Chapter : 1 - Analog Integrated Circuit Design : An Overview

Example - 1.1 : Page No - 35

In [1]:
#Given data
V_EE = 10 # in V
R2 = 2.4 # in k ohm
R1 = 2.4 # in k ohm
R3 = 1 # in k ohm
V_BE3 = 0.7 # in V
I = (V_EE - ((R2*V_EE)/(R1+R2)) - V_BE3)/R3 # in mA
print "The constant current = %0.1f mA" %I

The constant current = 4.3 mA

Example - 1.2 : Page No - 39

In [3]:
from math import log
#Given data
V_CC = 50 # in V
V_BE2 = 0.7 # in V
R = 50 # in k ohm
R = 50 * 10**3 # in ohm
I_C1 = 10 # in µA
I_C1 =I_C1 * 10**-6 # in A
V_T = 26 # in mV
V_T = V_T * 10**-3 # in V
I_C2 = (V_CC - V_BE2)/R # in A
R_E = (V_T*log(I_C2/I_C1))/I_C1 # in ohm
R_E = R_E * 10**-3 # in k ohm
print "The value of R_E = %0.3f kΩ" %R_E

The value of R_E = 11.937 kΩ

Example - 1.3 : Page No - 43

In [6]:
from __future__ import division
#Given data
V = 10 # in V
V_BE = 0.715 # in V
V_R = 0-(V_BE - V) # in V
R = 5.6 # in  k ohm
I_R = V_R/R # in  mA
bita = 100 
I_C = I_R * (bita/(1+bita)) # in mA
print "For transistor Q1, the collector current = %0.4f mA" %I_C
I_C2 = I_R # in mA
print "For transistor Q2, the collector current = %0.3f mA" %I_C2
I_C3 = I_R # in mA
print "For transistor Q3, the collector current = %0.3f mA " %I_C3
I_C4 = I_R # in mA
print "For transistor Q4, the collector current = %0.3f mA " %I_C4

For transistor Q1, the collector current = 1.6416 mA
For transistor Q2, the collector current = 1.658 mA
For transistor Q3, the collector current = 1.658 mA 
For transistor Q4, the collector current = 1.658 mA

Example - 1.4 : Page No - 44

In [7]:
#Given data
V = 10 # in V
V_BE = 0.715 # in V
R = 5.6 # in k ohm
I = (V-V_BE)/(R) # in mA
bita = 100 
I_C1 = (bita/(4+bita))*I # in mA
print "For transistor Q1, the collector current = %0.4f mA" %I_C1
I_C2 = I_C1 # in mA
print "For transistor Q2, the collector current = %0.4f mA" %I_C2
I_C3 = I_C1 # in mA
print "For transistor Q3, the collector current = %0.4f mA" %I_C3
I_C4 = I_C1 # in mA
print "For transistor Q4, the collector current = %0.4f mA" %I_C4

For transistor Q1, the collector current = 1.5943 mA
For transistor Q2, the collector current = 1.5943 mA
For transistor Q3, the collector current = 1.5943 mA
For transistor Q4, the collector current = 1.5943 mA

Example - 1.5 : Page No - 46

In [10]:
from math import sqrt
#Given data
I_D1 = 100 # in µA
k_n = 200 # in µA/V**2
W = 10 # in µm
l = 1 # in µm
V_A = 20 # in V
V_ov = sqrt((I_D1*2)/(k_n*(W/l))) # in V
V_t = 0.7 # in V
V_GS = V_t + V_ov # in V
V_GS = round(V_GS) # in V
V_DD = 3 # in V
I_REF = 100 # in µA
I_REF = I_REF * 10**-3 # in mA
R = (V_DD - V_GS)/I_REF # in k ohm
print "The value of R = %0.f kΩ" %R 
V_ov_min = V_ov  # in volt
print "The lowest possible value of V_o = %0.1f V" %V_ov_min 
r_o2 = V_A/I_D1 # in M ohm
print "The output resistance % 0.1f MΩ" %r_o2 
V_O = V_GS # in V
del_Io = V_O/r_o2 # in µA
print "The change in output current del_Io = %0.f µA" %del_Io 

The value of R = 20 kΩ
The lowest possible value of V_o = 0.3 V
The output resistance  0.2 MΩ
The change in output current del_Io = 5 µA