Chapter 6: Bipolar junction Transistors (BJTs)

Example 6.1 page No.215

In [2]:
#Exa 6.1
#find the Base current

#given data
Ic=9.95			#in mA
Ie=10    		#in mA

#Calculation
Ib=Ie-Ic		#in mA

#result
print"Emitter current is ",Ib,"mA"
Emitter current is  0.05 mA

Example 6.2 page No. 216

In [13]:
#Exa 6.2
#Find (i)Emitter current (ii)Current amplification factor (iii)Current gain factor 

#given data
IC=0.98			#in mA
IB=20.0			#in uA
IB=IB*10**-3		#in mA

#Calculation
#part (i)
IE=IB+IC		#in mA

#part (ii)
alpha=IC/IE		#unitless
#part (iii)
Beta=IC/IB		#unitless

#Result
print"Emitter current is",IE,"mA"
print"Current amplification factor is  ",alpha
print"Current gain factor is ",Beta
Emitter current is 1.0 mA
Current amplification factor is   0.98
Current gain factor is  49.0

Example 6.3 page No.216

In [10]:
#Exa 6.3
#Emitter current and Collector current

#given data
alfaDC=0.98			#unitless
ICBO=4				#in uA
ICBO=ICBO*10**-3		#in mA
IB=50				#in uA
IB=IB*10**-3			#in mA

#calculation
#Formula : IC=alfaDC*(IB+IC)+ICBO
IC=alfaDC*IB/(1-alfaDC)+ICBO/(1-alfaDC)	#in mA
IE=IC+IB			#in mA

#Result
print"Emitter current is ",IE,"mA"
print"Collector current is  ",IC,"mA"
Emitter current is  2.7 mA
Collector current is   2.65 mA

Example 6.4 page No. 216

In [9]:
#Exa 6.4
#Find the collector current

#given data
IB=10			#in uA
IB=IB*10**-3		#in mA
Beta=99			#Unitless
ICO=1			#in uA
ICO=ICO*10**-3		#in mA

#calculation
#Formula : IC=alfa*(IB+IC)+ICO
IC=Beta*IB+(1+Beta)*ICO	#in mA

#Result
print"Collector current in mA : ",IC,"mA"
Collector current in mA :  1.09 mA

Example 6.5 Page No.216

In [8]:
#Example 6.5
#Find (i) alpha , beta  and Ie 
#(ii)New level of Ib

#Given
Ic=5*10**-3        #mA collector current
Ic_=10*10**-3        #mA collector current
Ib=50*10**-6       #mA, Base current
Icbo=1*10**-6      #micro A, Current to base open current

#Calculation
beta=(Ic-Icbo)/(Ib+Icbo)
alpha=(beta/(1+beta))
Ie=Ib+Ic

Ib=(Ic_-(beta+1)*Icbo)/(beta)

#Result
print"(i) Current gain factor is",round(beta,0)
print" Current amplification factor is",round(alpha,2)
print" Emitter Current  is",Ie*1000,"mA"
print"(ii)New level of Ib  is",round(Ib*10**6,0),"micro A"
(i) Current gain factor is 98.0
 Current amplification factor is 0.99
 Emitter Current  is 5.05 mA
(ii)New level of Ib  is 101.0 micro A

Example 6.6 page No. 222

In [18]:
#Exa 6.6
#Find the dynamic input resistance

#given data
delVEB=200			#in Volts
delIE=5				#in mA

#calculation
rin=delVEB/delIE		#in ohm

#Result
print"Dynamic input resistance is ",rin,"mohm"
Dynamic input resistance is  40 mohm

Example 6.7 page No. 222

In [32]:
#Exa 6.7
#Determine Current gain and base current


#given data
ICBO=12.5 			#in uA
ICBO=ICBO*10**-3 		#in mA
IE=2 				#in mA
IC=1.97 			#in mA

#calculation
alfa=(IC-ICBO)/IE 		#unitless
IB=IE-IC 			#in mA

#result
print"Current gain : ",round(alfa,3)
print"Base current is ",IB,"mA"
Current gain :  0.979
Base current is  0.03 mA

Example 6.8 page No. 222

In [37]:
#Exa 6.8
#given data
RL=4.0 			#in Kohm
VL=3.0			#in volt
alfa=0.96 		#unitless
IC=VL/RL 		#in mA

#calculation
IE=IC/alfa 		#in mA
IB=IE-IC 		#in mA

#result
print"Base current ia",round(IB,2),"mA"
Base current ia 0.03 mA

Example 6.9 page No.227

In [41]:
#Exa 6.9
#Determine Collector emitter voltage and base current

#given data
VCC=10			 #in volt
RL=800			 #in ohm
VL=0.8			 #in volt
alfa=0.96		 #unitless

#calculation
#VR=IC*RL
VCE=VCC-VL 		#in Volt
IC=VL*1000/RL 		#in mA
Beta=alfa/(1-alfa) 	#unitless
IB=IC/Beta 		#in mA

#Result
print"Collector-emitter Voltage is ",VCE,"V"
print"Base current in uA : ",round(IB*1000,2),"microA"
Collector-emitter Voltage is  9.2 V
Base current in uA :  41.67 microA

Example 6.10 page No. 227

In [45]:
#Exa 6.10
#Determine Collector Current

#given data
alfao=0.98 		#unitless
ICO=10 			#in uA
ICO=ICO*10**-3 		#in mA
IB=0.22 		#in mA

#calculation
IC=(alfao*IB+ICO)/(1-alfao) 	#in mA

#result
print"Collector current is",IC,"mA"
Collector current is 11.28 mA

Example 6.11 page No. 228

In [47]:
#Exa 6.11
#determine Dynamic input resistance 

#given data
delVEB=250 		#in mVolts
delIE=1 		#in mA

#calculation
rin=delVEB/delIE 	#in ohm

#result
print"Dynamic input resistance is",rin,"ohm"
Dynamic input resistance is 250 ohm

Example 6.12 page No. 228

In [49]:
#Exa 6.12
#Determine Dynamic output resistance

#given data
delVCE=10-5 		#in Volts
delIC=5.8-5	 	#in mA

#calculation
rin=delVCE/delIC 	#in Kohm

#result
print"Dynamic output resistance is ",rin,"kohm"
Dynamic output resistance is  6.25 kohm

Example 6.13 page No.232

In [3]:
#Exa 6.13
#Determine operating point
%matplotlib inline
import matplotlib.pyplot as plt
#given data
VCC=10 			#in volt
RC=8 			#in Kohm
Beta=40 		#unitless
IB=15 			#in uA
IB=IB*10**-3 		#in mA

#calculation
# For VCE = 0 Volts
IC=VCC/RC 		#in mA
#For IC=0 VCE=VCC=10V :
IC=Beta*IB 		#in mA
VCE=VCC-IC*RC 		#in Volts

#result
print"Operating point Q is (",VCE,"V,",IC,"mA)"

#Plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)

Vce=[0,10]
Ic=[1.25,0]
plt.xlabel('Vce,V')
plt.ylabel('Ic,mA')
ax.plot([5.2], [0.6], 'o')
ax.annotate('(5.2V,0.6 mA)', xy=(5.4,0.7))

a=plt.plot(Vce,Ic)
plt.show(a)
Operating point Q is ( 5.2 V, 0.6 mA)

Example 6.14 page No. 232

In [65]:
#Exa 6.14
#How will the Q point change when load resistance will be change

#given data 
Vcc=12 		#in Volt collector supply voltage
Ic=1.2         #A, collector current
Rl=5            #kohm load resistance

#calculation
Vce=Vcc-Ic*Rl    #Collector emitter voltage
Rl1=7.5
Vce1=Vcc-Ic*Rl1

#result
print"Operating point at load resistance 5 kohm is (",Vce,"V,",Ic,"mA)"
print"Operating point at load resistance 7.5 kohm is (",Vce1,"V,",Ic,"mA)"
Operating point at load resistance 5 kohm is ( 6.0 V, 1.2 mA)
Operating point at load resistance 7.5 kohm is ( 3.0 V, 1.2 mA)

Example 6.15 Page No.233

In [6]:
#Example 6.15
#Given
Vcc=20         # V,  collector voltage
Rc=3.3*10**3

#calculation
#Appling kirchoff's Voltage Law
Ic=0          #for cut off point
Vce=Vcc
Ic=Vcc/Rc
print "Collector to emitter voltage is (Vce)",Vce,"V"
print "Collector current at saturation point is (Ic)",round(Ic*1000,0),"mA"

#Plot
%matplotlib inline
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)

Vce=[0,20]
Ic=[6,0]
plt.xlabel("Vce  (V)") 
plt.ylabel("Ic  (mA)") 
plt.xlim((0,25))
plt.ylim((0,8))
ax.plot([0], [6], 'o')
ax.annotate('(0,6mA)', xy=(0,6))

ax.plot([20], [0], 'o')
ax.annotate('(20V,0)', xy=(20,0))
a=plt.plot(Vce,Ic)
plt.show(a)
Collector to emitter voltage is (Vce) 20 V
Collector current at saturation point is (Ic) 6.0 mA

Example 6.16 page No. 233

In [7]:
#Exa 6.16
#find collector voltage and base voltage

#given data 
Beta=45 			#Unitless
VBE=0.7 			#in Volt
VCC=0 				#in Volt
RB=10**5 			#in ohm
RC=1.2*10**3 			#in ohm
VEE=-9 				#in Volt

#calculation
#Applying Kirchoffs Voltage Law in input loop we have
#IB*RB+VBE+VEE=0
IB=-(VBE+VEE)/RB 		#in mA
IC=Beta*IB 			#in mA
VC=VCC-IC*RC 			#in Volts
VB=VBE+VEE 			#in Volts

#Result
print"collector voltage is ",round(VC,1),"V"
print"Base voltage is ",VB,"V"
collector voltage is  -4.5 V
Base voltage is  -8.3 V