# 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);
# 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);
# 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);
# 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);
# 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);
# 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);
# 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);
# 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);
# 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);
# 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.
# 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);
# 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.
# 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);
# 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);