Scilab Textbook Companion forBasic Electronics And Linear Circuits by N. N. Bhargava, D. C. Kulshreshtha And S. C. Gupta1 Created by Mohammad Faisal Siddiqi B.Tech (pursuing) Electronics Engineering Jamia Millia Islamia, New Delhi College Teacher Dr. Sajad A. Loan Cross-Checked by TechPassion 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: Basic Electronics And Linear Circuits Author: N. N. Bhargava, D. C. Kulshreshtha And S. C. Gupta Publisher: Tata McGraw - Hill Education, New Delhi Edition: 1 Year: 2008 ISBN: 0074519654 1 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. 2 Contents List of Scilab Codes 4 1 INTRODUCTION TO ELECTRONICS 8 2 CURRENT AND VOLTAGE SOURCES 10 4 SEMICONDUCTOR DIODE 14 5 TRANSISTORS 20 6 VACUUM TUBES 27 7 TRANSISTOR BIASING AND STABILIZATION OF OPERATING POINT 32 8 SMALL SIGNAL AMPLIFIERS 46 9 MULTI STAGE AMPLIFIERS 53 10 POWER AMPLIFIERS 61 11 TUNED VOLTAGE AMPLIFIERS 64 12 FEEDBACK IN AMPLIFIERS 69 13 OSCILLATORS 73 14 ELECTRONIC INSTRUMENTS 75 3 . . . . . . . .3. .e 4.4. . .4 4. . . . . . . . . . . . . . . . . . . . Dynamic Input Resistance Determination . . . . Capacitance Determination on changing Bias Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 1. . . .a 4. . . . . . . RMS Value of Current Calculation .1 Resistor Range Calculation using Colour Band Sequence Resistor Range Calculation using Colour Band Sequence Equivalent Current Source Representation . .c 4.d 4. . . . . . . .8 Exa 6. Peak Value of Current Calculation . . . . . . . . Q Point Determination . DC or Average Value of Current Calculation . . . Collector and Base Currents Calculation . .2 5. . . . . . . Common Base Short Circuit Current Gain Calculation Common Emitter Short Circuit Current Gain Calculation . . Current Determination using Voltage Source and Current Source Representations . . . .5 5. Short Circuit Current Gain Determination . . Determination of Dynamic Output Resistance and AC and DC Current Gains . . . . . . . . . .3. Calculation of Dynamic Drain Resistance of JFET . .4. . .b Exa 5. .3. . .2 2. . . . . .3 2. . . . . . . DC Voltage and PIV Calculation . . . . DC Current Gain in Common Base Configuration .6 Exa 5. . . . .3. . . . .1 4. . . . . . . . . . DC Voltage and PIV Calculation . . . Maximum Permissible Current Determination . . . . .3 5. . . .3. Ripple Factor Determination .4 4. .List of Scilab Codes Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa 1. . . .5 Exa 5. .1 2. . .b 4.7 Exa 5. . . . . . . Dynamic Plate Resistance of the Diode Determination 4 8 9 10 11 11 12 14 15 15 16 16 17 17 18 19 20 20 21 22 22 23 23 24 25 27 . Rectification Efficiency Calculation . . . . . . . . . . . . . . . . . Equivalent Voltage Source Representation . Output Voltage Determination . . . .1 5. . .2 2.a 5. .2 4. . . . .4 8. Calculate Ic and Vce for given Circuit . . . . . . . . .4. . . Calculate Minimum and Maximum Ie and corresponding Vce . .c 8. . . . . Calculate Voltage Gain of Two Stage RC Coupled Amplifier . . . . . . . . . . . . . .10 7.1 9. . . Calculate Ic and Vce for given Circuit . . . . . . .2. . . . . . .6 9. . . . . .b 8. .4. . . . .b Exa 9. . Calculation of Voltage Gain of Amplifier . . . . . . .14 8. .b 8. . . . . . . . . . Calculation of Voltage Gain of Amplifier . . . Calculate DC bias Voltages and Currents . .c 8.11 7. . . . . .2 7.1 7. . . . . Calculate Ic and Vce for given Circuit . Calculate Maximum Voltage Gain and Bandwidth of Triode Amplifier . Calculation of Gain of Single Stage Amplifier . . . . . . . . . . . . .3 9. . . Calculation of Input Impedance of Amplifier .13 7. . . . . . . . . . . . . . . Calculate Input Impedance of Two Stage RC Coupled Amplifier . . . . . . . . . . . . . . . . .3. . . Calculate Minimum and Maximum Collector Currents Calculate Values of the three Currents . . . . 5 29 32 33 34 35 36 36 37 38 38 40 41 42 42 43 44 46 47 47 48 48 49 49 50 51 51 53 53 54 57 57 58 59 .12 7. . . . .2 7.7 7. . . . .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 Exa 6. . . . Calculate the value of Rb . .4. . . . . . . . . . . . . . . . . . . . . Calculation of Collector Current Ic . . . . . . . .a 8. .1 8. . . . Determine the new Q Points . . . . . To Plot the Frequency Response Curve . . . . . . . Calculation of Output Signal Voltage of FET Amplifier Calculate overall Voltage Gain in dB . . . . . . . . Calculate Re and Vce in the Circuit . . .a Exa 9.5 8. . . . . . . . Calculate Voltage at the Output Terminal . .5 Plotting of Static Plate Characteristics .6 7. . . . . . .3 7. . . . . . . . . . Calculation of Input Impedance of Amplifier . . . . . . . . . . .9 7. .4. Calculation of Ie and Vc in the Circuit . . . . . . . . . .8 7. . Calculate Ouput Impedance of Two Stage RC Coupled Amplifier . . Quiescent Operating Point Determination . . . . . .a 8. Calculate coordinates of Operating Point . . . . . . . . . . . . Determination of Hybrid Parameters .2. . . Calculate value of Resistance Rb . . . . . . .a 7. . . .2.2 9. . . . . . . . . . . . . . . Calculation of Voltage Gain of Amplifier . . .b 7. . . . .3. . . . . . . . . . . . Calculation of Current Gain of Amplifier . .4. .c Exa 9. . . . .5 7. . . . . .3. . . . Calculation of Q Point Parameters of Amplifier . . . . . . . . . Calculation of Voltage across each Component . . . . . . Percentage Increase in Power because of Distortion . . . . . Calculate Frequency of Oscillation of Phase Shift Oscillator . Designing of a Universal Shunt for making a Multi Range Milliammeter . . . . . . . . . . Calculation of Feedback Factor and Percent change in overall Gain . . . . . . . . Calculation of 2nd 3rd and 4th Harmonic Distortions . . Calculation of Impedance at Resonance . . . . . . . . . . . . . . . . . . . . . . . . .b 11. . . . Calculation of Effective Resistance seen at Primary . . . . . Determination of Peak and RMS AC Voltage . . . . . .5 Calculation of Transformer Turns Ratio . Calculation of Parameters of the Resonant Circuit at Resonance . .Exa Exa Exa Exa Exa Exa Exa Exa Exa 10. . . . . . . . . . .2 Exa 13. . Calculate Frequency of Oscillation of Tuned Collector Oscillator . . . . . Calculation of Gain of Negative Feedback Amplifier . . . Caculation of Series Resistance for coversion to Voltmeter Calculation of Shunt Resistance . . . . . . . . . . . .2 Exa 11. . . . . . . . . . . Calculate Frequency of Oscillation of Wein Bridge Oscillator . Calculation of Current at Resonance . .1. . . . . . . . . . .3 Exa 12. Determination of Magnitude and Frequency of Voltage Fed to Y Input . . .c 11. . . . . . .b 11. . . . . . . .1 Exa 12. . . . . . . . . . . . .1. . .2 10. . . . . . . . . . . . . . . . . . .a 10. . . . . . . . . . .1 Exa 14. .3.4 Exa 14. . .4 Exa 12. . .3 Exa 14. . . . . . . . . . 6 61 61 62 63 64 64 65 66 66 67 69 69 70 71 71 73 73 74 75 76 76 78 78 . .d 11. . . . . . . . Calculation of Resonant Frequency . . . . . . . . . . . . . Calculation of Impedance Q and Bandwidth of Resonant Circuit . . . . . . . . . Calculation of Input Impedance of the Feedback Amplifier . . . . . . . . . . . . . .a 11. .1. . . .1 10. . .3. . . . Calculation of Internal Gain and Feedback Gain . . . . . . . . . .1. . Calculation of change in overall Gain of Feedback Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . .3 Exa 14. . . . . . . . . . . .2 Exa 14. . . . . .2 Exa 12.5 Exa 13. . . . . . . . . . . .1 Exa 13.3 Exa 12. . . 1 To Plot the Frequency Response Curve . . . 28 29 9. . . . 79 7 . . . . . . . . . . . .2 Dynamic Plate Resistance of the Diode Determination . . . . . 55 14. . . . . Plotting of Static Plate Characteristics . .1 6. . .1 Determination of Magnitude and Frequency of Voltage Fed to Y Input . . . . . . . . . . . . . . .List of Figures 6. . . . . . . . . . . . clc . close . //NUMERICAL CODE FOR BAND YELLOW B =7. VIOLET . // D i s p l a y i n g The R e s u l t s i n Command Window 8 . e .T . ORANGE. A =4. 1 // Program t o f i n d Range o f a R e s i s t o r s o a s t o s a t i s f y manufacturer ’ s Tolerances // C o l o u r Band S e q u e n c e : YELLOW. //NUMERICAL CODE FOR BAND VIOLET C =3. 5% // R e s i s t o r V a l u e C a l c u l a t i o n R =( A *10+ B ) *10^ C . R1 =R . GOLD clear all .Chapter 1 INTRODUCTION TO ELECTRONICS Scilab code Exa 1. //NUMERICAL CODE FOR BAND ORANGE D =5. //TOLERANCE VALUE FOR BAND GOLD i .1 Resistor Range Calculation using Colour Band Sequence 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 // Example 1 . // T o l e r a n c e V a l u e C a l u l a t i o n T = D * R /100. R2 = R + T . close . D ) . A =8. 2 // Program t o f i n d Range o f a R e s i s t o r s o a s t o s a t i s f y manufacturer ’ s Tolerances // C o l o u r Band S e q u e n c e : GRAY. ” . T ) .R . R2 /1000) . //NUMERICAL CODE FOR BAND GOLD D =5. printf ( ” \n\n\ t R e s i s t o r V a l u e i s %f Ohms +− %f Ohms . e . // T o l e r a n c e V a l u e C a l u l a t i o n T = D * R /100. T /1000) . R2 ) . 9 . 20 printf ( ” \n\n\ t Range o f V a l u e s o f t h e R e s i s t o r i s %f kOhms & %f kOhms . 19 printf ( ” \n\n\ t R e s i s t o r V a l u e i s %f kOhms +− %f kOhms . //TOLERANCE VALUE FOR BAND GOLD i . R1 =R . 18 Scilab code Exa 1.printf ( ” \n\n\ t R e s i s t o r V a l u e i s %f kOhms +− %f p e r c e n t . ” . clc . GOLD. 5% // R e s i s t o r V a l u e C a l c u l a t i o n R =( A *10+ B ) *10^ C . ” . printf ( ” \n\n\ t Range o f V a l u e s o f t h e R e s i s t o r i s %f Ohms & %f Ohms . BLUE.T . //NUMERICAL CODE FOR BAND GRAY B =6. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\n\ t R e s i s t o r V a l u e i s %f Ohms +− %f p e r c e n t .R .R /1000 . R1 /1000 . ” . GOLD clear all . D ) .2 Resistor Range Calculation using Colour Band Sequence 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 // Example 1 .R /1000 . ” .R1 . //NUMERICAL CODE FOR BAND BLUE C = -1. R2 = R + T . ” . // V o l t a g e S o u r c e o r Thevenin ’ s R e p r e s e n t a i o n ( S e r i e s Voltage Source & R e s i s t o r ) Vs =2. // Amperes // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\n\ t The S h o r t C i r c u i t C u r r e n t V a l u e i s %f Amperes . ” . // V o l t s Rs =1.1 Equivalent Current Source Representation 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 2 . //Ohm // C u r r e n t S o u r c e o r Norton ’ s R e p r e s e n t a i o n ( P a r a l l e l Current Source & R e s i s t o r ) Is = Vs / Rs . Rs ) . printf ( ” \n\n\ t The S o u r c e I m p e d e n c e V a l u e i s %f Ohm . close . ” . Is ) . 10 . clc . 1 // Program t o O b t a i n E q u i v a l e n t C u r r e n t S o u r c e R e p r e s e n t a i o n from Given V o l t a g e S o u r c e Representation clear all .Chapter 2 CURRENT AND VOLTAGE SOURCES Scilab code Exa 2. ” .3 Current Determination using Voltage Source and Current Source Representations 1 2 // Example 2 . close .2 Equivalent Voltage Source Representation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 // Example 2 . // V o l t s // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\n\ t The Open C i r c u i t V o l t a g e i s %f V o l t s . ”). 2 // Program t o O b t a i n E q u i v a l e n t V o l t a g e S o u r c e R e p r e s e n t a i o n from Given C u r r e n t S o u r c e Representation clear all . ” . Scilab code Exa 2. printf ( ” \n\n\ t The S o u r c e I m p e d e n c e V a l u e i s %f Ohms . 3 // Program t o C a l c u l a t e C u r r e n t i n a Branch by U s i n g Current Source Representation 11 . Scilab code Exa 2. // C u r r e n t S o u r c e o r Norton ’ s R e p r e s e n t a i o n ( P a r a l l e l Current Source & R e s i s t o r ) Is =0.14 printf ( ” \n\n\ t The C u r r e n t S o u r c e & S o u r c e Impedance a r e connected i n P a r a l l e l . // Amperes Zs =100. ”). //Ohms // V o l t a g e S o u r c e o r Thevenin ’ s R e p r e s e n t a i o n ( S e r i e s Voltage Source & R e s i s t o r ) Vs = Is * Zs .2. printf ( ” \n\n\ t The V o l t a g e S o u r c e & S o u r c e Impedance a r e connected i n S e r i e s . clc . Vs ) . Zs ) . // Open C i r c u i t V o l a t g e I = Vs /( Zs + Zl ) . ” . //Ohms Z1 =10*10^3. I4V ) .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 // V e r i f y t h e C i r c u i t ’ s R e s u l t f o r i t s e q u i v a l e n c e with Voltage Source R e p r e s e n t a t i o n clear all . 4 // Program t o O b t a i n Output V o l t a g e Vo from Given A . I2 = Is * Zs /( Zs + Zl ) . I4I ) . ” . close .5*10^( -3) . // U s i n g C u r r e n t D i v i d e r Rule // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\n\ t The Load C u r r e n t u s i n g C u r r e n t S o u r c e R e p r e s e n t a i o n i s I 4 I = %f Amperes . else printf ( ” \n\n\ t Both R e s u l t s a r e Not E q u i v a l e n t . I4I = I2 * Z1 /( Z1 + Z2 ) . Scilab code Exa 2. E q u i v a l e n t o f an A m p l i f i e r u s i n g a T r a n s i s t o r 3 clear all .4 Output Voltage Determination // Example 2 . printf ( ” \n\n\ t The Load C u r r e n t u s i n g V o l t a g e S o u r c e R e p r e s e n t a i o n i s I4V = %f Amperes . // Given C i r c u i t Data Is =1. C . I4V = I * Z1 /( Z1 + Z2 ) . 1 2 12 . clc . // Amperes Zs =2*10^3. ” ) . //Ohms Z2 =40*10^3. //Ohms // C a l c u l a t i o n f o r C u r r e n t S o u r c e R e p r e s e n t a t i o n Zl = Z1 * Z2 /( Z1 + Z2 ) . end . // U s i n g C u r r e n t D i v i d e r Rule // C a l c u l a t i o n f o r C u r r e n t S o u r c e R e p r e s e n t a t i o n Vs = Is * Zs . ” ) . if I4I == I4V then printf ( ” \n\n\ t Both R e s u l t s a r e E q u i v a l e n t . // i . 10 mV Rs =1*10^3. close . // I n p u t C u r r e n t Io =100* i . // U s i n g C u r r e n t D i v i d e r Rule Vo = Il * Ro2 . // i . e .4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 clc . 1 kOhms // Output S i d e Ro1 =20*10^3. e . // i . // i . e . 2 kOhms // C a l c u l a t i o n i = Vs / Rs . // Output V o l a t g e // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Output V o l t a g e Vo = %f V o l t s . ” . Vo ) . 13 . e . // Output C u r r e n t Il = Io * Ro1 /( Ro1 + Ro2 ) . 20 kOhms Ro2 =2*10^3. // Given C i r c u i t Data // I n p u t S i d e Vs =10*10^( -3) . // V o l t s n2 =1. ” . ” . Vm ) . 1 // Program t o d e t e r m i n e DC V o l t a g e a c r o s s t h e l o a d and PIV o f t h e Diode clear all . 14 . // Assumption n1 =12* n2 . clc . //Maximum S e c o n d a r y V o l t a g e Vdc = Vm / %pi . Vdc ) .1 DC Voltage and PIV Calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 // Example 4 . close . printf ( ” \n\ t The Peak I n v e r s e V o l t a g e ( PIV ) i s = %f V . // Given C i r c u i t Data Vrms =220. //Maximum ( Peak ) Primary V o l t a g e Vm = n2 * Vp / n1 . //DC l o a d V o l t a g e // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The DC l o a d V o l t a g e i s = %f V . // Turns R a t i o // C a l c u l a t i o n Vp = sqrt (2) * Vrms .Chapter 4 SEMICONDUCTOR DIODE Scilab code Exa 4. Scilab code Exa 4. ” . // V o l t s n2 =1. close . clc . // Assumption n1 =12* n2 . // V o l t s ( Peak V a l u e o f V o l t a g e ) 15 .Scilab code Exa 4. ” . // Given C i r c u i t Data Rl =1*10^(3) . 2 // Program t o d e t e r m i n e DC V o l t a g e a c r o s s t h e l o a d and PIV o f t h e // C e n t r e Tap R e c t i f i e r and B r i d g e R e c t i f i e r clear all . //Ohms rd =10. //DC l o a d V o l t a g e // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The DC l o a d V o l t a g e i s = %f V .a Peak Value of Current Calculation 1 2 3 4 5 6 7 8 9 // Example 4 . ” .3. printf ( ” \n\ t The Peak I n v e r s e V o l t a g e ( PIV ) o f B r i d g e R e c t i f i e r i s = %f V .2 DC Voltage and PIV Calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 // Example 4 . printf ( ” \n\ t The Peak I n v e r s e V o l t a g e ( PIV ) o f C e n t r e −t a p R e c t i f i e r i s = %f V . clc .2* Vm ) . //Ohms Vm =220. //Maximum ( Peak ) Primary V o l t a g e Vm = n2 * Vp / n1 . 3 ( a ) // Program t o d e t e r m i n e t h e Peak V a l u e o f C u r r e n t clear all . // Given C i r c u i t Data Vrms =220. close . // Turns R a t i o // C a l c u l a t i o n Vp = sqrt (2) * Vrms . Vdc ) . //Maximum S e c o n d a r y V o l t a g e Vdc =2* Vm / %pi . Vm ) . //DC V a l u e o f C u r r e n t // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The DC o r A v e r a g e V a l u e o f C u r r e n t i s = %f mA . // Peak V a l u e o f C u r r e n t 12 // D i s p l a y i n g The R e s u l t s i n Command Window 13 printf ( ” \n\ t The Peak V a l u e o f C u r r e n t i s = %f mA .10 // C a l c u l a t i o n 11 Im = Vm /( rd + Rl ) . //Ohms rd =10. 3 ( c ) 2 // Program t o d e t e r m i n e t h e RMS V a l u e o f C u r r e n t 3 clear all . 5 close .c RMS Value of Current Calculation 1 // Example 4 . clc .3. // Peak V a l u e o f C u r r e n t Idc =2* Im / %pi . // Given C i r c u i t Data Rl =1*10^(3) . Im /10^( -3) ) .3. // V o l t s ( Peak V a l u e o f V o l t a g e ) // C a l c u l a t i o n Im = Vm /( rd + Rl ) . Scilab code Exa 4. Scilab code Exa 4. //Ohms Vm =220. ” . close . 3 ( b ) // Program t o d e t e r m i n e t h e DC o r A v e r a g e V a l u e o f Current clear all . Idc /10^( -3) ) . 4 clc .b DC or Average Value of Current Calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 // Example 4 . 6 // Given C i r c u i t Data 16 . ” . clc . //Ohms Vm =220.3. Scilab code Exa 4. //Ohms Vm =220. //RMS V a l u e o f C u r r e n t r = sqrt (( Irms / Idc ) ^2 -1) // R i p p l e F a c t o r // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The R i p p l e F a c t o r r = %f . 3 ( d ) // Program t o d e t e r m i n e t h e R i p p l e F a c t o r o f C e n t r e − t a p F u l l Wave R e c t i f i e r clear all . //RMS V a l u e o f C u r r e n t // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The RMS V a l u e o f C u r r e n t i s = %f mA . ” . // Peak V a l u e o f C u r r e n t Idc =2* Im / %pi . //DC V a l u e o f C u r r e n t Irms = Im / sqrt (2) .3. Irms /10^( -3) ) .e Rectification Efficiency Calculation 1 // Example 4 . Scilab code Exa 4. ” .r ) . close . // V o l t s ( Peak V a l u e o f V o l t a g e ) // C a l c u l a t i o n Im = Vm /( rd + Rl ) . //Ohms rd =10. // Given C i r c u i t Data Rl =1*10^(3) . // V o l t s ( Peak V a l u e o f V o l t a g e ) // C a l c u l a t i o n Im = Vm /( rd + Rl ) . // Peak V a l u e o f C u r r e n t Irms = Im / sqrt (2) . 3 ( e ) 17 . //Ohms rd =10.d Ripple Factor Determination 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 // Example 4 .7 8 9 10 11 12 13 14 Rl =1*10^(3) . ” .4 Maximum Permissible Current Determination 1 2 3 4 5 6 7 8 9 10 11 12 // Example 4 . //Ohms Vm =220. // V o l t s P =364*10^( -3) . Pac = Irms ^2*( rd + Rl ) . close . Scilab code Exa 4. clc . //Ohms rd =10.n *100) . // Peak V a l u e o f C u r r e n t Idc =2* Im / %pi . //RMS V a l u e o f C u r r e n t Pdc = Idc ^2* Rl . 4 // Program t o d e t e r m i n e Maximum C u r r e n t t h e Given Z e n e r Diode can h a n d l e clear all .1. // Given C i r c u i t Data Vz =9. // Given C i r c u i t Data Rl =1*10^(3) . // R e c t i f i c a t i o n E f f i c i e n c y // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The R e c t i f i c a t i o n E F f i c i e n c y n ( e e t a ) = %f p e r c e n t . //DC V a l u e o f C u r r e n t Irms = Im / sqrt (2) . close . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Maximum p e r m i s s i b l e C u r r e n t i s I z ( 18 .2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 // Program t o d e t e r m i n e t h e R e c t i f i c a t i o n E f f i c i e n c y o f C e n t r e −t a p F u l l Wave R e c t i f i e r clear all . // Watts // C a l c u l a t i o n Iz = P / Vz . // V o l t s ( Peak V a l u e o f V o l t a g e ) // C a l c u l a t i o n Im = Vm /( rd + Rl ) . n = Pdc / Pac . clc . Scilab code Exa 4. Cf = K / sqrt ( Vf ) . 18 pF Vi =4. // i . 5 // Program t o d e t e r m i n e C a p a c i t a n c e o f V a r a c t o r Diode i f the // R e v e r s e −B i a s V o l t a g e i s i n c r e a s e d from 4V t o 8V clear all . e .max ) = %f mA . 19 . ” . // Given C i r c u i t Data Ci =18*10^( -12) . // V o l t s // C a l c u l a t i o n K = Ci * sqrt ( Vi ) .5 Capacitance Determination on changing Bias Voltage 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 // Example 4 . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The F i n a l V a l u e o f C a p a c i t a n c e i s C = %f pF . clc . Iz /10^( -3) ) . ” . Cf /10^( -12) ) . // V o l t s Vf =8. close . // Given C i r c u i t Data alpha =0.2 Dynamic Input Resistance Determination 20 . Ib /10^( -6) ) . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Scilab code Exa 5.Chapter 5 TRANSISTORS Scilab code Exa 5.1 Collector and Base Currents Calculation // Example 5 . // Ampere // C a l c u l a t i o n Ic = alpha * Ie + Ico . ” . close . Ic /10^( -3) ) .98. // Ampere Ie =1*10^( -3) . ” . // Base C u r r e n t // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The C o l l e c t o r C u r r e n t i s I c= %f mA . // a l p h a ( dc ) Ico =1*10^( -6) . // C o l l e c t o r C u r r e n t Ib = Ie .Ic . 15 printf ( ” \n\ t The Base C u r r e n t i s I b= %f uA . clc . 1 // Program t o C a l c u l a t e C o l l e c t o r and Base C u r r e n t s clear all . 3 Short Circuit Current Gain Determination 1 2 3 4 5 6 7 8 9 10 11 12 // Example 5 . // S h o r t C i r c u i t C u r r e n t Gain // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The S h o r t C i r c u i t C u r r e n t Gain i s a l p h a o r h f b= %f . // Dynamic I n p u t R e s i s t a n c e a t Vcb= −10 V // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Dynamic I n p u t R e s i s t a n c e i s r i = %f Ohms . ” . clc . // From t h e I n p u t C h a r a c t e r i s t i c s dIe =(0.5 mA and Vcb= −10 V . clc .7 -0. ri ) .3) *10^( -3) . close .7 -0. //V // C a l c u l a t i o n ri = dVeb / dIe .62) .99*10^( -3) . 2 // Program t o D e t e r m i n e Dynamic I n p u t R e s i s t a n c e o f t h e T r a n s i s t o r a t // t h e p o i n t : I e =0. close . hfb ) .1 2 3 4 5 6 7 8 9 10 11 12 // Example 5 . 21 . clear all . // Given Data dIe =1*10^( -3) . Scilab code Exa 5. //A dVeb =(0. //A // C a l c u l a t i o n hfb = dIc / dIe . 3 // Program t o D e t e r m i n e S h o r t C i r c u i t C u r r e n t Gain o f the Tr a n s is t o r clear all . ” . //A dIc =0. 4 ( b ) // Program t o D e t e r m i n e Common E m i t t e r S h o r t C i r c u i t C u r r e n t Gain ( b e e t a ) // o f t h e T r a n s i s t o r clear all . //A dIc =0.b Common Emitter Short Circuit Current Gain Calculation 1 2 3 4 5 6 7 8 // Example 5 . close .4. ” . Scilab code Exa 5. 4 ( a ) // Program t o D e t e r m i n e Common Base S h o r t C i r c u i t C u r r e n t Gain ( a l p h a ) // o f t h e T r a n s i s t o r clear all . clc . alpha ) .995*10^( -3) . //A // C a l c u l a t i o n alpha = dIc / dIe . // Given Data dIe =1*10^( -3) .a Common Base Short Circuit Current Gain Calculation 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 5 .4. clc . // Given Data dIe =1*10^( -3) . //Common Base S h o r t C i r c u i t C u r r e n t Gain // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Common Base S h o r t C i r c u i t C u r r e n t Gain i s a l p h a= %f . close .Scilab code Exa 5. //A 22 . Alpha ) . 12 Scilab code Exa 5. //DC C u r r e n t Gain i n Common Base C o n f i g u r a t i o n // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The DC C u r r e n t Gain i n Common Base C o n f i g u r a t i o n i s Alpha= %f . clc . //A 10 // C a l c u l a t i o n 11 alpha = dIc / dIe .6 Determination of Dynamic Output Resistance and AC and DC Current Gains 1 2 // Example 5 . //Common Base S h o r t C i r c u i t C u r r e n t Gain beeta = alpha /(1 .alpha ) . close . Scilab code Exa 5. //Common E m i t t e r S h o r t C i r c u i t C u r r e n t Gain 13 // D i s p l a y i n g The R e s u l t s i n Command Window 14 printf ( ” \n\ t The Common E m i t t e r S h o r t C i r c u i t C u r r e n t Gain i s b e e t a= %f . ” . 6 // R e f e r F i g u r e 5 . 5 // Program t o D e t e r m i n e DC C u r r e n t Gain i n Common Base C o n f i g u r a t i o n clear all . beeta ) .995*10^( -3) . // C a l c u l a t i o n Alpha = Beeta /( Beeta +1) .5 DC Current Gain in Common Base Configuration 1 2 3 4 5 6 7 8 9 10 11 // Example 5 . ” . 2 0 i n t h e Textbook 23 .9 dIc =0. // Given Data Beeta =100. 5) . Beeta_dc ) .7 -3. Beeta_ac ) . close . 1 2 3 24 . //V Ib =30*10^( -6) . r o = %f kOhms” .6*10^( -3) . 5 clc .5 -7. // Given Data Vce =10. //A ro = dVce / dic . From Graph .3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 // Program t o D e t e r m i n e t h e Dynamic Output R e s i s t a n c e . Bac=d e l t a ( i c ) / d e l t a ( i b ) f o r g i v e n Vce // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t Dynamic Output R e s i s t a n c e . printf ( ” \n\ t DC C u r r e n t Gain .6) *10^( -3) ) /((40 -30) *10^( -6) ) . clc . // DC C u r r e n t Gain Beeta_ac =((4. Scilab code Exa 5.7 -3. //A // C a l c u l a t i o n from Given Output C h a r a c t e r i s t i c s a t I b = 30uA dVce =(12. //A Ic =3. 6 close . Bdc = %f ” . // Dynamic Output R e s i s t a n c e Beeta_dc = Ic / Ib . //AC C u r r e n t Gain . ro /10^(3) ) . //V dic =(3. //DC C u r r e n t Gain & AC C u r r e n t Gain from g i v e n output c h a r a c t e r i s t i c s clear all . Bac = %f ” . 7 // R e f e r F i g u r e 5 . printf ( ” \n\ t AC C u r r e n t Gain .7 Q Point Determination // Example 5 . 2 7 i n t h e Textbook // Program t o D e t e r m i n e t h e Q p o i n t from g i v e n collector characteristics 4 clear all .5) *10^( -3) . At t h e P o i n t o f I n t e r s e c t i o n : Vce =6. 25 . //V Rb =200*10^(3) . rd /10^(3) ) . A d o t t e d Curve i s drawn f o r // I b =40uA and I b =60uA . // Dynamic D r a i n R e s i s t a n c e // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Dynamic D r a i n R e s i s t a n c e o f JFET i s r d= %f kOhms . clc . ” . //V Rc =1*10^(3) . //V Ic =6*10^( -3) . printf ( ” \n\ t Vce = %f V” . Scilab code Exa 5. printf ( ” \n\ t I c = %f mA” . Vce ) . Ic /10^( -3) ) . // S . // Given Data u =80. Ib /10^( -6) ) .Vbe ) / Rb . //V // C a l c u l a t i o n Ib =( Vbb . //A // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t Q p o i n t : \n\n\ t I b = %f uA” .8 Calculation of Dynamic Drain Resistance of JFET 1 2 3 4 5 6 7 8 9 10 11 12 // Example 5 . close . //Ohms Vbb =10. //Ohms Vbe =0.7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 // Given Data Vcc =12. // A m p l i f i c a t i o n F a c t o r gm =200*10^( -6) . // V a l u e o f I b comes o u t t o be 50uA .7. T r a n s c o n d u c t a n c e // C a l c u l a t i o n rd = u / gm .7. 8 // Program t o C a l c u l a t e Dynamic D r a i n R e s i s t a n c e o f JFET clear all . 26 . Chapter 6 VACUUM TUBES Scilab code Exa 6.6 4 6. // C a l c u l a t i o n // V a l u e s from C h a r a c t e r i s t i c P l o t dVp =0. // Given C i r c u i t Data V =[0 0. ylabel ( ’ P l a t e C u r r e n t ( i n mA) ’ ) . clc . 1 // Program t o P l o t t h e C h a r a c t e r i s t i c s and // D e t e r m i n e Dynamic P l a t e R e s i s t a n c e clear all . close .5 2]. I ) . a = gca () .7 9. xlabel ( ’ P l a t e V o l t a g e ( i n V) ’ ) . //V 27 .5 1 1. title ( ’ STATIC CHARACTERISTIC CURVE OF THE DIODE ’ ) .8].1 Dynamic Plate Resistance of the Diode Determination 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 // Example 6 .5. //V I =[0 1. //mA // P l o t t i n g plot (V . 1: Dynamic Plate Resistance of the Diode Determination 28 .Figure 6. Figure 6. Mutual 29 .2: Plotting of Static Plate Characteristics 19 dIp =2. // Dynamic P l a t e R e s i s t a n c e 21 // D i s p l a y i n g The R e s u l t s i n Command Window 22 printf ( ” \n\ t The Dynamic P l a t e R e s i s t a n c e i s r p= %f Ohms . 2 // Program t o P l o t t h e S t a t i c P l a t e C h a r a c t e r i s t i c s and D e t e r m i n e // P l a t e AC R e s i s t a n c e . ” .7*10^( -3) . //A 20 rp = dVp / dIp . Scilab code Exa 6. rp ) .2 Plotting of Static Plate Characteristics 1 2 // Example 6 . 3]. //V rp = dvp / dip . //V V3 =[177 200 250]. plot ( V2 . //V V4 =[235 250 300]. plot ( V3 . ylabel ( ’ P l a t e C u r r e n t ( i n mA) ’ ) . close . u = gm * rp .4 -5. // Given C i r c u i t Data // A l l V a l u e s E x t r a p o l a t e d t o Touch x−a x i s V0 =[20 50 100 150].5 11. ut =(192 -150) /1.2 20]. I4 ) . title ( ’ STATIC PLATE CHARACTERISTIC CURVE OF THE TRIODE ’ ) . clc . // C a l c u l a t i o n // V a l u e s from C h a r a c t e r i s t i c P l o t dip =(14.4 14. //mA I1 =[0 4 12.3) *10^( -3) . // D i s p l a y i n g The R e s u l t s i n Command Window 30 . I3 ) .0 -10. plot ( V1 . plot ( V4 .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 31 32 33 34 35 36 37 38 Conductance & A m p l i f i c a t i o n Factor clear all .5 11.7) *10^( -3) . a = gca () .4 12. //V gm = diP / dvG . I2 ) .5]. //A dvp =20. I1 ) . //mA I3 =[0 3. //A dvG =1. diP =(12. I0 ) . //mA I4 =[0 2. //mA I2 =[0 5. //V V1 =[70 100 150 200]. //V V2 =[112 150 200].1]. //mA // P l o t t i n g plot ( V0 .4].4 21. //V I0 =[0 3. xlabel ( ’ P l a t e V o l t a g e ( i n V) ’ ) . ” . gm /10^( -3) ) .u ) . 41 printf ( ” \n\ t The G r a p h i c a l A m p l i f i c a t i o n F a c t o r i s u = %f . rp /10^(3) ) . ” . ” . 39 31 . ut ) . 42 printf ( ” \n\ t The T h e o r e t i c a l A m p l i f i c a t i o n F a c t o r i s u t= %f . ” . 40 printf ( ” \n\ t The Mutual C o n d u c t a n c e i s gm= %f mS .printf ( ” \n\ t The P l a t e AC R e s i s t a n c e i s r p= %f kOhms . Chapter 7 TRANSISTOR BIASING AND STABILIZATION OF OPERATING POINT Scilab code Exa 7. // Given C i r c u i t Data Vcc =9. // C a l c u l a t i o n Ib =( Vcc ) / Rb . Vce = Vcc . 1 // Program t o C a l c u l a t e // ( a ) C o l l e c t o r C u r r e n t // ( b ) C o l l e c t o r −to −E m i t t e r V o l t a g e clear all . //V Rb =300*10^3. //Ohms Beeta =50.1 Calculate Ic and Vce for given Circuit 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 // Example 7 . clc . 32 .Ic * Rc . //Ohms Rc =2*10^3. Ic = Beeta * Ib . close . Icsat = Vcc / Rc . Ib /10^( -6) ) . ” .0) . // C a l c u l a t i o n Ib =( Vcc ) / Rb . ” . if Ic < Icsat then disp ( ” T r a n s i s t o r i s n o t i n S a t u r a t i o n ” ) . ” . //Ohms Beeta =60. //Ohms Rc =1*10^3. end Scilab code Exa 7. else disp ( ” T r a n s i s t o r i s i n S a t u r a t i o n ” ) . Vce ) . Ic /10^( -3) ) . ” . Icsat = Vcc / Rc . 2 // Program t o C a l c u l a t e O p e r a t i n g P o i n t C o o r d i n a t e s of the C i r c u i t clear all . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The O p e r a t i n g P o i n t C o o r d i n a t e s o f t h e C i r c u i t a r e : \ n\ t I b = %f uA . Icsat /10^( -3) ) . printf ( ” \n\ t Vce = %f V . clc . ” . printf ( ” \n\ t I c = %f mA . Ib /10^( -6) ) . ” . // Given C i r c u i t Data Vcc =10. Ic = Beeta * Ib . Vce = Vcc . if Ic < Icsat then 33 .2 Calculate coordinates of Operating Point 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 // Example 7 . close .18 19 20 21 22 23 24 25 26 27 28 // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The d i f f e r e n t P a r a m e t e r s a r e \n\ t I b = %f uA . //V Rb =100*10^3. printf ( ” \n\ t I c = %f mA .Ic * Rc . printf ( ” \n\ t Vce = %f V . // Given C i r c u i t Data Vcc =10. printf ( ” \n\ t I c = %f mA . Vce = Vcc . 26 end Scilab code Exa 7. 22 else 23 disp ( ” T r a n s i s t o r i s i n S a t u r a t i o n ” ) . printf ( ” \n\ t Vce = %f V . ” .19 disp ( ” T r a n s i s t o r i s n o t i n S a t u r a t i o n ” ) . ” .Ic * Rc . 24 printf ( ” \n\ t I c = %f mA . Ic /10^( -3) ) . ” . Icsat = Vcc / Rc . clc . //Ohms Beeta =150. //V Rb =100*10^3. ” . //Ohms Rc =1*10^3. if Ic < Icsat then disp ( ” T r a n s i s t o r i s n o t i n S a t u r a t i o n ” ) . Vce ) . Ic /10^( -3) ) . close . Ic = Beeta * Ib .0) . ” . 21 printf ( ” \n\ t Vce = %f V . ” . ” . 3 // Program t o C a l c u l a t e O p e r a t i n g P o i n t C o o r d i n a t e s of the C i r c u i t clear all . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The O p e r a t i n g P o i n t C o o r d i n a t e s o f t h e C i r c u i t a r e : \ n\ t I b = %f uA . 25 printf ( ” \n\ t Vce = %f V . Ib /10^( -6) ) . Vce ) . // C a l c u l a t i o n Ib =( Vcc ) / Rb .3 Quiescent Operating Point Determination 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 // Example 7 . else 34 . 20 printf ( ” \n\ t I c = %f mA . Icsat /10^( -3) ) . 4. // Given C i r c u i t Data Vcc =6. //V Icbo =2*10^( -6) . ” . //V Vbe =0. Rb = %f kOhms . //A Ic =1*10^( -3) .3. printf ( ” \n\ t The Base R e s i s t a n c e ( N e g l e c t i n g I c b o and Vbe ) i s .23 disp ( ” T r a n s i s t o r i s i n S a t u r a t i o n ” ) . Rb1 /10^3) . // C a l c u l a t i o n // Case 1 : C o n s i d e r i n g I c b o and Vbe i n t h e calculations Ib =( Ic -( Beeta +1) * Icbo ) / Beeta .a Calculate value of Resistance Rb 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 // Example 7 . //A Beeta =20. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Base R e s i s t a n c e i s . clc .Rb1 ) / Rb1 *100. 35 . Icsat /10^( -3) ) . close . 24 printf ( ” \n\ t I c = %f mA . Rb = %f kOhms . 25 printf ( ” \n\ t Vce = %f V .0) . 26 end Scilab code Exa 7.Vbe ) / Ib . Rb2 = Vcc / Ib . // Case 2 : N e g l e c t i n g I c b o and Vbe i n t h e calculations Ib = Ic / Beeta . ” . Rb1 =( Vcc . ” . // P e r c e n t a g e E r r o r E =( Rb2 . Rb2 /10^3) . ” . 4 ( a ) // Program t o C a l c u l a t e V a l u e o f Rb i n t h e B i a s i n g Circuit clear all . //A 9 Beeta =25. clc . 5 // Program t o C a l c u l a t e // ( a ) I e // ( b ) Vc clear all . //V Rc =500. // Given C i r c u i t Data Vcc =10.5 Calculation of Ie and Vc in the Circuit 1 2 3 4 5 6 7 8 9 10 11 12 // Example 7 . 12 // D i s p l a y i n g The R e s u l t s i n Command Window 13 printf ( ” The C o l l e c t o r C u r r e n t i s : ” ) .24 printf ( ” \n\ t P e r c e n t a g e E r r o r i s = %f p e r c e n t . 14 printf ( ” \n\ t I c = %f mA . 6 // Given C i r c u i t Data 7 Icbo =10*10^( -6) . 4 ( b ) 2 // Program t o C a l c u l a t e Rb i n t h e B i a s i n g C i r c u i t 3 clear all .b Calculation of Collector Current Ic 1 // Example 7 . //Ohms Rb =500*10^3. Scilab code Exa 7.4. //A 8 Ib =47. 36 . close . 4 clc .E ) . Ic /10^( -3) ) . 10 // C a l c u l a t i o n 11 Ic = Beeta * Ib +( Beeta +1) * Icbo . ” .9*10^( -6) . ” . 5 close . //Ohms Beeta =100. Scilab code Exa 7. Scilab code Exa 7. Icmin = Beeta1 * Ibmin . Vc = Vce . clc . Vce = Vcc . close . Vc ) . // C a l c u l a t i o n CASE−2: Maximum C o l l e c t o r C u r r e n t Ibmax = Vcc /( Rb + Beeta2 * Rc ) .Ic * Rc . Icmax = Beeta2 * Ibmax . Beeta2 =200. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The D i f f e r e n t P a r a m e t e r s a r e : ” ) . printf ( ” \n\ t I e = %f mA . // C a l c u l a t i o n CASE−1: Minimum C o l l e c t o r C u r r e n t Ibmin = Vcc /( Rb + Beeta1 * Rc ) .13 14 15 16 17 18 19 20 21 22 // C a l c u l a t i o n Ib = Vcc /( Rb + Beeta * Rc ) . //V Rc =2*10^3. ” . //Ohms Beeta1 =50. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Minimum C o l l e c t o r C u r r e n t I c ( min ) = 37 . Ie = Ic . ” .6 Calculate Minimum and Maximum Collector Currents 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 // Example 7 . 6 // Program t o C a l c u l a t e // ( a ) Minimum C o l l e c t o r C u r r e n t // ( b ) Maximum C o l l e c t o r C u r r e n t clear all . Ic = Beeta * Ib . printf ( ” \n\ t Vc = %f V . // Given C i r c u i t Data Vcc =20. //Ohms Rb =200*10^3. Ie /10^( -3) ) . 22 printf ( ” \n\ t The Maximum C o l l e c t o r C u r r e n t I c ( max ) = %f mA . Ie = Ic + Ib . 21 printf ( ” \n\ t The Base C u r r e n t I b = %f uA . Ic /10^( -3) ) . clc . Ie /10^( -3) ) . ” . Scilab code Exa 7.7 Calculate Values of the three Currents // Example 7 . //Ohms Rb =1*10^6. ” .%f mA . Icmax /10^( -3) ) . Ic = Beeta * Ib . 7 // Program t o C a l c u l a t e // ( a ) I b // ( b ) I c // ( c ) I e clear all . //Ohms Beeta =100. //Ohms Re =1*10^3. ” . ” . close . // C a l c u l a t i o n Ib = Vcc /( Rb +( Beeta +1) * Re ) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The C o l l e c t o r C u r r e n t I c = %f mA . Icmin /10^( -3) ) . //V Rc =2*10^3. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 38 . ” . // Given C i r c u i t Data Vcc =10. Ib /10^( -6) ) . 22 printf ( ” \n\ t The E m i t t e r C u r r e n t I e = %f mA . ” . ” . // Given C i r c u i t Data Vcc =6. printf ( ” \n\ t The Maximum E m i t t e r C u r r e n t I e ( max ) = %f mA .Scilab code Exa 7. ” . clc . Vcemax = Vcc -( Rc + Re ) * Iemax . //V Rc =50. Vcemin ) . printf ( ” \n\ t The C o r r e s p o n d i n g Vce = %f V . Beeta2 =200. //V Vbe =0. close . Iemin /10^( -3) ) . Vcemin = Vcc -( Rc + Re ) * Iemin . printf ( ” \n\ t The C o r r e s p o n d i n g Vce = %f V . //Ohms Rb =10*10^3. ” . Vcemax ) .Vbe ) *( Beeta2 +1) /( Rb +( Beeta2 +1) * Re ) . 8 // Program t o C a l c u l a t e // ( a ) Minimum E m i t t e r C u r r e n t & c o r r e s p o n d i n g Vce // ( b ) Maximum E m i t t e r C u r r e n t & c o r r e s p o n d i n g Vce clear all . //Ohms Beeta1 =50. // C a l c u l a t i o n CASE−1: Minimum E m i t t e r C u r r e n t & c o r r e s p o n d i n g Vce Iemin =( Vcc . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Minimum E m i t t e r C u r r e n t I e ( min ) = %f mA .8 Calculate Minimum and Maximum Ie and corresponding Vce 1 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 // Example 7 .3. //Ohms Re =100. 39 . // C a l c u l a t i o e n CASE−2: Maximum E m i t t e r C u r r e n t & c o r r e s p o n d i n g Vce Iemax =( Vcc .Vbe ) *( Beeta1 +1) /( Rb +( Beeta1 +1) * Re ) . Iemax /10^( -3) ) . 9 // Program t o C a l c u l a t e new Q p o i n t s f o r // Minimum and Maximum v a l u e o f B e e t a clear all . //Ohms Rb =10*10^3. Vcemin = Vcc -( Rc + Re ) * Iemin . //Ohms Re =100. Icmin /10^( -3) ) . // C a l c u l a t i o n CASE−1: Minimum E m i t t e r C u r r e n t & c o r r e s p o n d i n g Vce Iemin =( Vcc . printf ( ” \n\ t Vc = %f V . Vcemax = Vcc -( Rc + Re ) * Iemax .Scilab code Exa 7. //V Rc =1*10^3. Vce ) .Vbe ) *( Beeta2 +1) /( Rb +( Beeta2 +1) * Re ) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” For B e e t a =50 : \ n\ t ” ) . 40 .Vbe ) *( Beeta1 +1) /( Rb +( Beeta1 +1) * Re ) . Beeta2 =200. ” . Icmin = Iemin .3. //V Vbe =0. // Given C i r c u i t Data Vcc =6.9 Determine the new Q Points 1 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 28 29 30 // Example 7 . if Icmin < Icsat then disp ( ” T r a n s i s t o r i s n o t i n S a t u r a t i o n ” ) . // D i s p l a y i n g The R e s u l t s i n Command Window Icsat = Vcc /( Rc + Re ) . Icmax = Iemax . //Ohms Beeta1 =50. printf ( ” \n\ t I c = %f mA . ” . close . clc . // C a l c u l a t i o e n CASE−2: Maximum E m i t t e r C u r r e n t & c o r r e s p o n d i n g Vce Iemax =( Vcc . 0) . ” . ” . 37 if Icmax < Icsat then 38 disp ( ” T r a n s i s t o r i s n o t i n S a t u r a t i o n ” ) . //A Beeta =80. 35 end 36 printf ( ” \ nFor B e e t a =200 : \ n\ t ” ) . // C a l c u l a t i o n Ib = Ic / Beeta . //V Re =500. 33 printf ( ” \n\ t I c ( s a t ) = %f mA . 41 else 42 disp ( ” T r a n s i s t o r i s i n S a t u r a t i o n ” ) . printf ( ” \n\ t Rb = %f kOhms . Rb =( Vcc -( Beeta +1) * Ib * Re ) / Ib . ” . ” . Icmax /10^( -3) ) . 39 printf ( ” \n\ t I c = %f mA . 43 printf ( ” \n\ t I c ( s a t ) = %f mA . close .31 else 32 disp ( ” T r a n s i s t o r i s i n S a t u r a t i o n ” ) .10 Calculate the value of Rb 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 // Example 7 . //Ohms Ic =8*10^( -3) . 1 0 // Program t o C a l c u l a t e Rb i n t h e B i a s i n g C i r c u i t clear all .0) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The Base R e s i s t a n c e i s : ” ) . 41 . 45 end Scilab code Exa 7. Vce ) . ” . // Given C i r c u i t Data Vcc =9. 44 printf ( ” \n\ t Vc ( s a t ) = %f V . ” . ” . //V Vce =3. clc . 34 printf ( ” \n\ t Vc ( s a t ) = %f V . 40 printf ( ” \n\ t Vc = %f V . Icsat /10^( -3) ) . Icsat /10^( -3) ) . Rb /10^3) . Ve ) . printf ( ” \n\ t Vc = %f V . //Ohms Re =1*10^3. Ve = Vb . Ie = Ve / Re . ” . printf ( ” \n\ t I c = %f mA . ” . Vc = Vcc . //Ohms Beeta =60. //Ohms Rc =5*10^3. // Given C i r c u i t Data Vcc =12. close . Scilab code Exa 7. printf ( ” \n\ t I e = %f mA . //V Vbe =0. clc . Ie /10^( -3) ) .3. // C a l c u l a t i o n Vb =( R2 /( R1 + R2 ) ) * Vcc . printf ( ” \n\ t Vce = %f V . //V R1 =40*10^3. printf ( ” \n\ t Ve = %f V . Vb ) .12 Calculate Re and Vce in the Circuit 42 . Ic /10^( -3) ) . ” . Vce ) .Ve . Vce = Vc . ” . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The D i f f e r e n t P a r a m e t e r s a r e : ” ) . //Ohms R2 =5*10^3. ” .11 Calculate DC bias Voltages and Currents 1 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 28 // Example 7 .Ic * Rc . Ic = Ie . Vc ) . 1 1 // Program t o C a l c u l a t e DC B i a s V o l t a g e s and C u r r e n t s clear all .Scilab code Exa 7.Vbe . printf ( ” \n\ t Vb = %f V . ” . printf ( ” \n\ t The C o l l e c t o r t o E m i t t e r V o l t a g e i s Vce = %f V . // N e g l e c t i n g Vbe Re = Ve / Ie . //V 43 . //V Vbe =0. clc .3. // Given C i r c u i t Data Vcc =15. 1 3 // / Program t o C a l c u l a t e I c and Vce o f t h e g i v e n Circuit Specifications clear all . clc .13 Calculate Ic and Vce for given Circuit 1 2 3 4 5 6 7 8 // Example 7 .1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 // Example 7 . Re ) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The E m i t t e r R e s i s t a n c e i s Re = %f Ohms . // Given C i r c u i t Data Vcc =12. //Ohms R2 =100. Vce ) . //Ohms Rc =20. 1 2 // Program t o C a l c u l a t e Re and Vce o f t h e g i v e n Circuit Specifications clear all . //A // C a l c u l a t i o n Ie = Ic . Ve = Vb . close . ” . //V R1 =200. Scilab code Exa 7. //Ohms Ic =100*10^( -3) . Vce = Vcc -( Rc + Re ) * Ic . ” . close . Vb =( R2 /( R1 + R2 ) ) * Vcc . Ic *( Rc + Re ) . //V Vee =15. //Ohms Rc =5*10^3. Vce ) . //Ohms Beeta =100.Vbe ) /( Rth + Beeta * Re ) . Rth = R1 * R2 /( R1 + R2 ) . //Ohms Re =10*10^3. printf ( ” \n\ t Vce = %f V . Ic = Ie . clc . printf ( ” \n\ t I c = %f mA . Vce = Vcc . //Ohms Re =1*10^3. //Ohms Beeta =60. ” . // Given C i r c u i t Data Vcc =12. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” The D i f f e r e n t P a r a m e t e r s a r e : ” ) . //V Rc =5*10^3. //Ohms Rb =10*10^3. // C a l c u l a t i o n Ie = Vee / Re . Ic /10^( -3) ) .14 Calculate Ic and Vce for given Circuit 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 // Example 7 .9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 R1 =40*10^3. ” . //Ohms R2 =5*10^3. close . 44 . Ib =( Vth . Scilab code Exa 7. Ic = Beeta * Ib . 1 4 // Program t o C a l c u l a t e // ( a ) I c // ( b ) Vce clear all . // C a l c u l a t i o n Vth =( R2 /( R1 + R2 ) ) * Vcc . 22 printf ( ” \n\ t Vce = %f V . ” . ” . 19 // D i s p l a y i n g The R e s u l t s i n Command Window 20 printf ( ” The P a r a m e t e r s a r e : ” ) . 21 printf ( ” \n\ t I c = %f mA . Ic /10^( -3) ) .18 Vce = Vcc .Ic * Rc . Vce ) . 45 . clc .72) /(20 -0) . Vce=c o n s t a n t hie =(0. 1 6 i n t h e Textbook // Program t o f i n d t h e H y b r i d P a r a m e t e r s from t h e given Transistor C h a r a c t e r i s t i c s clear all . // Given C i r c u i t Data Ic =2*10^( -3) .7 -1. //V // C a l c u l a t i o n // h f e=d e l t a ( i c ) / d e l t a ( i b ) . //A Vce =8. 1 // R e f e r F i g u r e 8 .715) /((20 -10) *10^( -6) ) . i b=c o n s t a n t hre =(0.1) *10^( -3) /(10 -7) .5.73 -0. // h r e=d e l t a ( Vbe ) / d e l t a ( Vce ) .Chapter 8 SMALL SIGNAL AMPLIFIERS Scilab code Exa 8. close . // h i e=d e l t a ( Vbe ) / d e l t a ( i b ) .1 Determination of Hybrid Parameters 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 // Example 8 . i b=c o n s t a n t hoe =(2.7) *10^( -3) /((20 -10) *10^( -6) ) . 1 5 and 8 . // hoe=d e l t a ( i c ) / d e l t a ( Vce ) . 46 . Vce=c o n s t a n t hfe =(2.73 -0.2 -2. Scilab code Exa 8. printf ( ” \n\ t h i e = %f kOhms” . printf ( ” \n\ t hoe = %f uS ” . Scilab code Exa 8. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The I n p u t Impedance o f t h e A m p l i f i e r i s Z i n = %f kOhms . 6 // Given C i r c u i t Data 7 Beeta =100. 2 ( a ) // Program t o f i n d t h e I n p u t Impedance o f t h e Amplifier clear all . Zin /10^3) . //Ohms Rb =150*10^3. // Given C i r c u i t Data ri =2*10^3. 2 ( b ) 2 // Program t o f i n d t h e V o l t a g e Gain o f t h e A m p l i f i e r 3 clear all .a Calculation of Input Impedance of Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 // Example 8 . hie /10^3) . hfe ) .19 20 21 22 23 24 // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The H y b r i d P a r a m e t e r s a r e : ” ) . 5 close . printf ( ” \n\n\ t h f e = %f ” . printf ( ” \n\ t h r e = %f ” .2. 47 . hoe /10^( -6) ) .b Calculation of Voltage Gain of Amplifier 1 // Example 8 . clc . 4 clc . //Ohms // C a l c u l a t i o n Zin = Rb * ri /( Rb + ri ) . ” . hre ) .2. close . Assumption io =100* ib .8 ri =2*10^3. // C a l c u l a t i o n Ai = io / ib . 2 ( c ) // Program t o f i n d t h e C u r r e n t Gain o f t h e A m p l i f i e r clear all .2. close . 4 clc .a Calculation of Voltage Gain of Amplifier 1 // Example 8 .c Calculation of Current Gain of Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 8 .3. Ai ) . //Ohms 10 // C a l c u l a t i o n 11 Av = Beeta * Rac / ri . clc . Scilab code Exa 8. Scilab code Exa 8. 6 // Given C i r c u i t Data 48 . ” . 5 close . Av ) . 12 // D i s p l a y i n g The R e s u l t s i n Command Window 13 printf ( ” \n\ t The V o l t a g e Gain o f t h e A m p l i f i e r i s Av = %f w i t h p h a s e o f 180 d e g r e e s . 3 ( a ) 2 // Program t o f i n d t h e V o l t a g e Gain o f t h e A m p l i f i e r 3 clear all . //Ohms 9 Rac =5*10^3. // C u r r e n t Gain // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The C u r r e n t Gain o f t h e A m p l i f i e r i s Ai = %f . //A. ” . // Given C i r c u i t Data // L e t i n p u t C u r r e n t i b =2A ib =2 . 3 ( c ) // Program t o f i n d t h e Q P o i n t o f t h e A m p l i f i e r 49 .b Calculation of Input Impedance of Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 8 .7 8 9 10 11 12 13 14 15 Bac =150.5*10^3. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V o l t a g e Gain o f t h e A m p l i f i e r i s Av = %f w i t h p h a s e o f 180 d e g r e e s .3. //Ohms R4 =7. rin =2*10^3.3.7*10^3. //Ohms // C a l c u l a t i o n Zin = R3 * R4 * rin /( R3 * R4 + R4 * rin + rin * R3 ) . Scilab code Exa 8. ” . clc . Av ) . // Given C i r c u i t Data rin =2*10^3. //Ohms // C a l c u l a t i o n Rac = R1 * R2 /( R1 + R2 ) . Av = Bac * Rac / rin . //Ohms R1 =4. 3 ( b ) // Program t o f i n d t h e I n p u t Impedance o f t h e Amplifier clear all . Scilab code Exa 8. //Ohms R2 =12*10^3. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The I n p u t Impedance o f t h e A m p l i f i e r i s Z i n = %f kOhms . ” . Zin /10^3) . close .c Calculation of Q Point Parameters of Amplifier 1 2 // Example 8 . //Ohms R3 =75*10^3. ” .7*10^3. 50 . 4 // Program t o f i n d t h e V o l t a g e Gain o f t h e A m p l i f i e r clear all . // Given C i r c u i t Data u =20. Rl =10*10^3. //Ohms Rc =4. //V R1 =75*10^3. close . Scilab code Exa 8. Ie /10^( -3) .2*10^3. Vce = Vcc -( Rc + Re ) * Ie . // Given C i r c u i t Data Vcc =15. //Ohms Re =1.5*10^3. //Ohms // C a l c u l a t i o n Vb = Vcc * R2 /( R1 + R2 ) .A ) . //Ohms R2 =7. clc . Ie = Ve / Re . Vce ) . close . clc .3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 clear all .4 Calculation of Voltage Gain of Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 8 . ” . Ve = Vb . //Ohms // C a l c u l a t i o n A = u * Rl /( rp + Rl ) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The D i f f e r e n t P a r a m e t e r s o f t h e A m p l i f i e r a r e I e = %f mA and Vce = %f V . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V o l t a g e Gain o f t h e A m p l i f i e r i s A = %f w i t h p h a s e o f 180 d e g r e e s . //Ohms rp =10*10^3. clc . // S Rl =22*10^3. //Ohms 51 .Scilab code Exa 8. For rp>>Rl we g e t A = gm * Rl . //Ohms rp =300*10^3. ” . //Ohms Rg =1*10^6. Scilab code Exa 8. // Given C i r c u i t Data Rl =12*10^3. // w i t h Phase o f 180 d e g r e e s // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Gain o f t h e A m p l i f i e r i s A = %f w i t h p h a s e o f 180 d e g r e e s . 5 // Program t o f i n d t h e Gain o f t h e A m p l i f i e r clear all . // Given C i r c u i t Data gm =3000*10^( -6) . close . //Ohms // C a l c u l a t i o n //A=−(gm∗ Rl /(1+( Rl / r p ) ) ) . close .5 Calculation of Gain of Single Stage Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 // Example 8 .A ) . //Ohms Rs =1*10^3.6 Calculation of Output Signal Voltage of FET Amplifier 1 2 3 4 5 6 7 8 9 // Example 8 . 6 // Program t o f i n d t h e Output S i g n a l V o l t a g e o f t h e Amplifier clear all . clc . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Output S i g n a l V o l t a g e o f t h e A m p l i f i e r i s vo = %f V . ” . //F u =20.10 11 12 13 14 15 16 17 18 19 20 21 Cs =25*10^( -6) . 52 . // Hz // C a l c u l a t i o n Xcs =1/(2* %pi * f * Cs ) . vo ) . vo = A * vi . Rs i s completely bypassed A = u * Rl /( Rl + rd ) . //Ohms vi =0. // As Xcs comes o u t t o be much s m a l l e r t h a n Rs . rd =100*10^3.1. //V f =1*10^3. A2 =50. // Given Data A1 =30. close . clc . 1 // Program t o C a l c u l a t e o v e r a l l V o l t a g e Gain o f a Multistage // A m p l i f i e r i n dB clear all . A3 =80. Scilab code Exa 9. Adb ) . // V o l t a g e Gain Adb =20* log10 ( A ) . // V o l t a g e Gain i n dB // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The o v e r a l l V o l t a g e Gain o f t h e M u l t i s t a g e A m p l i f i e r Adb = %f dB” . // C a l c u l a t i o n A = A1 * A2 * A3 .1 Calculate overall Voltage Gain in dB 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 // Example 9 .Chapter 9 MULTI STAGE AMPLIFIERS Scilab code Exa 9.2 Calculate Voltage at the Output Terminal 53 . // Given Data Vcc =30. Vb2 = Vc1 . Scilab code Exa 9. //V Vi =1. Vo ).Ic2 * R3 .7.3 To Plot the Frequency Response Curve 54 . Ic2 = Ie2 .4. Ie1 = Vbe / R2 . Ve2 = Vb2 . clc .1 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 28 // Example 9 .Ic1 * R1 .Vbe . Vo = %f V” .Vbe . //Ohms // C a l c u l a t i o n Ve = Vi . close . // B e e t a R1 =27*10^3.4*10^3. Vc2 = Vcc . //Ohms R3 =24*10^3. Ie2 = Ve2 / R4 . Ic1 = Ie1 . 2 // Program t o C a l c u l a t e V o l t a g e a t t h e Output Terminal o f //Two S t a g e D i r e c t Coupled A m p l i f i e r clear all . //V Vbe =0. //Ohms R2 =680. Vo = Vc2 . //V B =300. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V o l t a g e a t t h e Output T e r m i n a l o f Two S t a g e D i r e c t Coupled A m p l i f i e r . Vc1 = Vcc . //Ohms R4 =2. 1: To Plot the Frequency Response Curve 55 .Figure 9. y = [ Adbc Adb Adb Adbc ]. Adbc = Adb -3. x_label . Adbc ) . a . // Given Data A =100. x1 ) . f3 =80. x1 = [1 1 1 1]. y_label . a . a . close . text = ’ f3 f1 F r e q u e n c y ( Hz ) f2 f4 ’ .1 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 28 29 30 31 // Example 9 . // P l o t t i n g t h e F r e q u e n c y R e s p o n s e Curve x = [ f3 f1 f2 f4 ]. title . a = gca () . a . q = [10 100000]. a . Adb ( a t C u t o f f F r e q u e n c i e s ) = %f dB” . r = [37 37]. y ) . 3 // Program t o C a l c u l a t e Gain i n dB a t C u t o f f F r e q u e n c i e s and // P l o t F r e q u e n c y R e s p o n s e Curve clear all . y_location = ’ l e f t ’ . text = ’FREQUENCY RESPONSE CURVE ’ . 56 . x_location = ’ bottom ’ . plot2d (x . // C a l c u l a t i o n Adb =20* log10 ( A ) . gainplot (y . text = ’ AdB 37 ’ . // Lower by 3dB // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Gain a t C u t o f f F r e q u e n c i e s i s . f4 =40*10^3. f1 =400. clc . f2 =25*10^3. b Calculate Ouput Impedance of Two Stage RC Coupled Amplifier 1 // Example 9 . 34 q2 = [ f3 f1 f2 f4 ].6*10^3. Z i = %f kOhms ” . 6 close .r . //Ohms // C a l c u l a t i o n Zi = R1 * R2 * R3 /( R1 * R2 + R2 * R3 + R3 * R1 ) .32 plot2d2 (q .1*10^3. 5 clc . //Ohms R2 =56*10^3. 57 . clc . 4 ( a ) // Program t o C a l c u l a t e I n p u t Impedance o f t h e g i v e n //Two S t a g e RC Coupled A m p l i f i e r clear all . Scilab code Exa 9. 35 plot2d3 ( q2 . Scilab code Exa 9.4.6) . r2 . //Ohms R3 =1.4. 33 r2 = [37 40 40 37].a Calculate Input Impedance of Two Stage RC Coupled Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 // Example 9 . close .6) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The I n p u t Impedance . 4 ( b ) 2 // Program t o C a l c u l a t e Output Impedance o f t h e g i v e n 3 //Two S t a g e RC Coupled A m p l i f i e r 4 clear all . Zi /10^3) . // Given Data R1 =5. Zo = %f kOhms ” . A2 = . 12 // D i s p l a y i n g The R e s u l t s i n Command Window 13 printf ( ” \n\ t The Output Impedance . A1 = .6*10^3. Zo /10^3) . //Ohms 9 Ro2 =2.3*10^3. close .1*10^3. //Ohms R3 =5.1*10^3.8*10^3.4. //Ohms Ro2 =2. //Ohms R4 =1. // Given Data Ro1 =3. Scilab code Exa 9.3*10^3.2*10^3.hfe * Rac1 / hie . // O v e r a l l Gain // D i s p l a y i n g The R e s u l t s i n Command Window 58 . hie =1. //Ohms R1 =6. //Ohms // C a l c u l a t i o n Rac2 = Ro1 * Ro2 /( Ro1 + Ro2 ) .hfe * Rac2 / hie .2*10^3. clc . //Ohms 10 // C a l c u l a t i o n 11 Zo = Ro1 * Ro2 /( Ro1 + Ro2 ) . A = A1 * A2 . //Ohms hfe =120. //Ohms R2 =56*10^3.7 // Given Data 8 Ro1 =3. 4 ( c ) // Program t o V o l t a g e Gain o f t h e g i v e n Two S t a g e RC Coupled A m p l i f i e r clear all . Rac1 = R1 * R2 * R3 * R4 /( R1 * R2 * R3 + R2 * R3 * R4 + R1 * R3 * R4 + R1 * R2 * R4 ) .c Calculate Voltage Gain of Two Stage RC Coupled Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 // Example 9 . //Ohms Cs =100*10^( -12) . A = %f . Avmd = Avm ^2. ” . clc .22 printf ( ” \n\ t The O v e r a l l Gain . //F u =25. // Upper C u t o f f F r e q u e n c y f2d = f2 * sqrt ( sqrt (2) -1) . // Lower C u t o f f F r e q u e n c y o f Two S t a g e s f2 =1/(2* %pi * Cs * Req ) . 59 . // Given Data Rl =10*10^3. BW = %f kHz ” . // Upper C u t o f f F r e q u e n c y o f Two S t a g e s BW = f2d . // Bandwidth // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V o l t a g e Gain o f Two S t a g e s . Avmd ) . printf ( ” \n\ t The Bandwidth . 5 // Program t o C a l c u l a t e Maximum V o l t a g e Gain & Bandwidth clear all . Avm = gm * Req . BW /10^3) . f1 =1/(2* %pi * Cc * Rd ) .f1d .5 Calculate Maximum Voltage Gain and Bandwidth of Triode Amplifier 1 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 // Example 9 . //Ohms Cc =0.01*10^( -6) . rp =8*10^3. // Lower C u t o f f F r e q u e n c y f1d = f1 / sqrt ( sqrt (2) -1) . Scilab code Exa 9. Avmd = %f ” . //F // C a l c u l a t i o n gm = u / rp . close . // V o l t a g e Gain o f Two S t a g e s Rd =( rp * Rl /( rp + Rl ) ) + Rg . //Ohms Rg =470*10^3.A ) . Req = rp * Rl * Rg /( rp * Rl + Rl * Rg + Rg * rp ) . 60 . close . // Ohms RLd =10*10^3. // Ohms // C a l c u l a t i o n N12 = sqrt ( RLd / RL ) .1 Calculation of Transformer Turns Ratio 1 2 3 4 5 6 7 8 9 10 11 12 // Example 1 0 .N12 . ” .Chapter 10 POWER AMPLIFIERS Scilab code Exa 10.2 Calculation of Effective Resistance seen at Primary 1 2 // Example 1 0 . Scilab code Exa 10. // Given C i r c u i t Data RL =16. 1 // Program t o D e t e r m i n e t h e T r a n s f o r m e r Turns R a t i o clear all . clc .1) . 2 // Program t o D e t e r m i n e t h e E f f e c t i v e R e s i s t a n c e s e e n looking into 61 . // N12=N1/N2 // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The T r a n s f o r m e r Turns R a t i o i s N1/N2 = %d : %d . Scilab code Exa 10. D3 =( I3 / I1 ) *100. clc . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The E f f e c t i v e R e s i s t a n c e s e e n l o o k i n g i n t o t h e Primary . T h i r d & F o u r t h Harmonic D i s t o r t i o n s clear all . Rld /10^3) . D4 =( I4 / I1 ) *100. I2 =1. Rld = %f kOhms .3.5∗ s i n ( 2 4 0 0 ∗ t ) I1 =15. I4 =0. close .2∗ s i n ( 1 8 0 0 ∗ t ) +0. ” . 3 ( a ) // Program t o D e t e r m i n e t h e Second . //Ohms N12 =15. // Given C i r c u i t Data Rl =8. D2 = 62 . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The S e c o n d Harmonic D i s t o r t i o n i s .5. // N12=N1/N2 // C a l c u l a t i o n Rld =( N12 ) ^2* Rl . // C a l c u l a t i o n D2 =( I2 / I1 ) *100.2.5∗ s i n ( 1 2 0 0 ∗ t ) +1. // Given C i r c u i t Data // i o =15∗ s i n ( 6 0 0 ∗ t ) +1. clc .3 4 5 6 7 8 9 10 11 12 13 // t h e Primary clear all .5. close .a Calculation of 2nd 3rd and 4th Harmonic Distortions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 // Example 1 0 . I3 =1. Pi =(( P . I4 =0. P1 = poly (0 .3. Pi ( i n p e r c e n t )= ” ) .5∗ s i n ( 2 4 0 0 ∗ t ) I1 =15. I2 =1. clc . 19 printf ( ” \n\ t The F o u r t h Harmonic D i s t o r t i o n i s . D4 =( I4 / I1 ) *100. // D i s p l a y i n g The R e s u l t s i n Command Window disp ( Pi . ” The P e r c e n t a g e I n c r e a s e i n Power b e c a u s e o f D i s t o r t i o n i s . 18 printf ( ” \n\ t The T h i r d Harmonic D i s t o r t i o n i s .2∗ s i n ( 1 8 0 0 ∗ t ) +0. I3 =1.5. D3 = %f p e r c e n t . ”P1” ) .5∗ s i n ( 1 2 0 0 ∗ t ) +1. // C a l c u l a t i o n D2 =( I2 / I1 ) *100. D2 ) . ” .5. 63 . D3 ) . D3 =( I3 / I1 ) *100. ” . ” . // Given C i r c u i t Data // i o =15∗ s i n ( 6 0 0 ∗ t ) +1.P1 ) / P1 ) *100. D = sqrt ( D2 ^2+ D3 ^2+ D4 ^2) . D4 = %f p e r c e n t . Scilab code Exa 10. D4 ) . close .2. 3 ( b ) // Program t o D e t e r m i n e t h e P e r c e n t a g e I n c r e a s e i n Power b e c a u s e o f D i s t o r t i o n clear all .%f p e r c e n t . // D i s t o r t i o n F a c t o r P =(1+( D /100) ^2) * P1 .b Percentage Increase in Power because of Distortion 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 // Example 1 0 . fr /10^3) . clc .1. ” .Chapter 11 TUNED VOLTAGE AMPLIFIERS Scilab code Exa 11. //H R =20. //F L =220*10^( -6) .a Calculation of Resonant Frequency 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 1 1 . close . 64 . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The R e s o n a n t F r e q u e n c y . f r = %f kHz . // Given C i r c u i t Data C =300*10^( -12) . 1 ( a ) // Program t o C a l c u l a t e R e s o n a n t F r e q u e n c y o f t h e given Circuit clear all . //Ohms // C a l c u l a t i o n fr =1/(2* %pi * sqrt ( L * C ) ) . close .1. ” .1. //Ohms // C a l c u l a t i o n Rr = R . // Given C i r c u i t Data V =10. //H R =20. ” . I = %f A . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Impedance a t Resonance . Rr ) . //V C =300*10^( -12) . Scilab code Exa 11. 1 ( c ) // Program t o C a l c u l a t e t h e C u r r e n t a t R e s o n a n c e o f the given C i r c u i t clear all . //F L =220*10^( -6) .I 65 .Scilab code Exa 11.b Calculation of Impedance at Resonance 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 1 1 . //Ohms // C a l c u l a t i o n I=V/R. clc . Rr = %f Ohms . //F L =220*10^( -6) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The C u r r e n t a t Resonance . //H R =20.c Calculation of Current at Resonance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 // Example 1 1 . close . // Given C i r c u i t Data C =300*10^( -12) . clc . 1 ( b ) // Program t o C a l c u l a t e Impedance a t R e s o n a n c e o f t h e given Circuit clear all . ” . Vc ) . Vc = I * Xc . Xc =1/(2* %pi * fr * C ) . Vr ) . printf ( ” \n\ t V o l t a g e a c r o s s t h e R e s i s t a n c e . // Given C i r c u i t Data V =10.1. Scilab code Exa 11. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t V o l t a g e a c r o s s t h e I n d u c t a n c e . clc . //V C =300*10^( -12) . Vr = %f V .). Vr = I * R . close . printf ( ” \n\ t V o l t a g e a c r o s s t h e C a p a c i t a n c e . //Ohms // C a l c u l a t i o n fr =1/(2* %pi * sqrt ( L * C ) ) . I=V/R. Scilab code Exa 11.d Calculation of Voltage across each Component 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 // Example 1 1 . //F L =220*10^( -6) . Vc = %f V . //H R =20. ” . Vl = I * Xl . ” . Xl =2* %pi * fr * L . 1 ( d ) // Program t o C a l c u l a t e t h e V o l t a g e a c r o s s e a c h Component o f t h e g i v e n C i r c u i t clear all .2 Calculation of Parameters of the Resonant Circuit at Resonance 66 . Vl ) . Vl = %f V . //Ohms V =100. // Given C i r c u i t Data C =100*10^( -12) . //V // C a l c u l a t i o n fr =1/(2* %pi * sqrt ( L * C ) ) . Ic = V / Xc . 2 // Program t o C a l c u l a t e f r . ” . Il ) . printf ( ” \n\ t I l = %f A . ” . Xl =2* %pi * fr * L . close . Zp ) .1 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 // Example 1 1 . printf ( ” \n\ t Zp= %f Ohms . Ic ) . Xc =1/(2* %pi * fr * C ) . //F L =100*10^( -6) . Q and Bandwidth o f the 3 // R e s o n a n t C i r c u i t 1 2 67 .I /10^( -3) ) . Scilab code Exa 11.3 Calculation of Impedance Q and Bandwidth of Resonant Circuit // Example 1 1 . Il = V / Xl . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The C a l c u l a t e d V a l u e s a r e : ” ) . ” . fr /10^3) . printf ( ” \n\ t I c= %f A . printf ( ” \n\ t f r = %f kHz . ” . //H R =10. I c . L i n e C u r r e n t & Impedance o f // t h e R e s o n a n t C i r c u i t a t R e s o n a n c e clear all . ” . printf ( ” \n\ t I= %f mA. 3 // Program t o C a l c u l a t e Impedance . I l . Zp = L /( R * C ) . clc . I = V / Zp . 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 clear all . printf ( ” \n\ t Q u a l i t y F a c t o r . Zp /10^3) .Q ) . // Bandwidth // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The C a l c u l a t e d V a l u e s a r e : ” ) . df /10^3) . d f= %f kHz . // Given C i r c u i t Data C =100*10^( -12) . //H R =15. //Ohms // C a l c u l a t i o n fr =1/(2* %pi * sqrt ( L * C ) ) . printf ( ” \n\ t Impedance . Q= %f . ” . ” . clc . Q =2* %pi * fr * L / R . Zp= %f kOhms . 68 . printf ( ” \n\ t Bandwidth . //F L =150*10^( -6) . df = fr / Q . Zp = L /( R * C ) . ” . close . // Feedback F a c t o r // C a l c u l a t i o n Af = A /(1+ A * B ) . ” . // Given C i r c u i t Data A =100.Chapter 12 FEEDBACK IN AMPLIFIERS Scilab code Exa 12.1 Calculation of Gain of Negative Feedback Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 1 2 . clc . // I n t e r n a l Gain B =1/10. close .2 Calculation of Internal Gain and Feedback Gain 69 . Scilab code Exa 12. Af ) . 1 // Program t o C a l c u l a t e t h e Gain o f a N e g a t i v e Feedb ack A m p l i f i e r w i t h // Given S p e c i f i c a t i o n s clear all . Af = %f . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V a l u e o f t h e Gain o f Feedba ck A m p l i f i e r i s . V o l t a g e Gain B =0.B *100) . clc .1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 // Example 1 2 . 3 // Program t o C a l c u l a t e t h e c h a n g e i n o v e r a l l Gain o f t h e Feedb ack // A m p l i f i e r w i t h g i v e n Gain reduction clear all . A = %f . B = %f p e r c e n t . B =(( A / Af ) -1) / A . // Given C i r c u i t Data Af =100.6. I n p u t S i g n a l w i t h o u t F e e d a b a c k Gain Vi =0. // V o l t a g e Gain Vin =50*10^( -3) . 2 // Program t o C a l c u l a t e t h e A( I n t e r n a l Gain ) and B e e t a ( Fee dback Gain ) o f // a N e g a t i v e Fe edback A m p l i f i e r with given S p e c i f i c a t i o n s clear all . // 60dB . close . // Given C i r c u i t Data A =1000. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V a l u e o f t h e I n t e r n a l Gain A i s . I n p u t S i g n a l w i t h F e e d a b a c k Gain // C a l c u l a t i o n Vo = Af * Vi .3 Calculation of change in overall Gain of Feedback Amplifier 1 2 3 4 5 6 7 8 // Example 1 2 . Scilab code Exa 12. ” . A = Vo / Vin .A ) . //V . close .005. clc . // N e g a t i v e Feedb ack 70 . printf ( ” \n\ t The V a l u e o f t h e Feedback Gain B i s . //V . ” . 9 10 11 12 13 dAbyA = -0.12; //dA/A = 12 % // C a l c u l a t i o n dAfbyAf =1/(1+ A * B ) * dAbyA ; // dAf / Af =1/(1+A∗B) ∗dA/A // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The c h a n g e i n o v e r a l l Gain o f t h e Feedb ack A m p l i f i e r i s , dAf / Af = %f which i s e q u i v a l e n t t o %f p e r c e n t . ” , dAfbyAf , dAfbyAf * -100) ; Scilab code Exa 12.4 Calculation of Input Impedance of the Feedback Amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 1 2 . 4 // Program t o C a l c u l a t e t h e I n p u t Impedance o f t h e Feedb ack A m p l i f i e r // w i t h g i v e n S p e c i f i c a t i o n s clear all ; clc ; close ; // Given C i r c u i t Data Zi =1*10^3; //Ohms A =1000; // V o l t a g e Gain B =0.01; // N e g a t i v e Feedb ack // C a l c u l a t i o n Zid =(1+ A * B ) * Zi ; // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V a l u e o f t h e I n p u t Impedance o f t h e Feedb ack A m p l i f i e r i s , Z i d = %f kOhms . ” , Zid /10^3) ; Scilab code Exa 12.5 Calculation of Feedback Factor and Percent change in overall Gain 1 // Example 1 2 . 5 71 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 // Program t o C a l c u l a t e t h e v a l u e o f Feedbac k F a c t o r and P e r c e n t a g e // c h a n g e i n o v e r a l l Gain o f t h e Internal Amplifier clear all ; clc ; close ; // Given C i r c u i t Data A =1000; // 60dB , V o l t a g e Gain Zo =12000; //Ohms Zod =600; //Ohms dAbyA =0.1; //dA/A = 10 % // C a l c u l a t i o n B =(( Zo / Zod ) -1) / A ; // Zod=Zo /(1+A∗B) dAfbyAf =1/(1+ A * B ) * dAbyA ; // dAf / Af =1/(1+A∗B) ∗dA/A // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Feedback F a c t o r o f t h e Fe edback A m p l i f i e r i s , B = %f p e r c e n t . ” ,B *100) ; printf ( ” \n\ t The c h a n g e i n o v e r a l l Gain o f t h e Feedb ack A m p l i f i e r i s , dAf / Af = %f p e r c e n t . ” , dAfbyAf *100) ; 72 Chapter 13 OSCILLATORS Scilab code Exa 13.1 Calculate Frequency of Oscillation of Tuned Collector Oscillator 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 1 3 . 1 // Program t o C a l c u l a t e F r e q u e n c y o f O s c i l l a t i o n o f // Tuned C o l l e c t o r O s c i l l a t o r clear all ; clc ; close ; // Given C i r c u i t Data L =58.6*10^( -6) ; // H C =300*10^( -12) ; // F // C a l c u l a t i o n fo =1/(2* %pi * sqrt ( L * C ) ) ; // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The F r e q u e n c y o f O s c i l l a t i o n o f Tuned C o l l e c t o r O s c i l l a t o r i s f o = %f kHz . ” , fo /10^3) ; Scilab code Exa 13.2 Calculate Frequency of Oscillation of Phase Shift Oscillator 73 1 2 3 4 5 6 7 8 9 10 11 12 13 // Example 1 3 . 2 // Program t o C a l c u l a t e F r e q u e n c y o f O s c i l l a t i o n o f // Vacuum Tube Phase S h i f t O s c i l l a t o r clear all ; clc ; close ; // Given C i r c u i t Data R =100*10^3; // Ohms C =0.01*10^( -6) ; //F // C a l c u l a t i o n fo =1/(2* %pi * R * C * sqrt (6) ) ; // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The F r e q u e n c y o f O s c i l l a t i o n o f Vacuum Tube Phase S h i f t O s c i l l a t o r i s f o = %f Hz . ” , fo ) ; Scilab code Exa 13.3 Calculate Frequency of Oscillation of Wein Bridge Oscillator 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 // Example 1 3 . 3 // Program t o C a l c u l a t e F r e q u e n c y o f O s c i l l a t i o n o f // Wein B r i d g e O s c i l l a t o r clear all ; clc ; close ; // Given C i r c u i t Data R1 =220*10^3; // Ohms R2 =220*10^3; // Ohms C1 =250*10^( -12) ; //F C2 =250*10^( -12) ; //F // C a l c u l a t i o n fo =1/(2* %pi * sqrt ( R1 * C1 * R2 * C2 ) ) ; // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The F r e q u e n c y o f O s c i l l a t i o n o f Wein B r i d g e O s c i l l a t o r i s f o = %f kHz . ” , fo /10^3) ; 74 //A Vr =100.Chapter 14 ELECTRONIC INSTRUMENTS Scilab code Exa 14. // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The S e r i e s R e s i s t a n c e t o C o n v e r t g i v e n d A r s o n v a l movement i n t o a V o l t m e t e r i s . clc . //V // C a l c u l a t i o n Rtotal = Vr / Is . close . Rs = Rtotal . // Given C i r c u i t Data Rm =100.Rm . 1 // Program t o D e t e r m i n e t h e S e r i e s R e s i s t a n c e t o Convert given // d ’ A r s o n v a l movement i n t o a V o l t m e t e r w i t h t h e s p e c i f i e d Range clear all .1 Caculation of Series Resistance for coversion to Voltmeter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 // Example 1 4 . Rs = %f 75 . //Ohms Is =100*10^( -6) . close . clc . //A // C a l c u l a t i o n Ish = Imax . Scilab code Exa 14. ” . //A R =100.kOhms . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The V a l u e o f Shunt R e s i s t a n c e i s . //Ohms CS =100*10^( -6) . Scilab code Exa 14.0 −1 A clear all . //A Imax =10*10^( -3) . Rsh ) .3 Designing of a Universal Shunt for making a Multi Range Milliammeter 1 2 3 4 5 6 7 8 // Example 1 4 . Rsh = %f Ohms .0 −100 mA. 3 // Program t o D e s i g n t h e U n i v e r s a l Shunt f o r making M u l t i −Range // M i l l i a m m e t e r w i t h Range 0−1 mA . // Given C i r c u i t Data CS =100*10^( -6) . //Ohms 76 . Rs /10^3) . // Given C i r c u i t Data Rm =100. clc .2 Calculation of Shunt Resistance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 // Example 1 4 . Rsh = Rm * CS / Ish . 2 // Program t o D e t e r m i n e t h e Shunt R e s i s t a n c e r e q u i r e d clear all . close .CS . ” .0 −500 mA. CS .CS ) . R1 = %f Ohms .CS . Rm1 = Rm . Rsh1 = %f Ohms .R1 . ” .R1 . Rsh ) . R1 ) .R3 ) * Ish5 . R1 =( R * Ish2 . //A Rsh = CS * R /( Imax1 . //A Ish2 = Imax2 .R1 .CS ) .CS . Rsh1 ) . ” .CS ) . //Ohms // ( a ) C a l c u l a t i o n Imax1 =1*10^( -3) . printf ( ” \n\ t \ t \ t R5 = %f Ohms . 77 . //A Ish4 = Imax4 . R4 = %f Ohms . R2 ) . R3 ) . Rsh1 = Rm1 * CS / Ish1 .9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Rm =900. printf ( ” \n\ t For Range s w i t c h a t 10 mA . R2 =(( R .CS ) . printf ( ” \n\ t For Range s w i t c h a t 100 mA. ” . R4 =(( R . ” . printf ( ” \n\ t For Range s w i t c h a t 500 mA.CS . printf ( ” \n\ t For Range s w i t c h a t 1 A . // ( d ) C a l c u l a t i o n Imax4 =500*10^( -3) . R3 = %f Ohms .Rm * CS ) /( Ish2 . R4 ) .R4 . R3 =(( R . //A Ish5 = Imax5 . // ( e ) C a l c u l a t i o n Imax5 =1. ” .Rm * CS ) /( Ish3 . R5 ) . R2 = %f Ohms . printf ( ” \n\ t For Range s w i t c h a t 1 mA .CS . //A Ish3 = Imax3 . ” . // ( b ) C a l c u l a t i o n Imax2 =10*10^( -3) . Ish1 = Imax1 . ” .R1 ) * Ish3 . R5 =R . // ( c ) C a l c u l a t i o n Imax3 =100*10^( -3) .R2 . Rsh = %f Ohms .Rm * CS ) /( Ish5 .Rm * CS ) /( Ish4 .R2 .R3 .CS ) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t Shunt R e s i s t a n c e .R2 ) * Ish4 . D e f l e c t i o n S e n s i t i v i t y l =10. Vm = Vp /2. Vm ) . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Peak AC V o l t a g e . // Given C i r c u i t Data DS =5. ” .5 Determination of Magnitude and Frequency of Voltage Fed to Y Input 1 2 3 4 5 6 7 // Example 1 4 . 4 // Program t o D e t e r m i n e t h e AC V o l t a g e clear all . // Given C i r c u i t Data 78 . close . ” . Scilab code Exa 14. close . V = %f V .Scilab code Exa 14.4 Determination of Peak and RMS AC Voltage 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 // Example 1 4 . clc . 5 // Program t o D e t e r m i n e t h e Magnitude and t h e Frequency o f the // wave V o l t a g e f e d t o t h e Y−i n p u t clear all .V ) . //V/cm . //cm . clc . Vm = %f V . printf ( ” \n\ t The RMS AC V o l t a g e . T r a c e Length // C a l c u l a t i o n Vp = DS * l . V = Vm / sqrt (2) . ” . // D i s p l a y i n g The R e s u l t s i n Command Window printf ( ” \n\ t The Magnitude o f Wave V o l t a g e . xlabel ( ’X A x i s ’ ) . y_location = ” o r i g i n ” . a .V ) . f =1/ T . V = Vm / sqrt (2) . // s e c o n d s TP =4. y ) . T = TP * tb . x_location = ” o r i g i n ” .1*10^( -3) . a . A m p l i t u d e tb =0.6* x ) . V = %f V .Figure 14. ” . y = Am * cos (1. a = gca () . 79 .f /10^3) .01:6. // Time P e r i o d // C a l c u l a t i o n Vm =2* Am . printf ( ” \n\ t The F r e q u e n c y o f Wave V o l t a g e . // P l o t o f t h e g i v e n Wave figure x = -6:0. //V.5. f = %f kHz .1: Determination of Magnitude and Frequency of Voltage Fed to Y Input 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Am =3. // Given Waveform plot (x . title ( ’CRO OUTPUT ’ ) .28 29 30 ylabel ( ’Y A x i s ’ ) . 80 . xgrid (6) .
Report "Basic Electronics and Linear Circuits_N. N. Bhargava, D. C. Kulshreshtha and S. C. Gupta"