CHAPTER07:BJT FUNDAMENTALS

Example E01 : Pg 352

In [1]:
# Ex_7_1
#clc;
#clear;
#close;
# given data : 
#format('v',6);
alfa=0.90;# current gain
ICO=15.;# micro A(reverse saturation currenrt)
IE=4.;# mA(Emitter currenrt)
IC=ICO*10.**-3.+alfa*IE;# mA
IB=IE-IC;# mA
IB=IB*1000.;# micro A
print"Collector Current(mA)",IC
print"Base Current(micro A)",IB

Collector Current(mA) 3.615
Base Current(micro A) 385.0

Example E02 : Pg 352

In [2]:
# Ex_7_2
#clc;
#clear;
#close;
# given data : 
#format('v',6);
Beta=90.;# unitless
IC=4.;# mA(Collector Current)
alfa=Beta/(1.+Beta);# current gain
IB=IC/Beta;# mA(Base Current)
IE=IC+IB;# mA(Emitter currenrt)
print"Value of alfa",alfa
print"Base Current(micro A)",IB*1000.
print"Emmiter Current(mA)",IE

Value of alfa 0.989010989011
Base Current(micro A) 44.4444444444
Emmiter Current(mA) 4.04444444444

Example E03 : Pg 362

In [3]:
# Ex_7_3
# given data : 
alfa=0.90;# current gain
ICO=15.;# micro A(reverse saturation currenrt)
IB=0.5;# /mA(Base Current)
Beta=alfa/(1.-alfa);# unitless
IC=Beta*IB+(1.+Beta)*ICO/1000.;# mA(Collector Current)
print"Collector Current(mA)",IC

Collector Current(mA) 4.65

Example E04 : Pg 362

In [4]:
# Ex_7_4
# given data : 
IB=20.;# /micro A(Base Current)
IC=5.;# mA(Collector Current)
Beta=IC*1000./IB;# unitless
print"Beta=",Beta

Beta= 250.0

Example E05 : Pg 364

In [5]:
# Ex_7_5
# given data : 
IB=50.;# /micro A(Base Current)
IC=5.;# mA(Collector Current)
IE=IC+IB/1000.;# mA
Beta=IC*1000./IB;# unitless
alfa=IC/IE;# current gain
print"Emitter Current(mA)",IE
print"\nBeta=",Beta
print"\nalfa=",alfa
print"\nVerify that alfa=Beta/(Beta+1)"
print alfa==Beta/(Beta+1);
print"\nVerify that Beta=alfa/(1-alfa)"
print Beta==round(alfa/(1-alfa));

Emitter Current(mA) 5.05

Beta= 100.0

alfa= 0.990099009901

Verify that alfa=Beta/(Beta+1)
True

Verify that Beta=alfa/(1-alfa)
True

Example E06 : Pg 365

In [6]:
# Ex_7_6
# given data : 
IE=10.;# mA
IB=5.;# /mA(Base Current)
IC=IE-IB;# mA(Collector Current)
BetaR=IC/IB;# unitless
alfaR=IC/IE;# current gain
print "BetaR=",BetaR
print "alfaR=",alfaR
# Answer is wrong in the book.

BetaR= 1.0
alfaR= 0.5

Example E07 : Pg 366

In [7]:
# Ex_7_7
# given data : 
Beta=100.;# unitless
VBE=0.7;# V
VCC=10.;# V
# (a) VE=-0.7;# V
print"For the circuit in fig(a)"
VE=-0.7;# V(Constant voltage)
R1=10.;# kohm
R2=10.;# kohm
IE=(VCC+VE)/R2;# mA
IB=IE/(Beta+1.);# mA
VC=VCC-R1*1000.*(IE-IB)/1000.;# V
print"Constant voltage fo the given transistor, VE(V)",VE
print"Emitter current(mA)",IE
#format('v',5);
IB=IB*1000;# /micro A
print"Base current(micro A)",IB
#format('v',6);
print"VC(V)",VC
# (b) VE=-0.7;# V
R1=5.;# kohm
R2=5.;# kohm
VEE=-15.;# V
print"For the circuit in fig(b)"
VE=-0.7;# V(Constant voltage)
R1=5.;# kohm
R2=5.;# kohm
IE=(VCC+VE)/R2;# mA
IC=IE*Beta/(Beta+1.);# mA
VC=VEE+R2*IC;# V
print"Constant voltage fo the given transistor, VE(V)",VE
print"Emitter current(mA)",IE
print"Base current(mA)",IC
#format('v',5);
print"VC(V)",VC

For the circuit in fig(a)
Constant voltage fo the given transistor, VE(V) -0.7
Emitter current(mA) 0.93
Base current(micro A) 9.20792079208
VC(V) 0.792079207921
For the circuit in fig(b)
Constant voltage fo the given transistor, VE(V) -0.7
Emitter current(mA) 1.86
Base current(mA) 1.84158415842
VC(V) -5.79207920792

Example E08 : Pg 372

In [8]:
# Ex_7_8
# given data : 
import math 
Beta=1.+0j;#math.inf;# Current gain
VBE=0.7;# # V
VB=0;# V(For large Beta)
VE=VB-VBE;# # V
print"(a) Value of VB(V) : ",VB
print"(a) Value of VE(V) : ",VE
# Part (b)
R1=5.;# /kohm
R2=5.;# /kohm
VCC=10.;# /V
VEE=-15.;# /V
VE=VBE;# # V
VC=VEE+R1/R2*(VCC-VBE);# V
print"(b) Value of VE(V) : ",VE
print"(b) Value of VC(V) : ",VC

(a) Value of VB(V) :  0
(a) Value of VE(V) :  -0.7
(b) Value of VE(V) :  0.7
(b) Value of VC(V) :  -5.7

Example E09 : Pg 373

In [9]:
# Ex_7_9
# given data :
VEE=5.;# # V
VCC=-5.;# # V
VE=1.;# # V
RB=20.;# kohm
RE=5.;# kohm
RC=5.;# kohm
VBE=0.7;# # V
VB=VE-VBE;# /V
IB=VB/RB;# /mA
IE=(VEE-VE)/RE;# mA
IC=IE-IB;# mA
VC=VCC+IC*RC;# V
Beta=IC/IB;# Current gain
Alfa=IC/IE;# Current gain
print"VB(V) : ",VB
print"IB(mA) : ",IB
print"IE(mA) : ",IE
print"IC(mA) : ",IC
#format('v',5);
print"Beta : ",Beta
print"Alfa : ",Alfa

VB(V) :  0.3
IB(mA) :  0.015
IE(mA) :  0.8
IC(mA) :  0.785
Beta :  52.3333333333
Alfa :  0.98125

Example E10 : Pg 374

In [10]:
# Ex_7_10
#clc;
#clear;
#close;
#format('v',6);
# given data :
delVB=0.4;# V
delVC=0;# V# No change
delVE=delVB;# V# Same change
print"delVE(V) : ",delVE
print"delVC(V) : ",delVC

delVE(V) :  0.4
delVC(V) :  0

Example E11 : Pg 374

In [11]:
# Ex_7_11
# given data :
VBE=0.7;# # V
Beta=100.;# /Current Gain
# Part (a)
VCC=6.;# # V
VEE=0;# # V
VB=2.;# # V
RE=18.;# kohm
RC=3.;# kohm
VE=VB-VBE;# V
print"(a) Emitter Voltage(V) : ",VE
IE=1.;# /mA
IC=IE*Beta/(1+Beta);# /mA
VC=VCC-IC*RC;# V
if VC>VE:
    print"Circuit is in active mode."
# Part (b)
VEE=6.;# # V
VCC=0;# # V
VB=1.;# # V
RE=10.;# kohm
RC=10.;# kohm
VE=VB+VBE;# V
print"(b) Emitter Voltage(V) : ",VE
IE=(VEE-VE)/RE;# /mA
IC=IE;# /mA(Assumed nearly equal)
VC=VCC+IC*RC;# V
if VC>VB :
    print"Circuit is in saturation mode."
# Part (c)
VEE=9.5;# # V
VCC=-50.;# # V
VB=-5.;# # V
RE=200.;# kohm
RC=20.;# kohm
VE=VB+VBE;# V
print"(c) Emitter Voltage(V) : ",VE
IE=(VEE-VE)/RE;# /mA;# /mA
IC=IE*Beta/(1+Beta);# /mA
VC=VCC-IC*RC;# V
if VC>VE :
    print"Circuit is in active mode."
else :
    print"Circuit is in reverse active mode."
# Part (d)
VEE=-30.;# # V
VCC=-10.;# # V
VB=-20.;# # V
RE=6.;# kohm
RC=2.;# kohm
VE=VB-VBE;# V
print"(d) Emitter Voltage(V) : ",VE
IE=(VEE-VE)/RE;# /mA;# /mA
IC=IE*Beta/(1.+Beta);# /mA
VC=VCC-IC*RC;# V
if VC>VE :
    print"Circuit is in active mode."
# Note : Printing error in part (a) in the textbook. Answer is also not accurate in the textbook for part(c)

(a) Emitter Voltage(V) :  1.3
Circuit is in active mode.
(b) Emitter Voltage(V) :  1.7
Circuit is in saturation mode.
(c) Emitter Voltage(V) :  -4.3
Circuit is in reverse active mode.
(d) Emitter Voltage(V) :  -20.7
Circuit is in active mode.