chapter 7 BIPOLAR JUNCTION TRANSISTORB

Example 7_1 pgno: 220

In [1]:
#exa 7.1
from math import exp
Dnb =20
print"Dnb = ",Dnb," cmˆ2/s" #initializiation the value of one of base parametre of NPN transistor .
nB =10**4
print"nB = ",nB," /cmˆ3" # initializiation the value of one of base parametre of NPN transistor .
xB =1*10**-6
print"xB = ",xB," m" # initializiation the value of one of base parametre of NPN transistor .
AB =10**-4
print"AB = ",AB," cmˆ2" #initializiation the value of one of base parametre of NPN transistor .
e=1.6*10**-19
print"e = ",e," columns" # initializiation the value of electronic charge .
Vbe=0.5
print"Vbe = ",Vbe," V" # initializiation the value of base emitter voltage of NPN transistor ..
VT=0.0259
print"VT = ",VT," V" # initializiation the value of threshold voltage .
WB=10**-4
print"WB = ",WB," cm" # initializiation the value of base width of NPN transistor .
Io=((e*AB*Dnb*nB)/(WB))
print"Magnitude of Io , Io=((e∗AB∗Dnb∗nB)/(WB)))=",Io," A"# calculation
Ic=Io*(exp(Vbe/VT)-1)
print"Collector current ,Ic=Io((exp(Vbe/VT))−1))=",Ic," A"# calculation

Dnb =  20  cmˆ2/s
nB =  10000  /cmˆ3
xB =  1e-06  m
AB =  0.0001  cmˆ2
e =  1.6e-19  columns
Vbe =  0.5  V
VT =  0.0259  V
WB =  0.0001  cm
Magnitude of Io , Io=((e∗AB∗Dnb∗nB)/(WB)))= 3.2e-14  A
Collector current ,Ic=Io((exp(Vbe/VT))−1))= 7.7484232166e-06  A

Example 7_2 pgno: 221

In [2]:
#exa 7.2
NE =5*10**17
print"NE = ",NE," /cmˆ3" # initializiation the value of doping concentration in the emitter
NB =10**16
print"NB = ",NB," /cmˆ3" # initializiation the value of doping concentration in the base.
NC =10**15
print"NC = ",NC," /cmˆ3" # initializiation the value of doping concentration in the collector .
WB =0.8*10**-4
print"WB = ",WB," cm" # initializiation the value of base width of NPN transistor .
no=1.5*10**10
print"no = ",no,"cmˆ−3" # initializing the intrinsic carrier concentration .
pEO=(no**2/NE)
print"Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )=",pEO," /cmˆ3"# calculationnBO=(no^2/NB)
nBO=(no**2/NB)
print"Number of Majority holes in the base,nBO=(no ˆ2/NB))=",nBO," /cmˆ3"#calculation
pCO=(no**2/NC)
print"Number of Majority holes in the collector ,pCO=(noˆ2/NC) )=",pCO," /cmˆ3"# calculation

NE =  500000000000000000  /cmˆ3
NB =  10000000000000000  /cmˆ3
NC =  1000000000000000  /cmˆ3
WB =  8e-05  cm
no =  15000000000.0 cmˆ−3
Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )= 450.0  /cmˆ3
Number of Majority holes in the base,nBO=(no ˆ2/NB))= 22500.0  /cmˆ3
Number of Majority holes in the collector ,pCO=(noˆ2/NC) )= 225000.0  /cmˆ3

Example 7_3 pgno: 221

In [3]:
#exa 7.3
from math import exp
NE =5*10**17
print"NE = ",NE," /cmˆ3" # initializiation of doping concentration in the emitter .
NB =10**16
print"NB = ",NB," /cmˆ3" # initializiation of doping concentration in the base .
NC =10**15
print"NC = ",NC," /cmˆ3" # initializiation of doping concentration in the collector .
WB =0.8*10**-4
print"WB = ",WB," cm" # initializiation the value of base width of NPN transistor .
no=1.5*10**10
print"no = ",no,"cmˆ−3" # initializing the value of intrinsic carrier concentration .
VT=0.0259
print"VT = ",VT," V" # initializiation the value of threshold voltage .
VJ=0.6258
print"VJ=Vbe = ",VJ," V" # initializiation the value of base emitter voltage .
pEO=(no**2/NE)
print"Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )=",pEO," /cmˆ3"# calculation
nBO=(no**2/NB)
print"Number of Majority holes in the base,nBO=(no ˆ2/NB))=",nBO," /cmˆ3"#calculation
pCO=(no**2/NC)
print"Number of Majority holes in the collector ,pCO=(noˆ2/NC) )=",pCO," /cmˆ3"# calculation
pE=pEO*(exp(VJ/VT))
print"pE(O)=pEO∗( exp (VJ/VT) ) )=",pE, "/cmˆ3 " # calculation
nB=nBO*(exp(VJ/VT))
print"nB=(nBO∗( exp (VJ/VT) ) ) )=",nB,"/cmˆ3" # calculation
#the answer provided in the book for pE,nB is some what different than actual calculated .

NE =  500000000000000000  /cmˆ3
NB =  10000000000000000  /cmˆ3
NC =  1000000000000000  /cmˆ3
WB =  8e-05  cm
no =  15000000000.0 cmˆ−3
VT =  0.0259  V
VJ=Vbe =  0.6258  V
Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )= 450.0  /cmˆ3
Number of Majority holes in the base,nBO=(no ˆ2/NB))= 22500.0  /cmˆ3
Number of Majority holes in the collector ,pCO=(noˆ2/NC) )= 225000.0  /cmˆ3
pE(O)=pEO∗( exp (VJ/VT) ) )= 1.40186506034e+13 /cmˆ3 
nB=(nBO∗( exp (VJ/VT) ) ) )= 7.00932530169e+14 /cmˆ3

Example 7_5 pgno: 222

In [4]:
#exa 7.5
from math import sqrt
from math import acosh
Db=10
print"Db = ",Db," cmˆ2/s" # initializiation the value of one of parametere of the transistor .
Bt =0.95
print"Bt = ",Bt # initializiation the value of base transport factor of the transistor.
tb =10**-7
print"tb = ",tb," s" # initializiation the value of one of parametere of the transistor.
Lp=(sqrt(Db*tb))
print"Lp=(sqrt(Db∗tb)))=",Lp," cm"# calculation
WB=(Lp*(acosh(1/Bt)))
print"WB=(Lp∗( acosh (1/Bt) ) )=",WB,"cm" #calculation,

Db =  10  cmˆ2/s
Bt =  0.95
tb =  1e-07  s
Lp=(sqrt(Db∗tb)))= 0.001  cm
WB=(Lp∗( acosh (1/Bt) ) )= 0.000323036439272 cm

Example 7_7 pgno: 224

In [5]:
#exa 7.7
from math import exp
Jro =10**-9
print"Jro = ",Jro," A/cmˆ2" # initializiation the value of recombination current density .
Jo =10**-12
print"Jo = ",Jo," A/cmˆ2" # initializiation the value of reverse saturation current density.
Vbe =0.5
print"Vbe = ",Vbe," V" # initializiation the value of base emitter voltage .
VT =0.0259
print"VT = ",VT," V" # initializiation the value of threshold voltage .
delta=(1+((Jro/Jo)*(exp((-Vbe)/(2*VT)))))**-1
print"delta (recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe) /(2∗VT) ) ) ) )ˆ−1)=",delta # calculation .

Jro =  1e-09  A/cmˆ2
Jo =  1e-12  A/cmˆ2
Vbe =  0.5  V
VT =  0.0259  V
delta (recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe) /(2∗VT) ) ) ) )ˆ−1)= 0.939616412003

Example 7_8 pgno: 224

In [6]:
#exa 7.8
from math import sqrt
from math import cosh
from math import tanh
NE =1*10**17
print"NE = ",NE," /cmˆ3" # initializiation the value of doping concentration of emitter in the NPN transistor .
NB =10**15
print"NB = ",NB," /cmˆ3" # initializiation the value of doping concentration of base in the NPN transistor .
WE =0.6*10**-4
print"WE = ",WE," cm" # initializiation the value of one of parametre of the transistor.
WB =0.8*10**-4
print"WB = ",WB," cm" # initializiation the value of one of parametre of the transistor.
no=1.5*10**10
print"no = ",no,"cmˆ−3" # initializing the value of intrinsic carrier concentration .
e=1.6*10**-19
print"e = ",e," columns" # initializiation the value of electronic charge
DE=15
print"DE = ",DE," cmˆ2/s" # initializiation the value of one of parametere of the transistor .
DB=20
print"DB = ",DB," cmˆ2/s" #initializiation the value of one of parametere of the transistor .
tE =0.2*10**-6
print"tE = ",tE," s" # initializiation the value of one of parametere of the transistor.
tB =0.1*10**-6
print"tB = ",tB," s" # initializiation the value of one of parametere of the transistor.
Vbe=0.60
print"Vbe = ",Vbe," V" # initializiation the value of base emitter voltage .
VT=0.0259
print"VT = ",VT," V" # initializiation the value of threshold voltage .
Jro =2*10**-8
print"Jro = ",Jro," A/cmˆ2" # initializiation the value of recombination current density .
LE=(sqrt(DE*tE))
print"LE=(sqrt(DE∗tE)))=",LE," cm"#calculation
LB=(sqrt(DB*tB))
print"LB=(sqrt(DB∗tB)))=",LB," cm"#calculation
pEO=(no**2/NE)
print"Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )=",pEO," /cmˆ3"# calculation
nBO=(no**2/NB)
print"Number of Majority holes in the base,nBO=(no ˆ2/NB))=",nBO," /cmˆ3"#calculation
Y=(1+(((NB*DE*LB)/(NE*DB*LE))*((tanh(WB/LB)/tanh(WE/ LE)))))**(-1)
print"Emitter injection efficiency ,Y=(1+((NB∗DE∗LB) /(NE∗DB∗LE)∗(tanh(WB/LB)/tanh(WE/LE)))) )=",Y # calculation
Bt=(cosh(WB/LB))**-1
print"Base transport factor ,Bt=(cosh(WB/LB))ˆ−1)=",Bt# calculation
Jo=((e*DB*nBO)/(LB*tanh(WB/LB)))
print"Reverse saturation current Density , Jro=((e∗DB∗n BO)/(LB∗tanh(WB/LB)))) = ",Jo, "A/cmˆ2"  # calculation
delta=(1+((Jro/Jo)*(exp((-Vbe)/(2*VT)))))**-1
print"delta(recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe)/(2∗VT)))))ˆ−1)=",delta," A"# calculation
a=Bt*delta*Y
print"common base current amplification factor ,(alpha=Bt∗delta∗Y)=",a # calculation
B=(a/(1-a))
print"common emitter current amplification factor ,Beta=(a/(1−a) ) )=",B # calculation
#the value of NE provided in the question is different than used in the solution .
#I have used the value (while solving) provided in the question ( i . e NE=10ˆ17/cmˆ3) 

NE =  100000000000000000  /cmˆ3
NB =  1000000000000000  /cmˆ3
WE =  6e-05  cm
WB =  8e-05  cm
no =  15000000000.0 cmˆ−3
e =  1.6e-19  columns
DE =  15  cmˆ2/s
DB =  20  cmˆ2/s
tE =  2e-07  s
tB =  1e-07  s
Vbe =  0.6  V
VT =  0.0259  V
Jro =  2e-08  A/cmˆ2
LE=(sqrt(DE∗tE)))= 0.00173205080757  cm
LB=(sqrt(DB∗tB)))= 0.00141421356237  cm
Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )= 2250.0  /cmˆ3
Number of Majority holes in the base,nBO=(no ˆ2/NB))= 225000.0  /cmˆ3
Emitter injection efficiency ,Y=(1+((NB∗DE∗LB) /(NE∗DB∗LE)∗(tanh(WB/LB)/tanh(WE/LE)))) )= 0.990105536375
Base transport factor ,Bt=(cosh(WB/LB))ˆ−1)= 0.998402130561
Reverse saturation current Density , Jro=((e∗DB∗n BO)/(LB∗tanh(WB/LB)))) =  9.00959795262e-09 A/cmˆ2
delta(recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe)/(2∗VT)))))ˆ−1)= 0.999979304421  A
common base current amplification factor ,(alpha=Bt∗delta∗Y)= 0.988503018931
common emitter current amplification factor ,Beta=(a/(1−a) ) )= 85.9793551861

Example 7_9 pgno: 225

In [1]:
#exa 7.9
from math import log
NB =5e16
print"NB = ",NB," /cmˆ3" # initializiation the doping concentration in the base .
NC =2e15
print"NC = ",NC," /cmˆ3" # initializiation the doping concentration in the collector .
WBm =0.6e-4
print"WBm = ",WBm," cm" # initializiation the value of actual base width .
e=1.6e-19
print"e = ",e," columns" # initializiation the value of electronic charge .
VCB1=1
print"VCB1 = ",VCB1," V" # initializiation the initial value of collector base voltage .
VCB2=4
print"VCB2 = ",VCB2," V" # initializiation the final value of collector base voltage.
Er=11.9
print"Er = ",Er  # initializing value of relative dielectric permittivity constant .
Eo=8.854e-14
print"Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
no=1.5e10
print"no = ",no,"cmˆ−3" # initializing the value of intrinsic charge carriers
VT=0.0259
print"VT = ",VT," V" # initializiation the value of threshold voltage .
VBI=VT*(log((NB*NC)/no**2))
print" VBI=VT∗(log((NB∗NC)/noˆ2))=",VBI," V" # calculation
WBS1=((2*Eo*Er*(VBI+VCB1)/e)*(NC/NB)*(1/(NC+NB)))**(1./2.)
print"WBS=((2∗Eo∗Er∗(VBI+VCB1)/e)∗(NC/NB)∗(1/(NC+NB)))ˆ(1/2))=",WBS1," cm"#calculation
Wb1=WBm-WBS1
print"Neutral base width for VCB1,WB( neutral )=WBm− WBS1=",Wb1," cm"# calculation
WBS2=((2*Eo*Er*(VBI+VCB2)/e)*(NC/NB)*(1/(NC+NB)))**(1/2)
print"WBS=((2∗Eo∗Er∗(VBI+VCB2)/e)∗(NC/NB)∗(1/(NC+NB) ))ˆ(1/2))=",WBS2," cm"#calculation
Wb2=WBm-WBS2
print"Neutral base width for VCB2,WB( neutral )=WBm−WBS2=",Wb2," cm"# calculation
deltaWbneutral=Wb1-Wb2
print"change in the neutal base width ,deltaWb(neutral )=Wb1−Wb2=",deltaWbneutral," cm" # calculation

NB =  5e+16  /cmˆ3
NC =  2e+15  /cmˆ3
WBm =  6e-05  cm
e =  1.6e-19  columns
VCB1 =  1  V
VCB2 =  4  V
Er =  11.9
Eo =  8.854e-14  F/cm
no =  15000000000.0 cmˆ−3
VT =  0.0259  V
 VBI=VT∗(log((NB∗NC)/noˆ2))= 0.694640354303  V
WBS=((2∗Eo∗Er∗(VBI+VCB1)/e)∗(NC/NB)∗(1/(NC+NB)))ˆ(1/2))= 4.14348090604e-06  cm
Neutral base width for VCB1,WB( neutral )=WBm− WBS1= 5.5856519094e-05  cm
WBS=((2∗Eo∗Er∗(VBI+VCB2)/e)∗(NC/NB)∗(1/(NC+NB) ))ˆ(1/2))= 1.0  cm
Neutral base width for VCB2,WB( neutral )=WBm−WBS2= -0.99994  cm
change in the neutal base width ,deltaWb(neutral )=Wb1−Wb2= 0.999995856519  cm

Example 7_10 pgno: 226

In [8]:
#exa 7.10
ro=500*10e3
print"ro = ",ro," ohm" # initializiation the value of output resistance .
Vce1 =7
print"Vce1 = ",Vce1," V" # initializiation the initial value of collector emitter voltage .
Vce2 =1
print"Vce2 = ",Vce2," V" # initializiation the final value of collector emitter voltage .
Vce=6
print"change in the collector −emitter voltage , Vce1−Vce2 = ",Vce," V" # calculation .
Ic=(Vce/ro)
print"change in the collector current , Ic=(Vce/ro))=" ,Ic," A"# calculation,

ro =  5000000.0  ohm
Vce1 =  7  V
Vce2 =  1  V
change in the collector −emitter voltage , Vce1−Vce2 =  6  V
change in the collector current , Ic=(Vce/ro))= 1.2e-06  A