Chapter 9 - Negative Feedback Amplifier using Op-Amp

Example E1 - Pg 475

In [1]:
#Ex9_1 Pg-475
#calculate Close loop Bandwidth
Aol= 250000. #open loop gain
fol=160. #open loop frequency in HZ
Acl=50. #close loop gain
fcl=Aol/Acl*fol #close loop frequency in Hz
print '%s %.0f' %("Close loop Bandwidth = %.0f kHz",fcl*10**-3) 

Close loop Bandwidth = %.0f kHz 800

Example E2 - 475

In [2]:
#Ex9_2 Pg-475
#calculate Close loop Bandwidth
Aol= 50000. #open loop gain
fol=14. #open loop frequency in HZ
fcl=(Aol+1)*fol #close loop frequency in Hz
print '%s %.0f' %("Close loop Bandwidth = kHz",fcl*10**-3)

Close loop Bandwidth = kHz 700

Example E3 - 475

In [3]:
#Ex9_3 Pg-475
#calculate Close loop Bandwidth
Aol_Beta_1= 2500. #open loop gain
fol=20. #open loop frequency in HZ
fcl=Aol_Beta_1*fol #close loop frequency in Hz
print '%s %.0f' %("Close loop Bandwidth = kHz",fcl*10**-3)

Close loop Bandwidth = kHz 50

Example E4 - 475

In [4]:
#Ex9_4 Pg-475
#calculate Close loop Bandwidth,Peak value of output
import math
funi=1.*10.**(6.) #unity frequency in Hz
Sr=0.5/10.**(-6.) #slew rate in V/sec
Acl=10. #close loop gain
fcl=funi/Acl #close loop frequency in Hz
print '%s %.0f' %("(1)Close loop Bandwidth = kHz",fcl*10**-3)
Vp_max=Sr/(2.*math.pi*fcl) #output peak value
print '%s %.3f' %("(2)Peak value of output = V",Vp_max)

(1)Close loop Bandwidth = kHz 100
(2)Peak value of output = V 0.796

Example E5 - 475

In [5]:
#Ex9_5 Pg-475
#calculate Feedback fraction,Ideal closed loop gain,Exact closed loop voltage gain
Aol= 88. #open loop gain in db
R1=2.7*10.**(3.) #resistor R1 in ohm
R2=68.*10.**(3.) #resistor R2 in ohm
Beta=R1/(R1+R2) #Feedback fraction
print '%s %.3f' %("Feedback fraction = ",Beta)
Acl=1./Beta #ideal closed loop gain
print '%s %.2f' %("Ideal closed loop gain = ",Acl)
Aol=10.**(88./20.) #open loop gain
Acl=Aol/(1.+Beta*Aol) #exact closed loop voltage gain
print '%s %.2f' %("Exact closed loop voltage gain = ",Acl)

Feedback fraction =  0.038
Ideal closed loop gain =  26.19
Exact closed loop voltage gain =  26.16

Example E6 - 476

In [6]:
#Ex9_6 Pg-476
#calculate Feedback fraction,Closed loop input impedance
Aol=20000.#open loop gain
R1=100. #resistor R1 in ohm
R2=7.5*10.**(3.) #resistor R2 in ohm
Rin=3.*10.**(6.) #input resistor in ohm
Rcm=500.*10.**(6.) 
Beta=R1/(R1+R2) #Feedback fraction
print '%s %.5f' %("Feedback fraction = 1\\76 =",Beta)
Req=(Rin*Rcm)/(Rin+Rcm) #equivalent resistance of Rin and Rcm
Zcl=(1.+Beta*Aol)*Req #closed loop input impedance (textbook answer is wrong)
print '%s %.0f %s' %("Closed loop input impedance =",Zcl*10**-6,"Mohm")

Feedback fraction = 1\76 = 0.01316
Closed loop input impedance = 788 Mohm

Example E7 - 477

In [7]:
#Ex9_7 Pg-477
#calculate Current gain,Output current
R1=1.8 #resistor R1 in ohm
R2=1.5*10.**(3.) #resistor R2 in ohm
Iin=1.*10.**(-3.) #input current in A
Ai=1.+(R2/R1) #Current gain
print '%s %.0f' %("Current gain =",Ai)
Il=Ai*Iin #Output current
print '%s %.0f' %("Output current = mA pp",Il*10**3)

Current gain = 834
Output current = mA pp 834

Example E8 - 477

In [8]:
#Ex9_8 Pg-477
#calculate Output current,Load power,Load power
R1=2.7 #resistor R1 in ohm
R2=1.#resistor R2 in ohm
Vin=0.5 #input voltage in V
Io=Vin/R1 #output current
print '%s %.0f' %("(1) Output current = mA",Io*10**3)
P=Io**2.*R2 #load power
print '%s %.1f' %("(2a) Load power = mW",P*10**3)
R2=2 # new load resistor R2 in ohm
P=Io**2*R2 #load power
print '%s %.1f' %("(2b) Load power = mW",P*10**3)

(1) Output current = mA 185
(2a) Load power = mW 34.3
(2b) Load power = mW 68.6