Chapter 06 : Ac analysis of BJT circuits

Example 6.1, Page No 240

In [1]:
import math
#initialisation of variables
Vcc=12.0
R2=15.0*10**3
R1=33.0*10**3
rs=600

#Calculations
print("with no signal source")
Vb=(Vcc*R2)/(R1+R2)
print(" base bais voltage when no signal source is present %3.2fV " %Vb)
print(" signal source directly connected")
Vb=(Vcc*((rs*R2)/(rs+R2))/(R1+((rs*R2)/(rs+R2))))

#Results
print("base bais voltage is %3.2fV " %Vb)

with no signal source
 base bais voltage when no signal source is present 3.75V 
 signal source directly connected
base bais voltage is 0.21V

Example 6.2, Page No 244

In [2]:
import math

#initialisation of variables

Rc=2.2*10**3
Re=2.7*10**3
R1=18.0*10**3
R2=8.2*10**3
Vbe=.7

#Calculations
print("drawing dc load line")
Rldc=Rc+Re
print(" for Vce")
Ic=0
Vcc=20
Vce=Vcc-Ic*(Rc+Re)
print("plot point A at")
Ic=Vcc/(Rc+Re)
print("plot point B Ic=4.08mA and Vce=0")
print(" draw dc laod line through point A nad B")
Vb=(Vcc*R2)/(R1+R2)
Ve=Vb-Vbe
Ic=Ve/Re
Ie=Ic
print("drawing the ac load line")
Rlac=Rc#when there is no external Rl
Vce=Ic*Rc


#Results
print("The voltage is %.2f v " %Vce)

drawing dc load line
 for Vce
plot point A at
plot point B Ic=4.08mA and Vce=0
 draw dc laod line through point A nad B
drawing the ac load line
The voltage is 4.53 v

Example 6.3 Page No 251

In [3]:
import math

#initialisation of variables
Vce=4.5
Ib=40.0*10**-6

#Calculations
print("from current characteristic at Vce=4.5V and Ib=40uA")
Ic=4.0*10**-3
Ib=30.0*10**-6
hFE=Ic/Ib
print(" the value of hFE is %d " %hFE)
print("from output characteristic at Vce=4.5 and Ib=40uA")
Ic=0.2
Vce=6
hoe=(Ic/Vce)
R=1/hoe

#Results
print("the value of hoe is %3.1fuS " %(hoe*10**3))
print("the value of 1/hoe is %3.1fuS " %(1/hoe))

from current characteristic at Vce=4.5V and Ib=40uA
 the value of hFE is 133 
from output characteristic at Vce=4.5 and Ib=40uA
the value of hoe is 33.3uS 
the value of 1/hoe is 30.0uS

Example 6.4, Page No 253

In [4]:
import math
#initialisation of variables
hfe=133.0
hoe=33.3*10**-6
hfc=1+hfe

#Calculations
hob=hoe/(1+hfe)
A=hfe/(1+hfe)

#Results
print("tye value of a is %3.1fuS " %(A))

tye value of a is 1.0uS

Example 6.5 Page No 253

In [5]:
import math
#initialisation of variables

Ib=20.0*10**-6
Ic=1.0*10**-3
Ie=Ic

#Calculations
re=(26*10**-3)/Ie
hfe=Ic/Ib
hie=(1+hfe)*re
r=hie
B=hfe

#Results
print("the value of b is %3.1f " %(B))

the value of b is 50.0

Example 6.6 Page No 258

In [6]:
import math

#initialisation of variables
hie=2.1*10**3
hfe=75.0
hoe=1*10**-6
R1=68.0*10**3
R2=56.0*10**3
Rc=3.9*10**3
Rl=82*10**3

#Calculations
print(" input impedance Zi=R1||R2||hie")
Zi=((R1*R2*hie)/(R1+R2+hie))*10**-3
print(" input impedance is %3.2fKohm " %Zi)
print("output impedance is Zo=Rc||(1/hoe)")
Zo=((Rc*(1/hoe))/(Rc+(1/hoe)))*10**-3
print(" output impadance is %f3.2fKohm " %Zo)
Av=-(hfe*((Rc*Rl)/(Rc+Rl)))/hie


#Results
print(" voltage gain is %d " %Av)

 input impedance Zi=R1||R2||hie
 input impedance is 63416.34Kohm 
output impedance is Zo=Rc||(1/hoe)
 output impadance is 3.8848493.2fKohm 
 voltage gain is -132

Example 6.7, Page No 259

In [7]:
import math
#initialisation of variables
Ic=1.5*10**-3
Rc=4.7*10**3
Rl=56.0*10**3

#Calculations
Ie=Ic
re=(26*10**-3)/Ie
Av=-(((Rc*Rl)/(Rc+Rl))/re)

#Results
print(" voltage gain is %d " %Av)

 voltage gain is -250

Example 6.8 Page No 262

In [8]:
import math 

#initialisation of variables
hie=2.1*10**3
hfe=75.0
hoe=1.0*10**-6
Re=4.7*10**3
R1=68.0*10**3
R2=56.0*10**3
Rc=3.9*10**3
Rl=82.0*10**3

#Calculations
Zb=hie+Re*(1+hfe)
print(" input impedance is Zi=R1||R2||Zb")
Zi=((R1*R2*Zb)/(R1+R2+Zb))
print(" input circuit resistance is %3.3fKohm " %Zi)
Zo=Rc
Av=-hfe*((Rc*Rl)/(Rc+Rl))/(hie+Re*(1+hfe))

#Results
print("voltage gain is %3.3f " %Av)

 input impedance is Zi=R1||R2||Zb
 input circuit resistance is 2830983654.045Kohm 
voltage gain is -0.777

Example 6.9 Page No 267

In [9]:
import math
#initialisation of variables
hie=2.1*10**3
hfe=75.0
R1=10.0*10**3
R2=10.0*10**3
Re=4.7*10**3
Rl=12.0*10**3
rs=1.0*10**3

#Calculations
print(" Rl is not connected")
hic=hie
hfc=1+hfe
Zb=hic+hfc*(Re)
Zi=(R1*R2*Zb)/(R1+R2+Zb)
Ze=(hic+(R1*R2*rs)/(R1+R2+rs))/hfc
Z0=(Ze*Re)/(Ze+Re)
print(" when Rl is connected")
Zb=hic+hfc*((Re*Rl)/(Re+Rl))
Zi=(R1*R2*Zb)/(R1+R2+Zb)
hib=hie/(1+hfe)
Av=((Re*Rl)/(Re+Rl))/(hib+((Re*Rl)/(Re+Rl)))

#Results
print("voltage gain is %3.3f " %Av)

 Rl is not connected
 when Rl is connected
voltage gain is 0.992

Example 6.10 Page No 273

In [10]:
import math

#initialisation of variables
hie=2.1*10**3
hfe=75.0
Re=4.7*10**3
Rc=3.9*10**3
Rl=82.0*10**3

#Calculations
hib=hie/(1+hfe)
hfb=hfe/(1+hfe)
Zi=(hib*Re)/(Re+hib)
print("input impedance is %3.2fohm " %Zi)
Zo=Rc
print(" output impedance is %3.2fohm " %Zo)
Av=(hfb*((Rc*Rl)/(Rc+Rl)))/hib

#Results
print(" voltage gain is %3.2f " %Av)

input impedance is 27.47ohm 
 output impedance is 3900.00ohm 
 voltage gain is 132.96

Example 6.11, Page No 273

In [11]:
import math

#initialisation of variables
hib=27.6
hfb=.987
R1=68.0*10**3
R2=56.0*10**3
Re=4.7*10**3
Rc=3.9*10**3
Rl=82.0*10**3

#Calculations
Rb=(R1*R2)/(R1+R2)
Ze=hib+Rb*(1-hfb)
Zi=(Ze*Re)/(Ze+Re)
Av=(hfb*((Rc*Rl)/(Rc+Rl)))/(hib+Rb*(1-hfb))

#Results
print("voltage gain is %3.3f " %Av)

voltage gain is 8.609

Example 6.12, Page No 277

In [12]:
import math
#initialisation of variables
Rc=5.6*10**3
Rl=33.0*10**3
rs=600.0
hfe=100
hie=1.5*10**3
vs=50.0*10**-3

#Calculations
print(" CE circuit operation with vs at transistor base and Re bypassed")
Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie
Zb=hie
Rb=(R1*R2)/(R1+R2)
Zi=(Rb*Zb)/(Rb+Zb)
vi=(vs*Zi)/(rs+Zi)
vo=Av*vi
print("Cb circuit operation with vs at emitter and the base resistor bypassed")
Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie
Ze=hie/(1+hfe)
Zi=(Ze*Re)/(Ze+Re)
vi=(vs*Zi)/(rs+Zi)
vo=Av*vi

#Results
print("voltage vo is %3.2f " %(vo*10**3))

 CE circuit operation with vs at transistor base and Re bypassed
Cb circuit operation with vs at emitter and the base resistor bypassed
voltage vo is 384.29

Example 6.13, Page No 279

In [13]:
import math
#initialisation of variables
Io=50.0*10**-9
Vbe=.7
Vbc=-10
Af=.995
Ar=.5
Vt=26.0*10**-3
n=2.0
Vd=-10.0

#Calculations
x=Vd/(n*Vt)
Idc=(Io*((2.73**-x)-1))*10**9
Idc=Io*(-1)
y=Vbe/(n*Vt)
Ide=Io*((2.73**y)-1)
I1=Af*Ide
I2=Ar*Idc
Ic=I1-Idc
Ie=Ide-I2
Ib=Ie-Ic

#Results
print("voltage gain is %3.3f " %(Ib*10**6))

voltage gain is 185.909