Chapter 4 : Biplolar Junction Transistors(BJTs)

Example 4.1: Page No 197

In [1]:
#New value of Ic

VA=100# # Early voltage in volts
VCE_old=1# # in volts
Ic_old=1e-3# # in amperes
VCE_new=11# # in volts
ro=VA/Ic_old# # Output resistance in ohms
Ic_new=(VCE_new-VCE_old+Ic_old*ro)/ro# # in amperes
Ic_new=Ic_new*1e3# # in miliamperes
print "New value of Ic = %0.2f mA"%Ic_new

New value of Ic = 1.10 mA

Example 4.2: Page No 201

In [1]:
#Region of operation, All the node voltages and currents

betaf=100# # Current gain
print "Let us assume that the transistor is in active region."
VBE_active=0.7# # in volts
# From the equivalent circuit in Fig. 4.18(b)
VCC=10# # in volts
VBB=4# # in volts
RE=3.3e3# # in ohms
RC=5e3# # in ohms
VE=VBB-VBE_active# # in volts
# Writing KVL for base emitter loop and putting Ic= βF*Ib
IB=VE/((1+betaf)*RE)# # in amperes
IB=IB*1e3# # in miliamperes
IC=betaf*IB# # in miliamperes
IE=IB+IC# # in miliamperes
VC=VCC-IC*RC*1e-3# # in volts
print "VC = %0.2f V"%VC
print "VE = %0.2f V"%VE
print "VB = %0.2f V"%VBB
print "IC = %0.2f mA"%IC
print "IE = %0.2f mA"%IE
print "IB = %0.4f mA"%IB
print "Since the base is at 4 V and the collector is at 5.05 V, so the collector junction is reverse biased by 1.05 V. The transistor is indeed in forward active region as assumed."

Let us assume that the transistor is in active region.
VC = 5.05 V
VE = 3.30 V
VB = 4.00 V
IC = 0.99 mA
IE = 1.00 mA
IB = 0.0099 mA
Since the base is at 4 V and the collector is at 5.05 V, so the collector junction is reverse biased by 1.05 V. The transistor is indeed in forward active region as assumed.

Example 4.3: Page No 203

In [2]:
#Region of operation, Node currents and voltages

betaf=100# # Current gain
print "Let us assume that the transistor is in active region."
VBE_active=0.7# # in volts
# From Fig. 4.19
VCC=10# # in volts
VBB=5# # in volts
RB=100e3# # in ohms
RE=2e3# # in ohms
RC=2e3# # in ohms
# Writing KVL to the base circuit and putting Ic= βF*Ib
IB=(VBB-VBE_active)/(RB+(1+betaf)*RE)# # in amperes
IB=IB*1e3# # in miliamperes
IC=betaf*IB# # in miliamperes
IE=IB+IC# # in miliamperes
VB=VBB-IB*RB*1e-3# # in volts
VE=IE*RE*1e-3# # in volts
VC=VCC-IC*RC*1e-3# # in volts
print "VC = %0.2f V"%VC
print "VE = %0.2f V"%VE
print "VB = %0.2f V"%VBB
print "IC = %0.2f mA"%IC
print "IE = %0.2f mA"%IE
print "IB = %0.2f mA"%IB
print "Since base voltage VB is 3.6 V and collector is at 7.2 V, so collector-base junction is reverse biased by 3.6 V. Thus our assumption that the transistor is in active region is valid."

Let us assume that the transistor is in active region.
VC = 7.15 V
VE = 2.88 V
VB = 5.00 V
IC = 1.42 mA
IE = 1.44 mA
IB = 0.01 mA
Since base voltage VB is 3.6 V and collector is at 7.2 V, so collector-base junction is reverse biased by 3.6 V. Thus our assumption that the transistor is in active region is valid.

Example 4.4: Page No 203

In [3]:
#Region of operation

betaf=100# # Current gain
print "Let us assume that the transistor is in saturation region."
VBE_sat=0.8# # in volts
VCE_sat=0.2# # in volts
# From Fig. 4.21
VCC=10# # in volts
VBB=5# # in volts
RB=50e3# # in ohms
RC=2e3# # in ohms
# From the base loop
IB=(VBB-VBE_sat)/RB# # in amperes
IB=IB*1e3# # in miliamperes
IC_sat=(VCC-VCE_sat)/RC# # in amperes
IC_sat=IC_sat*1e3# # in miliamperes
IB_min=IC_sat/betaf# # in miliamperes
print "Minimum IB required to saturate the transistor = %0.3f mA"%IB_min
print "IB in the circuit = %0.3f mA"%IB
print "Since IB in the circuit is calculated as 0.084 mA, so it is greater than IB,min. Thus the transistor is indeed in saturation mode."

Let us assume that the transistor is in saturation region.
Minimum IB required to saturate the transistor = 0.049 mA
IB in the circuit = 0.084 mA
Since IB in the circuit is calculated as 0.084 mA, so it is greater than IB,min. Thus the transistor is indeed in saturation mode.

Example 4.5: Page No 205

In [4]:
#Value of RB so as to drive the transistor into saturation

bta=50# # Current gain
VBE_sat=0.8# # in volts
VCE_sat=0.2# # in volts
# From Fig. 4.22
VCC=10# # in volts
VBB=5# # in volts
RC=1e3# # in ohms
IC_sat=(VCC-VCE_sat)/RC# # in amperes
IB_min=IC_sat/bta# # Minimum base current in amperes to saturate the transistor
# Then base current can be taken as
IB=10*IB_min# # in amperes
RB=(VBB-VBE_sat)/IB# # in ohms
RB=RB*1e-3# # in kilo-ohms
print "Value of RB so as to drive the transistor into saturation = %0.2f kΩ"%RB

Value of RB so as to drive the transistor into saturation = 2.14 kΩ

Example 4.6: Page No 205

In [5]:
#Vo1, Vo2

betaf=100# # Current gain
print "Let us assume that the transistor is in active region."
VBE_active=-0.7# # in volts
# From Fig. 4.23
VCC=-10# # in volts
VEE=10# # in volts
VBB=2.5# # in volts
RE=6.8e3# # in ohms
RB=100e3# # in ohms
RC=10e3# # in ohms
# Writing KVL for base-emitter circuit and putting Ic= βF*Ib
IB=(VEE-VBB+VBE_active)/(RB+(1+betaf)*RE)# # in amperes

IC=betaf*IB# # in amperes
IE=IB+IC# # in amperes
Vo1=VCC+IC*RC# # in volts
Vo2=VEE-IE*RE# # in volts
VB=VBB+IB*RB# # in volts
print "Vo1 = %0.2f V"%Vo1
print "Vo2 = %0.2f V"%Vo2
print "Voltage at base = %0.2f V"%VB
print "As base voltage, VB is 3.36 V and voltage at collector is -1.4 V, collector base junction is reverse biased. Thus the transistor is indeed in active region as assumed."

Let us assume that the transistor is in active region.
Vo1 = -1.36 V
Vo2 = 4.06 V
Voltage at base = 3.36 V
As base voltage, VB is 3.36 V and voltage at collector is -1.4 V, collector base junction is reverse biased. Thus the transistor is indeed in active region as assumed.

Example 4.7: Page No 206

In [6]:
#Value of RC to obtain VC = +5 V

betaf=50# # Current gain
print "Let us assume that the transistor is in active region."
print "When current gain = 50"
VBE_active=-0.7# # in volts
# From Fig. 4.24
VC=5# # in volts
VEE=10# # in volts
RB=100e3# # in ohms
# Writing KVL for base circuit and putting Ic= βF*Ib
IB=(VEE+VBE_active)/RB# # in amperes
IC=IB*betaf# # in amperes
RC=VC/IC# # in ohms
RC=RC*1e-3# # in kilo-ohms
print "Value of RC to obtain VC = +5 V = %0.2f kΩ"%RC,
print "When current gain = 100"
IC=IB*100# # in amperes
VC=IC*RC*1e3# # in volts
print "Collector voltage = %0.2f V"%VC
print "Since collector voltage is greater than the base voltage, the transistor goes into saturation as collector junction gets forward biased."

Let us assume that the transistor is in active region.
When current gain = 50
Value of RC to obtain VC = +5 V = 1.08 kΩ When current gain = 100
Collector voltage = 10.00 V
Since collector voltage is greater than the base voltage, the transistor goes into saturation as collector junction gets forward biased.

Example 4.8: Page No 207

In [7]:
#:Labelled voltages and currents

betaf=100# # Current gain
print "Let us assume that the transistor is in active region."
VBE_active=-0.7# # in volts
# From Fig. 4.25(a)
VCC=-10# # in volts
VEE=10# # in volts
RE=6.8e3# # in ohms
RC=10e3# # in ohms
R1=300e3# # in ohms
R2=180e3# # in ohms
# Applying Thevnin's theorem at point B
R_th=R1*R2/(R1+R2)# # in ohms
V_th=VEE-(R2*(VEE-VCC)/(R1+R2))# # in volts
# From the Thevnin equivalent circuit in Fig. 4.25(b)
# Writing KVL for base-emitter circuit and putting Ic= βF*Ib
IB=(VEE-V_th+VBE_active)/(R_th+(1+betaf)*RE)# # in amperes
IB=IB*1e3# # in miliamperes
IC=betaf*IB# # in miliamperes
IE=IB+IC# # in miliamperes
VC=VCC+IC*RC*1e-3# # in volts
VE=VEE-IE*RE*1e-3# # in volts
VB=V_th+IB*R_th*1e-3# # in volts
I1=(VEE-VB)/R2# # in amperes
I1=I1*1e3# # in miliamperes
I2=I1+IB# # in miliamperes
print "IC = %0.2f mA"%IC
print "IE = %0.2f mA"%IE
print "IB = %0.2f mA"%IB
print "I1 = %0.2f mA"%I1
print "I2 = %0.2f mA"%I2
print "VC = %0.2f V"%VC
print "VE = %0.2f V"%VE
print "VB = %0.2f V"%VBB

Let us assume that the transistor is in active region.
IC = 0.85 mA
IE = 0.86 mA
IB = 0.01 mA
I1 = 0.04 mA
I2 = 0.04 mA
VC = -1.49 V
VE = 4.16 V
VB = 2.50 V