Introduction. Electronics

Transcription

1 R C R Introduction C to i b1 o1 o2 i b2 Electronics r π βi b1 βi b2 r π id /2 id /2 X An (β1)i b1 Online Text (β1)i b2 R EB Bob Zulinski R C R Associate Professor C od of Electrical Engineering i b1 o1 o2 i Michigan Technological Uniersity b2 id /2 r π βi b1 βi b2 r π id /2 X (β1)i b1 (β1)i b2 R EB Version 2.0

2 Introduction to Electronics ii Dedication Human beings are a delightful and complex amalgam of the spiritual, the emotional, the intellectual, and the physical. This is dedicated to all of them; especially to those who honor and nurture me with their friendship and loe.

3 Introduction to Electronics iii Table of Contents Preface xi Philosophy of an Online Text... xi Notes for Printing This Document... xiii Copyright Notice and Information... xiii Reiew of Linear Circuit Techniques 1 Resistors in Series... 1 Resistors in Parallel... 1 Product Oer Sum 1 Inerse of Inerses 1 Ideal Voltage Sources... 2 Ideal Current Sources... 2 Real Sources... 2 Voltage Diiders... 3 Current Diiders... 4 Superposition... 4 A quick exercise 4 What s missing from this reiew???... 5 You ll still need Ohm s and Kirchoff s Laws 5 Basic Amplifier Concepts 6 Signal Source... 6 Amplifier... 6 Load... 7 Ground Terminal... 7 To work with (analyze and design) amplifiers... 7 Voltage Amplifier Model 8 Signal Source... 8 Amplifier Input... 8 Amplifier Output... 8 Load... 8 OpenCircuit Voltage Gain... 9 Voltage Gain... 9 Current Gain Power Gain... 10

4 Introduction to Electronics i Power Supplies, Power Conseration, and Efficiency 11 DC Input Power Conseration of Power Efficiency Amplifier Cascades 13 Decibel Notation 14 Power Gain Cascaded Amplifiers Voltage Gain Current Gain Using Decibels to Indicate Specific Magnitudes Voltage leels: 15 Power leels 16 Other Amplifier Models 17 Current Amplifier Model Transconductance Amplifier Model Transresistance Amplifier Model Amplifier Resistances and Ideal Amplifiers 20 Ideal Voltage Amplifier Ideal Current Amplifier Ideal Transconductance Amplifier Ideal Transresistance Amplifier Uniqueness of Ideal Amplifiers Frequency Response of Amplifiers 24 Terms and Definitions Magnitude Response 24 Phase Response 24 Frequency Response 24 Amplifier Gain 24 The Magnitude Response Causes of Reduced Gain at Higher Frequencies Causes of Reduced Gain at Lower Frequencies... 26

5 Introduction to Electronics Differential Amplifiers 27 Example: 27 Modeling Differential and CommonMode Signals Amplifying Differential and CommonMode Signals CommonMode Rejection Ratio Ideal Operational Amplifiers 29 Ideal Operational Amplifier Operation Op Amp Operation with Negatie Feedback Slew Rate Op Amp Circuits The Inerting Amplifier 31 Voltage Gain Input Resistance Output Resistance Op Amp Circuits The Noninerting Amplifier 33 Voltage Gain Input and Output Resistance Op Amp Circuits The Voltage Follower 34 Voltage Gain Input and Output Resistance Op Amp Circuits The Inerting Summer 35 Voltage Gain Op Amp Circuits Another Inerting Amplifier 36 Voltage Gain Op Amp Circuits Differential Amplifier 38 Voltage Gain Op Amp Circuits Integrators and Differentiators 40 The Integrator The Differentiator... 41

6 Introduction to Electronics i Op Amp Circuits Designing with Real Op Amps 42 Resistor Values Source Resistance and Resistor Tolerances Graphical Solution of Simultaneous Equations 43 Diodes 46 Graphical Analysis of Diode Circuits 48 Examples of LoadLine Analysis Diode Models 50 The Shockley Equation Forward Bias Approximation 51 Reerse Bias Approximation 51 At High Currents 51 The Ideal Diode An Ideal Diode Example 53 PiecewiseLinear Diode Models A PiecewiseLinear Diode Example 57 Other PiecewiseLinear Models Diode Applications The Zener Diode Voltage Regulator 59 Introduction LoadLine Analysis of Zener Regulators Numerical Analysis of Zener Regulators Circuit Analysis 62 Zener Regulators with Attached Load Example Graphical Analysis of Loaded Regulator 64 Diode Applications The HalfWae Rectifier 66 Introduction A Typical Battery Charging Circuit The Filtered HalfWae Rectifier Relating Capacitance to Ripple Voltage 70

7 Introduction to Electronics ii Diode Applications The FullWae Rectifier 72 Operation st (Positie) HalfCycle 72 2 nd (Negatie) HalfCycle 72 Diode Peak Inerse Voltage Diode Applications The Bridge Rectifier 74 Operation st (Positie) HalfCycle 74 2 nd (Negatie) HalfCycle 74 Peak Inerse Voltage Diode Applications FullWae/Bridge Rectifier Features 75 Bridge Rectifier FullWae Rectifier Filtered FullWae and Bridge Rectifiers Bipolar Junction Transistors (BJTs) 76 Introduction Qualitatie Description of BJT ActieRegion Operation Quantitatie Description of BJT ActieRegion Operation BJT CommonEmitter Characteristics 80 Introduction Input Characteristic Output Characteristics Actie Region 81 Cutoff 82 Saturation 82 The pnp BJT 83 BJT Characteristics Secondary Effects 85

8 Introduction to Electronics iii The nchannel Junction FET (JFET) 86 Description of Operation Equations Goerning nchannel JFET Operation Cutoff Region 89 Triode Region 89 PinchOff Region 89 The Triode PinchOff Boundary The Transfer Characteristic MetalOxideSemiconductor FETs (MOSFETs) 92 The nchannel Depletion MOSFET The nchannel Enhancement MOSFET Comparison of nchannel FETs 94 pchannel JFETs and MOSFETs 96 Cutoff Region 98 Triode Region 98 PinchOff Region 98 Other FET Considerations 99 FET Gate Protection The Body Terminal Basic BJT Amplifier Structure 100 Circuit Diagram and Equations LoadLine Analysis Input Side LoadLine Analysis Output Side A Numerical Example Basic FET Amplifier Structure 107 Amplifier Distortion 110 Biasing and Bias Stability 112

9 Introduction to Electronics ix Biasing BJTs The Fixed Bias Circuit 113 Example For b = For b = Biasing BJTs The Constant Base Bias Circuit 114 Example For b = For b = Biasing BJTs The FourResistor Bias Circuit 115 Introduction Circuit Analysis Bias Stability To maximize bias stability 117 Example For b = 100 (and V BE = 0.7 V) 118 For b = Biasing FETs The Fixed Bias Circuit 119 Biasing FETs The Self Bias Circuit 120 Biasing FETs The Fixed Self Bias Circuit 121 Design of Discrete BJT Bias Circuits 123 Concepts of Biasing Design of the FourResistor BJT Bias Circuit Design Procedure 124 Design of the DualSupply BJT Bias Circuit Design Procedure 125 Design of the GroundedEmitter BJT Bias Circuit Design Procedure 126 Analysis of the GroundedEmitter BJT Bias Circuit

10 Introduction to Electronics x Bipolar IC Bias Circuits 129 Introduction The DiodeBiased Current Mirror Current Ratio 130 Reference Current 131 Output Resistance 131 Compliance Range Using a Mirror to Bias an Amplifier Wilson Current Mirror Current Ratio 133 Reference Current 134 Output Resistance 134 Widlar Current Mirror Current Relationship 135 Multiple Current Mirrors FET Current Mirrors Linear SmallSignal Equialent Circuits 138 Diode SmallSignal Equialent Circuit 139 The Concept The Equations Diode SmallSignal Resistance Notation 142 BJT SmallSignal Equialent Circuit 143 The CommonEmitter Amplifier 145 Introduction Constructing the SmallSignal Equialent Circuit Voltage Gain Input Resistance Output Resistance

11 Introduction to Electronics xi The Emitter Follower (Common Collector Amplifier) 149 Introduction Voltage Gain Input Resistance Output Resistance Reiew of Small Signal Analysis 153 FET SmallSignal Equialent Circuit 154 The SmallSignal Equialent Transconductance FET Output Resistance The Common Source Amplifier 157 The SmallSignal Equialent Circuit Voltage Gain Input Resistance Output Resistance The Source Follower 159 SmallSignal Equialent Circuit Voltage Gain Input Resistance Output Resistance Reiew of Bode Plots 164 Introduction The Bode Magnitude Response The Bode Phase Response SinglePole LowPass RC Gain Magnitude in db 167 Bode Magnitude Plot 168 Bode Phase Plot 169 SinglePole HighPass RC Bode Magnitude Plot 170 Bode Phase Plot 171

12 Introduction to Electronics xii Coupling Capacitors 172 Effect on Frequency Response Constructing the Bode Magnitude Plot for an Amplifier Design Considerations for RCCoupled Amplifiers 175 Low & MidFrequency Performance of CE Amplifier 176 Introduction Midband Performance Design Considerations The Effect of the Coupling Capacitors The Effect of the Emitter Bypass Capacitor C E The Miller Effect 183 Introduction Deriing the Equations The Hybridp BJT Model 185 The Model Effect of C p and C m HighFrequency Performance of CE Amplifier 189 The SmallSignal Equialent Circuit HighFrequency Performance The CE Amplifier Magnitude Response Nonideal Operational Amplifiers 193 Linear Imperfections Input and Output Impedance 193 Gain and Bandwidth 193 Nonlinear Imperfections Output Voltage Swing 194 Output Current Limits 194 SlewRate Limiting 194 FullPower Bandwidth 195

13 Introduction to Electronics xiii DC Imperfections Input Offset Voltage, V IO 195 Input Currents 195 Modeling the DC Imperfections Using the DC Error Model DC Output Error Example Finding WorstCase DC Output Error 201 Canceling the Effect of the Bias Currents Instrumentation Amplifier 204 Introduction Simplified Analysis Noise 206 Johnson Noise Johnson Noise Model 207 Shot Noise /f Noise (Flicker Noise) Other mechanisms producing 1/f noise 209 Interference Amplifier Noise Performance 211 Terms, Definitions, Conentions Amplifier Noise Voltage 211 Amplifier Noise Current 212 SignaltoNoise Ratio 212 Noise Figure 213 Noise Temperature 213 Conerting NF to/from T n 214 Adding and Subtracting Uncorrelated Quantities Amplifier Noise Calculations 215 Introduction Calculating Noise Figure Typical Manufacturer s Noise Data 217 Introduction Example # Example #

14 Introduction to Electronics xi Noise References and Credits 220 Introduction to Logic Gates 221 The Inerter The Ideal Case 221 The Actual Case 221 Manufacturer s Voltage Specifications Noise Margin Manufacturer s Current Specifications FanOut Power Consumption Static Power Consumption 224 Dynamic Power Consumption 224 Rise Time, Fall Time, and Propagation Delay SpeedPower Product TTL Logic Families & Characteristics CMOS Logic Families & Characteristics MOSFET Logic Inerters 230 NMOS Inerter with Resistie PullUp Circuit Operation 230 Drawbacks 231 CMOS Inerter Circuit Operation 232 Differential Amplifier 239 Modeling Differential and CommonMode Signals Basic Differential Amplifier Circuit Case #1 CommonMode Input 240 Case #2A Differential Input 241 Case #2B Differential Input 241 LargeSignal Analysis of Differential Amplifier 242

15 Introduction to Electronics x SmallSignal Analysis of Differential Amplifier 246 Differential Input Only Analysis of Differential HalfCircuit Differential Input Resistance 250 Differential Output Resistance 250 CommonMode Input Only Analysis of CommonMode HalfCircuit Commonmode input resistance 253 Commonmode output resistance 253 CommonMode Rejection Ratio

16 Introduction to Electronics xi Preface Philosophy of an Online Text I think of myself as an educator rather than an engineer. And it has long seemed to me that, as educators, we should endeaor to bring to the student not only as much information as possible, but we should strie to make that information as accessible as possible, and as inexpensie as possible. The technology of the Internet and the World Wide Web now allows us to irtually gie away knowledge! Yet, we don t, choosing instead to write another conentional text book, and print, sell, and use it in the conentional manner. The whys are undoubtedly intricate and many; I offer only a few obserations: Any change is difficult and resisted. This is true in the habits we form, the tasks we perform, the relationships we engage. It is simply easier not to change than it is to change. Though change is ineitable, it is not wellsuited to the behaior of any organism. The proper reward structure is not in place. Faculty are supposedly rewarded for writing textbooks, thereby bringing fame and immortality to the institution of their employ. 1 The recognition and reward structure are simply not there for a text that is simply posted on the web. No economic incentie exists to create and maintain a 1 I use the word supposedly because, in my iew, the official rewards for textbook authoring fall far short of what is appropriate and what is achieable through an equialent research effort, despite all the administratie lip serice to the contrary. These arguments, though, are more appropriately left to a different soapbox.

17 Introduction to Electronics xii structure that allows all authors to publish in this manner; that allows students easy access to all such material, and that rigorously ensures the material will exceed a minimum acceptable quality. If I were to do this the way I think it ought to be done, I would hae prepared the course material in two formats. The first would be a text, identical to the textbooks with which you are familiar, but aailable online, and intended to be used in printed form. The second would be a slide presentation, à la Corel Presentations or Microsoft PowerPoint, intended for use in the classroom or in an independent study. But, alas, I am still on that journey, so what I offer you is a hybrid of these two concepts: an online text somewhat less erbose than a conentional text, but one that can also sere as classroom oerhead transparencies. Other compromises hae been made. It would be adantageous to produce two online ersions one intended for use in printed form, and a second optimized for iewing on a computer screen. The two would carry identical information, but would be formatted with different page and font sizes. Also, to minimize file size, and therefore download times, font selection and ariations are somewhat limited when compared to those normally encountered in a conentional textbook. You may also note that exercise problems are not included with this text. By their ery nature problems quickly can become worn out. I beliee it is best to include problems in a separate document. Until all of these enhancements exist, I hope you will find this a suitable and worthwhile compromise. Enough of this; let s get on with it...

18 Introduction to Electronics xiii Notes for Printing This Document This document can be printed directly from the Acrobat Reader see the Acrobat Reader help files for details. If you wish to print the entire document, do so in two sections, as most printer driers will only spool a maximum of 255 pages at one time. Copyright Notice and Information This entire document is 1999 by Bob Zulinski. All rights resered. I copyrighted this online text because it required a lot of work, and because I hold a faint hope that I may use it to acquire immeasurable wealth, thereby supporting the insatiable, salacious lifestyle that I e always dreamed of. Thus, you will need my permission to print it. You may obtain that permission simply by asking: tell me who you are and what you want it for. Route your requests ia to rzulinsk@mtu.edu, or by USPS mail to Bob Zulinski, Dept. of Electrical Engineering, Michigan Technological Uniersity, Houghton MI Generous monetary donations included with your request will be looked upon with great faor.

19 Reiew of Linear Circuit Techniques Introduction to Electronics 1 Reiew of Linear Circuit Techniques R 1 R 2 Fig. 1. R s in series. Resistors in Series This is the simple one!!! Rtotal = R1 R2 R3 (1) Resistors must carry the same current!!! L s is series and C s in parallel hae same form. R 1 R 2 Resistors in Parallel Resistors must hae the same oltage!!! Equation takes either of two forms: Fig. 2. R s in parallel. Product Oer Sum: R total = RR 1 2 R R 1 2 (2) Only alid for two resistors. Not calculatorefficient!!! Inerse of Inerses: R total = R R R (3) Always alid for multiple resistors. Very calculatorefficient!!! L s in parallel and C s in series hae same forms.

20 Reiew of Linear Circuit Techniques Introduction to Electronics 2 3 V 5 V Fig. 3. Ideal oltage sources in parallel??? Ideal Voltage Sources Cannot be connected in parallel!!! Real oltage sources include a series resistance ( Theenin equialent ), and can be paralleled. Ideal Current Sources Cannot be connected in series!!! Real current sources include a parallel resistance ( Norton equialent ), and can be connected in series. Fig. 4. Ideal current sources in series??? I SC i 1/R TH V OC Fig. 5. Typical linear i characteristic of a real source. Real Sources All sources we obsere in nature exhibit a decreasing oltage as they supply increasing current. We presume that i relationship to be linear, so we can write the equations: = VOC irth i = ISC R or (4) TH

21 Reiew of Linear Circuit Techniques Introduction to Electronics 3 The linear equations help us isualize what might be inside of a real source: i i R TH V OC = V TH I SC R TH Note that: R TH V I OC = (5) SC Fig. 6. Theenin equialent circuit. Fig. 7. Norton equialent circuit. We can generalize this any linear resistie circuit can be represented as in Figs. 6 and 7. V X R A R B R C Fig. 8. Example of a oltage diider. V A V B V C Voltage Diiders Example finding the oltage across R B : V B RB = R R R V A B C Resistors must be in series, i.e., they must carry the same current!!! X (6) (Sometimes we cheat a little, and use the diider equation if the currents through the resistors are almost the same we ll note this in class if that is the case)

22 Reiew of Linear Circuit Techniques Introduction to Electronics 4 I X I B R A R B R C Fig. 9. Example of a current diider. Current Diiders I B 1 RB = R R R A B C I X (7) Resistors must be in parallel, i.e., hae the same oltage!!! Superposition Superposition applies to any linear circuit in fact, this is the definition of a linear circuit!!! An example of finding a response using superposition: I I A I B Fig. 10. The total response current I... Fig is the sum of the response I A... Fig and the response I B... A quick exercise: Use superposition and oltage diision to show that V X = 6 V: 10 kω 30 kω 4 V 12 V V X Fig. 13. A quick exercise...

23 Reiew of Linear Circuit Techniques Introduction to Electronics 5 What s missing from this reiew??? Node oltages / mesh currents... For the kinds of problems you ll encounter in this course, I think you should forget about these analysis methods!!! If there is any other way to sole a circuit problem, do it that other way... you ll arrie at the answer more efficiently, and with more insight. You ll still need Ohm s and Kirchoff s Laws: KVL: KCL: Sum of oltages around a closed loop is zero. We ll more often use a different form: Sum of oltages from point A to point B is the same regardless of the path taken. Sum of currents into a node (or area) is zero. I won t insult you by repeating Ohm s Law here...

24 Basic Amplifier Concepts Introduction to Electronics 6 Basic Amplifier Concepts Signal Source i (t) Amplifier o (t) Ground Load Signal Source Fig. 14. Block diagram of basic amplifier. A signal source is anything that proides the signal, e.g., the carbon microphone in a telephone handset the fuelleel sensor in an automobile gas tank... Amplifier An amplifier is a system that proides gain sometimes oltage gain (illustrated below), sometimes current gain, always power gain. i o o t t t Fig. 15. Generic input signal oltage. Fig. 16. Output oltage of noninerting amplifier. Fig. 17. Output oltage of inerting amplifier.

25 Basic Amplifier Concepts Introduction to Electronics 7 Signal Source i (t) Amplifier o (t) Ground Load Fig. 18. Block diagram of basic amplifier (Fig. 14 repeated). Load The load is anything we delier the amplified signal to, e.g., loudspeaker the leg of lamb in a microwae oen... Ground Terminal Usually there is a ground connection usually common to input and output maybe connected to a metal chassis maybe connected to powerline ground maybe connected to both maybe connected to neither... use caution!!! To work with (analyze and design) amplifiers we need to isualize what might be inside all three blocks of Fig. 18, i.e., we need models!!!

26 Voltage Amplifier Model Introduction to Electronics 8 Voltage Amplifier Model This is usually the one we hae the most intuition about... R S i i R o i o s i R A oc i i o R L Source Amplifier Load Fig. 19. Modeling the source, amplifier, and load with the emphasis on oltage. Signal Source Our emphasis is oltage... source oltage decreases as source current increases, as with any real source so we use a Theenin equialent. Amplifier Input When the source is connected to the amplifier, current flows the amplifier must hae an input resistance, R i. Amplifier Output Output oltage decreases as load current increases again we use a Theenin equialent. Load Load current flows... the load appears as a resistance, R L.

27 Voltage Amplifier Model Introduction to Electronics 9 R S i i R o i o s i R A oc i i o R L Source Amplifier Load Fig. 20. Voltage amplifier model (Fig. 19 repeated). OpenCircuit Voltage Gain If we remoe R L (i.e., with R L = ) the oltage of the Theenin source in the amplifier output is the opencircuit output oltage of the amplifier. Thus, A oc is called the opencircuit oltage gain: A oc = o i R L = (8) Voltage Gain With a load in place our concept of oltage gain changes slightly: A V o = o = i RL R R A A A RL oc i = oc R R o L o L (9) We can think of this as the amplifier oltage gain if the source were ideal: i i R o i o i i R A oc i i o R L Amplifier Fig. 21. A = o / i illustrated. Load

28 Voltage Amplifier Model Introduction to Electronics 10 R S i i R o i o s i R A oc i i o R L Source Amplifier Load Fig. 22. Voltage amplifier model (Fig. 19 repeated). With our real source model we define another useful oltage gain: A s o = i = s Ri R R A A Ri RL s s = oc R R R R S i S i o L (10) Notice that A and A s are both less than A oc, due to loading effects. Current Gain We can also define the amplifier current gain: A i o i R R A R o L o i i = = = = (11) i i i i RL RL R i Power Gain Because the amplifier input and load are resistances, we hae P o = V o I o, and P i = V i I i (rms alues). Thus: G P VI AA A R o o o i = = = i = = Ai P VI R i i i 2 2 L R R L i (12)

29 Power Supplies, Power Conseration, and Efficiency Introduction to Electronics 11 Power Supplies, Power Conseration, and Efficiency I A V AA V AA R S i i R o i o s i R A oc i i o R L Source Amplifier Load I B V BB V BB Fig. 23. Our oltage amplifier model showing power supply and ground connections. The signal power deliered to the load is conerted from the dc power proided by the power supplies. DC Input Power PS = VAAIA VBBIB (13) This is sometimes noted as P IN. Use care not to confuse this with the signal input power P i. Conseration of Power Signal power is deliered to the load Power is dissipated within the amplifier as heat The total input power must equal the total output power: P o P D PS Pi = Po PD (14) Virtually always P i << P S and is neglected.

30 Power Supplies, Power Conseration, and Efficiency Introduction to Electronics 12 I A V AA V AA R S i i R o i o s i R A oc i i o R L Source Amplifier Load I B V BB V BB Fig. 24. Our oltage amplifier model showing power supply and ground connections (Fig. 23 repeated). Efficiency Efficiency is a figure of merit describing amplifier performance: Po η = P S 100% (15)

31 Amplifier Cascades Introduction to Electronics 13 Amplifier Cascades Amplifier stages may be connected together (cascaded) : i i1 R o1 i i2 R o2 i o2 i1 R i1 A oc1 i1 o1 = i2 R i2 A oc2 i2 o2 Amplifier 1 Amplifier 2 Fig. 25. A twoamplifier cascade. Notice that stage 1 is loaded by the input resistance of stage 2. Gain of stage 1: A 1 o1 = (16) i1 Gain of stage 2: A 2 o2 o2 = = (17) i 2 o1 Gain of cascade: A oc o1 o2 = = i1 o1 A A 1 2 (18) We can replace the two models by a single model (remember, the model is just a isualization of what might be inside): i i1 R o2 i o2 i1 R i1 A oc i1 o2 Fig. 26. Model of cascade.

32 Decibel Notation Introduction to Electronics 14 Decibel Notation Amplifier gains are often not expressed as simple ratios... rather they are mapped into a logarithmic scale. The fundamental definition begins with a power ratio. Power Gain Recall that G = P o /P i, and define: GdB = 10log G (19) G db is expressed in units of decibels, abbreiated db. Cascaded Amplifiers We know that G total = G 1 G 2. Thus: G = 10logGG = 10logG 10logG = G G total, db , db 2, db (20) Thus, the product of gains becomes the sum of gains in decibels. Voltage Gain To derie the expression for oltage gain in decibels, we begin by recalling from eq. (12) that G = A 2 (R i /R L ). Thus: 10logG= 10log A R R 2 i L 2 i L = 10log A 10logR 10logR (21) = 20log A 10logR 10logR i L

33 Decibel Notation Introduction to Electronics 15 Een though R i may not equal R L in most cases, we define: A db = 20log A (22) Only when R i does equal R L, will the numerical alues of G db and A db be the same. In all other cases they will differ. From eq. (22) we can see that in an amplifier cascade the product of oltage gains becomes the sum of oltage gains in decibels. Current Gain In a manner similar to the preceding oltagegain deriation, we can arrie at a similar definition for current gain: A idb = 20log A (23) i Using Decibels to Indicate Specific Magnitudes Decibels are defined in terms of ratios, but are often used to indicate a specific magnitude of oltage or power. This is done by defining a reference and referring to it in the units notation: Voltage leels: dbv, decibels with respect to 1 V... for example, log. V V= 1 V = 10 dbv (24)

34 Decibel Notation Introduction to Electronics 16 Power leels: dbm, decibels with respect to 1 mw... for example 5 mw 5 mw = 10log 699. dbm 1 mw = (25) dbw, decibels with respect to 1 W... for example 5 mw = 10log 5 mw = 230. dbw (26) 1 W There is a 30 db difference between the two preious examples because 1 mw = 30 dbw and 1 W = 30 dbm.

35 Other Amplifier Models Introduction to Electronics 17 Other Amplifier Models Recall, our oltage amplifier model arose from our isualization of what might be inside a real amplifier: R S i i R o i o s i R A oc i i o R L Source Amplifier Load Fig. 27. Modeling the source, amplifier, and load with the emphasis on oltage (Fig. 19 repeated). Current Amplifier Model Suppose we choose to emphasize current. In this case we use Norton equialents for the signal source and the amplifier: i i i o i s R S i R i R o o R L A isc i i Source Current Amplifier Load Fig. 28. Modeling the source, amplifier, and load with the emphasis on current. The shortcircuit current gain is gien by: A isc = i i o i R L = 0 (27)

36 Other Amplifier Models Introduction to Electronics 18 Transconductance Amplifier Model Or, we could emphasize input oltage and output current: R S i i i o s i R i G msc i R o o R L Source Transconductance Amplifier Load Fig. 29. The transconductance amplifier model. The shortcircuit transconductance gain is gien by: G msc = i o i R L = 0 (siemens, S) (28) Transresistance Amplifier Model Our last choice emphasizes input current and output oltage: i i R o i o i s R S i R R moc i i i o R L Source Transresistance Amplifier Load Fig. 30. The transresistance amplifier model. The opencircuit transresistance gain is gien by: R moc = i o i R L = (ohms, Ω ) (29)

37 Other Amplifier Models Introduction to Electronics 19 Any of these four models can be used to represent what might be inside of a real amplifier. Any of the four can be used to model the same amplifier!!! Models obiously will be different inside the amplifier. If the model parameters are chosen properly, they will behae identically at the amplifier terminals!!! We can change from any kind of model to any other kind: Change Norton equialent to Theenin equialent (if necessary). Change the dependent source s ariable of dependency with Ohm s Law i = i i R i (if necessary). Try it!!! Pick some alues and practice!!!

38 Amplifier Resistances and Ideal Amplifiers Introduction to Electronics 20 Amplifier Resistances and Ideal Amplifiers Ideal Voltage Amplifier Let s reisit our oltage amplifier model: R S i i R o i o s i R A oc i i o R L Source Voltage Amplifier Load Fig. 31. Voltage amplifier model. We re thinking oltage, and we re thinking amplifier... so how can we maximize the oltage that gets deliered to the load? We can get the most oltage out of the signal source if R i >> R S, i.e., if the amplifier can measure the signal oltage with a high input resistance, like a oltmeter does. In fact, if R i, we won t hae to worry about the alue of R S at all!!! We can get the most oltage out of the amplifier if R o << R L, i.e., if the amplifier can look as much like a oltage source as possible. In fact, if at all!!! R o 0, we won t hae to worry about the alue of R L So, in an ideal world, we could hae an ideal amplifier!!!

39 Amplifier Resistances and Ideal Amplifiers Introduction to Electronics 21 i A oc i Fig. 32. Ideal oltage amplifier. Signal source and load are omitted for clarity. An ideal amplifier is only a concept; we cannot build one. But an amplifier may approach the ideal, and we may use the model, if only for its simplicity. Ideal Current Amplifier Now let s reisit our current amplifier model: i i i o i s R S i R i R o o R L A isc i i Source Current Amplifier Load Fig. 33. Current amplifier model (Fig. 28 repeated). How can we maximize the current that gets deliered to the load? We can get the most current out of the signal source if R i << R S, i.e., if the amplifier can measure the signal current with a low input resistance, like an ammeter does. In fact, if at all!!! R i 0, we won t hae to worry about the alue of R S

40 Amplifier Resistances and Ideal Amplifiers Introduction to Electronics 22 We can get the most current out of the amplifier if R o >> R L, i.e., if the amplifier can look as much like a current source as possible. In fact, if R o, we won t hae to worry about the alue of R L at all!!! This leads us to our conceptual ideal current amplifier: i i A isc i i Fig. 34. Ideal current amplifier. Ideal Transconductance Amplifier With a mixture of the preious concepts we can conceptualize an ideal transconductance amplifier. This amplifier ideally measures the input oltage and produces an output current: i G msc i Fig. 35. Ideal transconductance amplifier.

41 Amplifier Resistances and Ideal Amplifiers Introduction to Electronics 23 Ideal Transresistance Amplifier Our final ideal amplifier concept measures input current and produces an output oltage: i i R moc i i Fig. 36. Ideal transresistance amplifier. Uniqueness of Ideal Amplifiers Unlike our models of real amplifiers, ideal amplifier models cannot be conerted from one type to another (try it...).

42 Frequency Response of Amplifiers Introduction to Electronics 24 Frequency Response of Amplifiers Terms and Definitions In real amplifiers, gain changes with frequency... Frequency implies sinusoidal excitation which, in turn, implies phasors... using oltage gain to illustrate the general case: A Vo = = V i Vo Vo A A V V = (30) i i Both A and A are functions of frequency and can be plotted. Magnitude Response: A plot of A s. f is called the magnitude response of the amplifier. Phase Response: A plot of A s. f is called the phase response of the amplifier. Frequency Response: Taken together the two responses are called the frequency response... though often in common usage the term frequency response is used to mean only the magnitude response. Amplifier Gain: The gain of an amplifier usually refers only to the magnitudes: A db = 20log A (31)

43 Frequency Response of Amplifiers Introduction to Electronics 25 The Magnitude Response Much terminology and measures of amplifier performance are deried from the magnitude response... A db A mid db midband region 3 db Bandwidth, B f H f (log scale) Fig. 37. Magnitude response of a dccoupled, or directcoupled amplifier. A db A mid db midband region 3 db f L Bandwidth, B f H f (log scale) Fig. 38. Magnitude response of an accoupled, or RCcoupled amplifier. A mid db is called the midband gain... f L and f H are the 3dB frequencies, the corner frequencies, or the halfpower frequencies (why this last one?)... B is the 3dB bandwidth, the halfpower bandwidth, or simply the bandwidth (of the midband region)...

44 Frequency Response of Amplifiers Introduction to Electronics 26 Causes of Reduced Gain at Higher Frequencies Stray wiring inductances... Stray capacitances... Capacitances in the amplifying deices (not yet included in our amplifier models)... The figure immediately below proides an example: Fig. 39. Twostage amplifier model including stray wiring inductance and stray capacitance between stages. These effects are also found within each amplifier stage. Causes of Reduced Gain at Lower Frequencies This decrease is due to capacitors placed between amplifier stages (in RCcoupled or capacitielycoupled amplifiers)... This preents dc oltages in one stage from affecting the next. Signal source and load are often coupled in this manner also. Fig. 40. Twostage amplifier model showing capacitie coupling between stages.

45 Differential Amplifiers Introduction to Electronics 27 Differential Amplifiers Many desired signals are weak, differential signals in the presence of much stronger, commonmode signals. Example: Telephone lines, which carry the desired oice signal between the green and red (called tip and ring) wires. The lines often run parallel to power lines for miles along highway rightofways... resulting in an induced 60 Hz oltage (as much as 30 V or so) from each wire to ground. We must extract and amplify the oltage difference between the wires, while ignoring the large oltage common to the wires. Modeling Differential and CommonMode Signals ID /2 ICM I1 I2 ID /2 Fig. 41. Representing two sources by their differential and commonmode components. As shown aboe, any two signals can be modeled by a differential component, ID, and a commonmode component, ICM, if: ID = ICM I = ICM 2 2 ID I1 and (32) 2 2

46 Differential Amplifiers Introduction to Electronics 28 Soling these simultaneous equations for ID and ICM : = = 2 I1 I2 ID I1 I2 and (33) ICM Note that the differential oltage ID is the difference between the signals I1 and I2, while the commonmode oltage ICM is the aerage of the two (a measure of how they are similar). Amplifying Differential and CommonMode Signals We can use superposition to describe the performance of an amplifier with these signals as inputs: icm id /2 Amplifier o = A d id A cm icm id /2 Fig. 42. Amplifier with differential and commonmode input signals. A differential amplifier is designed so that A d is ery large and A cm is ery small, preferably zero. Differential amplifier circuits are quite cleer they are the basic building block of all operational amplifiers CommonMode Rejection Ratio A figure of merit for diff amps, CMRR is expressed in decibels: CMRR db Ad = 20log (34) A cm

47 Ideal Operational Amplifiers Introduction to Electronics 29 Ideal Operational Amplifiers The ideal operational amplifier is an ideal differential amplifier: O O = A 0 ( ) Fig. 43. The ideal operational amplifier: schematic symbol, input and output oltages, and inputoutput relationship. A 0 = A d = A cm = 0 R i = R o = 0 B = The input marked is called the noninerting input... The input marked is called the inerting input... The model, just a oltagedependent oltage source with the gain A 0 ( ), is so simple that you should get used to analyzing circuits with just the schematic symbol. Ideal Operational Amplifier Operation With A 0 =, we can conceie of three rules of operation: 1. If > then o increases If < then o decreases If = then o does not change... In a real op amp o cannot exceed the dc power supply oltages, which are not shown in Fig. 43. In normal use as an amplifier, an operational amplifier circuit employs negatie feedback a fraction of the output oltage is applied to the inerting input.

48 Ideal Operational Amplifiers Introduction to Electronics 30 Op Amp Operation with Negatie Feedback Consider the effect of negatie feedback: If > then o increases... Because a fraction of o is applied to the inerting input, increases... The gap between and is reduced and will eentually become zero... Thus, o takes on the alue that causes = 0!!! If < then o decreases... Because a fraction of o is applied to the inerting input, decreases... The gap between and is reduced and will eentually become zero... Thus, o takes on the alue that causes = 0!!! In either case, the output oltage takes on whateer alue that causes = 0!!! In analyzing circuits, then, we need only determine the alue of o which will cause = 0. Slew Rate So far we hae said nothing about the rate at which o increases or decreases... this is called the slew rate. In our ideal op amp, we ll presume the slew rate is as fast as we need it to be (i.e., infinitely fast).

49 Op Amp Circuits The Inerting Amplifier Introduction to Electronics 31 Op Amp Circuits The Inerting Amplifier Let s put our ideal op amp concepts to work in this basic circuit: 0 i 1 i 2 R 1 R 2 i O Voltage Gain Fig. 44. Inerting amplifier circuit. Because the ideal op amp has R i = will be zero., the current into the inputs This means i 1 = i 2, i.e., resistors R 1 and R 2 form a oltage diideriii Therefore, we can use superposition to find the oltage. (Remember the quick exercise on p. 4??? This is the identical problem!!!): = R i R or R (35) Now, because there is negatie feedback, o takes on whateer alue that causes = 0, and = 0!!! Thus, setting eq. (35) to zero, we can sole for o : R2 R R R A R2 i 2 o 1= 0 o = i = (36) R 1 1

50 Op Amp Circuits The Inerting Amplifier Introduction to Electronics 32 0 i 1 i 2 R 1 R 2 i O Fig. 45. Inerting amplifier circuit (Fig. 44 repeated). Input Resistance This means resistance seen by the signal source i, not the input resistance of the op amp, which is infinite. Because = 0, the oltage across R 1 is i. Thus: i 1 i i i = Rin = = = R 1 (37) R i 1 1 R i 1 Output Resistance This is the Theenin resistance which would be seen by a load looking back into the circuit (Fig. 45 does not show a load attached). Our op amp is ideal; its Theenin output resistance is zero: R O = 0 (38)

51 Op Amp Circuits The Noninerting Amplifier Introduction to Electronics 33 Op Amp Circuits The Noninerting Amplifier If we switch the i and ground connections on the inerting amplifier, we obtain the noninerting amplifier: i 0 O i 1 i 2 R 1 R 2 Fig. 46. Noninerting amplifier circuit. Voltage Gain This time our rules of operation and a oltage diider equation lead to: R1 R R i = = = o (39) from which: R = R R 1 2 R2 R A R2 = 1 = 1 (40) R 1 2 o i i Input and Output Resistance The source is connected directly to the ideal op amp, so: R in = R = (41) i A load sees the same ideal Theenin resistance as in the inerting case: R O = 0 (42)

52 Op Amp Circuits The Voltage Follower Introduction to Electronics 34 Op Amp Circuits The Voltage Follower i o Fig. 47. The oltage follower. Voltage Gain This one is easy: = = = A = 1 (43) i o i.e., the output oltage follows the input oltage. Input and Output Resistance By inspection, we should see that these alues are the same as for the noninerting amplifier... R in = and R = 0 (44) O In fact, the follower is just a special case of the noninerting amplifier, with R 1 = and R 2 = 0!!!

53 Op Amp Circuits The Inerting Summer Introduction to Electronics 35 Op Amp Circuits The Inerting Summer This is a ariation of the inerting amplifier: i A R A i B A R B R F B i F O Voltage Gain i A Fig. 48. The inerting summer. We could use the superposition approach as we did for the standard inerter, but with three sources the equations become unnecessarily complicated... so let s try this instead... Recall... O takes on the alue that causes = = 0... So the oltage across R A is A and the oltage across R B is B : A = ib = R R A Because the current into the op amp is zero: and B (45) B i i i R ( i i ) R A F = A B R = F F A B = F R R B and (46) A B Finally, the oltage rise to O equals the drop across R F : O RF R RF = R A A B B (47)

54 Op Amp Circuits Another Inerting Amplifier Introduction to Electronics 36 Op Amp Circuits Another Inerting Amplifier If we want ery large gains with the standard inerting amplifier of Fig. 44, one of the resistors will be unacceptably large or unacceptably small... We sole this problem with the following circuit: i 1 R 2 R 4 i R 1 i 2 R 3 O Voltage Gain Fig. 49. An inerting amplifier with a resistie Tnetwork for the feedback element. One common approach to a solution begins with a KCL equation at the R 2 R 3 R 4 junction we ll use the superposition & oltage diider approach, after we apply some network reduction techniques. Notice that R 3, R 4 and the op amp output oltage source can be replaced with a Theenin equialent: R 3 R 4 O R TH TH Fig. 50. Replacing part of the original circuit with a Theenin equialent

55 Op Amp Circuits Another Inerting Amplifier Introduction to Electronics 37 The alues of the Theenin elements in Fig. 50 are: R3 = R R and R = R R 3 4 (48) TH O TH 3 4 With the substitution of Fig. 50 we can simplify the original circuit: i R 1 R EQ = R 2 R TH TH = 0 Fig. 51. Equialent circuit to original amplifier. Again, O, and therefore TH, takes on the alue necessary to make = 0... We e now soled this problem twice before (the quick exercise on p. 4, and the standard inerting amplifier analysis of p. 31): TH R = R Substituting for TH and R EQ, and soling for O and A : EQ 1 i (49) ( ) R R R R R R R2 R R = = R R R O i i (50) A O R = = R i R R R 2 R R (51)

56 Op Amp Circuits The Differential Amplifier Introduction to Electronics 38 Op Amp Circuits Differential Amplifier The op amp is a differential amplifier to begin with, so of course we can build one of these!!! 2 R 1 R 2 i 1 i 2 R 1 R 2 1 Fig. 52. The differential amplifier. O Voltage Gain Again, O takes on the alue required to make =. Thus: R2 = R R = 2 (52) 1 2 We can now find the current i 1, which must equal the current i 2 : i 1 = 1 R 1 1 R2 = R R R R = i 1 ( ) (53) Knowing i 2, we can calculate the oltage across R 2... R2 = i R = R 1 R RR 2 2 R R R ( ) (54) Then we sum oltage rises to the output terminal: = = O R2 R2 R R R2 R RR 2 2 R R R ( ) (55)

57 Op Amp Circuits The Differential Amplifier Introduction to Electronics 39 Working with just the 2 terms from eq. (55)... R2 R R RR 2 2 R R R RR 1 2 R R R RR 2 2 R R R 2 = ( ) ( ) ( ) (56) = ( ) ( ) ( ) RR RR = = R R R R R R R R R R2 R (57) And, finally, returning the resulting term to eq. (55): R = 2 R R2 R = R2 R O ( ) 2 1 (58) So, under the conditions that we can hae identical resistors (and an ideal op amp) we truly hae a differential amplifier!!!

58 Op Amp Circuits Integrators and Differentiators Introduction to Electronics 40 Op Amp Circuits Integrators and Differentiators Op amp circuits are not limited to resistie elements!!! The Integrator i R R C i i C Fig. 53. Op amp integrator. O From our rules and preious experience we know that = 0 and i R = i C, so... i R i = = R i (59) C From the i relationship of a capacitor: C t 1 C idt 1 = C idt C = C t 0 C () 0 (60) Combining the two preious equations, and recognizing that O = C : O t 1 i C R dt 1 = = RC dt () 0 () 0 C i C 0 0 t (61) Normally C (0) = 0 (but not always). Thus the output is the integral of i, inerted, and scaled by 1/RC.

59 Op Amp Circuits Integrators and Differentiators Introduction to Electronics 41 The Differentiator i i C C R Fig. 54. The op amp differentiator. i R O This analysis proceeds in the same fashion as the preious analysis. From our rules and preious experience we know that = 0 and i C = i R... From the i relationship of a capacitor: i C C d C d C i = = = i (62) R dt dt Recognizing that O = R : i R RC d i O = R = R = (63) dt

60 Op Amp Circuits Designing with Real Op Amps Introduction to Electronics 42 Op Amp Circuits Designing with Real Op Amps i Resistor Values Our ideal op amp can supply unlimited current; real ones can t... R 1 R 2 i F i L R L Fig. 55. Noninerting amplifier with load. O To limit i F i L to a reasonable alue, we adopt the rule of thumb that resistances should be greater than approx. 100 Ω. Of course this is highly dependent of the type of op amp to be used in a design. Larger resistances render circuits more susceptible to noise and more susceptible to enironmental factors. To limit these problems we adopt the rule of thumb that resistances should be less than approximately 1 MΩ. Source Resistance and Resistor Tolerances i 1 i 2 R S R 1 R 2 i O In some designs R S will affect desired gain. Resistor tolerances will also affect gain. Fig. 56. Inerting amplifier including source resistance. If we wish to ignore source resistance effects, resistances must be much larger than R S (if possible). Resistor tolerances must also be selected carefully.

61 Graphical Solution of Simultaneous Equations Introduction to Electronics 43 Graphical Solution of Simultaneous Equations Let s reisit some 7 th grade algebra...we can find the solution of two simultaneous equations by plotting them on the same set of axes. Here s a triial example: We plot both equations: y = x and y = 4 (64) Fig. 57. Simple example of obtaining the solution to simultaneous equations using a graphical method. Obiously, the solution is where the two plots intersect, at x = 4, y = 4...

62 Graphical Solution of Simultaneous Equations Introduction to Electronics 44 Let s try another one: y 0, for x < 0 = 2 0.4x, for x 0 (65) and y x = (66) Fig. 58. Another example of graphically finding the solution to simultaneous equations. Here we see that the solution is approximately at x = 3.6, y = 5.2. Note that we lose some accuracy with a graphical method, but, we gain the insight that comes with the picture.

63 Graphical Solution of Simultaneous Equations Introduction to Electronics 45 If we change the preious example slightly, we ll see that we can t arbitrarily neglect the other quadrants: and y = 04. x 2, for all x (67) y x = (68) Fig. 59. Graphically finding multiple solutions. Now we hae two solutions the first one we found before, at x = 3.6, y = the second solution is at x = 5.5, y = In the pages and weeks to come, we will often use a graphical method to find current and oltage in a circuit. This technique is especially wellsuited to circuits with nonlinear elements.

64 Diodes Introduction to Electronics 46 Diodes When we place ptype semiconductor adjacent to ntype semiconductor, the result is an element that easily allows current to flow in one direction, but restricts current flow in the opposite direction... this is our first nonlinear element: free holes Anode Cathode ptype ntype free electrons i D D Fig. 60. Simplified physical construction and schematic symbol of a diode. The free holes wish to combine with the free electrons... When we apply an external oltage that facilitates this combination (a forward oltage, D > 0), current flows easily. When we apply an external oltage that opposes this combination, (a reerse oltage, D < 0), current flow is essentially zero. Of course, we can apply a large enough reerse oltage to force current to flow...this is not necessarily destructie.

65 Diodes Introduction to Electronics 47 Thus, the typical diode i characteristic: Fig. 61. PSpicegenerated i characteristic for a 1N750 diode showing the arious regions of operation. V F is called the forward knee oltage, or simply, the forward oltage. It is typically approximately 0.7 V, and has a temperature coefficient of approximately 2 mv/k V B is called the breakdown oltage. It ranges from 3.3 V to kv, and is usually gien as a positie alue. Diodes intended for use in the breakdown region are called zener diodes (or, less often, aalanche diodes). In the reerse bias region, i D diodes. 1 na for lowpower ( signal )

66 Graphical Analysis of Diode Circuits Introduction to Electronics 48 Graphical Analysis of Diode Circuits V S We can analyze simple diode circuits using the graphical method described preiously: R Fig. 62. Example circuit to illustrate graphical diode circuit analysis. i D D We need two equations to find the two unknowns i D and D. The first equation is proided by the diode i characteristic. The second equation comes from the circuit to which the diode is connected. V OC R TH (=V S ) (=R) Fig. 63. Theenin eq. of Fig. 62 identified. i This is just a standard Theenin equialent circuit and we already know its i characteristic... from Fig. 5 and eq. (4) on p. 2: = VOC irth i = ISC R or (69) TH I SC i=i D 1/R TH V OC = D... where V OC and I SC are the opencircuit oltage and the shortcircuit current, respectiely. A plot of this line is called the load line, and the graphical procedure is called loadline analysis. Fig. 64. Graphical solution.

67 Graphical Analysis of Diode Circuits Introduction to Electronics 49 Examples of LoadLine Analysis R V S i D D Case 1: V S = 2.5 V and R = 125 Ω Case 2: V S = 1 V and R = 25 Ω Case 3: V S = 10 V and R = 1 kω Fig. 65. Example circuit (Fig. 62 repeated). Case 1: V OC = V S = 2.5 V and I SC = 2.5 V / 125 Ω = 20 ma. We locate the intercepts, and draw the line. The solution is at D 0.71 V, i D 14.3 ma Case 2: V OC = V S = 1 V and I SC = 1 V / 25 Ω = 40 ma I SC is not on scale, so we use the slope: 1 = ma = Ω The solution is at D 0.70 V, i D 12.0 ma ma V 05. V Case 3: V OC = V S = 10 V I SC = 10 V / 1 kω = 10 ma V OC not on scale, use slope: 1 1k 1mA 25mA V 25V Ω = =.. Fig. 66. Example solutions. The solution is at: D 0.68 V, i D 9.3 ma