In [2]:

```
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
```

In [3]:

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

In [4]:

```
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
```

In [6]:

```
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
```

In [7]:

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

In [8]:

```
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
```

In [9]:

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

In [10]:

```
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"
```

In [1]:

```
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"
```