# 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
ICBO=4				#in uA
ICBO=ICBO*10**-3		#in mA
IB=50				#in uA
IB=IB*10**-3			#in mA

#calculation
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()

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

#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()

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