Chapter 15 : Operational amplifier frequency Response and compensation

Example 15.2, Page No 648

In [9]:
import math
#initialisation of variables

R2=1.0*10**6
Acl=4.5

#Calculations
R1=R2/Acl
R1=220*10**3#use standard value
R3=(R1*R2)/(R1+R2)
Cf=((R1*30*10**-12)/(R1+R2))*10**12

#Results
print(" suitable value of capacitor is %.2fpF " %Cf)

 suitable value of capacitor is 5.41pF

Example 15.3, Page No 649

In [10]:
import math

#initialisation of variables

f=35.0*10**3
Rf=68.0*10**3

#Calculations
Cf=(1.0/(2*3.14*f*Rf))*10**12

#Results
print(" suitable miller effect capacitor is %.2f pF " %Cf)

 suitable miller effect capacitor is 66.91 pF

Example 15.5 Page No 652

In [11]:
import math

#initialisation of variables

Acl=100.0
Av=10.0
print(" for Cf=30pF")
GBW=800.0*10**3

#Calculations
F2=GBW/Acl
print(" for Cf=3pF")
GBW=(800*10**3)*Av
f2=GBW/Acl

#Results
print(" The value of f2 is %.2f pF " %(f2/1000))

 for Cf=30pF
 for Cf=3pF
 The value of f2 is 80.00 pF

Example 15.6, Page No 654

In [12]:
import math
#initialisation of variables

Vip=1.0
R2=39.0*10**3
R3=4.7*10**3
SR=250.0/10**-6
f=100.0*10**3

#Calculations
print(" for the AD843")
Vop=((R2+R3)/R3)*Vip
fp=SR/(2*3.14*Vop)
print("full power bandwidth is %dHz " %fp)
print(" for a 741")
SR=0.5/10**-6
Vp=SR/(2*3.14*f)

#Results
print(" maximum peak output voltage is %3.2fV " %Vp)

 for the AD843
full power bandwidth is 4281508Hz 
 for a 741
 maximum peak output voltage is 0.80V

Example 15.7 Page No 656

In [13]:
import math
#initialisation of variables

rs=600.0
R1=1.0*10**3
R2=10.0*10**3
f=800.0*10**3

#Calculations
print(" stray capacitance")
Cs=1/(2*3.14*f*10*(((rs+R1)*R2)/(rs+R1+R2)))
print("compensation capacitor")
C2=((Cs*(rs+R1))/R2)*10**12

#Results
print("compensation capacitor is %.2fpF " %C2)

 stray capacitance
compensation capacitor
compensation capacitor is 2.31pF

Example 15.8 Page No 659

In [14]:
import math

#initialisation of variables
ro=25.0
f=2.0*10**6
R2=10.0*10**3
Rx=25.0

#Calculations
Cl=(1.0/(2.0*3.14*f*(10*ro)))*10**+12
print(" load capacitance is %3.2fpF " %Cl)
Cl=0.1*10**-6
C2=((Cl*(ro+Rx))/R2)*10**12

#Results
print(" compensation capacitance is %.2f pF " %C2)

 load capacitance is 318.47pF 
 compensation capacitance is 500.00 pF