Chapter 4 - Bipolar Junction Transistor

PageNumber 201 example 1

In [1]:
alpha=0.98#
vbe=0.7##base emitter voltage volt
ie=-4*10**-3##emitter current
vc=12##colector voltage volt
colr=3.3*10**3##ohms
colCurrent=ie*(-alpha)#
baseCurrent=0.02*ie#
vbn=vbe+(-4*10**-3*100)#
i2=-vbn/(10*10**3)#
i1=-(baseCurrent+i2)#
vcn=(vc-((colCurrent+i1)*colr))#
v1=vcn-0.9#
r1=v1/i1#
print "r1   =   %0.2f"%(abs(r1)),"ohm"

r1   =   19990.91 ohm

PageNumber 202 example 2

In [2]:
colvoltag=12##volts
vbe=5##volts
colcur=10*10**-3##ampere
vce=5##volts
beta1=50#
ib=colcur/beta1#
rb=(vbe-0.7)/ib#
rc=(12-vbe)/colcur#
#when 100ohm included
print "rb   =   %0.2f"%(rb),"ohm"
print "rc   =   %0.2f"%(rc),"ohm"
rb=(vce-0.7-(colcur+ib)*beta1)/ib#

print "rb at emitter resistance 100ohm   =   %0.2f"%(rb),"ohm"#correction in the book

rb   =   21500.00 ohm
rc   =   700.00 ohm
rb at emitter resistance 100ohm   =   18950.00 ohm

PageNumber 205 example 5

In [3]:
from math import log
#given
reveri=2*10**-6##ampere at 25
icb=2*10**-6*2**5##ampere at 75
basevoltag=5##volt
#(1)
rb=(-0.1+basevoltag)/(icb)#
print "max resistance   =   %0.2f"%((rb)),"ohm"#correction in the book
#(2)
basevoltag=1#
rb=100*10**3#
reveri=(-0.1+basevoltag)/rb#
q=reveri/(2*10**-6)#
w=q**10#
u=log(w)
t=25+(u/log((2)))#
print "baseresistance   =   %0.2f"%((rb)),"ohm"
print "temperature   =   %0.2f"%((t)),"celsius"

max resistance   =   76562.50 ohm
baseresistance   =   100000.00 ohm
temperature   =   46.70 celsius

PageNumber 205 example 6

In [4]:
#given
vbe=0.8##volt
beta1=100#
vce=0.2##volt
rb=200*10**3##ohm
bascur=(6-vbe)/rb#
colres=(10-vce)/(beta1*bascur)#
print "min resistance   =   %0.2f"%((colres)),"ohm"

min resistance   =   3769.23 ohm

PageNumber 206 example 7

In [5]:
beta1=100#
colres=3*10**3##collector resistance #ohm
rb=8*10**3##ohm
r1=500##ohm
voltag=5##volt
#(1)
ib=(-voltag+0.7)/((1+beta1)*r1+(rb))#
ic=beta1*ib#
vce=(-10-ic*(colres)+r1*(ib+ic))#
vcb=vce+0.7#
#(2)
volmin=-0.2+abs(ib+ic)*r1#
re=-(0.7+rb*ib+voltag)/((1+(beta1))*ib)#
print "in saturation mode"
print "vo   =   %0.2f"%((volmin)),"volt"#correction in the book
print "emitter resistance   <   %0.2f"%((re)),"ohm"

in saturation mode
vo   =   3.51 volt
emitter resistance   <   688.58 ohm

PageNumber 207 example 9

In [6]:
vcc=12##volt
rb=12*10**3##ohm
colres=2*10**3##ohm
beta1=100#
vb=0.7##volt
vce=0.1##volt

for q in range(1,3):
    if q==1:
        vbb=1
    else:
        vbb=12
    
    ib=(vbb-vb)/rb
    ic=beta1*ib
    ie=ic+ib
    vce=vcc-ic*colres
    if q==2 :
        ic=(vcc-0.1)/colres
    

    print "the operating point at vbb   =   %0.2f"%((vbb)),"volt ic   =   %0.2e"%((ic)),"ampere vce   =   %0.2f"%((vce))," volt"

beta1=ic/ib#

print "beta at saturation   =   %0.2f"%((beta1))

the operating point at vbb   =   1.00 volt ic   =   2.50e-03 ampere vce   =   7.00  volt
the operating point at vbb   =   12.00 volt ic   =   5.95e-03 ampere vce   =   -176.33  volt
beta at saturation   =   6.32

PageNumber 208 example 11

In [7]:
vbe=0.65##volt
colres=2*10**3##ohm
voltag=10##volt
i1=voltag/10#
q=(1.65-vbe)/(1*10**3)#


print "current   =   %0.2e"%((q)),"ampere"

current   =   1.00e-03 ampere

PageNumber 208 example 12

In [8]:
vcc=12##volt
r1=10*10**3##ohm
colres=1*10**3##ohm
re=5*10**3##ohm
rb=5*10**3##ohm
beta1=100#
vbe=0.7##volt
basvol=vcc*10/20#
ib=((basvol-vbe)/(rb+beta1*rb))#
ic=beta1*ib#
vce=vcc-ic*(colres+re)#
print "vce   =   %0.2f"%((vce)),"volt"
print "collector current   =   %0.2e"%((ic)),"ampere"

vce   =   5.70 volt
collector current   =   1.05e-03 ampere

PageNumber 209 example 13

In [9]:
colres=330##ohm
re=0.1*10**3##ohm
vcc=12##volt
vce=0.2##volt
revcur=18*10**-3#ampere
ib=0.3*10**-3##ampere
stability=10#
beta1=100#
colres=0.330##ohm
re=0.1*10**3##ohm
vbe=0.2#
rb=(((1+beta1)*re)/10-((1+beta1)*re))/(1-10.1)#
vb=2+ib*rb#
w=vcc/vb#
q=w-1#
r1=1.2*10**3#
r=q*1.2*10**3#
print "r1   =   %0.2f"%((q)),"times r2"
print "if r2 is 1200ohm"
print "r1   =   %0.2f"%((r)),"ohm"

print "r2   =   %0.2f"%((r1)),"ohm"

r1   =   4.22 times r2
if r2 is 1200ohm
r1   =   5061.77 ohm
r2   =   1200.00 ohm

PageNumber 210 example 14

In [10]:
alpha1=0.99#
ib=25*10**-6##ampere
icb=200*10**-9##ampere
beta1=alpha1/(1-alpha1)#
ic=beta1*ib+(beta1+1)*icb#
print "collector current   =   %0.2e"%((ic)),"ampere"
ie1=(ic-icb)/alpha1#
print "emitter current   =   %0.2e"%((ie1)),"ampere"
ic=beta1*ib#
print "collector current with ib   =   %0.2e"%((ic)),"ampere"
ie=ic/alpha1#
print "emitter current   =   %0.2e"%((ie)),"ampere"
w=(ie1-ie)/ie1#
print "error   =   %0.2e"%((w))

collector current   =   2.49e-03 ampere
emitter current   =   2.52e-03 ampere
collector current with ib   =   2.47e-03 ampere
emitter current   =   2.50e-03 ampere
error   =   7.94e-03

PageNumber 211 example 15

In [11]:
from __future__ import division
vcc=26##volt
colres=20*10**3##ohm
re=470##ohm
beta1=45#
vce=8##volt
ib=(vcc-vce)/((1+beta1)*(colres+re))#
ic=beta1*ib#
r1=((vcc-colres*(ib+ic)-re*(ib+ic)-(0.7)))/ib#
print "resistance   =   %0.2f"%((r1)),"ohm"
stability=(1+beta1)/(1+(beta1*re)/(re+colres))#
print "stability   =   %0.2f"%((stability))
#correction required in the book

resistance   =   381879.22 ohm
stability   =   22.62

PageNumber 211 example 16

In [12]:
vcc=1.5#volt in book should be changed as 1.5
colres=1.5*10**3##ohm
emresi=0.27*10**3##ohm
r1=2.7*10**3##ohm
r=2.7*10**3##ohm
beta1=45#
basre1=690##ohm
voltag=r*vcc/(r*r1)#
basres=(r*r1)/(r+r1)#
vbe=0.2#
for q in range (1,3):
    if q==2 :
        print "resistance   =   %0.2e"%((basre1)),"ohm"
        basres=basres+basre1
    
    bascur=(((voltag+vbe)))/(basres+(45*(emresi)))
    colcur=beta1*bascur
    vce=(vcc+colcur*colres+(bascur+colcur)*emresi)
    print "current   =   %0.2e"%((colcur)),"ampere"
    print "vce   =   %0.2f"%((vce)),"volt"

current   =   6.69e-04 ampere
vce   =   2.69 volt
resistance   =   6.90e+02 ohm
current   =   6.36e-04 ampere
vce   =   2.63 volt

PageNumber 212 example 17

In [13]:
beta1=25#
colres=2.5*10**3##ohm
vcc=10##volt
vce=-5##volt
ic=-(vcc+vce)/colres#
ib=ic/beta1#
rb=vce/ib#
stability=(1+beta1)/((1+beta1)*((colres)/(colres+rb)))#
print "base resistance   =   %0.2f"%((rb)),"ohm"#correction in book
print "stability   =   %0.2f"%((stability))

base resistance   =   62500.00 ohm
stability   =   26.00

PageNumber 212 example 18

In [14]:
therre=8##celsius per watts
tepera=27##celsius ambient temperature
potran=3##watt
tejunc=tepera+(therre*potran)#
print "junction temperature   =   %0.2f"%((tejunc)),"celsius"

junction temperature   =   51.00 celsius

PageNumber 213 example 19

In [15]:
from __future__ import division
ambtep=40##celsius
juntep=160##celsius
hs_a=8#
j_c=5#
c_a=85#
j_a=(j_c)+(c_a*hs_a)/(c_a+hs_a)#
podiss=(juntep-ambtep)/j_a#
print "dissipation   =   %0.2f"%((podiss)),"watt"

dissipation   =   9.75 watt

PageNumber 213 example 21

In [16]:
from __future__ import division
emicur=1*10**-3##ampere
colcur=0.995*10**-3##ampere
alpha1=colcur/emicur#
beta1=alpha1/(1-alpha1)#
print "alpha   =   %0.2f"%((alpha1))
print "beta   =   %0.2f"%((beta1))

alpha   =   0.99
beta   =   199.00

PageNumber 213 example 22

In [17]:
from __future__ import division
beta1=100#
alpha1=beta1/(beta1+1)#

print "alpha   =   %0.2f"%((alpha1))

alpha   =   0.99

PageNumber 213 example.23

In [18]:
from __future__ import division
rb=200*10**3##ohm
rc=2*10**3##ohm
vcc=20##volt
ib=(vcc)/(rb+200*rc)#
ic=200*ib#
print "ic   =   %0.4f"%((ic)),"ampere"
#correction required in book

ic   =   0.0067 ampere

PageNumber 214 example 24

In [19]:
from __future__ import division
alpha1=0.98#
revcur=1*10**-6##ampere
emicur=1*10**-3##ampere
colcur=alpha1*emicur+revcur#
bascur=emicur-colcur#
print "collector current   =   %0.2e"%((colcur)),"ampere"
print "base current   =   %0.2e"%((bascur)),"ampere"

collector current   =   9.81e-04 ampere
base current   =   1.90e-05 ampere

PageNumber 214 example 25

In [20]:
from __future__ import division
colcur=100*10**-3##ampere
ouresi=20##ohm
r=200##ohm
r1=100##ohm
vcc=15##volt
basvol=((r1)/(r+r1))*vcc#
em1res=basvol/colcur#
vce=vcc-(ouresi+em1res)*colcur#
print "vce   =   %0.2f"%((vce)),"volt"
print "emitter resistance   =   %0.2f"%((em1res)),"ohm"

vce   =   8.00 volt
emitter resistance   =   50.00 ohm

PageNumber 214 example 26

In [21]:
from __future__ import division
colres=1*10**3##ohm
beta1=50#
vbe=0.3##volt
vcc=6##volt
rb=10*10**3##ohm
re=100##ohm
em1cur=((vcc-vbe)*(beta1+1))/((rb+((beta1+1)*re)))#
for q in range(1,3):
    if q==2 :
        colres=1*10**3#
        vce=vcc-(colres+re)*em1cur#
        ic=vcc/(colres+re)#
        print "collector to emitter   =   %0.2f"%((vce)),"volt"
        print "collector current   =   %0.3f"%((ic)),"ampere"
    
    if q==1 :
        colres=50#
        rb=100#
        vce=vcc-(colres+rb)*em1cur#
        print "emitter current   =   %0.3f"%((em1cur)),"ampere"
        print "collector to emitter   =   %0.3f"%((vce)),"volt"
    

emitter current   =   0.019 ampere
collector to emitter   =   3.112 volt
collector to emitter   =   -15.18 volt
collector current   =   0.005 ampere

PageNumber 216 example 27

In [22]:
from __future__ import division
beta1=99#
stability=5#
vbe=0.2##volt
colres=2.5*10**3##ohm
vce=6##volt
ven=5.5##volt
vcc=15##volt
vcn=vce+ven#
colvol=vcc-vcn##voltage across collector resistance
ic=colvol/colres#
ib=ic/beta1#
colre1=ven/ic#
rb=stability*colre1/(1-(stability/(1+beta1)))##correction in the book taken collector resistance as 3.13*10**3ohm but it is 3.93*10**3ohm
v1=(ib*rb)+(vbe)+((ib+ic)*colre1)#
r=rb*vcc/v1#
r1=r*v1/(vcc-v1)#
print "resistance   =   %0.2f"%((colre1)),"ohm"
print "resistance r1    =   %0.2f"%((r)),"ohm"
print "resistance r2   =   %0.2f"%((r1)),"ohm"

resistance   =   3928.57 ohm
resistance r1    =   51281.87 ohm
resistance r2   =   34645.75 ohm

PageNumber 216 example 28

In [23]:
from __future__ import division
beta1=50#
vbb=5##volt
rb=10*10**3##ohm
colres=800##ohm
re=1.8*10**3##ohm
vcc=5##volt
ib=(0.7-vbb)/((rb)+(beta1+1)*re)##correction in book
re=beta1*ib#
ie=(ib+re)#
vce=vcc-colres*re-re*ie#
vcb=(vce-0.7)#
print "base current   =   %0.2e"%((ib)),"ampere"
print "collector current   =   %0.2e"%((re)),"ampere"
print "emitter current   =   %0.2e"%((ie)),"ampere"
print "vcb   =   %0.2f"%((vcb)),"volt"#correction in book
print "the collector base junction is reverse biased the transistor in active region"

base current   =   -4.22e-05 ampere
collector current   =   -2.11e-03 ampere
emitter current   =   -2.15e-03 ampere
vcb   =   5.99 volt
the collector base junction is reverse biased the transistor in active region

PageNumber 217 example 29

In [24]:
from __future__ import division
r=40*10**3##ohm
r1=5*10**3##ohm
colres=r1#
beta1=50#
em1res=1*10**3##ohm
vcc=12##volt
rth=r*r1/(r+r1)#
v1=r1*vcc/(r1+r)#
bascur=(v1-0.3)/(rth+(beta1*em1res))#
colcur=beta1*bascur#
vce=vcc-(colres+em1res)*colcur#
print "collector current   =   %0.2e"%((colcur)),"ampere"
print "collector emitter voltage   =   %0.2f"%((vce)),"volt"

collector current   =   9.49e-04 ampere
collector emitter voltage   =   6.31 volt

PageNumber 217 example 30

In [25]:
from __future__ import division
colcur=8*10**-3##ampere
re=500##ohm
vce=3##volt
beta1=80#
vcc=9##volt
ib=colcur/beta1#
rb=(vcc-(1+beta1)*(ib*re))/ib#
print " base resistance   =   %0.f"%((rb)),"ohm"

 base resistance   =   49500 ohm

PageNumber 217 example 31

In [26]:
from __future__ import division
vcc=10##volt
basres=1*10**6##ohm
colres=2*10**3##ohm
em1res=1*10**3##ohm
beta1=100#
bascur=vcc/(basres+(beta1+1)*(em1res))#
colcur=beta1*bascur#
em1cur=colcur+bascur#
print "base current   =   %0.2e"%((bascur)),"ampere"
print "collector current   =   %0.2e"%((colcur)),"ampere"#correction in book
print "emitter current   =   %0.2e"%((em1cur)),"ampere"#correction in book

base current   =   9.08e-06 ampere
collector current   =   9.08e-04 ampere
emitter current   =   9.17e-04 ampere

PageNumber 218 example 32

In [27]:
from __future__ import division
alpha1=0.99#
rebacu=1*10**-11##ampere
colres=2*10**3##ohm
vcc=10##volt
bascur=20*10**-6##ampere
beta1=alpha1/(1-alpha1)#
i1=(1+beta1)*rebacu#
colcur=beta1*bascur+i1#
em1cur=-(bascur+colcur)#
vcb=vcc-colcur*colres#
vce=vcb-0.7#
print "collector current   =   %0.2e"%((colcur)),"ampere"
print "emitter current   =   %0.2e"%((em1cur)),"ampere"
print "collector emitter voltage   =   %0.2f"%((vce)),"volt"

collector current   =   1.98e-03 ampere
emitter current   =   -2.00e-03 ampere
collector emitter voltage   =   5.34 volt

PageNumber 220 example 33

In [28]:
from __future__ import division
beta1=100#
revcur=20*10**-9##ampere
colres=3*10**3##ohm
rb=200*10**3##ohm
vbb=5##volt
vcc=11##volt
em1res=2*10**3##ohm
ib=(vbb-0.7)/rb#
ic=beta1*ib#
ie=ib+ic#
print "base current   =   %0.2e"%((ib)),"ampere"
print "collector current   =   %0.2e"%((ic)),"ampere"
print "emitter current   =   %0.2e"%((ie)),"ampere"#question asked only currents
#2*10**3 ohm added to emitter
ib=-(0.7-vcc)/(rb+((1+beta1)*em1res))#
ic=beta1*ib#
ie=ib+ic#
print "base current   =   %0.2e"%((ib)),"ampere"#correction in book
print "collector current   =   %0.2e"%((ic)),"ampere"
print "emitter current   =   %0.2e"%((ie)),"ampere"#question asked only currents

base current   =   2.15e-05 ampere
collector current   =   2.15e-03 ampere
emitter current   =   2.17e-03 ampere
base current   =   2.56e-05 ampere
collector current   =   2.56e-03 ampere
emitter current   =   2.59e-03 ampere

PageNumber 221 example 34

In [29]:
from __future__ import division
em1cur=2*10**-3##ampere
v1=12##volt
vcc=12##volt
format(12)#
colres=5*10**3##ohm
em1res=v1/em1cur#
colcur=em1cur#
voltag=colcur*colres##ic*r
v1=vcc-(colres*colcur)#
print "emitter current   =   %0.2e"%((em1cur)),"ampere"
print "collector current   =   %0.2e"%((colcur)),"ampere"
print "voltage   =   %0.2f"%((voltag)),"volt"
print "vcb   =   %0.2f"%(abs(v1)),"volt"
print "emitter resistance   =   %0.2f"%((em1res)),"ohm"

emitter current   =   2.00e-03 ampere
collector current   =   2.00e-03 ampere
voltage   =   10.00 volt
vcb   =   2.00 volt
emitter resistance   =   6000.00 ohm

PageNumber 221 example 35

In [30]:
from __future__ import division
vbb=4##volt
ib=50*10**-6##ampere
for q in [0, 0.7, 4, 12]:
    if q==0 :
        rb=(vbb-q)/ib#
        print "resistance at %0.1f"%((q)),"volt   %0.2f"%((rb)),"ohm"
    elif q==0.7:
        rb=(vbb-q)/ib
        print "resistance at %0.2f"%((q)),"volt   %0.2f"%((rb)),"ohm"
    elif q==4:
        print "vbb at 12volt"
        q=0
        vbb=12
        rb=(vbb-q)/ib
        print "resistance at %0.2f"%((q)),"volt   %0.2f"%((rb)),"ohm"
    else:
        q=0.7#
        vbb=12#
        rb=(vbb-q)/ib#
        
        
        print "resistance at %0.2f"%((q)),"volt   %0.2f"%((rb)),"ohm"

resistance at 0.0 volt   80000.00 ohm
resistance at 0.70 volt   66000.00 ohm
vbb at 12volt
resistance at 0.00 volt   240000.00 ohm
resistance at 0.70 volt   226000.00 ohm

PageNumber 222 example 36

In [31]:
from __future__ import division
ic=5.2*10**-3##ampere
ib=50*10**-6##ampere
icb=2*10**-6##ampere
beta1=(ic-icb)/(ib+icb)#
print "beta   =   %0.2f"%((beta1))
ie=ib+ic#

print "ie   =   %0.3f"%((ie)),"ampere"
alpha1=(ic-icb)/ic#
print "alpha   =   %0.2f"%((alpha1))



ic=10*10**-3##ampere
ib=(ic-(beta1+1)*(icb))/beta1#


print "ib   =   %0.2e"%((ib)),"ampere"
#correction required in the book

beta   =   99.96
ie   =   0.005 ampere
alpha   =   1.00
ib   =   9.80e-05 ampere

PageNumber 222 example 37

In [32]:
from __future__ import division
beta1=160
vb=-0.8##volt
re=2.5*10**3##ohm
vcc=10##volt
for q in [160, 80]:
    ib=(vcc-vb)*10**2/((re)*(1+q)*400)#
    ic=q*ib#
    colres=1.5*10**3##ohm
    print "collector current at beta %0.2f"%((q)),"   =   %0.2e"%((ic)),"ampere"
    #correction required in the book
    ie=(1+beta1)*ib#
    vce=-(vcc-colres*ic-re*ie)#
    print "vce at beta %0.2f"%((q)),"   =   %0.2f"%((vce)),"volt"
    #correction required in the book

collector current at beta 160.00    =   1.07e-03 ampere
vce at beta 160.00    =   -5.69 volt
collector current at beta 80.00    =   1.07e-03 ampere
vce at beta 80.00    =   -3.03 volt

PageNumber 222 example 38

In [33]:
from __future__ import division
vb=0.7##volt
vce=7##volt
ic=1*10**-3##ampere
vcc=12##volt
beta1=100#
colres=(vcc-vce)/ic#
ib=ic/beta1#
#rb
rb=(vcc-vb-ic*colres)/ib#
print "rb   =   %0.2f"%((rb))," ohm"
#stability
stability=(1+beta1)/(1+beta1*(colres/(colres+rb)))#
print "stability   =   %0.2f"%((stability))
#beta=50
beta1=50#
print "new point"
ib=(vcc-vb)/(beta1*colres+rb)#
ic=beta1*ib#
print "ic   =   %0.2e"%((ic))," ampere"
vce=vcc-(ic*colres)#
print "vce   =   %0.2f"%((vce))," volt"

rb   =   630000.00  ohm
stability   =   56.51
new point
ic   =   6.42e-04  ampere
vce   =   8.79  volt

PageNumber 223 example 39

In [34]:
from __future__ import division
vcc=16##volt
colres=3*10**3##ohm
re=2*10**3##ohm
r1=56*10**3##ohm
r2=20*10**3##ohm
alpha1=0.985#
vb=0.3##volt
#coordinates
beta1=alpha1/(1-alpha1)#
v1=vcc*r2/(r1+r2)#
rb=r2/(r1+r2)#
ic=(v1-vb)/((rb/beta1)+(re/beta1)+re)#
print "new point"
print "vce   =   %0.2f"%((v1))," volt"
print "ic   =   %0.2e"%((ic))," ampere"

new point
vce   =   4.21  volt
ic   =   1.93e-03  ampere

PageNumber 224 example 40

In [35]:
from __future__ import division
vce=12##volt
ic=2*10**-3##ampere
vcc=24##volt
vb=0.7##volt
beta1=50#
colres=4.7*10**3##ohm
#re
re=((vcc-vce)/(ic))-colres#
print "re   =   %0.2f"%((re))," ohm"
#r1
ib=ic/beta1#
v1=ib*3.25*10**3+vb+(ib+1.5*10**3)#
r1=3.25*18*10**3/2.23#
print "r1   =   %0.2f"%((r1))," ohm"
#r2
r2=26.23*2.23*10**3/(18-2.3)#
print "r2   =   %0.2f"%((r2))," ohm"

re   =   1300.00  ohm
r1   =   26233.18  ohm
r2   =   3725.66  ohm

PageNumber 225 example 41

In [36]:
from __future__ import division
colres=3*10**3##ohm
rb=150*10**3##ohm
beta1=125#
vcc=10##volt
v1=5##volt
vb=0.7##volt
ib=(v1-vb)/rb#
print "ib   =   %0.2e"%((ib))," ampere"
ic=beta1*ib#
ie=ic+ib#
print "ic   =   %0.2e"%((ic))," ampere"
print "ie   =   %0.2e"%((ie))," ampere"#correction in the book in question to find only currents

ib   =   2.87e-05  ampere
ic   =   3.58e-03  ampere
ie   =   3.61e-03  ampere

PageNumber 226 example 42

In [37]:
from __future__ import division
beta1=50#
vb=0.6##volt
vcc=18##volt
colres=4.3*10**3##ohm
ic=1.5*10**-3##ampere
vce=10##volt
stability=4#
r1=(vcc-vce)/ic#
re=r1-colres#
w=(beta1+1)*(stability)*re/(1+beta1-stability)#
print "re   =   %0.2f"%((re)),"ohm"
print "rb   =   %0.2f"%((w)),"ohm"#correction in the book

re   =   1033.33 ohm
rb   =   4485.11 ohm

PageNumber 226 example 43

In [38]:
from __future__ import division
re=100##ohm
beta1=100#
rb=1*10**3##ohm
stability=(1+beta1)/(1+beta1*(re/(re+rb)))#
r1=3.8#r2
print "r1   =   3.8*r2"#correction in the book not given in question

r1   =   3.8*r2

PageNumber 228 example 45

In [39]:
from __future__ import division
from math import log10
icb=2*10**-6##ampere
vbb=1##volt
r1=50*10**3##ohm
#current increases every 10celsius rb at 75celsius
vb=-0.1##volt
icb=2**6*10**-6##at 75celsius
rb=(vb+vbb)/icb#
print "rb at 75 celsius   =   %0.2f"%((rb)),"ohm"
icb=(vb+vbb)/r1#
print "icb   =   %0.2e"%((icb)),"ampere"
w=(log10(icb*10**6)*20/log10(2))-25#
print "temperature at which current till max   =   %0.2f"%((w)),"celsius"

rb at 75 celsius   =   14062.50 ohm
icb   =   1.80e-05 ampere
temperature at which current till max   =   58.40 celsius

PageNumber 228 example 46

In [40]:
from __future__ import division
vb=0.8##volt
beta1=100#
vce=0.2##volt
vcc=10##volt
rb=200*10**3##ohm
#collector resistance
ib=(5-0.7)/rb#
colres=(vcc-vce)/(beta1*ib)#
print "min collector resistance   =   %0.2f"%((colres)),"ohm"

min collector resistance   =   4558.14 ohm

PageNumber 229 example 47

In [41]:
from __future__ import division
alpha1=0.98#
alph11=0.96#
vcc=24##volt
colres=120##ohm
ie=100*10**-3##ampere
beta1=alpha1/(1-alpha1)#
bet11=alph11/(1-alph11)#
ib2=ie/(1+bet11)#
ie1=-ib2#
print "ib2   =   %0.2e"%((ib2)),"ampere"
print "ie1   =   %0.2e"%((ie1)),"ampere"


ic2=bet11*ib2#
ib1=ib2/(1+beta1)#
ic1=beta1*ib1#
print "ic2   =   %0.2e"%((ic2)),"ampere"
print "ib1   =   %0.2e"%((ib1)),"ampere"
print "ic1   =   %0.2e"%((ic1)),"ampere"
ic=ic1+ic2#
vce=vcc-ic*colres#
ib=ib1#
w=ic/ib#
q=-ic/ie#
print "ic   =   %0.2e"%((ic)),"ampere"
print "ic/ib   =   %0.2f"%((w))
print "ic/ie   =   %0.2f"%((q))
#correction required in the book
print "vce   =   %0.2f"%((vce)),"volt"

ib2   =   4.00e-03 ampere
ie1   =   -4.00e-03 ampere
ic2   =   9.60e-02 ampere
ib1   =   8.00e-05 ampere
ic1   =   3.92e-03 ampere
ic   =   9.99e-02 ampere
ic/ib   =   1249.00
ic/ie   =   -1.00
vce   =   12.01 volt