# chapter06:Transistor Biasing and Stabilization¶

## Example E1 - Pg 191¶

In :
#Determine the Q point
#given
B=50.;            #dc beta
Rc=2.2*10.**3.;#ohm      #resistor connected to collector
Rb=270.*10.**3.;#ohm      #resistor connected to base
Vcc=9.;#V            #Voltage supply across the collector resistor
Vbe=0.7;#V          #base to emitter voltage
Ib=(Vcc-Vbe)/Rb;         #Base current
Ic=B*Ib;         #Colletor current
Ics=Vcc/Rc;       #Colletor saturation current

#Actual Ic is the smaller of the above two values
Vce=Vcc-Ic*Rc;
print '%s %.1f %s %.1f %s' %("The Q point is =",Vce,'V',Ic*1000,'mA');

#Note--In book Vce = 5.7 V because of approaximation

The Q point is = 5.6 V 1.5 mA


## Example E2 - Pg 192¶

In :
#Determine the Q point
#given
B=150.;            #dc beta
Rc=1.*10.**3.;#ohm      #resistor connected to collector
Rb=100.*10.**3.;#ohm      #resistor connected to base
Vcc=10.;#V            #Voltage supply across the collector resistor
Vbe=0.7;#V          #base to emitter voltage
Ib=(Vcc-Vbe)/Rb;         #Base current
Ic=B*Ib;         #Colletor current
Ics=Vcc/Rc;       #Colletor saturation current

#Actual Ic is the smaller of the above two values i.e. Ic(sat) and since the transistor is in saturation mode therefore Vce will become 0

Vce=0;
print '%s %.f %s %.f %s' %("The Q point is =",Vce,"V",Ics*1000,"mA");

The Q point is = 0 V 10 mA


## Example E3 - Pg 192¶

In :
#Determine Rb and percentage change in collector current due to temperature rise
#given

#Calculating the base resistance
B=20.;            #dc beta
Rc=1.*10.**3.;#ohm      #resistor connected to collector
Ic=1.*10.**-3.;#A       #collector current
Vcc=6.;#V            #Voltage supply across the collector resistor
Vbe=0.3;#V       #for germanium
Icbo=2.*10.**-6.;#A       #collector  to base leakage current

Ib=(Ic-(1.+B)*Icbo)/B;
Rb=(Vcc-Vbe)/Ib;

print '%s %.f %s' %("The value of resistor Ib is =",120,'kohm');

Rb=120.*10.**3.;#ohm approax

#Now when temperature rise
Icbo=10.*10.**-6.;#A #collector  to base leakage current
B=25.;#dc beta
Ic1=B*Ib+(B+1)*Icbo;# #changed collector current
perc=(Ic1-Ic)*100./Ic;#percentage increase
print '%s %.f %s' %("The percentage change in collector current is =",perc,"percent");

The value of resistor Ib is = 120 kohm
The percentage change in collector current is = 46 percent


## Example E4 - Pg 193¶

In :
#Determine the Q point at two different B
#given

#At B=50

B=50.;            #dc beta
Rc=2.*10.**3.;#ohm      #resistor connected to collector
Rb=300.*10.**3.;#ohm      #resistor connected to base
Vcc=9.;#V            #Voltage supply across the collector resistor
Ib=Vcc/Rb;         #Base current
Ic=B*Ib;         #Colletor current
Ics=Vcc/Rc;       #Colletor saturation current

#Actual Ic is the smaller of the above two values

Vce=Vcc-Ic*Rc;
print '%s %.2f %s %.1f %s' %("The Q point (At B=50) =",Vce,"V",Ic*1000,"mA");

#At B=150

B1=150.;            #dc beta
Ic1=B*Ib;         #Colletor current
Ics1=Vcc/Rc;       #Colletor saturation current

#Actual Ic is the smaller of the above two values i.e. Ic(sat) and since the transistor is in saturation mode therefore Vce will become 0

Vce=0;
print '%s %.f %s %.1f %s' %("\nThe Q point (At B=150) is =",Vce,"V",Ics*1000,"mA");

print '%s %.f' %("\nThe factor at which collector current increases =",Ics1/Ic);

The Q point (At B=50) = 6.00 V 1.5 mA

The Q point (At B=150) is = 0 V 4.5 mA

The factor at which collector current increases = 3


## Example E5 - Pg 196¶

In :
#determine Q point in collector to base bias circuit
#given
B=100.;            #dc beta
Rc=500.;#ohm      #resistor connected to collector
Rb=500.*10.**3.;#ohm      #resistor connected to base
Vcc=10.;#V            #Voltage supply across the collector resistor
Ib=Vcc/(Rb+B*Rc);         #Base current
Ic=B*Ib;         #Colletor current
Ics=Vcc/Rc;       #Colletor saturation current

#Actual Ic is the smaller of the above two values

Vce=Vcc-(Ic+Ib)*Rc;
print '%s %.1f %s %.1f %s' %("The Q point is =",Vce,"V",Ic*1000,"mA");

The Q point is = 9.1 V 1.8 mA


## Example E6 - Pg 196¶

In :
#Calculate the collector current and change in it if B is changed by three times of previous B
#given
B=50.;            #dc beta
Rc=2.*10.**3.;#ohm      #resistor connected to collector
Rb=300.*10.**3.;#ohm      #resistor connected to base
Vcc=9.;#V            #Voltage supply across the collector as it is PNP so taking positive
Ib=Vcc/(Rb+B*Rc);         #Base current
Ic=B*Ib;         #Colletor current
print '%s %.3f %s' %("Collector current (B=50)=",Ic*1000,"mA\n");
#Now B=150
B=3.*B;      #three times of previous B
Ib1=Vcc/(Rb+B*Rc);         #Base current
Ic1=B*Ib1;         #Colletor current
print '%s %.2f %s' %("Collector current (B=150)=",Ic1*1000,"mA\n");
print '%s %.f' %("The factor at which collector current increases =",Ic1/Ic);

Collector current (B=50)= 1.125 mA

Collector current (B=150)= 2.25 mA

The factor at which collector current increases = 2


## Example E7 - Pg 199¶

In :
#Calculate the value of all three current Ie and Ic and Ib
#given
B=90.;            #dc beta
Rc=1.*10.**3.;#ohm      #resistor connected to collector
Rb=500.*10.**3.;#ohm      #resistor connected to base
Re=500.;#ohm      #resistor connected to emitter
Vcc=9.;#V            #Voltage supply across the collector resistor
Ib=Vcc/(Rb+B*Re);         #Base current
Ic=B*Ib;         #Colletor current
Ie=Ic+Ib;        #Emitter current
print '%s %.1f %s %s %.3f %s %s %.3f %s' %("Base current =",Ib*10**6,"uA\n","\nCollector current =",Ic*10**3,"mA\n","\nEmitter current =",Ie*10**3,"mA");


Base current = 16.5 uA

Collector current = 1.486 mA

Emitter current = 1.503 mA


## Example E8 - Pg 199¶

In :
#Calculate max and min value of emitter current
#given

#At B=50

B=50.;            #dc beta
Rc=75.;#ohm      #resistor connected to collector
Re=100.;#ohm      #resistor connected to emitter
Rb=10.*10.**3.;#ohm      #resistor connected to base
Vcc=6.;#V            #Voltage supply across the collector resistor
Vbe=0.3;#V       #for germanium
Ib=(Vcc-Vbe)/(Rb+(1.+B)*Re);         #Base current
Ie=(1.+B)*Ib;
Vce=Vcc-(Rc+Re)*Ie
print '%s %.2f %s' %("Minimum emitter current =",Ie*10**3,"mA\n");
print '%s %.2f %s' %("The collector to emitter volatge =",Vce,"V\n");

#At B=300

B1=300.; #dc beta
Ib1=(Vcc-Vbe)/(Rb+(1.+B1)*Re);#Base current
Ie1=(1.+B1)*Ib1;
Vce1=Vcc-(Rc+Re)*Ie1
#Here Vce1= -1.4874 V but can never have negative voltage because Ie1 is wrong as it cant be more than saturation value therefore
Ie1=Vcc/(Rc+Re);

#And Vce=0 V

Vce1=0;#V
print '%s %.2f %s' %("Maximum emitter current =",Ie1*10**3,"mA\n");
print '%s %.f %s' %("The collector to emitter volatge(saturation) =",Vce1,"V");

Minimum emitter current = 19.25 mA

The collector to emitter volatge = 2.63 V

Maximum emitter current = 34.29 mA

The collector to emitter volatge(saturation) = 0 V


## Example E9 - Pg 200¶

In :
#Determine the value of base resistance
#given

B=100.;            #dc beta
Rc=200.;#ohm      #resistor connected to collector
Re=500.;#ohm      #resistor connected to emitter
Vcc=9.;#V         #Voltage supply across the collector as it is PNP so taking positive
Vce=4.5;#V       #Collector to emitter voltage
Ic=(Vcc-Vce)/(Rc+Re);
Ib=Ic/B;
Rb=(Vcc-B*Re*Ib)/Ib;
print '%s %.f %s' %("The value of base resistance is =",Rb/1000,"kohm");

The value of base resistance is = 90 kohm


## Example E10 - Pg 200¶

In :
#Determine the collector current at two different B
#given

#At B=50

B=50.;#dc beta
Rc=2.;#ohm #resistor connected to collector
Re=1000.;#ohm #resistor connected to emitter
Rb=300.*10.**3.;#ohm #resistor connected to base
Vcc=9.;#V #Voltage supply across the collector resistor
Ib=Vcc/(Rb+B*Re);         #Base current
Ic=B*Ib;         #Colletor current
print '%s %.2f %s' %("The collector current at (B=50)=",Ic*1000,"mA\n");

#At B=150

B1=150.;#dc beta
Ib1=Vcc/(Rb+B1*Re);         #Base current
Ic1=B1*Ib1;         #Colletor current
print '%s %.1f %s' %("The collector current at (B=150)=",Ic1*1000,"mA\n");
print '%s %.1f' %("The factor at which collector current increases=",Ic1/Ic);

#IN BOOK Ic(AT B=50)= 1.25 mA and Ic1/Ic=2.4 DUE TO APPROAXIMATION

The collector current at (B=50)= 1.29 mA

The collector current at (B=150)= 3.0 mA

The factor at which collector current increases= 2.3


## Example E11 - Pg 205¶

In :
#Calculate Q point in voltage divider
#given
B=100.;            #dc beta
Rc=2.*10.**3.;#ohm      #resistor connected to collector
R1=10.*10.**3.;#ohm      #voltage divider resistor 1
R2=1.*10.**3.;#ohm      #voltage divider resistor 2
Re=200.;#ohm      #resistor connected to emitter
Vcc=10.;#V            #Voltage supply across the collector resistor
Vbe=0.3;#V          #base to emitter voltage
I=Vcc/(R1+R2);       #current through voltage divider
Vb=I*R2;        #voltage at base
Ve=Vb-Vbe;
Ie=Ve/Re;
Ic=Ie        #approaximating Ib is nearly equal to 0
Vc=Vcc-Ic*Rc;
Vce=(Vc)-Ve;
print '%s %.1f %s %.f %s' %("The Q point is =",Vce,"V",Ic*1000,"mA");

Ibc=I/20.;     #critical value of base current
Ib=Ic/B;      #actual base current

#Since Ib < Ibc, hence assumption is alright

The Q point is = 3.3 V 3 mA


## Example E12 - Pg 207¶

In :
#Solve the voltage divider accurately by applying thevenin's theorem
#given
B=100.;            #dc beta
Rc=2.*10.**3.;#ohm      #resistor connected to collector
R1=10.;#ohm      #voltage divider resistor 1
R2=1.;#ohm      #voltage divider resistor 2
Re=200.;#ohm      #resistor connected to emitter
Vcc=10.;#V            #Voltage supply across the collector resistor
Vbe=0.3;#V          #base to emitter voltage

Vth=Vcc*R2/(R1+R2);
Rth=R1*R2/(R1+R2);

print '%s %.1f %s' %("\nThevenin equivalent voltage Vth =",Vth,"V");
print '%s %.1f %s' %("\nThevenin equivalent resistance Rth =",Rth,"kohm");

Ib=(Vth-Vbe)/(Rth+(1.+B)*Re);
Ic=B*Ib;
Ie=Ic+Ib;
Vce=Vcc-Ic*Rc-Ie*Re;
print '%s %.4f %s' %("\nThe accurate value of Ic =",Ic*10**3,"mA");
print '%s %.5f %s' %("\nThe accurate value of Vce =",Vce,"V");
Icp=3.*10.**-3.; # Current calculated by voltage divider in previous example
Vcep=3.4; # Voltage calculated by voltage divider in previous example
Err_Ic=(Ic-Icp)*100./Ic;
Err_Vce=(Vce-Vcep)*100./Vce;
print '%s %.1f %s' %("\nError in Ic =",Err_Ic,"percent\n");
print '%s %.1f %s' %("\nError in Vce =",Err_Vce,"percent");

# The errors and The accurate values are different
# because of the approaximation in Vth and Rth in book

# In Book Ic = 2.8436 mA and Vce = 3.73839 V
# Error in Ic = -5.5%
# Error in Vce = +9%

Thevenin equivalent voltage Vth = 0.9 V

Thevenin equivalent resistance Rth = 0.9 kohm

The accurate value of Ic = 3.0152 mA

The accurate value of Vce = 3.36060 V

Error in Ic = 0.5 percent

Error in Vce = -1.2 percent


## Example E13 - Pg 209¶

In :
#determine the Q point for the emitter bias circuit
#given
B=100.;               #dc beta
Rc=5.*10.**3.;#ohm      #resistor connected to collector
Rb=10.*10.**3.;#ohm      #resistor connected to base
Re=10.*10.**3.;#ohm      #resistor connected to emitter
Vcc=12.;#V #Voltage supply across the collector resistor
Vee=15;#V #supply at emitter
Ie=Vee/Re;
Ic=Ie;
Vce=Vcc-Ic*Rc;
print '%s %.1f %s %.1f %s' %("The Q point is =",Vce,"V",Ic*1000,"mA");

The Q point is = 4.5 V 1.5 mA


## Example E14 - Pg 211¶

In :
#Calculate Vgs and Rs
#given
import math
Vp=2.;#V
Idss=1.75*10.**-3.;#A      #drain current at Vgs=0
Vdd=24.;#V             #drain to supply source
Id=1.*10.**-3.;#A       #drain current
Vgs=(-Vp)*(1-math.sqrt(Id/Idss));
Rs=abs(Vgs)/Id;
print '%s %.3f %s' %("Vgs =",Vgs,"V\n");
print '%s %.f %s' %("Rs =",Rs,"ohm");


Vgs = -0.488 V

Rs = 488 ohm