# Chapter 5 : BJT biasing (Q point) and stability¶

## Example 5.1: Page No 221¶

In [1]:
#RB, RC

IB=40e-6# # in amperes
VCE=6# # in volts
VCC=12# # in volts
betaf=80#
VBE=0.7# # in volts
RB=(VCC-VBE)/IB# # in ohms
RC=(VCC-VCE)/(betaf*IB)# # in ohms
RB=RB*1e-3# # in kilo-ohms
RC=RC*1e-3# # in kilo-ohms
print "RB = %0.2f kΩ "%RB
print "RC = %0.3f kΩ "%RC
RB = 282.50 kΩ
RC = 1.875 kΩ

## Example 5.2: Page No 227¶

In [2]:
#VCEQ, ICQ

VBE=0.7# # in volts
betaf=50#
# From Fig. 5.11(a)
VCC=18# # in volts
R1=82e3# # in ohms
R2=22e3# # in ohms
RC=5.6e3# # in ohms
RE=1.2e3# # in ohms
# Using Thevnin's theorem to obtain equivalent circuit given in Fig. 5.11(b)
VBB=R2*VCC/(R1+R2)# # in volts
RB=R1*R2/(R1+R2)# # in ohms
IB=(VBB-VBE)/(RB+(1+betaf)*RE)# # in amperes
IC=betaf*IB# # in amperes
VCE=VCC-IC*(RC+RE)-IB*RE# # in volts
IC=IC*1e3# # in mili-amperes
print "VCEQ = %0.2f V "%VCE
print "ICQ = %0.2f mA "%IC
VCEQ = 4.50 V
ICQ = 1.98 mA

## Example 5.3: Page No 228¶

In [3]:
#R1, R2, RC, RE

IC=1e-3# # in amperes
VCC=12# # in volts
betaf=100#
VBE=0.7# # in volts
# As suggested in the design constraints, allocate 1/3VCC to RC, another 1/3VCC to R2 leaving 1/3VCC for VCEQ.
VB=4# # in volts
VE=VB-VBE# # in volts
# Neglecting base current,
RE=VE/IC# # in ohms
# Select the current through R1R2 equal to 0.1IC
R1_plus_R2=VCC/(0.1*IC)# # in ohms
R2=VB*R1_plus_R2/VCC# # in ohms
R1=R1_plus_R2-R2# # in ohms
RC=VCC/(3*IC)# # in ohms
R1=R1*1e-3# # in kilo-ohms
R2=R2*1e-3# # in kilo-ohms
RC=RC*1e-3# # in kilo-ohms
RE=RE*1e-3# # in kilo-ohms
print "R1 = %0.2f kΩ "%R1
print "R2 = %0.2f kΩ "%R2
print "RC = %0.2f kΩ "%RC
print "RE = %0.2f kΩ "%RE
R1 = 80.00 kΩ
R2 = 40.00 kΩ
RC = 4.00 kΩ
RE = 3.30 kΩ

## Example 5.4: Page No 231¶

In [4]:
#VCEQ, ICQ

VBE=0.7# # in volts
betaf=45#
# From Fig. 5.14
VEE=9# # in volts
RB=100e3# # in ohms
RC=1.2e3# # in ohms
# Applying KVL in the clockwise direction base emitter loop
IB=(VEE-VBE)/RB# # in amperes
IC=betaf*IB# # in amperes
# Writing KVL for the collector loop
VCE=VEE-IC*RC# # in volts
IC=IC*1e3# # in mili-amperes
print "VCEQ = %0.2f V "%VCE
print "ICQ = %0.3f mA "%IC
VCEQ = 4.52 V
ICQ = 3.735 mA

## Example 5.5: Page No 232¶

In [5]:
#VCEQ, ICQ

VBE=0.7# # in volts
betaf=120#
# From Fig. 5.15
VCC=20# # in volts
VEE=20# # in volts
R1=8.2e3# # in ohms
R2=2.2e3# # in ohms
RC=2.7e3# # in ohms
RE=1.8e3# # in ohms
# Using Thevnin's theorem to obtain equivalent circuit given in Fig. 5.16(b)
RB=R1*R2/(R1+R2)# # in ohms
# From Fig. 5.16(a)
I=(VCC+VEE)/(R1+R2)# # in amperes
VBB=I*R2-VEE# # in volts
# Writing KVL for the base emitter loop and putting Ic= βF*Ib gives
IB=(VEE+VBB-VBE)/(RB+(1+betaf)*RE)# # in amperes
IC=betaf*IB# # in amperes
# KVL for the collector loop gives
VCE=VCC+VEE-IC*(RC+RE)-IB*RE# # in volts
IC=IC*1e3# # in mili-amperes
print "VCEQ = %0.2f V "%VCE
print "ICQ = %0.2f mA "%IC
VCEQ = 20.84 V
ICQ = 4.24 mA

## Example 5.6: Page No 233¶

In [6]:
#RF so that IE=+2 mA

IE=2e-3# # in amperes
VBE=0.7# # in volts
betaf=49#
# From Fig. 5.17
VCC=12# # in volts
RB=25e3# # in ohms
RC=2e3# # in ohms
I1=VBE/RB# # in amperes
IB=IE/(1+betaf)# # in amperes
# KVL for the indicated loop gives
RF=(VCC-RC*(I1+(1+betaf)*IB)-VBE)/(I1+IB)# # in ohms
RF=RF*1e-3# # in kilo-ohms
print "RF so that IE=+2 mA = %0.2f kΩ "%RF
RF so that IE=+2 mA = 106.53 kΩ

## Example 5.7: Page No 234¶

In [7]:
#RCQ, RE

VCEQ=3# # in volts
VBE=0.7# # in volts
betaf=200#
# From Fig. 5.18(a)
VCC=6# # in volts
VEE=6# # in volts
R1=90e3# # in ohms
R2=90e3# # in ohms
# Using Thevnin's theorem to obtain equivalent circuit given in Fig. 5.18(b)
RB=R1*R2/(R1+R2)# # in ohms
VBB=R2*(VCC+VEE)/(R1+R2)# # in volts
# In the output loop
x=VEE-VCEQ# # x = (IC+IB)RE in volts
# Applying KVL in the base emitter loop
IB=(VEE-VBE-x)/RB# # in amperes
IC=betaf*IB# # in amperes
# In the output loop
RC=VCC/IC# # in ohms
RE=x/(IC+IB)# # in ohms
RC=RC*1e-3# # in kilo-ohms
RE=RE*1e-3# # in kilo-ohms
print "RC = %0.3f kΩ "%RC
print "RE = %0.3f kΩ "%RE
RC = 0.587 kΩ
RE = 0.292 kΩ

## Example 5.8: Page No 235¶

In [8]:
#VCEQ

VBE=-0.7# # in volts
betaf=120#
# From Fig. 5.19(a)
VCC=18# # in volts
R1=47e3# # in ohms
R2=10e3# # in ohms
RC=2.4e3# # in ohms
RE=1.1e3# # in ohms
# Using Thevnin's theorem to obtain equivalent circuit given in Fig. 5.19(b)
VBB=R2*VCC/(R1+R2)# # in volts
RB=R1*R2/(R1+R2)# # in ohms
# Applying KVL in the base emitter loop and putting Ic= βF*Ib
IB=(VBB+VBE)/(RB+(1+betaf)*RE)# # in amperes
IC=betaf*IB# # in amperes
# In the collector emitter loop
VCE=-VCC+IC*(RC+RE)+IB*RE# # in volts
print "VCEQ = %0.1f V "%VCE
VCEQ = -10.7 V

## Example 5.9 : Page No 241¶

In [9]:
#              (i) RB
#             (ii) Stability factor
#            (iii) IC at 100°C

bta=50#
VBE=0.7# # in volts
VCE=5# # in volts
# From Fig. 5.21
VCC=24# # in volts
RC=10e3# # in ohms
RE=500# # in ohms

print "Part (i)"
# Applying KVL to the collector emitter circuit and putting Ic= βF*Ib
IB=(VCC-VCE)/((RC+RE)*(bta+1))# # in amperes
IC=bta*IB# #  at 25°C in amperes
RB=(VCE-VBE)/IB# # in ohms
RB=RB*1e-3# # in kilo-ohms
print "RB = %0.2f kΩ "%RB

print "Part (ii)"
S=(1+bta)/(1+bta*(RC+RE)/(RC+RE+RB*1e3))# # Stability factor
print "Stability factor = %0.2f "%S

print "Part (iii)"
# From Table 5.1
del_ICO=(20-0.1)*1e-9# # in amperes
del_IC=S*del_ICO# # in amperes
IC=IC+del_IC# #  at 100°C in amperes
IC=IC*1e3# #  at 100°C in mili-amperes
print "IC at 100°C = %0.2f mA "%IC
Part (i)
RB = 121.19 kΩ
Part (ii)
Stability factor = 10.23
Part (iii)
IC at 100°C = 1.77 mA

## Example 5.10: Page No 242¶

In [10]:
#               (i) S(ICO) for RB/RE=10 and change in IC
#              (ii) S(VBE) for RB = 240 kΩ, RE = 1 kΩ and change in IC

bta=100#

print "Part (i)"
RB_RE=10# # RB/RE
S_ICO=(1+bta)*(1+RB_RE)/(1+bta+RB_RE)#
# From Table 5.1
del_ICO=(20-0.1)*1e-9# # in amperes
del_IC=S_ICO*del_ICO# # in amperes
del_IC=del_IC*1e6# # in micro-amperes
print "S(ICO) for RB/RE=10, ",round(S_ICO,2)
print "Change in IC = %0.3f μA "%del_IC

print "Part (ii)"
RB=240e3# # in kilo-ohms
RE=1e3# # in kilo-ohms
S_VBE=-bta/(RB+(1+bta)*RE)#
# From Table 5.1
del_VBE=0.48-0.65# # in volts
del_IC=S_VBE*del_VBE# # in amperes
del_IC=del_IC*1e6# # in micro-amperes
print "S(VBE) for (RB = 240 kΩ, RE = 1 kΩ) = %0.3e "%S_VBE
print "Change in IC = %0.f μA "%del_IC
Part (i)
S(ICO) for RB/RE=10,  10.0
Change in IC = 0.199 μA
Part (ii)
S(VBE) for (RB = 240 kΩ, RE = 1 kΩ) = -2.933e-04
Change in IC = 50 μA

## Example 5.11: Page No 243¶

In [11]:
#S(β), IC at 100°C

IC=2e-3# #  at 25°C in amperes
# From Table 5.1
bta1=50# # at 25°C
bta2=80# # at 100°C
RB_RE=10# # RB/RE
S=IC*(1+RB_RE)/(bta1*(1+bta2+RB_RE))#
del_bta=bta2-bta1#
del_IC=S*del_bta# # in amperes
IC=IC+del_IC# #  at 100°C in amperes
IC=IC*1e3# #  at 100°C in mili-amperes
print "S(β) = %0.2e "%S
print "IC at 100°C = %0.3f mA "%IC
S(β) = 4.84e-06
IC at 100°C = 2.145 mA

## Example 5.12: Page No 244¶

In [12]:
#Variation of IC over the temperature range -65°C to 175°C

RB_RE=2# # RB/RE
RE=4.7e3# # in ohms
IC=2e-3# # at 25°C in amperes
# From Table 5.1
bta=50# # at 25°C
S_ICO=(1+bta)*(1+RB_RE)/(1+bta+RB_RE)#
S_VBE=-bta/(RE*(1+bta+RB_RE))#
# From Table 5.1
bta1=20# # at -65°C
bta2=120# # at 175°C
S_bta1=IC*(1+RB_RE)/(bta*(1+bta1+RB_RE))# # For 25°C to -65°C
S_bta2=IC*(1+RB_RE)/(bta*(1+bta2+RB_RE))# # For 25°C to 175°C
# From Table 5.1

# For 25°C to -65°C
del_ICO=(0.2e-3-0.1)*1e-9# # in amperes
del_VBE=0.85-0.65# # in volts
del_bta=bta1-bta#
del_IC=S_ICO*del_ICO+S_VBE*del_VBE+S_bta1*del_bta# # in amperes
IC1=IC+del_IC# # at -65°C in amperes
IC1=IC1*1e3# #  at -65°C in mili-amperes
print "IC at -65°C  = %0.3f mA "%IC1

# For 25°C to 175°C
del_ICO=(3.3e3-0.1)*1e-9# # in amperes
del_VBE=0.30-0.65# # in volts
del_bta=bta2-bta#
del_IC=S_ICO*del_ICO+S_VBE*del_VBE+S_bta2*del_bta# # in amperes
IC2=IC+del_IC# # at 175°C in amperes
IC2=IC2*1e3# #  at 175°C in mili-amperes
print "IC at 175°C = %0.2f mA "%IC2
IC at -65°C  = 1.803 mA
IC at 175°C = 2.15 mA

## Example 5.13: Page No 247¶

In [13]:
#               (i) R1
#              (ii) R1 for IC = 10 μA

IC=1e-3# # in amperes
VCC=10# # in volts
bta=125#
VBE=0.7# # in volts

print "Part (i)"
R1=bta*(VCC-VBE)/((bta+2)*IC)# # in ohms
R1=R1*1e-3# # in kilo-ohms
print "R1 = %0.2f kΩ "%R1

print "Part (ii)"
IC=10e-6# # in amperes
R1=bta*(VCC-VBE)/((bta+2)*IC)# # in ohms
R1=R1*1e-3# # in kilo-ohms
print "R1 for (IC = 10 μA) = %0.f kΩ "%R1
Part (i)
R1 = 9.15 kΩ
Part (ii)
R1 for (IC = 10 μA) = 915 kΩ

## Example 5.14: Page No 248¶

In [14]:
from math import log
#R1, RE

Io=10e-6# # in amperes
VCC=10# # in volts
bta=125#
VBE=0.7# # in volts
VT=25e-3# # in volts
# Let
I_ref=1e-3# # in amperes
R1=(VCC-VBE)/I_ref# # in ohms
R1=R1*1e-3# # in kilo-ohms
RE=VT*log(I_ref/Io)/((1+1/bta)*Io)# # in ohms
RE=RE*1e-3# # in kilo-ohms
print "R1 = %0.2f kΩ "%R1
print "RE = %0.2f kΩ "%RE
R1 = 9.30 kΩ
RE = 11.51 kΩ

## Example 5.15: Page No 250¶

In [1]:
from numpy import arange,log
#IC1, IC2, IC3

bta=125#
VBE=0.7# # in volts
VT=25e-3# # Voltage equivalent to temperatue at room temperature in volts
# From Fig. 5.27
VC=9# # in volts
RC=30# # in kilo-ohms
RE=1.94# # in kilo-ohms
I_ref=(VC-VBE)/RC# # in mili-amperes
IC=I_ref*bta/(3+bta)# # in mili-amperes

for i in arange(0.01,0.5,0.001):
if abs(VT*log(IC/i)/(i*(1+1/bta))-RE)<=0.1:
break#

print "IC1 = %0.3f mA "%IC
print "IC2 = %0.3f mA "%IC
print "IC3  = %0.4f mA "%i
IC1 = 0.270 mA
IC2 = 0.270 mA
IC3  = 0.0280 mA

## Example 5.16: Page No 254¶

In [1]:
#Io

bta=100#
VBE=0.7# # in volts
# From Fig. 5.30
# Writing KVL for the indicated loop
I_ref=(10-VBE)/10# # in mili-amperes
Io=bta*I_ref/(2*(1+bta))# # in mili-amperes
print "Io = %0.2f mA "%Io
Io = 0.46 mA

## Example 5.17: Page No 254¶

In [2]:
#               (i) IC1 and IC2
#              (ii) RC so that Vo = 6 V

bta=200#
# From Fig. 5.31

print "Part (i)"
I_ref=(12-0.7)/15# # in amperes
I1=0.7/2.8# # in amperes
IC=(I_ref-I1)*bta/(bta+2)# # in mili-amperes
print "IC1 = %0.2f mA "%IC
print "IC2 = %0.2f mA "%IC

print "Part (ii)"
Vo=6# # in volts
RC=(12-Vo)/IC# # in kilo-ohms
print "RC so that (Vo = 6 V)  = %0.2f kΩ "%RC
Part (i)
IC1 = 0.50 mA
IC2 = 0.50 mA
Part (ii)
RC so that (Vo = 6 V)  = 12.04 kΩ

## Example 5.18: Page No 255¶

In [3]:
#Emitter current in transistor Q3

bta=100#
VBE=0.75# # in volts
# From Fig. 5.32
I=(10-VBE)/4.7# # in mili-amperes
IE=I/2# # in mili-amperes
print "Emitter current in transistor Q3 = %0.2f mA "%IE
Emitter current in transistor Q3 = 0.98 mA