Chapter 5 Transistor Bias Considerations

Example 5.1 , Page no:143

In [1]:
import math
from __future__ import division

#CALCULATIONS
Icbo=(500*(2**((90-25)/10)))/1000

#RESULTS
print"The value of Icbo=",round(Icbo,3),"mA";

The value of Icbo= 45.255 mA

Example 5.6 , Page no:145

In [2]:
import math
from __future__ import division

#initialisation of variables
Vee=4 #v
Vbeq=0.7 #v
b=50 #Beta
Vcc=18 #v
Rc=6 #kΩ
Re=2*10**3     #kΩ
b1=100
Re1=2 #*10^3     //kΩ

#CALCULATIONS
Rb=25*10**3 #kΩ
Icq=((Vee-Vbeq)/((Rb/b)+((b+1)/b)*Re))*1000
Vceq=Vcc+Vee-(Rc+((b+1)/b)*Re1)*Icq  
Icq1=((Vee-Vbeq)/((Rb/b1)+((b1+1)/b1)*Re))*1000
Vceq1=Vcc+Vee-(Rc+((b1+1)/b1)*Re1)*Icq1

#RESULTS
print"The value of Icq=",round(Icq,3),"mA";
print"For beta=100";
print"The value of Vceq=",round(Vceq,3),"V";
print"For beta=100";
print"The value of Icq=",round(Icq1,3),"mA";
print"The value of Vceq=",round(Vceq1,3),"V";

The value of Icq= 1.299 mA
For beta=100
The value of Vceq= 11.554 V
For beta=100
The value of Icq= 1.454 mA
The value of Vceq= 10.341 V

Example 5.8 , Page no:146

In [3]:
import math
from __future__ import division

#initialisation of variables
Vcc=15 #v
Vee=4 #v 
Vbeq=0.7 #v
b=50 #Beta
Re=3*10**3 #kΩ
Rc=7 #kΩ
Vee1=5
Re1=3 #*10^3
b1=100
Vee2=5
Re2=3 #*10^3

#CALCULATIONS
Ieq=(Vee-Vbeq)/Re*1000
Icq=(b/(b+1))*Ieq
Ibq=Icq/b
Vceq=Vcc+Vee1-(Ieq*Re1)-(Icq*Rc)
Icq1=(b1/(b1+1))*Ieq
Ibq1=Icq/b1
Vceq1=Vcc+Vee2-(Ieq*Re2)-(Icq1*Rc)

#RESULTS
print"For beta=50";
print"The value of Ieq=",round(Ieq,3),"mA";
print"The value of Icq=",round(Icq,3),"mA";
print"The value of Ibq=",round(Ibq,3),"mA";
print"The value of Vceq=",round(Vceq,3),"V";
print"For beta=100";
print"The value of Ieq=",round(Ieq,3),"mA";
print"The value of Icq=",round(Icq1,3),"mA";
print"The value of Ibq=",round(Ibq1,3),"mA";
print"The value of Vceq=",round(Vceq1,3),"V";

For beta=50
The value of Ieq= 1.1 mA
The value of Icq= 1.078 mA
The value of Ibq= 0.022 mA
The value of Vceq= 9.151 V
For beta=100
The value of Ieq= 1.1 mA
The value of Icq= 1.089 mA
The value of Ibq= 0.011 mA
The value of Vceq= 9.076 V

Example 5.9 , Page no:147

In [4]:
import math
from __future__ import division

#initialisation of variables
Vcc=15
Vee=4
Vbeq=0.7
Rb=500

#CALCULATIONS
Sb=((Vcc-Vbeq)/Rb)*10**3
Icq=(Sb*(100-50))/1000

#RESULTS
print"The value of Sb=",round(Sb,3);
print"The value of Icq=",round(Icq,3),"mA";

The value of Sb= 28.6
The value of Icq= 1.43 mA

Example 5.11 , Page no:148

In [5]:
import math
from __future__ import division

#initialisation of variables
Vbb=6
Vbeq1=0.7
Icbo1=0.5
Rb=50
Re=1
B=75 #Beta

#CALCULATIONS
Icq1=((Vbb-Vbeq1+Icbo1*(0.5*51*10**-3))/((Rb*10**3/B)+Re*10**3))*10**3
Icbo2=(Icbo1*10**-6*2**2)*10**6
Vbeq=(-2*10**-3)*20
Vbeq2=Vbeq1+Vbeq
Icq2=((Vbb-Vbeq2+Icbo2*(2*51*10**-3))/((Rb*10**3/B)+Re*10**3))*10**3

#RESULTS
print"The value of Icq1=",round(Icq1,3),"mA";
print"The value of Icbo=",round(Icbo2,3),"mA";
print"The value of Vbeq=",round(Vbeq,3),"V";
print"The value of Vbeq2=",round(Vbeq2,3),"V";
print"The value of Icq2=",round(Icq2,3),"mA";

The value of Icq1= 3.188 mA
The value of Icbo= 2.0 mA
The value of Vbeq= -0.04 V
The value of Vbeq2= 0.66 V
The value of Icq2= 3.326 mA

Example 5.16 , Page no:150

In [6]:
import math
from __future__ import division

#initialisation of variables
B=50 #beta
Vee=5
Vbeq1=0.7
T2=125
Re=3*10**3
Icbo1=0.5 #*10^-6

#CALCULATIONS
Icq2=(((B+1)/B)*((Vee-Vbeq1+0.002*(T2-25))/Re)+(2**((T2-25)/10))*Icbo1*10**-6)*10**3

#RESULTS
print"The value of Icq2=",round(Icq2,3),"mA";

The value of Icq2= 2.042 mA

Example 5.19 , Page no:152

In [7]:
import math
from __future__ import division

#initialisation of variables
B=75 #beta
Rb=454.5 #kΩ
Vbb=1.818
Vbeq=0.7
Re=90

#CALCULATIONS
Icbo=0.2*10**-6
deltaRe=110-90
Sre=((B*Rb*Icbo-B**2*(Vbb-Vbeq+Icbo*Rb))/((Rb+B*Re)**2))*10**4
Icq=(Sre*deltaRe)/10

#RESULTS
print"The value of Sre=",round(Sre,3),"* 10^-4 A/Ohm";
print"The value of Icq=",round(Icq,3),"* 10^-4 mA";

The value of Sre= -1.212 * 10^-4 A/Ohm
The value of Icq= -2.423 * 10^-4 mA

Example 5.25 , Page no:156

In [8]:
import math
from __future__ import division

#initialisation of variables
Vdd=15
Idqmax=5.5
Idqmin=1.3
Rd=2.5 #kΩ

#CALCULATIONS
Vdsqmax=Vdd-Idqmax*Rd
Vdsqmin=Vdd-Idqmin*Rd

#RESULTS
print"The value of Vdsqmax=",round(Vdsqmax,3),"V";
print"The value of Vdsqmin=",round(Vdsqmin,3),"V";

The value of Vdsqmax= 1.25 V
The value of Vdsqmin= 11.75 V

Example 5.26 , Page no:157

In [9]:
import math
from __future__ import division

#initialisation of variables
Vdd=24 #V
Idqmax=2.5
Idqmin=1.2
Rs=1 #kΩ
Rd=3 #kΩ

#CALCULATIONS
Vdsqmax=Vdd-Idqmax*(Rs+Rd)
Vdsqmin=Vdd-Idqmin*(Rs+Rd)

#RESULTS
print"The value of Vdsqmax=",round(Vdsqmax,3),"V";
print"The value of Vdsqmin=",round(Vdsqmin,3),"V";

The value of Vdsqmax= 14.0 V
The value of Vdsqmin= 19.2 V

Example 5.28 , Page no:159

In [10]:
import math
from __future__ import division

#initialisation of variables
Vdd=24
Idqmax=4
Idqmin=2.8
Rs=2 #MΩ
Rd=1 #MΩ

#CALCULATIONS
Vdsqmax=Vdd-Idqmax*(Rs+Rd)
Vdsqmin=Vdd-Idqmin*(Rs+Rd)

#RESULTS
print"The value of Vdsqmax=",round(Vdsqmax,3),"V";
print"The value of Vdsqmin=",round(Vdsqmin,3),"V";

The value of Vdsqmax= 12.0 V
The value of Vdsqmin= 15.6 V