Chapter 6: Bipolar junction Transistors (BJTs)

Example 6.1 page No.215

In [2]:
Ic=9.95			#in mA
Ie=10    		#in mA

Ib=Ie-Ic		#in mA

print"Emitter current is ",Ib,"mA"

Emitter current is  0.05 mA

Example 6.2 page No. 216

In [13]:
IC=0.98			#in mA
IB=20.0			#in uA
IB=IB*10**-3		#in mA

IE=IB+IC		#in mA

alpha=IC/IE		#unitless
Beta=IC/IB		#unitless

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 [16]:
alfaDC=0.98			#unitless
ICBO=4				#in uA
ICBO=ICBO*10**-3		#in mA
IB=50				#in uA
IB=IB*10**-3			#in mA

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

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 [17]:
IB=10			#in uA
IB=IB*10**-3		#in mA
Beta=99			#Unitless
ICO=1			#in uA
ICO=ICO*10**-3		#in mA

IC=Beta*IB+(1+Beta)*ICO	#in mA

print"Collector current in mA : ",IC,"mA"

Collector current in mA :  1.09 mA

Example 6.5 Page No.216

In [6]:
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

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

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

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]:
delVEB=200			#in Volts
delIE=5				#in mA

rin=delVEB/delIE		#in ohm

print"Dynamic input resistance is ",rin,"mohm"

Dynamic input resistance is  40 mohm

Example 6.7 page No. 222

In [32]:
ICBO=12.5 			#in uA
ICBO=ICBO*10**-3 		#in mA
IE=2 				#in mA
IC=1.97 			#in mA

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

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]:
RL=4.0 			#in Kohm
VL=3.0			#in volt
alfa=0.96 		#unitless
IC=VL/RL 		#in mA

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

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

Base current ia 0.03 mA

Example 6.9 page No.227

In [41]:
VCC=10			 #in volt
RL=800			 #in ohm
VL=0.8			 #in volt
alfa=0.96		 #unitless

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

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]:
alfao=0.98 		#unitless
ICO=10 			#in uA
ICO=ICO*10**-3 		#in mA
IB=0.22 		#in mA

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

print"Collector current is",IC,"mA"

Collector current is 11.28 mA

Example 6.11 page No. 228

In [47]:
delVEB=250 		#in mVolts
delIE=1 		#in mA

rin=delVEB/delIE 	#in ohm

print"Dynamic input resistance is",rin,"ohm"

Dynamic input resistance is 250 ohm

Example 6.12 page No. 228

In [49]:
delVCE=10-5 		#in Volts
delIC=5.8-5	 	#in mA

rin=delVCE/delIC 	#in Kohm

print"Dynamic output resistance is ",rin,"kohm"

Dynamic output resistance is  6.25 kohm

Example 6.13 page No.232

In [ ]:
%matplotlib inline
In [17]:
VCC=10 			#in volt
RC=8 			#in Kohm
Beta=40 		#unitless
IB=15 			#in uA
IB=IB*10**-3 		#in mA

IC=VCC/RC 		#in mA
IC=Beta*IB 		#in mA
VCE=VCC-IC*RC 		#in Volts

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

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=plot(Vce,Ic)
show(a)

Operating point Q is ( 5.2 V, 0.6 mA)

Example 6.14 page No. 232

In [65]:
Vcc=12 		#in Volt collector supply voltage
Ic=1.2         #A, collector current
Rl=5            #kohm load resistance

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

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 [1]:
Vcc=20         # V,  collector voltage
Rc=3.3*10**3

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"

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

Vce=[0,20]
Ic=[6,0]
xlabel("Vce  (V)") 
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=plot(Vce,Ic)
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 [60]:
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

IB=-(VBE+VEE)/RB 		#in mA
IC=Beta*IB 			#in mA
VC=VCC-IC*RC 			#in Volts
VB=VBE+VEE 			#in Volts

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