Integrated Circuits K. R. Botkar

Scilab Textbook Companion for Integrated Circuits by K. R. Botkar1 Created by Tushar Kashyap B.TECH Electronics Engineering Model Engineering College College Teacher Ms.Vineetha George E, Model Engineering College Cross-Checked by Ms. Vineetha George E August 10, 2013

1 Funded

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

Book Description Title: Integrated Circuits Author: K. R. Botkar Publisher: Khanna Publishers Edition: 5 Year: 2010 ISBN: 81-7409-208-0

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Scilab numbering policy used in this document and the relation to the above 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 particular Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.

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Contents List of Scilab Codes

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2 Thick Film And Thin Film Hybrid ICs

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3 Semoconductor Devices Fundamentals

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5 Monolithic Components

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7 Operational Amplifier Characteristics

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8 Applications of Operational Amplifier

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9 Active Filters

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10 Special Purpose Amplifiers

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11 Nonlinear Circuit Application

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12 Signal Generators

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13 Voltage Regulators

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15 Phase Locked Loops

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16 Bipolar and MOS Digital Gate Circuits

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17 Light Emitting Diodes and Liquid Crystal Displays

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List of Scilab Codes Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

2.1 2.2 2.3 2.4 2.6 2.8 2.9 2.10 2.11 3.2 3.3 3.4 3.5 3.6 5.1 5.2 5.3 5.4 7.1 7.2 7.4 7.5 7.14 7.15 7.16 7.17 8.1 8.2

Resistance . . . . . . . . . . . . . . . . Resistance Calculation . . . . . . . . . . Sheet Resistivity . . . . . . . . . . . . . Design Capacitor . . . . . . . . . . . . . Capacitance . . . . . . . . . . . . . . . Thickness . . . . . . . . . . . . . . . . . Length . . . . . . . . . . . . . . . . . . Absolute Coefficient . . . . . . . . . . . Ratio . . . . . . . . . . . . . . . . . . . Resistivity . . . . . . . . . . . . . . . . Resistivity of Intrinsic Ge . . . . . . . . Hole Concentration . . . . . . . . . . . Resistivity of Cu . . . . . . . . . . . . . Bipolar Transistor Parameters . . . . . Transit Time . . . . . . . . . . . . . . . Unit gain frequency . . . . . . . . . . . Resistance and Sheet resistance . . . . . Capacitance per unit area . . . . . . . . Bipolar Differential Amplifier Parameter Rc and Re . . . . . . . . . . . . . . . . Offset Voltage Change . . . . . . . . . . Temperature Coefficient . . . . . . . . . Effect on Output Voltage . . . . . . . . Slew rate and Fmax . . . . . . . . . . . Largest Amplitude . . . . . . . . . . . . Maximum allowable frequency . . . . . Device Temperature . . . . . . . . . . . Device Temperature . . . . . . . . . . . 4

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6 7 8 8 9 10 10 11 12 14 15 15 16 16 19 20 20 21 23 24 25 26 27 28 28 29 31 32

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

8.3 8.4 8.7 8.8 9.6 9.12 10.3 10.4 10.8 10.9 11.4 11.5 11.11 12.6 12.11 12.12 12.20 13.3 13.14 15.2 15.3 15.4 15.5 16.2 16.3 16.12 16.13 16.14 17.2

Device Temperature . . . . . . . . . . . . . . . . . Device Temperature . . . . . . . . . . . . . . . . . Output Voltage . . . . . . . . . . . . . . . . . . . . Vp and Vo . . . . . . . . . . . . . . . . . . . . . . Determine Q Fl and Fh . . . . . . . . . . . . . . . Unity gain frequency and Capacitor determination Class B Power Amplifier . . . . . . . . . . . . . . . Power Output . . . . . . . . . . . . . . . . . . . . LM4250 Parameters . . . . . . . . . . . . . . . . . Common Emitter Amplifier Parameters . . . . . . Time taken . . . . . . . . . . . . . . . . . . . . . . Rise Time . . . . . . . . . . . . . . . . . . . . . . . Design Peak Detector . . . . . . . . . . . . . . . . 555 Timer . . . . . . . . . . . . . . . . . . . . . . . Design . . . . . . . . . . . . . . . . . . . . . . . . . Generating pulse by 555 timer . . . . . . . . . . . Waveform Generator . . . . . . . . . . . . . . . . . Maximum Efficiency and Power . . . . . . . . . . . Inductor and Capacitor . . . . . . . . . . . . . . . Output Signal Frequency . . . . . . . . . . . . . . VCO and Phase detector . . . . . . . . . . . . . . Second Order Butterworth Filter . . . . . . . . . . Lock Range . . . . . . . . . . . . . . . . . . . . . . Noise Margin . . . . . . . . . . . . . . . . . . . . . Fanouts . . . . . . . . . . . . . . . . . . . . . . . . NMOS operating region . . . . . . . . . . . . . . . Power Dissipation . . . . . . . . . . . . . . . . . . AC Power . . . . . . . . . . . . . . . . . . . . . . . Viewing distance . . . . . . . . . . . . . . . . . . .

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32 33 33 34 36 36 38 39 40 41 44 44 45 47 48 48 50 52 53 55 56 57 58 60 61 62 63 64 66

Chapter 2 Thick Film And Thin Film Hybrid ICs

Scilab code Exa 2.1 Resistance 1 2 3

4

5 6 7 8 9 10 11 12 13 14

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : R e s i s t a n c e // Example2 . 1 : a ) A r e s i s t o r h a s an a s p e c t r a t i o o f 2 0 : 1 and s h e e t r e s i s t a n c e o f 200 ohm/ s q u a r e . Find out the value o f r e s i s t a n c e . // b ) Find o u t t h e number o f s q u a r e s c o n t a i n e d i n a 2 k i l o o h m r e s i s t o r whose s h e e t r e s i s t a n c e i s 200 ohm p e r s q u a r e . // S o l u t i o n : a ) clear ; clc ; function y = myfunction (x , z ) // y : r e s i s t a n c e , x : s h e e t r e s i s t a n c e =200ohm/ s q u a r e , z : a s p e c t r a t i o =20:1 y = x * z // s i n c e , r e s i a t a n c e =s h e e t r e s i s t a n c e disp ( ’ r e s i s t a n c e i s =”) d i s p ( ’ ohm ’ , y ) endfunction // S o l u t i o n : b ) 6

15 16

17 18 19 20 21

// we have t o f i n d number o f s q u a r e which i s t o f i n d aspect ratio . function s = myfunction1 (r , p ) // r : r e s i s t o r =2kohm ( o r 2 0 0 0 ohm ) , p : s h e e t r e s i s t a n c e =200 Ohm/ s q u a r e , s : number o f s q u a r e ( a s p e c t r a t i o ) s = r / p // s i n c e , number o f s q u a r e =( r e s i s t o r / s h e e t resistance ) disp ( ’ number o f s q u a r e s a r e =”) // i n c l u d e ” ; ” a t t h e time o f c a l l i n g the f u n c t i o n d i s p ( ’ squares ’ , s ) endfunction // m y f u n c t i o n ( 2 0 0 , 2 0 / 1 ) ; m y f u n c t i o n 1 ( 2 0 0 0 , 2 0 0 ) ;

Scilab code Exa 2.2 Resistance Calculation 1 2 3

4 5 6 7

8 9

10 11 12

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : R e s i s t a n c e c a l c u l a t i o n // Example2 . 2 : A t h i c k f i l m r e s i s t o r i s s c r e e n e d w i t h a p a s t e o f f s h e e t r e s i s t i v i t y 1 0 0 0 0 ohm/ s q u a r e , and t h e r e s i s t o r i s d e f i n e d a s 0 . 2 4 cm l o n g and 0 . 0 6 cm w i d e . C a l c u l a t e t h e r e s i s t a n c e R . // S o l u t i o n : clear ; clc ; function R = myfunction3 (p ,l , w ) // r : r e s i s t o r , p=s h e e t r e s i s t a n c e =10000 ohm/ s q u a r e , l : l e n g t h o f r e s i s t o r =0.24cm , w : w i d t h o f t h e r e s i s t o r =0.06cm R = p *( l / w ) // s i n c e , r e s i s t a n c e =s h e e t r e s i s t a n c e ∗ ( l e n g t h o f r e s i s t o r / width o f the r e s i s t o r ) disp ( ’ r e s i s t a n c e o f t h e t h i c k f i l m r e s i s t o r i s = ’ ) // a t t h e t i m e o f c a l l i n g t h e f u n c t i o n include ”;” after it disp ( ’ ohm ’ ,R ) endfunction // m y f u n c t i o n 3 ( 1 0 0 0 0 , 0 . 2 4 , 0 . 0 6 ) ; 7

Scilab code Exa 2.3 Sheet Resistivity 1 2 3

4 5 6 7

8

9 10 11 12

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : S h e e t R e s i s t i v i t y // Example2 . 3 : C a l c u l a t e t h e s h e e t r e s i s t i v i t y o f a square o f t h i c k f i l m r e s i s t o r m a t e r i a l with the f o l l o w i n g p r o p e r t i e s : b u l k r e s i s t i v i t y =10ˆ−1ohm− cm and t h i c k f i l m t h i c k n e s s =10 m i c r o m e t e r . // S o l u t i o n : clear ; clc ; function Ps = myfunction4 (p , t ) // Ps : s h e e t r e s i s t a n c e , p : b u l k r e s i s t i v i t y o f t h i c k f i l m =10ˆ−1( o r 0 . 0 1 ) , t : t h i c k n e s s o f t h i c k f i l m =10 m i c r o m e t e r (=10∗10ˆ −4=0.001) Ps = p / t // s i n c e , s h e e t r e s i s t a n c e o f t h e f i l m= bulk r e s i s t a n c e of the f i l m / t h i c k n e s s of the film disp ( ’ s h e e t r e s i s t i v i t y i s = ’ ) // i n c l u d e ” ; ” a t a t the time o f c a l l i n g the f u n c t i o n disp ( ’ ohm p e r s q u a r e ’ , Ps ) endfunction // m y f u n c t i o n 4 (10ˆ −1 ,10∗10ˆ −4) ;

Scilab code Exa 2.4 Design Capacitor 1 2 3

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : D e s i g n C a p a c i t o r // Example2 . 4 : D e s i g n a c i r c u l a r 100 pF c a p a c i t o r w i t h the thick f i l m d i e l e c t r i c having d i e l e c t r i c f i l m t h i c k n e s s =0.02mm( o r 0 . 0 0 2 cm ) , assume Er =100 8

4 5 6 7 8 9 10

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// S o l u t i o n : We have t o f i n d t h e r a d i u s o f c r c u l a r c a p a c i t o r i n o r d e r to design i t . clear ; clc ; function r1 = myfunction5 (c , t ) r1 = c * t // c o n s t a n t=c a p a c i t o r ∗ t h i c k n e s s o f t h i c k film Er =100 // g i v e n r e l a t i v e p e r m e a b i l i t y o f t h i c k film r = sqrt ( r1 /( Er * %pi *8.85*10^ -12) ) // r a d i u s o f c i r c u l a r c a p a c i t o r , Eo =8 ,85∗10ˆ −12( d i e l e c t r i c constant of f r e e space ) disp ( ’ r a d i u s o f c i r c u l a r c a p a c i t o r i s = ’ ) // i n c l u d e ” ; ” at the time o f c a l l i n g the function disp ( ’ m e t e r ’ ,r ) endfunction // m y f u n c t i o n 5 ( 1 0 0 ∗ 1 0 ˆ − 1 2 , 0 . 0 0 2 ) ;

Scilab code Exa 2.6 Capacitance 1 2 3

4 5 6 7 8

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : C a p a c i t a n c e // Example2 . 6 : Find o u t t h e c a p a c i t a n c e o f a t h i c k f i l m c a p a c i t o r , i f t h e d i e l e c t r i c c o n s t a n t Er =100 , d i e l e c t r i c f i l m t h i c k n e s s =25 m i c r o m e t e r and a r e a A = 0 . 0 6 2 5 cm s q u a r e . // S o l u t i o n : clear ; clc ; function c = capacitance ( Er ,A , t ) c =8.8*10^ -12* Er * A /(10^ -12* t ) // c a p a c i t a n c e v a l u e w i l l be 2.2∗10ˆ −10 o r 220 pF , Eo : d i e l e c t r i c c o n s t a n t o f f r e e s p a c e =8.8∗10ˆ −12 , Er : d i e l e c t r i c c o n s t a n t o f t h i c k f i l m= 1 0 0 ( g i v e n ) 9

9 10

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, A : a r e a o f t h i c k f i l m = 0 . 0 6 2 6 cm s q u a r e ( o r 0 . 0 6 2 5 ∗ 1 0 ˆ − 4m s q u a r e ) , t : t h i c k n e s s o f t h e t h i c k f i l m= 25∗10ˆ −6m) // c a p a c i t a n c e=Eo∗ Er ∗A/ t disp ( ’ c a p a c i t a n c e i s = ’ ) // c = 2 . 2 0 0D −10(=2.2∗10ˆ −10)F , i n c l u d e ” ; ” a t l a s t a t t h e time o f c a l l i n g the f u n c t i o n disp ( ’ pF ’ ,c ) // pF : p i c o Farad endfunction // c a p a c i t a n c e ( 1 0 0 , 0 . 0 6 2 5 ∗ 1 0 ˆ − 4 , 2 5 ∗ 1 0 ˆ − 6 ) ;

Scilab code Exa 2.8 Thickness 1 2 3

4 5

6 7 8 9 10

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : T h i c k n e s s // Example2 . 8 : The b u l k r e s i s t i v i t y o f n i c h r o m i s 120 uohm−cm . C a l c u l a t e t h e t h i c k n e s s T i n a n g s t r o m s o f a f i l m w i t h s h e e t r e s i s t i v i t y o f 100ohm/ s q u a r e . // S o l u t i o n : function T = thickness ( Ps , p ) // Ps : s h e e t r e s i s t i v i t y o f n i c h r o m =100ohm/ s q u a r e , p : b u l k r e s i s t i v i t y o f n i c h r o m =120uohm−cm T = p /( Ps *10^ -8) // s i n c e Ps=p/T and 1 a n g s t r o m =10ˆ−8cm , s o d i v i d i n g by 10ˆ −8 h e r e disp ( ’ t h i c k n e s s i s = ’ ) // i n c l u d e ” ; ” a t t h e t i m e of ca ll im g the function disp ( ’ a n g s t r o m ’ ,T ) endfunction // t h i c k n e s s ( 1 0 0 , 1 2 0 ∗ 1 0 ˆ − 6 ) ;

Scilab code Exa 2.9 Length 10

1 2 3

4 5 6 7

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// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : Length // Rxample2 . 9 : C a l c u l a t e t h e l e n g t h o f a 400ohm t h i n f i l m r e s i s t o r . Given a s h e e t r e s i s t i v i t y o o f 100 ohm/ s q u a r e and a r e s i s t o r w i d t h o f 100um // S o l u t i o n : clear ; clc ; function L = extent (r ,w , Ps ) // L : l e n g t h o f t h i n f i l m , r : r e s i s t a n c e o f t h i n f i l m =400ohm , w : w i d t h o f r e s i s t o r =100um , Ps : s h e e t r e s i s t a n c e =100ohm/ s q u a r e L = r * w /(10^ -6* Ps ) // s i n c e , r=Ps ∗L/w and l e n g t h i n m i c r o m e t e r s o d i v i d i n g by 10ˆ −6. disp ( ’ l e n g t h o f t h i n f i l m i s = ’ ) // i n c l u d e ” ; ” a t the time o f c a l l i n g the f u n c t i o n at l a s t disp ( ’ m i c r o m e t e r ’ ,L ) endfunction // e x t e n t ( 4 0 0 , 1 0 0 ∗ 1 0 ˆ − 6 , 1 0 0 ) ;

Scilab code Exa 2.10 Absolute Coefficient 1 2 3

4 5 6 7

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : A b s o l u t e C o e f f i c i e n t // Example2 . 1 0 : A t h i n f i l m r e s i s t o r m e a s u r e s 150 ohmat 25 d e g r e e c e l c i u s and 1 5 1 . 5 ohm a t 100 d e g r e e c e l ci u s . Calculate i t s absolute c o e f f i c i e n t of r e s i s t a n c e i n p a r t s p e r m i l l i o n ( ppm ) p e r d e g r e celcius . // S o l u t i o n : clear ; clc ; function TCR = absresistor ( Rt1 , Rt2 , T1 , T2 ) // TCR : absolute temperature c o e f f i c i e n t of resist ance , Rt1 : r e s i s t a n c e a t 100 d e g r e e c e l c i u s =150ohm , Rt2 : r e s i s t a n c e a t 25 d e g r e e c e l c i u s = 1 5 1 . 5 ohm , T1 : 11

8 9

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t e m p e r a t u r e =100 d e g r e e c e l c i u s , T2 : t e m p e r a t u r e =25 degree c e l c i u s TCR =( Rt2 - Rt1 ) *10^6/( Rt1 *( T1 - T2 ) ) disp ( ’ a b s o l u t e c o e f f i c i e n t o f r e s i s t a n c e i s =”) // i n c l u d e ” ; ” at the time o f c a l l i n g the f u n c t i o n at l a s t d i s p ( ’ ppm / degree Celsius ’ , TCR ) // ppm : p a r t p e r million endfunction // a b s r e s i s t o r ( 1 5 0 , 1 5 1 . 5 , 1 0 0 , 2 5 ) ;

Scilab code Exa 2.11 Ratio 1 2 3

4 5 6 7 8 9 10 11 12 13 14

// C h a p t e r 2 T h i c k Film and Thin Film H y b r i d I C s // C a p t i o n : R a t i o // Example2 . 1 1 : Two t h i n r e s i s t o r a r e m e a s u r e d a t 50 d e g r e e c e l c i u s and 100 d e g r e e c e l s i u s and a r e f o u n d t o have t h e f o l l o w i n g v a l u e s : // Temperatur ( d e g r e e C) Ra ( ohm ) Rb ( ohm ) // 50 50 100 // 100 51 102.1 // C a l c u l l a t e t h e r a t i o TCR i n ppm/ d e g r e e c e l c i u s . // S o l u t i o n : function TCR = ratio ( Rat1 , Rbt1 , Rat2 , Rbt2 , T1 , T2 ) TCR =( Rat2 / Rbt2 - Rat1 / Rbt1 ) *10^6/(( Rat1 / Rbt1 ) *( T1 T2 ) ) disp ( ’ r a t i o TCR i s =”) // i i n c l u d e ” ; ” a t t h e t i m e of c a l l i n g the f u n c t i o n at l a s t d i s p ( ’ ppm / degree Celsius ’ , TCR ) //ppm : p a r t p e r million endfunction // r a t i o ( 1 0 0 , 5 0 , 1 0 2 . 1 , 5 1 , 1 0 0 , 5 0 ) ; 12

13

Chapter 3 Semoconductor Devices Fundamentals

Scilab code Exa 3.2 Resistivity 1 2 3

4 5 6 7

8 9 10

11 12 13

// C h a p t e r 3 S e m o c o n d u c t o r D e v i c e s F u n d a m e n t a l s // C a p t i o n : R e s i s t i v i t y // Example3 . 2 : A Sample o f S i i s doped w i t h 1 0 ˆ 1 7 p h o s p h o r u s atoms / c u b i c cm . What i s i t s r e s i s t i v i t y ? Given Un=700 s q u a r e cm/v−s e c . // S o l u t i o n : clear ; clc ; function Res = resistivity (u , n ) // n : doped c o n c e n t r a t i o n =10ˆ17 atoms / c u b i c cm , u : m o b i l i t y o f e l e c t r o n s =700 s q u a r e cm/v−s e c . q =1.6*10^ -19 // q : c h a r g e Res =1/( q * u * n ) // s i n c e P i s n e g l e g i b l e . disp ( ’ r e s i s t i v i t y o f t h e s i doped w i t h n− d o p a n t i s : ’ ) // i n c l u d e ” ; ” a t t h e t i m e o f calling disp ( ’ ohm−cm ’ , Res ) endfunction // a f t e r e x e c u t i n g c a l l i n g r e s i t i v i t y ( u=700 and n 14

=10ˆ17) i . e . , resistivity (10ˆ17 ,700) ; 14 // R e s u l t : R e s i s t i v i t y o f t h e S i doped w i t h n−d o p a n t i s : 0 . 0 8 9 ohm−cm ( a p p r o x )

Scilab code Exa 3.3 Resistivity of Intrinsic Ge 1 2 3

4 5 6 7 8 9 10 11 12

// C h a p t e r 3 S e m o c o n d u c t o r D e v i c e s F u n d a m e n t a l s // C a p t i o n : R e s i s t i v i t y o f I n t r i n s i c Ge // Example3 . 3 : Find t h e r e s i s t i v i t y o f i n t r i n s i c Ge a t 300K . Given un =3900 , and up =1900 cmˆ2/N s e c . and n i = 2 . 5 ∗ 1 0 ˆ 1 3 cmˆ−3 f o r i n t r i n s i c Ge . // S o l u t i o n : function RES = resistivity ( un , up ) // un : e l e c t r o n c o n c e n t r a t i o n , up : h o l e c o n c e n t r a t i o n q =1.6*10^ -19; // i n coulumb ni =2.5*10^13; // c o n c e n t r a t i o n i n cmˆ−3 RES =1/( q * ni *( un + up ) ) // s i n c e n=p=n i disp ( ’ r e s i s t i v i t y o f i n t r i n s i c Ge i s : ’ ) disp ( ’ ohm−cm ’ , RES ) endfunction // r e s i s t i v i t y ( 3 9 0 0 , 1 9 0 0 ) ;

Scilab code Exa 3.4 Hole Concentration // C h a p t e r 3 S e m o c o n d u c t o r D e v i c e s F u n d a m e n t a l s // C a p t i o n : H o l e C o n c e n t r a t i o n // Example3 . 4 : A s e m i c o n d u c t o r i s doped w i t h a c o n c e n t r a t i o n o f 1 0 ˆ 1 7 atoms /cmˆ3 o f r s e n i c . What i s t h e e q u i l i b r i u m h o l e c o n c e n t r a t i o n p a t 300K . Given n i = 1 . 5 ∗ 1 0 ˆ 1 0 cmˆ−3 4 // S o l u t i o n : 5 clear ;

1 2 3

15

6 clc ; 7 function p = holeconcentration ( ni , Nd ) // n i= i n t r i n s i i c

8 9 10 11 12

c o n c e n t r a t i o n = 1 . 5 ∗ 1 0 ˆ 1 0 cmˆ −3 , Nd : d o n a r c o n c e n t r a t i o n ; s i n c e , Nd>>n i , s o Nd=n =10ˆ17 atoms /cm ˆ 3 . p = ni ^2/ Nd disp ( ’ h o l e c o n c e n t r a r t i o n a t 300K i s : ’ ) disp ( ’ p e r c u b i c cm ’ ,p ) endfunction // h o l e c o n c e n t r a t i o n ( 1 . 5 ∗ 1 0 ˆ 1 0 , 1 0 ˆ 1 7 ) ;

Scilab code Exa 3.5 Resistivity of Cu 1 2 3

4 5 6 7 8 9 10 11

// C h a p t e r 3 S e m o c o n d u c t o r D e v i c e s F u n d a m e n t a l s // C a p t i o n : R e s i s t i v i t y o f Cu // Example3 . 5 : The r e s i s t i v i t y o f m e t a l i s g i v e n by p =1/nqu , where n i s number o f e l e c t r o n s p e r c u b i c meter , u i s m o b i l i t y , a nd q i s e l e c t r o n i c c h a r g e . D e t e r m i n e t h e r e s i s t i v i t o f c o p p e r a t room t e m p e r a t u r e . Given n = 8 . 5 ∗ 1 0 ˆ 2 8 p e r c u b i c meter , u =3.2∗10ˆ −3 mˆ2/V−s e c , a t room t e m p e r a t u r e . // S o l u t i o n : q =1.6*10^ -19; n =8.5*10^28; u =3.2*10^ -3; p =1/( n * q * u ) ; disp ( ’ r e s i s t i v i t y o f t h e c o p p e r i s : ’ ) disp ( ’ ohm−m e t e r ’ ,p ) // 2 . 2 9 8D−08 means 2 . 2 9 8 ∗ 1 0 ˆ − 8

Scilab code Exa 3.6 Bipolar Transistor Parameters 1

// C h a p t e r 3 S e m o c o n d u c t o r D e v i c e s F u n d a m e n t a l s 16

2 3

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

// C a p t i o n : B i p o l a r T r a n s i s t o r P a r a m e t e r s // Example3 . 6 : D e t e r m i n e Cu , Ccs ,Gm, C1 , R1 , R0 and Ru f o r a b i p o l a r t r a n s i s i t o r . Given : I c =0.2 mA, Vcb =10V, Vcs =15V, Cuo =0.25 pF , C j e=1 pF , Ccso =1.5 pF , Bo =2000 , Tf =0.3 ns , n=2∗10ˆ−4 and Vo =0.55V f o r a l l junctions . // S o l u t i o n : clear ; clc ; Cuo =0.25; // c o l l e c t o r −b a s e d e p l e t i o n r e g i o n c a p a c i t a n c e i n p i c o Farad ( pF ) f o r z e r o b i a s Ccso =1.5 ; // c o l l e c t o r −s u b s t r a t e j u n c t i o n c a p a c i t a n c e i n p i c o Farad ( pF ) f o r z e r o b i a s q =1.6*10^ -19 ; // e l e c t r o n c h a r g e i n coulomb Ic =0.2 ; // c o l l e c t o r c u r r e n t i n ampere (A) k =8.6*10^ -5; // i n eV/K, where 1eV =1.6∗10ˆ −19 T =300; // a b s o l u t e t e m p e r a t u r e i n k e l v i n (K) Vcb =10 ; // f o r w a r d b i a s on t h e j u n c t i o n i n v o l t ( v ) Vcs =15 ; // c o l l e c t o r −s u b s t r a t e b i a s i n v o l t (V) Cje =1 ; // d e p l e t i o n r e g i o n c a p a c i t a n c e i n p i c o Farad ( pF ) Bo =200; // s m a l l s i g n a l c u r r e n t g a i n Tf =0.3; // t r a n s i t t i m e i n f o r w a r d d i r e c t i o n i n nano s e c o n d s ( nS ) n =2*10^ -4; // p r o p o r t i o n a l i t y c o n s t a n t f o r Ro and gm Vo =0.55; // b i a s v o l t a g e i n v o l t (V) Cu = Cuo / sqrt (1+( Vcb / Vo ) ) ; // c o l l e c t o r −b a s e capacitance disp ( ’ Cu i s : ’ ) disp ( ’ pF ’ , Cu ) Ccs = Ccso / sqrt (1+( Vcs / Vo ) ) ; // c o l l e c t o r −s u b s t r a t e capacitance disp ( ’ Ccs i s : ’ ) disp ( ’ pF ’ , Ccs ) gm = q * Ic /( k * T *1.6*10^ -19) ; // s i n c e k i s i n eV s o c o n v e r t i n g i t i n Coulomb / K e l v i n disp ( ’gm i s : ’ ) // t r a n s c o n d u c t a n c e o f t h e b i p o l a r t r a n s i s t o r here 17

28 disp ( ’mA/V ’ , gm ) 29 Cb = Tf * gm ; // d i f f u s i o n c a p a c i t a n c e i n p i c o Farad ( pF ) 30 C1 = Cb + Cje ; // s m a l l s i g n a l c a p a c i t a n c e o f b i p o l a r

transistor 31 disp ( ”C1 i s : ” ) 32 disp ( ”pF” , C1 ) 33 R1 = Bo / gm ; // s m a l l 34 35 36 37 38 39 40 41

s i g n a l input r e s i s t a n c e of bipolar transistor disp ( ’ R1 i s : ’ ) disp ( ’ k i l o Ohm ’ , R1 ) Ro =1/( n * gm ) ; // s m a l l s i g n a l o u t p u t r e s i s t a n c e disp ( ’ R0 i s ’ ) disp ( ’ k i l o Ohm ’ , Ro ) Ru =10* Bo * Ro /10^3; // c o l l e c t o r −b a s e r e s i s t a n c e disp ( ’ Ru i s : ’ ) disp ( ’ Mega Ohm ’ , Ru )

18

Chapter 5 Monolithic Components

Scilab code Exa 5.1 Transit Time 1 2 3

4 5 6 7 8 9 10 11 12 13

// C h a p t e r 5 M o n o l i t h i c Components // C a p t i o n : T r a n s i t Time // Example5 . 1 : A l a t e r a l pnp d e v i c e b a s e w i d t h i s 8 um and t h e d i f f u s i o n c o f f i c i e n t f o r b a s e r e g i o n i s 10 cmˆ2/ s e c . C a l c u l a t e t h e b a s e t r a n s i t t i m e and t h e u n i t y g a i n f r e q u e n c y . // S o l u t i o n : function T = transittime (W , D ) //W: b a s e w i d t h =8um ; D : b a s e d i f f u s i o n c o f f i c i e n t =10 s q cm/ s e c . T = W ^2/(2* D ) ; // s i n c e f ( t r a n s i t f r e q u e n c y r e s p o n s e ) =2∗D/ (Wˆ 2 ) disp ( ’ b a s e t r a n s i t t i m e i s : ’ ) disp ( ’ n s ’ ,T *10^9) // i n n a n o s e c o n d s ( n s ) F =1/(2* %pi * T ) // where F=u n i t y g a i n f r e q u e n c y =1/(2∗ %pi ∗ t r a n s i t t i m e ) disp ( ’ u n i t y g a i n f r e q u e n c y i s : ’ ) disp ( ’MHz ’ ,F /10^6) // i n Mega H e r t z endfunction // t r a n s i t t i m e ( ( 8 ∗ 1 0 ˆ − 6 ) ,1 0 ∗ 1 0ˆ − 4 ) ;

19

Scilab code Exa 5.2 Unit gain frequency 1 2 3

4

5 6 7 8 9 10 11 12 13 14 15 16 17 18

// C h a p t e r 5 M o n o l i t h i c Components // C a p t i o n : U n i t g a i n f r e q u e n c y // Example5 . 2 : a ) Find F l ( u n i t g a i n f r e q u e n c y ) f o r t h e l a t e r a l pnp d e v i c e . Assume d i f f u s i o n c o f f i c i e n t o f h o l e s i n t h e b a s e o f 0 . 5 s q cm/ s e c and b a s e w i d t h o f 10um . // b ) Find t h e Fs ( u n i t g a i n f r e q u e n c y ) f o r t h e s u b s t r a t e pnp d e v i c e . Assume D=20 sqcm / s e c and W=8 um . clear ; clc ; // a ) S o l u t i o n : f o r t h e l a t e r a l pnp d e v i c e Wl =10*10^ -4; // b a s e w i d t h i n m i c r o c e n t i m e t e r ( ucm ) Dl =0.5; // b a s e d i f f u s i o n c o f f i c i e n t i n s q cm/ s e c Fl =2* Dl /( Wl ^2) ; disp ( ’ u n i t g a i n f r e q u e n c y f o r l a t e r a l pnp d e v i c e i s ; ’ ) disp ( ’MHz ’ , Fl /10^6) // b ) S o l u t i o n : f o r s u b s t r a t e pnp d e v i c e Ws =8*10^ -4; // b a s e i n ucm Ds =20; // b a s e d i f f u s i o n c o f f i c i e n t i n s q cm/ s e c Fs = Ds /( Ws ^2) ; disp ( ’ u n i t g a i n f r e q u e n c y f o r s u b s t r a t e pnp d e v i c e is ; ’) disp ( ’MHz ’ , Fs /10^6)

Scilab code Exa 5.3 Resistance and Sheet resistance 1 2

// C h a p t e r 5 M o n o l i t h i c Components // C a p a t i o n : R e s i s t a n c e and S h e e t r e s i s t a n c e 20

3

4 5 6 7 8 9 10 11 12 13 14 15 16 17

// Example5 . 3 : a )A b a s e d i f f u s i o n l a y e r l e n g t h i s 100 um and i t ’ s w i d t h i s 10um . The s h e e t r e s i s t a n c e o f t h e l a y e r i s 100 ohm/ s q u a r e . C a l c u l a t e i t s resistance . // b ) C a l c u l a t e t h e s h e e t r e s i s t a n c e o f a 20um t h i c k , 5 ohm−cm ep−l a y e r . // a ) S o l u t i o n : L =100; // b a s e d i f f u s i o n l a y e r i n um W =10; // b a s e d i f f u s i o n w i d t h i n um Rs =100; // s h e e t r e s i s t a n c e i n ohm/ s q u a r e R = L * Rs / W ; disp ( ’ r e s i s t a n c e o f b a s e d i f f u s i o n l a y e r i s : ’ ) disp ( ’Ohm ’ ,R ) // b ) S o l u t i o n : Pe =5*10^ -2; // ep−l a y e r r e s i s t i v i t y i n ucm t =20*10^ -6; // t h i c k n e s s o f t h e l a y e r i n um Rse = Pe / t ; // s h e e t r e s i t i v i t y o f ep−l a y e r disp ( ’ s h e e t r e s i s t a n c e o f ep−l a y e r i s : ’ ) disp ( ’Ohm ’ , Rse )

Scilab code Exa 5.4 Capacitance per unit area 1 2 3

4 5 6 7 8 9 10

// C h a p t e r 5 M o n o l i t h i c Components // C a p t i o n : C a p a c i t a n c e p e r u n i t a r e a // Example5 . 4 : D e t e r m i n e t h e c a p a c i t a n c e p e r u n i t a r e a o f t h e 400 a r m s t r o n g g a t e o x i d e o f a MOSFET device r e l a t i v e permittivity of s i l i c o n dioxide =3.9. // S o l u t i o n : clear ; clc ; Eo =8.86*10^ -14; // p e r m i t t i v i t y o f f r e e s p a c e i n F/cm Er =3.9; // r e l a t i v e p e r m i t t i v i t y o f MOSFET d e v i c e t =0.4*10^ -5; // t h i c k n e s s o f t h e g a t e o x i d e i n cm Co = Eo * Er / t ; // s i n c e c a p o a c i t a n c e (C)= p e r m i t t i v i t y (E) ∗ 21

a r e a (A) / t h i c k n e s ( t ) ; s o C/A=e / t 11 disp ( ’ c a p a c i t a n c e p e r u n i t a r e a o f g a t e o x i d e i s : ’ ) 12 disp ( ’ F/cmˆ2 ’ , Co )

22

Chapter 7 Operational Amplifier Characteristics

Scilab code Exa 7.1 Bipolar Differential Amplifier Parameter 1 2 3

4 5 6 7 8 9 10 11 12 13 14

// C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : B i p o l a r D i f f e r e n t i a l A m p l i f i e r P a r a m e t e r // Example7 . 1 : The f o l l o w i n g s p e c i f i c a t i o n a r e g i v e n f o r the dual input , balanced output b i p o l a r diferential amplifier : // Rc =2.2 kOhm , Re=4/7kOhm , Rs=50 ohm , Vcc=10V, Vee=−10V and Bf=Bo =100. Assume Vbe =0.7V . // D e t e r m i n e // a ) I c q and Vceq // b ) D i f f e r e n t i a l −mode v o l t a g e f g a i n , and // c ) I n p u t and o u t p u t r e s i s t a n c e s . clear ; clc ; // a ) S o l u t i o n : \ Rc =2.2*10^3; // c o l l e c t o r r e s i s t a n c e i n one t r a n s i s t o r i n ohm Re =4.7*10^3; // e m i t t e r e s i t a n c e o f one t r a n s i s t o r i n ohm Rs =50; // b a s e o r s o u r c e r e s i t a n c e o f one t r a n s i s t o r 23

15 16 17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32 33 34 35

i n ohm Vcc =10; // c o l l e c t o r power s u p p l y i n V o l t Vee = -10; // n e g a t i v e e m i t t e r power s u p p l y Bf =100; // g a i n o f t h e t r a n s i s t o r Bo = Bf ; Vbe =0.7; // b a s e e m i t t e r v o l t a g e o f one t r a n s i a t o r Icq =( abs ( Vee ) - Vbe ) /(2* Re +( Rs / Bf ) ) ; Vceq = Vcc + Vbe - Rc * Icq ; // b ) S o l u t i o n : gm = Icq /(25*10^ -3) ; // where t r a n s c o n d u c t a n c e gm=I c /Vt , Vt=25mV a t room t e m p e r a t u r e , s o gm =I c /25 Ad = - gm * Rc ; // d i f f e r e n t i a l mode v o l t a g e g a i n I c q h e r e w i l l be t a k e n a s f o u n d a b o v e n o t a p p r o x i m a t e d t o a s g i v e n book // c ) S o l u t i o n : r = Bo / gm ; // i n p u t r e s i s t a n c e o f one t r a n s i s t o r Ri =2* r ; // d i f f e r e n t i a l mode i n p u t r e s i s t a n c e Ro = Rc ; // d i f f e r e n t i a l mode o u t p u t r e s i s t a n c e disp ( ’A ’ , Icq *10^3 , ’ o p e r a t i n g p o i n t c o l l e c t o r c u r r e n t ’) disp ( ’V ’ , Vceq , ’ c o l l e c t o r −to −e m i t t e r v o l t a g e i s : ’ ) disp ( abs ( Ad ) , ’ D i f f e r e n t i a l −mode v o l t a g e g a i n ’ ) disp ( ’ k i l o Ohm ’ , Ri /10^3 , ’ I n p u t R e s i s t a n c e ’ ) disp ( ’ k i l o Ohm ’ , Ro /10^3 , ’ Output R e s i s t a n c e ’ ) // Note : // v a l u e o f I c q i s t a k e n a s 0 . 0 0 0 9 8 9 3 A o r 0 . 9 8 9 3 mA n o t a p p r o x i m a t e d t o 0 . 9 8 mA

Scilab code Exa 7.2 Rc and Re 1 2 3

// C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : Rc and Re // Example7 . 2 : A b i p o l a r d i f f e r e n t a i l a m p l i f i e r u s e s a t r a n s i s t o r h a v i n g Bo=200 and b i a s e d a t I c q =100 uA . D e t e r m i n e Rc and Re s o t h a t a b s ( Ad ) =500 and 24

4 5 6 7 8 9 10 11 12 13 14 15 16

CMRR=80 dB . // S o l u t i o n : clear ; clc ; //CMRR i n dB i s e x p r e s s e d a s 20logCMRR , s o 80=20 logCMRR o r CMRR =10^(80/20) ; Icq =100*10^ -6; // c o l l e c t o r c u r r e n t Vt =25*10^ -3; // s t a n d a r d v a l u e o f t h r e s h o l d v o l t a g e a t room t e m p e r a t u r e gm = Icq / Vt ; Re = CMRR /(2* gm ) ; // s i n c e CMRR=2∗gm∗Re ( a p p r o x ) Ad =500; // a b s o l u t e v a l u e o f d i f f e r e n t i a l mode voltage gain Rc = - Ad / gm ; // C o l l e c t o r r e s i s t a n c e disp ( ’ Mega Ohm ’ , Re /10^6 , ’ e m i t t e r r e s i t a n c e ( Re ) o f bipolar differential amplifier is : ’) disp ( ’ K i l o Ohm ’ , abs ( Rc ) /10^3 , ’ c o l l e c t o r r e s i s t a n c e ( Rc ) o f b i p o l a r d i f f e r e n t i a l a m p l i f i e r i s : ’ )

Scilab code Exa 7.4 Offset Voltage Change 1 2 3

4 5 6 7 8

// C h a p t e r 7 O p e r a t i o n A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : O f f s e t V o l t a g e Change // Example7 . 4 : What i s t h e c h a n g e i n t h e o f f s e t voltage of a bipolar transistor amplifier for a d i f f e r e n c e o f 10V i n t h e c o l l e c t o r −to −e m i t t e r v o l t a g e and Va=250 V . Assume room t e m p e r a t u r e . // S o l u t i o n : clear ; clc ; Vt =25*10^ -3; // t h r e s h o l d v o l t a g e a t room t e m p e r a t u r e in Volt Va =250; // e a r l y v o l t a g e o f t h e b i p o l a r t r a n s i s t o r i n volt 25

9 10 11 12 13 14 15 16

17 18

deltaVce =1; // l e t u s assume 1V o f c h a n g e i n Vce ( c o l l e c t o r −to −e m i t t e r v o l t a g e ) deltaVos1 = Vt *( - deltaVce / Va ) ; disp ( ’mV ’ , abs ( deltaVos1 ) *10^3 , ’ c h a n g e i n o f f s e t v o l t a g e f o r 1 V c h a n g e i n Vce i s : ’ ) for i =1:1 , if i ==1 then deltaVce =10; // i n v o l t deltaVos = deltaVce * deltaVos1 ; disp ( ’mV ’ , abs ( deltaVos ) *10^3 , ’ c h a n g e i n o f f s e t voltage of bipolar transistor for 10V c o l l e c t o r −to −e m i t t e r v o l t a g e ( Vce ) difference is : ’) end end

Scilab code Exa 7.5 Temperature Coefficient 1 2 3

4 5 6 7

8 9 10

// C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : T e m p e r a t u r e C o e f f i c i e n t // Example7 . 5 : D e t e r m i n e t h e t e m p e r a t u r e c o e f f i c i e n t of the input o f f s e t voltage f o r the b i p o l a r d i f f e r e n t i a l a m p l i f i e r h a v i n g Vos =1.5 mV. What i s t h e p e r c e n t a g e c h a n g e i n t h e Vos p e r d e g r e e temperature change . // S o l u t i o n : clear ; clc ; // t e m p e r a t u r e c o f f i c i e n t o f t h e i n p u t o f f s e t voltage f o r the b i p o l a r d i f f e r e n t i a l a m p l i f i e r Vos i s =dVos /dT=Vos /T ; Vos =1.5*10^ -3; // i n p u t o f f s e t v o l t a g e f o r b i p o l a r differential transistor amplifier T =300; // a s s u m i n g room t e m p e r a t u r e TC = Vos / T ; // t e m p e r a t u r e c o f f i c i e n t o f Vos 26

// p e r c e n t a g e c h a n g e i n t h e Vos p e r d e g r e e t e m p e r a t u r e c h a n g e w i l l be g i v e n by a s f o l l o w : 12 PC =( TC / Vos ) *100; // p e r c e n t a g e c h a n g e (PC) i n t h e Vos per degree temperature change 13 disp ( ’ %per d e g r e e c e l c i u s ’ ,PC , ’ p e r c e n t a g e c h a n g e i n t h e Vos per degree temperature change i s : ’ ) 11

Scilab code Exa 7.14 Effect on Output Voltage 1 2 3

4 5 6 7 8 9 10 11 12 13 14

15 16

// C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : E f f e c t on Output V o l t a g e // Example7 . 1 4 : For t h e n o n i n v e r t i n g OP−Amp w i t h i n p u t r e s i s t a n c e R1 nad f e e d b a c k r e s i s t a n c e R2 f i n d t h e e f f e c t on o u t p u t v o l t a g e Vo b e c a u s e o f t h e common mode v o l t a g e Vcm when t h e i n p u t v o l t a g e Vs c h a n g e s by 1V . Given CMRR=70 dB . // S o l u t i o n : clear ; clc ; CMRR =70; //Common Mode R e j e c t i o n R a t i o i n dB // s i n c e CMRR=20∗ l o g (Vcm/Vdm) dB // s o Vdm=Vcm/ 1 0 ˆ (CMRR/ 2 0 ) // s i n c e o u t p u t v o l t a g e o f OP−Amp i s Vo=(R1+R2 ) ∗Vdm/ R1=(R1+R2 ) ∗Vcm/ ( R1 ∗ 1 0 ˆ (CMRR/ 2 0 ) ) R1 =100; // a s s u m i n g i n p u t r e s i s t a n c e s t a n d a r d v a l u e i n k i l o Ohm R2 =900; // a s s u m i n g f e e d b a c k r e s i s t a n c e s t a n d a r d v a l u e i n k i l o Ohm Vs =1; // c h a n g e i n i n p u t v o l t a g e g i v e n i n q u e s t i o n Vcm = Vs ; // s i n c e c h a n g e i n i n p u t v o l t a g e i s a p p l i e d t o n o n i n v e r t i n g i n p u t and t h r o u g h t h e f e e d b a c k t o t h e i n v e r t i n g i p u t o f t h e Op−Amp a s w e l l . Vo =( R1 + R2 ) * Vcm /( R1 *10^( CMRR /20) ) disp ( ’mV ’ , abs ( Vo ) *10^3 , ’ c h a n g e i n o u t p u t v o l t a g e due t o common mode V o l t a g e (Vcm) i s : ’ ) 27

17 18

// Note : // CMRR, Vdm, Vo may be o f e i t h e r p o l a r i t y . Here absolute value i s calculated

Scilab code Exa 7.15 Slew rate and Fmax 1 2 3

4 5 6 7 8 9 10 11 12 13 14 15 16

// C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : S l e w r a t e and Fmax // Example7 . 1 5 : For t y p e 741 Op−Amp f o l l o w i n g parameter are given . Quiescent c o l l e c t o r current I c =9.5 uA , Cc=30 pF . Peak a m p l i t u d e o f i n p u t v o l t a g e Vm=15V . // a ) D e t e r m i n e t h e s l e w r a t e // b ) D e t e r m i n e f u l l power bandwidth Fmax f o r t h e s l e w r a t e a s o b t a i n e d from p a r t ( a ) . clear ; clc ; // a ) S o l u t i o n : Ic =9.5*10^ -6; // o p e r a t i n g c o l l e c t o r c u r r e n t i n A Cc =30*10^ -12; // p a r a s i t i c c a p a c i t a n c e SlewRate =2* Ic / Cc ; disp ( ’V/ u s ’ , SlewRate /10^6 , ’ S l e w r a t e i s : ’ ) // b ) S o l u t i o n : Vm =15; // a m p l i t u d e o f i n p u t v o l t a g e i n V o l t Fmax = SlewRate /(2* %pi * Vm ) ; // f u l l power bandwidth disp ( ’ kHz ’ , Fmax /10^3 , ’ f u l l power bandwidth Fmax f o r t h e S l e w Rate o b t a i n e d a b o v e i s : ’ )

Scilab code Exa 7.16 Largest Amplitude 1 2

// C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : L a r g e s t A m p l i t u d e 28

3

4 5 6 7 8 9 10

11 12

// Example7 . 1 6 : An a m p l i f i e r h a s a 10 kHz s i n e w a v e i n p u t s i g n a l . Find t h e l a r g e s t a m p l i t u d e t h a t t h e o u t p u t o f t h e a m p l i f i e r can be , w i t h o u t d i s t o r t i o n owing t o s l e w r a t e l i m i t i n g . Given s l e w r a t e =0.5V/ u s e c . // S o l u t i o n : clear ; clc ; Fmax =10*10^3; // f r e q u e n c y o f s i n e w a v e i n p u t s i g n a l i n Hz SlewRate =0.5*10^6; // g i v e n i n q u e s t i o n i n V/ s e c Vm = SlewRate /(2* %pi * Fmax ) ; // S i n c e Fmax=s l e w r a t e / ( 2 ∗ %pi ∗Vm) disp ( ’V( peak ) ’ ,Vm , ’ l a r g e s t a m p l i t u d e t h a t t h e o u t p u t o f t h e a m p l i f i e r can be w i t h o u t d i s t o r t i o n owing to slew rate l i m i t a t i o n i s : ’ ) // Note : // c a l c u l a t e d a m p l i t u d e i s 7 . 9 5 7 7 V, which can be approximated to 8 V

Scilab code Exa 7.17 Maximum allowable frequency // C h a p t e r 7 O p e r a t i o n a l A m p l i f i e r C h a r a c t e r i s t i c s // C a p t i o n : Maximum a l l o w a b l e f r e q u e n c y // Example7 . 1 7 : When a low f r e q u e n c y s i n u s o i d a l waveform i s a p p l i e d t o an i n p u t o f t h e n o n i n v e r t i n g Op−Amp t h e a m p l i f i e r r e s p o n d s l i n e a r l y o v e r an o u t p u t r a n g e from −10V t o +10V . I f R1=R2 and t h e s l e w r a t e o f t h e a m p l i f i e r i s 50 V/u s e c , what i s t h e maximum a l l o w a b l e f r e q u e n c y o f an i n p u t s i n u s o i d i f t h e o u t p u t s i g n a l s w i n g i s t o be m a i n t a i n e d from −10V t o +10V w i t h o u t d i s t o r t i o n ? r e s i s t a n c e and R2 i s f e e d b a c k resitance . 4 // S o l u t i o n 1 2 3

29

5 clear ; 6 clc ; 7 SlewRate =50/10^ -6; // i n V/ s e c 8 Vo =10 -( -10) ; // from q u e s t i o n o u t p u t 9 10 11 12

13 14

i s r a n g i n g from −10V t o +10V Vom = Vo ; // where Vom i s t h e maximum v a l u e o f Vo // t h e r e f o r e Fmax = SlewRate /(2* %pi * Vom ) ; disp ( ’ kHz ’ , Fmax /10^3 , ’ maximum a l l o w a b l e f r e q u e n c y o f an i n p u t s i n u s o i d a l f o r o u t p u t s w i n g m a i n t a i n e d from −10V t o +10v i s : ’ ) // Note : // o b t a i n e d maximum a l l o w a b l e a m p l i t u d e i s 3 9 7 . 8 8 7 3 6 kHz which can be a p p r o x i m a t e d t o 400 kHz

30

Chapter 8 Applications of Operational Amplifier

Scilab code Exa 8.1 Device Temperature 1 2 3

4 5 6 7 8 9 10 11 12

// C h a p t e r 8 A p p l i c a t i o n s o f O p e r a t i o n a l A m p l i f i e r // C a p t i o n : D e v i c e T e m p e r a t u r e // Example8 . 1 : The Heat g e n e r a t e d by a l i n e a r IC , uA 741 i s 200 mW. I f t h e t h e r m a l r e s i s t a n c e i s 150 d e g r e e C e l s i u s / Watt and t h e a m b i e n t t e m p e r a t u r e i s 25 d e g r e e c e l s i u s . c a l c u l a t e t h e d e v i c e temperature . // S o l u t i o n : clear ; clc ; Pd =200*10^ -3; // h e a t g e n e r a t e d Rt =150; // t h e r m a l r e s i s t a n c e Ta =25; // a m b i e n t t e m p e r a t u r e i n d e g r e e c e l s i u s // a s s u m i n g t h e r m a l e q u i l i b r i u m c o n d i t o n Td = Pd * Rt + Ta ; disp ( ’ d e g r e e c e l s i u s ’ ,Td , ’ The d e v i c e t e m p e r a t u r e i s : ’)

31

Scilab code Exa 8.2 Device Temperature 1 2 3

4 5 6 7 8 9 10 11 12

// C h a p t e r 8 A p p l i c a t i o n s o f O p e r a t i o n a l A m p l i f i e r // C a p t i o n : D e v i c e T e m p e r a t u r e // Example8 . 2 : For t h e d e v i c e i n Example8 . 1 , Pdmax =500 mW. D e t e r m i n e t h e d e v i c e t e m p e r a t u r e a f t e r e q u i l i b r i u m i s a t t a i n e d f o r an a m b i e n t t e m p e r a t u r e o f 75 d e g r e e c e l s i u s and i f t h e d e v i c e i s s u b j e c t e d t o maximum h e a t g e n e r a t i o n . Maximum a l l o w a b l e d e v i c e t e m p e r a t u r e i s 150 degree Celsius . // S o l u t i o n : clear ; clc ; Pmax =500*10^ -3; Pd = Pmax ; // s i n c e d e v i c e i s s u b j e c t e d t o maximum h e a t generation Rt =150; // t h e r m a l r e s i t a n c e Ta =75; // a m b i e n t t e m p e r a t u r e Td = Pd * Rt + Ta ; disp ( ’ d e g r e e c e l s i u s ’ ,Td , ’ d e v i c e t e m p e r a t u r e i s : ’ )

Scilab code Exa 8.3 Device Temperature // C h a p t e r 8 A p p l i c a t i o n s o f O p e r a t i o n a l A m p l i f i e r // C a p t i o n : D e v i c e T e m p e r a t u r e // Example8 . 3 : a ) The a m b i e n t t e m p e r a t u r e o f t h e d e v i c e o f Example8 . 2 r i s e s a b o v e 90 d e g r e e c e l s i u s . What i s t h e new v a l u e o f Td i f i t s t i l l g e n e r a t e s 500 mW? 4 // a ) S o l u t i o n : 5 clear ; 1 2 3

32

6 7 8 9 10 11

clc ; Pd =500*10^ -3; Rt =150; // t h e r m a l r e s i s t a n c e Ta =90; // a m b i e n t t e m p e r a t u r e Td = Pd * Rt + Ta ; disp ( ’ d e g r e e c e l s i u s ’ ,Td , ’ New v a l u e o f d e v i c e temperature i s : ’)

Scilab code Exa 8.4 Device Temperature 1 2 3

4 5 6 7 8 9 10 11

// C h a p t e r 8 A p p l i c a t i o n s o f O p e r a t i o n a l A m p l i f i e r // C a p t i o n : D e v i c e T e m p e r a t u r e // Example8 . 4 : F o r c e d a i r c o o l i n g p r o v i d e d f o r t h e d e v i c e i n Example8 . 3 l o w e r s t h e a m b i e n t t e m p e r a t u r e a t 60 d e g r e e c e l s i u s . What i s temperature of the device ? // S o l u t i o n : clear ; clc ; Pd =500*10^ -3; Rt =150; // t h e r m a l r e s i s t a n c e Ta =60; // a m b i e n t t e m p e r a t u r e Td = Pd * Rt + Ta ; disp ( ’ d e g r e e c e l s i u s ’ ,Td , ’ T e m p e r a t u r e o f t h e d e v i c e is : ’)

Scilab code Exa 8.7 Output Voltage 1 2 3

// C h a p t e r 8 A p p l i c a t i o n s o f O p e r a t i o n a l A m p l i f i e r // C a p t i o n : Output V o l t a g e // Example8 . 7 : I n t h e summing a m p l i f i e r ( i n v e r t i n g mode ) t h e s i g n a l s t o be combined a r e V1=3V, V2=2v , and V3=1V . The i n p u t r e s i s t o r a r e R1=R2=R3=3 33

4 5 6 7 8 9 10 11 12 13 14 15

k i l o ohm . The f e e d d b a c k r e s i s t o r Rf=1 k i l o ohm . C o n s i d e r i d e a l Op−Amp, d e t e r m i n e Vo . // S o l u t i o n : clear ; clc ; V1 =3; // i n p u t s i g n a l V2 =2; // i n p u t s i g n a l V3 =1; // i n p u t s i g n a l Rf =1*10^3; // f e e d b a c k r e s i t o r R1 =3*10^3; // i n p u t r e s i s t o r i n ohm R2 = R1 ; // i n p u t r e s i s t o r i n ohm R3 = R2 ; // i n p u t r e s i s t o r i n ohm Vo = -( Rf / R1 * V1 + Rf / R2 * V2 + Rf / R3 * V3 ) ; disp ( ’V ’ ,Vo , ’ Output V o l t a g e o f summing a m p l i f i e r i s : ’)

Scilab code Exa 8.8 Vp and Vo 1 2 3

4 5 6 7 8 9 10 11

// C h a p t e r 8 A p p l i c a t i o n s o f O p e r a t i o n a l A m p l i f i e r // C a p t i o n : Vp and Vo // Example8 . 8 : I n t h e c i r c u i t o f non−i n v e r t i n g summing Op−Amp, V1=+2V, V2=−4V, V3=+5V . i n p u t r e s i s t o r s f o r a l l t h e t h r e e i n p u t s i g n a l a r e same and a r e e q u a l t o 1 k i l o Ohm . The f e e d b a c k r e s i s t o r Rf i s 2 k i l o ohm . D e t e r m i n e t h e v o l t a g e Vp a t t h e n o n i n v e r t i n g p i n o f t h e Op−Amp and t h e o u t p u t Vo . Assume i d e a l Op=Amp . // S o l u t i o n : clear ; clc ; Rf =2*10^3; // f e e d b a c k r e s i s t o r R1 =1*10^3; R2 = R1 ; R3 = R2 ; V1 =2; 34

12 13 14 15 16 17 18

V2 = -4; V3 =5; n =3; // no o f i n p u t s Vp =( Rf / R1 * V1 + Rf / R2 * V2 + Rf / R3 * V3 ) / n ; Vo =(1+ Rf / R1 ) * Vp ; disp ( ’V ’ ,Vp , ’ v o l t a g e a t n o n i n v e r t i n g p i n i s : ’ ) disp ( ’V ’ ,Vo , ’ o u t p u t v o l t a g e v o l t a g e o f n o n i n v e r t i n g summing Op−Amp i s : ’ )

35

Chapter 9 Active Filters

Scilab code Exa 9.6 Determine Q Fl and Fh 1 2 3

4 5 6 7 8 9 10 11 12 13

// C h a p t e r 9 A c t i v e F i l t e r s // C a p t i o n : D e t e r m i n e Q F l and Fh // Example9 . 6 : A c e r t a i n two−p o l e band p a s s f i l t e r r e s p o n s e i s r e q u i r e d with a c e n t r e frequency o f 2 kHz and a 3 dB bandwidth o f 400 Hz . D e t e r m i n e Q, F l and Fh . // S o l u t i o n : clear ; clc ; Fo =2*10^3; // c e n t r e f r e q u e n c y i n Hz BW =400; // 3 dB bandwidth Q = Fo / BW ; // Q−f a c t o r o f band p a s s f i l t e r Fl = Fo * sqrt (1+1/(4* Q ^2) ) - Fo /(2* Q ) ; Fh = Fo * sqrt (1+1/(4* Q ^2) ) + Fo /(2* Q ) ; disp ( ’ Hz ’ ,Fl , ’ l o w e r c u t t o f f f r e q u e n c y i s : ’ ) disp ( ’ Hz ’ ,Fh , ’ H i g h e r c u t t o f f f r e q u e n c y i s : ’ )

Scilab code Exa 9.12 Unity gain frequency and Capacitor determination 36

1 2 3

4

5 6 7 8 9 10 11 12 13 14 15 16 17 18

// C h a p t e r 9 A c t i v e F i l t e r s // C a p t i o n : U n i t y g a i n f r e q u e n c y and C a p a c i t o r determination // Example9 . 1 2 : a ) D e t e r m i n e t h e u n i t y g a i n f r e q u e n c y , Fo , o f a s w i t c h e d c a p a c i t o r i n t e g r a t o r h a v i n g f o l l o w i n g s p e c i f i c a t i o n s : F c l k =1 kHz , C1=1 pF , and C2 =15.9 pF // b ) What i s t h e v a l u e o f c a p a c i t o r f o r an RC i n t e g r a t o r h a v i n g R=1.6 mega Ohm and Fo a s obtained in part ( a ) . // a ) S o l u t i o n : clear ; clc ; C1 =1*10^ -12; // s o u r c e c a p a c i t o r i n F C2 =15.9*10^ -12; // f e e d b a c k c a p a c i t o r Fclk =1*10^3; // c l o c k f r e q u e n c y o r s w i t c h i n g f r e q u e n c y Fo =1*( C1 / C2 ) * Fclk /(2* %pi ) ; disp ( ’ Hz ’ ,Fo , ’ u n i t y g a i n f r e q u e n c y i s : ’ ) // b ) S o l u t i o n : R =1.6*10^6; // r e s i s t o r o f RC i n t e g r a t o r i n Ohm C =1/(2* %pi * Fo * R ) ; disp ( ’ nF ’ ,C *10^9 , ’ f o r Rc i n t e g r a t o r v a l u e o f c a p a c i t o r needed i s : ’ ) // Note : // O b t a i n e d r e s u l t s a r e a p p r o x i m a t e d t o n e a r e s t v a l u e s , t h u s Fo=10 Hz and C=10 nF

37

Chapter 10 Special Purpose Amplifiers

Scilab code Exa 10.3 Class B Power Amplifier 1 2 3

4 5 6 7 8 9 10 11 12 13 14 15 16 17

// C h a p t e r 10 S p e c i a l P u r p o s e A m p l i f i e r s // C a p t i o n : C l a s s B Power A m p l i f i e r // Example10 . 3 : A c l a s s −B a u d i o power a m p l i f i e r h a s a s u p p l y v o l t a g e o f a b s ( Vcc ) =15V . The c l o s e d l o o p g a i n Av=50 and t h e a m p l i f i e r h a s t o d e l i v e r 10W o f power i n t o an 8 ohm l o a d . Find : // a ) t h e peak o u t p u t v o l t a g e s w i n g // b ) t h e peak o u t p u t c u r r e n t s w i n g // c ) t h e i n p u t s i g n a l r e q u i r e d ( rms ) // d ) t h e t o t a l power from t h e power s u p p l y // e ) t h e power d i s s i p a t e d i n t h e a m p l i f i e r // f ) t h e power c o n v e r s i o n e f f i c i e n c y clear ; clc ; // a ) S o l u t i o n : Po =10; // power i n Watt Rl =8; // l o a d r e s i s t a n c e i n Ohm ; Vorms = sqrt ( Po * Rl ) ; // s i n c e o u t p u t power Po=Vorms ˆ2/ Rl Vom = sqrt (2) * Vorms ; // peak o u t p u t v o l t a g e s w i n g disp ( ’V ’ , abs ( Vom ) , ’ The peak o u t p u t V o l t a g e s w i n g : ’ ) 38

18 // b ) S o l u t i o n : 19 Iom = Vom / Rl ; 20 disp ( ’A ’ , abs ( Iom ) , ’ The peak o u t p u t c u r r e n t s w i n g

is :

’) 21 // c ) S o l u t i o n : 22 Av =50; // c l o s e d l o o p g a i n 23 Vsrms = Vorms / Av ; 24 disp ( ’V ’ , Vsrms , ’ The i n p u t rms s i g n a l r e q u i r e d i s : ’ ) 25 // d ) S o l u t i o n : 26 Vcc =15; // a b s o l u t e v a l u e o f p o e r s u p p l y i n v o l t 27 Pin =2* Vcc * Iom / %pi ; // s i n c e I o r m s ∗ 2 ˆ ( 1 / 2 )=Iom 28 disp ( ’W’ ,Pin , ’ The t o t a l power from power s u p p l y i s : ’ 29 30 31 32 33 34 35 36

) // e ) S o l u t i o n : Pd =(2/ %pi ) * Vcc * sqrt (2* Po / Rl ) - Po ; disp ( ’W’ ,Pd , ’ The power d i s s i p a t e d i n t h e a m p l i f i e r is : ’) // f ) S o l u t i o n : n =( Po / Pin ) *100; disp ( ’% ’ ,n , ’ The power c o n v e r s i o n e f f i c i e n c y i s : ’ ) // Note : // Vcc , Vom and Iom can be o f e i t h e r p o l a r i t y but h e r e o n l y a b s o l u t e v a l u e i s c o n s i d e r e d and c a l c u l a t e d

Scilab code Exa 10.4 Power Output // C h a p t e r 10 S p e c i a l P u r p o s e A m p l i f i e r // C a p t i o n : Power Output // Example10 . 4 : For t h e a m p l i f i e r o f Example10 . 3 , f i n d t h e power o u t p u t l e v e l a t which t h e power d i s s i p a t i o n w i l l bw maximum and t h e maximum power dissipation . 4 // S o l u t i o n : 5 clear ; 6 clc ; 1 2 3

39

7 Vcc =15; // power s u p p l y i n v o l t 8 Rl =8; // l o a d r e s i s t a n c e i n ohm 9 // s i n c e Pd=2∗Vcc / %pi ∗ s q r t ( 2 ∗ Po/ Rl )−Po 10 // t o d e t e r m i n e t h e v a l u e o f Po a t which Pd i s

11 12 13 14 15

maximum we d i f f e r e n t i a t e a b o v e e q u a t i o n and e q u a t e t o z e r o , we f i n d Po a s Po =2* Vcc ^2/(( %pi ) ^2* Rl ) ; // t h e r e f o r e maximum power d i s s i p a t e d i s Pdmax =2* Vcc / %pi * sqrt (2* Po / Rl ) - Po ; disp ( ’W’ ,Po , ’ The power o u t p u t l e v e l f o r maximum power d i s s i p a t i o n i s : ’ ) disp ( ’W’ , Pdmax , ’ Maximum power d i s s i p a t i o n f o r c o r r e s p o n d i n g o u t p u t power l e v e l i s ; ’ )

Scilab code Exa 10.8 LM4250 Parameters 1 2 3

4 5 6 7 8 9 10 11 12 13

// C h a p t e r 10 S p e c i a l P u r p o s e A m p l i f i e r s // C a p t i o n : LM4250 P a r a m e t e r s // Example10 . 8 : The m i c r o p o w e r programmable Op−Amp LM 4 2 5 0 i s s u p p l i e d by 3 v s o u r s e ( a b s o l u t e v a l u e ) s o u r c e . Determine the value o f s e t r e s i s t o r f o r I s e t =0.1 uA i f R s e t i s c o n n e c t e d t o ( a ) Vee and ( b ) ground . ( c ) det er mi ne the q u i e s c e n t supply c u r r e n t and t h e q u i e s c e n t power d i s s i p a t i o n . clear ; clc ; // a ) S o l u t i o n : Vcc =3; // power s u p p l y i n V o l t Vee = - Vcc ; // n e g a t i v e power s u p p l y i n V o l t Iset =0.1*10^ -6; // b i a s s e t t i n g c u r r e n t i n A ; Rset =( Vcc + abs ( Vee ) -0.5) / Iset ; disp ( ’ mega Ohm ’ , Rset /10^6 , ’ The b i a s s e t t i n g c u r r e n t r e s i s t o r f o r Vee=−10 V i s : ’ ) // b ) S o l u t i o n : clear Vee ; 40

14 Vee =0; // s i n c e R s e t i s c o n n e c t e d t o g r o u n d 15 Rset =( Vcc + abs ( Vee ) -0.5) / Iset ; 16 disp ( ’ mega Ohm ’ , Rset /10^6 , ’ The b i a s s e t t i n g

current

r e s i s t o r f o r Vee=0 V i s : ’ ) // c ) S o l u t i o n : Qcurrent =5* Iset ; Qpower =( Vcc +3) * Qcurrent ; // where a b s ( Vee ) =3 V disp ( ’ uA ’ , Qcurrent *10^6 , ’ The q u i e s c e n t c u r r e n t supply i s : ’ ) 21 disp ( ’uW ’ , Qpower *10^6 , ’ The q u i e s c e n t power dissipated is : ’) 17 18 19 20

Scilab code Exa 10.9 Common Emitter Amplifier Parameters 1 2 3

4 5 6 7 8 9 10 11 12 13 14 15 16

// C h a p t e r 10 S p e c i a l P u r p o s e A m p l i f i e r s // C a p t i o n : Common E m i t t e r A m p l i f i e r P a r a m e t e r s // Example10 . 9 : A s i n g l e common e m i t t e r a m p l i f i e r h a s f o l l o w i n g d e v i c e and c i r c u i t p a r a m e t e r s : Rb=60 Ohm, Rs=40 Ohm, Cu=1.5 pF , Cl=1 pF , f t =1.6 GHz a t I c =2.5 mA q u i e s c e n t c u r r e n t . D e t e r m i n e e a c h o f t h e f o l l o w i n g f o r two v a l u e s o f Rl : 30 Ohm and 100 Ohm . a ) f 1 b ) F2 ( c )BW ( d ) Avmid ( e ) avmid ∗Bw . clear ; clc ; Ft =1.6*10^9; // r e d u c e d u n i t y g a i n f r e q u e n c y i n Hz Ic =2.5*10^ -3; // c o l l e c t o r c u r r e n t i n A Vt =25*10^ -3; // t h r e s h o l d v o l t a g e a t room t e m p e r a t u r e gm = Ic / Vt ; // t r a n s c o n d u c t a n c e Cu =1.5*10^ -12; Cl =1*10^ -12; Rs =40; Rb =60; C2 = gm /(2* %pi * Ft ) - Cu for i =1:2 , if i ==1 then 41

Rl =30; // l o a d r e s i s t a n c e F1 =1/(2* %pi *( Rs + Rb ) *( C2 + Cu *(1+ gm * Rl ) ) ) ; // f i r s t break frequency F2 =1/(2* %pi * Rl *( Cu + Cl ) ) ; // s e c o n d b r e a k frequency BW = F1 ; // s i n c e s i n g l e common e m i t t e r a m p l i f i e r s o n=1 t h u s BW=F1∗ s q r t ( 2 ˆ ( 1 / n ) −1) , i . e . ,BW=F1 Avmid = - gm * Rl ; // mid f r e q u e n c y g a i n GBW = Avmid * BW ; // g a i n −bandwidth p r o d u c t disp ( ’ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ For Rl =30 Ohm∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ’ ) disp ( ’MHz ’ , F1 /10^6 , ’ f i r s t b r e a k f r e q u e n c y i s : ’) disp ( ’MHz ’ , F2 /10^6 , ’ s e c o n d b r e a k f r e q u e n c y is : ’) disp ( ’MHz ’ , BW /10^6 , ’ Bandwidth i s : ’ ) disp ( abs ( Avmid ) , ’ mid f r e q u e n c y g a i n i s : ’ ) disp ( ’MHz ’ , abs ( GBW ) /10^6 , ’ g a i n −bandwidth product i s : ’ )

17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33

34 35 36 37 38 39 40 41

else Rl =100; // l o a d r e s i s t a n c e i n ohm F1 =1/(2* %pi *( Rs + Rb ) *( C2 + Cu *(1+ gm * Rl ) ) ) ; // f i r s t break frequency F2 =1/(2* %pi * Rl *( Cu + Cl ) ) ; // s e c o n d b r e a k frequency BW = F1 ; // s i n c e s i n g l e common e m i t t e r a m p l i f i e r s o n=1 t h u s BW=F1∗ s q r t ( 2 ˆ ( 1 / n ) −1) , i . e . ,BW=F1 Avmid = - gm * Rl ; // mid f r e q u e n c y g a i n GBW = Avmid * BW ; // g a i n −bandwidth p r o d u c t disp ( ’ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ For Rl =100 Ohm∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ’ ) disp ( ’MHz ’ , F1 /10^6 , ’ f i r s t b r e a k f r e q u e n c y i s : ’) disp ( ’MHz ’ , F2 /10^6 , ’ s e c o n d b r e a k f r e q u e n c y is : ’) disp ( ’MHz ’ , BW /10^6 , ’ Bandwidth i s : ’ ) disp ( abs ( Avmid ) , ’ mid f r e q u e n c y g a i n i s : ’ ) disp ( ’MHz ’ , abs ( GBW ) /10^6 , ’ g a i n −bandwidth 42

product i s : ’ ) 42 end 43 end

43

Chapter 11 Nonlinear Circuit Application

Scilab code Exa 11.4 Time taken 1 2 3

4 5 6 7 8 9 10

// C h a p t e r 11 N o n l i n e a r C i r c u i t A p p l i c a t i o n // C a p t i o n : Time t a k e n // Example11 . 4 : b ) Type 741 Op−amp i s u s e d a s a c o m p a r a t o r and i t s s l e w r a t e i s 0 . 5V/ u s . How l o n g w i l l i t c h a n g e from +10 V t o −10v ? // b ) S o l u t i o n : clear ; clc ; deltaVo =10 -( -10) ; SlewRate =0.5*10^ -6; t = deltaVo / SlewRate ; disp ( ’ u s ’ ,t /10^6 , ’ t i m e t a k e n by t h e o u t p u t v o l t a g e t o c h a n g e from +10 V t o −10 V i s : ’ )

Scilab code Exa 11.5 Rise Time 1 2

// C h a p t e r 11 N o n l i n e a r C i r c u i t A p p l i c a t i o n // C a p t i o n : R i s e Time 44

3

4 5 6 7 8 9 10 11 12

// Example11 . 5 : The u p p e r 3−dB f r e q u e n c y o f an Op−Amp i s 1MHz . C a l c u l a t e t h e r i s e t i m e o f t h e o u t p u t . I f t h e u p p e r 3−dB f r e q u e n c y o f t h e Op−Amp i s i n c r e a s e d t o 50 MHz by r e d u c i n g t h e g a i n s u c h t h a t g a i n bandwidth p r o d u c t r e m a i n s c o n s t a n t , t h e n f i n d o u t t h e new r i s e t i m e . D i s c u s s t h e e f f e c t o f i n c r e a s i n g bandwidth on a c c u r a c y o f c o m p a r a t o r . // S o l u t i o n : clear ; clc ; F3dB =1*10^6; // u p p e r 3−dB f r e q u e n c y o f Op−Amp Tr =0.35/ F3dB ; // from d e f i n i t i o n o f r i s e t i m e disp ( ’ n s e c ’ , Tr *10^9 , ’ R i s e t i m e o f t h e o u t p u t i s : ’ ) F3dB1 =50*10^6; Tr1 =0.35/ F3dB1 ; disp ( ’ n s e c ’ , Tr1 *10^9 , ’ R i s e t i m e o f t h e o u t p u t i s : ’ )

Scilab code Exa 11.11 Design Peak Detector 1 2 3

4 5 6 7 8 9 10

// C h a p t e r 11 N o n l i n e a r C i r c u i t A p p l i c a t i o n // C a p t i o n : D e s i g n Peak D e t e c t o r // Example11 . 1 1 : D e s i g n a p o s i t i v e peak d e t e c t o r u s i n g t y p e uA 760 c o m p a r a t o r t h a t can r e s p o n d t o a 100 mV( pp ) , 5 MHz s i n u s o i d a l i n p u t s i g n a l . The d e v i c e has f o l l o w i n g s p e c i f i c a t i o n s . Response t i m e =25 ns , p r o p a g a t i o n t i m e =12 ns , and I n p u t b i a s c u r r e n t =8uA . // S o l u t i o n : clear ; clc ; Vp =50*10^ -3; // s i n c e peak−peak v o l t a g e i s 100 mV f =5*10^6; T =200*10^ -9; t =15*10^ -9 // s i n c e r i s e t i m e ( t ) s h o u l d be g r e a t e r than p r o p a g a t i o n d e l a y (12 ns ) 45

11 deltaVc = Vp *(1 - cos (4* t / T *90*( %pi ) /180) ) ; 12 Ib =8*10^ -6; // i n p u t b i a s c u r r e n t 13 C = Ib /( deltaVc / T ) ; 14 disp ( ’mV ’ , deltaVc *10^3 , ’ v o l t a g e c h a n g e i s : ’ ) 15 disp ( ’ pF ’ ,C *10^12 , ’ c a p a c i t o r v a l u e i s : ’ ) 16 // Note : 17 // t h e Exact v a l u e a s c a l c u l a t e d i s t a k e n t o

c a l c u l a t e C , s o C= 2 9 3 . 5 9 5 5 5 pF . I f a p p r o x v a l u e o f d e l t a V c i s t a k e n a s 5 mV t h e n C=320 pF

46

Chapter 12 Signal Generators

Scilab code Exa 12.6 555 Timer 1 2 3

4 5 6 7 8 9 10 11 12 13 14 15 16

// C h a p t e r 12 S i g n a l G e n e r a t o r s // C a p t i o n : 555 Timer // Example12 . 6 : C a l c u l a t e ( a ) Tc ( b ) Td , and ( c ) t h e f r e e r u n n i n g f r e q u e n c y f o r t h e t i m e r 555 c o n n e c t e d i n a s t a b l e mode . Given Ra=6.8 k i l o Ohm ; Rb=3.3 k i l o Ohm ; C=0.1 uF . What i s t h e duty c y c l e ,d , of the c i r c u i t ? // S o l u t i o n : clear ; clc ; Ra =6.8*10^3; Rb =3.3*10^3; C =0.1*10^ -6; // U s i n g e q u a t i o n f o r a s t a b l e m u l t i v i b r a t o r we have Tc =0.69*( Ra + Rb ) * C ; // c h a r g i n g t i m e Td =0.69* Rb * C ; // d i s c h a r g i n g t i m e f =1.44/(( Ra +2* Rb ) * C ) ; // f r e e r u n n i n g f r e q u e n c y d = Rb /( Ra +2* Rb ) ; // duty c y c l e disp ( ’ ms ’ , Tc *10^3 , ’ c h a r g i n g t i m e o f 555 t i m e r i n a s t a b l e mode i s : ’ ) disp ( ’ ms ’ , Td *10^3 , ’ d i s c h a r g i n g t i m e o f 555 t i m e r i n 47

a s t a b l e mode i s : ’ ) 17 disp ( ’ kHz ’ ,f /10^3 , ’ f r e e r u n n i n g f r e q u e n c y o f 555 t i m e r i n a s t a b l e mode i s : ’ ) 18 disp (d , ’ duty c y c l e o f 555 t i m e r i n a s t a b l e mode i s : ’ )

Scilab code Exa 12.11 Design 1 2 3

4 5 6 7 8 9 10

11 12 13 14 15

// C h a p t e r 12 S i g n a l G e n e r a t o r s // C a p t i o n : D e s i g n // Example12 . 1 1 : A 555 one s h o t c i r c u i t w i t h Vcc=16 V i s t o have a 2 ms o u t p u t p u l s e w i d t h . D e s i g n a s u i t a b l e C i r c u i t . I t h r e s =0.25 uA ( max . ) from d a t a sheet of the device . // S o l u t i o n : clear ; clc ; Ithres =0.25*10^ -6; T =2*10^ -3 // o u t p u t p u l s e w i d t h Vcc =16; // power s u p p l y t o 555 // The v a l u e o f minimum c a p a c i t o r c h a r g i n g c u r r e n t I c s h o u l d be much g r e a t e r t h a n t h e t h r e s h o l d Current I t h r e s Icmin =1000* Ithres ; // s i n c e Icmin >> I t h r e s Ra = Vcc /(3* Icmin ) ; C = T /(1.1* Ra ) ; disp ( ’ k i l o Ohm ’ , Ra /10^3 , ’ r e s i t a n c e d e s i g n i s : ’ ) disp ( ’ uF ’ ,C *10^6 , ’ C a p a c i t o r d e s i g n i s : ’ )

Scilab code Exa 12.12 Generating pulse by 555 timer 1 2

// C h a p t e r 12 S i g n a l G e n e r a t o r s // C a p t i o n : G e n e r a t i n g p u l s e by 555 t i m e r 48

3

4

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

// Example12 . 1 2 : ( a ) D e s i g n a 555 a s t a b l e m u l t i v i b r a t o r t o g e n e r a t e an o u t p u t p u l s e w i t h p u l s e r e p e t i t i o n f r e q u e n c y (PRF) =4 kHz and a duty c y c l e o f 60%. Given Vcc=15V . // ( b ) A n a l y s e t h e c i r c u i t d e s i g n e d i n p a r t ( a ) t o d e t e r m i n e t h e a c t u a l PRF and duty c y c l e . Given I t h r e s =25 uA ( max . ) f o r t i m e r 5 5 5 . clear ; clc ; // a ) S o l u t i o n : d =60*10^ -2; // duty c y c l e g i v e n PRF =4*10^3; Vcc =15; // power s u p p l y T =1/ PRF ; // where T=Tc+Td Tc = d * T ; Td =T - Tc ; Ithres =25*10^ -6; Icmin =1*10^ -3; // s i n c e Icmin >>I t h r e s , s o a s s u m i n g I c m i n =1 mA R = Vcc /(3* Icmin ) ; // where R=Ra+Rb C = Tc /(0.7* R ) ; Rb = Td /(0.7* C ) ; Ra =R - Rb ; disp ( ’ k i l o Ohm ’ , Ra /10^3 , ’ D e s i g n e d r e s i s t o r ( Ra ) f o r 555 t i m e r i n a s t a b l e mode i s : ’ ) disp ( ’ k i l o Ohm ’ , Rb /10^3 , ’ D e s i g n e d r e s i s t o r ( Rb ) f o r 555 t i m e r i n a s t a b l e mode i s : ’ ) disp ( ’ uF ’ ,C *10^6 , ’ D e s i g n e d C a p a c i t o r f o r 555 t i m e r i n a s t a b l e mode i s : ’ ) // b ) S o l u t i o n : // from e q u a t i o n o f c h a r g i n g Tc1 =0.7* R * C ; Td1 =0.7* Rb * C ; T1 = Tc1 + Td1 ; PRFa =1/ T1 ; da = Tc1 /( Tc1 + Td1 ) *100; disp ( ’ kHz ’ , PRFa /10^3 , ’ a c t u a l P u l s e R e p e t i t i o n Frequency i s : ’ ) 49

31

disp ( ’% ’ ,da , ’ a c t u a l duty c y c l e i s : ’ )

Scilab code Exa 12.20 Waveform Generator 1 2 3

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

// C h a p t e r 12 S i g n a l G e n e r a t o r s // C a p t i o n : Waveform G e n e r a t o r // Example12 . 2 0 : D e s i g n a waveform g e n e r a t o r u s i n g t y p e 8 0 3 8 IC . The f r e q u e n c y o f O s c i l l a t i o n i s 5 kHz and t h e duty c y c l e s i s 50%. From d a t a s h e e t , t y p i c a l v a l u e s f o r t h e d e v i c e a t Vcc=5 V a r e a s follws : // Voh =3.6 V ; Vol =0.2 V ; I l l =−1.6 mA and I l h =40 uA . // S o l u t i o n : clear ; clc ; Fo =5*10^3; // f o r 50% duty c y c l e Tp=Tn Vcc =5; // i n v o l t Vol =0.2; // i n V o l t Voh =3.6; // i n v o l t Ill = -1.6*10^ -3; Ilh =40*10^ -6; Tp =1/(2* Fo ) ; C =0.01; // a s s u m i n g t h e C a p a c i t o r v a l u e i n uF f o r optimum d e s i g n Ra = Tp /(1.66* C ) ; Rb =2* Ra * Tp /(1.66* Ra * C + Tp ) ; R2min =( Vcc - Vol ) /(2*10^ -3 - abs ( Ill ) ) ; // s i n c e I l l i s negative R2max =( Vcc - Voh ) /(1*10^ -6+ Ilh ) ; // s i n c e I l h i s positive disp ( ’ k i l o Ohm ’ , Ra *10^3 , ’ d e s i g n e d v a l u e o f Ra i s : ’ ) disp ( ’ k i l o Ohm ’ , Rb *10^3 , ’ d e s i g n e d v a l u e o f Rb i s : ’ ) disp ( ’ k i l o Ohm ’ , R2min /10^3 , ’ minimum p u l l −up r e s i s t o r is : ’) 50

24

disp ( ’ k i l o Ohm ’ , R2max /10^3 , ’ maximum p u l l −up r e s i s t o r is : ’)

51

Chapter 13 Voltage Regulators

Scilab code Exa 13.3 Maximum Efficiency and Power 1 2 3

4 5 6 7 8

9 10 11 12 13

// C h a p t e r 13 V o l t a g e R e g u l a t o r s // C a p t i o n : Maximum E f f i c i e n c y and Power // Example13 . 3 : C a l c u l a t e t h e maximum e f f i c i e n c y and a s s o c i a t e d power d i s s i p a t i o n f o r t h e 5 V MC7805 s e r i e s r e g u l a t o r . The i n p u t r i p p l e i s 10 V and t h e l o a d c u r r e n t i s 1 A . The o u t p u t i s b e t w e e n 4 . 7 5 t o 5 . 2 5 f o r &v BW then disp ( ’ Both Sum and D i f f e r e n c e f r e q u e n c y c o m p o n e n t s a r e o u t s i d e t h e p a s s b a n d o f low− pass f i l t e r ’) disp ( ’ Loop w i l l n o t a c q u i r e l o c k ’ ) disp ( ’VCO f r e q u e n c y w i l l be i t s f r e e r u n n i n g frequency ’) end end

Scilab code Exa 15.4 Second Order Butterworth Filter 1 2 3

4 5 6 7 8 9 10 11 12 13

// C h a p t e r 15 P h a s e Locked Loops // C a p t i o n : S e c o n d Order B u t t e r w o r t h F i l t e r // Example 1 5 . 4 : A S y n t h e s i z e r u s i n g PLL h a s Kv=5∗%pi r a d / s . What v a l u e o f low−p a s s f i l t e r bandwidth s h o u l d be u s e d s o t h a t t h e c l o s e d −l o o p s y s t e m a p p r o x i m a t e s a s e c o n d −o r d e r B u t t e r w o r t h f i l t e r ? // S o l u t i o n : clear ; clc ; // For B u t t e r w o r t h f i l t e r t h e damping r a t i o ( Dr ) i s Dr =0.707; Kv =5* %pi ; Wl = Kv *(2* Dr ) ^2; // s i n c e ( Wl/Kv ) ˆ2=2∗Dr disp ( ’ r a d / s e c ’ ,Wl , ’ low p a s s f i l t e r bandwidth ’ ) // BW f o r c l o s e d l o o p s y s t e m i s BW = sqrt ( Kv * Wl ) ; // s i n c e BW=Wn, where Wn=n a t u r a l 57

14 15 16 17

f r e q u e n c y ,BW=bandwidth o f c l o s e d l o o p s y s t e m Wn = real ( BW ) ; t =2.2/ Wn ; disp ( ’ r a d / s e c ’ ,BW , ’ bandwidth o f c l o s e d l o o p s y s t e m is : ’) disp ( ’ s e c ’ ,t , ’ c o r r e s p o n d i n g s y s t e m r i s e t i m e i s : ’ )

Scilab code Exa 15.5 Lock Range 1 2 3

4 5 6 7 8 9 10 11 12 13 14 15 16

// C h a p t e r 15 P h a s e Locked Loops // C a p t i o n : Lock Range // Example15 . 5 : A PLL h a s a VCO w i t h Ko=25kHz /V and Fc=50kHz . The a m p l i f i e r g a i n i s A=2 and t h e p h a s e d e t e c t o r h a s a maximum o u t p u t v o l t a g e s w i n g o f +0.7V and −0.7V . Find t h e l o c k r a n g e o f t h e PLL . Assume f i l t e r g a i n e q u a l t o u n i t y . // S o l u t i o n : clear ; clc ; k1 =2*0.7/ %pi ; // p o s i t i v e maximum g a i n v a l u e o f p h a s e detector k2 = - k1 ; // n e g a t i v e maximum g a i n v a l u e o f p h a s e detector A =2; // a m p l i f i e r g a i n Ko =25; // VCO g a i n i n kHz // p o s i t i v e maximum o u t p u t v o l t a g e s w i n g o f p h a s e detector is V1 = k1 * %pi /2; // N e g a t i v e maximum o u t p u t v o l t a g e s w i n g o f p h a s e detector is V2 = k2 * %pi /2; Vf1 = k1 * A * %pi /2; // P o s i t i v e maximum c o n t r o l v o l t a g e a v a i l a b l e t o d r i v e VCO Vf2 = k2 * A * %pi /2; // n e g a t i v e maximum c o n t r o l v o l t a g e a v a i l a b l e t o d r i v e VCO 58

17 18 19 20 21 22

//maximum VCO f r e q u e n c y s w i n g t h a t can be o b t a i n e d is Fh = Ko * Vf1 ; // p o s i t i v e maximum VCO f r e q u e n c y s w i n g Fl = Ko * Vf2 ; // N e g a t i v e maximum VCO f r e q u e n c y s w i n g // s o l o c k r a n g e o f PLL i s f = Fh - Fl ; disp ( ’ kHz ’ ,f , ’ The l o c k r a n g e o f t h e PLL i s : ’ )

59

Chapter 16 Bipolar and MOS Digital Gate Circuits

Scilab code Exa 16.2 Noise Margin 1 2 3 4 5 6 7 8 9 10 11 12 13

// C h a p t e r 16 B i p o l a r and MOS D i g i t a l Gate C i r c u i t s // C a p t i o n : N o i s e Margin // Example 1 6 . 2 : An RTL g a t e h a s t h e w o r s t c a s e v o l t a g e s l i s t e d below : // Temp ( d e g r e e C) Voh (V) Vih (V) V i l (V) Vol (V) // −55 1.014 1.01 0.718 0.710 // 25 0.844 0.815 0.565 0.300 // 125 0.673 0.67 0.325 0.320 // C a l c u l t e t h e w o r s t c a s e NMl and NMh n o i s e m a r g i n s . // S o l u t i o n : clear ; clc ; T =[ -55;25;125]; // t e m p e r a t u r e s i n d e g r e e c e l s i u s given in table for j =1:3 , 60

14 if j ==1 then 15 disp ( ’ N o i s e m a r g i n s f o r T=−55 d e g r e e 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

c e l s i u s are : ’) NMl =0.718 -0.710; // s i n c e NMl=V i l −Vol NMh =1.014 -1.01; // s i n c e NMh=Vih−Voh disp ( ’ V o l t ’ ,NMl , ’ l o w e r l i m i t o f n o i s e m a r g i n a t −55 d e g r e e c e l s i u s i s : ’ ) disp ( ’ v o l t ’ ,NMh , ’ u p p e r l i m i t o f n o i s e l i m i t a t −55 d e g r e e c e l s i u s i s : ’ ) elseif j ==2 then disp ( ’ N o i s e m a r g i n f o r T=25 d e g r e e c e l s i u s a r e : ’ ) NMl =0.565 -0.300; NMh =0.844 -0.815; disp ( ’ V o l t ’ ,NMl , ’ l o w e r l i m i t o f n o i s e m a r g i n a t 25 d e g r e e c e l s i u s i s : ’ ) disp ( ’ V o l t ’ ,NMh , ’ u p p e r l i m i t o f n o i s e m a r g i n a t 25 d e g r e e c e l s i u s i s : ’ ) elseif j ==3 then disp ( ’ N o i s e m a r g i n f o r T=125 d e g r e e c e l s i u s a r e : ’) NMl =0.325 -0.320; NMh =0.673 -0.670; disp ( ’ V o l t ’ ,NMl , ’ l o w e r l i m i t o f n o i s e m a r g i n a t 125 d e g r e e c e l s i u s i s : ’ ) disp ( ’ V o l t ’ ,NMh , ’ uppwr l i m i t o f n o i s e m a r g i n a t 125 d e g r e e c e l s i u s i s : ’ ) end end

Scilab code Exa 16.3 Fanouts 1 2 3

// C h a p t e r 16 B i p o l a r and MOS D i g i t a l Gate C i r c u i t s // C a p t i o n : F a n o u t s // Example 1 6 . 3 : A TTL g a t e i s g u a r t n t e e d t o s i n k 10 61

4 5 6 7 8

9 10 11 12 13 14 15 16 17 18

mA w i t h o u t e x c e e d i n g ann o u t p u t v o l t a g e Vol =0.4V and t o s o u r c e 5mA w i t h o u t d r o p p i n g b e l o w Voh =2.4V . I f Tih =100uA a t 2 . 4V and I i l =1mA a t 0 . 4 V, c a l c u l a t e t h e low−s t a t e and h i g h −s t a t e f a n −o u t s . // S o l u t i o n : clear ; clc ; // f o r TTL g a t e // f a n o u t a t low o u t p u t i s = c o l l e c t o r s a t u r a t i o n c ur r e nt of output t r a n s i t o r / load c u r re n t of the driven gate . // f a n o u t f o r h i g h o u t p u t i s =s o u r c e c u r r e n t i n driving gatte / input current of load gate // from q u e s t i o n g i v e n Ic3 =10*10^ -3; // c o l l e c t o r s a t u r a t i o o n c u r r e n t a t output t r a n s i s t o r Ie =1*10^ -3; // l o a d c u r r e n t o f d r i v e n g a t e Ie4 =5*10^ -3; // s o u r c e c u r r e n t i n d r i v i n g g a t e Ic1 =100*10^ -6; // i n p u t c u r r e n t o f l o a d g a t e Fl = Ic3 / Ie ; disp ( Fl , ’ f a n o u t a t low o u t p u t s t a t e i s : ’ ) Fh = Ie4 / Ic1 ; disp ( Fh , ’ f a n o u t a t h i g h o u t p u t s t a t e i s : ’ )

Scilab code Exa 16.12 NMOS operating region // C h a p t e r 16 B i p o l a r and MOS D i g i t a l Gate C i r c u i t s // C a p t i o n : NMOS o p e r a t i n g r e g i o n // Example 1 6 . 1 2 : A NMOS t r a n s i s t o r w i t h K=20uA/Vˆ2 and Vth =1.5V i s o p e r a t e d a t Vgs=5V and I d s =100uA . D e t e r m i n e t h e r e g i o n o f t h e o p e r a t i o n on I−V c h a r a c t e r i s t i c s and f i n d Vds . 4 // S o l u t i o n : 5 clear ; 6 clc ;

1 2 3

62

7 8 9 10 11 12 13 14 15 16

17 18 19 20 21 22 23 24 25 26 27

K =20*10^ -6; Vgs =5; Vth =1.5; Ids =100*10^ -6; Id =( K /2) *( Vgs - Vth ) ^2; disp ( ’ uA ’ , Id /10^ -6 , ’ d r a i n c u r r e n t i n s a t u r a t i o n region ’) if Id > Ids then disp ( ’ r e g i o n o f o p e r a t i o n o f NMOS t r a n s i s t o r on I−V c h a r a c t e r i s t i c s i s LINEAR REGION ’ ) end // s i n c e NMOS l i e s i n LINEAR REGION s o I d s =(K/ 2 ) ∗ ( 2 ∗ ( Vgs−Vth ) ∗Vds−Vds ˆ 2 ) ; t h u s s u b s t i t u t i n g t h e v a l u e s we have // 100∗10ˆ −6==(20∗10ˆ −6/2) ∗ ( 2 ∗ ( 5 − 1 . 5 ) ∗Vds−Vds ˆ 2 ) ; // s o Vdsˆ2−7∗Vds +10=0; e q u i v a l e n t t o q u a d r a t t i c e q u a t i o n o f form aXˆ2+b∗X+c=0 Vds = poly (0 , ’ Vds ’ ) ; p = Vds ^2 -7* Vds +10; // e q u a t i o n whose r o o t s h a s t o be found z = roots ( p ) ; z = real ( z ) if ( z (1)