, providing the highest value of IB. The test setup is the same as that used in Project 3.1; however, the value of IB ...

  • Page 145

    121UNDERSTANDING TRANSISTORStransistor is fully ON (saturated) and IC can’t increase further. This agrees with the data sheet published by Fairchild Semiconductor for the 2N3904 transistor, which indicates that the transistor saturates at VC 5 0.2 volts.VCICIBTo multimeterset to voltsTo multime...

  • Page 146

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR122Because the base circuit is broken (that is, it is not complete), there is no base current flowing.QUESTIONSA. How much collector current is flowing? B. What is the collector voltage? C. What is the voltage at the point VC in the mechanical switch circ...

  • Page 147

    THE JUNCTION FIELD EFFECT TRANSISTOR JFET123100 kΩ1 kΩβ = 50VC10 VFIGURE 3.27QUESTIONSA. IB 5 IC 5 B. VR 5 C. VC 5 ANSWERSA. IvoltskmAB5V51010001 ImAmAC535500 15B. VR 5 1 kΩ 3 5 mA 5 5 voltsC. VC 5 10 volts–5 volts 5 5 voltsNOTE The output voltage in this problem is half of the ...

  • Page 148

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR124The BJT has a relatively low input impedance as compared to the JFET. Like the BJT, the JFET is a three-terminal device. The terminals are called the source, drain, and gate. They are similar in function to the emitter, collector, and base, respectively....

  • Page 149

    THE JUNCTION FIELD EFFECT TRANSISTOR JFET125ssggddssgg++ + +++++++++++++ ++ + ++ + + ++ + + + +++++++++ +−−−−−−− −−−−−−−+++++ + + + ++ +++−−−−−−−−−−− −−− −−− −−− −−− − −−− −− −− − −−−−−−...

  • Page 150

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR126Figure 3.29 shows a typical biasing circuit for the N-channel JFET. For a P-channel JFET, you must reverse the polarity of the bias supplies.RGRDVGSVDS+VDD−VGGFIGURE 3.29QUESTIONHow does the ON-OFF operation of a JFET compare to that of a BJT?ANSWERThe...

  • Page 151

    THE JUNCTION FIELD EFFECT TRANSISTOR JFET127B. As the gate becomes more negative with respect to the source, the resistance of the N-channel increases until the cutoff point is reached. At this point, the resistance of the channel is assumed to be infinite. What condition does this repres...

  • Page 152

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR128Further increasing the negative voltage on the gate repels additional electrons, increasing the width of the depletion region and decreasing the width of the channel. The narrower the channel, the higher its resistance. When you apply high enough negativ...

  • Page 153

    SELFTEST129SUMMARYAt this point, it’s useful to compare the properties of a mechanical switch with the prop-erties of both types of transistors, as summarized in the following table. SwitchBJTJFETOFF (or open)No current.No collector current.No drain current.Full voltage across terminals.Full...

  • Page 154

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR130+−RBVSRCVOβFIGURE 3.303. What causes the collector current to flow?4. What is meant by the term current gain? What symbol is used for this? What is its algebraic formula? Use the circuit in Figure 3.30 to answer questions 5 through 10. 5. Assume that...

  • Page 155

    SELFTEST13112. What controls the flow of current in both a JFET and a BJT?13. In the JFET common source circuit shown in Figure 3.31, add the correct polari-ties of the power supplies, and draw the current path taken by the drain current.RGVGGVDDIDsdgRDFIGURE 3.3114. When a base current is re...

  • Page 156

    CHAPTER 3 INTRODUCTION TO THE TRANSISTOR132ANSWERS TO SELFTESTIf your answers do not agree with those that follow, review the problems indicated in parentheses before you go to Chapter 4. 1.See Figure 3.33.NPNBBCECEPNPFIGURE 3.33 (problems 4 and 5 )2.See Figure 3.34.+−ICIBFIGURE 3.34 (probl...

  • Page 157

    SELFTEST13311.See Figure 3.35.dsN-channelP-channelgdsgFIGURE 3.35 (problem 29)12.The voltage on the gate controls the flow of drain current, which is similar to the base voltage controlling the collector current in a BJT.(problem 29)13.See Figure 3.36.VGGVDDID+_+_FIGURE 3.36 (problem 30)14.Th...

  • Page 158

    Book Authorc03V107/04/2012 2:56 PM c03.indd 1347/4/2012 3:05:21 PM

  • Page 159

    4The Transistor SwitchTransistors are everywhere. You can’t avoid them as you move through your daily tasks. For example, almost all industrial controls, and even your MP3 player, stereo, and television may use transistors as switches.In Chapter 3, “Introduction to the Transistor,” you saw ...

  • Page 160

    CHAPTER 4 THE TRANSISTOR SWITCH136When you complete this chapter, you will be able to do the following: ■ Calculate the base resistance, which turns a transistor ON and OFF. ■ Explain how one transistor turns another ON and OFF. ■ Calculate various currents and resistances in simple transis...

  • Page 161

    TURNING THE TRANSISTOR ON1372  In this problem circuit, a lamp can be substituted for the collector resistor. In this case, RC(the resistance of the lamp) is referred to as the load, and IC (the current through the lamp) is called the load current.QUESTIONSA. Is load current equivalent to base o...

  • Page 162

    CHAPTER 4 THE TRANSISTOR SWITCH138B. What other component does an ON transistor resemble? ANSWERSA. The same as the emitter voltage, which, in this circuit, is 0 voltsB. A closed mechanical switchNOTE In actual practice, there is a small voltage drop across the transistor from the collector to...

  • Page 163

    TURNING THE TRANSISTOR ON139QUESTIONSA. Why do you need base current? B. How can you make base current flow? ANSWERSA. To enable collector current to flow so that the lamp lights upB. By closing the mechanical switch in the base circuit 5  You can calculate the amount of base current fl...

  • Page 164

    CHAPTER 4 THE TRANSISTOR SWITCH140QUESTIONSA. What is the voltage drop across the base-emitter diode? B. What is the voltage drop across RB? ANSWERSA. 0.7 volt because it is a silicon transistorB. 24 volts if the 0.7 is ignored; 23.3 volts if it is not 7  The next step is to calculate RB. ...

  • Page 165

    TURNING THE TRANSISTOR ON141 8  Use the following steps to calculate the values of IB and RB needed to turn a transistor ON:1. Determine the required collector current.2. Determine the value of β.3. Calculate the required value of IB from the results of steps 1 and 2.4. Calculate the required v...

  • Page 166

    CHAPTER 4 THE TRANSISTOR SWITCH142ANSWERRB5V31 1kIn this calculation, VBE is included. 10  In practice, if the supply voltage is much larger than the 0.7-volt drop across the base-emitter junction, you can simplify your calculations by ignoring the 0.7-volt drop, and assume that all the supply v...

  • Page 167

    TURNING OFF THE TRANSISTOR 143an open mechanical switch. If the transistor is OFF, no current flows through the load (that is, no collector current flows).QUESTIONSA. When a switch is open, are the two terminals at different voltages or at the same voltage? B. When a switch is open, does curre...

  • Page 168

    CHAPTER 4 THE TRANSISTOR SWITCH144+−RCR2FIGURE 4.7QUESTIONSA. Why doesn’t current flow from the supply voltage to the base-emitter junction? B. How much current flows from collector to base? C. Why doesn’t current flow from collector to base through R2 ground?D. Why is the transistor bas...

  • Page 169

    TURNING OFF THE TRANSISTOR 145QUESTIONWhich of the following resistor values would you use to keep a transistor turned off? 1 ohm, 2 kΩ, 10 kΩ, 20 kΩ, 50 kΩ, 100 kΩ, 250 kΩ, and 500 kΩ. ANSWERThey would all be suitable except the 1 ohm because the rest are all above 1 kΩ and below 1 MΩ....

  • Page 170

    CHAPTER 4 THE TRANSISTOR SWITCH146WHY TRANSISTORS ARE USED AS SWITCHES15  You can use the transistor as a switch (as you saw in the previous problems) to per-form simple operations such as turning a lamp current on and off. Although often used between a mechanical switch and a lamp, there are o...

  • Page 171

    WHY TRANSISTORS ARE USED AS SWITCHES147QUESTIONWhat features mentioned in these examples make using transistors as switches desir-able? ANSWERThe switching action of a transistor can be directly controlled by an electrical signal, as well as by a mechanical switch in the base circuit. This provid...

  • Page 172

    CHAPTER 4 THE TRANSISTOR SWITCH148mass-produced electronic device and are the reason that electronic devices con-tinue to get smaller and lighter.QUESTIONWhat other features, besides the ones mentioned in the previous problem, are demon-strated in the examples given here?ANSWERTransistors can be ...

  • Page 173

    WHY TRANSISTORS ARE USED AS SWITCHES149ANSWERSA. Less current flows through the main switch than through the load.B.1. It increases safety and allows the operator to stay isolated from dangerous situations.2. Switches can be placed conveniently close together on a panel, or in the best place for...

  • Page 174

    CHAPTER 4 THE TRANSISTOR SWITCH150PROJECT 4.1: The Transistor SwitchOBJECTIVEThe objective of this project is to demonstrate how light can switch a transistor ON or OFF to control a device. GENERAL INSTRUCTIONSThis project uses two breadboarded circuits. The circuit shown on the left side of Figu...

  • Page 175

    WHY TRANSISTORS ARE USED AS SWITCHES151In this project, a photodiode detects the infrared light. When light strikes a PN junc-tion in a photodiode (or any diode), a current is generated. Infrared photodiodes also have a transparent case and junction material that produces a large current when it ...

  • Page 176

    CHAPTER 4 THE TRANSISTOR SWITCH152PN2222EmitterBaseCollectorFIGURE 4.12STEPBYSTEP INSTRUCTIONSSet up the circuits shown in Figure 4.9. 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 t...

  • Page 177

    WHY TRANSISTORS ARE USED AS SWITCHES153BuzzerPhotodiodeCathode(short lead)Anode(long lead)BaseCollectorPN2222transistorEmitter1 kΩresistor10kΩresistorFIGURE 4.13Figure 4.14 shows the breadboarded LED circuit for this project.LEDCathode(short lead)Anode(long lead)100Ωresistor.FIGURE 4.14Book Au...

  • Page 178

    CHAPTER 4 THE TRANSISTOR SWITCH154Figure 4.15 shows the test setup for this project with the rounded top of the LED and photodiode aligned toward each other.FIGURE 4.15The photodiode is connected to the base of a transistor. Therefore, current gener-ated by the photodiode turns the transistor ON....

  • Page 179

    WHY TRANSISTORS ARE USED AS SWITCHES155+−AR1IB1VSIC1Q1LampQ2R3R2BFIGURE 4.16When the switch is in position A, the base-emitter junction of Q1 is forward-biased. Therefore, base current (IB1) flows through R1 and through the base-emitter diode of Q1, turning the transistor ON. This causes the c...

  • Page 180

    CHAPTER 4 THE TRANSISTOR SWITCH156QUESTIONSA. What effect does IB1 have on transistor Q1?B. What effect does turning Q1 ON have on the following?1. Collector current IC12. Collector voltage VC1C. What effect does the change to VC1 covered in the previous question have on the following?1. The b...

  • Page 181

    WHY TRANSISTORS ARE USED AS SWITCHES157QUESTIONSA. How much base current IB1 flows into Q1?B. Is Q1 ON or OFF? C. What current flows through R3?D. Is Q2 ON or OFF? E. Is the lamp on or off? ANSWERSA. NoneB. OFFC. IB2D. ONE. On21  Refer to the circuit in Figures 4.16 and 4.17. Now, answer thes...

  • Page 182

    CHAPTER 4 THE TRANSISTOR SWITCH158C. What is the collector voltage of Q1 with the switch in each position? ANSWERSA. No. If Q1 is ON, all the current flows through it to ground as collector current. If Q1 is OFF, all the current flows through the base of Q2 as base current.B. In position A, 10 ...

  • Page 183

    WHY TRANSISTORS ARE USED AS SWITCHES159+−R110 VQ1Lamp1AQ2R3R2FIGURE 4.18 ■ A 10-volt lamp that draws 1 ampere; therefore VS = 10 volts, IC2 = 1 A. ■ β2 = 20, β1 = 100Ignore any voltage drops across the transistors.QUESTIONSCalculate the following:A. Find IB2 as in step 3.IB2 =B. Find R3 a...

  • Page 184

    CHAPTER 4 THE TRANSISTOR SWITCH160ANSWERSThe following answers correspond to the steps.A.1. IC2 is given as 1 ampere.2. β2 = 20 (given). This is a typical value for a transistor that would handle 1 ampere.3. I amperemAB21205055B. 4. R volts mA3105020055V Note that the 0.7 volt base-emitter drop ...

  • Page 185

    THE THREETRANSISTOR SWITCH161B. R3 = C. IC1 = D. IB1 = E. R1 = F. R2 = ANSWERSA. 56 mAB. 500 ohmsC. 56 mAD. 0.56 mAE. 50 kΩF. 50 kΩ by choiceTHE THREETRANSISTOR SWITCH24  The circuit shown in Figure 4.19 uses three transistors to switch a load on and off. In this circuit, Q1 is used to ...

  • Page 186

    CHAPTER 4 THE TRANSISTOR SWITCH162QUESTIONSAssume that the switch is in position A.A. Is Q1 ON or OFF? B. Is Q2 ON or OFF?C. Where is current through R4 flowing?D. Is Q3 ON or OFF? ANSWERSA. ONB. OFFC. Into the base of Q3D. ON25  Now use the same circuit as in problem 24.QUESTIONSAssume that th...

  • Page 187

    THE THREETRANSISTOR SWITCH163E. Which switch position turns on the lamp?F. How do the ON/OFF positions for the switch in the three-transistor switch differ from the ON/OFF positions for the switch in the two-transistor switch circuit?ANSWERSA. OFF.B. ON.C. Through Q2 to ground.D. OFF.E. Posit...

  • Page 188

    CHAPTER 4 THE TRANSISTOR SWITCH16411. Calculate IB1. Use IB1 = IC1/β1.12. Calculate R1. Use R1 = Vs/IB1.13. Choose R2.For this example, use a 10-volt lamp that draws 10 amperes. Assume that the βs of the transistors are given in the manufacturer’s data sheets as β1 = 100, β2 = 50, and β3 =...

  • Page 189

    THE THREETRANSISTOR SWITCH1656. β2 is given as 50.7. IImAmABC22250050105b55D.8. R voltsmA31010155 VkE.9. IImACB12105510. β1 is given as 100.11. IImAmABC11110100015b55F.12. R voltsmA1100110055VkG.13. R2 can be chosen to be 100 kΩ also. 27  Determine the values in the same circuit for a 75-v...

  • Page 190

    CHAPTER 4 THE TRANSISTOR SWITCH166ANSWERSA. 200 mAB. 375 ΩC. 2 mAD. 37.5 kΩE. 16.7 μAF. 4.5 MΩG. Choose R2 = 1 MΩALTERNATIVE BASE SWITCHING28  In the examples of transistor switching, the actual switching was performed using a small mechanical switch placed in the base circuit of the first...

  • Page 191

    ALTERNATIVE BASE SWITCHING167AI2IB+VSI1R2R1FIGURE 4.20QUESTIONSA. When the switch is open, is Q1 ON or OFF? B. When the switch is closed, is the lamp ON or Off? ANSWERSA. OFFB. ON 30  When the switch is closed, current flows through R1. However, at point A in Figure 4.20, the current div...

  • Page 192

    CHAPTER 4 THE TRANSISTOR SWITCH168 31  The problem now is to choose the values of both R1 and R2 so that when the current divides, there is sufficient base current to turn Q1 ON.QUESTIONConsider this simple example. Assume the load is a 10-volt lamp that needs 100 mA of current and β = 100. Cal...

  • Page 193

    ALTERNATIVE BASE SWITCHING169 33  Now you can calculate the value of R2. The voltage across R2 is the same as the voltage drop across the base-emitter junction of Q1. Assume that the circuit uses a silicon transistor, so this voltage is 0.7 volt.QUESTIONSA. What is the value of R2? B. What is...

  • Page 194

    CHAPTER 4 THE TRANSISTOR SWITCH170QUESTIONSFor each of the following lamps, perform the same calculations you used in the last few problems to find the values of R1 and R2.A. A 28-volt lamp that draws 56 mA. β = 100 B. A 12-volt lamp that draws 140 mA. β = 50 ANSWERSA. ImAmAImARvoltmAoh...

  • Page 195

    ALTERNATIVE BASE SWITCHING171The choice of I2 = 10IB is one such rule of thumb. Is it the only choice that works? Of course not. Almost any value of I2 that is at least 5 times larger than IB can work. Choos-ing 10 times the value is a good option for three reasons: ■ It is a good practical cho...

  • Page 196

    CHAPTER 4 THE TRANSISTOR SWITCH172ANSWERSA. The transistor is much faster.B. The transistor.C. The transistor.D. The transistor.E. Because transistors have no moving parts, they have a much longer operating lifetime than a mechanical switch. A mechanical switch will fail after several thousand op...

  • Page 197

    SWITCHING THE JFET173ANSWERSA. Drain, source, and gate, with the gate acting as the control.B. When the gate voltage is zero (at the same potential as the source), the JFET is ON. When the gate to source voltage difference is high, the JFET is OFF.PROJECT 4.2: The JFET OBJECTIVEThe objective of ...

  • Page 198

    CHAPTER 4 THE TRANSISTOR SWITCH174Parts ListYou need the following equipment and supplies:❑ One 6-volt battery pack (4 AA batteries)❑ One 12-volt battery pack (8 AA batteries)❑ One multimeter set to mA❑ One multimeter set to measure DC voltage❑ One 10 kΩ potentiometer❑ One breadboard...

  • Page 199

    SWITCHING THE JFET1755. Repeat steps 3 and 4 until ID drops to 0 mA.VGS (Volts)ID (mA)6. Graph the points recorded in the table, using the blank graph in Figure 4.24. Draw a curve through the points. Your curve should look like the one in Figure 4.22.ID−VGS4563211214108642mAFIGURE 4.24EXPECTED ...

  • Page 200

    CHAPTER 4 THE TRANSISTOR SWITCH176DrainGateSource2N3819 JFETFIGURE 4.25Figure 4.26 shows the test setup for this project.12-voltbattery packTo multimeterset to mATo multimeterset to voltsVGSID6-voltbattery pack10kΩpotentiometerFIGURE 4.26Book Authorc04V107/04/2012 2:59 PM c04.indd 1767/4/2012 ...

  • Page 201

    SWITCHING THE JFET177Compare your measurements with the ones shown in the following table. You should see a similar trend in the measured values, not exactly the same values.VGS (Volts)ID (mA)012.70.410.70.69.80.88.91.08.11.36.81.56.01.84.92.04.12.33.12.52.52.71.93.01.13.30.53.50.23.70.14.00Figur...

  • Page 202

    CHAPTER 4 THE TRANSISTOR SWITCH178With the potentiometer set to 0 ohms (point A in Figure 4.22), the voltage from the gate to the source is zero (VGS = 0). The current that flows between the drain and source terminals of the JFET at this time is at its maximum value and is called the saturation ...

  • Page 203

    SWITCHING THE JFET179C. What is the gate to source cutoff voltage for the curve shown? D. Why is this called a cutoff voltage? ANSWERSA. 12 mA on the graph.B. The word “saturation” is used to indicate that the current is at its maximum.C. Approximately –4.2 V on the graph.D. It is t...

  • Page 204

    CHAPTER 4 THE TRANSISTOR SWITCH180QUESTIONWhat should the value of RD be for the IDSS shown at point A in the curve? ANSWERRvoltsmAD55V2012167k 40  For the JFET circuit shown in Figure 4.29, assume that VDS = 1 volt when the ID is at saturation.QUESTIONSA. What is the required value of RD? ...

  • Page 205

    SUMMARY181in the “hard OFF” state. The purpose of resistor RG is to ensure that the gate is connected to ground while you flip the switch between terminals, changing the gate voltage from one level to the other. Use a large value of 1 MΩ here to avoid drawing any appreciable current from th...

  • Page 206

    CHAPTER 4 THE TRANSISTOR SWITCH182SELFTESTThese questions test your understanding of the concepts introduced in this chapter. Use a separate sheet of paper for your diagrams or calculations. Compare your answers with the answers provided.For the first three questions, use the circuit shown in...

  • Page 207

    SELFTEST183R1Q1Q2R3R4R210 VFIGURE 4.31NOTE Rounding off throughout a problem, or rounding off the final answer, could produce slightly different results.4. R4 = 100 ohms, β1 = 100, β2 = 20.R3 = R1 = R2 = 5. R4 = 10 ohms, β1 = 50, β2 = 20.R3 = R1 =R2 = 6. R4 = 250 ohms, β1 = 75, β2 = ...

  • Page 208

    CHAPTER 4 THE TRANSISTOR SWITCH184R1Q1Q2RCVCQ3R3R4R225 VFIGURE 4.327. RC = 10 ohms, β3 = 20, β2 = 50, β1 = 100.R4 = R2 = R3 = R1 = 8. RC = 28 ohms, β3 = 10, β2 = 75, β1 = 75.R4 = R2 = R3 = R1 = 9. RC = 1 ohm, β3 = 10, β2 = 50, β1 = 75.R4 = R2 = R3 = R1 = Questions 10–12 use the circuit...

  • Page 209

    SELFTEST18510 VRCVCR2R1FIGURE 4.3310. RC = 1 kΩ, β = 100.R1 = R2 = 11. RC = 22 kΩ, β = 75.R1 = R2 = 12. RC = 100 Ω, β = 30.R1 = R2 = 13. An N-channel JFET has a transfer curve with the following characteristics. When VGS = 0 volt, the saturation current (IDSS) is 10.5 mA, and the cutoff...

  • Page 210

    CHAPTER 4 THE TRANSISTOR SWITCH1861.100 kΩ(problem 8)2.235 kΩ. Choose 240 kΩ as a standard value.(problem 8)3.1.65 MΩ. Choose 1.6 MΩ as a standard value.(problem 8)4.R3 = 2 kΩ; R1 = 200 kΩ; R2 = 200 kΩ. Use these values.(problem 22)5.R3 = 200 ohms; R1 = 10 kΩ; R2 = 10 kΩ. Use these valu...

  • Page 211

    5AC Pre-Test and ReviewYou need to have some basic knowledge of alternating current (AC) to study electronics. To understand AC, you must understand sine waves.A sine wave is simply a shape, like waves in the ocean. Sine waves in electronics are used to represent voltage or current moving up and ...

  • Page 212

    CHAPTER 5 AC PRETEST AND REVIEW188This chapter discusses the following: ■ Generators ■ Sine waves ■ Peak-to-peak and root mean square voltages ■ Resistors in AC circuits ■ Capacitive and inductive reactance ■ ResonanceTHE GENERATOR1  In electronic circuits powered by direct curre...

  • Page 213

    THE GENERATOR189The symbol shown in Figure 5.2 represents a generator. Note that a sine wave shown within a circle designates an AC sine wave source.FIGURE 5.2QUESTIONSA. What is the most popular instrument used in the lab to produce waveforms? B. What does the term AC mean? C. What does the sine...

  • Page 214

    CHAPTER 5 AC PRETEST AND REVIEW190The zero axis is the reference point from which all voltage measurements are made.QUESTIONSA. What is the purpose of the zero axis? B. What is the usual point for making time measurements? ANSWERSA. It is the reference point from which all voltage measur...

  • Page 215

    THE GENERATOR191QUESTIONIf the pp voltage of a sine wave is 10 volts, find the rms voltage. ANSWERVVVrmspp535351220 7071023 535 5  Calculate the following for a sine wave.QUESTIONIf the rms voltage is 2 volts, find the pp voltage. ANSWERVVVpprms5335 33 52221 41425 656 6  Calculate the f...

  • Page 216

    CHAPTER 5 AC PRETEST AND REVIEW1927  There is a primary time measurement for sine waves. The duration of the complete sine wave is shown in Figure 5.4 and referred to as a cycle. All other time measurements are frac-tions or multiples of a cycle. TFIGURE 5.4QUESTIONSA. What is one complete si...

  • Page 217

    RESISTORS IN AC CIRCUITS193ANSWERSC. f5 1/TD. Hertz (Hz) is the standard unit for frequency. One Hertz equals one cycle per second.E. 2 kHz, 25 kHzF. 16.7 ms, 80 μsec, 1 μsec8  Choose all answers that apply.QUESTIONSWhich of the following could represent electrical AC signals? A. Simple sine w...

  • Page 218

    CHAPTER 5 AC PRETEST AND REVIEW194ANSWERUse Ohm’s law:I VRVohmsApppp5 5 5 10101Because the voltage is given in pp, the current is a pp current. 10  An AC signal of 10 Vrms is connected across a 20-ohm resistor.QUESTIONFind the current. ANSWERI VohmsArmsrms5 5 102005Because the voltage...

  • Page 219

    CAPACITORS IN AC CIRCUITS195QUESTIONFind Vout. ANSWERVVRRRkkkVoutinpp5 3 5353521210282102102()()++ΩΩΩCAPACITORS IN AC CIRCUITS 12  A capacitor opposes the flow of an AC current.QUESTIONSA. What is this opposition to the current flow called? B. What is this similar to in DC circuits? ...

  • Page 220

    CHAPTER 5 AC PRETEST AND REVIEW196B. 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? ANSWERSA. XfCC512B. XC 5 the reactance of the capacitor in ohms. f 5 the frequency of the signal in hertz. C 5 the...

  • Page 221

    CAPACITORS IN AC CIRCUITS19715  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).QUESTIONSFind XC1 and XC2:A. C5 0.1 μF, f 5 100 Hz. B. C5 100 μF, f 5 2 kHz. ...

  • Page 222

    CHAPTER 5 AC PRETEST AND REVIEW198USING THE OSCILLOSCOPEYou 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 ...

  • Page 223

    CAPACITORS IN AC CIRCUITS199Vout 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/DIV control to adjust the vertical scale and the TIME/DIV control to adjust t...

  • Page 224

    CHAPTER 5 AC PRETEST AND REVIEW200(continued)3.3 divisionsper cycleBecause the TIME/DIV knob 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:f Ts Hz30.3 kHz555551131000033303033.secYou can also measure the effect of ...

  • Page 225

    CAPACITORS IN AC CIRCUITS201divider circuit Vout, and the ground clip is clipped to a jumper wire connected to the ground bus. Red leadfrom functiongeneratorBlack leadfrom functiongeneratorChannel 2ground clipChannel 1ground clipChannel 2oscilloscopeprobeChannel 1oscilloscopeprobeThe function gen...

  • Page 226

    CHAPTER 5 AC PRETEST AND REVIEW202for 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.2 divisionspeak-to-peakPhase shift3 divisionspeak-to-...

  • Page 227

    THE INDUCTOR IN AN AC CIRCUIT203B. 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? ANSWERSA. Its AC reactance (XL), which can be quite high, is a result of the electromagnetic ...

  • Page 228

    CHAPTER 5 AC PRETEST AND REVIEW204QUESTIONSA. L5 1 mH (0.001 H), f 5 10 kHz B. L5 0.01 mH, f 5 5 MHz ANSWERSA. XL1 5 6.28 3 1033 0.001 5 6.28 ohmsXL2 5 6.28 310 3 1033 0.001 5 62.8 ohmsB. XL1 5 6.28 3 1033 0.01 3 10−35 0.0628 ohms XL2 5 6.28 3 5 3 106 × 0.01 3 10−35 314 ohmsA circuit cont...

  • Page 229

    RESONANCE205ANSWERSA. The resonant frequencyB. 2πfL 5 1/(2πfC). Rearranging the terms in this equation to solve for f yields the following formula for the resonant frequency ( fr ): fLCr512 20  If a capacitor and an inductor are connected in parallel, there is also a resonant fre-quency. Anal...

  • Page 230

    CHAPTER 5 AC PRETEST AND REVIEW206 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.QUESTIONSDetermine fr for the following: A. C 5 1 μF, L 5 1 henry B. C 5 0.2 μF, L 5 3.3 mH ...

  • Page 231

    SELFTEST207Filters 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, televi-sion, and other communications applications. Oscillators are electronic circuits that gener-ate a ...

  • Page 232

    CHAPTER 5 AC PRETEST AND REVIEW208B. Vp 5 80 mVVrms 5C. Vpp 5 100 VVrms 52. Convert the following rms values to the required values shown: A. Vrms 5 120 VVp 5B. Vrms 5 100 mVVp 5C. Vrms 5 12 VVpp 53. For the given value, find the period or frequency: A. T5 16.7 msf5B. f5 15 kHzT54. For the ci...

  • Page 233

    SELFTEST2096. Find the frequency necessary to cause each reactance shown: A. C5 1 μF, XC 5 200 ohmsf5B. L5 50 mH, XL 5 320 ohmsf57. What would be the resonant frequency for the capacitor and inductor values given in A and B of question 5 if they were connected in series? 8. What would be the ...

  • Page 234

    Book Authorc05V107/04/2012 3:03 PM c05.indd 2107/4/2012 3:04:35 PM

  • Page 235

    6FiltersCertain 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). The...

  • Page 236

    CHAPTER 6 FILTERS212When 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. ■ Ca...

  • Page 237

    CAPACITORS IN AC CIRCUITS213B. Does a capacitor appear as a short or an open circuit to an AC signal? ANSWERSA. 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...

  • Page 238

    CHAPTER 6 FILTERS214ANSWERIt decreases.If you had difficulty with these first three problems, you should review the examples in Chapter 5.CAPACITORS AND RESISTORS IN SERIES4  For simplicity, consider all inputs at this time to be pure sine waves. The circuit shown in Figure 6.2 shows a sine wav...

  • Page 239

    CAPACITORS AND RESISTORS IN SERIES215flow 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.QUESTIONWith an AC input to a simple circuit lik...

  • Page 240

    CHAPTER 6 FILTERS216ANSWERVVRRRoutin531212 7  You can calculate a total resistance to the flow of electric current for a circuit containing two resistors in series.QUESTIONWhat is the formula for this total resistance? ANSWERRRRT5112 8  You can also calculate the total opposition to the fl...

  • Page 241

    CAPACITORS AND RESISTORS IN SERIES217VinVoutCR = 300 Ωf = 1 kHz0.4 μF10 VppFIGURE 6.4A. XfCC5p512 B. ZXRC51 522 C. IVZ55 ANSWERSA. 400 ohmsB. 500 ohmsC. 20 mApp 9  Now, for the circuit shown in Figure 6.4, calculate the impedance and current using the values provided.QUESTIONSA. C = 5...

  • Page 242

    CHAPTER 6 FILTERS218ANSWERSA. Z = 13 ohms, I = 2 AppB. 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.VinVout = VRCRFIGURE 6.5The form...

  • Page 243

    CAPACITORS AND RESISTORS IN SERIES219A. Find XC:B. Find Z: C. Use the formula to find Vout:ANSWERSA. XC = 500 ohms (rounded off)B. Z= 1120 ohms (rounded off)C. Vout = 8.9 Vpp11  Now, find Vout for the circuit in Figure 6.5 using the given component values, signal volt-age, and frequency.QUE...

  • Page 244

    CHAPTER 6 FILTERS22012  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 prob-lems 10 and 11.QUESTIONWhat does attenuated mean? ANSWERTo reduce in amplitude or magnitude (...

  • Page 245

    CAPACITORS AND RESISTORS IN SERIES221QUESTIONSCalculate the output voltage for the circuit shown in Figure 6.8 for frequencies of 100 Hz, 1 kHz, 10 kHz, and 100 kHz.VinVoutCR1kΩ0.016μF10 VppFIGURE 6.8A. 100 Hz: B. 1 kHz: C. 10 kHz: D. 100 kHz: E. Plot these values for Vout against f, and draw a...

  • Page 246

    CHAPTER 6 FILTERS222100 Hz1V5V7V10 V1 kHz10 kHz(Note that this is a logarithmic frequency scale.)100 kHzFIGURE 6.9NOTE 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 pas...

  • Page 247

    223CAPACITORS AND RESISTORS IN SERIESmaterial than the more typical ceramic capacitor but performs the same func-tion. 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.) ...

  • Page 248

    CHAPTER 6 FILTERS2243. 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 ...

  • Page 249

    225CAPACITORS AND RESISTORS IN SERIES101001000100001000001000000024681012Vout(volts)fin (Hz)FIGURE 6.11EXPECTED RESULTSFigure 6.12 shows the breadboarded circuit for this project.1 kΩ resistor0.016μF capacitorFIGURE 6.12Book Authorc06V107/04/2012 3:08 PM c06.indd 2257/4/2012 3:08:59 PM

  • Page 250

    CHAPTER 6 FILTERS226Figure 6.13 shows a function generator and oscilloscope attached to the circuit.Channel 2ground clipChannel1ground clipChannel 2oscilloscopeprobeChannel1oscilloscopeprobeBlack leadfrom functiongeneratorRed leadfrom functiongeneratorFIGURE 6.13The input signal is represented by...

  • Page 251

    227CAPACITORS AND RESISTORS IN SERIESAs 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. Time/div control forboth channel 1 andchannel 2Channel 2volts/div controlChannel...

  • Page 252

    CHAPTER 6 FILTERS228finXCVout1 kHz10 kΩ1 volts3 kHz3.3 kΩ2.9 volts5 kHz2 kΩ4.5 volts7 kHz1.4 kΩ5.6 volts10 kHz1 kΩ7.1 volts20 kHz500 Ω8.9 volts30 kHz330 Ω9.5 volts50 kHz200 Ω9.8 volts100 kHz100 Ω10 volts200 kHz50 Ω10 volts500 kHz20 Ω10 volts1 MHz10 Ω10 voltsNotice the relatio...

  • Page 253

    229CAPACITORS AND RESISTORS IN SERIES14  Refer to the curve you drew in Project 6.1 for the following question.QUESTIONWhat would cause your curve to be moved slightly to the right or the left of the curve shown in Figure 6.16? ANSWERSlightly different values for the resistor and capacitor that...

  • Page 254

    CHAPTER 6 FILTERS230QUESTIONSA. What is the impedance formula for the circuit? B. What is the formula for the output voltage? ANSWERSA. ZXRC5122B. VVXZoutinC53 16  Refer to the circuit shown in Figure 6.17 and the following values:VVf kHzCF Rkinpp55 5m5V102011,.,QUESTIONSFind the following...

  • Page 255

    231CAPACITORS AND RESISTORS IN SERIES 17  Again, refer to the circuit shown in Figure 6.17 to answer the following question.QUESTIONCalculate the voltage across the resistor using the values given in problem 16, along with the calculated impedance value. ANSWERVVRZVRinpp555××.1010001277783 ...

  • Page 256

    CHAPTER 6 FILTERS232QUESTIONWhat parameters determine f1 and f2?ANSWERThe values of the capacitor and the resistorNOTE You can see in Figure 6.18 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 wil...

  • Page 257

    233CAPACITORS AND RESISTORS IN SERIES❑ 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 INSTRUCTIONSSet up the circuit shown in Figure 6.19. If you have some experi...

  • Page 258

    CHAPTER 6 FILTERS2346. 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 Pro...

  • Page 259

    235CAPACITORS AND RESISTORS IN SERIES101001000100001000001000000024681012Vout(volts)fin (Hz)FIGURE 6.20EXPECTED RESULTSFigure 6.21 shows the breadboarded circuit for this project.1 kΩ resistor0.016μF capacitorFIGURE 6.21Book Authorc06V107/04/2012 3:08 PM c06.indd 2357/4/2012 3:09:32 PM

  • Page 260

    CHAPTER 6 FILTERS236Figure 6.22 shows a function generator and oscilloscope attached to the circuit.Channel 2ground clipChannel1ground clipChannel 2oscilloscopeprobeChannel1oscilloscopeprobeBlack leadfrom functiongeneratorRed leadfrom functiongeneratorFIGURE 6.22The input signal is represented by...

  • Page 261

    237CAPACITORS AND RESISTORS IN SERIES2 divisionspeak-to-peak4.5 divisionspeak-to-peakFIGURE 6.23Time/div control forboth channel 1 andchannel 2Channel 2volts/div controlChannel1volts/div controlHorizontalposition knobChannel 2 verticalposition knobChannel 1 verticalposition knobFIGURE 6.24Book Au...

  • Page 262

    CHAPTER 6 FILTERS238Your values should be close to those shown in the following table, and the curve should be similar to Figure 6.25.finXCVout25 Hz400 kΩ10 volts50 Hz200 kΩ10 volts100 Hz100 kΩ10 volts250 Hz40 kΩ10 volts500 Hz20 kΩ10 volts1 kHz10 kΩ10 volts3 kHz3.3 kΩ9.4 volts5 kHz2...

  • Page 263

    239PHASE SHIFT OF AN RC CIRCUIT101001000100001000001000000024681012Vout(volts)fin (Hz)Low-pass filterFIGURE 6.25PHASE SHIFT OF AN RC CIRCUIT20  In both of the circuits shown in Figure 6.26, the output voltage is different from the input voltage.VinVout = VRCRVinVout = VCCR(1)(2)FIGURE 6.26Book ...

  • Page 264

    CHAPTER 6 FILTERS240QUESTIONIn what ways do they differ? ANSWERThe 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 whil...

  • Page 265

    PHASE SHIFT OF AN RC CIRCUIT241ANSWERSA. To the leftB. To the right22  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 deg...

  • Page 266

    CHAPTER 6 FILTERS242Figure 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.IVRVCVinθφFIGURE 6.28NOTE Although the voltage across a resistor is in phase with the current through the resistor, both are ...

  • Page 267

    PHASE SHIFT OF AN RC CIRCUIT243ANSWERSee Figure 6.29. Note that the phasor diagram shows that the magnitude of VC is greater than VR.IVRVCVinθθ = 68.7°φφ = 21.3°FIGURE 6.2924  Using the component values and input signal shown in Figure 6.30, answer the following questions.VinVout6Ω330μF1...

  • Page 268

    CHAPTER 6 FILTERS244D. VR: E. The current flowing through the circuit: F. The phase angle: ANSWERSA. XfCohmsC5p5128B. Zohms51 5861022C. VVVXZ voltsoutCinC5535 8D. VVRZ voltsRin53 5 6E. IVZV amperepp55V510101F. tanu55 VV5XRC86133 Therefore, θ = 53.13 degrees. 25  Use the circuit shown...

  • Page 269

    PHASE SHIFT OF AN RC CIRCUIT245QUESTIONSCalculate the following parameters:A. XC: B. Z: C. Vout: D. VR: E. The current flowing through the circuit: F. The phase angle: ANSWERSA. XC = 265 ohmsB. Z5 15V175265317 5722C. VC = 125 voltsD. VR = 83 voltsE. I = 0.472 ampereF. tanu5VV52...

  • Page 270

    CHAPTER 6 FILTERS246RESISTOR AND CAPACITOR IN PARALLEL26  The circuit shown in Figure 6.32 is a common variation on the low-pass filter circuit introduced in problem 15.VinVoutCR2R1FIGURE 6.32Because a DC signal will not pass through the capacitor, this circuit functions like the circuit shown ...

  • Page 271

    RESISTOR AND CAPACITOR IN PARALLEL247Calculating 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...

  • Page 272

    CHAPTER 6 FILTERS248ANSWERSA. 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.Vkkk voltsout53VV1 V52011110Vout1 kΩ1 kΩ20 VFIGURE 6.36C. Figure 6.37 shows the portion of the circuit that an...

  • Page 273

    RESISTOR AND CAPACITOR IN PARALLEL24927  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 proble...

  • Page 274

    CHAPTER 6 FILTERS250ANSWERS2.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 CIRCUITS28  Figure 6.40 shows a voltage divider circuit using an inductor, rat...

  • Page 275

    INDUCTORS IN AC CIRCUITS251ANSWERSA. ImpedanceB. XL = 2πfL.C. ZXRL5122In 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 s...

  • Page 276

    CHAPTER 6 FILTERS252 30  Find the output voltage for the circuit shown in Figure 6.42.VinVoutLR11 V10 V9 V1 kHz1 kΩ160 mH0 VFIGURE 6.42Use the steps in the following questions to perform the calculation.QUESTIONSA. Find the DC output voltage. Use the DC voltage divider formula.DC Vout5 B. Find...

  • Page 277

    INDUCTORS IN AC CIRCUITS253ANSWERSA. outDC V voltskk volts53VV151011010B. XL = 1 kΩ (approximately).C. Zk51 55V1121 41422D. outppppAC VVkkV53VV5211 4141 414E. The output waveform is shown in Figure 6.44.10.710 VVout9.3FIGURE 6.44 31  For the circuit shown in Figure 6.45, the DC resistance of ...

  • Page 278

    CHAPTER 6 FILTERS254C. 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 6.46ANSWERSA. outDC V voltskkvolts55VV1V5 1011500667()NOTE The 500 Ω DC resistance of the inductor has been added to the 1 kΩ resistor va...

  • Page 279

    255INDUCTORS IN AC CIRCUITS32  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 Voutalso change with the frequency of the input signal. If you plot the output voltage Vou...

  • Page 280

    CHAPTER 6 FILTERS256VinRLVoutFIGURE 6.49QUESTIONSA. 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. ANSWERSA. VVXZoutinL53B. See Figure 6.50.VoutfFIGUR...

  • Page 281

    257INDUCTORS IN AC CIRCUITSHIGHERORDER FILTERSFilter circuits that contain one capacitor or inductor are called first-order filters. Filter order numbers reflect the number of capacitors, inductors, or operational ampli-fiers (a component discussed in Chapter 8, “Transistor Amplifiers...

  • Page 282

    CHAPTER 6 FILTERS258PHASE SHIFT FOR AN RL CIRCUIT34  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 th...

  • Page 283

    PHASE SHIFT FOR AN RL CIRCUIT259VLVRVinIθFIGURE 6.52The phase angle is easily found:tanu555 pVVXRfLRLRL2QUESTIONCalculate the phase angle for the circuit discussed in problem 30. ANSWER45 degrees 36  Refer to the circuit discussed in problem 31.QUESTIONCalculate the phase angle. ANSWERtan...

  • Page 284

    CHAPTER 6 FILTERS260SUMMARYThis 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 ...

  • Page 285

    SELFTEST261A. B. C. D. E. 2. Use the circuit shown in Figure 6.54.VinVoutR= 30 ΩC= 0.4 μF100 Vpp10 kHzFIGURE 6.54A. B. C. D. E. 3. Use the circuit shown in Figure 6.55.VinVoutC= 32 μFR= 12 Ω26 Vpp1 kHzFIGURE 6.55A. B. C. Book Authorc06V107/04/2012 3:08 PM c06.indd 2617/4/201...

  • Page 286

    CHAPTER 6 FILTERS262D. E. For questions 4–6, calculate the following parameters for the circuit shown in each question.A. XCB. AC VoutC. DC Vout4. Use the circuit shown in Figure 6.56.AC 10 VppVinVoutC= 8 μF100Ω100Ω2 kHz20 VFIGURE 6.56A. B. C. 5. Use the circuit shown in Figure 6.57.AC ...

  • Page 287

    SELFTEST2636. Use the circuit shown in Figure 6.58.AC 10 VppVinC= 0.25 μF100Ω1kΩ10 kHz10 VFIGURE 6.58A. B. C. For questions 7–9, calculate the following parameters for the circuit shown in each question.A. DC VoutB. XLC. ZD. AC VoutE. tan θ and θ7. Use the circuit shown in Figure 6.5...

  • Page 288

    CHAPTER 6 FILTERS264D. E. 8. Use the circuit shown in Figure 6.60.AC 9.1 VppVinVoutL= 72 mHR= 100 Ωr= 1 Ω2 kHz10 VFIGURE 6.60A. B. C. D. E. 9. Use the circuit shown in Figure 6.61.AC 10 VppVinVoutR= 1 kΩL= 40 mHr= 04 kHz10 VFIGURE 6.61A. B. C. D. E. Book Authorc06V107/04/2012 3:08 ...

  • Page 289

    SELFTEST265ANSWERS TO SELFTESTIf your answers do not agree with those provided here, review the applicable problems in this chapter before you go to Chapter 7.1A.3 kΩproblems 8, 9, 10, 231B.5 kΩ1C.8 volts1D.2 amperes1E.36.87 degrees2A.40 ohmsproblems 8, 9, 232B.50 ohms2C.60 volts2D.2 am...

  • Page 290

    CHAPTER 6 FILTERS2667E.16.7 degrees8A.10 voltsproblems 28–30, 358B.904 ohms8C.910 ohms8D.1 volt8E.83.69 degrees9A.0 voltsproblems 28–30, 359B.1 kΩ9C.1.414 kΩ9D.7 volts9E.45 degrees(continued)Book Authorc06V107/04/2012 3:08 PM c06.indd 2667/4/2012 3:10:31 PM

  • Page 291

    7Resonant CircuitsYou have seen how the inductor and the capacitoreach present an opposition to the flow of an AC current, and how the magnitude of this reactance depends upon the frequency of the applied signal.When inductors and capacitors are used together in a circuit (referred to as an LC c...

  • Page 292

    CHAPTER 7 RESONANT CIRCUITS268After completing this chapter, you will be able to do the following: ■ Find the impedance of a series LC circuit. ■ Calculate the series LC circuit’s resonant frequency. ■ Sketch a graph of the series LC circuit’s output voltage. ■ Find the impedance of a...

  • Page 293

    THE CAPACITOR AND INDUCTOR IN SERIES269To simplify your calculations in the next few problems, you can assume that the small DC resistance of the inductor is much less than the resistance of the resistor R, and you can, therefore, ignore DC resistance in your calculations.When you apply an AC sig...

  • Page 294

    CHAPTER 7 RESONANT CIRCUITS270QUESTIONSFollow these steps to calculate the following:A. Find XL: B. Find XC: C. Use X = XL − XC to find the net reactance: D. Use ZX + R22 to find the impedance: ANSWERSA. XL = 628 ohmsB. XC = 160 ohmsC. X = 468 ohms (inductive)D. Z = 685 ohms 3  C...

  • Page 295

    THE CAPACITOR AND INDUCTOR IN SERIES271ANSWERSA. XL = 314 ohmsB. XC = 318 ohmsC. X = −4 ohms (capacitive)D. Z = 9 ohmsBy convention, the net reactance is negative when it is capacitive.4  Calculate the net reactance and impedance for an RLC series circuit, such as those shown in Figure 7.2, us...

  • Page 296

    CHAPTER 7 RESONANT CIRCUITS272In problems 1 through 4, the net reactance of the series inductor and capacitor changes as the frequency changes. Therefore, as the frequency changes, the voltage drop across the resistor changes and so does the amplitude of the output voltage Vout.If you plot Vout a...

  • Page 297

    THE CAPACITOR AND INDUCTOR IN SERIES273 6  You can find the frequency at which XL − XC = 0 by setting the formula for XL equal to the formula for XC and solving for f:212ppfLfCTherefore,fLCr12pwhere fr is the resonant frequency of the circuit.QUESTIONWhat effect does the value of the resista...

  • Page 298

    CHAPTER 7 RESONANT CIRCUITS2748  Calculate the resonant frequency for the circuit shown in Figure 7.3 using the capaci-tor and inductor values given in the following questions.QUESTIONSA. C = 0.1 μF, L = 1 mH fr = B. C = 1 μF, L = 2 mH fr = ANSWERSA. fr = 16 kHzB. fr = 3.6 kHz9  For the RLC c...

  • Page 299

    275THE CAPACITOR AND INDUCTOR IN SERIESThe output voltage drops to its minimum value at the resonant frequency for the cir-cuit, which you can calculate with the formula provided in problem 6. At the resonant frequency, the net reactance of the inductor and capacitor in series is at a minimum. Th...

  • Page 300

    CHAPTER 7 RESONANT CIRCUITS276and second significant digits of the inductance value. The third number is the multiplier, and the units are μH. Therefore, an inductor marked with 101 has a value of 100 μH.) ❑ One function generator.❑ One oscilloscope. ❑ One breadboard.STEPBYSTEP INS...

  • Page 301

    277THE CAPACITOR AND INDUCTOR IN SERIES5. Adjust the function generator to the frequency shown in the next row of the table (labeled 150 kHz in this instance). Each time you change the frequency, check Vinand adjust the amplitude knob on the function generator to maintain Vin at 5 Vpp if needed. ...

  • Page 302

    CHAPTER 7 RESONANT CIRCUITS27810020030040050060070080090012345f (kHz)Vout(volts)FIGURE 7.8EXPECTED RESULTSFigure 7.9 shows the breadboarded circuit for this project.100Ω resistor100μH inductor1000 pF capacitorFIGURE 7.9Book Authorc07V107/04/2012 3:13 PM c07.indd 2787/4/2012 3:13:36 PM

  • Page 303

    279THE CAPACITOR AND INDUCTOR IN SERIESFigure 7.10 shows a function generator and oscilloscope attached to the circuit.Channel 2ground clipChannel1ground clipChannel 2oscilloscopeprobeChannel1oscilloscopeprobeBlack leadfrom functiongeneratorRed leadfrom functiongeneratorFIGURE 7.10The input signa...

  • Page 304

    CHAPTER 7 RESONANT CIRCUITS2801 divisionpeak-to-peak5.4 divisionspeak-to-peakFIGURE 7.11Time/div control setto 1 μsec/divChannel 2 set to5 volts/divChannel 1 set to0.1 volts/divChannel 2 verticalposition knobChannel 1 verticalposition knobFIGURE 7.12Your values should be close to those shown in ...

  • Page 305

    281THE CAPACITOR AND INDUCTOR IN SERIESfin (kHz)Vout (volts)1004.91504.92004.92504.83004.73504.44004.04502.85000.55502.36003.66504.27004.47504.78004.78504.89004.810020030040050060070080090012345f (kHz)Vout(volts)FIGURE 7.13Book Authorc07V107/04/2012 3:13 PM c07.indd 2817/4/2012 3:14:21 PM

  • Page 306

    CHAPTER 7 RESONANT CIRCUITS282Notice the extra data points shown in the graph near the minimum Vout. These extra data points help you to determine the frequency at which the minimum Vout occurs. In this graph, the minimum Vout occurs at a frequency of 505 kHz, which is close to the cal-culated re...

  • Page 307

    283THE CAPACITOR AND INDUCTOR IN SERIESANSWERSA. fLCr12pB. fLCrCLr=−1212pNOTE Here is another version of the resonance frequency formula that is help-ful when Q is known:fLCQQr=+12122p 11  You can calculate the total opposition (impedance) of an inductor and capacitor con-nected in parallel to...

  • Page 308

    CHAPTER 7 RESONANT CIRCUITS284frVpVinVoutFIGURE 7.16QUESTIONSA. What would be the total impedance formula for the voltage divider circuit at reso-nance? B. What is the frequency called at the point where the curve is at its lowest point? C. Why is the output voltage at a minimum value at resonanc...

  • Page 309

    285THE CAPACITOR AND INDUCTOR IN SERIESCLRVoutVinFIGURE 7.17If Vout is plotted on a graph against frequency for the circuit shown in Figure 7.17, the curve looks like that shown in Figure 7.18. At the resonance frequency, the impedance of the parallel inductor and capacitor is at its maximum valu...

  • Page 310

    CHAPTER 7 RESONANT CIRCUITS28613  Find the resonant frequency in these two examples, where the capacitor and the inductor are in parallel. (Q is greater than 10.)QUESTIONSA. L = 5 mH, C = 5 μFfr = B. L = 1 mH, C = 10 μFfr = ANSWERSA. fr = 1 kHz (approximately)B. fr = 1600 Hz (approximately)THE...

  • Page 311

    THE OUTPUT CURVE287An input signal at the resonant frequency, fr, passes through a circuit with minimum attenuation, and with its output voltage equal to the peak output voltage, Vp, shown on this curve.The two frequencies f1 and f2 are “passed” almost as well as fr is passed. That is, signal...

  • Page 312

    CHAPTER 7 RESONANT CIRCUITS28815  Somewhere between fr and f3, and between fr and f4, there is a point at which frequencies are said to be either passed or reduced to such a level that they are effectively blocked. The dividing line is at the level at which the power output of the circuit is ha...

  • Page 313

    THE OUTPUT CURVE289 17  In these examples, find the voltage level at the half power points.QUESTIONSA. Vp = 20 volts B. Vp = 100 volts C. Vp = 3.2 volts ANSWERSA. 14.14 voltsB. 70.70 voltsC. 2.262 volts 18  Although this discussion started off by talking about the resonance frequency...

  • Page 314

    CHAPTER 7 RESONANT CIRCUITS290ANSWERThe bandwidth is wider for the frequencies given in A.When playing a radio, you listen to one station at a time, not to the adjacent stations on the dial. Thus, your radio tuner must have a narrow bandwidth so that it can select only the frequency of that one s...

  • Page 315

    291THE OUTPUT CURVEQUESTIONSA. What points on the curve shown in Figure 7.21 would you use to determine the circuit’s bandwidth? B. Would the output voltage at the resonant frequency be above or below these points? ANSWERSA. The half power points (0.707 Vout(max)).B. The output voltage at the r...

  • Page 316

    CHAPTER 7 RESONANT CIRCUITS292❑ One function generator.❑ One oscilloscope. ❑ One breadboard.STEPBYSTEP INSTRUCTIONSSet up the circuit shown in Figure 7.22. If you have some experience in building circuits, this schematic (along with the previous parts list) should provide all the info...

  • Page 317

    293THE OUTPUT CURVEand adjust the amplitude knob on the function generator to maintain Vin at 5 Vpp if needed. (If you leave the amplitude knob in one position, the voltage of the signal provided by the function generator will change as the net reactance of the circuit changes.) 6. Measure and re...

  • Page 318

    CHAPTER 7 RESONANT CIRCUITS29410020030040050060070080090012345f (kHz)Vout(volts)FIGURE 7.23EXPECTED RESULTSFigure 7.24 shows the breadboarded circuit for this project.100Ω resistor100μH inductor1000 pF capacitorFIGURE 7.24Book Authorc07V107/04/2012 3:13 PM c07.indd 2947/4/2012 3:14:34 PM

  • Page 319

    295THE OUTPUT CURVEFigure 7.25 shows a function generator and oscilloscope attached to the circuit.Channel 2ground clipChannel1ground clipChannel 2oscilloscopeprobeChannel1oscilloscopeprobeBlack leadfrom functiongeneratorRed leadfrom functiongeneratorFIGURE 7.251 divisionpeak-to-peak4.6 divisions...

  • Page 320

    CHAPTER 7 RESONANT CIRCUITS296The input signal is represented by the upper sine wave shown in Figure 7.26, and the output signal is represented by the lower sine wave. Read the number of divisions for the peak-to-peak output sine wave, and multiply it by the corresponding VOLTS/DIV setting to det...

  • Page 321

    297THE OUTPUT CURVEfin (kHz)Vout (volts)1000.31500.52000.72501.03001.43501.84002.64503.65004.65504.26003.46502.77002.27501.88001.68501.49001.310020030040050060070080090012345f (kHz)Vout(volts)FIGURE 7.28Because Q = 3.2 (well below 10), the curve for this circuit is not perfectly symmetrical. Book...

  • Page 322

    CHAPTER 7 RESONANT CIRCUITS298 20  You can find the bandwidth of a circuit by measuring the frequencies (f1 and f2) at which the half power points occur and then using the following formula:BWff=−21Or you can calculate the bandwidth of a circuit using this formula:BWfQrwhere:QXRLThe formula u...

  • Page 323

    299THE OUTPUT CURVEr= DC resistanceof inductorrLR (external resistor)FIGURE 7.30RLRLFIGURE 7.31QUESTIONSFor the circuit shown in Figure 7.32, all the component values are provided in the dia-gram. Find fr, Q, and BW.250μH160 pF12.6ΩFIGURE 7.32A. fr = B. Q = C. BW = Book Authorc07V107/04/2012 3:...

  • Page 324

    CHAPTER 7 RESONANT CIRCUITS300ANSWERSA. fLCkHZr==×××=−−12122501016010796612ppB. QXRfLRkHzHL===××.=.2279625012 699 2ppmVC. BWfQkHzkHzr==.=79699 28 21  Use the circuit and component values shown in Figure 7.33 to answer the following questions.10 mH1 μF100 ΩFIGURE 7.33QUESTIONSFind fr, ...

  • Page 325

    301THE OUTPUT CURVEANSWERSfr = 1590 Hz; Q = 1; BW = 1590 HzThe output curve is shown in Figure 7.34.795159015902385fVoutFIGURE 7.3422  Use the circuit and component values shown in Figure 7.35 to answer the following questions.100 mH1μF10ΩFIGURE 7.35QUESTIONSFind fr, Q, and BW for this circuit...

  • Page 326

    CHAPTER 7 RESONANT CIRCUITS302ANSWERSfr = 500 Hz; Q = 31.4; BW = 16 HzThe output curve is shown in Figure 7.36.500fVout16FIGURE 7.3623  Use the circuit shown in Figure 7.37 for this problem. In this case, the resistor value is 10 ohms. However, the inductor and capacitor values are not given.LC1...

  • Page 327

    303THE OUTPUT CURVEANSWERSA. BW = 15 HzB. L = 106 mHC. C = 0.166 μFYou can check these values by using the values of L and C to find fr.24  Use the circuit shown in Figure 7.37 for this problem. In this case, the resistor value is given as 10 ohms. However, the inductor and capacitor values ar...

  • Page 328

    CHAPTER 7 RESONANT CIRCUITS304Because of the narrow range of frequencies it passes, a high Q circuit is said to be selective in the frequencies it passes.A circuit that passes (or blocks) a wide range of frequencies is called a low Q circuit.Figure 7.39 shows the output curve for a low Q circuit....

  • Page 329

    305THE OUTPUT CURVErLC0.1Ω0.01μF1μHFIGURE 7.40QUESTIONSFind fr, Q, and BW for the circuit shown in Figure 7.40.A. fr = B. Q = C. BW = ANSWERSA. fr = 1.6 MHzB. XL = 10 ohms, so Q = 10/0.1 = 100 (The only resistance here is the small DC resis-tance of the inductor.)C. BW = 16 kHz (This is a fair...

  • Page 330

    CHAPTER 7 RESONANT CIRCUITS3061. Assume the peak output voltage Vp at the resonant frequency fr to be 100 percent. This is point A on the curve shown in Figure 7.41.2. The output voltage at f1 and f2 is 0.707 of 100 percent. On the graph, these are the two points labeled B in Figure 7.41. Note th...

  • Page 331

    307THE OUTPUT CURVEQUESTIONSCalculate fr, XL, Q, and BW for the circuit shown in Figure 7.42.100 pF256μH16ΩFIGURE 7.42A. fr = B. XL = C. Q = D. BW = ANSWERSA. fr = 1 MHzB. XL = 1607 ohmsC. Q = 100D. BW = 10 kHz28  Now, calculate the frequencies that correspond with each percentage of the peak ...

  • Page 332

    CHAPTER 7 RESONANT CIRCUITS308C. At what frequencies will the output level be 45 percent of Vp?D. At what frequencies will the output level be 32 percent of Vp?E. At what frequencies will the output level be 24 percent of Vp?F. At what frequencies will the output level be 13 percent of Vp?ANSWERS...

  • Page 333

    INTRODUCTION TO OSCILLATORS309ANSWERSA. V = 5 volts × 0.70 = 3.5 voltsB. V = 5 volts × 0.24 = 1.2 voltsFigure 7.43 shows the output curve generated by plotting the frequencies calculated in problem 28 and the corresponding output voltages calculated in this problem.%Vout980960990100010101020985...

  • Page 334

    CHAPTER 7 RESONANT CIRCUITS31030  When the switch in the circuit shown in drawing (1) of Figure 7.44 is closed, current flows through both sides of the parallel LC circuit in the direction shown.(1)(2)(3)FIGURE 7.44It is difficult for the current to flow through the inductor initially because ...

  • Page 335

    INTRODUCTION TO OSCILLATORS311QUESTIONWhen the capacitor is fully discharged, how much current is flowing through the inductor? ANSWERNone.32  Because there is no current in the inductor, its magnetic field collapses. The collaps-ing of the magnetic field induces a current to flow in the ind...

  • Page 336

    CHAPTER 7 RESONANT CIRCUITS312QUESTIONWhat do you think the current generated by the magnetic field in the inductor will do to the capacitor? ANSWERIt charges it to the original polarity.34  When the field has fully collapsed, the capacitor stops charging. It now begins to discharge again, cau...

  • Page 337

    INTRODUCTION TO OSCILLATORS313QUESTIONHow might you prevent this fade-out? ANSWERBy replacing a small amount of energy in each cycle.This lost energy can be injected into the circuit by momentarily closing and opening the switch at the correct time. (See drawing [1] of Figure 7.44.) This would...

  • Page 338

    CHAPTER 7 RESONANT CIRCUITS314The principles you learned in the last few problems are used in practical oscillator cir-cuits, such as those presented in Chapter 9.SUMMARYIn this chapter, you learned about the following topics related to resonant circuits: ■ How the impedance of a series LC circ...

  • Page 339

    SELFTEST3155. What is the voltage across a resistor in series with a parallel LC circuit at the reso-nant frequency? 6. What is the impedance of a series circuit at resonance? 7. What is the formula for the impedance of a parallel circuit at resonance? 8. What is the formula for the resonant f...

  • Page 340

    CHAPTER 7 RESONANT CIRCUITS31616. Use the output curve shown in Figure 7.47 to answer the following questions.111098765VoltskHz1001201401601802004321VfFIGURE 7.47A. What is the peak value of the output curve? B. What is the resonant frequency? C. What is the voltage level at the half power points...

  • Page 341

    SELFTEST317ANSWERS TO SELFTESTIf your answers do not agree with those given here, review the problems indicated in parentheses before you go on to Chapter 8, “Transistor Amplifiers.”1.Z = XL − XC(problem 2)2.ZXXRLC=−+()22(problem 2)3.XL = XC(problem 5)4.Maximum output(problem 5)5.M...

  • Page 342

    CHAPTER 7 RESONANT CIRCUITS31813.fr = 400 Hz, XL = XC = 40 ohms, Q = 2, BW = 200 Hz, Z = 20 ohms(problems 21−29)14.fr = 600 Hz, XL = 24 ohms, XC = 26.5, Q = 3, BW = 200 Hz, Z = 80 ohms(problems 21−29)Because Q is not given, you should use the more compli-cated of the two formulas shown in pro...

  • Page 343

    8Transistor AmplifiersMany of the AC signals you’ll work with in electron-ics are small. For example, the signal that an opti-cal detector reads from a DVD disk cannot drive a speaker, and the signal from a microphone’s output is too weak to send out as a radio signal. In cases such as these...

  • Page 344

    CHAPTER 8 TRANSISTOR AMPLIFIERS320amplifier, you can grasp the building block that makes up amplifier circuits used in elec-tronic devices such as cellphones, MP3 players, and home entertainment centers.Many amplifier circuit configurations are possible. The simplest and most basic of amplify...

  • Page 345

    WORKING WITH TRANSISTOR AMPLIFIERS321VSRBRC1 kΩβ = 100VR = VS − VCVCVCFIGURE 8.1Use the following steps to find the value of RB that will set the collector DC voltage (VC) to half the supply voltage (VS):1. Find IC by using the following equation:ICRCSCCVRRVV5522. Find IB by using the follow...

  • Page 346

    CHAPTER 8 TRANSISTOR AMPLIFIERS322ANSWERSA. IvoltskmAC5V5515B. ImAB55 5100005C. RvoltsmAkB55V10005200 2  You have seen that using a 200 kΩ resistor for RB gives an output level of 5 volts at the col-lector. This procedure of setting the output DC level is called biasing. In problem 1, you biase...

  • Page 347

    WORKING WITH TRANSISTOR AMPLIFIERS323QUESTIONSFor the circuit shown in Figure 8.1, calculate the following parameters when RB 5 168 kΩ and VS 5 10 volts:A. IRBCB55V B. IC 5 βIB 5 C. VR 5 ICRC 5 D. VC 5 VS2VR 5 ANSWERSA. I10 volts168 k0.059 mAB5V5B. IC 5 100 3 0.059 5 5.9 mAC. VR 5 1 kΩ 3 5....

  • Page 348

    CHAPTER 8 TRANSISTOR AMPLIFIERS324ANSWERSC. IB 5 0.075 mA, IC 5 7.5 mA, VC 5 2.5 voltsD. IB 5 0.025 mA, IC 5 2.5 mA, VC 5 7.5 volts5  The values of IC and VC that you calculated in problems 1 and 4 are plotted on the graph on the left side of Figure 8.2. The straight line connecting these points...

  • Page 349

    WORKING WITH TRANSISTOR AMPLIFIERS325B. OFF because essentially no current flows, and the transistor acts like an open cir-cuit. The voltage drop across the transistor is at its maximum (10 volts, in this case).6  Point A on the graph shown in Figure 8.2 is called the saturated point (or the sa...

  • Page 350

    CHAPTER 8 TRANSISTOR AMPLIFIERS326ANSWERC is the only one. A and B fall into nonlinear regions.8  If you apply a small AC signal to the base of the transistor after it has been biased, the small voltage variations of the AC signal (shown in Figure 8.3 as a sine wave) cause small variations in th...

  • Page 351

    WORKING WITH TRANSISTOR AMPLIFIERS327QUESTIONSA. If the input signal decreases, what happens to the collector voltage? B. If you apply a sine wave to the input, what waveform would you expect at the col-lector? ANSWERSA. The collector voltage, VC, increases.B. A sine wave, but inverted as shown i...

  • Page 352

    CHAPTER 8 TRANSISTOR AMPLIFIERS328QUESTIONSA. What is meant by VC? B. What is meant by Vout? ANSWERSA. Collector DC voltage, or the bias pointB. AC output voltageThe ratio of the output voltage to the input voltage is called the voltage gain of the amplifier.Voltage gainAVVVoutin55To calcu...

  • Page 353

    WORKING WITH TRANSISTOR AMPLIFIERS329Solving this for Vout results in the following equation. Here, the values of Rin 5 1 kΩ, Vin 5 1 mV, RC 5 1 kΩ, and β 5 100 were used to perform this sample calculation.VVRR1mV1001k1k100 mVoutinLin53 b 3533VV5QUESTIONSA. Calculate Vout if Rin 5 2 kΩ, Vin 5...

  • Page 354

    CHAPTER 8 TRANSISTOR AMPLIFIERS330NOTE An amplifier whose gain and DC level bias point change as described in this problem is said to be unstable. For reliable operation, an amplifier should be as stable as possible. In later problems, you see how to design a stable amplifier.A STABLE AMPLIFIE...

  • Page 355

    A STABLE AMPLIFIER331QUESTIONIn designing an amplifier circuit and choosing the resistor values, there are two goals. What are they? ANSWERA stable DC bias point, and a stable AC gain 12  Look at the gain first. The gain formula for the circuit shown in Figure 8.6 is as follows:AVVRRVoutinC...

  • Page 356

    CHAPTER 8 TRANSISTOR AMPLIFIERS332QUESTIONSA. Calculate the voltage gain (AV) of the amplifier circuit shown in Figure 8.6 if RC 5 10 kΩ and RE 5 1 kΩ. Then, use AV to calculate the output voltage if the input sig-nal is 2 mVpp. B. Calculate the voltage gain if RC 5 1 kΩ and RE 5 250 ohms....

  • Page 357

    A STABLE AMPLIFIER333QUESTIONSCalculate VC, VE, and AV for the circuit shown in Figure 8.7 with VS 5 10 volts, IC 5 1 mA, RC 5 1 kΩ, and RE 5 100 ohms. RCVCVEREFIGURE 8.7ANSWERV1k1mA1 voltVVV1019 voltsV100 ohms1mA 0.1 RCSRE5V 3552 5 2 5535vvoltARR1k100 ohms10VCE55V5 15  For this problem, use...

  • Page 358

    CHAPTER 8 TRANSISTOR AMPLIFIERS334ANSWERSVVVARCEV5V 3552 55V 355 2 k1mA2 volts1028 volts1k1mA1 voltRRCE55VV52k1k2 16  For this problem, use the circuit shown in Figure 8.7 with VS 5 10 volts and IC 5 1 mA.QUESTIONSFind VC, VE, and AV for the following values of RC and RE:A. RC 5 5 kΩ, RE 5 1 k...

  • Page 359

    BIASING335R1R2RCVSVCREFIGURE 8.8Read the following procedure and the relevant formulas first, and then you will work through an example.1. Find RE by using the following:ARR.VCE52. Find VE by using the following:AVVVVV.VRESCE55 23. Find VB by using the following:VV0.7 voltBE514. Find IC by using...

  • Page 360

    CHAPTER 8 TRANSISTOR AMPLIFIERS3368. Find R1 by using the following:RVVII.1SB2B5219. Steps 7 and 8 might produce nonstandard values for the resistors, so choose the nearest standard values.10. Use the voltage divider formula to see if the standard values you chose in step 9 result in a voltage le...

  • Page 361

    BIASING3374. IC 55. IB 56. I2 57. R2 58. R1 59. Choose the standard resistance values that are closest to the calculated values for R1 and R2.R1 5R2 510.Using the standard resistance values for R1 and R2, find VB.VB 5ANSWERSYou should have found values close to the following:1. 100 ohms2. 0.5 vo...

  • Page 362

    CHAPTER 8 TRANSISTOR AMPLIFIERS3381kΩ16 kΩ0.1μF2.4 kΩVin100ΩFIGURE 8.1018  Follow the steps in problem 17 to answer the following questions.QUESTIONSFind the values of the parameters specified in each question for the circuit shown in Figure 8.9 if AV 5 15, VC 5 6 volts, β 5 100, RC 5 3.3...

  • Page 363

    339BIASINGANSWERSFollowing are the values you should have found:1. 220 ohms2. 0.27 volt3. 0.97 volt (You can use 1 volt if you want.)4. 1.2 mA5. 0.012 mA6. 0.12 mA7. 8.3 kΩ8. 68.2 kΩ9. These are close to the standard values of 8.2 kΩ and 68 kΩ.10.1.08 volts using the standard values. This is ...

  • Page 364

    CHAPTER 8 TRANSISTOR AMPLIFIERS340❑ One 100 Ω, 0.25-watt resistor❑ One 15 kΩ, 0.25-watt resistor❑ One 2.2 kΩ, 0.25-watt resistor❑ One 3.3 kΩ, 0.25-watt resistor❑ One 220 Ω, 0.25-watt resistor❑ One 68 kΩ, 0.25-watt resistor❑ One 8.2 kΩ, 0.25-watt resistor❑ One 0.1 μF capaci...

  • Page 365

    341BIASING0.1μFFunctiongeneratorVoutVinVS = 10VR1R2RCREFIGURE 8.12Circuit #TransistorRCRER1R21PN22221 kΩ100 Ω15 kΩ2.2 kΩ22N39041 kΩ100 Ω15 kΩ2.2 kΩ3PN22223.3 kΩ220 Ω68 kΩ8.2 kΩ42N39043.3 kΩ220 Ω68 kΩ8.2 kΩCarefully check your circuit against the diagram.After you check your circu...

  • Page 366

    CHAPTER 8TRANSISTOR AMPLIFIERS3427. Repeat steps 5 and 6 until you have recorded Vin and Vout in the last row of the table.8. Determine β for each of the transistors used in this project. Insert the transistors one at a time into the circuit you built in Project 3-1 to take this measurement.9. F...

  • Page 367

    343BIASING3.3 kΩ resistor220 kΩ resistor68 kΩ resistor8.2 kΩ resistorBaseEmitterCollector0.1μF capacitorFIGURE 8.14Figure 8.15 shows a function generator and oscilloscope attached to the circuit.The input signal is represented by the upper sine wave shown in Figure 8.16, and the output signa...

  • Page 368

    CHAPTER 8 TRANSISTOR AMPLIFIERS344Black leadfrom powersupplyChannel 2ground clipChannel1ground clipChannel 2oscilloscopeprobeChannel1oscilloscopeprobeRed leadfrom powersupplyRed leadfrom functiongeneratorBlack leadfrom functiongeneratorFIGURE 8.152.2 divisionspeak-to-peak4.2 divisionspeak-to-peak...

  • Page 369

    345BIASINGChannel 2 set to0.1 volts/divChannel 2 verticalposition knobChannel 1 verticalposition knobChannel 1 set to0.5 volts/divTime/div control setto 50 μsec/divFIGURE 8.17The measured values of AV are quite close to the calculated values of AV, well within variations that could be caused by ...

  • Page 370

    CHAPTER 8 TRANSISTOR AMPLIFIERS346RECEFIGURE 8.18If the reactance of this capacitor for an AC signal is significantly smaller than RE, the AC signal passes through the capacitor rather than the resistor. Therefore, the capacitor is called an emitter bypass capacitor. The AC signal “sees” a d...

  • Page 371

    BIASING347ANSWERSA. It makes the emitter look like a ground and effectively turns the circuit into the circuit shown in Figure 8.4.B. It increases the gain.C. The same formula used in problem 10:ARRVCin5b3 20  You can use the circuit shown in Figure 8.18 when you need as much AC voltage gain as...

  • Page 372

    CHAPTER 8 TRANSISTOR AMPLIFIERS348QUESTIONSFollow the previous steps to calculate the value of CE required if the lowest operating frequency of the amplifier is 50 Hz.1. 50 Hz is the lowest frequency at which the amplifier must operate.2. XC 5 3. CE 5 ANSWERSXC 5 10 ohmsCE 5 320 μF (approxim...

  • Page 373

    BIASING349 ■ Circuit 1—Here, RE 5 zero, so the AC voltage gain formula is as follows:ARRVCin5b3. ■ Circuit 2—Here, RE 5 zero for an AC signal because the AC signal is grounded by the capacitor, and RE is out of the AC circuit. Thus, the AC voltage gain formula is as follows:ARRVCin5b 3. 2...

  • Page 374

    CHAPTER 8 TRANSISTOR AMPLIFIERS350ANSWERSA. 375B. 22122  Two-stage amplifiers can achieve large AC voltage gains if each amplifier uses an emitter bypass capacitor.QUESTIONWhat is the total AC voltage gain if each stage of a two-transistor amplifier has a gain of 100? ANSWER10,000THE EMITTER ...

  • Page 375

    THE EMITTER FOLLOWER351QUESTIONHow is the circuit shown in Figure 8.23 different from the amplifier circuit discussed in problems 11218? ANSWERThere is no collector resistor, and the output signal is taken from the emitter.24  The circuit shown in Figure 8.23 is called an emitter follower ampl...

  • Page 376

    CHAPTER 8 TRANSISTOR AMPLIFIERS352ANSWERSA. 1B. NoC. HighD. Low25  The example in this problem demonstrates the importance of the emitter follower circuit. The circuit shown in Figure 8.24 contains a small AC motor with 100 ohms resistance that is driven by a 10 Vpp signal from a generator. The ...

  • Page 377

    THE EMITTER FOLLOWER353You can use the following formula to calculate the approximate input resistance of the transistor:RR10010010,000(assuming that100)inE5b353V 5Vb5The 10 Vpp from the generator is divided between the 10,000-ohm input resistance of the transistor and the 50-ohm internal resista...

  • Page 378

    CHAPTER 8 TRANSISTOR AMPLIFIERS354QUESTIONSA. What is the approximate gain of an emitter follower circuit? B. What is the phase of the output signal compared to the phase of the input signal? C. Which has the higher value, the input resistance or the output resistance? D. Is the emitter ...

  • Page 379

    THE EMITTER FOLLOWER3557. Find R2 by using the following formula:RVI2B258. Find R1 by using the following formula:RVVII1SB2B521Usually, IB is small enough to be dropped from this formula.9. Choose the nearest standard values for R1 and R2.10. Check that these standard values give a voltage close ...

  • Page 380

    CHAPTER 8 TRANSISTOR AMPLIFIERS3568. R1 5 9. The nearest standard values are as follows: R1 5 R2 5 10. VB 5 ANSWERSYour answers should be close to the following values:1. 5 volts (This was given in Figure 8.26.)2. 5.7 volts3. 1 kΩ (This was given in Figure 8.26.)4. 5 mA5. 0.05 mA6. 0.5 mA7....

  • Page 381

    ANALYZING AN AMPLIFIER357In this case, to “analyze” means to calculate the collector DC voltage (the bias point) and find the AC gain. This procedure is basically the reverse of the design procedure.Start with the circuit shown in Figure 8.27.R1VS VCRECERCR2FIGURE 8.27Following are the steps...

  • Page 382

    CHAPTER 8 TRANSISTOR AMPLIFIERS35810 V 160 kΩ22 kΩ1 kΩ10 kΩ50 μFFIGURE 8.28QUESTIONSCalculate VB, VE, IC, VR, VC, and AV using Steps 126 of this problem.1. VB 5 2. VE 5 3. IC 5 4. VR 5 5. VC 5 6. AV 5 ANSWERS1. V1022 k160 k22 k1.2 voltsB53VV1V52. VE 5 1.2 2 0.7 5 0.5 volt3. I0.5 ...

  • Page 383

    ANALYZING AN AMPLIFIER359Without the capacitor: A10 k1k10 (a small gain)V5VV5 29  You can determine the lowest frequency the amplifier will satisfactorily pass by following these simple steps:1. Determine the value of RE.2. Calculate the frequency at which XC 5 RE/10. Use the capacitor reactanc...

  • Page 384

    CHAPTER 8 TRANSISTOR AMPLIFIERS36030  For the circuit shown in Figure 8.29, follow the steps given in problems 28 and 29 to answer the following questions.10 V 820 kΩ110 kΩ470Ω4.7 kΩ60μFβ= 120Rin = 1.5 kΩFIGURE 8.29QUESTIONS1. VB 52. VE 53. IC 54. VR 55. VC 56. With capacitor:AV 5Without ...

  • Page 385

    THE JFET AS AN AMPLIFIER3613. 1 mA4. 4.7 volts5. 5.3 volts (bias point)6. With capacitor: 376Without capacitor: 107. 57 Hz (approximately)THE JFET AS AN AMPLIFIER31  Chapter 3 discussed the JFET in problems 28231, and Chapter 4 discussed the JFET in problems 37241. You may want to review these p...

  • Page 386

    CHAPTER 8 TRANSISTOR AMPLIFIERS362D. What value of VGS would you need to turn the JFET completely OFF? E. When a JFET is alternately turned completely ON and OFF in a circuit, what type of component are you using the JFET as? ANSWERSA. N-channel JFET.B. VGS 5 0 V to turn the JFET completely O...

  • Page 387

    THE JFET AS AN AMPLIFIER363For the transfer curve shown in Figure 8.31, IDSS 5 12 mA and VGS(off) 5 −4 volts. Set-ting the bias voltage at VGS 5 −2 volts returns the following value for the drain current:I12 mA12412 mA(0.5)3mAD2253 2 22555⎛⎝⎜⎜⎞⎠⎟⎟⎟QUESTIONSCalculate the dra...

  • Page 388

    CHAPTER 8 TRANSISTOR AMPLIFIERS364ANSWERR(VV )I(24 volts 10 volts)3mA14 volts3mA4.67 kDDDDSD525555V 34  The circuit shown in Figure 8.32 (which is referred to as a JFET common source amplifier) applies a 0.5 Vpp sine wave to the gate of the JFET and produces an amplified sine wave output from ...

  • Page 389

    THE JFET AS AN AMPLIFIER365ANSWERFor I3.8 mA, V3.8 mA4.67 k17.7 voltsFor I2.3mA, V2.3mDRDDRD553V 555AA4.67 k10.7 volts5V 5This corresponds to a 7 Vpp sine wave. 36  As the voltage drop across RD changes, the output voltage also changes.QUESTIONFor the circuit shown in Figure 8.32, calculate the ...

  • Page 390

    CHAPTER 8 TRANSISTOR AMPLIFIERS366ANSWERThe output signal is a 7 Vpp sine wave with the same frequency as the input sine wave. As the input voltage on VGS increases (toward 0 volts), the output decreases. As the input voltage decreases (becomes more negative), the output voltage increases. This m...

  • Page 391

    THE JFET AS AN AMPLIFIER367In this equation, Δ indicates the change or variation in VGS and the corresponding drain current. The unit for transconductance is mhos.QUESTIONSA. Using the data from Table 8-1, what is the value of gm for the JFET used in the amplifier? B. What is the correspondi...

  • Page 392

    CHAPTER 8 TRANSISTOR AMPLIFIERS368E. Calculate the maximum and minimum values of Vout that result from the input signal using the procedures in problems 35 and 36. F. Calculate the gain of the amplifier. ANSWERSA. VGS 5 21.6 voltsB. ID 5 3.7 mAC. For VDS 5 10 volts,R14 volts3.7 mA3780 ohms...

  • Page 393

    THE JFET AS AN AMPLIFIER369 42  Figure 8.33 shows a JFET amplifier circuit that uses one power supply, rather than sepa-rate power supplies for the drain and gate used in the amplifier discussed in problems 34241.+24 V(VDD)+VinVoutRGRDIDRSVGSCSFIGURE 8.33The DC voltage level of the gate is zer...

  • Page 394

    CHAPTER 8 TRANSISTOR AMPLIFIERS370NOTE Choose the value of CS so that its reactance is less than 10 percent of RS at the lowest frequency you need to amplify. The DC load for the JFET is RD plus RS. The AC load is RD only because CS bypasses the AC signal around RS, which keeps the DC operating p...

  • Page 395

    THE OPERATIONAL AMPLIFIER371on a small silicon chip. These IC op-amps are much smaller and, therefore, more practical than an amplifier with equivalent performance that is made with discrete components.You can purchase op-amps in different case configurations. Some of these configura-tions ar...

  • Page 396

    CHAPTER 8 TRANSISTOR AMPLIFIERS372data sheet will specify their values for the particular op-amp you use. Datasheets usually contain circuit diagrams showing how you should connect external components to the op-amp for specific applications. These circuit diagrams (showing how a particular op-am...

  • Page 397

    THE OPERATIONAL AMPLIFIER373AC VoutRinRFAC Vin−RL10 kΩ10 kΩ10 kΩ+15741−15+−FIGURE 8.35Resistor RF is called a feedback resistor because it forms a feedback path from the output to the input. Many op-amp circuits use a feedback loop. Because the op-amp has such a high gain, it is easy to ...

  • Page 398

    CHAPTER 8 TRANSISTOR AMPLIFIERS374QUESTIONSA. Calculate the value of RF. B. Calculate the value of Vin required to produce the output voltage specified earlier. ANSWERSA. R506.8 k340 kF53V 5VB. V12 V500.24 Vor 0.168 Vinpppprms55PROJECT 8.2: The Operational AmplifierOBJECTIVEThe objective o...

  • Page 399

    375THE OPERATIONAL AMPLIFIER❑ One 220 kΩ, 0.25-watt resistor.❑ One 270 kΩ, 0.25-watt resistor.❑ One 330 kΩ, 0.25-watt resistor.❑ One 380 kΩ, 0.25-watt resistor.❑ Two terminal blocks.❑ Two 6-volt battery packs (4 AA batteries each).❑ One function generator.❑ One oscilloscope. ...

  • Page 400

    CHAPTER 8 TRANSISTOR AMPLIFIERS376VinRinRFVoutFunctiongenerator0.1μF10 kΩ10 kΩOPA134+−+ 6V− 6VFIGURE 8.37+−+−6VTo +V pinof OPA134To−V pinof OPA1346VFIGURE 8.38Carefully check your circuit against the diagram. After you check your circuit, follow these steps, and record your measureme...

  • Page 401

    377THE OPERATIONAL AMPLIFIER6. Measure and record Vout and Vin.7. Repeat steps 5 and 6 until you have recorded Vout and Vin for the last row of the table.RFCalculated AV (RF/Rin)Vin (volts) Vout (volts)Measured AV(Vout/Vin )51 kΩ100 kΩ150 kΩ220 kΩ270 kΩ330 kΩ380 kΩ8. Determine the calculat...

  • Page 402

    CHAPTER 8 TRANSISTOR AMPLIFIERS378Figure 8.40 shows a function generator and oscilloscope attached to the circuit.Channel 2oscilloscopeprobeChannel 2ground clipRed leadfrom functiongeneratorChannel1oscilloscopeprobeChannel1ground clip6 voltbattery pack6 voltbattery packBlack leadfrom functiongene...

  • Page 403

    379THE OPERATIONAL AMPLIFIERTime/div controlset to 50 μsec/divChannel 1 set to2 volts/divChannel 1 verticalposition knobChannel 2 verticalposition knobChannel 2 setto 0.1 volts/divFIGURE 8.42As you measure Vin and Vout, you may need to adjust the TIME/DIV control, the VOLTS/DIV control, and vert...

  • Page 404

    CHAPTER 8 TRANSISTOR AMPLIFIERS380SUMMARYThis chapter introduced the most common types of amplifiers in use today: the com-mon emitter BJT, the common source JFET, and the op-amp. At best, this chapter has scratched only the surface of the world of amplifiers. Actually, there are many variation...

  • Page 405

    SELFTEST381R1VSVER2VCRCREFIGURE 8.436. Repeat question 4 with these values: VS 5 14 volts, β 5 250, Rin 5 1 kΩ, and RC 515 kΩ. The bias point should be 7 volts and the AC voltage gain 50.7. Design an emitter follower amplifier given that VS 5 12 volts, RE 5 100 ohms, β 535, VE 5 7 volts, ...

  • Page 406

    CHAPTER 8 TRANSISTOR AMPLIFIERS38213. Design a JFET amplifier using the circuit shown in Figure 8.32. The characteristics of the JFET are IDSS 5 20 mA and VGS(off) 5 24.2 volts. The desired value of VDS is 14 volts. Find the value of RD. 14. If the transconductance of the JFET used in questio...

  • Page 407

    SELFTEST3834.R1 5 29 kΩ, R2 5 3.82 kΩ, RE 5 160 ohms, CE 5 200 μF, AV 5 120(problems 17222)5.R1 5 138 kΩ, R2 5 8 kΩ, RE 5 500 ohms, CE 5 64 μF, AV 5 800(problems 17222)6.R1 5 640 kΩ, R2 5 45 kΩ, RE 5 300 ohms, CE 5 107 μF, AV 5 750(problems 17222)7.R1 5 8 kΩ; R2 5 11.2 kΩ(problem...

  • Page 408

    Book Authorc08V107/04/2012 3:18 PM c08.indd 3847/4/2012 3:20:55 PM

  • Page 409

    9OscillatorsThis chapter introduces you to oscillators. An oscillator is a circuit that produces a continuous output signal. There are many types of oscillator circuits used extensively in electronic devices. Oscillators can produce a variety of different output signals, such as sine waves, squa...

  • Page 410

    CHAPTER 9 OSCILLATORS386 ■ Differentiate between positive and negative feedback. ■ Specify the type of feedback that causes a circuit to oscillate. ■ Specify at least two methods of obtaining feedback in an oscillator circuit. ■ Understand how resonant LC circuits set the frequency of an...

  • Page 411

    UNDERSTANDING OSCILLATORS3872  When you connect the output of an amplifier to its input, you get feedback. If the feedback is “out of phase” with the input, as shown in Figure 9.2, then the feedback is negative.VinVoutFIGURE 9.2When the signal from the collector is fed back to the base of t...

  • Page 412

    CHAPTER 9 OSCILLATORS388ANSWERSA. To reduce distortionB. Negative feedback3  If the feedback from the output is in phase with the input, as shown in Figure 9.4, the cir-cuit’s feedback is positive.VinVoutFIGURE 9.4In the circuit shown in Figure 9.5, the collector of the second transistor is co...

  • Page 413

    UNDERSTANDING OSCILLATORS389B. What type of feedback is used in oscillators? C. What parts of an amplifier do you connect to produce feedback? ANSWERSA. Negative feedback.B. Positive feedback.C. Connect the output of an amplifier to its input.4  The amplifier shown in Figure 9.6 is the same t...

  • Page 414

    CHAPTER 9 OSCILLATORS390 5  In the circuit shown in Figure 9.6, an input signal applied to the base will be amplified.QUESTIONSA. What is the basic formula for an amplifier’s voltage gain? B. What is the voltage gain formula for the amplifier circuit shown in Figure 9.6? ANSWERSA. ARR...

  • Page 415

    UNDERSTANDING OSCILLATORS391The voltage gain formula for this type of amplifier can be simplified because the input impedance to the amplifier is so low when the signal is fed into the emitter that you can discount it. This results in the following voltage gain formula for the common base ampl...

  • Page 416

    CHAPTER 9 OSCILLATORS392ANSWERSA. Amplified and not invertedB. LowC. ARRRRVLSCS55 8  Figure 9.8 shows an amplifier circuit with a parallel inductor and capacitor connected between the collector of the transistor and ground. A parallel inductor and capacitor circuit is sometimes called a tuned ...

  • Page 417

    UNDERSTANDING OSCILLATORS393NOTE The circuit shown in Figure 9.8 does not have an input signal either to the emitter or to the base. By adding a feedback connection to a parallel LC circuit, you provide an input signal to the emitter or base, as explained later in this chapter. 9  Write the volt...

  • Page 418

    CHAPTER 9 OSCILLATORS394The gain is increased to the point where you can consider it “large enough” to use the amplifier as an oscillator. When these capacitors are used in either a common emitter or common base amplifier, it is not usually necessary to calculate the gain of the amplifier....

  • Page 419

    UNDERSTANDING OSCILLATORS395You can use this circuit and the circuit shown in Figure 9.8 to selectively amplify one frequency far more than others.QUESTIONSA. What would you expect this one frequency to be? B. Write the formula for the impedance of the circuit at the resonance frequency. C....

  • Page 420

    CHAPTER 9 OSCILLATORS396QUESTIONIndicate which of the following is an accurate description of the circuit in Figure 9.10:A. OscillatorB. Tuned amplifierC. Common base circuitD. Common emitter circuitANSWERB. Tuned amplifierFEEDBACK14  To convert an amplifier into an oscillator, you must conne...

  • Page 421

    FEEDBACK397than the main coil). In all three of these methods, between one-tenth and one-half of the output must be used as feedback.QUESTIONSA. Where is the feedback taken from in a Colpitts oscillator? B. What type of oscillator uses a tap on the coil for the feedback voltage? C. What type does...

  • Page 422

    CHAPTER 9 OSCILLATORS398ANSWERVVX(XX)foutC2C1C251which becomesVVC(CC)fout11251 16  To find the resonance frequency in this circuit, first find the equivalent total capacitance CT of the two series capacitors. You then use CT in the resonance frequency formula.QUESTIONSA. What is the formula f...

  • Page 423

    FEEDBACK399 17  Figure 9.13 shows a parallel LC circuit in which the feedback voltage is taken from a tap N1 turns from one end of a coil, and N2 turns from the other end.N2VoutN1VfFIGURE 9.13You can calculate the feedback voltage with a voltage divider formula that uses the number of turns in e...

  • Page 424

    CHAPTER 9 OSCILLATORS400VfVoutN1 : N2FIGURE 9.14QUESTIONWho invented this type of oscillator? ANSWERArmstrong 19  For each of the feedback methods described in the last few problems, the voltage fed back from the output to the input is a fraction of the total output voltage ranging between on...

  • Page 425

    FEEDBACK401QUESTIONSA. What makes an amplifier into an oscillator? B. What input does an amplifier need to become an oscillator? ANSWERA. A resonant LC circuit with feedback of the correct phase and amount.B. None. Oscillations happen spontaneously if the feedback is correct.INSIDE THE INDUCTOR...

  • Page 426

    CHAPTER 9 OSCILLATORS402THE COLPITTS OSCILLATOR20  Figure 9.15 shows a Colpitts oscillator circuit, the simplest of the LC oscillators to build.The value of each color used in the bands is shown in the following table (with units in μH):ColorSignificant DigitsMultiplierToleranceBlack01Brown110...

  • Page 427

    THE COLPITTS OSCILLATOR403Feedback10 kΩ1 μF82 kΩ9 V0.85 VconnectionR18.2 kΩR2C10.1 μFC20.2 μFRCCCCB1 μFRE510 ΩL 0.5 mH(20 Ω DC)FIGURE 9.15The feedback signal is taken from the capacitive voltage divider and fed to the emitter. This connection provides a feedback signal to the emitter in ...

  • Page 428

    CHAPTER 9 OSCILLATORS404 21  Use the Colpitts oscillator component values shown in Figure 9.15 to answer the follow-ing questions.QUESTIONSA. What is the effective total capacitance of the two series capacitors in the tuned circuit? CT 5 B. What is the oscillator frequency? fr 5 C. What is ...

  • Page 429

    THE COLPITTS OSCILLATOR40522  Figure 9.16 shows a Colpitts oscillator circuit that uses a different method for making feedback connections between the parallel LC circuit and the transistor.C1C2CCLRECERCFIGURE 9.16QUESTIONList the differences between this circuit and the one shown in Figure 9....

  • Page 430

    CHAPTER 9 OSCILLATORS406ANSWERX1012fC0.1610CCE3E55p53510So, CE 5 3.2 μF. Thus, you should use a capacitor larger than 3 μF.PROJECT 9.1: The Colpitts OscillatorOBJECTIVEThe objective of this project is to demonstrate that an oscillator generates a sine wave when feedback is applied to either th...

  • Page 431

    407THE COLPITTS OSCILLATOR❑ One 0.5 mH inductor. (Suppliers may also refer to this value as 500 μH.)❑ One 9-volt battery pack.❑ One breadboard.❑ One oscilloscope. ❑ One PN2222 transistor. Figure 9.17 shows the pinout diagram for PN2222 transistors.PN2222EmitterBaseCollectorFIGURE 9.17S...

  • Page 432

    CHAPTER 9 OSCILLATORS408Carefully 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 1 to a jumper wire connected to Vout.Connect the groun...

  • Page 433

    409THE COLPITTS OSCILLATOR0.5 mHinductor510Ωresistor8.2 kΩresistor82 kΩresistor10kΩresistorPN2222transistor1μFcapacitor1μFcapacitor0.1μFcapacitor0.22μFcapacitorFIGURE 9.20Channel 1 ground clipChannel 1 oscilloscope probeFIGURE 9.21Book Authorc09V107/04/2012 3:23 PM c09.indd 4097/4/2012 ...

  • Page 434

    CHAPTER 9 OSCILLATORS410Figure 9.22 shows the sine wave generated by the Colpitts oscillator with feedback to the emitter. 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 divi...

  • Page 435

    411THE COLPITTS OSCILLATORHorizontalposition knobTime/div control set to 10 μsec/divChannel 1 vertical position knobChannel 1 set to 0.2 volts/divFIGURE 9.23Feedbackto base4.7μFcapacitorFIGURE 9.24Book Authorc09V107/04/2012 3:23 PM c09.indd 4117/4/2012 3:24:29 PM

  • Page 436

    CHAPTER 9 OSCILLATORS4123.4 divisionsper cycleFIGURE 9.25Your values should be close to those shown in the following table:Circuit #Period (μsec)Frequency (kHz)13429.423429.4Notice that the frequency of the sine wave generated by both circuits is the same. This demonstrates that an oscillator ca...

  • Page 437

    413THE COLPITTS OSCILLATORL = 160 mH DC resistance = 500 Ωapproximately9VC2C1R1R2CBRELFIGURE 9.26The following table shows possible values you might use for C1 and C2 in the circuit shown in Figure 9.26:C1C2CTfr0.01 μF0.1 μF0.01 μF0.2 μF0.01 μF0.3 μF0.1 μF1 μF0.2 μF1 μFQUESTIONSA. Calc...

  • Page 438

    CHAPTER 9 OSCILLATORS414D. What is the condition that results in the highest possible resonance frequency? E. What would be the highest resonance frequency if C1 is fixed at 0.01 μF, and C2 can vary from 0.005 μF to 0.5 μF? ANSWERSA. The following table shows the values of CT and fr:C1C2CTfr0...

  • Page 439

    THE HARTLEY OSCILLATOR415R2R1CCCLC2μFL= 2HDC resistance = 130 ΩRERC10μF10μFFIGURE 9.27Capacitor CL stops the emitter DC voltage from being pulled down to 0 volts through the coil. CL should have a reactance of less than RE/10, or less than 160 ohms at the oscil-lator frequency.QUESTIONSWork t...

  • Page 440

    CHAPTER 9 OSCILLATORS416Figure 9.28 shows a Hartley oscillator with the parallel LC circuit connected between the collector and the supply voltage. As with the circuit shown in Figure 9.27, this circuit provides a feedback signal to the emitter from a tap in the coil, in the correct phase to prov...

  • Page 441

    417THE HARTLEY OSCILLATORtotal inductance is found by adding the individual inductance values, using the following equation:LT 5 L1 1 L2Parts ListYou need the following equipment and supplies:❑ One 10 kΩ, 0.25-watt resistor.❑ One 510 Ω, 0.25-watt resistor.❑ One 82 kΩ, 0.25-watt resistor....

  • Page 442

    CHAPTER 9 OSCILLATORS418information 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.10 kΩ1 μF82 kΩR18.2 kΩR2C0.01 μFL23.3 mHL16.8 mHRCCC1 μFCCCB1 μFRE510 Ω9V+−VoutFIGURE 9....

  • Page 443

    419THE HARTLEY OSCILLATOR510Ωresistor6.8 mH inductor3.3 mH inductor8.2 kΩresistor82 kΩresistor10kΩresistor1μFcapacitor1μFcapacitor1μFcapacitor0.01μFcapacitorPN2222transistorFIGURE 9.31Channel 1 ground clipChannel 1 oscilloscope probeFIGURE 9.32Book Authorc09V107/04/2012 3:23 PM c09.indd ...

  • Page 444

    CHAPTER 9 OSCILLATORS420Figure 9.33 shows the sine wave generated by the Hartley oscillator. You can deter-mine 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/DIV setting...

  • Page 445

    421THE ARMSTRONG OSCILLATORHorizontalposition knobTime/div control set to 20 μsec/divChannel 1 vertical position knobChannel 1 set to 0.2 volts/divFIGURE 9.34THE ARMSTRONG OSCILLATORThe Armstrong oscillator shown in Figure 9.35 is somewhat more difficult to design and build. Here, the oscillatio...

  • Page 446

    CHAPTER 9 OSCILLATORS422Because of the large variety of transformers and coils available, it is almost impossible to give you a simple procedure for designing an Armstrong oscillator. Instead, the manu-facturer specifies the number of turns required on the coils, which guarantees that the oscill...

  • Page 447

    SIMPLE OSCILLATOR DESIGN PROCEDURE423D. Three: the Colpitts, Hartley, and the Armstrong.E. Nothing: The oscillations should start spontaneously if the component values in the circuit are correct.The main practical problem with building oscillators is selecting the coil. For mass production, a man...

  • Page 448

    CHAPTER 9 OSCILLATORS424Follow these steps:1. Choose the frequency of the oscillator output signal.2. Choose a suitable coil. This step presents the greatest practical difficulty. Some values of coil are often not available, so you must use whatever is readily available. Fortunately, you can use ...

  • Page 449

    SIMPLE OSCILLATOR DESIGN PROCEDURE425ANSWERSA. C1 5 0.016 μFB. C2 5 0.048 μF to 0.16 μF 28  Now, continue with the design procedure by following the next steps.6. Design an amplifier with a common emitter gain of about 20. Choose a collector DC voltage that is about half the supply voltage...

  • Page 450

    CHAPTER 9 OSCILLATORS426QUESTIONWhat is the value of CB? ANSWERC0.1 FB5m 30  Continue the design procedure steps.10. After you build an oscillator, apply power to the circuit and look at the output sig-nal on an oscilloscope. If the output signal is oscillating, check the frequency. If the fr...

  • Page 451

    OSCILLATOR TROUBLESHOOTING CHECKLIST4276. Check the DC voltage level of the collector, base, and emitter.7. Check the capacitor values of the LC circuit. If necessary, try some other values until the circuit oscillates.8. If none of the previous actions produce oscillations, check to see if any o...

  • Page 452

    CHAPTER 9 OSCILLATORS4288. Find CC.CC 59. Find CB.CB 5ANSWERSC1 5 0.01 μFC2 5 0.1 μFCC 5 0.047 μF (use 0.1 μF)CB 5 0.047 μF (use 0.1 μF)Steps 10–11 are the procedure you use to ensure that the oscillator works. If you built this circuit, go through steps 10–11. You don’t need to...

  • Page 453

    OSCILLATOR TROUBLESHOOTING CHECKLIST429Measurements of the output signal of this oscillator confirm a frequency close to 25 kHz.QUESTIONFind the impedance of the LC circuit at resonance. Note that r (the DC resistance of the inductor) is 12 ohms. ANSWERZLCr4100.011012=33 k (approximately)-3-6...

  • Page 454

    CHAPTER 9 OSCILLATORS4308. Find CC.CC 59. Find CB.CB 5ANSWERC1 5 0.0008 μF; therefore, choose a standard value of 0.001 μF.C2 5 0.0047 μF, which is a standard value.CB 5 CC 5 0.004 μF (minimum).34  The circuit you designed in problem 33 is shown in Figure 9.38.R1R2C20.0047μFC10.001μF...

  • Page 455

    OSCILLATOR TROUBLESHOOTING CHECKLIST431ANSWERZ 5 30 kΩThis is about 3 times the value of RC, rather than 10 times the value of RC, as suggested in step 6 of problem 28.35  Figure 9.39 shows several other oscillator circuits. Calculate the expected output fre-quency for each circuit and build a...

  • Page 456

    CHAPTER 9 OSCILLATORS432B. f 5C. f 5D. f 5ANSWERA. 8.8 kHzB. 10 kHzC. 3 kHzD. 1 kHzSUMMARY AND APPLICATIONSThis chapter covered the following topics related to oscillators: ■ The main elements that make up an oscillator ■ How to differentiate between positive and negative feedback ■ The...

  • Page 457

    SELFTEST4332. What is the difference between positive and negative feedback? 3. What type of feedback is required in an oscillator? 4. What is the formula for the frequency of an oscillator? 5. Draw the circuit for a Colpitts oscillator.6. Draw the circuit for a Hartley oscillator.7. Dr...

  • Page 458

    CHAPTER 9 OSCILLATORS4346.See Figure 9.27.(problem 25)7.See Figure 9.35.(problem 25)8A.V0.0470.147f5(problems 27–30)AV cannot be calculated.C1/C2  5 0.047/0.1  5 0.47Z cannot be calculated because r is unknown.fr  5 8.8 kHz (approximately).8B.V0.150.62f5AV  5 2.2 (approximately).C1/C2 ...

  • Page 459

    10The TransformerTransformers are used to “transform” an AC voltage to a higher or lower level. When you charge your cell-phone, 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 ...

  • Page 460

    CHAPTER 10 THE TRANSFORMER436You can also use transformers to increase voltage. For example, some of the equip-ment 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...

  • Page 461

    TRANSFORMER BASICS437C. If you apply an AC voltage to the terminals of the primary coil, what occurs in the secondary coil? ANSWERSA. 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 tran...

  • Page 462

    CHAPTER 10 THE TRANSFORMER438ANSWERSA. 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  Y...

  • Page 463

    TRANSFORMER BASICS439ANSWERThe 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.Center tapFIGURE 10.5QUESTIONWhat is the difference between t...

  • Page 464

    CHAPTER 10 THE TRANSFORMER440QUESTIONHow does increasing the number of turns of wire in a secondary coil affect the output voltage across the secondary coil? ANSWERIt increases the output voltage across the secondary coil. 6  Figure 10.6 shows the number of turns in the primary and secondary...

  • Page 465

    TRANSFORMER BASICS441 7  Use the formula from problem 6 to answer the following question.QUESTIONCalculate 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. ANSWERVVNNTRinoutps55VVNNVTRoutinspin55 31VVTRVoutinpp535 3 511...

  • Page 466

    CHAPTER 10 THE TRANSFORMER442ANSWERSA. 4 Vpp (This is a step-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...

  • Page 467

    TRANSFORMER BASICS443FIGURE 10.8+40 V−28 V−40 V+28 V0VQUESTIONSA. 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? ANSWERSA. rmsB. 23 1.414 3 28 5 79.184 volts11  Like the 28-volt transformer output value, the 120-volt w...

  • Page 468

    CHAPTER 10 THE TRANSFORMER44412  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.(1)(2)10 V20 V10 VFIGURE 10...

  • Page 469

    TRANSFORMER BASICS445ANSWERSA. 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 5 2 3 1.414 3 30 5 84.84 volts. 13  When the magnetic field induces an AC signal on the...

  • Page 470

    CHAPTER 10 THE TRANSFORMER446ANSWERSIn AC power calculations, you must use the rms values of current and voltage.A. IPVAinininrms=== .1212001B. VVTRVoutinrms555120524C. Iout 5 Iin (TR) 5 0.1 3 5 5 0.5 ArmsD. Pout 5 VinIout 5 24 3 0.5 5 12 watts (same as the power in)INSIDE THE TRANSFORMERIn addit...

  • Page 471

    TRANSFORMERS IN COMMUNICATIONS CIRCUITS447TRANSFORMERS IN COMMUNICATIONS CIRCUITS14  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...

  • Page 472

    CHAPTER 10 THE TRANSFORMER448ANSWER 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 conne...

  • Page 473

    TRANSFORMERS IN COMMUNICATIONS CIRCUITS449can determine the turns ratio for a transformer that matches impedances between a gen-erator and a load. In this way, the generator “sees” an impedance equal to its own imped-ance, and the load also “sees” an impedance equal to its own impedance.F...

  • Page 474

    CHAPTER 10 THE TRANSFORMER45010 VppZG10 kΩZL600Ω2.45 VFIGURE 10.12NOTE 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 p...

  • Page 475

    SUMMARY AND APPLICATIONS451 18  In this problem you use a transformer to match a generator to a 2 kΩ load.QUESTIONSA. What turns ratio is required to match a 2 kΩ load with a source that has an output impedance of 5 kΩ? B. If the load requires a power of 20 mW, what should the source be? (...

  • Page 476

    CHAPTER 10 THE TRANSFORMER452 ■ 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 signalSELFTESTThese questions test your understanding of the material in this chapter. Use...

  • Page 477

    SELFTEST4536. Vin 5 10 Vpp and Vout 5 7 Vpp, what is the turns ratio?TR57. In the center-tapped secondary winding shown in Figure 10.14, the voltage between points A and B may be expressed as VA-B 5 28 Vpp. What is the voltage between C and A? ACBFIGURE 10.148. In the center-tapped secondary w...

  • Page 478

    CHAPTER 10 THE TRANSFORMER454ANSWERS TO SELFTESTIf 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....

  • Page 479

    11Power Supply CircuitsA 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 be...

  • Page 480

    CHAPTER 11 POWER SUPPLY CIRCUITS456When 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 ...

  • Page 481

    DIODES IN AC CIRCUITS PRODUCE PULSATING DC4572  Figure 11.2 shows the circuit in Figure 11.1 with a 20 Vpp AC input signal centered at 120 volts DC.20 V+20 VFIGURE 11.2QUESTIONSA. What are the positive and negative peak voltages of the input signal? B. What is the output waveform of this circuit...

  • Page 482

    CHAPTER 11 POWER SUPPLY CIRCUITS458QUESTIONSA. 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. Output0VFIGURE 11.4ANSWERSA. Positive peak voltage is 110 volts. Ne...

  • Page 483

    DIODES IN AC CIRCUITS PRODUCE PULSATING DC459InputOutput+10 V−10 V0VFIGURE 11.6ANSWERSee Figure 11.7.+10 V0VFIGURE 11.75  Figure 11.7 shows the output waveform of the circuit shown in Figure 11.3, for one com-plete cycle of the input waveform.QUESTIONNow, draw the output waveform for three com...

  • Page 484

    CHAPTER 11 POWER SUPPLY CIRCUITS4606  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 inv...

  • Page 485

    DIODES IN AC CIRCUITS PRODUCE PULSATING DC461B. What is the output voltage? ANSWERSA. Never because the voltage that results from adding the AC and DC signals ranges from 210 volts to 230 volts. Therefore, the diode is always reverse-biased.B. A constant 0 volts.8  As you have seen, a diode pass...

  • Page 486

    CHAPTER 11 POWER SUPPLY CIRCUITS462QUESTIONSA. 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 o...

  • Page 487

    DIODES IN AC CIRCUITS PRODUCE PULSATING DC463C. 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.ANSWERSA. During the first half of the cycle, D1 is forward-biased and conducts current. D2 is reve...

  • Page 488

    CHAPTER 11 POWER SUPPLY CIRCUITS464QUESTIONOn 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 wave-forms at points C and D shown in Figure 11.14.)ANSWERSee Figure 11.16.ACBDEThis is c...

  • Page 489

    465DIODES IN AC CIRCUITS PRODUCE PULSATING DCQUESTIONHow does this circuit differ from the circuit shown in Figure 11.15? ANSWERThis 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 posi...

  • Page 490

    CHAPTER 11 POWER SUPPLY CIRCUITS466ANSWERSA. Two diodes in each case.B. See Figure 11.20.FIGURE 11.20PROJECT 11.1: The Full-Wave RectifierOBJECTIVEThe objective of this project is to compare the outputs of the two types of full-wave rectifiers. GENERAL INSTRUCTIONSYou set up two circuits. One of...

  • Page 491

    467DIODES IN AC CIRCUITS PRODUCE PULSATING DC❑ Two breadboards.❑ One function generator.❑ One oscilloscope. STEPBYSTEP INSTRUCTIONSSet 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 t...

  • Page 492

    CHAPTER 11 POWER SUPPLY CIRCUITS4681N40011N40011N40011N400110 kΩFunctiongeneratorFIGURE 11.226. Connect the oscilloscope probe for Channel 1 to a jumper wire connected to one end of the resistor. Connect the ground clip for Channel 1 to a jumper wire con-nected to the other end of the resistor.7...

  • Page 493

    469DIODES IN AC CIRCUITS PRODUCE PULSATING DCFigure 11.24 shows a function generator and oscilloscope attached to Circuit #1.Channel 2ground clipChannel1oscilloscopeprobeChannel1ground clipRed leadfrom functiongeneratorChannel 2oscilloscopeprobeBlack leadfrom functiongeneratorFIGURE 11.24In Figur...

  • Page 494

    CHAPTER 11 POWER SUPPLY CIRCUITS4704 divisionspeak-to-peak2.2 divisionsto peakFIGURE 11.25Time/div control setto 0.2 msec/divChannel 1 set to2 volts/divChannel1verticalposition knobChannel 2verticalposition knobChannel 2 set to5 volts/divFIGURE 11.26Book Authorc11V107/04/2012 3:27 PM c11.indd 4...

  • Page 495

    471DIODES IN AC CIRCUITS PRODUCE PULSATING DCFigure 11.27 shows the breadboarded bridge rectifier (Circuit # 2).1N40011N4001Transformer10kΩresistorFIGURE 11.27Figure 11.28 shows a function generator and oscilloscope attached to Circuit #2.In Figure 11.29, the signal across the primary coil is r...

  • Page 496

    CHAPTER 11 POWER SUPPLY CIRCUITS472Black leadfrom functiongeneratorRed leadfrom functiongeneratorChannel 2oscilloscopeprobeChannel 2group clipChannel1oscilloscopeprobeChannel1ground clipFIGURE 11.284 divisionspeak-to-peak1.75 divisionsto peakFIGURE 11.29Book Authorc11V107/04/2012 3:27 PM c11.indd...

  • Page 497

    473DIODES IN AC CIRCUITS PRODUCE PULSATING DCChannel 1 set to5 volts/divChannel 1 verticalposition knobTime/div control setto 0.2 msec/divChannel 2 verticalposition knobChannel 2 set to5 volts/divFIGURE 11.30Your values should be close to those shown in the following table.Circuit #Vpp (Primary C...

  • Page 498

    CHAPTER 11 POWER SUPPLY CIRCUITS474LEVEL DC SMOOTHING PULSATING DC14  A basic power supply circuit can be divided into four sections, as shown in Figure 11.31.RectifierSmoothing sectionTransformerLoadFIGURE 11.31QUESTIONSA. If you use a center-tap transformer in a power supply, how many di...

  • Page 499

    LEVEL DC SMOOTHING PULSATING DC4750VPulsating DCDC with AC rippleFIGURE 11.32The load (which is “driven” by the power supply) can be a simple lamp or a complex electronic circuit. Whatever load you use, it requires a certain voltage across its terminals and draws a current. Therefore, t...

  • Page 500

    CHAPTER 11 POWER SUPPLY CIRCUITS476QUESTIONSLook at the circuit shown in Figure 11.33 and answer the following questions:A. What type of secondary coil is used? B. What type of rectifier is used? C. What components make up the smoothing section? D. What output would you expect from the rectifie...

  • Page 501

    LEVEL DC SMOOTHING PULSATING DC477As the voltage level of the DC pulse rises to its peak, the capacitor C1 is charged to the peak voltage of the DC pulse.When the input DC pulse drops from its peak voltage back to 0 volts, the electrons stored on capacitor C1 discharge through the circuit. ...

  • Page 502

    CHAPTER 11 POWER SUPPLY CIRCUITS478However, if another pulse passes through the diode before the capacitor is discharged, the resulting waveform looks like that shown in Figure 11.38.C1 rechargesFIGURE 11.38The capacitor discharges only briefly before the second pulse recharges it to peak value....

  • Page 503

    LEVEL DC SMOOTHING PULSATING DC479by the capacitor that is primarily responsible for leveling out the pulsating DC. For that reason, this discussion uses the term smoothing section.QUESTIONEstimate the average DC output level of the waveform shown in Figure 11.39. ANSWERApproximately 90 p...

  • Page 504

    CHAPTER 11 POWER SUPPLY CIRCUITS480 20  Figure 11.41 shows the waveform after the diode has rectified the sine wave for the half-way rectifier circuit shown in Figure 11.40.40 V28 VFIGURE 11.41QUESTIONCalculate the duration of one pulse. ANSWER60 Hz represents 60 cycles (that is, wavelengths...

  • Page 505

    LEVEL DC SMOOTHING PULSATING DC4811036 1001001531()RTherefore, R1 5 260 ohms 22  Figure 11.42 shows the half-wave rectifier circuit with the 260-ohm value you calculated for R1.C1C2100 Ω260 ΩFIGURE 11.42Now, choose a value for C1 that produces a discharge time through the two resistors ...

  • Page 506

    CHAPTER 11 POWER SUPPLY CIRCUITS48223  Figure 11.43 shows voltage waveforms at various points in the half-wave rectifier circuit.BAC100Ω260Ω230μF40 V0VABC11.11 V8.89 V0V40 V32 V0VDC level of 36 VDC level of 10 VFIGURE 11.43QUESTIONSA. What happens to the DC output voltage between points B an...

  • Page 507

    LEVEL DC SMOOTHING PULSATING DC483 24  In most cases, the level of the AC ripple is still too high, and further smoothing is required. Figure 11.44 shows the portion of the half-wave rectifier circuit that forms a voltage divider using R1, and the parallel combination of RL and C2. This v...

  • Page 508

    CHAPTER 11 POWER SUPPLY CIRCUITS484 25  Figure 11.45 shows the half-wave rectifier circuit with all capacitor and resistor values.RL260 Ω100 Ω265 μF230 μFFIGURE 11.45Because XC2 is one-tenth of RL, you can ignore RL in AC voltage divider calculations. Figure 11.46 shows the resulting AC vol...

  • Page 509

    LEVEL DC SMOOTHING PULSATING DC485NOTE This result means that the addition of C2 lowers the AC ripple shown by curve C in Figure 11.43, with peak values of 11.11 and 8.89, to values of 10.155 and 9.845 volts. This represents a lower ripple at the output. Hence, C2 aids the smoothing of the ...

  • Page 510

    CHAPTER 11 POWER SUPPLY CIRCUITS486B. What is the average DC level for the waveform in Figure 11.48 given that Vp 5 40 volts? C. Why does a full-wave rectifier have a higher average DC level than a half-wave rec-tifier? ANSWERSA. 36 volts, which is 90 percent of Vp.B. 38 volts, which is 9...

  • Page 511

    LEVEL DC SMOOTHING PULSATING DC487QUESTIONCalculate the value of C1. ANSWERWith a time constant of τ 5 83.3 ms. and a discharge resistance of R1 1 RL 5 380 ohms, C1 5 220 μF. 29  You can use the voltage divider equation to find the amount of AC ripple across the load resistor for a fu...

  • Page 512

    CHAPTER 11 POWER SUPPLY CIRCUITS488QUESTIONCalculate the upper and lower values of the AC ripple at the output if you use a second capacitor with a value of 135 μF in parallel with the load resistor. You can use the same formulas as those for the half-wave rectifier in problem 25. XC2 is 10 Ω ...

  • Page 513

    LEVEL DC SMOOTHING PULSATING DC489B. Calculate the value of R1 required to make the DC level at the output 5 volts. C. Calculate the value of C1. D. Calculate the value of C2. E. What is the amount of AC ripple at the output? F. Draw the final circuit showing the calculated val...

  • Page 514

    CHAPTER 11 POWER SUPPLY CIRCUITS490SUMMARYThis chapter introduced the following concepts and calculations related to power supplies: ■ The effects of diodes on AC signals ■ Methods of rectifying an AC signal ■ Half-wave and full-wave rectifier circuit designs ■ The calculations you can ...

  • Page 515

    SELFTEST4913. See Figure 11.53.VoutFIGURE 11.534. See Figure 11.54.VoutFIGURE 11.545. See Figure 11.55.VoutFIGURE 11.556. In the circuit shown in Figure 11.56, 100 Vrms at 60 Hz appears at the secondary coil of the transformer; 28 volts DC with as little AC ripple as possible is required acros...

  • Page 516

    CHAPTER 11 POWER SUPPLY CIRCUITS492ANSWERS TO SELFTESTIf your answers do not agree with those given here, review the problems indicated in parentheses.1.See Figure 11.57.10 V0 VFIGURE 11.57 (problems 1–5)2.See Figure 11.58.45 V40 V35 VFIGURE 11.58 (problem 2)3.See Figure 11.59.FIGURE 11.59 (...

  • Page 517

    12Conclusion and Final Self-TestIn this book, you have discovered basic concepts and formulas that provide a foundation for your studiesin modern electronics, whether you become a dedicated hobbyist or study electrical or electronics engineering.CONCLUSIONHaving read this book, you should now kno...

  • Page 518

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST494pursue electronics to whatever depth and for whatever reason you want. Specifically, you should now be able to do the following: ■ Recognize all the important, discrete electronics components in a schematic diagram. ■ Understand how circuits that...

  • Page 519

    FINAL SELFTEST495FINAL SELFTESTThis final test allows you to assess your overall knowledge of electronics. Answers and review references follow the test. Use a separate sheet of paper for your calculations and drawings.1. If R5 1 MV and I5 2 μA, find the voltage. 2. If V5 5 volts and R5 ...

  • Page 520

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST49611. If R5 10 kV and C5 1 μF, find the time constant. 12. If R5 1 MV and C5 250 μF, find the time constant. 13. Three capacitors of 1 μF, 2 μF, and 3 μF are connected in parallel. Find the total capacitance. 14. Three capacitors of 100 μF, 220 ...

  • Page 521

    FINAL SELFTEST497VSR1R2VZFIGURE 12.221. For the circuit in Figure 12.2, VS 5 10 volts, R1 5 1 kV, R2 5 10 kV, and VZ 5 6.3 volts. Find IZ.22. Using the circuit shown in Figure 12.3, find the DC collector voltage, VC, if VS 528 volts, β5 10, RB 5 200 kV, and RC 5 10 kV.VSRBRCVCFIGURE 12.323. ...

  • Page 522

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST498VSRBRCVCFIGURE 12.427. Again, using the circuit shown in Figure 12.4, find the value of RB required to turn the transistor ON if VS 5 5 volts, RC 5 4.7 kV, and β5 100. 28. Using the circuit shown in Figure 12.5,find the values of R1, R2, and R3 tha...

  • Page 523

    FINAL SELFTEST49933. If Vpp 5 10 volts, find Vrms.34. If Vrms 5 120 volts, find Vpp.35. If the frequency of a sine wave is 14.5 kHz, what is the period of the waveform? 36. Find the reactance XC for a 200 μF capacitor when the frequency is 60 Hz. 37. Find the value of the capacitance that g...

  • Page 524

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST50043. Using the circuit shown in Figure 12.7, find XC, AC Vout, and DC Vout, if Vin 5 1 Vpp AC, riding on a 5-volt DC level; f5 10 kHz; R1 5 10 kV; R2 5 10 kV; and C5 0.2 μF. R1R2CVoutVin0VFIGURE 12.744. Again, using the circuit shown in Figure 12.7, ...

  • Page 525

    FINAL SELFTEST50147. In the circuit shown in Figure 12.10, L5 10 mH, C5 0.02 μF, and r5 7 ohms.Find fr, XL, XC, Z, Q, and the bandwidth. LrCFIGURE 12.1048. If the voltage across the resonant circuit of question 47 is at a peak value of 8 volts at the resonant frequency, what is the voltage at...

  • Page 526

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST50251. Using the circuit shown in Figure 12.11, how would you modify the amplifier in question 50 to obtain a maximum gain? Assume that the lowest frequency it has to pass is 50 Hz. 52. Using the JFET amplifier circuit shown in problem 42 of Chapter 8,...

  • Page 527

    FINAL SELFTEST50364. Name the two main coils used on a transformer. 65. What is the equation that shows the relationship between the input voltage, the output voltage, and the number of turns in each coil of a transformer? 66. What is the equation that shows the relationship between the turns ...

  • Page 528

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST504ANSWERS TO FINAL SELFTESTThe references in parentheses to the right of the answers give you the chapter and prob-lem number where the material is introduced so that you can easily review any concepts covered in the test.1.V5 2 volts(Chapter 1, prob...

  • Page 529

    FINAL SELFTEST50526.RB 5 500 kΩ(Chapter 4, problems 8)27.RB 5 470 kΩ(Chapter 4, problems 4–8)28.R3 5 44 kΩ, R1 5 2.2 kΩ, R2 5 2.2 kΩ(Chapter 4, problems 19–23)29.R3 5 2.2 kΩ, R1 5 66 kΩ, R2 5 66 kΩ(Chapter 4, problems 19–23)30.RD 5 2 kΩ(Chapter 4, problem 39)31.See Figu...

  • Page 530

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST50643.XC 5 80 ohms, AC Vout 5 8 mV, DC Vout 5 2.5 volts(Chapter 6, problem 26)44.XC 5 1.33 ohms, AC Vout 5 8.3 mV, DC Vout 5 2 volts(Chapter 6, problem 26)45.XL 5 62.8 ohms, Z5 89 ohms, AC Vout 5 6.07 volts, DC Vout 5 4.3 volts, tan θ5 1, θ5 45 degrees...

  • Page 531

    FINAL SELFTEST50758.A capacitive voltage divider(Chapter 9, problem 14)59.An inductive voltage divider(Chapter 9, problem 14)60.See Figure 12.15.VSCBREC1R1R2C2LFIGURE 12.15(Chapter 9, problem 24)61.See Figure 12.16.VSCBRER1R2CCCLFIGURE 12.16(Chapter 9, problem 25)62.f12LCr 5p(Chapter 9, proble...

  • Page 532

    CHAPTER 12 CONCLUSION AND FINAL SELFTEST50863.See Figure 12.17.PriSeccenter tapFIGURE 12.17(Chapter 10, problem 4)64.Primary and secondary(Chapter 10, problem 2)65.Vin/Vout 5 VP/VS 5 NP/NS 5 TR(Chapter 10, problem 6)66.Iout/Iin 5 IS/IP 5 NP/NS 5 TR(Chapter 10, problem 13)67.Zin/Zout 5 (NP/NS)2...

  • Page 533

    Appendix AGlossaryAmpere (A)The unit of measurement of electric current.AmplifierElectronic device or circuit that produces an output signal with greater power, voltage, or current than that provided by the input signal.Capacitance (C)The capability of a component to store an electric charge whe...

  • Page 534

    APPENDIX A GLOSSARY510Farad (F)The unit of measurement of capacitance.FeedbackA connection from the output of an amplifier back to the input. In some instances, a portion of the output voltage is used to control, stabilize, or modify the operation of the amplifier. However, in some instances, u...

  • Page 535

    APPENDIX A GLOSSARY511OscillatorAn electronic circuit that produces a continuous output signal such as a sine wave or square wave.Phase angleFor a signal, the angle of lead or lag between the current waveform and the voltage waveform, expressed in degrees.Phase shiftThe change in a phase of a sig...

  • Page 536

    APPENDIX A GLOSSARY512Transistor, JFETA junction field effect transistor (JFET), which, like the bipolar junc-tion transistor, can be used either as a switch or an amplifier.Transistor, MOSFETLike the BJT and JFET, a metal oxide silicon field effect transistor (MOSFET) that can be used eithe...

  • Page 537

    Appendix BList of Symbols and AbbreviationsThe following table lists common symbols and abbreviations.Symbol/AbbreviationMeaningAAmpereACAlternating currentAppPeak-to-peak amperesArmsRoot mean square amperesAVAC voltage gainβ (beta)Current gainBWBandwidthCCapacitorDCDirect currentFFaradgmTransc...

  • Page 538

    514APPENDIX B LIST OF SYMBOLS AND ABBREVIATIONSSymbol/AbbreviationMeaningfrResonant frequencyHHenryHzHertzIElectric currentIBBase currentICCollector currentIDDrain current of a field effect transistor (FET); also current through a diodeIDSSSaturation currentIppPeak-to-peak currentIrmsRoot mean ...

  • Page 539

    515APPENDIX B LIST OF SYMBOLS AND ABBREVIATIONSSymbol/AbbreviationMeaningrDC resistance of an inductorTPeriod of a waveformτTime constantTRTurns ratioθPhase angleμVMicrovoltVVoltageVCVoltage at the collector of a transistorVDDDrain supply voltageVEVoltage at the emitter of a transistorVGGGate ...

  • Page 540

    bapp02.indd 5166/14/2012 9:06:18 AM

  • Page 541

    Appendix CPowers of Ten and Engineering PrefixesThe following table shows powers of the number 10, decimal equivalents, prefixes used to denote the value, symbols used, and typical usages.PowerDecimalPrefixSymbolTypical Uses1091,000,000,000Giga-GGHz1061,000,000Mega-MMΩ, MHz, MV1031,000Kilo-kK...

  • Page 542

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  • Page 543

    Appendix DStandard Composition Resistor ValuesThe most commonly used type of resistor is the carbon film resistor with a ± 5 percent tolerance and either a 1/4 or 1/2 watt power rating. The standard resistance values for this type of resistor are listed in the following table (in ohms). You sho...

  • Page 544

    520APPENDIX D STANDARD COMPOSITION RESISTOR VALUES3.9434705.1 k56 k620 k4.3475105.6 k62 k680 k4.7515606.2 k68 k750 k5.1566206.8 k75 k820 k5.6626807.5 k82 k910 k6.2687508.2 k91 k1.0 M6.8758209.1 k100 k1.2 M7.58291010 k110 k1.5 M8.2911.0 k11 k120 k1.8 M9.11001.1 k12 k130 k2.2 M101101.2 k13 k150 k2....

  • Page 545

    Appendix ESupplemental ResourcesThis appendix provides a list of websites, books, magazines, tutorials, and electronics suppliers that should be of interest if you want more knowledge about basic electronics concepts, reference material for circuit design, or the supplies needed to build circuits...

  • Page 546

    522APPENDIX E SUPPLEMENTAL RESOURCES ■ All About Circuits actionURI(http://www.allaboutcircuits.com/)%E2%80%94):(www.allaboutcircuits.com/)—This site includes an online book on electronics theory and circuits, as well as discussion forums on electronics projects, microcontrollers, and general...

  • Page 547

    523SUPPLIERS ■ Nuts and Volts Magazine actionURI(http://www.nutsvolts.com/)%E2%80%94):(www.nutsvolts.com/)—This magazine provides information on new components for hobbyists and projects, focusing on circuits using microcontrollers. ■ EDN Magazine actionURI(http://www.edn.com/)%E2%80%94):(w...

  • Page 548

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  • Page 549

    Appendix FEquation ReferenceThe following table provides a quick reference to common equations.ParameterEquationChapter ReferenceBandwidthBWQ5frChapter 7, problem 20CapacitanceParallel CapafitanceCCC CT12N511 1...Chapter 1, SummarySeries Capacitance 1C1C1C...1C,orC11C1C1C...1C,orCCCCT12NT123NT1...

  • Page 550

    526APPENDIX F EQUATION REFERENCEParameterEquationChapter ReferenceFrequencyf1T5Chapter 5, problem 7Resonance Frequency (series LC circuit)fr12LC5pChapter 7, problem 6Resonance Frequency (parallel LC circuit)fr12LC1rCLfr12LCif Q1025p25p$if Q is less thanor10,,Chapter 7, problem 10GainVoltage Ga...

  • Page 551

    527APPENDIX F EQUATION REFERENCEParameterEquationChapter ReferenceResistanceParallel Resistance1111111111123123RRRRRorRRRRRorRTRTNTN511551155......,,11212RRRfor two resistorsChapter 1, SummarySeries ResistanceR R R RT12N511 1...Chapter 1, SummaryPowerPVIorPI RorPVR555,,22Chapter 1, SummaryReactan...

  • Page 552

    528APPENDIX F EQUATION REFERENCEParameterEquationChapter ReferenceVoltageOhm’s law (DC)V = IRChapter 1, SummaryOhm’s law (AC)V = IZChapter 6, problem 8Voltage dividerVVRRST115Chapter 1, SummaryPeak-to-Peak Volt-age (sine wave)VVVVppprmsrms55 33532222 828.Chapter 5, problem 4RMS Volt-age (sine...

  • Page 553

    Appendix GSchematic Symbols Used in This BookThe following table shows schematic symbols used in this book.ComponentSymbolBatteryBuzzerCapacitorsCapacitorVariable CapacitorContinuedbapp07.indd 5296/15/2012 5:42:48 PM

  • Page 554

    530APPENDIX G SCHEMATIC SYMBOLS USED IN THIS BOOKComponentSymbolDiodesDiodeZener DiodeLight Emitting DiodePhotodiodeGenerator (DC)GGroundInductorsInductorTapped InductorLamp(continued)bapp07.indd 5306/15/2012 5:42:49 PM

  • Page 555

    APPENDIX G SCHEMATIC SYMBOLS USED IN THIS BOOK531ComponentSymbolMetersMeterAmmeterAVoltmeterVMotorMOperational Amplifier−+ResistorsResistorTwo-Contact Variable Resis-tor (Potentiometer)Three-Contact Variable Resis-tor (Potentiometer)Signal Generator (Sine Wave)SwitchesSingle-pole, Single-Thro...

  • Page 556

    APPENDIX G SCHEMATIC SYMBOLS USED IN THIS BOOK532ComponentSymbolTest PointTransformersTransformerCenter Tap TransformerTransistorsNPN BJTPNP BJTN-Channel JFETP-Channel JFET(continued)bapp07.indd 5326/15/2012 5:42:51 PM

  • Page 557

    533IndexAA (amperes)calculating power, 12–14defined, 509microamperes, 15VA (volts × amperes) rating, 446abbreviations, 513–515AC (alternating current)answers, 209biasing. See biasingcapacitors and, 33capacitors in AC circuits, 195–197capacitors in AC filter circuits, 212–214diodes in AC ...

  • Page 558

    INDEX534bandwidth (BW). See BW (bandwidth)base amplifier, 390–391base currentalternative switching, 166–172biasing and, 322–323calculating from b, 108–109collector current and, 105current gain, 105–106defined, 97–99path of, 103–104in PNP transistor, 102relationships in amplifier cir...

  • Page 559

    INDEX535bridge rectifierdefined, 464project, 466–473Building Gadgets website, 494, 521burnout, diode, 72buzzersschematic symbol, 529transistor switch project, 150–154BW (bandwidth)equation, 525general resonance curve, 306output curves, 300–305for parallel LC circuits, 305resonant circuits, ...

  • Page 560

    INDEX536circuits (continued)switches, 30–32transformers in communications, 447–451transistors. See transistorsusing breadboards, 21–23closed switchescomparing bipolar junction transistor (BJT) to, 118–119comparing junction field effect transistor (JFET) to, 127comparing transistors to, 12...

  • Page 561

    INDEX537path of collector, 104–105preventing excessive, 61relationships in amplifier circuit, 332–333saturation, 178understanding diodes, 67–69using multimeter, 28–29using transistors as switches, 146in zener diodes, 77–78current divider, 24–27current flowdefined, 2–4, 41diodes in A...

  • Page 562

    INDEX538diodes (continued)breakdown, 70–72defined, 509finding current for, 67–69imperfect diodes, 61inside the diode, 72–74limiting voltage, 60–61overview, 47–48perfect diodes, 64–65power dissipation, 65–66project, 53–60schematic symbol, 530self-test, 87–89summary, 86–87transi...

  • Page 563

    INDEX539emitter follower, 354–355graph of resistance, 18impedance, 216, 230inductor impedance, 251inductor reactance, 251junction field effect transistor (JFET) amplifier voltage gain, 366–367Ohm’s law, 5–8op-amp voltage gain, 372output voltage, 230phase angle, 259power for resistors, 65r...

  • Page 564

    INDEX540filters (continued)defined, 510high-order, 257high-pass project, 222–228inductors in AC circuits, 250–256low-pass, 478low-pass project, 232–239notch project, 275–281overview, 211–212phase shift for resistors and inductors (RL) circuits, 258–259phase shift of resistors and capa...

  • Page 565

    INDEX541schematic symbol, 530ground bus, 22HH (henrys)defined, 510measuring inductance, 203half power frequencies, 288half power pointsbandwidth and, 291defined, 288measuring bandwidth, 298output curve, 289half-wave rectifier circuitsdefined, 461level DC, 484–486smoothing pulsating DC, 474volta...

  • Page 566

    INDEX542inductors (continued)in oscillators, 309–314practical oscillator design, 423resonance frequency, 205–206schematic symbol, 530in series, 416–421input currentefficiency of transformer, 445–446Kirchhoff’s Current Law, 26input resistancecommon base amplifier, 391defined, 328in emitt...

  • Page 567

    INDEX543labeling of inductors, 401–402lag the output voltage waveform, 241lampsschematic symbol, 530using transistors as switches, 146, 154–161zener diode voltage regulator, 82–86lawsKirchhoff’s, 510Kirchhoff’s Current Law, 26Kirchhoff’s Voltage Law, 21Ohm’s law, 5–8LC (inductance...

  • Page 568

    INDEX544meters, 531microampere (μA), 15microfarad (μF), 39microprocessors, 134milliampere (mA), 15mobile charge, 74mobile devices, 146motorin emitter follower circuit, 352–353schematic symbol, 531Mouser Electronics, 523multimetersinside, 27–29test setup for diode project, 58test setup for t...

  • Page 569

    INDEX545comparing junction field effect transistor (JFET) to, 127comparing transistors to, 129diodes and, 53oscillatorsArmstrong, 421–422Colpitts, 402–406, 412–414Colpitts oscillator project, 406–412defined, 506, 511feedback, 396–401Hartley, 414–416Hartley oscillator project, 416–42...

  • Page 570

    INDEX546output voltage (continued)phase shift of resistors and capacitors (RC) circuit, 239–241for resistors in series, 216resonant frequency, 291resistance, inductors, and capacitors (RLC) circuit, 271–272, 274–275smoothing pulsating DC, 482transformer basics, 439–440transistor amplifier...

  • Page 571

    INDEX547defined, 52–53pF (picofarad), 39phasedefined, 197negative feedback, 387positive feedback, 388transformers, 438phase angledefined, 511equation, 526for resistors and inductors (RL) circuits, 259phase difference, 241phase shiftdefined, 241, 511equation, 526of resistors and capacitors (RC) ...

  • Page 572

    INDEX548projects (continued)transistor amplifiers, 339–345transistor switch, 150–154transistors, 110–114zener diode voltage regulator, 82–86PRV (peak reverse voltage), 72pulsating DC (PDC), 456–466. See alsolevel DCpulse duration, 480QQ (quality factor)calculating bandwidth, 298–300de...

  • Page 573

    INDEX549full-wave rectifier circuit project, 466–473half-wave rectifier circuit. See half-wave rectifier circuitsresistance (R). See R (resistance)resistorsin AC circuits, 193–195alternative base switching, 169–170biasing, 322, 334–339calculating Q, 298–299creating stable amplifier, 330...

  • Page 574

    INDEX550RLC (resistors, inductors and capacitors) circuits, 268rms (root mean square) voltageequations, 528sine wave measurements, 190–191Ssafety power ratings, 65–66saturated transistorsdefined, 511project, 119–121saturationcurrent, 178defined, 117–118drain current, 362–363op-amp, 373p...

  • Page 575

    INDEX551alternative base switching, 166–167defined, 30–32schematic symbol, 532stable amplifiers, 330–334step-down transformers, 442step-up transformers, 442supplemental resources, 521–523suppliers, 523supply voltageturning off transistor, 143turning on transistor, 142using transistors as ...

  • Page 576

    INDEX552transfer curve (continued)switching junction field effect transistor (JFET), 178–179switching junction field effect transistor (JFET) project, 177transformer output voltage equation, 528transformersbasics, 436–446in communications circuits, 447–451defined, 511inside, 446overview, 43...

  • Page 577

    INDEX553Uunits of capacitance, 39units of resistance, 15unstable amplifiers, 330VV (volt), 512VA (volts × amperes) rating, 446vacuum tube, 47variable capacitors, 529variable resistors, 8V-I curve (characteristic curve)defined, 16–18diode project, 54–60reversed diode, 70switching junction fie...

  • Page 578

    INDEX554voltage gain (continued)transistor amplifiers, 328–329unstable amplifiers, 330voltmeter, 531volts × amperes (VA) rating, 446VOLTS/DIV control, 199–202WW (watt)calculating power, 12–14defined, 512waveformsinductors in AC circuits, 252–256oscillator, 309output. See output waveforms...