# Chapter 16 Op-Amp Negative Feedback¶

## Example 16.1 Page No 385¶

In [3]:
# given data
A=100000.0##unit less
R1= 98.0*10**3##  Ω
R2= 2.0*10**3##  Ω
Vin= 1.*10**-3##  V
B= R2/(R1+R2)## unit less
A_CL= 1/B## unit less
A_CL= A/(1+A*B)## unit less
# The output voltage
Vout= Vin*A_CL##  V
# The error voltage
Verror= Vout/A##  V
Vout= Vout*10**3##  mV
Verror= Verror*10**6##  µV
print "The output voltage = %.2f mV"%Vout
print "The error voltage = %.2f µV"%Verror

The output voltage = 49.98 mV
The error voltage = 0.50 µV


## Example 16.2 Page No 386¶

In [4]:
# given data
A=20000#
B= 0.02#
Vin= 1##  mV
Vin= Vin*10**-3##  V
# The closed loop voltage gain,
A_CL= A/(1+A*B)#
# The output voltage,
Vout= Vin*A_CL##  V
# The error voltage,
Verror= Vout/A##  V
Vout= Vout*10**3##  mV
Verror= Verror*10**6##  µV
print "The value of A_CL = %.2f"%A_CL
print "The value of Vout = %.2f mV"%Vout
print "The value of Verror = %.2f µV"%Verror

The value of A_CL = 49.88
The value of Vout = 49.88 mV
The value of Verror = 2.49 µV


## Example 16.3 Page No 389¶

In [5]:
# given data
A=100000.0#
R1= 100.0*10**3##  Ω
R2= 100.0##  Ω
r_in= 2.0*10**6##  Ω
r_out= 75.0##  Ω
B= R2/(R1+R2)## unit less
# The closed loop input impedence
r_in_CL= (1+A*B)*r_in##  Ω
# The closed loop output impedence
r_out_CL= r_out/(1+A*B)##  Ω
r_in_CL=r_in_CL*10**-6##  Mohm
print "The closed loop input impedence = %.2f MΩ"%r_in_CL
print "The closed loop output  impedence = %.2f Ω"%r_out_CL

The closed loop input impedence = 201.80 MΩ
The closed loop output  impedence = 0.74 Ω


## Example 16.4 Page No 389¶

In [11]:
# given data
A=100.0#
R_B= 39.0*10**3##  Ω
r_in= 2.0*10**6##  Ω
r_out= 75.0##  Ω
Vin_off= 2.0*10**-3##  V
I_B1= 90.0*10**-9##  A
I_in_off= 20.0*10**-9##  A
# The closed loop voltage gain
B=1## unit less
# The closed-loop input impedance
r_in_CL= (1.0+A*B)*r_in##  Ω
r_in_CL= r_in_CL*10**-6##  Mohm
print "The closed loop voltage gain = %.2f"%B
print "The closed-loop input impedance = %.2f MΩ"%r_in_CL
A=100000.0#
# The closed-loop output impedance
r_out_CL= r_out/A##  Ω
print "The closed-loop output impedance = %.2e Ω"%r_out_CL
#Let V= V1-V2 = Vin_off+I_B1*R_B
V= Vin_off+I_B1*R_B##  A
# The output offset voltage
Voo_CL= A*V/A##  V
Voo_CL= Voo_CL*10**3##  mV
print "The output offset voltage = %.2f mV"%Voo_CL

The closed loop voltage gain = 1.00
The closed-loop input impedance = 202.00 MΩ
The closed-loop output impedance = 7.50e-04 Ω
The output offset voltage = 5.51 mV


## Example 16.5 Page No 393¶

In [12]:
# given data
R_F= 22.0*10**3##  Ω
R_S= 1.0*10**3##  Ω
A= 100000.0## unit less
# The closed-loop voltage gain
A_CL= R_F/R_S#
# The desensitivity
desensitivity= A/A_CL#
print "The closed-loop voltage gain = %.2f"%A_CL
print "The desensitivity = %.2f"%desensitivity

The closed-loop voltage gain = 22.00
The desensitivity = 4545.45


## Example 16.6 Page No 396¶

In [13]:
# given data
f_unity= 1.0*10**6##  Hz
# For A_CL= 1000, The value of f_CL
A_CL= 1000.0#
f_CL= f_unity/A_CL##  Hz
f_CL= f_CL*10**-3##  kHz
print "For A_CL= 1000, The value of f_CL = %.2f kHz"%f_CL
# For A_CL= 100, The value of f_CL
A_CL= 100.0#
f_CL= f_unity/A_CL##  Hz
f_CL= f_CL*10**-3##  kHz
print "For A_CL= 100, The value of f_CL = %.2f kHz"%f_CL
# For A_CL= 10, The value of f_CL
A_CL= 10.0#
f_CL= f_unity/A_CL##  Hz
f_CL= f_CL*10**-3##  kHz
print "For A_CL= 10, The value of f_CL = %.2f kHz"%f_CL
# For A_CL= 1, The value of f_CL
A_CL= 1.0#
f_CL= f_unity/A_CL##  Hz
f_CL= f_CL*10**-6##  MHz
print "For A_CL= 1, The value of f_CL = %.2f MHz"%f_CL

For A_CL= 1000, The value of f_CL = 1.00 kHz
For A_CL= 100, The value of f_CL = 10.00 kHz
For A_CL= 10, The value of f_CL = 100.00 kHz
For A_CL= 1, The value of f_CL = 1.00 MHz