Chapter 14 : Ic operational Amplifier and basic Op amp circuits¶

Example 14.1, Page No 597¶

In [1]:
import math
#initialisation of variables
Vbe=0.7
Ib=500.0*10**-9

#Calculations
R1=Vbe/(10.0*Ib)
R1=120.0*10**3#use standard value
R2=R1
I2=100.0*Ib
Vr1=15.0
Vr2=Vr1
R1=Vr1/I2
R1=270.0*10**3#use satndard value
R2=R1
R3=(R1*R2)/(R1+R2)

#Results
print('The value of R3= %.2f kohm ' %(R3/1000))

The value of R3= 135.00 kohm


Example 14.2, Page No 599¶

In [2]:
import math

#initialisation of variables

R2=1.0*10**6
Vb=3.0
Vo=3.0
Vee=9.0

#Calculations
Vr2=Vb-(-Vee)
Vr1=Vee-Vb
I2=Vr2/R2
R1=Vr1/I2
R3=0

#Results
print('The value of R1= %.2f kohm ' %(R1/1000))

The value of R1= 500.00 kohm


Example 14.3 Page No 601¶

In [3]:
import math

#initialisation of variables

Av=200000.0
ri=2.0*10**6
ro=75.0
Vo=1.0
B=1.0

#Calculations
Vd=Vo/Av
Zi=(1+Av*B)*ri
Zo=ro/(1+Av*B)

#Results
print('The value of Zo= %.2f X 10^-3 kohm ' %(Zo*10**3))

The value of Zo= 0.37 X 10^-3 kohm


Example 14.4, Page No 603¶

In [4]:
import math
#initialisation of variables

f=70.0
Rl=4.0*10**3
Ib=500.0*10**-9
Vbe=0.7

#Calculations
R1=Vbe/(10*Ib)
R1=120*10**3#use standard value
R2=R1
print(" desire value of capacitor is C=1/2*3.14*f*R")
C2=1/(2*3.14*f*Rl)
C1=1/(2*3.14*f*(R1/10))

#Results
print('The value of C1= %.2f mF ' %(C1*10**6))

 desire value of capacitor is C=1/2*3.14*f*R
The value of C1= 0.19 mF


Example 14.5 Page No 605¶

In [5]:
import math
#initialisation of variables

Ib=500.0*10**-9
Vi=50.0*10**-3
Vo=2.0

#Calculations
I2=100.0*Ib
R3=Vi/I2
R2=(Vo/I2)-R3
R1=(R2*R3)/(R2+R3)

#Results
print('The value of R1= %.2f kohm ' %(R1/1000))

The value of R1= 0.97 kohm


Example 14.6 Page No 606¶

In [6]:
import math

#initialisation of variables

Av=200000.0
ri=2.0*10**6
ro=75.0
R3=1.0*10**3
R2=39*10**3

#Calculations
B=R3/(R2+R3)
Zi=(1+Av*B)*ri

#Results
print(" typical input impedance for non-inverting amplifier is %.2f ohm " %Zi)
Zo=ro/(1+Av*B)
print('The value of Zo= %.2f kohm ' %(Zo*10))

 typical input impedance for non-inverting amplifier is 10002000000.00 ohm
The value of Zo= 0.15 kohm


Example 14.7, Page No 607¶

In [7]:
import math
#initialisation of variables

R2=50.0*10**3
R3=2.2*10**3
C2=8.2*10**-6
Rl=600.0

#Calculations
print("voltage gain ")
Acl=(R3+R2)/R3

#Results
print("lower cuttoff frequency ")
f=1/(2*3.14*C2*Rl)
print('The value of f= %.2f kohm ' %(f))

voltage gain
lower cuttoff frequency
The value of f= 32.36 kohm


Example 14.8 Page No 610¶

In [8]:
import math

#initialisation of variables

Acl=144.0
Vi=20.0*10**-3
Ib=500.0*10**-9

#Calculations
I1=100.0*Ib
R1=Vi/I1
R1=390.0   #use standard value
R2=Acl*R1
R3=(R1*R2)/(R1+R2)

#Results
print('The value of R3= %.2f kohm ' %(R3))

The value of R3= 387.31 kohm


Example 14.9 Page No 612¶

In [9]:
import math
#initialisation of variables
Acl=3.0
R4=1.0*10**6
Vi=1.0

#Calculations
R1=R4/Acl
R1=330.0*10**3#use standard value
R2=R1
R3=R1
I1=Vi/R1
I2=I1
I3=I1
I4=I1+I2+I3
Vo=-I4*R4

#Results
print('The value of Vo= %.2f v ' %(Vo))

The value of Vo= -9.09 v


Example 14.10 Page No 615¶

In [10]:
import math

#initialisation of variables

Ib=500.0*10**-9
Vi=1.0
Acl=10.0

#Calculations
I1=100*Ib
R1=Vi/I1
R1=18*10**3#use standard value
R2=Acl*R1
R4=R1
R3=R1/Acl

#Results
print('The value of R3= %.2f kohm ' %(R3/1000))

The value of R3= 1.80 kohm


Example 14.11, Page No 619¶

In [11]:
import math

#initialisation of variables

Vi=10*10**-3
Vn=1.0
R1=33.0*10**3
R2=300.0
R5=15.0*10**3
R4=15.0*10**3
Vi2=-10.0*10**-3
R3=R1
R6=15.0*10**3

#Calculations
R7=R6
Acl=((2*R1+R2)/R2)*(R5/R4)
print("at junction of R1 and R2")
Vb=Vi+Vn
print("at junction of R2 and R3")
Vc=Vi2+Vn
print(" current through R2")
I2=(Vb-Vc)/R2
print("at the output of A1")
Va=Vb+(I2*R1)
print("at output of A2")
Vd=Vc-(I2*R3)
print("at junction of R6 and R7")
Vf=Vd*(R7/(R6+R7))
print("at junction of R4 and R5")
Ve=Vf
print("current through R4")
I4=(Va-Ve)/R4

#Results
print("at output of A3")
Vg=Ve-(I4*R5)
print('The value of Vg= %.2f kohm ' %(Vg))

at junction of R1 and R2
at junction of R2 and R3
current through R2
at the output of A1
at output of A2
at junction of R6 and R7
at junction of R4 and R5
current through R4
at output of A3
The value of Vg= -4.42 kohm


Example 14.12, Page No 623¶

In [12]:
import math
#initialisation of variables

Vcc=15.0
Vee=-15.0
Av=200000.0
SR=0.5/10**-6
Vo=14.0

#Calculations
V=(Vcc-1)-(Vee+1)
Vi=Vo/Av
print("rise time of output is ")
t=(V/SR)*10**6

#Results
print("rise time of output is %d ms " %t)

rise time of output is
rise time of output is 56 ms


Example 14.13, Page No 627¶

In [13]:
#initialisation of variables
Ib=500.0*10**-9
UTP=5.0
Vcc=15.0

#Calculations
I1=100.0*Ib
R2=UTP/I1
R1=((Vcc-1)-5)/I1

#Results
print('The value of R1= %.2f kohm ' %(R1/1000))

The value of R1= 180.00 kohm


Example 14.14, Page No 630¶

In [14]:
import math
#initialisation of variables

Vcc=15.0
Vsat=Vcc
R2=150.0*10**3
Vf=0.7
R1=27.0*10**3
R3=120.0*10**3

#Calculations
I2=(Vsat-Vf)/R2
UTP=I2*R1

#Results
print(" LTP calculation including Vf")
I3=(Vsat-Vf)/R3
LTP=-I3*R1
print('The value of LTP= %.2f kohm ' %(LTP))

 LTP calculation including Vf
The value of LTP= -3.22 kohm