Chapter 05 : Transistor Characteristic

Example 5.1a, Page No 133

In [1]:
import math
#initialisation of variables
B=100.0  #Beta
Ico=20.0  #in nA 
Rc=3.0
Rb=200.0
Vbb=5.0   #in V
Vcc=10      #in V
Vbe=0.7      #in Active region

#Applying KVL to base circuit

#Vbb+Rb*Ib+Vbe=0

#Calculations
Ib=(Vbb-Vbe)/Rb      #in mA
#Ico<<Ib
Ic=B*Ib  #in mA
#To verify the Active region Assumption
#Vcc+Rc*Ic+Vcb+Vbe=0

Vcb=(-Rc*Ic)+Vcc-Vbe       #in V

#Results
print("Vcb =  %.2f V " %Vcb)
if Vcb>0 :
  print('Positive value of Vcb represents reversed biased collector junction and Transistor in active region')


print("Current in transistor(Ic) is %.2f mA " %Ic)
print("Current in transistor(Ib) is %.2f mA " %Ib)

Vcb =  2.85 V 
Positive value of Vcb represents reversed biased collector junction and Transistor in active region
Current in transistor(Ic) is 2.15 mA 
Current in transistor(Ib) is 0.02 mA

Example 5.1b, Page No 133

In [2]:
import math

#initialisation of variables
B=100.0 #Beta
Ico=20.0    #in nA 
Rc=3.0
Ico=20    #in nA
Rb=200.0
Re=2.0
Vbb=5.0      #in V
Vcc=10.0    #in V
Vbe=0.7   #in Active region

#Ico<<Ib  Assuming
#Itot=Ib+Ic=Ib+B*Ib=(B+1)*Ib
#Applying KVL to base circuit
#Vbb+Rb*Ib+Vbe+Re*Itot=0

#Calculations
Ib=(Vbb-Vbe)/(Rb+(Re*(B+1)))     #in mA

Ic=B*Ib     #in mA

#Hence Ico<<Ib
#To verify the Active region Assumption
#Vcc+Rc*Ic+Vcb+Vbe=0

Vcb=(-Rc*Ic)+Vcc-Vbe-(Re*(B+1)*Ib)   #in V

print("Vcb =  %.2f V " %Vcb)

if Vcb>0 :
  print('Positive value of Vcb represents reversed biased collector junction and Transistor in active region')

#Results
print("Current in transistor(Ic) is %.2f mA " %Ic)
print("Current in transistor(Ib) is %.2f mA " %Ib)

Vcb =  3.93 V 
Positive value of Vcb represents reversed biased collector junction and Transistor in active region
Current in transistor(Ic) is 1.07 mA 
Current in transistor(Ib) is 0.01 mA

Example 5.2 Page No 134

In [3]:
import math

#initialisation of variables
Rc=3.0
Rb=50.0 
Vbb=5.0       #in V
Vcc=10.0 #in V
Vce=0.2    #in V
Vbe=0.8      #in Active region
hFE=100.0 

#Assuming transistor in saturated region
#Applying KVL to base circuit
#Vbb+Rb*Ib+Vbe=0

#Calculations
Ib=(Vbb-Vbe)/Rb    #in mA

#Applying KVL to Collector circuit
#Vcc+Rc*Ic+Vce=0

Ic=(Vcc-Vce)/Rc     #in mA

Ib_min=Ic/hFE

print("Minimum Ib = %.2f mA " %Ib_min)

if Ib>Ib_min :
  print('Transistor in saturated Region')

#Results
print("Current in transistor(Ic) is %.2f mA " %Ic)
print("Current in transistor(Ib) is %.2f mA " %Ib)

Minimum Ib = 0.03 mA 
Transistor in saturated Region
Current in transistor(Ic) is 3.27 mA 
Current in transistor(Ib) is 0.08 mA

Example 5.3, Page No 134

In [4]:
import math
#initialisation of variables
Ib=0.01     #mA
Ic=100.0*Ib


#Calculations
Vcb=5-Ic-0.7-(101*Ib)


#Results
print("The value of Vcb is= %.2f V " %(Vcb))
print("Since Vcb is positive, the transistor is in the active region ")

The value of Vcb is= 2.29 V 
Since Vcb is positive, the transistor is in the active region

Example 5.4, Page No 135

In [5]:
import math
#initialisation of variables
Ve= 4.0     #V
Ie=2        #mA
Vc=12-(2*2)
beta=19.0

#Calculations
Rb=2*(1.0+beta)

#Results
print("The value of Vcb is= %.2f V " %(Rb))

The value of Vcb is= 40.00 V

Example 5.5, Page No 135

In [6]:
import math
#initialisation of variables
Ve= 4.0     #V
Ie=2        #mA
Vc=12-(2*2)
beta=100.0

#Calculations
Ib=(2.7-0.7)/beta
Ic=beta*Ib
Vc=(10.0-Ic)/2

#Results
print("The value of Vcb is= %.2f V " %(Vc))

The value of Vcb is= 4.00 V

Example 5.6, Page No 135

In [7]:
import math
#initialisation of variables
Ie1=1.0     #mA
Vc1=10.7*Ie1-10
Vbe2=0.7

#Calculations
Ie2=(10+Vc1-Vbe2)/10.0
Vc2=10-Ie1
Vcb2=Vc2-Vbe2

#Results
print("The value of Vcb is= %.2f V " %(Vcb2))

The value of Vcb is= 8.30 V

Example 5.7a, Page No 142

In [8]:
import math
#initialisation of variables
hfe=100
Ib=4.2/50     #mA
Ic=9.8/3
Ib=Ic/hfe

#Results
print("The value of Ib is= %.3f V " %(Ib))

The value of Ib is= 0.033 V

Example 5.9a, Page No 144

In [9]:
import math
#initialisation of variables
Ie1=1.0     #mA
Ic1=0.99    #mA
Vcb1=12-(5*Ic1)-5
Ve=5-0.7
Vbe2=2.5-Ve

#Results
print("The value of Vbe2 is= %.2f V " %(Vbe2))

The value of Vbe2 is= -1.80 V

Example 5.9b, Page No 144

In [10]:
import math
#initialisation of variables
Ie2=1.0        #mA
Ic2=0.99     #mA
Vcb2=12-(5*Ic2)-2.5

#Calculations
Ve2=2.5-0.7
Vbe1=0-Ve
V0=12-(5*Ic2)

#Results
print("The value of V0 is= %.2f V " %(V0))

The value of V0 is= 7.05 V