Ch-20 : Operational Amplifiers

Page No. 620 Example 20.1.

print "    CMRR = Ad / Acm = 10**5"
acm=(1.0*10**5)/(10**5)
print "    Therefore, the common-mode gain, Acm = Ad / CMRR =%0.2f"%acm

    CMRR = Ad / Acm = 10**5
    Therefore, the common-mode gain, Acm = Ad / CMRR =1.00

Page No. 620 Example 20.2.

sr=20./(4) # in V/us
print "    The slew rate,        SR = dVo / dt"
print "    SR =%0.2f V/us"%sr

    The slew rate,        SR = dVo / dt
    SR =5.00 V/us

Page No. 621 Example 20.3.

from math import sqrt,pi
print "The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,"
print "      SR = 2*pi*f*Vm / 10**6 = 0.5 V/us"
vm=50.*(20*10**-3) # in volts
print "The maximum output voltage,  Vm = A*Vid =%0.2f V"%vm
print "The maximum frequency of the input for which undistorted output is obtained is given by,"
f=(0.5*10**6)/(2*pi*1) # in kHz
x1=f*10**-3
print "     fmax = SR*10**6 / 2*pi*Vm =%0.2f"%x1

The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,
      SR = 2*pi*f*Vm / 10**6 = 0.5 V/us
The maximum output voltage,  Vm = A*Vid =1.00 V
The maximum frequency of the input for which undistorted output is obtained is given by,
     fmax = SR*10**6 / 2*pi*Vm =79.58

Page No. 622 Example 20.4.

from math import pi,sqrt
print "The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,"
print "      SR = 2*pi*f*Vm / 10**6 = 0.5 V/us"
vm=(0.5*10**6)/(2*pi*(40*10**3)) # in volts
print "The maximum output voltage,  Vm(V peak-to-peak) = SR*10**6 / 2*pi*f =%0.2f"%vm,"    = 3.98 V peak-to-peak"
print "The maximum peak-to-peak input voltage for undistorted output is,"
vid=3.98/10 # in volts
print "      Vid(V peak-to-peak) = Vm/A = %0.2f"%vid

The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,
      SR = 2*pi*f*Vm / 10**6 = 0.5 V/us
The maximum output voltage,  Vm(V peak-to-peak) = SR*10**6 / 2*pi*f =1.99     = 3.98 V peak-to-peak
The maximum peak-to-peak input voltage for undistorted output is,
      Vid(V peak-to-peak) = Vm/A = 0.40

Page No. 623 Example 20.5.

af=-10./1
print "  The closed-loop voltage gain Af = -RF / R1 =%0.2f"%af

  The closed-loop voltage gain Af = -RF / R1 =-10.00

Page No. 624 Example 20.6.

af=1.+(10./1)
print "  The closed-loop voltage gain, AF = 1 + RF/R1 =%0.2f"%af
beta=1/(1+10.)
print "  The feedback factor,  beta = R1 / R1+RF =%0.2f"%beta

  The closed-loop voltage gain, AF = 1 + RF/R1 =11.00
  The feedback factor,  beta = R1 / R1+RF =0.09

Page No. 625 Example 20.7.

v=-(2+3+4) # in volts
print "The output voltage is given by,"
print "    Vo = -Rf/R * (V1+V2+...+Vn) =%0.2f V"%v

The output voltage is given by,
    Vo = -Rf/R * (V1+V2+...+Vn) =-9.00 V

Page No. 626 Example 20.8.

from math import pi
print "1.  Given: fL = 1 kHz"
print "2.  Since R and C values are not given, let assume C = 0.01 uF"
r=1/(2*pi*(10**3)*(0.01*10**-6))
x1=r*10**-3  # in k-ohm
print "3.  Therefore, R(k-ohm) = 1 / 2*pi*fL*C =%0.2f kohm"%x1
print "4.  Given pass band gain A = 1 + Rf/Ri = 2 i.e. the value of Rf = Ri"
print "Let Rf = Ri = 10 k-ohm. The high pass circuit values are shown in Fig.20.31"

1.  Given: fL = 1 kHz
2.  Since R and C values are not given, let assume C = 0.01 uF
3.  Therefore, R(k-ohm) = 1 / 2*pi*fL*C =15.92 kohm
4.  Given pass band gain A = 1 + Rf/Ri = 2 i.e. the value of Rf = Ri
Let Rf = Ri = 10 k-ohm. The high pass circuit values are shown in Fig.20.31

Page No. 627 Example 20.9.

r=(0.5)/0.01 # in k-ohm
print "(b) With      C = 0.01 uF,  R = 0.5*10**-3/0.01*10**-6 =%0.2f kohm"%r
print "(c) Maximum value of differential input voltage is"
x=2.*14*(100/(100+116))
print "          2*Vsat*(R2 / R1+R2) =%0.2f"%x
print "Therefore, the peak values for the differential input voltage just exceed +-2 x 6.48 V"

(b) With      C = 0.01 uF,  R = 0.5*10**-3/0.01*10**-6 =50.00 kohm
(c) Maximum value of differential input voltage is
          2*Vsat*(R2 / R1+R2) =0.00
Therefore, the peak values for the differential input voltage just exceed +-2 x 6.48 V