Chapter 9 Class A Power Amplifiers

Example 9.1 Page No 220

In [1]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import plot,xlabel,ylabel,show,title,subplot
# given data
V_CC= 10##  V
V_BE= 0.7##  V
R1= 2.2##  kΩ
R2= 10##  kΩ
R_E= 1##  kΩ
R_C= 3.6##  kΩ
R= 1.5##  kΩ
# The base voltage
V_B= R1*V_CC/(R1+R2)##  V
# The emitter current,
I_E= (V_B-V_BE)/R_E##  mA
# The collector current,
I_CQ= I_E##  mA
# The collector emitter voltage,
V_CE= V_CC-I_E*(R_C+R_E)##  V
V_CEQ= V_CE##  V
# The saturation current,
I_Csat= V_CC/(R_C+R_E)##  mA
V_CEcutoff= V_CC##  V
V_CE= arange(0,0.1+V_CEcutoff,0.1)  #  V
I_C= (V_CC-V_CE)/(R_C+R_E)##  mA
# The dc and ac load line
subplot(121)
plot(V_CE,I_C)
xlabel("V_CE in volts")
ylabel("I_C in mA")
title("DC load line")
r_L= R_C*R/(R_C+R)##  kΩ
I_Csat= I_CQ+V_CEQ/r_L##  mA
Vce_cutoff= V_CEQ+I_CQ*r_L##  V
x=[0,Vce_cutoff]#
y=[I_Csat, 0]
subplot(122)
plot(x,y)
xlabel("V_CE in volts")
ylabel("I_C in mA")
title("AC load line")
show()
print "DC and AC load line shown in figure."

DC and AC load line shown in figure.

Example 9.2 Page No 221

In [6]:
# given data
V_BE= 0.7##  V
V_CC= 30##  V
R_E= 8.2##  Ω
R1= 22##  Ω
R2= 47##  Ω
R_C= 10##  Ω
R_L= 30##in Ω
# The base to ground voltage,
V_B= R1*V_CC/(R1+R2)##  V
# The emitter current,
I_E= (V_B-V_BE)/R_E##  A
# The collector current,
I_CQ= I_E##  A
# The collector emitter voltage,
V_CEQ= V_CC-I_E*(R_E+R_C)##  V
# The load resistance,
r_L= R_C*R_L/(R_C+R_L)##  Ω
I_Csat= I_E+V_CEQ/r_L##  A
Vce_cutoff= V_CEQ+I_CQ*r_L##  V
print "The cut off value of V_CE = %.2f volts"%Vce_cutoff

The cut off value of V_CE = 18.66 volts

Example 9.3 Page No 223

In [7]:
# given data
V_BE= 0.7##  V
V_CC= 20##  V
V_B= 10##  V
R_E= 50##  Ω
# The collector current,
I_CQ= (V_B-V_BE)/R_E##  A
# The collector emitter voltage,
V_CEQ= V_CC-I_CQ*R_E##  V
R1= 50##  Ω
R2= 50##  Ω
# The load resistance,
r_L= R1*R2/(R1+R2)##  Ω
I_Csat= I_CQ+V_CEQ/r_L##  A
Vce_cutoff= V_CEQ+I_CQ*r_L##  V
print "The cut off value of V_CE = %.2f volts"%Vce_cutoff

The cut off value of V_CE = 15.35 volts

Example 9.4 Page No 227

In [12]:
# given data
V_E= 1.0##  V
R_E=1*10**3##  Ω
V_CC= 10.0##  V
R_C= 4.0*10**3##  Ω
R_L= 10.0*10**3##  Ω
# The collector current,
I_CQ= V_E/R_E##  A
I_C= I_CQ##  A
# The collector emitter voltage,
V_CEQ= V_CC-I_C*(R_C+R_E)##  V
# The load resistance,
r_L= R_L*R_C/(R_L+R_C)##  Ω
#The ac compliance,
PP= 2*I_CQ*r_L##  V
print "The ac compliance = %.2f volts"%PP

The ac compliance = 5.71 volts

Example 9.5 Page No 227

In [13]:
# given data
V_E= 1.0##  V
R_E=1*10**3##  Ω
R_C= 4.0*10**3##  Ω
V_CC= 10.0##  V
I_CQ= V_E/R_E##  A
I_C= I_CQ##  A
V_CEQ= V_CC-I_C*(R_C+R_E)##  V
# (i) when R_L = 1 MΩ, the value of 2I_CQrL
R_L= 1*10**6##  Ω
r_L= R_L*R_C/(R_L+R_C)##  Ω
I_CQrL= I_CQ*r_L##in A
print "When R_L = 1 MΩ, the value of 2I_CQrL = %.2f volts"%(2*I_CQrL)
# (ii) when R_L = 100 kΩ, the value of 2I_CQrL
R_L= 100*10**3##  Ω
r_L= R_L*R_C/(R_L+R_C)##  Ω
I_CQrL= I_CQ*r_L##in A
print "When R_L = 100 kΩ, the value of 2I_CQrL = %.2f volts"%(2*I_CQrL)
# (iii) when R_L = 10 kΩ, the value of 2I_CQrL
R_L= 10*10**3##  Ω
r_L= R_L*R_C/(R_L+R_C)##  Ω
I_CQrL= I_CQ*r_L##in A
print "When R_L = 10 kΩ, the value of 2I_CQrL = %.2f volts"%(2*I_CQrL)
# (iv) when R_L = 1 kΩ, the value of 2I_CQrL
R_L= 1*10**3##  Ω
r_L= R_L*R_C/(R_L+R_C)##  Ω
I_CQrL= I_CQ*r_L##in A
print "When R_L = 1 kΩ, the value of 2I_CQrL = %.2f volts"%(2*I_CQrL)
# (v) when R_L = 100 Ω, the value of 2I_CQrL
R_L= 100##  Ω
r_L= R_L*R_C/(R_L+R_C)##  Ω
I_CQrL= I_CQ*r_L##in A
print "When R_L = 100 Ω, the value of 2I_CQrL = %.2f volts"%(2*I_CQrL)

When R_L = 1 MΩ, the value of 2I_CQrL = 7.97 volts
When R_L = 100 kΩ, the value of 2I_CQrL = 7.69 volts
When R_L = 10 kΩ, the value of 2I_CQrL = 5.71 volts
When R_L = 1 kΩ, the value of 2I_CQrL = 1.60 volts
When R_L = 100 Ω, the value of 2I_CQrL = 0.20 volts

Example 9.6 Page No 230

In [18]:
# given data
V_CC= 12##  V
V_BE= 0.7##  V
I_CQ= 5*10**-3##  A
bita= 200## unit less
# The emitter voltage,
V_E= 0.1*V_CC##  V
# The emitter current,
I_E= I_CQ##  A
# The emitter resistance,
R_E= V_E/I_E##  Ω
# The collector resistance,
R_C= 4*R_E##  Ω
# The base voltage,
V_B= V_E+V_BE##  V
I_C= I_CQ##  A
I_B= I_C/bita##  A
R= V_CC/(10*I_B)##  Ω
R2= V_B/(10*I_B)##  Ω
R1= R-R2##  Ω
R1= R1*10**-3##  k ohm
R2= R2*10**-3##  k ohm
R_C= R_C*10**-3##  k ohm
print "The value of R1 is    : %.2f"%(R1),"  kΩ  (standard value : 39 kΩ)"
print "The value of R2 is    : %.2f"%(R2),"   kΩ  (standard value : 7.5 kΩ)"
print "The value of R_E is   : %.2f"%(R_E),"  Ω  (standard value : 240 Ω)"
print "The value of R_C is   : %.2f"%(R_C),"   kΩ  (standard value : 1 kΩ)"

The value of R1 is    : 40.40   kΩ  (standard value : 39 kΩ)
The value of R2 is    : 7.60    kΩ  (standard value : 7.5 kΩ)
The value of R_E is   : 240.00   Ω  (standard value : 240 Ω)
The value of R_C is   : 0.96    kΩ  (standard value : 1 kΩ)

Example 9.7 Page No 231

In [19]:
# given data
I_CQ= 5*10**-3##  A
R_C= 1*10**3##  Ω
R_L= 1*10**3##  Ω
# The load resistance
r_L= R_C*R_L/(R_C+R_L)##  Ω
# The ac compliance,
PP= 2*I_CQ*r_L##  V
I_CQ= 5.15*10**-3##  A
PP= 2*I_CQ*r_L##  V
print "The ac compliance = %.2f volts"%PP

The ac compliance = 5.15 volts

Example 9.9 Page No 234

In [21]:
# given data
V_CC= 12.0##  V
V_BE= 0.7##  V
R_C= 1*10**3##  Ω
R_E= 240.0##  Ω
r_L= 500.0##  Ω
bita= 200.0## unit less
# The required collector current,
I_CQ= V_CC/(R_C+R_E+r_L)##  A
# The emitter voltage,
V_E= I_CQ*R_E##  V
# The base voltage,
V_B= V_E+V_BE##  V
I_C= I_CQ##  A
I_B= I_C/bita##  A
# The total resistance of the voltage divider,
R= V_CC/(10*I_B)##  Ω
R2= V_B/(10*I_B)##  Ω
R1= R-R2##  Ω
R1= R1*10**-3##  k ohm
R2= R2*10**-3##  k ohm
R_C= R_C*10**-3##  k ohm
print "The value of R1 is    : %.2f"%(R1),"  kΩ  (standard value : 39 kΩ)"
print "The value of R2 is    : %.2f"%(R2),"   kΩ  (standard value : 7.5 kΩ)"
print "The value of R_E is   : %.2f"%(R_E),"  Ω  (standard value : 240 Ω)"
print "The value of R_C is   : %.2f"%(R_C),"   kΩ  (standard value : 1 kΩ)"

The value of R1 is    : 27.97   kΩ  (standard value : 39 kΩ)
The value of R2 is    : 6.83    kΩ  (standard value : 7.5 kΩ)
The value of R_E is   : 240.00   Ω  (standard value : 240 Ω)
The value of R_C is   : 1.00    kΩ  (standard value : 1 kΩ)

Example 9.10 Page No 236

In [22]:
# given data
R_C= 3.6##  kΩ
R_L= 1.5##  kΩ
V_CEQ= 4.94##  V
I_CQ= 1.1##  mA
# The quiescent power dissipation of the transistor,
P_DQ= V_CEQ*I_CQ##  mW
r_L= R_C*R_L/(R_C+R_L)##  kΩ
PP= 2*I_CQ*r_L##  V
# The maximum ac load power,
P_Lmax= PP**2/(8*R_L)##  mW
print "The maximum ac load power = %.2f mW"%P_Lmax

The maximum ac load power = 0.45 mW

Example 9.11 Page No 240

In [23]:
# given data
V_E= 1.71##  V
R_E= 240##  Ω
V_CC= 12##  V
R_C= 1*10**3##  Ω
R_L= 1*10**3##  Ω
I= 0.355*10**-3##  A
I_CQ= V_E/R_E##  A
I_C= I_CQ##  A
# The collector emitter voltage,
V_CEQ= V_CC-I_C*(R_C+R_E)##  V
r_L= R_C*R_L/(R_C+R_L)##  Ω
PP= 2*V_CEQ##  V
# The maximum ac load power,
P_Lmax= PP**2/(8*R_L)##  W
I_CC= I_C+I##  A
P_CC= V_CC*I_CC##  W
# The efficiency 
Eta= P_Lmax/P_CC*100##  %
print "The efficiency = %.2f %%"%Eta

The efficiency = 5.58 %

Example 9.12 Page No 244

In [24]:
# given data
Ta= 70## ambient temperature = %.2f °C
P= 30##  power dissipation = %.2f W
theta_CS= 0.5##  °C/W
theta_SA= 1.5##  °C/W
# The case temperature
Tc= Ta+P*(theta_CS+theta_SA)##  °C
# The power rating
P_Dmax= 60##  W
print "The case temperature = %.2f °C"%Tc
print "The power rating = %.2f watt"%P_Dmax

The case temperature = 130.00 °C
The power rating = 60.00 watt