chapter10:Feedback in Amplifiers

Example E1 - Pg 368

#Determine the gain of feedback amplifier
#given
A=100.;#internal gain
B=0.1;#feedback factor
Af=A/(1.+A*B);
print '%s %.2f %s' %("The gain of feedback amplifier =",Af,"\n");

The gain of feedback amplifier = 9.09 

Example E2 - Pg 368

#Determine the gain of feedback amplifier in dB
#given
import math
Ad=60.;#dB      #internal gain in dB
A=10.**(Ad/20.);    #internal gain
B=1./20.;#feedback factor
Af=A/(1.+A*B);
Afd=20.*math.log10(Af);
print '%s %.2f %s' %("The gain of feedback amplifier =",Afd,"dB");

The gain of feedback amplifier = 25.85 dB

Example E3 - Pg 368

#Calculate the percentage of output fed back to input
#given
A=600.;#internal gain
Af=50.;#gain of feedback amplifier
B=(A/Af-1.)/A;
print '%s %.3f %s' %("The percentage of output fed back to input =",B*100,"percent");

The percentage of output fed back to input = 1.833 percent

Example E4 - Pg 368

#Calculate the internal gain and percentage of output fed back to input
#given
Af=80.;#gain of feedback amplifier
Vi=0.05;#V#input with feedback
Vi_=4.*10.**-3.;#V#input without feedback
Vo_=Af*Vi;
A=Vo_/Vi_;
print '%s %.f %s' %("The internal gain is =",A,"\n");
B=(A/Af-1.)/A;
print '%s %.2f %s' %("The percentage of output fed back to input =",B*100,"percent\n");

The internal gain is = 1000 

The percentage of output fed back to input = 1.15 percent

Example E5 - Pg 369

#Calculate the gain with and without feedback and feedback factor
#given
Vo_=5.;#V         #output voltage
Vi=0.2;#V    #input with feedback
Vi_=0.05;#V   #input without feedback
A=Vo_/Vi_;
Af=Vo_/Vi;
print '%s %.f %s' %("The gain without feedback is =",A,"\n");
print '%s %.f %s' %("The gain with feedback is =",Af,"\n");
B=(A/Af-1.)/A;
print '%s %.f %s' %("The feedback factor =",B*100,"percent\n");

The gain without feedback is = 100 

The gain with feedback is = 25 

The feedback factor = 3 percent

Example E6 - Pg 369

#Calculate the gain of feedback amplifier and feedback factor
#given
A=100.;      #internal gain
N=20.;#dB   #negative feedback
B=(10.**(-N/(-20.))-1.)/A;      #taking antilog
Af=A/(1.+A*B);
print '%s %.f %s' %("The feedback factor =",B*100,"percent\n");
print '%s %.f %s' %("The gain of the feedback amplifier is =",Af,"\n");

The feedback factor = 9 percent

The gain of the feedback amplifier is = 10 

Example E7 - Pg 371

#Calculate percentage change in the overall gain
#given
A=1000.;#internal gain
N=40.;#dB#negative feedback
D=10.**((-N)/-20.);#D=(1+AB)desensitivity
dA_A=10.;#percent#dA/A
dAf_Af=dA_A/D;#dAf/Af
print '%s %.1f %s' %("The percentage change in the overall gain =",dAf_Af,"percent");

The percentage change in the overall gain = 0.1 percent

Example E8 - Pg 371

#Calculate percentage change in the overall gain
#given
Adb=60.;#dB#internal gain in dB
B=0.005;#feedback factor
A=10.**(Adb/(20.));#taking antilog
dA_A=-12.;#percent #dA/A
D=(1.+A*B);#D=(1+AB)desensitivity
dAf_Af=dA_A/D;#dAf/Af
print '%s %.f %s' %("The percentage change in the overall gain reduces by =",-dAf_Af,"percent");

The percentage change in the overall gain reduces by = 2 percent

Example E9 - Pg 374

#Determine the input resistance of feedback amplifier
#given
A=250.;#internal gain
B=0.1;#feedback factor
Ri=1.1*10.**3.;#ohm    #input resistance
Rif=Ri*(1.+A*B);
print '%s %.1f %s' %("The input resistance of feedback amplifier =",Rif/1000,"kohm");

#The ans in book is incorrect due to use of (2+A*B) instead of (1+A*B) the ans in book is 29.7 kohm

The input resistance of feedback amplifier = 28.6 kohm

Example E10 - Pg 374

#Calculate the percentage of negative feedback to input
#given
Adb=60.;#dB      #internal gain in dB
A=10.**(Adb/(20.));      #taking antilog
Ro=12.*10.**3.;#ohm      #output resistance
Rof=600.;#ohm
B=(Ro/Rof-1.)/A;
print '%s %.1f %s' %("The percentage of negative feedback to input =",B*100,"percent");

The percentage of negative feedback to input = 1.9 percent