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Library o f Congress Cataloging-in-Publication Data C o u g h lin , R o b e rt F. O p e ra tio n a l a m p lifie rs and lin e a r in te g ra te d c irc u its / R o b e rt F. C o u g h lin , F re d e ric k F. D risc o ll. — 6th ed. p. cm . In c lu d e s b ib lio g ra p h ic a l re fe re n c e s and index. ISB N 0 -1 3 -0 1 4 9 9 1 -8 1. O p e ra tio n a l am p lifie rs. 2. L in e a r in te g ra te d c irc u its. I. D risc o ll, F re d e ric k F., II. T itle. T K 7 8 7 1.5 8 .0 6 C 6 8 2001 6 2 1 .3 8 1 5 — dc21

0 0 -0 4 0 6 3 3 C IP

Vice President and Publisher: Dave Garza Editor in Chief: Stephen Helba Acquisitions Editor: Scott J. Sambucci Production Editor: Rex Davidson Design C oordinator: Karrie Converse-Jones C over Designer: Thomas Mack Cover art: M arjory Dressier Production M anager: Pat Tonneman M arketing M anager: Ben Leonard

T his book was set in Times Roman by York Graphic Services, Inc. It was printed and bound by R. R. Donnelley & Sons Company. The cover was printed by Phoenix Color Corp.

Copyright © 2001, 1998, 1991, 1987, 1982, 1977 by Prentice-Hall, Inc., Upper Saddle River, New Jersey 07458. All rights reserved. Printed in the United States of America. This publication is pro­ tected by Copyright and permission should be obtained from the publisher prior to any prohibited re­ production, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

Prentice IIall

To O ur Partners in Ballroom Dane and O ur L ifetim e Partners, B arbara and Jean As We Grow O lder We Grow C loser

Contents

PREFACE

XXV

INTRODUCTION TO OP AMPS Learning Objectives

1

1-0

Introduction

1-1

Is There Still a Need for Analog Circuitry? 7-7.7 7-7.2 1-1.3

1-2

2

Analog and Digital Systems, 2 Op Amp Development, 3 Op Amps Become Specialized, 3

741 General-Purpose Op Amp 1-2.1 1-2.2

2

4

Circuit- Symbol and Terminals, 4 Simplified Internal Circuitry o f a General-Purpose Op Amp, 5

Contents

VI 1-2.3 1-2.4 1-2.5

1-3

Packaging and Pinouts 1-3.1 1-3.2

1-4

Input Stage—Differential Amplifier, 6 Intermediate Stage—Level Shifter, 6 Output Stage—Push-Pull, 6

7

Packaging, 7 Combining Symbol and Pinout, 8

How to Identify or Order an Op Amp 1-4.1 1-4.2

The Identification Code, 9 Order Number Example, 10

1-5

Second Sources

1-6

Breadboarding Op Amp Circuits 1-6.1 1-6.2

9

10 11

The Power Supply, 11 Breadboarding Suggestions, 11

Problems

12

FIRST EXPERIENCES WITH AN OP AMP Learning Objectives 2-0

Introduction

2-1

Op Amp Terminals 2-1.1 2-1.2 2-1.3 2-1.4

2-2

2-3

14

Power Supply Terminals, 15 Output Terminal, 16 Input Terminals, 16 Input Bias Currents and Offset Voltage, 17

18

Definition, 18 Differential Input Voltage, Ed, 18 Conclusions, 19

20

Noninverting Zero-Crossing Detector, 20 Inverting Zero-Crossing Detector, 21

Positive- and Negative-Voltage-Level Detectors 2-4.1 2-4.2

2-5

14

Zero-Crossing Detectors 2-3.1 2-3.2

2-4

13

Open-Loop Voltage Gain 2-2.1 2-2.2 2-2.3

21

Positive-Level Detectors, 21 Negative-Level Detectors, 21

Typical Applications of Voltage-Level Detectors 2-5.1

13

Adjustable Reference Voltage, 21

21

Contents 2-5.2 2-5.3 2-5.4

2-6

Voltage Reference ICs 2-6.1 2-6.2 2-6.3

2-7

Introduction, 27 Ref-02, 27 Ref-02/Voltage Level Detector Applications, 27

Introduction, 29 Sine-to-Square Wave Converter, 29 Sawtooth-to-Pulse Wave Converter, 29 Quad Voltage Comparator, LM339, 30

32

Introduction, 32 Pulse-Width Modulator, Noninverting, 33 Inverting and Noninverting Pulse-Width Modulators, 35

A Pulse-Width Modulator Interface to a Microcontroller

2-10 Op Amp Comparator Circuit Simulation 2-10.1 2-10.2

38

Introduction, 38 Creating, Initializing, and Simulating a Circuit, 38

Problems

41

INVERTING AND NONINVERTING AMPLIFIERS Learning Objectives 3-0

Introduction

3-1

The Inverting Amplifier 3-1.1 3-1.2 3-1.3 3-1.4 3-1.5 3-1.6 3-1.7

3-2

29

Computer Interfacing with Voltage-Level Detectors 2-8.1 2-8.2 2-8.3

2-9

27

Signal Processing with Voltage-Level Detectors 2-7.1 2-7.2 2-7.3 2-7.4

2-8

Sound-Activated Switch, 22 Light Column Voltmeter, 24 Smoke Detector, 26

44

45 45

Introduction, 45 Positive Voltage Applied to the Inverting Input, 45 Load and Output Currents, 47 Negative Voltage Applied to the Inverting Input, 48 Voltage Applied to the Inverting Input, 49 Design Procedure, 51 Analysis Procedure, 51

Inverting Adder and Audio Mixer 3-2.1 3-2.2 3-2.3

Inverting Adder, 52 Audio Mixer, 53 DC Offsetting an AC Signal, 53

52

3-3

Multichannel Amplifier 3-3.1 3-3.2 3-3.3

55

The Need fo r a Multichannel Amplifier, 55 Circuit Analysis, 55 Design Procedure, 56

3-4

Inverting Averaging Amplifier

3-5

Noninverting Amplifier 3-5.1 3-5.2

3-6

3-7

57

Circuit Analysis, 57 Design Procedure, 59

Voltage Follower 3-6.1 3-6.2

61

Introduction, 61 Using the Voltage Follower, 62

The "Ideal" Voltage Source 3-7.1 3-7.2 3-7.3 3-7.4

Noninverting Adder

3-9

Single-Supply Operation

66

3-10 Difference Amplifiers 3-10.1 3-10.2

64

Definition and Awareness, 64 The Unrecognized Ideal Voltage Source, 64 The Practical Ideal Voltage Source, 65 Precise Voltage Sources, 66

3-8

3-11

56

67

69

The Subtractor, 70 Inverting-Noninverting Amplifier, 71

Designing a Signal Conditioning Circuit

3-12 PSpice Simulation 3-12.1 3-12.2 3-12.3 3-12.4

76

Inverting Amplifier—DC Input, 76 Inverting Amplifier—AC Input, 77 Inverting Adder, 78 Noninverting Adder, 79

Problems

80

COMPARATORS AND CONTROLS Learning Objectives 4-0 4-1

Introduction

84

85

Effect of Noise on Comparator Circuits

Contents OP AMPS WITH DIODES Learning Objectives

187

7-0

Introduction to Precision Rectifiers

7-1

Linear Half-Wave Rectifiers 7-1.1 7-1.2 7-1.3 7-1.4

7-2

198

Positive Peak Follower and Hold, 198 Negative Peak Follower and Hold, 200

200

AC-to-DC Conversion or MAV Circuit, 200 Precision Rectifier with Grounded Summing Inputs, 202 AC-to-DC Converter, 203

Dead-Zone Circuits 7-5.1 7-5.2 7-5.3 7-5.4

203

Introduction, 203 Dead-Zone Circuit with Negative Output, 203 Dead-Zone Circuit with Positive Output, 205 Bipolar-Output Dead-Zone Circuit, 208

7-6

Precision Clipper

7-7

Triangular-to-Sine Wave Converter

7-8

PSpice Simulation of Op Amps with Diodes 7-8.1 7-8.2 7-8.3

194

Introduction, 194 Types o f Precision Full-Wave Rectifiers, 195

AC-to-DC Converter 7-4.1 7-4.2 7-4.3

7-5

Introduction, 189 Inverting Linear Half-Wave Rectifier, Positive Output, 190 Inverting Linear Half-Wave Rectifier, Negative Output, 192 Signal Polarity Separator, 193

Peak Detectors 7-3.1 7-3.2

7-4

189

Precision Rectifiers: The Absolute-Value Circuit 7-2.1 7-2.2

7-3

188

208 208 209

Linear Half-Wave Rectifier, 209 Precision Full-Wave Rectifier, 211 Mean-Absolute-Value Amplifier, 213

Problems

215

DIFFERENTIAL, INSTRUMENTATION, AND BRIDGE AMPLIFIERS Learning Objectives

216

Introduction

217

Basic Differential Amplifier 8-1.1 8-1.2 8-1.3

217

Introduction, 217 Common-Mode Voltage, 219 Common-Mode Rejection, 220

Differential versus Single-Input Amplifiers 8-2.1 8-2.2

Measurement with a Single-Input Amplifier, 221 Measurement with a Differential Amplifier, 222

Improving the Basic Differential Amplifier 8-3.1 8-3.2

223

Increasing Input Resistance, 223 Adjustable Gain, 223

Instrumentation Amplifier 8-4.1 8-4.2

221

226

Circuit Operation, 226 Referencing Output Voltage, 228

Sensing and Measuring with the Instrumentation Amplifier 229 8-5.1 8-5.2 8-5.3

Sense Terminal 229 Differential Voltage Measurements, 230 Differential Voltage-to-Current Converter, 231

The Instrumentation Amplifier as a Signal Conditioning Circuit 233 8-6.1 8-6.2 8-6.3 8-6.4 8-6.5

Introduction to the Strain Gage, 233 Strain-Gage Material, 233 Using Strain-Gage Data, 234 Strain-Gage Mounting, 235 Strain-Gage Resistance Changes, 235

Measurement of Small Resistance Changes 8-7.1 8-7.2 8-7.3

Need fo r a Resistance Bridge, 235 Basic Resistance Bridge, 236 Thermal Effect on Bridge Balance, 237

Balancing a Strain-Gage Bridge 8-8.1 8-8.2

235

238

The Obvious Technique, 238 The Better Technique, 238

Increasing Strain-Gage Bridge Output Practical Strain-Gage Application

239

241

Measurement of Pressure, Force, and Weight

243

Contents 8-12

Basic Bridge Amplifier 8-12.1 8-12.2 8-12.3 8-12.4

8-13

. 243

Introduction, 243 Basic Bridge Circuit Operations. 244 Temperature Measurement with a Bridge Circuit, 245 Bndge Amplifiers and Computers, 248

Adding Versatility to the Bridge Amplifier 8-13.1 8-13.2

248

Grounded Transducers, 248 High-Current Transducers, 248

Problems

249

DC PERFORMANCE: BIAS, OFFSETS, AND DRIFT Learning Objectives

252

9-0

Introduction

9-1

Input Bias Currents

9-2

Input Offset Current

9-3

Effect of Bias Currents on OutputVoltage 9-3.1 9-3.2 9-3.3

9-4

9-5

255 256

Simplification, 256 Effect o f ( —) Input Bias Current, 256 Effect o f (+) Input Bias Current, 258

259

9-4.1 9-4.2 9-4.3

260

Current-Compensating the Voltage Follower, 259 Current-Compensating Other Amplifiers, 260 Summary on Bias-Current Compensation,

Input Offset Voltage

261

Definition and Model, 261 Effect o f Input Offset Voltage on Output Voltage, 262 Measurement of Input Offset Voltage, 262

Input Offset Voltage for the Adder Circuit 9-6.1 9-6.2

9-7

254

Effect of Offset Current on Output Voltage

9-5.1 9-5.2 9-5.3

9-6

253

Comparison o f Signal Gain and Offset Voltage Gain. 264 How Not to Eliminate the Effects o f Offset Voltage, 265

Nulling-Out Effect of Offset Voltage and Bias Currents 265 9-7.1

264

Design or Analysis Sequence, 265

9-7.2 9-7.3

Null Circuits fo r Offset Voltage, 266 Nulling Procedure fo r Output Voltage, 267

9-8

Drift

267

9-9

Measurement of Offset Voltage and Bias Currents

9-10

Common-Mode Rejection Ratio

9-11

Power Supply Rejection Ratio Problems

270 271

272

AC PERFORMANCE: BANDWIDTH, SLEW RATE, NOISE Learning Objectives

274

10-0

Introduction

10-1

Frequency Response of the Op Amp 10-1.1 10-1.2 10-1.3 10-1.4

10-2

JO-2.2 10-2.3 10-2.4 JO-2.5

279

Effect o f Open-Loop Gain on Closed-Loop Gain o f an Amplifier, DC Operation, 279 Small-Signal Bandwidth, Low- and High-Frequency Limits, 28J Measuring Frequency Response, 282 Bandwidth o f Inverting and Noninverting Amplifiers, 282 Finding Bandwidth by a Graphical Method, 283

284

Definition o f Slew Rate, 284 Cause o f Slew-Rate Limiting, 285 Slew-Rate Limiting o f Sine Waves, 285 Slew Rate Made Easy, 288

Noise in the Output Voltage 10-4.1 JO-4.2 J 0-4.3 JO-4.4 10-4.5

10-5

Internal Frequency Compensation, 275 Frequency-Response Curve, 276 Unity-Gain Bandwidth, 277 Rise Time, 278

Slew Rate and Output Voltage 10-3.1 10-3.2 JO-3.3 10-3.4

10-4

275

Amplifier Gain and Frequency Response 10-2.1

10-3

275

289

Introduction, 289 Noise in Op Amp Circuits, 289 Noise Gain, 290 Noise in the Inverting Adder, 290 Summary, 290

Loop Gain Problems

291 292

Contents ACTIVE FILTERS Learning Objectives 11-0

Introduction

11-1

Basic Low-Pass Filter 11-1.1 11-1.2 11-1.3

294

295 296

Introduction, 296 Designing the Filter, 297 Filter Response, 299

11-2

Introduction to the Butterworth Filter

11-3

-40-dB/Decade Low-Pass Butterworth Filter 11-3.1 11-3.2

11-4

Simplified Design Procedure, 300 Filter Response, 302

-60-dB/Decade Low-Pass Butterworth Filter 11-4.1 11-4.2

Simplified Design Procedure, 302 Filter Response, 304

11-5 High-Pass Butterworth Filters 11-5.1 11-5.2 11-5.3 11-5.4 11-5.5

11-6

315

Cascading, 315 Wideband Filter Circuit, 315 Frequency Response, 315

316

Narrowband Filter Circuit, 317 Performance, 317 Stereo-Equalizer Octave Filter, 318

Notch Filters 11-9.1 11-9.2

312

Frequency Response, 312 Bandwidth, 313 Quality Factor, 314 Narrowband and Wideband Filters, 314

Narrowband Bandpass Filters 11-8.1 11-8.2 11-8.3

11-9

Introduction, 305 20-dB/Decade Filter, 306 40-dB/Dec,ade Filter, 308 60-dB/Decade Filter, 309 Comparison o f Magnitudes and Phase Angles, 311

Basic Wideband Filter 11-7.1 11-7.2 11-7.3

11-8

305

Introduction to Bandpass Filters 11-6.1 11-6.2 11-6.3 11-6.4

11-7

299

319

Introduction, 319 Notch Filter Theory, 320

11-10

120-Hz Notch Filter 11-10.1 11-10.2 11-10.3 11-10.4 11-10.5

11-11

320

Need fo r a Notch Filter 320 Statement o f the Problem, 321 Procedure to Make a Notch Filter, 321 Bandpass Filler Components, 321 Final Assembly, 322

Simulation of Active Filter Circuits Using PSpice 11-1 L I 11-11.2 11-11.3

322

Low-Pass Filter 323 High-Pass Filter, 325 Bandpass Filter 326

Problems

328

MODULATING, DEMODULATING, AND FREQUENCY CHANGING WITH THE MULTIPLIER Learning Objectives

330

12-0

Introduction

12-1

Multiplying DC Voltages 12-1.1 12-1.2

331 331

Multiplier Scale Factor, 331 Multiplier Quadrants, 332

12-2

Squaring a Number or DC Voltage

12-3

Frequency Doubling 12-3.1 12-3.2

12-4

337

Basic Theory, 337 Phase-Angle Meter, 339 Phase Angles Greater than ±90°, 340

12-5

Analog Divider

12-6

Finding Square Roots

12-7

334

Principle of the Frequency Doubler, 334 Squaring a Sinusoidal Voltage, 335

Phase-Angle Detection 12-4.1 12-4.2 12-4.3

334

340 342

Introduction to Amplitude Modulation 12-7.1 12-7.2 12-7.3 12-7.4 12-7.5 12-7.6

Need fo r Amplitude Modulation, 342 Defining Amplitude Modulation, 343 The Multiplier Used as a Modulator, 343 Mathematics o f a Balanced Modulator, 343 Sum and Difference Frequencies, 345 Side Frequencies and Sidebands, 347

342

Contents

12-8

Standard Amplitude Modulation 12-8.1 12-82 12-8.3

348

Amplitude Modulator Circuit, 348 Frequency Spectrum o f a Standard AM Modulator, 351 Comparison o f Standard AM Modulators and Balanced Modulators, 352

12-9

Demodulating an AM Voltage

12-10

Demodulating a Balanced Modulator Voltage

12-11

Single-Sideband Modulation and Demodulation 356

12-12

Frequency Shifting

12-13

Universal Amplitude Modulation Receiver 12-13.1 12-13.2 12-13.3 12-13.4

352

356

Tuning and Mixing, 358 Intermediate-Frequency Amplifier, 360 Detection Process, 360 Universal AM Receiver, 360

Problem s

361

INTEGRATED-CIRCUIT TIMERS Learning Objectives

362

13-0

Introduction

13-1

Operating Modes of the 555 Timer

13-2

Terminals of the 555 13-2.1 13-2.2 13-2.3 13-2.4 13-2.5 13-2.6 13-2.7

13-3

363 364

365

Packaging and Power Supply Terminals, 365 Output Terminal, 366 Reset Terminal, 366 Discharge Terminal, 366 Control Voltage Terminal, 366 Trigger and Threshold Terminals, 366 Power-on Tune Delays, 368

Free-Running or Astable Operation 13-3.1 13-3.2 13-3.3 13-3.4

3!

Circuit Operation, 371 Frequency o f Oscillation, 371 Duty Cycle, 373 Extending the Duty Cycle, 374

371

358

13-4

Applications of the 555 as an Astable Multivibrator 375 13-4.1 13-4.2

13-5

One-Shot or Monostable Operation 13-5.1 13-5.2

13-6

Tone-Burst Oscillator, 375 Voltage-Controlled Frequency Shifter, 377

378

Introduction, 378 Input Pulse Circuit, 380

Applications of the 555 as a One-Shot Multivibrator 381 13-6.1 13-6.2 13-6.3 13-6.4

Water-Level Fill Control, 381 Touch Switch, 381 Frequency Divider, 382 Missing Pulse Detector, 383

13-7

Introduction to Counter Tinners

13-8

The XR 2240 Programmable Timer/Counter 13-8.1 13-8.2 13-8.3

13-9

13-11

389

Timing Applications, 389 Free-Running Oscillator, Synchronized Outputs, 390 Binary Pattern Signal Generator, 391 Frequency Synthesizer, 392

13-10 Switch Programmable Timer 13-10.1 13-10.2

394

Timing Intervals, 394 Circuit Operation, 394

PSpice Simulation of 555 Timer 13-1 L I 13-11.2

3

Circuit Description, 385 Counter Operation, 386 Programming the Outputs, 388

Timer/Counter Applications 13-9.1 13-9.2 13-9.3 13-9.4

384

394

Astable or Free-Running Multivibrator, 394 Tone-Burst-Control Circuit, 397

Problems

399

DIGITAL-TO-ANALOG CONVERTERS Learning Objectives 14-0

Introduction

14-1

DAC Characteristics

400

401 401

ix

Contents 4-2

Positive Feedback 4-2.1 4-2.2 4-2.3

4-3

87

Introduction, 87 Upper-Threshold Voltage, 58 Lower-Threshold Voltage, 88

Zero-Crossing Detector with Hysteresis

90

4-3.1 Defining Hysteresis, 90 4-3.2 Zero-Crossing Detector with Hysteresis as a Memory Element, 91

4-4

Voltage-Level Detectors with Hysteresis

91

Introduction, 97 4-4.2 Noninverting Voltage-Level Detector with Hysteresis, 92 4-4.3 Inverting Voltage-Level Detector with Hysteresis, 94 4-4.1

4-5

Voltage-Level Detector with Independent Adjustment of Hysteresis and Center Voltage 96 4-5.7 Introduction, 96 4-5.2

4-6

Battery-Charger Control Circuit, 9is positive, from Eq. (2-2).

(a) Noninverting: W hen E x is above Vref, Vt) = + 1 ^ .

(b) Inverting: W hen E, is above Vrcf> V0 = - V 5al.

FIG U RE 2-4 Zero-crossing detectors, noninverting in (a) and inverting in (b). If the sig­ nal Ej is applied to the ( + ) input, the circuit action is noninverting. If the signal £, is ap­ plied to the ( - ) input, the circuit action is inverting.

T he polarity o f V0 “tells’’ if E , is above or below Vref. T he transition o f V0 tells when Ej crossed the reference and in w hat direction. For exam ple, w hen Va m akes a positivegoing transition from —Vsat to + K sal, it indicates that £*, ju st crossed 0 in the positive d i­ rection. T he circuit o f Fig. 2-4(a) is a noninverting zero-crossing detector.

First Experiences with an Op Am p

21

2-3.2 Inverting Zero-Crossing Detector T he op am p ’s ( - ) input in Fig. 2-4(b) com pares E f w ith a reference voltage o f 0 V (Vref = 0 V). T his circuit is an inverting zero-crossing detector. T he w aveshapes o f VQ versus tim e and Va versus E f in Fig. 2-4(b) can be explained by the follow ing sum m ary: 1. If E, is m ore positive than Vref, then V0 equals - Vsat. 2. W here Et crosses the reference going positive, Va m akes a negative-going transition from + Vsat to - Vsat.

Summary. If the signal or voltage to be m onitored is connected to the ( + ) in­ put, a noninverting com parator results. If the signal or voltage to be m onitored is co n ­ nected to the ( —) input, an inverting com parator results. W hen V0 = + Vsa(, the signal is above (m ore positive than) Vref in a noninverting com parator and below (m ore negative than) Vref in an inverting com parator. 2-4 POSITIVE- AND NEGATIVE-VOLTAGE-LEVEL DETECTO RS

_______________________

2-4.1 Positive-Level Detectors In Fig. 2-5 a positive reference voltage Vref is applied to one o f the op a m p ’s inputs. T his m eans that the op am p is set up as a com parator to detect a positive voltage. If the volt­ age to be sensed, E h is applied to the op a m p ’s ( + ) input, the result is a noninverting p o s ­ itive-level detector. Its operation is show n by the w aveshapes in Fig. 2-5(a). W hen E x is above Vref, V0 equals + Vsat. W hen £, is below Vref, V0 equals —Vsat. If Ej is applied to the inverting input as in Fig. 2-5(b), the circuit is an inverting p o s­ itive-level detector. Its operation can be sum m arized by the statem ent: W hen E-, is above Kef, V0 equals —Vsat. This circuit action can be seen m ore clearly by observing the plot o f Ej and Vref versus tim e in Fig. 2-5(b).

2-4.2 Negative-Level Detectors Figure 2-6(a) is a noninverting negative-level detector This circuit detects when input signal Ej crosses the negative voltage - Vref. W hen Et is above - Vref, VQ equals + Vsat. W hen E , is below —Vref, V0 = —Vsaj. The circuit of Fig. 2-6(b) is an inverting negative-level detector W hen E ( is above - Vref, V0 equals - Vsat> and when Ej is below - Vref, V0 equals + Vsat.

2-5 TYPICAL APPLICATIONS OF VOLTAGE-LEVEL D ETECTO RS

______________________

2-5.1 Adjustable Reference Voltage ICs are available to set p recise voltage references. T hese referen ce chips w ill be in tro ­ duced in the next section. In this section a resistive divider netw ork is used to set Vref. F igure 2-7 show s how to m ake an adjustable reference voltage. Two iO -kH resistors and

22

Chapter 2

(a) Noninverting: W hen E , is above

+v

+K,

+vQ

(b) Inverting: When £,- is above

FIG U RE 2-5 Positive-voltage-level detector, noninverting in (a) and inverting in (b). If the signal E, is applied to the ( + ) input, the circuit action is noninverting. If the signal E-t is applied to the ( —) input, the circuit action is inverting.

a 10-kfi potentiom eter are connected in series to m ake a 1-mA voltage divider. Each kilohm o f resistance corresponds to a voltage drop o f 1 V. Vref can be set to any value betw een - 5 and + 5 V. R em ove the — V connection to the bottom 10-kf2 resistor and substitute a ground. You now have a 0.5-m A divider, and Vref can be adjusted from 5 to 10 V.

2-5.2 Sound-Activated Switch Figure 2-8 first shows how to make an adjustable reference voltage o f 0 to 100 mV. Pick a 10-kH pot, 5-kH resistor, and + I5 - V supply to generate a convenient large adjustable voltage of 0 to 10 V. N ext connect a 100:1 (approxim ately) voltage divider that divides the 0-to-10-V adjustm ent dow n to the desired 0-to-100-m V adjustable reference voltage. {Note: Pick the large 100-kH divider resistor to be 10 tim es the potentiom eter resistance; this avoids loading dow n the 0-to -10-V adjustm ent.)

24

Chapter 2

+ 15 V

+V

A 5 kn

Sonalert alarm or lamp No. 1893

+ 15 V

-6

SCR C106B

0 tn m n m V

0 to +10 V

o - io k n

} }

Reset switch

Ej I -V

v M icrophone

FIG U RE 2-8 A sound-activated switch is made by connecting the output of a noninverting voltage-level detector to an alarm circuit. 1. O pen the reset sw itch to turn off both SCR and alarm. 2. In a quiet environm ent, adjust the sensitivity control until Va ju st sw ings to ~ V SM. 3. C lose the reset sw itch. The alarm should rem ain off. A ny noise signal w ill now generate an ac voltage and be picked up by the m icro­ phone as an input. T he first positive sw ing o f E t above Vrel drives Va to + Vsat. T he diode now conducts a current pulse of about 1 mA into the gate, G, o f the silicon-controlled rectifier (SCR). Norm ally, the S C R ’s anode, A , and cathode, K, term inals act like an open sw itch. How ever, the gate current pulse m akes the SCR turn on, and now the anode and cathode term inals act like a closed sw itch. The audible or visual alarm is now activated. Furtherm ore, the alarm stays on because once an SCR has been turned on, it stays on u n ­ til its an o d e -cath o d e circuit is opened. T he circuit of Fig. 2-8 can be m odified to photograph high-speed events such as a bullet penetrating a glass bulb. Som e cam eras have m echanical sw itch contacts that close to activate a stroboscopic flash. To build this sound-activated flash circuit, rem ove the alarm and connect anode and cathode term inals to the strobe input in place o f the cam ­ era sw itch. Turn off the room lights. O pen the cam era shutter and fire the rifle at the glass bulb. T he rifle ’s sound activates the sw itch. T he strobe does the w ork o f apparently sto p ­ ping the bullet in midair. C lose the shutter. T he position o f the bullet in relation to the bulb in the picture can be adjusted experim entally by m oving the m icrophone closer to or farther aw ay from the rifle.

2-5.3 Light-Column Voltmeter A light-colum n voltm eter displays a colum n o f light w hose height is proportional to volt­ age. M anufacturers o f audio and m edical eq u ip m en t m ay replace analog m eter panels with light-colum n voltm eters because they are easier to read at a distance.

First Exp eriences with an Op Am p

25

A light-colum n voltm eter is constructed in the circuit o f Fig. 2-9. R cal is adjusted so that 1 m A flow s through the equal resistor divider netw ork to R ]0. Ten separate ref­ erence voltages are established in 1-V steps from 1 V to 10 V. W hen Ei = 0 V or is less than 1 V, the outputs o f all op am ps are at - Vsat. T he sil­ icon diodes protect the light-em itting diodes against excessive reverse bias voltage. W hen + 15 V = +V

A

FIGURE 2-9 Light-column voltmeter. Reference voltages to each op amp are in steps of 1 V. As Et is increased from 1 V to 10 V, LED1 through LED 10 light in sequence. R { to R [0 are 1% resistors. The op amps are 741 8-pin mini-DIPS.

26

Chapter 2

Ej is increased to a value betw een I and 2 V, only the output o f op am p 1 goes positive to light L E D I. Note that the op am p ’s output current is autom atically lim ited by the op am p to its short-circuit value approxim ately 20 to 25 mA. T he 220-H output resistors d i­ vert heat away from the op amp. As Ei is increased, the L ED s light in num erical order. This circuit can also be built using tw o-an d -o n e-h alf L M 324 quad op am ps; som e m anufacturers have designed IC packages for this particular application, such as N ational S em ico n d u cto r’s LM 3914.

2-5.4 Sm oke Detector A nother practical application of a voltage-level detector is a sm oke or dust particle d e­ tector, as show n in Fig. 2-10. The lam p and photoconductive cell are m ounted in an en ­ closed cham ber that adm its sm oke or dust but not external light. T he p hotoconductor is a light-sensitive resistor. In the absence o f sm oke or dust, very little light strikes the pho­ toconductor and its resistance stays at som e high value, typically several hundred kilohms. T he 10-kH sensitivity control is adjusted until the alarm turns off. Smoke particles •

Photoconductor

FIGURE 2-10 With no smoke or dust present the 10-kH sensitivity control is adjusted until the alarm stops. Light reflected off any smoke or dust particles causes the alarm to sound. Any particles entering the cham ber cause light to reflect off the particles and strike the photoconductor. This, in turn, causes the photoconductor’s resistance to decrease and the voltage across R i to increase. As Ei increases above Vref, V0 sw itches from —Vsat to + y sat, causing the alarm to sound. The alarm circuit o f Fig. 2-10 does not include an SCR. T herefore, w hen the particles leave the cham ber, the photoconductor’s resistance increases and the alarm turns off. If you w ant the alarm to stay on, use the SCR alarm circuit show n in Fig. 2-8. The lam p and photoresistor m ust be m ounted in a flat black, lightproof box that adm its smoke. A m bient (room ) light prevents proper operation. T he resistive netw ork

First Experiences with an Op Am p

27

at the input of the op am p form s a W heatstone bridge. This circuit can be used to m onitor the level of dust particles in a clean room environm ent.

2-6 VOLTAGE R EFER EN C E ICs

_____________________________________________________________

2-6.1 Introduction V oltage reference ICs are used to provide a precise voltage for circuit and system de­ signers, especially w hen setting the reference voltage for com parator circuits as well as A/D or D/A converters. Any fluctuation on the reference pin(s) o f converter devices pro ­ duces an inaccuracy in the conversion. Fluctuations on the reference input to a com para­ tor can result in data being lost or erroneous data being sent to a com puter system . In Figs. 2-7 and 2-8, we needed a reference voltage, Vref, and used a resistor divider netw ork connected between the supply voltages or the positive supply and ground. A lthough this circuit may be adequate for som e quick-testing or low -cost designs, a better solution is to use a precision voltage reference chip. M any o f these chips are inexpensive (less than a dollar), set a constant output voltage independent o f tem perature, and can be operated from a w ide range of input pow er supply voltages. Variations in pow er supply voltages do not affect their output reference voltage. Som e o f these chips use the bandgap diode principle to produce a constant voltage o f 1.2 V. This tem perature-independent voltage is followed by an am plifier and buffer (am plifiers and buffers are topics covered in C hapter 3) to provide standard output voltages such as 2.5 V, 5 V, or 10 V. O ther reference chips use a Z ener diode as the reference follow ed by a buffer and am plifier to provide output voltages such as + 5 V and - 1 0 V, as well as ± 5 V and ± 1 0 V tracking outputs. Som e of the m ost com m only used IC voltage reference chips are the REF-01 (4- 10 V), REF-02 (5 V), and REF-03 (2.5 V). W e’ll use the REF-02 IC as an introduction to precision volt­ age reference devices.

2-6.2 REF-02 The R EF-02 IC outputs a stable + 5 .0 V, w hich can be adjusted by ± 6 % ( ± 3 0 0 mV) u s­ ing one external potentiom eter as show n in Fig. 2 - 11(a). The 1 0 -k fl potentiom eter al­ lows the actual output voltage to be adjusted from 4.7 to 5.3 V. T hus for an 8-bit A/D converter the reference voltage can be set to 5.12 V, creating a resolution o f 20 m V /bit. (R esolution of A/D converters is discussed in detail in C hapter 14.) T he R EF-02 can o p ­ erate from an input supply voltage o f from 7 to 40 V, m aking it an ideal voltage refer­ ence device for a w ide range o f applications. Two com m on package styles are show n in Figs. 2 - 1 1(b) and (c).

2-6.3 REF-02/Voltage Level Detector Applications Figure 2 -1 1(d) shows how the REF-02 can be connected to an op am p com parator to set the reference voltage at 5.0 V. In this circuit, the adjustm ent potentiom eter is not used and the REF-02 is used in its basic configuration. In this application, Vref for the com parator

First Experiences with an Op Am p

29

If your application requires a stable but variable reference voltage o f 0 to 5.0 V you still may use the R E F-02 w ith a potentiom eter connected betw een the R E F -0 2 ’s output term inal and com m on, as show n in Fig. 2 -1 1(e). {Note: T he 5-kH potentiom eter in this figure allow s us to vary for the com parator.) T his po ten tio m eter is not being used to adjust the output voltage o f the R EF-02 but rather the input reference to the com parator so that Vref can be varied from 0 to 5.0 V. You still may use the circuit o f Fig. 2 -1 1(a) if you need to set the R E F -0 2 ’s m axim um output voltage to 5.0 V ± 3 0 0 mV. T he tem p er­ ature pin (pin 3) is used if the R E F-02 is being used as a tem perature sensor. For an ex ­ am ple, refer to A nalog D evices’ Web Site, specifically the R E F -0 2 ’s data sheet, to see this device as a sensor in a tem perature controller application.

2-7 SIGNAL PROCESSIN G WITH VOLTAGE LEV EL DETECTO RS

_____________________

2-7.1 Introduction Arm ed with only the know ledge gained thus far, we will m ake a sine-to-square w ave co n ­ verter, an analog-to-digital converter, and a pulse-w idth m odulator out o f the versatile op amp. T hese open-loop com parator (or voltage-level detector) applications are offered to show how easy it is to use op am ps.

2-7.2 Sine-to-Square Wave Converter The zero-crossing detector o f Fig. 2-4 will convert the output o f a sine-w ave from a func­ tion generator into a variable-frequency square wave. If E t is a sine wave, triangular wave, or a wave o f any other shape that is sym m etrical around zero, the zero-crossing d etecto r’s output will be square. T he frequency of £ , should be below 100 Hz, for reasons that are explained in C hapter 10.

2-7.3 Sawtooth-to-Pulse Wave Converter Zener Diode Method: The voltage level detector circuits o f Section 2-4 can be used to convert a saw tooth wave to a pulse w ave provided that the output o f the op am p is modified to create only a positive pulse. T his m odification is show n in Fig. 2.12(a) and consists o f a silicon diode, a resistor, and a zener diode in series. W hen the output volt­ age o f the op am p is at + V sa!, the resistor lim its the current to appro x im ately 5 mA, enough current to cause zener breakdow n. For this condition the output voltage o f the cir­ cuit, V09 equals the zener voltage. Vz 5 4.7 V in Fig. 2.12(a). W hen the op a m p ’s output voltage is at —Vsal, diode D1 is reversed biased and the op a m p ’s output current is zero, hence V0 = 0 V. This circuit is a quick way o f converting a saw tooth-to-pulse wave that is T T L com patible. T he input and output w aveform s are show n in Figs. 2.12(b) and (c), respectively. For this application, however, a better m ethod is to use an integrated circuit called a com parator because we can get V0 to sw ing betw een 0 and 5 V w ithout the ex ­ ternal diodes.

31

First Experiences with an Op Am p

o 14

lOo o 13

Il o

- V or gnd

FIG U RE 2-13 Connection diagram for the LM339 quad comparator. Four voltage comparators are contained in one 14-pin dual-in-line package.

Power supply terminals. Pins 3 and 12 are positive and negative supply volt­ age term inals, respectively, for all four com parators. M axim um supply voltage betw een pins 3 and 12 is ± 1 8 V. In m ost applications, the negative supply term inal, pin 12, is grounded. T hen pin 3 can be any voltage from 2 to 36 V dc. T he L M 339 is used p rim ar­ ily for single-supply operation. Output terminals. T he output term inal o f each op am p is an o p en-collector npn transistor. Each transistor collector is connected to the respective output term inals 2, 1, 14, and 13. All em itters are connected together and then to pin 12. If pin 12 is grounded, the output term inal acts like a sw itch. A closed sw itch extends the ground from pin 12 to the output term inal [see Fig. 2-14(b)]. If you w ant the output to go high w hen the sw itch is open, you m ust install a pullup resistor and an external voltage source. As show n in Fig. 2 - 14(a), this feature allow s easy interfacing betw een a ± 1 5 -V analog system and a 5-V digital system . T he output term inal should not sink more than 16 mA. Input terminals. T he input term inals are differential. Use Eq. (2-1) to determ ine the sign for Ed. If Ed is p ositive, the output sw itch is o pen, as in Fig. 2 - 14(a). If Ed is neg ­ ative, the output sw itch is closed, as in Fig. 2 - 14(b). U nlike many other op am ps, the in ­ put term inals can be brought dow n to ground potential w hen pin 12 is grounded.

146

Chapter 5

Design Exam ple 5-18 (a) D esign the circuit in Fig. 5 -17(b) to differentiate signals in the range from 500 Hz to 1000 Hz. C hoose C, = 0.1 fi¥. (b) If the input signal is e in = 0.4 sin 2 tt (1000 )t volts, w hat is the expression of the output voltage?

Solution

(a) R earranging Eq. (5-16) 1 *,■ =

277(1 X 103 Hz)(0.1 X 10-6 F)

= 1.59 kft

Let T2 = 207’| and solve for f y fro m Eq. (5-17).

20 Rf

277(1 X 103 H z)(0.1 X 10~ 6 F)

= 31.8 kH

U se Eq. (5-16) to solve for Cf . 1 C/

277(1 X 103 H z )(3 1 .8 k fi)

= 0.005 /jlF

(b) F rom Fig. 5 - 17(a), the output voltage is = -R /C r

d etn dt

T hen,

vc = -(3 1 .8 kn)(0.1 /i.F)— [0.4 sin 2tt(1000)«] dt

= - ( 3 1 .8 kflX O .l / aF )(0.4)(27 t)( 1000)cos[27t( 1000)t] = ~ 8 cos[27r(1000)t] V Note: W ith large values o f Rf , the output o f the op am p may go into saturation because it is m ultiplied by the input sig n al’s peak value, frequency, and the constant 2 tt.

5-13 PSPICE SIMULATION Ln this section, we w ill use PSpice and sim ulate the perform ance o f the differential volt­ age-to-current converter w ith a grounded load as show n in Fig. 5-5. U se an IPROBE to m easure IL. Place the follow ing parts on the right-hand side o f the w ork area:

Selected Applications of Op Am ps

147

Draw => Get New Part

Part = = = = = =

>uA 741 > VDC > R > GLOBAL > AGND > IP R O B E

Num ber

Library

1 4 5 4 5 1

eval»slb source.slb analog.slb p o r ts lb p o rt.slb special.sib

A rrange the parts as show n in the schem atic o f Fig. 5-5 but include the IPROBE to m ea­ sure IL. Set the four resistors labeled R as /?, = R 2 = R 3 = #4 = 10 k f l, E x = 5 V, E 2 = 3 V, and R L = 1 k fl. N ow select Analysis = > Setup = > DC Sweep. O pen DC Sweep and select

Swept Van Type = > Voltage Source

Sweep TVpe = > Linear Now set

Name = > V I, Start Value = > 5V, End Value = > 5V, Increment = > IV Select

Nested Sweep = > Swept Var. Type = > Voltage Source = > Sweep Type = > Linear N ow set

Name = > V2, Start Value = > 3V, End Value = > 3V, Increment = > IV In this sim ulation you do not w ant to run Probe. Rem ove Probe execution by

Analysis = > Probe Setup = > Click

Do Not Auto-Run Probe Save the file w ith an .SCH extension and click Analysis = > Simulate. T he com pleted circuit w ith current value l L is show n in Fig. 5-18.

148

Chapter 5

15 V - 7

Vo

15 V

VO

FIGURE 5-18

PSpice model of Fig. 5-5.

PROBLEMS 5-1. Refer to Example 5-1 and Fig. 5-1. Assume that 1FS = 1 mA and meter winding resistance Rm = 1 k f l If Ei = ~ 1.0 V and /?, = Ik ft, find (a) (b) V0. 5-2. A 1-mA movement, with R„, = 1 kfl, is to be substituted in thecircuit inFig. 5-2. Redesign the R; resistors for full-scale meter deflection when (a) Ej = ± 6 V dc; (b) Et = 6 V rms; (c) E, = 6 V p-p; (d) E, = 6 V peak. 5-3. In Fig. P5-3 complete the schematic wiring between op amp,diodes, and milliammeter.The current through the meter must be steered from right to left. 5-4. Calculate a value for Ridc in Fig. P5-3 so that the meter readsfull scale whenEt—5Vand the range switch is on the 5-V position. 5-5. Consider that the range switch is in the 5-V position in Fig. P5-3.Calculatevalues for the following resistors to give a full-scale meter deflection of 5 V: (a)Rirms for Et= 5 Vrms; (b) Rj p-p for Ej = 5 V p-p; (c) R, peak for E, = 5 V peak. 5-6. With the circuit conditions shown in Problem 5-4, (a) which diodes are conducting? (b) Find Va. Assume that diode drops are 0.6 V. 5-7. For the constant-current source shown in Fig. P5-7, (a) draw the current direction, the emit­ ter arrow, and state if the transistor is npn or pnp\ (b) find /; (c) find VL. 5-8. If V0 = 11 V and Et = 5 V in Fig. 5-3, find Vz. 5-9. / 1 must equal 20 mA in Fig. 5-4 when £ f- = - 10 V. Find Rh 5-10. Define a floating load. 5-11. In Fig. 5-5, E2 = 0 V, R = 10 kH, and RL = 5 k i l Find lL> VL, and V0 for (a) Et = - 2 V; (b) £, = + 2 V.

150

Chapter 5

5-12. In Fig. 5-5, E ] = 0 V, R = 10 kft, and R L = 1 kCl. Find l u VLt and V0 for (a) E2 = - 2 V; (b) E2 = +2 V. 5-13. In Fig. 5-5, E, = E2 = - 5 V, and R = RL = 5 k f l Find IL, VL, and V*. 5-14. Replace Vz in Fig. 5-6 with a 900-H resistor. Find lL. 5-15. Sketch an op amp circuit that will draw short-circuit current from a signal source and con­ vert the short-circuit current to a voltage. 5-16. A CL5M9M photocell has a resistance of about 10 k il under an illumination of 2 fc. If Et —10 V in Fig. 5-9, calculate Rf for an output Va of 0.2 V when the photoconductive cell is illuminated by 2 fc. 5-17. Change multiplier resistor mR in Fig. 5-10 to 49 kfi. Find lL. 5-18. A solar cell that has a maximum short-circuit current of 0.1 A = ISc is installed in the cir­ cuit in Fig. 5-12. (a) Select RF to give Va = 10 V when Isc = 0.1 A. (b) A 50-[xA meter movement is to indicate full scale when Isc = 0.1 A. Find /?scaie if RM = 5 kfl. 5-19. Resistor Rj is changed to 10 kfl in Example 5-14. Find the phase angle 6. 5-20. Design a phase shifter to give a —90° shift at 1 Hz. Choose C, from 0.001, 0.01, 0 .1, or 1.0 / j F . R, must lie between 2 and 100 kfl. 5-21. Design a -9 0 ° phase shift at 1590 Hz. Then for your design, calculate (a) 6 at 15 Hz; (b) 6 at 15 kHz. 5-22. Calculate the net current through Rf in Fig. 5 -14(a) if the AD590 temperature is 100°C. Then find V.. 5-23. Calculate the net current through Rf in Fig. 5 -14(b) when the temperature is 100°F. Find V0. 5-24. Calculate the value of fl/in Fig. 5 -14(a) to design a signal conditioning circuit that interfaces with a microcontroller’s A/D converter. The voltage range of the converter is 0 to 5 V. 5-25. Use a simulation program and design an integrating circuit. The input sugnal is ein = t sin 20007rt V 5-26. Use a simulation program and design a differentiating circuit. The input signal is (a) sine wave of 500 Hz and a peak value of 0.2 V; (b) square wave of 500 Hz and a peak value of 0.2 V; (c) cosine wave of 500 Hz and a peak value of 0.2 V.

CHAPTER 6 Signal Generators

LEARNING

OBJECTIVES

_________________________________________________

U pon com pletion o f this chapter on signal generators, you w ill be able to: • E xplain the operation o f a m ultivibrator circuit, sketch its output voltage w aveshapes, and calculate its frequency of oscillation. • M ake a one-shot m ultivibrator and explain the purpose o f this circuit. • Show how tw o op am ps, three resistors, and one capacitor can be connected to form an inexpensive triangle/square-w ave generator. • Predict the frequency o f oscillation and am plitude o f the voltages in a bipolar or unipo­ lar triangle-w ave generator and identify its disadvantages. • Build a saw tooth wave generator and tell how it can be used as a voltage-to-frequency converter, frequency m odulator, or frequency shift key circuit. • C on n ect an A D 630 balanced m o d u la to r/d em o d u lato r to o p erate as a sw itch ed gain am plifier.

151

152

Chapter 6

C onnect the A D 630 to an op am p circuit to m ake a precision triangle-w ave generator w hose output voltage am plitude can be adjusted independently o f the oscillating fre­ quency, and vice versa. B uild, test, m easure, and explain the operation o f an A D 639 universal trigonom etric function generator w hen it is w ired to generate sine functions. C onnect the A D 639 to the triangle-w ave generator to m ake a superb p recisio n sine-w ave generator. Its oscillating frequency can be adjusted over a w ide frequency range by a single resistor, w ithout changing am plitude. Know about the operation o f a single IC function generator.

6-0 INTRODUCTION Up to now our m ain concern has been to use the op am p in circuits that process signals. In this chapter w e concentrate on op am p circuits that generate signals. Four o f the m ost com m on and useful signals are described by their shape w hen viewed on an oscilloscope. T hey are the square wave, triangular wave, saw tooth wave, and sine wave. A ccordingly, the signal generator is classified by the shape o f the wave it generates. Som e circuits are so w idely used that they have been assigned a special name. For exam ple, the first circuit presented in Section 6-1 is a m ultivibrator that generates prim arily square waves and ex ­ ponential waves. Som e ICs that generate these w aveform s from a single IC are available. How ever, you may need a w aveform quickly and not have on hand one o f these function generator ICs.

6-1 FREE-RUNNING MULTIVIBRATOR 6-1.1 Multivibrator Action A fre e-ru n n in g or astable m ultivibrator is a square-w ave generator. T he circuit o f Fig. 6 - 1 is a m u ltiv ib rato r circu it and looks so m eth in g like a co m p arato r w ith h y steresis (C hapter 4), except that the input voltage is replaced by a capacitor. R esistors R i and R 2 form a voltage divider to feed back a fraction o f the output to the ( + ) input. W hen V0 is at + Vsat, as show n in Fig. 6 -l(a ), the feedback voltage is called the upper-threshold volt­ age VUT. VUT is given in Eq. ( 4 -1) and repeated here for convenience: V,rr =

( + y sa|)

R esistor Rf provides a feedback path to the ( —) input. W hen V0 is at + Vsat, current I + flows through Rf to charge capacitor C tow ard VUT. A s long as the capacitor voltage Vc is less than VUTt the output voltage rem ains at + Vsal. W hen Vc charges to a value slightly greater than VUT, the ( —) input goes positive with respect to the ( + ) input. This sw itches the output from + Vsat to —Vsal. The ( + ) input is now held negative with respect to ground because the feedback voltage is negative and given by

154

Chapter 6

lished except that C now has an initial charge equal to VLT. T he cap acito r w ill discharge from VLT to 0 V and then recharge to VUT> and the process is repeating. F ree-running m ul­ tivibrator action is sum m arized as follow s: 1. W hen V0 — — Vsal, C discharges from VUT to VLT and sw itches V0 to + Vsat. 2. W hen V0 = + V sat, C charges from VLT to VUT and sw itches V0 to —Vsat. T he tim e needed for C to charge and discharge determ ines the frequency o f the m ultivi­ brator.

6-1.2 Frequency of Oscillation T he capacitor and output-voltage w aveform s for the free-running m ultivibrator are show n in Fig. 6-2. R esistor R 2 is chosen to equal 0 .86/?i to sim plify calculation o f cap acito r charge time. Tim e intervals /] and t2 show how Vc and VD change w ith tim e for Figs. 6 - 1(a) and (b), respectively. Tim e intervals t x and t2 are equal to the product o f /fy-and C. T he period o f oscillation, T, is the tim e needed for one com plete cycle. S ince T is the sum o f t, and t2, T = 2Rf C

for R2 = O M R

(6-3a)

v» (volts)

A V = ^+vv sai vo

10

T=2RC=y

Tim e

10

/, - RjC

15

FIGURE 6-2

Voltage waveshapes for the multivibrator of Fig. 6-1.

155

Signal G enerators T he frequency of o s c illa tio n /is the reciprocal o f period T and is expressed by

(6-3b) w here T is in s e c o n d s ,/in hertz, / ^ i n ohm s, and C in farads.

Example 6-1 In Fig. 6-1, if R { = 100 k fl, R 2 = 86 k fl, + V sat = + 1 5 V, and - V sat = - 1 5 V, find (a) VuT>'

(b) V l t -

Solution

(a) By Eq. (6-1),

(b) By Eq. (6-2), 86 k fl

Example 6-2 Find the period o f the m ultivibrator in E xam ple 6-1 if Rf = 100 k f i and C = 0.1 fxF.

Solution

U sing Eq. (6-3a), T - (2)(100 k H )(0 .1 pF ) = 0.020 s = 20 ms.

Example 6-3 Find the frequency o f oscillation for the m ultivibrator o f E xam ple 6-2.

Solution

From Eq. (6-3b),

Example 6-4 Show w hy 7 = 2 Rf C w hen R 2 = 0.86R, as stated in Eq. (6-3a).

Solution T he tim e required for a capacitor C to charge through a resistor Rf from som e starting capacitor voltage toward som e aim ing voltage to a stop voltage is expressed g en ­ erally as aim — start aim — stop

156

Chapter 6

Applying the equation to Fig. 6-2 yields 11 = Rf C In

+ V/sat ~ VLT + Vsat - V u r

If the m agnitudes of + V sat and — VsiLX are equal, the term in parentheses sim plifies to

In

Since In 2.718 = 1, the In term can be reduced to I if R| 2R 2 ----- —------= 2.718

or

R 2 = 0.86/?,

Now /, = R jC and t2 = R fC if R 2 = 0.86/? |. Therefore, T = t x + t2 = 2Rf C.

6-2 ONE-SHOT MULTIVIBRATOR

___________________________________________________________

6-2.1 Introduction A o n e-sh o t m u ltivib ra to r generates a single o u tp u t pulse in response to an input sig ­ nal. T he length o f the o u tp u t pulse depends only on external co m p o n en ts (resistors and cap acito rs) co n nected to the op am p. As show n in Fig. 6-3, the o n e-sh o t generates a sin g le o u tpu t pulse on the neg ative-going edge o f £,. The d uration o f the input pulse can be longer or sh o rter than the ex pected output pulse. T he duratio n o f the o u tp u t p ulse is represen ted by r in Fig. 6-3. S ince r can be changed only by ch an g ing re sis­ tors or capacitors, the o n e-sh o t can be con sid ered a p u lse stretch er T his is b ecause the w idth o f the pulse can be lon ger than the input pulse. M oreover, the o n e-sh o t in tro ­ duces an idea o f an ad ju stab le delay, that is, the delay betw een the tim e w hen E, goes negative and the tim e for VQ to go positive again. O peration o f the o n e-sh o t w ill be stu d ied in three parts: (1) the stable state, (2) tran sition to the tim ing state, and (3) the tim ing state.

6-2.2 Stable State In Fig. 6-4(a), VQ is at + Vsat. Voltage divider /?, and R2 feeds back VUT to the ( + ) input. V u r is given by Eq. (6-1). The diode D x clam ps the ( —) input at approxim ately + 0 .5 V. T he ( + ) input is positive with respect to the ( —) input, and the high open-loop gain tim es the differential input voltage (Ed = 2.1 - 0.5 = 1.6 V) holds V0 at + K sat.

Chapter 6

O,

Dt

FIGURE 6-4

Monostable or one-shot multivibrator.

163

Signal G enerators

Next we select and C. B egin by m aking a trial choice for C = 0.05 /xF. T hen ca l­ culate a value for /?, to see if R t is greater than 10 k fl. F rom Eq. (6-6),

2.8 4/C

4(1000 H z)(0.05

jjlF)

= 14 k a

In practice it w ould be prudent for to be a 12 -k fl resistor in series w ith a 0 to 5 -k fl pot. T he 5 - k fl pot m ay then be adjusted for an oscillation frequency o f precisely 1.00 kHz.

6-3.3 Unipolar Triangle-Wave Generator T he bipolar triangle-w ave generator circuit o f Fig. 6-6 can be changed to produce a u nipo­ lar triangle w ave output. S im ply add a diode in series w ith p R as show n in Fig. 6-7. C ircuit operation is studied by reference to the w aveshapes in Fig. 6-7(b). W hen VB is at + Vsal, the diode stops current flow through p R and sets V lt at 0 V. W hen VB is at —ysat, the diode allow s current flow through p R and sets VUT at a value o f V ur = —

~ ysat + 0-6 V

(6-7 a)

P

Frequency of oscillation is then given approxim ately by (6-7b)

2 R,C

Example 6-7 Find the approxim ate peak voltage and frequency for the unipolar triangle-w ave g enerator in Fig. 6-7.

Solution

C alculate _ p R _ 28 k f l _ P ~

R ~ 10 k f l _

Find the peak value of VA from Eq. (6-7a): -1/ t + 0 .6 V

- 1 3 .8 V + 0.6 V

VUT ~

2.8

From Eq. (6-7b),

2.8 f =

2R £

2(28 kH )(0.05 jjlF )

= 1000 Hz

N ote the change in R , value from Fig. 6-6(a) to Fig. 6-7(a).

- 4.7 V

Chapter 6

FIGURE 6-7 Diode D in (a) converts the bipolar triangle-wave generator into a unipolar triangle-wave generator. Waveshapes are shown in (b). (a) Basic unipolar triangle-wave generator; oscillating frequency is 1000 Hz. (b) Outputvoltage waveshapes.

165

Signal G enerators

6-4 SAWTOOTH-WAVE GENERATOR

_____________________________________________________

6-4.1 Circuit Operation A low -parts-count saw tooth-w ave generator circuit is show n in Fig. 6-8(a). Op am p A is a ram p generator. Since E t is negative, Voramp can only ram p up. T he rate o f rise of the ram p voltage is constant at

t

R tC

(6- 8)

The ram p voltage is m onitored by the ( + ) input o f com parator 301B. If Voramp is below Vref, the com parator’s output is negative. D iodes protect the transistors against ex ­ cessive reverse bias. W hen Voramp rises to ju st exceed Vref, the output Vocomp goes to positive saturation. This forward biases “dum p’' transistor Q D into saturation. T he saturated transistor acts as a short circuit across the integrating capacitor C. C discharges quickly through QD to es­ sentially 0 V. W hen Vocomp goes positive, it turns on Q x to short-circuit the lO-kO p o ­ tentiom eter. This drops Vref to alm ost zero volts. As C discharges toward 0 V, it drives Vorarnp rapidly toward 0 V. Voramp drops be­ low Vref, causing l^ocomp to go negative and turn off Q D. C begins charging linearly, and generation o f a new saw tooth wave begins.

6-4.2 Sawtooth Waveshape Analysis The ram p voltage rises at a rate of 1 V per m illisecond in Fig. 6-8(b). M eanw hile, Vocomp is shown to be negative. W hen the ram p crosses Vref, VOCOmp snaps positive to drive the ram p voltage quickly toward 0 V. As Voramp snaps to 0 V, the co m p arator’s output is re­ set to negative saturation. R am p operation is sum m arized in Fig. 6-8(c).

6-4.3 Design Procedure T he tim e for one saw tooth-w ave period can be derived m ost efficiently by analogy with a fam iliar experience. tim e (of rise) =

distance (of rise) speed (of rise)

(6-9a)

(6-9b) Since frequency is the reciprocal o f the period (6-9c)

Signal G enerators

167

Design Example 6-8 D esig n a sa w to o th -w av e g e n e ra to r to have a 10-V p ea k o u tp u t and a fre q u en c y o f 100 Hz. L et Ei = 1 V.

Design Procedure 1. D esign a voltage divider to give a reference voltage Vref = + 1 0 V for op am p B in Fig. 6-8(a). 2. L e t’s select a ram p rate rise o f 1 V /m s. P ick any R tC com b in atio n to give 1.0 ms. T herefore, let’s select /?,* = 10 k f I and C = 0.1 fiF. ■ 3. T he resulting circuit is show n in Fig. 6-8(a). 4. Ei may be m ade from a voltage divider and voltage follow er to m ake an ideal voltage source (see Section 3.7). 5. A lternatively, you could pick a trial value for R tC and solve for £ , in Eq. (6-9b) 6. C heck the design values in Eq. 6-9c.

6-4.4 Voltage-to-Frequency Converter T here are tw o w ays to change or m odulate the o scillating frequency o f Fig. 6-8. We see from Eq. (6-9c) that the frequency is directly pro p o rtio n al to E L and inversely p ro p o r­ tional to Vref. T he advantages and disadvantages o f each m ethod are exam ined w ith an exam ple. T his type o f frequency m odulation by Vref has tw o disadvantages w ith respect to control o f frequency by E t. First, the relationship betw een input voltage Vrcf and output frequency is not linear. Second, the saw too th ’s peak output voltage is not constant, since it varies directly w ith Vref.

6-4.5 Frequency Modulation and Frequency Shift Keying E xam ples 6-9 and 6-10 indicate one way o f achieving fre q u en c y m odulation (FM ). T hus, if the am plitude o f Ei varies, the frequency o f the saw tooth oscillator w ill be changed or m odulated. If E { is keyed betw een tw o voltage levels, the saw tooth oscillator changes fre­ quencies. T his type o f application is called fre q u en c y shift keying (FSK ) and is used for data transm ission. T hese tw o preset frequencies correspond to “0 ” and “ 1” states (co m ­ m only called space and m ark) in binary.

168

Chapter 6

Example 6-9 If Ei is doubled to - 2 V in Fig. 6-8, find the new frequency o f oscillation. S o lu tio n

In Eq. (6 —9c) use | E, 1 (10 X 103 n x o . l X 10“ 6 F) 10 V ---------1

= ^

1.0 X 10-3 s

10 V

= M00Hz\ I

V

For E, = - 2 V ,/ = (2 V )(100 H z/V) = 200 Hz. T hus as £ , changes from 0 V to - 1 0 V, fre­ quency changes from 0 Hz to 1 kHz. The peak am plitude o f the saw tooth w ave rem ains equal to Vref (10 V) for all frequencies.

Example 6-10 K eep E h and C at their value show n in Fig. 6-8(a). R educe Vref by o n e-h alf to 5 V. Is the frequency doubled or halved?

Solution

From Exam ple 6-8 and Eq. (6-9c), 1V f

(ms) Vref

_ (1000 H z)/V Kef

For Vrei = 10 V , / = 100 Hz. For Vref = 5 V the frequency is doubled to 200 Hz. As Vref is reduced from 10 V to 0 V, the frequency is increased from 100 H z to a very high value.

6-4.6 Disadvantages T he triangle-w ave generators o f Section 6-3 are inexpensive and reliable. However, they have tw o disadvantages. T he rates o f rise and fall o f the triangle w ave are unequal. A lso, the peak values o f both triangle-w ave and square-w ave outputs are unequal, because the m agnitudes o f + V sat and - V sat are unequal. In the next section w e substitute an A D 630 for the com parator. T his will give the equivalent o f precisely equal square-w ave ± voltages that w ill a lso be equal to the ± peak values o f triangle-w ave voltage. O nce we have m ade a precision triangle-w ave g en ­ erator, we w ill use it to drive a new state-of-the-art trigonom etric function g enerator to make a precision sine-w ave generator.

170

Chapter 6

6-5 BALANCED MODULATOR/DEMODULATOR, THE AD630

_________________________

6-5.1 Introduction T he A D 630 is an advanced integrated circuit. It has 20 pins, w hich allow s this versatile sw itched voltage gain IC to act as a m odulator, dem odulator, phase detector, and m ulti­ plexer, as well as perform other signal conditioning tasks. We connect the A D 630, as in Fig. 6-9(a), as a controlled sw itched gain ( + 1 or —1) am plifier. T his p articu lar applica­ tion w ill be exam ined by discussing the role perform ed by the d om inant term inals.

6-5.2 Input and Output Terminals T he input signal Vref is connected to m odulation pins 16 and 17 in Fig. 6-9, and thus to the inputs of two am plifiers, A and B. The gain o f A is program m ed for —1 and B for + 1 by shorting term inals (1) 13 to 14, (2) 15 to 19 to 20, (3) 16 to 17, and (4) grounding pin 1. T he carrier input term inal, pin 9 (in this application), determ ines w hich am plifier, A or B , is connected to the o u tp u t term in al. If pin 9 is a b o ve the v o ltag e at pin 10 (ground), am plifier B is selected. Voltage at output pin 13 then equals Vref tim es (4-1). If pin 9 voltage is below ground (negative), am plifier A is selected and output pin 13 equals l/ref times ( —1). (Note that in com m unication circuits, Vref is called the analog data or signal voltage, Vc is called a chopper or carrier voltage, and VQ is the m odulated output. T hat is, the am plitude o f the low -frequency signal voltage is impressed upon the higherfrequency carrier wave— hence the names selected for the A D 630’s input and output terminals.)

6-5.3 Input-Output Waveforms Vrcf is a dc voltage of 5.0 V in Fig. 6-9(b). Vc is a 100-Hz square w ave w ith peak am ­ plitudes that m ust exceed ± 1 mV. O utput voltage V0 is show n in Fig. 6-9(c) to sw itch synchronously with Vc from + l/ref to —\/ref> and vice versa. We are going to replace the unpredictable ± Vsat o f the 301 com parator in Fig. 6-6 with precisely + or —Vref. M oreover, Vref can be adjusted easily to any required value. As show n in the next sectio n , Vref w ill set the positive and negative peak values o f both triangle-w ave and square-w ave generators.

6-6 PRECISION TRIANGLE/SQUARE-WAVE GENERATOR

_____________________________

6-6.1 Circuit Operation O nly six parts plus a voltage source, Vref, m ake up the versatile precision triangle- and squarewave generator in Fig. 6 -10(a). Circuit operation is explained by referencing the waveshapes in Fig. 6 - 10(b). We begin at time zero. Square-wave output Vos begins at —Vref or —5 V. This forces the triangle wave Vot to go positive from a starting point of — = —5 V. D uring this time, pin 9 is below ground to select an AD630 gain o f —1 and holds Vos at —5 V.

172

Chapter 6

6-6.2 Frequency of Oscillation T he easiest way to find the frequency o f oscillation is to begin w ith the rate o f rise o f the triangle wave, Vol/ t y in volts per second. T he rate o f rise o f the triangle wave, from 0 to 0.5 ms in Fig. 6-10(b), is found from

v„ t

R,C

(6- ,0 )

T he tim e t for a half-cycle is 772, and during this tim e, Vol changes by 2Vref. S ubstituting these for t and Vot into Eq. (6-10), w e obtain 2 V re f

=

772

Vn f

R tC

(6-11)

and solve for both period T and frequency o f o sc illa tio n /: T = 4 RfC

and

(6-12)

N ote that Vref cancels out in Eqs. (6-11) and (6-12). This is a very im portant ad ­ vantage. T he peak output voltages of both square- and triangle-w ave signals are set by + Vref. As Vref is adjusted, the frequency o f oscillation does not change.

Example 6-11 M ake a triangle/square-w ave generator that has peak voltages o f ± 5 V and oscillates at a frequency o f 1.0 kHz.

Solution

C hoose Vref = 5.0 V. For low im pedance, Vref should be the output o f an op am p. A rbitrarily choose C = 0.01 /xF. F rom Eq. (6-12), 1 */ =

4 /C

1 4(1000)(0.01

jjlF)

For a fine adjustm ent of the output frequency, m ake /?; from a 2 2 -k fl resistor in series w ith a 5- or 10-kH variable resistor.

6-7 SINE-W AVE GENERATION SU R V EY C om m ercial function generators produce triangular, square, and sinusoidal signals w hose frequency and am plitude can be changed by the user. To obtain a sine-w ave output, the triangle w ave is passed through a shaping netw ork m ade o f carefully selected resistors

Signal Generators

173

and diodes (see Fig. 7-19). T he sine waves thus produced are reasonably good. H owever, there is inevitably som e distortion, particularly at the peaks o f the sine wave. A nother so­ lution is to use an IC function generator chip such as M A X IM ’s M A X 038, w hich is ca­ pable of producing sine, square, and triangle w aveform s fro m 1 H z to 20 M Hz. W hen an application requires a single-frequency sine wave, conventional oscillators use phase-shifting techniques that usually em ploy (1) tw o R C tuning netw orks, and (2) com plex am plitude lim iting circuitry. To m inim ize distortion, the lim it circuit m ust usu ­ ally be custom -adjusted for each oscillator. T he frequency o f this oscillator is difficult to vary because tw o R C netw orks m ust be varied and their values m ust track w ithin ± 1%. W aveform s m ay also be generated by using the A D 630 w ith a universal trig o n o ­ metric function generator, the A D 639. The A D 630 has already been used to generate a precision triangle wave w hose frequency and am plitudes are precise and easy to adjust. We w ill co n n ect the triangle-w ave outp u t Vol o f Fig. 6-10(a) to an A D 639 universal trigonom etric function generator. The resulting circuit w ill have the best qualities o f a p re ­ cision sine-w ave generator whose fre q u en c y will be easily a djustable.

6-8 UNIVERSAL TRIGONOMETRIC FUNCTION GENERATOR, THEAD639 _________________________________________________________________________________ 6-8.1 Introduction T he A D 639 is a state -o f-th e-art trig o n o m etric fu n ctio n generator. It w ill perfo rm all trigonom etric functions in real tim e, including sin, cos, tan, cosec, sec, and cotan. W hen a calculator perform s a trig function, the operator punches in a num ber corresponding to the num ber of angular degrees and punches SIN. T he calculator pauses, then displays a num ber indicating the sine o f the angle. T hat is, a num ber for angle 0 is entered and the calculator produces a num ber for sin 0. T he A D 639 accepts an input voltage that represents the angle. It is called the angle voltage, Vang. For the A D 639, the angle voltage is found from (6-13) F our input term inals are available. However, we shall look at only the single active input that generates sin functions. T he output voltage will equal sin 0 or 10 sin 0, d e­ pending how the internal gain control is pin program m ed.

6-8.2 Sine Function Operation T he A D 639 is w ired to output VQ = 1 sin 0 in Fig. 6-11. T here are four input term inals: 1, 2, 7, and 8. W ired as show n, the chip perform s a sine function. Pins 3, 4, and 10 co n ­ trol gain. N orm ally, 3 and 4 are grounded so that pin 10 can activate the internal gain co n ­ trol. A gain o f 1 results w hen pin 10 is w ired to — Vs or pin 9. W ire pin 10 to + Vs or pin

174

Chapter 6

-v,

+V*

+ 15 V

-15 V

9

10

16

Angle voltage

Vn = 1 sin 0 AJ3639A

5

7 12

8

T (a)

(a) °a

,(V)

FIG U RE 6-11 The AD639 is pin-programmed in (a) to act as a sine function genera­ tor. Each ± 20 mV of input angle voltage corresponds to an input angle of 0 = ±1°. Output Va equals 1 X sin 0. (a) The AD639A is pin-programmed to output the sine of the angle voltage; (b) output voltage V0 equals the sine of 8 if 9 is represented by an angle voltage of 20 mV per angular degree.

175

Signal G enerators

16 and obtain a gain of + 1 0 . Then VQ — 10 sin 0. Pin 6 is a precision 1.80-V reference voltage that corresponds to an angle voltage of 90° (see Eq. (6-13)). We analyze sine func­ tion operation by an exam ple.

Example 6-12 C a lc u late the req u ired in p u t angle voltage and resu ltan t o u tp u t v o ltag e fo r an g les o f (a) ± 4 5 °; (b) ± 9 0 °; (c) ± 2 2 5 °; (d) ± 4 0 5 °.

Solution (a)

From Eq. (6-13) and Fig. 6-11,

— 2

0

( ±45° ) = ± 0 .9 0 V,

= 1 sin (± 4 5 ° ) = ± 0 .7 0 7 V.

(b) Vang =

(± 9 0 ° ) = ± 1 .8 0 V, V0 = 1 sin (± 9 0 ° ) = ± 1 .0 V.

(c) Vane =

( ± 2 2 5 ° ) = ± 4 .5 0 V, V„ = 1 sin (± 2 2 5 ° ) = ± 0 .7 0 7 V.

(d) Vang =

20

(± 4 0 5 ° ) = ± 8 .1 0 V,

= 1 sin (± 4 0 5 ° ) = ± 0 .7 0 7 V.

E xam ple 6-12 clearly illustrates that the A D 639, rem arkable as it is, cannot output the sine of, for exam ple, 36,000°. This w ould require an angle voltage o f 720 V T he nor­ mal ± 1 5 -V supply lim its the guaranteed usable input angle to ± 5 0 0 ° , or ± 1 0 .0 0 0 V. We extend the results o f E xam ple 6-12 to sum m arize briefly the perform ance o f the sine fu n c­ tion generator in Table 6-1 and Fig. 6 -1 1(b). In Fig. 6 -1 1(b), VQ is plotted against both Vang and 0. A study o f this figure shows that if Vang could be varied linearly by a triangle wave, VD w ould vary sinusoidally. Further, if the frequency o f the triangle wave could be varied easily, the sine-w ave frequency could easily be tuned, adjusted, or varied. We pursue this observation in the next section.

6-9 PRECISION SINE-WAVE GENERATOR

________________________________________________

6-9.1 Circuit Operation C onnect the precision triangle-w ave oscillator in Fig. 6-10 to the sine function generator in Fig. 6-11 to construct the precision sine-w ave generator in Fig. 6-12. As a bonus, we also have precision triangle-w ave and square-w ave outputs. The 1.80-V reference voltage o f the A D 639 is connected to m odulation inputs 16 and 17 o f the A D 630 m odulator (Fig. 6-9). C ircuit operation is now exam ined by reference to Fig. 6-12.

176

Chapter 6

TABLE 6-1

AD639 Sine Functions3 Output (V)

Input 0 (angular degrees) 0 ±45 ±90 ±135 ±180 ±225 ±270 ±315 ±360 ±405 ±450 ±495 ±500

= 10 sin 6 (wire pin

V.„fi (V)

V0 - 1 sin 6 (wire pin 10 to 9)

0.00 ± 0 .9 0 ± 1 .8 0 ± 2 .7 0 ± 3 .6 0 ± 4 .5 0 ± 5 .4 0 ± 6 .3 0 ± 7 .2 0 ± 8 .1 0 ± 9 .0 0 ± 9 .9 0 ± 1 0 .0 0

0.000 ± 0.707 ± 1.000 ± 0.707 0.000 ± 0.707 ± 1.000 ± 0.707 0.000 ± 0.707 ± 1.000 ± 0.707 ± 0.643

0.000 ± 7.07 ± 10.07 ± 7.07 0.000 ± 7 .0 7 ± 1 0 .0 0 ± 7 .0 7 0.00 ± 7 .0 7 ± 1 0 .0 0 ± 7.07 ± 6 .4 3

Angle voltage,

10 to 16)

a C onnect terminal 10 to 9 to pin program V0 = sin 0; or connect pin 10 to 16 to pin program Va = 10 sin 0. Input angle voltage Vang = (20 m V /l°C ) 0.

Triangle-wave rise time, 0 to T/2 in Fig. 6-12(b) 1. A D 63 0 a. Pin 13 is at —Vref = —1.8 V, causing b. Pin 9 to select gain = - 1 to hold 13 at - 1 . 8 V and c. O p am p output voltage to ram p up. 2. Op am p a. Pin 6 ram ps from —Vref = —1.8 V tow ard + V ref = 1.8 V to b. H old pin 9 of the A D 630 negative and c. D rive input 1 o f the A D 639 w ith an angle voltage linearly from - 1 . 8 to 1.8 V. 3. A D 639 a. Pin l ’s input angle voltage corresponds to an input angle that varies linearly from - 9 0 ° to + 9 0 °. b. Pin 13 outputs VQ — 10 sin 6 from —10 to + 1 0 V. W hen op am p pin 6 reaches + 1 .8 V, pin 9 o f the A D 630 goes positive to select a gain o f + 1 . Its output, in 13, snaps to + 1 .8 V. T his begins the fa ll time.

Triangle-wave fall time, T/2 to T in Fig. 6-12(b) l . A D 630: C auses the triangle w ave to ram p down from + 1 .8 V to - 1 .8 V. A t - 1 . 8 V, gain is sw itched to —1 and a new cycle begins.

177

Signal Generators

(a) Precision sine-wave generator circuit.

( a ) £/\ pus '

FIG U RE 6-12 Frequency of the precision sine-square-triangle-wave generator in (a) can be easily changed by adjusting Output waveshapes are shown in (b). Their am­ plitudes are independent of frequency.

178

Chapter 6

2. Op am p: A pplies an angle voltage to input pin 1 o f the A D 639 that varies linearly from + 1 .8 to ~ 1 .8 V. 3. A D 639: Its input angle voltage corresponds to an input angle o f 0 = + 9 0 ° to - 9 0 ° . Pin 13 outputs a sine w ave that varies from + 1 0 to —10 V.

6-9.2 Frequency of Oscillation T he frequency of o s c illa tio n ,/ is determ ined by R if C, and the op am p in Fig. 6 - 12(a) from

( 6 - | 4 )

Peak am plitudes of the triangle wave and square w ave are precisely equal to ± 1 .8 V. T he sine w ave has peak am plitude o f ± 1 0 V and is synchronized to the triangle wave (for the ± 1-V peak, change the A D 639 pin 10 connection to —Vs).

E x a m p le 6-13 Let C = 0.025 |jlF in Fig. 6 - 12(a) (two 0.05-fjiF capacitors in series). How does frequency change as Ri is changed from 10 k f l to 100 k fl? S o lu tio n

F rom Eq. (6-14),

/*_ l __ * 1 | | J ~ 4(10 kfl)(0 .0 2 5 |xF) ” Z

/»_ l t _ *Q Q |^| ^ 1 ~ 4(100 k fl)(0 .0 2 5 julF) ~~ Z

E xam ple 6-13 show s the overw helm ing superiority o f this m ulti w ave generator. Frequency is tuned easily an d with precision. A lthough we have used the A D 639 to gen ­ erate a sine wave, this chip is a universal trigonom etric function g enerator and could be rew ired to produce other trigonom etric w aveform s.

6-9.3 High-Frequency Waveform Generator M a x im ’s M A C 038 is a 20-pin, high-frequency, precision function g enerator w hose fre­ quency can be controlled over a w ide range from 0.1 Hz to 20 M H z. It can produce sine, trian g le, or sq uare w aves at o utputs that are selected by an ap p ro p riate co d e at tw o (transistor-transistor logic) T T L -com patible select inputs. It can also produce saw tooth or pulse outputs as well as a synchronizing output. A pplications for this versatile chip include function or FSK generation, V C O s (volt­ age controlled oscillator), frequency m odulators, and synthesizers, as well as pulse-w idth m odulation.

179

Signal Generators 6-10 PSPiCE SIMULATION O F SIGNAL GENERATOR CIRCUITS

______________________

In this section, we will use PSpice to m odel and sim ulate the perform ance o f four signal generator circuits studied in this chapter: the free-running m ultivibrator, the one-shot m ul­ tivibrator, the bipolar triangle-w ave generator, and the unipolar triangle-w ave generator.

6-10.1 Free-Running Multivibrator R efer to Fig. 6-1 and create the PSpice model o f the circuit. Set the resistor and capacitor values as given in E xam ples 6-1 and 6-2. Use the 741 op am p if you are using the evalu­ ation softw are package o f PSpice. O btain a plot o f Vc and VQ versus tim e. For the circuit to begin oscillating in a sim ulation it is necessary to provide a sudden im pulse at the be­ ginning of the sim ulation. This stim ulus can be generated by using tw o pulse sources in­ stead o f two dc sources to pow er the op amp. T he pulse w idth will be set to a m uch longer tim e interval than the period o f oscillation and will have a fast rise tim e to sim ulate sud­ denly applying pow er to the circuit. To begin, place the follow ing parts in the w ork area. Draw = > Get New Part Part = = = = = =

> > > > > >

uA741 VPULSE R C G LO BAL AGND

Num ber

Library

1 2 3 J 4 4

eval.slb source.sib analog.slb analog.sJb port.sib port.sib

{Note: We are using VPULSE instead o f VDC for the op am p supplies.) A rrange the parts and w ire the circuit as show n in Fig. 6-1. C hange the attributes o f the parts as given in E xam ples 6-1 and 6-2. Set up each o f the VPULSE attributes by double-clicking on the sym bol. In the pop-up box, set the values for VI (m inim um input voltage), V2 (m axim um input voltage), TD (tim e delay), TR (rise tim e), TF (fall tim e), PW (pulse w idth), and PER (period):

VI = > OV = > Save Attr V2 = > 15 V = > Save Attr = > Change Display = > Both name and value TD = > 0 = > Save Attr TR = > Ins = > Save Attr TF = > Ins = > Save Attr PW = > 50s = > Save Attr PER = > 51s = > Save Attr E ach pulse attribute pop-up box is set up the sam e because the —V supply has been ro ­ tated 180 degrees, as we have done w ith the dc supplies connected to pin 4 o f other op am ps. D ouble-click on the lead from the capacitor and label it Vc. D ouble-click on the lead from the output term inal of the op am p and label it Vo (see Fig. 6-13).

182

Chapter 6

In order to obtain a plot o f E , and V0 versus tim e, w e m ust initialize the transient menu.

Analysis = > Setup = > Select Transient Click Transient = > Print Step: = > lOOus = > Final Time: = > 10ms Save the circuit as a file w ith the .SCH extension. Run the sim ulation

Analysis = > Simulate In the P robe w'indow, select

Trace = > Add = > V[Ei] = > V[Vo] Label the plots and obtain a printout as show n in Fig. 6-16.

15 v r

10 v 5V

Input _ pulse,

0 V

^

V0

/

-5 V

-10 V -1 5 V Os

Period -► 1 2 ms

1---------------1 4 ms

6 ms

1 ------------1 10 ms

8.ms

Tim e

FIGURE 6-16 Waveforms for the one-shot multivibrator modeled in PSpice in Fig. 6-15.

6-10.3 Bipolar Triangle-Wave Generator C reate the P Spice m odel o f the bipolar triangle-w ave generator show n in Fig. 6-6. Use tw o 741 op am ps. O btain a plot o f VA and V0 versus tim e. Place the follow ing parts in the w ork area. Draw => Qet New Part Part = = = = = =

> > > > > >

uA741 VPULSE R C G LOBAL AGND

Num ber

Library

2 2 4 1 6 5

eval.slb source.sib analog.sib analog.slb port, sib port.slb

183

Signal Generators

A rrange the parts, change the attributes, and w ire the circuit as show n in Fig. 6-6. T he VPULSE attributes are set the sam e as they are in Section 6-10.1 (see Fig. 6-17).

FIG U RE 6-17

Bipolar triangle-wave generator modeled in PSpice.

In order to obtain a plot of VA and VQ versus tim e, set the transient m enu.

Analysis = > Setup = > Select Transient C lick Transient = > Print Step: = > 0.01ms = > Final Time: = > 3ms Save the circuit as a file with the .SCH extension. Run the sim ulation

Analysis = > Simulate In the Probe w indow, select

Trace = > Add = > V[VA] = > V[Vo] Label the plots and obtain a printout as show n in Fig. 6-18.

6-10.4 Unipolar Triangle-Wave Generator M odify the P S pice m odel o f Fig. 6-17 to create the un ip o lar trian g le-w av e g en erato r show n in Fig. 6-7. From the P Spice parts list, obtain a 1N4002 diode and place it in se ­ ries w ith pR (see Fig. 6-19). Save the circuit w ith the .SCH extension.

185

Signal Generators 15 V

10 V

5V

0V

-5 V

-10 V -1 5 V Os

0.5 ms

1.0 ms

1.5 ms

2.0 ms

2.5 ms

3.0 ms

Tim e

FIGURE 6-20 Waveforms VA and V(, for the unipolar triangle-wave genera­ tor of Fig. 6-19.

PROBLEMS

__________________________________________________________________________

6-1. Make two drawings of a multivibrator circuit with R\ = 100 kH, R2 = 86 kfl, Rf = 10 kfl, and C = 0.01 /xF. Show the direction of current through C and calculate both VUT and VLT for (a) = +Ksa( = 15 V; (b) V, = - V sa( = - 1 5 V. 6-2. Calculate the frequency of oscillation for the multivibrator circuit in Problem 6-1. 6-3. In Problem 6-1, if C is changed to 0.1 ^F, do you expect the output frequency to oscillate at 500 Hz? (See Example 6-3.) What could you do to Rf to increase frequency to 1000 Hz? 6-4. The monostable multivibrator of Figs. 6-4 and 6-5 generates a negative output pulse in re­ sponse to a negative-going input signal. How would you change these circuits to get a posi­ tive output pulse for a positive-going input edge? 6-5. Explain what is meant by monostable recovery time. 6-6. Sketch a one-shot multivibrator circuit whose output will deliver a negative pulse lasting 1 ms with a recovery time of about 0.1 ms. 6 -7 .

Assume for simplicity that saturation voltages in the triangle-wave oscillator of Fig. 6-6 are ± 15 V, Rj = R = 10 kH, C = 0.1 fiF, and pR = 50 kfl. Find the peak triangle-wave voltages and oscillating frequency.

6-8. Refer to the triangular-wave oscillator circuit of Fig. 6-6. What happens to peak output volt­ ages and oscillating frequency if you (a) double pR only; (b) double R, only; (c) double ca­ pacitor C only? 6-9. Change pR to 14 kfl and C to 0.1 jjlF in the unipolar triangle-wave generator of Fig. 6-7. Find the resulting peak output voltage and frequency of oscillation. (See Example 6-7.)

186

Chapter 6

6-10. In the sawtooth-wave generator of Fig. 6-8(a), let Vref = 1 V, R, = 10 kfl, and C = 0.1 jjlF. (a) Find an expression for frequency / i n terms of Ev (b) C alculate/for Et = I V and E, = 2 V. 6-11. These questions refer to the AD630 balanced modulator circuit in Fig. 6-9. (a) Name the application for which the A.D630 is wired. (b) When pin 9 is at a positive voltage, which amplifier is selected, and what is the value of \/»o?• (c) Suppose that Vrcl. is a ± l-V-peak sine wave and pin 9 is at 1 V; what happens at V0 when pin 9 is changed to - 1 V? 6-12. Figure 6-10 shows a precision triangle/square-wave oscillator.Threecomponentscontrol peak output voltages and oscillating frequency, R h C, and Vref. (a) Which does what? (b) Can the oscillating frequency be adjusted independent of peak outputs, and vice versa? (c) What must be done to change the frequency from 100 to 500 Hz and the peak voltages from ±5 V to ±1 V? 6-13. V0 ~ 0.866 V in the sine function generator circuit of Fig. 6-11. (a) What angle does this represent? (b) What is the value of the input angle voltage? 6-14. Calculate V0 in Fig. 6-11 when the input angle is 30° and pin 10 is wired to (a) pin 9; (b) pin 16. 6-15. Design a sine-wave oscillator whose frequency can be varied from 0.5 Hz to 50 Hz with just a single variable resistor.

CHAPTER 7 Op Amps with Diodes

LEARNING

OBJECTIVES

____________________________________________________

U pon com pleting this chapter on op am ps w ith diodes, you w ill be able to: - D raw the circuit for a precision (or linear) half-w ave rectifier. • Show current flow and circuit voltages in a precision half-w ave rectifier for either p o s­ itive or negative inputs. • D o the sam e for precision full-w ave rectifiers. • Sketch tw o types o f precision full-w ave rectifier circuits. • E xplain the operation o f a peak detector circuit. • A dd one cap acito r to a p rec isio n half-w ave rectifier to m ake an ac-to -d c co n v erter (m ean-average-value) circuit. • Explain the operation o f dead-zone circuits. • D raw the circuit for and explain the operation o f precision clipper circuits.

187

diode can be designed using an op am p and tw o ordinary diodes. T he result is a pow er­ ful circuit capable o f rectifying input signals o f only a few m illivolts. T he low cost o f this equivalent ideal diode circuit allow s it to be used routinely for many applications. They can be grouped loosely into the follow ing classifications: linear half-w ave rectifiers and precision full-w ave rectifiers. 1. L inear half-w ave rectifiers. T he linear half-w ave rectifier circuit delivers an output that depends on the m agnitude and p o la rity o f the input voltage. It w ill be show n that the linear half-w ave rectifier circuit can be m odified to perform a variety o f sig­ nal-processing applications. T he linear half-w ave rectifier is also called a precision half-w ave rectifier and acts as an ideal diode. 2. P recision fu ll-w a v e rectifiers. T he precision full-w ave rectifier circuit delivers an output proportional to the m agnitude but not the polarity o f the input. For exam ple, the output can be positive at 2 V for inputs o f either + 2 V or - 2 V. Since the ab ­ solute value o f + 2 V and —2 V is equal to + 2 V, the precision full-w ave rectifier is also called an absolute-value circuit. A pplications for both linear half-w ave and precision full-w ave rectifiers includes: 1. 2. 3. 4. 5. 6. 7.

8. 9. 10. 1 1.

D etection o f am plitude-m odulated signals D ead-zone circuits Precision bound circuits or clippers C urrent sw itches W aveshapers Peak-value indicators Sam ple-and-hold circuits A bsolute-value circuits A veraging circuits Signal polarity detectors A c-to-dc converters

T he functions listed are often necessary to condition signals before they are applied to an input of a m icrocontroller.

7-1 LINEAR HALF-WAVE RECTIFIERS ______________________________________________________ 7-1.1 Introduction L inear half-w ave rectifier circuits transm it only one-half cycle o f a signal and elim inate the other by bounding the output to 0 V. The input half-cycle that is transm itted can be either inverted or noninverted. It can also experience gain or attenuation, or rem ain un­ changed in m agnitude, depending on the choice o f resistors and placem ent o f diodes in the op am p circuit.

In Fig. 7-2(b), negative input E t forces the op am p output VOA to go positive. This causes D 2 to conduct. T he circ u it then acts like an inverter, since Rf = /?, and VD = - ( - £ , ) = + £ ,. Since the ( - ) input is at ground potential, diode D , is reverse biased. Input current is set by £,//?, and gain by - R f /Ri. R em em ber that this gain equation ap ­ plies only for negative inputs, and V0 can only be positive or zero. C ircuit operation is sum m arized by the w aveshapes in Fig. 7-3. V0 can only go p o s­ itive in a linear response to negative inputs. The m ost im portant property o f this linear half-w ave rectifier w ill now be exam ined. An ordinary silicon diode or even a hot-carrier diode requires a few tenths o f volts to becom e forw ard biased. A ny signal voltage below this threshold voltage cannot be rectified. However, by connecting the diode in the feed ­ back loop o f an op am p, the threshold voltage o f the diode is essentially elim inated. For exam ple, in Fig. 7-2(b) let £, be a low voltage o f - 0 .1 V. E i and /?, convert this low v o lt­ age to a current that is conducted through D 2. VOA goes to w hatever voltage is required to supply the necessary diode drop plus the voltage drop across Rf . T hus m illivolts o f in ­ put voltage can be rectified, since the d io d e’s forw ard bias is supplied autom atically by the negative feedback action o f the op amp.

and V0A

FIGURE 7-3 Input, output, and transfer characteristics of a positiveoutput, ideal, inverting half-wave rectifier.

194

Chapter 7

£ , and V

FIGURE 7-6

Input and output voltages for the polarity separator of Fig. 7-5

7-2 PRECISION RECTIFIERS: THE A BSO LU TE-VALU E CIRCUIT

_______________________

7-2.1 Introduction The precision full-w ave rectifier transm its one polarity o f the input signal and inverts the other. T hus both half-cycles o f an alternating voltage are transm itted but are converted to a single polarity o f the circu it’s output. T he precision full-w ave rectifier can rectify input voltages with m illivolt am plitudes. T his type o f circuit is useful to prepare signals for m ultiplication, averaging, or d e­ m odulation. T he characteristics o f an ideal precision rectifier are show n in Fig. 7-7. T he precision rectifier is also called an absolute-value circuit. The absolute value of a num ber (or voltage) is equal to its m agnitude regardless o f sign. For exam ple, the absolute values of | +2 | and |—2 | are +2. T he sym bol 11 m eans “absolute value of.” Figure 1-1 shows that the output equals the absolute value of the input. In a precision rectifier circuit the o ut­ put is either negative or positive, depending on how the diodes are installed.

198

Chapter 7

T he w aveshapes in Fig. 7-8(c) show that V0 is alw ays o f positive polarity and equal to the absolute value of the input voltage. To obtain negative outputs fo r either polarity o f E h sim ply reverse the diodes.

High-impedance precision full-wave rectifier. T he second type o f p reci­ sion rectifier is show n in Fig. 7-9. T he input signal is connected to the noninverting op am p inputs to obtain high input im pedance. Figure 7-9(a) show s w hat happens for p o si­ tive inputs. Ej and set the current through diode D P. T he ( —) inputs o f both op am ps are at a potential equal to E L so that no current flow s through R 2>R 3, and R 4. T herefore, V0 = Ei for all positive input voltages. W hen Ei goes negative in Fig. 7-9(b), E, and R j set the current through both R ] and R 2 to turn on diode D /V. Since R { = R 2 = R> the anode o f D N goes to 2E f or 2 ( - £ , ) = —4 V. The ( - ) input of op amp B is at ~ E r The voltage drop across R 3 is 2 E{ —E { or ( - 4 V) — ( —2) = - 2 V. This voltage drop and R 3 establishes a current / 3 through both /?3 and R4 equal to the input current /. Consequently, V0 is positive when E t is negative. Thus VQ is always positive despite the polarity of E h so VG = lE t |. T he w aveshapes for this circuit are the sam e as in Fig. 7-8(c). N ote that the m axi­ m um value o f Ei is lim ited by the negative saturation voltage o f the op am ps.

7-3 PEAK D ETECTO RS

_______________________________________________________________________

In addition to rectifying a signal precisely, diodes and op am ps can be interconnected to build a peak detector circuit. This circuit follow s the voltage peaks o f a signal and stores the hig h est value (alm ost indefinitely) on a capacitor. If a higher peak signal value com es along, this new value is stored. The highest peak voltage is stored until the capacitor is discharged by a m echanical or electronic sw itch. This peak detector circuit is also called a fo llo w -a n d -h o ld circuit or pea k follow er. We shall also see that reversing tw o diodes changes this circuit from a peak to a valley follower.

7-3.1 Positive Peak Follower and Hold T he peak follow er-and-hold circuit is shown in Fig. 7-10. It consists o f tw o op am ps, two diodes, a resistor, a hold capacitor, and a reset switch. Op am p A is a precision half-wave rectifier that charges C only when input voltage E t exceeds capacitor voltage Vc . Op am p B is a voltage follow er w hose output signal is equal to Vc . The follow er's high input im ­ pedance does not allow the capacitor to discharge appreciably. To analyze circuit operation, let us begin with Fig. 7 - 10(a). W hen E t exceeds Vc, diode D P is forw ard biased to charge hold capacitor C. As long as Et is greater than Vc , C charges tow ard £ :. Consequently, Vc follow s £, as long as E t exceeds Vc . W hen E { drops below Vc , diode D N turns on as show n in Fig. 7 - 10(b). D iode D P turns off and d is­ connects C from the output o f op am p A. D iode D P m ust be a low -leakage-type diode or the capacitor voltage will discharge (droop). To m inim ize droop, op am p B should require

Op Am ps with Diodes

203

7-4.3 AC-to-DC Converter A large-value low -leakage capacitor (10-/xF tantalum ) is added to the absolute-value cir­ cuit of Fig. 7-13. The resultant circuit is the MAV am plifier or ac-to-dc converter show n in Fig. 7-14. C apacitor C does the averaging o f the rectified output o f op am p B. It takes about 50 to 500 cycles of input voltage before the capacitor voltage settles dow n to its fi­ nal reading. If the w aveshapes of Fig. 7-12 are applied to the ac-to-dc converter, its o u t­ put will be the MAV o f each input signal.

FIGURE 7-14 Add one capacitor to the absolute-value amplifier of Fig. 7-13 to get this ac-to-dc converter or mean-absolute-value amplifier.

7-5 DEAD ZONE CIRCUITS 7-5.1 Introduction C om parator circuits tell i f a signal is below or above a particular reference voltage. In contrast w ith the com parator, a dead-zone circuit tells by how m uch a signal is below or above a reference voltage.

7-5.2 Dead-Zone Circuit with Negative Output A nalysis of a dead-zone circuit begins w ith the circuit o f Fig. 7-15. A convenient regulated supply voltage + V and resistor m R establish a reference voltage Uref. Vre( is found from the equation Vrc{ — + V /m . As will be shown, the negative o f Vref, —Vref, will establish the dead zone. In Fig. 7-15(a), current I is determ ined by + V and resistor m R at / = + V /m R D iode D n will conduct for all positive values o f E h clam ping VOA and V qb t0 0 V. T herefore, all positive inputs are elim inated from affecting the output. In order to get any

205

Op A m ps with Diodes

output at VOA, Ei m ust go negative, as show n in Fig. 7 - 15(b). D iode D P will conduct when the loop current E t/R through £, exceeds the loop current V/mR through resistor mR. T he value o f Ei necessary to turn on D P in Fig. 7-15(b) is equal to - Vref. T his co n ­ clusion is found by equating the currents

= +v R

mR

and solving for E (: +V E i = ---------- = - V ref m

(7 -la )

+ \/ ^ re f= -----m

( 7 - lb)

w here

T hus all values o f E t above - Vref will lie in a dead zone w here they will not be tran s­ m itted [see Fig. 7-1 5(c)]. O utputs VOA and VOB will be zero. W hen Ei is below Vref, £, and VTCf are added and their sum is inverted at output VOA. VOA is reinverted by op am p B. T hus VOB only has an output w hen E, goes below Vref. VOB tells you by how many volts £, lies below Vref. Circuit operation is sum m arized by the w aveshapes o f Fig. 7 - 15(c) and illustrated by an exam ple.

E x a m p le 7-1 In the circuit o f Fig. 7-15, 4- V = + 15 V, m R = 30 k fi, and R = 10 k f), so that m = 3. Find (a) Vref; (b) VOA w hen E, = - 1 0 V; (c) w hen E, = - 1 0 V. S o lu tio n (a) From Eq. (7 -lb ), Vref = + 1 5 V/3 = 5 V. (b) VOA and VOB will all values o f E { above —Vref = - 5 V, from Eq. (7 -la ). T h erefo re, VOA = - ( - 1 0 V) - 5 V = + 5 V. (c) Op am p B inverts the output o f VOA so that T hus, VOB indicates how m uch E { goes below —Vref. All input signals above dead zone and are elim inated from the output.

equal zero for - E j - Vr&f = VOB = - 5 V. — Vref fall in a

7-5.3 Dead-Zone Circuit with Positive Output If the diodes in Fig. 7-15 are reversed, the result is a positive-output dead-zone circuit as show n in Fig. 7-16. R eference voltage Vref is found from Eq. (7 -lb ): Vrer = —15 V/3 = —5 V. W henever Ej goes above —Vref = - ( - 5 V) = + 5 V, the output VOB tells by how m uch Ei exceeds —Vref. T he dead zone exists for all values o f below — Vrcf.

2R

+

+V0

(b) W aveshapes for precision clipper.

FIGURE 7-18 A precision clipper is made from a bipolar dead-zone circuit plus an added resistor Rc.

208

Chapter 7

7-5.4 Bipolar-Output Dead-Zone Circuit T he positive and negative output dead-zone circuits can be com bined as show n in Fig. 7-17 and discussed in Fig. 7-18. The VOA outputs from Figs. 7-15 and 7-16 are connected to an inverting adder. The adder output VOB tells how much Ej goes above one positive ref­ erence voltage and also how m uch E( goes below a different negative reference voltage.

7-6 PRECISION CLIPPER

_____________________________________________________________________

A clipper or am plitude lim iter circuit clips off all signals above a positive reference volt­ age and all signals below a negative reference voltage. The reference voltages can be m ade sym m etrical or nonsym m etrical around zero. C onstruction o f a precision clipper cir­ cuit is accom plished by adding a single resistor, R c, to a bipolar output dead-zone circuit as show n in Fig. 7-18. The outputs o f op am ps A and B are each connected to the input o f the inverting adder. Input signal £, is connected to a third input o f the inverting adder, via resistor R c . If R c is rem oved, the circuit would act as a dead-zone circuit. However, w hen R c is present, input voltage Et is subtracted from the dead-zone circu it’s output and the result is an inverting precision clipper. C ircuit operation is sum m arized by the w aveshapes in Fig. 7 - 18(b). O utputs VQa and VOB are inverted and added to ~ E r The plot of V0 versus tim e show s by solid lines how the clipped output appears. The dashed lines show how the circuit acts as a deadzone circuit if R c is removed.

7-7 TRIANGULAR TO-SINE-WAVE CONVERTER

_________________________________________

V ariable-frequency sine-w ave oscillators are m uch harder to build than variable-frequency triangular-w ave generators. The circuit of Fig. 7-19 converts the output o f a triangularwave generator into a sine wave that can be adjusted for less than 5% distortion. The triangle-to-sine-w ave converter is an am plifier w hose gain varies inversely with am plitude o f the output voltage. R { and R3 set the slope of V0 at low am plitudes near the zero crossings. As V0 in­ creases, the voltage across R 3 increases to begin forw ard biasing D, and Z)3 for positive outputs, or D 2 and D 4 for negative outputs. W hen these diodes conduct, they shunt feed­ back resistance /?3, lowering the gain. This tends to shape the triangular output above about 0.4 V into a sine wave. In order to get rounded tops for the sine-wave output, R 2 and diodes D5 and D 6 are adjusted to make am plifier gain approach zero at the peaks o f V0. T he circu it is adjusted by co m paring a 1-kHz sine wave and the o u tp ut o f the triangle/sine-w ave converter on a dual-trace CRO. /?,, R 2, R 3, and the peak am plitude of Ej are adjusted in sequence for best sinusoidal shape. The adjustm ents interact, so they should be repeated as necessary. {Note: A lthough the circuit of Fig. 7-19 will shape a tri­ angular wave to a sine wave, the parts count is high, but you may need to generate such a w aveform with readily available parts. A better solution is to purchase an IC chip that generates triangle, square, and sine waves in a single package.)

210

Chapter 7

A rrange the parts as show n and change the attributes o f the parts as given in Fig. 7-2. Set up the sine-w ave attributes by double-clicking the sym bol. In the pop-up w indow change VOFF, VAMPL, and FREQ

VOFF = > 0 = > Save Attr VAMPL = > 2V = > Save Attr —> Change Display —> Both name and value FREQ = > 1kHz = > Save Attr = > Change Display = > Both name and value C lose the pop-up box. D ouble-click on the lead from the sine wave to R x and label it E h D ouble-click on the lead from the cathode term inal of diode D 2 and label it V0 (see Fig. 7-20).

_y2 15V

VO -V Vam pl = 2V freq = lk “ l5V

Vo FIGURE 7-20

PSpice model of Fig. 7-

In order to obtain a plot o f E, and VG versus tim e, w e m ust initialize the T ransient menu.

Analysis = > Setup = > S elect Transient C lick Transient = > Print Step: = > 1/us = > Final Time: = > 2ms Save the circuit as a file w ith the .SCH extension. Run the sim ulation

Analysis = > Simulate In the P robe window, select

Trace = > Add = > V[Ei]

= > V[Vo] L abel the plots and obtain a printout as show n in Fig. 7-21.

215

Op Am ps with Diodes

FIG U RE 7-25 in Fig. 7-24.

PROBLEMS

Plot of Vr, versus time for the mean-absolute-value amplifier

_____________________________________________________________________

7-1. What is the absolute value of +3 V and - 3 V? 7-2. If the peak value of E{ — 0.5 V in Fig. 7-1, sketch the waveshapes of VQ vs. t and Va vs. £, for both a silicon and an ideal diode. 7-3. If E{ is a sine wave with a peak value of 1 V in Figs. 7-2 and 7-3, sketch the waveshapes of V0 vs. l and VQ vs. Er 7-4. If diodes D\ and D2 are reversed in Fig. 7-2, sketch V0 vs. £, and Va vs. t. 7-5. Sketch the circuit for a signal polarity separator. 7-6. Let both diodes be reversed in Fig. 7-8. What is the value of Vn if £,• = +1 V or E, - - 1 V? 7-7. What is the name of a circuit that follows the voltage peaks of a signal and stores the highest value? 7-8. How do you reset the hold capacitor’s voltage to zero in a peak follower-and-hold circuit? 7-9. How do you convert the absolute-value amplifier of Fig. 7-13 to an ac-to-dc converter? 7-10. If resistor mR is changed to 50 kH in Example 7-1, find (a) Vref; (b) VOA when E, = 10 V; (c) when Et = 10 V. 7 - 11. If resistor Rc is removed in Fig. 7-18, sketch Va vs. Et.

CHAPTER 8 Differential, Instrumentation, and Bridge Amplifiers

LEARNING O B JEC TIVES

_________________________________________________

W hen you com plete this chapter on differential, instrum entation, and bridge am plifiers, you w ill be able to: • D raw the circuit for a basic differential am plifier, state its o u tp u t-in p u t equation, and ex ­ plain why it is superior to a single-input am plifier. •D e fin e com m on-m ode and differential input voltage. •D raw the circuit for a differential input to differential output voltage am plifier and add a differential am plifier to m ake a three-op-am p instrum entation am plifier (IA). • C alculate the output voltage o f a three-op-am p instrum entation am plifier if you are given the input voltages and resistance values. • U se com m ercially available instrum entation am plifiers. • Explain how the sense and reference term inals o f an IA allow you to (1) elim inate the effects o f connecting-w ire resistance on load voltage, (2) obtain load current boost, or (3) m ake a differential voltage-to-current converter (ac current source).

Differential, Instrumentation, and Bridge Amplifiers

217

• E xplain how a strain gage converts tension or com pression forces into a change in re­ sistance. • C onnect strain gages into a passive bridge resistance netw ork to convert gage resistance change into an output voltage. •A m p lify the strain gage b rid g e’s differential output w ith an instrum entation am plifier. • M easure pressure, force, or weight. • Draw the circuit for a bridge am plifier and show how it converts a change in transducer resistance to an output voltage. • U se the bridge am plifier to m ake a tem perature-to-voltage converter.

8-0 INTRODUCTION

__________________________________________________________________________

The m ost useful am plifier for m easurem ent, instrum entation, or control is the in stru m en ­ tation amplifier. It is designed w ith several op am ps and precision resistors, w hich m ake the circuit extrem ely stable and useful w here accuracy is im portant. T here are now many integrated circuits available in single packages. A lthough these packages are m ore ex ­ pensive than a single op am p, w hen perform ance and precision are required, the in stru ­ m entation am plifier is well w orth the price, because its perform ance cannot be m atched by the average op amp. A first cousin and basic block w ithin the instrum entation am plifier is the differen­ tial amplifier, also referred to as a subtractor circu it T his chapter begins with the differ­ ential am plifier, show ing the applications in w hich it is superior to the ordinary inverting or noninverting am plifier. T he differential am plifier, w ith som e additions, leads into the instrum entation am plifier, w hich is discussed in the second part o f this chapter. T he final sections consider bridge am plifiers, w hich involve both instrum entation and basic differ­ ential am plifiers.

8-1 BASIC DIFFERENTIAL AMPLIFIER

______________________________________________________

8-1.1 Introduction T he differential am plifier can m easure as well as am plify sm all signals that are buried in m uch larger signals. How the differential am plifier accom plishes this task w ill be studied in Section 8-2, but first, let us build and analyze the circuit perform ance o f the basic d if­ ferential am plifier. Four precision (1% ) resistors and an op am p m ake up a differential am plifier, as show n in Fig. 8-1. T here are two input term inals, labeled ( - ) input and ( + ) input, co r­ responding to the closest op am p term inal. If E v is replaced by a short circuit, E 2 sees an inverting am plifier w ith a gain o f —m. T herefore, the output voltage due to E 2 is —m E 2. N ow Jet E 2 be short-circuited; £ , divides betw'een R and m R to apply a voltage o f E xm! (1 + m) at the op a m p ’s ( + ) input. This divided voltage sees a noninverting am plifier with a gain of {m + 1). T he output voltage due to E\ is the divided voltage, E vm!{ 1 + m), tim es the noninverting am plifier gain, (1 + m), w hich yields m E x. T herefore, E { is am plified at

219

Differential, Instrum entation, and Bridge A m plifiers

8-1.2 Common-Mode Voltage The output o f the differential am plifier should be 0 w hen E x — E 2. T he sim plest w ay to apply equal voltages is to w ire the inputs together and connect them to the voltage source (see Fig. 8-2a). For such a connection, the input voltage is called the com m on-m ode in ­ p u t voltage, E Cm > Now VQ w ill be 0 if the resistor ratios are equal (m R to R for the in ­ verting am plifier gain equals m R to R o f the voltage-divider netw ork). Practically, the re­ sistor ratios are equalized by installing a p o ten tio m eter in series w ith one resistor, as show n in Fig. 8-2a. T he potentiom eter is trim m ed until V0 is reduced to a negligible value. This causes the com m on-m ode voltage gain, Vn/E CMi to approach 0. It is this ch aracter­ istic o f a differential am plifier that allow s a sm all signal voltage to be picked out o f a

mR = 100 k£2

------------V W -------------

+v

(a) + 10V

*4

(b) FIG U RE 8-2 (a) The common-mode voltage gain should be zero, (b) Basic differential amplifier connected to a Wheatstone bridge.

220

Chapter 8

larger noise voltage. It may be possible to arrange the circuit so that the larger undesired signal is the com m on-m ode input voltage and the sm all signal is the differential input voltage. T hen the differential am plifier’s output voltage will contain only an am plified version o f the differential input voltage. O perational am plifier circuits, as w ell as op am ps, have com m on m ode error, and the com m on m ode voltage is different for the circuit and the op am p. In this section, we will introduce com m on m ode voltage and com m on m ode rejection for a basic differential am plifier circuit, w hich is the fundam ental building block for the instrum entation am pli­ fier. In C hapter 10, com m on m ode rejection is revisited, but for the op am p only. R efer to the basic differential am plifier circuit in Fig. 8-2(b). T he difference v o lt­ age, £ , - E2, may be defined as input difference signal voltage, Eds, and the average o f the tw o inputs, ( E l + E 2)/2 , is defined as the com m on m ode voltage, E cm. In this circuit, the differential am plifier is being used to m easure the voltage difference betw een the two output nodes o f a W heatstone bridge.

E x a m p le 8-2 For the circuit o f Fig. 8-2(b), the bridge is slightly unbalanced and the input voltages with respect to ground are £ , = 5 V 4- 2 mV and E2 = 5 V — 2 mV. D eterm ine (a) the input signal voltage and (b) the com m on m ode voltage. S o lu tio n

(a) T he difference input signal voltage

is

£ ds = E x - E 2 = (5 V + 2 m V) - (5 V - 2 mV) = 4 mV (b) T he com m on m ode voltage is Cm

_ E l + E2 2

(5 V + 2 mV)

Ideally, the circuit design is to reject the com m on signal voltage.

+ (5V - 2 mV) 2

_

m ode voltage and am plify the input

8-1.3 Common M ode Rejection Introduction C ircuits or op am ps are com pared to one another by their com m on-m ode rejection (CM R ) or com m on m ode rejection ratio (CM R R). C M R is a m easure o f the change in output voltage w hen both inputs are changed by an equal am ount. CMR_R is a ratio expression,

Differential, Instrum entation, and Bridge A m plifiers

221

w hile com m on-m ode rejection is a logarithm o f that ratio. For exam ple, if C M R R is 10,000, the C M R is calculated by C M R = 20 log jo CM R R

( 8- 2 )

For this exam ple, CM R - 20 lo g ,o (10,000) = 80 dB

CMR of a Differential Amplifier Circuit R eferring to the circuit in Fig. 8-2(b) and using the superposition principle discussed in Section 8-1, the general equation for output voltage is Vn = E>

R?

7?3 + R4

Ri + R 2

(8-3)

For this circuit to w ork properly all resistors have to be carefully ratio-m atched to m aintain excellent com m on-m ode rejection. For exam ple, if all the resistors are equal, then the output voltage is the difference o f the input signals, V0 = E\ ~ E 2>and if the in ­ put voltages are equal, then VQ is 0. T his results in an infinite com m on-m ode rejection. How ever, let's consider the situation w here one resistor is m ism atched and the circu it val­ ues are E y — E 2 — E and R x = R 3 — R4 = R. T he m ism atched resisto r is R 2 w here R 2 = R + 0A % R . A pplying Eq. (8.3) yields Vn =

1.0017?\ / 2R 2.001 R ) \ R

X

E = 0.0005 X E

A lthough this expression shows the input voltage is reduced by 2000 (1/0.0005), the com m on-m ode rejection o f the circuit has been degraded to CM R = 20 log jo (2000) = 66 dB In this application, we have discussed only the com m on-m ode rejection o f the cir­ cuit and considered the op am p to be ideal. A disadvantage w ith the basic differential am ­ plifier circuit in Fig. 8.2(b) is that a slight m ism atch o f the resistors causes a degradation of the CM R. In this exam ple, the m ism atch was R2, but a m ore com m on m ism atch is the input resistors R\ or R 3 because o f loading effect(s). A solution to this problem will be covered in Section 8-3.

8-2 DIFFERENTIAL V ER SU S SINGLE-INPUT AM PLIFIERS 8-2.1 Measurem ent with a Single-Input Amplifier A sim plified w iring diagram o f an inverting am plifier is show n in Fig. 8-3. T he pow er com m on term inal is show n connected to earth ground. Earth ground com es from a co n ­ nection to a w ater pipe on the street side o f the w ater meter. G round is extended via co n ­ duit or a bare Rom ex w ire to the third (green) w ire o f the instrum ent line cord and finally to the chassis o f the am plifier. T his equipm ent or chassis ground is m ade to ensure the

Am plifier

ground

ground

FIGURE 8-3 Noise voltages act as if they are in series with the input signal Consequently, both are amplified equally. This arrangement is unworkable if En is equal or greater than £,. safety o f hum an operators. It also helps to drain off static charges or any capacitive co u ­ pled noise currents to earth. T he signal source is also show n in Fig. 8-3 to be connected to earth ground. Even if it w ere not grounded, there would be a leakage resistance or capacitance coupling to earth, to com plete a ground loop. Inevitably, noise currents and noise voltages abound from a variety o f sources that are often not easily identifiable. T he net effect o f all this noise is m odeled by noise volt­ age source En in Fig. 8-3. U sing the pow er supply com m on as the reference, it is evident that E n is in series with signal voltage E h so that both are am plified by a factor o f - 1 0 0 due to the inverting am plifier. En may be m uch larger than For exam ple, the skin sig ­ nal voltage due to heart beats is less than 1 mV, w hereas the bod y ’s noise voltage m ay be tenths o f volts or m ore; it w ould be im possible to m ake an EK G m easu rem en t w ith a sin ­ gle-input am plifier. W hat is needed is an am plifier that can distinguish betw een Et and E n and am plify only E h Such a circuit is the differential am plifier.

8-2.2 Measurem ent with a Differential Amplifier A differential am plifier is used to measure only the signal voltage (see Fig. 8-4). The signal voltage Ej is connected across the ( + ) and ( —) inputs of the differential amplifier. Therefore, Ei is am plified by a gain o f —100. Noise voltage En becom es the com m on-m ode voltage in­ put voltage to the differential am plifier as shown in Fig. 8-2. (Note: Apply superposition.) Therefore, the noise voltage is not am plified and has been effectively elim inated from hav­ ing any significant effect on the output Vot as long as the resistors are matched as shown.

223

Differential, Instrum entation, and Bridge A m p lifiers A m plifier

Third line ' cord wire

Earth g ro u n d

En noise voltage becom es the com m on-m ode voltage of the diff. amp and is not am plified

FIGURE 8-4 The differential amplifier is connected so that noise voltage be­ comes the common-mode voltage and is not amplified. Only the signal voltage Ej is amplified because it has been connected as the differential input voltage.

8-3 IMPROVING THE BASIC DIFFERENTIAL AMPLIFIER 8-3.1 Increasing Input Resistance T here are tw o disadvantages to the basic differential am plifier studied thus far: It has low input resistance, and changing gain is difficult because the resistor ratios m ust be closely m atched. T he first disadvantage is elim inated by buffering or isolating the inputs with volt­ age followers. This is accom plished w ith two op am ps connected as voltage follow ers in Fig. 8-5(a). The output o f op am p A, with respect to ground is E u and the output o f op am p A 2 w ith respect to ground is E2. The differential output voltage V0 is developed across the load resistor R L. V0 equals the difference between E\ and E 2 (V0 = E x — E 2). N ote that the output o f the basic differential am plifier of Fig. 8-1 is a single-ended output; that is, one side o f R L is connected to ground, and V0 is m easured from the output pin o f the op am p to ground. The buffered differentia] am plifier o f Fig. 8-5(a) is a differential output; that is, neither side of R L is connected to ground, and V0 is m easured only across R L.

8-3.2 Adjustable Gain T he second disadvantage o f the basic differential am plifier is the lack o f adjustable gain. T his problem is elim inated by adding three m ore resistors to the buffered am plifier. T he resulting buffered, differential-input to differential-output am plifier w ith adjustable gain is show n in Fig. 8-5(b). T he high input resistance is preserved by the voltage follow ers.

(a) B u ffered d iffer en tia l-in p u t to d iffe r e n tia l-o u tp u l a m p lifier.

+V

FIG U RE 8-5

Improving the basic differentia] amplifier.

226

Chapter 8

8-4 INSTRUMENTATION AMPLIFIER

_______________________________________________________

8-4.1 Circuit Operation T he instrum entation am plifier (IA ) is one o f the m ost useful, precise, and versatile am ­ plifiers available today. Youwill find at least one in every data acquisition unit. The b a­ sic IA is m ade from three op am ps and seven resistors, as show n in Fig. 8-6. To sim plify circuit analysis, note that the instrum entation am plifier is actually m ade by connecting a buffered am plifier [Fig. 8-5(b)] to a basic differential am plifier (Fig. 8-1). O p am p A 3 and its four equal resistors, R, form a differential am plifier w ith a gain o f 1. O nly the A 3 re­ sistors have to be m atched. T he prim ed resistor, R \ can be m ade variable to balance out any com m on-m ode voltage, as show n in Fig. 8-2. O nly one resistor, aR, is used to set the gain according to Eq. (8-6), repeated here for convenience: V 2 ^ — = 1+ E ]- E2 a

(8-6)

w here a = aR /R . £ ] is applied to the ( + ) input and E2 to the ( —) input. V0 is proportional to the dif­ ference betw een input voltages. C haracteristics o f the instrum entation am plifier are su m ­ m arized as follow s:

FIGURE 8-6

Basic instrumentation amplifier model.

Differential, Instrum entation, and Bridge A m p lifiers

227

1. T he voltage gain, from differential input ( E x — E2) to single-ended output, is set by one resistor. 2. T he input resistance o f both inputs is very high and does not change as the gain is varied. 3. VD does not depend on the voltage com m on to both E x and E 2 (com m on-m ode volt­ age), only on their difference.

E x a m p le 8-3 In Fig. 8-6, R = 25 k f l and aR = 50 ( 1 C alculate the voltage gain. S o lu tio n

From Eq. (8-6), aR

50

1

R

25,000

500

= a

1+ - = 1+ = 1 + (2 X 500) = 1001 a 1/500

£, ~ E 7 E x a m p le 8-4

If aR is rem oved in Fig. 8-6 so that aR = S o lu tio n

w hat is the voltage gain?

a = «>, so 1/ e

{-

e2

2 = 1+ — = 1

E x a m p le 8-5 In Fig. 8-6, the follow ing voltages are applied to the inputs. E ach voltage polarity is given w ith respect to ground. A ssum ing the gain o f 1001 from E xam ple 8-3, find V0 fo r (a) E } = 5.001 V and E 2 = 5.002 V; (b) £ , = 5.001 V and E 2 = 5.000 V; (c) E x = - 1 .0 0 1 V, E 2 = - 1 .0 0 2 V. S o lu tio n

(a) V0 = 1001(£, - E 2) = 1001(5.001 - 5.002) V -

lO O l(-O .O O l) V = —1.001 V

(b) V0 = 1001(5.001 - 5.000) V = 1001(0.001) V = 1.001 V (c) VQ = 1001[—1.001 - ( -1 .0 0 2 ) ] V = 1001(0.001) V = 1.001 V

229

Differential, Instrum entation, and Bridge A m plifiers

An offset voltage or reference voltage Kref is inserted in series w ith reference ter­ m inal R. Vref is divided by 2 and applied to the A3 op am p ’s ( + ) input. T hen the nonin­ verting am plifier gives a gain o f 2 so that V0 equals Vref. N ow V0 can be set to any d e­ sired offset value by adjusting Vref. In practice Vref is the output o f a voltage-follow er circuit as show n in Fig. 8-7(b).

8-5 SEN SIN G AND M EASURIN G WITH THE INSTRUMENTATION AMPLIFIER

______________________________________________

8-5.1 Sen se Terminal T he versatility and perform ance o f the instrum entation am plifier in Fig. 8-6 can be im ­ proved by breaking the negative feedback loop around op am p A3 and bringing out three term inals. As show n in Fig. 8-8, these term inals are output term inal 0, sense term inal Sf and reference term inal R. If long w ires or a current-boost transistor are required betw een the instrum entation am plifier and load, there will be voltage drops across the connecting w ires. To elim inate these voltage drops, the sense term inal and reference term inal are w ired directly to the load. Now, w ire resistance is added equally to resistors in series w ith the sense and reference term inals to m inim ize any m ism atch. Even m ore im portant, by sensing voltage at the load term inals and not at the am p lifier’s output term inal, feedback acts to hold load voltage constant. If only the basic differential am plifier is used, the out-

+v A R

Sense

R AAA---- o­ 2

Connecting wire resistance FIGURE 8-8 Extending the sense and reference terminals to the load termi­ nals makes Va depend on the amplifier gain and the input voltages, not on the load current or load resistance.

231

Differential, Instrum entation, and Bridge A m p lifiers 49,400 \

G ain = 1 +

Rc

(8-7)

I

For gain values of 1, 10, 100, and 1000, a table listing the R G values is given in Fig. 8-9(b). The usual w ay to m easure VCE o f a w orking com m on-em itter am plifier circuit is to (1) m easure collector voltage (w ith respect to ground), (2) m easure em itter voltage (w ith respect to ground), and (3) calculate the difference. The IA allow s you to m ake the m ea­ surem ent in one step, as show n in Fig. 8-9(b). Since £ , = Vcolleclor and E 2 = Vemi(ter, = ( ! ) ( £ , - E2) = (l)(V collect0r - ^emitter) =

(8-8)

Example 8-6 G iven V0 = 5 V in Fig. 8-9(b), find VC£.

Solution

From Eq. (8-8), 5 V = (£ , - E 2) =

Example 8-7 E xtend E xam ple 8-6 as follows. C onnect + In to the em itter and - In to ground, assum e V0 m easures 1.2 V, and calculate (a) em itter current Z£ ; (b) the voltage across R L or VRL.

Solution

(a) Since V0 — 1.2 V, E { — E 2 = 1.2 V, and therefore VRE — 1.2 V. U se O h m ’s law to find IE.

/£ = ^ Re

= i ^

i kn

= 1.2 mA

(b)

V,c o l l e c t o r = V « + V „ = 5 V + 1 .2 V = 6 .2 V

Vrl = Vrr - V,c o l l e c t o r = 15 V - 6.2 V = 8.8 V C C

v

Part (a) o f this exam ple show s how to m easure current in a w orking circuit by m easuring the voltage drop across a know n resistor.

8-5.3 Differential Voltage-to-Current Converter T he A D 620 instrum entation am plifier does not have a sense term inal. T herefore, if your application requires this term inal, choose another IA such as the A D 524 or A D 624. Figure 8-10 show s how to m ake an excellent current source that can sink or source dc current into a grounded load. It can also be an ac current source.

232

Chapter 8 + 15 V

-1 5 V

FIGURE 8-10 A differential voltage-to-current converter is made from an IA, op amp, and resistor. To understand how this circuit operates, one m ust understand that the IA’s output voltage at pin 9 depends on load current, / L, load resistor, R Ly and current set resistor, R s. In equation form Vg = I^R S + I lJRl

(8-9a)

T he output voltage o f an IA can also be expressed generally by V9 = K ef + g a ln (£ | - E 2)

(8-9b)

T he A D 547 voltage follow er forces the reference voltage to equal load voltage or VrCf = I l R l . Since the IA’s gain is set for 10 in Fig. 8-10, we can rew rite Eq. (8-9b) as V9 = l LR L + 10(£, - E 2)

(8-9c)

E quate Eqs. (8-9a) and (8-9c) to solve for 1L, w hich yields

E quation (8-9d) indicates that load resistor, R L , does not control load current; this is true as long as neither am plifier is forced to saturation. 1L is controlled by R , and the difference betw een £ , and E 2.

Example 8-8 In the circuit o f Fig. 8-10, Rs = 1 kH , E\ = 100 mV, E 2 = 0 V, and R L = 5 kCl. Find (a) (b) VR - (c) Vref; (d) V9.

Differential, Instrum entation, and Bridge A m plifiers

Solution

233

(a) F rom Eq. (8-9d), 4 =

10

IlRs = (1 m A )(l (c)vj= IlRl (1 m A )(5 (b) VRr =

k ft) = 1 V k ft) = 5 V From Eq. (8-9a) or (8-9c), =

(d)

V9 = I l R s + I l R l = 1 V + 5 V = 6 V or Vg = Vref 4- g a in (£ 1 - E2) = 5 V + 10(0.1 V) = 6 V

8-6 THE INSTRUMENTATION AMPLIFIER A S A SIGNAL CONDITIONING CIRCUIT _____________________________ In C hapter 3, we m entioned that som e sensor circuits are designed with a differential o u t­ put. However, the input voltage to a m icrocontroller is w ith respect to ground and hence is single ended. T herefore we need a signal conditioning circuit (SC C) that has a differ­ ential input and a single-ended output— the instrum entation am plifier. We begin by intro­ ducing a sensor circuit with a differential output.

8-6.1 Introduction to the Strain Gage A strain gage is a conducting w ire w hose resistance changes by a sm all am ount w hen it is lengthened or shortened. T he change in length is sm all, a few m illionths o f an inch. T he strain gage is bonded to a structure so that the percent change in length o f the strain gage and structure are identical. A foil-type gage is show n in Fig. 8 -1 1(a). T he active length o f the gage lies along the transverse axis. T he strain gage m ust be m ounted so that its transverse axis lies in the sam e direction as the structure m otion that is to be m easured [see Figs. 8 -1 1(b) and (c)]. L engthening the bar by tension lengthens the strain gage conductor and increases its re­ sistance. C om pression reduces the g ag e’s resistan ce b ecause the norm al length o f the strain gage is reduced.

8-6.2 Strain-Gage Material S train gages are m ade from m etal alloy such as co n stan tan , N ich ro m e V, D ynaloy, Stabiloy, or platinum alloy. For high-tem perature w ork they are m ade o f wire. For m od­ erate tem perature, strain gages are m ade by form ing the m etal alloy into very thin sheets by a photoetching process. The resultant product is called a foil-type strain gage and a typical exam ple is show n in Fig. 8 -1 1(a).

No. 30 insulated instrument wire

Lateral axis ^

I _______ A c tiv e __ length (a) Metal foil-type strain gage. Hi A D

1kor

(b) Tension lengthens bar and gage to increase gage resistance by AR.

(c) Com pression shortens bar and gage to reduce gage resistance by AR.

FIG U RE 8-11 structure.

Using a strain gage to measure the change in length of a

8-6.3 Using Strain-Gage Data In the next section, we show that our instrum entation m easures only the g ag e’s change in resistance AR. The m anufacturer specifies the unstrained g ag e’s resistance R. O nce AR has been m easured, the ratio AR /R can be calculated. The m anufacturer also furnishes a spec­ ified gage fa c to r (GF) for each gage. T he gage factor is the ratio o f the percent change in resistance of a gage to its percent change in length. T hese percent changes m ay also be ex ­ pressed as decim als. If the ratio AR /R is divided by gage factor G, the result is the ratio o f the change in length o f the gage AL to its original length L O f course the structure w here the gage is m ounted has the sam e AL /L An exam ple will show how gage factor is used.

E x a m p le 8-9 A 1 2 0 strain gage w ith a gage factor o f 2 is affixed to a m etal bar. T he bar is stretched and causes a A R o f 0.001 Cl. Find A L/L.

234

Differential, Instrum entation, and Bridge A m p lifiers

235

Solution AL

A R/R

o.ooi a/120 a

L

GF

2

— 4.1 m icroinches per inch

T he ratio AL/L has a nam e. It is cal led unit strain. It is the unit strain data (we have developed from a m easurem ent o f AR) that m echanical engineers need. They can use this unit strain data together w ith know n characteristics o f the structural m aterial (m odulus o f elasticity) to find the stress on the beam. Stress is the am ount o f fo rc e acting on a unit area. T he unit for stress is pounds per square inch (psi). If the bar in E xam ple 8-9 w ere m ade of m ild steel, its stress w ould be about 125 psi. Strain is the deform ation o f a ma~

B efore m ounting a strain gage the surface o f the m ounting beam m ust be cleaned, sanded, and rinsed w ith alcohol, Freon, or m ethyl ethyl ketone (M EK ). The gage is then fastened perm anently to the cleaned surface by E astm an 910, epoxy, polym ide adhesive, or ce­ ram ic cem ent. T he m anufacturer’s procedures should be follow ed carefully.

8-6.5 Strain-Gage Resistance Changes It is the change o f resistance in a strain gage AR that m ust be m easured and this change is small. AJ? has values of a few m illiohm s. T he technique em ployed to m easure small re-

8-7 M EA SU R EM EN T O F SM A LL RESISTA N CE CHANGES

_____________________________

To m easure resistance, w e m ust first find a technique to convert the resistance change to a current or voltage for display on an am m eter or voltm eter. If we m ust m easure a sm all change of resistance, we will obtain a very sm all voltage change. For exam ple, if we passed 5 m A of current through a 1 2 0 -0 strain gage, the voltage across the gage w ould be 0.600 V. If the resistance changed by 1 m ft, the voltage change w ould be 5 fiV . To display the 5-/llV change, w e w ould need to am plify it by a factor of, for exam ple, 1000 to 5 mV. H ow ever, we w ould also am plify the 0.6 V by 1000 to obtain 600 V plus 5 mV. It is difficult to detect a 5-m V difference in a 600-V signal. T herefore, we need a circuit that allow s us to am plify only the difference in voltage across the strain gage caused by a change in resistance. T he solution is found in the W heatstone bridge circuit.

238

Chapter 8

8-8 BALANCING A STRAIN-GAGE BRIDGE 8-8.1 The Obvious Technique S uppose that you had a w orking gage and tem perature-com pensation gage in Fig. 8-14 that are equal to w ithin 1 m ft. To com plete the bridge, you install two 1%, 1 2 0 -ft resis­ tors. O ne is high by 1% at 121.200 f t and one is low by 1% at 118.800 ft. T hey m ust be equalized to balance the bridge. To do so, a 5 -ft, 20-turn balancing pot is installed, as show n in Fig. 8-14. Theoretically, the pot should be set as show n to equalize resistances in the top branches of the bridge at 122.500 ft. F urther assum e that an instrum entation am plifier w ith a gain o f 1000 is connected to the bridge of Fig. 8-14. F rom E xam ple 8-10, the o u t p u t o f the instrum entation am pli­ fier (IA ) w ill be about 22 mV per m illiohm o f unbalance. This m eans th at the 5 -fl pot m ust be adjusted to w ithin 1 m ft o f the values show n, so that E x — E2 and consequently VQ o f the IA will equal 0 V ± 22 mV. U nfortunately, it is very difficult in practice to adjust for balance. T his is because each turn o f the pot is worth 5 ft/2 0 turns = 250 m ft. W hen you adjust the pot it is normal to expect a backlash o f o f a turn. Therefore, your best efforts result in an unbalance at the pot o f about ± 5 m ft. You observe this unbalance at the IA’s output, w here V0 changes by ± 0.1 V on either side of zero as you fine-tune the 20-tum pot. It turns out there is a better technique that uses an ordinary linear potentiom eter ( | turn) and a single resistor. £=10 V Set for 3.700 Cl

R2 118.800 Cl

^ y

Set for 1.300 a

*3 = 121.200 Cl £j to differential amp (+) in J 20.000 Cl w orking gage

120.000 Cl tem perature com pensation gage

FIGURE 8-14

Balance pot RB is adjusted in an attempt to make £, - E2 — 0 V.

8-8.2 The Better Technique To analyze operation of the balance netw ork in Fig. 8-15, assum e that the R 2 and R 3 bridge resistors are reasonably equal, to w ithin ± I %. The strain g ag e’s resistance should have equal resistances w ithin several m illiohm s if the w orking gage is not under strain.

239

Differential, Instrum entation, and Bridge A m plifiers

Balance

network

E= 10 V I

f= 1.0$ RB] = 10 kQ. pot

- w hich is found by adding the m agnitudes of / fl+ and IB- and dividing this sum by 2. In equation form , Ib

(9-1)

w here | IB+ | is the m agnitude o f IB+ and | /# _ | is the m agnitude o f IB- . T he range of IB is from 1 /j l A or m ore for general-purpose op am ps to 1 pA or less for op am ps that

256

Chapter 9

Example 9-1 C onsider a 741 op am p that has IB+ = 0.4 /xA and IB- — 0.3 bias current JB\ (b) the offset current l os. S o lu tio n

Find (a) the average

(a) By Eq. (9-1), r (0.4 + 0.3) MA aoc IB = ---------- o-----------= °*35

a

(b) By Eq. (9-2), I os = (0.4 — 0.3) fi A = 0.1 /xA

A general-purpose op am p such as the O P -177 has typical values o f IB = 2.4 nA and Ios = 0.5 nA. N ew er versions o f the 741 op am p have typical values low er than those given in E xam ple 9-1.

9-3 E FFEC T O F BIAS CURREN TS ON OUTPUT VOLTAGE 9-3.1 Simplification In this section it is assum ed that bias currents are the only op am p characteristic that will cause an undesired com ponent in the output voltage. T he effects o f other op am p ch arac­ teristics on VQ will be dealt w ith individually.

9-3.2 Effect of c C = (6.28)(2

X

103)(5 X 10- 9 ) = 15'9 kft

Exam ple 11-3 Calculate R in Fig. 1 l-2(a) for a cutoff frequency of 30 krad/s and C = 0.01 fjF. Solution

From Eq. (1 l-2b),

R ~ wc C ~ (30

X

103)(1 X 10- 8 )

- 3,3 ^

Design Procedure The design o f a low-pass filter similar to Fig. 11-2(a) is accom­ plished in three steps: 1. Choose the cutoff frequency— either c. To satisfy the Butterworth criteria, the frequency response must be 0.707 at o>c and be 0 dB in the pass band. These condi­ tions will be met if the following design procedure is followed:

Design procedure for 40-dB/decade high-pass 1. Choose a cutoff frequency, u)c or / c. 2. Let Ci = C2 = C and choose a convenient value.

(b) Frequency response for circuit of part (a).

FIGURE 11-8 Circuit and frequency response for a 40-dB/decade high-pass Butterworth filter.

309

Active Filters 3. Calculate R\ from 1.414 = -L±7T

(H -9 )

a>c C

4. Select

R2 = \Ri

(11-10)

5. To minimize dc offset, let Rf = R\.

Exam ple 11-8 In Fig. 1 l-8(a), let C\ = C2 = 0.01 fiF. Calculate (a) R\ and (b) R2 for a cutoff frequency o f 1 kHz. Solution

(a) From Eq. (11-9),

n _ “

1.414 „„ (6.28)(1 X 103)(0.01 X 10“6) “ 22-5

^

(b) R2 = ^(22.5 k fl) ■ 11.3 kfl. Exam ple 11-9 Calculate (a) Ri and (b) R2 in Fig. 11 -8(a) for a cutoff frequency o f 80 krad/s. Ci = C2 = 125 pF. Solution

(a) From (11-9), /?l ~

(80 X 103)(125 X 10“ 12) ~ 140

(b)

r 2 = i( i4 0 kft) = 70 m .

11-5.4 60-dB/Decade Filter As with the low-pass filter o f Fig. 11-5, a high-pass filter o f + 6 0 dB/decade can be con­ structed by cascading a +40-dB/decade filter with a +20-dB/decade filter. This circuit (like the other high- and low-pass filters) is designed as a Butterworth filter to have the frequency response in Fig. ll-9 (b ). The design steps for Fig. 11 -9(a) are as follows:

Design procedure for 60-dB/decade high-pass 1. Choose the cutoff frequency, wc or fc. 2. Let Ci = C2 = C3 = C and choose a convenient value between 100 pF and 0.1 /iF.

40 dB/decade

20 dB/decade

1^1 E,

lirl

(b) Frequency-response for the circuit of part (a).

FIGURE 11-9 Circuit and frequency response for a 60-dB/decade Butterworth high-pass filter.

3. Calculate R3 from

n C

4. Select

Rx = 2 R 3

310

( 11- 12)

312

Chapter 11 TABLE 11-5 COMPARISON OF \A c l \ FOR FIGS. 117(a), 118(a), AND 11 -9(a)

O)

20 dB/decade; Fig. 1l-7(a)

40 dB/decade; Fig. 11-8(a)

0A(dc 0.25 o)c 0.5o)c (Or 2a>c 4 (t)c 10o)c

0.1 0.25 0.445 0.707 0.89 0.97 1.0

0.01 0.053 0.24 0.707 0.97 0.998 1.0

60 dB/decade; Fig. 11-9(a) 0.001 0.022 0.124 0.707 0.992 0.999 1.0

TABLE 11-6 COMPARISON OF PHASE AN GLES FOR FIGS. 11-7(a), 11-8(a), AND 11 -9(a)

a)

20 dB/decade; Fig. 11-7(a)

40 dB/decade; Fig. 1l-8(a)

60 dB/decade; Fig. 11-9(a)

0.1 U)c 0.25 0.5wc Cl) 2o,t. 4(oc 10coc

84° 76° 63° 45° IT 14° 6°

172° 143° 137° 90° 43° 21° 8°

256° 226° 210° 135° 60° 29° 12°

11-6 INTRODUCTION TO BANDPASS FILTERS ___________________________________________ 11-6.1 Frequency Response A bandpass filter is a frequency selector. It allow s one to select or pass only one p artic­ ular band o f frequencies from all other frequencies that may be present in a circuit. Its norm alized frequency response is show n in Fig. 11-10. T his type o f filter has a m axim um gain at a resonant frequency f r. In this chapter all bandpass filters will have a gain o f 1. or 0 dB at/,.. T here is one frequency b elow /,, w here the gain falls to 0.707. It is the low er c u to ff frequency, f h At the higher cutofffrequency, f h the gain also equals 0.707, as in Fig. 11 - 1 0 .

313

Active Filters

Resonant frequency f r 1.0

A

— Bandwidth « = /* - //

0.707

-------- ^---------------- ► Frequency fr

f,

FIGURE 11-10 A bandpass filter has a maximum gain at resonant frequency f r . The band of frequencies transmitted lies between / and//,.

11-6.2 Bandwidth The range of frequencies betw een / and f h is called bandw idth Bf or B = fh ~ ft

(I I -14)

T he bandw idth is not exactly centered on the resonant frequency. (It is for this reason that we use the historical nam e “ resonant frequency” rather than “center frequency” to de­ s c rib e /..) If you know the values for / and / /7, the resonant frequency can be found from fr = 'S J J h

(11-15)

If you know the resonant frequency, f r >and bandw idth, B, cutoff frequencies can be found from (I I-16a) L = f i

+ B

(I I -16b)

Example 11-12 A bandpass voice filter has low er and upper cutoff frequencies o f 300 and 3000 Hz. Find (a) the bandw idth; (b) the resonant frequency.

Chapter 11

314 Solution

(a) From Eq. (11-14), B = f h ~ f i = (3000 - 300) = 2700 Hz

(b) From Eq. (11-15), f r = y /J J h = V (3 0 0 )(3 0 0 0 ) = 948.7 H z Note: f r is alw ays below the center frequency o f (3000 + 300)/2 = 1650 Hz.

Example 11-13 A bandpass filter has a resonant frequency o f 950 Hz and a bandw idth o f 2700 Hz. Find its low er and upper cutoff frequencies.

Solution

F rom Eq. (11 -16a),

= 1650 - 1350 = 300 Hz From Eq. ( 1 1- I 6b) t f h = 300 + 2700 = 3000 Hz.

11-6.3 Quality Factor T he quality fa c to r Q is defined as the ratio o f resonant frequency to bandw idth, or ( 11- 1 7 )

Q is a m easure o f the bandpass filter's selectivity. A high Q indicates that a filter selects a sm aller band of frequencies (m ore selective).

11-6.4 Narrowband and Wideband Filters A w ideband filter has a bandw idth that is tw o or m ore tim es the resonant frequency. T hat is, Q < 0.5 for w ideband filters. In general, w ideband filters are m ade by cascading a lowpass filter circuit with a high-pass filter circuit. T his topic is covered in the next section, A narrow band filter (Q > 0.5) can usually be m ade w ith a single stage. T his type o f fil­ ter is presented in Section 11-8.

Example 11-14 Find the quality factor o f a voice filter that has a bandw idth o f 2700 H z and a resonant fre­ quency o f 950 H z (see E xam ples 11-12 and 11-13).

315

Active Filters

Solution

From Eq. (11-7), £> = — = = 0.35 “ B 2700

This filter is classified as w ideband because Q < 0.5.

11-7 BASIC WIDEBAND FILTER 11-7.1 Cascading W hen the output of one circuit is connected in series w ith the input o f a second circuit, the process is called cascading gain stages. In Fig. 11-11, the first stage is a 3000-H z lowpass filter (Section 11-3). Its output is connected to the input o f a 300-H z high-pass fil­ ter (Section 11-5.3). The cascaded pair o f active filters now form a bandpass filter from input Et to output Va. N ote that it m akes no difference if the high-pass is connected to the low -pass, or vice versa. Note: Each op am p circuit in Fig. 11-11 has unity gain.

11-7.2 Wideband Filter Circuit In general, a w ideband filter (Q < 0.5) is m ade by cascading a low- and a high-pass fil­ ter (see Fig. 11-11). C utoff frequencies o f the low- and high-pass sections m ust n o t over­ lap, and each m ust have the sam e passband gain. F urtherm ore, the low -pass filter's cu t­ off frequency m ust be 10 or m ore tim es the high-pass filter’s cutoff frequency. For cascaded low- and high-pass filters, the resulting w ideband filter will have the follow ing characteristics: 1. T he low er cutoff frequency,//, will be determ ined only by the high-pass filter. 2. The high cutoff frequency,/*, will be set only by the low -pass filter. 3. G ain will be m axim um at resonant fre q u e n c y ,/,, and equal to the passband gain o f either filter. T hese principles are illustrated next.

11-7.3 Frequency Response In Fig. 11-11 the frequency response o f a basic —40-dB /decade 3000-H z low -pass filter is plotted as a dashed line. T he frequency response o f a 300-H z high-pass filter is plotted as a solid line. T he 40-dB /decade roll-off o f the high-pass filter is seen to d e te r m in e /. T he -4 0 -d B /d e c a d e roll-off of the low -pass s e ts /,. B oth roll-off curves m ake up the fre­ quency response o f the bandpass filter, V0 versus / O bserve that the resonant, low, and high c u to ff fre q u en c ies plus b an d w id th agree ex actly w ith the v alues ca lc u lated in

320

Chapter 11

11-9.2 Notch Filter Theory As show n in Fig. 11-14, a notch filter is m ade by subtracting the output o f a bandpass fil­ ter from the original signal. For frequencies in the notch filter’s passband, the output o f the bandpass filter section approaches zero. T herefore, input £, is transm itted via adder input resistor /?, to drive V0 to a value equal to —E h Thus VQ = in both low er and upper passbands of the notch filte r Suppose that the frequency of E-, is adjusted to resonant frequency f r o f the narrow bandpass filter com ponent. (Note: f r o f the bandpass sets the notch frequency.) £, will exit from the bandpass as — £, and then is inverted by R x and R to drive V0 to + E t. However, E t is transm itted via R 2 to drive V0 to —E {. T hus VQ responds to both inputs o f the adder and becom es Va = - E t = 0 V at f n In practice, VG approaches zero only a t / r . T he depth o f the notch depends on how closely the resistors and capacitors are m atched in the bandpass filter and ju d icio u s fine adjustm ent o f resistor /?, at the inverting ad d er’s onput. T his procedure is explained in Section 11-10.3.

R AAAt

Fig. 11-12

Narrow bandpass filter

V FIGURE 11-14 A notch filter is made by a circuit that subtracts the output of a bandpass filter from the original signal.

11-10 120-HZ NOTCH FILTER

________________________________________________________________

11-10.1 Need for a Notch Filter In applications w here low -level signals m ust be am plified, there may be present one or m ore o f an assortm ent o f unw anted noise signals. E xam ples are 50-, 60-, or 400-H z fre­ quencies from pow er lines, 120-Hz ripple from full-w ave rectifiers, or even higher fre­ quencies from regulated sw itching-type pow er supplies or clock oscillators. If both sig­ nals and a signal-frequency noise com ponent are passed through a notch filter, only the desired signals will exit from the filter. T he noise frequency is “notched out.” As an ex ­ am ple, let us m ake a notch filter to elim inate 120-Hz hum.

321

Active Filters

11-10.2 Statement of the Problem T h e p ro b lem is to m ake a n o tch filte r w ith a n o tch (re so n a n t) fre q u e n c y o f f r = 120 Hz. L et us select a stopband o f B = 12 Hz. T he gain o f the notch filter in the passband w ill be unity (0 dB ) so that the desired sig n als w ill be tran sm itted w ith o u t at­ tenuations. We use Eq. (1 1-17) to d eterm in e a value for Q th at is req u ired by the notch filter:

This high value of Q m eans that (1) the notch and com ponent bandpass filter w ill have narrow bands with very sharp frequency -resp o n se curves, and (2) the bandw idth is es­ sentially centered on the resonant frequency. A ccordingly, this filter will transm it all fre­ quencies from 0 to (120 - 6) = 114 Hz and all frequencies above (120 + 6) = 126 Hz. The notch filter will stop all frequencies betw een 114 and 126 Hz.

11-10.3 Procedure to Make a Notch Filter A notch filter is m ade in two steps: 1. M ake a bandpass filter that has the sam e resonant frequency, bandw idth, and conse­ quently Q as the notch filter. 2. C o n n e ct the in v e rtin g ad d e r o f Fig. 11-15 by se le c tin g eq u al re sis to rs for R . U sually, R — 10 k fl. (A practical fine-tuning p ro ced u re is p resented in the next section.)

11-10.4 Bandpass Filter Components T he first step in m aking a 120-Hz notch filter is best illustrated by an exam ple (see Fig. 11-15).

D esign E x a m p le 11-17 D esign a bandpass filter w ith a resonant frequency o f f r = 120 Hz and a bandw idth o f 12 H z so that 2 = 1 0 . T hus gain of the bandpass section will be 1 at f r and approach zero at the output o f the notch labeled V0.

Solution

C hoose C = 0.33 /xF. From Eq. (11-18a), 0.1591 (12)(0.33 X 10“ 6)

= 40.2 k f l

323

Active Filters

11-11.1 Low-Pass Filter R efer to Fig. ll- 4 ( a ) and create the PSpice m odel o f the circuit using a 741 op am p if you are using the evaluation softw are package. The input voltage source will be VAC and w ill be set for a 1-V m agnitude. We w ant a plot o f V0 versus frequency. To begin, place the follow ing parts in the w ork area. Qraw => get New Part Part = = = = = = =

> > > > > > >

uA741 VAC VDC R C GLOBAL AGND

Number

Library

1

eval.slb source.sib source^lb analog.slb analog.slb port.slb porLslb

1

2 4 2 4 5

Note: We are using VAC as the input source instead o f VSIN as we have in previ­ ous chapters. The VAC sym bol requires only that m agnitude and phase be set. T he fre­ quency range will be set in the A nalysis Setup m enu. A rrange the parts as show n in Fig. 11-4(a). C hange the attributes o f the parts to those values given in Exam ple 11-4. Set up the VAC sine w ave attributes by d o u b le-click in g the sym bol; in the p o p-up w indow change phase and m agnitude.

ACPHASE = > 0 = > Save Attr ACMAG = > IV = > Save Attr —> Change Display = > Both name and value D ouble-click on the lead from the output term inal of the op am p and label it Vo (see Fig. 11-16). To obtain a plot of V0 versus frequency, we m ust initialize the AC Sw eep m enu.

Analysis = > Setup = > Open AC Sweep = > => => =>

Enable AC Sweep Decade Pts/Decade = > 1 0 Start Freq = > 10Hz End Freq = > 10kHz

Save the circuit as a file with the .SCH extension. Run the sim ulation

Analysis = > Simulate In the Probe window, we need to select both P lot and Trace options from the m enu bar.

Plot = > Y Axis Settings = > Scale = > Log Trace = > Add = > V[Vo] Label the plots and obtain a printout as show n in Fig. 11-17.

325

A ctive Filters

11-11.2 High-Pass Filter T he procedure for m odeling and sim ulating a high-pass filter is sim ilar to that for the lowpass filter previously described. R efer to Fig. ll- 8 ( a ) and create the PSpice m odel o f the circuit using a 741 op amp. T he input voltage source w ill be VAC and will be set for a 1-V m agnitude. O btain a plot o f V0 versus frequency. To begin, place the follow ing parts in the w ork area. Draw => Get New Part

=> => => => => => =>

Part

Number

Library

uA741 VAC VDC

1 I 2 4 2 4 5

evalslb source.slb source.slb analog.slb analog.s!b portslb port.slb

R

C GLOBAL AGND

As previously m entioned w e are using VAC as the input source instead o f VSIN so that we may vary frequency through a range, because the VAC sym bol requires only m agni­ tude and phase to be set. T he frequency range is set in the A nalysis Setup m enu. A rrange the parts as show n in Fig. 1 l-8(a). Change the attributes o f the parts to those values given in Exam ple 11-8. Set up the VAC sine wave attributes by double-clicking the sym bol; in the pop-up w indow change phase and m agnitude.

ACPHASE = > 0 = > Save Attr ACMAG = > IV = > Save Attr —> Change Display —> Both name and value D ouble-click on the lead from the output term inal o f the op am p and label it Vo (see Fig. 11-18). To obtain a plot o f V0 versus frequency, we m ust initialize the AC Sw eep menu.

Analysis = > Setup = > Open AC Sweep = > => => =>

Enable AC Sweep Decade Pts/Decade = > 1 0 Start Freq = > 100Hz End Freq = > 100kHz

Save the circuit as a file w ith the .SCH extension. Run the sim ulation

Analysis = > Simulate In the Probe window, we need to select both Plot and T race options from the m enu bar.

Plot = > Y Axis Settings = > Scale = > Log Trace = > Add = > V[Vo] Label the plots and obtain a printout as shown in Fig. 11-19.

328 PROBLEMS

Chapter 11

_____________________________________________________________________

11-1. List the four types of filters. 11-2. What type of filter has a constant output voltage from dc up to the cutoff frequency? 11-3. What is a filter called that passes a band of frequencies while attenuating all frequencies outside the band? 11-4. In Fig. 11-2(a), if R = 100 kO and C = 0.02 /zF, what is the cutoff frequency? 11-5. The low-pass filter of Fig. 11 -2(a) is to be designed for a cutoff frequency of 4.5 kHz. If C = 0.005 /aF, calculate R. 11-6. Calculate the cutoff frequency for each value of C in Fig. PI 1-6.

Rf = 10 kQ AAAr

FIG U RE PI 1-6

11-7. What are the two characteristics of a Butterworth filter? 11-8. Design a -40-dB /decade low-pass filter at a cutoff frequency of 10 krad/s. Let C | = 0.02 fxF. 11-9. In Fig. 11-4(a), if off frequency f c.

= R2 = 10 kll, C} = 0.01 /jlF, and C2 = 0.002 /jlF, calculate the cut­

11-10. Calculate (a) /?3, (b) Ru and (c) R2 in Fig. 1 l-5(a) for a cutoff frequency of 10 krad/s. Let C3 = 0.005 mF. 11-11. If R y = R2 = R3 = 20 kfl, C, = 0.002 fjuF, C2 = 0.008 ^ and C3 = 0.004 ^ in Fig. 115(a), determine the cutoff frequency (oc. 11-12. In Fig. 11-5(a), C{ = 0.01 /xF, C2 = 0.04 fjuF, and C3 = 0.02 y,F. Calculate R for a cutoff frequency of 1 kHz. 11-13. Calculate R in Fig. 1 l-7(a) if C = 0.04 fjuF a n d /c = 500 Hz.

329

Active Filters 11-14. In Fig. 1 l-7(a) calculate (a) a)c and (b) f c if R = 10 k ft and C = 0.01 /xF 11-15. Design a 40-dB/decade high-pass filter for o)c = 5 krad/s. G = C2 = 0.02

jjlF.

11-16. Calculate (a) R ] and (b) /?2 in Fig. 11 -8(a) for a cutoff frequency of 40 krad/s. CA = C2 = 250 pF. 11-17. For Fig. I I-9(a), let C{ = C2 = C3 = 0.05 /xF. Determine (a) /?3, (b) /?,, and (c) R2 for a cutoff frequency of 500 Hz. 11-18. The circuit of Fig. 11-9(a) is designed with thevalues CA = C2 = C3 = 400 pF, 100 kH, /?2 = 25 kH, and /?3 = 50 kfi. Calculate the cutoff frequency/,.

=

11-19. Find the (a) bandwidth, (b) resonant frequency, and (c) quality factor of a bandpass filter with lower and upper cutoff frequencies of 55 and 65 Hz. 11-20. A bandpass filter has a resonant frequency of the lower and upper cutoff frequencies.

1000 Hz and a bandwidth of 2500Hz.Find

11-21. Use the capacitor and resistor values of the high-pass filter in Fig. 11-11 to prove f c = 3000 Hz. 11-22. Use the capacitor and resistor values of the high-pass filter in Fig. 11-11 to prove that f c = 300 Hz. 11-23. Find Q for the bandpass filter of Fig. 11-11. 11-24. Design a narrow bandpass filter using one op amp.The resonantfrequency Q = 1.5. Select C = 0.1 /jlF in Fig. 10-12. 11-25. (a) How would you convert the bandpass filter of Problem 11-24 into same resonant frequency and Q1 (b) Calculate / and f h for the notch

is 128Hz and a notchfilter with the filter.

CHAPTER 12 Modulating, Demodulating, and Frequency Changing with the Multiplier

LEARNING

O B J E C T I V E S _____________________________________________________

U pon com pletion o f this chapter on m ultiplier ICs, you will be able to: • W rite the output-input equation of a m ultiplier IC and state the value o f its scale factor. • M ultiply tw o dc voltages or divide one dc voltage by another. • Square the value of a dc voltage or take its square root. • D ouble the frequency o f any sine wave. • M easure the phase angle betw een tw o sine w aves o f equal frequency. • Show that am plitude m odulation is actually a m ultiplication process. • M ultiply a carrier sine w ave by a m odulating sine wave and express the output voltage either by a product term or by a term containing sum and difference frequencies. • C alculate the am plitude and frequency o f each output frequency term. • M ake either a balanced am plitude m odulator or a standard am plitude m odulator. • Show how a m ultiplier can be used to shift frequencies.

Modulating, Demodulating, and Frequency Changing with the Multiplier

12-0 INTRODUCTION

331

_________________________________________________________________

A nalog m ultipliers are arrangem ents o f op am ps and other circuit elem ents available as an integrated circuit. M ultipliers are easy to use; som e o f their applications are (1) m ea­ surem ent o f power, (2) frequency doubling and shifting, (3) detecting phase-angle differ­ ence betw een two signals o f equal frequency, (4) m ultiplying tw o signals, (5) dividing one signal by another, (6) taking the square root o f a signal, (7) squaring a signal, and (8) designing nonlinear signal conditioning circuits. A nother use for m ultipliers is to dem on­ strate the principles o f am plitude m odulation and dem odulation. T he schem atic o f an A D 633 m ultiplier is show n in Fig. 12-l(a). The device is a four-quadrant analog m ulti­ plier. It has high input im pedance, w hich makes signal source loading negligible. Pow er supply voltages can range from ± 8 V to ± 1 8 V. No external com ponents or user calibra­ tion are required. T he output voltage is a scaled version o f the x and y inputs. T he scale fa c to r is explained in Section 12-1.

8-Pin plastic DIP (N) package W

FIG U RE 12-1 multiplier.

AD633JN

Introduction to the

12-1 MULTIPLYING DC VOLTAGES __________________________________________________ 12-1.1 Multiplier Scale Factor T he 8-pin m ini-D IP housing and internal schem atic o f the A D 633 m ultiplier is show n in Fig. 12-1(a). In general term s, the output voltage V0 is expressed by, V0 = — ~

~

+ ;

(12-1)

w here V0 is the output voltage m easured at term inal W w ith respect to ground. T he fac­ tor of is called a scale fa c to r and is typical o f m ultipliers, because m ultipliers are d e­ signed for the sam e type o f pow er supplies used for op am ps, nam ely ± 1 5 V. For best re­ sults, the voltages applied to either x or y inputs should not exceed + 1 0 V or —10 V w ith respect to ground. T his ± 1 0 -V lim it also holds for the output, so the scale factor is u su ­ ally the reciprocal o f the voltage lim it, or 1/10 V. If both input voltages are at their posi­ tive lim its o f + 1 0 V, the output w ill be at its positive lim it o f 10 V.

332

Chapter 12

12-1.2 Multiplier Quadrants M ultipliers are classified by quadrants; for exam ple, there are one-quadrant, tw o-quadrant, and four-quadrant m ultipliers. T he classification is explained in tw o w ays in Fig. 12-2. In Fig. 12-2(a), the input voltages can have four possible polarity com binations. C onsider *2 ~ )>2 = z = 0 , then both x v and y x are positive and operation is in quadrant 1, since x, is the horizontal and y x the vertical axis. If x x is positive and y ] is negative, quadrant 4 o p ­ eration results, and so forth.

Example 12-1 Find V0 for the follow ing com bination o f inputs if x 2 = y2 = z — 0: (a) x x = 10 V, y x = 10 V; ( b )x , - - 1 0 V ,y, = 10 V; (c) jca = 10 V, y x - - 1 0 V; (d) jc, = - 1 0 V, y x = - 1 0 V.

Solution

From Eq. (1 2 -lb ),

(a) VB = ( 1° X 10) = 10

V, quadrant 1

(b) V0

= ( - 1|q ^ 1Q) =

- 1 0 V, quadrant

2

(c) V0

=

- 1 0 V, quadrant

4

(d) V0 = —-—

^

=

^ = 10 V, quadrant 3

In Fig. 12-2(b), V0 is plotted on the vertical axis and x, on the horizontal axis. If w e apply 10 V to the y input and vary jc from - 1 0 V to + 1 0 V, we plot the line ab, la­ beled y = 10 V. If y x is changed to - 1 0 V, the line cdy labeled 3^1 = —10 V, results. T hese lines can be seen on an oscilloscope by connecting V0 o f the m ultiplier to the y x input o f the oscilloscope and X\ of the m ultiplier to the + x x input o f the oscilloscope. F or accu­ racy, V0 should be 0 V w hen either m ultiplier input is 0 V. M ultiplier errors are prim ar­ ily due to input and output offsets, scale factor error, and/or nonlinearity o f the core m ul­ tiplying unit. T hese errors are only fractions o f a percen t error, and if they need to be elim inated refer to the m anufacturer’s data sheet.

Modulating, Demodulating, and Frequency Changing with the Multiplier

12-4 PHASE-ANGLE DETECTION

337

___________________________________________________________

12-4.1 Basic Theory If two sine w aves o f the sam e frequency are applied to the m ultiplier inputs in Fig. 125(a), the output voltage V0 has a dc voltage com ponent and an ac com ponent w hose fre­ quency is tw ice that o f the input frequency. This conclusion was developed in Section 12­ 3.2. T he dc voltage is actually proportional to the difference in phase angle 0 betw een EX] and Eyx. For exam ple, in Fig. 12-5, 0 = 0°, because there was no phase difference be­ tw een EXi and Eyr Figure 12-5(b) shows two sine waves o f identical frequency but a phase difference o f 90°; therefore, 0 = 90°. If one input sine w ave differs in phase angle from the other, it is possible to calcu­ late or m easure the phase-angle difference from the dc voltage com ponent in VD. This dc com ponent VQ dc is given by* ( 12-4a) w here Exp and Eyp are peak am plitudes o f EX] and Eyr For exam ple, if Exp = 10 V, Eyp = 5 V, and they are in p h a se, then Va dc would indicate 2.5 V on a dc voltm eter. This volt­ m eter point w ould be m arked as a phase angle o f 0° (cos 0° = 1). If 0 = 45° (cos 45° = 0.707), the dc m eter w ould read 0.707 X 2.5 V — 1.75 V. O ur dc voltm eter can be cal­ ibrated as a phase-angle m eter 0° at 2.5 V, 45° at 1.75 V, and 90° at 0 V. E quation (12-4a) m ay also be expressed by* ( 12-4b) •xp

L *yp

If we could arrange for the product Exp Eyp to equal 20, we could use a 0 -to -l-V dc voltm eter to read cos 0 directly from the m eter face and calibrate the m eter face in d e­ grees from a cosine table. T hat is, Eq. (12-4b) reduces to V0 dc = cos 0

for Exp = Eyp = 4.47 V

T his point is explored further in Section 12-4.2.

^Trigonometric identity: sin A sin B = {[cos (A - B) - cos (A + £)] For equal frequencies, different phase angle: A = l u f i + 6 for E„

B = 27rft for Ey

Therefore, [sin (27r/r + 0)][sin 2irft] = {[cos 0 — cos (477/if + 0)J = y(dc - double frequency term)

(] 2-4c)

+ 15 V

x2 = y2 = z = 0 (a) Phase-angle m easurem ent.

(b) Input voltage for 0 = 90°.

VQ— -s in 2rc2000/

FIGURE 12-5 Multiplier used to measure the phase-angle difference between two equal frequencies.

340

Chapter 12

0 (deg)

cos 0

V„dc(V)

±30 ±45 ±60 ± 0 ±90

0.866 0.707 0.500 1.000 0.000

0.866 0707 0.500 1.000 0.000

(The last tw o rows of this table com e from E xam ples 12-5 and 12-6.)

T he 0 -to -l-V voltm eter scale can now be calibrated in degrees, 0 V for a 90° phase angle and 1.0 V for 0° phase angle. A t 0.866 V, 0 = 30°, and so forth. T he phase-angle m eter does not indicate w hether 0 is a leading or lagging phase angle but only the phase difference betw een EXx and Eyr

12-4.3 Phase Angles Greater Than ±90° T he cosine o f phase angles greater than + 9 0 ° or - 9 0 ° is a negative value. T herefore, V0 will be negative. This extends the capability o f the phase angle m eter in E xam ple 12-7.

Example 12-8 C alculate VQ dc for phase angles o f (a) 0 = ± 9 0 ° ; (b) 0 = ± 1 2 0 °; (c) 0 = ± 135°; (d) 6 ± 1 5 0 °; (e) 0 = ± 1 8 0 °.

Solution

U sing Eq. (12-4a) and tabulating results, we have ±90°

± 120°

± 135°

±150°

+1

vodc

0V

- 0 .5 V

- 0 .7 0 V

- 0 .8 6 6 V

-1 V

o

OO

O

0

F rom the results o f E xam ples 12-7 and 12-8, a ± 1-V voltm eter can be calibrated to read from 0 to ± 1 8 0 °.

72-5 ANALOG DIVIDER

______________________________________________________________________

A n analog divider gives the ratio o f two signals or provides gain control. It is constructed as show n in Fig. 12-6 by inserting a m ultiplier in the feedback loop o f an op am p. Since the op a m p ’s ( - ) input draw s negligible current, the sam e value o f current I flow s through

Modulating, Demodulating, and Frequency Changing with the Multiplier

341

resistors R. Therefore, the output voltage o f the m ultiplier Vm is equal in m agnitude but opposite in polarity (with respect to ground) to E in or £ in = - V m

( 12-5a)

B ut Vm is also equal to one-tenth (scale factor) o f the product o f input Ex and output of the op am p V0. Substituting for Vm yields V E 3 " = -“

U 2-5b)

S olving for Voy we obtain

E quation (12-5c) show s that the d iv id e r's o u tp u t VQ is p ro p o rtio n al to the ratio o f inputs £ in and Ex. Ex should never be allow ed to go to 0 V or to a negative voltage, b ecause the op am p w ill saturate. E m can be positive, negative, or 0 V. N ote th at the d ivider can be view ed as a voltage gain 10!EX acting on £ in. So if Ex is ch an g ed , the gain w ill change. T his voltage control o f the gain is useful in au to m atic g ain -co n tro l circuits.

342

Chapter 12

12-6 FINDING SQ U A RE ROOTS

____________________________________________________________

A divider can be m ade to find square roots by connecting both inputs o f the m ultiplier to the output o f the op am p (see Fig. 12-7). E quation (l2 -5 a ) also pertains to Fig. 12-7. But now Vm is one-tenth (scale factor) o f V0 X V0 or V2 ~E-m = V m = - £

( 12-6a)

Solving for V0 (elim inate V - T ) yields

Vc = V l O l E j

(12-6b)

Equation (12-6b) states that V0 equals the square root o f 10 tim es the m agnitude o f £ in. E in m ust be a negative voltage, or else the op am p saturates. T he range o f E m is betw een — 1 and —10 V. Voltages sm aller than —1 V w ill cause inaccuracies. T he diode prevents ( - ) saturation for positive E m. If E [n has positive values, reverse the diode.

Square root finder

FIGURE 12-7

Square rooting with an op amp and a multiplier (x2 = y i

z = 0).

12-7 INTRODUCTION TO AMPLITUDE MODULATION

__________________________________

12-7.1 Need for Amplitude Modulation L ow -frequency audio or data signals cannot be transm itted from antennas o f reasonable size. A udio signals can be transm itted by changing or m odulating som e characteristic o f a higher-frequency carrier wave. If the am plitude o f the carrier w ave is changed in pro-

33

First Experiences with an Op Am p Input = tem perature 0 to 50 °C ■N W ■ > W’ —

115 to 323 (iA w W

Transducer

Current-tovoltage converter

output 0 to 5 V W

C

(a) Block diagram of a temperatureto-voltage converter.

0

10

20

30

40

50

Tem perature (°C) (b) In put-output characteristic of a tem perature-to-voltage converter.

FIG U RE 2-15 An example of how room or a process temperature is mea­ sured electronically. A ssum e that you have available a circu it that gives 0 to 5 V out fo r a ro o m -tem p erature change o f 0° to 50°C (see Fig. 2-15). T he o u tput, Vtemp, can now be used as a m easu rem en t o f tem perature, or it can be used to co n tro l tem p eratu re. S u p p o se that you w ant to send this te m p eratu re in fo rm atio n to a co m p u ter so th at the co m p u te r could m onitor, control, or chan g e room tem p eratu re. A voltag e-lev el d etec to r can a c ­ com plish this task. To u nderstand how this can be done, w e p resen t a p u lse -w id th m o d ­ ula to r using the L M 339 com parator.

2-8.2 Pulse-Width Modulator; Noninverting T he L M 339 co m p arato r in Fig. 2 - 16(a) co m p ares tw o input v o ltag es, Ec and Vtemp. [Figure 2-16(b) is sim ilar to Fig. 2 - 12(a).] A saw tooth wave, Ec, w ith constant frequency is connected to the ( - ) input, it is called a carrier w ave. Vlemp is a tem perature-controlled voltage. Its rate o f change must be m uch less than that o f E c. In this design, Vtemp is the signal from the tem perature transducer. It can be treated as a variable reference voltage when Fig. 2 - 16(a) is com pared to Fig. 2 - 12(a) or Fig. 2-5(b). In this circuit the input signal is defined as Viemp. The output is defined as the high tim e, T h o f Va. In Fig. 2 - 16(b), the output stays high for 2 ms w hen Vtemp = 1 V.If Vtemp increases to 4 V, high tim e TH increases to 8 ms as in Fig. 2 - 16(c).

(a) N o n i n v e r t i n g p u l s e - w i d t h - m o d u l a t o r circuit.

( t » ^ m P = ' V , r „ = 2 m s.

(c ) ^lemp = 4 V ,

= 8 ms.

r H= r -

(d) Input voltage Viemv vs. output high time Th -

FIGURE 2-16 V,emp is defined as the input signal in (a). As V'(emp increases from 0 to 5 V, the high time of output voltage V0 increases from 0 to 10 ms. The circuit is called a noninverting pulsewidth modulator

35

First Experiences with an Op Am p

O peration of the circuit is sum m arized by the in p u t-o u tp u t characteristics in Fig. 216(d). T he w idth of output pulse TH is changed (m odulated) by Vtemp. T he constant p e­ riod o f the output wave is set by Ec. Thus Ec carries the inform ation co n ta in ed in Vtemp. V0 is then said to be a pulse-w idth-m odulated wave. The in p u t-o u tp u t equation is output T h = (V^erop) - J —

(2-3)

w here T = period of saw tooth carrier wave E cm = m axim um peak voltage o f a saw tooth carrier Exam ple 2-2 shows that the pulse-w idth m odulator can also be called a duty-cycle controller.

Example 2-2 A 10-V, 50-H z saw tooth wave is pulse-w idth m odulated by a 4-V signal. Find the o u tp u t’s (a) high tim e; (b) duty cycle. S o lu tio n

Period T is found from the reciprocal o f the frequency: = 20 ms T = ~ = f 50 Hz

(a) From Eq. (2-3), 20 ms T„ = (4 V)

10 V

= o ms

(b) D uty cycle is defined as the ratio o f high tim e to the period and is expressed in percent: duty cycle =

X 100 8 ms 20 ms

(2-4)

X 100

Thus the output stays high for 40% o f each signal.

2-8.3 Inverting and Noninverting Pulse- Width Modulators F igure 2-17 show s the difference betw een noninverting and inverting pulse-w idth m odu­ lators. If signal Vlemp is applied to the ( + ) input, the circuit is defined as noninverting [see Figs. 2 - 17(a), (b), and (c)]. T he slope of TH versus Vlemp rises to the right and is p o sitive or noninverting.

(a) N o n i n v e r t i n g P W M .

(d) Inve rting P W M .

(A )°3 (A) °A (b) Input and output wave forms for Vref = I V.

(e) Input and output wave forms for Vref = I V.

lemp V v t

(c )

Output T h

vs.

\/[emp 7 = 1 0 ms.

(f) Output T h

vs.

Vle[np T = 10 ms.

FIGURE 2-17 Output high time increases as input Vlemp increases in a noninverting pulse-width modulator [see (a), (b), and (c)]. Output high time decreases as Vicmp increases in an inverting pulse-width modulator

37

First Experiences with an Op Am p

Vtemp is applied to the ( - ) input in Fig. 2 - 17(d). As Vtemp increases, TH decreases, The slope of TH versus Vternp is show n in Fig. 2 -1 7(f) and is negative. T he inverting p er­ form ance equation is (2-5)

Example 2-3 C alculate the output high tim e if Vternp = 4 V in Fig. 2 - 17(d).

Solution

From Eq. (2-5), 4 V

2-9 A PULSE-WIDTH MODULATOR INTERFACE TO A MICROCONTROLLER __________________ Either circuit of Fig. 2-17 can be used to transm it tem perature inform ation as a pulse-w idth m odulated signal to a com puter. The advantage o f such an analog interface circuit is to elim inate a voltage drop over distances o f several hundred feet. Thus the pulse-w idth m od­ ulator can interface an analog signal with an input port o f a m icrocontroller (see Fig. 2 ­ 18). T he tem perature is first converted to a voltage by the sensor. A noninverting pulsew idth m odulator then converts this analog voltage to an output that is digital in nature; that is, its output is either high or low and the high tim e is directly proportional to tem perature.

^T em p erature 20° C ^

w W

( ^ 2 V = Vltmp ^

Tem perature-tovoltage converter

FIGURE 2-18

TH = 40 ms ^

Noninverting pulsewidth m odulator

Internal counter of a m icrocontroller

Block diagram of a computerized temperature measurement.

T he com puter program m er can perform the analog-to-digital conversion o f the high tim e to a digital code. T his may be done by using a 1-ms tim ing loop and counting the num ber o f tim es that the loop is executed. A nother and m ore efficient m ethod is to use the internal counter designed into m ost m icrocontrollers. T he O-to-5-V transition o f V0 is

38

Chapter 2

used to start the m icrocontroller’s counter and the 5-to-O-V transition stops the co u n ter T he count, w hich is autom atically stored in one o f the m icro co n tro ller’s internal registers, is directly proportional to the tem perature.

2-10 OP AM P COMPARATOR CIRCUIT SIMULATION 2-10.1 Introduction PSpice is a softw are package for analog and digital design analysis. S tudents w ho are studying op am ps usually have aLready used PSpice in previous courses, so all the PSpice fundam entals are not introduced; however, enough introduction steps are included th ro u g h ­ out this text to allow first-tim e users to create and analyze their circuits.

2-10.2 Creating, Initializing, and Simulating a Circuit L et us create and analyze the noninverting positive-level detector circuit show n in Fig. 2-5(a). We will use a sine wave for the input signal because it is easy to obtain from the basic parts list. (Note: T he parts list does not contain a triangular w aveform although one can be created, w hich we w ill do in a later chapter.) To create and sim ulate Fig. 2-5(a), open a new w orksheet either by clicking on File = > New, or if the PSpice w indow is not open, double-click on the Schem atics icon in the window, [f necessary, enlarge the w ork area to fill the entire screen. The basic parts list browser may be obtained by clicking Draw from the Menu bar and then clicking Get New Part from the drop-down menu. These steps will be represented by

Draw = > Get New Part A shortcut for obtaining the parts list is to click the icon on the toolbar. (Note: T he icon sym bol is different for different versions o f PSpice.) E ith er m ethod produces a p o p ­ up m enu that contains the Parts Browser’s basic list. Click Advanced > > and the b a­ sic m enu expands to include a w indow to show you the part before you place it in the w ork area. O ther libraries o f parts can be obtained by clicking on the Libraries button. The general guidelines for creating and sim ulating a circuit in P Spice are: 1. O pen a new work area. 2. O btain each part from the parts list and place it in the w ork area. T hen close the parts list. 3. A rrange the parts the way they appear in the circuit schem atic. 4. Interconnect the parts. 5. C hange any attribute value(s) for a part if necessary. 6. Initialize setup param eters— Analysis = > Setup. 7. Initialize probe setup if you w ant a plot— Analysis = > Probe Setup 8. Save the schem atic as a file with the .SCH extension. 9. E nsure there are no w iring errors— Analysis = > Create Netlist. 10. E xecute the program to observe the results— Analysis = > Simulate.

39

First Experiences with an Op Am p

Let us create the noninverting positive-level detector circuit o f Fig. 2-5(a) by call­ ing up the follow ing parts and placing them in the w ork area. It is easier if you get all the parts at once and place them in the right section o f the w ork area, close the parts list, and then arrange the parts as they appear in the circuit schem atic. For this application, w e will use three dc supplies for + V, - V, and Vref. Draw => Get New Part Part

=> => => => => =>

UA741 VDC VSIN GLOBAL AGND R

Num ber

Version 6.2 Library

pins 1 and 5 are shown but not used place three for + V, - Vr and Vre( sine wave place six analog ground, place five resistor for RL

eval. sib source.slb source.slb portslb port.slb analog.slb

Close the parts list and arrange the parts as in Fig. 2-5(a). (Note: T he op am p PSpice m odel com es from the parts list w ith the inverting term inal at the bottom and the nonin­ verting term inal at the top o f the diagram . For now w e will leave it w ith this orientation. T he term inals can be sw itched if the op am p is rotated tw ice and then flipped. In this new orientation, however, + V is at the bottom and - V is at the top.) To interconnect the parts, click D ra w = > W ire or click the thin Pencil icon in the toolbar. F igure 2-19 show s how the parts can be interconnected.

+V

ref

V4

v2 — 6V -

+V

15V

15V -

VO

Vo

Vo

. 5

VampJ = 10 V Freq = 100 Hz

Rl = uA741

10 k

V

FIG U RE 2-19

PSpice model of a noninverting comparator circuit.

The parts in this circuit that require setting new values (attributes) are the three dc supply voltages; the six globals; the sine w ave’s am plitude, frequency, and offset; and the value o f R l . C hanging a p art’s attributes is done by first double-clicking on the part or value to be changed and then entering the new value. D ouble-clicking highlights the part or value in red and then opens an attribute box that allow s you to en ter the new value.

40

Chapter 2 O ne at a tim e, double-click on 0 V and set the voltages at the supplies as:

+ V = 15 V —V = 15 V (N ote the orientation o f this supply.)

V ref = 6 V O ne at a tim e, set the six G L O B A L labels as: + V — at pin 7 of the op am p and that connected to + 1 5 V —V— at pin 4 of the op am p and that connected to —15 V V ref — at pin 2 o f the op am p and the source used for Vref. Sim ilarly the label o f the resistor and its value can be changed to R L and 10 k fl, re­ spectively. To change the attributes of the input sine-w ave signal, double-click the sym ­ bol and a VS1N attribute box appears. O ne at a time, change each attribute by dou b le­ clicking the attribute and setting the new value in the window. For this circuit, am plitude, frequency, and offset values have to be changed as show n:

AM PL = to 10 V = > Save Attr = > Change Display = > Both name and value FREQ - to 100 Hz = > Save Attr = > Change Display = > Both name and value VOFF = to O V = > Save Attr (not necessary to change display for this application) In this application, we w ant a plot o f E n Vref, and VQ versus tim e sim ilar to w hat is show n in Fig. 2-5(a). In order to do this, we first m ust add the location o f Et and VDto the op a m p ’s inverting and output term inals, respectively. T he location o f Vref is already show n on the circuit diagram . This step is done by double-clicking the “w ire” connection at the point o f interest and entering the label in the w indow o f the pop-up box. Figure 2­ 19 show s the com pleted schem atic ready for analysis. To obtain these plots, open Analysis = > Probe Setup and click A u to m a tic ally Run P robe A fter S im u la tio n . N ow open Analysis = > Setup and click the box next to Transient. An x appears indicating it has been selected. Now click on Transient and set Print Step to 0.05 ms and Final T im e to 20 ms. This will allow P robe to display tw o com plete cycles o f a 100-H z sine wave. Save the file by File = > Save or by clicking the Disk icon in the toolbar. You may use any file nam e but be sure to use extension .SCH. A check o f the w iring corrections is done by creating a netlist— Analysis = > Create Netlist. A w arning appears if there are any w iring errors. C lick O K and a list o f the error location(s) is obtained. If there are no errors, then the circuit is ready to run the sim ulation program . T his step is done by Analysis = > Simulate or using the hot key FI 1. T he Probe window (a b lack screen) ap ­ pears. To plot the graphs, use Trace = > Add and click V[Ei], V[Vref], and V[Vo] and then OK. T he three w aveform s should now be plotted as show n in Fig. 2-20. To add la­ bels to the graphs, use Tools = > Label = > Text and a text box appears. Type in the la­ bel you w ish to place on the graph and then click OK. Use the m ouse to place the label w here you w ant it and repeat the procedure for any new labels. To add arrow s, use Tools = > Label = > Arrow. Use the m ouse to place the tail o f the arrow at the starting point and draw out the arrow. C lick the left m ouse button to stop and the com pleted arrow is draw n.

42

Chapter 2

2-7. Ej is applied to the ( - ) input and ground to the ( + ) input of a 741 in Fig. P2-7. Sketch ac­ curately (a) Va vs. / and (b) V0 vs. E-r

FIGURE P2-7

2-8. Swap the input connections to E, and ground in Fig. P2-7. Sketch (a) Vn vs. t and (b) V0 vs. E, 2-9. Refer to Problems 2-7 and 2-8. Which circuit is the noninverting zero-crossing detector, and which is the inverting zero-crossing detector? 2-10. To which input would you connect a reference voltage to make an inverting level detector? 2-11. You need a 741 noninverting voltage-level detector, (a) Will the output be at +V sa, or ~ V sM when the signal voltage is above the reference voltage? (b) To which input do you connect the signal? 2-12. Design a reference voltage that can be varied from 0 to - 5 V. Assume that the negative sup­ ply voltage is - 1 5 V. 2-13. Design a 0 to +50 mV adjustable reference voltage. Derive it from the + 15-V supply. 2-14. The frequency of carrier wave Ec is constant at 50 Hz in Fig. P2-14. If Vlemp = 5 V, (a) cal­ culate high time TH\ (b) plot V0 vs. time.

FIGURE P2-14

2-15. Assume that Vtemp is varied from 0 V to + 10 V in Problem 2-14. Plot TH vs. Vlemp

43

First Experiences with an Op Am p

2-16. In Fig. P2-16, £ in is a triangle wave. The amplitude is - 5 V to + 5 V and the frequency is 100 Hz. Sketch accurately the graphs of (a) Va vs. £ in; (b) V0 vs. /. + 15 V

A

+5 V

FIGURE P2-16 2-17. Draw the schematic of a circuit whose output voltage will go positive to + Vsa„ when the in­ put signal crosses +5 V in the positive direction. 2-18. Is the solution of Problem 2 -1 7 classified as an inverting or noninverting comparator? 2-19. Draw a circuit whose output goes to + Vsat when the input signal is below - 4 V. The output should be at —Vsa, when the input is above - 4 V. 2-20. Does the solution circuit for Problem 2-19 represent an (a) inverting or noninverting, (b) pos­ itive- or negative-voltage-level detector?

CHAPTER 3 Inverting and Noninverting Amplifiers

LEARNING

OBJECTIVES

___________________________________________________

U pon com pleting this chapter on inverting and noninverting am plifiers, you will be able to: • D raw the circuit for an inverting am plifier and calculate all voltages and currents for a given input signal. • Draw the circuit for a noninverting am plifier and calculate all voltages and currents. • Plot the output voltage w aveshape and o u tp u t-in p u t characteristics o f eith er an invert­ ing or a noninverting am plifier for any input voltage w aveshape. • D esign an am plifier to m eet a gain and input resistance specification. • Build an inverting or noninverting adder and audio mixer. • U se a voltage follow er to m ake an ideal voltage source. • C reate a negative output voltage from a positive reference voltage. • Add a dc offset voltage to an ac signal voltage.

Inverting and Noninverting A m plifiers

45

• M easure the average value of several signals. • D esign with single-supply op am ps. • Build a subtractor. • D esign a signal conditioning circuit for a tem perature sensor. • A nalyze inverting and noninverting am plifier circuits using PSpice.

3-0 INTRODUCTION

_________________________________________________________________________

This chapter uses the op am p in one of its m ost im portant applications— m aking an am ­ plifier. An am plifier is a circuit that receives a signal at its input and delivers an undis­ torted larger version o f the signal at its output. All circuits in this chapter have one fea­ ture in com m on: An external feedback resistor is connected betw een the output term inal and ( —) input term inal. T his type of circuit is called a negative fe ed b a c k circuit. T here are many advantages obtained with negative feedback, all based on the fact that circuit perform ance no longer depends on the open-loop gain o f the op am p, A OL. By adding the feedback resistor, we form a loop from output to ( —) input. T he resulting cir­ cuit now has a closed-loop gain or am plifier gain, A CL, w hich is independent o f A OL (pro­ vided that A 0 l is m uch larger than ACL). As will be shown, the closed-loop gain, A CL, depends only on external resistors. For best results 1% resistors should be used, and A CL will be know n within 1%. N ote that adding external resistors does not change the open-loop gain A OL. A OL still varies from op am p to op am p, so adding negative feedback will allow us to ignore changes in A o L as long as A OL is large. We begin with the inverting am plifier to show that A CL depends sim ply on the ratio o f two resistors.

3-1 THE INVERTING AMPLIFIER

____________________________________________________________

3-1.1 Introduction T he circuit of Fig. 3 - 1 is one of the most w idely used op am p circuits. It is an am plifier w hose closed-loop gain from Et to V0 is set by Rf and /?,. It can am plify ac or dc signals. To understand how this circuit operates, we make two realistic sim plifying assum ptions that w ere introduced in C hapter 2. 1. T he voltage E d betw een the ( + ) and ( —) inputs is essen tially 0 if VQ is not in saturation. 2. The current drawn by either the ( + ) or the ( - ) input term inal is negligible.

3-1.2 Positive Voltage Applied to the Inverting Input In Fig. 3-1, positive voltage E, is applied through input resistor R t to the op am p ’s ( —) in­ put. N egative feedback is provided by feedback resistor Rf . T he voltage betw een the ( + ) and ( - ) inputs is essentially equal to 0 V. T herefore, the ( - ) input term inal is also at

46

Chapter 3 Voltage across R f equals V(>.

— Rf = 100 kft Voltage across Rj equals Et

+V

— W v ----------R; = 10 kQ

-0V

V

FIG U RE 3-1 A positive input voltage is applied to the ( —) input of an in­ verting amplifier, ft, converts this voltage to a current, /; /^converts / back into an amplified version of Er

0 V, so ground potential is at the ( —) input. For this reason, the ( - ) input is said to be at virtual ground. S ince one side of /?, is at E, and the other is at 0 V, the voltage drop across /?, is E h T he current I through /?, is found from O h m 's law: (3 -la ) Rj includes the resistance of the signal generator. This point is discussed further in Section 3-5.2. All of the input current I flow s through Rf> since a negligible am ount is draw n by the ( —) input term inal. N ote that the current through Rf is set by /?, and not by Rft V0, or the op amp. T he voltage drop across R f is sim ply I (Rf), or (3 -lb ) As shown in Fig. 3-1, one side of R f and one side o f load R L are connected. The voltage from this connection to ground is V0. The other sides o f R f and o f R L are at ground potential. Therefore, V0 equals VRj (the voltage across RJ). To obtain the polarity o f VQ, note that the left side of R f is at ground potential. The current direction established by Ef forces the right side of R f to go negative. Therefore, VQ is negative when Et is positive. Equating V0 with VRj and adding a minus sign to signify that VG goes negative when Ex goes positive, we have

47

Inverting and Noninverting A m plifiers

0

(3-2a)

'/?,■

Now, introducing the definition that the closed-loop gain o f the am plifier is A cL, we rew rite Eq. (3-2a) as A

C L

—

(3-2b)

R,

T he m inus sign in Eq. (3-2b) show s that the polarity o f the output Va is inverted w ith respect to E h For this reason, the circuit o f Fig. 3-1 is called an inverting amplifier.

3-1.3 Load and Output Currents The load current 1L that flow s through R L is determ ined only by R L and V0 and is fu r­ nished from the op am p ’s output term inal. T hus 1L = V J R L. The current / through Rf m ust also be furnished by the output term inal. T herefore, the op am p output current l0 is (3-3)

E x a m p le 3-1 For Fig. 3-1, let Rf = 100 k fl, R, = 10 k fl, and £, = 1 V. C alculate (a) /; (b) V*; (c) A b ­ s o lu tio n

(a) From Eq. (3 -la), F 1V I = — = “4 ^ 7 7 = 0.1 mA

/?,

io k n

(b) From Eq. (3-2a), Rr

v’ = - %

x e >-

(iv»-

- |ov

(c) U sing Eq. (3-2b), we obtain R>r

Ri

_

100 k f l =

10 k f l

-10

This answ er m ay be checked by taking the ratio o f Vv to AcL = ^ = ^ y C E, 1V

= - l0

E x a m p le 3-2 U sing the values given in Exam ple 3-1 and R L = 25 k fl, determ ine (a) IL\ (b) the total cu r­ rent into the output pin of the op amp.

49

Inverting and Noninverting A m plifiers

E x a m p le 3-3 For Fig. 3-2, let R j = 250 k age across R /y (c) V0. S o lu tio n

R t = 10 kCl, and

= - 0 . 5 V. C alculate (a) /; (b) the volt­

(a) From Eq. (3 -la ), 1=

Ej_ = 0.5 V R,

10 k a

= 50 /jlA = 0.05 mA

(b) From Eq. (3 -lb ), VR f = I X R f = (50 /iA )(2 5 0 k ft) = 12.5 V (c) From Eq. (3-2a), 250 kH

Rr

io k n

( - 0 . 5 V) = + 1 2 .5 V

Thus the m agnitude of the output voltage does equal the voltage across Rf, and A C L = —25. E x a m p le 3-4 U sing the values in Exam ple 3-3, determ ine (a) R L for a load current o f 2 mA; (b) I0\ (c) the circu it’s input resistance. S o lu tio n

(a) U sing O h m ’s law and VD from E xam ple 3-3, Vn IL

12.5 V 2 mA

(b) From Eq. (3-3) and E xam ple 3-3, IQ = 1 + IL = 0.05 mA + 2 mA = 2.05 mA (c) The circu it’s input resistance, or the resistance seen by , is /?, = 10 kH. A PSpice model and sim ulation results are given in Section 3-13.

3-1.5 Voltage Applied to the Inverting Input Figure 3-3(a) shows an ac signal voltage E t applied via R , to the inverting input. For the positive half-cycle, the voltage polarities and the direction o f currents axe the sam e as in Fig. 3-1. For the negative half-cycle voltage, the polarities and direction o f currents are the sam e as in Fig. 3-2. T he output waveform is the negative (or 180° out o f phase) o f the

Inverting and Noninverting Amplifiers

51

E x a m p le 3-6 If the input voltage in Exam ple 3-5 is - 5 V, determ ine the output voltage. S o lu tio n

U sing Eq. (3-2a) or rearranging Eq. (3-2b), we obtain V, = —

X E, = A cl E, = ( —2 )(—5 V) = 10 V

See tim e 0 in Figs. 3-3(b) and (c). T he frequency o f the output and input signals is the same. A P Spice m odel and sim ulation are given in Section 3-13. The sim ulation uses a 5-Vpeak sine w ave w ith a frequency set at 500 H z as the input signal.

3-1.6 Design Procedure Follow ing is an exam ple o f the design procedure for an inverting am plifier.

D esign E x a m p le 3-7 D esign an am plifier w ith a gain of —25. The input resistance R {n should equal or exceed

10 m . D esign P ro c e d u re 1. C hoose the circuit type illustrated in Figs. 3-1 to 3-3. 2. Pick Ri = 10 k f l (safe, prudent choice). 3. C alculate R f from Rf ~ (gain)(/?>). (For this calculation, use the m agnitude o f gain.)

3-1.7 Analysis Procedure You are interview ing for a jo b in the electronics field. T he technical interview er asks you to analyze the circuit. A ssum e that you recognize the circuit as that o f an inverting am ­ plifier. Then, 1. Look at R h State that the input resistance o f the circuit equals the resistance o f R 2. D ivide the value of R f by the value o f R h State that the m agnitude o f gain equals R f/R h A lso, the output voltage will be negative w hen the input voltage is positive.

53

Inverting and Noninverting A m plifiers

Solution From Eq. (3-4), VQ = - ( 2 V + 3 V + 1 V) = —6 V. T he PSpice model for this circuit is shown in Fig. 3-24. Example 3-9 If the polarity of E 3 is reversed in Fig. 3-4 but the values are the sam e as in E xam ple 3-8, find V0.

Solution

From Eq. (3-4), V , = - ( 2 V + 3 V -

1 V) = - 4 V.

If only two input signals, E x and E 2, are needed, sim ply replace E 3 w ith a short cir­ cuit to ground. If four signals m ust be added, sim ply add another equal resistor R betw een the fourth signal and the sum m ing point S . E quation (3-4) can be changed to include any num ber o f input voltages.

3-2.2 Audio Mixer In the adder of Fig. 3-4, all the input currents flow through feedback resistor Rf . T his m eans that /, does not affect / 2 or / 3. M ore generally, the input currents do not affect one another because each “sees” ground potential at the sum m ing node. T herefore, the input currents— and consequently the input voltages £ ], E 2, and E 3— do not interact. T his feature is especially desirable in an audio mixer. For exam ple, let E u E2, and E 3 be replaced by m icrophones. The ac voltages from each m icrophone w ill be added or m ixed at every instant. Then if one m icrophone is carrying guitar m usic, it will not com e out of a second m icrophone facing the singer. If a lOO-kfl volum e control is installed be­ tw een each m icrophone and associated input resistor, their relative volum es can be ad ­ justed and added. A w eak singer can then be heard above a very loud guitar.

3-2.3 DC Offsetting an AC Signal Som e applications require that you add a dc offset voltage or current to an ac signal. Suppose that you m ust transm it an audio signal via an infrared em itting diode (IR ED ) or light-em itting diode. It is first necessary to bias the LRED on w ith a dc current. T hen the audio signal can be superim posed as an ac current that rides on or m odulates the dc cu r­ rent. T he result is a light or infrared beam w hose intensity changes directly w ith the au ­ dio signal. We illustrate this principle by an exam ple.

Example 3-10 D esign a circuit that allow s you to add a dc voltage to a triangle wave.

Solution S elect a tw o-channel adder circuit as in Fig. 3-5(a). A variable dc offset voltage Edc, is connected to one channel. The ac signal, £ ac, is connected to the other.

56

Chapter 3 Rf

E 1\„n

Rr

A1

Rr

+ E2 D-7T + E 3 -Z-D A2

A3

(3-7a)

E quation (3-7a) shows that the gain of each channel can be changed independently o f the others by sim ply changing its input resistor. Rf_

CL-S

(3-7b)

R?

/?>’ or

V0 — E ]A Cl } + E 2A CL2 + E?A c L)

3-3.3 Design Procedure Follow ing is an exam ple of the design procedure for a m ultichannel am plifier.

D esign E x a m p le 3-11 D esign a three-channel inverting am plifier. T he gains for each channel will be

Channel number

Voltage gain

1 2 3

-10 -5 -2

D esign P ro c e d u re 1. S elect a 10-kfi resistor for the input resistance o f the channel with the highest gain. C hoose /?! = 10 k f l since A CLi is the largest. 2. C alculate feedback resistor Rf from Eq. (3-7b): A

C L ,

= _K l

.

10

R, =

10 k f l ’

R f = 100 k n

3. C alculate the rem aining input resistors from Eq. (3-7b) to get R 2 — 20 k H and R 3 = 50 kfl.

3-4 INVERTING AVERAGING AMPLIFIER Suppose that you had to m easure the average tem perature at three locations in a dw elling. First m ake three tem perature-to-voltage converters (shown in Section 5-14). T hen connect their outputs to an averaging amplifier. An averaging am plifier gives an output voltage

57

Inverting and Noninverting A m plifiers

proportional to the average o f ail the input voltages. If there are three input voltages, the averager should add the input voltages and divide the sum by 3. The averager is the sam e circuit arrangem ent as the inverting adder in Fig. 3-4 or the inverting adder w ith gain in Fig. 3-6. T he difference is that the input resistors are m ade equal to som e convenient value R and the feedback resistor is made equal to R divided by the num ber of inputs. Let n equal the num ber of inputs. Then for a three-input averager, n = 3 and Rf = R!3. Proof is found by substituting into Eq. (3-7a), for Rf = R!3 and R x = R 2 — R^ = R to show that (3-8)

E x a m p le 3-12 In Fig. 3-4, = R2 = R 3 = R = 100 kH and Rf = 100 kH /3 ~ 33 k f l If E x = + 5 V, E 2 = + 5 V, and E 3 = - IV , find V0. S o lu tio n have

Since R f = R B y the am plifier is an averager, and from Eq. (3-8) with n — 3, we

5 V + 5 V + ( - 1 V) 3

Up to now we have dealt w ith am plifiers w hose input signals were applied via R{ to the op am p ’s inverting input. We turn our attention next to am plifiers in w hich E-, is ap ­ plied directly to the op am p ’s noninverting input.

3-5 NONINVERTING AMPLIFIER

___________________________________________________________

3-5.1 Circuit Analysis Figure 3-7 is a noninverting am plifier; that is, the output voltage, VD, is the sam e polar­ ity as the input voltage, T he input resistance o f the inverting am plifier (Section 3-1) is R h but the input resistance o f the noninverting am plifier is extrem ely large, typically exceeding 100 M fl. Since there is practically 0 voltage between the ( + ) and ( - ) pins of the op am p, both pins are at the sam e potential E r Therefore, Et appears across R\> causes current I to flow as given by (3-9a) T he direction o f / depends on the polarity of E r C om pare Figs. 3-7(a) and (b). The input current to the op am p ’s ( —) term inal is negligible. T herefore, I flows through f y a n d the voltage drop across Rf is represented by VR and expressed as

59

Inverting and Noninverting A m plifiers Rf Va = Ei + - ^ - E, R\ or Vo ~ [ 1 +

Ri

E,-

(3-IOa)

R earranging Eq. (3 -10a) to express voltage gain, we get V A cl = - ^ = Ej

Rf R f + R] I+ ~ J ~ = / p R\ Ri

(3-IOb)

E quation ( 3 - 10b) shows that the voltage gain o f anoninverting am plifier is alw ays greater than 1. The load current 1L is given by V0/R l and therefore depends only on V0 and R L. I0 is the op am p ’s output current and is given by Eq. (3-3).

Example 3-13 (a) Find the voltage gain for the noninverting am plifier o f Fig. 3-8. If E , is a 100-Hz trian ­ gle w ave w ith a 2-V peak, plot (b) V0 vs. t\ (c) VD vs. E h S o lu tio n

(a) From Eq. ( 3 - 10b), R f+ R t CL~

R]

( 4 0 + 1 0 ) k fl ~

10 kH

“

(b) See Fig. 3-8(b). T hese are the w aveshapes that w ould be seen on a dc-coupled, d u al­ trace oscilloscope. (c) See Fig. 3-8(c). S et an oscillo sco p e for an x - y display, vertical 5 V /div, h o rizo n tal 1 V/div. N ote that the slope rises to the right and is positive. R ise over run gives you the gain m agnitude of + 5 .

3-5.2 Design Procedure Follow ing is an exam ple o f the design procedure for a noninverting am plifier.

Design Example 3-14 D esign an am plifier with a gain of + 10.

Chapter 3

R/+ R \

■E, = 5E:

(a) Noninverting am plifier circuit with gain of +5.

K>(V) (A) !3 (A) '3 PU« °A (c) Input-output characteristic o f a noninverting amplifier.

(b) W aveshape of VQ and £ , vs. /.

FIGURE 3-8

Noninverting amplifier circuit analysis for Example 3-13.

61

Inverting and Noninverting A m plifiers

Design Procedure 1. Since the gain is positive, select a noninverting am plifier. T hat is, we apply E, to the op a m p ’s ( + ) input. 2. Choose R { = 10 kft. 3. C alculate Rf from Eq. (3-10b). 10 = 1 + 777777’ 10 k i l

3 6 VOLTAGE FOLLOW ER

Rf = 9(10 k fl) = 90 left

___________________________________________________________________

3-6.1 Introduction The circuit o f Fig. 3-9 is called a voltage follow er, but it is also referred to as a source follower, unity-gain amplifier, buffer amplifier, or isolation amplifier. It is a special case o f the noninverting amplifier. The input voltage, £, , is applied directly to the ( + ) input. Since the voltage betw een ( + ) and ( —) pins of the op am p can be considered 0, ( 3 - II a) Note that the output voltage equals the input voltage in both m agnitude and sign. T herefore, as the nam e o f the circuit im plies, the output voltage fo llo w s the input or source voltage. The voltage gain is 1 (or unity), as shown by (3-1 lb)

FIG U RE 3-9

Voltage follower.

64

Chapter 3

Now let us consider the sam e signal source connected to an inverting am p lifier w hose gain is - 1 [see Fig. 3 - 1 1(b)]. As stated in Section 3-1.3, the input resistan ce to an inverting am plifier is R,. T his causes the g en erato r voltage £ gen to divide betw een R ini and /?,. U sing the voltage division law to find the g en erato r term inal voltage E { yields R,

E ,= R\nl

+ Ri

X Earn =

io k n 10 k n + 90 k.Q

x (1.0 V) = 0.1 V

T hus it is this 0.1 V that becom es the input voltage to the inverting am plifier. If the in­ verting am plifier has a gain of only —1, the output voltage V0 is —0.1 V. In conclusion, if a high-im pedance source is connected to an inverting am plifier, the voltage gain from V0 to Egen is not set by /fy-and R , as given in Eq. (3-2b). T he actual gain m ust include R inl, as R71 Ri

+ R \,

io kn ioo k n

o .i

If you m ust am plify and invert a signal from a high-im pedance source and w ish to draw no signal current, first buffer the source w ith a voltage follower. Then feed the fo l­ low er’s output into an inverter. If you need buffering and do not w ant to invert the input signal, use the noninverting amplifier.

3-7 THE "IDEAL" VOLTAGE SO U RC E 3-7.1 Definition and Awareness T he ideal voltage source is first encountered in textbooks concerned w ith fundam entals. By definition, the voltage does not vary regardless o f how m uch current is draw n from it. You m ay not be aw are of the fact that you create a perfect voltage source w hen you m e a­ sure the frequency response o f an am plifier or filter. We explain how this apparently per­ fect perform ance com es about in the next section.

3-7.2 The Unrecognized Ideal Voltage Source The lab or field procedure typically goes like this: Set the input signal am plitude at 0.2 V rm s and frequency at the low est lim it. M easure output V0. Hold E in at 0.2 V rm s for each m easurem ent. Plot VQ or V0/E [u versus frequency. As you dial higher frequencies, E m b e­ gins to decrease (because of input capacitance loading). You autom atically increase the function g en erato r’s volum e control to hold E lu at 0.2 V. You have ju st, by definition, cre­ ated an “ideal” voltage source. £ in never varied throughout the test sequence no m atter how m uch current was draw n from it. This is an exam ple o f the unrecognized ideal volt­ age source.

66

Chapter 3

To preserve the value of any reference voltage, sim ply buffer it with a voltage fol­ lower. T he 7.5-V reference voltage is connected to a voltage follow er in Fig. 3-12(c). The output of the follow er equals K e f - You can extract up to 5 mA from the fo llo w er's output w ith no change in Vref. T he buffer m akes an excellent clandestine bug. You can m onitor w hat is going on at any circuit point. Since a follow er has a high input im pedance, it draw s negligible cu r­ rent from the circuit. T herefore, it is nearly im possible to detect.

3-7.4 Precise Voltage Sources Section 2-6 introduced precision voltage reference ICs such as the R E F -02 (a precision + 5-V reference chip). You can use these voltage reference chips w ith an inverting am ­ plifier to create precise negative voltages as well as positive and negative voltages. The circuit o f F igure 3 - 13(a) show s how a negative voltage o f —5 V can be created using the R E F -02 and an inverting am plifier. T his circu it has a low er parts count and m ore p reci­ sion than the circuit o f Figure 3-12(c). T he parts co u n t is obvious, and the precision is obtained by the R E F -02 in place o f /?, and Rf . C onsider R t and Rjr to be 1% resistors. T hen there is a possibility that one resistor could be + 1 % w hile the other is —1%. T his w ill produce an o utput voltage w ith a 2% error, w hich may not be acceptable for your design. A n o th er ap p licatio n using the R E F -02 w ith an inverting am p lifier is show n in F igure 3 - 13(b). T his circuit creates a ± 5 -V source from a single R E F-02 chip and an in­ verting am plifier.

3-8 NONINVERTING ADDER

________________________________________________________________

A three-input noninverting adder is constructed with a passive averager and noninverting am plifier as shown in Fig. 3 - 14(a). The passive averager circuit consists o f three equal re­ sistors R a and the three voltages to be added. The output o f the passive averager is £ in, w here £ in is the average o f E lt E2, and £ 3, or E in — (E , 4- E 2 + E3)/3. C onnect a voltage follow er to E in if you need a noninverting averager (in contrast with Sec. 3-4). O utput V0 results from am plifying £ in by a gain equal to the num ber o f inputs rt. In Fig. 3 - 14(a), n — 3. D esign the am plifier by choosing a convenient value for resistor R. T hen find Rf from Rf = R ( n - 1)

(3-12)

As shown in Fig. 3-14(a), the value for /^ s h o u ld be Rf = 10 kH (3 — 1) = 20 kH . If E 2, and E 3 are not ideal voltage sources, such as a battery or output of an op am p, buffer them w ith follow ers as in Fig. 3 - 14(b).

67

Inverting and Noninverting A m plifiers + 15 V

1

2

V

-1 5 V

SJ

(b) FIGURE 3-13

(a) A - 5 V precision output voltage; (b) a ±5 V output voltage.

3-9 SIN GLE-SUPPLY OPERATION

__________________________________________________________

Som e am plifier designs require battery-pow ered operation and output voltage sw ings to w ithin m illivolts o f the supply voltages. Thus, you w ant an op am p described by m anu­ facturers as a device capable o f single-supply and rail-to-rail operation. Two such devices are the A D 820 and the O P-90 from A nalog D evices. T h ese op am ps can operate from either a single or dual supply. For exam ple, the A D 820 can operate from a dual supply

.

sources, simply buffer each one with a voltage follower.

FIG U RE 3-14 All resistors of an n-input noninverting adder are equal except the feedback resistor; choose R = 10 kfl and RA = 10 kfl. Then /^eq u als R times the number of inputs minus one: Rf = R(n — I).

(± 1 .5 V to ± 1 8 V) or from a single supply ( + 3 V to + 3 6 V). T he input signal applied to these devices can be brought to ground, and the output can sw ing to w ithin 10 mV of either supply voltage. They are available in an 8-pin m ini-D IP package w ith the sam e pinouts as the 741 or O P-177 op am ps. Single-supply op am ps are often used in batterypow ered applications, portable instrum ents, m edical instrum entation, and data acquisition units. O ften they are used to am plify positive signals com ing from sensors such as strain

69

Inverting and Noninverting A m plifiers Rf = 90 k Q

---------W V -------+5 V

A

FIG U R E 3-15 The AD820 can operate for single-supply applications as shown in (a) and (b).

gages or therm ocouples. Figure 3 -15(a) show s the A D 820 w ired as a noninverting am ­ plifier w ith a gain o f 10. If you are operating the A D 820 from a single supply and w ant to am plify an ac signal, then the input ac signal has to have a dc offset or be com bined w ith a dc voltage as show n in F igure 3 -15(b). (Note: T his circuit is sim ilar to the invert­ ing adder studied in F igure 3-5(a) but now operated from a single supply.)

3-10 DIFFERENCE AM PLIFIERS

_____________________________________________________________

T he differential am plifier and its more pow erful relative, the instrum entation am plifier, will be studied in C hapter 8. However, as other applications of inverting and noninvert­ ing am plifiers, we offer two exam ples o f the difference am plifier in this section and the design o f a signal conditioning circuit for a tem perature sensor in the next.

70

Chapter 3

3-10.1 The Subtractor A circuit that takes the difference betw een tw o signals is called a su b tra cto r [see Fig. 316(a)]. It is m ade by connecting an inverting am plifier to a tw o-input inverting averager. To analyze this circuit, note that E, is transm itted through op am p A w ith a gain o f — 1 and appears as V0] = —E x. VDi is then inverted (tim es - 1 ) by the top channel o f the in ­ verting am plifier B. Thus E { is inverted once by op am p A and again by op am p B to ap ­ pear at VQ as E ,. E 2 is inverted by the bottom channel o f op am p B and drives V0 to —E 2. T hus V0 responds to the difference betw een E { and E2, or V0 = E l - E 2

Inverting am plifier

(3-13a)

T w o-input inverting adder

10 kO

J5 V (a) An inverting am plifier and a two-input inverting adder make a subtractor. VQ= E { - £ 2-

Rr= JOkft

------- ^A/V---------

15 V (b) Both am plifier inputs are used to make an am plifier that calculates the difference between 2E, and E 2.

FIG U RE 3-16 Two examples of difference amplifiers are the subtractor in (a) and us­ ing the op amp as both an inverting and a noninverting amplifier in (b).

71

Inverting and Noninverting A m plifiers

As show n in Fig. 3-16(a), for £ , = 2 V and E2 = 3 V, V0 = 2 - 3 = - 1 V. If the value o f R f is m ade larger than R h the subtractor w ill have gain

Va = ^

(£, - E2)

( 3 - 13b)

K,

3-10.2 Inverting-Noninverting Amplifier In Fig. 3-16(b), signal E\ is applied to the am p lifier’s noninverting input and signal E2 is applied to the inverting input. We w ill use superposition to analyze this circuit. First re­ m ove E2 and replace it by a ground. E { sees a noninverting am plifier w ith a gain o f (Rf + Ri)IRit or 2. T hus E x alone drives V0 to 2E {. N ext reconnect E2 and replace E } by a ground. E2 sees an inverting am plifier with a gain o f - 1 . E2 drives V0 to —E2. W hen both E x and E2 are connected, V0 is given by

V„ = 2E[ - E2

(3-14)

As show n in Fig. 3 - 16(b), VQ = 1 V w hen - 2 V and E2 = 3 V. W e will now show how to design the subtractor circuit o f Fig. 3 - 16(a) to be the an a­ log interface circuitry connected betw een a tem perature sensor and the analog—digital converter o f a m icrocontroller. T his analog interface circuit is also know n as a signal co n ­ ditioning circuit.

3-77 DESIGNING A SIGNAL CONDITIONING CIRCUIT* A nother way of viewing the circuit in Fig. 3-5(a) (redraw n in Figure 3-17 for convenience) is that it allows us to design a signal conditioning circuit (SCC) for a m icrocontroller ap­ plication that satisfies the equation of a straight line, y = mx + b. This equation occurs quite often w hen designing SCCs. Com paring the equation o f y = mx + b to the circuit o f Fig. 3-17, the y term is the output voltage, V0; the x term is the input signal voltage, E the m term is the gain o f the circuit, R f/R {\ and the b term is Rf /R2 tim es Edc. Therefore, if your application uses a sensor that generates an output signal m easured with respect to ground, w hich m ust be am plified and offset, then an SCC sim ilar to Fig. 3-17 may be used. (Note: T he outputs o f som e sensors generate a differential output, and these devices require an SCC capable o f m easuring a differential voltage. Such circuits are studied in C hapter 8.) The de­ sign of any SCC unit requires obtaining the equation o f the circuit. This equation is obtained from w hat y o u ’ve got (the output conditions o f the sensor) to what you w ant (the input con­ ditions o f the m icrocontroller’s A /D converter). Let us study this topic.

Statement of the problem. D esign a signal conditioning circuit to interface betw een a tem perature sensor and the A /D converter o f a m icrocontroller. T he tem pera­ * For more exam ples of the linear circuil design procedure refer to D ata Acquisition and Process Control with the M 6 8 H C I1 Microcontroller, 2nd Ed. by F. Driscoll, R. Coughin, and R. Villanucci, Prentice Hall (2000).

72

Chapter 3

AA/V

R

+ 15 V

A

FIGURE 3-17 The inverting summer can be designed to satisfy the equation of a straight line, y = mx + b

ture range to be m easured is 0° to 50°C, and the range o f the A /D converter is from 0 to 5 V. You w ant the output o f the SCC to be linear; that is, w hen the sensor is m easuring 0°C , the output o f the SCC is 0 V; w hen the sensor is m easuring 10°C, the S C C ’s output is to be 1 V; and so forth up to 50°C, at w hich tem perature the SC C outputs 5 V. Solution. A lthough our goal is to design the SCC, w hich is an op amp circuit, our starting point is the sensor and w riting an equation for it, because the output o f the sensor is the input to the SCC. Therefore, once the sensor is picked, this is w hat y o u ’ve got. W hat you want is the output o f the SCC to fit the range o f the m icrocontroller’s A/D converter. Therefore, the SCC design is being squeezed between w hat y o u ’ve got and w hat you want. L et’s first learn about one type of tem perature sensor and how to w rite the equation for it.

Introduction to a temperature sensor. For this application, we shall choose the LM 335, which is a solid-state temperature sensor that belongs to a family o f devices that has a sensitivity of 10 mV/°K. It is used in applications that require m easuring tem peratures from —10° to 100°C and is m odeled as a two-terminal zener. The package style and model are shown in Figures 3-18(a) and (b), respectively. This device is capable o f operating over a current range of from 400 fiA to 5 mA. The data sheet for the LM 335 gives the device’s sensitivity as 10 mV/°K. However, our application is to measure degrees Celsius. The rela­ tionship between degrees Kelvin and degrees Celsius is: A 1-degree rise in Kelvin equals a 1-degree rise in Celsius, and the freezing point o f water is 0°C, which equals 273°K. Writing an equation that describes the sensor. A plot o f the output volt­ age of the JLM335 versus tem perature is given in Figure 3-18(c). The slope o f the line is the d ev ice’s sensitivity— 10 m V /°K . T herefore, in term s o f °K the output voltage is Vj = (10 mV/°K) (7 inOK) w here T is the tem perature in °K. At 273°K (0°C), the sen so r’s output voltage is VV = (10 m V /°K ) (273°K ) - 2.73 V as show n in Figure 3 - 18(c).

(3-15)

TO-92 Plastic package

TO-46 Metal can package Current-limiting extemaj resistor, R Model of the LM335

Voltage (VT)

(c)

FIGURE 3-18 The LM335 (a) package styles, (b) model, and (c) voltage ver­ sus temperature characteristic.

N ow the sensor’s output voltage can be w ritten in term s o f degrees C elsius as Vr - (10 m V /°C ) (7 inoC) + 2.73 V

(3-16)

w here T is the tem perature in degrees Celsius. For our application, at 0°C VT = 2.73 V and at 50°C VT = 3.23 V. This is the input voltage range for the SCC. T he output range o f the SC C is the input range o f the A /D converter, w hich is 0 V to 5 V. F igure 3-19 show s a block diagram o f this data acquisition system for m easuring tem perature.

Writing an equation that describes the SCC. F rom the inform ation we know about the sensor and the A /D converter, we can plot the output/input characteristics o f the SCC. F igure 3-19 includes such a plot. T he output values o f the SCC are plotted on the y-axis. R em em ber, these values are the voltage range o f the A /D converter— 0 V to 5 V. T he input values to the SCC are plotted on the je-axis. T hese values are the volt­ age range o f the sensor— 2.73 V to 3.23 V for this application.

74

Chapter 3

■Ar

o°c

2.73 V LM335

/

50°C

3.23 V

Signal conditioning circuit

0 V M icrocontroller 5V

VV(V)

(3.23 - 2.73) = 0.5 V -2 7 .3

FIGURE 3-19 Block diagram of a temperature measuring system and the de­ sired output-input characteristics of the SCC.

T he slope o f the line is (5 ~ 0) V (3.23 - 2.73) V

10

A vy

(3-17)

T his value of 10 is the gain that VT m ust be m ultiplied by. T he dc offset is found from ch o o sin g a point on the line and su b stitu tin g into the eq u atio n o f a stra ig h t lin e— y = m x + b. C hoosing the coordinate pair (2.73, 0), we obtain 0 = (10) (2.73) + b S olving for b yields b = - 2 7 .3 V T hus, the equation of the S C C ’s output voltage is V0 = (10) (V T ) - 27.3 V

(3-18)

Note: A lthough the dc offset is - 2 7 .3 V, the output voltage, Vot never goes to this value be­ cause the range of VT is from 2.73 V to 3.23 V. This range o f VT limits V0 from 0 to 5 V.

Designing the signal conditioning circuit. N ow that we know the equation for the SC C and it is in the form o f y = m x + b, we w ant a circuit in w hich the gain o f 10 and the offset o f —27.3 V can be set independently. A noninverting sum m er is not the

77

Inverting and Noninverting A m plifiers

3-12.2 Inverting Amplifier—A C Input R efer to Fig. 3-3 and create the PSpice model o f the circuit. Set the input voltage to a sine w ave with a peak value o f 5 V and a frequency o f 500 Hz. O btain a plot o f £, and V0 versus tim e. To begin, place the follow ing parts in the w ork area. Draw = > Get New Part Part = = = = = =

> > > > > >

uA741 VSIN VDC R GLOBAL AGND

Num ber

Library

1 1 2 3 4 5

eval.slb source.slb source.slb analog.sib port.slb p ortslb

N ote that we are using a sine wave as the input signal instead o f a triangular wave as show n in Fig. 3-3. A rrange the parts as shown in Fig. 3-3. C hange the attributes o f the parts as given in Fig. 3-3. Set up the sine-w ave attributes by double-clicking on the sym ­ bol. In the pop-up w indow change VOFF, VAMPL, and FREQ.

VOFF = > 0 = > Save Attr VAMPL = > 5V — > Save Attr — > Change Display — > Both name and value FREQ = > 500Hz = > Save Attr = > Change Display = > Both name and value D ouble-click on the lead from the sine w ave generator to Ri and label it Ei. D ouble-click on the lead from the output term inal o f the op am p and label it Vo. See Fig. 3-22.

*/

- A /W -

20 kQ.

FIGURE 3-22

PSpice model for Fig. 3-3

In order to obtain a plot o f E, and V0 versus tim e, we m ust initialize the T ransient m enu

79

Inverting and Noninverting A m plifiers Draw =>

G e t

New Part Part = = = = = = =

> > > > > > >

uA741 VDC R G LOBAL AGND IPROBE V IEW PO INT

Num ber

Library

1 5 5 4 7 4 2

evai.slb source.slb analog.slb port.slb port.slb special.sib special.slb

A rrange and w ire the parts as show in Fig. 3-4. Place the IP R O B E s to m easure /,, / 2, / 3, and If, and place the V IE W P O IN T S to m easure the voltage at the sum m ing point, S , and V0. C hange the parts attributes to correspond to Fig. 3-4. Save the circuit in a file and run the sim ulation A n aly sis = > S im u la te . T he results are show n in Fig. 3-24.

2.000E-03

Jo FIGURE 3-24

PSpice model of Fig. 3-4— the inverting adder.

3-12A Noninverting Amplifier C reate a P S pice m odel o f the noninverting am p lifier show n in Fig. 3-7(a) w ith Rf = 20 k fl, /?, = 10 k f l, and £,■ = 2 V. U se IP R O B E to m easure l 0, If , ILy and / ( - ) . Use V IE W P O I N T to m easure VL and the v o ltag e at the ( - ) input. If you are using the

80

Chapter 3

evaluation softw are package, build the circuit with a 741 op am p instead o f the O P - 177 as show n in Fig. 3-7(a). Place the follow ing parts in the w ork area. Draw => Get New Part Part

=> => => => => => =>

uA741 VDC R GLOBAL AGND IPROBE VIEWPOINT

N um ber 1 3 3 4 5 4 2

Library

eval.slb source.slb analog.slb

port.slb portslb special.sib speciahslb

A rrange the parts and include the IP R O B E s and V IE W P O IN T S . Save the file and run the sim ulation A nalysis = > S im u la te. The results are show n in Fig. 3-25.

FIGURE 3-25

PRO BLEM S

PSpice model for the noninverting amplifier of Fig. 3-7(a).

___________________________________________________________________________

3-1. What type of feedback is applied to an op amp when an external component is connected between the output terminal and the inverting input? 3-2. If the open-loop gain is very large, does the closed-loop gain depend on the external com­ ponents or the op amp? 3-3. What two assumptions have been used to analyze the circuits in this chapter?

Modulating, Demodulating, and Frequency Changing with the Multiplier

343

portion to the audio signal, the process is called am plitude m odulation (A M ). C hanging the frequency or the phase angle of the carrier wave results in freq u en cy m odulation (FM ) and p hase-angle m odulation (PM ), respectively. O f course, the original audio signal m u st eventually be recovered by a pro cess called dem odulation or detection. T he rem ainder o f this section concentrates on using the m ultiplier for am plitude m odulation. (“M o d u late” is from the ancient G reek language m eaning to “change.” Curiously, it is the Latin prefix “d e” that converts the m eaning to “change back.” )

12-7.2 Defining Amplitude Modulation T he introduction to am plitude m odulation begins w ith the am plifier in Fig. 12-8(a). T he input voltage E c is am plified by a constant gain A. A m plifier output VQ is the product gain o f A and E c. N ow suppose that the am p lifier’s gain is varied. T his co n cep t is rep ­ resented by an arrow through A in Fig. 12-8(b). A ssum e that A is varied from 0 to a m ax ­ im um and back to 0 as show n in Fig. 12-8(b) by the plot o f A versus t. T his m eans that the am plifier m ultiplies the input voltage E c by a different value (gain) over a period o f tim e. VD is now the am plitude o f input E c varied or m ultiplied by an am plitude o f A. This process is an exam ple o f am plitude m odulation, and the output voltage VQ is called the am plitude m odulated signal. T herefore, to obtain an am plitude-m odulated signal (V0), the am plitude o f a high-frequency carrier signal (Ec) is varied by an intelligence or data signal A .

12-7.3 The Multiplier Used as a Modulator F rom S ection 12-7.2 and Fig. 12-8(b), V0 equals E c m ultiplied by A. T h erefo re, am p li­ tude m odulation is a m ultiplication p ro c ess. As show n in Fig. 12-8(c), E c is applied to a m u ltip lie r’s x input. E m [having the sam e shape as A in Fig. 12-8(b)] is applied to the m u ltip lier’s y input. E c is m ultiplied by a voltage th at varies from 0 th rough a m ax i­ m um and back to 0. So VQ has the sam e envelope as Em. T he m u ltip lier can be co n sid ­ ered a voltage-controlled gain device as w ell as an am plitude m odulator. T he w aveshape show n is th a t o f a b a la n c e d m odulator. T h e reaso n fo r th is nam e w ill be given in S ection 12-8.3. N ote carefully in Fig. 12-8(c) that VQ is not a sine wave; that is, the peak values o f successive half-cycles are different. This principle is used in Section 12-12 to show how a frequen cy-sh ifter (heterodyne) circuit w orks, but first, w e exam ine am plitude m odula­ tion in greater detail.

72-7.4 Mathematics of a Balanced Modulator A high-frequency sinusoidal carrier wave Ec is applied to one input o f a m ultiplier. A low er-frequency audio or data signal is applied to the second input o f a m odulator and w ill be called the m odulating w ave, Em. For test and analysis, both E c and Em w ill be sine w aves described as follows.

345

Modulating, Dem odulating, and Frequency Changing with the M ultiplier M odulating wave, Em: = Emp sin l u f j

( l2-7b)

w here E mp is the peak value of the m odulating wave and f m is the m odulating frequency. Now let the carrier voltage Ec be applied to the x input o f a m ultiplier as Ex, and let the m odulating voltage E m be applied to the y input o f a m ultiplier as Ey. The m ulti­ p lier’s output voltage VD is expressed as a product term from Eq. (12-16) as

=

Emp\ Q cp (sin 27* > ,)(sin 2jrf[l)

( 12' 8)

Equation (12-8) is called the product term , because it represents the product of two sine waves w ith different frequencies. However, it is not in the form used by ham radio o per­ ators or com m unications personnel. They prefer the form obtained by applying to Eq. (12­ 8) the trigonom etric identity (sin y4)(sin B) = i[co s (A - B) - cos (A + 5)]

( 12-9)

Substituting Eq. (12-9) into Eq. (12-8), w here A - E c and B - Ern, we have COS 2 M / C -

va =

fjt ~

cos 2 -nifc + f m)t

(12-10)

Equation (12-10) is analyzed in Section 12-7.5.

12-7.5 Sum and Difference Frequencies R ecall from Section 12-7.3 that E cis a sine w ave and E,„ is a sine w ave, but no part o f V0 is a sine wave. V0in Fig. 12-8(c) is expressed m athem atically by either Eq. (12-8) or (12-10). B ut Eq. (12-10) show s that VD is m ade up o f tw o cosine w aves w ith fre­ q uencies d ifferent from either E m or E c. T hey are the sum fre q u e n c y f c -I- f m and the d iffe re n c e fr e q u e n c y f c —f m. T he sum and d iffe re n c e fre q u e n c ie s are ev a lu a ted in E xam ple 12-9.

Example 12-9 In Fig. 12-9, carrier signal Ec has a peak voltage o f Ecp = 5 V and a frequency o f f c = 10,000 Hz. The m odulating signal Em has a peak voltage o f Em = 5 V and a frequency of f m = 1000 Hz. C alculate the peak voltage and frequency o f (a) the sum frequency; (b) the difference frequency. S o lu tio n

From Eq. (12-10), the peak value o f both sum and difference voltages is Emp ECp _ 5 V X 5 V — | 25 V

20

_

20

~

*

Modulating, Dem odulating, and Frequency Changing with the M ultiplier

347

T he sum frequency is f c + f m = 10,000 Hz + 1000 Hz = 11,000 Hz; the difference fre­ quency is f c — f m = 10,000 Hz - 1000 Hz = 9000 Hz. T hus V0 is m ade up o f the difference o f two cosine waves: V0 = 1.25 cos 2tt9 0 0 0 / - 1.25 cos 2tt1 1,0001 This result can be verified by connecting a wave or spectrum analyzer to the m ulti­ p lier’s output; a 1.25-V deflection occurs at 11,000 Hz and at 9000 Hz. T he original input signals o f 1 kH z and 10 kH z do not exist at the output.

An oscilloscope can be used to show input and output voltages o f the m ultiplier o f E xam ple 12-9. The product term for VQ is found from Eq. (12-8): V0 = 2.5 V (sin 2 7 rl0 ,0 0 0 0 (sin 2 tt1 0 0 0 0 2.5 X

Ec

X

Em

VQ is show n w ith Em in the top draw ing and with E c in the bottom draw ing o f Fig. 12-9. O bserve that Em and E c have peak voltages o f 5 V. The peak value o f VD is 2.5 V. N ote that the upper and low er envelopes o f V0 are not the sam e shape as Em. T herefore, w e cannot rectify and filter Va to recover Em. T his characteristic distinguishes the balance moduJator.

12-7.6 Side Frequencies and Sidebands A nother way o f displaying the output o f a m odulator is by a graph show ing the peak am ­ plitude as a vertical line for each frequency. T he resulting fre q u en c y spectrum is show n in Fig. 12-10(a). The sum and difference frequencies in V0 are called upper and low er side frequencies because they are above and below the carrier frequency on the graph. W hen m ore than one m odulating signal is applied to the m odulator ( y input) in Fig. 12­ 9, each generates a sum and difference frequency in the output. Thus, there will be tw o side frequencies for each y input frequency, placed sym m etrically on either side o f the carrier. If the expected range of m odulating frequencies is known, the resulting range o f side frequencies can be predicted. For exam ple, if the m odulating frequencies range b e­ tw een 1 and 4 kHz, the low er side frequencies fall in a band betw een (10 — 4) kH z = 6 kH z and ( 1 0 — 1) kH z = 9 kHz. The band betw een 6 and 9 kH z is called the low er sid e b a n d . F or this sam e exam ple, the u p p er sid e b a n d ran g es from ( 1 0 + 1 ) k Hz = 11 kH z to (10 + 4) kH z = 14 kHz. Both upper and low er sidebands are show n in Fig 1 2-10(b).

M odulating, Dem odulating, and Frequency Changing with the M ultiplier

349

standard am plitude m odulator (A M ) adds the carrier term to the output. The A M car ra­ dio uses standard AM . O ne way of adding the carrier term to generate a standard AM out­ put is show n in Fig. 12-11(a). The m odulating signal is fed into one input o f an adder. A dc voltage equal to the peak value o f the carrier voltage E cp is fed into the other input. The output of the adder is then fed into the y input o f a m ultiplier, as show n in Fig. 1211(b). T he carrier signal is fed into the x input. T he m ultiplier m ultiplies Ex by Ey, and its output voltage is the standard AM voltage given by either o f the follow ing equations: ' E2

sin 27jfct

10

(carrier term )

( 12- 1

+

E E —^ — — (sin 277/c/)(sin 2

ttf mt)

(product term )

10

or E 2cp 10

■

sin 2.'nfct

(carrier term ) +

V0 = ^ £