Chapter 8 Common-Base Approximations¶

Example 8.1 Page No 209¶

In [1]:
# given data
V_EE= 10##  V
V_BE= 0.7##  V
R_E= 20*10**3##  Ω
V_CC= 25##  V
R_C= 10*10**3##  Ω
# The emitter current
I_E= (V_EE-V_BE)/R_E##  A
I_C= I_E##  A
# The collector to base voltage,
V_CB= V_CC-I_C*R_C##  V
print "The value of V_CB = %.2f volts"%V_CB

The value of V_CB = 20.35 volts


Example 8.2 Page No 209¶

In [2]:
# given data
V_EE= 12##  V
V_BE= 0.7##  V
R_E= 5.6*10**3##  Ω
V_CC= 15##  V
R_C= 6.8*10**3##  Ω
# The emitter current,
I_E= (V_EE-V_BE)/R_E##  A
I_C= I_E##  A
# The collector to base voltage
V_CB= V_CC-I_C*R_C##  V
print "The value of V_CB = %.2f volts"%V_CB

# Note : The answer in the book is not accurate.

The value of V_CB = 1.28 volts


Example 8.3 Page No 211¶

In [3]:
# given data
V_EE= 15##  V
V_BE= 0.7##  V
R_E= 22*10**3##  Ω
Vin= 2*10**-3##  V
V= 25*10**-3##  V
R1= 10*10**3##  Ω
R2= 30*10**3##  Ω
I_E= (V_EE-V_BE)/R_E##  A
# The ac resistance of emitter diode,
r_desh_e= V/I_E##  Ω
r_L= R1*R2/(R1+R2)##  Ω
# The voltage gain
A= r_L/r_desh_e#
# The output voltage
Vout= A*Vin##  V
Vout= Vout*10**3##  mV
print "The output voltage = %.2f mV"%Vout

The output voltage = 390.00 mV


Example 8.4 Page No 212¶

In [4]:
# given data
V_EE= 10##  V
V_BE= 0.7##  V
R_E= 6.8*10**3##  Ω
Rs= 100##  Ω
R1= 3.3*10**3##  Ω
R2= 1.5*10**3##  Ω
V= 25*10**-3##  V
Vs= 1*10**-3##  V
I_E= (V_EE-V_BE)/R_E##  A
r_desh_e= V/I_E##  Ω
Zin= r_desh_e##  Ω
# The input voltage to the emitter,
Vin= Zin*Vs/(Rs+Zin)##  V
r_L= R1*R2/(R1+R2)##  Ω
# The voltage gain,
A= r_L/r_desh_e#
# The output voltage
Vout= A*Vin##  V
Vout= Vout*10**3##  mV
print "The output voltage = %.2f mV"%Vout

The output voltage = 8.72 mV