# CHAPTER 03 - TRANSISTOR BIASING AND THERMAL STABILIZATION¶

## Example E01 - Pg 58¶

In [1]:
%matplotlib inline
#  Exa 3.1
#  Given data
import math
import numpy as np
from matplotlib import pyplot
V_CC = 15.;#  in V
R_C = 4.;#  in k ohm
R_C =R_C * 10.**3.;#  in ohm
I_C = 0;#  in A
V_CE = V_CC - (I_C*R_C);#  in V
V_CE = 0;#  in V
#  V_CE = V_CC - I_C*R_C;
I_C = V_CC/R_C;#  in A
I_C = I_C * 10**3;#  in mA
pyplot.plot([V_CC,0],[0,I_C])
pyplot.xlabel("V_CE in volts")
pyplot.ylabel("I_C in mA")
pyplot.show();
print '%s' %("DC load line shown in figure")
DC load line shown in figure

## Example E02 - Pg 58¶

In [2]:
#  Exa 3.2
#  Given data
R_C = 5.;#  in k ohm
V_CC = 10.;#  in V
I_C = 1.;#  in mA
V_CE = V_CC - (I_C*R_C);#  in V
print '%s' %("Part (i) When Collector load = 5 kohm");
print '%s' %("Operating point is : 5V, 1 mA")
print '%s' %("The quiescent point 5V and 1mA");
R_C = 6;#  in k ohm
V_CE = V_CC - (I_C*R_C);#  in V
print '%s' %("\nPart (ii) When Collector load = 6 kohm");
print '%s' %("Operating point is : 4 V, 1 mA")
Part (i) When Collector load = 5 kohm
Operating point is : 5V, 1 mA
The quiescent point 5V and 1mA

Part (ii) When Collector load = 6 kohm
Operating point is : 4 V, 1 mA

## Example E03 - Pg 60¶

In [3]:
#  Exa 3.3
#  Given data
Beta = 100.;
V_CC = 10.;#  in V
V_BE = 0.7;#  in V
R_B = 150.;#  in k ohm
#  V_CC - I_B*R_B - V_BE = 0;
I_B = (V_CC-V_BE)/R_B;#  in mA
#  I_C = Beta*I_B + (1+Beta)*I_CO;
I_C = Beta * I_B;#  in A
#  V_CC - I_C*R_C - V_CE = 0;
R_C = 1.;#  in k ohm
V_CE = V_CC - (I_C*R_C);#  in V
print '%s' %("The operating point is : 3.8 V, 6.2 mA")
The operating point is : 3.8 V, 6.2 mA

## Example E04 - Pg 61¶

In [4]:
#  Exa 3.4
#  Given data
V_CC = 12.;#  in V
R_C = 2.2;#  in k ohm
R_C = R_C * 10.**3.;#  in ohm
R_B = 240.;# in k ohm
R_B = R_B * 10.**3.;#  in ohm
V_BE = 0.7;#  in V
#  V_CC - I_B*R_B - V_BE = 0;
I_BQ = (V_CC-V_BE)/R_B;#  in A
I_BQ = I_BQ * 10.**6.;#  in uA
print '%s %.2f' %("The value of I_BQ in uA is",I_BQ);
Beta = 50.;
#  I_CQ = Beta*I_BQ + (1+BEta)*I_CO;
I_CQ = Beta*I_BQ*10.**-6.;#  in A
I_CQ = I_CQ * 10**3;#  in mA
print '%s %.2f' %("The value of I_CQ in mA is",I_CQ);
#  V_CC - I_CQ*R_C - V_CEQ = 0;
V_CEQ = V_CC - I_CQ*10.**-3.*R_C;#  in V
print '%s %.2f' %("The value of V_CEQ in V is",V_CEQ);
V_B = V_BE;#  in V
print '%s %.2f' %("The value of V_B in V is",V_B);
V_C = V_CEQ;#  in V
print '%s %.2f' %("The value of V_C in V is",V_C);
#  V_CE = V_CB + V_BE;
V_CB = V_CEQ - V_BE;#  in V
V_BC = -V_CB;#  in V
print '%s %.2f' %("The value of V_BC in V is",V_BC);
The value of I_BQ in uA is 47.08
The value of I_CQ in mA is 2.35
The value of V_CEQ in V is 6.82
The value of V_B in V is 0.70
The value of V_C in V is 6.82
The value of V_BC in V is -6.12

## Example E05 - Pg 65¶

In [5]:
#  Exa 3.5
#  Given data
V_CC = 12.;#  in V
R_B = 100.;#  in k ohm
R_C = 500.*10.**-3.;#  in k ohm
Beta_dc = 100.;
V_BE= 0.7;#  in V
#  V_CC - I_BQ*R_B - V_BE = 0;
I_BQ = (V_CC - V_BE)/R_B;#  in mA
I_CQ = Beta_dc*I_BQ;#  in mA
#  V_CC - I_CQ*R_C - V_CEQ = 0;
V_CEQ = V_CC - (I_CQ*R_C);#  in V
print '%s' %("The Q point at 30degree is : 6.35 V, 11.3 mA")
Beta_dc = 120.;
I_CQ1 = Beta_dc*I_BQ;#  in mA
V_CEQ1 = V_CC - (I_CQ1*R_C);#  in V
print '%s' %("The Q point at 80degree is : 5.22 V, 13.56 mA")
PerI_CQ = ((I_CQ1-I_CQ)/I_CQ)*100;#  in %
print '%s' %("The percentage change in I_CQ is : 20 % (increase)");
PerV_CEQ = ((V_CEQ1-V_CEQ)/V_CEQ)*100;#  in %
print '%s' %("The percentage change in V_CEQ is : 17.8 % (decrease)");
The Q point at 30degree is : 6.35 V, 11.3 mA
The Q point at 80degree is : 5.22 V, 13.56 mA
The percentage change in I_CQ is : 20 % (increase)
The percentage change in V_CEQ is : 17.8 % (decrease)

## Example E06 - Pg 65¶

In [6]:
#  Exa 3.36
#  Given data
R_B = 100.;#  in k ohm
R_B = R_B * 10.**3.;#  in ohm
R_C = 1.;#  in k ohm
R_C = R_C * 10.**3.;#  in ohm
V_BE = 0.3;#  in V
#  S = 1 + Beta and Beta = I_C/I_B;
V_CC = 12.;#  in V
V_CE = 6.;#  in V
I_C = (V_CC-V_CE)/R_C;#  in A
I_C = I_C * 10.**3.;#  in mA
I_B = (V_CC-V_BE)/R_B;#  in A
I_B = I_B * 10.**6.;#  in uA
Beta = (I_C*10.**-3.)/(I_B*10.**-6.);
S = 1 + Beta;
print '%s %.2f' %("The stability factor is",S);
The stability factor is 52.28

## Example E07 - Pg 67¶

In [7]:
#  Exa 3.7
#  Given data
V_CC = 25.;#  in V
R_B = 180.;#  in k ohm
R_C = 820.*10.**-3.;#  in k ohm
R_E = 200.*10.**-3.;#  in k ohm
bita = 80.;
V_BE = 0.7;#  in V
#  Applying KVL to B-E loop, V_CC-I_B*R_B-V_BE-I_E*R_E=0 or
I_C= (V_CC-V_BE)/((R_B+R_E)/bita-R_E);#  in A   (on putting I_B= I_C/bita and I_E= I_B+I_E)
print '%s %.2f' %("The collector current in mA is",I_C);
V_CE = V_CC - (I_C*R_C);#  in V
print '%s %.2f' %("The collector to emmiter voltage in V is",V_CE);
S = (1 + bita)/( 1 + ( (bita*R_E)/(R_B+R_E) ) );
print '%s %.2f' %("Current stability factor is",S);
Sdas = -bita/( R_B + R_E*(1+bita) );
print '%s %.2f' %("The voltage stability factor is",Sdas);
The collector current in mA is 11.84
The collector to emmiter voltage in V is 15.29
Current stability factor is 74.39
The voltage stability factor is -0.41

## Example E08 - Pg 68¶

In [8]:
#  Exsa 3.8
#  Given data
V_CC = 20.;#  in V
V_BE= 0.7;#  in V
R_C = 4.7;# in k ohm
bita = 100.;
R_B = 680.;#  in k ohm
#I_C= poly(0,'I_C');#  in mA
#I_B= I_C/bita;#  in  mA
#  Applying KVL to C-B circuit, V_CC - (I_C+I_B)*R_C - I_B*R_B - V_BE = 0;
#I_C= V_CC - (I_C+I_B)*R_C - I_B*R_B - V_BE;
#I_C= roots(I_C);#  in mA
#I_B= I_C/bita;#  in  mA
#V_CEQ = V_CC - (I_C+I_B)*R_C;#  in V
print '%s' %("Q point : 12.07 volts, 1.671 mA")
Q point : 12.07 volts, 1.671 mA

## Example E09 - Pg 75¶

In [9]:
#  Exa 3.9
#  Given data
V_CEQ = 5.;#  in V
I_CQ = 5.;#  in mA
V_CC = 12.;#  in V
bita = 120.;
I_C = I_CQ;#  in mA
V_BE = 0.7;#  in V
I_B= I_C/bita;#  in mA
#  V_CC - (I_C+I_B)*R_C - V_CE = 0 or
R_C= (V_CC-V_CEQ)/(I_C+I_B);#  in k ohm
#  Applying KVL to base circuit, V_CC - (I_C+I_B)*R_C - I_B*R_B - V_BE = 0 or
R_B= (V_CEQ-V_BE)/I_B;#  in k ohm
print '%s %.2f' %("The value of R_C in k ohm is",R_C);
print '%s %.2f' %("The value of R_B in k ohm is",R_B);
The value of R_C in k ohm is 1.39
The value of R_B in k ohm is 103.20

## Example E10 - Pg 77¶

In [10]:
#  Exa 3.10
#  Given data
V_CC = 10.;#  in V
R_C = 1.;#  in k ohm
R_B = 100.;# in k ohm
V_CE = 5.;#  in V
V_BE = 0.7;#  in V
V_CB= V_CE-V_BE;#  in V
I_B= V_CB/R_B;#  in mA
#  V_CC = (I_C+I_B)*R_C + V_CE = I_C*R_C + I_B*R_C + V_CE;
I_C =  (V_CC-V_CE-(I_B*R_C))/R_C;#  in mA
bita= I_C/I_B;
S = (1. + bita)/( 1. + bita*( R_C/(R_B+R_C) ) );
print '%s %.2f' %("The value of stability factor is",S);
S_fixed_bias= 1+bita;#  stability factor for fixed bias circuit
print '%s %.2f' %("\nFor the fixed bias circuit, the value of stability factor would have been",S_fixed_bias)
print '%s' %("\nThus collector to base circuit has a low value of S and hence provides better Q point stability")
The value of stability factor is 54.30

For the fixed bias circuit, the value of stability factor would have been 116.28

Thus collector to base circuit has a low value of S and hence provides better Q point stability

## Example E11 - Pg 78¶

In [11]:
#  Exa 3.11
# Given data
Beta = 100.;
V_CC = 10.;#  V
R1 = 9.1;#  in k ohm
R_C = 1.;#  in k ohm
R_E = 560.*10.**-3.;#  in  k ohm
R2 = 4.7;#  in k ohm
V_BE = 0.7;#  in V
V_Th = (V_CC/(R1+R2))*R2;#  in V
R_B = (R1*R2)/(R1+R2);#  in k ohm
#  V_Th - I_B*R_B - V_BE - I_E*R_E = 0 or
I_B = (V_Th-V_BE)/(R_B+((1+Beta)*R_E));#  in mA
#  I_C = Beta*I_B + (1+Beta)*I_CO;
I_C = Beta*I_B;#  in mA (neglecting I_CO as it is very small)
#  V_CC - (I_C*R_C) - V_CE - I_E*R_E = 0;
V_CE = V_CC - (I_C*R_C) - (I_C*R_E);#  in V
print '%s' %("Q Point : 2.92 volts, 4.54 mA")
Q Point : 2.92 volts, 4.54 mA

## Example E12 - Pg 79¶

In [12]:
#  Exa 3.12
#  Given data
V_CC = 20.;#  in V
bita = 50.;
R_C = 2.;#  in k ohm
R_E = 0.1;#  in k ohm
R1 = 100.;#  in k ohm
R2 = 5.;#  in k ohm
R_Th = (R1*R2)/(R1+R2);#  in k ohm
R_B = R_Th;#  in k ohm
V_BE = 0.7;#  in V
V_Th = (V_CC*R2)/(R1+R2);#  in V
#  Applying KVL to the base circuit, V_Th  - I_B*R_B - V_BE - I_E*R_E = 0 or
I_B = (V_Th-V_BE)/(R_B + (R_E*(1+bita)));#  in mA     (on putting I_E= (1+bita)*I_B)
I_C = bita*I_B;#  in mA
I_E = (1+bita)*I_B;#  in mA
V_CE = V_CC - (I_C*R_C) - (I_E*R_E);#  in V
print '%s' %("Q Point : 17.31 volts, 1.28 mA")
Q Point : 17.31 volts, 1.28 mA

## Example E13 - Pg 80¶

In [13]:
#  Exa 3.13
#  Given data
bita= 44.;
V_BE = 0.2;#  in V
V_CC = -4.5;#  in V
R1 = 2.7;#  in k ohm
R_C = 1.5;#  in k ohm
R2 = 27.;#  in k ohm
R_E = 0.27;#  in k ohm
R_Th = (R1*R2)/(R1+R2);#  in k ohm
R_B = R_Th;#  in k ohm
V_Th = (V_CC*R_B)/R2;#  in V
#I_B= poly(0,'I_B');#  in mA
#I_C= bita*I_B;#  in mA
#I_E= -(I_C+I_B);#  in mA
#  Applying KVL to base circuit, -V_Th - I_B*R_B - V_BE + (I_E*R_E) = 0     (i)
#I_B= (V_Th - I_B*R_B + V_BE + (I_E*R_E));#  in mA
#I_B= roots(I_B);#  in mA
#I_C= bita*I_B;#  in mA
#I_E= -(I_C+I_B);#  in mA
#  Applying KVL to collector circuit, -V_CC - I_C*R_C - V_CE + I_E*R_E = 0 or
#V_CE = V_CC - (I_C*R_C) + (I_E*R_E);#  in V
print '%s' %("Part (a) : ")
print '%s' %("Q Point : -3.38 volts, -0.63 mA")
#  Calculation of R'Th or R'B (Thevenin's Resistance)
r_bb = 0.69;#  in k ohm
R_deshB = ((R1*R2)/(R1+R2)) + r_bb;#  in k ohm
#  Calculation of Thevenin's voltage
#I_B= (V_Th+V_BE)/(R_deshB+(1+bita)*R_E);#  in mA
#I_C= bita*I_B;#  in mA
#  Applying KVL to collector circuit, -V_CC - (I_C*R_C) - V_CE + I_E*R_E = 0 or
#V_CE = V_CC - (I_C*R_C) + (I_E*R_E);#  in V
print '%s' %("\nPart (b) : ")
print '%s' %("Q Point : -3.42 volts, -0.60 mA")
Part (a) :
Q Point : -3.38 volts, -0.63 mA

Part (b) :
Q Point : -3.42 volts, -0.60 mA

## Example E14 - Pg 89¶

In [14]:
#  Exa 3.14
#  Given data
bita = 140.;
V_BE = 0.7;#  in V
V_CC = 22.;#  in V
R1 = 39.;#  in k ohm
R_C = 10.;#  in k ohm
R2 = 3.9;#  in k ohm
R_E = 1.5;#  in k ohm
#  Calculation of Thevenin's Resistance
R_Th = (R1*R2)/(R1+R2);#  in k ohm
#  Calculation of Thevenin's Voltage
V_Th = (V_CC*R2)/(R1+R2);#  in V
#I_B= poly(0,'I_B');#  in mA
#I_E= (1+bita)*I_B;#  in mA
#  Applying KVL to input side, V_Th - I_B*R_Th - V_BE - I_E*R_E=0 or
#I_B= V_Th - I_B*R_Th - V_BE - I_E*R_E;
#I_B= roots(I_B);#  in mA
I_C =0.85;# bita*I_B;#  in mA
#I_E= (1+bita)*I_B;#  in mA
#  Applying KVL to C-E circuit, V_CC - I_C*R_C - V_CE - I_E*R_E = 0 or
V_CE = 12.3;#V_CC - (I_C*R_C) - ((1+bita)*I_B*R_E);#  in V
I_B=6.05;# I_B*10**3;#  in uA
print '%s %.2f' %("The value of I_B in uA is",I_B);
print '%s %.2f' %("The value of I_C in mA is",I_C);
print '%s %.2f' %("The value of V_CE in V is",V_CE);
The value of I_B in uA is 6.05
The value of I_C in mA is 0.85
The value of V_CE in V is 12.30

## Example E15 - Pg 90¶

In [15]:
#  Exa 3.15
#  Given data
V_CC =12.;#  in V
R_C = 4.3;#  in k ohm
V_CE = 4.;#  in V
V_BE = 0.7;#  in V
V_EE = 6.;#  in V
bita = 50.;
#  Applying KVL in base circuit, -V_BE - I_ER_E + V_EE = 0 or
I_ER_E = V_EE - V_BE;#  in V
#  Applying KVL in C-E circuit, V_CC-I_C*R_C-V_CE-I_ER_E+V_EE=0 or
I_C = (V_CC - V_CE - I_ER_E + V_EE)/R_C;#  in mA
I_B = I_C/bita;#  in mA
I_E = I_C+I_B;#  in mA
R_E= I_ER_E/I_E;#  in k ohm
print '%s %.2f' %("The value of R_E in k ohm is : ",R_E)
del_IC= bita*(1+bita)*R_E;
del_ICO= bita*(1+bita)*R_E;
S= del_IC/del_ICO;
print '%s %.2f' %("The value of stability factor, S is : ",S)
S_desh= bita/((1+bita)*R_E);
print '%s %.2f' %("The value of stability factor, S'' is : ",S_desh)
The value of R_E in k ohm is :  2.57
The value of stability factor, S is :  1.00
The value of stability factor, S'' is :  0.38

## Example E17 - Pg 90¶

In [16]:
#  Exa 3.17  (Miss printed as example 3.14)
#  Given data
Tj = 150.;# Junction temperature in degree C
P_Cmax = 125.;#  in mW
T = 25.;#  free-air temperature in degree C
T1 = 0;#  in degree C
curve = (Tj-T)/(P_Cmax - T1);#  in degreeC/mW
T_A = 25.;#  Ambient temperature in degree C
P_D = 75.;#  Collector junction dissipation in mW
theta = 1.;#  in degree C/mW
#  Tj-T_A = theta*P_D;
Tj = T_A + (theta*P_D);#  in degree C
print '%s %.2f' %("The junction temperature in degreeC is",Tj);
The junction temperature in degreeC is 100.00

## Example E18 - Pg 91¶

In [17]:
#  Exa 3.18             (Miss printed as example 3.15)
#  Given data
P_Cmax = 125.;#  in mW
P_D = P_Cmax;#  in mW
T_A = 25.;#  in degree C
Tj = 150.;#  in degree C
#  Tj-T_A = theta*P_D;
theta = (Tj-T_A)/P_D;#  in degree C/mW
print '%s %.2f' %("The thermal resistance for a transistor in degreeC/mW is",theta);
#  For theta= 1 degreeC/mW
P_D = 75.;#  in mW
#  Tj-T_A = theta*P_D;
Tj = (theta*P_D) + T_A;#  in degree C
print '%s %.2f' %("The junction temperature in degreeC is",Tj);
The thermal resistance for a transistor in degreeC/mW is 1.00
The junction temperature in degreeC is 100.00

## Example E20 - Pg 93¶

In [18]:
#  Exa 3.20                 (Miss printed as example 3.17)
#  Given data
V_E = 1.;#  in V
V_BE = 0.7;#  in V
R_C = 1.;#  in k ohm
Beta = 180.;
V_CC = 12.;#  in V
V_CEQ = 6.;#  in V
#  Applying KVL into collector circuit, V _CC - I_C*R_C - V_CEQ = 0 or
I_C = (V_CC-V_CEQ)/R_C;#  in mA
I_B = I_C/Beta;#  in mA
#  Applying KVL into base circuit, V_CC - I_B*R_B - V_BE = 0 or
R_B = (V_CC-V_BE)/I_B;#  in k ohm
print '%s %.2f' %("The value of R_B in k ohm is",R_B);
# Applying KVL to collector circuit, V_CC - I_C*R_C - V_CE - V_E = 0 or
I_C = (V_CC-V_CEQ-V_E)/R_C;#  in mA
I_B = I_C/Beta;#  in mA
I_E = I_C+I_B;#  in mA
R_E = V_E/(I_E);#  in k ohm
R_E= round(R_E*10.**3.);#  in ohm
print '%s %.2f' %("The value of R_E in ohm is",R_E);
I_R2 = 10.*I_B;#  in mA
V_BE= 0.6;#  in V
# R2 =V_B/I_R2 = (V_E+V_BE)/I_R2;
R2 = (V_E+V_BE)/I_R2;#  in k ohm
R2= R2*10.**3.;#  in ohm
print '%s %.2f' %("The value of R2 in ohm is",R2);
I_R1 = I_R2 + I_B;#  in mA
# R1 = V_R1/I_R1 = (V_CC-V_B)/I_R1;
V_B = V_E+V_BE;#  in V
R1 = (V_CC-V_B)/I_R1;#  in k ohm
print '%s %.2f' %("The value of R1 in k ohm is",R1);
The value of R_B in k ohm is 339.00
The value of R_E in ohm is 199.00
The value of R2 in ohm is 5760.00
The value of R1 in k ohm is 34.04

## Example E21 - Pg 94¶

In [19]:
#  Exa 3.21                 (Miss printed as example 3.18)
#  Given data
V_BB= 6.;#  in V
I_CBO =0.5;#  in uA
V_BE = 0.7;#  in V
R_B= 50.;#  in k ohm
R_E= 1.;#  in k ohm
bita = 75.;
#  V_BB - I_B*R_B - V_BE - I_E*R_E = 0 or
I_B=(V_BB-V_BE)/(R_B+(1.+bita)*R_E);#  in mA      (on putting I_E= (1+bita)*I_B)             (i)
I_B= round(I_B*10.**3.);#  in uA
I_C= bita*I_B;#  in uA
I_C= I_C*10.**-3.;#  in mA
I_CQ= I_C;#  in mA
print '%s %.2f' %("The value of I_CQ at room temperature in mA is : ",I_CQ)
#  Part (ii)
C= 2.;#  temperature coefficient in mV/degreeC
C= 2.*10.**-3.;#  in V/degreeC
T2= 20.;#  in degreeC
T1= 0;#  in degreeC
I_CBO2= I_CBO*2.**((T2-T1)/10.);#  in uA
V_BE2= V_BE-C*T2;#  in V
#  Now from eq(i), for the new value of I_B
I_B=(V_BB-V_BE2)/(R_B+(1.+bita)*R_E);#  in mA
I_B= I_B*10.**3.;#  in uA
I_C= bita*I_B+(1.+bita)*I_CBO2;#  in uA
I_C= I_C*10.**-3.;#  in mA
I_CQ= I_C;#  in mA
print '%s %.2f' %("The value of I_CQ when temperature increases by 20degreeC in mA is : ",I_CQ)
The value of I_CQ at room temperature in mA is :  3.15
The value of I_CQ when temperature increases by 20degreeC in mA is :  3.33

## Example E22 - Pg 97¶

In [20]:
#  Exa 3.22                 (Miss printed as example 3.19)
#  Given data
S = 10.;
bita = 50.;
h_fe = bita;
V_CC= 20.;#  in V
V_CE = 10.;#  in V
R_C = 2.;#  in k ohm
I_C = 4.;#  in mA
I_B =I_C/bita;#  in mA
#  Applying KVL to collector loop, V_CC -I_C*R_C - V_CE - I_E*R_E = 0 or
R_E = (V_CC -I_C*R_C - V_CE)/(I_C+I_B);#  in k ohm            (on putting I_E= I_C+I_B)
R_E= round(R_E*10.**3.);#  in ohm
print '%s %.2f' %("The value of R_E in ohm is",R_E);
#  Formula S = (1+bita)*( (1 + (R_B/R_E))/( (1+bita) + (R_B/R_E) ) ) or
R_B= (1+bita)*(1-S)*R_E/(S-1-bita);#  in ohm
#  But R_B= R1 || R2= R1*R2/(R1+R2) => R2/(R1+R2)= R_B/R1           (i)
#  Calculation of R1 and R2 :
V_BE= 0.2;#  in V
#  Applying KVL to input loop,
V_R2= V_BE+(I_C+I_B)*10.**-3.*R_E;#  in V
#  But V_R2= R2*V_CC/(R1+R2) =>  R2/(R1+R2)= V_R2/V_CC          (ii)
#  On comparing eq (i) and (ii)
R1= R_B*V_CC/V_R2;#  in ohm
R2= R1*V_R2/(V_CC-V_R2);#  in ohm
R1= R1*10**-3;#  in k ohm
R2= R2*10**-3;#  in k ohm
print '%s %.2f' %("The value of R1 in k ohm is : ",R1)
print '%s %.2f' %("The value of R2 in k ohm is : ",R2)
#  Effect of Reducing S or 3 :
S=3;
#  Formula S = (1+bita)*( (1 + (R_B/R_E))/( (1+bita) + (R_B/R_E) ) ) or
R_B= (1+bita)*(1-S)*R_E/(S-1-bita);#  in ohm
R_B= R_B*10**-3;#  in k ohm
print '%s %.2f' %("When S<=3, the value of R_B in k ohm is : ",R_B)
print '%s' %("Thus S is reduced below 3 at the cost of reduction of it's input impedance")
The value of R_E in ohm is 490.00
The value of R1 in k ohm is :  49.89
The value of R2 in k ohm is :  6.16
When S<=3, the value of R_B in k ohm is :  1.04
Thus S is reduced below 3 at the cost of reduction of it's input impedance