# Chapter 10 - Other Power Amplifiers¶

## Example 10.1 Page No 253¶

In [1]:
# given data
V_CEQ= 7.5##  V
R_L= 50##  Ω
I_Csat= V_CEQ/R_L##  A
I_CQ= 0.01*I_Csat##  A
P_DQ= V_CEQ*I_CQ##  W
PP= 2*V_CEQ##  V
P_Dmax= PP**2/(40*R_L)##  W
P_Lmax= PP**2/(8*R_L)##  W
# The value of P_DQ
P_DQ= P_DQ*10**3##  mW
# The value of P_Dmax
P_Dmax= P_Dmax*10**3##  mW
# The value of P_Lmax
P_Lmax= P_Lmax*10**3##  mW
print "The value of P_DQ = %.2f mW"%P_DQ
print "The value of P_Dmax = %.2f mW"%P_Dmax
print "The value of P_Lmax = %.2f mW"%P_Lmax

The value of P_DQ = 11.25 mW
The value of P_Dmax = 112.50 mW
The value of P_Lmax = 562.50 mW


## Example 10.2 Page No 255¶

In [2]:
# given data
V_CC= 15##  V
I_Csat= 150##  mA
P_Lmax= 563##  mW
I= 0.02*I_Csat##  mA
Idc= 0.318*I_Csat##  mA
I_CC= I+Idc##  mA
P_CC= V_CC*I_CC##  mW
# The efficiency of amplifier
Eta= P_Lmax/P_CC*100##  %
print "The efficiency of amplifier = %.2f %%"%Eta

# Note: The answer in the book is not accurate

The efficiency of amplifier = 74.03 %


## Example 10.3 Page No 260¶

In [17]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import plot,xlabel,ylabel,title,show
# given data
V_CC= 40.0##  V
V_CEQ= 20.0##  V
R_L= 10.0##  Ω
I_Csat= V_CEQ/R_L##  A
V_CEcutoff= V_CEQ##  V
V_CE= arange(0,0.1+V_CEcutoff,0.1) #  V
I_C= (V_CEQ-V_CE)/R_L##  A
# The plot of ac load line,
plot(V_CE,I_C)
xlabel("VCE in volts")
ylabel("IC in A")
show()
print "AC load line shown in figure"

AC load line shown in figure


## Example 10.4 Page No 260¶

In [5]:
# given data
V_CC= 40##  V
V_BE= 0.7##  V
R= 1*10**3##  Ω
R_L= 10##  Ω
V_CEQ= 20##  V
I_CQ= (V_CC-2*V_BE)/(2*R)##  A
# The value of P_DQ
P_DQ= V_CEQ*I_CQ##  W
print "The value of P_DQ = %.2f W"%P_DQ
PP= 2*V_CEQ##  V
# The value of P_Lmax
P_Lmax= PP**2/(8*R_L)##  W
# The value of P_Dmax
P_Dmax= PP**2/(40*R_L)##  W
print "The value of P_Lmax = %.2f W"%P_Lmax
print "The value of P_Dmax = %.2f W"%P_Dmax

The value of P_DQ = 0.39 W
The value of P_Lmax = 20.00 W
The value of P_Dmax = 4.00 W


## Example 10.5 Page No 263¶

In [9]:
# given data
V_E= 1.43##  V
R_E= 100##  Ω
R_L= 100##  Ω
R_C= 1*10**3##  Ω
bita= 200#
Vt= 25*10**-3##  V
I_E= V_E/R_E##  A
I_CQ= I_E##  A
Zin= bita*R_L##  Ω
r_desh_e= Vt/I_CQ##  Ω
# The voltage gain of the driver stage
A= (R_C*Zin/(R_C+Zin))/(R_E+r_desh_e)#
print "The voltage gain of the driver stage = %.2f"%A
# On ignoring Zin and r_desh_e,
A= R_C/R_E#
print "On ignoring the value of Zin and r'e, the voltage gain = %.2f"%A

The voltage gain of the driver stage = 9.36
On ignoring the value of Zin and r'e, the voltage gain = 10.00


## Example 10.6 Page No 264¶

In [11]:
# given data
V_CC= 30.0##  V
PP= V_CC##  V
R_L= 100.0##  Ω
# The value of P_Lmax
P_Lmax= PP**2/(8*R_L)##  W
print "The value of PP = %.2f volts"%PP
print "The value of P_Lmax = %.2f W"%P_Lmax

The value of PP = 30.00 volts
The value of P_Lmax = 1.12 W


## Example 10.7 Page No 264¶

In [13]:
# given data
R_C= 1*10**3##  Ω
r_desh_e= 2.5##in Ω
Zin= 1.0*10**3##  Ω
A2= 10## unit less
A3= 1## unit less
A1= (R_C*Zin/(R_C+Zin))/r_desh_e## unit less
# The overall voltage gain
A= A1*A2*A3#
print "The overall voltage gain = %.2f"%A

The overall voltage gain = 2000.00


## Example 10.8 Page No 266¶

In [15]:
# given data
V_CC= 50.0##  V
V_CEsat= 1.0##  V
R_L= 5##  Ω
bita_dc= 90## unit less
I_Csat= (V_CC-V_CEsat)/R_L##  A
# The minimum base current that produces saturation
I_Bsat= I_Csat/bita_dc##  A
I_Bsat= I_Bsat*10**3##  mA
print "The minimum base current that produces saturation = %.2f mA"%I_Bsat

The minimum base current that produces saturation = 108.89 mA


## Example 10.9 Page No 267¶

In [16]:
# given data
I_Csat= 109*10**-3##  A
bita_dc= 200#
R_B= 1*10**3##  Ω
V_BE1= 0.7##  V
V_BE2= 1.6##  V
# The base current,
I_Bsat= I_Csat/bita_dc##  A
# The input voltage
Vin= I_Bsat*R_B+V_BE1+V_BE2##  V
print "The input voltage = %.2f volts"%Vin

The input voltage = 2.85 volts