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Contents

OperationalAmplifiers andLinear IntegratedCircuits

Table of Contents

Index

Copyright

Dedication

FOREWORD

PREFACE

ACKNOWLEDGEMENTS

Ch. 1. Operational Amplifiers—Basics

1.1. INTRODUCTION

1.2. OPERATIONALAMPLIFIERS

1.3. CLASSIFICATION OFOPERATIONAL AMPLIFIERS

1.4. SYMBOL

1.5. OPERATIONALAMPLIFIER PARAMETERS

1.6. FREQUENCY ROLL OFF

1.7. OPERATIONAL-AMPLIFIER IN OPEN LOOPCONFIGURATION

1.8. OPERATIONAL-AMPLIFIER GOING TOSATURATION

1.9. VIRTUAL GROUND

1.10. OP-AMP—A DIRECTCOUPLED HIGH GAINAMPLIFIER

1.11. INVERTINGAMPLIFIER

1.12. BASIC LINEARCIRCUITS USINGOPERATIONAL AMPLIFIERS

1.13. IDEAL OPERATIONALAMPLIFIER AS ASUBTRACTOR

1.14. IDEAL OPERATIONALAMPLIFIER AS ANINTEGRATOR

1.15. IDEAL OPERATIONALAMPLIFIER AS ADIFFERENTIATOR

1.16. OPERATIONALAMPLIFIER DESIGNTECHNIQUES

1.17. MEASUREMENT OFOPERATIONAL AMPLIFIERPARAMETERS

1.18. MEASUREMENT OFPOWER SUPPLY REJECTIONRATIO (PSRR)

1.19. MEASUREMENT OFSLEW RATE

1.20. MEASUREMENT OFOPEN LOOP GAIN

1.21. FREQUENCYRESPONSE

1.22. Summary

1.23. SOLVED EXAMPLES

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Next(Key:n)1.23. SOLVED EXAMPLES

Example 1.1: Example 1.1

For an op-amp integrator with R = 100 MΩ and C = 1 μF, an input of 2 sin1000 t is applied. Determine the value of v0.

Figure 1.42. Circuit for Ex. 1.1

1.23. SOLVED EXAMPLES

1.24. ESSAY-TYPEQUESTIONS

1.25. OBJECTIVE-TYPEQUESTIONS

1.26. PROBLEMS

1.27. SELF-ASSESSMENTQUESTIONS

1.28. ASSIGNMENT

1.29. UNSOLVEDPROBLEMS

Ch. 2. Op-amp Applications

Ch. 3. Active Filters andOscillators

Ch. 4. Timer and PhaseLocked Loop ICs

Ch. 5. Digital to AnalogConverters (DACs) andAnalog to Digital Converters(ADCs)

Ch. 6. Voltage Regulators

Appx. A. Appendix A

Appx. B. Appendix B

Appx. C. Appendix C

Appx. D. Appendix D

Bibliography

Index

Example 1.2: Example 1.2

For an op-amp differentiator with R = 1 kΩ and C = 0.01 μF, a square waveinput of 200 Hz is applied. Sketch the output waveform giving reasons.

Figure 1.43. Waveforms for Ex. 1.2 Differentiator Circuit, the +ve and−ve slopes of Square Wave are Given as Output Waveform

The reason is T = RC << T. So near ideal differentiation is done. The slope ofthe square wave (output of differentiation) is ∞ for vertical line and zero forhorizontal zone.

Differentiation means the slope of the line. For a perfect vertical line, the slopeis ∞. For a perfect horizontal line, the slope is O. Therefore, if a square waveinput is given to ideal differentitor circuit, the output waveform is as shown inFig. for V0 Vst, a series of spikes.

Example 1.3: Example 1.3

(a) Give the specifications and typical values of op-amp 741.

Minimum Typical Maximum

Input bias current 200 nA

Input offset current 30 nA

Input offset voltage 1 mV

Input requistance 1 MV

Open loop voltage gain 100,000

Common-mode rejectionratio

100 db

Slew rate

0.7 V/μsec.

Unity gain bandwidth 1 MHz

Occupy voltage rejectionratio

90 db

PSRR 15 μV/V 15 μ/V

(b) Give the pin configuration of 741 IC How do you identify thepins of the IC?

Figure 1.44. Pin Configuration of IC741

Example 1.4: Example 1.4

Explain different IC packages.

The IC packages are

1. Ceramic flint packages

2. TO-5 package (with straight leads)

3. TO-5 package (with dual-on-line format)

4. DIP (dual-in-line) package

The tab in the IC package indicates pin 8. The other pins, starting from 1, arecounted in an anticlockwise direction for the TO-5 package. The dot on thenotch side indicates pin 1 in a DIP package. The first pin on the ridge side ispin 1 in some packages.

The package types are indicated by the code

I: Mini Dip

P: Plastic Dip

F: Flat Pack

Example 1.5: Example 1.5

(a) A triangular wave form of (5x + 3) with a peak to peak valueof 2 V is applied to the inverting terminal of an op-amp in openloop configuration. Sketch the input and output wave format tothe same time scale.

Figure 1.45. Waveforms for Ex. 1.6

(b) An output v0 = vi2 is to be realised using op-amps where v1 is

an electrical signal. Give the schematic.

Figure 1.46. Realisation of Expression V0 = Vi2

Example 1.6: Example 1.6

An op-amp has R1 and L1 in series on the input side, connected to the

inverting terminal and R2, C2 connected in parallel in the feedback path. The

non-inverting terminal is grounded. Derive the transfer function.

Figure 1.47. Circuit for Ex. 1.6

For the circuit shown in Fig. 1.48, obtain the expression for e0.

Example 1.7: Example 1.7

For the Fig. 1.48, obtain the expression for e0

The input impedance of an op-amp is very high. Therefore, the currents intothe op-amp are negligible.

Hence I1 = I2

and V1 = V2

Applying Kirchhoff's law of currents at points 1 and 2

Figure 1.48. Circuit for Ex. 1.7

Example 1.8: Example 1.8

For the circuit shown in Fig. 1.49, show that current through R2 is

independent of R2 and is equal to ,

Applying Kirchhoff's law of currents at point 1

The current through the resistor R2

The negative sign indicates that the direction of current is opposite to whatwas assumed. current through resistor

Figure 1.49. Circuit for Ex. 1.8

Example 1.9: Example 1.9

For the circuit shown in Fig. 1.50, obtain the expression for current throughR4. Find the current through R4. Assume R2 >> R3.

Figure 1.50. Circuit for Ex. 1.9

We know that I1 = I2 Since op-amp does not draw any current.

e0 is the drop across R2. Non-inverting terminal is at virtual ground point.

Assuming R2 >> R2

The current through

Example 1.10: Example 1.10

Show that V0 = 2(V1 − V2).

Figure 1.51. Circuit for Ex. 1.10

The circuit can be drawn as given below.

Figure 1.52. Redrawn Circuit for Ex. 1.10

Example 1.11: Example 1.11

Write a short note on frequency compensation techniques used in op-amps.

The essential idea of compensation is to reshape the magnitude and phaseplots of βA so that |βA| < 1 when the angle of βA is 180°. There are threegeneral methods of accomplishing this goal.

(i) Dominant pole or lag compensation

This method exerts an extra pole into the transfer function at alater frequency than the existing poles. Such a circuit introduced aphone log into the amp. The loop gain drops 0 dB with the slope of6 dB/octave at a frequency where the poles of Av contribute

negligible phase shift. The only disadvantage is it wastes theworkable bandwidth.

(ii) Lead compensation

The amplifier or the feedback network is modified so as to add azero to the transfer function, thereby increasing the phase.

Figure 1.53. Frequency Compensation Circuit

Figure 1.54. Frequency Compensation Circuit

(iii) Pole-zero or lead-lead compensation

This technique adds both a pole (a lag) and zero (lead) to thetransfer gain. The zero is to cancel the lowest pole. The transferfunction of the phase network is found to be

f1 is lower cuf-off frequency and f2 is upper cut-off frequency

Compensation of three techniques.

Figure 1.55. Frequency Response

Example 1.12: Example 1.12

Explain the advantages and disadvantages of ICs. List typical specifications ofan IC op-amp.

All the components in each integrated circuit are fabricated on the same chip.IC's have become a vital part of modern electronics circuit design. They areused in the computer industry, automobile industry, home appliances,

communications, and central systems where they permit minimisation andsuperior performance not possible with discrete components. They providelong, trouble-free service and are economical. Digital ICs are used to formcircuits as gates, counters, multipliers, shift registers, and so on. Linear ICs areequivalent to discrete transistor networks. They are used for amplifier filters,modulations, integrators, timers, and other special purposes.

ICs are (a) small in size, (b) are low in cost, (c) have a low offset voltage, (d)low offset current (e) high reliability, and (f) good temperature tracking.

Disadvantages

(a) Fabrication of Inductor with large value of quality factor Q inICs has not been sucessfull. Building of inductor with reasonablevalue and by IC has not been successful.

(b) Integrated resistor and capacitor have limited moderate valuesand are available with wide tolerance.

(c) Circuit adjustments are difficult.

Typical specifications of an IC amplifier are as follows.

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