Scilab Textbook Companion for
Analog And Digital Electronics
by U. A. Bakshi And A. P. Godse1
Created by
Priyank Bangar
B.Tech
Electronics Engineering
NMIMS
College Teacher
No Teacher
Cross-Checked by
May 23, 2016
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: Analog And Digital Electronics
Author: U. A. Bakshi And A. P. Godse
Publisher: Technical Publications, Pune
Edition: 1
Year: 2009
ISBN: 9788184316902
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 Special Diodes
6
2 Frequency Response
8
3 Feedback Amplifiers
14
4 Oscillators
37
5 Combinational Logic Circuits
56
6 Sequential Logic Circuits
80
7 Shift Registers
82
8 Counters
83
9 Op amp Applications
115
10 Voltage Regulators
134
3
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
1.1
1.2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
current through LED . . . . . . . . . . . . . . .
transition capacitance and constant K . . . . . .
maximum voltage gain . . . . . . . . . . . . . . .
gain of the amplifier . . . . . . . . . . . . . . . .
gain of an amplifier . . . . . . . . . . . . . . . .
low frequency response of the amplifier . . . . . .
low frequency response of the FET amplifier . . .
high frequency response of the amplifier . . . . .
high frequency response of the amplifier . . . . .
gain fL and fH . . . . . . . . . . . . . . . . . . .
Vo and second harmonic distortion with feedback
beta and Af . . . . . . . . . . . . . . . . . . . . .
beta and Av and Avf and Rif and Rof . . . . . .
Avf and Rif and Rof . . . . . . . . . . . . . . . .
Avf and Rif and Rof . . . . . . . . . . . . . . . .
D and Avf and Rif and Rof . . . . . . . . . . . .
voltage gain . . . . . . . . . . . . . . . . . . . . .
Avf . . . . . . . . . . . . . . . . . . . . . . . . .
Avf and Rif and Rof . . . . . . . . . . . . . . . .
voltage gain and input and output resistance . .
Ai and beta and Aif . . . . . . . . . . . . . . . .
beta and Av and Avf and Rif and Rof . . . . . .
gain and new bandwidth . . . . . . . . . . . . . .
show voltage gain with feedback . . . . . . . . .
GMf and Rif and Rof . . . . . . . . . . . . . . .
feedback factor and Rif and Rof . . . . . . . . .
gain with feedback and new bandwidth . . . . .
overall voltage gain and bandwidth . . . . . . . .
4
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6
6
8
8
9
9
10
11
12
14
14
15
16
18
19
21
23
24
25
27
28
30
32
32
33
34
35
36
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
Exa
Exa
Exa
Exa
Exa
Exa
Exa
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
4.15
4.16
4.17
4.18
4.19
4.20
4.21
4.22
4.23
4.24
4.25
4.26
4.27
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
C and hfe . . . . . . . . . . . . . . . . . . . . . . . . .
frequency of oscillation . . . . . . . . . . . . . . . . .
R and C . . . . . . . . . . . . . . . . . . . . . . . . . .
C and RD . . . . . . . . . . . . . . . . . . . . . . . . .
minimum and maximum R2 . . . . . . . . . . . . . . .
range over capacitor is varied . . . . . . . . . . . . . .
frequency of oscillation . . . . . . . . . . . . . . . . .
frequency of oscillation . . . . . . . . . . . . . . . . .
calculate C . . . . . . . . . . . . . . . . . . . . . . . .
series and parallel resonant freqency . . . . . . . . . .
series and parallel resonant freqency . . . . . . . . . .
Varify Barkhausen criterion and find frequency of oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . .
minimum and maximum values of R2 . . . . . . . . .
frequency of oscillation and minimum hfe . . . . . . .
range of frequency of oscillation . . . . . . . . . . . . .
frequency of oscillation . . . . . . . . . . . . . . . . .
gain of the transistor . . . . . . . . . . . . . . . . . . .
new frequency and inductance . . . . . . . . . . . . .
new frequency of oscillation . . . . . . . . . . . . . . .
R and hfe . . . . . . . . . . . . . . . . . . . . . . . . .
component values of wien bridge . . . . . . . . . . . .
values of C2 and new frequency of oscillation . . . . .
series and parallel resonant freqency and Qfactor . . .
change in frequency and trimmer capacitance . . . . .
design RC phase shift oscillator . . . . . . . . . . . . .
range of capacitor . . . . . . . . . . . . . . . . . . . .
change in frequency of oscillation . . . . . . . . . . . .
design a combinational logic circuit . . . . . . . . . . .
design a circuit with control line C and data lines . . .
design combinational circuit . . . . . . . . . . . . . . .
design logic circuit . . . . . . . . . . . . . . . . . . . .
design circuit to detect invalid BCD number . . . . . .
design two level combinational circuit . . . . . . . . .
design 32 to 1 multiplexer . . . . . . . . . . . . . . . .
design a 32 to 1 multiplexer . . . . . . . . . . . . . . .
implement boolean function using 8to1 multiplexer . .
implement boolean function using 4to1 multiplexer . .
5
37
38
38
38
39
40
40
41
41
42
42
43
45
45
46
46
47
47
48
48
49
50
51
51
53
54
54
56
57
58
59
60
61
62
62
63
63
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
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
5.11
5.12
5.13
5.14
5.15
5.16
5.17
5.18
5.19
5.20
5.21
5.22
5.23
5.24
5.25
5.26
5.27
5.28
5.29
5.31
6.4
7.1
8.1
8.2
8.4
8.5
8.6
8.8
8.9
8.10
8.11
8.12
8.13
8.14
8.15
8.16
8.17
8.18
implement boolean function using 8to1 MUX
implement boolean function using 4to1 MUX
implement boolean function using 8to1 MUX
implement boolean function using 8to1 MUX
implement boolean function using 8to1 MUX
determine boolean expression . . . . . . . . .
realize using 4 to 1 MUX . . . . . . . . . . .
design 1 to 8 DEMUX . . . . . . . . . . . . .
implement full subtractor . . . . . . . . . . .
construst 1 to 32 DEMUX . . . . . . . . . . .
design 3 to 8 decoder . . . . . . . . . . . . .
design 5 to 32 decoder . . . . . . . . . . . . .
design 4 line to 16 line decoder . . . . . . . .
implement using 74LS138 . . . . . . . . . . .
implement full substractor . . . . . . . . . . .
implement gray to binary code converter . . .
design 2 bit comparator . . . . . . . . . . . .
design full adder circuit . . . . . . . . . . . .
implement BCD to 7 segment decoder . . . .
implement 32 input to 5 output encoder . . .
analyze the circuit . . . . . . . . . . . . . . .
determine number of flip flops needed . . . .
count after 12 pulse . . . . . . . . . . . . . .
count in binary . . . . . . . . . . . . . . . . .
draw logic diagram . . . . . . . . . . . . . . .
output frequency . . . . . . . . . . . . . . . .
maximum operating frequency . . . . . . . .
design 4 bit up down ripple counter . . . . .
design divide by 9 counter . . . . . . . . . . .
design a divide by 128 counter . . . . . . . .
design divide by 78 counter . . . . . . . . . .
design divide by 6 counter . . . . . . . . . . .
find fmax . . . . . . . . . . . . . . . . . . . .
determine states . . . . . . . . . . . . . . . .
design MOD 10 counter . . . . . . . . . . . .
design down counter . . . . . . . . . . . . . .
design programmable frequency divider . . .
design a counter . . . . . . . . . . . . . . . .
6
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65
65
66
67
68
68
69
70
70
71
72
73
74
74
75
75
76
77
78
78
80
82
83
83
84
84
84
85
85
86
86
87
87
88
88
89
90
90
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
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
8.19
8.20
8.21
8.22
8.23
8.24
8.25
8.26
8.28
8.30
8.31
8.32
8.33
8.34
8.35
8.36
8.37
8.38
8.39
8.40
8.41
8.42
8.44
8.45
9.1
9.3
9.4
9.5
9.6
9.8
9.9
9.10
9.11
9.12
9.13
9.17
9.18
9.19
programmable frequency divider . . . . . .
design divide by 2 and divide by 5 counter .
design mod 9 counter . . . . . . . . . . . .
determine MOD number and counter range
design a divide by 20 counter . . . . . . . .
design divide by 96 counter . . . . . . . . .
design divide by 93 counter . . . . . . . . .
design divide by 78 counter . . . . . . . . .
7490 IC . . . . . . . . . . . . . . . . . . . .
sketch output waveforms of counter . . . . .
explain operation of circuit . . . . . . . . .
design divide by 40000 counter . . . . . . .
design modulo 11 counter . . . . . . . . . .
design excess 3 decimal counter . . . . . . .
modulus greater than 16 . . . . . . . . . . .
modulo 8 counter . . . . . . . . . . . . . . .
synchronous decade counter . . . . . . . . .
flip flops . . . . . . . . . . . . . . . . . . . .
design mod 5 synchronous counter . . . . .
design MOD 4 down counter . . . . . . . .
design MOD 12 synchronous counter . . . .
design 4 bit 4 state ring counter . . . . . .
design 4 bit 8 state johnson counter . . . .
johnson counter . . . . . . . . . . . . . . .
output voltage . . . . . . . . . . . . . . . .
input bias current . . . . . . . . . . . . . .
design inverting schmitt trigger . . . . . . .
threshold voltage . . . . . . . . . . . . . . .
tripping voltage . . . . . . . . . . . . . . .
time duration . . . . . . . . . . . . . . . . .
calculate R1 and R2 . . . . . . . . . . . . .
calculate trip point and hysteresis . . . . .
design schmitt trigger . . . . . . . . . . . .
design op amp schmitt trigger . . . . . . . .
VUT and VLT and frequency of oscillation
design op amp circuit . . . . . . . . . . . .
design monostable using IC 555 . . . . . . .
design a timer . . . . . . . . . . . . . . . .
7
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92
92
94
94
94
95
96
96
96
97
100
101
102
102
103
104
104
106
108
110
111
113
113
114
115
116
116
117
117
118
119
119
120
121
122
123
123
124
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
9.20
9.21
9.26
9.27
9.28
9.32
9.33
9.34
9.35
9.36
9.38
9.39
9.40
9.41
9.42
9.43
10.1
10.2
10.4
10.7
10.8
draw timer using IC 555 . . . . . . . . . . . . . .
frequency of output and duty cycle . . . . . . . .
design astable multivibrator . . . . . . . . . . . .
design astable multivibrator . . . . . . . . . . . .
design astable mode to generate square wave . .
define resolution . . . . . . . . . . . . . . . . . .
output voltage . . . . . . . . . . . . . . . . . . .
find VoFS . . . . . . . . . . . . . . . . . . . . . .
find out stepsize and analog output . . . . . . . .
find output voltage . . . . . . . . . . . . . . . . .
find resolution and digital output . . . . . . . . .
calculate quantizing error . . . . . . . . . . . . .
dual scope ADC . . . . . . . . . . . . . . . . . .
find digital output of ADC . . . . . . . . . . . .
find conversion time . . . . . . . . . . . . . . . .
find maximum frequency of input sine wave . . .
find line and load regulation and ripple refection
design op amp series voltage regulator . . . . . .
calculate output voltage . . . . . . . . . . . . . .
determine regulated output voltage . . . . . . . .
find the range . . . . . . . . . . . . . . . . . . . .
8
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125
125
126
127
128
129
129
130
130
130
131
132
132
132
133
133
134
135
135
136
136
List of Figures
8.1
sketch output waveforms of counter . . . . . . . . . . . . . .
9
100
Chapter 1
Special Diodes
Scilab code Exa 1.1 current through LED
1 // Example 1 . 1
2 clc
3 format (5)
4 disp ( ” Assume t h e d r o p a c r o s s t h e LED a s 2 V . ” )
5 disp ( ” T h e r e f o r e ,
VD = 2 V” )
6 disp ( ”From f i g . 1 . 1 1 ,
RS = 2 . 2 k−ohm and VS = 15 V” )
7 is =(15 -2) /(2.2)
// i n mA
8 disp ( is , ” T h e r e f o r e ,
I S (mA) = VS−VD / RS =” )
Scilab code Exa 1.2 transition capacitance and constant K
1 // Example 1 . 2
2 clc
3 disp ( ” The t r a n s i s t o r c a p a c i t a n c e i s g i v e n by , ” )
4 disp ( ” CT = C ( 0 ) / [ 1 + |VR/VJ | ˆ n ] ” )
5 disp ( ”Now C ( 0 ) = 80pF , n = 1/3 a s d i f f u s e d j u n c t i o n
6
”)
disp ( ”
VR = 4 . 2 V,
VJ = 0 . 7 V” )
10
7 ct =((80*10^ -12) /((1+(4.2/0.7) ) ^(1/3) ) ) *10^12
// i n
pF
8 format (6)
9 disp ( ct , ” T h e r e f o r e , CT( pF ) = ” )
10 disp ( ” t h e t r a n s i s t o r c a p a c i t a n c e i s
11 disp ( ” CT = K / [VR+VJ ] ˆ n” )
12 format (10)
13 k =(41.82*10^ -12) *((4.2+0.7) ^(1/3) )
14 disp (k , ” T h e r e f o r e , K = ” )
11
a l s o g i v e n by , ” )
Chapter 2
Frequency Response
Scilab code Exa 2.1 maximum voltage gain
1 // Example 2 . 1
2 clc
3 format (7)
4 disp ( ”We know t h a t maximam v o l t a g e g a i n o f
voltage
a m p l i f i e r i s given as ”)
5 mv =200* sqrt (2)
6 disp ( mv , ” T h e r e f o r e ,
Maximum v o l t a g e g a i n = Gain a t
cut − o f f x s q r t ( 2 ) =” )
Scilab code Exa 2.2 gain of the amplifier
1 // Example 2 . 2
2 clc
3 format (6)
4 disp ( ”We know t h a t , ” )
5 a =100/ sqrt (1+((1000/20) ^2) )
6 disp (a , ” Below midband :
/ f ) ˆ 2 ) =” )
12
A = A mid / s q r t (1+( f 1
Scilab code Exa 2.3 gain of an amplifier
1 // Example 2 . 3
2 clc
3 format (7)
4 a =200* sqrt (2)
5 disp (a , ”We know t h a t
A mid = 3dB g a i n x s q r t ( 2 )
=” )
6 am =282.84/( sqrt (1+(((10/2) ^2) ) ) )
7 format (6)
8 disp ( am , ” Above midband :
A = A mid / s q r t (1+(
f 1 / f ) ˆ 2 ) =” ) // a n s w e r i n t e x t b o o k i s wrong
Scilab code Exa 2.4 low frequency response of the amplifier
1 // Example 2 . 4
2 clc
3 format (6)
4 disp ( ” I t i s n e c e s s a r y t o a n a l y z e e a c h n e t w o r k t o
5
6
7
8
9
10
11
12
determine the c r i t i c a l frequency of the a m p l i f i e r
”)
disp ( ” ( a ) I n p u t RC n e t w o r k ” )
fc1 =1/(2* %pi *[680+1031.7]*(0.1*10^ -6) )
disp ( fc1 , ”
f c ( i n p u t ) ( i n Hz ) = 1 / 2∗ p i ∗ [ RS+(
R1 | | R2 | | h i e ) ] C1 =” ) // i n Hz
disp ( ” ( b ) Output RC n e t w o r k ” )
format (7)
fc2 =1/(2* %pi *((2.2+10) *10^3) *(0.1*10^ -6) )
disp ( fc2 , ”
f c ( o u t p u t ) ( i n Hz ) = 1 / 2∗ p i ∗ (RC+
RL) ∗C2 =” ) // i n Hz
disp ( ” ( c ) Bypass RC n e t w o r k ” )
13
13 rth =((68*22*0.680) /((22*0.680) +(68*0.680) +(68*22) ) )
*10^3
14 disp ( rth , ” R th ( i n ohm ) = R1 | | R2 | | RS =” )
15 format (6)
16 fc3 =1/(2* %pi *17.23*10*10^ -6)
17 disp ( fc3 , ”
f c ( b y p a s s ) ( i n Hz ) = 1 / 2∗ p i ∗ [ (
R th+h i e / b e t a ) | | RE ] ∗ CE” )
disp ( ”We have c a l c u l a t e d a l l t h e t h r e e c r i t i c a l
f r e q u e n c i e s : ”)
19 disp ( ” ( a ) f c ( i n p u t ) = 9 2 9 . 8 Hz” )
20 disp ( ” ( b ) f c ( o u t p u t ) = 1 3 0 . 4 5 Hz” )
21 disp ( ” ( c ) f c ( b y p a s s ) = 9 2 3 . 7 Hz” )
18
Scilab code Exa 2.5 low frequency response of the FET amplifier
1 // Example 2 . 5
2 clc
3 disp ( ” I t i s n e c e s s a r y t o a n a l y z e e a c h n e t w o r k t o
4
5
6
7
8
9
10
11
12
13
14
15
16
17
determine the c r i t i c a l frequency of the a m p l i f i e r
”)
disp ( ” ( a ) I n p u t RC Network ” )
disp ( ”
f c = 1 / 2∗ p i ∗ R i n ∗C1” )
format (6)
rin =(100*100) /(100+100)
disp ( rin , ” where R i n ( i n M−ohm ) = RG | | R i n ( g a t e ) =
RG | | | VGS/ IGSS | =” )
format (5)
fc1 =1/(2* %pi *50*10^6*0.001*10^ -6)
disp ( fc1 , ” T h e r e f o r e ,
f c ( i n Hz ) =” )
disp ( ” ( b ) Output RC Network ” )
format (6)
fc2 =1/(2* %pi *(24.2*10^3) *(1*10^ -6) )
disp ( fc2 , ”
f c ( i n Hz ) = 1 / 2∗ p i ∗ (RD+RL) ∗C2 =” )
disp ( ”We have c a l c u l a t e d two c r i t i c a l f r e q u e n c i e s ” )
disp ( ” ( a ) f c ( i n p u t ) = 3 . 1 8 Hz” )
14
18
disp ( ” ( b ) f c ( o u t p u t ) = 6 . 5 7 7 Hz” )
Scilab code Exa 2.6 high frequency response of the amplifier
1 // Example 2 . 6
2 clc
3 disp ( ” B e f o r e c a l c u l a t i n g
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
critical frequencies it is
n e c e s s a r y t o c a l c u l a t e mid f r e q u e n c y g a i n o f t h e
given c i r c u i t . This i s r e q u i r e d t o c a l c u l a t e C in (
m i l l e r ) and C o u t ( m i l l e r ) ” )
disp ( ”
Av = −h f e ∗Ro / Ri ” )
disp ( ” where Ri = h i e | | R1 | | R2” )
disp ( ” and
Ro = RC | | RL” )
format (6)
av =( -100*1.8) /1.032
disp ( av , ” T h e r e f o r e , Av = −h f e (RC | | RL) / h i e | | R1 | | R2
=” )
disp ( ” N e g a t i v e s i g n i n d i c a t e s 180 d e g r e e s h i f t
b e t w e e n i n p u t and o u t p u t ” )
format (7)
cin =(4*(174.4+1) ) *10^ -3 // i n nF
disp ( cin , ”
C i n ( m i l l e r ) ( i n nF ) = C bc ∗ ( Av+1) =” )
cout =(4*175.4) /(174.4) // i n pF
format (4)
disp ( cout , ”
C o u t ( m i l l e r ) ( i n pF ) = C bc ∗ ( Av+1) /
Av =” )
disp ( ”We now a n a l y z e i n p u t and o u t p u t n e t w o r k f o r
c r i t i c a l frequency . ”)
format (8)
fci =(1/(2* %pi *410*0.7216*10^ -9) ) *10^ -3 // i n kHz
disp ( fci , ”
f c ( i n p u t ) ( i n kHz ) = 1 / 2∗ p i ∗ ( Rs | | R1
| | R2 | | h i e ) ∗ ( C be+C i n ( m i l l e r ) ) =” )
format (5)
fco =(1/(2* %pi *((22*10^6) /(12.2*10^3) ) *(4*10^ -12) ) )
*10^ -6 // i n MHz
15
disp ( fco , ”
f c ( o u t p u t ) ( i n MHz) = 1 / 2∗ p i ∗ (RC | | RL
) ∗ C o u t ( m i l l e r ) =” )
24 disp ( ”We have c a l c u l a t e d b o t h t h e c r i t i c a l
f r e q u e n c i e s ”)
25 disp ( ” ( a ) f c ( i n p u t ) = 5 3 7 . 9 4 7 kHz ” )
26 disp ( ” ( b ) f c ( o u t p u t ) = 2 2 . 1 MHz” )
23
Scilab code Exa 2.7 high frequency response of the amplifier
1 // Example 2 . 7
2 clc
3 disp ( ” B e f o r e c a l c u l a t i n g
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
critical frequencies it is
n e c e s s a r y t o c a l c u l a t e mid f r e q u e n c y g a i n o f t h e
given a m p l i f i e r c i r c u i t . This i s r e q u i r e d to
c a l c u l a t e C i n ( m i l l e r ) and C o u t ( m i l l e r ) ” )
disp ( ”
Av = −gm ∗ RD” )
disp ( ” Here , RD s h o u l d b r r e p l a c e d by RD | | RL” )
av = -6*2
disp ( av , ” T h e r e f o r e ,
Av = −gm∗ (RD | | RL) =” )
cin =2*(12+1) // i n pF
disp ( cin , ” C i n ( m i l l e r ) ( i n pF ) = C gd ∗ ( Av+1) = C r s s
∗ ( Av+1) =” )
format (6)
cout =(2*13) /12 // i n pF
disp ( cout , ” C o u t ( m i l l e r ) ( i n pF ) = C gd ∗ ( Av+1) / Av =
”)
disp ( ” G gs = C i s s − C r s s = 4 pF” )
disp ( ”We know a n a l y z e i n p u t and o u t p u t n e t w o r k f o r
c r i t i c a l frequency ”)
disp ( ”
f c ( i n p u t ) = 1 / 2∗ p i ∗RS∗CT” )
disp ( ”
= 1 / 2∗ p i ∗RS ∗ [ C g s+C i n ( m i l l e r
) ] ”)
format (4)
fc1 =(1/(2* %pi *100*30*10^ -12) ) *10^ -6 // i n MHz
disp ( fc1 , ”
f c ( i n p u t ) ( i n MHz)= ” )
16
20 fc2 =(1/(2* %pi *((48.4*10^6) /(24.2*10^3) )
*(2.166*10^ -12) ) ) *10^ -6 // i n MHz
format (6)
disp ( fc2 , ”
f c ( o u t p u t ) ( i n MHz) = 1 / 2∗ p i ∗ (RD | | RL
) ∗ C o u t ( m i l l e r ) =” )
23 disp ( ”We have c a l c u l a t e d b o t h t h e c r i t i c a l
f r e q u e n c i e s : ”)
24 disp ( ” ( a ) f c ( i n p u t ) = 53 MHz” )
25 disp ( ” ( b ) f c ( o u t p u t ) = 3 6 . 7 4 MHz” )
21
22
17
Chapter 3
Feedback Amplifiers
Scilab code Exa 3.1 gain fL and fH
1 // Example 3 . 1
2 clc
3 disp ( ” ( a ) Gain w i t h f e e d b a c k ” )
4 format (5)
5 av =1000/(1+(0.05*1000) )
6 disp ( av , ”
AV mid = Av mid / 1+ b e t a ∗ Av mid =” )
7 flf =50/(1+(0.05*1000) )
// i n Hz
8 disp ( flf , ” ( b )
f L f ( i n Hz ) = f L / 1+ b e t a ∗ Av mid
=” )
9 fhf =((50*10^3) *(1+(0.05*1000) ) ) *10^ -6
// i n MHz
10 disp ( fhf , ” ( c )
f H f ( i n MHz) = f H ∗ (1+ b e t a ∗
Av mid ) =” )
Scilab code Exa 3.2 Vo and second harmonic distortion with feedback
1 // Example 3 . 2
2 clc
3 disp ( ” ( a ) b e t a :
−40 = 20∗ l o g [1+ b e t a ∗A ] ” )
18
4 disp ( ” T h e r e f o r e ,
1+ b e t a ∗A = 100 ” )
5 b =99/1000
6 format (6)
7 disp (b , ” T h e r e f o r e ,
b e t a =” )
8 disp ( ” Gain o f t h e a m p l i f i e r w i t h f e e d b a c k
9
10
11
12
13
14
15
16
i s given
as ”)
avf =1000/100
disp ( avf , ”
A Vf = A V / 1+ b e t a ∗A V =” )
disp ( ” ( b ) To m a i n t a i n o u t p u t power 10 W, we s h o u l d
m a i n t a i n o u t p u t v o l t a g e c o n s t a n t and t o m a i n t a i n
o u t p u t c o n s t a n t w i t h f e e d b a c k g a i n r e q u i r e d Vs i s
”)
vsf =10*100*10^ -3 // i n V
disp ( vsf , ”
V s f ( i n V) = Vs ∗ 100 =” )
disp ( ” ( c ) S e c o n d h a r m o n i c d i s t o r t i o n i s r e d u c e d by
f a c t o r 1 + b e t a ∗A” )
d2f =(0.1/100) *100 // i n p e r c e n t a g e
disp ( d2f , ”
D 2 f ( i n p e r c e n t a g e ) = D 2 / 1+ b e t a ∗A
=” )
Scilab code Exa 3.3 beta and Af
1 // Example 3 . 3
2 clc
3 disp ( ” ( a ) We know t h a t ” )
4 disp ( ”
dAf / Af = 0 . 1 / 1 + b e t a ∗A ∗ dA/A” )
5 disp ( ” T h e r e f o r e ,
1+ b e t a ∗A = 3 7 . 5 ” )
6 b =(36.5/2000) *100 // i n p e r c e n t a g e
7 format (6)
8 disp (b , ” T h e r e f o r e ,
b e t a ( i n p e r c e n t a g e ) =” )
9 af =2000/(1+(0.01825*2000) )
10 disp ( af , ” ( b )
Af = A / 1+ b e t a ∗A =” )
19
Scilab code Exa 3.4 beta and Av and Avf and Rif and Rof
1 // Example 3 . 4
2 clc
3 disp ( ” S t e p 1 : I d e n t i t y t o p o l o g y ” )
4 disp ( ” The f e e d b a c k v o l t a g e i s a p p l i e d
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
a c r o s s the
r e s i s t a n c e R e1 and i t i s i n s e r i e s w i t h i n p u t
s i g n a l . Hence f e e d b a c k i s v o l t a g e s e r i e s f e e d b a c k
. ”)
disp ( ” ” )
disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t . ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 (
c o n n e c t i n g C2 t o g r o u n d ) , which g i v e s p a r a l l e l
c o m b i n a t i o n o f Re w i t h Rf a t E1 . To f i n d o u t p u t
c i r c u i t , s e t I i = 0 ( o p e n i n g t h e i n p u t node E1 a t
e m i t t e r o f Q1 ) , which g i v e s s e r i e s c o m b i n a t i o n
o f Rf and Re1 a c r o s s t h e o u t p u t . The r e s u l t a n t
c i r c u i t i s shown i n F i g . 3 . 3 2 ” )
disp ( ” ” )
disp ( ” S t e p 4 : Find open l o o p v o l t a g e g a i n ( A v ) ” )
format (5)
rl2 =(4.7*10.1) /(4.7+10.1) // i n k−ohm
disp ( rl2 , ” R L2 ( i n k−ohm ) = R c2 | | ( R e1+Rf ) =” )
disp ( ” A i 2 = −h f e = −100 ” )
disp ( ” R i 2 = h i e = 1 1 0 0 ohm” )
format (7)
av2 =( -100*3.21*10^3) /1100
disp ( av2 , ” A v2 = A i 2 ∗ R L2 / R i 2 =” )
disp ( ” A i 1 = −h f e = −100 ” )
format (5)
rl1 =(22*220*22*1.100) /((220*22*1.100) +(22*22*1.100)
+(22*220*1.100) +(22*220*22) ) // i n ohm
disp ( rl1 *10^3 , ” R L1 ( i n ohm ) = R c1 | | R3 | | R4 | |
R i 2 =” )
ri1 =1.1+(101*((0.1*10) /(0.1+10) ) ) // i n k−ohm
format (5)
disp ( ri1 , ” R i 1 ( i n k−ohm ) = h i e + (1+ h f e ) ∗ R e 1 e f f =
20
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
where R e 1 e f f = ( R e1 | | Rf ) ” )
av1 =( -100*995) /(11.099*10^3)
disp ( av1 , ” T h e r e f o r e , A v1 = A i 1 ∗RL1 / Ri1 =” )
disp ( ” The o v e r a l l v o l t a g e g a i n w i t h o u t f e e d b a c k i s
g i v e n as , ” )
av = -291.82* -8.96
format (7)
disp ( av , ” Av = A v1 ∗ A v2 =” )
disp ( ” The o v e r a l l v o l t a g e g a i n t a k i n g Rs i n a c c o u n t
i s g i v e n as , ” )
aV =(2614.7*11.099*10^3) /((11.099*10^3) +100)
format (8)
disp ( aV , ” Av = Vo / Vs = Av∗ R i 1 / R i 1+Rs =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e b e t a ” )
disp ( ” L o o k i n g a t F i g . 3 . 3 3 . ” )
beta =100/(100+(10*10^3) )
format (7)
disp ( beta , ” b e t a = Vf / Vo =” )
d =1+(0.0099*2591.35)
format (6)
disp (d , ” D = 1 + b e t a ∗Av =” )
avf =2591.35/26.65
disp ( avf , ” A v f = Av/D =” )
rif =26.65*11.099 // i n k−ohm
format (8)
disp ( rif , ” R i f ( i n k−ohm ) = R i 1 ∗ D =” )
riff =(295.788*220*22) /((220*22) +(295.788*22)
+(295.788*220) ) // i n k−ohm
format (6)
disp ( riff , ” R ’ ’ i f ( i n k−ohm ) = R i f | | R1 | | R2 =” )
disp ( ” R o f = Ro / D = i n f i n i t y / D = i n f i n i t y ” )
disp ( ” T h e r e f o r e , R ’ ’ o f = R ’ ’ o / D
where
R ’ ’ o = R L2 ” )
roff =(3.21*10^3) /26.65 // i n omh
format (7)
disp ( roff , ” T h e r e f o r e , R ’ ’ o f ( i n ohm ) = ” )
21
Scilab code Exa 3.5 Avf and Rif and Rof
1 // Example 3 . 5
2 clc
3 disp ( ” S t e p 1 : I d e n t i t y t o p o l o g y ” )
4 disp ( ” The f e e d b a c k v o l t a g e i s a p p l i e d
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
a c r o s s R1
( 1 0 0 ohm ) , which i s i n s e r i e s w i t h i n p u t s i g n a l .
Hence f e e d b a c k i s v o l t a g e s e r i e s f e e d b a c k . ” )
disp ( ” ” )
disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 , which
g i v e s p a r a l l e l c o m b i n a t i o n o f R1 w i t h R2 a t E1 a s
shown i n t h e f i g . 3 . 4 5 . To f i n d o u t p u t c i r c u i t ,
s e t I i = 0 by o p e n i n g t h e i n p u t node , E1 a t
e m i t t e r o f Q1 , which g i v e s t h e s e r i e s c o m b i n a t i o n
o f R2 and R1 a c r o s s t h e o u t p u t . The r e s u l t a n t
c i r c u i t i s shown i n f i g . 3 . 4 5 ” )
disp ( ” ” )
disp ( ” S t e p 4 : Find t h e open l o o p v o l t a g e g a i n ( Av ) ” )
rl2 =(4.7*4.8) /(4.7+4.8) // i n k−ohm
format (5)
disp ( rl2 , ” R L2 ( i n k−ohm ) =” )
disp ( ” S i n c e
h o e = h r e = 0 we can u s e a p p r o x i m a t e
a n a l y s i s ”)
disp ( ” A i 2 = −h f e = −50” )
disp ( ” R i 2 = h i e = 1 . 1 k−ohm” )
av2 =( -50*2.37) /1.1
format (7)
disp ( av2 , ” A v2 = A i 2 ∗ R L2 / R i 2 =” )
rl1 =(10*47*33*1.1) /((47*33*1.1) +(10*33*1.1)
+(10*47*1.1) +(10*47*33) ) // i n ohm
format (5)
disp ( rl1 *10^3 , ” R L1 ( i n ohm ) =” )
22
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
disp ( ” A i 1 = −h f e = −50” )
ri1 =1.1+(51*((0.1*4.7) /(4.8) ) ) // i n k−ohm
format (6)
disp ( ri1 , ” R i 1 ( i n k−ohm ) = h i e + (1+ h f e ) ∗Re =” )
av1 =( -50*942) /(6.093*10^3)
format (5)
disp ( av1 , ” A v1 = A i 1 ∗ R L1 / R i 1 =” )
av = -7.73* -107.73
format (7)
disp ( av , ” T h e r e f o r e , A v = A v1 ∗ A v2 =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e b e t a and D” )
disp ( ” b e t a = R1 / R1+R2 = 1 / 4 8 ” )
d =1+(832.75/48) // i n ohm
format (6)
disp (d , ” D( i n ohm ) = 1 + A∗ b e t a =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e A vf , R o f and R i f ” )
avf =832.75/18.35
disp ( avf , ” A v f = A v / D =” )
rif =6.093*18.35 // i n k−ohm
disp ( rif , ” R i f ( i n k−ohm ) = R i 1 ∗ D =” )
rof =(2.37*10^3) /18.35 // i n ohm
format (7)
disp ( rof , ” R o f ( i n ohm ) = R o / D =” )
Scilab code Exa 3.6 Avf and Rif and Rof
1 // Example 3 . 6
2 clc
3 disp ( ” S t e p 1 : I d e n t i f y t o p o l o g y ” )
4 disp ( ” The f e e d b a c k v o l t a g e i s a p p l i e d
a c r o s s R e1
= 1 . 5 k−ohm , which i s i n s e r i e s w i t h i n p u t s i g n a l
. Hence f e e d b a c k i s v o l t a g e s e r i e s f e e d b a c k ” )
5 disp ( ” ” )
23
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
disp ( ” S t e p 2 and s t e p 3 : Find i n p u t and o u t p u t
c i r c u i t ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 , which
g i v e s p a r a l l e l c o m b i n a t i o n o f R e1 w i t h R f a t E1
a s shown i n f i g . 3 . 4 7 . To f i n d ouput c i r c u i t , s e t
I i = 0 by o p e n i n g t h e i n p u t node , E1 a t e m i t t e r
o f Q1 , which g i v e s t h e s e r i e s c o m b i n a t i o n o f R f
and R e1 a c r o s s t h e o u t p u t . The r e s u l t a n t
c i r c u i t i s shown i n f i g . 3 . 4 7 ” )
disp ( ” ” )
disp ( ” S t e p 4 : Find t h e open l o o p v o l t a g e g a i n ( Av ) ” )
rl2 =(2.2*57.5) /(2.2+57.5) // i n k−ohm
format (6)
disp ( rl2 , ” R L2 ( i n k−ohm ) = R c2 | | ( Rf + R e1 ) =” )
disp ( ” S i n c e hoe ∗ R L2 = 1 0 ˆ − 6 ∗ 2 . 1 1 9 k−ohm = 0 . 0 0 2 1 1 9
i s l e s s t h a n 0 . 1 we u s e a p p r o x i m a t e a n a l y s i s . ” )
disp ( ” A i 2 = − h f e = −200 ” )
disp ( ” R i 2 = h i e = 2 k−ohm” )
av2 =( -200*2.119) /2
disp ( av2 , ” A v2 = A i 2 ∗ R L2 / R i 2 =” )
rl1 =(120*2) /(122) // i n k−ohm
disp ( rl1 , ” R L1 ( i n k−ohm ) = R C1 | | R i 2 =” )
disp ( ” S i n c e hoe ∗ R L1 = 1 0 ˆ − 6 ∗ 1 . 9 6 7 = 0 . 0 0 1 9 6 7 i s
l e s s t h a n 0 . 1 we u s e a p p r o x i m a t e a n a l y s i s . ” )
disp ( ” A i 1 = −h f e = −200 ” )
ri1 =2+(201*((1.5*56) /(57.5) ) ) // i n k−ohm
format (7)
disp ( ri1 , ” R i 1 ( i n k−ohm ) = h i e + (1+ h f e ) ∗Re =” )
av1 =( -200*1.967) /295.63
format (5)
disp ( av1 , ” T h e r e f o r e , A v1 = A i 1 ∗ R L1 / R i 1 =” )
disp ( ” The o v e r a l l g a i n w i t h o u t f e e d b a c k i s ” )
av = -1.33* -211.9
format (7)
disp ( av , ” Av = A v1 ∗ A v2 =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e b e t a ” )
beta =1.5/57.5
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51
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format (6)
disp ( beta , ” b e t a = Vf / Vo =” )
disp ( ” ” )
disp ( ” S t e p 6 : c a l c u l a t e D, A vf , R i f , R o f ” )
d =1+(0.026*281.82)
disp (d , ” D = 1 + Av∗ b e t a =” )
avf =281.82/8.327
disp ( avf , ” T h e r e f o r e ,
A v f = Av / D =” )
ri =(295.63*150) /(295.63+150) // i n k−ohm
format (5)
disp ( ri , ” Ri ( i n k−ohm ) = R i 1 | | R =” )
rif =99.5*8.327 // i n k−ohm
format (7)
disp ( rif , ” R i f ( i n k−ohm ) = Ri ∗D =” )
disp ( ” Ro = 1/ hoe = 1 M−ohm” )
rof =((1*10^6) /8.327) *10^ -3 // i n k−ohm
format (4)
disp ( rof , ” R o f ( i n k−ohm ) = Ro / D =” )
ro =(1000*2.119) /(2.119+1000) // i n k−ohm
format (7)
disp ( ro , ” R ’ ’ o ( i n k−ohm ) = Ro | | R c2 | | ( Rf+R e1 )
= Ro | | R L2 =” )
56 rof =(2.1145*10^3) /8.327
// i n ohm
57 format (4)
58 disp ( rof , ” R ’ ’ o f ( i n ohm ) = R ’ ’ o / D =” )
Scilab code Exa 3.7 D and Avf and Rif and Rof
1 // Example 3 . 7
2 clc
3 disp ( ” S t e p 1 : I d e n t i t y t o p o l o g y ” )
4 disp ( ” By s h o r t i n g o u t p u t v o l t a g e ( Vo = 0 ) ,
f e e d b a c k v o l t a g e Vf becomes z e r o and h e n c e i t i s
v o l t a g e s a m p l i n g . The f e e d b a c k v o l t a g e i s a a p l i e d
in s e r i e s with input v o l t a g e hence the topology
25
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i s v o l t a g e s e r i e s feedback . ”)
disp ( ” ” )
disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t . ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 . T h i s
p l a c e s t h e p a r a l l e l c o m b i n a t i o n o f r e s i s t o r 10 K
and 200 ohm a t f i r s t s o u r c e . To f i n d o u t p u t
c i r c u i t , s e t I i = 0 . T h i s p l a c e s t h e r e s i s t o r 10
K and 200 ohm i n s e r i e s a c r o s s t h e o u t p u t . The
r e s u l t a n t c i r c u i t i s shown i n f i g . 3 . 5 0 . ” )
disp ( ” ” )
disp ( ” S t e p 4 : R e p l a c e FET w i t h i t s e q u i v a l e n t
c i r c u i t a s shown i n f i g . 3 . 5 1 ” )
disp ( ” ” )
disp ( ” S t e p 5 : Find open l o o p t r a n s f e r g a i n . ” )
disp ( ” Av = Vo / Vs = A v1 ∗ A v2 ” )
disp ( ” A v2 = −u∗ R L2 / R L2+r d ” )
rl2 =(10.2*47) /(10.2+47) // i n k−ohm
format (5)
disp ( rl2 , ” where R L2 ( i n k−ohm ) =” )
av2 =( -40*8.38) /(8.38+10)
format (7)
disp ( av2 , ” T h e r e f o r e , A v2 =” )
disp ( ” A v1 = u∗ R D e f f / r d+R D e f f +(1+u ) ∗ R s e f f ” )
rdeff =(47*1000) /(47+1000) // i n k−ohm
format (6)
disp ( rdeff , ” where R D e f f ( i n k−ohm ) = R D | | R G2 =”
)
disp ( ” R s e f f = 200 | | 10 K” )
av1 =( -40*44.98*10^3) /((10*10^3) +(44.89*10^3)
+(41*((10*0.2) /(10.2) ) ) )
// a n s w e r i n t e x t b o o k i s
disp ( av1 , ” A v1 =” )
wrong
oav = -28.59* -18.237
format (7)
disp ( oav , ” T h e r e f o r e ,
O v e r a l l Av =” )
disp ( ” ” )
disp ( ” S t e p 6 : C a l c u l a t e b e t a ” )
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beta =200/(10.2*10^3)
disp ( beta , ” b e t a = Vf / Vo =” )
disp ( ” ” )
disp ( ” S t e p 7 : C a l c u l a t e D, A vf , R i f , R ’ ’ o f ” )
d =1+(0.0196*521.39)
format (6)
disp (d , ” D = 1 + Av∗ b e t a =” )
avf =521.39/11.22
disp ( avf , ” A v f = Av / D =” )
disp ( ” Ri = R G = 1 M−ohm” )
rif =11.22
disp ( rif , ” R i f ( i n M−ohm ) = Ri ∗ D =” )
disp ( ” Ro = r d = 10 k−ohm” )
ro =(10*8.38) /(18.38) // i n k−ohm
disp ( ro , ” R ’ ’ o ( i n k−ohm ) = r d | | R L2 =” )
rof =(4.559*10^3) /11.22 // i n ohm
format (4)
disp ( rof , ” R ’ ’ o f ( i n ohm ) = R ’ ’ o / D =” )
Scilab code Exa 3.8 voltage gain
1 // Example 3 . 8
2 clc
3 disp ( ” Here , o u t p u t v o l t a g e
4
5
6
7
8
9
i s s a m p l e d and f e d i n
s e r i e s w i t h t h e i n p u t s i g n a l . Hence t h e t o p o l o g y
i s v o l t a g e s e r i e s feedback . ”)
disp ( ” The open l o o p v o l t a g e g a i n f o r one s t a g e i s
g i v e n as , ” )
disp ( ”
Av = −gm∗ R eq ” )
req =(8*40*1000) /((40*1000) +(8*1000) +(8*40) ) // i n k
−ohm
format (5)
disp ( req , ” R eq ( i n k−ohm ) = r d | | R d | | ( R i 1+R 2
) =” )
av = -5*6.62
27
10 format (6)
11 disp ( av , ” Av =” )
12 avm = -33.11^3
13 disp ( avm , ”Av = O v e r a l l
14
15
16
17
18
19
20
21
v o l t a g e g a i n = | A vmid | ˆ 3 =” )
// a n s w e r i n t e x t b o o k i s wrong
beta =50/(10^6)
format (7)
disp ( beta , ” b e t a = Vf / Vo = −R 1 / R g = −R 1 /
R 1+R 2 =” )
d =1+(( -5*10^ -5) * -36306)
format (6)
disp (d , ” D = 1 + | Av | ∗ b e t a =” )
avf = -36306/2.8153
disp ( avf , ” A v f = Av / D =” )
Scilab code Exa 3.9 Avf
1 // Example 3 . 9
2 clc
3 disp ( ” Here , o u t p u t t e r m i n a l s a r e B and ground ,
4
5
6
7
8
9
10
11
thus
t h e f o r w a r d g a i n i s t h e g a i n o f Q1 and i t i s , ” )
disp ( ” A BN = −33.11 ” )
disp ( ” However , Q2 and Q3 must be c o n s i d e r e d a s a
part of feedback loop ”)
disp ( ” Here beta BN = V f / V B = V f / V o ∗ V o /V C
∗ V C/V B” )
disp ( ” where V B and V C a r e v o l t a g e s a t p o i n t B and
C, r e s p e c t i v e l y . ”)
disp ( ” T h e r e f o r e , beta BN = V f / V o ∗ A v3 ∗ A v2
b e c a u s e V o /V C = A v3 and V C/V B = A v2
”)
bbn = -(5*10^ -5) *(33.11^2)
format (7)
disp ( bbn , ” T h e r e f o r e , beta BN = −R1/ R g ∗ A v3 ∗
A v2 =” )
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15
16
disp ( ” Note t h a t t h e l o o p g a i n − beta BN ∗ A BN = Aˆ3
Vo ∗ R1/Rg = −1.815 = −A∗ b e t a ” )
disp ( ” I t s h o u l d be c l e a r t h t r e g a r d l e s s o f w h e r e t h e
output t e r m i n a l s a r e taken , the l o o p g a i n i s
unchanged . ” )
avf = -33.11/2.815
format (6)
disp ( avf , ” T h e r e f o r e ,
A v f = A BN / 1+A∗ b e t a =” )
Scilab code Exa 3.10 Avf and Rif and Rof
1
2
3
4
5
6
7
8
9
10
11
12
13
// Example 3 . 1 0
disp ( ” S t e p 1 : I d e n t i f y t o p o l o g y ” )
disp ( ” By s h o r t i n g o u t p u t v o l t a g e ( Vo = 0 ) ,
f e e d b a c k v o l t a g e Vf becomes z e r o and h e n c e i t i s
v o l t a g e s a m p l i n g . The f e e d b a c k v o l t a g e i s a p p l i e d
in s e r i e s with the input v o l t a g e hence the
topology i s v o l t a g e s e r i e s feedback . ”)
disp ( ” ” )
disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t . ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 . T h i s
p l a c e s t h e p a r a l l e l c o m b i n a t i o n o f r e s i s t o r 10 K
and 300 ohm a t f i r s t s o u r c e . To f i n d o u t p u t
c i r c u i t , s e t I i = 0 . T h i s p l a c e s t h e r e s i s t o r 10K
and 300 ohm i n s e r i e s a c r o s s t h e o u t p u t . The
r e s u l t a n t c i r c u i t i s shown i n f i g . 3 . 5 4 . ” )
disp ( ” ” )
disp ( ” S t e p 4 : R e p l a c e FET w i t h i t s e q u i v a l e n t
c i r c u i t a s shown i n f i g . 3 . 5 5 . ” )
disp ( ” ” )
disp ( ” S t e p 5 : Find open l o o p t r a n s f e r g a i n . ” )
disp ( ” Av = Vo / Vs = A v1 ∗ A v2 ” )
disp ( ” A v2 = −u∗ R L2 / R L2+r d ” )
rl2 =(10.3*22) /(10.3+22) // i n k−ohm
29
14 format (3)
15 disp ( rl2 , ” where
R L2 ( i n k−ohm ) =” )
16 av2 =( -50*7) /17
17 format (6)
18 disp ( av2 , ”
A v2 =” )
19 disp ( ”
A v1 = u∗ R D e f f / r d+R D e f f +(1+u ) ∗ R s e f f ” )
20 rdeff =(22*1000) /(22+1000)
// i n k−ohm
21 disp ( rdeff , ”
R D e f f ( i n k−ohm ) = R D | | R G2 =” )
22 disp ( ”
R s e f f = 330 | | 10K” )
23 av1 =( -50*21.53) /(10+21.53+(51*((0.33*10) /(10+0.33) ) )
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)
disp ( av1 , ” T h e r e f o r e , A v1 =” )
av = -20.59* -22.51
disp ( av , ” O v e r a l l Av = A v1 ∗ A v2 =” )
disp ( ” ” )
disp ( ” S t e p 6 : C a l c u l a t e b e t a ” )
beta =330/(330+10000)
format (7)
disp ( beta , ” b e t a = Vf / Vo = Rs / Rs+Rf =” )
disp ( ” ” )
disp ( ” s t e p 7 : C a l c u l a t e D, A vf , R i f , R ’ ’ o f ” )
d =1+(0.0319*463.5)
disp (d , ” D = 1 + Av∗ b e t a =” )
avf =463.5/15.785
format (6)
disp ( avf , ” A v f = Av / D =” )
disp ( ” Ri = R G = 1 M−ohm” )
rif =15.785
format (7)
disp ( rif , ” R i f ( i n k−ohm ) = Ri ∗ D =” )
ro =(10*7) /(10+7) // i n k−ohm
format (6)
disp ( ro , ” R ’ ’ o ( i n k−ohm ) = r d | | R L2 =” )
rof =(4.118*10^3) /15.785 // i n ohm
format (4)
disp ( rof , ” R ’ ’ o f ( i n ohm ) = R ’ ’ o / D =” )
30
Scilab code Exa 3.11 voltage gain and input and output resistance
1 // Example 3 . 1 1
2 clc
3 disp ( ” S t e p 1 : I d e n t i f y t o p o l o g y ” )
4 disp ( ” The f e e d b a c k v o l t a g e i s a p p l i e d
5
6
7
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9
10
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12
13
14
15
16
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18
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20
21
a c r o s s the
r e s i s t a n c e R e1 and i t i s i n s e r i e s w i t h i n p u t
s i g n a l . Hence f e e d b a c k i s v o l t a g e s e r i e s f e e d b a c k
. ”)
disp ( ” ” )
disp ( ” s t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t . ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 (
c o n n e c t i n g C2 t o g r o u n d ) , which g i v e s p a r l l e l
c o m b i n a t i o n o f Re w i t h Rf a t E1 . To f i n d o u t p u t
c i r u i t , s e t I i = 0 ( o p e n i n g t h e i n p u t node E1 a t
e m i t t e r o f Q1 ) , which g i v e s s e r i e s c o m b i n a t i o n od
Rf and R e1 a c r o s s t h e o u t p u t . The r e s u l t a n t
c i r c u i t i s shown i n f i g . 3 . 5 7 ” )
disp ( ” ” )
disp ( ” S t e p 4 : Find open l o o p v o l t a g e g a i n ( Av ) ” )
rl2 =(4.7*3.42) /(4.7+3.42) // i n k−ohm
format (5)
disp ( rl2 , ” R L2 ( i n k−ohm ) = R c2 | | ( Rs+R) =” )
disp ( ” A i 2 = −h f e = −50” )
disp ( ” R i 2 = h i e = 1 0 0 0 ohm = 1 k−ohm” )
av2 = -50*1.98
format (3)
disp ( av2 , ” A v2 = A i 2 ∗ R L2 / R i 2 =” )
disp ( ” A i 1 = −h f e = −50” )
format (7)
rl1 =((10*100*22*1) /((100*22) +(10*22) +(10*100)
+(10*100*22) ) ) *10^3 // i n ohm
disp ( rl1 , ” R L1 ( i n ohm ) = R c1 | | R3 | | R4 | | R i 2
31
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23
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49
50
51
=” )
disp ( ” R i 1 = h i e + (1+ h f e ) ∗ R e 1 e f f ” )
re1 =1+(51*((3.3*0.12) /(3.42) ) ) // i n k−ohm
format (4)
disp ( re1 , ” where R e 1 e f f ( i n k−ohm ) = Rs | | R =” )
av1 =( -50*865.46) /6900
format (5)
disp ( av1 , ” A v1 = A i 1 ∗ R L1 / R i 1 =” )
disp ( ” The o v e r a l l v o l t a g e g a i n , ” )
av = -6.27* -99
format (7)
disp ( av , ” Av = A v1 ∗ A v2 =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e b e t a ” )
beta =120/(120+3300)
format (6)
disp ( beta , ” b e t a = Vf / Vo = Rs / Rs+R =” )
disp ( ” ” )
disp ( ” S t e p 6 : C a l c u l a t e D, A vf , R i f , R o f and R ’ ’
o f ”)
d =1+(0.035*620.73)
format (7)
disp (d , ” D = 1 + Av∗ b e t a =” )
avf =620.73/22.725
format (5)
disp ( avf , ” A v f = Av / D =” )
rif =6.9*22.725 // i n k−ohm
format (6)
disp ( rif , ” R i f ( i n k−ohm ) = R i 1 ∗ D =” )
disp ( ” R o f = Ro / D = i n f i n i t y ” )
rof =(1.98*10^3) /22.725 // i n ohm
disp ( rof , ” R ’ ’ o f ( i n ohm ) = R ’ ’ o / D = R L2 / D =” )
Scilab code Exa 3.12 Ai and beta and Aif
32
1 // Example 3 . 1 2
2 clc
3 disp ( ” S t e p 1 : I d e n t i f y t o p o l o g y ” )
4 disp ( ” The f e e b d a c k i s g i v e n from e m i t t e r
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
o f Q2 t o
t h e b a s e o f Q2 . I f I o = 0 t h e n f e e d b a c k c u r r e n t
through 5 K r e g i s t e r i s zero , hence i t i s c u r r e n t
s a m p l i n g . As f e e d b a c k s i g n a l i s mixed i n s h u n t
with input , the a m p l i f i e r i s c u r r e n t shunt
feedback a m p l i f i e r . ”)
disp ( ” ” )
disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t ” )
disp ( ” The i n p u t c i r c u i t o f t h e a m p l i f i e r w i t h o u t
f e e d b a c k i s o b t a i n e d by o p e n i n g t h e o u t p u t l o o p
a t t h e e m i t t e r o f Q2 ( I o = 0 ) . T h i s p l a c e s R ’ ’ ( 5 K
) i n s e r i e s w i t h Re from b a s e t o e m i t t e r o f Q1 .
The o u t p u t c i r c u i t i s f o u n d by s h o r t i n g t h e i n p u t
node , i . e . making Vi = 0 . T h i s p l a c e s R ’ ’ ( 5 K)
i n p a r a l l e l w i t h Re . The r e s u l t a n t e q u i v a l e n t
c i r c u i t i s shown i n f i g . 3 . 5 9 ” )
disp ( ” ” )
disp ( ” S t e p 4 : Find open c i r c u i t t r a n s f e r g a i n . ” )
disp ( ” A I = I o / I s = −I c / I b 2 ∗ I b 2 / I c 1 ∗ I c 1 /
I b1 ∗ I b1 / I s ”)
disp ( ”We know t h a t − I c 2 / I b 2 = A i 2 = −h f e = −50
and ” )
disp ( ” − I c 1 / I b 1 = A i 1 = −h f e = 50 ” )
disp ( ” I c 1 / I b 1 = 50 ” )
disp ( ” L o o k i n g a t f i g . 3 . 5 9 we can w r i t e , ” )
disp ( ” I b 2 / I c 1 = −R c1 / R c1+R i 2 ” )
ri2 =1.5+(51*((5*0.5) /(5.5) ) ) // i n k−ohm
format (8)
disp ( ri2 , ” where R i 2 ( i n k−ohm ) = h i e + (1+ h f e ) ∗ (
R e2 | | R ’ ’ ) =” )
x1 = -2/(2+24.6818)
disp ( x1 , ” I b 2 / I c 1 =” )
disp ( ” I b 1 / I s = R / R+R i 1
where R = Rs | | ( R
’ ’ + R e2 ) ” )
r =((1*5.5) /(1+5.5) ) *10^3 // i n ohm
33
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36
37
38
39
40
41
format (9)
disp (r , ” T h e r e f o r e , R( i n ohm ) =” )
disp ( ” and R i 1 = h i e + (1+ h f e ) ∗ R e1 = 1 6 . 8 k−ohm”
)
x1 =846.1538/(846.1538+(16.8*10^3) )
format (8)
disp ( x1 , ” T h e r e f o r e ,
I b 1 / I s =” )
ai =50*0.07495*50*0.04795
format (7)
disp ( ai , ” A I =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e b e t a ” )
beta =500/(500+(5*10^3) )
disp ( beta , ” b e t a = I f / I o = R e2 / R e2 | R ’ ’ =” )
disp ( ” ” )
disp ( ” S t e p 6 : C a l c u l a t e D, A I f ” )
d =1+(0.0909*8.9848)
disp (d , ” D = 1 + A I ∗ b e t a =” )
aif =8.9848/1.8168
disp ( aif , ” A I f = A I / D =” )
Scilab code Exa 3.13 beta and Av and Avf and Rif and Rof
1 // Example 3 . 1 3
2 clc
3 disp ( ” S t e p 1 : I d e n t i f y t o p o l o g y ” )
4 disp ( ” The f e e d b a c k v o l t a g e i s a p p l i e d
a c r o s s R1
( 1 5 0 ohm ) , which i s i n s e r i e s w i t h i n p u t s i g n a l .
Hence f e e d b a c k i s v o l t a g e s e r i e s f e e d b a c k . ” )
5 disp ( ” ” )
6 disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t ”)
7 disp ( ” To f i n d i n p u t c i r c u i t , s e t Vo = 0 , which
g i v e s p a r a l l e l c o m b i n a t i o n o f R1 w i t h R2 a t E1 a s
shown i n t h e f i g . 3 . 6 1 . To f i n d o u t p u t c i r c u i t ,
34
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
s e t I i = 0 by o p e n i n g t h e i n p u t node , E1 a t
e m i t t e r o f Q1 , which g i v e s t h e s e r i e s c o m b i n a t i o n
o f R2 and R1 a c r o s s t h e o u t p u t . The r e s u l t a n t
c i r c u i t i s shown i n f i g . 3 . 6 1 . ” )
disp ( ” ” )
disp ( ” S t e p 4 : Find t h e open l o o p v o l t a g e g a i n ( Av ) ” )
rl2 =(4.7*15.15) /(4.7+15.15) // i n k−ohm
format (5)
disp ( rl2 , ” RL2 ( i n k−ohm ) =” )
disp ( ” S i n c e hoe = h r e = 0 , we can u s e a p p r o x i m a t e
a n a l y s i s . ”)
disp ( ” A i 2 = − h f e = −500 ” )
disp ( ” R i 2 = h i e = 1 1 0 0 ohm” )
av2 =( -500*3.59*10^3) /1100
disp ( av2 , ” A v2 = A i 2 ∗ R L2 / R i 2 =” )
rl1 =((10*47*33*1.1) /((47*33*1.1) +(10*33*1.1)
+(10*47*1.1) +(10*47*33) ) ) *10^3 // i n ohm
disp ( rl1 , ” R L1 ( i n ohm ) = 10K | | 47K | | 33K | | R i 2
=” )
disp ( ” A i 1 = − h f e = −500 ” )
ri1 =1.1+(501*((0.15*15) /(0.15+15) ) ) // i n k−ohm
disp ( ri1 , ” R i 1 ( i n k−ohm ) = h i e + (1+ h f e ) ∗Re =” )
av1 =( -500*942) /(75.5*10^3)
format (6)
disp ( av1 , ” A v1 = A i 1 ∗ R L1 / R i 1 =” )
av = -6.238* -1632
disp ( av , ” Av = A v1 ∗ A v2 =” )
disp ( ” ” )
disp ( ” S t e p 5 : C a l c u l a t e b e t a and D” )
beta =150/(150+15000)
format (7)
disp ( beta , ” b e t a = R1 / R1+R2 =” )
d =1+(10180*0.0099)
format (8)
disp (d , ” D = 1 + A∗ b e t a =” )
disp ( ” ” )
disp ( ” S t e p 6 : C a l c u l a t e A vf , R o f and R i f ” )
avf =10180/101.782
35
39 format (4)
40 disp ( avf , ”
A v f = Av / D =” )
41 rif =75.5*101.782*10^ -3
// i n M−ohm
42 format (6)
43 disp ( rif , ”
R i f ( i n M−ohm ) = R i 1 ∗ D =” )
44 rof =(3.59*10^3) /101.782
45 disp ( rof , ”
R o f ( i n ohm ) = Ro / D = R L2 / D =” )
Scilab code Exa 3.14 gain and new bandwidth
1 // Example 3 . 1 4
2 clc
3 disp ( ” Given :
A vmid = 5 0 0 , f L = 100 kHz , f H = 20
4
5
6
7
8
9
10
11
12
13
kHz and b e t a = 0 . 0 1 ” )
avf =500/(1+(0.01*500) )
format (6)
disp ( avf , ” A v f = A vmid / 1+ b e t a ∗ A vmid =” )
flf =100/(1+(0.01*500) ) // i n Hz
disp ( flf , ” f L f ( i n Hz ) = f L / 1+ b e t a ∗ A vmid =” )
fhf =20*(1+(0.01*500) ) // i n kHz
disp ( fhf , ” f H f ( i n kHZ ) = f H ∗ ( 1 + b e t a ∗ A vmid ) =
”)
bw =120 -0.01667 // i n kHZ
format (9)
disp ( bw , ” BW f ( i n kHz ) = f H f − f L f =” )
Scilab code Exa 3.15 show voltage gain with feedback
1 // Example 3 . 1 5
2 clc
3 disp ( ” S t e p 1 : I d e n t i f y t o p o l o g y ” )
4 disp ( ” By s h o r t i n g o u t p u t ( Vo = 0 ) , f e e d b a c k v o l t a g e
d o e s n o t become z e r o . By o p e n i n g t h e o u t p u t l o o p
36
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
f e e d b a c k becomes z e r o and h e n c e i t i s c u r r e n t
s a m p l i n g . The f e e d b a c k i s a p p l i e d i n s e r i e s w i t h
the input s i g n a l , hence topology used i s c u r r e n t
s e r i e s feedback . ”)
disp ( ” ” )
disp ( ” S t e p 2 and S t e p 3 : Find i n p u t and o u t p u t
c i r c u i t . ”)
disp ( ” To f i n d i n p u t c i r c u i t , s e t I o = 0 . T h i s
p l a c e s Re i n s e r i e s w i t h i n p u t . To f i n d o u t p u t
c i r c u i t I i = 0 . T h i s p l a c e s Re i n o u t p u t s i d e .
The r e s u l t a n t c i r c u i t i s shown i n f i g . 3 . 6 3 . ” )
disp ( ” ” )
disp ( ” S t e p 4 : R e p l a c e t r a n s i s t o r w i t h i t s h−
p a r a m e t e r e q u i v a l e n t a s shown i n f i g . 3 . 6 4 . ” )
disp ( ” ” )
disp ( ” S t e p 5 : Find open l o o p t r a n s f e r g a i n . ” )
disp ( ” From q u a t i o n ( 5 ) o f s e c t i o n 3 . 9 . 1 we have ” )
disp ( ” A v f = I o ∗ R L / Vs = G Mf ∗ R L ” )
disp ( ”
= − h f e ∗ R L / R ’ ’ s+ h i e +(1+ h f e ) ∗Re” )
disp ( ” Here R ’ ’ s = Rs | | R1 | | R2” )
disp ( ”
= Rs | | Rb
b e c a u s e R b = R1
| | R2” )
disp ( ” T h e r e f o r e , Vo / Vs = Vo/ Vi ∗ Vi / Vs ” )
disp ( ” where
Vi / Vs = Rb / Rs+Rb” )
disp ( ” T h e r e f o r e , Vo / Vs = (− h f e ∗ R L / R ’ ’ s+ h i e
+(1+ h f e ) ∗Re ) ∗ ( Rb / Rs+Rb ) ” )
disp ( ” D i v i d i n g b o t h n u m e r a t o r and d e n o m i n a t o r by Rs+
Rb we g e t , ” )
disp ( ” A v f = Vo / Vs = [− h f e ∗Rc ∗ ( Rb/Rb+Rs ) ] / R ’ ’
s+ h i e +(1+ h f e ) ∗Re
b e c a u s e RL = Rc” )
disp ( ”
= − h f e ∗Rc ∗ [ 1 / 1 + ( Rs /Rb ) ] / R ’ ’ s+ h i e
+(1+ h f e ) ∗Re” )
Scilab code Exa 3.16 GMf and Rif and Rof
37
1 // Example 3 . 1 6
2 clc
3 disp ( ” R e f e r e x a m p l e 3 . 1 5 ” )
4 disp ( ”
A v f = − h f e ∗Rc ∗ [ 1 / 1 + ( Rs /Rb ) ] / R ’ ’ s+ h i e
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
+(1+ h f e ) ∗Re
where R ’ ’ s = Rs | | R1 | | R2” )
avf =( -50*(1.8*10^3) *[1/(1+(1000/4272) ) ])
/(810+1000+((1+50) *1000) )
format (5)
disp ( avf , ” A v f =” )
gmf = -1.38/(1.8*10^3)
format (9)
disp ( gmf , ” G Mf = A v f / R L =” )
disp ( ” b e t a = Vf / I o = I e ∗Re / I o = −I o ∗Re / I o =
−Re = −1 K” )
disp ( ” G Mf = G M / 1+ b e t a ∗G M” )
gm =1/((1/( -7.66*10^ -4) ) +1000)
format (10)
disp ( gm , ” T h e r e f o r e , G M =” )
d =1+( -1000* -3.2735*10^ -3)
format (7)
disp (d , ” D = 1 + G M∗ b e t a =” )
ri =(1+1.36) // i n k−ohm
format (5)
disp ( ri , ” R i ( i n k−ohm ) = Rs+( h i e +Re ) | | R D =” )
rif =2.36*4.2735 // i n k−ohm
format (3)
disp ( rif , ” R i f ( i n k−ohm ) = R i ∗ D =” )
disp ( ” R o = i n f i n i t y ” )
disp ( ” R o f = R o ∗ D = i n f i n i t y ” )
disp ( ” R ’ ’ o f = R o f | | R L = R L = 1 . 8 k−ohm” )
Scilab code Exa 3.17 feedback factor and Rif and Rof
1 // Example 3 . 1 7
2 clc
38
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
disp ( ” R e f e r e x a m p l e 3 . 1 5 ” )
disp ( ” A v f = − h f e ∗Rc ∗ [ 1 / 1 + ( Rs /Rb ) ] / R ’ ’ s+ h i e
+(1+ h f e ) ∗Re
where R ’ ’ s = Rs | | R1 | | R2” )
avf =( -50*(4*10^3) *[1/(1+(1000/9000) ) ])
/(900+1000+((1+150) *1000) )
format (6)
disp ( avf , ” A v f =” )
gmf = -1.177/(4*10^3)
format (9)
disp ( gmf , ” G Mf = A v f / R L =” )
disp ( ” b e t a = Vf / I o = I e ∗Re / I o = −I o ∗Re / I o =
−Re = −1 K” )
disp ( ” G Mf = G M / 1+ b e t a ∗G M” )
gm =1/((1/( -2.943*10^ -4) ) +1000)
format (9)
disp ( gm , ” T h e r e f o r e , G M =” )
d =1+( -1000* -4.17*10^ -4)
format (6)
disp (d , ” D = 1 + G M∗ b e t a =” )
ri =1+((2*9) /(2+9) ) // i n k−ohm
disp ( ri , ” R i ( i n k−ohm ) = Rs+( h i e +Re ) | | R D =” )
rif =2.636*1.417 // i n k−ohm
format (6)
disp ( rif , ” R i f ( i n k−ohm ) = R i ∗ D =” )
disp ( ” R o = i n f i n i t y ” )
disp ( ” R o f = R o ∗ D = i n f i n i t y ” )
disp ( ” R ’ ’ o f = R o f | | R L = R L = 4 k−ohm” )
Scilab code Exa 3.18 gain with feedback and new bandwidth
1 // Example 3 . 1 8 .
2 clc
3 disp ( ” Given : A v mid = 4 0 , f L = 100 Hz , f H = 15
kHz and b e t a = 0 . 0 1 ” )
4 avf =400/(1+(0.01*400) )
39
5 format (3)
6 disp ( avf , ”
A v f = A v mid / 1+ b e t a ∗ A v mid =” )
7 flf =100/(1+(0.01*400) )
8 disp ( flf , ”
f L f = f L / 1+ b e t a ∗ A v mid =” )
9 fhf =(15) *(1+(0.01*400) )
// i n kHz
10 disp ( fhf , ”
f H f ( i n kHz ) = f H ∗ (1+ b e t a ∗ A v mid ) =”
)
11 bw =75 -0.02
// i n kHz
12 format (6)
13 disp ( bw , ” BW f ( i n kHz ) = f H f − f L f =” )
Scilab code Exa 3.19 overall voltage gain and bandwidth
1 // Example 3 . 1 9
2 clc
3 disp ( ” Given : Av = 1 0 , BW = 1 ∗ 1 0 ˆ 3 , n =3” )
4 disp ( ” ( i ) O v e r a l l v o l t a g e g a i n ” )
5 disp ( ” The g a i n o f c a s c a d e d a m p l i f i e r w i t h o u t
f e e d b a c k = 10∗10∗10 = 1000 ”)
6 avf =1000/(1+(0.1*1000) )
7 format (4)
8 disp ( avf , ” A v f = Av / 1+Av∗ b e t a =” )
9 disp ( ” ( i i ) Bandwidth o f c a s c a d e d s t a g e ” )
10 disp ( ” Bandwidth o f c a s c a d e d a m p l i f i e r w i t h o u t
11
12
13
14
15
16
feedback ”)
bw =((1*10^6) * sqrt ((2^(1/3) ) -1) ) *10^ -3 // i n kHz
format (7)
disp ( bw , ” BW( c a s c a d e ) ( i n kHz ) = BW∗ s q r t ( 2 ˆ ( 1 / n ) −
1 ) =” )
bwf =(509.82*10^3*(1+(0.1*1000) ) ) *10^ -6 // i n MHz
format (6)
disp ( bwf , ” BW f ( i n MHz) = BW ∗ ( 1 + b e t a ∗ A v mid ) =
”)
40
Chapter 4
Oscillators
Scilab code Exa 4.1 C and hfe
1 // Example 4 . 1
2 clc
3 disp ( ” R e f e r i n g t o e q u a t i o n ( 1 ) , ” )
4 ri =(25*57*1.8) /((57*1.8) +(25*1.8) +(25*57) )
// i n k−
ohm
5 format (6)
6 disp ( ri , ” R ’ ’ i ( i n k−ohm ) = R1 | | R2 | | h i e =” )
7 disp ( ”Now R ’ ’ i + R3 = R” )
8 r3 =7.1 -1.631
// i n k−ohm
9 format (5)
10 disp ( r3 , ” T h e r e f o r e ,
R3 ( i n k−ohm ) = R − R ’ ’ i =” )
11 k =20/7.1
12 format (6)
13 disp (k , ” K = R C / R =” )
14 disp ( ”Now
f = 1 / 2∗ p i ∗R∗C∗ s q r t (6+4K) ” )
15 c =(1/( sqrt (6+(4*2.816) ) *2* %pi *7.1*10*10^6) ) *10^12
16
17
18
19
// i n pF
format (8)
disp (c , ” T h e r e f o r e , C( i n pF ) =” )
disp ( ” h f e >= 4K + 23 + 29/K” )
hfe =(4*2.816) +23+(29/2.816)
41
20
21
format (7)
disp ( hfe , ”
h f e >=” )
Scilab code Exa 4.2 frequency of oscillation
1 // Example 4 . 2
2 clc
3 disp ( ” The g i v e n v a l u e s a r e , R = 4 . 7 k−ohm and C =
0 . 4 7 uF” )
4 f =1/(2* %pi * sqrt (6) *(4.7*10^3) *(0.47*10^ -6) )
// i n
Hz
5 format (7)
6 disp (f , ”
f ( i n Hz ) = 1 / 2∗ p i ∗ s q r t ( 6 ) ∗R∗C =” )
Scilab code Exa 4.3 R and C
1 // Example 4 . 3
2 clc
3 disp ( ” f = 1 kHz ” )
4 disp ( ”Now
f = 1 / 2∗ p i ∗ s q r t ( 6 ) ∗R∗C” )
5 disp ( ” Choose C = 0 . 1 uF” )
6 r =1/( sqrt (6) *2* %pi *0.1*1*10^ -3)
// i n ohm
7 format (8)
8 disp (r , ” T h e r e f o r e , R( i n ohm ) = ” )
9 disp ( ” Choose R = 680 ohm
standard value ”)
Scilab code Exa 4.4 C and RD
1 // Example 4 . 4
2 clc
42
3 disp ( ” U s i n g t h e e x p r e s s i o n f o r t h e f r e q u e n c y ” )
4 disp ( ”Now ,
f = 1 / 2∗ p i ∗R∗C∗ s q r t ( 6 ) ” )
5 f =(1/( sqrt (6) *2* %pi *9.7*5*10^6) ) *10^9
// i n nF
6 format (5)
7 disp (f , ” T h e r e f o r e , C( i n nF ) =” )
8 disp ( ”Now u s i n g t h e e q u a t i o n ( 2 7 ) ” )
9 disp ( ”
|A| = g m ∗ R L”)
10 disp ( ” T h e r e f o r e ,
| A | >= 29 ” )
11 disp ( ” T h e r e f o r e ,
g m ∗ R L >= 29 ” )
12 rl =(29/(5000*10^ -6) ) *10^ -3
// i n k−ohm
13 format (4)
14 disp ( rl , ” T h e r e f o r e ,
R L ( i n k−ohm ) >= 29 / g m =” )
15 disp ( ” R L = R D∗ r d / R D+r d ” )
16 rd =(40) /4.8823
17 format (5)
18 disp ( rd , ”
T h e r e f o r e , R D ( i n k−ohm ) = ” )
19 disp ( ” While f o r minimum v a l u e o f R L = 5 . 8 k−ohm” )
20 disp ( ”
R D = 6 . 7 8 k−ohm” )
Scilab code Exa 4.5 minimum and maximum R2
1 // Example 4 . 5
2 clc
3 disp ( ” The f r e q u e n c y o f t h e o s c i l l a t o r i s g i v e n by , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗ s q r t ( R1∗R2∗C1∗C2 ) ” )
5 disp ( ” For
f = 10 kHz , ” )
6 r2 =(1/(4*( %pi ^2) *(100*10^6) *(10*10^3) *(0.001*10^ -12)
7
8
9
10
11
12
) ) // i n k−ohm
format (6)
disp ( r2 , ” T h e r e f o r e , R2 ( i n k−ohm ) =” )
disp ( ” For
f = 50 kHz , ” )
r2 =(1/(4*( %pi ^2) *(2500*10^6) *(10*10^3)
*(0.001*10^ -12) ) ) // i n k−ohm
format (6)
disp ( r2 , ” T h e r e f o r e , R2 ( i n k−ohm ) =” )
43
13
disp ( ” So minimum v a l u e o f R2 i s 1 . 0 1 3 k−ohm w h i l e
t h e maximum v a l u e o f R2 i s 2 5 . 3 3 k−ohm” )
Scilab code Exa 4.6 range over capacitor is varied
1 // Example 4 . 6
2 clc
3 disp ( ” The f r e q u e n c y i s g i v e n by , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗ s q r t (C∗ L e q ) ” )
5 leq =(2*10^ -3) +(20*10^ -6)
6 format (8)
7 disp ( leq , ” where
L e q = L1 + L2 =” )
8 disp ( ” For
f = f m a x = 2 0 5 0 kHz ” )
9 format (5)
10 c =(1/(4*( %pi ^2) *((2050*10^3) ^2) *0.00202) ) *10^12
11
12
13
14
15
16
//
i n pF
disp (c , ” T h e r e f o r e , C( i n pF ) =” )
disp ( ” For f = f m i n = 950 kHz ” )
c =(1/(4*( %pi ^2) *((950*10^3) ^2) *0.00202) ) *10^12 //
i n pF
format (6)
disp (c , ” T h e r e f o r e , C( i n pF ) =” )
disp ( ” Hence C must be v a r i e d from 2 . 9 8 pF t o 1 3 . 8 9
pF , t o g e t t h e r e q u i r e d f r e q u e n c y v a r i a t i o n . ” )
Scilab code Exa 4.7 frequency of oscillation
1 // Example 4 . 7
2 clc
3 disp ( ” The g i v e n v a l u e s a r e , ” )
4 disp ( ” L1 = 0 . 5 mH,
L2 = 1 mH, C = 0 . 2 uF” )
5 disp ( ”Now
f = 1 / 2∗ p i ∗ s q r t (C∗ L e q ) ” )
6 leq =0.5+1
// i n mH
44
7 disp ( leq , ” and
L e q ( i n mH) = L1 + L2 =” )
8 f =(1/(2* %pi * sqrt (1.5*0.2*10^ -9) ) ) *10^ -3
// i n kHz
9 format (5)
10 disp (f , ” T h e r e f o r e ,
f ( i n kHz ) =” )
Scilab code Exa 4.8 frequency of oscillation
1 // Example 4 . 8
2 clc
3 disp ( ” The e q u i v a l e n t c a p a c i t a n c e i s g i v e n by , ” )
4 ceq =(150*1.5*10^ -21) /((150*10^ -12) +(1.5*10^ -9) )
5
6
7
8
9
10
in F
format (12)
disp ( ceq , ” C eq ( i n F ) = C1∗C2 / C1+C2 =” )
disp ( ”Now ,
f = 1 / 2∗ p i ∗ s q r t ( L∗ C eq ) ” )
f =(1/(2* %pi * sqrt (50*136.363*10^ -18) ) ) *10^ -6
MHz
format (6)
disp (f , ” f ( i n MHz) =” )
//
// i n
Scilab code Exa 4.9 calculate C
1 // Example 4 . 9
2 clc
3 disp ( ” The g i v e n v a l u e s a r e , ” )
4 disp ( ”
L = 100 uH ,
C1 = C2 = C
5
6
7
8
and
f = 500
kHz ” )
disp ( ”Now ,
f = 1 / 2∗ p i ∗ s q r t ( L∗ C eq ) ” )
ceq =1/(4*( %pi ^2) *(100*10^ -6) *((500*10^3) ^2) )
F
format (11)
disp ( ceq , ” T h e r e f o r e , C eq ( i n F ) =” )
45
// i n
9
10
11
12
13
disp ( ” but
C eq = C1∗C2 / C1+C2
”)
disp ( ” T h e r e f o r e , C eq = C / 2 ” )
c =1.0132*2
format (6)
disp (c , ” T h e r e f o r e , C( i n nF ) =” )
and C1 = C2 = C
Scilab code Exa 4.10 series and parallel resonant freqency
1 // Example 4 . 1 0
2 clc
3 fs =(1/(2* %pi * sqrt (0.4*0.085*10^ -12) ) ) *10^ -6
4
5
6
7
8
9
10
11
12
13
14
// i n
MHz
format (6)
disp ( fs , ” ( i )
f s ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗C) =” )
ceq =0.085/1.085 // i n pF
disp ( ceq , ” ( i i ) C eq ( i n pF ) = C∗C M / C+C M =” )
fp =(1/(2* %pi * sqrt (0.4*0.078*10^ -12) ) ) *10^ -6 // i n
MHz ( t h e a n s w e r i n t e x t b o o k i s wrong )
disp ( fp , ” T h e r e f o r e ,
f p ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗
C eq ) =” )
inc =((0.899 -0.856) /0.856) *100 // i n p e r c e n t a g e
disp ( inc , ” ( i i i ) % i n c r e a s e =” )
q =(2* %pi *0.4*0.856*10^6) /(5*10^3)
format (8)
disp (q , ” ( i v ) Q = o m e g a s ∗L / R = 2∗ p i ∗ f s ∗L / R =” )
Scilab code Exa 4.11 series and parallel resonant freqency
1 // Example 4 . 1 1
2 clc
3 disp ( ”
C M = 2 pF” )
4 fs =(1/(2* %pi * sqrt (2*0.01*10^ -12) ) ) *10^ -6
46
// i n MHz
5 format (6)
6 disp ( fs , ”Now
f s ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗C) =” )
7 ceq =(2*0.01*10^ -24) /(2.01*10^ -12)
// i n F
8 format (9)
9 disp ( ceq , ”
C eq ( i n F ) = C M∗C / C M+C =” )
10 fp =(1/(2* %pi * sqrt (2*9.95*10^ -15) ) ) *10^ -6
// i n MHz
11 format (6)
12 disp ( fp , ”
f p = 1 / 2∗ p i ∗ s q r t ( L∗ C eq ) =” )
13 disp ( ” So f s a n d f p v a l u e s a r e a l m o s t same . ” )
Scilab code Exa 4.12 Varify Barkhausen criterion and find frequency of
oscillation
1 // e x a m p l e 4 . 1 2 .
2 clc
3 disp ( ”From t h e g i v e n i n f o r m a t i o n we can w r i t e , ” )
4 disp ( ”
A = −16∗10ˆ6/ j ∗ omega and b e t a =
5
6
7
8
9
10
11
1 0 ˆ 3 / [ 2 ∗ 1 0 ˆ 3 + j ∗ omega ] ˆ 2 ” )
disp ( ”To v e r i f y t h e B a r k h a u s e n c o n d i t i o n means t o
v e r i f y w h e t h e r | A∗ b e t a | = 1 a t a f r e q u e n c y f o r
which A∗ b e t a = 0 d e g r e e . L e t u s e x p r e s s , A∗ b e t a
i n i t s r e c t a n g l u a r form . ” )
disp ( ” A∗ b e t a = −16∗10ˆ6∗10ˆ3 / j ∗ omega ∗ [ 2 ∗ 1 0 ˆ 3 + j ∗
omega ] ˆ 2 = −16∗10ˆ9 / j ∗ omega ∗ [ 4 ∗ 1 0 ˆ 6 + 4 ∗ 1 0 ˆ 3 ∗ j ∗
omega+( j ∗ omega ) ˆ 2 ] ” )
disp ( ”
= −16∗10ˆ9 / j ∗ omega ∗ [ 4 ∗ 1 0 ˆ 6 + 4 ∗ 1 0 ˆ 3 ∗ j
∗ omega−omega ˆ 2 ]
a s j ∗2 = −1” )
disp ( ”
= −16∗10ˆ9 / 4 ∗ 1 0 ˆ 6 ∗ j ∗ omega +4∗10ˆ3∗ j
ˆ2∗ omegaˆ2− j ∗ omega ˆ 3 ] ” )
disp ( ”
= −16∗10ˆ9 / j ∗ omega ∗[4∗10ˆ6 − omega
ˆ 2 ] − [ omega ˆ 2 ∗ 4 ∗ 1 0 ˆ 3 ] ” )
disp ( ” R a t i o n a l i s i n g t h e d e n o m i n a t o r f u n c t i o n we g e t ,
”)
disp ( ” A∗ b e t a = −16∗10ˆ9[ − omega ˆ 2 ∗ 4 ∗ 1 0 ˆ 3 − j ∗ omega
∗[4∗10ˆ6 − omega ˆ 2 ] ] / [ − [ omega ˆ 2 ∗ 4 ∗ 1 0 ˆ 3 ] − j ∗ omega
47
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
∗[4∗10ˆ6 − omega ˆ 2 ] ] ∗ [ − omega ˆ 2 ∗ 4 ∗ 1 0 ˆ 3 − j ∗ omega
∗[4∗10ˆ6 − omega ˆ 2 ] ] ” )
disp ( ” U s i n g ( a−b ) ( a+b ) = a ˆ2 − b ˆ2 i n t h e
denominator , ”)
disp ( ” A∗ b e t a = 1 6 ∗ 1 0 ˆ 9 [ omega ˆ2∗4∗10ˆ3+ j ∗ omega
∗[4∗10ˆ6 − omega ˆ 2 ] ] / [− omega ˆ 2 ∗ 4 ∗ 1 0 ˆ 3 ] ˆ 2 − [ j ∗
omega ∗[4∗10ˆ6 − omega ˆ 2 ] ˆ 2 ” )
disp ( ” A∗ b e t a = 1 6 ∗ 1 0 ˆ 9 [ omega ˆ2∗4∗10ˆ3+ j ∗ omega
∗[4∗10ˆ6 − omega ˆ 2 ] ] / 1 6 ∗ 1 0 ˆ 6 ∗ omega ˆ4 + omega
ˆ2(4∗10ˆ6 − omega ˆ 2 ) ˆ2 ” )
disp ( ”Now t o have A∗ b e t a = 0 d e g r e e , t h e i m a g i n a r y
p a r t o f A∗ b e t a must be z e r o . T h i s i s p o s s i b l e
when , ” )
disp ( ” T h e r e f o r e , omega ∗ ( 4 ∗ 1 0 ˆ 6 − omega ˆ 2 ) = 0 ” )
disp ( ” T h e r e f o r e , omega = 0 o r 4 ∗ 1 0 ˆ 6 − omega ˆ2 =
0”)
disp ( ” T h e r e f o r e , omega ˆ2 = 4 ∗ 1 0 ˆ 6
N e g l e c t i n g zero value of frequency ”)
disp ( ” T h e r e f o r e , omega = 2 ∗ 1 0 ˆ 3 r a d / s e c ” )
disp ( ” At t h i s f r e q u e n c y | A∗ b e t a | can be o b t a i n e d as ,
”)
disp ( ” | A∗ b e t a | = 1 6 ∗ 1 0 ˆ 9 [ 4 ∗ 1 0 ˆ 3 ∗ omega ˆ 2 ] /
1 6 ∗ 1 0 ˆ 6 ∗ omegaˆ4+omega ˆ2[4∗10ˆ6 − omega ˆ 2 ] ˆ 2
a t omega = 2 ∗ 1 0 ˆ 3 ” )
ab =(2.56*10^20) /(2.56*10^20)
disp ( ab , ” | A∗ b e t a | =” )
disp ( ” T h e r e f o r e , At omega = 2 ∗ 1 0 ˆ 3 r a d / s e c , A∗ b e t a
= 0 d e g r e e a s i m a g i n a r y p a r t i s z e r o w h i l e | A∗
b e t a | = 1 . Thus B a r k h a u s e n C r i t e r i o n i s s a t i s f i e d
. ”)
disp ( ” The f r e q u e n c y a t which c i r c u i t w i l l o s c i l l a t e
i s t h e v a l u e o f omega f o r which | A∗ b e t a | = 1 and
A∗ b e t a = 0 d e g r e e a t t h e same t i m e ” )
disp ( ” i . e .
omega = 2 ∗ 1 0 ˆ 3 r a d / s e c ” )
disp ( ” But
omega = 2∗ p i ∗ f ” )
f =(2*10^3) /(2* %pi ) // i n Hz
format (9)
disp (f , ” T h e r e f o r e ,
f ( i n Hz ) = omega / 2 p i =” )
48
Scilab code Exa 4.13 minimum and maximum values of R2
1 // Example 4 . 1 3
2 clc
3 disp ( ” The f r e q u e n c y o f t h e o s c i l l a t o r i s g i v e n by , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗ s q r t ( R1∗R2∗C1∗C2 ) ” )
5 disp ( ” For
f = 20 kHz , ” )
6 r2 =(1/(4*( %pi ^2) *((20*10^3) ^2) *(10*10^3)
7
8
9
10
11
12
13
*((0.001*10^ -6) ^2) ) ) *10^ -3
format (5)
disp ( r2 , ” T h e r e f o r e , R2 ( i n k−ohm ) =” )
disp ( ” For f = 70 kHz , ” )
r2 =(1/(4*( %pi ^2) *((70*10^3) ^2) *(10*10^3)
*((0.001*10^ -6) ^2) ) ) *10^ -3
format (6)
disp ( r2 , ” T h e r e f o r e , R2 ( i n k−ohm ) =” )
disp ( ” So minimum v a l u e o f R2 i s 0 . 5 1 7 k−ohm w h i l e
t h e maximum v a l u e o f R2 i s 6 . 3 3 k−ohm” )
Scilab code Exa 4.14 frequency of oscillation and minimum hfe
1 // Example 4 . 1 4 .
2 clc
3 disp ( ”R = 6 k−ohm , C = 1 5 0 0 pF ,
R C = 18 k−ohm” )
4 k =18/6
5 disp (k , ”Now
K = R C / R =” )
6 disp ( ” T h e r e f o r e ,
f = 1 / 2∗ p i ∗R∗C∗ s q r t (6+4K) ” )
7 f =(1/(2* %pi *(6*10^3) *(1500*10^ -12) * sqrt (6+12) ) )
8
9
*10^ -3 // i n kHZ
format (6)
disp (f , ” f ( i n kHz ) =” )
49
10 hfe =(4*3) +23+(29/3)
11 disp ( hfe , ”
( h f e ) min = 4K + 23 + 29/K =” )
Scilab code Exa 4.15 range of frequency of oscillation
1 // Example 4 . 1 5 .
2 clc
3 format (6)
4 disp ( ” For a Wien b r i d g e o s c i l l a t o r , ” )
5 disp ( ”
f = 1 / 2∗ p i ∗R∗C” )
6 fm =(1/(2* %pi *(100*10^3) *(50*10^ -12) ) ) *10^ -3
// i n
kHz
7 disp ( fm , ” T h e r e f o r e ,
f m a x ( i n kHz ) =” )
8 fmi =(1/(2* %pi *(100*10^3) *(500*10^ -12) ) ) *10^ -3
9 disp ( fmi , ” and
f m i n ( i n kHz ) =” )
10 fn =31.83+50
11 disp ( fn , ”Now f n e w ( i n kHz ) = f m a x + 5 0 ∗ 1 0 ˆ 3 =” )
12 disp ( ” The c o r r e s p o n d i n g R = R ’ ’ w i t h an a d d i t i o n a l
13
14
15
16
17
18
r e s i s t a n c e R x in p a r a l l e l ”)
disp ( ” T h e r e f o r e ,
f = 1 / 2∗ p i ∗R’ ’ ∗ C” )
r =(1/(2* %pi *(50*10^ -12) *(81.83*10^3) ) ) *10^ -3
k−ohm
disp (r , ” T h e r e f o r e , R ’ ’ ( i n k−ohm ) =” )
rx =1/((1/38.89) -(1/100) ) // i n k−ohm
disp ( ” T h e r e f o r e , R ’ ’ = R∗ R x / R+R x ” )
disp ( rx , ” T h e r e f o r e , R x ( i n k−ohm ) =
i n p a r a l l e l w i t h 100 k−ohm” )
Scilab code Exa 4.16 frequency of oscillation
1 // Example 4 . 1 6 .
2 clc
3 format (6)
50
// i n
4
5
6
7
8
9
disp ( ” For a H a r t l e y o s c i l l a t o r t h e f r e q u e n c y i s
g i v e n by , ” )
disp ( ” f = 1 / 2∗ p i ∗ s q r t ( L e q ∗C)
where L e q
= L1+L2 ” )
leq =20+5 // i n mH
disp ( leq , ” T h e r e f o r e ,
L e q ( i n mH) = 20+5 =” )
f =(1/(2* %pi * sqrt (25*500*10^ -15) ) ) *10^ -3 // i n kHz
disp (f , ” T h e r e f o r e ,
f ( i n kHz ) =” )
Scilab code Exa 4.17 gain of the transistor
1 // Example 1 4 . 7
2 clc
3 disp ( ” For a H a r t l e y o s c i l l a t o r , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗ s q r t ( L e q ∗C)
5
6
7
8
9
10
11
12
where L e q
= L1 + L2 + 2M” )
leq =(1/(4*( %pi ^2) *((168*10^3) ^2) *(50*10^ -12) ) ) *10^3
// i n mH
format (6)
disp ( leq , ” T h e r e f o r e ,
L e q ( i n mH) =” )
l2 =((17.95*10^ -3) -(15*10^ -3) -(5*10^ -6) ) *10^3 // i n
mH
disp ( l2 , ” T h e r e f o r e , L2 ( i n mH) =” )
hfe =((15*10^ -3) +(5*10^ -6) ) /((2.945*10^ -3) +(5*10^ -6) )
format (5)
disp ( hfe , ”Now
h f e = L1+M / L2+M =” )
Scilab code Exa 4.18 new frequency and inductance
1 // Example 4 . 1 8
2 clc
3 disp ( ” For a C o l p i t t s o s c i l l a t o r , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗ s q r t ( L∗ C eq ) ” )
51
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
disp ( ” where C eq = C1∗C2 / C1+C2 but C1 = C2 =
0 . 0 0 1 uF” )
ceq =((0.001*10^ -6) ^2) /(2*(0.001*10^ -6) ) // i n F
format (7)
disp ( ceq , ” T h e r e f o r e , C eq ( i n F ) =” )
disp ( ” L = 5∗10ˆ −6 H” )
f =(1/(2* %pi * sqrt (25*10^ -16) ) ) *10^ -6 // i n MHz
format (6)
disp (f , ” T h e r e f o r e ,
f ( i n MHz) =” )
disp ( ”Now L i s d o u b l e d i . e . 10 uH” )
f1 =(1/(2* %pi * sqrt (50*10^ -16) ) ) *10^ -6 // i n MHz
format (5)
disp ( f1 , ” T h e r e f o r e ,
f ( i n MHz) =” )
nf = 2*3.183
format (6)
disp ( nf , ”New f r e q u e n c y ( i n MHz) = 2 ∗ 3 . 1 8 3 =” )
l =(1/(4*( %pi ^2) *((6.366*10^6) ^2) *(5*10^ -10) ) ) *10^6
// i n uH
format (5)
disp (l , ” T h e r e f o r e , L ( i n uH ) =” )
Scilab code Exa 4.19 new frequency of oscillation
1 // Example 4 . 1 9
2 clc
3 disp ( ” For a Clapp o s c i l l a t o r , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗ s q r t ( L∗C3 ) ” )
5 disp ( ” where
C3 = 63 pF” )
6 f =(1/(2* %pi * sqrt (315*10^ -18) ) ) *10^ -6
7 format (6)
8 disp (f , ” T h e r e f o r e ,
f ( i n MHz) =” )
Scilab code Exa 4.20 R and hfe
52
// i n MHz
1 // Example 4 . 2 0
2 clc
3 disp ( ” R e f e r i n g t o e q u a t i o n ( 1 )
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
of s e c t i o n 4 . 5 . 3 , the
i n p u t i m p e d a n c e i s g i v e n by , ” )
disp ( ”R ’ ’ i = R1 | | R2 | | h i e ” )
disp ( ”Now R1 = 25 k−ohm , R2 = 47 k−ohm , and h i e
= 2 k−ohm” )
format (7)
ri =(25*47*2) /((47*2) +(25*2) +(25*47) ) // i n k−ohm
disp ( ri , ” T h e r e f o r e , R ’ ’ i ( i n k−ohm ) =” )
disp ( ” K = R C / R” )
. . . given ”)
disp ( ”Now R C = 10 k−ohm
disp ( ”Now f = 1 / 2∗ p i ∗R∗C∗ s q r t (6+4K) ” )
disp ( ” T h e r e f o r e , R∗ s q r t (6+4K) = 3 1 8 3 0 . 9 8 9 ” )
disp ( ”Now K = R C / R = 1 0 ∗ 1 0 ˆ 3 / R” )
disp ( ” T h e r e f o r e , R∗ s q r t ( 6 + ( 4 0 ∗ 1 0 ∗ 1 0 ˆ 3 /R) ) =
31830.989 ”)
disp ( ” T h e r e f o r e , Rˆ 2 ∗ ( 6 + ( 4 0 ∗ 1 0 ∗ 1 0 ˆ 3 /R) ) =
( 3 1 8 3 0 . 9 8 9 ) ˆ2 ” )
R = poly (0 , ’R ’ )
p1 =6* R ^2+(40*10^3) *R -(31830.989) ^2
t1 = roots ( p1 )
ans1 = t1 (1)
format (6)
disp (( - ans1 ) *10^ -3 , ” T h e r e f o r e , R( i n k−ohm )=
N e g l e c t i n g n e g a t i v e value ”)
k =10/16.74
format (7)
disp (k , ” T h e r e f o r e , K = R C / R =” )
disp ( ” T h e r e f o r e ,
h f e >= 4K + 23 + 29/K” )
hfe =(4*0.5973) +23+(29/0.5973)
format (6)
disp ( hfe , ” h f e >=” )
Scilab code Exa 4.21 component values of wien bridge
53
1 // Example 4 . 2 1
2 clc
3 disp ( ” The f r e q u e n c y i s g i v e n by , ” )
4 disp ( ”
f = 1 / 2∗ p i ∗R∗C” )
5 disp ( ” L e t t h e r e s i s t a n c e v a l u e t o be s e l e c t e d as , ” )
6 disp ( ” R1 = R2 = R = 50 k−ohm” )
7 disp ( ”
f = 1 / 2∗ p i ∗ 5 0 ∗ 1 0 ˆ 3 ∗C” )
8 f =(1/(2* %pi *(50*10^3) *100) ) *10^9
// i n nF
9 format (6)
10 disp (f , ”
f ( i n nF ) =” )
11 disp ( ” and
f m a x = 1 / 2∗ p i ∗ 5 0 ∗ 1 0 ˆ 3 ∗C” )
12 c =(1/(2* %pi *(50*10^3) *(10*10^3) ) ) *10^9
// i n pF
13 disp (c , ” C( i n nF ) =” )
14 disp ( ” Thus t o v a r y t h e f r e q u e n c y from 100 Hz t o 10
kHz , t h e c a p a c i t o r r a n g e s h o u l d be s e l e c t e d a s
0 . 3 1 8 nF t o 3 1 . 8 3 nF” )
15 disp ( ” S i m i l a r l y k e e p i n g t h e c a p a c i t o r v a l u e c o n s t a n t
, t h e r a n g e o f t h e r e s i s t a n c e v a l u e s can be
obtained . ”)
Scilab code Exa 4.22 values of C2 and new frequency of oscillation
1 // Example 4 . 2 2
2 clc
3 disp ( ”
f = 2 . 5 MHz, L = 10 uH , C1 = 0 . 0 2 uF” )
4 disp ( ” For C o l p i t t s o s c i l l a t o r , t h e f r e q u e n c y i s
5
6
7
8
9
10
11
g i v e n by , ” )
disp ( ” f = 1 / 2 p i ∗ s q r t ( L∗ C eq ) ” )
ceq =(1/(4*( %pi ^2) *((2.5*10^6) ^2) *(10*10^ -6) ) ) *10^12
// i n pF
format (8)
disp ( ceq , ” T h e r e f o r e , C eq ( i n pF ) =” )
disp ( ” ( i ) But
C eq = C1∗C2 / C1+C2” )
c2 =((0.02*10^ -6) /49.348) *10^9 // i n nF
format (7)
54
12
13
14
15
16
disp ( c2 , ” T h e r e f o r e , C2 ( i n nF ) =” ) // a n s w e r i n
t e x t b o o k i s wrong
disp ( ” ( i i )
L = 2 ∗ 10 = 20 uH” )
disp ( ” and C eq = 4 0 5 . 2 8 4 pF” )
f =(1/(2* %pi * sqrt (20*405.284*10^ -18) ) ) *10^ -6 // i n
MHz
disp (f , ” f ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗ C eq ) =” )
Scilab code Exa 4.23 series and parallel resonant freqency and Qfactor
1 // Example 4 . 2 3 .
2 clc
3 f =(1/(2* %pi * sqrt (0.33*0.065*10^ -12) ) ) *10^ -6
4
5
6
7
8
9
10
11
12
13
14
// i n
MHz
format (6)
disp (f , ” ( i )
f ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗C) =” )
ceq =0.065/1.065 // i n pF
disp ( ceq , ” ( i i ) C eq ( i n pF ) = C∗C M / C+C M =” )
fp =(1/(2* %pi * sqrt (0.33*0.061*10^ -12) ) ) *10^ -6 // i n
MHz
disp ( fp , ” ( i )
f p ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗ C eq ) =”
)
pi =((1.121 -1.087) /1.087) *100 // i n p e r c e n t a g e
disp ( pi , ” ( i i i ) % i n c r e a s e =” )
q =(2* %pi *1.087*0.33*10^6) /(5.5*10^3)
format (8)
disp (q , ” ( i v ) Q = omega x ∗L / R = 2∗ p i ∗ f s ∗L / R =” )
Scilab code Exa 4.24 change in frequency and trimmer capacitance
1 // Examle 4 . 2 4
2 clc
55
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
disp ( ” ( i ) Assume one p e r t i c u l a r c o u p l i n g d i r e c t i o n
f o r which , ” )
disp ( ” L e q = L1 + L2 + 2M = 0 . 2 5 mH” )
format (8)
f =(1/(2* %pi * sqrt (0.25*100*10^ -15) ) ) *10^ -6 // i n MHz
disp (f , ” T h e r e f o r e ,
f ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L e q ∗C
) =” )
disp ( ” L e t t h e d i r e c t i o n o f c o u p l i n g i s r e v e r s e d , ” )
disp ( ” L e q = L1 + L2 − 2M = 0 . 1 5 mH” )
fd =(1/(2* %pi * sqrt (0.15*100*10^ -15) ) ) *10^ -6 // i n
MHz
format (7)
disp ( fd , ” T h e r e f o r e ,
f ’ ’ ( i n MHz) = 1 / 2∗ p i ∗ s q r t (
L e q ∗C) =” )
pc =((1.2994 -1.00658) /1.00658) *100 // i n p e r c e n t a g e
format (6)
disp ( pc , ” T h e r e f o r e , % c h a n g e = f ’ ’ − f / f ∗ 100 =” )
disp ( ” ( i i ) L e t u s assume d i r e c t i o n o f c o u p l i n g s u c h
that , ”)
disp ( ” L e q = L1 + L2 + 2M = 0 . 2 5 mH” )
disp ( ”
C t = Trim c a p a c i t o r = 100 pF” )
disp ( ” T h e r e f o r e , C eq = C∗ C t / C+C t = 50 pF” )
f1 =(1/(2* %pi * sqrt (0.25*50*10^ -15) ) ) *10^ -6 // i n MHz
format (7)
disp ( f1 , ” T h e r e f o r e ,
f = 1 / 2∗ p i ∗ s q r t ( L e q ∗ C eq ) =”
)
disp ( ” I f now d i r e c t i o n o f c o u p l i n g i s r e v e r s e d , ” )
disp ( ” L e q = L1 + L2 − 2M = 0 . 1 5 mH” )
f2 =(1/(2* %pi * sqrt (0.15*50*10^ -15) ) ) *10^ -6 // i n MHz
format (8)
disp ( f2 , ” T h e r e f o r e ,
f ’ ’ = 1 / 2∗ p i ∗ s q r t ( L e q ∗ C eq )
=” )
pc1 =((1.83776 -1.4235) /1.4235) *100
format (7)
disp ( pc1 , ” T h e r e f o r e , % c h a n g e = f ’ ’ − f / f ∗ 100 =” )
56
Scilab code Exa 4.25 design RC phase shift oscillator
1 // Example 4 . 2 5
2 clc
3 disp ( ” For RC p h a s e s h h i f t o s c i l l a t o r , ” )
4 disp ( ”
h f e = 4K + 23 + 29/K
. . . given
h f e = 150 ” )
5 disp ( ” T h e r e f o r e ,
150 = 4K + 23 + 29/K” )
6 disp ( ” T h e r e f o r e ,
4Kˆ2 − 127K + 29 = 0 ” )
7 K = poly (0 , ’K ’ )
8 p1 =4* K ^2 -127* K +29
9 t1 = roots ( p1 )
10 format (6)
11 disp ( t1 , ” T h e r e f o r e , K =” )
12 disp ( ”
f = 1 / 2∗ p i ∗R∗C∗ s q r t (6+4K)
13
14
15
16
17
18
19
20
21
22
23
. . . given
f = 5 kHz ” )
disp ( ” T h e r e f o r e , Choose
C = 100 pF” )
r =(1/(2* %pi *(1000*10^ -12) *(5*10^3) * sqrt (6+(4*0.23) ) )
) *10^ -3 // i n k−ohm
format (3)
disp (r , ” T h e r e f o r e , R( i n k−ohm ) =” )
disp ( ” K = R C / R i . e . R C = KR = 2 . 7 k−ohm” )
disp ( ” N e g l e c t i n g e f f e c t o f b i a s i n g r e s i s t a n c e s
a s s u m i n g them t o be l a r g e and s e l e c t i n g
t r a n s i s t o r w i t h h i e = 2 k−ohm” )
disp ( ” R ’ ’ i = h i e = 2 k−ohm” )
disp ( ” T h e r e f o r e , L a s t r e s i s t a n c e i n p h a s e n e t w o r k ” )
r3 =12 -2
disp ( r3 , ” R3 = R − R ’ ’ i =” )
disp ( ” U s i n g t h e back t o back c o n n e c t e d z e n e r d i o d e s
o f 9 . 3 V ( Vz ) e a c h a t t h e o u t p u t o f e m i t t e r
f o l l o w e r and u s i n g t h i s a t t h e o u t p u t o f t h e
o s c i l l a t o r , t h e o u t p u t a m p l i t u d e can be
c o n t r o l l e d t o 10 V i . e . 20 V peak t o peak . The
57
z e n e r d i o d e 9 . 3V and f o r w a r d b i a s e d d i o d e o f 0 . 7
V g i v e s t o t a l 10 V” )
24 disp ( ” The d e s i g n e d c i r c u i t i s shown i n f i g . 4 . 4 9 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 4.26 range of capacitor
1 // Example 4 . 2 6
2 clc
3 disp ( ” L1 = 20 uH ,
L2 = 2 mH” )
4 leq =((20*10^ -6) +(2*10^ -3) ) *10^3
// i n mH
5 format (7)
6 disp ( leq , ” T h e r e f o r e ,
L e q ( i n mH) = L1 + L2 =” )
7 disp ( ”
For f = f m a x = 2 . 5 MHz” )
8 disp ( ”
f = 1 / 2∗ p i ∗ s q r t (C∗ L e q ) ” )
9 c =(1/(4*( %pi ^2) *((2.5*10^6) ^2) *(2.002*10^ -3) ) ) *10^12
10
11
12
13
14
15
16
17
// i n pF
format (7)
disp (c , ” T h e r e f o r e , C( i n pF ) =” )
disp ( ” For f = f m i n = 1 MHz” )
disp ( ” f = 1 / 2∗ p i ∗ s q r t (C∗ L e q ) ” )
c1 =(1/(4*( %pi ^2) *((1*10^6) ^2) *(2.002*10^ -3) ) ) *10^12
// i n pF
format (8)
disp ( c1 , ” T h e r e f o r e , C( i n pF ) =” )
disp ( ” Thus C must be v a r i e d from 2 . 0 2 4 4 pF t o
1 2 . 6 5 2 5 pF” )
Scilab code Exa 4.27 change in frequency of oscillation
1
// Example 4 . 2 7
58
2 clc
3 ceq =((0.02*12*10^ -24) /(12.02*10^ -12) ) *10^12
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
// i n
pF
format (8)
disp ( ceq , ” C eq ( i n pF ) = C1∗C2 / C1+C2 =” )
fs =(1/(2* %pi * sqrt (50*0.02*10^ -15) ) ) *10^ -6 // i n MHz
format (7)
disp ( fs , ” T h e r e f o r e ,
f s ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗
C1 ) =” )
fp =(1/(2* %pi * sqrt (50*0.01996*10^ -15) ) ) *10^ -6 // i n
MHz
format (7)
disp ( fp , ” T h e r e f o r e ,
f p ( i n MHz) = 1 / 2∗ p i ∗ s q r t ( L∗
C eq ) =” )
disp ( ” L e t C s = 5 pF c o n n e c t e d a c r o s s t h e c r y s t a l ” )
c2 =12+5
disp ( c2 , ” T h e r e f o r e , C ’ ’ 2 ( i n pF ) = C2 + C x =” )
format (10)
ceq1 =0.019976
disp ( ceq1 , ” T h e r e f o r e , C ’ ’ e q ( i n pF ) = C1∗C ’ ’ 2 / C1+
C ’ ’ 2 =” )
fp1 =5.03588
disp ( fp1 , ” T h e r e f o r e ,
f ’ ’ p ( i n MHz) = 1 / 2∗ p i ∗ s q r t (
L∗ C eq ) =” )
disp ( ”New C x = 6 pF i s c o n n e c t e d then , ” )
c21 =12+6
disp ( c21 , ” C ’ ’ ’ ’ 2 ( i n pF ) = C2 + C x =” )
ceq2 =0.0199778
disp ( ceq2 , ” T h e r e f o r e , C ’ ’ ’ ’ e q ( i n pF ) = C1∗C ’ ’ ’ ’ 2 /
C1+C ’ ’ ’ ’ 2 =” )
fp2 =5.035716
disp ( fp2 , ” T h e r e f o r e ,
f ’ ’ ’ ’ p ( i n MHz) = 1 / 2∗ p i ∗
s q r t ( L∗C ’ ’ ’ ’ e q ) =” )
c =(5.03588 -5.035716) *10^6
disp (c , ” T h e r e f o r e , Change ( i n Hz ) = f ’ ’ p − f ’ ’ ’ ’ p
=” )
59
Chapter 5
Combinational Logic Circuits
Scilab code Exa 5.1 design a combinational logic circuit
1 // Example 5 . 1
2 clc
3 disp ( ” Given p r o b l e m
4
5
6
7
8
9
10
11
12
13
s p e c i f i c that there are three
i n p u t v a r i a b l e s and one o u t p u t v a r i a b l e . We
a s s i g n A, B and C l e t t e r s y m b o l s t o t h r e e i n p u t
v a r i a b l e s and a s s i g n Y l e t t e r symbol t o one
o u t p u t v a r i a b l e . The r e l a t i o n s h i p b e t w e e n i n p u t
v a r i a b l e s and o u t p u t v a r i a b l e can be t a b u l a t e d a s
shown i n t r u t h t a b l e 5 . 1 ” )
disp ( ” A
B
C
Y” )
disp ( ” 0
0
0
0”)
disp ( ” 0
0
1
0”)
disp ( ” 0
1
0
0”)
disp ( ” 0
1
1
1”)
disp ( ” 1
0
0
0”)
disp ( ” 1
0
1
1”)
disp ( ” 1
1
0
1”)
disp ( ” 1
1
1
1”)
disp ( ”Now we o b t a i n t h e s i m p l i f i e d B o o l e a n
e x p r e s s i o n f o r o u t p u t v a r i a b l e Y u s i n g K−map
s i m p l i f i c a t i o n . ”)
60
14
15
16
17
disp ( ”
BC
BC’ ’
B’ ’C’ ’
disp ( ”A
0
0
1
disp ( ”A’ ’ 0
1
1
disp ( ” Y = AC + BC + AB” )
B ’ ’ C” )
0”)
1”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.2 design a circuit with control line C and data lines
1 // Example 5 . 2
2 clc
3 disp ( ” The t r u t h
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
t a b l e f o r the g i v e n problem i s as
shown b e l o w . ” )
disp ( ” C
D3
D2
D1
Output ” )
disp ( ” 0
x
x
x
0”)
disp ( ” 0
0
0
0
0”)
disp ( ” 0
0
0
1
1”)
disp ( ” 0
0
1
0
1”)
disp ( ” 0
1
0
0
1”)
disp ( ” ” )
disp ( ”K−map s i m p l i f i c a t i o n ” )
disp ( ”
D1 ’ ’ D2 ’ ’
D1 ’ ’ D2
D1D2
D1D2
’ ’ ”)
disp ( ”C ’ ’ D3 ’ ’
0
0
0
0”)
disp ( ”C ’ ’ D3
0
0
0
0”)
disp ( ”CD3
1
X
X
X” )
disp ( ”CD3 ’ ’
0
1
X
1”)
disp ( ” ” )
disp ( ” T h e r e f o r e ,
Y = CD3 + CD2 + CD1” )
This code can be downloaded from the website wwww.scilab.in
61
Scilab code Exa 5.3 design combinational circuit
1 // Example 5 . 3
2 clc
3 disp ( ” Truth t a b l e ” )
4 disp ( ”
Input
5 disp ( ”
Decimal D i g i t
6
7
8
9
10
11
12
13
14
15
16
17
18
)
disp ( ”
A B C
Y0” )
disp ( ”
0 0 0
0”)
disp ( ”
0 0 0
1”)
disp ( ”
0 0 1
0”)
disp ( ”
0 0 1
1”)
disp ( ”
0 1 0
0”)
disp ( ”
0 1 0
1”)
disp ( ”
0 1 1
0”)
disp ( ”
0 1 1
1”)
disp ( ”
1 0 0
0”)
disp ( ”
1 0 0
1”)
disp ( ” ” )
disp ( ” Here Y0 =
C , Y5 = B ,
given c i r c u i t
l i n e s without
Output ” )
Digit 1
Digit 0”
D
Y7
Y6
Y5
Y4
Y3
Y2
Y1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
1
0
0
0
0
1
0
0
0
0
1
0
0
1
0
0
1
0
0
0
0
1
1
0
0
0
1
0
0
1
1
0
1
0
0
0
1
0
0
0
0
0
1
0
1
0
0
0
1
0
D, Y1 = 0 , Y2
Y6 = A and Y7 =
can be o b t a i n e d
u s i n g any l o g i c
62
= D, Y3 = 0 , Y4 =
0. Therefore , the
from t h e i n p u t
g a t e s ”)
Scilab code Exa 5.4 design logic circuit
1 // Example 5 . 4
2 clc
3 disp ( ” L e t u s c o n s i d e r D f o r Door ,
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
I for ignition , L
f o r L i g h t . Then c o n d i t i o n s t o a c t i v a t e t h e a l a r m
are : ”)
disp ( ” 1 . The h e a d l i g h t s a r e ON w h i l e t h e i g n i t i o n i s
OFF . ” )
disp ( ”
i . e . L = 1 , I = 0 and D may be a n y t h i n g . ” )
disp ( ” 2 . The d d o r i s open w h i l e t h e i g n i t i o n i s ON” )
disp ( ”
i . e . D = 1 , I = 1 , L may be a n y t h i n g . ” )
disp ( ” A l s o a l a r m w i l l sound i f l o g i c c i r c u i t o u t p u t
i s zero . ”)
disp ( ” T h e r e f o r e , o u t p u t f o r a b o v e c o n d i t i o n i s z e r o
and f o r r e s t o f t h e c o n d i t i o n i t i s 1 which i s
summarized i n t h e f o l l o w i n g t a b l e . ” )
disp ( ” D
I
L
Y” )
disp ( ” 0
0
0
1”)
disp ( ” X
0
1
0”)
disp ( ” 0
1
0
1”)
disp ( ” 0
1
1
1”)
disp ( ” 1
0
0
1”)
disp ( ” 1
1
X
0”)
disp ( ” T h e r e f o r e , K−map f o r l o g i c c i r c u i t . ” )
disp ( ”
I ’ ’L’ ’
I ’ ’L
IL
IL ’ ’ ” )
disp ( ”D’ ’
1
0
1
1”)
disp ( ”D
1
0
0
0”)
disp ( ” Output = Y = I ’ ’ L ’ ’ + D’ ’ I ” )
disp ( ” As AND−OR l o g i c can be d i r e c t l y r e p l a c e d by
NAND−NAND, l o g i c c i r c u i t u s i n g o n l y NAND g a t e s i s
a s shown i n f i g . 5 . 9 and f i g . 5 . 1 0 ” )
63
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.5 design circuit to detect invalid BCD number
1 // Example 5 . 5
2 clc
3 disp ( ” Truth t a b l e ” )
4 disp ( ” Dec
A B C D
Output Y” )
5 disp ( ”
0
0 0 0 0
0”)
6 disp ( ”
1
0 0 0 1
0”)
7 disp ( ”
2
0 0 1 0
0”)
8 disp ( ”
3
0 0 1 1
0”)
9 disp ( ”
4
0 1 0 0
0”)
10 disp ( ”
5
0 1 0 1
0”)
11 disp ( ”
6
0 1 1 0
0”)
12 disp ( ”
7
0 1 1 1
0”)
13 disp ( ”
8
1 0 0 0
0”)
14 disp ( ”
9
1 0 0 1
0”)
15 disp ( ”
10
1 0 1 0
1”)
16 disp ( ”
11
1 0 1 1
1”)
17 disp ( ”
12
1 1 0 0
1”)
18 disp ( ”
13
1 1 0 1
1”)
19 disp ( ”
14
1 1 1 0
1”)
20 disp ( ”
15
1 1 1 1
1”)
21 disp ( ” ” )
22 disp ( ”K−map s i m p l i f i c a t i o n ” )
23 disp ( ”
C ’ ’ D’ ’
C’ ’D
CD
24
25
)
disp ( ”A’ ’ B ’ ’
disp ( ”A’ ’ B
0
0
0
0
64
0
0
CD’ ’ ”
0”)
0”)
26
27
28
disp ( ”AB
1
disp ( ”AB’ ’
0
disp ( ” Y = AB + AC” )
1
1
0
1
1”)
1”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.6 design two level combinational circuit
1 // Example 5 . 6
2 clc
3 disp ( ”
I n p u t 1 −> P r e s s u r e i n f u e l t a n k ” )
4 disp ( ”
I n p u t 2 −> P r e s s u r e i n o x i d i z e r t a n k ” )
5 disp ( ”
Input = 1 I n d i c a t e s p r e s s u r e i s equal to or
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
a b o v e t h e r e q u i r e d minimum” )
disp ( ”
= 0 Otherwise ”)
disp ( ”
I n p u t 3 −> From t i m e r ” )
disp ( ” i f i n p u t 3 = 1 I n d i c a t e s t h a t t h e r e a r e l e s s
t h a n o r e x a c t l y 10 m i n u t e s f o r l i f t o f f ” )
disp ( ”
= 0 Otherwise ”)
disp ( ”
Output −> P a n e l l i g h t , i f l i g h t g o e s on
then ”)
disp ( ”
Output = 1 ” )
disp ( ” e l s e Output = 0 ” )
disp ( ” ” )
disp ( ” Truth t a b l e ” )
disp ( ” L e t i n p u t 1 = A,
input 2 = B,
input 3 = C. ”)
disp ( ”
Inputs
Output ” )
disp ( ” A
B
C
Y” )
disp ( ” 0
0
0
1”)
disp ( ” 0
0
1
0”)
disp ( ” 0
1
0
1”)
disp ( ” 0
1
1
0”)
disp ( ” 1
0
0
1”)
disp ( ” 1
0
1
0”)
65
disp ( ” 1
disp ( ” 1
disp ( ” ” )
disp ( ”K−map
disp ( ”
disp ( ”A’ ’
disp ( ”A
disp ( ” Y =
disp ( ”
=
disp ( ”
=
A + B] ”)
34 disp ( ”
=
35 disp ( ”
=
24
25
26
27
28
29
30
31
32
33
1
1
0
1
0”)
1”)
s i m p l i f i c a t i o n ”)
B’ ’C’ ’
B’ ’C
BC
BC’ ’ ” )
1
0
0
1”)
1
0
1
0”)
ABC + A’ ’ B ’ ’ C ’ ’ + B ’ ’ C ’ ’ ” )
ABC + C ’ ’ ( B’ ’ +A’ ’ B) ” )
ABC + C ’ ’ ( B’ ’ +A ’ ’ )
[ A’ ’ +A’ ’ B =
ABC + C ’ ’ ( A’ ’ B ’ ’ ) ” )
A’ ’ B ’ ’ XOR C ’ ’ ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.7 design 32 to 1 multiplexer
1 // Example 5 . 7
2 clc
3 disp ( ” F i g . 5 . 2 0 shows a 32 t o 1 m u l t i p l e x e r
u s i n g 74
LS150 I C s . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.8 design a 32 to 1 multiplexer
1 // Example 5 . 8
2 clc
3 disp ( ” F i g . 5 . 2 1 shows a 32 t o 1 m u l t i p l e x e r
using
f o u r 8 t o 1 m u l t i p l x e r e s and 2 t o 4 d e c o d e r . . ” )
66
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.9 implement boolean function using 8to1 multiplexer
1 // Example 5 . 9
2 clc
3 disp ( ” The f u n c t i o n can be i m p l e m e n t e d w i t h a 8 t o 1
m u l t i p l e x e r , a s shown i n f i g . 5 . 2 2 . Three
v a r i a b l e s A, B and C a r e a p p l i e d t o t h e s e l e c t
l i n e s . The m i n t e r m s t o be i n c l u d e d ( 1 , 3 , 5 and
6 ) a r e c h o s e n by making t h e i r c o r r e s p o n d i n g i n p u t
l i n e s e q u a l t o 1 . Mintems 0 , 2 , 4 and 7 a r e n o t
i n c l u d e d by making t h e i r i n p u t l i n e s e q u a l t o 0 . ”
)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.10 implement boolean function using 4to1 multiplexer
1 // Example 5 . 1 0
2 clc
3 disp ( ” F i g . 5 . 2 3 shows t h e i m p l e m e n t a t i o n o f
function
w i t h 4 t o 1 m u l t i p l e x e r . Two o f t h e v a r i a b l e s , B
and C , a r e a p p l i e d t o t h e s e l e c t i o n l i n e s . B i s
c o n n e c t e d t o S1 and C i s c o n n e c t e d t o S0 . The
i n p u t s f o r m u l t i p l e x e r a r e d e r i v e d from t h e
implementation t a b l e . ”)
4 disp ( ” Truth t a b l e ” )
5 disp ( ” Minterm
A B C
F” )
67
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
disp ( ”
0
0 0 0
0”)
disp ( ”
1
0 0 1
1”)
disp ( ”
2
0 1 0
0”)
disp ( ”
3
0 1 1
1”)
disp ( ”
4
1 0 0
0”)
disp ( ”
5
1 0 1
1”)
disp ( ”
6
1 1 0
1”)
disp ( ”
7
1 1 1
0”)
disp ( ” ” )
disp ( ” I m p l e m e n t a t i o n t a b l e ” )
disp ( ”
D0 D1 D2 D3” )
disp ( ”A’ ’
0
1
2
3
Row 1 ” )
disp ( ”A
4
5
6
7
Row 2 ” )
disp ( ”
0
1
A
A’ ’ ” )
disp ( ” ” )
disp ( ” As shown i n f i g . 5 . 2 3 ( c ) t h e i m p l e m e n t a t i o n
t a b l e i s n o t h i n g but t h e l i s t o f t h e i n p u t s o f
t h e m i l t i p l e x e r s and u n d e r them l i s t o f a l l t h e
m i n t e r m s i n two r o w s . The f i r s t row l i s t s a l l
t h o s e m i n t e r m s where A i s complemented , and t h e
s e c o n d row l i s t s a l l t h e m i n t e r m s w i t h A
uncomplemented . The m i n t e r m s g i v e n i n t h e
f u n c t i o n a r e c i r c l e d and t h e n e a c h column i s
i n s e r t e d s e p a r a t e l y as f o l l o w s . ”)
disp ( ” 1 . I f t h e two m i n t e r m s i n a column a r e n o t
c i r c l e d , O i s applied to the corresponding
m u l t i p l e x e r i n p u t ( s e e column 1 ) . ” )
disp ( ” 2 . I f t h e two m i n t e r m s i n a column a r e c i r c l e d
, 1 i s applied to the corresponding m u l t i p l e x e r
i n p u t ( s e e column 2 ) . ” )
disp ( ” 3 . I f t h e minterm i n t h e s e c o n d row i s c i r c l e d
and m i n t e r m s i n t h e f i r s t row i s n o t c i r c l e d , A
i s applied to the corresponding m u l t i p l e x e r input
( s e e column 3 ) . ” )
disp ( ” 4 . I f t h e minterm i n t h e f i r s t row i s c i r c l e d
and minterm i n t h e s e c o n d row i s n o t c i r c l e d , A’ ’
i s applied to the corresponding m u l t i p l e x e r
i n p u t ( s e e column 4 ) . ” )
68
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.11 implement boolean function using 8to1 MUX
1 // Example 5 . 1 1
2 clc
3 disp ( ” F i g 5 . 2 5 shows t h e i m p l e m e n t a t i o n o f g i v e n
4
5
6
7
8
Booolean f u n c t i o n with 8 : 1 m i l t i p l e x e r . ”)
disp ( ” I m p l e m e n t a t i o n t a b l e ” )
disp ( ”
D0 D1 D2 D3 D4 D5 D6 D7” )
disp ( ”A’ ’
0
1
2
3
4
5
6
7”)
disp ( ”A
8
9
10 11 12 13 14 15 ” )
disp ( ”
1
1
0
A’ ’ A’ ’ 0
0
A” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.12 implement boolean function using 4to1 MUX
1 // Example 5 . 1 2
2 clc
3 disp ( ” The f u n c t i o n h a s f o u r
v a r i a b l e s . To i m p l e m e n t
t h i s f u n c t i o n we r e q u i r e 8 : 1 m u l t i p l e x e r . i . e . ,
two 4 : 1 m u l t i p l e x e r s . We have a l r e a d y s e e n how
t o c o n s t r u c t 8 : 1 m u l t i p l e x e r u s i n g two 4 : 1
m u l t i p l e x e r s . The same c o n c e p t i s u s e d h e r e t o
implement g i v e n Boolean f u n c t i o n . ”)
4 disp ( ” ” )
5 disp ( ” I m p l e m e n t a t i o n t a b l e ” )
6 disp ( ”
D0 D1 D2 D3 D4 D5 D6 D7” )
69
7
8
9
disp ( ”A’ ’
disp ( ”A
disp ( ”
0
1
8
9
A’ ’ 1
2
3
4
5
6
7”)
10 11 12 13 14 15 ” )
A’ ’ 0
1
0
1
0”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.13 implement boolean function using 8to1 MUX
1 // Example 5 . 1 3
2 clc
3 disp ( ” The g i v e n B o o l e a n e x p r e s s i o n
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
i s not in
s t a n d a r d SOP form . L e t u s f i r s t c o n v e r t t h i s i n
s t a n d a r d form . ” )
disp ( ” F (A, B , C , D) = A’ ’ BD’ ’ ( C+C ’ ’ ) + ACD(B+B ’ ’ )
+ B ’ ’CD(A+A ’ ’ ) + A’ ’ C ’ ’ D(B+B ’ ’ ) ” )
disp ( ”
= A’ ’ BCD’ ’ + A’ ’ BC’ ’ D’ ’ + ABCD
+ AB’ ’CD + AB’ ’CD + A’ ’ B ’ ’CD + A’ ’ BC’ ’ D + A’ ’ B ’ ’
C ’ ’ D” )
disp ( ”
= A’ ’ BCD’ ’ + A’ ’ BC’ ’ D’ ’ + ABCD
+ AB’ ’CD + A’ ’ B ’ ’CD + A’ ’ BC’ ’ D + A’ ’ B ’ ’ C ’ ’ D” )
disp ( ” ” )
disp ( ” The t r u t h t a b l e f o r t h i s s t a n d a r d SOP form can
be g i v e n a s ” )
disp ( ” No .
Minterms
A B C D Y” )
disp ( ” 0
0 0 0 0 0”)
disp ( ” 1
A’ ’ B ’ ’ C ’ ’ D
0 0 0 1 1”)
disp ( ” 2
0 0 1 0 0”)
disp ( ” 3
A’ ’ B ’ ’CD
0 0 1 1 1”)
disp ( ” 4
A’ ’ BC’ ’ D’ ’
0 1 0 0 1”)
disp ( ” 5
A’ ’ BC’ ’ D
0 1 0 1 1”)
disp ( ” 6
A’ ’ BCD’ ’
0 1 1 0 1”)
disp ( ” 7
0 1 1 1 0”)
disp ( ” 8
1 0 0 0 0”)
disp ( ” 9
1 0 0 1 0”)
70
20
21
22
23
24
25
26
27
28
29
30
31
32
disp ( ” 10
1 0 1 0 0”)
disp ( ” 11
AB’ ’CD
1 0 1 1 1”)
disp ( ” 12
1 1 0 0 0”)
disp ( ” 13
1 1 0 1 0”)
disp ( ” 14
1 1 1 0 0”)
disp ( ” 15
ABCD
1 1 1 1 1”)
disp ( ” ” )
disp ( ”From t h e t r u t h t a b l e B o o l e a n f u n c t i o n can be
implemented u s i n g 8 : 1 m u l t i p l e x e r as f o l l o w s : ”
)
disp ( ” I m p l e m e n t a t i o n t a b l e : ” )
disp ( ”
D0 D1 D2 D3 D4 D5 D6 D7” )
disp ( ”A’ ’
0
1
2
3
4
5
6
7”)
disp ( ”A
8
9
10 11 12 13 14 15 ” )
disp ( ”
0
A’ ’ 0
1
A’ ’ A’ ’ A’ ’ A” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.14 implement boolean function using 8to1 MUX
1 // Example 5 . 1 4 .
2 clc
3 disp ( ” Here , i n s t e a d
4
5
6
7
8
9
o f minterms , maxterms a r e
s p e c i f i e d . Thus , we have t o c i r c l e maxterms which
are not i n c l u d e d in the Boolean f u n c t i o n . Fig .
5 . 2 8 shows t h e i m p l e m e n t a t i o n o f B o o l e a n f u n c t i o n
with 8 : 1 m u l t i p l e x e r . ”)
disp ( ” ” )
disp ( ” I m p l e m e n t a t i o n t a b l e : ” )
disp ( ”
D0 D1 D2 D3 D4 D5 D6 D7” )
disp ( ”A’ ’
0
1
2
3
4
5
6
7”)
disp ( ”A
8
9
10 11 12 13 14 15 ” )
disp ( ”
0
A’ ’ A’ ’ A
A’ ’ A
0
1”)
71
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.15 implement boolean function using 8to1 MUX
1 // Example 5 . 1 5
2 clc
3 disp ( ” I n t h e g i v e n B o o l e a n f u n c t i o n
4
5
6
7
8
9
10
11
t h r e e don ’ ’ t
c a r e c o n d i t i o n s a r e a l s o s p e c i f i e d . We know t h a t
don . . t c a r e c o n d i t i o n s can be t r e a t e d a s e i t h e r 0
s o r 1 s . F i g . 5 . 2 9 shows t h e i m p l e m e n t a t i o n o f
given Boolean f u n c t i o n using 8 : 1 m u l t i p l e x e r . ”)
disp ( ” ” )
disp ( ” I m p l e m e n t a t i o n t a b l e : ” )
disp ( ”
D0 D1 D2 D3 D4 D5 D6 D7” )
disp ( ”A’ ’
0
1
2
3
4
5
6
7”)
disp ( ”A
8
9
10 11 12 13 14 15 ” )
disp ( ”
1
0
1
1
A
A
1
0”)
disp ( ” ” )
disp ( ” I n t h i s example , by t a k i n g don ’ ’ t c a r e
c o n d i t i o n s 8 and 14 we have e l i m i n a t e d A’ ’ term
and h e n c e t h e i n v e r t e r . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.16 determine boolean expression
1 // Example 1 5 . 6
2 clc
3 disp ( ”
D’ ’
D” )
4 disp ( ”D
0
1”)
72
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
disp ( ” 0
2
3”)
disp ( ” 0
4
5”)
disp ( ” 1
6
7”)
disp ( ”D
8
9”)
disp ( ” 1
10
11 ” )
disp ( ”D’ ’ 12
13 ” )
disp ( ” 0
14
15 ” )
disp ( ” ” )
disp ( ” Here , i m p l e m e n t a t i o n t a b l e i s l i s t e d f o r l e a s t
s i g n i f i c a n t b i t i . e . D . The f i r s t column l i s t
a l l m i n t e r m s w i t h D i s c o m p l e m e n t a t e d and t h e
s e c o n d column l i s t s a l l t h e m i n t e r m s w i t h D
uncomplemented , a s shown i n f i g . 5 . 3 0 ( a ) . Then
a c c o r d i n g to data i n p u t s given to the m u l t i p l e x e r
minterms a r e c i r c l e d a p p l y i n g f o l l o w i n g r u l e s . ”)
disp ( ” 1 . I f m u l t i p l e x e r i n p u t i s 0 , don ’ ’ t c i r c l e
any minterm i n t h e c o r r e s p o n d i n g row . ” )
disp ( ” 2 . I f m u l t i p l e x e r i n p u t 1 , c i r c l e b o t h t h e
m i n t e r m s i n t h e c o r r e s p o n d i n g row . ” )
disp ( ” 3 . I f m u l t i p l e x e r i n p u t i s D, c i r c l e t h e
minterm b e l o n g s t o cloumn D i n t h e c o r r e s p o n d i n g
row . ” )
disp ( ” 4 . I f m u l t i p l e x e r i n p u t i s D’ ’ , c i r c l e t h e
minterm b e l o n g s t o column D’ ’ i n t h e
c o r r e s p o n d i n g row . ” )
disp ( ” T h i s i s i l l u s t r a t e d i n f i g . 5 . 3 0 ( b ) . Now
c i r c l e d m i n t e r m s can be w r i t t e n t o g e t B o o l e a n
e x p r e s s i o n as f o l l o w s : ”)
disp ( ” Y = A’ ’ B ’ ’ C ’ ’ D + A’ ’ BCD’ ’ + A’ ’BCD + AB’ ’ C ’ ’
D + AB’ ’ CD’ ’ + AB’ ’CD + ABC’ ’ D’ ’ ” )
Scilab code Exa 5.17 realize using 4 to 1 MUX
1 // Example 5 . 1 7
2 clc
73
3
4
5
6
7
8
9
10
11
12
disp ( ”
disp ( ”w ’ ’ x ’ ’
disp ( ”w ’ ’ x
disp ( ”wx ’ ’
disp ( ”wx
disp ( ” ” )
disp ( ”D0 = w ’
disp ( ”D1 = w ’
disp ( ”D2 = w ’
disp ( ”D3 = w ’
D0
D3” )
0
1
2
3”)
4
5
6
7”)
8
9
10 11 ” )
12 13 14 15 ” )
’x +
’x’ ’
’x +
’x +
D1
D2
wx ’ ’ =
+ wx ’ ’
wx ’ ’ =
wx ’ ’ +
w XOR x ” )
= x ’ ’ ”)
w XOR x ” )
wx = x + wx ’ ’ = w + x ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.18 design 1 to 8 DEMUX
1 // Example 5 . 1 8
2 clc
3 disp ( ” The c a s c a d i n g
of d e m u l t i p l e x e r s i s s i m i l a r to
t h e c a s c a d i n g o f d e c o d e r . F i g . 5 . 3 3 shows
c a s c a d i n g o f two 1 : 4 d e m u l t i p l e x e r s t o form 1 :
8 d e m u l t i p l e x e r . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.19 implement full subtractor
1 // Example 5 . 1 9
2 clc
3 disp ( ” L e t u s s e e t h e t r u t h
”)
4 disp ( ”
A
B
Bin
D
table of f u l l subtractor .
Bout ” )
74
disp ( ” 0 0
0
0
0”)
disp ( ” 0 0
1
1
1”)
disp ( ” 0 1
0
1
1”)
disp ( ” 0 1
1
0
1”)
disp ( ” 1 0
0
1
0”)
disp ( ” 1 0
1
0
0”)
disp ( ” 1 1
0
0
0”)
disp ( ” 1 1
1
1
1”)
disp ( ” ” )
disp ( ” For f u l l s u b t r a c t o r d i f f e r e n c e D f u n c t i o n can
be w r i t t e n a s D = f (A, B , C) = summation m( 1 , 2 ,
4 , 7 ) and Bout f u n c t i o n can be w r i t t e n a s ” )
15 disp ( ”
Bout = F (A, B , C) = summation m( 1 , 2 , 3 ,
7) ”)
16 disp ( ” With Din i n p u t 1 , d e m u l t i p l e x e r g i v e s m i n t e r m s
a t t h e o u t p u t s o by l o g i c a l l y ORing r e q u i r e d
m i n t e r m s we can i m p l e m e n t B o o l e a n f u n c t i o n s f o r
f u l l s u b t r a c t o r . F i g . 5 . 3 4 shows t h e
implementation of f u l l subtractor using
d e m u l t i p l e x e r . ”)
5
6
7
8
9
10
11
12
13
14
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.20 construst 1 to 32 DEMUX
1 // Exmaple 5 . 2 0
2 clc
3 disp ( ” The f i g . 5 . 3 7 shows t h e i m p l e m e n t a t i o n o f 1 t o
32 d e m u l t i p l e x e r u s i n g two 74 X154 I C s . Here , t h e
most s i g n i f i c a n t b i t o f s e l e c t s i g n a l ( A4 ) i s
u s e d t o e n a b l e e i t h e r u p p e r 1 t o 16 d e m u l t i p l e x e r
o r l o w e r 1 t o 16 d e m u l t i p l e x e r . The d a t a i n p u t
and o t h e r s e l e c t s i g n a l s a r e c o n n e c t e d p a r a l l e l
t o b o t h t h e d e m u l t i p l e x e r I C s . When A4 = 0 , u p p e r
75
d e m u l t i p l e x e r i s e n a b l e d and t h e d a t a i n p u t i s
routed to the output corresponds to the s t a t u s of
A0 A1 A2 and A3 l i n e s . When A4 = 1 , l o w e r
m i l t i p l e x e r i s e n a b l e d and t h e d a t a i n p u t i s
routed to the output corresponds to the s t a t u s of
A0 A1 A2 and A3 l i n e s . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.21 design 3 to 8 decoder
1 // Example 5 . 2 1
2 clc
3 disp ( ” F i g . 5 . 4 0 shows 3 t o 8 l i n e
4
5
6
7
8
9
10
11
d e c o d e r . Here , 3
i n p u t s are decoded i n t o e i g h t outputs , each
o u t p u t r e p r e s e n t one o f t h e m i n t e r m s o f t h e 3
i n p u t v a r i a b l e s . The t h r e e i n v e r t e r s p r o v i d e t h e
complement o f t h e i n p u t s , and e a c h one o f t h e
e i g h t AND g a t e s g e n e r a t e s one o f t h e m i n t e r m s .
Enable input i s provided to a c t i v a t e decoded
o u t p u t b a s e d on d a t a i n p u t s A, B and C . The t a b l e
shows t h e t r u t h t a b l e f o r 3 t o 8 d e c o d e r . ” )
disp ( ” ” )
disp ( ” Truth t a b l e f o r a 3 t o 8 d e c o d e r ” )
disp ( ”
Inputs
|
Outputs ” )
disp ( ”EN A B C | Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0
”)
disp ( ” 0
X X X | 0
0
0
0
0
0
0
0”
)
disp ( ” 1
0 0 0 | 0
0
0
0
0
0
0
1”
)
disp ( ” 1
0 0 1 | 0
0
0
0
0
0
1
0”
)
disp ( ” 1
0 1 0 | 0
0
0
0
0
1
0
0”
76
12
13
14
15
16
)
disp ( ” 1
)
disp ( ” 1
)
disp ( ” 1
)
disp ( ” 1
)
disp ( ” 1
)
0
1
1
|
0
0
0
0
1
0
0
0”
1
0
0
|
0
0
0
1
0
0
0
0”
1
0
1
|
0
0
1
0
0
0
0
0”
1
1
0
|
0
1
0
0
0
0
0
0”
1
1
1
|
1
0
0
0
0
0
0
0”
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.22 design 5 to 32 decoder
1 // Example 5 . 2 2
2 clc
3 disp ( ” The F i g . 5 . 4 5 shows t h e c o n s t r u c t i o n
of 5 to
32 d e c o d e r u s i n g f o u r 74 LS138s and h a l f 74 LS139 .
The h a l f s e c t i o n o f 74 LS139 IC u s e d a s a 2 t o 4
d e c o d e r t o d e c o d e t h e two h i g h e r o r d e r i n p u t s , D
and E . The f o u r o u t p u t s o f t h i s d e c o d e r a r e u s e d
t o e n a b l e one o f t h e f o u r 3 t o 8 d e c o d e r s . The
t h r e e l o w e r i n p u t s A, B and C a r e c o n n e c t e d i n
p a r a l l e l t o f o u r 3 t o 8 d e c o d e r s . T h i s means t h a t
t h e same o u t p u t p i n o f e a c h o f t h e f o u r 3 t o 8
d e c o d e r s i s s e l e c t e d but o n l y one i s e n a b l e . The
remaining enables s i g n a l s of four 3 to 8 decoders
ICs a r e co nn ect ed i n p a r a l l e l to c o n s t r u c t
e n a b l e s i g n a l s f o r 5 t o 32 d e c o d e r . ” )
77
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.23 design 4 line to 16 line decoder
1 // Example 5 . 2 3
2 clc
3 disp ( ” 4 l i n e t o 16 l i n e
4
decoder using 1 l i n e to 4
l i n e decoder ”)
disp ( ” As shown i n f i g . 5 . 4 6 f i v e numbers o f 2 : 4
d e c o d e r a r e r e q u i r e d t o d e s i g n 4 : 16 d e c o d e r .
D e c o d e r 1 i s u s e d t o e n a b l e one o f t h e d e c o d e r 2 ,
3 , 4 and 5 . I n p u t s o f f i r s t d e c o d e r a r e t h e A
and B MSB i n p u t s o f 4 : 16 d e c o d e r . The i n p u t s o f
d e c o d e r a r e c o n n e c t e d t o g e t h e r f o r m i n g C and D
LSB i n p u t s o f 4 : 16 d e c o d e r . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.24 implement using 74LS138
1 // Example 5 . 2 4
2 clc
3 disp ( ” I n t h i s example , we u s e IC 74 LS138 , 3 : 8
decoder to implement m u l t i p l e output f u n c t i o n .
The o u t p u t s o f 74 LS138 a r e a c t i v e low , t h e r e f o r e ,
SOP f u n c t i o n ( f u n c t i o n F1 ) can be i m p l e m e n t e d
u s i n g NAND g a t e and POS f u n c t i o n ( f u n c t i o n F2 )
can be i m p l e m e n t e d u s i n g AND g a t e , a s shown i n
f i g . 5 . 5 0 ”)
78
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.25 implement full substractor
1 // Example 5 . 2 5
2 clc
3 disp ( ” The t r u t h
4
5
6
7
8
9
10
11
12
13
14
15
16
table for
f u l l s u b t r a c t o r i s as
shown b e l o w ” )
disp ( ” ” )
disp ( ” I n p u t s
Outputs ” )
disp ( ”A B Bin
D Bout ” )
disp ( ” 0 0
0
0
0”)
disp ( ” 0 0
1
1
1”)
disp ( ” 0 1
0
1
1”)
disp ( ” 0 1
1
0
1”)
disp ( ” 1 0
0
1
0”)
disp ( ” 1 0
1
0
0”)
disp ( ” 1 1
0
0
0”)
disp ( ” 1 1
1
1
1”)
disp ( ” ” )
disp ( ” I m p l e m e n t a t i o n o f f u l l s u b t r a c t o r u s i n g 3 : 8
d e c o d e r i s shown i n f i g . 5 . 5 1 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.26 implement gray to binary code converter
1 // Example 5 . 2 6
2 clc
79
3
4
5
6
7
8
9
10
11
12
13
14
15
disp ( ” The t r u t h t a b l e f o r 3− b i t b i n a r y t o g r a y c o d e
c o n v e r t e r i s a s shown b e l o w ” )
disp ( ” ” )
disp ( ”A B C
G2 G1 G0” )
disp ( ” 0 0 0
0
0
0”)
disp ( ” 0 0 1
0
0
1”)
disp ( ” 0 1 0
0
1
1”)
disp ( ” 0 1 1
0
1
0”)
disp ( ” 1 0 0
1
1
0”)
disp ( ” 1 0 1
1
1
1”)
disp ( ” 1 1 0
1
0
1”)
disp ( ” 1 1 1
1
0
0”)
disp ( ” ” )
disp ( ” The f i g . 5 . 5 2 shows t h e i m p l e m e n t a t i o n o f 3−
b i t binary to gray code c o n v e r t e r using 3 : 8
d e c o d e r . As o u t p u t s o f 7 4 1 3 8 a r e a c t i v e low we
have t o u s e NAND g a t e i n s t e a d o f OR g a t e . The
a c t i v e low o u t p u t from t h e d e c o d e r f o r c e s o u t p u t (
s ) o f c o n n e c t e d NAND g a t e ( s ) t o become HIGH , t h u s
implementing the f u n c t i o n . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.27 design 2 bit comparator
1 // Example
2 clc
3 disp ( ” ” )
4 disp ( ”A1
5 disp ( ” 0
6 disp ( ” 0
7 disp ( ” 0
8 disp ( ” 0
9 disp ( ” 0
5.27
A0
0
0
0
0
1
B1
0
0
1
1
0
B0
0
1
0
1
0
A>B
0
0
0
0
1
80
A=B
0
0
0
0
0
A<B” )
0”)
1”)
1”)
1”)
0”)
10
11
12
13
14
15
16
17
18
19
20
disp ( ” 0
disp ( ” 0
disp ( ” 0
disp ( ” 1
disp ( ” 1
disp ( ” 1
disp ( ” 1
disp ( ” 1
disp ( ” 1
disp ( ” 1
disp ( ” 1
1
1
1
0
0
0
0
1
1
1
1
0
1
1
0
0
1
1
0
0
1
1
1
0
1
0
1
0
1
0
1
0
1
0
0
0
1
1
0
0
1
1
1
0
1
0
0
0
0
1
0
0
0
0
1
0”)
1”)
1”)
0”)
0”)
0”)
1”)
0”)
0”)
0”)
0”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.28 design full adder circuit
1 // Example 5 . 2 8
2 clc
3 disp ( ” Truth t a b l e
4
5
6
7
8
9
10
11
12
13
)
disp ( ” I n p u t s
disp ( ”A B Cin
disp ( ” 0 0
0
disp ( ” 0 0
1
disp ( ” 0 1
0
disp ( ” 0 1
1
disp ( ” 1 0
0
disp ( ” 1 0
1
disp ( ” 1 1
0
disp ( ” 1 1
1
for
f u l l a d d e r i s a s shown b e l o w . ”
Outputs ” )
C a r r y Sum” )
0
0”)
0
1”)
0
1”)
1
0”)
0
1”)
1
0”)
1
0”)
1
1”)
81
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.29 implement BCD to 7 segment decoder
1 // Example 5 . 2 9
2 clc
3 disp ( ”BCD−to −common anode 7−s e g m e n t d e c o d e r ” )
4 disp ( ” D i g i t
A B C D
a b c d e f
5 disp ( ” 0
0 0 0 0
0 0 0 0 0 0
6 disp ( ” 1
0 0 0 1
1 0 0 1 1 1
7 disp ( ” 2
0 0 1 0
0 0 1 0 0 1
8 disp ( ” 3
0 0 1 1
0 0 0 0 1 1
9 disp ( ” 4
0 1 0 0
1 0 0 0 1 1
10 disp ( ” 5
0 1 0 1
0 1 0 0 1 0
11 disp ( ” 6
0 1 1 0
0 1 0 0 0 0
12 disp ( ” 7
0 1 1 1
0 0 0 1 1 1
13 disp ( ” 8
1 0 0 0
0 0 0 0 0 0
14 disp ( ” 9
1 0 0 1
0 0 0 0 1 0
g”)
1”)
1”)
0”)
0”)
0”)
0”)
0”)
1”)
0”)
0”)
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
Scilab code Exa 5.31 implement 32 input to 5 output encoder
1 // Example 5 . 3 1
2 clc
3 disp ( ” F i g . 5 . 7 5 shows how f o u r 74 LS148 can be
c o n n e c t e d t o a c c e p t 32 i n p u t s and p r o d u c e a 5− b i t
82
e n c o d e d o u t p u t , A0 − A4 . EO’ ’ s i g n a l i s
c o n n e c t e d t o t h e EI ’ ’ i n p u t o f t h e n e x t l o w e r
p r i o r i t y e n c o d e r and EI ’ ’ i n p u t o f t h e h i g h e s t
p r i o r i t y e n c o d e r i s g r o u n d e d . T h e r e f o r e , a t any
t i m e o n l y one e n c o d e r i s e n a b l e d . S i n c e , t h e A2 −
A0 o u t p u t s o f a t t h e most one 74 LS148 w i l l be
e n a b l e d a t a time , t h e o u t p u t s o f t h e i n d i v i d u a l
74 LS148s can be ORed t o p r o d u c e A2 − A0 . L i k e w i s e
, t h e i n d i v i d u a l GS ’ ’ o u t p u t s can be combined i n
a 4 t o 2 e n c o d e r t o p r o d u c e A4 and A3 . The GS
o u t p u t f o r 32− b i t e n c o d e r i s p r o d u c e d b y ORing GS
’ ’ outputs of a l l encoders . ”)
This code can be downloaded from the website wwww.scilab.in
83
Chapter 6
Sequential Logic Circuits
Scilab code Exa 6.4 analyze the circuit
1 // Example 6 . 4
2 clc
3 disp ( ”To a n a l y z e t h e
4
5
6
7
8
9
10
11
12
c i r c u i t means t o d r i v e t h e
truth t a b l e f o r i t . ”)
disp ( ”We have , D = I n p u t XOR Q n ” )
disp ( ” ” )
disp ( ”CLK
Input
Q n
D = i n p u t XOR Q n
Q n+1” )
disp ( ” down
0
0
0
0”)
disp ( ” down
0
1
1
1”)
disp ( ” down
1
0
1
1”)
disp ( ” down
1
1
0
0”)
disp ( ” ” )
disp ( ” I n t h e c i r c u i t f i g . 6 . 5 3 , o u t p u t d o e s n o t
c h a n g e when i n p u t i s 0 and i t t o g g l e s when i n p u t
i s 1 . T h i s i s t h e c h a r a c t e r i s t i c s o f T f l i p −f l o p .
Hence , t h e g i v e n c i r c u i i s T f l i p −f l o p
84
c o n s t r u c t e d u s i n g D f l i p −f l o p . ” )
This code can be downloaded from the website wwww.scilab.in
85
Chapter 7
Shift Registers
Scilab code Exa 7.1 determine number of flip flops needed
1 // Example 7 . 1
2 clc
3 disp ( ” ( i ) A 6− b i t
4
5
6
7
8
9
b i n a r y number r e q u i r e s r e g i s t e r
with 6 f l i p −f l o p s . ”)
disp ( ” ” )
disp ( ” ( i i ) ( 3 2 ) 1 0 = ( 1 0 0 0 0 0 ) 2 . The number o f b i t s
r e q u i r e d t o r e p r e s e n t 32 i n b i n a r y a r e s i x ,
t h e r e f o r e , 6 f l i p −f l o p s a r e needed to c o n s t r u c t a
r e g i s t e r c a p a b l e od s t o r i n g 32 d e c i m a l . ” )
disp ( ” ” )
disp ( ” ( i i i ) ( F ) 1 6 = ( 1 1 1 1 ) 2 . The number o f b i t s
r e q u i r e d to r e p r e s e n t (F) 16 in binary are four ,
t h e r e f o r e f o u r f l i p −f l o p s a r e needed to c o n s t r u c t
a r e g i s t e r c apable o f s t o r i n g (F) 16 ”)
disp ( ” ” )
disp ( ” ( i v ) ( 1 0 ) 8 = ( 1 0 0 0 ) 2 . The number o f b i t s
r e q u i r e d to r e p r e s e n t (10) 8 in binary are four ,
t h e r e f o r e , f o u r f l i p −f l o p s a r e needed to
c o n s t r u c t a r e g i s t e r capable of s t o r i n g (10) 8 . ”)
86
Chapter 8
Counters
Scilab code Exa 8.1 count after 12 pulse
1 // Example 8 . 1
2 clc
3 disp ( ” A f t e r 12 p u l s e s ,
t h e c o u n t w i l l be ( 1 1 0 0 ) 2 , i
. e . 12 i n d e c i m a l . ” )
Scilab code Exa 8.2 count in binary
1 // Example 8 . 2
2 clc
3 a = dec2bin (144)
4 disp (a , ” d e c i m a l ( 1 4 4 ) =” )
5 disp ( ” S i n c e c o u n t e r i s a 5− b i t c o u n t e r ,
it resets
a f t e r 2ˆ5 = 32 c l o c k p u l s e s . ” )
6 disp ( ” D i v i d i n g 144 by 32 we g e t q u o t i e n t 2 and
remainder 6”)
7 disp ( ” T h e r e f o r e , c o u n t e r r e s e t s f o u r t i m e s and t h e n
i t c o u n t s r e m a i n i n g 16 c l o c k p u l s e s . Thus , t h e
c o u n t w i l l be b i n a r y ( 1 1 0 0 0 0 ) , i . e . , 16 i n
decimal ”)
87
Scilab code Exa 8.4 draw logic diagram
1 // Example 8 . 4
2 clc
3 disp ( ”When f l i p − f l o p s
are negatively
the Q output of pre vi ous stage i s
the c l o c k input of the next stage .
3− s t a g e a s y n c h r o n o u s c o u n t e r w i t h
t r i g g e r e d f l i p −f l o p s . ”)
edge t r i g g e r e d ,
connected to
F i g . 8 . 5 shows
n e g a t i v e edge
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.5 output frequency
1 // e x a m p l e 8 . 5
2 clc
3 of =50/14
4 format (5)
5 disp ( of , ” Output f r e q u e n c y = 50 kHz / 14 =” )
Scilab code Exa 8.6 maximum operating frequency
1 // Example 8 . 6
2 clc
3 disp ( ”We know t h a t MOD−32 u s e s
f i v e f l i p − f l o p s . With
t p d = 50 ns , t h e f m a x f o r r i p p l e c o u n t e r can
be g i v e n as , ” )
4 fm =(1/(250*10^ -9) ) *10^ -6
88
5
disp ( fm , ” f m a x ( r i p p l e ) = ” )
Scilab code Exa 8.8 design 4 bit up down ripple counter
1 // Example 8 . 8
2 clc
3 disp ( ” The 4− b i t
c o u n t e r n e e d s f o u r f l i p − f l o p s . The
c i r c u i t f o r 4− b i t up / down r i p p l e c o u n t e r i s
s i m i l a r t o 3− b i t up / down r i p p l e c o u n t e r e x c e p t
t h a t 4− b i t c o u n t e r h a s one more f l i p −f l o p and i t s
c l o c k d r i v i n g c i r c u i t i n g . ”)
4 disp ( ” The f i g . 8 . 1 4 shows t h e 4− b i t up / down r i p p l e
counter . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.9 design divide by 9 counter
1 // Example 8 . 9
2 clc
3 disp ( ” I n t e r n a l
s t r u c t u r e o f 7 4 9 2 i s a s shown i n f i g
. 8 . 1 6 . ”)
4 disp ( ” ” )
5 disp ( ” The c i r c u i t d i a g r a m f o r d i v i d e −by−9 c o u n t e r i s
a s shown i n f i g . 8 . 1 7 . ” )
This code can be downloaded from the website wwww.scilab.in This code
89
can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.10 design a divide by 128 counter
1 // Example 8 . 1 0
2 clc
3 disp ( ” S i n c e 128 = 16 x 8 , a d i v i d e −by −16 c o u n t e r
f o l l o w e d by a d i v i d e −by−8 c o u n t e r w i l l become a
d i v i d e −by −128 c o u n t e r . IC 7 4 9 3 i s a 4− b i t b i n a r y
c o u n t e r ( i . e . mod−16 o r d i v i d e −by −16) , t h e r e f o r e ,
two IC p a c k a g e s w i l l be r e q u i r e d . ” )
4 disp ( ” The c i r c u i t d i a g r a m i s a s shown i n f i g . 8 . 1 8 . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.11 design divide by 78 counter
1 // Example 8 . 1 1
2 clc
3 disp ( ” S i n c e 78 = 13 x 6 , we have t o u s e 7 4 9 3 a s mod−
a3 and 7 4 9 2 a s mod−6 c o u n t e r s . For t h e mod−13
c o u n t e r QD, QC and QA o u t p u t s o f 7 4 9 3 a n s ANDed
and u s e d t o c l e a r t h e c o u n t when t h e c o u n t
r e a c h e s 1 1 0 1 . For t h e mod−6 c o u n t e r , c l o c k i s
a p p l i e d to B input of 7492. ”)
4 disp ( ” The c i r c u i t d i a g r a m i s a s shown i n t h e f i g .
8.19 ”)
This code can be downloaded from the website wwww.scilab.in
90
Scilab code Exa 8.12 design divide by 6 counter
1 // Example 8 . 1 2
2 clc
3 disp ( ” The f i g . 8 . 2 0 shows d i v i d e d −by−6 (MOD 6 )
c o u n t e r u s i n g 7 4 9 3 . As shown i n t h e f i g . 8 . 2 0 , t h e
c l o c k i s a p p l i e d t o i n o u t B o f IC 7 4 9 3 and t h e
o u t p u t c o u n t s e q u e n c e i s t a k e n from QD, QC and QB.
As s o o n a s c o u n t i s 1 1 0 , i . e . QD and QC = 1 , t h e
i n t e r n a l NAND g a t e o u t p u t g o e s low and i t r e s e t s
the counter . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.13 find fmax
1 // Example 8 . 1 3
2 clc
3 disp ( ” For a s y n c h r o n o u s c o u n t e r t h e
4
5
6
7
8
9
10
t o t a l delay that
must be a l l o w e d b e t w e e n i n p u t c l o c k p u l s e s i s
e q u a l t o f l i p −f l o p t p d + AND g a t e t p d . Thus
T c l o c k >= 50 + 20 = 70 n s and s o t h e c o u n t e r h a s
”)
fm =(1/(70*10^ -9) ) *10^ -6
format (5)
disp ( fm , ” f m a x ( i n MHz) =” )
disp ( ”We know t h a t MOD−16 r i p p l e c o u n t e r u s e d f o u r
f l i p − f l o p s . With f l i p −f l o p t p d = 50 ns , t h e
f m a x f o r r i p p l e c o u n t e r can be g i v e n as , ” )
fma =(1/(4*(50*10^ -9) ) ) *10^ -6
format (3)
disp ( fma , ” f m a x ( r i p p l e ) ( i n MHz) =” )
91
Scilab code Exa 8.14 determine states
1 // Example 8 . 1 4
2 clc
3 disp ( ” IC 7 4 1 9 1
i s a 4− b i t c o u n t e r . Thus i t i s MOD−16
c o u n t e r . However , we r e q u i r e MOD−11 c o u n t e r . The
d i f f e r e n c e b e t w e e n 16 and 11 i s 5 . Hence 5 s t e p s
must be s k i p p e d from t h e f u l l modulus s e q u e n c e .
T h i s can be a c h i e v e d by p r e s e t t i n g c o u n t e r t o
v a l u e 5 . Each t i m e when c o u n t e r r e c y c l e s i t
s t a r t s c o u n t i n g from 5 u p t o 16 on e a c h f u l l c y c l e
. Therefore , each f u l l c y c l e o f the counter
c o n s i s t s o f 11 s t a t e s . ” )
Scilab code Exa 8.15 design MOD 10 counter
1 // Example 8 . 1 5
2 clc
3 disp ( ” IC 7 4 1 9 1
i s a 4− b i t c o u n t e r . Thus i t i s MOD−16
c o u n t e r . However , we r e q u i r e MOD−10 c o u n t e r . The
d i f f e r e n c e b e t w e e n 16 and 10 i s 6 . Hence 6 s t e p s
must be s k i p p e d from t h e f u l l modulus s e q u e n c e .
T h i s can be a c h i e v e d by p r e s e t t i n g c o u n t e r t o
v a l u e 6 . Each t i m e when c o u n t e r r e c y c l e s i t
s t a r t s c o u n t i n g from 6 u p t o 16 on e a c h f u l l c y c l e
. Therefore , each f u l l c y c l e o f the counter
c o n s i s t s o f 10 s t a t e s . ” )
This code can be downloaded from the website wwww.scilab.in
92
Scilab code Exa 8.16 design down counter
1 // Example 8 . 1 6
2 clc
3 disp ( ” The f i g 8 . 3 6 shows t h e c o n n e c t i o n s
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
f o r 74 LS191
t o g e t d e s i r e o p e r a t i o n . We can d e s i g n t h e
c o m b i n a t i o n a l c i r c u i t f o r s u c h c o u n t e r from t h e
t r u t h t a b l e shown b e l o w . ” )
disp ( ” ” )
disp ( ”Q3 Q2 Q1 Q0
Y” )
disp ( ” 0
0
0
0
0”)
disp ( ” 0
0
0
1
0”)
disp ( ” 0
0
1
0
0”)
disp ( ” 0
0
1
1
1”)
disp ( ” 0
1
0
0
1”)
disp ( ” 0
1
0
1
1”)
disp ( ” 0
1
1
0
1”)
disp ( ” 0
1
1
1
1”)
disp ( ” 1
0
0
0
1”)
disp ( ” 1
0
0
1
1”)
disp ( ” 1
0
1
0
1”)
disp ( ” 1
0
1
1
1”)
disp ( ” 1
1
0
0
1”)
disp ( ” 1
1
0
1
1”)
disp ( ” 1
1
1
0
0”)
disp ( ” 1
1
1
1
0”)
disp ( ” ” )
disp ( ”K=map s i m p l i f i c a t i o n ” )
disp ( ”
Q1 ’ ’ Q0 ’ ’ Q1 ’ ’ Q0 Q1Q0 Q1Q0 ’ ’ ” )
disp ( ”Q3 ’ ’ Q2 ’ ’
0
0
1
0”)
disp ( ”Q3 ’ ’ Q2
1
1
1
1”)
disp ( ”Q3Q2
1
1
0
0”)
disp ( ”Q3Q2 ’ ’
1
1
1
1”)
disp ( ” ” )
disp ( ” T h e r e f o r e , PL ’ ’ = Y = Q3 ’ ’ Q1Q0 + Q3 ’ ’ Q2 +
Q3Q1 ’ ’ + Q3Q2 ’ ’ ” )
disp ( ” A f t e r s w i t c h ON, i f t h e c o u n t e r o u t p u t i s
o t h e r t h a n 1 1 0 1 t h r o u g h 0 0 1 1 , t h e PL ’ ’ g o e s low
93
and c o u n t 1 1 0 1 i s l o a d e d i n t h e c o u n t e r . The
c o u n t e r i s t h e n d e c r e m e n t e d on t h e o c c u r r e n c e o f
c l o c k p u l s e s . When c o u n t e r r e a c h e s 0 0 1 0 , t h e PL ’ ’
a g a i n g o e s low and c o u n t 1 1 0 1 i s l o a d e d i n t h e
counter ”)
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.17 design programmable frequency divider
1 // Example 8 . 1 7
2 clc
3 disp ( ” The IC 7 4 1 9 1
i s a 4− b i t b i n a r y c o u n t e r ,
t h e r e f o r e f o u t = f CLK / 16 i n up and down
c o u n t e r mode . I f f CLK = 500 Hz and f o u t = 50 Hz
we n e e d mod 10 ( 5 0 0 / 5 0 ) c o u n t e r . The f i g . 8 . 3 9
shows t h e mod−10 c o u n t e r u s i n g IC 7 4 1 9 1 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.18 design a counter
1 // Example 8 . 1 8
2 clc
3 disp ( ” The f i g shows t h e c o n n e c t i o n s
f o r 74 LS191 t o
g e t d e s i r e o p e r a t i o n . We can d e s i g n t h e
c o m b i n a t i o n a l c i r c u i t f o r s u c h c o u n t e r from t h e
t r u t h t a b l e shown b e l o w . ” )
94
disp ( ” ” )
disp ( ”Q3 Q2 Q1 Q0
Y” )
disp ( ” 0
0
0
0
0”)
disp ( ” 0
0
0
1
0”)
disp ( ” 0
0
1
0
0”)
disp ( ” 0
0
1
1
1”)
disp ( ” 0
1
0
0
1”)
disp ( ” 0
1
0
1
1”)
disp ( ” 0
1
1
0
1”)
disp ( ” 0
1
1
1
1”)
disp ( ” 1
0
0
0
1”)
disp ( ” 1
0
0
1
1”)
disp ( ” 1
0
1
0
1”)
disp ( ” 1
0
1
1
1”)
disp ( ” 1
1
0
0
1”)
disp ( ” 1
1
0
1
1”)
disp ( ” 1
1
1
0
1”)
disp ( ” 1
1
1
1
0”)
disp ( ” ” )
disp ( ”K=map s i m p l i f i c a t i o n ” )
disp ( ”
Q1 ’ ’ Q0 ’ ’ Q1 ’ ’ Q0 Q1Q0 Q1Q0 ’ ’ ” )
disp ( ”Q3 ’ ’ Q2 ’ ’
0
0
1
0”)
disp ( ”Q3 ’ ’ Q2
1
1
1
1”)
disp ( ”Q3Q2
1
1
0
1”)
disp ( ”Q3Q2 ’ ’
1
1
1
1”)
disp ( ” ” )
disp ( ” T h e r e f o r e , PL ’ ’ = Y = Q3 ’ ’ Q1Q0 + Q3 ’ ’ Q2 +
Q3Q1 ’ ’ + Q3Q2 ’ ’ + Q2Q1Q0 ’ ’ ” )
31 disp ( ” A f t e r s w i t c h ON, i f t h e c o u n t e r o u t p u t i s
o t h e r t h a n 1 1 1 0 t h r o u g h 0 0 1 1 , t h e PL ’ ’ g o e s low
and c o u n t 1 1 1 0 i s l o a d e d i n t h e c o u n t e r . The
c o u n t e r i s t h e n d e c r e m e n t e d on t h e o c c u r r e n c e o f
c l o c k p u l s e s . When c o u n t e r r e a c h e s 0 0 1 0 , t h e PL ’ ’
a g a i n g o e s low and c o u n t 1 1 1 0 i s l o a d e d i n t h e
counter ”)
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
95
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.19 programmable frequency divider
1 // Example 8 . 1 9
2 clc
3 disp ( ” IC 7 4 1 9 1
i s a 4− b i t b i n a r y c o u n t e r . Thus i t
d i v i d e s t h e i n p u t f r e q u e n c y by 1 6 . However , we
can d e s i g n MOD−N c o u n t e r u s i n g IC 7 4 1 9 1 . For MOD−
N c o u n t e r t h e o u t p u t f r e q u e n c y w i l l be f o u t =
f i n / N . Thus by c h a n g i n g N we can c h a n g e t h e
o u t p u t f r e q u e n c y . The f i g . 8 . 4 0 shows t h e
programmable f r e q u n c y d i v i d e r u s i n g IC 7 4 1 9 1 . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.20 design divide by 2 and divide by 5 counter
1 // Example 8 . 2 0
2 clc
3 disp ( ” F i g . 8 . 4 1 shows D i v i d i n g −by−2 f o r up c o u n t i n g ”
)
4 disp ( ” D i v i d e −by−2 i s a mod−2 c o u n t e r . S i n c e , a f t e r
p r e s e t above c o u n t e r g o e s through 2 s t a t e s 1110
and 1 1 1 1 , i t i s a mod−2 c o u n t e r . Thus , a b o v e
c i r c u i t i s a d i v i d e −by−2 c o u n t e r f o r up c o u n t i n g
mode . ” )
96
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
disp ( ” ” )
disp ( ” D i v i d e −by−5 f o r down c o u n t i n g mode : ” )
disp ( ” ” )
disp ( ”Q3 Q2 Q1 Q0
Y” )
disp ( ” 0
0
0
0
0”)
disp ( ” 0
0
0
1
0”)
disp ( ” 0
0
1
0
0”)
disp ( ” 0
0
1
1
0”)
disp ( ” 0
1
0
0
0”)
disp ( ” 0
1
0
1
0”)
disp ( ” 0
1
1
0
0”)
disp ( ” 0
1
1
1
0”)
disp ( ” 1
0
0
0
0”)
disp ( ” 1
0
0
1
0”)
disp ( ” 1
0
1
0
0”)
disp ( ” 1
0
1
1
1”)
disp ( ” 1
1
0
0
1”)
disp ( ” 1
1
0
1
1”)
disp ( ” 1
1
1
0
1”)
disp ( ” 1
1
1
1
1”)
disp ( ” ” )
disp ( ”K=map s i m p l i f i c a t i o n ” )
disp ( ”
Q1 ’ ’ Q0 ’ ’ Q1 ’ ’ Q0 Q1Q0 Q1Q0 ’ ’ ” )
disp ( ”Q3 ’ ’ Q2 ’ ’
0
0
0
0”)
disp ( ”Q3 ’ ’ Q2
0
0
0
0”)
disp ( ”Q3Q2
1
1
1
1”)
disp ( ”Q3Q2 ’ ’
0
0
1
0”)
disp ( ” ” )
disp ( ” T h e r e f o r e , Y = Q3Q2 + Q3Q1Q0” )
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
97
Scilab code Exa 8.21 design mod 9 counter
1 // Example 8 . 2 1
2 clc
3 disp ( ” IC 7 4 1 9 1
i s a 4− b i t c o u n t e r . Thus i t i s MOD−16
c o u n t e r . However , we r e q u i r e MOD−9 c o u n t e r . The
d i f f e r e n c e b e t w e e n 16 and 9 i s 7 . Hence 7 s t e p s
must be s k i p p e d from t h e f u l l modulus s e q u e n c e .
T h i s can be a c h i e v e d by p r e s e t t i n g c o u n t e r t o
v a l u e 7 . Each t i m e when c o u n t e r r e c y c l e s i t
s t a r t s c o u n t i n g from 7 u p t o 16 on e a c h f u l l c y c l e
. Therefore , each f u l l c y c l e o f the counter
c o n s i s t s of 9 s t a t e s . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.22 determine MOD number and counter range
1 // Example 8 . 2 2
2 clc
3 disp ( ” C l o c k f r e q u e n c y = 256 kHz ” )
4 disp ( ” Output f r e q u e n c y = 2 kHz ” )
5 format (4)
6 mn =256/2
7 disp ( mn , ” T h e r e f o r e ,
Mod number = n =” )
8 disp ( ” T h e r e f o r e ,
C o u n t e r i s MOD−128 c o u n t e r ” )
9 disp ( ”Mod−128 c o u n t e r can c o u n t t h e numbers from 0
to 127. ”)
Scilab code Exa 8.23 design a divide by 20 counter
1
// Example 8 . 2 3
98
2 clc
3 disp ( ” I n t e r n a l
4
5
6
7
8
9
10
s t r u c t u r e o f 7 4 9 0 r i p p l e c o u n t e r IC
i s a s shown i n f i g . 8 . 5 0 ” )
disp ( ” ” )
disp ( ”We know t h a t , one IC can work a s mod−10 (BCD)
c o u n t e r . T h e r e f o r e , we n e e d two I C s . The c o u n t e r
w i l l go t h r o u g h s t a t e s 0−19 and s h o u l d be r e s e t
on s t a t e 2 0 . i . e . ” )
disp ( ”
QD QC QB QA
QD QC QB QA” )
disp ( ”
0
0
1
0
0
0
0
0”)
disp ( ”
7490(2)
7490(1) ”)
disp ( ” ” )
disp ( ” The d i a g r a m o f d i v i d e −by −20 c o u n t e r u s i n g IC
7 4 9 0 i s a s shown i n f i g . 8 . 5 1 ” )
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.24 design divide by 96 counter
1 // Example 8 . 2 4
2 clc
3 disp ( ” IC 7 4 9 0 i s a d e c a d e c o u n t e r . When two s u c h I C s
a r e c a s c a d e d , i t becomes a d i v i d e −by −100 c o u n t e r
. To g e t a d i v i d e −by −96 c o u n t e r , t h e c o u n t e r i s
r e s e t a s s o o n a s i t becomes 1 0 0 1 0 1 1 0 . The
d i a g r a m i s shown i n f i g . 8 . 5 2 . ” )
This code can be downloaded from the website wwww.scilab.in
99
Scilab code Exa 8.25 design divide by 93 counter
1 // Example 8 . 2 5 .
2 clc
3 disp ( ” IC 7 4 9 0 i s a d e c a d e c o u n t e r . Whentwo s u c h I C s
a r e c a s c a d e d , i t becomes a d i v i d e −by −100 c o u n t e r .
To g e t a d i v i d e −by −93 c o u n t e r , t h e c o u n t e r i s
r e s e t a s s o o n a s o t becomes 1 0 0 1 0 0 1 1 . The
d i a g r a m i s a s shown i n f i g . 8 . 5 3 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.26 design divide by 78 counter
1 // Example 8 . 2 6
2 clc
3 disp ( ” IC 7 4 9 0 i s a d e c a d e c o u n t e r . When two s u c h I C s
a r e c a s c a d e d , i t becomes a d i v i d e −by −100 c o u n t e r
. To g e t a d i v i d e by 78 o r MOD−78 c o u n t e r , t h e
c o u n t e r i s r e s e t a s s o o n a s o t becomes 0 1 1 1 1 0 0 0
a s shown i n f i g . 8 . 5 4 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.28 7490 IC
1 // Example 8 . 2 8
2 clc
3 disp ( ” I f t h e QD o u t p u t
i s connected to A input of
7 4 9 0 IC and , i n p u t c l o c k i s a p p l i e d t o B i n p u t
100
4
5
6
7
8
9
10
11
12
13
14
15
d i v i d e by t e n s q u a r e wave i s o b t a i n e d a t o u t p u t
QA. ” )
disp ( ” C l o c k
Outputs ” )
disp ( ”
QA QD QC QB” )
disp ( ” 0
L
L
L
L” )
disp ( ” 1
L
L
L
H” )
disp ( ” 2
L
L
H
L” )
disp ( ” 3
L
L
H
H” )
disp ( ” 4
L
H
L
L” )
disp ( ” 5
H
L
L
L” )
disp ( ” 6
H
L
L
H” )
disp ( ” 7
H
L
H
L” )
disp ( ” 8
H
L
H
H” )
disp ( ” 9
H
H
L
L” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.30 sketch output waveforms of counter
1 // e x a m p l e 1 0 . 9
2
3 clc ;
4 clear ;
5 close ;
6 c = [0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
7
8
9
10
11
0];
q = [0
0];
a = [0
0];
b = [0
0];
y1 = q ;
y2 = a ;
// t a k i n g t h e v a l u e s f o r a mod −6 c o u n t e r
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0
0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1
101
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
41
42
43
44
45
46
47
48
49
y3 = b ;
y11p =1;
y22p =1;
y33p =1;
y44p =1;
cp =1;
yf1p =1;
for i =1:25
// making a r r a y s t o draw t h e o u t p u t
if y1 ( i ) ==1 then
for o =1:100
y11 ( y11p ) =1;
y11p = y11p +1;
end
else
for o =1:100
y11 ( y11p ) =0;
y11p = y11p +1;
end
end
if y2 ( i ) ==1 then
for o =1:100
y21 ( y22p ) =1;
y22p = y22p +1;
end
else
for o =1:100
y21 ( y22p ) =0;
y22p = y22p +1;
end
end
if y3 ( i ) ==1 then
for o =1:100
y31 ( y33p ) =1;
y33p = y33p +1;
end
else
102
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
for o =1:100
y31 ( y33p ) =0;
y33p = y33p +1;
end
end
if c ( i ) ==1 then
for o =1:100
c1 ( cp ) =1;
cp = cp +1;
end
else
for o =1:100
c1 ( cp ) =0;
cp = cp +1;
end
end
end
z =[2 2];
subplot (4 ,1 ,1) ; // p l o t i n g t h e o u t put
title ( ’ Timing Diagram ’ ) ;
plot ( c1 ) ;
plot ( z ) ;
ylabel ( ’QA ’ ) ;
subplot (4 ,1 ,2) ;
plot ( y11 ) ;
ylabel ( ’QB ’ ) ;
plot ( z ) ;
subplot (4 ,1 ,3) ;
plot ( y21 ) ;
ylabel ( ’QC ’ ) ;
plot ( z ) ;
subplot (4 ,1 ,4) ;
plot ( z ) ;
ylabel ( ’QD ’ ) ;
plot ( y31 ) ;
disp ( ” The c o u n t e r g o e s t h r o u g h s t a t e s 0 0 0 0 ( D e c i m a l
103
Figure 8.1: sketch output waveforms of counter
0 ) t o 1 0 1 1 ( D e c i m a l 1 1 ) , i . e . , t h r o u g h 12 s t a t e s .
Thus i t i s a MOD−12 c o u n t e r . ” )
Scilab code Exa 8.31 explain operation of circuit
1 // Example 8 . 3 1
2 clc
3 disp ( ” The f i g . 8 . 6 3 shows t h e c a s c a d e d c o n n e c t i o n o f
4− b i t b i n a r y c o u n t e r s . L e t u s s e e t h e c i r c u i t
o p e r a t i o n . The c o u n t e r IC1 o p e r a t e s a s a c o u n t e r
f o r c o u n t i o n i n t h e UP d i r e c t i o n s i n c e CLEAR =
104
LOAD = 1 . When t h e c o u n t r e a c h e s t h e maximum
v a l u e ( 1 1 1 1 ) i t s RC ( R i p p l e C a r r y Output ) g o e s
HIGH which makes P and T ( E n a b l e ) i n p u t s o f IC2
HIGH f o r one c l o c k c y c l e a d v a n c i n g i t s o u t p u t by
1 and making Q o u t p u t s o f IC1 , 0 a t t h e n e x t
clock cycle . After this clock cycle P = T = 0 for
IC2 and IC1 w i l l go on c o u n t i n g t h e p u l s e s . When
t h e o u t p u t s o f IC1 and IC2 b o t h r e a c h t h e
maximum count , RC o u t p u t s o f b o t h o f t h e s e I C s
w i l l go HIGH . T h i s w i l l make P = T o f IC3 HIGH
and t h e r e f o r e , i n t h e n e x t c l o c k c y c l e IC3 c o u n t
w i l l be i n c r e m e n t e d and s i m u l t a n e o u s l y IC1 and
IC2 w i l l be c l e a r e d . T h i s way t h e c o u n t i n g w i l l
continue . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.32 design divide by 40000 counter
1 // Example 8 . 3 2
2 clc
3 disp ( ” C a s c a d i n g f o u r 7 4 1 6 1 ( e a c h 4− b i t ) c o u n t e r s we
g e t 16 ( 4 x 4 ) b i t c o u n t e r a s shown i n f i g 8 . 6 3 . ”
)
4 disp ( ” T h e r e f o r e , we g e t 2 ˆ 1 6 = 6 5 , 5 3 6 modulus
counter ”)
5 disp ( ” However , we r e q u i r e d i v i d e −by −40 ,000 c o u n t e r .
The d i f f e r e n c e b e t w e e n 6 5 , 5 3 6 and 4 0 , 0 0 0 i s
2 5 , 5 3 6 , which i s t h e number o f s t a t e s t h o s e must
be s k i p p e d from t h e f u l l modulus s e q u e n c e . T h i s
can be a c h i e v e d by p r e s e t t i n g t h e c o u n t i n g from
2 5 , 5 3 6 u p t o 6 5 , 5 3 6 on e a c h f u l c y c l e . T h e r e f o r e ,
each f u l l c y c l e o f the counter c o n s i s t s o f 40 ,000
s t a t e s . ”)
105
Scilab code Exa 8.33 design modulo 11 counter
1 // Example 8 . 3 3
2 clc
3 disp ( ” A l t h o u g h t h e 74 X163 i s a modulo −16 c o u n t e r ,
it
can be made t o c o u n t i n a modulus l e s s t h a n 16
by u s i n g t h e CLR’ ’ o r LD ’ ’ i n p u t t o s h o r t e n t h e
n o r m a l c o u n t i n g s e q u e n c e . The f i g . 8 . 6 9 shows
c i r c u i t c o n n e c t i o n s f o r modulo −11 c o u n t e r . Here ,
l o a d i n p u t i s a c t i v a t e d upon a c t i v a t i o n o f RCO (
r i p p l e −c a r r y −o u t p u t ) . S i n c e l o a d i n p u t i s
a d j u s t e d t o s t a t e 5 , c o u n t e r c o u n t s from 5 t o 15
and t h e n s t a r t s a t 5 a g a i n , f o r a t o t a l o f 11
s t a t e s per counting c y c l e . ”)
4 disp ( ” ” )
5 disp ( ”We can a l s o d e s i g n modulo −11 c o u n t e r u s i n g CLR
’ ’ i n p u t a s shown i n f i g . 8 . 7 0 . h e r e , NAND g a t e i s
u s e d t o d e t e c t s t a t e 10 and f o r c e t h e n e x t s t a t e
t o 0 . A 2− i n p u t g a t e i s u s e d t o d e t e c t s t a t e 10
( b i n a r y 1 0 1 0 ) by c o n n e c t i n g Q1 and Q3 t o t h e
i n p u t s o f t h e NAND g a t e . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.34 design excess 3 decimal counter
1 // Example 8 . 3 4
2 clc
3 disp ( ”An e x c e s s −3 d e c i m a l c o u n t e r s h o u l d
start
c o u n t i n g from c o u n t 3 ( b i n a r y 0 0 1 1 ) and c o u n t
106
u p t o c o u n t 12 ( b i n a r y 1 1 0 0 ) . S t a r t i n g c o u n t i s
a d j u s t e d by l o a d i n g 0 0 1 1 a t l o a d i n p u t s . To
r e c y c l e c o u n t from 1 1 0 0 t o 0 0 1 1 , Q3 and Q2 o u t p u t
a r e c o n n e c t e d a s i n p u t s f o r 2− i n p u t NAND g a t e .
Thus , NAND g a t e d e t e c t s s t a t e 1 1 0 0 and f o r c e s
0 0 1 1 t o be l o a d e d a s t h e n e x t s t a t e . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.35 modulus greater than 16
1 // Example 8 . 3 5
2 clc
3 disp ( ”A b i n a r y c o u n t e r w i t h a modulus g r e a t e r t h a n
16 can be b u i l t by c a s c a d i n g 74 X163s . When
c o u n t e r s a r e c a s c a d e d , CLK, CLR’ ’ and LD ’ ’ o f a l l
t h e 74 X163s a r e c o n n e c t e d i n p a r a l l e l , s o t h a t
a l l o f them c o u n t o r a r e c l e a r e d o r l o a d e d a t t h e
same t i m e . The RCO s i g n a l d r i v e s t h e ENT i n p u t
o f t h e n e x t c o u n t e r . The f i g . 8 . 7 3 shows modulo −60
c o u n t e r . To have a modulo 60 c o u n t we n e e d a t
l e a s t 6− b i t c o u n t e r , t h u s two 74 X163s a r e
c a s c a d e d . C o u n t e r i s d e s i g n e d t o c o u n t from 196
t o 2 5 5 . The MAXCNT s i g n a l d e t e c t s t h e s t a t e 255
and s t o p s t h e c o u n t e r u t i l GO’ ’ i s a s s e r t e d . When
GO’ ’ i s a s s e r t e d t h e c o u n t e r i s r e l o a d e d w i t h
196 ( b i n a r y 1 1 0 0 0 1 0 0 ) and c o u n t s u p t o 2 5 5 . To
e n a b l e c o u n t i n g , CNTEN i s c o n n e c t e d t o t h e ENP
i n p u t s i n p a r a l l e l . A NAND g a t e a s s e t s RELOAD’ ’
t o g e t back t o s t a t e 196 o n l y i f GO’ ’ i s a s s e r t e d
and t h e c o u n t e r i s i n s t a t e 2 5 5 . ” )
107
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.36 modulo 8 counter
1 // Example 8 . 3 6
2 clc
3 disp ( ”A b i n a r y c o u n t e r may be combined w i t h a
d e c o d e r t o o b t a i n a s e t o f 1−out−o f −M c o d e d
s i g n a l s , where one s i g n a l i s a s s e r t e d i n e a c h
c o u n t s t a t e . T h i s i s u s e f u l when c o u n t e r s a r e
u s e d t o c o n t r o l a s e t o f d e v i c e s , where a
d i f f e r e n t d e v i c e s i s enabled in each counter
s t a t e . ”)
4 disp ( ” The f i g . 8 . 7 4 shows how a 74 X163 c o n n e c t e d a s a
modulo −8 c o u n t e r can be combined w i t h a 74 X138
3−8 d e c o d e r t o p r o v i d e e i g h t s i g n a l s , e a c h one
r e p r e s e n t i n g a counter s t a t e . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.37 synchronous decade counter
1 // Example 8 . 3 7
2 clc
3 disp ( ” E x c i t a t i o n t a b l e ” )
4 disp ( ” P r e s e n t S t a t e
Next S t a t e
F l i p −f l o p I n p u t s ” )
5 disp ( ”QD QC QB QA
Q D+1 Q C+1 Q B+1
JK D JK C JK B JK A” )
108
Q A+1
6
disp ( ” 0
7
disp ( ” 0
8
disp ( ” 0
0
0
0
0
0
0
9
disp ( ” 0
0
0
10
disp ( ” 0
1
0
11
disp ( ” 0
12
disp ( ” 0
13
disp ( ” 0
1
0
1
0
1
1
14
disp ( ” 1
0
0
15
disp ( ” 1
16
disp ( ” 1
17
disp ( ” 1
0
1
0
X
0
X
18
disp ( ” 1
1
X
19
disp ( ” 1
1
X
20
disp ( ” 1
21
disp ( ” 1
1
X
1
X
22
23
24
25
26
27
0
0
0
0
1
0
1
1
0
0
0
0
1
0
1
1
0
0
0
0
1
X
1
X
0
X
0
X
1
X
1
X
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
0
0
X
1
X
0
X
1
X
0
X
1
X
0
1”)
0
1”)
0
1”)
0
1”)
0
1”)
0
1”)
0
1”)
1
1”)
1
1”)
0
1”)
X
1”)
X
1”)
X
X” )
X
X” )
X
X” )
X
X” )
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
0
0
0
0
0
1
0
0
0
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
disp ( ” ” )
disp ( ”K−map S i m p l i f i c a t i o n ” )
disp ( ”
For JK D” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA
disp ( ”QD’ ’ QC’ ’
0
0
0
disp ( ”QD’ ’QC
0
0
1
109
QBQA’ ’ ” )
0”)
0”)
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
0
1
X
X” )
disp ( ”JK D = QA QD + QA QB QC” )
disp ( ” ” )
disp ( ”
For JK C” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
0
0
1
0”)
disp ( ”QD’ ’QC
0
0
1
0”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
0
0
X
X” )
disp ( ”JK C = QA QB” )
disp ( ” ” )
disp ( ”
For JK B” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
0
1
1
0”)
disp ( ”QD’ ’QC
0
1
1
0”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
0
0
X
X” )
disp ( ”JK B = QA QD’ ’ ” )
disp ( ” ” )
disp ( ”
For JK A” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
1
1
1
1”)
disp ( ”QD’ ’QC
1
1
1
1”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
1
1
X
X” )
disp ( ”JK A = 1 ” )
disp ( ” ” )
disp ( ” F i g shows t h e l o g i c d i a g r a m f o r t h e
s y n c h r o n o u s d e c a d e c o u n t e r u s i n g JK f l i p −f l o p ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.38 flip flops
110
1 // Example 8 . 3 8
2 clc
3 disp ( ” E x c i t a t i o n
4 disp ( ” I n p u t
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
t a b l e ”)
Present State
F l i p −f l o p I n p u t s ” )
disp ( ”UP/DOWN’ ’
QC QB QA
JK C JK B JK A” )
Q A+1
disp ( ” UD” )
disp ( ” 0
0
0
0
1
1
1
1”)
disp ( ” 0
0
0
1
0
0
0
1”)
disp ( ” 0
0
1
0
1
0
1
1”)
disp ( ” 0
0
1
1
0
0
0
1”)
disp ( ” 0
1
0
0
1
1
1
1”)
disp ( ” 0
1
0
1
0
0
0
1”)
disp ( ” 0
1
1
0
1
0
1
1”)
disp ( ” 0
1
1
1
0
0
0
1”)
disp ( ” 1
0
0
0
1
0
0
1”)
disp ( ” 1
0
0
1
0
0
1
1”)
disp ( ” 1
0
1
0
1
0
0
1”)
disp ( ” 1
0
1
1
0
1
1
1”)
disp ( ” 1
1
0
0
1
0
0
1”)
disp ( ” 1
1
0
1
0
0
1
1”)
disp ( ” 1
1
1
0
1
0
0
1”)
111
Next S t a t e
Q C+1
Q B+1
1
1
0
0
0
0
0
1
0
1
1
0
1
0
1
1
0
0
0
1
0
1
1
0
1
0
1
1
1
1
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
disp ( ”
1
1
1
1
0
0
0
1
1
1”)
disp ( ” ” )
disp ( ”K−map S i m p l i f i c a t i o n ” )
disp ( ”
For JK C” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
1
0
0
0”)
disp ( ”QD’ ’QC
1
0
0
0”)
disp ( ”QDQC
0
0
1
0”)
disp ( ”QDQC’ ’
0
0
1
0”)
disp ( ”JK C =UD’ ’ QB’ ’ QB’ ’ + UD QB QA” )
disp ( ” ” )
disp ( ”
For JK B” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
1
0
0
1”)
disp ( ”QD’ ’QC
1
0
0
1”)
disp ( ”QDQC
0
1
1
0”)
disp ( ”QDQC’ ’
0
1
1
0”)
disp ( ”TB =UD’ ’ QA’ ’ + UD QA” )
disp ( ” ” )
disp ( ”
For JK A” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
1
1
1
1”)
disp ( ”QD’ ’QC
1
1
1
1”)
disp ( ”QDQC
1
1
1
1”)
disp ( ”QDQC’ ’
1
1
1
1”)
disp ( ”TA = 1 ” )
disp ( ” ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.39 design mod 5 synchronous counter
1
// Example 8 . 3 9
112
2 clc
3 disp ( ” For mod−5 c o u n t e r we r e q u i r e 3 f l i p − f l o p s . ” )
4 disp ( ” E x c i t a t i o n t a b l e ” )
5 disp ( ”
Present State
Next S t a t e
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
F l i p −f l o p I n p u t s ” )
QC QB QA
TA
T B
T C” )
0
0
0
0
0
1”)
0
0
1
0
1
1”)
0
1
0
0
0
1”)
0
1
1
1
1
1”)
1
0
0
1
0
0”)
disp ( ”
Q C+1
disp ( ” 0
1
disp ( ” 1
0
disp ( ” 2
1
disp ( ” 3
0
disp ( ” 4
0
disp ( ” ” )
disp ( ”K−map
disp ( ”
disp ( ”QA’ ’
disp ( ”QA
disp ( ”T A =
disp ( ” ” )
disp ( ”
disp ( ”QA’ ’
disp ( ”QA
disp ( ”T B =
disp ( ” ” )
disp ( ”
disp ( ”QA’ ’
disp ( ”QA
disp ( ”T C =
disp ( ” ” )
S i m p l i f i c a t i o n ”)
QB’ ’ QC’ ’ QB’ ’QC
0
0
1
X
QA + QB QC” )
Q A+1
Q B+1
0
0
0
1
0
1
1
0
0
0
QBQC
1
X
QBQC’ ’ ” )
0”)
X” )
QB’ ’ QC’ ’
0
0
QC” )
QB’ ’QC
1
X
QBQC
1
X
QBQC’ ’ ” )
0”)
X” )
QB’ ’ QC’ ’
1
0
QA’ ’ ” )
QB’ ’QC
1
X
QBQC
1
X
QBQC’ ’ ” )
1”)
X” )
113
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.40 design MOD 4 down counter
1 // Example 8 . 4 0
2 clc
3 disp ( ” E x c i t a t i o n t a b l e ” )
4 disp ( ” P r e s e n t S t a t e
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Next S t a t e
f l o p Inputs ”)
disp ( ”
QC QB
A+
B+
K B” )
J B
disp ( ”
0
0
1
1
1
X” )
disp ( ”
0
1
0
0
X
1”)
disp ( ”
1
0
0
1
1
X” )
disp ( ”
1
1
1
0
X
1”)
disp ( ” ” )
disp ( ”K−map S i m p l i f i c a t i o n ” )
disp ( ” For J A ” )
disp ( ”
B’ ’
B” )
disp ( ”A’ ’
1
0”)
disp ( ”A
X
X” )
disp ( ” J A = B ’ ’ ” )
disp ( ” ” )
disp ( ” For K A” )
disp ( ”
B’ ’
B” )
disp ( ”A’ ’
X
X” )
disp ( ”A
1
0”)
disp ( ”K A = B ’ ’ ” )
disp ( ” ” )
114
Flip −
J A
KA
1
X
0
X
X
1
X
0
24
25
26
27
28
29
30
31
32
33
34
35
disp ( ” For J B ” )
disp ( ”
B’ ’
B” )
disp ( ”A’ ’
1
X” )
disp ( ”A
1
X” )
disp ( ” J B = 1 ” )
disp ( ” ” )
disp ( ” For K B” )
disp ( ”
B’ ’
B” )
disp ( ”A’ ’
X
1”)
disp ( ”A
X
1”)
disp ( ”K B = 1 ” )
disp ( ” ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.41 design MOD 12 synchronous counter
1 // Example 8 . 4 1
2 clc
3 disp ( ”Mod−12 s y n c h r o n o u s c o u n t e r u s i n g D f l i p −f l o p
4
5
6
7
8
9
10
11
12
13
14
15
16
:
”)
disp ( ” L e t
Number o f f l i p −f l o p r e q u i r e d = n” )
disp ( ”
2ˆ n >= 12 ” )
disp ( ”
n = 4”)
disp ( ” E x c i t a t i o n t a b l e ” )
disp ( ” P r e s e n t S t a t e
Next S t a t e
”)
disp ( ”QD QC QB QA
Q D+1 Q C+1 Q B+1 Q A+1” )
disp ( ” 0
0
0
0
0
0
0
1”)
disp ( ” 0
0
0
1
0
0
1
0”)
disp ( ” 0
0
1
0
0
0
1
1”)
disp ( ” 0
0
1
1
0
1
0
0”)
disp ( ” 0
1
0
0
0
1
0
1”)
disp ( ” 0
1
0
1
0
1
1
0”)
disp ( ” 0
1
1
0
0
1
1
1”)
115
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
42
43
44
45
46
47
48
49
50
51
52
53
54
disp ( ” 0
1
1
1
1
0
0
0”)
disp ( ” 1
0
0
0
1
0
0
1”)
disp ( ” 1
0
0
1
1
0
1
0”)
disp ( ” 1
0
1
0
1
0
1
1”)
disp ( ” 1
0
1
1
0
0
0
0”)
disp ( ” ” )
disp ( ”K−map S i m p l i f i c a t i o n ” )
disp ( ”
For D A” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
1
0
0
1”)
disp ( ”QD’ ’QC
1
0
0
1”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
1
0
0
1”)
disp ( ”D A = QA’ ’ ” )
disp ( ” ” )
disp ( ”
For D B” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
0
1
0
1”)
disp ( ”QD’ ’QC
0
1
0
1”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
0
1
0
1”)
disp ( ”D B = QB’ ’ QaA + QA’ ’ QB” )
disp ( ”
= QA XOR QB” )
disp ( ” ” )
disp ( ”
For D C” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
0
0
1
0”)
disp ( ”QD’ ’QC
1
1
0
1”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
0
0
0
0”)
disp ( ”D C = QC QB’ ’ + QC QA’ ’ + QD’ ’ QC’ ’ QB QA” )
disp ( ” ” )
disp ( ”
For D D” )
disp ( ”
QB’ ’ QA’ ’ QB’ ’QA QBQA QBQA’ ’ ” )
disp ( ”QD’ ’ QC’ ’
0
0
0
0”)
disp ( ”QD’ ’QC
0
0
1
0”)
disp ( ”QDQC
X
X
X
X” )
disp ( ”QDQC’ ’
1
1
0
1”)
116
55
disp ( ”D D = QD QB’ ’ + QC QB QA + QD QA’ ’ ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.42 design 4 bit 4 state ring counter
1 // Example 8 . 4 2
2 clc
3 disp ( ” The f i g . 8 . 9 9 shows t h e
c i r c u i t diagram f o r a
4− b i t , 4− s t a t e r i n g c o u n t e r w i t h a s i n g l e
c i r c u l a t i n g 1 . Here , 74 X194 u n i v e r s a l s h i f t
r e g i s t e r i s connected so that i t normally
p r e f o r m s a l e f t − s h i f t . However , when RESET i s
a s s e r t e d i t l o a d s 0 0 0 1 . Once RESET i s n e g a t e d ,
t h e 7 4 1 9 4 s h i f t s l e f t on e a c h c l o c k p u l s e . The
D SL s e r i a l i n p u t i s c o n n e c t e d t o t h e l e f t m o s t
o u t p u t ( Q3 : MSB) , s o t h e n e x t s t a t e s a r e 0 0 1 0 ,
0 1 0 0 , 1 0 0 0 , 0 0 0 1 , 0 0 1 0 , . . . . . Thus t h e c o u n t e r
counter v i s i t s four unique s t a t e s b e f o r e
r e p e a t i n g . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.44 design 4 bit 8 state johnson counter
1 // Example 8 . 4 4
2 clc
3 disp ( ” J o h n s o n c o u n t e r
is basically a twisted ring
c o u n t e r . The f i g . 8 . 1 0 4 ( a ) shows t h e b a s i c c i r c u i t
f o r a J o h n s o n c o u n t e r . The t a b l e shows t h e
s t a t e s o f a 4− b i t J o h n s o n c o u n t e r . ” )
117
4
5
6
7
8
9
10
11
12
13
14
15
16
disp ( ” ” )
disp ( ” S t a t e s o f 4− b i t J o h n s o n c o u n t e r ” )
disp ( ” S t a t e name
Q3 Q2 Q1 Q0” )
disp ( ”
S1
0
0
0
0”)
disp ( ”
S2
0
0
0
1”)
disp ( ”
S3
0
0
1
1”)
disp ( ”
S4
0
1
1
1”)
disp ( ”
S5
1
1
1
1”)
disp ( ”
S6
1
1
1
0”)
disp ( ”
S7
1
1
0
0”)
disp ( ”
S8
1
0
0
0”)
disp ( ” ” )
disp ( ” T h i s c o u n t e r can be m o d i f i e d t o have s e l f
c o r r e c t i n g J o h n s o n c o u n t e r a s shown i n f i g . 8 . 1 0 4 (
c ) . Here , t h e c o n n e c t i o n s a r e made s u c h t h a t
c i r c u i t oads 0001 as the next s t a t e whenever the
c u r r e n t s t a t e i s 0XX0 . ” )
This code can be downloaded from the website wwww.scilab.in This code
can be downloaded from the website wwww.scilab.in
Scilab code Exa 8.45 johnson counter
1 // Example 8 . 4 5
2 clc
3 disp ( ” J o h n s o n c o u n t e r
w i l l p r o d u c e a modulus o f 2 xn
where n i s t h e number o f s t a g e s ( i . e . f l i p − f l o p s )
i n t h e c o u n t e r . T h e r e f o r e , Mod 10 r e q u i r e s 5
f l i p − f l o p s and Mod 16 r e q u i r e s 8 f l i p − f l o p s . ” )
118
Chapter 9
Op amp Applications
Scilab code Exa 9.1 output voltage
1 // Example 9 . 1
2 clc
3 disp ( ” The d i f f e r e n t i a l
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
a m p l i f i e r i s r e p r e s e n t e d as
shown i n f i g . 9 . 5 . ” )
disp ( ” ( i ) CMRR = 100 ” )
vd =300 -240
disp ( vd , ”
Vd ( i n uV ) = V1 − V2 = ” )
vc =(300+240) /2
disp ( vc , ”
Vc ( i n uV ) = V1+V2 / 2 =” )
disp ( ”CMRR = Ad / Ac” )
ac =5000/100
disp ( ac , ” T h e r e f o r e , Ac =” )
format (6)
vo =((5000*60) +(50*270) ) *10^ -3
disp ( vo , ” T h e r e f o r e , Vo ( i n mV) = Ad∗Vd + Ac∗Vc =” )
disp ( ” ( i i ) CMRR = 1 0 ˆ 5 ” )
ac =5000/(10^5)
disp ( ac , ” T h e r e f o r e , Ac = Ad / CMRR =” )
vo =((5000*60) +(0.05*270) ) *10^ -3
format (9)
disp ( vo , ” T h e r e f o r e , Vo ( i n mV) = Ad∗Vd + Ac∗Vc =” )
119
21
disp ( ” I d e a l l y Ac must be z e r o and o u t p u t s h o u l d be
o n l y Ad∗Vd which i s 5000∗60∗10ˆ −6 i . e . 300 mV. I t
can be s e e n t h a t h i g h e r t h e v a l u e o f CMRR, t h e
output i s almost p r o p o r t i o n a l to the d i f f e r e n c e
v o l t a g e Vd , r e j e c t i n g t h e common mode s i g n a l . So
i d e a l v a l u e o f CMRR f o r a d i f f e r e n t i a l a m p l i f i e r
i s i n f i n i t y . ”)
Scilab code Exa 9.3 input bias current
1 // Example 9 . 3
2 clc
I b = 60 nA” )
3 disp ( ” I i O S = 20 nA ,
4 disp ( ”Now I i O S = I b 1 − I b 2 = 20 ” )
5 disp ( ”
I b = I b 1+I b 2 / 2 = 60 ” )
6 disp ( ” T h e r e f o r e ,
I b 1 + I b 2 = 120 ” )
7 disp ( ” T h e r e f o r e ,
2∗ I b 1 = 140 ” )
I b 2 = 50 nA” )
8 disp ( ” T h e r e f o r e ,
I b 1 = 70 nA ,
Scilab code Exa 9.4 design inverting schmitt trigger
1
2
3
4
5
6
7
8
9
10
11
12
// Example 9 . 4
clc
disp ( ”V UT = +4 V, V LT = −4 V,
S u p p l y = +− 15 V” )
disp ( ”+− V s a t = 0 . 9 x [ S u p p l y ] = +− 1 3 . 5 V = Vo” )
disp ( ” For op−amp 7 4 1 , I B ( max ) = 500 nA” )
disp ( ” T h e r e f o r e ,
I 2 = 100 I B ( max ) = 50 uA” )
r2 =(4/(50*10^ -6) ) *10^ -3
disp ( r2 , ” T h e r e f o r e , R2 ( i n k−ohm ) = V UT / I 2 =” )
i2 =(4/(82*10^3) ) *10^6
format (6)
disp ( i2 , ” R e c a l c u l a t i n g I2 ,
I 2 = V UT / R2 =” )
r1 =((13.5 -4) /(48.78*10^ -6) ) *10^ -3
120
13
14
format (7)
disp ( r1 , ” T h e r e f o r e , R1 = Vo−V UT / I 2 = +V s a t −V UT
/ I 2 =” )
15 disp ( ” The d e s i g n e d c i r c u i t i s shown i n f i g ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.5 threshold voltage
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
// Example 9 . 5
clc
disp ( ”V CC = +15 V” )
vsat =0.9*15
format (5)
disp ( vsat , ” T h e r e f o r e ,
V s a t ( i n V) = 0 . 9 V CC =” )
disp ( ” R1 = 51 k−ohm , R2 = 120 ohm” )
vut =(13.5*120) /((51*10^3) +120)
format (8)
disp ( vut , ”V UT ( i n V) = +V s a t ∗R2 / R1+R2 =” )
vlt =( -13.5*120) /((51*10^3) +120)
disp ( vlt , ”V LT ( i n V) = −V s a t ∗R2 / R1+R2 =” )
h =(0.03169*2) *10^3
format (6)
disp (h , ”H( i n mV) = V UT − V LT =” )
Scilab code Exa 9.6 tripping voltage
1 // Example 9 . 6
2 clc
3 disp ( ” As i n p u t
i s a p p l i e d t o t h e non−i n v e r t i n g
t e r m i n a l , t h e c i r c u i t i s non−i n v e r t i n g S c h m i t t
t r i g g e r . ”)
121
4 disp ( ” R1 = 100 k−ohm ,
R2 = 1 k−ohm” )
5 vut =13.5*(1/100)
6 format (6)
7 disp ( vut , ” T h e r e f o r e ,
V UT ( i n V) = +V s a t ∗ R2/R1 =”
)
8 vlt = -13.5*(1/100)
9 disp ( vlt , ” T h e r e f o r e ,
)
V LT ( i n V) = −V s a t ∗ R2/R1 =”
Scilab code Exa 9.8 time duration
1 // Example 9 . 8
2 clc
3 disp ( ”LTP = −1.5 V and H = 2 V” )
4 disp ( ”Now
H = UTP − LTP” )
5 disp ( ” T h e r e f o r e ,
2 = UTP − ( − 1 . 5 ) ” )
6 disp ( ” T h e r e f o r e , UTP = 0 . 5 V” )
7 disp ( ” I n t h e f i g . 9 . 4 7 , t h e a n g l e t h e t a can be
8
9
10
11
12
13
14
15
16
17
18
19
20
21
o b t a i n e d from e q u a t i o n o f s i n e wave . S i n e wave i s
r e p r e s e n t e d as , ” )
when p i < omega ∗ t
disp ( ” V i n = V p ∗ s i n ( p i+t h a t a )
< 2 pi ”)
disp ( ” At LTP ,
−1.5 = 5∗ s i n ( p i+t h e t a ) ” )
disp ( ”
= − 5∗ s i n ( t h e t a ) ” )
disp ( ” T h e r e f o r e ,
s i n ( theta ) = 0 . 3 ”)
t = asind (0.3)
format (6)
disp (t , ” T h e r e f o r e ,
t h e t a ( i n d e g r e e ) =” )
disp ( ” The t i m e p e r i o d o f s i n e wave i s , ” )
T =1
disp (T , ”
T( i n ms ) = 1/ f =” )
disp ( ” At UTP,
0 . 5 = V p∗ s i n ( theta ) ”)
disp ( ” T h e r e f o r e , 0 . 5 = 5 ∗ s i n ( t h e t a ) ” )
disp ( ” T h e r e f o r e , s i n ( t h e t a ) = 0 . 1 ” )
t = asind (0.1)
122
22
23
24
25
26
27
28
29
30
disp (t , ” T h e r e f o r e , t h e t a ( i n d e g r e e ) =” )
disp ( ” The t i m e T1 f o r o u t p u t i s from 5 . 7 3 9 d e g r e e t o
(180 degree + 17.45 degree ) ”)
t1 =197.45 -5.739
format (7)
disp ( t1 , ” T h e r e f o r e , T1 ( i n d e g r e e ) =” )
T1 =(191.71/360)
disp ( T1 , ” i . e . T1 ( ms ) =” )
t2 =1 -0.5325
disp ( t2 , ” and
T2 ( i n ms ) = T − T1 =” )
Scilab code Exa 9.9 calculate R1 and R2
1 // Example 9 . 9
2 clc
3 disp ( ” Given +V s a t = 12 V,
−V s a t = −12 V,
V H = 6 V” )
4 disp ( ”We know t h a t h y s t e r e s i s w i d t h i s g i v e n a s ” )
5 disp ( ” V H = ( R2/R1+R2 ) [+ V s a t −V s a t ] ” )
6 disp ( ” T h e r e f o r e ,
R2 / R1+R2 = V H / +V s a t −V s a t ” )
7 r =6/(24)
8 disp (r , ” T h e r e f o r e ,
R2 / R1+R2 =” )
9 disp ( ” T h e r e f o r e ,
R2 = 0 . 2 5 R1 + 0 . 2 5 R2” )
10 disp ( ” T h e r e f o r e ,
0 . 7 5 R2 = 0 . 2 5 R1” )
11 r2 =0.25/0.75
12 format (7)
13 disp ( r2 , ” T h e r e f o r e ,
R2 / R1 =” )
14 disp ( ” Assuming R2 = 10 k−ohm” )
15 r1 =(10000/0.3333) *10^ -3
16 format (3)
17 disp ( r1 , ” R1 ( i n k−ohm ) =” )
Scilab code Exa 9.10 calculate trip point and hysteresis
123
1 // Example 9 . 1 0
2 clc
3 disp ( ”From f i g
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
9 . 4 5 , R1 = 68 k−ohm , R2 = 1 . 5 k−ohm
and V s a t = 1 3 . 5 V” )
vut =(1.5/(1.5+68) ) *13.5
format (7)
disp ( vut , ”V UT ( i n V) = R2/R1+R2 ∗ V s a t =” )
vlt =( -1.5/(1.5+68) ) *13.5
disp ( vlt , ”V LT ( i n V) = −R2/R1+R2 ∗ V s a t =” )
h =2*0.2913
disp (h , ” T h e r e f o r e , H( i n V) = V UT − V LT =” )
disp ( ”Now
H = ( 2 ∗ R2 / R1+R2 ) ∗ V s a t ” )
disp ( ” For minimum H, R2 must be minimum and R1 must
be maximum” )
r2min =((1.5) -(0.05*1.5) )
format (6)
disp ( r2min , ” T h e r e f o r e , R2 min ( i n k−ohm ) = R2 − 5%∗
R2 =” )
r2max =((68) +(0.05*68) )
disp ( r2max , ” T h e r e f o r e , R1 max ( i n k−ohm ) = R1 + 5%∗
R1 =” )
hm =((2*1.425) /(71.4+1.425) ) *13.5
disp ( hm , ” T h e r e f o r e , H min ( i n V) =” )
Scilab code Exa 9.11 design schmitt trigger
1 // Example 9 . 1 1
2 clc
3 disp ( ” Choose op−amp LM318 w i t h V s a t a s +− 1 3 . 5 V
4
5
6
7
8
w i t h s u p p l y v o l t a g e +− 15 V” )
disp ( ” V UT = + 5 V” )
disp ( ”Now V UT = ( R2 / R1+R2 ) ∗ V s a t ” )
disp ( ” T h e r e f o r e , 5 = ( R2 / R1+R2 ) ∗ 1 3 . 5 ” )
disp ( ” T h e r e f o r e , R1 + R2 = 2 . 7 ∗ R2” )
disp ( ” T h e r e f o r e , R1 = 1 . 7 ∗ R2” )
124
9 disp ( ” Choose R2 = 10 k−ohm” )
10 r1 =1.7*10
11 disp ( r1 , ” T h e r e f o r e ,
R1 ( i n k−ohm ) =” )
12 disp ( ” The d e s i g n e d c i r c u i t i s shown i n
f i g . 9 . 4 6 ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.12 design op amp schmitt trigger
1 // Example 9 . 1 2
2 clc
3 disp ( ” For t h e S c h m i t t t r i g g e r ” )
V LT = −4 V,
4 disp ( ”V UT = 2 V,
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
+−V s a t = +−10
V” )
disp ( ” For u n e q u a l UTP and LTP v a l u e s , a m o d i f i e d
c i r c u i t i s r e q u i r e d a s shown i n t h e f i g . 9 . 5 2 . ” )
disp ( ” The v o l t a g e V1 d e c i d e s t h e UTP and LTP l e v e l s .
A p p l y i n g KVL t o t h e o u t p u t c i r c u i t and
n e g l e c t i n g op−amp i n p u t c u r r e n t we can w r i t e , ” )
disp ( ”−IR2 − IR1 − x + V0 = 0 ” )
disp ( ” T h e r e f o r e ,
I = V0−x / R1+R2” )
disp ( ”And V1 = IR1 + x ” )
disp ( ” T h e r e f o r e , V1 = ( V0−x /R1+R2 ) ∗R1 + x ” )
disp ( ” For
+V s a t = 10 V, ” )
disp ( ”V1 = V UT = 2 V, ” )
disp ( ”V0 = 10 V” )
disp ( ” T h e r e f o r e , 2 = (10 − x /R1+R2 ) ∗R1 + x
( 1) ”)
disp ( ” For −V s a t = −10 V, ” )
disp ( ”V1 = V LT = −4 V, ” )
disp ( ”V0 = −10 V” )
disp ( ” T h e r e f o r e , −4 = (−10−x /R1+R2 ) ∗R1 + x
(2) ”
)
disp ( ” S u b t r a c t i n g e q u a t i o n s ( 2 ) and ( 1 ) , ” )
disp ( ” T h e r e f o r e , 6 = 20∗ R1 / R1+R2” )
125
21 disp ( ” T h e r e f o r e ,
R1+R2 = 3 . 3 3 3 ∗ R1” )
22 disp ( ” T h e r e f o r e ,
R2 = 2 . 3 3 3 ∗ R1
( 3 ) ”)
23 disp ( ” S u b s t i t u t i n g ( 3 ) i n e q u a t i o n ( 1 ) ” )
24 disp ( ” 2 = ((10 − x ) ∗R1 / 3 . 3 3 3 ∗ R1 ) + x ” )
25 disp ( ” T h e r e f o r e ,
2 . 3 3 3 ∗ x = −3.3334 ” )
26 x = -3.3334/2.333
27 format (7)
28 disp (x , ” T h e r e f o r e ,
x =” )
29 disp ( ” So a c t u a l l y p o l a r i t y o f t h e v o l t a g e s o u r c e
’ ’x
’ ’ must be o p p o s i t e t o what i s assumed e a r l i e r a s
shown i n f i g . 9 . 5 2 . ” )
30 disp ( ” Choose
R1 = 1 k−ohm
h e n c e R2 = 2 . 3 3 3 k−
ohm” )
31 disp ( ” T h e r e f o r e ,
R comp = R1 | | R2 = 0 . 7 k−ohm” )
32 disp ( ”Now a s l o n g a s V i n i s l e s s t h a n V UT , t h e
o u t p u t i s a t +V s a t = 10 V and when V i n > V UT ,
t h e o u t p u t s w i t c h e s from +V s a t t o −V s a t . While
a s l o n g a s V i n > V LT , t h e o u t p u t i s a t −V s a t =
−10 V and when V i n < V LT , t h e o u t p u t s w i t c h e s
from −V s a t t o +V s a t . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.13 VUT and VLT and frequency of oscillation
1 // Example 9 . 1 3
2 clc
3 disp ( ” ( a ) We know t h a t
4 vut =(86*15) /(86+100)
5 format (5)
6 disp ( vut , ” V UT ( i n V)
7 vlt =(86* -15) /(86+100)
8 disp ( ” ( b ) We know t h a t
9 disp ( vlt , ” V LT ( i n V)
, ”)
= R1∗+ V s a t / R1+R2 =” )
, ”)
= R1∗− V s a t / R1+R2 =” )
126
10 disp ( ” ( c ) We know t h a t , ” )
11 f0 =1/0.02
12 disp ( f0 , ” f 0 ( i n Hz ) = 1 / 2∗ Rf ∗ C i n ∗[+ V s a t −V LT/+
V s a t −V UT ] =” )
Scilab code Exa 9.17 design op amp circuit
1 // Example 9 . 1 7
2 clc
3 disp ( ” The m o n o s t a b l e m u l t i v i b r a t o r
4
5
6
7
8
9
10
11
12
13
14
u s i n g op−amp
p r o d u c e s t h e p u l s e waveform . The p u l s e w i d t h i s
g i v e n by , ” )
disp ( ” T = RC∗ l n [1+ V D1/ V s a t / 1− b e t a ] ” )
disp ( ” where V D1 = 0 . 7 V, +v s a t = +−12 V f o r op−
amp 741 ” )
disp ( ” b e t a = R2 / R1+R2 = 0 . 5 w i t h R1 = R2” )
t =1/(2*10^3)
format (6)
disp (t , ”T( i n s e c ) = 1/ f ” )
disp ( ” Choose C = 0 . 1 uF” )
disp ( ” T h e r e f o r e , 5∗10ˆ −4 = R∗ 0. 1 ∗ 10 ˆ − 6 ∗ l n
[1+(0.7/12) /1 −0.5] ”)
disp ( ” T h e r e f o r e , R = 6 . 7 k−ohm” )
disp ( ” Choose R1 = R2 = 10 k−ohm” )
disp ( ” The d e s i g n e d c i r c u i t i s shown i n f i g . 9 . 6 3 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.18 design monostable using IC 555
1 // Example 9 . 1 8
2 clc
127
3
4
5
6
7
8
9
10
11
disp ( ” The r e q u i r e d p u l s e w i d t h i s , ” )
disp ( ” W = 10 ms” )
disp ( ” The p u l s e w i d t h i s g i v e n by , ” )
disp ( ” W = 1 . 1 ∗R∗C” )
disp ( ” T h e r e f o r e , 10∗10ˆ −3 = 1 . 1 ∗R∗C” )
disp ( ” T h e r e f o r e , RC = 9 . 0 9 0 9 ∗ 1 0 ˆ − 3 ” )
disp ( ” Choose C = 0 . 1 uF” )
disp ( ” T h e r e f o r e , R = 9 0 . 9 0 9 k−ohm ˜ 91 k−ohm” )
disp ( ” The d e s i g n e d c i r c u i t i s shown i n f i g . 9 . 7 8 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.19 design a timer
1 // Example 9 . 1 9
2 clc
3 disp ( ” F i g . 9 . 7 9 shows m o n o s t a b l e
4
5
6
7
8
9
10
11
12
13
14
c i r c u i t used to
d r i v e the r e l a y . ”)
disp ( ” T h i s r e l a y s h o u l d be e n e r g i z e d f o r 5 s e c o n d t o
h o l d h e a t e r ’ ’ON’ ’ f o r 5 s e c o n d s . Thus , T ON f o r
monostable i s 5 seconds . ”)
disp ( ”We know t h a t t h e p u l s e w i d t h i s g i v e n by , ” )
disp ( ”
W = 1 . 1 RC” )
disp ( ” T h e r e f o r e , 5 = 1 . 1 RC” )
disp ( ”Now , t h e r e a r e two unknowns . I n t h i s c a s e , we
have t o s e l e c t v a l u e f o r c a p a c i t o r and w i t h t h e
s e l e c t e d v a l u e we have t o f i n d t h e v a l u e o f
r e s i s t a n c e from t h e f o r m u l a . ” )
disp ( ” T h e r e f o r e ,
I f c a p a c i t o r v a l u e i s 10 uF” )
disp ( ” t h e n
5 = 1 . 1 ∗R∗10 uF” )
r =(5/(1.1*10*10^ -6) ) *10^ -3
format (7)
disp (r , ” T h e r e f o r e , R( i n k−ohm ) =” )
disp ( ” The c a l c u l a t e d v a l u e i s n o t s t a n d a r d v a l u e ,
128
but we can a d j u s t t h i s v a l u e by c o n n e c t i n g
v a r i a b l e r e s i s t a n c e i . e . potentiometer . ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.20 draw timer using IC 555
1 // Example 9 . 2 0
2 clc
3 disp ( ” The r e q u i r e m e n t
4
5
6
7
8
9
10
11
12
i s t h a t t h e d o o r must be open
f o r 15 s e c a f t e r r e c e i v i n g a t r i g g e r s i g n a l and
then g e t s shut door a u t o m a t i c a l l y . This r e q u i r e s
IC 555 i n a m o n o s t a b l e mode w i t h a p u l s e w i d t h o f
15 s e c . ” )
disp ( ” T h e r e f o r e , W = 15 s e c ” )
disp ( ”Now
W = 1 . 1 RC” )
disp ( ” T h e r e f o r e , 15 = 1 . 1 RC” )
disp ( ” Choose
C = 100 uF” )
r =(15/(1.1*100*10^ -6) ) *10^ -3
format (8)
disp (r , ” T h e r e f o r e , R( i n k−ohm ) =” )
disp ( ” The d e s i g n e d c i r c u i t i s shown i n t h e f i g . 9 . 8 0
”)
disp ( ” The s u p p l y v o l t a g e 10 o r 15 V h a s no e f f e c t on
the operation of the c i r c u i t or the values of R
and C s e l e c t e d . ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.21 frequency of output and duty cycle
129
1 // / Example 9 . 2 1
2 clc
3 disp ( ” The f r e q u e n c y o f o u t p u t i s g i v e n by , ” )
4 f =(1.44/(12*0.01*10^ -3) ) *10^ -3
5 format (3)
6 disp (f , ”
f ( i n kHz ) = 1 . 4 4 / ( R A+2∗R B ) ∗C =” )
7 disp ( ” The duty c y c l e i s g i v e n by , ” )
8 d =8/12
9 format (7)
10 disp (d , ” D = R A+R B / R A+2∗R B =” )
11 disp ( ” Thus t h e duty c y c l e 6 6 . 6 7%” )
Scilab code Exa 9.26 design astable multivibrator
1 // Example 9 . 2 6
2 clc
3 disp ( ” f = 1 kHz ” )
4 disp ( ”D = 75% = 0 . 7 5 ” )
5 disp ( ”Now
f = 1 . 4 4 / ( R A+2∗R B ) ∗C” )
6 disp ( ” T h e r e f o r e ,
1 ∗ 1 0 ˆ 3 = 1 . 4 4 / ( R A+2∗R B ) ∗C” )
7 disp ( ” T h e r e f o r e ,
( R A+2∗R B ) ∗C = 1.44 ∗10ˆ −3
8
9
10
11
12
13
14
15
16
17
. . . . ( 1 ) ”)
disp ( ” T h e r e f o r e , w h i l e %D = ( ( R A+R B ) / ( R A+2∗R B ) )
∗ 100 ” )
disp ( ” T h e r e f o r e ,
0 . 7 5 = R A+R B / R A+2∗R B” )
disp ( ” T h e r e f o r e , R A+2∗R B = ( R A+R B ) / 0 . 7 5 ” )
disp ( ” T h e r e f o r e , R A+2∗R B = 1 . 3 3 ∗ ( R A+R B ) ” )
disp ( ” T h e r e f o r e ,
0 . 6 6 ∗ R B = 0 . 3 3 ∗ R A” )
disp ( ” T h e r e f o r e , R B = 0 . 5 ∗ R A
. . . . ( 2 ) ”)
disp ( ” Choose C = 0 . 1 uF” )
disp ( ” S u b s t i t u t i n g i n ( 1 ) , ” )
disp ( ” ( R A+2∗R B ) ∗0.1∗10ˆ −6 = 1.44 ∗10ˆ −3 ” )
disp ( ” T h e r e f o r e , R A+2∗R B = 1 4 4 0 0
. . . . ( 3 ) ”)
130
18 disp ( ” S u b s t i t u t i n g ( 2 ) i n ( 3 ) , ” )
19 disp ( ”R A + 2 ( 0 . 5 ∗ R A ) = 1 4 4 0 0 ” )
20 ra =(14400/2) *10^ -3
21 format (4)
22 disp ( ra , ” T h e r e f o r e ,
R A ( i n k−ohm ) =” )
23 rb =0.5*7.2
24 disp ( rb , ” T h e r e f o r e ,
R B ( i n k−ohm ) =” )
25 disp ( ” and C = 0 . 1 uF” )
26 disp ( ” Hence t h e c i r c u i t d i a g r a m i s a s shown i n
fig
. 9 . 1 0 0 ”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.27 design astable multivibrator
1 // Example 9 . 2 7
2 clc
3 disp ( ”T ON = 0 . 6 ms , T = 1 ms” )
4 d =0.6*100
5 disp (d , ” T h e r e f o r e , D( i n p e r c e n t a g e ) = t ON / T =” )
6 disp ( ”Now D = R A+R B / R A+2∗R B = 0 . 6 ” )
7 disp ( ” T h e r e f o r e ,
R A+R B = 0 . 6 ∗ R A + 1 . 2 ∗ R B” )
8 disp ( ” T h e r e f o r e ,
0 . 4 ∗ R B = 0 . 2 ∗ R B” )
9 disp ( ” T h e r e f o r e ,
R B = 2∗R A
. . . . ( 1 ) ”)
10 disp ( ”
f = 1 . 4 4 / ( R A+2∗R B ) ∗C = 1/T = 1 0 0 0 ”
11
12
13
14
15
16
17
18
)
disp ( ” Choose C = 0 . 1 uF” )
disp ( ” T h e r e f o r e , R A+2∗R B = 1 4 4 0 0 ” )
disp ( ” U s i n g ( 1 ) ,
5∗R A = 1 4 4 0 0 ” )
ra =(14400/5) *10^ -3
format (5)
disp ( ra , ” T h e r e f o r e ,
R A ( i n k−ohm ) =” )
rb =2.88*2
disp ( rb , ” R B ( i n k−ohm ) =” )
131
19
disp ( ” The c i r c u i t
i s shown i n t h e f i g . 9 . 1 0 1 ” )
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.28 design astable mode to generate square wave
1 // Example 9 . 2 8
2 clc
3 disp ( ”T ON = T OFF = 0 . 5 ms” )
4 disp ( ” T h e r e f o r e , T = T ON + T OFF = 1 ms” )
5 disp ( ” i . e .
f = 1/T = 1 kHz ” )
6 disp ( ”Now T d = T OFF = 0 . 6 9 ∗ R B∗C” )
7 disp ( ” Choose
C = 0 . 1 uF” )
8 rb =((0.5*10^ -3) /(0.69*0.1*10^ -6) ) *10^ -3
9 format (6)
10 disp ( rb , ” T h e r e f o r e ,
R B ( i n k−ohm ) =” )
11 disp ( ”Now duty c y c l e i s 50% s o R A = R B = 7 . 2 4 6 k−
12
13
14
15
16
17
18
19
ohm” )
disp ( ” P r a c t i c a l l y a m o d i f i e d c i r c u i t i s r e q u i r e d f o r
50% duty c y c l e where d i o d e i s c o n n e c t e d a c r o s s
R B and c h a r g i n g t a k e s p l a c e t h r o u g h R A and
d i o d e . And R B must be e q u a l t o sum o f R A and
d i o d e f o r w a r d r e s i s t a n c e . So t o have p e r f e c t
s q u a r e wave , R A i s k e p t v a r i a b l e i . e . p o t o f s a y
10 k−ohm i n t h i s c a s e . I t i s t h e n a d j u s t e d t o
o b t a i n p r e c i s e s q u a r e wave . The r e s i s t a n c e
r e q u i r e d i n s e r i e s w i t h LED t o be c o n n e c t e d i s , ” )
disp ( ”R = V 0−V LED / I LED ” )
disp ( ” Assuming V LED = 0 . 7 V” )
r =(5 -0.7) /(50*10^ -3)
format (3)
disp (r , ” C u r r e n t l i m i t i n g R( ohm ) = ” )
disp ( ” The v o l t a g e o f R i s ” )
disp ( ”P = (50∗10ˆ −3) ˆ2 ∗ 100 ” )
132
20 p = ((50*10^ -3) ^2) *100
21 disp (p , ”P( i n W) =” )
22 disp ( ” Both r e s i s t o r s R can be o f 1/4 W” )
23 disp ( ” The r e q u i r e d c i r c u i t i s shown i n t h e
fig .9.102
”)
This code can be downloaded from the website wwww.scilab.in
Scilab code Exa 9.32 define resolution
1 // Example 9 . 3 2
2 clc
3 disp ( ” For t h e g i v e n DAC, ” )
4 disp ( ” n = Number o f b i t s = 8 ” )
5 disp ( ” ( i ) R e s o l u t i o n = 2ˆ n = 2ˆ8 = 256 ” )
6 disp ( ” i . e . t h e o u t p u t v o l t a g e can have 256 d i f f e r e n t
v a l u e s i n c l u d i n g zero . ”)
disp ( ” ( i i ) V 0FS = F u l l s c a l e o u t p u t v o l t a g e ” )
disp ( ”
= 2 . 5 5 V” )
disp ( ” T h e r e f o r e ,
R e s o l u t i o n = V 0FS / 2ˆ n − 1 =
2 . 5 5 / 2ˆ8 − 1 = 10mV / 1LSB” )
10 disp ( ” Thus an i n p u t c h a n g e o f 1LSB c a u s e s t h e o u t p u t
t o c h a n g e by 10mV” )
7
8
9
Scilab code Exa 9.33 output voltage
1 // Example 9 . 3 3
2 clc
3 disp ( ” For g i v e n DAC, ” )
4 disp ( ” n = 4 ” )
5 disp ( ” T h e r e f o r e ,
V 0FS = 15 V” )
133
disp ( ” T h e r e f o r e
/ LSB” )
7 disp ( ” T h e r e f o r e
8 disp ( ”Now D =
9 disp ( ” T h e r e f o r e
6
,
R e s o l u t i o n = V 0FS / 2ˆ n − 1 = 1V
, V0 = R e s o l u t i o n ∗ D” )
Decimal o f 0110 = 6”)
, V0 = 1V / LSB ∗ 6 = 6 V” )
Scilab code Exa 9.34 find VoFS
1 // Example 9 . 3 4
2 clc
3 disp ( ” R e s o l u t i o n = V 0FS / 2ˆ n − 1 ” )
4 disp ( ” T h e r e f o r e ,
20 = V 0FS / 2ˆ n − 1 ” )
5 disp ( ” T h e r e f o r e ,
V 0FS = 5 . 1 V” )
6 disp ( ” D = E q u i v a l e n t o f 1 0 0 0 0 0 0 0 = 128 ” )
7 disp ( ” T h e r e f o r e ,
V0 = R e s o l u t i o n ∗ D = 20 ∗ 128 =
2 . 5 6 V” )
Scilab code Exa 9.35 find out stepsize and analog output
1 // Example 9 . 3 5
2 clc
3 disp ( ” For g i v e n DAC,
n = 4 , V 0FS = +5 V” )
4 disp ( ” R e s o l u t i o n = V 0FS / 2ˆ n − 1 = 1/3 V/LSB” )
5 disp ( ” T h e r e f o r e ,
V0 = R e s o l u t i o n ∗ D” )
6 disp ( ” For D = D e c i m a l od 1 0 0 0 0 = 8 ” )
7 disp ( ” V0 = 1/3 ∗ 8 = 2 . 6 6 6 7 V” )
8 disp ( ” For D = D e c i m a l o f 1 1 1 1 = 15 ” )
9 disp ( ” V0 = 1/3 ∗ 15 = 5 V” )
Scilab code Exa 9.36 find output voltage
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5
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8
9
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11
12
// Example 9 . 3 6
clc
disp ( ” For 12− b i t DAC, s t e p s i z e i s 8 mV” )
v =(8*10^ -3) *((2^12) -1)
format (6)
disp (v , ” V 0FS = 8 mV ∗ 2 ˆ 1 2 − 1 =” )
r =((8*10^ -3) /32.76) *100
format (8)
disp (r , ”% R e s o l u t i o n = 8mV/ 3 2 . 7 6V ∗ 100 =” )
q =(8*10^ -3) *1389
format (7)
disp (q , ” The o u t p u t v o l t a g e f o r t h e i n p u t
0 1 0 1 0 1 1 0 1 1 0 1 i s = 8mV ∗ 1 3 8 9 =” )
Scilab code Exa 9.38 find resolution and digital output
1 // Example 9 . 3 8 .
2 clc
3 disp ( ” ( a ) From e q u a t i o n ( 1 ) we have , ” )
4 r =2^8
5 format (4)
6 disp (r , ” R e s o l u t i o n = 2ˆ8 =” )
7 disp ( ” and from e q u a t i o n ( 2 ) we have , ” )
8 disp ( ” R e s o l u t i o n = 5 . 1V/ ( 2 ˆ 8 − 1 ) = 20 mV/LSB” )
9 disp ( ” T h e r e f o r e , we can s a y t h a t t o c h a n g e o u t p u t by
10
11
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14
1 LSB we have t o c h a n g e
disp ( ” ( b ) For 1 . 2 8 V a n a l o g
can be c a l c u l a t e d as , ” )
d =1.28/(20*10^ -3)
format (3)
disp (d , ”D ( i n LSBs ) = 1 . 2 8V
disp ( ” The b i n a r y e q u i v a l e n t
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i n p u t by 20 mV” )
input , d i g i t a l output
/ 20 mV/LSB =” )
o f 64 i s 0 1 0 0 0 0 0 0 ” )
Scilab code Exa 9.39 calculate quantizing error
1 // Example 9 . 3 9
2 clc
3 disp ( ”From e q u a t i o n ( 3 ) we g e t ” )
4 qe =(4.095/(4095*2) ) *10^3
5 format (4)
6 disp ( qe , ”Q E ( i n mV) = 4 . 0 9 5 / (4096 −1) ∗2 =” )
Scilab code Exa 9.40 dual scope ADC
1 // Example 9 . 4 0
2 clc
3 disp ( ”We know t h a t , ” )
4 disp ( ” t 2 = ( V1/VR) ∗ t 1 ” )
5 t2 =83.33
6 format (6)
7 disp ( t2 , ” ( i ) t 2 ( i n ms ) = ( 1 0 0 / 1 0 0 ) ∗ 8 3 . 3 3 =” )
8 disp ( ” ( i i ) V1 = 200 mV” )
9 t2 =83.33*2
10 disp ( t2 , ” T h e r e f o r e , t 2 ( i n ms ) = ( 2 0 0 / 1 0 0 ) ∗ 8 3 . 3 3 =” )
Scilab code Exa 9.41 find digital output of ADC
1 // Example 8 . 4 1
2 clc
3 disp ( ” The d i g i t a l o u t p u t i s g i v e n as , ” )
4 disp ( ” D i g i t a l o u t p u t = ( Counts / S e c o n d ) ∗ t 1 ∗ ( V i /V R ) ”
5
6
7
8
)
disp ( ”Now C l o c k f r e q u e n c y = 12 kHz ” )
disp ( ”
i . e . = 12000 counts / second ”)
d =12000*83.33*(100/100) *10^ -3
format (5)
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9
disp (d , ” T h e r e f o r e ,
D i g i t a l output ( in counts ) =
1 2 0 0 0 ∗ 8 3 . 3 3 ∗ ( 1 0 0 / 1 0 0 ) ∗10ˆ −3 =” )
Scilab code Exa 9.42 find conversion time
1 // Example 9 . 4 2 .
2 clc
3 disp ( ”
f = 1 MHz” )
4 disp ( ” T h e r e f o r e , T = 1/ f = 1 / 1 ∗ 1 0 ˆ 6 = 1 u s e c ” )
5 disp ( ”
n = 8”)
6 tc =1*(8+1)
7 format (2)
8 disp ( tc , ” T h e r e f o r e ,
T C ( i n u s e c ) = T∗ ( n+1) =” )
Scilab code Exa 9.43 find maximum frequency of input sine wave
1 // Example 9 . 4 3
2 clc
3 disp ( ” The maximum f r e q u e n c y i s g i v e n by , ” )
4 f =1/(2* %pi *(9*10^ -6) *2^8)
5 format (6)
6 disp (f , ” f m a x ( i n Hz ) = 1 / 2∗ p i ∗ ( T C ) ∗2ˆ n =” )
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Chapter 10
Voltage Regulators
Scilab code Exa 10.1 find line and load regulation and ripple refection
1 // Example 1 0 . 1
2 clc
3 disp ( ” Z Z = 7 ohm ,
4
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17
R3 = 330 ohm ,
V 0 = 4 . 7 V,
V i n = 15 V” )
disp ( ” The s p e c i f i e d c h a n g e i n V i n i s 10%, ” )
vin =0.1*15
format (4)
disp ( vin , ” T h e r e f o r e ,
d e l t a V i n ( i n V) = 10% o f V i n
=” )
vo =(1.5*7) /330
format (8)
disp ( vo , ” T h e r e f o r e ,
d e l t a V 0 ( i n V) = d e l t a V i n ∗ Z Z
/ R3 =” )
lr =0.03181*100/4.7
format (6)
disp ( lr , ” T h e r e f o r e ,
Line r e g u l a t i o n ( in percentage )
= d e l t a V 0 ∗ 100 / V 0 =” )
disp ( ” For I L ( max ) = 50 mA, ” )
dvo =(20*7*50*10^ -3) /330
format (8)
disp ( dvo , ” T h e r e f o r e ,
d e l t a V 0 ( i n V) = I L ( max ) ∗ R S ∗
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Z Z / R3 =” )
lr =0.02121*100/4.7
format (7)
disp ( lr , ” T h e r e f o r e ,
Line r e g u l a t i o n ( in precentage )
= d e l t a V 0 ∗ 100 / V 0 =” )
disp ( ”Now
V R ( o u t ) = V R ( i n ) ∗ Z Z / R3” )
zz =7/330
format (8)
disp ( zz , ” T h e r e f o r e , V R ( o u t ) /V R ( i n ) = Z Z /R3 =” )
rr =20* log10 (0.02121)
format (6)
disp ( rr , ” T h e r e f o r e , RR( i n dB ) = 20∗ l o g ( 0 . 0 2 1 2 1 ) = ”
)
Scilab code Exa 10.2 design op amp series voltage regulator
1 // Example 1 0 . 4
2 clc
3 disp ( ”R1 = 5 k−ohm ,
R2 = 10 k−ohm” )
4 disp ( ” The IC i s 7 8 0 8 i . e . V r e g = +8 V” )
5 vt =8*(3)
6 format (3)
7 disp ( vt , ” T h e r e f o r e ,
V out ( i n V) = V r e g ∗ [ 1 + R2/R1 ]
=” )
8 disp ( ”Now R2 = 1 k−ohm then , ” )
9 vo =8*(1+(1/5) )
10 format (4)
11 disp ( vo , ” V out ( i n V) = 8 ∗ [ 1 + 1 / 5 ] =” )
12 disp ( ” Thus t h e V out can be v a r i e d from 9 . 6 V t o 24
V, by v a r i n g R2 from 1 k−ohm t o 10 k−ohm . ” )
Scilab code Exa 10.4 calculate output voltage
139
1 // Example 1 0 . 4
2 clc
3 disp ( ”R1 = 5 k−ohm ,
R2 = 10 k−ohm” )
4 disp ( ” The IC i s 7 8 0 8 i . e . V r e g = +8 V” )
5 vo =8*3
6 format (3)
7 disp ( vo , ” T h e r e f o r e ,
V out ( i n V) = V r e g ∗ [ 1 + R2/R1 ]
8
9
10
11
12
=” )
disp ( ”Now R2 = 1 k−ohm then , ” )
vou =8*(1+(1/5) )
format (4)
disp ( vou , ” V out ( i n V) =” )
disp ( ” Thus t h e V out can be v a r i e d from 9 . 6 V t o 24
V, by v a r i n g R2 from 1 k−ohm t o 10 k−ohm” )
Scilab code Exa 10.7 determine regulated output voltage
1 // Example 1 0 . 7
2 clc
3 disp ( ” The r e s i s t a n c e u s e d a r e , ” )
4 disp ( ” R1 = 220 ohm and R2 = 1 . 5 k−ohm” )
5 disp ( ” w h i l e f o r LM 3 1 7 ,
I ADJ = 100 uA” )
6 disp ( ” T h e r e f o r e ,
V 0 = 1 . 2 5 ∗ [ 1 + R2/R1 ] + I ADJ ∗R2” )
7 vo =(1.25*(1+((1.5*10^3) /220) ) ) +(100*1.5*10^ -3)
8 format (5)
9 disp ( vo , ” T h e r e f o r e ,
V 0 ( i n V) =” )
Scilab code Exa 10.8 find the range
1 // Example 1 0 . 8
2 clc
3 disp ( ” For LM 3 1 7 , t h e c u r r e n t I ADJ = 100 uA” )
4 disp ( ”When R2 i s maximum i . e . R2 = 0 then , ” )
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5 disp ( ” V 0 = 1 . 2 5 ∗ [ 1 + R2/R1 ] + I ADJ ∗R2 = 1 . 2 5 V” )
6 disp ( ”When R2 i s maximum , i . e . R2 = 10 k−ohm t h e n ” )
7 vo =(1.25*(1+((10*10^3) /820) ) ) +(100*10*10^ -3)
8 format (6)
9 disp ( vo , ” V 0 ( i n V) = ” )
10 disp ( ” Thus t h e o u t p u t v o l t a g e can be v a r i e d i n t h e
r a n g e 1 . 2 5 V t o 1 7 . 4 9 V” )
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