# Chapter 6 : BJT amplifiers¶

## Example 6.2: Page No 275¶

In :
#rπ, gm

IBQ=7.6e-6 # in amperes
bta=104#
VT=25e-3 # Voltage equivalent to temperatue at room temperature in volts
ICQ=IBQ*bta # in amperes
gm=ICQ/VT # in ampere per volt
gm=gm*1e3 # in mili-ampere per volt
r_pi=bta/gm # in kilo-ohms
print "rπ = %0.2f kΩ "%r_pi
print "gm = %0.2f mA/V "%gm

rπ = 3.29 kΩ
gm = 31.62 mA/V


## Example 6.3: Page No 287¶

In :
#AI, Ri, AV, AVs, Ro, Ro'

hie=1e3 # in ohms
hfe=100#
hre=2e-4#
hoe=20e-6 # in amperes per volt
RC=5e3 # in ohms
Rs=1e3 # in ohms
# From Table 6.3
AI=-hfe/(1+hoe*RC)#
Ri=hie+hre*AI*RC # in ohms
AV=AI*RC/Ri#
AVs=AV*Ri/(Ri+Rs)#
Yo=hoe-hfe*hre/(hie+Rs) # in ohms inverse
Ro=1/Yo # in ohms
Ro_dash=Ro*RC/(Ro+RC) # in ohms
Ri=Ri*1e-3 # in kilo-ohms
Ro=Ro*1e-3 # in kilo-ohms
Ro_dash=Ro_dash*1e-3 # in kilo-ohms
print "AI = %0.2f "%AI
print "Ri = %0.2f kΩ "%Ri
print "AV = %0.2f "%AV
print "AVs = %0.2f "%AVs
print "Ro = %0.2f kΩ "%Ro
print "Ro'' = %0.2f kΩ "%Ro_dash

AI = -90.91
Ri = 0.91 kΩ
AV = -500.00
AVs = -238.10
Ro = 100.00 kΩ
Ro'' = 4.76 kΩ


## Example 6.4: Page No 288¶

In :
#AI', AVs, Ri,eff, Ro, Ro'

hie=2e3 # in ohms
hfe=50#
hre=2e-4#
hoe=20e-6 # in amperes per volt
# From Fig. 6.22(a)
Rs=2e3 # in ohms
R1=90e3 # in ohms
R2=10e3 # in ohms
RC=5e3 # in ohms
# From the Table 6.3
RB=R1*R2/(R1+R2) # in ohms
AI=-hfe/(1+hoe*RC)#
Ri=hie+hre*AI*RC # in ohms
Ri_eff=RB*Ri/(RB+Ri) # in ohms
AI_dash=AI*RB/(RB+Ri)#
AVs=AI*RC*Ri_eff/(Ri*(Rs+Ri_eff))#
Rs_eff=Rs*RB/(Rs+RB) # in ohms
Yo=hoe-hfe*hre/(hie+Rs_eff) # in ohms inverse
Ro=1/Yo # in ohms
Ro_dash=Ro*RC/(Ro+RC) # in ohms
Ri_eff=Ri_eff*1e-3 # in kilo-ohms
Ro=Ro*1e-3 # in kilo-ohms
Ro_dash=Ro_dash*1e-3 # in kilo-ohms
print "AI'' = %0.2f "%AI_dash
print "AVs = %0.2f "%AVs
print "Ri,eff = %0.2f kΩ "%Ri_eff
print "Ro = %0.2f kΩ "%Ro
print "Ro'' = %0.2f kΩ "%Ro_dash

AI'' = -37.34
AVs = -51.78
Ri,eff = 1.61 kΩ
Ro = 57.97 kΩ
Ro'' = 4.60 kΩ


## Example 6.5: Page No 295¶

In :
#AI, AVs, Ri, Ro'

hie=4e3 # in ohms
hfe=200#
# From Fig. 6.27(a)
Rs=5e3 # in ohms
R1=90e3 # in ohms
R2=10e3 # in ohms
RC=5e3 # in ohms
RE=1e3 # in ohms
# From Fig 6.27(b)
RB=R1*R2/(R1+R2) # in ohms
Ri=hie+(1+hfe)*RE # in ohms
Ri_eff=RB*Ri/(RB+Ri) # in ohms
AI=-hfe*RB/(RB+Ri)#
AVs=-hfe*RC*Ri_eff/(Ri*(Rs+Ri_eff))#
Ro_dash=RC # in ohms
Ri=Ri*1e-3 # in kilo-ohms
Ro_dash=Ro_dash*1e-3 # in kilo-ohms
print "AI = %0.2f "%AI
print "AVs = %0.2f "%AVs
print "Ri = %0.2f kΩ "%Ri
print "Ro'' = %0.2f kΩ "%Ro_dash

AI = -8.41
AVs = -3.09
Ri = 205.00 kΩ
Ro'' = 5.00 kΩ


## Example 6.6: Page No 305¶

In :
#AI, Ri, AVs

bta=100#
VBE=0.7 # Cut-in voltage in volts
VT=25e-3 # Voltage equivalent to temperatue at room temperature in volts
# From Fig. 6.33
RB=100e3 # in ohms
RC=3e3 # in ohms
VBB=3 # in volts

# DC analysis
# From dc equivalent circuit in Fig. 6.34(a)
IBQ=(VBB-VBE)/RB # in amperes
ICQ=bta*IBQ # in amperes
gm=ICQ/VT # in ampere per volt
r_pi=bta/gm # in ohms

# AC analysis
# From ac equivalent circuit using approximate hybrid-π model in Fig. 6.34(b)
AI=-bta#
Ri=RB+r_pi # in ohms
AVs=-bta*RC/(RB+r_pi)#
Ri=Ri*1e-3 # in kilo-ohms
print "AI = %0.2f "%AI
print "Ri = %0.2f kΩ "% Ri
print "AVs = %0.2f "%AVs

AI = -100.00
Ri = 101.09 kΩ
AVs = -2.97


## Example 6.7: Page No 309¶

In :
#              (a) Load resistance RE to make Ri ≥ 500 kΩ
#              (b) AV, Ro, Ro'

IC=2e-3 # in amperes
Rs=5e3 # Source resistance in ohms
bta=125#
VT=25e-3 # Voltage equivalent to temperatue at room temperature in volts

print "Part (a)"
Ri=500e3 # in ohms
gm=IC/VT # in mho
r_pi=bta/gm # in ohms
RE=(Ri-r_pi)/(1+bta) # in ohms
REk=RE*1e-3 # in kilo-ohms
print "RE = %0.2f kΩ "%REk

print "Part (b)"
AV=(1+bta)*RE/(Rs+Ri)#
Ro=(Rs+r_pi)/(1+bta) # in ohms
Ro_dash=Ro*RE/(Ro+RE) # in ohms
print "Ro = %0.2f Ω "%Ro
print "Ro''  = %0.2f Ω "%Ro_dash

Part (a)
RE = 3.96 kΩ
Part (b)
Ro = 52.08 Ω
Ro''  = 51.41 Ω


## Example 6.8: Page No 313¶

In :
#Ri, AVs

IC=0.2e-3 # in amperes
bta=125#
Rs=2e3 # in ohms
RE=100 # in ohms
RC=5e3 # in ohms
VT=25e-3 # Voltage equivalent to temperatue at room temperature in volts
gm=IC/VT # in mho
r_pi=bta/gm # in ohms
Ri=r_pi+(1+bta)*RE # in ohms
AVs=-bta*RC/(Rs+r_pi+(1+bta)*RE)#
Ri=Ri*1e-3 # in kilo-ohms
print "Ri = %0.1f kΩ "%Ri
print "AVs = %0.1f "%AVs

Ri = 28.2 kΩ
AVs = -20.7


## Example 6.9: Page No 314¶

In :
from numpy import inf
#rπ, AI, Ri, AVs, Ro, Ro'

bta=200#
VT=25e-3 # Voltage equivalent to temperatue at room temperature in volts
# From Fig. 6.39
VBE=0.7 # Cut-in voltage in volts
VCC=9 # in volts
RB=200e3 # in ohms
RC=2e3 # in ohms

# DC analysis
# From dc equivalent circuit in Fig. 6.40(a)
# Writing KVL from collector to base loop
IB=(VCC-VBE)/(RB+(1+bta)*RC) # in amperes
ICQ=bta*IB#  # in amperes
gm=ICQ/VT # in mho
r_pi=bta/gm # in ohms

# AC analysis
# From ac equivalent circuit using Miller's theorem in Fig. 6.40(b)
# Assuming AV >> 1
RL=RB*RC/(RB+RC) # Effective load resistance in ohms
# Using hybrid-π model and approximate resulta given in Table 6.5 for CE amplifier stage, we have
AI=-bta#
AV=-bta*RL/r_pi#
Ro=inf#
r_pi=r_pi*1e-3 # in kilo-ohms
RL=RL*1e-3 # in kilo-ohms
print "rπ  = %0.2f kΩ "%r_pi
print "AI = %0.2f "%AI
print "AVs = %0.2f "%AV
print "Ro = %0.2f "%Ro
print "Ro'' = %0.2f kΩ "%RL

rπ  = 1.81 kΩ
AI = -200.00
AVs = -218.41
Ro = inf
Ro'' = 1.98 kΩ


## Example 6.10: Page No 323¶

In :
#Ri,eff, Ro, AV, AI

bta=200#
ro=50e3 # in ohms
VBE=0.7 # Cut-in voltage in volts
VT=25e-3 # Voltage equivalent to temperatue at room temperature in volts
# From Fig. 6.44
VCC=16 # in volts
R1=90e3 # in ohms
R2=10e3 # in ohms
RC=2.2e3 # in ohms
RE=0.68e3 # in ohms

# DC analysis
# From the Thevnin's equivalent circuit in Fig. 6.45(a)
RB=R1*R2/(R1+R2) # in ohms
VBB=VCC*R2/(R1+R2) # in volts
# From the base loop
IB=(VBB-VBE)/(RB+(1+bta)*RE) # in amperes
IE=(1+bta)*IB # in amperes
re=VT/IE # in ohms

# AC analysis
Ri=bta*re+(1+bta)*RE # in ohms
Ri_eff=RB*Ri/(RB+Ri) # in ohms
AI=-bta*RB/(RB+bta*(re+RE))#
AV=-RC/RE#
Ri_eff=Ri_eff*1e-3 # in kilo-ohms
print "Ri,eff = %0.2f kΩ "%Ri_eff
print "Ro = %0.2f "%inf
print "AI = %0.2f "%AI
print "AVs = %0.2f "%AV

Ri,eff = 8.46 kΩ
Ro = inf
AI = -12.08
AVs = -3.24