op-amps and linear integrated circuits_r. a. gayakwad_2

8/20/2019 Op-Amps and Linear Integrated Circuits_R. a. Gayakwad_2

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Scilab Textbook Companion for

Op-Amps And Linear Integrated Circuits

by R. A. Gayakwad1

Created byAmar Ds

BEInstrumentation Engineering

Sri Jayachamarajendra College Of EngineeringCollege Teacher

Rathnnakara SCross-Checked by

TechPassion

August 10, 2013

1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes written in it can be downloaded from the ”Textbook Companion Project”section at the website http://scilab.in


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Book Description

Title: Op-Amps And Linear Integrated Circuits

Author: R. A. Gayakwad

Publisher: PHI Learning Pvt. Ltd., New Delhi

Edition: 4

Year: 2004

ISBN: 978-81-203-2058-1

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Scilab numbering policy used in this document and the relation to theabove book.

Exa Example (Solved example)

Eqn Equation (Particular equation of the above book)

AP Appendix to Example(Scilab Code that is an Appednix to a particularExample of the above book)

For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 meansa scilab code whose theory is explained in Section 2.3 of the book.

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Contents

List of Scilab Codes 4

1 Introduction to Operational Amplifier 7

2 Interpretation of Data Sheets and Characteristics of an Opamp 11

3 An Opamp with Negative Feedback 14

4 The Practical Opamp 21

5 Frequency Response of an Opamp 37

6 General Linear Applications 40

7 Active Filters and Oscillators 53

8 Comparators and Converters 74

9 Specialized IC Applications 80

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List of Scilab Codes

Exa 1.1.a Collector current and dc voltage . . . . . . . . . . . . 7Exa 1.1.b Voltage gain of the opamp . . . . . . . . . . . . . . . . 9Exa 1.1.c Input resistance of the opamp . . . . . . . . . . . . . . 9Exa 2.1.a Output voltage for Openloop differential amplifier . . 11Exa 2.1.b Output voltage for openloop differential amplifier . . . 12Exa 2.2.a Output voltage for inverting amplifier . . . . . . . . . 12Exa 2.2.b Output voltage for inverting amplifier . . . . . . . . . 13Exa 3.1 Parameters Of voltageseries feedback amplifier . . . . 14Exa 3.2 Parameters Of voltageseries feedback amplifier . . . . 15Exa 3.3 Parameters of Voltageshunt feedback amplifier . . . . 16Exa 3.4 Output voltage of voltage shunt feedback amplifier . . 17Exa 3.5.a Gain input resistance of the amplifier . . . . . . . . . 18Exa 3.5.b Output voltage of an Opamp . . . . . . . . . . . . . . 18

Exa 3.6.a Voltage gain and input resistance of the Opamp . . . . 19Exa 3.6.b Output voltage of the Opamp . . . . . . . . . . . . . . 20Exa 4.1 Design of Compensating Network . . . . . . . . . . . . 21Exa 4.2 Max Output offset voltage . . . . . . . . . . . . . . . . 22Exa 4.3 Design of input offset voltage compensating network . 22Exa 4.4.a Max Output offset voltage . . . . . . . . . . . . . . . . 23Exa 4.4.b Effect of input bias current . . . . . . . . . . . . . . . 24Exa 4.5.a Max Output offset voltage . . . . . . . . . . . . . . . . 24Exa 4.5.b Effect of input bias current . . . . . . . . . . . . . . . 25Exa 4.6 Max Output offset voltage . . . . . . . . . . . . . . . . 25Exa 4.7 Total Output offset voltage . . . . . . . . . . . . . . . 26

Exa 4.8.a Error voltage and output voltage . . . . . . . . . . . . 27Exa 4.8.b Error voltage and output voltage . . . . . . . . . . . . 28Exa 4.9.a Error voltage and output voltage . . . . . . . . . . . . 29Exa 4.9.b Output waveform . . . . . . . . . . . . . . . . . . . . 30

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Exa 4.10.a Error voltage and output voltage . . . . . . . . . . . . 30

Exa 4.10.b Error voltage and output voltage . . . . . . . . . . . . 31Exa 4.11.a Output offset voltage . . . . . . . . . . . . . . . . . . 32Exa 4.11.b Output offset voltage . . . . . . . . . . . . . . . . . . 33Exa 4.12 Output ripple voltage . . . . . . . . . . . . . . . . . . 34Exa 4.13 Change in output offset voltage . . . . . . . . . . . . . 34Exa 4.14.a Output voltage . . . . . . . . . . . . . . . . . . . . . . 35Exa 4.14.b Output common mode voltage . . . . . . . . . . . . . 35Exa 5.1 Maximum gain . . . . . . . . . . . . . . . . . . . . . . 37Exa 5.2 Gain equation and break frequencies . . . . . . . . . . 38Exa 5.3 Stability of voltage follower . . . . . . . . . . . . . . . 38Exa 6.1 Bandwidth of the amplifier . . . . . . . . . . . . . . . 40

Exa 6.2.a Bandwidth of the amplifier . . . . . . . . . . . . . . . 41Exa 6.2.b Max output voltage swing . . . . . . . . . . . . . . . . 42Exa 6.3 Components of peak amplifier . . . . . . . . . . . . . . 42Exa 6.4 Output voltage . . . . . . . . . . . . . . . . . . . . . . 43Exa 6.5 Output voltage . . . . . . . . . . . . . . . . . . . . . . 44Exa 6.6 Output voltage . . . . . . . . . . . . . . . . . . . . . . 44Exa 6.7 Output voltage . . . . . . . . . . . . . . . . . . . . . . 45Exa 6.8 Change in resistance in straingage . . . . . . . . . . . 46Exa 6.9 Gain of the amplifier . . . . . . . . . . . . . . . . . . . 46Exa 6.10 Range of input voltage . . . . . . . . . . . . . . . . . . 47

Exa 6.11 Current and voltage drop . . . . . . . . . . . . . . . . 47Exa 6.12.a Load current . . . . . . . . . . . . . . . . . . . . . . . 48Exa 6.12.b Output voltage . . . . . . . . . . . . . . . . . . . . . . 48Exa 6.13 Range of output voltage . . . . . . . . . . . . . . . . . 49Exa 6.14 Change in output voltage . . . . . . . . . . . . . . . . 49Exa 6.15 Output voltage of an integrator . . . . . . . . . . . . . 50Exa 6.16.a Design of differentiator . . . . . . . . . . . . . . . . . 51Exa 6.16.b Output waveform of differentiator . . . . . . . . . . . 52Exa 7.1 Design of low pass filter . . . . . . . . . . . . . . . . . 53Exa 7.2 Design of low pass filter . . . . . . . . . . . . . . . . . 54Exa 7.3 Frequency response of low pass filter . . . . . . . . . . 54

Exa 7.4.a Design of second order low pass filter . . . . . . . . . . 56Exa 7.4.b Frequency response of second order highpass filter . . 57Exa 7.5.a Design of highpass filter . . . . . . . . . . . . . . . . . 58Exa 7.5.b Frequency response of highpass filter . . . . . . . . . . 59Exa 7.6.a Determination of low cutoff frequency . . . . . . . . . 61

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Exa 7.6.b Frequency response of second order highpass filter . . 61

Exa 7.7.a Design of wide bandpass filter . . . . . . . . . . . . . . 63Exa 7.7.b Frequency response of bandpass filter . . . . . . . . . 64Exa 7.7.c Calculation of quality factor . . . . . . . . . . . . . . . 65Exa 7.8.a Design of narrow bandpass filter . . . . . . . . . . . . 66Exa 7.8.b Design of narrow bandpass filter . . . . . . . . . . . . 67Exa 7.9 Design of wide bandreject filter . . . . . . . . . . . . . 67Exa 7.10 Design of notch filter . . . . . . . . . . . . . . . . . . . 68Exa 7.11 Phase angle . . . . . . . . . . . . . . . . . . . . . . . . 68Exa 7.12 Design of phase shift oscillator . . . . . . . . . . . . . 69Exa 7.13 Design of wein bridge oscillator . . . . . . . . . . . . . 70Exa 7.14 Design of quadrature oscillator . . . . . . . . . . . . . 70

Exa 7.15 Design of squarewave oscillator . . . . . . . . . . . . . 71Exa 7.16 Design of triangular wave generator . . . . . . . . . . 71Exa 7.17.a Nominal frequency . . . . . . . . . . . . . . . . . . . . 72Exa 7.17.b Modulation in output frequency . . . . . . . . . . . . 73Exa 8.1 Threshold voltage . . . . . . . . . . . . . . . . . . . . 74Exa 8.2 Output voltage swing . . . . . . . . . . . . . . . . . . 75Exa 8.3 Output frequencies . . . . . . . . . . . . . . . . . . . . 76Exa 8.4 Output voltage . . . . . . . . . . . . . . . . . . . . . . 78Exa 8.5 Output voltage . . . . . . . . . . . . . . . . . . . . . . 78Exa 9.1 Second order inverting butterworth lowpass filter . . . 80

Exa 9.2 Second order inverting butterworth bandpass filter . . 81Exa 9.3 Design of notch filter . . . . . . . . . . . . . . . . . . . 82Exa 9.4 Second order butterworth lowpass filter . . . . . . . . 83Exa 9.5 Value of capacitor . . . . . . . . . . . . . . . . . . . . 83Exa 9.6 Value of resistor . . . . . . . . . . . . . . . . . . . . . 84Exa 9.7 Value of tc td and f0 . . . . . . . . . . . . . . . . . . . 84Exa 9.8 Freq of free running ramp generator . . . . . . . . . . 85Exa 9.9 Value of fout fl fc . . . . . . . . . . . . . . . . . . . . . 86Exa 9.10 Design of current source . . . . . . . . . . . . . . . . . 86Exa 9.11 Design of voltage regulator . . . . . . . . . . . . . . . 87Exa 9.12 Design of stepdown switching regulator . . . . . . . . 88

Exa 9.13 Design of stepdown switching regulator . . . . . . . . 90

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Chapter 1

Introduction to Operational

Amplifier

Scilab code Exa 1.1.a Collector current and dc voltage

1 / / C h a p t e r 12 // Page . No−4 ,3 / / E x a mp l e 1 1 a , F i g u r e . No−1.24 / / C o l l e c t o r c u r r e nt and dc v o l t a g e5 // Given :6 clear ; clc ;

7 V c c = 6 ; V b e 5 = 0 . 7 ; V e e = 6 ; V b e 3 = 6 . 7 ; V b e 6 = 0 . 7 ; V b e 7 = 0 . 7 ; //V ol ta ge i n v o l t s

8 R c 1 = 6 . 7 * 1 0 ^ 3 ; // R e s i s t a n ce i n ohms9 I c 1 = rand () ;

10 V c 1 = V c c - R c 1 * I c 1 ;

11 V e 4 = V c 1 - V b e 5 ;

12 I 4 = ( V e 4 + V e e ) / ( 9 . 1 * 1 0 ^ 3 + 5 . 5 * 1 0 ^ 3 ) ;

13 V b 3 = 5 . 5 * 1 0 ^ 3 * I 4 - V e e ;

14 V e 3 = V b 3 - V b e 3 ;15 I e 3 = ( V e 3 + V b e 3 ) / 3 . 3 * 1 0 ^ 3 ;

16 I c 1 = 1 . 0 8 * 1 0 ^ - 3 / 2 . 7 6 5 ; // S i n ce I e 3 =2∗ I c 1 ,s u b s t i t u t i n g i n above e qu a ti o n a nd s i m p l i f y i n g

17 printf ( ” \n C o l l e c t o r c u rr e n t I c 1 i s = %. 5 f A \n” ,Ic1

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) // R e su l t

18 V c 1 = V c c - R c 1 * I c 1 ;19 printf ( ” \n V o lt ag e Vc1 i s = %. 2 f V \n” , V c 1 ) //R e s u l t

20 V e 4 = V c 1 - V b e 5 ;

21 printf ( ” \n V o lt ag e Ve4 i s = %. 2 f V \n” , V e 4 ) //R e s u l t

22 I e 4 = ( V e 4 + V e e ) / ( 2 9 . 2 * 1 0 ^ 3 ) ;

23 printf ( ” \n C ur re nt I e 4 i s = %. 6 f A \n” , I e 4 ) //R e s u l t

24 I c 5 = I e 4 ;

25 printf ( ” \n C ur re nt I c 5 i s = %. 6 f A \n” , I c 5 ) //

R e s u l t26 Vc5=Vcc -3.8*10^3* Ic5;

27 printf ( ” \n V o lt ag e Vc5 i s = %. 2 f V \n” , V c 5 ) //R e s u l t

28 V e 6 = V c 5 - V b e 6 ;


30 I e 6 = ( V e 6 + V e e ) / ( 1 5 * 1 0 ^ 3 ) ;


32 V e 7 = V e 6 + V b e 7 ;


34 I 1 = ( V c c - V e 7 ) / 4 0 0 ;

35 printf ( ” \n C ur re nt I 1 i s = %. 6 f A \n” , I 1 ) // R e su l t36 I e 8 = I 1 ;


38 V e 8 = - V e e + 2 * 1 0 ^ 3 * I e 8 ;

39 printf ( ” \n V ol ta ge Ve8 a t t he o ut p ut t e rm i na l i s = %. 2 f V \n” , V e 8 ) // R e su l t

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Scilab code Exa 1.1.b Voltage gain of the opamp

1 / / C h a p t e r 12 // Page . No−4 ,3 / / E x a mp l e 1 1 b , F i g u r e . No−1.24 / / V o l t a g e g a i n o f t h e opamp5 // Given :6 clear ; clc ;

7 I e 1 = 0 . 3 9 * 1 0 ^ - 3 ; I e 4 = 0 . 2 9 8 * 1 0 ^ - 3 ; I e 6 = 0 . 6 7 8 * 1 0 ^ - 3 ; //C u rr e nt i n amps

8 R c 1 = 6 . 7 * 1 0 ^ 3 ; R c 5 = 3 . 8 * 1 0 ^ 3 ; // R e s i s t a n ce i n ohms9 b e t a _ a c = 1 5 0 ;

10 r e 1 = ( 2 5 * 1 0 ^ - 3 ) / I e 1 ;11 r e 2 = r e 1 ;

12 r e 4 = ( 2 5 * 1 0 ^ - 3 ) / I e 4 ;

13 r e 5 = r e 4 ;

14 r e 6 = ( 2 5 * 1 0 ^ - 3 ) / I e 6 ;

15 k = ( R c 1 * 2 * b e t a _ a c * r e 4 ) / ( R c 1 + 2 * b e t a _ a c * r e 4 ) ;

16 A d 1 = k / r e 1 ;

17 printf ( ” \n V ol ta ge g ai n o f t he dual −i n p u t , b a l a n c e dout put − d i f f e r e n t i a l a m p l i f i e r i s = % . 2 f \n” , A d 1 )

// R e su l t18 k 1 = ( R c 5 * b e t a _ a c * ( r e 6 + 1 5 * 1 0 ^ 3 ) ) / ( R c 5 + b e t a _ a c * ( r e 6

+ 1 5 * 1 0 ^ 3 ) ) ;

19 A d 2 = k 1 / ( 2 * r e 5 ) ;

20 printf ( ” \n V ol ta ge g ai n o f t he dual −i n p u t , u n b a l a n c e dout put − d i f f e r e n t i a l a m p l i f i e r i s = % . 1 f \n” ,Ad2

) // R e su l t21 A d = 8 2 . 5 5 * 2 2 . 6 ; // Us ing Ad=Ad1∗Ad222 printf ( ” \n O v e r a l l g ai n o f t he o p−amp i s = %. 2 f \n”

, A d ) // R e su l t

Scilab code Exa 1.1.c Input resistance of the opamp

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1 / / C h a p t e r 1

2 // Page . No−4 ,3 / / E x a m pl e 1 1 c , F i g u r e . No−1.24 // I n pu t r e s i s t a n c e o f t he opamp5 // Given :6 clear ; clc ;

7 b e t a _ a c = 1 5 0 ;

8 r e 1 = 6 4 . 1 ; // R e s i s t a n ce i n ohms9 R i = 2 * b e t a _ a c * r e 1 ;

10 printf ( ” \n I np ut r e s i s t a n c e Ri i s = %. 1 f ohm \n” , R i )// R e su l t

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Chapter 2

Interpretation of Data Sheets

and Characteristics of an

Opamp

Scilab code Exa 2.1.a Output voltage for Openloop differential amplifier

1 / / C h a p t e r 22 // Page . No−4 5 , F i g u r e . No−2.93 / / E x a m p l e 2 1 a4 / / Output v o l t a g e f o r open−l o o p d i f f e r e n t i a l

a m p l i f i e r5 // Given :6 clear ; clc ;

7 v i n 1 = 5 * 1 0 ^ - 6 ; v i n 2 = - 7 * 1 0 ^ - 6 ; // Both i n p u t v o l t a g e sa re i n v o l t s

8 A = 2 0 0 0 0 0 ; // V o lt ag e g a in9 v o = A * ( v i n 1 - v i n 2 ) ; // Output v o l t a g e i n v o l t s

10 printf ( ” \n Output v o l t a g e i s vo = %. 1 f V d c \n” , v o )

// R e su l t

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Scilab code Exa 2.1.b Output voltage for openloop differential amplifier

1 / / C h a p t e r 22 // Page . No−4 5 , F i g u r e . No−2.93 //EXAMPLE 2 1 b4 / / Output v o l t a g e f o r open−l o o p d i f f e r e n t i a l

a m p l i f i e r5 // Given :6 clear ; clc ;

7 v i n 1 = 1 0 * 1 0 ^ - 3 ; v i n 2 = 2 0 * 1 0 ^ - 3 ; // Both i n pu t v o l t a g e sa re i n v o l t s

8 A = 2 0 0 0 0 0 ; // V o lt ag e g a in9 v o = A * ( v i n 1 - v i n 2 ) ; // Output v o l t ag e i n v o l t s

10 printf ( ” \n Output v o l t a g e i s vo = %. f V rms \n” , v o )// R e su l t

Scilab code Exa 2.2.a Output voltage for inverting amplifier

1 / / C h a p t e r 22 // Page . No−4 6 , F i g u r e . No−2.103 / / E x a m p l e 2 2 a4 // Output v o l t a g e f o r i n v e r t i n g a m p l i f i e r5 // Given :

6 clear ; clc ;7 v i n = 2 0 * 1 0 ^ - 3 ; // I np u t v o l t a g e i n v o l t s8 A = 2 0 0 0 0 0 ; // V o lt ag e g a in9 v o = - ( A * v i n ) ; // Output v o l t ag e i n v o l t s

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10 printf ( ” \n Output v o l t a g e i s vo = %. f V \n” , v o ) //

R e s u l t

Scilab code Exa 2.2.b Output voltage for inverting amplifier

1 / / C h a p t e r 22 // Page . No−4 6 , F i g u r e . No−2.10

3 / / E x a m p l e 2 2 b4 // Output v o l t a g e f o r i n v e r t i n g a m p l i f i e r5 // Given :6 clear ; clc ;

7 v i n = - 5 0 * 1 0 ^ - 6 ; // I np u t v o l t a g e i n v o l t s8 A = 2 0 0 0 0 0 ; // V o lt ag e g a in9 v o = - ( A * v i n ) ; // Output v o l t ag e i n v o l t s

10 printf ( ” \n Output v o l t a g e i s vo = %. f V \n” , v o ) //R e s u l t

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Chapter 3

An Opamp with Negative

Feedback

Scilab code Exa 3.1 Parameters Of voltageseries feedback amplifier

1 / / C h a pt e r32 // P age . No−7 5 , F i g u r e . No−3.23 / / E xa mp le 3 14 / / P ar am et er s o f v o lt a ge − s e r i e s f ee db ac k a m p l i f i e r5 / / G iv en6 clear ; clc ;

7 R 1 = 1 0 0 0 ; R f = 1 0 0 0 0 ;

8 A = 2 0 0 0 0 0 ; // Open−l oo p v o l t ag e g ai n9 R i = 2 * 1 0 ^ 6 ; // I np ut r e s i s t a n c e w it ho ut f ee db ac k

10 R o = 7 5 ; // Output r e s i s t a n c e w it ho ut f e ed b ac k11 f o = 5 ; // B reak f r e qu e n cy o f an Op−amp12 V s a t = 1 3 ; // S a t u r a ti o n v o l t ag e13 B = R 1 / ( R 1 + R f ) ; // Gain o f t he f ee db ac k c i r c u i t14 A f = A / ( 1 + A * B ) ; / / C l os e d−l oo p v o l t a g e g ai n

15 printf ( ” \n C l os ed −l oo p v o l t ag e g ai n i s Af = %. 2 f \n”, A f ) // R e su l t16 R i F = R i * ( 1 + A * B ) ; // I np ut r e s i s t a n c e w i th f ee db ac k17 printf ( ” \n I np u t r e s i s t a n c e w i t h f ee db ac k i s RiF = %

. 2 f ohms \n” , R i F ) // R e su l t

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18 R o F = R o / ( 1 + A * B ) ; // Output r e s i s t a n c e w it h f e ed b ac k

19 printf ( ” \n O utput r e s i s t a n c e w it h f e ed b ac k i s RoF =%f ohms \n” , R o F ) // R e su l t20 f F = f o * ( 1 + A * B ) ; / / B an dwi dt h w i th f e e d b a c k21 printf ( ” \n B an dwidth w it h f e e d ba c k i s vo = %. 1 f Hz \

n” , f F ) // R e su l t22 V o o T = V s a t / ( 1 + A * B ) ; // T ot a l o ut p u t o f f s e t v o l t a ge

w it h f e e d ba c k23 printf ( ” \n T o ta l o ut pu t o f f s e t v o l t a g e w i th f ee db ac k

i s VooT = % f V \n” , V o o T ) // R e su l t

Scilab code Exa 3.2 Parameters Of voltageseries feedback amplifier

1 / / C h a pt e r32 // P age . No−8 3 , F i g u r e . No−3.73 / / E xa mp le 3 24 / / P ar am et er s o f v o lt a ge − s e r i e s f ee db ac k a m p l i f i e r5 / / G iv en

6 clear ; clc ;7 R 1 = 1 0 0 0 ; R f = 1 0 0 0 0 ;

8 A = 2 0 0 0 0 0 ; // Open−l oo p v o l t ag e g ai n9 R i = 2 * 1 0 ^ 6 ; // I np ut r e s i s t a n c e w it ho ut f ee db ac k

10 R o = 7 5 ; // Output r e s i s t a n c e w it ho ut f e ed b ac k11 f o = 5 ; // B reak f r e qu e n cy o f an Op−amp12 V s a t = 1 3 ; // S a t u r a ti o n v o l t ag e13 B = 1 ; // Gain o f t h e f e ed ba ck c i r c u i t and i t i s e q ua l

t o 1 f o r v o l t a g e f o l l o w e r14 A f = A / ( 1 + A * B ) ; / / C l os e d−l oo p v o l t a g e g ai n

15 printf ( ” \n C l os ed −l oo p v o l t ag e g ai n i s Af = %. f \n” ,Af ) // R e su l t16 R i F = R i * ( 1 + A * B ) ; // I np ut r e s i s t a n c e w i th f ee db ac k17 printf ( ” \n I np u t r e s i s t a n c e w i t h f ee db ac k i s RiF = %

. 1 f ohms \n” , R i F ) // R e su l t

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18 R o F = R o / ( 1 + A * B ) ; // Output r e s i s t a n c e w it h f e ed b ac k

19 printf ( ” \n O utput r e s i s t a n c e w it h f e ed b ac k i s RoF =%f ohms \n” , R o F ) // R e su l t20 f F = f o * ( 1 + A * B ) ; / / B an dwi dt h w i th f e e d b a c k21 printf ( ” \n B an dwidth w it h f e e d ba c k i s vo = %. 1 f Hz \




Scilab code Exa 3.3 Parameters of Voltageshunt feedback amplifier

1 / / C h a pt e r32 // P age . No−8 6 , F i g u r e . No−3.83 / / E xa mp le 3 34 / / P ar am et er s o f v o lt a ge −s hu nt f ee db ac k a m p l i f i e r5 / / G iv en

6 clear ; clc ;7 R 1 = 4 7 0 ; R f = 4 . 7 * 1 0 ^ 3 ;

8 A = 2 0 0 0 0 0 ; // Open−l oo p v o l t ag e g ai n9 R i = 2 * 1 0 ^ 6 ; // I np ut r e s i s t a n c e w it ho ut f ee db a ck

10 R o = 7 5 ; // Output r e s i s t a n c e w it ho ut f e ed b ac k11 f o = 5 ; // B reak f r e qu e n cy o f an Op−amp12 V s a t = 1 3 ; // S a t u r a ti o n v o l t ag e13 K = R f / ( R 1 + R f ) ; // V ol ta ge a t t en u a ti o n f a c t o r14 B = R 1 / ( R 1 + R f ) ; // Gain o f t he f ee db ac k c i r c u i t15 A f = - ( A * K ) / ( 1 + A * B ) ; / / C l os e d−l oo p v o l t a g e g ai n

16 printf ( ” \n C l os ed −l oo p v o l t ag e g ai n i s Af = %. f \n” ,Af ) // R e su l t17 X = R f / ( 1 + A ) ;

18 R i F = R 1 + ( X * R i ) / ( X + R i ) ; // I np ut r e s i s t a n c e w i t hf e e d b a c k

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19 printf ( ” \n I np u t r e s i s t a n c e w i t h f ee db ac k i s RiF = %

. 1 f ohms \n” , R i F ) // R e su l t20 R o F = R o / ( 1 + A * B ) ; // Output r e s i s t a n c e w it h f e ed b ac k21 printf ( ” \n O utput r e s i s t a n c e w it h f e ed b ac k i s RoF =

%f ohms \n” , R o F ) // R e su l t22 f F = f o * ( 1 + A * B ) / K ; // B an dw id th w i th f e e d b a c k23 printf ( ” \n B an dwidth w it h f e e d ba c k i s vo = %. 2 f Hz \




Scilab code Exa 3.4 Output voltage of voltage shunt feedback amplifier

1 / / C h a pt e r32 // P age . No−8 6 , F i g u r e . No−3.83 / / E xa mp le 3 4

4 / / O utput v o l t a g e o f v o lt a g e −s hu nt f e e db a c ka m p l i f i e r

5 / / G iv en6 clear ; clc ;

7 R 1 = 4 7 0 ; R f = 4 . 7 * 1 0 ^ 3 ;

8 A = 2 0 0 0 0 0 ; // Open−l oo p v o l t ag e g ai n9 v i n = 1 ; // I np ut v o lt a g e i n v o l t s

10 K = R f / ( R 1 + R f ) ; // V ol ta ge a t t en u a ti o n f a c t o r11 B = R 1 / ( R 1 + R f ) ; // Gain o f t he f ee db ac k c i r c u i t12 A f = - ( A * K ) / ( 1 + A * B ) ; / / C l os e d−l oo p v o l t a g e g ai n

13 v o = A f * v i n ; / / Output v o l t a g e14 printf ( ” \n Output v o l t a g e i s vo = %. f V \n” , v o ) //R e s u l t

15 t = 0 : 0 . 1 : 2 * % p i ;

16 v o = - 1 0 * sin ( t ) ;

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17 plot ( t , v o ) ;

18 t i t l e ( ’ O ut pu t V o l t a g e ’ ) ;19 x l a b e l ( ’ t ’ ) ;20 y l a b e l ( ’ Vo ’ ) ;

Scilab code Exa 3.5.a Gain input resistance of the amplifier

1 / / C h a pt e r32 // P age . No−9 6 , F i g u r e . No−3. 143 // E xa mp le 3 5 a4 // G ain I np ut r e s i s t a n c e o f t h e a m p l i f i e r5 / / G iv en6 clear ; clc ;

7 R 1 = 1 0 0 0 ; R 2 = 1 0 0 0 ;

8 R f = 1 0 * 1 0 ^ 3 ; R 3 = 1 0 * 1 0 ^ 3 ;

9 A D = - R f / R 1 ; // V o lt a ge g a in10 printf ( ” \n V o lt ag e g a in i s AD = %. f \n” , A D ) //

R e s u l t

11 R i F x = R 1 ; // I n p u t r e s i s t a n c e o f i n v e r t i n g a m p l i f i e r12 printf ( ” \n I np u t r e s i s t a n c e o f i n v e r t i n g a m p l i f i e r

i s Ri Fx = % . f ohms \n” , R i F x ) // R e su l t13 R i F y = R 2 + R 3 ; // I np ut r e s i s t a n c e o f non−i n v e r t i n g

a m p l i f i e r14 printf ( ” \n I n p u t r e s i s t a n c e o f non− i n v e r t i n g

a m p l i f i e r i s RiFy = %. f ohms \n” , R i F y ) // R e su l t

Scilab code Exa 3.5.b Output voltage of an Opamp

1 / / C h a pt e r3

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2 // P age . No−9 6 , F i g u r e . No−3. 14

3 // E xa mp le 3 5 b4 / / Output v o l t a g e o f an Op−amp5 / / G iv en6 clear ; clc ;

7 v x = 2 . 7 ; v y = 3 ; // Both i np ut v o l t a g e s a re i n v o l t s8 R f = 1 0 * 1 0 ^ 3 ; R 1 = 1 0 0 0 ; // Both a r e i n ohms9 A D = - R f / R 1 ; // V o lt a ge g a in

10 v x y = v x - v y ;

11 v o = A D * v x y ; / / Output v o l t a g e12 printf ( ” \n Output v o l t a g e i s vo = %. f V \n” , v o ) //

R e s u l t

Scilab code Exa 3.6.a Voltage gain and input resistance of the Opamp

1 / / C h a pt e r32 // P age . No−9 9 , F i g u r e . No−3. 163 // E xa mp le 3 6 a

4 // V ol ta ge g ai n and i np ut r e s i s t a n c e o f Op−amp5 / / G iv en6 clear ; clc ;

7 R 1 = 6 8 0 ; R 3 = 6 8 0 ; // Both a r e i n ohms8 R F = 6 8 0 0 ; R 2 = 6 8 0 0 ; // Both a r e i n ohms9 R i = 2 * 1 0 ^ 6 ; // Open−l oo p i np ut r e s i s t a n c e o f t h e op−

amp10 v x = - 1 . 5 ; v y = - 2 ; // Both a re i n v o l t s11 A = 2 0 0 0 0 0 ; // Open−l o o p Ga in12 A D = 1 + R F / R 1 ; // V ol t ag e g a in

13 printf ( ” \n V o lt ag e g a in i s AD = %. f \n” , A D ) //R e s u l t14 B = R 2 / ( R 2 + R 3 ) ;

15 R i F y = R i * ( 1 + A * B ) ; / / I np ut r e s i s t a n c e o f f i r s t s t a gea m p l i f i e r

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16 printf ( ” \n I np u t r e s i s t a n c e o f f i r s t s t a g e a m p l i f i e r

i s RiFy = %. 1 f ohms \n” , R i F y ) // R e su l t17 B = R 1 / ( R 1 + R F ) ;18 R i F x = R i * ( 1 + A * B ) ; // I np ut r e s i s t a n c e o f s e co nd s t a g e

a m p l i f i e r19 printf ( ” \n I np u t r e s i s t a n c e o f s ec o nd s t a g e

a m p l i f i e r i s RiFx = %. 1 f ohms \n” , R i F x ) // R e su l t

Scilab code Exa 3.6.b Output voltage of the Opamp

1 / / C h a pt e r32 // P age . No−9 9 , F i g u r e . No−3. 163 // E xa mp le 3 6 b4 / / Output v o l t a g e o f t he Op−amp5 / / G iv en6 clear ; clc ;

7 R 1 = 6 8 0 ; R F = 6 8 0 0 // Both a r e i n ohms8 v x = - 1 . 5 ; v y = - 2 ; // Both i np ut v o l t a g e s a re i n v o l t s

9 A D = 1 + R F / R 1 ; // V ol t ag e g a in10 v x y = v x - v y ;

11 v o = A D * v x y ; / / Output v o l t a g e12 printf ( ” \n Output v o l t a g e i s vo = %. 1 f V \n” , v o ) //

R e s u l t

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Chapter 4

The Practical Opamp

Scilab code Exa 4.1 Design of Compensating Network

1 / / C h a pt e r42 // P age . No−1143 / / E xa mp le 4 14 / / D e s ig n o f C om pe ns at in g N et wo rk5 / / G iv en6 clear ; clc ;

7 V = 1 0 // S up pl y v o l t a g e8 V i o = 1 0 * 1 0 ^ - 3 ; // I np ut o f f s e t v o l t a ge9 R c = 1 0 ; / / A s s um p t io n

10 R b = ( V / V i o ) * R c ;

11 printf ( ” \n R e s i s t a n ce Rb i s = %. f ohms \n” , R b ) //R e s u l t

12 R a = R b / 2 . 5 ; / / S i n ce Rb>Rmax , l e t u s c h o o s e Rb=1 0∗Rmaxwh er e Rmax=Ra/4

13 printf ( ” \n R e s i s t a n ce Ra i s = %. f ohms \n” , R a ) //R e s u l t

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This code can be downloaded from the website wwww.scilab.in

Scilab code Exa 4.2 Max Output offset voltage

1 / / C h a pt e r42 // P age . No−1 21 , F i g u r e . No−4.133 / / E xa mp le 4 24 // Max O utput o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 * 1 0 ^ 3 ;

8 V i o = 1 0 * 1 0 ^ - 3 ; // I np ut o f f s e t v o l t a ge9 A o o = 1 + R f / R 1 ; / / To f i n d max v a l u e o f Voo , we r e d uc e

i np ut v o l t a ge v in t o z er o .10 V o o = A o o * V i o ; // Max o ut p ut o f f s e t v o l t a g e11 printf ( ” \n Max o ut p ut o f f s e t v o l t a g e i s = %. 3 f V dc

\n” , V o o ) // R e su l t

Scilab code Exa 4.3 Design of input offset voltage compensating network

1 / / C h a pt e r42 // P age . No−1 21 , F i g u r e . No−4.14

3 / / E xa mp le 4 34 // D e s ig n o f i np ut o f f s e t v ol ta ge −c o m p e n s a t i n g

ne t w or k5 / / G iv en6 clear ; clc ;

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7 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 * 1 0 ^ 3 ; R c = 1 0 ;

8 A f = 1 + R f / ( R 1 + R c ) ; // C lo se d l oo p g ai n o f non−i n v e r t i n g a m p l i f i e r9 printf ( ” \n C lo se d l oo p g ai n o f non− i n v e r t i n g

a m p l i f i e r i s = %. 1 f \n” , A f ) // R e su l t


Scilab code Exa 4.4.a Max Output offset voltage

1 / / C h a pt e r42 // P age . No−1 27 , F i g u r e . No−4.193 // E xa mp le 4 4 a4 // Max O utput o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 4 7 0 ; R f = 4 7 * 1 0 ^ 3 ;

8 V i o = 6 * 1 0 ^ - 3 ;9 I b = 5 0 0 * 1 0 ^ - 9 ;

10 V s = 1 5 ;

11 / / Max o ut p u t o f f s e t v o l t a ge due t o i np ut o f f s e tv o l ta g e , Vio i s :

12 V o o = ( 1 + R f / R 1 ) * V i o ; // Max o ut pu t o f f s e t v o l t ag e13 printf ( ” \n Max o ut p ut o f f s e t v o l t a g e i s = %. 3 f V dc

\n” , V o o ) // R e su l t14 / / Max o ut p u t o f f s e t v o l t a ge due t o i np ut o f f s e t

v o lt ag e , I b i s :15 V o I b = R f * I b ;

// Max o ut pu t o f f s e t v o l t ag e16 printf ( ” \n Max o ut p ut o f f s e t v o l t a g e due t o i np uto f f s e t c ur re nt , I b i s = %. 6 f V dc \n” , V o I b ) //R e s u l t

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Scilab code Exa 4.4.b Effect of input bias current

1 / / C h a pt e r42 // P age . No−1 27 , F i g u r e . No−4.193 // E xa mp le 4 4 b4 // E f f e c t o f i np ut b i as c u r r e n t5 / / G iv en6 clear ; clc ;

7 R 1 = 4 7 0 ; R f = 4 7 * 1 0 ^ 3 ;

8 R O M = R 1 * R f / ( R 1 + R f ) ; // P a r a l l e l c om bi na ti on o f R1 andRf

9 printf ( ” \n P a r a l l e l c o m b in a t i o n o f R1 and Rf , i . e ROMi s = %. 1 f ohm \n” , R O M ) // A p p ro x im a t el y t h e

v a l ue i s 47 ohm

Scilab code Exa 4.5.a Max Output offset voltage

1 / / C h a pt e r42 // P age . No−1 27 , F i g u r e . No−4.193 // E xa mp le 4 5 a4 // Max O utput o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ;8 V i o = 6 * 1 0 ^ - 3 ;

9 I b = 5 0 0 * 1 0 ^ - 9 ;

10 V s = 1 5 ;

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11 / / Max o ut p u t o f f s e t v o l t a ge due t o i np ut o f f s e t

v o l ta g e , Vio i s :12 V o o = ( 1 + R f / R 1 ) * V i o ; // Max o ut pu t o f f s e t v o l t ag e13 printf ( ” \n Max o ut p ut o f f s e t v o l t a g e due t o i np ut

o f f s e t v o l ta g e , Vio i s = %. 4 f V dc \n” , V o o ) //R e s u l t

14 / / Max o ut p u t o f f s e t v o l t a ge due t o i np ut o f f s e tv o lt ag e , I b i s :

15 V o I b = R f * I b ; // Max o ut pu t o f f s e t v o l t ag e16 printf ( ” \n Max o ut p ut o f f s e t v o l t a g e due t o i np ut

o f f s e t c ur re nt , I b i s = %. 4 f V dc \n” , V o I b ) //R e s u l t

Scilab code Exa 4.5.b Effect of input bias current

1 / / C h a pt e r42 // P age . No−1 27 , F i g u r e . No−4.193 // E xa mp le 4 4 b

4 // E f f e c t o f i np ut b i as c u r r e n t5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ;

8 R O M = R 1 * R f / ( R 1 + R f ) ; // P a r a l l e l c om bi na ti on o f R1 andRf

9 printf ( ” \n P a r a l l e l c o m b in a t i o n o f R1 and Rf , i . e ROMi s = % . f ohm \n” , R O M )

Scilab code Exa 4.6 Max Output offset voltage

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1 / / C h a pt e r4

2 // P age . No−1 30 , F i g u r e . No−4.213 / / E xa mp le 4 64 // Max O utput o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 I i o = 2 0 0 * 1 0 ^ - 9 ; // I np u t o f f s e t c u r r en t8 R f = 1 0 0 * 1 0 ^ 3 ;

9 V o I i o = R f * I i o ; // Max o ut pu t o f f s e t v o l t a g e10 printf ( ” \n Max o ut p ut o f f s e t v o l t a g e due t o i np ut

o f f s e t c ur re nt , I b i s = %. 4 f V dc \n” , V o I i o ) //R e s u l t

Scilab code Exa 4.7 Total Output offset voltage

1 / / C h a pt e r42 // P age . No−1 32 , F i g u r e . No −4 . 2 2 ( a )3 / / E xa mp le 4 7

4 / / T ot al Output o f f s e t v o l t ag e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 * 1 0 ^ 3 ;

8 V i o = 7 . 5 * 1 0 ^ - 3 ; // Max i n p ut o f f s e t v o l t ag e9 I i o = 5 0 * 1 0 ^ - 9 ; // Max i n p ut o f f s e t c u r r e nt

10 I b = 2 5 0 * 1 0 ^ - 9 ; // Max i n pu t b i a s c u r r e nt11 // For f i g u r e 4 . 2 2( a )12 V o o T = ( 1 + R f / R 1 ) * V i o + ( R f * I b ) ; // S i nc e t he c u r re n t

g en e r at ed o u t p ut o f f s e t v o l t a ge i s due t o i np ut

b i as c u r r e n t I b13 printf ( ” \n Max t o t a l o ut p u t o f f s e t v o l t a ge due t oi np ut o f f s e t c ur re nt , I b i s = %. 4 f V \n” , V o o T ) //

R e s u l t14

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15 // F or f i g u r e 4 . 2 2( b )

16 V o o T = ( 1 + R f / R 1 ) * V i o + ( R f * I i o ) ; // S i nc e t he c u r r e ntg en e r at ed o u t p ut o f f s e t v o l t a ge i s due t o i np uto f f s e t c ur r e nt I b

17 printf ( ” \n Max t o t a l o ut p u t o f f s e t v o l t a ge due t oi np ut o f f s e t c ur re nt , I b i s = %. 4 f V \n” , V o o T ) //

R e s u l t

Scilab code Exa 4.8.a Error voltage and output voltage

1 / / C h a pt e r42 // P age . No−1 36 , F i g u r e . No−4.243 // E xa mp le 4 8 a4 / / E rr or v o l t a g e and o ut pu t v o l t ag e5 / / G iv en6 clear ; clc ;

7 d e l t a _ V i o = ( 3 0 * 1 0 ^ - 6 ) ; // Change i n i np ut o f f s e tv o l t a g e

8 d e l t a _ T = 1 ; // U ni t c ha ng e i n t e mp e ra t u re9 d e l t a _ I i o = ( 3 0 0 * 1 0 ^ - 1 2 ) ; // Change i n i np ut o f f s e t

c u r r e n t10 V s = 1 5 ;

11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ; R l = 1 0 * 1 0 ^ 3 ;

12 V i n = 1 * 1 0 ^ - 3 ; // I np ut v o l t a g e13 k = 2 5 ; // A mp l i f i e r i s n u l l e d a t 25 deg14 T = 3 5 - k ; // Chang e i n t e m p e r a tu r e15 E v = ( 1+ R f / R 1 ) * ( d e l t a_ V i o / d e l ta _ T ) * T + R f * ( d e l ta _ I io /

d e l t a _ T ) * T ; // E rr or v o l t ag e

16 printf ( ” \n E rr or v o l t a ge i s = %. 4 f V \n” , E v ) //R e s u l t17 V o = - ( R f / R 1 ) * V i n + E v ; / / Output v o l t a g e18 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //

R e s u l t

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19 // (OR)

20 V o = - ( R f / R 1 ) * V i n - E v ; / / Output v o l t a g e21 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //R e s u l t

Scilab code Exa 4.8.b Error voltage and output voltage

1 / / C h a pt e r42 // P age . No−1 36 , F i g u r e . No−4.243 // E xa mp le 4 8 b4 / / E rr or v o l t a g e and o ut pu t v o l t ag e5 / / G iv en6 clear ; clc ;



c u r r e n t

10 V s = 1 5 ;11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ; R l = 1 0 * 1 0 ^ 3 ;

12 V i n = 1 0 * 1 0 ^ - 3 ; // I np ut v o l t ag e13 k = 2 5 ; // A mp l i f i e r i s n u l l e d a t 25 deg14 T = 3 5 - k ; // Chang e i n t e m p e r a tu r e15 E v = ( 1+ R f / R 1 ) * ( d e l t a_ V i o / d e l ta _ T ) * T + R f * ( d e l ta _ I io /

d e l t a _ T ) * T ; // E rr or v o l t ag e16 printf ( ” \n E rr or v o l t a ge i s = %. 4 f V \n” , E v ) //

R e s u l t17 V o = - ( R f / R 1 ) * V i n + E v ; / / Output v o l t a g e

18 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //R e s u l t19 // (OR)20 V o = - ( R f / R 1 ) * V i n - E v ; / / Output v o l t a g e21 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //

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R e s u l t


1 / / C h a pt e r42 // P age . No−1 36 , F i g u r e . No−4.243 // E xa mp le 4 9 a

4 / / E rr or v o l t a g e and o ut pu t v o l t ag e5 / / G iv en6 clear ; clc ;



c u r r e n t10 V s = 1 5 ;

11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ; R l = 1 0 * 1 0 ^ 3 ;

12 V i n = 1 0 * 1 0 ^ - 3 ; // I np ut v o l t ag e

13 k = 2 5 ; // A mp l i f i e r i s n u l l e d a t 25 deg14 T = 5 5 - k ; // Chang e i n t e m p e r a tu r e15 E v = ( 1+ R f / R 1 ) * ( d e l t a_ V i o / d e l ta _ T ) * T + R f * ( d e l ta _ I io /

d e l t a _ T ) * T ; // E rr or v o l t ag e16 printf ( ” \n E rr or v o l t a ge i s = %. 4 f V d c \n” , E v ) //

R e s u l t17 V o = - ( R f / R 1 ) * V i n + E v ; / / Output v o l t a g e18 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //

R e s u l t19 // (OR)

20 V o = - ( R f / R 1 ) * V i n - E v ; / / Output v o l t a g e21 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //R e s u l t

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Scilab code Exa 4.9.b Output waveform

1 / / C h a pt e r42 // P age . No−1 36 , F i g u r e . No−4.243 // E xa mp le 4 9 b4 / / O ut pu t w a ve fo r m5 / / G iv en6 clear ; clc ;

7 t = 0 : 0. 1 : 2* % pi ;

8 y = - 10 00 * sin ( t ) + 9 1 . 8 ;

9 a = gca () ;

10 a . x _ l a b el . t e x t = ’ Time ’ ;11 a . y _ l a b el . t e x t = ’ V ol t ag e ’ ;12 a . t it le . t e xt = ’ O utput w av ef or m ’ ;13 plot2d ( t , y ) ;

14 t 1 = 0 : 0 . 1 : 2 * % p i ;

15 y 1 = 9 1 . 8 * ( t 1 > = 0 ) ;

16 b = gca () ;17 b . l i n e _ s t y l e = 3 ;

18 plot2d ( t 1 , y 1 ) ;



3 // E xa mp le 4 1 0 a4 / / E rr or v o l t a g e and o ut pu t v o l t ag e5 / / G iv en6 clear ; clc ;

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7 d e l t a _ V i o = ( 3 0 * 1 0 ^ - 6 ) ; // Change i n i np ut o f f s e t

v o l t a g e8 d e l t a _ T = 1 ; // U ni t c ha ng e i n t e mp e ra t u re9 d e l t a _ I i o = ( 3 0 0 * 1 0 ^ - 1 2 ) ; // Change i n i np ut o f f s e t

c u r r e n t10 V s = 1 5 ;

11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ;

12 V i n = 1 * 1 0 ^ - 3 ; // I np ut v o l t a g e13 k = 2 5 ; // A mp l i f i e r i s n u l l e d a t 25 deg14 T = 3 5 - k ; // Chang e i n t e m p e r a tu r e15 E v = ( 1+ R f / R 1 ) * ( d e l t a_ V i o / d e l ta _ T ) * T + R f * ( d e l ta _ I io /

d e l t a _ T ) * T ; // E rr or v o l t ag e

16 printf ( ” \n E rr or v o l t a ge i s = %. 4 f V d c \n” , E v ) //R e s u l t

17 V o = ( 1 + R f / R 1 ) * V i n + E v ; / / Output v o l t a g e18 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //

R e s u l t19 // (OR)20 V o = ( 1 + R f / R 1 ) * V i n - E v ; / / Output v o l t a g e21 printf ( ” \n Output v o l t a g e i s = %. 4 f V \n” , V o ) //

R e s u l t

Scilab code Exa 4.10.b Error voltage and output voltage

1 / / C h a pt e r42 // P age . No−1 41 , F i g u r e . No−4.263 // E xa mp le 4 1 0 b4 / / E rr or v o l t a g e and o ut pu t v o l t ag e

5 / / G iv en6 clear ; clc ;7 d e l t a _ V i o = ( 3 0 * 1 0 ^ - 6 ) ; // Change i n i np ut o f f s e t

v o l t a g e8 d e l t a _ T = 1 ; // U ni t c ha ng e i n t e mp e ra t u re

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9 d e l t a _ I i o = ( 3 0 0 * 1 0 ^ - 1 2 ) ; // Change i n i np ut o f f s e t

c u r r e n t10 V s = 1 5 ;11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ; R l = 1 0 * 1 0 ^ 3 ;

12 V i n = 1 0 * 1 0 ^ - 3 ; // I np ut v o l t ag e13 k = 2 5 ; // A mp l i f i e r i s n u l l e d a t 25 deg14 T = 3 5 - k ; // Chang e i n t e m p e r a tu r e15 E v = ( 1+ R f / R 1 ) * ( d e l t a_ V i o / d e l ta _ T ) * T + R f * ( d e l ta _ I io /

d e l t a _ T ) * T ; // E rr or v o l t ag e16 printf ( ” \n E rr or v o l t a ge i s = %. 4 f V d c \n” , E v ) //

R e s u l t17 V o = ( 1 + R f / R 1 ) * V i n + E v ; / / Output v o l t a g e

18 printf ( ” \n Output v o l t a g e i s = %. 4 f V d c \n” , V o ) //R e s u l t

19 // (OR)20 V o = ( 1 + R f / R 1 ) * V i n - E v ; / / Output v o l t a g e21 printf ( ” \n Output v o l t a g e i s = %. 4 f V d c \n” , V o ) //

R e s u l t

Scilab code Exa 4.11.a Output offset voltage

1 / / C h a pt e r42 // P age . No−1 41 , F i g u r e . No−4.283 // E xa mp le 4 1 1 a4 // Output o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 d e l t a _ V i o = 1 5 . 8 5 * 1 0 ^ - 6 ; // Change i n i np ut o f f s e t

v o l t a g e8 d e l t a _ V = 1 ; // U ni t c ha ng e i n s up pl y v o l t a g e9 V = 2 ; // Change i n s up p ly v o l t a g e

10 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ;

11 d e l t a _ V o o = ( 1 + R f / R 1 ) * ( d e l t a _ V i o / d e l t a _ V ) * V ; // C hange

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i n o u t p u t o f f s e t v o l t a ge

12 printf ( ” \n Change i n o ut pu t o f f s e t v o l t a g e i s = %. 4 f V \n” , d e l t a _ V o o ) // R e su l t

Scilab code Exa 4.11.b Output offset voltage

1 / / C h a pt e r4

2 // P age . No−1 41 , F i g u r e . No−4.283 // E xa mp le 4 1 1 b4 // Output o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 d e l t a _ V i o = 1 5 . 8 5 * 1 0 ^ - 6 ; // Change i n i np ut o f f s e tv o l t a g e

8 d e l t a _ V = 1 ; // U ni t c ha ng e i n s up pl y v o l t a g e9 V = 2 ; // Change i n s up p ly v o l t a g e

10 V i n = 1 0 * 1 0 ^ - 3 ;

11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ;

12 d e l t a _ V o o = ( 1 + R f / R 1 ) * ( d e l t a _ V i o / d e l t a _ V ) * V ; / / O ut pu to f f s e t v o l t a g e

13 V o = ( - R f / R 1 ) * V i n + d e l t a _ V o o ; // T ot al o ut pu t o f f s e tv o l t a g e

14 printf ( ” \n T o ta l o u t p ut o f f s e t v o l t a ge i s = %. 4 f V \n” , V o ) // R e su l t

15 // (OR)16 V o = ( - R f / R 1 ) * V i n - d e l t a _ V o o ; // T ot al o ut pu t o f f s e t

v o l t a g e17 printf ( ” \n T o ta l o u t p ut o f f s e t v o l t a ge i s = %. 4 f V \

n” , V o ) // R e su l t

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Scilab code Exa 4.12 Output ripple voltage

1 / / C h a pt e r42 // P age . No−1 41 , F i g u r e . No −4 . 2 8 ( b )3 / / E xa m pl e 4 124 // Output r i p p l e v o l t a g e5 / / G iv en6 clear ; clc ;

7 d e l t a _ V i o = 1 5 . 8 5 * 1 0 ^ - 6 ; // Change i n i np ut o f f s e tv o l t a g e

8 d e l t a _ V = 1 ; // U ni t c ha ng e i n s up pl y v o l t a g e9 V = 1 0 * 1 0 ^ - 3 ; // Change i n s up p ly v o l t a g e

10 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ;

11 d e l t a _ V o o = ( 1 + R f / R 1 ) * ( d e l t a _ V i o / d e l t a _ V ) * V ; // C hangei n o u t p u t o f f s e t v o l t a ge

12 printf ( ” \n Change i n o ut pu t o f f s e t v o l t a g e i s = %. 6 f V \n” , d e l t a _ V o o ) // R e su l t

Scilab code Exa 4.13 Change in output offset voltage

1 / / C h a pt e r42 // P age . No−1 36 , F i g u r e . No−4.243 / / E xa m pl e 4 13

4 / / Change i n o ut pu t o f f s e t v o l t a g e5 / / G iv en6 clear ; clc ;

7 d e l t a _ V i o = 5 * 1 0 ^ - 6 ; // Change i n i np ut o f f s e t v o l t ag e8 d e l t a _ t = 1 ; // U ni t c ha ng e i n t im e

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9 d e l t a _ I i o = 2 * 1 0 ^ - 9 ; // Change i n i np ut o f f s e t c u r r e nt

10 t = 4 ; / / Ti me e l a p s e d ( w e e ks )11 R 1 = 1 * 1 0 ^ 3 ; R f = 1 0 0 * 1 0 ^ 3 ; R l = 1 0 * 1 0 ^ 3 ;12 d e l t a _ V o o t = ( 1 + R f / R 1 ) * ( d e l t a _ V i o / d e l t a _ t ) * t + R f * (

d e l t a _ I i o / d e l t a _ t ) * t ; // Change i n o ut pu t o f f s e tv o l t a g e

13 printf ( ” \n Change i n o ut pu t o f f s e t v o l t a g e i s = %. 4 f V \n” , d e l t a _ V o o t ) // R e su l t

Scilab code Exa 4.14.a Output voltage

1 / / C h a pt e r42 // P age . No−1 53 , F i g u r e . No−4.323 // E xa mp le 4 1 4 a4 / / Output v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R 2 = 1 * 1 0 ^ 3 ; R f = 1 0 * 1 0 ^ 3 ; R 3 = 1 0 * 1 0 ^ 3 ;

8 v d = 5 * 1 0 ^ - 3 ; // D i f f e r e n t i a l v o l t a g e9 v c m = 2 * 1 0 ^ - 3 ; / / Common−mode v o l t a g e

10 A d = R f / R 1 ; / / C l os e d−lo op d i f f e r e n t i a l ga in11 v o = A d * v d ; / / Output v o l t a g e12 printf ( ” \n Output v o l t a g e i s = %. 3 f V \n” , v o ) //

R e s u l t

Scilab code Exa 4.14.b Output common mode voltage


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3 // E xa mp le 4 1 4 b

4 // Output common−mode v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R 2 = 1 * 1 0 ^ 3 ; R f = 1 0 * 1 0 ^ 3 ; R 3 = 1 0 * 1 0 ^ 3 ;

8 v d = 5 * 1 0 ^ - 3 ; // D i f f e r e n t i a l v o l t a g e9 v c m = 2 * 1 0 ^ - 3 ; / / Common−mode v o l t a g e

10 A d = R f / R 1 ; / / C l os e d−lo op d i f f e r e n t i a l ga in11 C M R R d b = 9 0

12 C M R R = 1 0 ^ ( 9 0 / 2 0 ) ; // Us in g CMRRdb=20∗ l o g 1 0 (CMRR) , toc o n v e r t t h e CMRR( dB ) v a l u e i n t o i t s e q u u i v a l e n tn u m e ri c a l v a l u e

13 printf ( ” \n CMRR i s = %.2 f \n” , C M R R ) // R e su l t14 v o c m = ( A d * v c m ) / C M R R ; // Output common−mode v o l t a g e15 printf ( ” \n Output common−mode v o l t a g e i s = %. 8 f V \n

” , v o c m ) // R e su l t

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Chapter 5

Frequency Response of an

Opamp

Scilab code Exa 5.1 Maximum gain

1 / / C ha p t er 52 // P age . No−1 71 , F i g u r e . No−5.43 / / E xa mp le 5 14 / / Maximum ga in5 / / G iv en6 clear ; clc ;

7 f o = 5 ; // Break f r e q o f t h e op−amp i n Hz8 s = % s ;

9 A = 2 0 0 0 0 0 ; // Gain o f t he op−amp a t 0 Hz10 H = syslin ( ’ c ’ , ( A * f o * 2 * % p i ) / ( ( f o * 2 * % p i ) + s ) ) ;11 f m i n = 1 ;

12 f m a x = 1 0 0 0 0 0 ;

13 bode ( H , f m i n , f m a x ) ;

14 A o l = 4 0 ;

15 printf ( ” \n Maximum g a i n i s = % . f dB \n ” , A o l ) ; //From t h e g r a ph

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Scilab code Exa 5.2 Gain equation and break frequencies

1 / / C h a p t e r 52 // Page . No−1 72 , F i g u r e . No−5.53 / / E x a m p l e 5 24 / / Gain e q ua t i on and b re ak f r e q u e n c i e s5 // Given :6 clear ; clc ;

7 p h a s e = - 1 5 7 . 5 ; // Phas e s h i f t a t a bo ut 3 MHz8 f = 3 * 1 0 ^ 6 ;

9 disp ( ” G a in e q u a t i o n i s A o l ( f )=A / ( ( 1 + ( f / f o 1 ) ∗ j ) ∗(1+( f / fo 2 ) ∗ j ) ) , where fo1 − f i r s t b r e ak f r e q u e n c y and f o 2−s e co n d b r ea k f r e q u e n cy ” ) // From t he f i g u r e

10 f o 1 = 6 ;

11 printf ( ” \n F i r s t b r ea k f r e qu e n c y f o1 i s = %. f Hz \n”, f o 1 ) / / F rom t h e g r a ph

12 k = - a t a n d ( f / f o 1 ) - p h a s e ;

13 f o 2 = f / t a n d ( k ) ;14 printf ( ” \n S eco nd b re ak f r eq u e nc y f o 2 i s = %. 1 f Hz \

n” , f o 2 ) // R e su l t

Scilab code Exa 5.3 Stability of voltage follower

1 / / C ha p t er 52 // P age . No−8 3 , F i g u r e . No−3.73 / / E xa mp le 5 34 // S t a b i l i t y o f v o l t a g e f o l l o w e r5 / / G iv en

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6 clear ; clc ;

7 f o = 5 ; // Break f r e q o f t h e op−amp i n Hz8 s = % s ;9 A = 2 0 0 0 0 0 ; // Gain o f t he op−amp a t 0 Hz

10 H = syslin ( ’ c ’ , ( A * f o * 2 * % p i ) / ( ( f o * 2 * % p i ) + s ) ) ;11 f m i n = 1 0 ;

12 f m a x = 1 0 0 0 0 0 0 ;

13 bode ( H , f m i n , f m a x ) ;

14 A o l = 0 ;

15 printf ( ” \n M agni tude a t whi ch v o l t a g e f o l l o w e r i ss t a b l e i s = %. f dB \n ” , A o l ) ; // From t h e g r a ph

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Chapter 6

General Linear Applications

Scilab code Exa 6.1 Bandwidth of the amplifier

1 / / C h a pt e r62 // P age . No−1 93 , F i g u r e . No − 6 . 3 ( a )3 / / E xa mp le 6 14 / / B andwidth o f t he a m p l i f i e r5 / / G iv en6 clear ; clc ;

7 R 1 = 1 0 0 ; R f = 1 * 1 0 ^ 3 ; R i n = 5 0 ; R l = 1 0 * 1 0 ^ 3 ;8 C i = 0 . 1 * 1 0 ^ - 6 ; // C ap a ci t an ce b /w 2 s t a g e s b e in g

c o u p l e d9 R i F = R 1 ; // ac i np ut r e s i s t a n c e o f t h e s ec o nd s t a g e

10 R o = R i n ; // ac o ut p u t r e s i s t a n c e o f t he 1 s t s t a g e11 U G B = 1 0 ^ 6 ; // U ni t y g a i n b an dw id th12 f l = 1 / ( 2 * % p i * C i * ( R i F + R o ) ) ; // Low−f r e q c u t o f f 13 printf ( ” \n Low−f r e q c u t o f f i s = %. 1 f Hz \n” , f l ) //

R e s u l t14 K = R f / ( R 1 + R f ) ;

15 A f = - R f / R 1 ; // c l o se d l oo p v o l t a ge g ai n16 f h = U G B * K / abs ( A f ) ; / / H ig h−f r e q c u t o f f 17 printf ( ” \n H ig h−f r e q c u t o f f i s = %. 1 f Hz \n” , f h ) //

R e s u l t18 B W = f h - f l ; // Bandw i dt h

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19 printf ( ” \n B an dw id th i s = %. 1 f Hz \n” , B W ) // R e su l t

Scilab code Exa 6.2.a Bandwidth of the amplifier

1 / / C h a pt e r62 // P age . No−1 93 , F i g u r e . No − 6. 4( c )3 // E xa mp le 6 2 a

4 / / B andwidth o f t he a m p l i f i e r5 / / G iv en6 clear ; clc ;

7 R 1 = 1 0 0 * 1 0 ^ 3 ; R 2 = 1 0 0 * 1 0 ^ 3 ; R 3 = 1 0 0 * 1 0 ^ 3 ; R f = 1 * 1 0 ^ 6 ; R i n

=50;

8 C i = 0 . 1 * 1 0 ^ - 6 ; // C ap a ci t an ce b /w 2 s t a g e s b e in gc o u p l e d

9 R o = R i n ; // ac o ut p u t r e s i s t a n c e o f t he 1 s t s t a g e10 V c c = 1 5 ;

11 U G B = 1 0 ^ 6 ; // U ni t y g a i n b an dw id th12 R i f = R 2 * R 3 / ( R 2 + R 3 ) ; // s i n c e R i ∗(1+A∗B)>>R2 o r R3

13 f l = 1 / ( 2 * % p i * C i * ( R i f + R o ) ) ; / / l ow−f r e q c u t o f f 14 printf ( ” \n Low−f r e q c u t o f f i s = %. 1 f Hz \n” , f l ) //

R e s u l t15 K = R f / ( R 1 + R f ) ;

16 A f = - R f / R 1 ; // c l o se d l oo p v o l t a ge g ai n17 f h = U G B * K / abs ( A f ) ; / / H ig h−f r e q c u t o f f 18 printf ( ” \n H ig h−f r e q c u t o f f i s = %. 1 f Hz \n” , f h ) //

R e s u l t19 B W = f h - f l ; // Bandw i dt h20 printf ( ” \n B an dw id th i s = %. 1 f Hz \n” , B W ) // R e su l t

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Scilab code Exa 6.2.b Max output voltage swing

1 / / C h a pt e r62 // P age . No−1 93 , F i g u r e . No − 6. 4( c )3 // E xa mp le 6 2 b4 / / Max o ut pu t v o l t a g e s wi ng5 / / G iv en6 clear ; clc ;

7 R 1 = 1 0 0 * 1 0 ^ 3 ; R 2 = 1 0 0 * 1 0 ^ 3 ; R 3 = 1 0 0 * 1 0 ^ 3 ; R f = 1 * 1 0 ^ 6 ; R i n

=50;

8 C i = 0 . 1 * 1 0 ^ - 6 ; // C ap a ci t an ce b /w 2 s t a g e s b e in gc o u p l e d

9 R o = R i n ; // ac o ut p u t r e s i s t a n c e o f t he 1 s t s t a g e10 U G B = 1 0 ^ 6 ; // U ni t y g a i n b an dw id th11 V c c = 1 5 ;

12 printf ( ” \n The i d e a l maximum o u tp u t v o l t a g e s wi ng i s= %. f V pp \n” , V c c )

Scilab code Exa 6.3 Components of peak amplifier

1 / / C h a pt e r62 // P age . No−1 93 , F i g u r e . No − 6 . 5 ( a )3 / / E xa mp le 6 34 / / C omponents o f p ea k a m p l i f i e r5 / / G iv en6 clear ; clc ;

7 f p = 1 6 * 1 0 ^ 3 ; / / Peak f r e q u e n c y8 A f = 1 0 ; // Gain a t p eak f r e qu e n cy

9 C = 0 . 0 1 * 1 0 ^ - 6 ; // Assume10 L = 1 / ( ( ( 2 * % p i * f p ) ^ 2 ) * 1 0 ^ - 8 ) ; / / S i m p l i f y i n g f p =1/ (2∗p i ∗ s q r t ( L∗C) )

11 printf ( ” \n I n du c ta n ce i s = %. 4 f H \n” ,L )12 L = 1 0 * 1 0 ^ - 3 ; / / A p p ro x i m at e

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13 R = 3 0 ; // A ssume t h e v al ue o f i n t e r n a l r e s i s t a n c e o f

t he i n d u c to r14 X l = 2 * % p i * f p * L ; // I n d uc t i v e r e a ct a n ce15 Q c o i l = X l / R ; // F ig ur e o f m er i t o f t h e c o i l16 printf ( ” \n F i gu re o f m e r it o f t h e c o i l i s = %. 1 f \n”

, Q c o i l )

17 R p = ( Q c o i l ) ^ 2 * R ; // P a r a l l e l r e s i s t a n c e o f t h e t an kc i r c u i t

18 printf ( ” \n P a r a l l e l r e s i s t a n c e o f t h e t a nk c i r c u i ti s = %. 1 f ohm \n” , R p )

19 R 1 = 1 0 0 ; // Assume t h e v al ue o f i n t e r n a l r e s i s r a n c eo f t h e c o i l

20 R f = - R p / ( 1 - ( R p / ( A f * R 1 ) ) ) ; // S i m p l i f y i n g Af =( Rf | | Rp) /R1

21 printf ( ” \n Feed ba ck r e s i s t a n c e i s = %. 1 f ohm \n” , R f )

Scilab code Exa 6.4 Output voltage

1 / / C h a pt e r62 // P age . No−2 00 , F i g u r e . No−6.63 / / E xa mp le 6 44 / / Output v o l t a g e5 / / G iv en6 clear ; clc ;

7 V a = 1 ; V b = 2 ; V c = 3 ; // I n pu t v o l t a g e s i n v o l t s8 R a = 3 * 1 0 ^ 3 ; R b = 3 * 1 0 ^ 3 ; R c = 3 * 1 0 ^ 3 ; R f = 1 * 1 0 ^ 3 ;

9 V o = - ( ( R f / R a ) * V a + ( R f / R b ) * V b + ( R f / R c ) * V c ) ; // O utputv o l t a g e

10 printf ( ” \n Output v o l t a g e i s = %. f V \n” , V o )

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1 / / C h a pt e r62 // P age . No−2 03 , F i g u r e . No−6.73 / / E xa mp le 6 54 / / Output v o l t a g e5 / / G iv en6 clear ; clc ;

7 V a = 2 ; V b = - 3 ; V c = 4 ; // I np ut v o l t a g e s i n v o l t s8 R 1 = 1 * 1 0 ^ 3 ; R f = 2 * 1 0 ^ 3 ;

9 V 1 = ( V a + V b + V c ) / 3 ; // V ol t ag e a t non− i n v e r t i n gt e r m i n a l

10 printf ( ” \n V o lt ag e a t non− i n v e r t i n g t e r m i n al i s = %.f V \n” , V 1 )

11 V o = ( 1 + R f / R 1 ) * V 1 ; // Output v o l t a g e12 printf ( ” \n Output v o l t a g e i s = %. f V \n” , V o )


1 / / C h a pt e r62 // P age . No−2 05 , F i g u r e . No−6.93 / / E xa mp le 6 64 / / Output v o l t a g e


7 V a = 2 ; V b = 3 ; V c = 4 ; V d = 5 ; // I n pu t v o l t a g e s i n v o l t s8 R = 1 * 1 0 ^ 3 ;

9 V o = - V a - V b + V c + V d ; // Output v o l t a g e

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10 printf ( ” \n Output v o l t a g e i s = %. f V \n” , V o )


1 / / C h a pt e r62 // P age . No−2 09 , F i g u r e . No−6.123 / / E xa mp le 6 7

4 / / Output v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R f = 4 . 7 * 1 0 ^ 3 ; R a = 1 0 0 * 1 0 ^ 3 ; R b = 1 0 0 * 1 0 ^ 3 ; R c

= 1 0 0 * 1 0 ^ 3 ;

8 V d c = 5 ;

9 R t = 1 0 0 * 1 0 ^ 3 ; // R e s i st a nc e o f a t h er m is t or10 t e m p _ c o e f f = 1 * 1 0 ^ 3 ;

11

12 / / O utput v o l t a g e a t 0 d e g re e13 d e l t a _ R = - t e m p _ c o e f f * ( 0 - 2 5 ) ; // Change i n r e s i s t a n c e

14 R = R a ; // Ra=Rb=Rc=R15 V o = ( ( R f * d e l t a _ R ) / ( R 1 * 4 * R ) ) * V d c ;

16 printf ( ” \n Output v o l t ag e a t 0 d e g r ee i s = %. 2 f V \n” , V o )

17

18 / / Output v o l t a g e a t 1 00 d e gr e e19 d e l t a _ R = - t e m p _ c o e f f * ( 1 0 0 - 2 5 ) ; / / C ha ng e i n

r e s i s t a n c e20 V o = ( ( R f * d e l t a _ R ) / ( R 1 * 4 * R ) ) * V d c ;

21 printf ( ” \n Output v o l t ag e a t 100 d e g r ee i s = %. 2 f V

\n” , V o )

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Scilab code Exa 6.8 Change in resistance in straingage

1 / / C h a pt e r62 // P age . No−2 09 , F i g u r e . No−6.123 / / E xa mp le 6 74 / / Change i n r e s i s t a n c e i n s t r ai n g a ge5 / / G iv en6 clear ; clc ;

7 A = - 1 0 0 ; / / G ai n o f t h e d i f f e r e n t i a l i n s t r u m e n t a t i o na m p l i f i e r

8 R a = 1 0 0 ; R b = 1 0 0 ; R c = 1 0 0 ;

9 V d c = 1 0 ; V o = 1 ;

10 R = R a ; // Ra=Rb=Rc=R11 d e l t a _ R = ( V o * R ) / ( V d c * abs ( A ) ) ; // Change i n r e s i s t a n c e12 printf ( ” \n Change i n r e s i s t a n c e i s = %. 1 f ohm \n” ,

d e l t a _ R )

Scilab code Exa 6.9 Gain of the amplifier

1 / / C h a pt e r62 // P age . No−2 16 , F i g u r e . No −6 . 1 4 ( a )3 / / E xa mp le 6 94 / / Gain o f t h e a m p l i f i e r


7 V o = 3 . 7 ; V i n = 1 0 0 * 1 0 ^ - 3 ;

8 R 1 = 1 0 0 ; // Assume9 R f = 0 . 5 * ( ( V o * R 1 ) / V i n - 1 ) ; // Fee db ack r e s i s r a n c e

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10 printf ( ” \n Feed ba ck r e s i s r a n c e i s = %. 1 f ohm \n” , R f )

11 A = ( 1 + 2 * R f / R 1 ) ; / / G ai n o f t h e d i f f e r e n t i a l a m p l i f i e r12 printf ( ” \n G ai n o f t h e d i f f e r e n t i a l a m p l i f i e r i s = %. 1 f \n” ,A )

Scilab code Exa 6.10 Range of input voltage

1 / / C h a pt e r62 // P age . No−2 20 , F i g u r e . No−6.173 / / E xa m pl e 6 104 / / Range o f i n pu t v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 m i n = 1 * 1 0 ^ 3 ; R 1 m a x = 6 . 8 * 1 0 ^ 3 ;

8 i o = 1 * 1 0 ^ - 3 ; // Meter c u rr e n t f o r f u l l −waver e c t i f i c a t i o n

9 v i n _ m i n = 1 . 1 * R 1 m i n * i o ; // Minimum i n p u t v o l t a g e10 printf ( ” \n Minimum i n p u t v o l t a g e i s = %. 1 f V \n” ,

v i n _ m i n )11 v i n _ m a x = 1 . 1 * R 1 m a x * i o ; / / Maximum i n p u t v o l t a g e12 printf ( ” \n Maximum i n p u t v o l t a g e i s = %. 2 f V \n” ,

v i n _ m a x )

Scilab code Exa 6.11 Current and voltage drop

1 / / C h a pt e r62 // P age . No−2 22 , F i g u r e . No−6.183 / / E xa m pl e 6 114 / / C ur re nt and v o l t a g e d ro p

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5 / / G iv en

6 clear ; clc ;7 V i n = 0 . 5 ; V o = 1 . 2 ;

8 R 1 = 1 0 0 ;

9 I o = V i n / R 1 ; // C ur re nt t hr o ug h d i o d e10 printf ( ” \n C ur re nt t hr ou gh d i od e i s = %. 4 f A \n” , I o )11 V d = V o - V i n ; // V ol ta ge dr op a c r o s s d io de12 printf ( ” \n V ol ta ge dr op a c r o s s d io de i s = %. 1 f V \n”

, V d )

Scilab code Exa 6.12.a Load current

1 / / C h a pt e r62 // P age . No−2 22 , F i g u r e . No−6.193 // E xa mp le 6 1 2 a4 / / Load c u r r e n t5 / / G iv en6 clear ; clc ;

7 V i n = 5 ; V 1 = 1 ;8 R = 1 0 * 1 0 ^ 3 ;

9 I l = V i n / R ; // Load c u r r e n t10 printf ( ” \n Load c u r re n t i s = %. 5 f A \n” , I l )

Scilab code Exa 6.12.b Output voltage

1 / / C h a pt e r62 // P age . No−2 22 , F i g u r e . No−6.193 // E xa mp le 6 1 2 b4 / / Output v o l t a g e

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5 / / G iv en

6 clear ; clc ;7 V i n = 5 ; V 1 = 1 ;

8 R = 1 0 * 1 0 ^ 3 ;

9 V o = 2 * V 1 ; // Output v o l t a g e10 printf ( ” \n Output v o l t a g e i s = %. f V \n” , V o )

Scilab code Exa 6.13 Range of output voltage

1 / / C h a pt e r62 // P age . No−2 22 , F i g u r e . No−6.203 / / E xa m pl e 6 134 / / Range o f o ut pu t v o l t a g e5 / / G iv en6 clear ; clc ;

7 R 1 = 1 * 1 0 ^ 3 ; R f = 2 . 7 * 1 0 ^ 3 ;

8 V r e f = 2 ;

9 I o = 0 ; // S i n ce a l l t h e b in ar y i np u t s D0 t o D7 a r e

l o g i c z er o10 V o _ m i n = I o * R f ; / / Minimum o u t p u t v o l t a g e11 printf ( ” \n Minimum o u t pu t v o l t a g e i s = % . f V \n” ,

V o _ m i n )

12 I o = ( V r e f / R 1 )

* ( 1 / 2 + 1 / 4 + 1 / 8 + 1 / 1 6 + 1 / 3 2 + 1 / 6 4 + 1 / 1 2 8 + 1 / 2 5 6 ) ;

13 V o _ m a x = I o * R f ; / / Maximum o u t p u t v o l t a g e14 printf ( ” \n Maximum o u t pu t v o l t a g e i s = %. 2 f V \n” ,

V o _ m a x )

Scilab code Exa 6.14 Change in output voltage

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1 / / C h a pt e r6

2 // P age . No−2 28 , F i g u r e . No−6.213 / / E xa m pl e 6 144 / / C hange i n o u tp ut v o l t a g e5 / / G iv en6 clear ; clc ;

7 R f = 3 * 1 0 ^ 3 ;

8 V d c = 5 ;

9 R t = 1 0 0 * 1 0 ^ 3 ; // R e s is t a nc e a t d a rk ne ss10 V o m i n = - ( V d c / R t ) * R f ; // Min o ut pu t v o l t a g e a t

d a r k n e s s11 printf ( ” \n Min o ut p ut v o l t a g e a t d a rk ne ss i s = %. 2 f

V \n” , V o m i n )12 R t = 1 . 5 * 1 0 ^ 3 ; // R e s i st a nc e a t I l l u mi n a t i o n13 V o m a x = - ( V d c / R t ) * R f ; // Max o ut p ut v o l t a g e a t

I l l u m i n a t i o n14 printf ( ” \n Max o u tp ut v o l t a g e a t I l l u m i n a t i o n i s = %

. f V \n” , V o m a x )

Scilab code Exa 6.15 Output voltage of an integrator

1 / / C h a pt e r62 // P age . No−2 30 , F i g u r e . No−6.233 / / E xa m pl e 6 154 / / O utput v o l t a g e o f an i n t e g r a t o r5 / / G iv en6 clear ; clc ;

7 V i n = 2 ; // I np ut v o lt a g e i n v o l t

8 V o 0 = 0 ;9 V o 1 = - i n t e g r a t e ( ’ 2 ’ , ’ t ’ ,0,1);10 disp ( V o 1 )

11 V o 2 = - i n t e g r a t e ( ’ 2 ’ , ’ t ’ , 1 , 2 ) + V o 1 ;12 disp ( V o 2 )

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13 V o 3 = - i n t e g r a t e ( ’ 2 ’ , ’ t ’ , 2 , 3 ) + V o 2 ;

14 disp ( V o 3 )15 V o 4 = - i n t e g r a t e ( ’ 2 ’ , ’ t ’ , 3 , 4 ) + V o 3 ;16 disp ( V o 4 )

17 V o = [ Vo 0 V o1 V o2 V o3 V o4 ] ;

18 t = [0 1 2 3 4];

19 plot ( t , V o ) ;


Scilab code Exa 6.16.a Design of differentiator

1 / / C h a pt e r62 // P age . No−2383 // E xa mp le 6 1 6 a4 // D e s i g n o f d i f f e r e n t i a t o r5 / / G iv en6 clear ; clc ;

7 C 1 = 0 . 1 * 1 0 ^ - 6 ; // Assume

8 f a = 1 * 1 0 ^ 3 ; // Freq a t whi ch g ai n i s 0 dB9 R f = 1 / ( 2 * % p i * f a * C 1 ) ; / / U s i n g f a = 1/ (2∗ %p i ∗ Rf ∗C1 )

10 printf ( ” \n Feed ba ck r e s i s t a n c e i s = %. 1 f ohm \n” , R f )11 R f = 1 . 5 * 1 0 ^ 3 ; / / A p p ro x i ma t i o n12 f b = 2 0 * 1 0 ^ 3 ; // Gain l i m i t i n g f r e q13 R 1 = 1 / ( 2 * % p i * f b * C 1 ) ;

14 printf ( ” \n R e s i s t a nc e , R1 i s = %. 1 f ohm \n” , R 1 )15 R 1 = 8 2 ; / / A p p ro x i m at i o n16 C f = R 1 * C 1 / R f ;

17 printf ( ” \n C a pa c it a nc e , Cf i s = %. 1 0 f f a r a d \n” , C f )

18 C f = 0 . 0 0 5 * 1 0 ^ - 6 ; / / A p p ro x i m at i o n

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Scilab code Exa 6.16.b Output waveform of differentiator

1 / / C h a pt e r62 // P age . No−2383 // E xa mp le 6 1 6 b4 // Output w aveform o f d i f f e r e n t i a t o r5 / / G iv en6 clear ; clc ;

7 s t e p = 0 . 0 1 ;

8 t = 0 : s t e p : 2 * % p i ;

9 dy = diff ( sin ( t ) ) / s t e p ; // a p pr ox im at e d i f f e r e n t i a t i o no f s i n e f u n c t i o n

10 V o = - 1 . 5 * 1 0 ^ 3 * 0 . 1 * 1 0 ^ - 6 * 2 * % p i * 1 0 ^ 3 * d y ;

11 plot ( V o ) ;


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Chapter 7

Active Filters and Oscillators

Scilab code Exa 7.1 Design of low pass filter

1 / / C h a pt e r72 // P age . No−2563 / / E xa mp le 7 14 / / D es i gn o f l ow p a s s f i l t e r5 / / G iv en6 clear ; clc ;

7 f h = 1 * 1 0 ^ 3 ; / / Cut− o f f f r eq u en c y8 C = 0 . 0 1 * 1 0 ^ - 6 ; / / A s s um p t io n9 R = 1 / ( 2 * % p i * f h * C ) ;

10 printf ( ” \n R e s i s t a n ce R i s = %. 1 f ohm \n” ,R ) //R e s u l t

11 printf ( ” \n U se 2 0 kohm POT a s R \n” )12 R 1 = 1 0 * 1 0 ^ 3 ; / / A s s um p t io n13 printf ( ” \n R e s i s t a n ce R1 i s = %. 1 f ohm \n” , R 1 )14 R f = R 1 ; // S i c e p as sb an d g a in i s 2 , R1 a nd Rf must be

e q u a l

15 printf ( ” \n R e s i s t a n ce Rf i s = %. 1 f ohm \n” , R f )

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Scilab code Exa 7.2 Design of low pass filter

1 / / C h a pt e r72 // P age . No−2563 / / E xa mp le 7 24 / / D es i gn o f l ow p a s s f i l t e r5 / / G iv en6 clear ; clc ;

7 f c 0 = 1 * 1 0 ^ 3 ; // O r i g i na l cut − o f f f r eq u en c y8 f c 1 = 1 . 6 * 1 0 ^ 3 ; // New c ut−o f f f r e q ue n cy9 R = 1 5 . 9 * 1 0 ^ 3 ; // O r i g i n a l r e s i s t a n c e v al ue

10 k = f c 0 / f c 1 ;

11 R n e w = R * k ;

12 printf ( ” \n New R e s i s t a n c e Rnew i s = %. 1 f ohm \n” ,

R n e w ) // R e su l t

Scilab code Exa 7.3 Frequency response of low pass filter

1 / / C h a pt e r72 // P age . No−257

3 / / E xa mp le 7 34 / / F re qu en cy r e s p o n s e o f l ow p a s s f i l t e r5 / / G iv en6 clear ; clc ;

7 A f = 2 ; / / P as sb an d g a i n o f t h e f i l t e r

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8 f h = 1 0 0 0 ; / / Cut− o f f f r eq u en c y

9 f 1 = 1 0 ; // I np ut f r e q i n Hz10 a v 1 = A f / sqrt ( 1 + ( f 1 / f h ) ^ 2 ) ;11 printf ( ” \n Gain m ag ni tu de av1 a t f 1 i s = %. 2 f \n” ,

a v 1 ) // R e su l t12 f 2 = 1 0 0 ; // Freq i n Hz13 a v 2 = A f / sqrt ( 1 + ( f 2 / f h ) ^ 2 ) ;

14 printf ( ” \n Gain m ag ni tu de av2 a t f 2 i s = %. 2 f \n” ,a v 2 ) // R e su l t

15 f 3 = 2 0 0 ; // Freq i n Hz16 a v 3 = A f / sqrt ( 1 + ( f 3 / f h ) ^ 2 ) ;

17 printf ( ” \n Gain m ag ni tu de av3 a t f 3 i s = %. 2 f \n” ,

a v 3 ) // R e su l t18 f 4 = 7 0 0 ; // Freq i n Hz19 a v 4 = A f / sqrt ( 1 + ( f 4 / f h ) ^ 2 ) ;


21 f 5 = 1 0 0 0 ; // Fre q i n Hz22 a v 5 = A f / sqrt ( 1 + ( f 5 / f h ) ^ 2 ) ;






30 f 8 = 1 0 0 0 0 ; // Fr eq i n Hz31 a v 8 = A f / sqrt ( 1 + ( f 8 / f h ) ^ 2 ) ;




36 f 1 0 = 1 0 0 0 0 0 ; // Fr eq i n Hz

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37 a v 1 0 = A f / sqrt ( 1 + ( f 1 0 / f h ) ^ 2 ) ;

38 printf ( ” \n Gain m ag ni tu de a v1 0 a t f 1 0 i s = %. 2 f \n” ,a v 1 0 ) // R e su l t39 x =[ f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 ];

40 y = [ av 1 a v2 a v3 a v4 a v5 a v6 a v7 a v8 a v9 a v1 0 ];

41 gainplot ( x , y ) ;

42 t i t l e ( ’ F r e q u e n c y R e s p o n s e ’ ) ;43 x l a b e l ( ’ Frequenc y (Hz) ’ ) ;44 y l a b e l ( ’ V o l t a g e g a i n ( dB ) ’ ) ;

Scilab code Exa 7.4.a Design of second order low pass filter

1 / / C h a pt e r72 // P age . No−2563 // E xa mp le 7 4 a4 / / D es i gn o f s e co n d o r d e r l ow p a s s f i l t e r5 / / G iv en6 clear ; clc ;

7 f h = 1 * 1 0 ^ 3 ; / / Cut− o f f f r eq u en c y8 C 2 = 0 . 0 0 4 7 * 1 0 ^ - 6 ; / / A s s um p t io n9 C 3 = C 2 ;

10 R 2 = 1 / ( 2 * % p i * f h * C 2 ) ;

11 printf ( ” \n R e s i s t a n ce R2 i s = %. 1 f ohm \n” , R 2 ) //R e s u l t

12 R 2 = 3 3 * 1 0 ^ 3 ; / / A p p ro x i m at i o n13 R 3 = R 2 ;

14 printf ( ” \n R e s i s t a n ce R3 i s = %. 1 f ohm \n” , R 3 ) //R e s u l t

15 R 1 = 2 7 * 1 0 ^ 3 ; / / A s s um p t io n16 R f = 0 . 5 8 6 * R 1 ;17 printf ( ” \n R e s i s t a n ce Rf i s = %. 1 f ohm \n” , R f ) //

R e s u l t18 printf ( ” \n U se 2 0 kohm POT a s R f \n” )

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Scilab code Exa 7.4.b Frequency response of second order highpass filter

1 / / C h a pt e r72 // P age . No−2603 // E xa mp le 7 4 b4 // F re qu en cy r e s p on s e o f s ec on d o r d er h i g h pa s s

f i l t e r5 / / G iv en6 clear ; clc ;

7 A f = 1 . 5 8 6 ; / / P as sb an d g a i n o f t h e f i l t e r8 f h = 1 0 0 0 ; / / Cut− o f f f r eq u en c y9 f 1 = 1 0 ; // I np ut f r e q i n Hz

10 a v 1 = A f / sqrt ( 1 + ( f 1 / f h ) ^ 4 ) ;










24 f 6 = 3 0 0 0 ; // Fre q i n Hz

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25 a v 6 = A f / sqrt ( 1 + ( f 6 / f h ) ^ 4 ) ;

26 printf ( ” \n Gain m ag ni tu de av6 a t f 6 i s = %. 2 f \n” ,a v 6 ) // R e su l t27 f 7 = 7 0 0 0 ; // Fre q i n Hz28 a v 7 = A f / sqrt ( 1 + ( f 7 / f h ) ^ 4 ) ;




33 f 9 = 3 0 0 0 0 ; // Fr eq i n Hz

34 a v 9 = A f / sqrt ( 1 + ( f 9 / f h ) ^ 4 ) ;35 printf ( ” \n Gain m ag ni tu de av9 a t f 9 i s = %. 5 f \n” ,

a v 9 ) // R e su l t36 f 1 0 = 1 0 0 0 0 0 ; // Fr eq i n Hz37 a v 1 0 = A f / sqrt ( 1 + ( f 1 0 / f h ) ^ 4 ) ;

38 printf ( ” \n Gain m ag ni tu de a v1 0 a t f 1 0 i s = %. 6 f \n” ,a v 1 0 ) // R e su l t

39 x =[ f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 ];

40 y = [ av 1 a v2 a v3 a v4 a v5 a v6 a v7 a v8 a v9 a v1 0 ];

41 gainplot ( x , y ) ;

42 t i t l e ( ’ F r e q u e n c y R e s p o n s e ’

) ;

43 x l a b e l ( ’ Frequenc y (Hz) ’ ) ;44 y l a b e l ( ’ V o l t a g e g a i n ( dB ) ’ ) ;

Scilab code Exa 7.5.a Design of highpass filter

1 / / C h a pt e r72 // P age . No−2633 // E xa mp le 7 5 a4 / / D es i gn o f h i g h p a s s f i l t e r5 / / G iv en

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6 clear ; clc ;

7 f h = 1 * 1 0 ^ 3 ; / / Cut− o f f f r eq u en c y8 A f = 2 ; / / P as sb an d g a i n o f t h e f i l t e r9 C = 0 . 0 1 * 1 0 ^ - 6 ; / / A s s um p t io n

10 R = 1 / ( 2 * % p i * f h * C ) ;

11 printf ( ” \n R e s i s t a n ce R i s = %. 1 f ohm \n” ,R ) //R e s u l t

12 printf ( ” \n U se 2 0 kohm POT a s R \n” )13 R 1 = 1 0 * 1 0 ^ 3 ; / / A s s um p t io n14 printf ( ” \n R e s i s t a n ce R1 i s = %. 1 f ohm \n” , R 1 )15 R f = R 1 ; // S i c e p as sb an d g a in i s 2 , R1 a nd Rf must be

e q u a l

16 printf ( ” \n R e s i s t a n ce Rf i s = %. 1 f ohm \n” , R f )

Scilab code Exa 7.5.b Frequency response of highpass filter

1 / / C h a pt e r72 // P age . No−263

3 // E xa mp le 7 5 b4 / / F re qu en cy r e s p o n s e o f h i g h p a s s f i l t e r5 / / G iv en6 clear ; clc ;

7 A f = 2 ; / / P as sb an d g a i n o f t h e f i l t e r8 f l = 1 0 0 0 ; / / Cut− o f f f r eq u en c y9 f 1 = 1 0 0 ; // I np u t f r e q i n Hz

10 a v 1 = ( A f * ( f 1 / f l ) ) / sqrt ( 1 + ( f 1 / f l ) ^ 2 ) ;


12 f 2 = 2 0 0 ; // Freq i n Hz13 a v 2 = ( A f * ( f 2 / f l ) ) / sqrt ( 1 + ( f 2 / f l ) ^ 2 ) ;14 printf ( ” \n Gain m ag ni tu de av2 a t f 2 i s = %. 2 f \n” ,

a v 2 ) // R e su l t15 f 3 = 4 0 0 ; // Freq i n Hz

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16 a v 3 = ( A f * ( f 3 / f l ) ) / sqrt ( 1 + ( f 3 / f l ) ^ 2 ) ;

17 printf ( ” \n Gain m ag ni tu de av3 a t f 3 i s = %. 2 f \n” ,a v 3 ) // R e su l t18 f 4 = 7 0 0 ; // Freq i n Hz19 a v 4 = ( A f * ( f 4 / f l ) ) / sqrt ( 1 + ( f 4 / f l ) ^ 2 ) ;


21 f 5 = 1 0 0 0 ; // Fre q i n Hz22 a v 5 = ( A f * ( f 5 / f l ) ) / sqrt ( 1 + ( f 5 / f l ) ^ 2 ) ;


24 f 6 = 3 0 0 0 ; // Fre q i n Hz

25 a v 6 = ( A f * ( f 6 / f l ) ) / sqrt ( 1 + ( f 6 / f l ) ^ 2 ) ;26 printf ( ” \n Gain m ag ni tu de av6 a t f 6 i s = %. 2 f \n” ,

a v 6 ) // R e su l t27 f 7 = 7 0 0 0 ; // Fre q i n Hz28 a v 7 = ( A f * ( f 7 / f l ) ) / sqrt ( 1 + ( f 7 / f l ) ^ 2 ) ;


30 f 8 = 1 0 0 0 0 ; // Fr eq i n Hz31 a v 8 = ( A f * ( f 8 / f l ) ) / sqrt ( 1 + ( f 8 / f l ) ^ 2 ) ;

32 printf ( ” \n Gain m ag ni tu de av8 a t f 8 i s = %. 2 f \n” ,a v 8 )

// R e su l t33 f 9 = 3 0 0 0 0 ; // Fr eq i n Hz34 a v 9 = ( A f * ( f 9 / f l ) ) / sqrt ( 1 + ( f 9 / f l ) ^ 2 ) ;

35 printf ( ” \n Ga

op-amps and linear integrated circuits_r. a. gayakwad_2

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