analog and digital electronics_u. a. bakshi and a. p. godse

8/10/2019 Analog and Digital Electronics_U. a. Bakshi and a. P. Godse

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Scilab Textbook Companion for

Analog And Digital Electronics

by U. A. Bakshi And A. P. Godse1

Created byPriyank Bangar

B.TechElectronics Engineering

NMIMSCollege Teacher

No TeacherCross-Checked by

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 Projectsection at the website http://scilab.in


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Book Description

Title: Analog And Digital Electronics

Author: U. A. Bakshi And A. P. Godse

Publisher: Technical Publications, Pune

Edition: 1

Year: 2009

ISBN: 9788184316902

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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 Special Diodes 10

2 Frequency Response 12

3 Feedback Amplifiers 18

4 Oscillators 41

5 Combinational Logic Circuits 60

6 Sequential Logic Circuits 84

7 Shift Registers 86

8 Counters 87

9 Op amp Applications 119

10 Voltage Regulators 138

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List of Scilab Codes

Exa 1.1 current through LED . . . . . . . . . . . . . . . . . . 10Exa 1.2 transition capacitance and constant K . . . . . . . . . 10Exa 2.1 maximum voltage gain. . . . . . . . . . . . . . . . . . 12Exa 2.2 gain of the amplifier . . . . . . . . . . . . . . . . . . . 12Exa 2.3 gain of an amplifier . . . . . . . . . . . . . . . . . . . 13Exa 2.4 low frequency response of the amplifier. . . . . . . . . 13Exa 2.5 low frequency response of the FET amplifier. . . . . . 14Exa 2.6 high frequency response of the amplifier . . . . . . . . 15Exa 2.7 high frequency response of the amplifier . . . . . . . . 16Exa 3.1 gain fL and fH . . . . . . . . . . . . . . . . . . . . . . 18Exa 3.2 Vo and second harmonic distortion with feedback . . . 18Exa 3.3 beta and Af. . . . . . . . . . . . . . . . . . . . . . . . 19Exa 3.4 beta and Av and Avf and Rif and Rof . . . . . . . . . 20

Exa 3.5 Avf and Rif and Rof . . . . . . . . . . . . . . . . . . . 22Exa 3.6 Avf and Rif and Rof . . . . . . . . . . . . . . . . . . . 23Exa 3.7 D and Avf and Rif and Rof . . . . . . . . . . . . . . . 25Exa 3.8 voltage gain. . . . . . . . . . . . . . . . . . . . . . . . 27Exa 3.9 Avf . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Exa 3.10 Avf and Rif and Rof . . . . . . . . . . . . . . . . . . . 29Exa 3.11 voltage gain and input and output resistance . . . . . 31Exa 3.12 Ai and beta and Aif . . . . . . . . . . . . . . . . . . . 32Exa 3.13 beta and Av and Avf and Rif and Rof . . . . . . . . . 34Exa 3.14 gain and new bandwidth. . . . . . . . . . . . . . . . . 36Exa 3.15 show voltage gain with feedback . . . . . . . . . . . . 36

Exa 3.16 GMf and Rif and Rof . . . . . . . . . . . . . . . . . . 37Exa 3.17 feedback factor and Rif and Rof . . . . . . . . . . . . 38Exa 3.18 gain with feedback and new bandwidth . . . . . . . . 39Exa 3.19 overall voltage gain and bandwidth . . . . . . . . . . . 40

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Exa 4.1 C and hfe . . . . . . . . . . . . . . . . . . . . . . . . . 41

Exa 4.2 frequency of oscillation . . . . . . . . . . . . . . . . . 42Exa 4.3 R and C. . . . . . . . . . . . . . . . . . . . . . . . . . 42Exa 4.4 C and RD. . . . . . . . . . . . . . . . . . . . . . . . . 42Exa 4.5 minimum and maximum R2. . . . . . . . . . . . . . . 43Exa 4.6 range over capacitor is varied . . . . . . . . . . . . . . 44Exa 4.7 frequency of oscillation . . . . . . . . . . . . . . . . . 44Exa 4.8 frequency of oscillation . . . . . . . . . . . . . . . . . 45Exa 4.9 calculate C . . . . . . . . . . . . . . . . . . . . . . . . 45Exa 4.10 series and parallel resonant freqency . . . . . . . . . . 46Exa 4.11 series and parallel resonant freqency . . . . . . . . . . 46Exa 4.12 Varify Barkhausen criterion and find frequency of oscil-

lation . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Exa 4.13 minimum and maximum values of R2 . . . . . . . . . 49Exa 4.14 frequency of oscillation and minimum hfe . . . . . . . 49Exa 4.15 range of frequency of oscillation. . . . . . . . . . . . . 50Exa 4.16 frequency of oscillation . . . . . . . . . . . . . . . . . 50Exa 4.17 gain of the transistor. . . . . . . . . . . . . . . . . . . 51Exa 4.18 new frequency and inductance . . . . . . . . . . . . . 51Exa 4.19 new frequency of oscillation . . . . . . . . . . . . . . . 52Exa 4.20 R and hfe . . . . . . . . . . . . . . . . . . . . . . . . . 52Exa 4.21 component values of wien bridge . . . . . . . . . . . . 53

Exa 4.22 values of C2 and new frequency of oscillation . . . . . 54Exa 4.23 series and parallel resonant freqency and Qfactor . . . 55Exa 4.24 change in frequency and trimmer capacitance . . . . . 55Exa 4.25 design RC phase shift oscillator . . . . . . . . . . . . . 57Exa 4.26 range of capacitor . . . . . . . . . . . . . . . . . . . . 58Exa 4.27 change in frequency of oscillation . . . . . . . . . . . . 58Exa 5.1 design a combinational logic circuit. . . . . . . . . . . 60Exa 5.2 design a circuit with control line C and data lines. . . 61Exa 5.3 design combinational circuit. . . . . . . . . . . . . . . 62Exa 5.4 design logic circuit . . . . . . . . . . . . . . . . . . . . 63Exa 5.5 design circuit to detect invalid BCD number. . . . . . 64

Exa 5.6 design two level combinational circuit . . . . . . . . . 65Exa 5.7 design 32 to 1 multiplexer . . . . . . . . . . . . . . . . 66Exa 5.8 design a 32 to 1 multiplexer . . . . . . . . . . . . . . . 66Exa 5.9 implement boolean function using 8to1 multiplexer . . 67Exa 5.10 implement boolean function using 4to1 multiplexer . . 67

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Exa 5.11 implement boolean function using 8to1 MUX . . . . . 69

Exa 5.12 implement boolean function using 4to1 MUX . . . . . 69Exa 5.13 implement boolean function using 8to1 MUX . . . . . 70Exa 5.14 implement boolean function using 8to1 MUX . . . . . 71Exa 5.15 implement boolean function using 8to1 MUX . . . . . 72Exa 5.16 determine boolean expression . . . . . . . . . . . . . . 72Exa 5.17 realize using 4 to 1 MUX . . . . . . . . . . . . . . . . 73Exa 5.18 design 1 to 8 DEMUX . . . . . . . . . . . . . . . . . . 74Exa 5.19 implement full subtractor . . . . . . . . . . . . . . . . 74Exa 5.20 construst 1 to 32 DEMUX. . . . . . . . . . . . . . . . 75Exa 5.21 design 3 to 8 decoder . . . . . . . . . . . . . . . . . . 76Exa 5.22 design 5 to 32 decoder . . . . . . . . . . . . . . . . . . 77

Exa 5.23 design 4 line to 16 line decoder . . . . . . . . . . . . . 78Exa 5.24 implement using 74LS138 . . . . . . . . . . . . . . . . 78Exa 5.25 implement full substractor. . . . . . . . . . . . . . . . 79Exa 5.26 implement gray to binary code converter. . . . . . . . 79Exa 5.27 design 2 bit comparator . . . . . . . . . . . . . . . . . 80Exa 5.28 design full adder circuit . . . . . . . . . . . . . . . . . 81Exa 5.29 implement BCD to 7 segment decoder . . . . . . . . . 82Exa 5.31 implement 32 input to 5 output encoder . . . . . . . . 82Exa 6.4 analyze the circuit . . . . . . . . . . . . . . . . . . . . 84Exa 7.1 determine number of flip flops needed . . . . . . . . . 86

Exa 8.1 count after 12 pulse . . . . . . . . . . . . . . . . . . . 87Exa 8.2 count in binary . . . . . . . . . . . . . . . . . . . . . . 87Exa 8.4 draw logic diagram. . . . . . . . . . . . . . . . . . . . 88Exa 8.5 output frequency . . . . . . . . . . . . . . . . . . . . . 88Exa 8.6 maximum operating frequency . . . . . . . . . . . . . 88Exa 8.8 design 4 bit up down ripple counter . . . . . . . . . . 89Exa 8.9 design divide by 9 counter . . . . . . . . . . . . . . . . 89Exa 8.10 design a divide by 128 counter . . . . . . . . . . . . . 90Exa 8.11 design divide by 78 counter . . . . . . . . . . . . . . . 90Exa 8.12 design divide by 6 counter . . . . . . . . . . . . . . . . 91Exa 8.13 find fmax . . . . . . . . . . . . . . . . . . . . . . . . . 91

Exa 8.14 determine states . . . . . . . . . . . . . . . . . . . . . 92Exa 8.15 design MOD 10 counter . . . . . . . . . . . . . . . . . 92Exa 8.16 design down counter . . . . . . . . . . . . . . . . . . . 93Exa 8.17 design programmable frequency divider . . . . . . . . 94Exa 8.18 design a counter . . . . . . . . . . . . . . . . . . . . . 94

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Exa 8.19 programmable frequency divider . . . . . . . . . . . . 96

Exa 8.20 design divide by 2 and divide by 5 counter. . . . . . . 96Exa 8.21 design mod 9 counter . . . . . . . . . . . . . . . . . . 98Exa 8.22 determine MOD number and counter range . . . . . . 98Exa 8.23 design a divide by 20 counter . . . . . . . . . . . . . . 98Exa 8.24 design divide by 96 counter . . . . . . . . . . . . . . . 99Exa 8.25 design divide by 93 counter . . . . . . . . . . . . . . . 100Exa 8.26 design divide by 78 counter . . . . . . . . . . . . . . . 100Exa 8.28 7490 IC . . . . . . . . . . . . . . . . . . . . . . . . . . 100Exa 8.30 sketch output waveforms of counter. . . . . . . . . . . 101Exa 8.31 explain operation of circuit . . . . . . . . . . . . . . . 104Exa 8.32 design divide by 40000 counter . . . . . . . . . . . . . 105

Exa 8.33 design modulo 11 counter . . . . . . . . . . . . . . . . 106Exa 8.34 design excess 3 decimal counter . . . . . . . . . . . . . 106Exa 8.35 modulus greater than 16. . . . . . . . . . . . . . . . . 107Exa 8.36 modulo 8 counter. . . . . . . . . . . . . . . . . . . . . 108Exa 8.37 synchronous decade counter . . . . . . . . . . . . . . . 108Exa 8.38 flip flops. . . . . . . . . . . . . . . . . . . . . . . . . . 110Exa 8.39 design mod 5 synchronous counter . . . . . . . . . . . 112Exa 8.40 design MOD 4 down counter . . . . . . . . . . . . . . 114Exa 8.41 design MOD 12 synchronous counter . . . . . . . . . . 115Exa 8.42 design 4 bit 4 state ring counter . . . . . . . . . . . . 117

Exa 8.44 design 4 bit 8 state johnson counter . . . . . . . . . . 117Exa 8.45 johnson counter . . . . . . . . . . . . . . . . . . . . . 118Exa 9.1 output voltage . . . . . . . . . . . . . . . . . . . . . . 119Exa 9.3 input bias current . . . . . . . . . . . . . . . . . . . . 120Exa 9.4 design inverting schmitt trigger . . . . . . . . . . . . . 120Exa 9.5 threshold voltage . . . . . . . . . . . . . . . . . . . . . 121Exa 9.6 tripping voltage . . . . . . . . . . . . . . . . . . . . . 121Exa 9.8 time duration. . . . . . . . . . . . . . . . . . . . . . . 122Exa 9.9 calculate R1 and R2 . . . . . . . . . . . . . . . . . . . 123Exa 9.10 calculate trip point and hysteresis . . . . . . . . . . . 123Exa 9.11 design schmitt trigger . . . . . . . . . . . . . . . . . . 124

Exa 9.12 design op amp schmitt trigger. . . . . . . . . . . . . . 125Exa 9.13 VUT and VLT and frequency of oscillation . . . . . . 126Exa 9.17 design op amp circuit . . . . . . . . . . . . . . . . . . 127Exa 9.18 design monostable using IC 555. . . . . . . . . . . . . 127Exa 9.19 design a timer . . . . . . . . . . . . . . . . . . . . . . 128

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List of Figures

8.1 sketch output waveforms of counter . . . . . . . . . . . . . . 104

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Chapter 1

Special Diodes

Scilab code Exa 1.1 current through LED

1 / / E xa mp le 1 . 12 clc

3 format (5)

4 disp ( Assume t h e d ro p a c r o s s t h e LED a s 2 V . )5 disp ( T he re fo re , VD = 2 V )6 disp ( From f i g . 1 . 1 1 , RS = 2 . 2 kohm and V S = 15 V )

7 i s = ( 1 5 - 2 ) / ( 2 . 2 ) / / i n mA8 disp ( i s , T h e re fo r e , I S (mA) = VSVD / RS =)

Scilab code Exa 1.2 transition capacitance and constant K

1 / / E xa mp le 1 . 22 clc

3 disp ( The t r a n s i s t o r c a p ac i t a n c e i s g i ve n by , )4 disp ( CT = C( 0 ) / [ 1+ |VR/VJ | n ] )5 disp ( Now C( 0 ) = 80pF , n = 1/3 a s d i f f u s e d j u n c ti o n

)6 disp ( VR = 4 . 2 V, VJ = 0 . 7 V )

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7 c t = ( ( 8 0 * 1 0 ^ - 1 2 ) / ( ( 1 + ( 4 . 2 / 0 . 7 ) ) ^ ( 1 / 3 ) ) ) * 1 0 ^ 1 2 // i n

pF8 format (6)9 disp ( c t , T h e r e fo r e , CT( pF ) = )

10 disp ( t he t r a n s i s t o r c a pa c it a nc e i s a l s o g i v en by , )11 disp ( CT = K / [ VR+VJ ] n )12 format (10)

13 k = ( 4 1 . 8 2 * 1 0 ^ - 1 2 ) * ( ( 4 . 2 + 0 . 7 ) ^ ( 1 / 3 ) )

14 disp (k , T h er ef or e , K = )

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Chapter 2

Frequency Response

Scilab code Exa 2.1 maximum voltage gain

1 / / E xa mp le 2 . 12 clc

3 format (7)

4 disp ( We know t h a t maximam v o l t a g e g a i n o f v o l t a g ea m p l i f i e r i s g iv en a s )

5 m v = 2 0 0 * sqrt (2)

6 disp ( m v , T h e r e fo r e , Maximum v o l t a g e g a i n = Ga in a tcut o f f x s q r t ( 2 ) = )

Scilab code Exa 2.2 gain of the amplifier

1 / / E xa mp le 2 . 22 clc

3 format (6)

4 disp ( We know tha t , )5 a = 1 0 0 / sqrt ( 1 + ( ( 1 0 0 0 / 2 0 ) ^ 2 ) )

6 disp (a , Below midband : A = A mid / s q r t (1+( f 1/ f ) 2 ) = )

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Scilab code Exa 2.3 gain of an amplifier

1 / / E xa mp le 2 . 32 clc

3 format (7)

4 a = 2 0 0 * sqrt (2)

5 disp (a , We know t ha t A mid = 3dB g a in x s q r t ( 2 )= )

6 a m = 2 8 2 . 8 4 / ( sqrt ( 1 + ( ( ( 1 0 / 2 ) ^ 2 ) ) ) )

7 format (6)

8 disp ( a m , Above midband : A = A mid / s q r t (1+(f 1 / f ) 2 ) = ) // a ns we r i n t ex tb o ok i s wrong

Scilab code Exa 2.4 low frequency response of the amplifier

1 / / E xa mp le 2 . 4

2 clc3 format (6)

4 disp ( I t i s n e c e s s ar y t o a n al y ze e a ch n et w or k t od e te r m in e t h e c r i t i c a l f r e q u e n c y o f t h e a m p l i f i e r )

5 disp ( ( a ) I n p ut RC n e tw o rk )6 f c 1 = 1 / ( 2 * % p i * [ 6 8 0 + 1 0 3 1 . 7 ] * ( 0 . 1 * 1 0 ^ - 6 ) )

7 disp ( f c 1 , f c ( i n p u t ) ( i n Hz ) = 1 / 2p i [RS+(R1| |R2| | h ie ) ] C1 = ) // i n Hz

8 disp ( ( b ) O ut pu t RC n e tw o r k )9 format (7)

10 f c 2 = 1 / ( 2 * % p i * ( ( 2 . 2 + 1 0 ) * 1 0 ^ 3 ) * ( 0 . 1 * 1 0 ^ - 6 ) )11 disp ( f c 2 , f c ( o u t p u t ) ( i n Hz ) = 1 / 2p i (RC+

RL) C2 = ) // i n Hz12 disp ( ( c ) B y p as s RC n e t w o r k )

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13 r t h = ( ( 6 8 * 2 2 * 0 . 6 8 0 ) / ( ( 2 2 * 0 . 6 8 0 ) + ( 6 8 * 0 . 6 8 0 ) + ( 6 8 * 2 2 ) ) )

*10^314 disp ( r t h , R t h ( i n ohm) = R1 | | R2 | | RS = )15 format (6)

16 f c 3 = 1 / ( 2 * % p i * 1 7 . 2 3 * 1 0 * 1 0 ^ - 6 )

17 disp ( f c 3 , f c ( b y p a s s ) ( i n Hz ) = 1 / 2p i [ (R t h+ hi e / be t a ) | | RE] CE)

18 disp ( We h av e c a l c u l a t e d a l l t h e t h r e e c r i t i c a lf r e q u e n c i e s : )

19 disp ( ( a ) f c ( i n p u t ) = 9 2 9 . 8 Hz )20 disp ( ( b ) f c ( o u t pu t ) = 1 3 0 . 4 5 Hz )21 disp ( ( c ) f c ( b y p a s s ) = 9 2 3 . 7 Hz )

Scilab code Exa 2.5 low frequency response of the FET amplifier

1 / / E xa mp le 2 . 52 clc

3 disp ( I t i s n e c e s s ar y t o a n al y ze e a ch n et w or k t od e te r m in e t h e c r i t i c a l f r e q u e n c y o f t h e a m p l i f i e r )

4 disp ( ( a ) I n p ut RC N et wo rk )5 disp ( f c = 1 / 2p i R i n C1 )6 format (6)

7 r i n = ( 1 0 0 * 1 0 0 ) / ( 1 0 0 + 1 0 0 )

8 disp ( r i n , wher e R i n ( i n Mohm) = RG | | R i n ( g a t e ) =RG | | | VGS/IGSS| = )

9 format (5)

10 f c 1 = 1 / ( 2 * % p i * 5 0 * 1 0 ^ 6 * 0 . 0 0 1 * 1 0 ^ - 6 )

11 disp ( f c 1 , T h e r e fo r e , f c ( i n Hz ) = )12 disp ( ( b ) O ut pu t RC N et wo rk )

13 format (6)14 f c 2 = 1 / ( 2 * % p i * ( 2 4 . 2 * 1 0 ^ 3 ) * ( 1 * 1 0 ^ - 6 ) )

15 disp ( f c 2 , f c ( i n Hz ) = 1 / 2p i (RD+RL) C2 = )16 disp ( We h av e c a l c u l a t e d two c r i t i c a l f r e q u e n c i e s )17 disp ( ( a ) f c ( i n p u t ) = 3 . 1 8 Hz )

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18 disp ( ( b ) f c ( o u t pu t ) = 6 . 5 7 7 Hz )

Scilab code Exa 2.6 high frequency response of the amplifier

1 / / E xa mp le 2 . 62 clc

3 disp ( B e f o r e c a l c u l a t i n g c r i t i c a l f r e q u e n c i e s i t i sn e c e s s a r y t o c a l c u l a t e mid f re q u e n cy g ai n o f t heg iv en c i r c u i t . T h i s i s r e q u i r e d t o c a l c u l a t e C in (

m i l l e r ) and C o ut ( m i l l e r ) )4 disp ( Av = h f e R o / R i )5 disp ( where Ri = h i e | | R1 | | R2 )6 disp ( and Ro = RC | | RL )7 format (6)

8 a v = ( - 1 0 0 * 1 . 8 ) / 1 . 0 3 2

9 disp ( a v , T h e re fo r e , Av = h f e (RC | | RL) / h i e | |R1| |R2= )

10 disp ( N eg at iv e s i gn i n d i c a t e s 180 d eg re e s h i f tb et we en i n p u t and o u t pu t )

11 format (7)

12 c i n = ( 4 * ( 1 7 4 . 4 + 1 ) ) * 1 0 ^ - 3 // i n nF13 disp ( c i n , C in ( m i l l e r ) ( i n nF ) = C bc (Av+1) = )14 c o u t = ( 4 * 1 7 5 . 4 ) / ( 1 7 4 . 4 ) // i n pF15 format (4)

16 disp ( c o u t , C out ( m i l l e r ) ( i n pF ) = C bc (Av+1) /Av = )

17 disp ( We now a n a l y z e i n p u t and o u tp ut n et wo rk f o rc r i t i c a l fr eq ue nc y . )

18 format (8)

19 f c i = ( 1 / ( 2 * % p i * 4 1 0 * 0 . 7 2 1 6 * 1 0 ^ - 9 ) ) * 1 0 ^ - 3 // i n kHz

20 disp ( f c i , f c ( i np ut ) ( i n kHz ) = 1 / 2p i ( Rs | |R1| |R2| | h i e ) ( C b e+C i n ( m i l l e r ) ) = )21 format (5)

22 f c o = ( 1 / ( 2 * % p i * ( ( 2 2 * 1 0 ^ 6 ) / ( 1 2 . 2 * 1 0 ^ 3 ) ) * ( 4 * 1 0 ^ - 1 2 ) ) )

* 1 0 ^ - 6 / / i n MHz

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23 disp ( f c o , f c ( o ut pu t ) ( i n MHz) = 1 / 2p i (RC | | RL

) C o u t ( m i l l e r ) = )24 disp ( We h a ve c a l c u l a t e d b o th t h e c r i t i c a lf r e q u e n c i e s )

25 disp ( ( a ) f c ( i n p u t ) = 5 3 7 . 9 4 7 kHz )26 disp ( ( b ) f c ( o u t p u t ) = 2 2 . 1 MHz )

Scilab code Exa 2.7 high frequency response of the amplifier

1 / / E xa mp le 2 . 72 clc

3 disp ( B e f o r e c a l c u l a t i n g c r i t i c a l f r e q u e n c i e s i t i sn e c e s s a r y t o c a l c u l a t e mid f re q u e n cy g ai n o f t heg iv en a m p l i f i e r c i r c u i t . T h i s i s r e q u i r e d t oc a l c u l a t e C i n ( m i l l e r ) and C o u t ( m i l l e r ) )

4 disp ( Av = gm RD )5 disp ( H er e , RD s h o u l d b r r e p l a c e d by RD | | RL )6 a v = - 6 * 2

7 disp ( a v , T h e r e f o r e , Av = gm (RD | | RL) = )8 c i n = 2 * ( 1 2 + 1 ) // i n pF

9 disp ( c i n , C i n ( m i l l e r ) ( i n pF ) = C g d ( A v + 1 ) = C r s s (Av+1) = )

10 format (6)

11 c o u t = ( 2 * 1 3 ) / 1 2 // i n pF12 disp ( c o u t , C o u t ( m i l l e r ) ( i n pF ) = C g d ( Av +1) / Av =

)13 disp ( G gs = C i s s C r s s = 4 p F )14 disp ( We know a n a l y z e i n p u t and o u tp u t n et wo r k f o r

c r i t i c a l fr eq ue nc y )15 disp ( f c ( i n p u t ) = 1 / 2 p i RSCT )

16 disp ( = 1 / 2p i RS [ C g s+C i n ( m i l l e r) ] )17 format (4)

18 f c 1 = ( 1 / ( 2 * % p i * 1 0 0 * 3 0 * 1 0 ^ - 1 2 ) ) * 1 0 ^ - 6 / / i n MHz19 disp ( f c 1 , f c ( i n p u t ) ( i n MHz)= )

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20 f c 2 = ( 1 / ( 2 * % p i * ( ( 4 8 . 4 * 1 0 ^ 6 ) / ( 2 4 . 2 * 1 0 ^ 3 ) )

* ( 2 . 1 6 6 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 6 / / i n MHz21 format (6)22 disp ( f c 2 , f c ( o ut pu t ) ( i n MHz) = 1 / 2p i (RD | | RL

) C o u t ( m i l l e r ) = )23 disp ( We h a ve c a l c u l a t e d b o th t h e c r i t i c a l

f r e q u e n c i e s : )24 disp ( ( a ) f c ( i n p u t ) = 5 3 MHz )25 disp ( ( b ) f c ( o u t p u t ) = 3 6 . 7 4 MHz )

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Chapter 3

Feedback Amplifiers

Scilab code Exa 3.1 gain fL and fH

1 / / E xa mp le 3 . 12 clc

3 disp ( ( a ) Gain w it h f e ed b a ck )4 format (5)

5 a v = 1 0 0 0 / ( 1 + ( 0 . 0 5 * 1 0 0 0 ) )

6 disp ( a v , AV mid = Av mid / 1+ b e t a Av mid = )

7 f l f = 5 0 / ( 1 + ( 0 . 0 5 * 1 0 0 0 ) ) // i n Hz8 disp ( f l f , (b ) f L f ( i n Hz ) = f L / 1+ b e t a Av mid

= )9 f h f = ( ( 5 0 * 1 0 ^ 3 ) * ( 1 + ( 0 . 0 5 * 1 0 0 0 ) ) ) * 1 0 ^ - 6 / / i n MHz

10 disp ( f h f , ( c ) f H f ( i n MHz) = f H ( 1+ be t a A v mi d) = )

Scilab code Exa 3.2 Vo and second harmonic distortion with feedback

1 / / E xa mp le 3 . 22 clc

3 disp ( ( a ) b et a : 4 0 = 2 0l o g [ 1+ b e t a A ] )

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4 disp ( T h e r e fo r e , 1+ be t a A = 1 0 0 )

5 b = 9 9 / 1 0 0 06 format (6)

7 disp (b , Th e r e f o r e , b e t a = )8 disp ( Gain o f t h e a m p l i f i e r w i t h f ee db ac k i s g iv en

a s )9 a v f = 1 0 0 0 / 1 0 0

10 disp ( a v f , A Vf = A V / 1+ b e t a A V = )11 disp ( ( b ) To m a in t ai n o ut p ut p ower 1 0 W, we s h o ul d

m a in ta in o ut pu t v o l t a g e c o n st a n t and t o m a in ta i no ut pu t c on s ta n t w i th f ee db a ck g ai n r e q u i r e d Vs i s )

12 v s f = 1 0 * 1 0 0 * 1 0 ^ - 3 // i n V13 disp ( v s f , V s f ( i n V) = Vs 100 = )14 disp ( ( c ) S eco nd h ar mo ni c d i s t o r t i o n i s r ed uc ed by

f a c t o r 1 + b et a A )15 d 2 f = ( 0 . 1 / 1 0 0 ) * 1 0 0 // i n p e rc e nt a ge16 disp ( d 2 f , D 2 f ( i n p er c en ta ge ) = D 2 / 1+ b e t a A

= )

Scilab code Exa 3.3 beta and Af

1 / / E xa mp le 3 . 32 clc

3 disp ( ( a ) We know t h a t )4 disp ( dAf / Af = 0. 1/1+ b e t a A dA/A )5 disp ( T h e r e fo r e , 1+ be t a A = 3 7 . 5 )6 b = ( 3 6 . 5 / 2 0 0 0 ) * 1 0 0 // i n p e rc e nt a ge7 format (6)

8 disp (b , T h e r e fo r e , b e ta ( i n p e r c e n t a g e ) = )

9 a f = 2 0 0 0 / ( 1 + ( 0 . 0 1 8 2 5 * 2 0 0 0 ) )10 disp ( a f , ( b ) Af = A / 1+ b e t a A = )

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Scilab code Exa 3.4 beta and Av and Avf and Rif and Rof

1 / / E xa mp le 3 . 42 clc

3 disp ( S t e p 1 : I d e n t i t y t op ol og y )4 disp ( The f e e db ac k v o l t a g e i s a p p l i e d a c r o s s t h e

r e s i s t a n c e R e1 and i t i s i n s e r i e s w i t h i n p u ts i g n a l . Hence f ee db ac k i s v o l t a ge s e r i e s f ee db ac k. )

5 disp ( )6 disp ( S te p 2 and S te p 3 : Fi nd i n pu t and o ut pu t

c i r c u i t . )

7 disp ( To f i n d i n p u t c i r c u i t , s e t Vo = 0 (c o n n ec t i ng C2 t o g ro un d ) , wh ich g i v e s p a r a l l e lc om bi na ti o n o f Re w it h Rf a t E1 . To f i n d o ut pu tc i r c u i t , s e t I i = 0 ( o pe n in g t h e i np ut node E1 a t

e m i tt e r o f Q1 ) , whi ch g i v e s s e r i e s c om bi na ti ono f Rf and Re1 a c r o s s t he o ut pu t . The r e s u l t a n tc i r c u i t i s shown i n F i g . 3 . 3 2 )

8 disp ( )9 disp ( S te p 4 : Find open l oo p v o l t a g e g ai n ( A v ) )

10 format (5)

11 r l 2 = ( 4 . 7 * 1 0 . 1 ) / ( 4 . 7 + 1 0 . 1 ) // i n kohm12 disp ( r l 2 , R L2 ( i n ko h m ) = R c 2 | | ( R e 1+Rf ) = )

13 disp ( A i 2 = h f e = 100 )14 disp ( R i 2 = h i e = 1100 ohm )15 format (7)

16 a v 2 = ( - 1 0 0 * 3 . 2 1 * 1 0 ^ 3 ) / 1 1 0 0

17 disp ( a v 2 , A v2 = A i 2 R L2 / R i 2 = )18 disp ( A i 1 = h f e = 100 )19 format (5)

20 r l 1 = ( 2 2 * 2 2 0 * 2 2 * 1 . 1 0 0 ) / ( ( 2 2 0 * 2 2 * 1 . 1 0 0 ) + ( 2 2 * 2 2 * 1 . 1 0 0 )

+ ( 2 2 * 2 2 0 * 1 . 1 0 0 ) + ( 2 2 * 2 2 0 * 2 2 ) ) / / i n ohm

21 disp ( r l 1 * 1 0 ^ 3 , R L1 ( i n ohm ) = R c1 | | R3 | | R4 | |R i 2 = )

22 r i 1 = 1 . 1 + ( 1 0 1 * ( ( 0 . 1 * 1 0 ) / ( 0 . 1 + 1 0 ) ) ) // i n kohm23 format (5)

24 disp ( r i 1 , R i 1 ( i n kohm ) = h i e + (1+ h f e ) R e 1 e f f =

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Scilab code Exa 3.5 Avf and Rif and Rof

1 / / E xa mp le 3 . 52 clc

3 disp ( S t e p 1 : I d e n t i t y t op ol og y )4 disp ( The f ee db ac k v o lt a g e i s a p pl i ed a c r o s s R1

( 10 0 ohm ) , which i s i n s e r i e s w it h i np ut s i g n a l .Hence f ee db a ck i s v o l t a g e s e r i e s f ee db a ck . )


c i r c u i t )7 disp ( To f i n d i np ut c i r c u i t , s e t Vo = 0 , which

g i v e s p a r a l l e l c om bi na ti on o f R1 w i t h R2 a t E1 a sshown i n t h e f i g . 3 . 4 5 . To f i n d o ut p u t c i r c u i t ,

s e t I i = 0 by o pe ni ng t he i np ut node , E1 a te m i t t e r o f Q1 , whi ch g i v e s t he s e r i e s c om bi na ti on

o f R2 a nd R1 a c r o s s t he o ut pu t . The r e s u l t a n tc i r c u i t i s shown i n f i g . 3 . 4 5 )

8 disp ( )

9 disp ( S te p 4 : Find t he open l oo p v o l t ag e g ai n ( Av ) )10 r l 2 = ( 4 . 7 * 4 . 8 ) / ( 4 . 7 + 4 . 8 ) // i n kohm11 format (5)

12 disp ( r l 2 , R L2 ( i n kohm) = )13 disp ( S i nc e h oe = h r e = 0 we can u s e a pp ro x i ma te

a n a l y s i s )14 disp ( A i 2 = h f e = 50 )15 disp ( R i 2 = h i e = 1 . 1 kohm )16 a v 2 = ( - 5 0 * 2 . 3 7 ) / 1 . 1

17 format (7)

18 disp ( a v 2 , A v2 = A i 2 R L2 / R i 2 = )19 r l 1 = ( 1 0 * 4 7 * 3 3 * 1 . 1 ) / ( ( 4 7 * 3 3 * 1 . 1 ) + ( 1 0 * 3 3 * 1 . 1 )

+ ( 1 0 * 4 7 * 1 . 1 ) + ( 1 0 * 4 7 * 3 3 ) ) / / i n ohm20 format (5)

21 disp ( r l 1 * 1 0 ^ 3 , R L1 ( i n ohm ) = )

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22 disp ( A i 1 = h f e = 50 )

23 r i 1 = 1 . 1 + ( 5 1 * ( ( 0 . 1 * 4 . 7 ) / ( 4 . 8 ) ) ) // i n kohm24 format (6)25 disp ( r i 1 , R i 1 ( i n kohm ) = h i e + (1+ h f e ) Re = )26 a v 1 = ( - 5 0 * 9 4 2 ) / ( 6 . 0 9 3 * 1 0 ^ 3 )

27 format (5)

28 disp ( a v 1 , A v1 = A i 1 R L1 / R i 1 = )29 a v = - 7 . 7 3 * - 1 0 7 . 7 3

30 format (7)

31 disp ( a v , T h er ef or e , A v = A v1 A v 2 = )32 disp ( )33 disp ( S te p 5 : C a l c u l a te b et a and D )

34 disp ( b e t a = R1 / R1+R2 = 1/ 48 )35 d = 1 + ( 8 3 2 . 7 5 / 4 8 ) / / i n ohm36 format (6)

37 disp (d , D( i n ohm ) = 1 + A b e t a = )38 disp ( )39 disp ( S te p 5 : C a l c u l at e A vf , R of and R i f )40 a v f = 8 3 2 . 7 5 / 1 8 . 3 5

41 disp ( a v f , A v f = A v / D = )42 r i f = 6 . 0 9 3 * 1 8 . 3 5 // i n kohm43 disp ( r i f , R i f ( i n ko h m ) = R i 1 D = )44 r o f = ( 2 . 3 7 * 1 0 ^ 3 ) / 1 8 . 3 5

/ / i n ohm45 format (7)46 disp ( r o f , R o f ( i n ohm ) = R o / D = )


1 / / E xa mp le 3 . 62 clc

3 disp ( S t e p 1 : I d e n t i f y t op ol og y )4 disp ( The f e e db ac k v o l t a g e i s a p p l i e d a c r o s s R e1= 1 .5 kohm , which i s i n s e r i e s w i t h i np ut s i g n a l. Hence f ee db ac k i s v o l t a ge s e r i e s f ee db ac k )

5 disp ( )

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6 disp ( S te p 2 and s t ep 3 : Fi nd i np ut and o ut p ut

c i r c u i t )7 disp ( To f i n d i np ut c i r c u i t , s e t Vo = 0 , whichg i v e s p a r a l l e l c om bi na ti on o f R e1 w i t h R f a t E1

a s shown i n f i g . 3 . 4 7 . To f i n d ouput c i r c u i t , s e tI i = 0 by o pe ni ng t he i np ut node , E1 a t e m i t t e ro f Q1 , whi ch g i v e s t he s e r i e s c o mb in at io n o f R f and R e1 a c r o s s t he o ut pu t . The r e s u l t a n t

c i r c u i t i s shown i n f i g . 3 . 4 7 )8 disp ( )9 disp ( S te p 4 : Find t he open l oo p v o l t ag e g ai n ( Av ) )

10 r l 2 = ( 2 . 2 * 5 7 . 5 ) / ( 2 . 2 + 5 7 . 5 ) // i n kohm

11 format (6)12 disp ( r l 2 , R L2 ( i n ko h m ) = R c 2 | | ( R f + R e 1 ) = )13 disp ( S i n c e ho e R L 2 = 1 0 62 .1 19 ko h m = 0 . 0 0 2 1 1 9

i s l e s s t han 0 . 1 we u se a pp ro xi ma te a n a l y s i s . )14 disp ( A i 2 = h f e = 200 )15 disp ( R i 2 = h i e = 2 kohm )16 a v 2 = ( - 2 0 0 * 2 . 1 1 9 ) / 2

17 disp ( a v 2 , A v2 = A i 2 R L2 / R i 2 = )18 r l 1 = ( 1 2 0 * 2 ) / ( 1 2 2 ) // i n kohm19 disp ( r l 1 , R L1 ( i n kohm) = R C1 | | R i 2 = )20 disp (

S i n c e ho e R L 1 = 1 0 61 .9 67 = 0 . 0 0 19 6 7 i sl e s s t ha n 0 . 1 we u se a pp ro xi ma te a n a l y s i s . )21 disp ( A i 1 = h f e = 200 )22 r i 1 = 2 + ( 2 0 1 * ( ( 1 . 5 * 5 6 ) / ( 5 7 . 5 ) ) ) // i n kohm23 format (7)

24 disp ( r i 1 , R i 1 ( i n kohm ) = h i e + (1+ h f e ) Re = )25 a v 1 = ( - 2 0 0 * 1 . 9 6 7 ) / 2 9 5 . 6 3

26 format (5)

27 disp ( a v 1 , T h er ef or e , A v1 = A i 1 R L1 / R i 1 = )28 disp ( The o v e r a l l g ai n w it ho ut f ee db ac k i s )29 a v = - 1 . 3 3 * - 2 1 1 . 9

30 format (7)31 disp ( a v , Av = A v1 A v 2 = )32 disp ( )33 disp ( S te p 5 : C a l c u l at e b et a )34 b e t a = 1 . 5 / 5 7 . 5

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35 format (6)

36 disp ( b e t a , b e ta = Vf / Vo = )37 disp ( )38 disp ( St e p 6 : c a l c u l a t e D , A vf , R i f , R of )39 d = 1 + ( 0 . 0 2 6 * 2 8 1 . 8 2 )

40 disp (d , D = 1 + Av b e t a = )41 a v f = 2 8 1 . 8 2 / 8 . 3 2 7

42 disp ( a v f , T h er e fo r e , A vf = Av / D = )43 r i = ( 2 9 5 . 6 3 * 1 5 0 ) / ( 2 9 5 . 6 3 + 1 5 0 ) // i n kohm44 format (5)

45 disp ( r i , Ri ( i n ko h m ) = R i 1 | | R =)46 r i f = 9 9 . 5 * 8 . 3 2 7 // i n kohm

47 format (7)48 disp ( r i f , R i f ( i n kohm) = R i D = )49 disp ( Ro = 1/ hoe = 1 Mohm )50 r o f = ( ( 1 * 1 0 ^ 6 ) / 8 . 3 2 7 ) * 1 0 ^ - 3 // i n kohm51 format (4)

52 disp ( r o f , R of ( i n kohm ) = Ro / D = )53 r o = ( 1 0 0 0 * 2 . 1 1 9 ) / ( 2 . 1 1 9 + 1 0 0 0 ) // i n kohm54 format (7)

55 disp ( r o , R o ( i n kohm) = Ro | | R c2 | | ( Rf+R e1 )= Ro | | R L2 = )

56 r o f = ( 2 . 1 1 4 5 * 1 0 ^ 3 ) / 8 . 3 2 7 / / i n ohm57 format (4)

58 disp ( r o f , R o f ( i n ohm ) = R o / D = )

Scilab code Exa 3.7 D and Avf and Rif and Rof

1 / / E xa mp le 3 . 72 clc

3 disp ( S t e p 1 : I d e n t i t y t op ol og y )4 disp ( By s h o r t i ng o ut p ut v o l t a g e ( Vo = 0) ,f ee db ac k v o l t a g e Vf becomes z e ro and h en ce i t i sv o l t a g e s am pl in g . The f ee db ac k v o l t ag e i s a a p l i e d

i n s e r i e s w i t h i np ut v o l t a ge h e n c e t h e t op ol og y

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i s v o l t a ge s e r i e s f ee db ac k . )

5 disp ( )6 disp ( S te p 2 and S te p 3 : Fi nd i n pu t and o ut pu tc i r c u i t . )

7 disp ( To f i n d i n pu t c i r c u i t , s e t Vo = 0 . T h i sp l a c es t h e p a r a l l e l c o m b i n a ti o n o f r e s i s t o r 10 Kand 2 00 ohm a t f i r s t s o u r c e . To f i n d o ut pu tc i r c u i t , s e t I i = 0 . T h i s p l a c e s t h e r e s i s t o r 10

K and 200 ohm i n s e r i e s a c r o s s t he o ut pu t . Ther e s u l t a n t c i r c u i t i s shown i n f i g . 3 . 5 0 . )

8 disp ( )9 disp ( S te p 4 : R ep la ce FET wi th i t s e q u i v a l e n t

c i r c u i t a s shown i n f i g . 3 . 5 1 )10 disp ( )11 disp ( S te p 5 : Find open l oo p t r a n s f e r g ai n . )12 disp ( Av = Vo / Vs = A v1 A v 2 )13 disp ( A v2 = u R L2 / R L2+ r d )14 r l 2 = ( 1 0 . 2 * 4 7 ) / ( 1 0 . 2 + 4 7 ) // i n kohm15 format (5)

16 disp ( r l 2 , wher e R L2 ( i n kohm) = )17 a v 2 = ( - 4 0 * 8 . 3 8 ) / ( 8 . 3 8 + 1 0 )

18 format (7)

19 disp ( a v 2 , T h er e fo r e , A v2 =

)

20 disp ( A v1 = u R D e f f / r d +R D e f f +(1+u ) R s e f f )21 r d e f f = ( 4 7 * 1 0 0 0 ) / ( 4 7 + 1 0 0 0 ) // i n kohm22 format (6)

23 disp ( r d e f f , w here R D ef f ( i n kohm) = R D | | R G2 =)

24 disp ( R s e f f = 200 | | 10 K )25 a v 1 = ( - 4 0 * 4 4 . 9 8 * 1 0 ^ 3 ) / ( ( 1 0 * 1 0 ^ 3 ) + ( 4 4 . 8 9 * 1 0 ^ 3 )

+ ( 4 1 * ( ( 1 0 * 0 . 2 ) / ( 1 0 . 2 ) ) ) )

26 disp ( a v 1 , A v1 = ) // a nsw e r i n t ex tb oo k i swrong

27 o a v = - 2 8 . 5 9 * - 1 8 . 2 3 728 format (7)

29 disp ( o a v , T h er e fo r e , O v e r al l Av = )30 disp ( )31 disp ( S te p 6 : C a l c u l at e b et a )

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32 b e t a = 2 0 0 / ( 1 0 . 2 * 1 0 ^ 3 )

33 disp ( b e t a , b e ta = Vf / Vo = )34 disp ( )35 disp ( S te p 7 : C a l c u l at e D, A vf , R i f , R o f )36 d = 1 + ( 0 . 0 1 9 6 * 5 2 1 . 3 9 )

37 format (6)

38 disp (d , D = 1 + Av b e t a = )39 a v f = 5 2 1 . 3 9 / 1 1 . 2 2

40 disp ( a v f , A vf = Av / D = )41 disp ( Ri = R G = 1 Mohm )42 r i f = 1 1 . 2 2

43 disp ( r i f , R i f ( i n Mohm) = R i D =)

44 disp ( Ro = r d = 10 kohm )45 r o = ( 1 0 * 8 . 3 8 ) / ( 1 8 . 3 8 ) // i n kohm46 disp ( r o , R o ( i n ko h m ) = r d | | R L2 = )47 r o f = ( 4 . 5 5 9 * 1 0 ^ 3 ) / 1 1 . 2 2 / / i n ohm48 format (4)


Scilab code Exa 3.8 voltage gain

1 / / E xa mp le 3 . 82 clc

3 disp ( Here , o ut pu t v o l t a g e i s s am pl ed and f e d i ns e r i e s w i th t he i np ut s i g n a l . Hence t he t op o lo g yi s v o l t a ge s e r i e s f ee db ac k . )

4 disp ( The open l oo p v o l t a g e g ai n f o r one s t a g e i sg i v e n a s , )

5 disp ( Av = gm R e q )6 r e q = ( 8 * 4 0 * 1 0 0 0 ) / ( ( 4 0 * 1 0 0 0 ) + ( 8 * 1 0 0 0 ) + ( 8 * 4 0 ) ) // i n k

ohm7 format (5)8 disp ( r e q , R eq ( i n ko h m ) = r d | | R d | | ( R i1+R 2

) =)9 a v = - 5 * 6 . 6 2

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10 format (6)

11 disp ( a v , Av = )12 a v m = - 3 3 . 1 1 ^ 313 disp ( a v m , Av = O v e r a l l v o l t a g e g a in = |A vmid| 3 = )

// a ns we r i n t ex tb o ok i s wrong14 b e t a = 5 0 / ( 1 0 ^ 6 )

15 format (7)

16 disp ( b e t a , b e t a = Vf / Vo = R 1 / R g = R 1 /R 1+R 2 = )

17 d = 1 + ( ( - 5 * 1 0 ^ - 5 ) * - 3 6 3 0 6 )

18 format (6)

19 disp (d , D = 1 + | Av | b e t a = )

20 a v f = - 3 6 3 0 6 / 2 . 8 1 5 321 disp ( a v f , A vf = Av / D = )

Scilab code Exa 3.9 Avf

1 / / E xa mp le 3 . 92 clc

3 disp ( Here , o ut p ut t e r m i n a l s a r e B a nd g ro un d , t h us

t h e f or wa rd g ai n i s t he g ai n o f Q1 and i t i s , )4 disp ( A BN = 33. 11 )5 disp ( However , Q2 and Q3 must b e c o n s i d e r e d a s a

p ar t o f f ee db a ck l oo p )6 disp ( Her e beta BN = V f / V B = V f / V o V o / V C

V C/V B )7 disp ( where V B and V C a r e v o l t a g e s a t p o in t B a nd

C , r e s p e c t i v e l y . )8 disp ( T h e re fo r e , beta BN = V f / V o A v3 A v2

b e c a u s e V o /V C = A v3 and V C/ V B = A v2

)9 b b n = - ( 5 * 1 0 ^ - 5 ) * ( 3 3 . 1 1 ^ 2 )10 format (7)

11 disp ( b b n , T h er e fo r e , beta BN = R1/R g A v3 A v 2 = )

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12 disp ( Note t ha t t he l oo p g ai n be t a BN A BN = A3

Vo R1/Rg = 1 . 8 1 5 = A b e t a )13 disp ( I t s ho ul d be c l e a r t ht r e g a r d l e s s o fw he re t h eo ut pu t t e r mi n a l s a r e ta ke n , t he l oo p g ai n i sunch ange d . )

14 a v f = - 3 3 . 1 1 / 2 . 8 1 5

15 format (6)

16 disp ( a v f , T h e r e fo r e , A v f = A BN / 1+A b e t a = )


1 / / E x am pl e 3 . 1 02 disp ( S t e p 1 : I d e n t i f y t op ol og y )3 disp ( By s h o r t i ng o ut p ut v o l t a g e ( Vo = 0) ,

f ee db ac k v o l t a g e Vf becomes z e ro and h en ce i t i sv o l t a g e s am pl in g . The f ee db ac k v o l t ag e i s a p p li e d

i n s e r i e s w i t h t h e i np ut v o lt a ge h e n c e t het op o lo g y i s v o l t a g e s e r i e s f ee db ac k . )


c i r c u i t . )6 disp ( To f i n d i n pu t c i r c u i t , s e t Vo = 0 . T h i s

p l a c es t h e p a r a l l e l c o m b i n a ti o n o f r e s i s t o r 10 Kand 3 00 ohm a t f i r s t s o u r c e . To f i n d o ut pu tc i r c u i t , s e t I i = 0 . T h i s p l a c e s t h e r e s i s t o r 10K

and 300 ohm i n s e r i e s a c r o s s t he o ut pu t . Ther e s u l t a n t c i r c u i t i s shown i n f i g . 3 . 5 4 . )

7 disp ( )8 disp ( S te p 4 : R ep la ce FET wi th i t s e q u i v a l e n t

c i r c u i t a s shown i n f i g . 3 . 5 5 . )

9 disp ( )10 disp ( S te p 5 : Find open l oo p t r a n s f e r g ai n . )11 disp ( Av = Vo / Vs = A v1 A v 2 )12 disp ( A v2 = u R L2 / R L2+ r d )13 r l 2 = ( 1 0 . 3 * 2 2 ) / ( 1 0 . 3 + 2 2 ) // i n kohm

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14 format (3)

15 disp ( r l 2 , wher e R L2 ( i n kohm) = )16 a v 2 = ( - 5 0 * 7 ) / 1 717 format (6)

18 disp ( a v 2 , A v2 = )19 disp ( A v1 = u R D e f f / r d +R D e f f +(1+u ) R s e f f )20 r d e f f = ( 2 2 * 1 0 0 0 ) / ( 2 2 + 1 0 0 0 ) // i n kohm21 disp ( r d e f f , R De f f ( i n kohm) = R D | | R G2 = )22 disp ( R s e f f = 330 | | 10K )23 a v 1 = ( - 5 0 * 2 1 . 5 3 ) / ( 1 0 + 2 1 . 5 3 + ( 5 1 * ( ( 0 . 3 3 * 1 0 ) / ( 1 0 + 0 . 3 3 ) ) )

)

24 disp ( a v 1 , T h er e fo r e , A v1 = )

25 a v = - 2 0 . 5 9 * - 2 2 . 5 126 disp ( a v , O v e r a l l Av = A v1 A v 2 = )27 disp ( )28 disp ( S te p 6 : C a l c u l at e b et a )29 b e t a = 3 3 0 / ( 3 3 0 + 1 0 0 0 0 )

30 format (7)

31 disp ( b e t a , b e ta = V f / Vo = Rs / Rs+Rf = )32 disp ( )33 disp ( s t ep 7 : C a l c u l at e D, A vf , R i f , R o f )34 d = 1 + ( 0 . 0 3 1 9 * 4 6 3 . 5 )

35 disp (d , D = 1 + Av b e t a =

)

36 a v f = 4 6 3 . 5 / 1 5 . 7 8 5

37 format (6)

38 disp ( a v f , A vf = Av / D = )39 disp ( Ri = R G = 1 Mohm )40 r i f = 1 5 . 7 8 5

41 format (7)

42 disp ( r i f , R i f ( i n kohm) = R i D =)43 r o = ( 1 0 * 7 ) / ( 1 0 + 7 ) // i n kohm44 format (6)

45 disp ( r o , R o ( i n kohm) = r d | | R L2 = )

46 r o f = ( 4 . 1 1 8 * 1 0 ^ 3 ) / 1 5 . 7 8 5 / / i n ohm47 format (4)


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Scilab code Exa 3.11 voltage gain and input and output resistance

1 / / E x am pl e 3 . 1 12 clc

3 disp ( S t e p 1 : I d e n t i f y t op ol og y )4 disp ( The f e e db ac k v o l t a g e i s a p p l i e d a c r o s s t h e

r e s i s t a n c e R e1 and i t i s i n s e r i e s w i t h i n p u ts i g n a l . Hence f ee db ac k i s v o l t a ge s e r i e s f ee db ac k. )

5 disp ( )6 disp ( s t e p 2 and S te p 3 : Fi nd i n pu t and o ut pu t

c i r c u i t . )7 disp ( To f i n d i n p u t c i r c u i t , s e t Vo = 0 (

c o n n e c t i n g C2 t o g ro un d ) , w hi ch g i v e s p a r l l e lc om bi na ti o n o f Re w it h Rf a t E1 . To f i n d o ut pu tc i r u i t , s e t I i = 0 ( o pe n in g t he i np ut node E1 a te m i t t e r o f Q1 ) , which g i v e s s e r i e s c om bi na ti on od

Rf and R e1 a c r o s s t he o ut pu t . The r e s u l t a n tc i r c u i t i s shown i n f i g . 3 . 5 7 )

8 disp ( )9 disp ( S te p 4 : Find open l oo p v o l t a g e g ai n ( Av ) )

10 r l 2 = ( 4 . 7 * 3 . 4 2 ) / ( 4 . 7 + 3 . 4 2 ) // i n kohm11 format (5)

12 disp ( r l 2 , R L2 ( i n ko h m ) = R c 2 | | ( Rs+R) = )13 disp ( A i 2 = h f e = 50 )14 disp ( R i 2 = h i e = 1 00 0 ohm = 1 kohm )15 a v 2 = - 5 0 * 1 . 9 8

16 format (3)

17 disp ( a v 2 , A v2 = A i 2 R L2 / R i 2 = )18 disp ( A i 1 = h f e = 50 )19 format (7)

20 r l 1 = ( ( 1 0 * 1 0 0 * 2 2 * 1 ) / ( ( 1 0 0 * 2 2 ) + ( 1 0 * 2 2 ) + ( 1 0 * 1 0 0 )

+ ( 1 0 * 1 0 0 * 2 2 ) ) ) * 1 0 ^ 3 / / i n ohm21 disp ( r l 1 , R L1 ( i n ohm ) = R c1 | | R3 | | R4 | | R i 2

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= )

22 disp ( R i 1 = h i e + (1+ h f e ) R e 1 e f f )23 r e 1 = 1 + ( 5 1 * ( ( 3 . 3 * 0 . 1 2 ) / ( 3 . 4 2 ) ) ) // i n kohm24 format (4)

25 disp ( r e 1 , wher e R e 1 e f f ( i n kohm) = Rs | | R = )26 a v 1 = ( - 5 0 * 8 6 5 . 4 6 ) / 6 9 0 0

27 format (5)

28 disp ( a v 1 , A v1 = A i 1 R L1 / R i 1 = )29 disp ( The o v e r a l l v o l t a g e g ai n , )30 a v = - 6 . 2 7 * - 9 9

31 format (7)

32 disp ( a v , Av = A v1 A v 2 = )

33 disp ( )34 disp ( S te p 5 : C a l c u l at e b et a )35 b e t a = 1 2 0 / ( 1 2 0 + 3 3 0 0 )

36 format (6)

37 disp ( b e t a , b e ta = Vf / Vo = Rs / Rs+R = )38 disp ( )39 disp ( S te p 6 : C a l c u l at e D, A vf , R i f , R of and R

o f )40 d = 1 + ( 0 . 0 3 5 * 6 2 0 . 7 3 )

41 format (7)

42 disp (d , D = 1 + Av b e t a =

)

43 a v f = 6 2 0 . 7 3 / 2 2 . 7 2 5

44 format (5)

45 disp ( a v f , A vf = Av / D = )46 r i f = 6 . 9 * 2 2 . 7 2 5 // i n kohm47 format (6)

48 disp ( r i f , R i f ( i n ko h m ) = R i 1 D = )49 disp ( R of = Ro / D = i n f i n i t y )50 r o f = ( 1 . 9 8 * 1 0 ^ 3 ) / 2 2 . 7 2 5 / / i n ohm51 disp ( r o f , R o f ( i n ohm ) = R o / D = R L2 / D = )

Scilab code Exa 3.12 Ai and beta and Aif

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1 / / E x am pl e 3 . 1 2

2 clc3 disp ( S t e p 1 : I d e n t i f y t op ol og y )4 disp ( The f e eb da ck i s g i v en from e m i t te r o f Q2 t o

t h e b a se o f Q2 . I f I o = 0 t h e n f ee db ac k c u r r e n tt h r o u g h 5 K r e g i s t e r i s z e r o , h e n c e i t i s c ur r e nt

s am pl in g . As f ee db ac k s i g n a l i s mixed i n s hu ntw i t h i n p u t , t h e a m p l i f i e r i s c u rr e n t s h u n tf e ed b ac k a m p l i f i e r . )

5 disp ( )6 disp ( S te p 2 and S te p 3 : Fi nd i np ut and o ut p ut )7 disp ( The i np u t c i r c u i t o f t h e a m p l i f i e r w i t h o u t

f ee db ac k i s o b ta i ne d by o pe ni ng t he o ut pu t l oo pa t t he e m i t t e r o f Q2 ( I o = 0 ) . T h is p l a c e s R ( 5 K) i n s e r i e s w i t h Re f rom b as e t o e m i t te r o f Q1 .The o ut p u t c i r c u i t i s f o u nd by s h o rt i n g t he i np ut

node , i . e . making Vi = 0 . T hi s p l a c e s R ( 5 K)i n p a r a l l e l w i t h Re . The r e s u l t a n t e q ui v a l e n tc i r c u i t i s shown i n f i g . 3 . 5 9 )

8 disp ( )9 disp ( S t e p 4 : Find open c i r c u i t t r a n s f e r g ai n . )

10 disp ( A I = I o / I s = Ic / I b2 I b 2 / I c 1 I c 1 /

I b 1 I b1 / Is )

11 disp ( We know th at I c 2 / I b 2 = A i2 = h f e = 50and )

12 disp ( I c 1 / I b 1 = A i1 = h f e = 5 0 )13 disp ( I c 1 / I b 1 = 50 )14 disp ( L oo ki ng a t f i g . 3 . 5 9 we ca n w ri te , )15 disp ( I b 2 / I c 1 = R c1 / R c1+R i 2 )16 r i 2 = 1 . 5 + ( 5 1 * ( ( 5 * 0 . 5 ) / ( 5 . 5 ) ) ) // i n kohm17 format (8)

18 disp ( r i 2 , wher e R i 2 ( i n kohm ) = h i e + (1+ h f e ) (R e2 | | R ) = )

19 x 1 = - 2 / ( 2 + 2 4 . 6 8 1 8 )20 disp ( x 1 , I b 2 / I c 1 = )21 disp ( I b 1 / I s = R / R+R i 1 where R = Rs| | ( R

+ R e 2 ) )22 r = ( ( 1 * 5 . 5 ) / ( 1 + 5 . 5 ) ) * 1 0 ^ 3 / / i n ohm

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23 format (9)

24 disp (r , T h e r e f o r e , R( i n ohm ) = )25 disp ( and R i 1 = h i e + (1+ h f e ) R e1 = 1 6 . 8 kohm)

26 x 1 = 8 4 6 . 1 5 3 8 / ( 8 4 6 . 1 5 3 8 + ( 1 6 . 8 * 1 0 ^ 3 ) )

27 format (8)

28 disp ( x 1 , T he r e f o re , I b1 / I s = )29 a i = 5 0 * 0 . 0 7 4 9 5 * 5 0 * 0 . 0 4 7 9 5

30 format (7)

31 disp ( a i , A I = )32 disp ( )33 disp ( S te p 5 : C a l c u l at e b et a )

34 b e t a = 5 0 0 / ( 5 0 0 + ( 5 * 1 0 ^ 3 ) )35 disp ( b e t a , b e t a = I f / I o = R e2 / R e2| R = )36 disp ( )37 disp ( S te p 6 : C a l c u l at e D , A I f )38 d = 1 + ( 0 . 0 9 0 9 * 8 . 9 8 4 8 )

39 disp (d , D = 1 + A I b e t a = )40 a i f = 8 . 9 8 4 8 / 1 . 8 1 6 8

41 disp ( a i f , A I f = A I / D = )

Scilab code Exa 3.13 beta and Av and Avf and Rif and Rof

1 / / E x am pl e 3 . 1 32 clc

3 disp ( S t e p 1 : I d e n t i f y t op ol og y )4 disp ( The f ee db ac k v o lt a g e i s a p pl i ed a c r o s s R1

( 15 0 ohm ) , which i s i n s e r i e s w it h i np ut s i g n a l .Hence f ee db a ck i s v o l t a g e s e r i e s f ee db a ck . )

5 disp ( )

6 disp ( S te p 2 and S te p 3 : Fi nd i n pu t and o ut pu tc i r c u i t )7 disp ( To f i n d i np ut c i r c u i t , s e t Vo = 0 , which

g i v e s p a r a l l e l c om bi na ti on o f R1 w i t h R2 a t E1 a sshown i n t h e f i g . 3 . 6 1 . To f i n d o ut p u t c i r c u i t ,

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s e t I i = 0 by o pe ni ng t he i np ut node , E1 a t

e m i t t e r o f Q1 , whi ch g i v e s t he s e r i e s c om bi na ti ono f R2 a nd R1 a c r o s s t he o ut pu t . The r e s u l t a n tc i r c u i t i s shown i n f i g . 3 . 6 1 . )

8 disp ( )9 disp ( S te p 4 : Find t he open l oo p v o l t ag e g ai n ( Av ) )

10 r l 2 = ( 4 . 7 * 1 5 . 1 5 ) / ( 4 . 7 + 1 5 . 1 5 ) // i n kohm11 format (5)

12 disp ( r l 2 , RL2 ( i n kohm) = )13 disp ( S i n c e ho e = h re = 0 , we ca n u se a pp ro xi ma te

a n a l y s i s . )14 disp ( A i 2 = h f e = 500 )

15 disp ( R i 2 = h i e = 1 1 00 ohm )16 a v 2 = ( - 5 0 0 * 3 . 5 9 * 1 0 ^ 3 ) / 1 1 0 0

17 disp ( a v 2 , A v2 = A i 2 R L2 / R i 2 = )18 r l 1 = ( ( 1 0 * 4 7 * 3 3 * 1 . 1 ) / ( ( 4 7 * 3 3 * 1 . 1 ) + ( 1 0 * 3 3 * 1 . 1 )

+ ( 1 0 * 4 7 * 1 . 1 ) + ( 1 0 * 4 7 * 3 3 ) ) ) * 1 0 ^ 3 / / i n ohm19 disp ( r l 1 , R L1 ( i n ohm ) = 10K | | 47K | | 33K | | R i 2

= )20 disp ( A i 1 = h f e = 500 )21 r i 1 = 1 . 1 + ( 5 0 1 * ( ( 0 . 1 5 * 1 5 ) / ( 0 . 1 5 + 1 5 ) ) ) // i n kohm22 disp ( r i 1 , R i 1 ( i n kohm ) = h i e + (1+ h f e ) Re = )23 a v 1 = ( - 5 0 0 * 9 4 2 ) / ( 7 5 . 5 * 1 0 ^ 3 )

24 format (6)

25 disp ( a v 1 , A v1 = A i 1 R L1 / R i 1 = )26 a v = - 6 . 2 3 8 * - 1 6 3 2

27 disp ( a v , Av = A v1 A v 2 = )28 disp ( )29 disp ( S te p 5 : C a l c u l a te b et a and D )30 b e t a = 1 5 0 / ( 1 5 0 + 1 5 0 0 0 )

31 format (7)

32 disp ( b e t a , b e t a = R1 / R1+R2 = )33 d = 1 + ( 1 0 1 8 0 * 0 . 0 0 9 9 )

34 format (8)35 disp (d , D = 1 + A b e t a = )36 disp ( )37 disp ( S te p 6 : C a l c u l at e A vf , R of and R i f )38 a v f = 1 0 1 8 0 / 1 0 1 . 7 8 2

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39 format (4)

40 disp ( a v f , A vf = Av / D = )41 r i f = 7 5 . 5 * 1 0 1 . 7 8 2 * 1 0 ^ - 3 // i n Mohm42 format (6)

43 disp ( r i f , R i f ( i n Mo h m ) = R i 1 D = )44 r o f = ( 3 . 5 9 * 1 0 ^ 3 ) / 1 0 1 . 7 8 2

45 disp ( r o f , R o f ( i n ohm ) = Ro / D = R L2 / D = )

Scilab code Exa 3.14 gain and new bandwidth

1 / / E x am pl e 3 . 1 42 clc

3 disp ( Given : A vmid = 5 00 , f L = 10 0 kHz , f H = 2 0kHz and b e t a = 0 . 0 1 )

4 a v f = 5 0 0 / ( 1 + ( 0 . 0 1 * 5 0 0 ) )

5 format (6)

6 disp ( a v f , A vf = A vmid / 1+ be ta A vmid = )7 f l f = 1 0 0 / ( 1 + ( 0 . 0 1 * 5 0 0 ) ) // i n Hz8 disp ( f l f , f L f ( i n Hz ) = f L / 1+ b et a A vmid = )9 f h f = 2 0 * ( 1 + ( 0 . 0 1 * 5 0 0 ) ) // i n kHz

10 disp ( f h f , f H f ( i n kHZ ) = f H ( 1 + b e t a A v m i d ) = )

11 b w = 1 2 0 - 0 . 0 1 6 6 7 / / i n kHZ12 format (9)

13 disp ( b w , BW f ( i n kHz ) = f H f f L f = )

Scilab code Exa 3.15 show voltage gain with feedback

1 / / E x am pl e 3 . 1 52 clc

3 disp ( S t e p 1 : I d e n t i f y t op ol og y )4 disp ( By s h o r t i n g o ut pu t ( Vo = 0 ) , f e ed b ac k v o l t a g e

d o es n ot become z e r o . By o p en i ng t h e o u tp ut l o o p

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f ee db a ck becomes z e ro and h en ce i t i s c u r re n t

s am pl in g . The f ee db ac k i s a p p l i e d i n s e r i e s w i tht he i np ut s i g n al , h en ce t o po l og y u se d i s c u r r e n ts e r i e s f e ed b ac k . )


c i r c u i t . )7 disp ( To f i n d i n p u t c i r c u i t , s e t I o = 0 . T h i s

p l a c e s Re i n s e r i e s w it h i np ut . To f i n d o ut pu tc i r c u i t I i = 0 . T h i s p l ac e s Re i n o u t p u t s i d e .The r e s u l t a n t c i r c u i t i s shown i n f i g . 3 . 6 3 . )

8 disp ( )

9 disp ( St e p 4 : R e p l a c e t r a n s i s t o r w i t h i t s hp ar am et er e q u i v al e n t a s shown i n f i g . 3 . 6 4 . )

10 disp ( )11 disp ( S te p 5 : Find open l oo p t r a n s f e r g ai n . )12 disp ( From q ua t i on ( 5 ) o f s e c t i o n 3 . 9 . 1 we h ave )13 disp ( A v f = I o R L / Vs = G Mf R L )14 disp ( = h f e R L / R s+ h i e +(1+ h f e ) Re )15 disp ( H er e R s = Rs | | R1 | | R2 )16 disp ( = Rs | | Rb b e c a u s e R b = R1

| | R2 )17 disp (

T h e r e fo r e , Vo / Vs = Vo/ Vi Vi/Vs)

18 disp ( where Vi / Vs = Rb / Rs+Rb )19 disp ( T h er ef or e , Vo / Vs = ( h f e R L / R s+ h i e

+(1+ h f e ) Re ) ( Rb / Rs+Rb) )20 disp ( D i v i d i n g b ot h n u me ra t or and d e n om i n at o r by Rs+

Rb we ge t , )21 disp ( A v f = Vo / Vs = [h f eRc (Rb/Rb+Rs ) ] / R

s+ h i e +(1+ h f e ) Re b e c a us e RL = Rc )22 disp ( = h f eRc [ 1 / 1 + ( R s /Rb ) ] / R s+ h i e

+(1+ h f e ) Re )

Scilab code Exa 3.16 GMf and Rif and Rof

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1 / / E x am pl e 3 . 1 6

2 clc3 disp ( R e f e r e xa mp le 3 . 1 5 )4 disp ( A v f = h f eRc [ 1 / 1 + ( R s /Rb ) ] / R s+ h i e

+(1+ h f e ) Re where R s = Rs| |R1| |R2 )5 a v f = ( - 5 0 * ( 1 . 8 * 1 0 ^ 3 ) * [ 1 / ( 1 + ( 1 0 0 0 / 4 2 7 2 ) ) ] )

/ ( 8 1 0 + 1 0 0 0 + ( ( 1 + 5 0 ) * 1 0 0 0 ) )

6 format (5)

7 disp ( a v f , A v f = )8 g m f = - 1 . 3 8 / ( 1 . 8 * 1 0 ^ 3 )

9 format (9)

10 disp ( g m f , G Mf = A v f / R L = )

11 disp ( b e t a = Vf / I o = I eRe / I o = I o Re / I o =Re = 1 K )

12 disp ( G Mf = G M / 1+ b et a G M )13 g m = 1 / ( ( 1 / ( - 7 . 6 6 * 1 0 ^ - 4 ) ) + 1 0 0 0 )

14 format (10)

15 disp ( g m , T h e r e fo r e , G M = )16 d = 1 + ( - 1 0 0 0 * - 3 . 2 7 3 5 * 1 0 ^ - 3 )

17 format (7)

18 disp (d , D = 1 + G M b e t a = )19 r i = ( 1 + 1 . 3 6 ) // i n kohm20 format (5)

21 disp ( r i , R i ( i n kohm) = Rs+( h i e +Re ) | | R D = )22 r i f = 2 . 3 6 * 4 . 2 7 3 5 // i n kohm23 format (3)

24 disp ( r i f , R i f ( i n ko h m ) = R i D = )25 disp ( R o = i n f i n i t y )26 disp ( R o f = R o D = i n f i n i t y )27 disp ( R o f = R o f | | R L = R L = 1 . 8 kohm )

Scilab code Exa 3.17 feedback factor and Rif and Rof

1 / / E x am pl e 3 . 1 72 clc

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3 disp ( R e f e r e xa mp le 3 . 1 5 )

4 disp ( A v f = h f eRc [ 1 / 1 + ( R s /Rb ) ] / R s+ h i e+(1+ h f e ) Re where R s = Rs| |R1| |R2 )5 a v f = ( - 5 0 * ( 4 * 1 0 ^ 3 ) * [ 1 / ( 1 + ( 1 0 0 0 / 9 0 0 0 ) ) ] )

/ ( 9 0 0 + 1 0 0 0 + ( ( 1 + 1 5 0 ) * 1 0 0 0 ) )

6 format (6)

7 disp ( a v f , A v f = )8 g m f = - 1 . 1 7 7 / ( 4 * 1 0 ^ 3 )

9 format (9)

10 disp ( g m f , G Mf = A v f / R L = )11 disp ( b e t a = Vf / I o = I eRe / I o = I o Re / I o =

Re = 1 K )

12 disp ( G Mf = G M / 1+ b et a G M )13 g m = 1 / ( ( 1 / ( - 2 . 9 4 3 * 1 0 ^ - 4 ) ) + 1 0 0 0 )

14 format (9)

15 disp ( g m , T h e r e fo r e , G M = )16 d = 1 + ( - 1 0 0 0 * - 4 . 1 7 * 1 0 ^ - 4 )

17 format (6)

18 disp (d , D = 1 + G M b e t a = )19 r i = 1 + ( ( 2 * 9 ) / ( 2 + 9 ) ) // i n kohm20 disp ( r i , R i ( i n kohm) = Rs+( h i e +Re ) | | R D = )21 r i f = 2 . 6 3 6 * 1 . 4 1 7 // i n kohm22 format (6)

23 disp ( r i f , R i f ( i n ko h m ) = R i D = )24 disp ( R o = i n f i n i t y )25 disp ( R o f = R o D = i n f i n i t y )26 disp ( R o f = R o f | | R L = R L = 4 kohm )

Scilab code Exa 3.18 gain with feedback and new bandwidth

1 / / E xam ple 3 . 1 8 .2 clc3 disp ( Given : A v mid = 4 0 , f L = 1 00 Hz , f H = 1 5

kHz and b e t a = 0 . 0 1 )4 a v f = 4 0 0 / ( 1 + ( 0 . 0 1 * 4 0 0 ) )

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5 format (3)

6 disp ( a v f , A vf = A v mid / 1+ be ta A v m id = )7 f l f = 1 0 0 / ( 1 + ( 0 . 0 1 * 4 0 0 ) )8 disp ( f l f , f L f = f L / 1+ b e t a A v m id = )9 f h f = ( 1 5 ) * ( 1 + ( 0 . 0 1 * 4 0 0 ) ) // i n kHz

10 disp ( f h f , f H f ( i n kHz ) = f H ( 1+ be t a A v mid ) =)

11 b w = 7 5 - 0 . 0 2 // i n kHz12 format (6)

13 disp ( b w , BW f ( i n kHz ) = f H f f L f = )

Scilab code Exa 3.19 overall voltage gain and bandwidth

1 / / E x am pl e 3 . 1 92 clc

3 disp ( Gi v e n : Av = 10 , BW = 1 1 0 3 , n =3 )4 disp ( ( i ) O v er a l l v o l t ag e g ai n )5 disp ( The g ai n o f c as ca de d a m p l i f i e r w it ho ut

f e e d b a c k = 1 0 1 0 1 0 = 1 0 0 0 )6 a v f = 1 0 0 0 / ( 1 + ( 0 . 1 * 1 0 0 0 ) )

7 format (4)8 disp ( a v f , A v f = A v / 1 + A v b e t a = )9 disp ( ( i i ) B an dw id th o f c a s c a d e d s t a g e )

10 disp ( Bandwidth o f c as c ad e d a m p l i f i e r w it ho utf e e d b a c k )

11 b w = ( ( 1 * 1 0 ^ 6 ) * sqrt ( ( 2 ^ ( 1 / 3 ) ) - 1 ) ) * 1 0 ^ - 3 // i n kHz12 format (7)

13 disp ( b w , BW( ca sc ad e ) ( in kHz ) = BWs q r t ( 2 ( 1 / n ) 1 ) = )

14 b w f = ( 5 0 9 . 8 2 * 1 0 ^ 3 * ( 1 + ( 0 . 1 * 1 0 0 0 ) ) ) * 1 0 ^ - 6 / / i n MHz

15 format (6)16 disp ( b w f , BW f ( i n MHz ) = BW ( 1 + b e t a A v mid ) = )

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Chapter 4

Oscillators

Scilab code Exa 4.1 C and hfe

1 / / E xa mp le 4 . 12 clc

3 disp ( R e f e r i n g t o e q u a t i o n ( 1 ) , )4 r i = ( 2 5 * 5 7 * 1 . 8 ) / ( ( 5 7 * 1 . 8 ) + ( 2 5 * 1 . 8 ) + ( 2 5 * 5 7 ) ) // i n k

ohm5 format (6)

6 disp ( r i , R i ( i n kohm) = R1 | | R2 | | h i e = )7 disp ( Now R i + R3 = R )8 r 3 = 7 . 1 - 1 . 6 3 1 // i n kohm9 format (5)

10 disp ( r 3 , T h e re fo r e , R3 ( i n kohm) = R R i = )11 k = 2 0 / 7 . 1

12 format (6)

13 disp (k , K = R C / R = )14 disp ( Now f = 1 / 2p i RCs q r t (6+4K) )15 c = ( 1 / ( sqrt ( 6 + ( 4 * 2 . 8 1 6 ) ) * 2 * % p i * 7 . 1 * 1 0 * 1 0 ^ 6 ) ) * 1 0 ^ 1 2

// i n pF16 format (8)17 disp (c , T h e r e fo r e , C( i n pF ) = )18 disp ( h f e >= 4 K + 2 3 + 2 9 / K )19 h f e = ( 4 * 2 . 8 1 6 ) + 2 3 + ( 2 9 / 2 . 8 1 6 )

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20 format (7)

21 disp ( h f e , h f e >

= )

Scilab code Exa 4.2 frequency of oscillation

1 / / E xa mp le 4 . 22 clc

3 disp ( The g i v en v a l ue s a re , R = 4 . 7 kohm and C =0 . 4 7 uF )

4 f = 1 / ( 2 * % p i * sqrt ( 6 ) * ( 4 . 7 * 1 0 ^ 3 ) * ( 0 . 4 7 * 1 0 ^ - 6 ) ) // i nHz5 format (7)

6 disp (f , f ( i n Hz ) = 1 / 2p i s q r t ( 6 )RC = )

Scilab code Exa 4.3 R and C

1 / / E xa mp le 4 . 3

2 clc3 disp ( f = 1 kHz )4 disp ( Now f = 1 / 2p i s q r t ( 6 )RC )5 disp ( Choos e C = 0 . 1 uF )6 r = 1 / ( sqrt ( 6 ) * 2 * % p i * 0 . 1 * 1 * 1 0 ^ - 3 ) / / i n ohm7 format (8)

8 disp (r , T h e r e fo r e , R( i n ohm ) = )9 disp ( Choose R = 680 ohm s ta nd ar d v al ue )

Scilab code Exa 4.4 C and RD

1 / / E xa mp le 4 . 42 clc

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3 disp ( U si n g t he e x p r e s s i o n f o r t he f r eq u en c y )

4 disp ( Now , f = 1 / 2 p i RCs q r t ( 6 ) )5 f = ( 1 / ( sqrt ( 6 ) * 2 * % p i * 9 . 7 * 5 * 1 0 ^ 6 ) ) * 1 0 ^ 9 // i n nF6 format (5)

7 disp (f , T h e r e fo r e , C( i n nF ) = )8 disp ( Now u s i n g t h e e q u a t i o n ( 2 7 ) )9 disp ( | A | = g m R L )

10 disp ( T h e r e f o r e , | A | >= 2 9 )11 disp ( T h er ef or e , g m R L >= 2 9 )12 r l = ( 2 9 / ( 5 0 0 0 * 1 0 ^ - 6 ) ) * 1 0 ^ - 3 // i n kohm13 format (4)

14 disp ( r l , T h er ef or e , R L ( i n kohm) >= 2 9 / g m = )

15 disp ( R L = R D r d / R D+r d )16 r d = ( 4 0 ) / 4 . 8 8 2 3

17 format (5)

18 disp ( r d , T h er e f o re , R D ( i n ko h m ) = )19 disp ( While f o r minimum v a lu e o f R L = 5 . 8 kohm )20 disp ( R D = 6 . 7 8 kohm )

Scilab code Exa 4.5 minimum and maximum R2

1 / / E xa mp le 4 . 52 clc

3 disp ( The f r e q u e n c y o f t h e o s c i l l a t o r i s g i v e n by , )4 disp ( f = 1 / 2p i sq r t ( R1R2C1C 2) )5 disp ( For f = 10 kHz , )6 r 2 = ( 1 / ( 4 * ( % p i ^ 2 ) * ( 1 0 0 * 1 0 ^ 6 ) * ( 1 0 * 1 0 ^ 3 ) * ( 0 . 0 0 1 * 1 0 ^ - 1 2 )

) ) // i n kohm7 format (6)

8 disp ( r 2 , T h e re fo r e , R2 ( i n kohm) = )

9 disp ( For f = 50 kHz , )10 r 2 = ( 1 / ( 4 * ( % p i ^ 2 ) * ( 2 5 0 0 * 1 0 ^ 6 ) * ( 1 0 * 1 0 ^ 3 )* ( 0 . 0 0 1 * 1 0 ^ - 1 2 ) ) ) // i n kohm

11 format (6)

12 disp ( r 2 , T h e re fo r e , R2 ( i n kohm) = )

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13 disp ( So minimum v a l ue o f R2 i s 1 . 0 1 3 kohm w h i l e

t h e maximum v a l u e o f R2 i s 2 5 . 3 3 kohm )

Scilab code Exa 4.6 range over capacitor is varied

1 / / E xa mp le 4 . 62 clc

3 disp ( The f r e q u e n c y i s g i v e n by , )4 disp ( f = 1 / 2p i s q r t (C L e q ) )

5 l e q = ( 2 * 1 0 ^ - 3 ) + ( 2 0 * 1 0 ^ - 6 )6 format (8)

7 disp ( l e q , wher e L eq = L1 + L2 = )8 disp ( For f = f max = 2050 kHz )9 format (5)

10 c = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 2 0 5 0 * 1 0 ^ 3 ) ^ 2 ) * 0 . 0 0 2 0 2 ) ) * 1 0 ^ 1 2 //i n pF

11 disp (c , T h e r e fo r e , C( i n pF ) = )12 disp ( For f = f m in = 950 kHz )13 c = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 9 5 0 * 1 0 ^ 3 ) ^ 2 ) * 0 . 0 0 2 0 2 ) ) * 1 0 ^ 1 2 //

i n pF

14 format (6)15 disp (c , T h e r e fo r e , C( i n pF ) = )16 disp ( Hence C must be v a r i e d from 2 . 9 8 pF t o 1 3 . 8 9

pF , t o g e t t he r e q u i r e d f r e q ue n cy v a r i a t i o n . )


1 / / E xa mp le 4 . 7

2 clc3 disp ( The g i v e n v a l u e s a re , )4 disp ( L1 = 0 . 5 mH, L2 = 1 mH, C = 0 . 2 uF )5 disp ( Now f = 1 / 2p i s q r t (C L e q ) )6 l e q = 0 . 5 + 1 / / i n mH

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7 disp ( l e q , a nd L e q ( i n mH) = L 1 + L2 = )

8 f = ( 1 / ( 2 * % p i * sqrt ( 1 . 5 * 0 . 2 * 1 0 ^ - 9 ) ) ) * 1 0 ^ - 3 // i n kHz9 format (5)10 disp (f , T h e r e fo r e , f ( i n kHz ) = )


1 / / E xa mp le 4 . 82 clc

3 disp ( The e q u i v a l e n t c a p a c i t a n ce i s g i ve n by , )4 c e q = ( 1 5 0 * 1 . 5 * 1 0 ^ - 2 1 ) / ( ( 1 5 0 * 1 0 ^ - 1 2 ) + ( 1 . 5 * 1 0 ^ - 9 ) ) //

i n F5 format (12)

6 disp ( c e q , C eq ( i n F) = C1C2 / C1+C2 =)7 disp ( Now , f = 1 / 2 p i s q r t ( L C e q ) )8 f = ( 1 / ( 2 * % p i * sqrt ( 5 0 * 1 3 6 . 3 6 3 * 1 0 ^ - 1 8 ) ) ) * 1 0 ^ - 6 // i n

MHz9 format (6)

10 disp (f , f ( i n MHz) = )

Scilab code Exa 4.9 calculate C

1 / / E xa mp le 4 . 92 clc

3 disp ( The g i v e n v a l u e s a re , )4 disp ( L = 100 uH , C1 = C2 = C and f = 500

kHz )5 disp ( Now , f = 1 / 2p i s q r t ( L C e q ) )6 c e q = 1 / ( 4 * ( % p i ^ 2 ) * ( 1 0 0 * 1 0 ^ - 6 ) * ( ( 5 0 0 * 1 0 ^ 3 ) ^ 2 ) ) // i n

F7 format (11)

8 disp ( c e q , T h e r e f o r e , C e q ( i n F ) = )

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9 disp ( but C eq = C1C2 / C1+C2 and C1 = C2 = C

)10 disp ( T he re fo re , C eq = C / 2 )11 c = 1 . 0 1 3 2 * 2

12 format (6)

13 disp (c , T h e r e fo r e , C( i n nF ) = )

Scilab code Exa 4.10 series and parallel resonant freqency

1 / / E x am pl e 4 . 1 02 clc3 f s = ( 1 / ( 2 * % p i * sqrt ( 0 . 4 * 0 . 0 8 5 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 6 // i n

MHz4 format (6)

5 disp ( f s , ( i ) f s ( i n MHz) = 1 / 2p i s q r t ( LC) = )6 c e q = 0 . 0 8 5 / 1 . 0 8 5 // i n pF7 disp ( c e q , ( i i ) C e q ( i n pF ) = CC M / C+C M = )8 f p = ( 1 / ( 2 * % p i * sqrt ( 0 . 4 * 0 . 0 7 8 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 6 // i n

MHz ( t he a ns we r i n t ex tb o ok i s wrong )9 disp ( f p , T h e re fo r e , f p ( i n MHz) = 1 / 2p i s q r t ( L

C eq ) = )10 i n c = ( ( 0 . 8 9 9 - 0 . 8 5 6 ) / 0 . 8 5 6 ) * 1 0 0 // i n p e rc e n t ag e11 disp ( i n c , ( i i i ) % i n c re a s e = )12 q = ( 2 * % p i * 0 . 4 * 0 . 8 5 6 * 1 0 ^ 6 ) / ( 5 * 1 0 ^ 3 )

13 format (8)

14 disp (q , ( i v ) Q = o me ga s L / R = 2p i f sL / R = )

Scilab code Exa 4.11 series and parallel resonant freqency

1 / / E x am pl e 4 . 1 12 clc

3 disp ( C M = 2 pF )4 f s = ( 1 / ( 2 * % p i * sqrt ( 2 * 0 . 0 1 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 6 / / i n MHz

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5 format (6)

6 disp ( f s , Now f s ( i n MHz) = 1 / 2p i s q r t ( LC) = )7 c e q = ( 2 * 0 . 0 1 * 1 0 ^ - 2 4 ) / ( 2 . 0 1 * 1 0 ^ - 1 2 ) // i n F8 format (9)

9 disp ( c e q , C eq ( i n F) = C MC / C M+C = )10 f p = ( 1 / ( 2 * % p i * sqrt ( 2 * 9 . 9 5 * 1 0 ^ - 1 5 ) ) ) * 1 0 ^ - 6 / / i n MHz11 format (6)

12 disp ( f p , f p = 1 / 2p i s q r t ( L C eq ) = )13 disp ( So f s a n d f p v a l ue s a r e a lm os t same . )

Scilab code Exa 4.12 Varify Barkhausen criterion and find frequency ofoscillation

1 / / ex am pl e 4 . 1 2 .2 clc

3 disp ( From t h e g i v e n i n f o r m a t i o n we c an w r it e , )4 disp ( A = 16106/ j omega and b et a =

1 0 3 / [ 2 103+ j omega ] 2 )5 disp ( To v e r i f y t he B ar kh au sen c o n d i t i o n means t o

v e r i f y w he th er | A b e t a| = 1 a t a f re qu en cy f o r

w h i c h A b et a = 0 d e gr e e . Le t us e xp r e ss , A b e t ai n i t s r e c t a ng l u a r form . )

6 disp ( A b e t a = 16106103 / j omega [ 2 103 + jome ga ] 2 = 16109 / j omega [ 4 1 0 6 + 4 1 0 3 jomega+(j omega ) 2 ] )

7 disp ( = 16109 / j omega [ 4 1 0 6 + 4 1 0 3 j omegaomega 2 ] a s j2 = 1 )

8 disp ( = 16109 / 4 1 0 6j omega+4103j2 omega2j omega 3 ] )

9 disp ( = 16109 / j omega [4106 omega

2 ] [ ome ga2 4 1 0 3 ] )10 disp ( R a t i o n a l i s i n g t he d en om in at or f u n c t i o n we g et , )

11 disp ( A b e t a = 16109[ omega2 4 1 0 3 j omega [ 4106 omega 2 ] ] / [ [omega2 4103] j omega

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[ 4106 o m e g a 2 ] ] [ omega2 4 1 0 3 j omega

[ 4106 omega 2 ] ] )12 disp ( U si ng ( ab ) ( a +b ) = a 2 b 2 i n t h ede nomi nat or , )

13 disp ( A b e t a = 1 6 1 0 9 [ o m e g a 24103+ j omega [ 4106 omega 2 ] ] / [omega2 4 1 0 3 ] 2 [ jomega [ 4106 o m e g a 2 ] 2 )

14 disp ( A b e t a = 1 6 1 0 9 [ o m e g a 24103+ j omega [ 4106 omega 2 ] ] / 1 6 1 0 6 omega4 + omega2(4106 omega 2) 2 )

15 disp ( Now t o h a ve A b et a = 0 d eg re e , t he i m ag i na r yp ar t o f A b et a must be z e ro . T h is i s p o s s i b l e

when , )16 disp ( T h e r e fo r e , omega ( 4 1 0 6 o m e g a 2 ) = 0 )17 disp ( T h er e f o re , omega = 0 o r 4 1 0 6 ome ga2 =

0 )18 disp ( T h e r e fo r e , omega 2 = 41 0 6

N e gl e c ti n g z e r o v al ue o f f r eq u en c y )19 disp ( T h e re fo r e , omega = 2 1 0 3 r a d / s e c )20 disp ( At t h i s f r eq u e nc y | A b e t a| ca n be o b ta i ne d as ,

)21 disp ( | A b e t a | = 1 6 1 0 9 [ 4 1 0 3 o me ga 2 ] /

1 6 1 0 6 omega4+omega2[4 106 o m e g a 2 ] 2a t omega = 2 1 0 3 )22 a b = ( 2 . 5 6 * 1 0 ^ 2 0 ) / ( 2 . 5 6 * 1 0 ^ 2 0 )

23 disp ( a b , | A b e t a| = )24 disp ( T h e r e fo r e , At omega = 2 1 0 3 r ad / s e c , A b e t a

= 0 d eg r e e a s i ma gi na ry p ar t i s z er o w hi l e | Ab e t a| = 1 . Thus B a rk ha u se n C r i t e r i o n i s s a t i s f i e d. )

25 disp ( The f r e q u e n c y a t w hi ch c i r c u i t w i l l o s c i l l a t ei s t he v al u e o f omega f o r which | A b e t a| = 1 a ndA b et a = 0 d e g re e a t t he same t im e )

26 disp ( i . e . omega = 2 1 0 3 r ad / s e c )27 disp ( But omega = 2p i f )28 f = ( 2 * 1 0 ^ 3 ) / ( 2 * % p i ) // i n Hz29 format (9)

30 disp (f , T h e r e fo r e , f ( i n Hz ) = o mega / 2 p i = )

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Scilab code Exa 4.13 minimum and maximum values of R2

1 / / E x am pl e 4 . 1 32 clc

3 disp ( The f r e q u e n c y o f t h e o s c i l l a t o r i s g i v e n by , )4 disp ( f = 1 / 2p i sq r t ( R1R2C1C 2) )5 disp ( For f = 2 0 kHz , )6 r 2 = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 2 0 * 1 0 ^ 3 ) ^ 2 ) * ( 1 0 * 1 0 ^ 3 )

* ( ( 0 . 0 0 1 * 1 0 ^ - 6 ) ^ 2 ) ) ) * 1 0 ^ - 37 format (5)

8 disp ( r 2 , T h e re fo r e , R2 ( i n kohm) = )9 disp ( For f = 7 0 kHz , )

10 r 2 = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 7 0 * 1 0 ^ 3 ) ^ 2 ) * ( 1 0 * 1 0 ^ 3 )

* ( ( 0 . 0 0 1 * 1 0 ^ - 6 ) ^ 2 ) ) ) * 1 0 ^ - 3

11 format (6)

12 disp ( r 2 , T h e re fo r e , R2 ( i n kohm) = )13 disp ( So minimum v a l ue o f R2 i s 0 . 5 1 7 kohm w h i l e

t h e maximum v a l u e o f R2 i s 6 . 3 3 kohm )

Scilab code Exa 4.14 frequency of oscillation and minimum hfe

1 / / E xam ple 4 . 1 4 .2 clc

3 disp ( R = 6 kohm , C = 150 0 pF , R C = 18 kohm )4 k = 1 8 / 6

5 disp (k , Now K = R C / R = )6 disp (

T h e r e f o r e , f = 1 / 2p i RCs q rt (6+4K))

7 f = ( 1 / ( 2 * % p i * ( 6 * 1 0 ^ 3 ) * ( 1 5 0 0 * 1 0 ^ - 1 2 ) * sqrt ( 6 + 1 2 ) ) )

* 1 0 ^ - 3 / / i n kHZ8 format (6)

9 disp (f , f ( i n kHz ) = )

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10 h f e = ( 4 * 3 ) + 2 3 + ( 2 9 / 3 )

11 disp ( h f e , ( h f e ) m in = 4K + 2 3 + 2 9/K = )

Scilab code Exa 4.15 range of frequency of oscillation

1 / / E xam ple 4 . 1 5 .2 clc

3 format (6)

4 disp ( For a Wien b r i d g e o s c i l l a t o r , )

5 disp ( f = 1 / 2p i RC )6 f m = ( 1 / ( 2 * % p i * ( 1 0 0 * 1 0 ^ 3 ) * ( 5 0 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 3 // i nkHz

7 disp ( f m , T h e r e fo r e , f m ax ( i n kHz ) = )8 f m i = ( 1 / ( 2 * % p i * ( 1 0 0 * 1 0 ^ 3 ) * ( 5 0 0 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 3

9 disp ( f m i , and f m i n ( i n kHz ) = )10 f n = 3 1 . 8 3 + 5 0

11 disp ( f n , Now f n e w ( i n kHz ) = f m ax + 5 0 1 0 3 = )12 disp ( The c o r r es p o n d i ng R = R w it h an a d d i t i o n a l

r e s i s t a n c e R x i n p a r a l l e l )13 disp ( T h e r e f o r e , f = 1 / 2p i R C )

14 r = ( 1 / ( 2 * % p i * ( 5 0 * 1 0 ^ - 1 2 ) * ( 8 1 . 8 3 * 1 0 ^ 3 ) ) ) * 1 0 ^ - 3 // i nkohm

15 disp (r , T h e re fo r e , R ( i n kohm) = )16 rx =1/((1/3 8.89) -(1/100)) // i n kohm17 disp ( T h e re fo r e , R = R R x / R+R x )18 disp ( r x , T h er ef or e , R x ( i n kohm) =

i n p a r a l l e l w i t h 100 kohm )


1 / / E xam ple 4 . 1 6 .2 clc

3 format (6)

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4 disp ( F or a H a r t l e y o s c i l l a t o r t h e f r e q u e n c y i s

g i v e n by , )5 disp ( f = 1 / 2p i s q r t ( L e q C) where L e q= L1+L2 )

6 l e q = 2 0 + 5 / / i n mH7 disp ( l e q , T h e r e f o r e , L e q ( i n mH) = 20+5 = )8 f = ( 1 / ( 2 * % p i * sqrt ( 2 5 * 5 0 0 * 1 0 ^ - 1 5 ) ) ) * 1 0 ^ - 3 // i n kHz9 disp (f , T h e r e fo r e , f ( i n kHz ) = )

Scilab code Exa 4.17 gain of the transistor

1 / / E x am pl e 1 4 . 72 clc

3 disp ( For a H ar tl e y o s c i l l a t o r , )4 disp ( f = 1 / 2p i s q r t ( L e q C) where L e q

= L1 + L2 + 2M )5 l e q = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 1 6 8 * 1 0 ^ 3 ) ^ 2 ) * ( 5 0 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ 3

/ / i n mH6 format (6)

7 disp ( l e q , T h e r e f o r e , L e q ( i n mH) = )

8 l2= ((17.95*10^ -3) -(15*10^-3) -(5*10^-6)) *10^3 // i nmH

9 disp ( l 2 , T h e r e fo r e , L2 ( i n mH) = )10 h f e = ( ( 1 5 * 1 0 ^ - 3 ) + ( 5 * 1 0 ^ - 6 ) ) / ( ( 2 . 9 4 5 * 1 0 ^ - 3 ) + ( 5 * 1 0 ^ - 6 ) )

11 format (5)

12 disp ( h f e , Now h f e = L1+M / L 2+M = )

Scilab code Exa 4.18 new frequency and inductance

1 / / E x am pl e 4 . 1 82 clc

3 disp ( For a C o l p i t t s o s c i l l a t o r , )4 disp ( f = 1 / 2p i s q r t ( L C e q ) )

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5 disp ( wh ere C eq = C1C2 / C1+C2 b u t C1 = C2 =

0 . 0 0 1 uF )6 c e q = ( ( 0 . 0 0 1 * 1 0 ^ - 6 ) ^ 2 ) / ( 2 * ( 0 . 0 0 1 * 1 0 ^ - 6 ) ) // i n F7 format (7)

8 disp ( c e q , T h e r e fo r e , C eq ( i n F ) = )9 disp ( L = 510 6 H )

10 f = ( 1 / ( 2 * % p i * sqrt ( 2 5 * 1 0 ^ - 1 6 ) ) ) * 1 0 ^ - 6 / / i n MHz11 format (6)

12 disp (f , T h e r e f o r e , f ( i n MHz) = )13 disp ( Now L i s d o u bl e d i . e . 1 0 uH )14 f 1 = ( 1 / ( 2 * % p i * sqrt ( 5 0 * 1 0 ^ - 1 6 ) ) ) * 1 0 ^ - 6 / / i n MHz15 format (5)

16 disp ( f 1 , T h e r e f o r e , f ( i n MHz ) = )17 n f = 2 * 3 .1 8 3

18 format (6)

19 disp ( n f , New f r e q ue nc y ( i n MHz ) = 2 3 . 1 8 3 = )20 l = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 6 . 3 6 6 * 1 0 ^ 6 ) ^ 2 ) * ( 5 * 1 0 ^ - 1 0 ) ) ) * 1 0 ^ 6

// i n uH21 format (5)

22 disp (l , T h e r e fo r e , L ( i n uH ) = )

Scilab code Exa 4.19 new frequency of oscillation

1 / / E x am pl e 4 . 1 92 clc

3 disp ( For a Clapp o s c i l l a t o r , )4 disp ( f = 1 / 2p i s q r t ( LC 3) )5 disp ( where C3 = 63 pF )6 f = ( 1 / ( 2 * % p i * sqrt ( 3 1 5 * 1 0 ^ - 1 8 ) ) ) * 1 0 ^ - 6 / / i n MHz7 format (6)

8 disp (f , T h e r e f o r e , f ( i n MHz) = )

Scilab code Exa 4.20 R and hfe

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1 / / E x am pl e 4 . 2 0

2 clc3 disp ( R e fe r i n g t o e q u a ti o n ( 1 ) o f s e c t i o n 4 . 5 . 3 , t hei n p u t i mp ed an ce i s g i v e n by , )

4 disp ( R i = R1 | | R2 | | h i e )5 disp ( Now R1 = 25 kohm , R2 = 47 kohm , and h i e

= 2 kohm )6 format (7)

7 r i = ( 2 5 * 4 7 * 2 ) / ( ( 4 7 * 2 ) + ( 2 5 * 2 ) + ( 2 5 * 4 7 ) ) // i n kohm8 disp ( r i , T h e re fo r e , R i ( i n kohm) = )9 disp ( K = R C / R )

10 disp ( Now R C = 10 kohm . . . g i v e n )

11 disp ( Now f = 1 / 2p i RCs q r t (6+4K) )12 disp ( T h er ef or e , Rs q r t ( 6+4K) = 3 1 8 3 0 . 9 8 9 )13 disp ( Now K = R C / R = 1 0 1 0 3 / R )14 disp ( T h er ef or e , Rs q r t ( 6 + ( 4 0 1 0 103/R ) ) =

3 1 8 3 0 . 9 8 9 )15 disp ( T h e re fo r e , R 2 ( 6 + ( 4 0 1 0 103 /R) ) =

( 3 1 8 3 0 . 9 8 9 ) 2 )16 R = poly (0 , R )17 p 1 = 6 * R ^ 2 + ( 4 0 * 1 0 ^ 3 ) * R - ( 3 1 8 3 0 . 9 8 9 ) ^ 2

18 t1 = roots ( p 1 )

19 a n s 1 = t 1 ( 1 )

20 format (6)

21 disp ( ( - a n s 1 ) * 1 0 ^ - 3 , T h er e fo r e , R( i n kohm)=N e g le c t i n g n e ga t i ve v al ue )

22 k = 1 0 / 1 6 . 7 4

23 format (7)

24 disp (k , T h e re fo r e , K = R C / R = )25 disp ( T he re fo re , h f e >= 4 K + 2 3 + 2 9 / K )26 h f e = ( 4 * 0 . 5 9 7 3 ) + 2 3 + ( 2 9 / 0 . 5 9 7 3 )

27 format (6)

28 disp ( h f e , h f e >= )

Scilab code Exa 4.21 component values of wien bridge

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1 / / E x am pl e 4 . 2 1

2 clc3 disp ( The f r e q u e n c y i s g i v e n by , )4 disp ( f = 1 / 2p i RC )5 disp ( Le t t he r e s i s t a n c e v al ue t o be s e l e c t e d as , )6 disp ( R1 = R2 = R = 50 kohm )7 disp ( f = 1 / 2p i 5 0 1 0 3 C )8 f = ( 1 / ( 2 * % p i * ( 5 0 * 1 0 ^ 3 ) * 1 0 0 ) ) * 1 0 ^ 9 // i n nF9 format (6)

10 disp (f , f ( i n nF ) = )11 disp ( and f max = 1 / 2p i 5 0 1 0 3 C )12 c = ( 1 / ( 2 * % p i * ( 5 0 * 1 0 ^ 3 ) * ( 1 0 * 1 0 ^ 3 ) ) ) * 1 0 ^ 9 // i n pF

13 disp (c , C( i n nF ) = )14 disp ( Thus t o v ar y t he f r e qu e n cy fro m 1 00 Hz t o 10

kHz , t he c a p a c i t o r r an ge s ho ul d be s e l e c t e d a s0 . 3 1 8 nF t o 3 1 . 83 nF )

15 disp ( S i m i l a r l y k ee pi ng t he c a p a c i t or v al ue c on st an t, t h e r a n g e o f t h e r e s i s t a n c e v al u e s can beo b t a i n e d . )

Scilab code Exa 4.22 values of C2 and new frequency of oscillation

1 / / E x am pl e 4 . 2 22 clc

3 disp ( f = 2. 5 MHz, L = 10 uH , C1 = 0 . 02 uF )4 disp ( For C o l p i t t s o s c i l l a t o r , t h e f r e qu e n c y i s

g i v e n by , )5 disp ( f = 1 / 2 p i s q r t ( L C e q ) )6 c e q = ( 1 / ( 4 * ( % p i ^ 2 ) * ( ( 2 . 5 * 1 0 ^ 6 ) ^ 2 ) * ( 1 0 * 1 0 ^ - 6 ) ) ) * 1 0 ^ 1 2

// i n pF

7 format (8)8 disp ( c e q , T h e r e fo r e , C eq ( i n pF ) = )9 disp ( ( i ) But C eq = C1C2 / C1+C2)

10 c 2 = ( ( 0 . 0 2 * 1 0 ^ - 6 ) / 4 9 . 3 4 8 ) * 1 0 ^ 9 // i n nF11 format (7)

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12 disp ( c 2 , T h e r e fo r e , C2 ( i n nF ) = ) // a ns we r i n

t e xt b o ok i s wrong13 disp ( ( i i ) L = 21 0 = 2 0 u H )14 disp ( and C eq = 4 05 . 2 84 pF )15 f = ( 1 / ( 2 * % p i * sqrt ( 2 0 * 4 0 5 . 2 8 4 * 1 0 ^ - 1 8 ) ) ) * 1 0 ^ - 6 // i n

MHz16 disp (f , f ( i n MHz) = 1 / 2p i s q r t ( L C eq ) = )

Scilab code Exa 4.23 series and parallel resonant freqency and Qfactor

1 / / E xam ple 4 . 2 3 .2 clc

3 f = ( 1 / ( 2 * % p i * sqrt ( 0 . 3 3 * 0 . 0 6 5 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 6 // i nMHz

4 format (6)

5 disp (f , ( i ) f ( i n MHz) = 1 / 2p i s q r t ( LC) = )6 c e q = 0 . 0 6 5 / 1 . 0 6 5 // i n pF7 disp ( c e q , ( i i ) C e q ( i n pF ) = CC M / C+C M = )8 f p = ( 1 / ( 2 * % p i * sqrt ( 0 . 3 3 * 0 . 0 6 1 * 1 0 ^ - 1 2 ) ) ) * 1 0 ^ - 6 // i n

MHz

9 disp ( f p , ( i ) f p ( i n MHz) = 1 / 2p i s q r t ( L C eq ) =)

10 p i = ( ( 1 . 1 2 1 - 1 . 0 8 7 ) / 1 . 0 8 7 ) * 1 0 0 // i n p e rc e nt a ge11 disp ( p i , ( i i i ) % i n c r e a s e = )12 q = ( 2 * % p i * 1 . 0 8 7 * 0 . 3 3 * 1 0 ^ 6 ) / ( 5 . 5 * 1 0 ^ 3 )

13 format (8)

14 disp (q , ( i v ) Q = o meg a x L / R = 2p i f sL / R = )

Scilab code Exa 4.24 change in frequency and trimmer capacitance

1 / / E xa ml e 4 . 2 42 clc

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3 disp ( ( i ) Assume one p e r t i c u l a r c o up l i n g d i r e c t i o n

f o r w hi ch , )4 disp ( L e q = L1 + L2 + 2M = 0 . 2 5 mH )5 format (8)

6 f = ( 1 / ( 2 * % p i * sqrt ( 0 . 2 5 * 1 0 0 * 1 0 ^ - 1 5 ) ) ) * 1 0 ^ - 6 / / i n MHz7 disp (f , T h e r e fo r e , f ( i n MHz) = 1 / 2p i s q r t ( L e q C

) =)8 disp ( Le t t he d i r e c t i o n o f c o u p li n g i s r ev er se d , )9 disp ( L e q = L1 + L2 2M = 0. 15 mH )

10 f d = ( 1 / ( 2 * % p i * sqrt ( 0 . 1 5 * 1 0 0 * 1 0 ^ - 1 5 ) ) ) * 1 0 ^ - 6 // i nMHz

11 format (7)

12 disp ( f d , T h e re fo r e , f ( i n MHz) = 1 / 2p i s q r t (L e q C) = )

13 p c = ( ( 1 . 2 9 9 4 - 1 . 0 0 6 5 8 ) / 1 . 0 0 6 5 8 ) * 1 0 0 // i n p e rc e nt a ge14 format (6)

15 disp ( p c , T h e re fo r e , % c h an ge = f f / f 100 = )16 disp ( ( i i ) Le t us assume d i r e c t i o n o f c o u p li n g s uc h

that , )17 disp ( L e q = L1 + L2 + 2M = 0 . 2 5 mH )18 disp ( C t = Trim c a p ac i t or = 100 pF )19 di

analog and digital electronics_u. a. bakshi and a. p. godse

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