# Chapter 8:Junction Transistors: Biasing and Amplification¶

## Example 1, Page 172¶

In :
#Variable declaration
b=99.;
Vbe=0.7;    #Volatge between base and emitter in V
Vcc=12;     #Volatge source applied at collector in V4
Rb=100*10**3;       #Resistance at base in ohms

#Calculations&Results
Ib=(12-0.7)/((100*Rl)+Rb);        #Base current in micro Ampere
print 'Ib=%.1f uA'%(Ib*10**6)

Ic=b*Ib;
print 'Ic=%.2f mA'%(Ic*10**3)
Vce=4.47;       #Voltage between collector and emitter in V

S=(b+1)/(1+b*Rl/(Rl+Rb));       #stabilty factor 1
print 'S=%.f'%S;
S1=b/(Rb+Rl*(1+b));      #stabilty factor 2 in A/V
print 'S1=%.1e A/V'%S1
S2=(Vcc-Vbe-(Ic*Rl))/(Rb+Rl*(1+b));        #stability factor 3 in A
print 'S2=%.2e A'%S2

Ib=37.7 uA
Ic=3.73 mA
S=34
S1=3.3e-04 A/V
S2=1.28e-05 A


## Example 2, Page 172¶

In :
#Variable declaration
b=49;#b=dc current gain of the common emitter transistor
Vbe=0.2;    #Volatge between base and emitter in V
Vcc=10.;     #Volatge source applied at collector in V4
Vce=5;      #Collector to emitter voltage in V
Ic=4.9;     #collector current in mA
Rl=1.;      #load resistance in kilo ohms
S=10;       #stability factor

#Calculations&Results
Ib=Ic/b;    #base current in mA
Re=((Vcc-Vce-(Ic*Rl))/(Ic+Ib))*1000;       #Resistance at emitter in ohms
print 'Re=%.f ohms'%Re;
#S=((1+b)*(1+(RT/Re)))/(1+b+(RT/Re))
RT=((S-1)*Re)/(1-(S/(1+b)))#RT=Thevenin resistance =(R1*R2)/(R1+R2)
VT=(Ib*(10**-3)*RT)+Vbe+((Ib+Ic)*(10**-3)*Re)#VT=Thevenin voltage=(R2*Vcc)/(R1+R2)
# R2/(R1+R2)=VT/Vcc
R1=(RT*Vcc)/VT
print "R1=%.3f k ohm"%(R1/10**3)
R2=((VT/Vcc)*R1)/(1-(VT/Vcc))
print "R2=%.1f ohm"%R2#incorrect answer in textbook

Re=20 ohms
R1=5.660 k ohm
R2=185.9 ohm


## Example 3, Page 173¶

In :
#Variable declaration
hib=30;     #h parameter of CB a transistor
hrb=4*10**-4;        #h parameter of CB a transistor
hfb=-0.99;        #h parameter of CB a transistor
hob=0.9*10**-6;        #h parameter of CB a transistor in S

#Calculations&Results
AI=-hfb/(1+(hob*Rl));        #Current gain
print 'AI=%.3f'%AI;

Ri=hib-((hfb*hrb*Rl)/(1+(hob*Rl)));    #Input resistance in ohms
print 'Ri=%.1f ohms'%Ri;

Ro=hib/((hib*hob)-(hfb*hrb));       #Output Resistance in kohms
print 'Ro=%.2f k ohms'%(Ro*10**-3);

AV=AI*Rl/Ri;        #Voltage gain
print 'AV=%.1f'%AV;

AP=AI*AV;       #Power gain
print 'AP=%.1f'%AP

AI=0.985
Ri=32.4 ohms
Ro=70.92 k ohms
AV=182.6
AP=179.8


## Example 4, Page 173¶

In :
#Variable declaration
Rg=1.*10**3;      #internal resistance in ohms
hie=1*10**3;     #h parameter of the transistor in terms of ohms
hre=2.5*10**-4;     #h parameter of the transistor
hfe=150.;     #h parameter of the transistor

#Calculations&Results
hoe=1./(40*10**3);      #h parameter of the transistor in terms of mho
AI=(-hfe)/(1+(hoe*Rl));      #Current gain
print 'AI=%.f'%AI;
Ri=hie+(AI*hre*Rl);     #input resistance in ohms
print 'Ri=%.f ohms'%Ri;
Ro=(Rg+hie)/((Rg*hoe)+(hie*hoe)-(hfe*hre));     #output resistance in ohms
print 'Ro=%.f k ohms'%(Ro*10**-3);

AI=-100
Ri=500 ohms
Ro=160 k ohms


## Example 5, Page 173¶

In :
#Variable declaration
hie=1*10**3;     #h parameter of the transistor in terms of ohms
hre=5*10**-4;     #h parameter of the transistor
hfe=100;     #h parameter of the transistor
hoe=25*10**-6;      #h parameter of the transistor in terms of mho
Rg=1*10**3;      #source reistance in ohms

#Calculations&Results
AI=(-hfe)/(1+(hoe*Rl));      #Current gain
print 'AI=%.2f'%AI;

Ri=hie+(AI*hre*Rl);     #input resistance in ohms
print 'Ri=%.1f ohms'%Ri;

AVo=AI*Rl/(Rg+Ri);      #Overall voltage gain including source resistance
print 'AVo=%.f'%AVo;

APo=AVo*AI;     #Overall voltage gain including source resistance
print 'APo=%.2e'%APo;

AI=-88.89
Ri=777.8 ohms
AVo=-250
APo=2.22e+04


## Example 6, Page 174¶

In :
#Variable declaration
hoe=25*10**-6;       #h parameter in A/V
hie=4000;       #h paramater in ohms
hfe=135;        #h paramater of transistor
hre=7*10**-4;        #h paramater of transistor
Re=100;     #emitter resistance in ohms

#Calculations&Results
#Here hoe*Rl is less than 0.1. So we can simplify the circuit and according to it the current gain is AI=Ic/Ib. here Ic=-hfe*Ib.

AI=-hfe;        #current gain
print 'AI=%.f'%AI;

Ri=hie+(1+hfe)*Re;      #input resistance in ohms
print 'Ri=%.1f k ohms'%(Ri*10**-3);

AV=AI*Rl/Ri;        #voltage gain
print 'AV=%.f'%AV;

print "The output resistance of the transistor excluding RL is infinite."
print "The output resistance of the transistor including RL is = %.f k ohms"%(Rl/10**3)

AI=-135
Ri=17.6 k ohms
AV=-24
The output resistance of the transistor excluding RL is infinite.
The output resistance of the transistor including RL is = 3 k ohms


## Example 7, Page 175¶

In :
#Variable declaration
hfe=100;        #h parameter of transistor
hie=560;        #h parameter of transistor in ohms
Rc=2*10**3;      #collector resistance in ohms
Re=10**3;        #emitter resistance in ohms
Rb=600*10**3;        #Base resistance in ohms

#Calculations&Results
#Since hoe is neglected we can use the simplified equivalent circuit hence the Ri is

Ri=hie+(1+hfe)*Re;      #Input resistance in ohms
print 'Ri=%.2f K ohms'%(Ri*10**-3);

Rib=(Ri*Rb)/(Ri+Rb);        #Input resistance including Rb in ohms
print 'Input resistance (including Rb)=%.2f k ohms'%(Rib*10**-3);

print "The output resistance excluding load is infinita"
Ro=Rc;
print "Output resistance including load = %.f k ohms"%(Ro*10**-3)

AV=-(hfe*Ro)/(hie+((1+hfe)*Re));        #voltage gain
print 'AV=%.2f'%AV;
print "\nSmall signals are used,since otherwise the output waveform will be distorted.Also,the equivalent circuit will not hold."

#Taking DC emitter current and collector current nearly equal

Ib=20./(Rb+Re*101);      #base current in mA
print 'Ib=%.4f mA'%(Ib*10**3);

print "\nThe Q-point is defined by"
Ic=hfe*Ib;      #collector current in mA
print 'Ic=%.2f mA'%(Ic*10**3);

VCE=20-(3*Ic*10**3)
print 'VCE=%.2f V'%VCE;

Ri=101.56 K ohms
Input resistance (including Rb)=86.86 k ohms
The output resistance excluding load is infinita
Output resistance including load = 2 k ohms
AV=-2.00

Small signals are used,since otherwise the output waveform will be distorted.Also,the equivalent circuit will not hold.
Ib=0.0285 mA

The Q-point is defined by
Ic=2.85 mA
VCE=11.44 V


## Example 8, Page 176¶

In :
import math

#Variable declaration
#For a CE transistor amplifier circuit with self-bias
f=1000#f=frequency in Hz
AV=-200#AV=voltage gain
hfe=100#hfe=current gain
hie=1#hie=input impedance in kilo ohms
Pcmax=75*10**-3#Pcmax=maximum collector dissipation in Watt
#hre and hoe are to be neglected
VCC=12#VCC=collector supply voltage

#Calculations&Results
#AV=-(hfe*RL)/hie where RL is the load resistance
RL=-(AV*hie)/hfe
print "The designed values of the components of a CE transistor amplifier are:"
print "The load resistance RL is = %.f K ohms"%RL
#For the amplifier to be linear,the quiescent point is chosen to lie in the middle of the DC load line
VCG=VCC/2  #VCG=DC collector to ground voltage
#VCC=(IC*RL)+VCG where IC=DC collector current
IC=(VCC-VCG)/RL
print "Ihe DC collector current is = %.f mA"%IC
Pr=(IC**2)*RL#Pr=power dissipation in RL
#Pc=the collector dissipation is set at 14.5 mW which is below the value of Pcmax
#Pc=VCE*IC
Pc=14.5
VCE=Pc/IC#VCE=collector-to-emitter voltage drop
VEG=VCG-VCE#VEG=DC voltage drop across resistance Re
IE=IC#IE=emitter current
Re=VEG/(IC)
print "The resistance Re is = %.f ohm"%(Re*1000)#Re is converted in terms of ohms
Pe=(IC**2)*Re#Pe=power dissipation in Re
VBE=0.7#VBE=assumed DC base-to-emitter voltage drop
VBG=VBE+(IE*Re)#VBG=DC voltage across resistance R2
#VT=(VCC*R2)/(R1+R2) where VT=Thevenin equivalent voltage
#RT=(R1*R2)/(R1+R2).............(1) where RT=Thevenin equivalent resistance
#VBG=VT-(IB*RT)
#VBG=((VCC*R2)/(R1+R2))-(IB*((R1*R2)/(R1+R2)))..................(2)
#Let (R2/(R1+R2))=x ..............(3)
x=VBG/VCC#neglecting the second term on the right hand side of equation (2)
a=(1-x)/x    #a=R1/R2
#S=((1+b)*(1+RT/Re))/(1+b+(RT/Re)) where S=stability factor and b=current gain=hfe
#b>>1 hence S=(hfe*(1+RT/Re))/(1+b+(RT/Re))
#For good stability we choose S=hfe/20
RT=((hfe-20.)/19)*Re
R1=RT/x#from equation (1) and (3)
print "The resistance R1 is=%.1f k ohm"%R1
R2=R1/5.33
print "The resistance R2 is = %.f k ohm"%R2
Pr2=(VBG**2)/R2#Pr2=power dissipation in R2
Pr1=((VCC-VBG)**2)/R1 #Pr1=power dissipation in R1
Ce=1./(2*math.pi*f*((Re*1000)/10))#Ce=bypass capacitor
print "The bypass capacitance Ce is %.f uF"%(Ce/10**-6)#Ce is converted in terms of micro farad
C1=2/(2*math.pi*f*100)#C1=coupling capacitor
print "The coupling capacitance C1 is %.1f uF"%(C1/10**-6)#C1 is converted in terms of micro farad
Rin=20*1000#Rin=assumed input impedance in ohms
C2=1/(2*math.pi*f*0.1*Rin)#C2=coupling capacitor
print "The coupling capacitance C2 is %.1f uF"%(C2/10**-6)#C2 is converted in terms of micro farad

The designed values of the components of a CE transistor amplifier are:
The load resistance RL is = 2 K ohms
Ihe DC collector current is = 3 mA
The resistance Re is = 389 ohm
The resistance R1 is=10.5 k ohm
The resistance R2 is = 2 k ohm
The bypass capacitance Ce is 4 uF
The coupling capacitance C1 is 3.2 uF
The coupling capacitance C2 is 0.1 uF


## Example 9, Page 177¶

In :
import math

#Variable declaration
VCC=12#VCC=collector supply voltage
a=0.98#a=dc current gain of the common base transistor
VBE=0.7#VBE=base emitter voltage
IE=2#IE=emitter current in mA

#Calculations&Results
#Ico is to be neglected
b=a/(1-a)#b=dc current gain of the common emitter transistor
#IC=b*IB where IC=collector current and IB=base current
#IE=IC+IB
#IE=(b+1)*IB
IB=IE/(b+1)
IC=b*IB
RE=0.1#RE=resistance in kilo ohms connected to the emitter terminal
R2=20#R2=resistance in kilo ohms
RC=3.3#RC=resistance in kilo ohms connected to the collector terminal
#Let I be the current in the resistance R2
#Applying Kirchhoff's voltage law in the base-emitter circuit
#VBE+(RE*IE)=R2*I
I=(1./R2)*(VBE+(RE*IE))
#Applying Kirchhoff's voltage law
#((I+IB+IC)*RC)+((I+IB)*R1)+(I*R2)=VCC
R1=(VCC-((I+IB+IC)*RC)-(I*R2))/(I+IB)

print "The resistance R1 is = %.1f k ohm"%R1

The resistance R1 is = 51.2 k ohm


## Example 10, Page 178¶

In :
#Variable declaration
VBE=0.7#VBE=base emitter voltage
b=90#b=dc current gain of the common emitter transistor
VCC=10#VCC=collector supply voltage
RE=1.2#RE=resistance in kilo ohms connected to the emitter terminal
RC=4.7#RC=resistance in kilo ohms connected to the collector terminal
RB=250.#RB=resistance in kilo ohms connected to the base terminal

#Calculations&Results
#Applying Kirchhoff's voltage law
#VCE=(RB*IB)+VBE where VCE=collector emitter voltage
#Also VCC=((IB+IC)*RC)+VCE+(IE*RE)
#IC=b*IB where IC=collector current and IB=base current
#IE=IC+IB where IE=emitter current
#IE=(b+1)*IB
IB=(VCC-VBE)/(((b+1)*(RC+RE))+RB)
IE=(b+1)*IB
VCE=(RB*IB)+VBE
IC=b*IB
print "The quiescent value of IE is = %.3f mA"%IE
print "The quiescent value of VCE is = %.2f V"%VCE
print "When dc current gain=90,IC= %.3f mA"%IC
#b is increased by 50%
b1=((50*b)/100)+b
IB1=(VCC-VBE)/(((b1+1)*(RC+RE))+RB)
IC1=b1*IB1
print "When dc current gain is increased by 50%%,IC=%.3f mA"%IC1
x=((IC1-IC)/IC)*100#x=increase in the collector current
print "The increase in the collector current IC is = %.1f %%"%x
print "\nThe percentage increase of IC being less than that of the dc current gain,the circuit provides some stabilization against the changes in the dc current gain."
print "VCE does not depend on dc current gain and hence it is not affected when the dc current gain changes."

The quiescent value of IE is = 1.075 mA
The quiescent value of VCE is = 3.65 V
When dc current gain=90,IC= 1.064 mA
When dc current gain is increased by 50%,IC=1.193 mA
The increase in the collector current IC is = 12.2 %

The percentage increase of IC being less than that of the dc current gain,the circuit provides some stabilization against the changes in the dc current gain.
VCE does not depend on dc current gain and hence it is not affected when the dc current gain changes.


## Example 11, Page 178¶

In :
#Variable declaration
VBE=0.7#VBE=base emitter voltage
b=99#b=dc current gain of the common emitter transistor
VCC=15#VCC=collector supply voltage
RE=7#RE=resistance in kilo ohms connected to the emitter terminal
RC=4#RC=resistance in kilo ohms connected to the collector terminal
RB=5#RB=resistance in kilo ohms connected to the base terminal
VEE=(-15)#VEE=emitter supply voltage

#Calculations&Results
#Applying Kirchhoff's voltage law in the base emitter loop
#-VEE=(RB*IB)+VBE +(IE*RE)
#IC=b*IB where IC=collector current and IB=base current
#IE=IC+IB where IE=emitter current
#IE=(b+1)*IB
IB=(-VEE-VBE)/(RB+((b+1)*RE))
print "The quiescent value of IB is = %.4f mA"%IB
IC=b*IB
print "The quiescent value of IC is = %.2f mA"%IC
IE=(b+1)*IB
print "The quiescent value of IE is = %.2f mA"%IE
#Applying Kirchhoff's voltage law in the output circuit
#(IC*RC)+VCE+(IE*RE)=VCC-VEE
VCE=(VCC-VEE)-(IE*RE)-(IC*RC)
print "The quiescent value of VCE is = %.2f V"%VCE
#b is increased by 20%
b1=((20*b)/100)+b
IB1=(-VEE-VBE)/(RB+((b1+1)*RE))
IC1=b1*IB1
print "When dc current gain is increased by 20%%,IC=%.2f mA"%IC1
x=((IC1-IC)/IC)*100#x=increase in the collector current
print "The increase in the collector current IC is = %.1f %%"%x
print "\nSince a 20% increase in current gain produces a mere 0.284% enhancement of IC,the circuit provides a good",\
"stabilization against the changes in the current gain"
#In textbook the increase in the collector current is given as 0.5% which is actually coming as 0.284% approximately

The quiescent value of IB is = 0.0203 mA
The quiescent value of IC is = 2.01 mA
The quiescent value of IE is = 2.03 mA
The quiescent value of VCE is = 7.77 V
When dc current gain is increased by 20%,IC=2.01 mA
The increase in the collector current IC is = 0.3 %

Since a 20% increase in current gain produces a mere 0.284% enhancement of IC,the circuit provides a good stabilization against the changes in the current gain


## Example 12, Page 179¶

In :
#Variable declaration
#For a self-bias circuit
VBE=0.7#VBE=base emitter voltage
b=100#b=dc current gain of the common emitter transistor
VCC=22#VCC=collector supply voltage
R1=82.#R1=resistance in kilo ohms
R2=16#R2=resistance in kilo ohms
Re=0.750#Re=resistance in kilo ohms connected to the emitter terminal

#Calculations&Results
#ICO is to be neglected
VT=(R2*VCC)/(R1+R2)#VT=Thevenin equivalent voltage
RT=(R1*R2)/(R1+R2)#RT=Thevenin equivalent resistance
#Applying Kirchhoff's voltage law to the base circuit
#(IB*(RT+Re))+(IC*Re)=VT-VBE
#IC=b*IB
IB=(VT-VBE)/(RT+Re+(b*Re))#IB=base current
IC=b*IB#IC=collector current
#Applying Kirchhoff's voltage law to the collector circuit
#(IC*(RL+Re))+(IB*Re)+VCE=VCC
VCE=VCC-((IC*(RL+Re))+(IB*Re))#VCE=collector emitter voltage
print "The operating point is specified by"
print "IC=%.2f mA"%IC
print "VCE=%.1f V"%VCE

The operating point is specified by
IC=3.24 mA
VCE=12.4 V


## Example 13, Page 180¶

In :
#Variable declaration
RE=0.680#RE=resistance in kilo ohms connected to the emitter terminal
RC=2.7#RC=resistance in kilo ohms connected to the collector terminal
#RB=resistance connected to the base terminal
VCE=7.3#VCE=collector emitter voltage
VBE=0.7#VBE=base emitter voltage
Vre=2.1#Vre=voltage across RE resistance
IB=0.02#IB=base current in mA

#Calculations&Results
IE=Vre/RE#IE=emitter current in mA
IC=IE-IB#IC=collector current in mA
b=IC/IB#b=current gain
print "The current gain β is = %.1f"%b
VCC=(IC*RC)+VCE+Vre#VCC=collector supply voltage
print "The collector supply voltage VCC is = %.1f V"%VCC
#Voltage across RB (Vrb)resistance is given by
Vrb=VCC-(VBE+Vre)
RB=Vrb/IB
print "The resistance RB is = %.f k ohm"%RB
#To draw the DC load line,we neglect the base current in RE resistance
#Equation for DC load line is:
#VCE=VCC-(RC+RE)*IC
print "\nFor the DC load line"
print "The intercept of the load line on the VCE-axis(X-axis) is = %.1f V"%VCC
print "The intercept of the load line on the IC axis(Y-axis) is = %.2f mA"%(VCC/(RC+RE))
print "The DC load line is the straight line joining above two intercepts."
print "The co-ordinates of the operating point Q on the load line are (7.3V,3.07mA)"

The current gain β is = 153.4
The collector supply voltage VCC is = 17.7 V
The resistance RB is = 744 k ohm

The intercept of the load line on the VCE-axis(X-axis) is = 17.7 V
The intercept of the load line on the IC axis(Y-axis) is = 5.23 mA
The DC load line is the straight line joining above two intercepts.
The co-ordinates of the operating point Q on the load line are (7.3V,3.07mA)


## Example 14, Page 180¶

In :
#Variable declaration
VBE=0.7#VBE=base emitter voltage
b=120#b=dc current gain of the common emitter transistor
VCC=15.#VCC=collector supply voltage
R1=72.#R1=resistance in kilo ohms
R2=8#R2=resistance in kilo ohms
Re=0.700#Re=resistance in kilo ohms connected to the emitter terminal
RC=2#RC=resistance in kilo ohms connected to the collector terminal
Rin=1.5#Rin=input resistance in kilo ohms of the amplifier
vi=1#vi=amplitude of the ac input signal in mV

#Calculations&Results
VT=(R2*VCC)/(R1+R2)#VT=Thevenin equivalent voltage
RT=(R1*R2)/(R1+R2)#RT=Thevenin equivalent resistance
#Applying Kirchhoff's voltage law to the base circuit
#(IB*(RT+Re))+(IC*Re)=VT-VBE
#IC=b*IB
IB=(VT-VBE)/(RT+Re+(b*Re))#IB=base current
IC=b*IB#IC=collector current
#Applying Kirchhoff's voltage law to the collector circuit
#(IC*(RL+Re))+(IB*Re)+VCE=VCC
VCE=VCC-((IC*(RL+Re)))#VCE=collector emitter voltage(neglecting small term IB*RE)
#Equation for DC load line is:
#VCE=VCC-(RL+Re)*IC
print "1. For the DC load line"
print "The operating point Q is specified by %.3f mA"%IC
print "VCE=%.1f V"%VCE
print "The intercept of the dc load line on the VCE-axis(X-axis) is = %.f V"%VCC
print "The intercept of the dc load line on the IC axis(Y-axis) is = %.2f mA"%(VCC/(RC+Re))
print "The DC load line is the straight line joining above two intercepts."
print "\n2. For the AC load line"
print "The intercept of the ac load line on the VCE-axis(X-axis) is = %.1f V"%(VCE+(IC*Rac))
print "The line joining the above intercept and the operating point Q extended to meet the IC axis(Y-axis) gives the AC load line"
AV=-(b*Rac)/Rin#AV=voltage gain of the amplifier
vo=abs(AV)*vi#vo=amplitude of the output voltage signal
print "\n3. The amplitude of the output voltage vo is = %.f mV"%vo

1. For the DC load line
The operating point Q is specified by 1.045 mA
VCE=12.2 V
The intercept of the dc load line on the VCE-axis(X-axis) is = 15 V
The intercept of the dc load line on the IC axis(Y-axis) is = 5.56 mA
The DC load line is the straight line joining above two intercepts.

2. For the AC load line
The intercept of the ac load line on the VCE-axis(X-axis) is = 13.2 V
The line joining the above intercept and the operating point Q extended to meet the IC axis(Y-axis) gives the AC load line

3. The amplitude of the output voltage vo is = 80 mV