Scilab Textbook Companion for
Microelectronic Circuits
by A. S. Sedra And K. C. Smith1
Created by
Shruthi S.H
B.Tech (pursuing)
Electronics Engineering
NIT, Surathkal
College Teacher
Mrs Rekha S, NIT Surathkal
Cross-Checked by
Giridharan, IITB
May 20, 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: Microelectronic Circuits
Author: A. S. Sedra And K. C. Smith
Publisher: Oxford University Press
Edition: 5
Year: 2004
ISBN: 0-19-514252-7
1
Scilab numbering policy used in this document and the relation to the
above book.
Exa Example (Solved example)
Eqn Equation (Particular equation of the above book)
AP Appendix to Example(Scilab Code that is an Appednix to a particular
Example of the above book)
For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means
a scilab code whose theory is explained in Section 2.3 of the book.
2
Contents
List of Scilab Codes
4
1 Introduction to Electronics
5
2 Operational Amplifiers
10
3 Diodes
12
4 MOS Field Effect Transistors
19
5 Bipolar Junction Transistor
28
6 single stage integrated circuit amplifiers
42
7 Differential and multistage amplifier
60
8 Feedback
69
9 Operational amplifier and data converter circuits
74
10 Digital CMOS logic circuits
81
11 Memory and advanced digital circuits
86
12 Filters and tuned amplifiers
89
14 Output Stages and amplifier
90
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
1.3
1.4
1.5
1.6
2.1
2.3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.9
4.10
4.11
4.12
Amplifier gain power and eficiency . . . . . . . . . .
Gain of transistor amplifier . . . . . . . . . . . . . .
Overall voltage gain of three stage amplifier . . . . .
Bipolar junction transistor . . . . . . . . . . . . . .
DC gain 3dB frequency and frequency at which gain
Evaluation of tPHL . . . . . . . . . . . . . . . . . .
Closed loop and open loop gain . . . . . . . . . . . .
Design instrumentation amplifier . . . . . . . . . . .
Peak value of diode current and maximum reverse vo
Values of Iand V for the circuit given . . . . . . . . .
Evaluating junction scaling constant . . . . . . . . .
To determine ID and VD . . . . . . . . . . . . . . .
Repeating example 4 using piecewise linear model .
Power supply ripple . . . . . . . . . . . . . . . . . .
Percentage change in regulated voltage . . . . . . . .
line regulation load regulation . . . . . . . . . . . . .
Value of capacitance C that will result in peak pe . .
To determine operating point parameters . . . . . .
Design of given circuit . . . . . . . . . . . . . . . . .
Design of given circuit . . . . . . . . . . . . . . . . .
Design of given circuit . . . . . . . . . . . . . . . . .
To determine all node voltages and currents throug .
Design of given circuit . . . . . . . . . . . . . . . . .
To determine drain currents and output voltage . . .
Design of given circuit . . . . . . . . . . . . . . . . .
Small signal analysis . . . . . . . . . . . . . . . . . .
To determine all parameters of transistor amplifie . .
Midband gain and upper 3dB frequency . . . . . . .
4
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12
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19
20
20
21
22
22
23
24
24
25
26
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
4.13
5.1
5.2
5.3
5.4
5.5
5.7
5.8
5.9
5.10
5.11
5.13
5.14
5.17
5.18
5.19
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
7.1
7.2
7.3
7.5
8.1
8.2
8.3
8.4
Coupling capacitor values . . . . . . . . . . . . . . . .
Design of given circuit with current 2mA . . . . . . .
Consider a common Emitter circuit . . . . . . . . . . .
Determine RB . . . . . . . . . . . . . . . . . . . . . .
Analyse the circuit to find node voltages and bran . .
Analyse the circuit to find node voltages and bran . .
Analyse the circuit to find node voltages and bran . .
Analyse the circuit to find node voltages and bran . .
Analyse the circuit to find node voltages and bran . .
Analyse the circuit to find node voltages and bran . .
Analyse the circuit to find node voltages and bran . .
Design of bias network of the amplifier . . . . . . . . .
Analysis of transistor amplifier . . . . . . . . . . . . .
Amplifier parameters . . . . . . . . . . . . . . . . . . .
Midband gain and 3dB frequency . . . . . . . . . . . .
To select values of capacitance required . . . . . . . .
To find the operating point of NMOS transistor . . . .
Comparison between NMOS transistor and npn transis
Comparison between NMOS transistor and npn transis
Design of the circuit with output current 100uA . . . .
Determine 3dB frequency . . . . . . . . . . . . . . . .
To determine midband gain and upper 3dB frequency
Application of miller theorem . . . . . . . . . . . . . .
Analysis of CMOS CS amplifier . . . . . . . . . . . . .
Analysis of CMOS CS amplifier . . . . . . . . . . . . .
To determine AM ft fZ f3dB . . . . . . . . . . . . . .
Avo Rin Rout Gi Gis Gv fH . . . . . . . . . . . . . . .
Comparison between Cascode amplifier and CS amplif
Analysis of CC CE amplifier . . . . . . . . . . . . . .
To determine required resistor values . . . . . . . . . .
Analysis of differential amplifier . . . . . . . . . . . . .
Analysis of Active loaded MOS differential amplifi . .
To determine all parameters for different transist . . .
Analysis of given circuit . . . . . . . . . . . . . . . . .
Analysis of op amp connected in an inverting conf . .
Feedback triple . . . . . . . . . . . . . . . . . . . . . .
Small signal analysis . . . . . . . . . . . . . . . . . . .
Small signal analysis . . . . . . . . . . . . . . . . . . .
5
27
28
29
30
30
31
32
33
33
34
35
36
37
38
39
40
42
43
44
47
48
48
49
49
51
52
53
54
57
58
60
61
63
67
69
70
71
72
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
9.1
9.2
9.3
10.1
10.2
10.3
10.4
11.1
11.2
11.3
12.4
14.1
14.2
14.3
14.4
14.5
Design of two stage CMOS op amp . . . . . . . . . . .
To determine Av ft fP SR and PD of folded casc . . .
To determine input offset voltage . . . . . . . . . . . .
To determine tPHL tPLH and tP . . . . . . . . . . . .
WbyL ratios for the logic circuit . . . . . . . . . . . .
To determine the parameters of pseudo NMOS inverte
To determine parameters for NMOS transistor . . . .
Min WbyL ratio to ensure flip flop will switch . . . . .
Design of two stage CMOS op amp . . . . . . . . . . .
Time required . . . . . . . . . . . . . . . . . . . . . .
To design tuned amplifier . . . . . . . . . . . . . . . .
To design a Class B Output Amplifier . . . . . . . . .
To determine quiescent current and power . . . . . . .
Redesign the output stage of Example 2 . . . . . . . .
To determine thermal resistance junction temperat . .
To determine the maximum power dissipated . . . . .
6
74
76
80
81
82
82
84
86
87
88
89
90
91
92
92
93
Chapter 1
Introduction to Electronics
Scilab code Exa 1.1 Amplifier gain power and eficiency
1 // Example1 . 1 : A m p l i f i e r g a i n , power and e f i c i e n c y
2 // A m p l i f i e r o p e r a t e s a t +10−V/−10−V power s u p p l y .
3 A_v =9/1; // s i n u s o i d a l v o l t a g e i n p u t o f 1V peak and
4
5
6
7
8
9
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11
12
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17
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19
20
s i n u s o i d a l o u t p u t v o l t a g e o f 9V peak
I_o =9/1000; // 1 k i l o ohms l o a d
disp ( A_v , ” V o l t a g e g a i n (V/V) =” )
disp (20* log10 ( A_v ) ,” V o l t a g e g a i n ( dB ) =” )
I_i =0.0001 // s i n u s o i d a l c u r r e n t i n p u t o f 0 . 1mA peak
A_i = I_o / I_i ;
disp ( A_i , ” C u r r e n t g a i n (A/A) =” )
disp (20* log10 ( A_i ) ,” C u r r e n t g a i n ( dB ) =” )
V_orms = 9/ sqrt (2) ;
I_orms = 9/ sqrt (2) ;
P_L = V_orms * I_orms ; // o u t p u t power i n mW
V_irms =1/ sqrt (2) ;
I_irms =0.1/ sqrt (2) ;
P_I = V_irms * I_irms ; // i n p u t power i n mW
A_p = P_L / P_I ;
disp ( A_p , ” Power g a i n (W/W) =” )
disp (10* log10 ( A_p ) ,” Power g a i n ( dB ) =” )
P_dc =10*9.5+10*9.5; // a m p l i f i e r draws a c u r r e n t o f
7
21
22
23
24
25
9 . 5mA from e a c h o f i t s two power s u p p l i e s
disp ( P_dc , ” Power drawn from t h e dc s u p p l i e s (mW) =” )
P_dissipated = P_dc + P_I - P_L ;
disp ( P_dissipated , ” Power d i s s i p a t e d i n t h e a m p l i f i e r
(mW) ” )
n = P_L / P_dc *100;
disp (n , ” A m p l i f i e r e f f i c i e n c y i n p e r c e n t a g e ” )
Scilab code Exa 1.2 Gain of transistor amplifier
1
2
3
4
5
6
7
8
9
10
11
12
13
// Example 1 . 2 : Gain o f t r a n s i s t o r a m p l i f i e r
// A m p l i f i e r h a s t r a n s f e r c h a r a c t e r i s t i c s v O
=10 −(10ˆ −11) ∗ ( exp ˆ40 ∗ v 1 ) a p p l i e s f o r v 1 i s
g r e a t e r t h a n o r e q u a l 0V and v o i s g r e a t e r t h a n
o r e q u a l t o 0 . 3V
L_l = 0.3; // l i m i t L −
disp ( L_l , ” The l i m i t L − (V) =” )
v_I =1/40* log ((10 -0.3) /10^ -11) ; // from t h e t r a n s f e r
c h a r a c t e r i s t i c s and v o =0.3V
disp ( v_I , ” v I i n v o l t s =” )
L_u =10 -10^ -11; // o b t a i n e d by v I =0 i n t r a n s f e r
characteristics
disp ( L_u , ” t h e l i m i t L + (V) =” )
V_I =1/40* log ((10 -5) /10^ -11) ; // V O=5V
disp ( V_I , ” The v a l u e o f t h e dc b i a s v o l t a g e t h a t
r e s u l t s i n V O=5V (V)=” )
A_v = -10^ -11* exp (40* V_I ) *40; // A v=dv O / d v I
disp ( A_v , ” Gain a t t h e o p e r a t i n g p o i n t (V/V) =” )
disp ( ”NOTE t h e g a i n i s n e g a t i v e t h a t i m p l i e s t h e
a m p l i f i e r i s an i n v e r t i n g a m p l i f i e r ” )
Scilab code Exa 1.3 Overall voltage gain of three stage amplifier
8
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
// Example 1 . 3 : O v e r a l l v o l t a g e g a i n o f c t h r e e −
stage amplifier
gainloss_in =10^6/(1*10^6+100*10^3) ; // f r a c t i o n o f
input s i g n a l i s obtained using voltage divider
r u l e , g a i n l o s s i n= v i 1 / v s
A_v1 =10*100000/(100000+1000) ; // A v1 = v i 2 / v i 1 i s
the voltage gain at f i r s t stage
A_v2 =100*10000/(10000+1000) ; // A v2 = v i 3 / v i 2 i s
the voltage gain at second stage
A_v3 =100/(100+10) ; // A v3 = v L / v i 3 i s t h e v o l t a g e
gain at the output stage
A_v = A_v1 * A_v2 * A_v3 ; // A v i s t h e t o t a l v o l t a g e g a i n
disp ( A_v , ” The o v e r a l l v o l t a g e g a i n (V/V) =” )
disp (20* log10 ( A_v ) ,” The o v e r a l l v o l t a g e g a i n ( dB ) =”
)
gain_src_ld = A_v * gainloss_in ;
disp ( gain_src_ld , ” The v o l t a g e g a i n from s o u r c e t o
g a i n (V/V) =” )
disp (20* log10 ( gain_src_ld ) ,” The v o l t a g e g a i n from
s o u r c e t o l o a d ( dB ) =” )
A_i =10^4* A_v ; // A i= i o / i i =( v L / 1 0 0 ) / ( v i 1 / 1 0 ˆ 6 )
disp ( A_i , ” The c u r r e n t g a i n (A/A)=” )
disp (20* log10 ( A_i ) ,” The c u r r e n t g a i n ( dB ) =” )
A_p =818*818*10^4; // A p=P L / P I=v L ∗ i o / v i 1 ∗ i i
disp ( A_p , ” The power g a i n (W/W) =” )
disp (10* log10 ( A_p ) ,” The power g a i n ( dB ) =” )
Scilab code Exa 1.4 Bipolar junction transistor
1
2
3
4
// Example1 . 4 : B i p o l a r j u n c t i o n t r a n s i s t o r
// 1 , 4 a
// u s i n g v o l t a g e d i v i d e r r u l e t h e f r a c t i o n o f i n p u t
s i g n a l v b e=v s ∗ r p i / ( r p i +R s )
5 // o u t p u t v o l t a g e v o=−g m v b e ( R L | | r o )
9
r_pi =2.5*10^3; // ( ohm )
R_s =5*10^3; // ( ohm )
R_L =5*10^3 // ( ohm )
g_m =40*10^ -3; // ( mho )
r_o =100*10^3; // ( ohm )
gain = -( r_pi * g_m *( R_L * r_o /( R_L + r_o ) ) ) /( r_pi + R_s ) ; //
g a i n=v o / v s
12 disp ( gain , ” The v o l t a g e g a i n (V/V) =” )
13 gain_negl_r_o = - r_pi * g_m * R_L /( r_pi + R_s ) ;
14 disp ( gain_negl_r_o , ” Gain n e g l e c t i n g t h e e f f e c t o f
r o (V/V) =” )
6
7
8
9
10
11
15
16 // 1 . 4 b
17 // B i b=g m ∗ v b e
18 // B i s s h o r t c i r c u i t g a i n
19 B = g_m * r_pi ;
20 disp (B , ” The s h o r t c i r c u i t g a i n (A/A) =” )
Scilab code Exa 1.5 DC gain 3dB frequency and frequency at which gain
1
2
3
4
5
6
7
8
9
10
11
12
13
14
// Example 1 . 5 : DC g a i n , 3dB f r e q u e n c y and
f r e q u e n c y a t which g a i n =0 o f v o l t a g e a m p l i f i e r
// 1 . 5 b
R_s =20*10^3; // ( ohm )
R_i =100*10^3; // ( ohm )
C_i =60*10^ -12; // ( ohm )
u = 144; // (V/V)
R_o = 200; // ( ohm )
R_L = 1000; // ( ohm )
K = u /((1+ R_s / R_i ) *(1+ R_o / R_L ) ) ;
disp (K , ” The dc g a i n (V/V)= ” )
disp (20* log10 ( K ) ,” The dc g a i n ( dB ) =” )
w_o =1/( C_i * R_s * R_i /( R_s + R_i ) ) ;
disp ( w_o , ” The 3−dB f r e q u e n c y ( r a d / s ) =” )
10
15 f_o = w_o /2/ %pi ;
16 disp ( f_o , ” F r e q u e n c y ( Hz ) =” )
17 disp (100* w_o , ” u n i t y g a i n f r e q u e n c y ( r a d / s )=” ,100* f_o
, ” U n i t y g a i n f r e q u e n c y ( Hz ) ” )
Scilab code Exa 1.6 Evaluation of tPHL
1
2
3
4
5
6
7
8
9
10
11
12
// Example 1 . 6 : Time f o r t h e o u t p u t t o r e a c h (V OH+
V OL ) /2
V_DD =5; // (V)
R =1000; // ( ohm )
R_on =100; // ( ohm )
V_offset =0.1; // (V)
C =10*10^ -12; // ( F )
V_OH =5; // (V)
V_OL = V_offset +( V_DD - V_offset ) * R_on /( R + R_on ) ;
T=R*C;
v_o_t_PLH =( V_OH + V_OL ) /2; // t o f i n d t PLH
t_PLH =0.69* T ; // t PLH i s low t o h i g h p r o p o g t i o n
delay
disp ( t_PLH , ” t i m e r e q u i r e d f o r he o u t p u t t o r e a c h (
V OH+V OL ) /2 ( s e c o n d s ) =” )
11
Chapter 2
Operational Amplifiers
Scilab code Exa 2.1 Closed loop and open loop gain
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3
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21
// Example 2 . 1 : C l o s e d l o o p and open l o o p g a i n
// C o n s i d e r i n v e r t i n g c o n f i g u r a t i o n
// 2 . 1 a
R_1 =1000; // ( ohm )
R_2 =100*10^3; // ( ohm )
A =10^3; // (V/V)
disp (A , ”A (V/V) ” )
G = - R_2 / R_1 /(1+(1+ R_2 / R_1 ) / A ) ;
disp ( -G , ”G” )
e =( -G -( R_2 / R_1 ) ) /( R_2 / R_1 ) *100;
disp (e , ” e (%) ” )
v_1 =0.1; // (V)
v_1 = G * v_1 / A ;
disp ( v_1 , ” v 1 (V) ” )
A =10^4; // (V/V)
disp (A , ”A (V/V) ” )
G = - R_2 / R_1 /(1+(1+ R_2 / R_1 ) / A ) ;
disp ( -G , ”G” )
e =( -G -( R_2 / R_1 ) ) /( R_2 / R_1 ) *100;
disp (e , ” e (%) ” )
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39
40
v_1 =0.1; // (V)
v_1 = G * v_1 / A ;
disp ( v_1 , ” v 1 (V) ” )
A =10^5; // (V/V)
disp (A , ”A (V/V) ” )
G = - R_2 / R_1 /(1+(1+ R_2 / R_1 ) / A ) ;
disp ( -G , ”G” )
e =( -G -( R_2 / R_1 ) ) /( R_2 / R_1 ) *100;
disp (e , ” e (%) ” )
v_1 =0.1; // (V)
v_1 = G * v_1 / A ;
disp ( v_1 , ” v 1 (V) ” )
// 2 . 1 b
A =50000; // (V/V)
disp (A , ”A (V/V) ” )
G = - R_2 / R_1 /(1+(1+ R_2 / R_1 ) / A ) ;
disp ( -G , ”G” )
disp ( ” Thus a −50% c h a n g e i n t h e open l o o p g a i n
r e s u l t s i n o n l y −0.1% i n t h e c l o s e d l o o p g a i n ” )
Scilab code Exa 2.3 Design instrumentation amplifier
1 // Example 2 . 3 : D e s i g n i n s t r u m e n t a t i o n
2 R_2 =1 -50000 -1/1000+50;
3 disp ( R_2 , ” R 2 ( ohm ) ” )
4 R_1 =2* R_2 /999;
5 disp ( R_1 , ” R 1 ( ohm ) ” )
13
amplifier
Chapter 3
Diodes
Scilab code Exa 3.1 Peak value of diode current and maximum reverse vo
1
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3
4
5
6
// Example 3 . 1 : Peak v a l u e o f d i o d e c u r r e n t and
maximum r e v e r s e v o l t a g e
// v s i s s i n u s o i d a l i n p u t v o l t a g e w i t h peak 24V
// b a t t e r y c h a r g e s t o 12V
I_d =(24 -12) /100
max_v_rev =24+12;
disp ( I_d , ” peak v a l u e o f d i o d e c u r r e n t (A) ” ,
max_v_rev , ”maximum r e v e r s e v o l t a g e a c r o s s t h e
d i o d e (V) ” )
Scilab code Exa 3.2 Values of Iand V for the circuit given
// Example 3 . 2 : V a l u e s o f I a n d V f o r t h e c i r c u i t
given
2 disp ( ” C o n s i d e r f i g 3 . 6 ( a ) . Assume b o t h d i o d e s a r e
conducting ”)
3 I_D2 =(10 -0) /10;
4 I =(0 -( -10) ) /5 - I_D2 ; // node e q u t i o n a t B f o r f i g
3.6( a)
1
14
5 V_B =0;
6 V =0;
7 disp (I , ” I (mA)=” , V , ”V (V) ” ,” D 1 i s
8
9
10
11
12
13
14
15
16
conducting as
assumed o r i g i n a l l y ” )
disp ( ” C o n s i d e r f i g 3 . 6 ( a ) . Assume b o t h d i o d e s a r e
conducting ”)
I_D2 =(10 -0) /5;
I =(0 -( -10) ) /10 -2; // node e q u t i o n a t B f o r f i g 3 . 6 ( b
)
disp (I , ” I (mA)=” , V ,( ”V (V) ” ) )
disp ( ” I m p l i e s a s s u m p t i o n i s wrong . l e t s assume D 1
i s o f f and D 2 i s on ” )
I_D2 =(10 -( -10) ) /15;
V_B = -10+10* I_D2 ;
I =0;
disp (I , ” I (mA) ” , V_B , ”V (V) ” ,” D 1 i s r e v e r s e b i a s e d
”)
Scilab code Exa 3.3 Evaluating junction scaling constant
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// Example 3 . 3 : E v a l u a t i n g j u n c t i o n s c a l i n g c o n s t a n t
// i −I S ∗ exp ( v / ( n∗V T ) ) i m p l i e s I S=i ∗ exp (−v / ( n∗V T ) )
n =1;
i =10^ -3; // (A)
v =700; // (V)
V_T =25; // (V)
I_S = i * exp ( - v /( n * V_T ) )
disp ( I_S , ” I S (A) f o r n=1” )
n =2;
I_S = i * exp ( - v /( n * V_T ) )
disp ( I_S , ” I S (A) f o r n=2” )
disp ( ” These v a l u e s i m p l i e s I S i s 1 0 0 0 t i m e s g r e a t e r
”)
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Scilab code Exa 3.4 To determine ID and VD
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// Example 3 . 4 : To d e t e r m i n e I D and V D
V_DD =5; // (V)
R =1000; // ( ohm )
I_1 =1*10^ -3; // (A)
V_D =0.7; // (V)
V_1 = V_D ;
I_D =( V_DD - V_D ) / R ;
I_2 = I_D ;
V_2 = V_1 +0.1* log10 ( I_2 / I_1 ) ;
I_D =( V_DD - V_2 ) / R ;
disp ( I_D , ” The d i o d e c u r r e n t (A) ” )
V_D = V_2 +0.1* log10 ( I_D / I_2 )
disp ( V_D , ” The d i o d e v o l a g e (V) ” )
Scilab code Exa 3.5 Repeating example 4 using piecewise linear model
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// Example 3 . 5 : R e p e a t i n g e x a m p l e 3 . 4 u s i n g
p i e c e w i s e l i n e a r model
V_D0 =0.65; // (V)
r_D =20; // ( ohm )
R =1000; // ( ohm )
V_DD =5; // (V)
I_D =( V_DD - V_D0 ) /( R + r_D ) ;
disp ( I_D , ” I D (A) ” )
V_D = V_D0 + I_D * r_D ;
disp ( V_D , ” The d i o d v o l t a g e (V) ” )
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Scilab code Exa 3.6 Power supply ripple
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// Example 3 . 6 : Power s u p p l y r i p p l e
V_S =10; // V S=V +
V_D =0.7; // (V)
R =10*10^3; // ( ohm )
n =2;
V_T =25*10^ -3; // (V)
V_s =1; // (V)
I_D =( V_S - V_D ) / R ;
r_D = n * V_T / I_D ;
v_d = V_s * r_D /( R + r_D ) ;
disp ( v_d , ” v d ( peak (V) ) ” )
Scilab code Exa 3.7 Percentage change in regulated voltage
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// Example 3 . 7 : P e r c e n t a g e c h a n g e i n r e g u l a t e d
voltage
V_DD =10; // (V)
V_D =0.7*3; // s t r i n g o f 3 d i o d e s p r o v i d e t h i s
constant voltage
R =1*10^3;
I_D =( V_DD - V_D ) / R ;
n =2;
V_T = 25*10^ -3; // (V)
r_d = n * V_T / I_D ; // i n c r e m e n t a l r e s i s t a n c e
r =3* r_d ; // t o t a l i n c r e m e n t a l r e s i s t a n c e
deltav_O =2* r /( r + R ) ; // d e l t a v i s peak t o peak c h a n g e
in output v o l t a g e
disp ( deltav_O , ” P e r c e n t a g e c h a n g e (V) i n r e g u l a t e d
v o l t a g e c a u s e d by 10% c h a n g e i n power s u p p l y ” )
I =2.1*10^ -3; // The c u r r e n t drawn from t h e d i o d e
string
deltav_O = - I * r ; // D e c r e a s e i n v o l t a g e a c r o s s d i o d e
string
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disp ( deltav_O , ” D e c r e a s e i n v o l t a g e a c r o s s d i o d e
s t r i n g (V) ” )
Scilab code Exa 3.8 line regulation load regulation
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// Example 3 . 8 : l i n e r e g u l a t i o n l o a d r e g u l a t i o n
V_Z =6.8; // (V)
I_Z =0.005; // (A)
r_Z =20; // ( ohm )
V =10; // V=V +
R =0.5*10^3; // ( ohm )
// 3 . 8 a
V_ZO = V_Z - r_Z * I_Z ;
I_Z =( V - V_ZO ) /( R + r_Z )
V_O = V_ZO + I_Z * r_Z ;
disp ( V_O , ”V O (V) ” )
// 3 . 8 b
deltaV =1; // c h a n g e i n V i s +1 and −1
deltaV_O = deltaV * r_Z /( R + r_Z ) ; // c o r r e s p o n d i n g c h a n g e
in output v o l t a g e
18 disp ( deltaV_O , ” L i n e r e g u l a t i o n (V/V) ” )
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20 // 3 . 8 c
21 I_L =1*10^ -3; // l o a d c u r r e n t
22 deltaI_L =1*10^ -3;
23 deltaI_Z = -1*10^ -3; // c h a n g e i n z e n e r c u r r e n t
24 deltaV_O = r_Z * deltaI_Z ;
25 disp ( deltaV_O , ” Load r e g u l a t i o n (V/A) ” )
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27 // 3 . 8 d
28 I_L =6.8/2000; // l o a d c u r r e n t w i t h l o a d r e s i s t a n c e
o f 2000
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deltaI_Z = - I_L ;
deltaV_O = r_Z * deltaI_Z ;
disp ( deltaV_O , ” C o r r e s p o n d i n g c h a n g e i n z e n e r v o l t a g e
(V) f o r z e n e r c u r r e n t c h a n g e o f −3.4mA” )
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33 // 3 . 8 e
34 R_L =0.5*10^3; // ( ohm )
35 V_O = V * R_L /( R + R_L ) ;
36 disp ( V_O , ”V O (V) f o r R L =0.5K ohm” )
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38 // 3 . 8 f
39 I_Z =0.2*10^ -3; // Z e n e r t be a t t h e e d g e o f
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breakdown I Z=I ZK
V_Z =6.7; // V Z=V ZK
I_Lmin =(9 -6.7) /0.5; // Lowest c u r r e n t s u p p l i e d t o R
I_L = I_Lmin - I_Z ; // l o a d c u r r e n t
R_L = V_Z / I_L ;
disp ( R_L , ” R L ( ohm ) ” )
Scilab code Exa 3.9 Value of capacitance C that will result in peak pe
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// Example 3 . 9 : V a l u e o f c a p a c i t a n c e C t h a t w i l l
r e s u l t i n peak−peak r i p p l e o f 2V
V_P =100; // (V)
V_r =2; // (V)
f =60; // ( Hz )
R =10*10^3; // ( ohm )
I_L = V_P / R ;
C = V_P /( V_r * f * R ) ;
disp (C , ”C (V) ” )
wdeltat = sqrt (2* V_r / V_P ) ;
disp ( wdeltat , ” C o n d u c t i o n a n g l e ( r a d ) ” )
i_Dav = I_L *(1+ %pi * sqrt (2* V_P / V_r ) ) ;
disp ( i_Dav , ” i D a v (A) ” )
i_Dmax = I_L *(1+2* %pi * sqrt (2* V_P / V_r ) ) ;
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disp ( i_Dmax , ” i Dmax (A) ” )
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Chapter 4
MOS Field Effect Transistors
Scilab code Exa 4.1 To determine operating point parameters
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// Example 4 . 1 : To d e t e r m i n e o p e r a t i n g p o i n t
parameters
L_min =0.4*10^ -6; // (m)
t_ox =8*10^ -9; // ( s )
u_n =450*10^ -4; // (A/Vˆ 2 )
V_t =0.7; // (V)
e_ox =3.45*10^ -11;
// 4 . 1 a
C_ox = e_ox / t_ox ;
disp ( C_ox , ” C ox ( F/mˆ 2 ) ” )
k_n = u_n * C_ox ;
disp ( k_n , ” k n (A/Vˆ 2 ) ” )
// 4 . 1 b
// O p e r a t i o n i n s a t u r a t i o n r e g i o n
W =8*10^ -6; // (m)
L =0.8*10^ -6; // (m)
i_D =100*10^ -6; // (A)
V_GS = sqrt (2* L * i_D /( k_n * W ) ) + V_t ;
disp ( V_GS , ”V GS (V) ” )
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V_DSmin = V_GS - V_t ;
disp ( V_DSmin , ” V DSmin (V) ” )
// 4 . 1 c
// MOSFET i n t r i o d e r e g i o n
r_DS =1000; // ( ohm )
V_GS =1/( k_n *( W / L ) * r_DS ) + V_t ;
disp ( V_GS , ”V GS (V) ” )
Scilab code Exa 4.2 Design of given circuit
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// Example 4 . 2 : D e s i g n o f g i v e n c i r c u i t t o o b t a i n
I D =0.4mA and V D=0.5V
// NMOS t r a n s i s t o r i s o p e r a t i n g i n s a t u r a t i o n r e g i o n
I_D =0.4*10^ -3; // (A)
V_D =0.5; // (V)
V_t =0.7; // (V)
uC_n =100*10^ -6; // (A/Vˆ 2 )
L =1*10^ -6; // (m)
W =32*10^ -6; // (m)
V_SS = -2.5; // (V)
V_DD =2.5; // (V)
V_OV = sqrt ( I_D *2* L /( uC_n * W ) ) ;
V_GS = V_t + V_OV ;
disp ( V_GS , ” V t (V) ” )
V_S = -1.2; // (V)
R_S =( V_S - V_SS ) / I_D ;
disp ( R_S , ” R S ( ohm ) ” )
V_D =0.5; // (V)
R_D =( V_DD - V_D ) / I_D ;
disp ( R_D , ”R D ( ohm ) ” )
Scilab code Exa 4.3 Design of given circuit
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// Example 4 . 3 : D e s i g n o f g i v e n c i r c u i t t o o b t a i n
I D =80uA
// FET i s o p e r a t i n g i n s a t u r a t i o n r e g i o n
I_D =80*10^ -6; // (A)
V_t =0.6; // (V)
uC_n =200*10^ -6; // (A/Vˆ 2 )
L =0.8*10^ -6; // (m)
W =4*10^ -6; // (m)
V_DD =3; // (V)
V_OV = sqrt (2* I_D /( uC_n *( W / L ) ) ) ;
V_GS = V_t + V_OV ;
V_DS = V_GS ;
V_D = V_DS ;
disp ( V_D , ”V D (V) ” )
R =( V_DD - V_D ) / I_D ;
disp (R , ”R ( ohm ) ” )
Scilab code Exa 4.4 Design of given circuit
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// Example 4 . 4 :
Design of given c i r c u i t to obtain
V D=0.1V
// MOSFET i s o p e r a t i n g i n t r i o d e r e g i o n
V_D =0.1; // (V)
V_DD =5; // (V)
V_t =1; // (V)
K =1*10^ -3; // K=k ’ n (W/L )
V_GS =5; // (V)
V_DS =0.1; // (V)
I_D = K *(( V_GS - V_t ) * V_DS -( V_DS ^2) /2) ;
disp ( I_D , ” I D (A) ” )
R_D =( V_DD - V_D ) / I_D ;
disp ( R_D , ”R D ( ohm ) ” )
r_DS = V_DS / I_D ;
disp ( r_DS , ” r DS ( ohm ) ” )
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Scilab code Exa 4.5 To determine all node voltages and currents throug
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// Example 4 . 5 : To d e t e r m i n e a l l node v o l t a g e s and
c u r r e n t s through a l l branches
V_t =1; // (V)
K =1*10^ -3; // K=k ’ n (W/L )
V_DD =10; // (V)
R_G1 =10*10^6; // ( ohm )
R_G2 =10*10^6; // ( ohm )
R_D =6*10^3; // ( ohm )
R_S =6*10^3; // ( ohm )
p = poly ([8 -25 18] , ’ I D ’ , ’ c o e f f ’ ) ;
I_D = roots ( p ) ;
// I D =0.89mA w i l l r e s u l t i n t r a n s i s t o r c u t o f f
h e n c e we t a k e t h e o t h e r r o o t o f t h e e q u a t i o n
V_G = V_DD * R_G2 /( R_G2 + R_G1 ) ;
I_D = I_D (1) *10^ -3;
disp ( I_D , ” I D (A) ” )
V_S = I_D * R_S ;
disp ( V_S , ” V S (V) ” )
V_GS = V_G - V_S ;
disp ( V_GS , ”V GS (V) ” )
V_D = V_DD - R_D * I_D ;
disp ( V_D , ”V D (V) ” )
// V D>V G−V t t h e t r a n s i s t o r i s o p e r a t i n g i n
s a t u r a t i o n a s i n i t i a l l y assumed
Scilab code Exa 4.6 Design of given circuit
// Example 4 . 6 ;
Design of given c i r c u i t to obtain
I D =0.5mA and V D=3V
2 // MOSFET i s i n s a t u r a t i o n
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V_DD =5; // (V)
V_D =3; // (V)
I_D =0.5*10^ -3; // (A)
V_t = -1; // (V)
K =1*10^ -3; // K=k ’ n (W/L )
V_OV = sqrt (2* I_D / K ) ;
V_GS = V_t +( - V_OV )
R_D = V_D / I_D ;
V_Dmax = V_D - V_t ; // − s i g n a s m a g n i t u d e o f V t i s
considered
12 R_D = V_Dmax / I_D ;
13 disp ( R_D , ”R D ( ohm ) ” )
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Scilab code Exa 4.7 To determine drain currents and output voltage
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// Example 4 . 7 : To d e t e r m i n e d r a i n c u r r e n t s and
output v o l t a g e
K_n =1*10^ -3; // K n=k n ∗W n/ L n (A/Vˆ 2 )
K_p = 1*10^ -3; // K p=k p ∗W p/ L p (A/Vˆ 2 )
V_tn = 1; // (V)
V_tp = -1; // (V)
V_I = -2.5:2.5:2.5; // (V)
V_DD =2.5; // (V)
R =10; // ( k i l o ohm )
// For V I=0
I_DP =( K_p *( V_DD - V_tn ) ^2) /2;
I_DN = I_DP ;
disp ( I_DP , I_DN , ” I DP (A) and I DN (A) f o r V I=0V” )
disp (0 , ”V O f o r V I =0V” )
// For V I =2.5V
// I DN=K N ( V GS−V tn ) V DS
// I DN=v O /R
// S o l v i n g t h e two e q u a t i o n s we g e t
I_DN =0.244*10^ -3; // (V)
V_O = -2.44; // (V)
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20 disp ( I_DN , V_O , ”V O and I DN f o r V I =2.5V” )
21 // For V I =−2.5V Q N i s c u t o f f
22 I_DP =2.44*10^ -3; // (A)
23 V_O =2.44; // (V)
24 disp (0 , I_DP , V_O , ”V O (V) , I DP (A) and I DN (A) f o r
V I =−2.5V” )
Scilab code Exa 4.9 Design of given circuit
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// Example 4 . 9 : D e s i g n o f g i v e n c i r c u i t t o o b t a i n
I D =0.5mA
I_d =0.5*10^ -3; // (A)
I_S =0.5*10^ -3; // (A)
V_t =1:0.5:1.5; // (V)
K_n =1*10^ -3; // K n=k n ∗W/L (A/Vˆ 2 )
V_DD =15; // (V)
V_D =10; // (V)
V_S =5; // (V)
R_D =( V_DD - V_D ) / I_d ;
R_S = V_S / I_S ;
V_OV = sqrt ( I_d *2/ K_n ) ;
V_GS = V_t + V_OV ;
V_G = V_S + V_GS ;
// V t =1.5V
// I D=K( V GS−V t ) ˆ 2 / 2
// 7=V GS+10 I D
// s o l v i n g a b o v e e q u a t i o n s
I_D =0.455*10^ -3;
deltaI_D = I_D - I_d ; // Change i n I D (A)
change = deltaI_D *100/ I_d ; // Change i n %
disp ( change , ” Change i n I D (%) ” )
Scilab code Exa 4.10 Small signal analysis
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// Example 4 . 1 0 : S m a l l s i g n a l a n a l y s i s
V_t =1.5; // (V)
K =0.00025; //K= k nW/L (A/Vˆ 2 )
V_A =50; // (V)
I_D =1.06*10^ -3; // (A)
V_D =4.4; // (V)
R_D =10000; // ( ohm )
R_L =10000; // ( ohm )
V_GS = V_D ;
g_m = K *( V_GS - V_t ) ;
r_o = V_A / I_D ;
A_v = - g_m *( R_L * R_D * r_o ) /( R_D * R_L + R_D * r_o + R_L * r_o ) ;
disp ( A_v , ” V o l t a g e g a i n (V/V) ” )
R_G =10*10^6; // ( ohm )
// i i = v i ∗(1 − A v ) /R G
R_in = R_G /(1 -( A_v ) ) ;
disp ( R_in , ” I n p u t r e s i s t a n c e ( ohm ) ” )
// v DSmin=v GSmin−V t
v_i = V_t /(1+( - A_v ) ) ; // − s i g n t o make A v p o s i t i v e
disp ( v_i , ” L a r g e s t a l l o w a b l e i n p u t s i g n a l (V) ” )
Scilab code Exa 4.11 To determine all parameters of transistor amplifie
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// Example 4 . 1 1 : To d e t e r m i n e a l l p a r a m e t e r s o f
transistor amplifier
v_o =90; // (V)
v_i =9; // (V)
R_sig =100*10^3; // ( ohm )
R_L =10*10^3; // ( ohm )
v_sig =10; // (V)
A_vo = v_o / v_i ;
disp ( A_vo , ” A vo (V/V) ” )
G_vo = v_o / A_vo ;
disp ( G_vo , ” G vo (V/V) ” )
R_i = G_vo * R_sig /( A_vo - G_vo )
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disp ( R_i , ” R i ” )
disp ( ” assume R L = 10 k i l o ohm i s c o n n e c t e d ” )
v_o =70; // (V)
v_i =8; // (V)
A_v = v_o / v_i ;
disp ( A_v , ” A v (V/V) ” )
G_v = v_o / A_vo ;
disp ( G_v , ” G v (V/V) ” )
R_o = R_L *( A_vo - A_v ) / A_v ;
disp ( R_o , ” R o ( ohm ) ” )
R_out = R_L *( G_vo - G_v ) / G_v ;
disp ( R_out , ” R out ( ohm ) ” )
R_in =( v_i *100) /( v_sig - v_i ) ;
disp ( R_in , ” R i n ( ohm ) ” )
G_m = A_vo / R_o ;
disp ( G_m , ”G m ( mho ) ” )
A_i = A_v * R_in / R_L ;
disp ( A_i , ” A i (V/V) ” )
R_inL0 = R_sig /((1+ R_sig / R_i ) *( R_out / R_o ) -1) ; // R i n |
R L=0 ( ohm )
31 disp ( R_inL0 , ” R i n a t R L=0” )
32 A_is = A_vo * R_inL0 / R_o ;
33 disp ( A_is , ” A i s (A/A) ” )
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Scilab code Exa 4.12 Midband gain and upper 3dB frequency
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// Example 4 . 1 2 : Midband g a i n and u p p e r 3dB
frequency
R_sig = 100*10^3; // ( ohm )
R_G =4.7*10^6; // ( ohm )
R_D =15*10^3; // ( ohm )
R_l =15*10^3; // ( ohm )
g_m =1*10^ -3; // ( mho )
r_o =150*10^3; // ( ohm )
C_gs =1*10^ -12; // ( F )
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9 C_gd =0.4*10^ -12; // ( F )
10 R_L = 1/(1/ r_o + 1/ R_D + 1/ R_l )
11 A_M = R_G /( R_G + R_sig ) * g_m * R_L ;
12 disp ( A_M , ” midband g a i n A M (V/V) ” )
13 C_eq =(1+ g_m * R_L ) * C_gd ;
14 C_in = C_gs + C_eq ;
15 f_H =( R_G + R_sig ) /(2* %pi * C_in * R_sig * R_G ) ;
16 disp ( f_H , ” f H ( Hz ) ” )
Scilab code Exa 4.13 Coupling capacitor values
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// Example 4 . 1 3 : C o u p l i n g c a p a c i t o r v a l u e s
R_G =4.7*10^6; // ( ohm )
R_D =15*10^3; // ( ohm )
R_L =15*10^3; // ( ohm )
R_sig =100*10^3; // ( ohm )
g_m =1*10^ -3; // ( mho )
f_L =100; // ( Hz )
C_S = g_m /(2* %pi * f_L )
disp ( C_S , ” C S ( F ) ” )
f_P2 =1/(2* %pi * C_S / g_m ) ;
f_P1 =10; // ( Hz )
f_P2 =10; // ( Hz )
C_C1 =1/(2* %pi *( R_G + R_sig ) *10)
disp ( C_C1 , ” C C1 ( F ) ” )
C_C2 =1/(2* %pi *( R_D + R_L ) *10)
disp ( C_C2 , ” C C2 ( F ) ” )
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Chapter 5
Bipolar Junction Transistor
Scilab code Exa 5.1 Design of given circuit with current 2mA
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// Example 5 . 1 : D e s i g n o f g i v e n c i r c u i t w i t h
c u r r e n t 2mA
// BJT w i l l be o p e r a t i n g i n a c t i v e mode
B =100; // B i s b e t a v a l u e
a = B /( B +1) ; // a i s a l p h a v a l u e
v_BE =0.7; // v BE (V) a t i C =1mA
i_C =1*10^ -3:1*10^ -3:2*10^ -3; // (A)
I_C =2*10^ -3; // (A)
V_T =25*10^ -3; // (V)
V_C =5; // (V)
V_CC =15; // (V)
V_B =0; // (V)
V_RC = V_CC - V_C ; // V RC i s t h e v o l t a g e d r o p a c r o s s
resistor R C
R_C = V_RC / I_C ;
disp ( R_C , ” C o l l e c t o r R e s i s t a n c e R C ( ohm ) ” )
V_BE = v_BE + V_T * log ( i_C (2) / i_C (1) ) ;
disp ( V_BE , ” Base e m i t t e r v o l t a g e V BE (V) a t i C =2mA”
)
V_E = V_B - V_BE ;
disp ( V_E , ” E m i t t e r v o l t a g e V E (V) ” )
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19 I_E = I_C / a ;
20 disp ( I_E , ” E m i t t e r c u r r e n t I E (A) ” )
21 R_E =( V_E -( - V_CC ) ) / I_E ;
22 disp ( R_E , ” E m i t t e r r e s i s t a n c e R C ( ohm ) ” )
Scilab code Exa 5.2 Consider a common Emitter circuit
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// Example 5 . 2 : C o n s i d e r a common E m i t t e r c i r c u i t
I_S =10^ -15; // (A)
R_C =6.8*10^3; // ( ohm )
V_CC =10; // (V)
V_CE =3.2; // (V)
V_T =25*10^ -3; // (V)
// 5 . 2 a
I_C =( V_CC - V_CE ) / R_C ;
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
V_BE = V_T * log ( I_C / I_S ) ;
disp ( V_BE , ”V BE (V) ” )
// 5 . 2 b
V_in =5*10^ -3; // s i n u o s o i d a l i n p u t 0 f peak a m p l i t i d e
5mv
A_v = -( V_CC - V_CE ) / V_T ;
disp ( A_v , ” V o l t a g e g a i n ” )
V_o = - A_v * V_in ; // n e g a t i v e s i g n t o make p o s i t i v e
value of voltage gain
disp ( V_o , ” A m p l i t u d e o f o u t p u t v o l t a g e (V) ” )
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21 // 5 . 2 c
22 v_CE =0.3 // (V)
23 i_C =( V_CC - v_CE ) / R_C ;
24 disp ( i_C , ” i C (A) ” )
25 v_be = V_T * log ( i_C / I_C ) ; // v BE i s
i n v BE
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p o s i t i v e increment
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disp ( v_be , ” r e q u i r e d i n c r e m e n t (V) ” )
// 5 . 2 d
v_O =0.99* V_CC ;
R_C =6.8*10^3; // ( ohm )
i_C =( V_CC - v_O ) / R_C ;
I_C =1*10^ -3; // (A)
disp ( i_C , ” i C (A) ” )
v_be = V_T * log ( i_C / I_C ) ;
disp ( v_be , ” n e g a t i v e i n c r e m e n t i n v BE (V) ” )
Scilab code Exa 5.3 Determine RB
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// Example 5 . 3 : D e t e r m i n e R B
// t r a n s i s t o r i s s p e c i f i e d t o have B v a l u e i n t h e
r a n g e o f 50 t o 150
V_C =0.2; // V C=V CEsat
V_CC =10; // (V)
R_C =10^3; // ( ohm )
V_BB =5; // (V)
V_BE =0.7; // (V)
bmin =50; // r a n g e o f b e t e i s 50 t o 150
I_Csat =( V_CC - V_C ) / R_C ;
I_BEOS = I_Csat / bmin ; // I B (EOS)=I BEOS
I_B =10* I_BEOS ; // b a s e c u r r e n t f o r an o v e r d r i v e
f a c t o r 10
R_B =( V_BB - V_BE ) / I_B ;
disp ( R_B , ” V a l u e o f R B ( ohm ) ” )
Scilab code Exa 5.4 Analyse the circuit to find node voltages and bran
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// Example 5 . 4 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
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V_BB = 4; // (V)
V_CC =10; // (V)
V_BE =0.7; // (V)
b =100; // b e t a = 100
R_E =3.3*10^3; // ( ohm )
R_C =4.7*10^3; // ( ohm )
V_E = V_BB - V_BE ;
disp ( V_E , ” E m i t t e r v o l t a g e (V) ” )
I_E =( V_E -0) / R_E ;
disp ( I_E , ” E m i t t e r c u r r e n t (A) ” )
a = b /( b +1) // a l p h a v a l u e
I_C = I_E * a ;
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
V_C = V_CC - I_C * R_C ; // A p p l y i n g ohm ’ s law
disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ” )
I_B = I_E /( b +1) ;
disp ( I_B , ” Base c u r r e n t (A) ” )
Scilab code Exa 5.5 Analyse the circuit to find node voltages and bran
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// Example 5 . 5 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
disp ( ” Assuming a c t i v e mode o p e r a t i o n ” )
V_CC =10; // (V)
R_C =4.7*10^3; // (V)
R_E =3.3*10^3; // ( ohm )
V_BE =0.7; // (V)
V_BB =6; // (V)
V_CEsat =0.2; // (V)
V_E = V_BB - V_BE ;
disp ( V_E , ” E m i t t e r v o l t a g e (V) ” )
I_E = V_E / R_E ;
disp ( I_E , ” E m i t t e r c u r r e n t (A) ” )
V_C = V_CC - I_E * R_C ; // I E=I C
disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ” )
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disp ( ” S i n c e V C < V B o u r a s s u m p t i o n i s wrong \n
Hence i t s s a t u r a t i o n mode o p e r a t i o n ” )
V_E = V_BB - V_BE ;
disp ( V_E , ” E m i t t e r v o l t a g e (V) ” )
I_E = V_E / R_E ;
disp ( I_E , ” E m i t t e r c u r r e n t (A) ” )
V_C = V_E + V_CEsat ;
disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ” )
I_C =( V_CC - V_C ) / R_C ;
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
I_B = I_E - I_C ;
disp ( I_B , ” Base c u r r e n t (A) ” )
Bforced = I_C / I_B ; // t r a n s i s t o r i s made t o o p e r a t e a t
a forced beta value
disp ( Bforced , ” f o r c e d b e t a ” )
Scilab code Exa 5.7 Analyse the circuit to find node voltages and bran
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// Example 5 . 7 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
V_CC = -10; // (V)
R_E =2000; // ( ohm )
R_C =1000; // ( ohm )
V_EE =10; // (V)
V_E =0.7; // (V) e m i t t e r b a s e j u n c t i o n w i l l be
f o r w a r d b i a s e d w i t h V E=V EB=0.7V
disp ( V_E , ” E m i t t e r b a s e j u n c t i o n i s f o r w a r d b i a s e d
w i t h V E (V) ” )
I_E =( V_EE - V_E ) / R_E ;
disp ( I_E , ” E m i t t e r c u r r e n t (A) ” )
B =100; // Assuming b e t a 100
a = B /( B +1) ;
I_C = a * I_E ; // Assuming t h e t r a n s i s t o r t o o p e r a t e i n
a c t i v e mode
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
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14 V_C = V_CC + I_C * R_C ;
15 disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ” )
16 I_B = I_E /( B +1) ;
17 disp ( I_B , ” Base c u r r e n t (A) ” )
Scilab code Exa 5.8 Analyse the circuit to find node voltages and bran
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// Example 5 . 8 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
V_CC = 10; // (V)
R_C =2000; // ( ohm )
V_BB =5; // (V)
R_B =100*10^3; // ( ohm )
B =100; // b e t a v a l u e
I_B =( V_BB - V_BE ) / R_B ;
disp ( I_B , ” Base c u r r e n t (A) ” )
I_C = B * I_B ;
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
V_C = V_CC - I_C * R_C ;
disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ” )
V_B =0.7 ; // V B=V BE
disp ( V_B , ” Base v o l t a g e (V) ” )
I_E =( B +1) * I_B ;
disp ( I_E , ” E m i t t e r c u r r e n t (A) ” )
Scilab code Exa 5.9 Analyse the circuit to find node voltages and bran
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// Example 5 . 9 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
// a s s u m i n g t h a t t h e t r a n s i s t o r i s s a t u r a t e d
V_CC = -5; // (V)
V_EE =5; // (V)
R_B =10000; // ( ohm )
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R_C =10000; // ( ohm )
R_E =1000; // ( ohm )
V_EB =0.7; // (V)
V_ECsat =0.2; // (V)
// u s i n g t h e r e l a t i o n I E=I C+I B
V_B =3.75/1.2; // (V)
disp ( V_B , ” Base v o l t a g e (V) ” )
V_E = V_B + V_EB ;
disp ( V_E , ” E m i t t e r v o l t a g e (V) ” )
V_C = V_E - V_ECsat ;
disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ” )
I_E =( V_EE - V_E ) / R_E ;
disp ( I_E , ” E m i t t e r c u r r e n t (A) ” )
I_B = V_B / R_B ;
disp ( I_B , ” Base c u r r e n t (A) ” )
I_C =( V_C - V_CC ) / R_C ;
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
Bforced = I_C / I_B ; // V a l u e o f f o r c e d b e t a
disp ( Bforced , ” F o r c e d Beta v a l u e ” )
Scilab code Exa 5.10 Analyse the circuit to find node voltages and bran
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// Exampe 5 . 1 0 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
V_CC =15; // (V)
R_C =5000; // ( ohm )
R_B1 =100*10^3; // ( ohm )
R_B2 =50*10^3; // ( ohm )
R_E =3000; // ( ohm )
V_BE =0.7; // (V)
B =100; // b e t a v a l u e
V_BB = V_CC * R_B2 /( R_B1 + R_B2 ) ;
disp ( V_BB , ”V BB (V) ” )
R_BB = R_B1 * R_B2 /( R_B1 + R_B2 ) ;
disp ( R_BB , ”R BB ( ohm ) ” )
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I_B = I_E /( B +1) ;
disp ( I_B , ” Base c u r r e n t (A) ” )
I_E =( V_BB - V_BE ) /( R_E +( R_BB /( B +1) ) )
disp ( I_E , ” E m i t e r c u r r e n t (A) ” )
I_B = I_E /( B +1)
disp ( I_B , ” Base c u r r e n t (A) ” )
V_B = V_BE + I_E * R_E ;
disp ( V_B , ” Base v o l t a g e (V) ” )
a = B /( B +1) ; // a l p h a v a l u e
I_C = a * I_E
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
V_C = V_CC - I_C * R_C ;
disp ( V_C , ” C o l l e c t o r v o l t a g e (V) ) ” )
Scilab code Exa 5.11 Analyse the circuit to find node voltages and bran
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// Example 5 . 1 1 : A n a l y s e t h e c i r c u i t t o f i n d node
v o l t a g e s and b r a n c h c u r r e n t s
V_CC =15; // (V)
R_C1 =5000; // ( ohm )
R_B1 =100*10^3; // ( ohm )
R_B2 =50*10^3; // ( ohm )
R_E =3000; // ( ohm )
V_BE =0.7; // (V)
R_E2 =2000; // ( ohm )
R_C2 =2700; // ( ohm )
V_EB =0.7; // (V)
B =100; // b e t a v a l u e
V_BB = V_CC * R_B2 /( R_B1 + R_B2 ) ;
R_BB = R_B1 * R_B2 /( R_B1 + R_B2 ) ;
I_E1 =( V_BB - V_BE ) /( R_E +( R_BB /( B +1) ) )
disp ( I_E1 , ” I E 1 (A) ” )
I_B1 = I_E1 /( B +1)
disp ( I_B1 , ” I B 1 (A) ” )
V_B1 = V_BE + I_E * R_E ;
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19 disp ( V_B1 , ” V B1 (V) ” )
20 a = B /( B +1) ; // a l p h a v a l u e
21 // b e t a and a l p h a v a l u e s a r e same f o r t h e two
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transistors
I_C1 = a * I_E
disp ( I_C1 , ” IC1 (A) ” )
V_C1 = V_CC - I_C1 * R_C1 ;
disp ( V_C1 , ” V C1 (V) ) ” )
V_E2 = V_C1 + V_EB ;
disp ( V_E2 , ” V E2 (V) ” )
I_E2 =( V_CC - V_E2 ) / R_E2 ;
disp ( I_E2 , ” I E 2 (A) ” )
I_C2 = a * I_E2 ;
disp ( I_C2 , ” I C 2 (A) ” )
V_C2 = I_C2 * R_C2 ;
disp ( V_C2 , ” V C2 (V) ” )
I_B2 = I_E2 /( B +1) ;
disp ( I_B2 , ” I B 2 (A) ” )
Scilab code Exa 5.13 Design of bias network of the amplifier
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// Example 5 . 1 3 : D e s i g n o f b i a s n e t w o r k o f t h e
amplifier
I_E =1*10^ -3; // (A)
V_CC =12; // (V)
B =100; // b e t a v a l u e
V_B =4; // (V)
V_BE =0.7; // (V)
R1 =80; // ( ohm )
R2 =40; // ( ohm )
V_C =8; // (V)
V_E = V_B - V_BE ;
disp ( V_E , ” E m i t t e r v o l t a g e (V) ” )
R_E = V_E / I_E ;
disp ( R_E , ” E m i t t e r r e s i s t a n c e ( ohm ) ” )
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14 I_E =( V_B - V_BE ) /( R_E +( R1 * R2 /( R1 + R2 ) ) /( B +1) ) ;
15 disp ( I_E , ” more a c c u r a t e v a l u e f o r I E (A) f o r R1=80
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ohm and R2=40 ohm” )
R1 =8; // ( ohm )
R2 =4; // ( ohm )
I_E =( V_B - V_BE ) /( R_E +( R1 * R2 /( R1 + R2 ) ) /( B +1) ) ;
disp ( I_E , ” more a c c u r a t e v a l u e f o r I E (A) f o r R1=8
ohm and R2=4 ohm” )
R_C =( V_CC - V_C ) / I_E ; // I E=I C
disp ( R_C , ” C o l l e c t o r r e s i s t o r ( ohm ) ” )
Scilab code Exa 5.14 Analysis of transistor amplifier
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// Example 5 . 1 4 : A n a l y s i s o f t r a n s i s t o r a m p l i f i e r
V_CC =10; // (V)
R_C =3000; // ( ohm )
R_BB =100*10^3; // ( ohm )
V_BB =3; // (V)
V_BE =0.7; // (V)
V_T =25*10^ -3; // (V)
I_B =( V_BB - V_BE ) / R_BB ;
disp ( I_B , ” Base c u r r e n t (A) ” )
I_C = B * I_B ;
disp ( I_C , ” C o l l e c t o r c u r r e n t (A) ” )
V_C = V_CC - I_C * R_C ;
disp ( V_C , ” C o l l e c o r v o l t a g e (V) ” )
I_E = B * I_C /( B +1) ;
r_e = V_T / I_E ;
disp ( r_e , ” r e ( ohm ) ” )
g_m = I_C / V_T ;
disp ( g_m , ” g m ( mho ) ” )
r_pi = B / g_m ;
disp ( r_pi , ” r p i ( ohm ) ” )
// v i i s i n p u t v o l t a g e l e t u s assume i t t o be 1 V
v_i =1;
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23 v_be = v_i * r_pi /( r_pi + R_BB )
24 disp ( v_be , ” v b e ” )
25 v_o = - g_m * R_C * v_be ;
26 disp ( v_o , ” Output v o l t a g e (V) ” )
27 A_v = v_o / v_i ;
28 disp ( A_v , ” V o l t a g e g a i n ” )
Scilab code Exa 5.17 Amplifier parameters
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// Example 5 . 1 7 : A m p l i f i e r p a r a m e t e r s
// T r a n s i s t o r a m p l i f i e r i s h a v i n g a open c i r c u i t
v o l t a g e o f v s i g o f 10mV
v_sig =10*10^ -3; // (V)
R_L =10*10^3; // ( ohm )
R_sig =100*10^3; // ( ohm )
disp ( ” C a l c u l a t i o n w i t h R L i n f i n i t e ” )
v_i =9; // (V)
v_o =90; // (V)
A_vo = v_o / v_i ;
disp ( A_vo , ” A vo (V/V) ” )
G_vo = v_o / A_vo ;
disp ( G_vo , ” G vo (V/V) ” )
R_i = G_vo * R_sig /( A_vo - G_vo )
disp ( R_i , ” R i ( ohm ) ” )
disp ( ” C a l c u l a t i o n s w i t h R L = 10 k ohm” )
v_o =70*10^ -3; // (V)
v_i =8*10^ -3; // (V)
A_v = v_o / v_i ;
disp ( A_v , ” V o l t a g e g a i n A v (V/V) ” )
G_v = v_o *10^3/10;
disp ( G_v , ” G v (V/V) ” )
R_o =( A_vo - A_v ) * R_L / A_v ;
disp ( R_o , ” R o ( ohm ) ” )
R_out =( G_vo - G_v ) * R_L / G_v ;
disp ( R_out , ” R out ( ohm ) ” )
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26 R_in = v_i * R_sig /( v_sig - v_i ) ;
27 disp ( R_in , ” R i n ( ohm ) ” )
28 G_m = A_vo / R_o ;
29 disp ( G_m , ”G m (A/V) ” )
30 A_i = A_v * R_in / R_L ;
31 disp ( A_i , ” A i (A/A) ” )
32 R_ino = R_sig /((1+ R_sig / R_i ) *( R_out / R_o ) -1) ; // R i n o
i s R i n a t R L=0
33 disp ( R_ino , ” R i n a t R L =0” )
34 A_is = A_vo * R_ino / R_o ;
35 disp ( A_is , ” A i s (A/A) ” )
Scilab code Exa 5.18 Midband gain and 3dB frequency
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// Example 5 . 1 8 : Midband g a i n and 3dB f r e q u e n c y
// T r a n s i s t o r i s b i a s e d a t I C =1mA
V_CC =10; // (V)
V_EE =10; // (V)
I =0.001; // (A)
R_B =100000; // ( ohm )
R_C =8000; // ( ohm )
R_sig =5000; // ( ohm )
R_L =5000; // ( ohm )
B =100; // b e t a v a l u e
V_A =100; // (V)
C_u =1*10^ -12; // ( F )
f_T =800*10^6; // ( Hz )
I_C =0.001; // (A)
r_x =50; // ( ohm )
// V a l u e s o f h y b r i d p i model p a r a m e t e r s
g_m = I_C / V_T ;
r_pi = B / g_m ;
r_o = V_A / I_C ;
w_T =2* %pi * f_T ;
CpiplusCu = g_m / w_T ; // C u+C p i
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A_M = R_B * r_pi * g_m * R_l /(( R_B + R_sig ) *( r_pi + r_x +( R_B *
R_sig /( R_B + R_sig ) ) ) ) ;
disp ( A_M , ” Midband g a i n (V/V) ” )
R_seff =( r_pi *( r_x + R_B * R_sig /( R_B + R_sig ) ) ) /( r_pi + r_x +
R_B * R_sig /( R_B + R_sig ) ) ; // E f f e c t i v e s o u r c e
r e s i s t a n c e R s e f f =R ’ s i g
C_in = C_pi + C_u *(1+ R_l * g_m ) ;
f_H =1/(2* %pi * C_in * R_seff ) ;
disp ( f_H , ” 3dB f r e q u e n c y ( Hz ) ” )
Scilab code Exa 5.19 To select values of capacitance required
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// Example 5 . 1 9 : To s e l e c t v a l u e s o f c a p a c i t a n c e
required
R_B =100000; // ( ohm )
r_pi =2500; // ( ohm )
R_C =8000; // ( ohm )
R_L =5000; // ( ohm )
R_sig =5000; // ( ohm )
B =100; // b e t a v a l u e
g_m =0.04; // (A/V)
r_pi =2500; // ( ohm )
f_L =100; // ( Hz )
r_e =25; // ( ohm )
R_C1 = R_B * r_pi /( R_B + r_pi ) + R_sig ; // R e s i s t a n c e s e e n
by C C1
R_E = r_e + R_B * R_sig /(( R_B + R_sig ) *( B +1) ) ; // R e s i s t a n c e
s e e n by C E
R_C2 = R_C + R_L ; // R e s i s t a n c e s e e n by C C2
w_L =2* %pi * f_L ;
C_E =1/( R_E *0.8* w_L ) ; // C E i s t o c o n t r i b u t e o n l y 80%
of the value of w L
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disp ( C_E , ”C E ( F ) ” )
C_C1 =1/( R_C1 *0.1* w_L ) ; // C C1 i s t o c o n t r i b u t e o n l y
10% o f t h e v a l u e o f f L
19 disp ( C_C1 , ” C C1 ( F ) ” )
20 C_C2 =1/( R_C2 *0.1* w_L ) ; // C C2 s h o u l d c o n t r i b u t e o n l y
10% o f t h e v a l u e o f f L
21 disp ( C_C2 , ” C C2 ( F ) ” )
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Chapter 6
single stage integrated circuit
amplifiers
Scilab code Exa 6.1 To find the operating point of NMOS transistor
// Example 6 . 1 : To f i n d t h e o p e r a t i n g p o i n t o f NMOS
transistor
2 // C o n s i d e r NMOS t r a n s i s t o r
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// 6 . 1 a
I_D =100*10^ -6; // (A)
K_n =387*10^ -6*10; // K n=u n ∗ C ox (W/L ) (A/Vˆ 2 )
V_th =0.48; // (V)
V_OV = sqrt (2* I_D / K_n ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_th + V_OV ;
disp ( V_GS , ”V GS (V) ” )
// 6 . 1 b
I_C =100*10^ -6; // (A)
I_S =6*10^ -18 // (A)
V_T =0.025; // (V)
V_BE = V_T * log ( I_C / I_S ) ;
disp ( V_BE , ”V BE (V) ” )
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Scilab code Exa 6.2 Comparison between NMOS transistor and npn transis
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// Example 6 . 2 : Comparison b e t w e e n NMOS t r a n s i s t o r
and npn t r a n s i s t o r
disp ( ” For NMOS t r a n s i s t o r ” )
I_D =100*10^ -6; // (A)
V_a =5; // V’ A=V a (A)
L =0.4; // (um)
K_n =267*4/0.4*10^ -6; // K n=u n ∗ C ox ∗ (W/L ) (A/Vˆ 2 )
V_OV = sqrt (2* I_D / K_n ) ;
g_m = sqrt (2* K_n * I_D )
disp ( g_m , ” g m (A/V) ” )
disp ( ” R i n i s i n f i n i t e ” )
r_o = V_a * L / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = g_m * r_o ;
disp ( A_O , ”A O (V/V) ” )
disp ( ” For npn t r a n s i s t o r ” )
I_C =0.1*10^ -3; // c o l l e c t o r c u r r e n t
B_o =100; // b e t a v a l u e
V_A =35; // (V)
g_m = I_C / V_T ;
disp ( g_m , ” g m (A/V) ” )
R_in = B_o / g_m ;
disp ( R_in , ” R i n ( ohm ) ” )
r_o = V_A / I_C ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = g_m * r_o ;
disp ( A_O , ”A O (V/V) ” )
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Scilab code Exa 6.3 Comparison between NMOS transistor and npn transis
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// Example 6 . 3 : Comparison b e t w e e n NMOS t r a n s i s t o r
and npn t r a n s i s t o r
// For npn t r a n s i s t o r
disp ( ” For npn t r a n s i s t o r ” )
I_C =10*10^ -6; // (A)
V_T =0.025; // (V)
V_A =35; // (V)
C_jeO =5*10^ -15; // ( F )
C_uO =5*10^ -15; // ( F )
C_L =1*10^ -12; // ( F )
disp ( ” The d a t a c a l c u l a t e d f o r I C =10uA” )
g_m = I_C / V_T ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_C ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = V_A / V_T ;
disp ( A_O , ”A O (V/V) ” )
T_F =10*10^ -12;
C_de = T_F * g_m ;
disp ( C_de , ” C de ( F ) ” )
C_je =2* C_jeO ;
disp ( C_je , ” C j e ( F ) ” )
C_pi = C_de + C_je ;
disp ( C_pi , ” C p i ( F ) ” )
C_u = C_uO ;
disp ( C_u , ” C u ( F ) ” )
f_T = g_m /(2* %pi *( C_pi + C_u ) ) ;
disp ( f_T , ” f T ( Hz ) ” )
f_t = g_m /(2* %pi * C_L ) ;
disp ( f_t , ” f t ( Hz ) ” )
disp ( ” The d a t a c a l c u l a t e d f o r I C =100uA” )
I_C =100*10^ -6;
g_m = I_C / V_T ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_C ;
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disp ( r_o , ” r o ( ohm ) ” )
A_O = V_A / V_T ;
disp ( A_O , ”A O (V/V) ” )
T_F =10*10^ -12;
C_de = T_F * g_m ;
disp ( C_de , ” C de ( F ) ” )
C_je =2* C_jeO ;
disp ( C_je , ” C j e ( F ) ” )
C_pi = C_de + C_je ;
disp ( C_pi , ” C p i ( F ) ” )
C_u = C_uO ;
disp ( C_u , ” C u ( F ) ” )
f_T = g_m /(2* %pi *( C_pi + C_u ) ) ;
disp ( f_T , ” f T ( Hz ) ” )
f_t = g_m /(2* %pi * C_L ) ;
disp ( f_t , ” f t ( Hz ) ” )
disp ( ” The d a t a c a l c u l a t e d f o r I C =1mA” )
I_C =1*10^ -3;
g_m = I_C / V_T ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_C ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = V_A / V_T ;
disp ( A_O , ”A O (V/V) ” )
T_F =10*10^ -12;
C_de = T_F * g_m ;
disp ( C_de , ” C de ( F ) ” )
C_je =2* C_jeO ;
disp ( C_je , ” C j e ( F ) ” )
C_pi = C_de + C_je ;
disp ( C_pi , ” C p i ( F ) ” )
C_u = C_uO ;
disp ( C_u , ” C u ( F ) ” )
f_T = g_m /(2* %pi *( C_pi + C_u ) ) ;
disp ( f_T , ” f T ( Hz ) ” )
f_t = g_m /(2* %pi * C_L ) ;
disp ( f_t , ” f t ( Hz ) ” )
// For NMOS t r a n s i s t o r
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L =0.4*10^ -6; // (m)
C_L =1*10^ -12; // ( F )
disp ( ” The d a t a c a l c u l a t e d f o r I D = 10uA” )
I_D =10*10^ -6; // (A)
WbyL =0.12* I_D ; // WbyL=(W/L )
disp ( WbyL *10^6 , ” (W/L ) ” )
g_m =8* I_D ;
disp ( g_m , ” g m (A/V) ” )
r_o =2/ I_D ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = g_m * r_o ;
disp ( A_O , ”A O (V/V) ” )
C_gs =(2/3) * WbyL *0.4*0.4*5.8+0.6* WbyL *0.4;
disp ( C_gs , ” C g s ( fF ) ” )
C_gd =0.6* WbyL *0.4;
disp ( C_gd , ” C gd ( fF ) ” )
f_T = g_m /(2* %pi *( C_gs *10^ -15+ C_gd *10^ -15) ) ;
disp ( f_T , ” f T ( Hz ) ” )
f_t = g_m /(2* %pi * C_L )
disp ( f_t , ” f t ( Hz ) ” )
disp ( ” The d a t a c a l c u l a t e d f o r I D = 100uA” )
I_D =100*10^ -6; // (A)
WbyL =0.12* I_D ; // WbyL=(W/L )
disp ( WbyL *10^6 , ” (W/L ) ” )
g_m =8* I_D ;
disp ( g_m , ” g m (A/V) ” )
r_o =2/ I_D ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = g_m * r_o ;
disp ( A_O , ”A O (V/V) ” )
C_gs =(2/3) * WbyL *0.4*0.4*5.8+0.6* WbyL *0.4;
disp ( C_gs , ” C g s ( fF ) ” )
C_gd =0.6* WbyL *0.4;
disp ( C_gd , ” C gd ( fF ) ” )
f_T = g_m /(2* %pi *( C_gs *10^ -15+ C_gd *10^ -15) ) ;
disp ( f_T , ” f T ( Hz ) ” )
f_t = g_m /(2* %pi * C_L )
disp ( f_t , ” f t ( Hz ) ” )
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disp ( ” The d a t a c a l c u l a t e d f o r I D = 1mA” )
I_D =1*10^ -3; // (A)
WbyL =0.12* I_D ; // WbyL=(W/L )
disp ( WbyL *10^6 , ” (W/L ) ” )
g_m =8* I_D ;
disp ( g_m , ” g m (A/V) ” )
r_o =2/ I_D ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = g_m * r_o ;
disp ( A_O , ”A O (V/V) ” )
C_gs =(2/3) * WbyL *0.4*0.4*5.8+0.6* WbyL *0.4;
disp ( C_gs , ” C g s ( fF ) ” )
C_gd =0.6* WbyL *0.4;
disp ( C_gd , ” C gd ( fF ) ” )
f_T = g_m /(2* %pi *( C_gs *10^ -15+ C_gd *10^ -15) ) ;
disp ( f_T , ” f T ( Hz ) ” )
f_t = g_m /(2* %pi * C_L )
disp ( f_t , ” f t ( Hz ) ” )
Scilab code Exa 6.4 Design of the circuit with output current 100uA
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// Example 6 . 4 : D e s i g n o f t h e c i r c u i t w i t h o u t p u t
c u r r e n t =100uA
V_DD =3; // (V)
I_REF =100*10^ -6; // (A)
I_D1 =100*10^ -6; // (A)
L =1*10^ -6; // (m)
W =10*10^ -6; // (m)
V_t =0.7; // (V)
k_n =200*10^ -6; // k n=k ’ n (A/Vˆ 2 )
V_A =20; // V A=V’ A (V)
V_OV = sqrt ( I_D1 *2* L /( k_n * W ) ) ;
V_GS = V_t + V_OV ;
R =( V_DD - V_GS ) / I_REF ;
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15 disp ( V_Omin , ” V min (V) ” )
16 r_o2 = V_A / I_REF ;
17 disp ( r_o2 , ” r o 2 ( ohm ) ” )
18 V_O = V_GS ;
19 deltaV_O =1; // Change i n V O (V)
20 deltaI_O = deltaV_O / r_o2 ; // C o r r e s p o n d i n g c h a n g e i n
I O (A)
21 disp ( deltaI_O , ” The c o r r e s p o n d n g c h a n g e i n I O (A) ” )
Scilab code Exa 6.5 Determine 3dB frequency
// Example 6 . 5 : D e t e r m i n e 3dB f r e q u e n c y
// High f r e q u e n c y r e s p o n s e o f an a m p l i f i e r can be
c h a r a c t e r i z e d by t h t r a n s f e r f u n c t i o n
3 // F H ( s ) =(1− s / 1 0 ˆ 5 ) /(1+ s / 1 0 ˆ 4 ) (1+ s / 4 ∗ 1 0 ˆ 4 )
4 w_H =1/ sqrt (1/10^8+1/(16*10^8) -2/10^10) ; // w H=1/
s q r t ( 1 / w P1 ˆ2+1/ w P2 ˆ2 −2/ w Z1 ˆ2−2 w Z2 ˆ 2 )
5 disp ( w_H , ”w H ( r a d / s ) ” )
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Scilab code Exa 6.6 To determine midband gain and upper 3dB frequency
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// Example 6 . 6 : To d e t e r m i n e midband g a i n and u p p e r
3dB f r e q u e n c y
R_in =420*10^3; // ( ohm )
R_sig =100*10^3; // ( ohm )
g_m =4*10^ -3; // ( mho )
R_L =3.33*10^3; // R L=R ’ L ( ohm )
C_gs =1*10^ -12; // F
C_gd = C_gs ;
A_M = - R_in * g_m * R_L /( R_in + R_sig )
disp ( A_M , ” Midband f r e q u e n c y g a i n A M (V/V) ” )
R_gs = R_in * R_sig /( R_in + R_sig ) ;
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12 T_gs = C_gs * R_gs ; // Oen c i r c u i t
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time c o n s t a n t o f C gs
(s)
R_gd = R + R_L + g_m * R_L * R ;
T_gd = R_gd * C_gd ; // open c i r c u i t t i m e c o n s t a n t o f
C gd ( s )
w_H =1/( T_gs + T_gd ) ; // u p p e r 3dB f r e q u e n c y w H
f_H = w_H /(2* %pi ) ;
disp ( f_H , ” Upper 3dB f r e q u e n c y f H ( Hz ) ” )
Scilab code Exa 6.7 Application of miller theorem
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// Example 6 . 7 : A p p l i c a t i o n o f m i l l e r ’ s t h e o r e m
// 6 . 7 a
// By m i l l e r ’ s t h e o r e m
Z =1000*10^3; // ( ohm )
K = -100; // (V/V)
R_sig =10*10^3; // ( ohm )
Z_1 = Z /(1 - K ) ;
disp ( Z_1 , ” Z 1 ( ohm ) ” )
Z_2 = Z /(1 -(1/ K ) ) ;
disp ( Z_2 , ” Z 2 ( ohm ) ” )
VobyVsig = -100* Z_1 /( Z_1 + R_sig ) ; // VobyVsig =(V o /
V sig )
13 disp ( VobyVsig , ” ( V o / V s i g ) (V/V) ” )
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// 6 . 7 b
// A p p l y i n g m i l l e r ’ s t h e o r e m
f_3dB =1/(2* %pi *1.01*10^ -6) ;
disp ( f_3dB , ” f 3 d B ( Hz ) ” )
Scilab code Exa 6.8 Analysis of CMOS CS amplifier
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// Example 6 . 8 : A n a l y s i s o f CMOS CS a m p l i f i e r
k_n =200*10^ -6; // (A/Vˆ 2 )
W =4*10^ -6; // (m)
L =0.4*10^ -6; // (m)
I_REF =100*10^ -6; // (A)
V_An =20; // (A)
I_D1 =0.1*10^ -3; // (A)
V_Ap =10; // (V)
V_DD =3; // (V)
I_D2 =0.1*10^ -3; // (A)
V_tp =0.6; // (V)
V_tn =0.6; // (V)
g_m1 = sqrt (2* k_n *( W / L ) * I_REF ) ;
disp ( g_m1 , ” g m1 (A/V) ” )
r_o1 = V_An / I_D1 ;
disp ( r_o1 , ” r o 1 ( ohm ) ” )
r_o2 = V_Ap / I_D2 ;
disp ( r_o2 , ” r o 2 ( ohm ) ” )
A_v = - g_m1 * r_o1 * r_o2 /( r_o1 + r_o2 ) ;
disp ( A_v , ” A v ( v /V) ” )
I_D =100*10^ -6; // (A)
k_n =65*10^ -6; // (A/Vˆ 2 )
V_OV3 =0.53; // (V)
V_SG = V_tp + V_OV3 ;
disp ( V_SG , ”V SG (V) ” )
V_OA = V_DD - V_OV3 ;
disp ( V_OA , ”V OA (V) ” )
V_IB =0.93; // (V)
V_IA =0.88; // (V)
disp ( V_IA , V_IB , ” C o o r d i n a t e s o f t h e e x t r e m i t i e s o f
t h e a m p l i f i e r V IB and V IA ” )
deltavI = V_IB - V_IA ; // w i d t h o f a m p l i f i e r r e g i o n
V_OB =0.33; // (V)
deltavO = V_OB - V_OA ; // c o r r e s p o n d i n g o u t p u t r a n g e (V)
deltavObydeltavI = - deltavO / deltavI ; // L a r g e s i g n a l
v o l t a g e g a i n (V/V)
disp ( deltavObydeltavI , ” L a r g e s i g n a l v o l t a g e g a i n (V/
V) ” )
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Scilab code Exa 6.9 Analysis of CMOS CS amplifier
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// Example 6 . 9 : A n a l y s i s o f CMOS CS a m p l i f i e r
// C o n s i d e r CMOS open s o u r c e a m p l i f i e r
I_D =100*10^ -6; // (A)
I_REF = I_D ;
uC_n =387*10^ -6; // u n ∗ C ox=uC n (A/Vˆ 2 )
uC_p =86*10^ -6; // u n ∗ C ox=uC n (A/Vˆ 2 )
W =7.2*10^ -6; // (m)
L =0.36*10^ -6; // (m)
V_An =5*10^ -6; // (A)
R_sig =10*10^3; // ( ohm )
V_OV = sqrt (2* I_D * L /( W * uC_n ) ) ;
g_m = I_D /( V_OV /2) ;
disp ( g_m , ” g m (A/V) ” )
r_o1 =5*0.36/(0.1*10^ -3) ;
disp ( r_o1 , ” r o 1 ( ohm ) ” )
r_o2 =6*0.36/(.1*10^ -3) ;
disp ( r_o2 , ” r o 2 ( ohm ) ” )
R_L = r_o1 * r_o2 /( r_o1 + r_o2 ) ;
disp ( R_L , ” R L ( ohm ) ” )
A_m = - g_m * R_L ;
disp ( A_m , ”A m (V/V) ” )
C_gs =20*10^ -15; // ( F )
C_gd =5*10^ -15; // ( F )
C_in = C_gs + C_gd *(1+ g_m * R_L ) ; // u s i n g m i l l e r
equivalence
disp ( C_in , ” C i n ( F ) ” )
f_H =1/(2* %pi * C_in * R_sig ) ;
disp ( f_H , ” f H ( Hz ) ” )
R_gs =10*10^3; // ( ohm ) u s i n g open c i r c u i t
time
c o n s t a n t s methods
R_L =9.82*10^3; // ( ohm )
R_gd = R_sig *(1+ g_m * R_L ) + R_L ;
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disp ( R_gd , ” R gd ( ohm ) ” )
R_CL = R_L ;
T_gs = C_gs * R_gs ;
disp ( T_gs , ” T g s ( s ) ” )
T_gd = C_gd * R_gd ;
disp ( T_gd , ” T gd ( s ) ” )
C_L =25*10^ -15;
T_CL = C_L * R_CL ;
disp ( T_CL , ”T CL ( s ) ” )
T_H = T_gs + T_gd + T_CL ;
disp ( T_H , ”T H ( s ) ” )
f_H =1/(2* %pi * T_H ) ; // 3dB f r e q u e n c y
disp ( f_H , ” f H ( Hz ) ” )
f_Z = g_m /(2* %pi * C_gd ) ; // f r e q u e n c y o f t h e z e r o
disp ( f_Z , ” f Z ( Hz ) ” )
// D enom inat or p o l y n o m i a l
p = poly ([1 1.16*10^ -9 0.0712*10^ -18] , ’ s ’ , ’ c o e f f ’ )
disp (p , ” De nomi nat or p o l y n o m i a l ” )
s = roots ( p ) ;
f_P2 = s (2) /( -2* %pi ) ;
f_P1 = s (1) /( -2* %pi )
disp ( f_P2 , f_P1 , ” The f r e q u e n c i e s f P 1 ( Hz ) and f P 2
( Hz ) a r e f o u n d a s t h e r o o t s o f t h e d e n o m i n a t o r
frequency ”)
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54 disp ( f_H , ” A n o t h e r e s t i m a t e f o r f H ( Hz ) ” )
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Scilab code Exa 6.10 To determine AM ft fZ f3dB
1 // Example 6 . 1 0 : To d e t e r m i n e A M , f t , f Z , f 3 d B
2 // C o n s i d e r t h e CS a m p l i f i e r
3 A_M = -12.3; // (V/V) f o u n d from Example 6 . 9
4 C_L =25*10^ -15; // ( F )
5 C_gd =5*10^ -15; // ( F )
6 R_L =9.82*10^3; // ( F )
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7 g_m =1.25*10^ -3; // ( mho )
8 f_H =1/(2* %pi *( C_L + C_gd ) * R_L ) ; // 3dB f r e q u e n c y
9 disp ( f_H , ” f H ( Hz ) ” )
10 f_t = - A_M * f_H ; // Unity −g a i n f r e q u e n c y − s i g n t o make
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disp ( f_t , ” f t ( Hz ) ” )
f_Z = g_m /(2* %pi * C_gd ) ; // f r e q u e n c y o f t h e z e r o
disp ( f_Z , ” f Z ( Hz ) ” )
I_D =400*10^ -6; // I D must be q u a d r u p l e d by c h a n g i n g
I REF t o 400 uF
V_OV =0.32;
g_m = I_D /( V_OV /2) ;
disp ( g_m , ” g m (A/V) ” )
r_o1 =5*0.36/(0.4*10^ -3) ;
disp ( r_o1 , ” r o 1 ( ohm ) ” )
r_o2 =6*0.36/(0.4*10^ -3) ;
disp ( r_o2 , ” r o 2 ( ohm ) ” )
R_L =( r_o1 * r_o2 ) /( r_o1 + r_o2 ) ;
disp ( R_L , ” R L ( ohm ) ” )
A_M = - g_m * R_L ;
disp ( A_M , ”A M (V/V) ” )
f_H =1/(2* %pi *( C_L + C_gd ) * R_L ) ;
disp ( f_H , ” f H ( Hz ) ” )
f_t = f_H * - A_M ; // U n i t y g a i n f r e q u e n c y
disp ( f_t , ” f t ( Hz ) ” )
Scilab code Exa 6.11 Avo Rin Rout Gi Gis Gv fH
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// Example 6 . 1 1 : Avo Rin Rout Gi G i s Gv fH
// C o n s i d e r t h e common g a t e a m p l i f i e r
g_m =1.25*10^ -3; // (A/V)
r_o =18000; // ( ohm )
I_D =100*10^ -6; // (A)
X =0.2;
R_S =10*10^3; // ( ohm )
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R_L =100*10^3; // ( ohm )
C_gs =20*10^ -15; // ( F )
C_gd =5*10^ -15; // ( F )
C_L =0; // ( F )
gmplusgmb = g_m +0.2* g_m ; // gmplusgmb=g m+g mb
A_vo =1+( gmplusgmb ) * r_o ;
disp ( A_vo , ” A vo (V/V) ” )
R_in =( r_o + R_L ) / A_vo ;
disp ( R_in , ” R i n ( ohm ) ” )
R_out = r_o + A_vo * R_S ;
disp ( R_out , ”ohm” )
G_v = A_vo * R_L /( R_L + R_out ) ;
disp ( G_v , ” G v (V/V) ” )
G_is = A_vo * R_S / R_out ;
disp ( G_is , ” G i s (A/A) ” )
G_i = G_is * R_out /( R_out + R_L )
disp ( G_i , ” G i (A/A) ” )
R_gs = R_S * R_in /( R_S + R_in ) ;
R_gd = R_L * R_out /( R_L + R_out ) ;
T_H = C_gs * R_gs + C_gd * R_gd ;
f_H =1/(2* %pi * T_H ) ;
disp ( f_H , ” f H ( Hz ) ” )
Scilab code Exa 6.12 Comparison between Cascode amplifier and CS amplif
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// Example 6 . 1 2 : Comparison b e t w e e n C a s c o d e
a m p l i f i e r and CS a m p l i f i e r
// 6 . 1 2 a
// CS a m p l i f i e r
g_m =1.25*10^ -3;
r_o =20*10^3;
R_L = r_o * r_o /( r_o + r_o ) ;
C_gs =20*10^ -15;
R_sig =10000;
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C_gd =5*10^ -15;
C_L =5*10^ -15;
C_db =5*10^ -15;
A_o = g_m * r_o ;
disp ( A_o , ” A o (V/V) ” )
A_v = - A_o /2;
disp ( A_v , ” A v (V/V) ” )
T_H = C_gs * R_sig + C_gd *[(1+ g_m * R_L ) * R_sig + R_L ]+( C_L +
C_db ) * R_L ;
disp ( T_H , ”T H ( s ) ” )
f_H =1/(2* %pi * T_H ) ;
disp ( f_H , ” f H ( Hz ) ” )
f_t = - A_v * f_H ;
disp ( f_t , ” f t ( Hz ) ” )
// C a s c o d e a m p l i f i e r
g_m1 =1.25*10^ -3;
r_o1 =20000;
X =0.2;
r_o2 =20000;
R_L =20000;
A_o1 = g_m1 * r_o1 ;
disp ( A_o1 , ” A o1 (V/V) ” )
gm2plusgmb2 = g_m1 + X * g_m ;
A_vo2 =1+( gm2plusgmb2 ) * r_o2 ;
disp ( A_vo2 , ” A vo2 (V/V) ” )
R_out1 = r_o1 ;
R_in2 =1/( gm2plusgmb2 ) + R_L / A_vo2 ;
disp ( R_in2 , ” R i n 2 ( ohm ) ” )
R_d1 = R_out1 * R_in2 /( R_out1 + R_in2 ) ;
disp ( R_d1 , ” R d1 ( ohm ) ” )
R_out = r_o2 + A_vo2 * r_o1 ;
disp ( R_out , ” R out ( ohm ) ” )
vo1byvi = - g_m1 * R_d1 ;
disp ( vo1byvi , ” ( v o 1 / v i ) (V/V) ” )
A_v = - A_o1 * A_vo2 * R_L /( R_L + R_out ) ;
disp ( A_v , ” A v (V/V) ” )
C_gs1 =20*10^ -15;
R_sig =10*10^3;
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50 C_gd2 =5*10^ -15;
51 C_db1 =5*10^ -15;
52 T_H = R_sig *[ C_gs1 + C_gd1 *(1+ gm1Rd1 ) ]+ R_d1 *( C_gd1 + C_db1
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+ C_gs2 ) +(( R_L * R_out ) /( R_L + R_out ) ) *( C_L + C_db2 +
C_gd2 ) ;
f_H =1/(2* %pi * T_H ) ;
disp ( T_H , ”T H ( s ) ” )
disp ( f_H , ” f H ( Hz ) ” )
f_t = - A_v * f_H ;
disp ( f_t , ” f t ( Hz ) ” )
// 6 . 1 2 b
// CS a m p l i f i e r
A_v = -12.5;
R_L =10*10^3;
disp ( A_v , ” A v (V/V) ” )
T_H =( C_gd + C_L + C_db ) * R_L ;
disp ( T_H , ”T H ( s ) ” )
f_H =1/(2* %pi * T_H ) ;
disp ( f_H , ”F H ( Hz ) ” )
f_t = - A_v * f_H ;
disp ( f_t , ” f t ( Hz ) ” )
// C a s c o d e a m p l i f i e r
R_L =640*10^3;
R_out =640*10^3;
R_out1 =20*10^3;
A_v = - A_o1 * A_vo2 * R_L /( R_L + R_out ) ;
disp ( A_v , ” A v (V/V) ” )
R_in2 =1/ gm2plusgmb2 + R_L / A_vo2 ;
disp ( R_in2 , ” R i n 2 ( ohm ) ” )
R_d1 = R_in2 * R_out1 /( R_in2 + R_out1 ) ;
disp ( R_d1 , ” R d1 ( ohm ) ” )
T_H = R_d1 *( C_gd1 + C_db1 + C_gs2 ) +( R_L * R_out /( R_L + R_out ) )
*( C_L + C_gd2 + C_db2 ) ;
disp ( T_H , ”T H ( s ) ” )
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81 f_H =1/(2* %pi * T_H ) ;
82 disp ( f_H , ” f H ( Hz ) ” )
83 f_t = - A_v * f_H ;
84 disp ( f_t , ” f t ( Hz ) ” )
Scilab code Exa 6.13 Analysis of CC CE amplifier
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// Example 6 . 1 3 : A n a l y s i s o f CC−CE a m p l i f i e r
// C o n s i d e r a CC−CE a m p l i f i e r
// a t an e m i t t e r b i a s c u r r e n t o f 1mA f o r Q 1 and Q 2
g_m =40*10^ -3; // (A/V)
r_e =25; // ( ohm )
B =100; // b e t a v a l u e
C_u =2*10^ -12; // ( F )
f_T =400*10^6 // ( Hz )
r_pi = B / g_m ;
disp ( r_pi , ” r p i ( ohm ) ” )
C_pi = g_m /(2* %pi * f_T ) - C_u ;
disp ( C_pi , ” C p i ( F ) ” )
R_in2 =2500; // ( ohm )
r_pi2 =2500; // ( ohm )
r_pi1 =2500; // ( ohm )
r_e1 =0.025; // ( ohm )
B_1 =100; // b e t a v a l u e
R_in =( B_1 +1) *( r_e1 + R_in2 ) ;
disp ( R_in , ” R i n ( ohm ) ” )
R_sig =4*10^3; // ( ohm )
R_L =4000; // ( ohm )
Vb1byVsig = R_in /( R_in + R_sig ) ; // ( V b1 / V s i g )
disp ( Vb1byVsig , ” ( V b1 / V s i g ) (V/V) ” )
Vb2plusVb1 = R_in2 /( R_in2 + r_e1 ) ; // ( V b2 / V b1 )
disp ( Vb2plusVb1 , ” ( V b2 / V b1 ) (V/V) ” )
VobyVb2 = - g_m * R_L ; // ( V o / V b2 )
disp ( VobyVb2 , ” ( V o / V b2 ) (V/V) ” )
A_M = VobyVb2 * Vb2plusVb1 * Vb2plusVb1 ;
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disp ( A_M , ”A M (V/V) ” )
R_u1 = R_sig * R_in /( R_sig + R_in ) ;
disp ( R_u1 , ” R u1 ( ohm ) ” )
R_pi1 =( R_sig + R_in2 ) /(1+( R_sig / r_pi1 ) +( R_in2 / r_e1 ) ) ;
// C p i 1 s e e s a r e s i s t a n c e R p i 1
disp ( R_pi1 , ” R p i 1 ( ohm ) ” )
R_out1 =25+4000/101;
R_pi2 = R_in2 * R_out1 /( R_in2 + R_out1 ) ; // C p i 2 s e e s a
r e s i s t a n c e R pi2
disp ( R_pi2 , ” R p i 2 ( ohm ) ” )
R_u2 =(1+ g_m * R_L ) * R_pi2 + R_L ;
disp ( R_u2 , ” R u2 ( ohm ) ” )
C_u1 =2*10^ -12; // ( F )
R_u1 =3940; // ( ohm )
C_pi1 =13.9*10^ -12; // ( F )
C_u2 =2*10^ -12; // ( F )
C_pi2 =13.9*10^ -12; // ( F )
T_H = C_u1 * R_u1 + C_pi1 * R_pi1 + C_u2 * R_u2 + C_pi2 * R_pi2 ;
disp ( T_H , ”T H ( s ) ” )
f_H =1/(2* %pi * T_H ) ;
disp ( f_H , ” f H ( Hz ) ” )
A_M = r_pi *( - g_m * R_L ) /( r_pi + R_sig ) ;
disp ( A_M , ”A M (V/V) ” )
R_pi = r_pi * R_sig /( r_pi + R_sig ) ;
disp ( R_pi , ” R p i ( ohm ) ” )
R_u =(1+ g_m * R_L ) * R_pi + R_L ;
disp ( R_u , ” R u ( ohm ) ” )
T_H = C_pi * R_pi + C_u * R_u ;
disp ( T_H , ”T H ( s ) ” )
f_H =1/(2* %pi * T_H ) ;
disp ( f_H , ” f H ( Hz ) ” )
Scilab code Exa 6.14 To determine required resistor values
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// Example 6 . 1 4 : To d e t e r m i n e r e q u i r e d r e s i s t o r
values
// The c i r c u i t s g e n e r a t e a c o n s t a n t c u r r e n t I D =10
uA which o p e r a t e a t a s u p p l y o f 10V
V_BE =0.7; // (V)
V_t =0.025; // (V)
I_REF =10*10^ -6; // (A)
V_DD =10; // (V)
I =1*10^ -3; // (A)
V_BE1 = V_BE + V_t * log ( I_REF / I ) ; // V o l t a g e d r o p a c r o s s
Q 1
disp ( V_BE1 , ”V BE1 (V) ” )
R_1 =( V_DD - V_BE1 ) /( I_REF ) ; // For t h e W i d l a r c i r c u i t
we d e c i d e I REF=1mA and V BE1 =0.7V
disp ( R_1 , ” R 1 ( ohm ) ” )
R_2 =( V_DD - V_BE ) / I ;
disp ( R_2 , ” R 2 ( ohm ) ” )
R_3 =( V_t / I_REF ) * log ( I / I_REF ) ;
disp ( R_3 , ” R 3 ( ohm ) ” )
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Chapter 7
Differential and multistage
amplifier
Scilab code Exa 7.1 Analysis of differential amplifier
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// Example 7 . 1 A n a l y s i s o f d i f f e r e n t i a l
// C o n s i d e r t h e d i f f e r e n t i a l a m p l i f i e r
B =100; // b e t a v a l u e
amplifier
// 7 . 1 a
V_T =0.025; // (V)
I_E =0.0005; // (A)
R_E =150; // ( ohm )
r_e1 = V_T / I_E ; // e m i t t e r r e s i s t a n c e ( ohm )
r_e2 = r_e1 ; // e m i t t e r r e s i s t a n c e ( ohm )
r_e = r_e1 ;
R_id =2*( B +1) *( r_e + R_E ) ;
disp ( R_id , ” The i n p u t d i f f e r e n t i a l r e s i s t a n c e R i d (
ohm ) ” )
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16 R_id =40000; // ( ohm )
17 R_sig =5000; // ( ohm )
18 R_C =10000; // ( ohm )
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R_E =150; // ( ohm )
A_v = R_id /( R_id + R_sig ) ; // A v= v o / v s i g (V/V)
A_V =2* R_C /(2*( r_e + R_E ) ) ; // A V= v o / v i d (V/ v )
A_d = A_v * A_V ; // A d=v o / v s i g (V/V)
disp ( A_d , ” O v e r a l l d i f f e r e n t i a l v o l t a g e g a i n (V/V) ” )
// 7 . 1 c
R_EE =200000; // ( ohm )
deltaR_C =0.02* R_C ; // i n t h e w o r s t c a s e
A_cm = R_C * deltaR_C /(2* R_EE * R_C )
disp ( A_cm , ” Worst c a s e common mode g a i n (V/V) ” )
// 7 . 1 d
CMRR =20* log10 ( A_d / A_cm )
disp ( CMRR , ”CMRR i n dB” )
// 7 . 1 e
r_o =200000; // ( ohm )
R_icm =( B +1) *( R_EE * r_o /2) /( R_EE + r_o /2) ;
disp ( R_icm , ” I n p u t common mode r e s i s t a n c e ( ohm ) ” )
Scilab code Exa 7.2 Analysis of Active loaded MOS differential amplifi
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// Example 7 . 2 : A n a l y s i s o f A c t i v e l o a d e d MOS
differential amplifier
W =7.2*10^ -6; // (m)
L =0.36*10^ -6; // (m)
C_gs =29*10^ -15; // ( F )
C_gd =5*10^ -15; // ( F )
C_db =5*10^ -15; // ( F )
uC_n =387*10^ -6; // uC n=u n C o x (A/Vˆ 2 )
uC_p =86*10^ -6; // uC p=u p C o x (A/Vˆ 2 )
V_an =5; // V an=V’ An (V/um) (V)
V_ap =6; // V ap=V’ Ap (V/um) (V)
I =0.2*10^ -3; // (A)
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R_SS =25000; // ( ohm )
C_SS =0.2*10^ -12; // ( F )
C_S =25*10^ -15; // ( F )
K_n = uC_n * W / L ;
I_D =100*10^ -6; // b i a s c u r r e n t (A)
V_OV = sqrt (2* I_D / K_n ) ;
g_m = I / V_OV ;
g_m1 = g_m ;
g_m2 = g_m ;
r_o1 = V_an *0.36/(0.1*10^ -3) ;
r_o2 = r_o1 ;
K_p = uC_p * W / L ;
V_OV34 = sqrt (2* I_D / K_p ) ; // V OV3 , 4
g_m3 =2*0.1*10^ -3/ V_OV34 ;
g_m4 = g_m3 ;
r_o3 = V_ap *0.36/(0.1*10^ -3) ;
r_o4 = r_o3 ;
A_d = g_m *( r_o2 * r_o4 ) /( r_o2 + r_o4 ) ;
disp ( A_d , ” A d (V/V) ” )
A_cm = -1/(2* g_m3 * R_SS ) ;
disp ( A_cm , ”A cm (V/V) ” )
CMRR =20* log10 ( - A_d / A_cm ) ; // n e g a t i v e s i g n t o make
A cm p o s i t i v e
disp ( CMRR , ”CMRR i n dB” )
C_gd1 =5*10^ -15; // ( F )
C_db1 =5*10^ -15; // ( F )
C_db3 =5*10^ -15; // ( F )
C_gs3 =20*10^ -15; // ( F )
C_gs4 =20*10^ -15; // ( F )
C_m = C_gd1 + C_db1 + C_db3 + C_gs3 + C_gs4 ;
C_gd2 =5*10^ -15; // ( F )
C_db2 =5*10^ -15; // ( F )
C_gd4 =5*10^ -15; // ( F )
C_db4 =5*10^ -15; // ( F )
C_x =25*10^ -15; // ( F )
C_L = C_gd2 + C_db2 + C_gd4 + C_db4 + C_x ;
disp ( ” p o l e s and z e r o e s o f A d ” )
R_o = r_o2 * r_o4 /( r_o2 + r_o4 )
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49 f_p1 =1/(2* %pi * C_L * R_o ) ;
50 disp ( f_p1 , ” f p 1 ( Hz ) ” )
51 f_p2 = g_m3 /(2* %pi * C_m ) ;
52 disp ( f_p2 , ” f p 2 ( Hz ) ” )
53 f_Z =2* f_p2 ;
54 disp ( f_Z , ” f Z ( Hz ) ” )
55 disp ( ” Dominant p o l e o f CMRR i s
a t l o c a t i o n o f commom
−mode g a i n z e r o ” )
56 f_Z =1/(2* %pi * C_SS * R_SS ) ;
57 disp ( f_Z , ” f Z ( Hz ) ” )
Scilab code Exa 7.3 To determine all parameters for different transist
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// Example 7 . 3 : To d e t e r m i n e a l l p a r a m e t e r s f o r
different transistor
I_REF =90*10^ -6; // (A)
V_tn =0.7; // (V)
V_tp =0.8; // Magnitude i s c c o n s i d e r e d
uC_n =160*10^ -6; // uC n=u n ∗ C ox
uC_p =40*10^ -6; // uC p=u p ∗ C ox
V_A =10; // (V)
V_DD =2.5; // (V)
V_SS =2.5; // (V)
L =0.8*10^ -6; // (m)
r_o2 =222; // ( ohm )
r_o4 =222; // ( ohm )
g_m1 =0.3; // ( mho )
A_1 = - g_m1 * r_o2 * r_o4 /( r_o2 + r_o4 ) ;
disp ( A_1 , ” A 1 (V/V) ” )
r_o6 =111; // ( ohm )
r_o7 =111; // ( ohm )
g_m6 =0.6; // ( mho )
A_2 = - g_m6 * r_o6 * r_o7 /( r_o6 + r_o7 ) ;
disp ( A_2 , ” A 2 (V/V) ” )
disp ( ” For Q 1 ” )
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W =20*10^ -6; // (m)
I_D = I_REF /2; // (A)
disp ( I_D , ” I D (A) ” )
K_p = uC_p * W / L ;
V_OV = sqrt (2* I_D / K_p ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tp + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
disp ( ” For Q 2 ” )
W =20*10^ -6; // (m)
I_D = I_REF /2; // (A)
disp ( I_D , ” I D (A) ” )
K_p = uC_p * W / L ;
V_OV = sqrt (2* I_D / K_p ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tp + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
disp ( ” For Q 3 ” )
W =5*10^ -6; // (m)
I_D = I_REF /2; // (A)
disp ( I_D , ” I D (A) ” )
K_n = uC_n * W / L ;
V_OV = sqrt (2* I_D / K_n ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tn + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
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disp ( ” For Q 4 ” )
W =5*10^ -6; // (m)
I_D = I_REF /2; // (A)
disp ( I_D , ” I D (A) ” )
K_n = uC_n * W / L ;
V_OV = sqrt (2* I_D / K_n ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tn + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
disp ( ” For Q 5 ” )
W =40*10^ -6; // (m)
I_D = I_REF ; // (A)
disp ( I_D , ” I D (A) ” )
K_p = uC_p * W / L ;
V_OV = sqrt (2* I_D / K_p ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tp + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
disp ( ” For Q 6 ” )
W =10*10^ -6; // (m)
I_D = I_REF ;
disp ( I_D , ” I D (A) ” )
K_n = uC_n * W / L ;
V_OV = sqrt (2* I_D / K_n ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tn + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
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disp ( r_o , ” r o ( ohm ) ” )
disp ( ” For Q 7 ” )
W =40*10^ -6; // (m)
I_D = I_REF ;
disp ( I_D , ” I D (A) ” )
K_p = uC_p * W / L ;
V_OV = sqrt (2* I_D / K_p ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tp + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
disp ( ” For Q 8 ” )
W =40*10^ -6; // (m)
I_D = I_REF ;
disp ( I_D , ” I D (A) ” )
K_p = uC_p * W / L ;
V_OV = sqrt (2* I_D / K_p ) ;
disp ( V_OV , ”V OV (V) ” )
V_GS = V_tp + V_OV ;
disp ( V_GS , ”V GS (V) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
A_O = A_1 * A_2 ;
disp (20* log10 ( A_O ) ,” The dc open l o o p g a i n i n dB” )
v_ICMmin = -2.5+1;
disp ( v_ICMmin , ” Lower l i m i t o f i n p u t common−mode (V) ”
)
v_ICMmax =2.2 -1.1;
disp ( v_ICMmax , ” Upper l i m i t o f i n p u t common−mode (V) ”
)
v_Omax = V_DD - V_OV ;
disp ( v_Omax , ” H i g h e s t a l l o w a b l e o u t p u t v o l t a g e (V) ” )
v_Omin = - V_SS + V_OV ;
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disp ( v_Omin , ” Lowest a l l o w a b l e o u t p u t v o l t a g e (V) ” )
Scilab code Exa 7.5 Analysis of given circuit
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// Example 7 . 5 : A n a l y s i s o f g i v e n c i r c u i t
B =100; // b e t a v a l u e
I_E =0.2510^ -3; // (A)
R_1 =20000; // ( ohm )
R_2 =20000; // ( ohm )
R_3 =3000; // ( ohm )
R_4 =2300; // ( ohm )
R_5 =15700; // ( ohm )
R_6 =3000; // ( ohm )
r_e1 =25/0.25; // ( ohm )
r_e2 = r_e1 ; // ( ohm )
r_pi1 =( B +1) * r_e1 ;
r_pi2 =( B +1) * r_e2 ;
R_id = r_pi1 + r_pi2 ;
disp ( R_id , ” I n p u t d i f f e r e n t i a l r e s i s t a n c e ( ohm ) ” )
I_E =1*10^ -3;
r_e4 =25/1;
r_e5 = r_e4 ;
r_pi4 =( B +1) * r_e4 ;
r_pi5 =( B +1) * r_e5 ;
R_i2 = r_pi4 + r_pi5 ;
disp ( R_i2 , ” I n p u t r e s i s t a n c e o f t h e s e c o n d s t a g e R i 2
( ohm ) ” )
A_1 =( R_i2 *( R_1 + R_2 ) /(( R_i2 + R_1 + R_2 ) *( r_e1 + r_e2 ) ) )
disp ( A_1 , ” V o l t a g e g a i n o f t h e f i r s t s t a g e (V/V) ” )
r_e7 =25/1;
R_i3 =( B +1) *( R_4 + r_e7 ) ;
disp ( R_i3 , ” I n p u t r e s i s t a n c e o f t h e t h i r d s t a g e R i 3
( ohm ) ” )
A_2 =( - R_3 * R_i3 ) /(( R_3 + R_i3 ) *( r_e4 + r_e5 ) ) ;
disp ( A_2 , ” V o l t a g e g a i n o f t h e s e c o n d s t a g e (V/V) ” )
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r_e8 =25/5;
R_i4 =( B +1) *( r_e8 + R_6 ) ;
disp ( R_i4 , ” I n p u t r e s i s t a n c e o f t h e t h i r d s t a g e R i 2
( ohm ) ” )
A_3 =( - R_5 * R_i4 ) /(( R_5 + R_i4 ) *( r_e7 + R_4 ) ) ;
disp ( A_3 , ” V o l t a g e g a i n o f t h e t h i r d s t a g e (V/V) ” )
A_4 = R_6 /( R_6 + r_e8 ) ;
disp ( A_4 , ” V o l t a g e g a i n o f t h e f o u r t h s t a g e (V/V) ” )
A = A_1 * A_2 * A_3 * A_4 ; // A=v o / v i d (V/V)
disp (A , ” O v e r a l l o u t p u t g a i n (V/V) ” )
disp (20* log10 ( A ) ,” O v e r a l l o u t p u t g a i n i n dB” )
R_o = R_6 *( r_e8 + R_5 /( B +1) ) /( R_6 + r_e8 + R_5 /( B +1) )
disp ( R_o , ” Output r e s i s t a n c e ( ohm ) ” )
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Chapter 8
Feedback
Scilab code Exa 8.1 Analysis of op amp connected in an inverting conf
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// Example 8 . 1 : A n a l y s i s o f op amp c o n n e c t e d i n an
inverting configuration
// By i n s p e c t i o n we can w r i t e down t h e e x p r e s s i o n s
f o r A, B , c l o s e d l o o p g a i n , t h e i n p u t
r e s i s t a n c e and t h e o u t p u t r e s i s t a n c e
u =10^4; // ( ohm )
R_id =100*10^3; // ( ohm )
r_o =1000; // ( ohm )
R_L =2000; // ( ohm )
R_1 =1000; // ( ohm )
R_2 =10^6; // ( ohm )
R_S =10000; // ( ohm )
A = u *( R_L *( R_1 + R_2 ) /( R_L + R_1 + R_2 ) ) * R_id /((( R_L *( R_1 +
R_2 ) ) /( R_L + R_1 + R_2 ) + r_o ) *( R_id + R_S +( R_1 * R_2 ) /( R_1
+ R_2 ) ) )
disp (A , ” 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 (V/V) ” )
B = R_1 /( R_1 + R_2 ) ; // Beta v a l u e
disp (B , ” Beta v a l u e ” )
A_f = A /(1+ A * B ) ;
disp ( A_f , ” V o l t a g e g a i n w i t h f e e d b a c k (V/V) ” )
R_i = R_S + R_id +( R_1 * R_2 /( R_1 + R_2 ) ) // I n p u t r e s i s t a n c e
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of the A c i r c u i t in f i g 8.12 a of textbook
R_if = R_i *7;
R_in = R_if - R_S ;
disp ( R_in , ” I n p u t r e s i s t a n c e ( ohm ) ” )
R_o =1/(1/ r_o +1/ R_L +1/( R_1 + R_2 ) ) ;
R_of = R_o /(1+ A * B ) ;
R_out = R_of * R_L /( R_L - R_of ) ;
disp ( R_out , ” t h e o u t p u t r e s i s t a n c e ( ohm ) ” )
Scilab code Exa 8.2 Feedback triple
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// Example 8 . 2 : Feedback t r i p l e
// C o n s i d e r t h e g i v e n t h r e e s t a g e s e r i e s − s e r i e s
feedback
h_fe =100;
g_m2 =40*10^ -3; // (A/V)
r_e1 =41.7; // ( ohm )
a_1 =0.99; // a l p h a v a l u e
R_C1 =9000; // ( ohm )
R_E1 =100; // ( ohm )
R_F =640; // ( ohm )
R_E2 =100; // ( ohm )
r_pi2 = h_fe / g_m2 ;
R_C2 =5000; // ( ohm )
r_e3 =6.25; // ( ohm )
R_C3 =800; // ( ohm )
// F i r s t s t a g e g a i n A 1=V c1 / V i
A_1 = - a_1 * R_C1 * r_pi2 /(( R_C1 + r_pi2 ) *( r_e1 +(( R_E1 *( R_F +
R_E2 ) ) /( R_E1 + R_F + R_E2 ) ) ) )
disp ( A_1 , ” The v o l t a g e g a i n o f t h e f i r s t s t a g e (V/V) ”
)
// Gain o f t h e s e c o n d s t a g e A 2=Vc2 / V c1
A_2 = - g_m2 *{( R_C2 *( h_fe +1) /( R_C2 + h_fe +1) ) *[ r_e3 +( R_E2
*( R_F + R_E1 ) ) /( R_E2 + R_F + R_E1 ) ]}
disp ( A_2 , ” The s e c o n d s t a g e g a i n (V/V) ” )
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// T h i r d s t a g e g a i n A 3 I O / V i
A_3 =1/( r_e3 +( R_E2 *( R_F + R_E1 ) /( R_E2 + R_F + R_E1 ) ) ) ;
disp ( A_3 , ” The t h i r d s t a g e g a i n (V/V) ” )
A = A_1 * A_2 * A_3 ; // combined g a i n
disp (A , ” Combined g a i n (V/V) ” )
B = R_E1 * R_E2 /( R_E2 + R_F + R_E1 ) ;
disp (B , ” Beta v a l u e ” )
A_f = A /(1+ A * B ) ;
disp ( A_f , ” C l o s e d l o o p g a i n (A/V) ” )
A_v = - A_f * R_C3 ; // V o l t a g e g a i n
disp ( A_v , ” V o l t a g e g a i n (V/V) ” )
R_i =( h_fe +1) *( r_e1 +( R_E1 *( R_F + R_E2 ) ) /( R_E1 + R_F + R_E2 )
);
R_if = R_i *(1+ A * B ) ;
disp ( R_if , ” I n p u t r e s i s t a n c e ( ohm ) ” )
R_o =( R_E2 *( R_F + R_E1 ) /( R_F + R_E1 + R_E2 ) ) + r_e3 + R_C2 /(
h_fe +1) ;
R_of = R_o *(1+ A * B ) ;
disp ( R_of , ” Output v o l t a g e ( ohm ) ” )
r_o =25000; // ( ohm )
g_m3 =160*10^ -3; // ( mho )
r_pi3 =625; // ( ohm )
R_out = r_o +(1+ g_m3 * r_o ) * R_of * r_pi3 /( R_of + r_pi3 ) ;
disp ( R_out , ” R out ( ohm ) ” )
Scilab code Exa 8.3 Small signal analysis
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// Example 8 . 3 : S m a l l s i g n a l a n a l y s i s
B =100; // b e t a v a l u e
I_B =0.015*10^ -3; // (A)
I_C =1.5*10^ -3; // (A)
V_C =4.7; // (V)
g_m =40*10^ -3;
R_f =47000;
R_S =10000;
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10 r_pi = B / g_m ;
11 A = -358.7*10^3; // V o / I i = −g m ( R f | | R C ) ( R S | | R F | |
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r pi )
R_i =1400; // R i=R S | | R f | | r p i ( ohm )
R_o = R_C * R_f /( R_C + R_f ) ;
B = -1/ R_f ;
A_f = A /(1+ A * B ) ; // V o / I s
A_v = A_f / R_S ; // V o / V s
disp ( A_v , ” The g a i n (V/V) ” )
R_if = R_i /(1+ A * B ) ;
disp ( R_if , ” R i f ( ohm ) ” )
R_of = R_o /(1+ A * B ) ;
disp ( R_of , ” R o f ( ohm ) ” )
Scilab code Exa 8.4 Small signal analysis
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// Example 8 . 4 : S m a l l s i g n a l a n a l y s i s
R_S =10*10^3; // ( ohm )
R_B1 =100*10^3; // ( ohm )
R_B2 =15*10^3; // ( ohm )
R_C1 =10*10^3; // ( ohm )
R_E1 =870; // ( ohm )
R_E2 =3400; // ( ohm )
R_C2 =8000; // ( ohm )
R_L =1000; // ( ohm )
R_f =10000; // ( ohm )
B =100; // b e t a v a l u e
V_A =75; // (V)
A = -201.45 // I o / I i (A/A)
R_i =1535; // ( ohm )
R_o =2690; // ( ohm )
B = - R_E2 /( R_E2 + R_f ) ;
R_if = R_i /(1+ A * B ) ;
disp ( R_if )
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19 R_in =1/((1/ R_if ) -(1/ R_S ) ) ;
20 disp ( R_in , ” R i n ( ohm ) ” )
21 A_f = A /(1+ A * B ) ; // I o / I S
22 gain = R_C2 * A_f /( R_C2 + R_L ) ; // I o / I S
23 disp ( gain , ” I o / I S (A/A) ” )
24 R_of = R_o *(1+ A * B ) ; // ( ohm )
25 r_o2 =75/0.0004; // ( ohm )
26 g_m2 =0.016; // (A/V)
27 r_pi2 =6250; // ( ohm )
28 R_out = r_o2 *[1+ g_m2 *( r_pi2 * R_of /( r_pi2 + R_of ) ) ]
29 disp ( R_out , ” R out ( ohm ) ” )
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Chapter 9
Operational amplifier and data
converter circuits
Scilab code Exa 9.1 Design of two stage CMOS op amp
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// Example 9 . 1 D e s i g n o f two−s t a g e CMOS op−amp
A_v =4000; // (V/V)
V_A =20; // (V)
k_p =80*10^ -6; // k ’ n=k n (A/Vˆ 2 )
k_n =200*10^ -6; // k ’ p=k P (A/Vˆ 2 )
V_SS =1.65; // (V)
V_DD =1.65; // (V)
V_tn =0.5; // (V)
V_tp =0.5; // (V)
C_1 =0.2*10^ -12; // ( F )
C_2 =0.8*10^ -12; // ( F )
I_D =100*10^ -6; // (A)
V_OV = sqrt ( V_A ^2/ A_v ) ;
WbyL_1 = I_D *2/( V_OV ^2* k_p ) ; // WbyL 1=(W/L ) 1
disp ( WbyL_1 , ” R e q u i r e d (W/L ) r a t i o f o r Q 1 ” )
WbyL_2 = WbyL_1 ; // WbyL 2=(W/L ) 2
disp ( WbyL_2 , ” R e q u i r e d (W/L ) r a t i o f o r Q 2 ” )
WbyL_3 = I_D *2/( V_OV ^2* k_n ) ; // WbyL 3=(W/L ) 3
disp ( WbyL_3 , ” R e q u i r e d (W/L ) r a t i o f o r Q 3 ” )
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20 WbyL_4 = WbyL_3 ; // WbyL 4=(W/L ) 4
21 disp ( WbyL_4 , ” R e q u i r e d (W/L ) r a t i o f o r Q 4 ” )
22 I_D =200*10^ -6;
23 WbyL_5 = I_D *2/( V_OV ^2* k_p ) ; // WbyL 5=(W/L ) 5
24 disp ( WbyL_5 , ” R e q u i r e d (W/L ) r a t i o f o r Q 5 ” )
25 I_D =500*10^ -6;
26 WbyL_7 =2.5* WbyL_5 ; // WbyL 7=(W/L ) 7
27 disp ( WbyL_7 , ” R e q u i r e d (W/L ) r a t i o f o r Q 7 ” )
28 WbyL_6 = I_D *2/( V_OV ^2* k_n ) ; // WbyL 6=(W/L ) 6
29 disp ( WbyL_6 , ” R e q u i r e d (W/L ) r a t i o f o r Q 6 ” )
30 WbyL_8 =0.1* WbyL_5 ; // WbyL 8=(W/L ) 8
31 disp ( WbyL_8 , ” R e q u i r e d (W/L ) r a t i o f o r Q 8 ” )
32 V_ICMmin = - V_SS + V_OV + V_tn - V_tp ;
33 disp ( V_ICMmin , ” The l o w e s t v a l u e o f i n p u t common mode
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v o l t a g e ”)
V_ICMmax = V_DD - V_OV - V_OV - V_tp ;
disp ( V_ICMmax , ” The h i g h e s t v a l u e o f i n p u t common
mode v o l t a g e ” )
v_omin = - V_SS + V_OV ;
disp ( v_omin , ” The l o w e s t v a l u e o f o u t p u t s w i n g
a l l o w a b l e ”)
v_omax = V_DD - V_OV ;
disp ( v_omax , ” The h i g h e s t v a l u e o f o u t p u t s w i n g
a l l o w a b l e ”)
R_o =20/(2*0.5) ;
disp ( R_o , ” I n p u t r e s i s t a n c e i s p r a c t i c a l l y i n f i n i t e
and o u t p u t r e i s t a n c e i s ( ohm ) ” )
G_m2 =2* I_D / V_OV ;
disp ( G_m2 , ”G m2 (A/V) ” )
f_P2 =3.2*10^ -3/(2* %pi * C_2 ) ;
disp ( f_P2 , ” f P 2 ( Hz ) ” )
R =1/ G_m2 ;
disp (R , ”To move t h e t r a n s m i s s i o n z e r o t o s= i n f i n i t e
, r v a l u e s e l e c t e d a s ( ohm ) ” )
f_t = f_P2 * tand (15) ; // Phase m a r g i n o f 75 d e g r e e s ,
t h u s p h a s e s h i f t due t o s e c c o n d p o l e must be 15
degrees
disp ( f_t , ” f t ( Hz ) ” )
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50 G_m1 =2*100*10^ -6/ V_OV ; // I D = 100uA
51 C_C1 = G_m1 /(2* %pi * f_t ) ;
52 disp ( C_C1 , ” C C1 ( F ) ” )
53 SR =2* %pi * f_t * V_OV ;
54 disp ( SR , ”SR (V/ s ) ” )
Scilab code Exa 9.2 To determine Av ft fP SR and PD of folded casc
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// Example 9 . 2 : To d e t e r m i n e A v , f t , f P , SR and P D
of folded cascode amplifier
// C o n s i d e r a d e s i g n o f t h e f o l d e d −c a s c o d e op amp
I =200*10^ -6; // (A)
I_B =250*10^ -6; // (A)
V_OV =0.25; // (V)
k_n =100*10^ -6; // k n=k ’ n (A/Vˆ 2 )
k_p =40*10^ -6; // k p=k ’ p (A/Vˆ 2 )
V_A =20; // V A=V’ A (V/um)
V_DD =2.5; // (V)
V_SS =2.5; // (V)
V_t =0.75; // (V)
L =1*10^ -6; // (m)
C_L =5*10^ -12; // ( F )
disp ( ” Data c a l c u l a t e d f o r Q1” )
I_D = I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL =W/L
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q2” )
I_D = I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
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disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q3” )
I_D = I_B - I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_p * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q4” )
I_D = I_B - I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_p * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q5” )
I_D = I_B - I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q6” )
I_D = I_B - I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
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=W/L
=W/L
=W/L
=W/L
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disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q7” )
I_D = I_B - I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q8” )
I_D = I_B - I /2;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q9” )
I_D = I_B ;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_p * V_OV ^2) ; // WbyL
disp ( WbyL , ”W/L” )
disp ( ” Data c a l c u l a t e d f o r Q10” )
I_D = I_B ;
disp ( I_D , ” I D (A) ” )
g_m =2* I_D / V_OV ;
disp ( g_m , ” g m (A/V) ” )
r_o = V_A / I_D ;
disp ( r_o , ” r o ( ohm ) ” )
WbyL =2* I_D /( k_p * V_OV ^2) ; // WbyL
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=W/L
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=W/L
103 disp ( WbyL , ”W/L” )
104 disp ( ” Data c a l c u l a t e d f o r Q11” )
105 I_D = I ;
106 disp ( I_D , ” I D (A) ” )
107 g_m =2* I_D / V_OV ;
108 disp ( g_m , ” g m (A/V) ” )
109 r_o = V_A / I_D ;
110 disp ( r_o , ” r o ( ohm ) ” )
111 WbyL =2* I_D /( k_n * V_OV ^2) ; // WbyL =W/L
112 disp ( WbyL , ”W/L” )
113 gmro =160; // gmro=g m ∗ r o
114 disp ( gmro , ” g m ∗ r o f o r a l l t r a n s i s t o r s i s (V/V) ” )
115 V_GS =1;
116 disp ( V_GS , ”V GS f o r a l l t r a n s i s t o r s i s (V) ” )
117 V_ICMmin = - V_SS + V_OV + V_OV + V_t ;
118 disp ( V_ICMmin , ” The l o w e s t v a l u e o f i n p u t common mode
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v o l t a g e (V) ” )
V_ICMmax = V_DD - V_OV + V_t ;
disp ( V_ICMmax , ” The h i g h e s t v a l u e o f i n p u t common
mode v o l t a g e (V) ” )
v_omin = - V_SS + V_OV + V_OV + V_t ;
disp ( v_omin , ” The l o w e s t v a l u e o f o u t p u t s w i n g
a l l o w a b l e (V) ” )
v_omax = V_DD - V_OV - V_OV ;
disp ( v_omax , ” The h i g h e s t v a l u e o f o u t p u t s w i n g
a l l o w a b l e (V) ” )
r_o2 =200*10^3; // r o c a l c u l a t e d f o r Q2
r_o10 =80*10^3; // r o c a l c u l a t e d f o r Q10
R_o4 = gmro *( r_o2 * r_o10 ) /( r_o2 + r_o10 ) ;
r_o8 =133333; // r o c a l c u l a t e d f o r Q8
R_o6 = gmro * r_o8 ;
R_o = R_o4 * R_o6 /( R_o4 + R_o6 ) ;
disp ( R_o , ” Output r e s i s t a n c e ( ohm ) ” )
G_M =0.0008;
A_v = G_M * R_o ;
disp ( A_v , ” V o l t a g e g a i n (V/V) ” )
f_t = G_M /(2* %pi * C_L ) ;
disp ( f_t , ” U n i t y g a i n bandwidth ( Hz ) ” )
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f_P = f_t / A_v ;
disp ( f_P , ” Dominant p o l e f r e q u e n c y ( Hz ) ” )
SR = I / C_L ;
disp ( SR , ” S l e w Rate (V/ s ) ” )
I_t =0.5*10^ -3; // t o t a l c u r r e n t
V_S =5; // S u p p l y v o l t a g e
P_D = I_t * V_S ;
disp ( P_D , ” Power d i s s i p a t e d (W) ” )
Scilab code Exa 9.3 To determine input offset voltage
1 // Example 9 . 3 : To d e t e r m i n e i n p u t o f f s e t v o l t a g e
2 r_e =2.63*10^3; // ( ohm )
3 R =1000; // ( ohm )
4 I =9.5*10^ -6; // (A)
5 deltaRbyR =0.02; // 2% mismatch b e t w e e n R 1 and R 2
6 G_m1 =10^ -3/5.26; // (A/V)
7 deltaI = deltaRbyR /(1+ deltaRbyR + r_e / R ) ; // Change o f
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d e l t a I i n I E (A)
V_OS = deltaI / G_m1 ;
disp ( V_OS , ” O f f s e t v o l t a g e (V) ” )
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Chapter 10
Digital CMOS logic circuits
Scilab code Exa 10.1 To determine tPHL tPLH and tP
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// Example 1 0 . 1 : To d e t e r m i n e t PHL , t PLH and t P
// C o n s i d e r CMOS i n v e r t e r
C_ox =6*10^ -15; // ( F/umˆ 2 )
uC_n =115*10^ -6; // uC n=u n ∗ C ox (A/Vˆ 2 )
uC_p =30*10^ -6; // uC p=u p ∗ C ox (A/Vˆ 2 )
V_tn =0.4; // (V)
V_tp = -0.4; // (V)
V_DD =2.5; // (V)
W_n =0.375*10^ -6; // W f o r Q N
L_n =0.25*10^ -6; // L f o r Q N
W_p =1.125*10^ -6; // W f o r Q P
L_p =0.25*10^ -6; // L f o r Q P
C_gd1 =0.3* W_n *10^ -9; // ( F )
C_gd2 =0.3* W_p *10^ -9; // ( F )
C_db1 =10^ -15; // ( F )
C_db2 =10^ -15; // ( F )
C_g3 = 0.375*0.25*6*10^ -15+2*0.3*0.375*10^ -15; // ( F )
C_g4 =1.125*0.25*6*10^ -15+2*0.3*1.125*10^ -15; // ( F )
C_w =0.2*10^ -15; // ( F )
C =2* C_gd1 +2* C_gd2 + C_db1 + C_db2 + C_g3 + C_g4 + C_w ; // ( F )
i_DN0 = uC_n * W_n *( V_DD - V_tn ) ^2/(2* L_n ) ; // i DN0 =
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i DN ( 0 ) (A)
i_DNtPHL = uC_n * W_n *(( V_DD - V_tn ) * V_DD /2 -(( V_DD /2) ^2)
/2) / L_n ; // i DNtPHL = i DN ( t PHL ) (A)
i_DNav =( i_DN0 + i_DNtPHL ) /2; // i DN | av (A)
t_PHL = C *( V_DD /2) / i_DNav ;
disp ( t_PHL , ” t PHL ( s ) ” )
t_PLH =1.3* t_PHL ; // S i n c e W p/W n=3 and u n / u p =3.83
t h u s t PLH i s g r e a t e r t h a n t PHL by 3 . 8 3 / 3
disp ( t_PLH , ” t PLH ( s ) ” )
t_P =( t_PHL + t_PLH ) /2;
disp ( t_P , ” t P ( s ) ” )
Scilab code Exa 10.2 WbyL ratios for the logic circuit
1 // Example 1 0 . 2 : W/L r a t i o s f o r t h e l o g i c
2 // For b a s i c i n v e r t e r
3 n =1.5;
4 p =5;
5 L =0.25*10^ -6; // (m)
6 WbyL =2* n ; // W/L f o r Q NB , Q NC , Q ND
7 disp ( WbyL , ”W/L r a t i o f o r Q NB” )
8 disp ( WbyL , ”W/L r a t i o f o r Q NC” )
9 disp ( WbyL , ”W/L r a t i o f o r Q ND” )
10 WbyL = n ; // W/L r a t i o f o r Q NA
11 disp ( WbyL , ”W/L r a t i o f o r Q NA” )
12 WbyL =3* p ; // W/L f o r Q PA , Q PC , Q PD
13 disp ( WbyL , ”W/L r a t i o f o r Q PA” )
14 disp ( WbyL , ”W/L r a t i o f o r Q PC” )
15 disp ( WbyL , ”W/L r a t i o f o r Q PD” )
circuit
Scilab code Exa 10.3 To determine the parameters of pseudo NMOS inverte
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// Example 1 0 . 3 : To d e t e r m i n e t h e p a r a m e t e r s o f
p s e u d o NMOS i n v e r t e r
// C o n s i d e r a p s e u d o NMOS i n v e r t e r
uC_n =115*10^ -6; // uC n=u n ∗ C ox (A/Vˆ 2 )
uC_p =30*10^ -6; // uC p=u p ∗ C ox (A/Vˆ 2 )
V_tn =0.4; // (V)
V_tp = -0.4; // (V)
V_DD =2.5; // (V)
W_n =0.375*10^ -6; // W f o r Q N (m)
L_n =0.25*10^ -6; // L f o r Q N (m)
r =9;
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12 // 1 0 . 3 a
13 V_OH = V_DD ;
14 disp ( V_OH , ”V OH (V) ” )
15 V_OL =( V_DD - V_tn ) *(1 - sqrt (1 -1/ r ) ) ;
16 disp ( V_OL , ”V OL (V) ” )
17 V_IL = V_tn +( V_DD - V_tn ) / sqrt ( r *( r +1) ) ;
18 disp ( V_IL , ” V IL (V) ” )
19 V_IH = V_tn +2*( V_DD - V_tn ) /( sqrt (3* r ) ) ;
20 disp ( V_IH , ” V IH (V) ” )
21 V_M = V_tn +( V_DD - V_tn ) / sqrt ( r +1) ;
22 disp ( V_M , ”V M (V) ” )
23 NM_H = V_OH - V_IH ;
24 NM_L = V_IL - V_OL ;
25 disp ( NM_L , NM_H , ” The h i g h e s t and t h e l o w e s t
values of
a l l o w a b l e n o i s e m a r g i n (V) ” )
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27 // 1 0 . 3 b
28 WbyL_p = uC_n *( W_n / L_n ) /( uC_p * r ) ; // WbyL p=(W/L ) p
29 disp ( WbyL_p , ” (W/L ) p ” )
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31 // 1 0 . 3 c
32 I_stat =( uC_p * WbyL_p *( V_DD - V_tn ) ^2) /2;
33 disp ( I_stat , ” I s t a t (A) ” )
34 P_D = I_stat * V_DD ;
35 disp ( P_D , ” S t a t i c power d i s s i p a t i o n P D (W) ” )
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37 // 1 0 . 3 d
38 C =7*10^ -15;
39 t_PLH =1.7* C /( uC_p * WbyL_p * V_DD ) ;
40 disp ( t_PLH , ” t PLH ( s ) ” )
41 t_PHL =1.7* C /( uC_n *( W_n / L_n ) * sqrt (1 -0.46/ r ) * V_DD ) ;
42 disp ( t_PHL ,( ” t PHL ( s ) ” ) )
43 t_p =( t_PHL + t_PLH ) /2;
44 disp ( t_p , ” t p ( s ) ” )
Scilab code Exa 10.4 To determine parameters for NMOS transistor
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// Example 1 0 . 4 : To d e t e r m i n e p a r a m e t e r s f o r NMOS
transistor
// C o n s i d e r NMOS t r a n s i s t o r s w i t c h
uC_n =50*10^ -6; // uC n=u n ∗ C ox (A/Vˆ 2 )
‘= u p ∗ C ox (A/Vˆ 2 )
uC_p =20*10^ -6; // uC px
V_tO =1; // (V)
y =0.5; // (Vˆ 1 / 2 )
fie_f =0.6/2; // (V)
V_DD =5; // (V)
W_n =4*10^ -6; // (m)
L_n =2*10^ -6; // (m)
C =50*10^ -15; // ( F )
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13 // 1 0 . 4 a
14 V_t =1.6; // (V)
15 V_OH = V_DD - V_t ; // V OH i s
t h e v a l u e o f v O a t which
Q s t o p s c o n d u c t i n g (V)
16 disp ( V_OH , ”V OH (V) ” )
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18 // 1 0 . 4 b
19 W_p =10*10^ -6; // (m)
20 L_p =2*10^ -6; // (m)
21 i_DP = uC_p * W_p *(( V_DD - V_OH - V_tO ) ^2) /(2* L_p ) ;
22 disp ( i_DP , ” S t a t i c c u r r e n t o f t h e i n v e r t e r (A) ” )
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23 P_D = V_DD * i_DP ;
24 disp ( P_D , ” Power d i s s i p a t e d (W) ” )
25 V_O =0.08; // Output v o l t a g e (V) f o u n d by e q u a t i n g
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t h e c u r r e n t o f Q N=18uA
disp ( V_O , ” The o u t p u t v o l t a g e o f t h e i n v e r t e r (V) ” )
// 1 0 . 4 c
i_D0 = uC_n * W_n *(( V_DD - V_tO ) ^2) /(2*2*10^ -6) ; // i D 0=
i D ( 0 ) (A) c u r r e n t i D a t t =0
v_O =2.5; // (V)
V_t = V_tO +0.5*( sqrt ( v_O +2* fie_f ) - sqrt (2* fie_f ) ) ; //
a t v O =2.5V
i_DtPLH =( uC_n * W_n *( V_DD - v_O - V_t ) ^2) /(2* L_n ) ; //
i DtPLH=i D ( t PLH ) (A) c u r r e n t i D a t t=t PLH
i_Dav =( i_D0 + i_DtPLH ) /2; // i D a v=i D | av (A) a v e r a g e
discharge current
t_PLH = C *( V_DD /2) / i_Dav ;
disp ( t_PLH , ” t PHL ( s ) ” )
// 1 0 . 4 d
// Case w i t h v t g o i n g low
i_D0 = uC_n * W_n *(( V_DD - V_tO ) ^2) /(2*2*10^ -6) ; // i D 0=
i D ( 0 ) (A) c u r r e n t i D a t t =0
i_DtPHL = uC_n * W_n *(( V_DD - V_tO ) * v_O -( v_O ^2) /2) /( L_n ) ;
// i DtPHL=i D ( t PHL ) (A) c u r r e n t i D a t t=T PHL
i_Dav =( i_D0 + i_DtPHL ) /2; // i D a v=i D | av (A) a v e r a g e
discarge current
t_PHL = C *( V_DD /2) / i_Dav ;
disp ( t_PHL , ” t PHL ( s ) ” )
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45 // 1 0 . 4 e
46 t_P =( t_PHL + t_PLH ) /2;
47 disp ( t_P , ” t P ( s ) ” )
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Chapter 11
Memory and advanced digital
circuits
Scilab code Exa 11.1 Min WbyL ratio to ensure flip flop will switch
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// Example 1 0 . 1 : To d e t e r m i n e t PHL , t PLH and t P
// C o n s i d e r CMOS i n v e r t e r
C_ox =6*10^ -15; // ( F/umˆ 2 )
uC_n =115*10^ -6; // uC n=u n ∗ C ox (A/Vˆ 2 )
uC_p =30*10^ -6; // uC p=u p ∗ C ox (A/Vˆ 2 )
V_tn =0.4; // (V)
V_tp = -0.4; // (V)
V_DD =2.5; // (V)
W_n =0.375*10^ -6; // W f o r Q N
L_n =0.25*10^ -6; // L f o r Q N
W_p =1.125*10^ -6; // W f o r Q P
L_p =0.25*10^ -6; // L f o r Q P
C_gd1 =0.3* W_n *10^ -9; // ( F )
C_gd2 =0.3* W_p *10^ -9; // ( F )
C_db1 =10^ -15; // ( F )
C_db2 =10^ -15; // ( F )
C_g3 = 0.375*0.25*6*10^ -15+2*0.3*0.375*10^ -15; // ( F )
C_g4 =1.125*0.25*6*10^ -15+2*0.3*1.125*10^ -15; // ( F )
C_w =0.2*10^ -15; // ( F )
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20 C =2* C_gd1 +2* C_gd2 + C_db1 + C_db2 + C_g3 + C_g4 + C_w ; // ( F )
21 i_DN0 = uC_n * W_n *( V_DD - V_tn ) ^2/(2* L_n ) ; // i DN0 =
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i DN ( 0 ) (A)
i_DNtPHL = uC_n * W_n *(( V_DD - V_tn ) * V_DD /2 -(( V_DD /2) ^2)
/2) / L_n ; // i DNtPHL = i DN ( t PHL ) (A)
i_DNav =( i_DN0 + i_DNtPHL ) /2; // i DN | av (A)
t_PHL = C *( V_DD /2) / i_DNav ;
disp ( t_PHL , ” t PHL ( s ) ” )
t_PLH =1.3* t_PHL ; // S i n c e W p/W n=3 and u n / u p =3.83
t h u s t PLH i s g r e a t e r t h a n t PHL by 3 . 8 3 / 3
disp ( t_PLH , ” t PLH ( s ) ” )
t_P =( t_PHL + t_PLH ) /2;
disp ( t_P , ” t P ( s ) ” )
Scilab code Exa 11.2 Design of two stage CMOS op amp
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// Example 1 1 . 2 D e s i g n o f two−s t a g e CMOS op−amp
uC_n =50*10^ -6; // u n ∗ C ox (A/Vˆ 2 )
uC_p =20*10^ -6; // u p ∗ C ox (A/Vˆ 2 )
V_tn0 =1; // (V)
V_tp0 = -1; // (V)
fie_f =0.6/2; // (V)
y =0.5; // (Vˆ 1 / 2 )
V_DD =5; // (V)
W_n =4*10^ -6; // (m)
L_n =2*10^ -6; // (m)
W_p =10*10^ -6; // (m)
L_p =2*10^ -6; // (m)
W =10*10^ -6; // (m)
L =10*10^ -6; // (m)
C_B =1*10^ -12; // b i t l i n e c a p a c i t a n c e ( F )
deltaV =0.2; // 0 . 2 V d e c r e m e n t
WbyL_eq =1/( L_p / W_p + L_n / W_n ) ; // WbyL eq=(W/L ) e q
// E q u i v a l e n t t r a n s i s t o r w i l l o p e r a t e i n s a t u r a t i o n
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20 I =( uC_n * WbyL_eq *( V_DD - V_tn0 ) ^2) /2
21 r_DS =1/( uC_n *( W_n / L_n ) *( V_DD - V_tn0 ) ) ;
22 v_Q = r_DS * I ; // v Q=r DS ∗ I
23 I_5 =0.5*10^ -3; // (A)
24 deltat = C_B * deltaV / I_5 ;
25 disp ( deltat , ” The t i m e ( s ) r e q u i r e d t o d e v e l o p an
o u t p u t v o l t a g e o f 0 . 2V” )
Scilab code Exa 11.3 Time required
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// Example 1 1 . 3 : Time r e q u i r e d f o r v B t o r e a c h 4 . 5
V
// C o n s i d e r s e n s e −a m p l i f i e r c i r c u i t
uC_n =50*10^ -6; // uC n=u n ∗ C ox (A/Vˆ 2 )
uC_p =20*10^ -6; // uC p=u p ∗ C ox (A/Vˆ 2 )
W_n =12*10^ -6; // (m)
L_n =4*10^ -6; // (m)
W_p =30*10^ -6; // (m)
L_p =4*10^ -6; // (m)
v_B =4.5; // (V)
C_B =1*10^ -12; // ( F )
V_GS =2.5; // (V)
V_t =1; // (V)
deltaV =0.1; // (V)
g_mn = uC_n *( W_n / L_n ) *( V_GS - V_t ) ; // (A/V)
g_mp = uC_p *( W_p / L_p ) *( V_GS - V_t ) ; // (A/V)
G_m = g_mn + g_mp ; // (A/V)
T = C_B / G_m ; // ( s )
deltat = T *( log ( v_B / V_GS ) - log ( deltaV ) ) ;
disp ( deltat , ” The t i m e f o r v B t o r e a c h 4 . 5V ( s ) ” )
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Chapter 12
Filters and tuned amplifiers
Scilab code Exa 12.4 To design tuned amplifier
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// Example 1 2 . 4 To d e s i g n t u n e d a m p l i f i e r
cfg = -10; // C e n t e r f r e q u e n c y g a i n (V/V)
g_m =0.005; // (A/V)
r_o =10000; // ( ohm )
f_o =1*10^6; // ( Hz )
B =2* %pi *10^4; // Bandwidth
R = - cfg / g_m ;
R_L = R * r_o /( r_o - R ) ;
disp ( R_L , ” R L ( ohm ) ” )
C =1/( R * B )
disp (C , ”C ( F ) ” )
w_o =2* %pi * f_o ;
L =1/( w_o ^2* C ) ;
disp (L , ”L (H) ” )
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Chapter 14
Output Stages and amplifier
Scilab code Exa 14.1 To design a Class B Output Amplifier
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// Example 1 4 . 1 To d e s i g n a C l a s s B Output A m p l i f i e r
P_L =20; // A v e r a g e power (W)
R_L =8; // Load r e s i s t a n c e ( ohm )
V_o = sqrt (2* P_L * R_L ) ;
disp ( V_o , ” S u p p l y v o l t a g e r e q u i r e d (V) ” )
V_CC =23; // We s e l e c t t h i s v o l t a g e (V)
I_o = V_o / R_L ;
disp ( I_o , ” Peak c u r r e n t drawn from e a c h s u p p l y (A) ” )
P_Sav = V_CC * I_o / %pi ; // P S+ = P S− = P Sav
P_S = P_Sav + P_Sav ; // T o t a l s u p p l y power
disp ( P_S , ” The t o t a l power s u p p l y (W) ” )
n = P_L / P_S ; // n i s power c o n v e r s i o n e f f i c i e n c y
disp ( n *100 , ” Power c o n v e r s i o n e f f i c i e n c y %” )
P_DPmax = V_CC ^2/( %pi ^2* R_L ) ;
P_DNmax = P_DPmax ;
disp ( P_DPmax , ”Maximun power d i s s i p a t e d i n e a c h
t r a n s i s t o r (W) ” )
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Scilab code Exa 14.2 To determine quiescent current and power
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// Example 1 4 . 2 To d e t e r m i n e q u i e s c e n t c u r r e n t and
power
// C o n s i d e r C l a s s AB A m p l i f i e r
V_CC =15; // (V)
R_L =100; // ( ohm )
v_O = -10:10:10; // A m p l i t u d e o f s i n u s o i d a l o u t p u t
v o l t a g e (V)
I_S =10^ -13; // (A)
V_T =25*10^ -3; // (V)
B =50; // Beta v a l u e
i_Lmax =10/(0.1*10^3) ; // Maximum c u r r e n t t h r o u g h Q N
(A)
// I m p l i e s max b a s e c u r e n t
in Q N i s approximately
2mA
I_BIAS =3*10^ -3; // We s e l e c t I BIAS=3mA i n o r d e r t o
m a i n t a i n a minimum o f 1mA t h r o u g h t h e d i o d e s
I_Q =9*10^ -3; // The a r e a r a t i o o f 3 y e i l d s q u i e s c e n t
c u r r e n t o f 9mA
P_DQ =2* V_CC * I_Q ;
disp ( P_DQ , ” Q u i e s c e n t power d i s s i p a t i o n (W) ” )
// For v O=0V b a s e c u r r e n t o f Q N i s 9 / 5 1 = 0 . 1 8 mA
// L e a v e s a c u r r e n t o f 3 −0.18=2.83mA t o f l o w t h r o u g h
the diodes
I_S = (10^ -13) /3; // D i o d e s have I S = (1 ∗10ˆ −13) /3
V_BB =2* V_T * log ((2.83*10^ -3) / I_S ) ;
disp ( V_BB , ”V BB (V) f o r v O = 0V” )
// For v O=+10V, c u r r e n t t h r o u g h t h e d i o d e s w i l l
d e c r e a s e t o 1mA
V_BB =2* V_T * log ((1*10^ -3) / I_S ) ;
disp ( V_BB , ”V BB (V) f o r v O = +10V” )
// For v O=−10V , Q N w i l l c o n d u c t v e r y s m a l l
current thus base current i s n e g l i g i b l e
// A l l o f t h e I BIAS ( 3mA) f l o w s t h r o u g h t h e d i o d e s
V_BB =2* V_T * log ((3*10^ -3) / I_S ) ;
disp ( V_BB , ”V BB (V) f o r v O = −10V” )
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Scilab code Exa 14.3 Redesign the output stage of Example 2
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// Example 1 4 . 3 R e d e s i g n t h e o u t p u t s t a g e o f Example
14.2
V_T =25*10^ -3; // (V)
I_S =10^ -14; // (A)
I_Q =2*10^ -3; // R e q u i r e d q u i e s c e n t c u r r e n t (A)
// We s e l e c t I BIAS=3mA which i s d i v i d e d b e t w e e n I R
and I C 1
// Thus we s e l e c t I R =0.5mA and I C 1 =2.5mA
V_BB =2* V_T * log ( I_Q /10^ -13) ;
disp ( V_BB , ”V BB (V) ” )
I_R =0.5*10^ -3;
R1plusR2 = V_BB / I_R ; // R1plusR2 = R 1+R 2
I_C1 =2.5*10^ -3;
V_BE1 = V_T * log ( I_C1 / I_S ) ;
disp ( V_BE1 , ”V BE1 (V) ” )
R_1 = V_BE1 / I_R ;
disp ( R_1 , ” R 1 ( ohm ) ” )
R_2 = R1plusR2 - R_1 ;
disp ( R_2 , ” R 2 ( ohm ) ” )
Scilab code Exa 14.4 To determine thermal resistance junction temperat
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// Example 1 4 . 4 To d e t e r m i n e t h e r m a l r e s i s t a n c e ,
junction temperature
// C o n s i d e r BJT w i t h f o l l o w i n g
specifications
P_D0 =2; // Maximum power d i s s i p a t i o n (W)
T_A0 =25; // Ambient t e m p e r a t u r e ( d e g r e e c e l c i u s )
T_Jmax =150; // maximum j u n c t i o n t e m p e r a t u r e ( d e g r e e
celcius )
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// 1 4 . 4 a
theta_JA =( T_Jmax - T_A0 ) / P_D0 ; // Thermal r e s i s t a n c e
disp ( theta_JA , ” The t h e r m a l r e s i s t a n c e ( d e g r e e
c e l s i u s /W) ” )
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11 // 1 4 . 4 b
12 T_A =50; // ( d e g r e e c e l c i u s )
13 P_Dmax =( T_Jmax - T_A ) / theta_JA ;
14 disp ( P_Dmax , ”Maximum power t h a t can be d i s s i p a t e d
at
an a m b i e n t t e m p e r a t u r e o f 50 d e g r e e c e l s i u s (W) ”
)
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16 // 1 4 . 4 c
17 T_A =25; // ( d e g r e e c e l c i u s )
18 P_D =1; // (W)
19 T_J = T_A + theta_JA * P_D ;
20 disp ( T_J , ” J u n c t i o n t e m p e r a t u r e ( d e g r e e
celcius ) if
t h e d e v i c e i s o p e r a t i n g a t T A=25 d e g r e e c e l s i u s
and i s d i s s i p a t i n g 1W” )
Scilab code Exa 14.5 To determine the maximum power dissipated
// Example 1 4 . 5 To d e t e r m i n e t h e maximum power
dissipated
2 // C o n s i d e r a BJT w i t h f o l l o w i n g s p e c i f i c a t i o n s
3 T_Jmax =150; // ( d e g r e e c e l c i u s )
4 T_A =50; // ( d e g r e e c e l c i u s )
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// 1 4 . 5 a
theta_JA =62.5; // ( d e g r e e c e l c i u s /W)
P_Dmax =( T_Jmax - T_A ) / theta_JA ;
disp ( P_Dmax , ” The maximum power (W) t h a t can be
d i s s i p a t e d s a f e l y by t h e t r a n s i s t o r when o p e r a t e d
in f r e e a i r ”)
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// 1 4 . 5 b
theta_CS =0.5; // ( d e g r e e c e l c i u s /W)
theta_SA =4; // ( d e g r e e c e l c i u s /W)
theta_JC =3.12; // ( d e g r e e c e l c i u s /W)
theta_JA = theta_JC + theta_CS + theta_SA ;
P_Dmax =( T_Jmax - T_A ) / theta_JA
disp ( P_Dmax , ” The maximum power (W) t h a t can be
d i s s i p a t e d s a f e l y by t h e t r a n s i s t o r when o p e r a t e d
a t an a m b i e n t t e m p e r a t u r e o f 50 d e g r e e c e l c i u s
but w i t h a h e a t s i n k f o r which t h e t a C S= 0 . 5 (
d e g r e e c e l c i u s /W) and t h e t a S A = 4 ( d e g r e e
c e l c i u s /W) (W) ” )
// 1 4 . 5 c
theta_CA =0 // s i n c e i n f i n i t e h e a t s i n k
P_Dmax =( T_Jmax - T_A ) / theta_JC ;
disp ( P_Dmax , ” The maximum power (W) t h a t can be
d i s s i p a t e d s a f e l y i f an i n f i n i t e h e a t s i n k i s
u s e d and T A=50 ( d e g r e e c e l c i u s ) ” )
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