Chapter- 1 : Differential And Cascode Amplifiers

Example : 1.1 - Page No 12

In [4]:
# Given data
V_CC= 10 # in volt
V_EE= V_CC 
V_BE= 0.715 # in volt
R_c1= 2.7 # in k ohm
R_c1= R_c1*10**3 # in ohm
R_c2= R_c1 # in ohm
R_E=3.9 # in k ohm
R_E= R_E*10**3 # in ohm
Bita_ac= 100 
Bita_dc= Bita_ac 
I_E= (V_EE-V_BE)/(2*R_E) # in amp
I_C= I_E # in amp
V_C= V_CC-I_C*R_c1 # in volt
V_E= 0-V_BE # in volt
V_CE= V_C-V_E # in volt
re_desh= 25*10**-3/I_E 
A_d= R_c1/re_desh 
print "Operating current = %0.2f mA" %(I_C*10**3)
print "Operating voltage = %0.1f V" %V_CE
print "Voltage gain = %0.1f" %A_d

Operating current = 1.19 mA
Operating voltage = 7.5 V
Voltage gain = 128.6

Example : 1.2 - Page No 12

In [5]:
# Given data
V_CC= 10 # in volt
V_EE= 10 # in volt
V_BE=0.7# in volt
I_C=0.5 # in mA
I_C=I_C*10**-3 # in amp
R_C= 10 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 9.3 # in k ohm
R_E= R_E*10**3 # in ohm
I_E= (V_EE-V_BE)/(2*R_E) # in amp
I_CQ= I_E # in amp
print "Quiescent collector current = %0.1f mA" %(I_CQ*10**3)
V_CEQ= V_CC+V_BE-I_C*R_C # in volt
print "Quiescent collector emitter voltage = %0.1f V" %V_CEQ

Quiescent collector current = 0.5 mA
Quiescent collector emitter voltage = 5.7 V

Example : 1.3 - Page No 13

In [8]:
from __future__ import division
# Given data
V_CC= 12 # in volt
V_EE= 12 # in volt
V_BE= 0.7 # in volt
R_C= 10 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 10 # in k ohm
R_E= R_E*10**3 # in ohm
R_B= 20 # in k ohm
R_B= R_B*10**3 # in ohm
Bita_dc= 75 
# Part (i)
#Ignoring V_BE
I_T= V_EE/R_E#in amp   
I_E= I_T/2 # in amp
I_C=I_E 
V_out= V_CC-I_C*R_C# in volt
print "Part : a"
print "Output Voltage = %0.f volt (Ignoring V_BE)" %V_out
#Considering V_BE
I_T= (V_EE-V_BE)/R_E#in amp   
I_E= I_T/2 # in amp
I_C=I_E 
V_out= V_CC-I_C*R_C# in volt
print "Output Voltage = %0.2f volt (Condidering V_BE)" %V_out
I_T= (V_EE-V_BE)/(R_E+R_B/(2*Bita_dc)) # in amp
I_E= I_T/2 # in amp
I_C=I_E 
V_out= V_CC-I_C*R_C# in volt
print "Output Voltage = %0.3f volt (With Bita_dc)" %V_out

# Part(ii)
I_C= 0.6 # in mA
I_C=I_C*10**-3 
I_B= I_C/Bita_dc # in amp
print "\nPart : b"
print "Base current = %0.f micro amphere" %(I_B*10**6)
V_B= -I_B*R_B # in volt
print "Base Voltage = %0.2f V" %V_B

# Part (iii)
Bita_dc= 60 
I_B1= I_C/Bita_dc # in amp
print "\nPart : c"
print "Base current for transistor Q1 = %0.f micro amphere" %(I_B1*10**6)
V_B1= -I_B1*R_B # in volt
print "Base Voltage for transistor Q1 = %0.1f V" %V_B1
Bita_dc= 80 
I_B2= I_C/Bita_dc # in amp
print "Base current for transistor Q2 = %0.1f micro amphere" %(I_B2*10**6)
V_B2= -I_B2*R_B # in volt
print "Base Voltage for transistor Q2 = %0.2f V" %V_B2

Part : a
Output Voltage = 6 volt (Ignoring V_BE)
Output Voltage = 6.35 volt (Condidering V_BE)
Output Voltage = 6.424 volt (With Bita_dc)

Part : b
Base current = 8 micro amphere
Base Voltage = -0.16 V

Part : c
Base current for transistor Q1 = 10 micro amphere
Base Voltage for transistor Q1 = -0.2 V
Base current for transistor Q2 = 7.5 micro amphere
Base Voltage for transistor Q2 = -0.15 V

Example : 1.4 - Page No 14

In [10]:
# Given data
V_CC= 10 # in volt
V_EE= 10 # in volt
V_BE= 0.7 # in volt
R_C= 2.2 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 4.7 # in k ohm
R_E= R_E*10**3 # in ohm
Ri_1= 50 # in ohm
Ri_2= Ri_1 # in ohm
Bita_dc= 100 
Bita_ac = Bita_dc 
# Part (a)
I_CQ = (V_EE-V_BE)/(2*R_E+Ri_1/Bita_dc) # in amp
I_E= I_CQ # in amp
print "Part : a"
print "Value of I_CQ = %0.4f mA" %(I_CQ*10**3)
V_CEQ= V_CC + V_BE - I_CQ*R_C # in volt
print "Value of V_CEQ = %0.4f volt" %V_CEQ

# Part(b)
re_desh= (26*10**-3)/I_E # in ohm
# A_d= V_out/V_ind = R_C/re_desh
A_d = R_C/re_desh 
print "\nPart : b"
print "Voltage gain = %0.2f " %A_d

# Part(c)
# R_in1= R_in2= 2*Bita_ac*re_desh
R_in1= 2*Bita_ac*re_desh # in ohm
print "\nPart : c"
print "Input resistance = %0.3f k ohm" %(R_in1*10**-3)

# Part(d)
# R_out1= R_out2= R_C
R_out1= R_C # in ohm
print "\nPart : d"
print "Output resistance = %0.1f k ohm" %(R_out1*10**-3)

Part : a
Value of I_CQ = 0.9893 mA
Value of V_CEQ = 8.5235 volt

Part : b
Voltage gain = 83.71 

Part : c
Input resistance = 5.256 k ohm

Part : d
Output resistance = 2.2 k ohm

Example : 1.5 - Page No 14

In [15]:
# Given data
V_CC= 15 # in volt
V_EE= 15 # in volt
V_BE= 0.7 # in volt
R_C= 1 # in M ohm
R_C= R_C*10**6 # in ohm
R_E= R_C # in ohm

Bita_ac= 100 
I_E = (V_EE-V_BE)/(2*R_E) # in amp
re_desh= (26*10**-3)/I_E # in ohm
A_d = R_C/re_desh 
print "Voltage gain = %0.f " %A_d
Z_in= 2*Bita_ac*re_desh # in ohm
print "Input impedence = %0.4f M ohm" %(Z_in*10**-6)
Z_out= R_C # in ohm
print "Output impedence = %0.f M ohm" %(Z_out*10**-6)
A_cm= R_C/(2*R_E+re_desh) 
CMRR= A_d/A_cm 
print "Common-mode rejection ratio = %0.f" %CMRR
I_C=I_E 
V_out= V_CC-I_C*R_C # in volt
print "Total output voltage at the quiescent value = %0.2f volt" %V_out
# when v_in = 1
v_in= 1 # in mV
v_in= v_in*10**-3 # in volt
v_out= A_d*v_in 
print "The ac output voltage = %0.3f volt" %v_out

# Note : Answer of CMRR in the book is wrong due to wrong calculation of A_cm

Voltage gain = 275 
Input impedence = 0.7273 M ohm
Output impedence = 1 M ohm
Common-mode rejection ratio = 551
Total output voltage at the quiescent value = 7.85 volt
The ac output voltage = 0.275 volt

Example : 1.6 - Page No 17

In [16]:
# Given data
V_EE= 5 # in volt
R_C= 2 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 4.3 # in k ohm
R_E= R_E*10**3 # in ohm
V_BE=0.7 # in volt (Assuming)
V_T= 26*10**-3 # in volt
I_E = (V_EE-V_BE)/(2*R_E) # in amp
re_desh= V_T/I_E # in ohm
A_d = R_C/(2*re_desh) 
print "Voltage gain = %0.2f " %A_d

Voltage gain = 19.23

Example : 1.7 - Page No 17

In [17]:
# Given data
V_EE= 5 # in volt
R_C= 2 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 4.3 # in k ohm
R_E= R_E*10**3 # in ohm
V_BE=0.7 # in volt (Assuming)
V_T= 26*10**-3 # in volt
I_E = (V_EE-V_BE)/(2*R_E) # in amp
re_desh= V_T/I_E # in ohm
A_d = R_C/(2*re_desh) 
A_cm= R_C/(2*R_E+re_desh) 
print "Common mode gain= %0.4f" %A_cm 
CMRR= A_d/A_cm 
print "Common mode rejection ratio = %0.1f" %CMRR

Common mode gain= 0.2312
Common mode rejection ratio = 83.2

Example : 1.8 - Page No 20

In [18]:
# Given data
V_CC= 9 # in volt
V_EE= 9 # in volt
V_BE= 0.7 # in volt (Assuming value)
R_C= 47 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 43 # in k ohm
R_E= R_E*10**3 # in ohm
Ri_1= 20 # in ohm
Ri_2= Ri_1 # in ohm
v_in1= 2.5 # in mv
v_in1=v_in1*10**-3 # in volt
Bita_1= 75 
Bita_2= Bita_1 
I_CQ = (V_EE-V_BE)/(2*R_E+Ri_1/Bita_1) # in amp
I_E= I_CQ # in amp
V_CEQ= V_CC + V_BE - I_CQ*R_C # in volt
re_desh= (26*10**-3)/I_E # in ohm
A_d = R_C/re_desh 
v_out= A_d*v_in1 # in volt
print "Output voltage = %0.3f volt" %v_out

Output voltage = 0.436 volt

Example : 1.9 - Page No 20

In [19]:
# Given data
RC= 2.2;#in kΩ
RE= 4.7;# in kΩ
Ri1= 50*10**-3;# in kΩ
Ri2= 50*10**-3;# in kΩ
VCC= 10;#in V
VEE= 10;# in V
VBE= 0.7;# in V
beta_dc= 100;
beta_ac= 100;

# Part (i)
# Formula Used : ICQ= IE= (VEE-VBE)/(2*RE+Ri/beta_dc)
ICQ= (VEE-VBE)/(2*RE+Ri1/beta_dc);#quiescent collector current in mA
IE= ICQ;# in mA
print "Part (i) : Dual-input, unbalanced output"
print "The value of ICQ = %0.4f mA" %ICQ
# Quiescent collector-emitter voltage,
VCEQ= VCC+VBE-ICQ*RC;# in V
print "The value of VCEQ = %0.4f V" %VCEQ
re_desh= 26/IE;# AC emitter resistance in Ω
Rin1= 2*beta_ac*re_desh;# input resistance in Ω
Rin1= Rin1*10**-3;#in kΩ
Rin2= Rin1;# in kΩ
print "The value of Rin1 = %0.3f kΩ " %Rin1
print "The value of Rin2 = %0.3f kΩ " %Rin1
Rout= RC;# in kΩ
print "The value of Rout = %0.1f kΩ" %Rout 
print "The value of RC = %0.1f kΩ" %RC
# Formula Used : Ad= Vout/Vind= RC/re_desh
Ad= RC*10**3/(re_desh*2);# voltage gain of dual input, unbalanced output
print "The value of Ad = %0.3f " %Ad

# Part (ii)
print "\nPart (ii) : Single-output, balanced output"
print "The value of ICQ = %0.4f mA" %ICQ
print "The value of VCEQ = %0.4f V" %VCEQ
print "The value of Rin = %0.3f kΩ" %Rin1
print "The value of Rout1 = %0.1f kΩ" %Rout
print "The value of Rout2 = %0.1f kΩ" %Rout
# Formula Used : Ad= Vout/Vind= RC/re_desh
Ad= RC*10**3/(re_desh);# voltage gain of dual input, unbalanced output
print "The value of Ad = %0.2f" %Ad

Part (i) : Dual-input, unbalanced output
The value of ICQ = 0.9893 mA
The value of VCEQ = 8.5235 V
The value of Rin1 = 5.256 kΩ 
The value of Rin2 = 5.256 kΩ 
The value of Rout = 2.2 kΩ
The value of RC = 2.2 kΩ
The value of Ad = 41.855 

Part (ii) : Single-output, balanced output
The value of ICQ = 0.9893 mA
The value of VCEQ = 8.5235 V
The value of Rin = 5.256 kΩ
The value of Rout1 = 2.2 kΩ
The value of Rout2 = 2.2 kΩ
The value of Ad = 83.71

Example : 1.10 - Page No 23

In [20]:
# Given data
VEE= 9;#in V
VCC= 9;#in V 
RC= 47*10**3;# collector resistance in Ω
RE= 43*10**3;# emitter resistance in Ω
vin1= 2.5*10**-3;# in V
Ri1= 20*10**3;# in Ω
Ri2= Ri1;# in Ω
VBE= 0.7;# in V
VT= 26*10**-3;# in V
beta1= 75;
beta2= 75;
IE= (VEE-VBE)/(2*RE+Ri1/beta1);#emitter current in A
ICQ= IE;# quiescent current in A
VCEQ= VCC+VBE-ICQ*RC;# quiescent collector voltage in V
re_desh= VT/IE;#AC emitter resistance in Ω
Ad= RC/(2*re_desh);# voltage gain
vout= Ad*vin1;# output voltage in V
print "The output voltage = %0.3f V" %vout

The output voltage = 0.217 V

Example : 1.11 - Page No 25

In [23]:
# Given data
R_E_desh= 200 # in ohm
V_CC= 10 # in volt
V_EE= 10 # in volt
V_BE= 0.7 # in volt
R_C= 2.2 # in k ohm
R_C= R_C*10**3 # in ohm
R_E= 4.7 # in k ohm
R_E= R_E*10**3 # in ohm
Ri_1= 50 # in ohm
Ri_2= Ri_1 # in ohm
Bita_dc= 100 
Bita_ac = Bita_dc 
I_CQ = (V_EE-V_BE)/(2*R_E+ R_E_desh+Ri_1/Bita_dc) # in amp
I_E= I_CQ # in amp
print "Value of I_CQ = %0.4f mA" %(I_CQ*10**3)
V_CEQ= V_CC + V_BE - I_CQ*R_C # in volt
print "Value of V_CEQ = %0.3f volt" %V_CEQ
re_desh= (26*10**-3)/I_E # in ohm
A_d = R_C/(re_desh+R_E_desh) 
print "Voltage gain= %0.2f" %A_d
R_in1= 2*Bita_ac*(re_desh+R_E_desh) # in ohm
print "Input resistance = %0.3f k ohm" %(R_in1*10**-3)
R_out1= R_C # in ohm
print "Output resistance = %0.1f k ohm" %(R_out1*10**-3)

Value of I_CQ = 0.9687 mA
Value of V_CEQ = 8.569 volt
Voltage gain= 9.70
Input resistance = 45.368 k ohm
Output resistance = 2.2 k ohm

Example : 1.12 - Page No 29

In [24]:
# Given data
VEE= 15 #in V
VD1= 0.7 # in V
VD2= 0.7 # in V
VBE= 0.7 # in V
Beta= 100 
VT= 26 # in mV
R3= 180 #in Ω
RC= 470 # in Ω
VB3= -VEE+VD1+VD2 #in V
VE3= VB3-VBE #voltage at emitter terminal of transistor Q3 in V
IE3= (VE3-(-VEE))/R3 #emitter current through transistor Q3 in A

#Part  (i) 
ICQ= IE3/2 #quiescent current in A
ICQ= round(ICQ*10**3) #in mA
IE= ICQ #emitter current in mA
print "Part (i) : Quiescent current = %0.f mA " %ICQ
VCEQ= VEE+VBE-ICQ*10**-3*RC #quiescent collector-emitter voltage in V
print "The quiescent collector-emitter voltage = %0.2f V" %VCEQ
re_desh= VT/IE #AC emitter resistance in Ω

# Part (ii)
Ad= RC/re_desh # differential voltage gain
print "Part (ii) : Differential voltage gain = %0.2f " %Ad

# Part (iii)
Rin1= 2*Beta*re_desh # in Ω
Rin1= Rin1*10**-3 # in kΩ
print "Part (iii) : The input resistance = %0.1f kΩ" %Rin1

Part (i) : Quiescent current = 2 mA 
The quiescent collector-emitter voltage = 14.76 V
Part (ii) : Differential voltage gain = 36.15 
Part (iii) : The input resistance = 2.6 kΩ

Example : 1.13 - Page No 30

In [25]:
# Given data
VEE= 10 #in V
VCC=10 # in V
VD1= 0.715 # in V
Vz= 6.2# in V
VBE= VD1 # in V
Izt= 41 # in mA
R3= 2.7 # in kΩ
RC= 4.7 # in kΩ
VT= 26 # in mV
beta_ac= 100 
beta_dc= 100 
VB3= -VEE+Vz+VD1 #voltage at the base of transistor Q3 in V
VE3= VB3-VBE # voltage at the emitter of transistor Q3 in V
IE3= (VE3-(-VEE))/R3 #emitter current through transistor Q3 in mA
ICQ= IE3/2 #quiescent current in mA
VCEQ= VCC+VBE-ICQ*RC # in V
print "Part (c) : The Q-point values : " 
print "The value of ICQ = %0.3f mA" %ICQ
print "The value of VCEQ = %0.2f V" %VCEQ
re_desh= VT/ICQ #dynamic emitter resistance in Ω
Ad= RC*10**3/re_desh # voltage gain
print "Part (a) : The voltage gain = %0.1f" %Ad
Rin= 2*beta_ac*re_desh # differential input resistance in Ω
Rin=Rin*10**-3 # in kΩ
print "Part (b) : The differential input resistance = %0.2f kΩ" %Rin

Part (c) : The Q-point values : 
The value of ICQ = 1.148 mA
The value of VCEQ = 5.32 V
Part (a) : The voltage gain = 207.5
Part (b) : The differential input resistance = 4.53 kΩ

Example : 1.14 - Page No 34

In [26]:
# Given data
V_CC=12 # in volt
V_BE=0.7 # in volt
R1= 25 # in k ohm
R1=R1*10**3 # in ohm
# I=I_REF= (V_CC-V_BE)/R1
I= (V_CC-V_BE)/R1  # in amp
print "Mirrored current = %0.3f mA" %(I*10**3)

Mirrored current = 0.452 mA

Example : 1.15 - Page No 34

In [27]:
# Given data
VCC= 10 # in V
VBE= 0.7 # in V
R1= 15 # in kΩ
Beta= 100 
I_REF= (VCC-VBE)/R1 #reference current in mA
print "The reference current = %0.2f mA" %I_REF
Iout= I_REF*Beta/(Beta+2) # output current in mA
print "The output current = %0.3f mA" %Iout

The reference current = 0.62 mA
The output current = 0.608 mA

Example : 1.16 - Page No 34

In [28]:
# Given data
VCC= 15 # in V
VBE= 0.7 # in V
R1= 2.2 # in kΩ
Beta= 220 
I_REF= (VCC-VBE)/R1 #reference current in mA
# Formula : I= IC= I_REF*(Beta/(Beta+2))
IC= I_REF*Beta/(Beta+2) # in mA
print "The value of current = %0.2f mA" %IC

The value of current = 6.44 mA

Example : 1.17 - Page No 35

In [29]:
# Given data
Vz= 1.8 # in V
VBE= 0.7 # in V
RE= 1 # in kΩ
Beta= 180 
VB= Vz-VBE # in V
IE= VB/RE #emitter current in mA
# Formula : I= IC= IE*(Beta/(Beta+1))
IC= IE*Beta/(Beta+1) # in mA
print "The value of current = %0.3f mA" %IC

The value of current = 1.094 mA

Example : 1.18 - Page No 35

In [30]:
# Given data
VCC= 9 # in V
R1= 12 # in kΩ
VBE= 0.7 # in V
Beta= 100 
I_REF= (VCC-2*VBE)/R1 #reference current in mA
print "The reference current = %0.4f mA" %I_REF
Iout= I_REF/(1+2/(Beta*(1+Beta))) #output current in mA
print "The output current = %0.4f mA" %Iout
IC2= Iout #collector current in mA
print "The collector current = %0.4f mA" %IC2
# IB3= I_REF-IC1= I_REF-IC2  (since IC1= IC2)
IB3= I_REF-IC2 #base current of transistor Q3 in mA
IB3= IB3*10**3 # in µA
print "The base current of transistor Q3 = %0.1f µA" %IB3
IB3= 0.1 # in µA
IE3= (1+Beta)*IB3 # emitter current of transistor Q3 in µA
print "The emitter current of transistor Q3 = %0.1f µA" %IE3
IB1= IE3/2 #base current in µA
IB2= IB1 # in µA
print "The base current = %0.2f µA" %IB1

The reference current = 0.6333 mA
The output current = 0.6332 mA
The collector current = 0.6332 mA
The base current of transistor Q3 = 0.1 µA
The emitter current of transistor Q3 = 10.1 µA
The base current = 5.05 µA

Example : 1.19 - Page No 36

In [31]:
# Given data
V_BE=0.715 # in volt
V_CC=9 # in volt
Bita_dc=100 
Bita_ac= Bita_dc 
V_EE= 10 # in volt
R=5.6 # in k ohm
R= R*10**3 # in ohm
I_REF= (V_EE-V_BE)/R # in amp
# From 2*I_B + I_C1 -I_REF =0
I_C1= I_REF*Bita_dc/(2+Bita_dc) # in amp
# By symmetry
I_C2= I_C1 
I_C3= I_C2 
I=3*I_C1 # current through R_C in amp
print "Current through R_C = %0.2f mA" %(I*10**3)

Current through R_C = 4.88 mA

Example : 1.20 - Page No 36

In [32]:
# Given data
V_BE=0.7 # in volt
V_CC=5 # in volt
V_EE=-5 # in volt
Bita=100 
R=18.6 # in k ohm
R= R*10**3 # in ohm
I2= (V_CC-V_BE-V_EE)/R # in amp
I_C3=I2 
I_E= I_C3/2 # in amp
re_desh= (26*10**-3)/I_E # in ohm
re1_desh=re_desh 
re2_desh=re1_desh 
R_in1= 2*Bita*re_desh # in ohm
R_in2= R_in1
print "Differential input resistance = %0.1f k ohm" %(R_in1*10**-3)

Differential input resistance = 20.8 k ohm

Example : 1.21 - Page No 44

In [33]:
# Given data
V_BE=0.7 # in volt
V_CC=18 # in volt
R_E=1.1 # in k ohm
R_C=1.8 # in k ohm
R_C=R_C*10**3 # in ohm
R1=4.7 # in k ohm
R2=5.6 # in k ohm
R3=6.8 # in k ohm
I_E1= (V_CC*R1/(R1+R2+R3)-V_BE)/R_E # in mA
re_desh= 26/I_E1 # in ohm
re2_desh=re_desh
Av= -R_C/re2_desh 
print "Voltage gain of the cascode amplifier = %0.1f" %Av



 

Voltage gain of the cascode amplifier = -267.3