# Chapter 8 - Feedback Amplifiers¶

## Example E1 - Pg 434¶

In [1]:
#Ex8_1 pg-434
#calculate Input signal,Gain after feedback,Feedback
import math
A=120. #amplification gain
Vi=50.*10.**(-3.) #input voltage
Beta=0.1 #feedback factor
Vo= A*Vi #output voltage without feedback
print '%s %.2f' %("Input signal = V",Vo)
Vs=Vi-Beta*Vo
#input signal +ve output because of -ve feedback (calue in texxtbook is wrong)
print '%s %.2f' %("Input signal = V",abs(Vs))
vg=A/(1.+Beta*A) #voltage gain
print '%s %.1f' %("Gain after feedback =",vg)
fb=(-1.)*20.*math.log10(1+(Beta*A))
print '%s %.3f' %("Feedback (db)=",fb)

Input signal = V 6.00
Input signal = V 0.55
Gain after feedback = 9.2
Feedback (db)= -22.279


## Example E2 - Pg 435¶

In [2]:
#Ex8_2 pg-435
#calculate
ff=4. #feedback factor
BW=6.*10.**(6.) #bandwidth in Hz
BW_fb=BW*(1.+ff) #Bandwidth with feedback factor(since Beta is +ve)
print '%s %.0f' %("Bandwidth with feedback factor = MHz",BW_fb*10**-6)

Bandwidth with feedback factor = MHz 30


## Example E3 - Pg 435¶

In [3]:
#Ex8_3 pg-435
#calculate Negative Feedback factor,Beta
openA=60000. #open loop gain
closeA=10000. #closed loop gain
Beta=((openA/closeA)-1.)/closeA
print '%s %.4f' %("Negative Feedback factor=",Beta)
BA=Beta*openA #value of Beta*A
print '%s %.0f' %("Beta*A=",BA)

Negative Feedback factor= 0.0005
Beta*A= 30


## Example E4 - Pg 435¶

In [4]:
#Ex8_4 pg-435
#calculate Feedback factor,Gain after negative feedback
Df=0.5/100. #distortion after negative feedback
D=8./100. #harmonic distortion
BA=D/Df-1 #value of Beta*A
A=200.
Beta=BA/A #feedback factor
print'%s %.3f'%("Feedback factor=",Beta)
Af=A/(1.+BA) #Gain after -ve feedback
print'%s %.1f'%("Gain after negative feedback=",Af)

Feedback factor= 0.075
Gain after negative feedback= 12.5


## Example E5 - Pg 436¶

In [5]:
#Ex8_5 pg-436
#calculate Decrement in distortion,Percentage change in distortion
A=100. #voltage gain
per=10./100. #percentage of negative feedback applied
BA=A*per #value of Beta*A
Af=A/(1.+BA) #gain after negative feedback
print '%s' %("Decrement in distortion,D-Df = D-(D/(1+Beta*A))")
print '%s %.1f' %("=",Af)
per_dis=(BA/(1.+BA))*100. #percentage change in distortion
print '%s %.0f' %("Percentage change in distortion=",per_dis)
red=100-per_dis #reduction
print '%s %.0f' %("Therefore reduction is=",red)

Decrement in distortion,D-Df = D-(D/(1+Beta*A))
= 9.1
Percentage change in distortion= 91
Therefore reduction is= 9


## Example E6 - Pg 436¶

In [6]:
#Ex8_6 pg-436
#calculate Voltage gain after negative feedback,Voltage gain after negative feedback,Beta
A=50. #voltage gain
per=10./100. #percentage of negative feedback applied
BA=per*A #value of Beta*A
Af=A/(1.+BA) #gain after negative feedback
print '%s' %("(1) Voltage gain after negative feedback")
print '%s %.2f' %("=",Af)
A=50. #voltage gain
per=5./100. #percentage of negative feedback applied
BA=per*A #value of Beta*A
Af1=A/(1.+BA) #gain after negative feedback
print '%s' %("(2) Voltage gain after negative feedback")
print '%s %.1f' %("=",Af1)
print '%s' %("\nSo the new gain is not double the previous case")
print '%s' %("The difference between expected value and actual value of gain obtained is")
diff_value=2.*Af-Af1
print '%s %.2f' %("=",diff_value)
print '%s' %("\n(3) To have the gain double of case(1) i.e 16.66,let the feedback introduced be Beta(assuming negative feedback)")
Af=16.66 #voltage gain with negative feedback
A=50. #voltage gain
Beta=((A/Af)-1.)/A #feedback in percentage
print '%s %.2f' %("Beta=",Beta)

(1) Voltage gain after negative feedback
= 8.33
(2) Voltage gain after negative feedback
= 14.3

So the new gain is not double the previous case
The difference between expected value and actual value of gain obtained is
= 2.38

(3) To have the gain double of case(1) i.e 16.66,let the feedback introduced be Beta(assuming negative feedback)
Beta= 0.04