Chapter 4: The Practical Op-Amp

Example 4.1

In [23]:
#Example 4.1
#Design a Compensating Network for the opamp LM307.
#The opamp uses +10 V and -10 V supply voltages.

#Variable declaration
V=10             #Supply voltage
Vio=10*10**-3    #Input offset voltage
Rc=10            #Assumption

#calculation
Rb=(V/Vio)*Rc
Ra=Rb/2.5        #Since Rb>Rmax,let us choose Rb=10*Rmax where Rmax=Ra/4

#Result
print "Resistance Rb is",Rb/10**3,"kilo ohms"
print "Resistance Ra is ",Ra/10**3,"kilo ohms"

Resistance Rb is 10.0 kilo ohms
Resistance Ra is  4.0 kilo ohms

Example 4.2

In [24]:
#Example 4.2
#The opamp in the circuit of figure 4-13 is the LM307 with Vio=10 mV dc maximum.
#What is the maximum possible output offset voltage, Voo, caused by
#the input offset voltage Vio?

#Variable declaration
R1=1*10**3
Rf=10*10**3
Vio=10*10**-3      #Input offset voltage

#calculation
Aoo=1+Rf/R1        #To find max value of Voo,we reduce input voltage vin to zero.
Voo=Aoo*Vio        #Max output offset voltage

#Result
print "Max output offset voltage is",Voo*10**3,"milli Volts"

Max output offset voltage is 110.0 milli Volts

Example 4.3

In [25]:
#Example 4.3
#Design an input offset voltage-compensating network for the circuit in
#figure 4-13

from __future__ import division #to perform decimal division

#Variable declaration
R1=1*10**3
Rf=10*10**3
Rc=10


#calculation
Af=1+Rf/(R1+Rc)    #Closed loop gain of non-inverting amplifier

#Result
print "Closed loop gain of non-inverting amplifier is",round(Af)

Closed loop gain of non-inverting amplifier is 11.0

Example 4.4

In [26]:
#Example 4.4
#a) For the inverting amplifier of Figure 4-19, determine the maximum possible
#output offset voltage due to input offset voltage Vio and input bias current Ib
#The opamp is a type 741.
#b) What value of ROM is needed to reduce the effect of input bias current Ib.

from __future__ import division #to perform decimal division

#Variable declaration
R1=470
Rf=47*10**3
Vio=6*10**-3          #Input offset voltage
Ib=500*10**-9         #Input bias current
Vs=15                 #Supply voltage

#calculation
Voo=(1+Rf/R1)*Vio     #Max output offset voltage due to input offset voltage Vio
VoIb=Rf*Ib            #Max output offset voltage due to input offset voltage Ib
ROM=R1*Rf/(R1+Rf)     #Parallel combination of R1 and Rf

#Result
print "Max output offset voltage due to Vio is",round(Voo*10**3),"milli Volts"
print "Max output offset voltage due to Ib is ",round(VoIb*10**3,1),"milli Volts"
print "Parallel combination of R1 and Rf,i.e ROM is ",round(ROM,2),"Ohms"

Max output offset voltage due to Vio is 606.0 milli Volts
Max output offset voltage due to Ib is  23.5 milli Volts
Parallel combination of R1 and Rf,i.e ROM is  465.35 Ohms

Example 4.5

In [27]:
#Example 4.5
#Repeat example 4.4 if R1 replaced by 1 kilo Ohm and Rf replaced by 100 kilo Ohm

from __future__ import division #to perform decimal division

#Variable declaration
R1=1*10**3
Rf=100*10**3
Vio=6*10**-3          #Input offset voltage
Ib=500*10**-9         #Input bias current
Vs=15                 #Supply voltage

#calculation
Voo=(1+Rf/R1)*Vio     #Max output offset voltage due to input offset voltage Vio
VoIb=Rf*Ib            #Max output offset voltage due to input offset voltage Ib
ROM=R1*Rf/(R1+Rf)     #Parallel combination of R1 and Rf

#Result
print "Max output offset voltage due to Vio is",round(Voo*10**3),"milli Volts"
print "Max output offset voltage due to Ib is ",round(VoIb*10**3,1),"milli Volts"
print "Parallel combination of R1 and Rf,i.e ROM is ",round(ROM,2),"Ohms"

Max output offset voltage due to Vio is 606.0 milli Volts
Max output offset voltage due to Ib is  50.0 milli Volts
Parallel combination of R1 and Rf,i.e ROM is  990.1 Ohms

Example 4.6

In [6]:
#Example 4.6
#For the inverting amplifier in figure 4-21, determine the maximum output offset
#voltage VoIio caused by the input offset current Iio.
#The opamp is a type 741

from __future__ import division #to perform decimal division

#Variable declaration
Iio=200*10**-9      #Input offset current
Rf=100*10**3

#calculation
VoIio=Rf*Iio        #Max output offset voltage due to input offset voltage Ib

#Result
print "Max output offset voltage due to Ib is ",round(VoIio*10**3),"milli Volts"

Max output offset voltage due to Ib is  20.0 milli Volts

Example 4.7

In [7]:
#Example 4.7
#Compute the maximum possible output offset voltages in the amplifier circuits
#shown in the figure 4-22. The opamp is MC1536 with the following specifications.
#Vio=7.5 mV maximum, Iio=50 nA maximum,Ib=250 nA maximum at TA=25 degree celcius

from __future__ import division #to perform decimal division

#Variable declaration
R1=1*10**3
Rf=10*10**3
Vio=7.5*10**-3 #Max input offset voltage
Iio=50*10**-9  #Max input offset current
Ib=250*10**-9  #Max input bias current

#calculation
# For figure 4.22(a)
VooT1=(1+Rf/R1)*Vio+(Rf*Ib)  #Since the current generated output offset voltage is due to input bias current Ib
# For figure 4.22(b)
VooT2=(1+Rf/R1)*Vio+(Rf*Iio) #Since the current generated output offset voltage is due to input offset current Ib

#Result
print "Max output offset voltage due to Ib is ",VooT1*10**3,"milli Volts"
print "Max output offset voltage due to Iio is ",VooT2*10**3,"milli Volts"

Max output offset voltage due to Ib is  85.0 milli Volts
Max output offset voltage due to Iio is  83.0 milli Volts

Example 4.8_a

In [29]:
#Example 4.8_a
#Refer to the inverting amplifier in figure 4-24. The opamp is the LM307 with
#the following specifications.
#delta_Vio/delta_T =30 microVolt/degree celcius maximum
#delta_Iio/delta_T= 300 pA/degree celcius
#Vs=15 V, R1=1 Kilo Ohm, Rf=100 Kilo Ohm,Rl=10 Kilo Ohm
#Assume that the amplifier is nulled at 25 degree celcius. Calculate the value
#of the error voltage Ev and output voltage at 35 degree celcius if
#a) Vin= 1 mV dc

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=30*10**-6        #Change in input offset voltage
delta_T=1                  #Unit change in temperature
delta_Iio=(300*10**-12)    #Change in input offset current
Vs=15
R1=1*10**3
Rf=100*10**3
Rl=10*10**3
Vin=1*10**-3               #Input voltage
k=25                       #Amplifier is nulled at 25 deg
T=35-k                     #Change in temperature

#calculation
Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T   #Error voltage
Vo1=-(Rf/R1)*Vin+Ev                                             #Output voltage
Vo2=-(Rf/R1)*Vin-Ev                                             #Output voltage

#Result
print "Error voltage is ",round(Ev*10**3,1),"mV"
print "Output voltage 1 is ",round(Vo1*10**3,1),"mV"
print "Output voltage 2 is ",round(Vo2*10**3,1),"mV"

Error voltage is  30.6 mV
Output voltage 1 is  -69.4 mV
Output voltage 2 is  -130.6 mV

Example 4.8_b

In [30]:
#Example 4.8_b
#Refer to the inverting amplifier in figure 4-24. The opamp is the LM307 with
#the following specifications.
#delta_Vio/delta_T =30 microVolt/degree celcius maximum
#delta_Iio/delta_T= 300 pA/degree celcius
#Vs=15 V, R1=1 Kilo Ohm, Rf=100 Kilo Ohm,Rl=10 Kilo Ohm
#Assume that the amplifier is nulled at 25 degree celcius. Calculate the value
#of the error voltage Ev and output voltage at 35 degree celcius if
#a) Vin= 10 mV dc

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=30*10**-6        #Change in input offset voltage
delta_T=1                  #Unit change in temperature
delta_Iio=(300*10**-12)    #Change in input offset current
Vs=15
R1=1*10**3
Rf=100*10**3
Rl=10*10**3
Vin=10*10**-3              #Input voltage
k=25                       #Amplifier is nulled at 25 deg
T=35-k                     #Change in temperature

#calculation
Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T   #Error voltage
Vo1=-(Rf/R1)*Vin+Ev                                             #Output voltage
Vo2=-(Rf/R1)*Vin-Ev                                             #Output voltage

#Result
print "Error voltage is ",round(Ev*10**3,1),"mV"
print "Output voltage 1 is ",round(Vo1*10**3,1),"mV"
print "Output voltage 2 is ",round(Vo2*10**3,1),"mV"

Error voltage is  30.6 mV
Output voltage 1 is  -969.4 mV
Output voltage 2 is  -1030.6 mV

Example 4.8_a

In [31]:
#Example 4.8_a
#Design of Compensating Network

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=30*10**-6        #Change in input offset voltage
delta_T=1                  #Unit change in temperature
delta_Iio=(300*10**-12)    #Change in input offset current
Vs=15
R1=1*10**3
Rf=100*10**3
Rl=10*10**3
Vin=1*10**-3                #Input voltage
k=25                        #Amplifier is nulled at 25 deg
T=35-k                      #Change in temperature

#calculation
Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T   #Error voltage
Vo1=-(Rf/R1)*Vin+Ev                                             #Output voltage
Vo2=-(Rf/R1)*Vin-Ev                                             #Output voltage

#Result
print "Error voltage is ",Ev,"Volts"
print "Output voltage 1 is ",Vo1,"Volts"
print "Output voltage 2 is ",Vo2,"Volts"

Error voltage is  0.0306 Volts
Output voltage 1 is  -0.0694 Volts
Output voltage 2 is  -0.1306 Volts

Example 4.9_a

In [34]:
#Example 4.9_a
#Refer again to the amplifier circuit in figure 4-24.Use the same circuit
#specifications that are given in example 4-8. Assume that the amplifier is
#nulled at 25 degree celcius. If Vin is a 10 mV peak sine wave at 1 kilo Hz
#Calculate Ev and Vo at 55 degree celcius.

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=30*10**-6        #Change in input offset voltage
delta_T=1                  #Unit change in temperature
delta_Iio=(300*10**-12)    #Change in input offset current
Vs=15
R1=1*10**3
Rf=100*10**3
Rl=10*10**3
Vin=10*10**-3               #Input voltage
k=25                        #Amplifier is nulled at 25 deg
T=55-k                      #Change in temperature

#calculation
Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T   #Error voltage
Vo1=-(Rf/R1)*Vin+Ev                                             #Output voltage
Vo2=-(Rf/R1)*Vin-Ev                                             #Output voltage

#Result
print "Error voltage is ",round(Ev*10**3,1),"mV"
print "Output voltage 1 is ",round(Vo1*10**3,1),"mV"
print "Output voltage 2 is ",round(Vo2*10**3,1),"mV"

Error voltage is  91.8 mV
Output voltage 1 is  -908.2 mV
Output voltage 2 is  -1091.8 mV

Example 4.9_b

In [4]:
#Example 4.9_b
#Refer again to the amplifier circuit in figure 4-24.Use the same circuit
#specifications that are given in example 4-8. Assume that the amplifier is
#nulled at 25 degree celcius. If Vin is a 10 mV peak sine wave at 1 kilo Hz
#Draw the output voltage waveform at 55 degree celcius.


from __future__ import division #to perform decimal division
%matplotlib inline

from matplotlib.pyplot import ylabel, xlabel, title, plot, show
import matplotlib.pyplot as plt
import math
import numpy as np



#calculation
x=np.arange(0,2*math.pi,0.1)   #x coordinate
y=-1000*np.sin(x)+91.8         #y coordinate

#result
plt.plot(x,y)
plt.ylabel('voltage')
plt.xlabel('time')
plt.title(r'$output waveform$')
plt.show()

Example 4.10_a

In [37]:
#Example 4.10_a
#Repeat example 4-8 for the noniverting amplifier shown in figure 4-26.
#Assume that Rc<<R1.


from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=30*10**-6        #Change in input offset voltage
delta_T=1                  #Unit change in temperature
delta_Iio=(300*10**-12)    #Change in input offset current
Vs=15
R1=1*10**3
Rf=100*10**3
Rl=10*10**3
Vin=1*10**-3               #Input voltage
k=25                       #Amplifier is nulled at 25 deg
T=35-k                     #Change in temperature

#calculation
Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T   #Error voltage
Vo1=(1+Rf/R1)*Vin+Ev                                            #Output voltage
Vo2=(1+Rf/R1)*Vin-Ev                                            #Output voltage

#Result
print "Error voltage is ",round(Ev*10**3,1),"mV"
print "Output voltage 1 is ",round(Vo1*10**3,1),"mV"
print "Output voltage 2 is ",round(Vo2*10**3,1),"mV"

Error voltage is  30.6 mV
Output voltage 1 is  131.6 mV
Output voltage 2 is  70.4 mV

Example 4.10_b

In [38]:
#Example 4.10_b
#Repeat example 4-8 for the noniverting amplifier shown in figure 4-26.
#Assume that Rc<<R1.

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=30*10**-6        #Change in input offset voltage
delta_T=1                  #Unit change in temperature
delta_Iio=(300*10**-12)    #Change in input offset current
Vs=15
R1=1*10**3
Rf=100*10**3
Rl=10*10**3
Vin=10*10**-3              #Input voltage
k=25                       #Amplifier is nulled at 25 deg
T=35-k                     #Change in temperature

#calculation
Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T   #Error voltage
Vo1=(1+Rf/R1)*Vin+Ev                                            #Output voltage
Vo2=(1+Rf/R1)*Vin-Ev                                            #Output voltage

#Result
print "Error voltage is ",round(Ev*10**3,1),"mV"
print "Output voltage 1 is ",round(Vo1*10**3,1),"mV"
print "Output voltage 2 is ",round(Vo2*10**3,1),"mV"

Error voltage is  30.6 mV
Output voltage 1 is  1040.6 mV
Output voltage 2 is  979.4 mV

Example 4.11_a

In [39]:
#Example 4.11_a
#The amplifier in figure 4-28(b) is nulled when the low dc supply voltage is
#20 mV.
#Because of poor regulation,low dc voltage varies with time from 18 V to 22 V.
#a) Determine the change in the output offset voltage caused by the change in
#supply voltages

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=15.85*10**-6  #Change in input offset voltage
delta_V=1               #Unit change in supply voltage
V=2                     #Change in supply voltage
R1=1*10**3
Rf=100*10**3

#calculation
delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V  #Change in output offset voltage
        

#Result
print "Change in Output offset voltage is ",round(delta_Voo*10**3,2),"mV"

Change in Output offset voltage is  3.2 mV

Example 4.11_b

In [41]:
#Example 4.11_b
#The amplifier in figure 4-28(b) is nulled when the low dc supply voltage is
#20 mV.
#Because of poor regulation,low dc voltage varies with time from 18 V to 22 V.
#b) Determine the output voltage Vo if Vin=10 mV dc. The opamp is the LM307
#with SVRR=96 dB.


from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=15.85*10**-6  #Change in input offset voltage
delta_V=1               #Unit change in supply voltage
V=2                     #Change in supply voltage
R1=1*10**3
Rf=100*10**3
Vin=10*10**-3

#calculation
delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V   #Output offset voltage
Vo1=(-Rf/R1)*Vin+delta_Voo                  #Total output offset voltage 1
Vo2=(-Rf/R1)*Vin-delta_Voo                  #Total output offset voltage 2
        

#Result
print "Total Output offset voltage 1 is ",round(Vo1*10**3,1),"mV"
print "Total Output offset voltage 2 is ",round(Vo2*10**3,1),"mV"

Total Output offset voltage 1 is  -996.8 mV
Total Output offset voltage 2 is  -1003.2 mV

Example 4.11_b

In [42]:
#Example 4.11_b
#Design of Compensating Network

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=15.85*10**-6  #Change in input offset voltage
delta_V=1               #Unit change in supply voltage
V=2                     #Change in supply voltage
R1=1*10**3
Rf=100*10**3
Vin=10*10**-3

#calculation
delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V   #Output offset voltage
Vo1=(-Rf/R1)*Vin+delta_Voo                  #Total output offset voltage 1
Vo2=(-Rf/R1)*Vin-delta_Voo                  #Total output offset voltage 2
        

#Result
print "Total Output offset voltage 1 is ",round(Vo1,4),"Volts"
print "Total Output offset voltage 2 is ",round(Vo2,4),"Volts"

Total Output offset voltage 1 is  -0.9968 Volts
Total Output offset voltage 2 is  -1.0032 Volts

Example 4.12

In [44]:
#Example 4.12
#Referring to figure 4-28(b), suppose that the circuit is nulled when the voltage
#across terminals +Vcc and -Vee measures 20 V dc. Also suppose that because of
#poor filtering, 10 mV rms ac ripple is measured across terminals +Vcc and -Vee.
#While the input signal Vin=0V, how much ripple voltage can we expect at the
#output if the opamp is the LM307.

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=15.85*10**-6  #Change in input offset voltage
delta_V=1               #Unit change in supply voltage
V=10*10**-3             #Change in supply voltage
R1=1*10**3
Rf=100*10**3


#calculation
delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V   #Output offset voltage

#Result
print "Change in Output offset voltage 1 is ",round(delta_Voo*10**6),"uV rms"

Change in Output offset voltage 1 is  16.0 uV rms

Example 4.13

In [5]:
#Example 4.13
#Suppose that the circuit in figure 4-24 is initially nulled. Assume also that
#room temperature and the voltage terminals +Vcc and -Vee remain constant.
#Determine the maximum possible chnage in output offset voltage after one month
#if the opamp is the LH0041C.
#Assume that R1=1kilo Ohm, Rf=100 kilo Ohm and Rl=10 kilo Ohm.

from __future__ import division #to perform decimal division

#Variable declaration
delta_Vio=5*10**-6  #Change in input offset voltage
delta_t=1           #Unit change in time
delta_Iio=2*10**-9  #Change in input offset current 
t=4                 #Time elapsed(weeks)
R1=1*10**3
Rf=100*10**3
Rl=10*10**3


#calculation
delta_Voot=(1+Rf/R1)*(delta_Vio/delta_t)*t+Rf*(delta_Iio/delta_t)*t  #Output offset voltage

#Result
print "Change in Output offset voltage 1 is ",round(delta_Voot*10**3,2),"mV"

Change in Output offset voltage 1 is  2.82 mV