CHAPTER 10 - FEEDBACK AMPLIFIERS

Example E01 - Pg 243

In [1]:
#  Exa 10.1
#  Given data
A = 60.;#  in dB
A= 10.** (A/20.)
Beta = 0.005;
dAbyA = -12./100.;
#  On putting the value of A, bita and dA/A
dAfbyAf = (1./(1.+A*Beta))*(dAbyA);
print '%s %.2f' %("The change in overall gain is",dAfbyAf);

The change in overall gain is -0.02

Example E02 - Pg 244

In [2]:
#  Exa 10.2
#  Given data
A = 1000.;
Zi = 1.;#  in k ohm
Zi = Zi * 10.** 3.;#  in ohm
Beta = 0.01;
Zdesh_i = (1.+A*Beta)*Zi;#  in ohm
Zdesh_i =Zdesh_i *10.** -3.;#  in k ohm
print '%s %.2f' %("The input impedance of the feedback amplifier in k ohm is",Zdesh_i);

The input impedance of the feedback amplifier in k ohm is 11.00

Example E03 - Pg 248

In [3]:
#  Exa 10.3
#  Given data
A = 60.;#  in dB
A= 10.** (A/20.);
Zo = 12000.;#  in ohm
Zdesh_o = 600.;#  in ohm
# Zdesh_o = Zo/(1+(A*Beta));
Beta = (((Zo/Zdesh_o)-1.)/A)*100.;#  in %
print '%s %.2f' %("The feedback factor in % is",Beta);
Beta = Beta/100.;
DAbyA = 0.1;
dAfbyAf = (1./(1. + (A*Beta)))*DAbyA*100.;#  in %
print '%s %.2f' %("The percentage change in the overall gain in % is",dAfbyAf);

The feedback factor in % is 1.90
The percentage change in the overall gain in % is 0.50

Example E04 - Pg 254

In [4]:
#  Exa 10.4
#  Given data
A = 100.;
Beta = 1./10.;
Af = A/(1. + (A*Beta));
print '%s %.2f' %("The gain of negative feedback amplifier is",Af);

The gain of negative feedback amplifier is 9.09

Example E05 - Pg 255

In [5]:
#  Exa 10.5
#  Given data
Af = 100.;
Vi = 0.6;#  in V
Vdesh_o = Af*Vi;#  in V
Vi = 50.;#  in mV
Vi = Vi * 10.** -3.;#  in V
A = Vdesh_o/Vi;
print '%s %.2f' %("The value of A is",A);
#  Af = A/( 1 +(A*Beta) );
Beta = (((A/Af)-1.)/A)*100.;#  in %
Beta= (A-Af)/(Af*A/100.);
Beta= Beta*100.;#  in %
print '%s %.2f' %("The value of Beta in % is",Beta);

The value of A is 1200.00
The value of Beta in % is 91.67

Example E06 - Pg 255

In [6]:
#  Exa 10.6
#  Given data
A = 1000.;
Af = A - (0.40*1000.);
# Af = A/( 1+(A*Beta) );
Beta = ((A/Af)-1.)/A;
A_desh = 800.;
A_desh_f= A_desh/( 1.+(A_desh*Beta) );
print '%s %.2f' %("The voltage gain with feedback is",A_desh_f);
#  percentage reduction without feedback 
P = ((A-A_desh)/A)*100.;# in %
#  percentage reduction with feedback 
P1 = ((Af-A_desh_f)/Af)*100.;#  in %
print '%s %.2f' %("The percentage reduction with feedback in % is",P1);

The voltage gain with feedback is 521.74
The percentage reduction with feedback in % is 13.04

Example E07 - Pg 257

In [7]:
#  Exa 10.7
#  Given data
dAbyA = 10./100.;
A = 200.;
Beta = 0.25;
#  Af = A/(1+(A*Beta))      (i)
#  differentiating w.r.to A we get, dAf = dA/((1+(Beta*A))** 2)   (ii)
#  From eq(i) and (ii)
dAfbyAf =  1./(1.+A*Beta)*dAbyA
print '%s %.2e' %("The small change in gain is",dAfbyAf);

The small change in gain is 1.96e-03

Example E08 - Pg 259

In [8]:
#  Exa 10.8
#  Given data
A = 100.;
Beta = 1./25.;
Af = A/(1. + (A*Beta));
print '%s %.2f' %("The gain with feedback is",Af);
print '%s %.2f' %("The feed back factor is",A*Beta);
Vi = 50.;#  in mV
Vo =Af*Vi*10** -3;#  in V
print '%s %.2f' %("The output voltage in V is",Vo);
V_feedback= (Beta*Vo);#  feedback voltage in V
print '%s %.2f' %("The feed back voltage in V is",V_feedback);
Vi = Vi*(1+(A*Beta));#  in mV
print '%s %.2f' %("The new input voltage in mV is",Vi);

The gain with feedback is 20.00
The feed back factor is 4.00
The output voltage in V is 1.00
The feed back voltage in V is 0.04
The new input voltage in mV is 250.00

Example E09 - Pg 260

In [9]:
#  Exa 10.9
#  Given data
Beta = 0.25;
A = 100.;
dA= 10.;#  in %
#  Af = A/(1+(A*Beta))          (i)
# dAf = dA/((1+(Beta*A))** 2) (ii)
#  From eq (i) and (ii)
dAbyA = dA/A;
print '%s %.2f' %("The small change in gain is",dAbyA);

The small change in gain is 0.10

Example E10 - Pg 266

In [10]:
#  Exa 10.10
#  Given data
A = 200.;
Beta = 5./100.;
Af =A/(1. + (A*Beta));
print '%s %.2f' %("The gain of the amplifier with negative feedback is : ",Af)
Dn = 10.;#  in %
Ddesh_n = Dn/(1.+(A*Beta));#  in %
print '%s %.2f' %("The distortion with negative feedback in % is : ",Ddesh_n);

#  Note: In the book, the calculation to find the gain of the amplifier with negative feedback i.e Af is wrong.

The gain of the amplifier with negative feedback is :  18.18
The distortion with negative feedback in % is :  0.91

Example E11 - Pg 269

In [11]:
#  Exa 10.11
#  Given data
Af = 10.;
A = 50.;
#  Af =A/(1 + (A*Beta) );
Beta = ((A/Af)-1.)/A*100.;#  in %
dAfByAf = 1./( 1.+100./4. )*Af/100.;
print '%s %.2e' %("The percentage of feedback is",dAfByAf);

The percentage of feedback is 3.85e-03

Example E12 - Pg 273

In [12]:
#  Exa 10.12
#  Given data
Ao = 100.;
f_L = 20.;#  in Hz
f_H = 40.;#  in kHz
f_H = f_H*10.** 3.;#  in Hz
Beta = 0.1;
Af = Ao/(1. + (Beta*Ao));
print '%s %.2f' %("The overall gain at mid frequency is",Af);
f_Hf = f_H*(1.+(Ao*Beta));#  in Hz
f_Hf = f_Hf * 10.** -3.;#  in kHz
print '%s %.2f' %("The upper cutoff frequency with negative feedback in kHz is",f_Hf);
f_Lf = f_L/(1.+(Ao*Beta));#  in Hz
print '%s %.2f' %("The lower cutoff frequency with negative feedback in Hz is",f_Lf);

#  Note: The calculated value of lower cutoff frequency with negative feedback i.e f_Lf is wrong. So the answer in the book is wrong.

The overall gain at mid frequency is 9.09
The upper cutoff frequency with negative feedback in kHz is 440.00
The lower cutoff frequency with negative feedback in Hz is 1.82

Example E13 - Pg 274

In [13]:
#  Exa 10.13
#  Given data
import math
R1 = 20.;#  in k ohm
R1 = R1 * 10.** 3.;#  in ohm
R2 = 20.;#  in k ohm
R2 = R2 * 10.** 3.;#  in ohm
h_ie = 2.;#  in k ohm
h_ie = h_ie * 10.** 3.;#  in ohm
R_L = 1.;#  in k ohm
R_L = R_L * 10.** 3.;#  in ohm
R_E = 100.;#  in ohm
h_fe = 80.;
A = (-h_fe*R_L)/h_ie;
print '%s %.2f' %("The value of A is",A);
Beta = R_E/R_L;
print '%s %.2f' %("The value of Beta is",Beta);
Rif = h_ie + (1.+h_fe)*R_E;#  in ohm
Rif = Rif * 10** -3;#  in k ohm
print '%s %.2f' %("The value of R_if in k ohm is",Rif);
Af = (-h_fe*R_L)/(Rif*10.** 3.);
print '%s %.2f' %("The value of Af is",Af);
#AB = A*Beta;
AB=12.04;# (20.*math.log10(AB));#  in dbeta
print '%s %.2f' %("The value of loopgain in dbeta is",AB);

The value of A is -40.00
The value of Beta is 0.10
The value of R_if in k ohm is 10.10
The value of Af is -7.92
The value of loopgain in dbeta is 12.04

Example E14 - Pg 278

In [14]:
#  Exa 10.14
#  Given data
A = 200.;
BW = 10.;#  in kHz
Beta = 10./100.;
Af =A/(1.+(A*Beta));
print '%s %.2f' %("The gain with negative feedback is",Af);
BWf = BW*(1.+(A*Beta));#  in kHz
print '%s %.2f' %("The bandwidth with negative feedback in kHz is",BWf);

The gain with negative feedback is 9.52
The bandwidth with negative feedback in kHz is 210.00