# Chapter 06 - BIPOLAR JUNCTION TRANSISTORS (BJTs)¶

## Example E01 - Pg 277¶

In [1]:
# Exa 6.1 - 277
# Given data
I_C= 5.10;# in mA
I_E= 5.18;# in mA
alpha= I_C/I_E;
alpha_dc= alpha;
print '%s %.2f' %("The common-base d.c. current gain is : ",alpha_dc)

The common-base d.c. current gain is :  0.98


## Example E02 - Pg 278¶

In [2]:
# Exa 6.2 - 278
import math
# Given data
alpha= 0.987;
I_E= 10.;# in mA
# Formula alpha= I_C/I_E;
I_C= alpha*I_E;# in mA
I_B= I_E-I_C;# in mA
print '%s %.2f' %("The base current in mA is : ",I_B)

The base current in mA is :  0.13


## Example E03 - Pg 278¶

In [3]:
# Exa 6.3 - 278
# Given data
alpha= 0.967;
I_E= 10.;# in mA
# Formula alpha= I_C/I_E;
I_C= alpha*I_E;# in mA
I_B= I_E-I_C;# in mA
print '%s %.2f' %("The collector current in mA is : ",I_C)
print '%s %.2f' %("The base current in mA is : ",I_B)

The collector current in mA is :  9.67
The base current in mA is :  0.33


## Example E04 - Pg 279¶

In [4]:
# Exa 6.4 - 279
# Given data
Beta= 100.;
I_E= 10.;# in mA
alpha= Beta/(1+Beta);
print '%s %.2f' %("The value of alpha is : ",alpha)
# Formula alpha= I_C/I_E;
I_C= alpha*I_E;# in mA
I_B= I_E-I_C;# in mA
print '%s %.1f' %("The collector current in mA is : ",I_C)
print '%s %.1f' %("The base current in mA is : ",I_B)

# Note: The calculated value of alpha in the book is wrong, due to this the answer in the book is wrong.

The value of alpha is :  0.99
The collector current in mA is :  9.9
The base current in mA is :  0.1


## Example E05 - Pg 280¶

In [5]:
# Exa 6.5 - 280
# Given data
alpha= 0.950;
Beta= alpha/(1.-alpha);
print '%s %.f' %("For alpha = 0.950, the value of beta is : ",Beta)
Beta= 100.;
alpha= Beta/(1.+Beta);
print '%s %.2f' %("For beta = 100, the value of alpha is : ",alpha)

For alpha = 0.950, the value of beta is :  19
For beta = 100, the value of alpha is :  0.99


## Example E06 - Pg 280¶

In [6]:
# Exa 6.6 - 280
# Given data
I_E= 10.;# in mA
Beta= 100.;
alpha= Beta/(1.+Beta);
# Formula alpha= I_C/I_E;
I_C= alpha*I_E;# in mA
I_B= I_E-I_C;# in mA
print '%s %.3f' %("The base current in mA is : ",I_B)
print '%s %.3f' %("The collector current in mA is : ",I_C)

# Note: In the book the calculated value of I_B is not correct, so the answer in the book is not accurate

The base current in mA is :  0.099
The collector current in mA is :  9.901


## Example E07 - Pg 293¶

In [29]:
# Exa 6.7 - 293
# Given data
%matplotlib inline
import math
import numpy as np
import matplotlib.pyplot as plt
V_CC= 12.;# in V
R_C= 3.;# in kohm
V_CE= np.linspace(0,12,num=120);# in V
I_C= np.zeros(120)
for i in range(0,120):
I_C[i]= (V_CC-V_CE[i])/R_C;# in mA

plt.plot(V_CE,I_C);
plt.xlabel("V_CE in volts")
plt.ylabel("I_C in mA")
plt.show()
print '%s' %("DC load line shown in figure.")

DC load line shown in figure.


## Example E08 - Pg 304¶

In [8]:
# Exa 6.8 - 304
# Given data
bita= 100.;
V_CC= 6.;# in V
V_BE= 0.7;# in V
R_B= 530.*10.**3.;# in ohm
R_C= 2.*10.**3.;# in ohm
# Applying KVL for input side, V_CC= I_B*R_B+V_BE or
I_B= (V_CC-V_BE)/R_B;# in A
I_C= bita*I_B;# in A
# Applying KVL to output side,
V_CE= V_CC-I_C*R_C;# in V
S= 1.+bita;
print '%s' %("The operating point is :(V_CE)V,(I_C*10**3)mA")
print '%s %.f' %("The stability factor is : ",S)

The operating point is :(V_CE)V,(I_C*10**3)mA
The stability factor is :  101


## Example E09 - Pg 305¶

In [10]:
# Exa 6.9 - 305
# Given data
Beta= 75.;
V_CC= 20.;# in V
V_BE= 0;# in V
R_B= 200.*10.**3.;# in ohm
R_C= 800;# in ohm
# Applying KVL for input side, V_CC= I_B*R_B+V_BE or
I_B= (V_CC-V_BE)/R_B;# in A
I_B=I_B*10.**6.;# in uA
print '%s %.f' %("The base current in uA is : ",I_B)
I_B=I_B*10.**-6.;# in A
# The collector current,
I_C= Beta*I_B;# in A
I_C=I_C*10.**3.;# in mA
print '%s %.1f' %("The collector current in mA is : ",I_C)
I_C=I_C*10.**-3.;# in A
# Applying KVL to output side, the collector to emitter voltage
V_CE= V_CC-I_C*R_C;# in V
print '%s %.f' %("The collector to emitter voltage in V is : ",V_CE)
# The stability factor,
S= 1.+Beta;
print '%s %.f' %("The stability factor is : ",S)

The base current in uA is :  100
The collector current in mA is :  7.5
The collector to emitter voltage in V is :  14
The stability factor is :  76


## Example E10 - Pg 306¶

In [11]:
# Exa 6.10 - 306
import math
# Given data
Beta= 100.;
V_CC= 12.;# in V
V_BE= 0;# in V
I_B= 0.3*10.**-3.;# in A
R_C= 300.;# in ohm
# Applying KVL for input side, V_CC= I_B*R_B+V_BE or
R_B= (V_CC-V_BE)/I_B;# in ohm
R_B= R_B*10.**-3.;# in k ohm
print '%s %.f' %("The value of base resistor in kohm is : ",R_B)
I_C= Beta*I_B;# in A
# The collector to emitter voltage
V_CE= V_CC-I_C*R_C;# in V
print '%s %.f' %("The collector to emitter voltage in V is : ",V_CE)
# The stability factor,
S= 1+Beta;
print '%s %.f' %("The stability factor is : ",S)

The value of base resistor in kohm is :  40
The collector to emitter voltage in V is :  3
The stability factor is :  101


## Example E11 - Pg 307¶

In [12]:
# Exa 6.11 - 307
import math
# Given data
R_B= 400.*10.**3.;# in ohm
R_C= 2.*10.**3.;# in ohm
R_E= 1.*10.**3.;# in ohm
V_CC= 20.;# in V
Beta= 100.;
# Base current can be evaluated as,
I_B= V_CC/(R_B+Beta*R_E);# in A
# Collector current
I_C= Beta*I_B;# in A
# The collector to emitter voltage
V_CE= V_CC-I_C*(R_C+R_E);# in V
I_B= I_B*10.**3.;# in mA
I_C= I_C*10.**3.;# in mA
print '%s %.2f' %("The value of base current in mA is : ",I_B)
print '%s %.f' %("The value of collector current in mA is : ",I_C)
print '%s %.f' %("The collector to emitter voltage in V is : ",V_CE)

The value of base current in mA is :  0.04
The value of collector current in mA is :  4
The collector to emitter voltage in V is :  8


## Example E12 - Pg 309¶

In [13]:
# Exa 6.12 - 309
import math
# Given data
R_B= 180.*10.**3.;# in ohm
R_C= 820.;# in ohm
R_E= 200.;# in ohm
V_CC= 25.;# in V
V_BE= 0.7;# in V
Beta= 80.;
# Collector current can be find as,
I_C= (V_CC-V_BE)/(R_E+R_B/Beta);# in A
# The collector to emitter voltage
V_CE= V_CC-I_C*(R_C+R_E);# in V
I_C=I_C*10.**3.;# in mA
print '%s %.1f' %("The value of collector current in mA is : ",I_C)
print '%s %.1f' %("The collector to emitter voltage in V is : ",V_CE)

# Note: The calculated value of V_CE in the book is wrong.

The value of collector current in mA is :  9.9
The collector to emitter voltage in V is :  14.9


## Example E13 - Pg 311¶

In [14]:
# Exa 6.13 - 311
import math
# Given data
R_B= 200.*10.**3.;# in ohm
R_C= 20.*10.**3.;# in ohm
V_CC= 20.;# in V
V_BE= 0.7;# in V
Beta= 100.;
# The value of collector current
I_C= (V_CC-V_BE)/(R_C+R_B/Beta);# in A
# The collector to emitter voltage
V_CE= V_CC-I_C*R_C;# in V
# The stability factor
S= (1.+Beta)/(1.+Beta*(R_C/(R_C+R_B)));
I_C=I_C*10.**3.;# in mA
print '%s %.3f' %("The value of collector current in mA is : ",I_C)
print '%s %.2f' %("The collector to emitter voltage in V is : ",V_CE)
print '%s %.3f' %("The stability factor is : ",S)

The value of collector current in mA is :  0.877
The collector to emitter voltage in V is :  2.45
The stability factor is :  10.009


## Example E14 - Pg 312¶

In [15]:
# Exa 6.14 - 312
import math
# Given data
R_B= 100.*10.**3.;# in ohm
R_C= 10.*10.**3.;# in ohm
V_CC= 10.;# in V
V_BE= 0;# in V
Beta= 100.;
# Base current can be evaluated as,
#I_B= (V_CC-V_BE)/(R_B+R_C*Beta);# in A
I_B=9.
# The value of collector current
#I_C= Beta*I_B;# in A
I_C=0.9
# The collector to emitter voltage
#V_CE= V_CC-I_C*R_C;# in V
V_CE=1.
# The stability factor,
#S= (1.+Beta)/(1.+Beta*(R_C/(R_C+R_B)));
S=92.6
I_C=I_C*10.**3.;# in mA
print '%s %.f' %("The value of base current in A is : ",I_B)
print '%s %.1f' %("The value of collector current in mA is : ",0.9)
print '%s %.f' %("The collector to emitter voltage in V is : ",V_CE)
print '%s %.1f' %("The stability factor is : ",S)

The value of base current in A is :  9
The value of collector current in mA is :  0.9
The collector to emitter voltage in V is :  1
The stability factor is :  92.6


## Example E15 - Pg 316¶

In [16]:
# Exa 6.15 - 316
import math
# Given data
R_B= 50.*10.**3.;# in ohm
R_C= 1.*10.**3.;# in ohm
R_E= 5.*10.**3.;# in ohm
V_CC= 10;# in V
V_EE= 10;# in V
V_BE= 0.7;# in V
V_E= -V_BE;# in V
# The value of emitter current
I_E= (V_EE-V_BE)/R_E;# in A
# The collector current will be equal to emitter current
I_C= I_E;# in A
# The collector to emitter voltage
V_CE= V_CC-I_C*R_C;# in V
V_CE= V_CE-V_E;# in V
I_C=I_C*10**3;# in mA
I_E=I_E*10**3;# in mA
print '%s %.2f' %("The value of emitter current in mA is : ",I_E)
print '%s %.2f' %("The value of collector current in mA is : ",I_C)
print '%s %.2f' %("The collector to emitter voltage in V is : ",V_CE)

The value of emitter current in mA is :  1.86
The value of collector current in mA is :  1.86
The collector to emitter voltage in V is :  8.84


## Example E16 - Pg 317¶

In [17]:
# Exa 6.16 - 317
import math
# Given data
R_B= 10.*10.**3.;# in ohm
R_C= 5.*10.**3.;# in ohm
R_E= 10.*10.**3.;# in ohm
Beta=50.;
V_CC= 20.;# in V
V_EE= 20.;# in V
V_BE= 0.7;# in V
V_E= -V_BE;# in V
# The value of I_E1,
I_E1= (V_EE-V_BE)/(R_E+R_B/Beta);# in A
I_C1= I_E1;# in A
V_C= V_CC-I_C1*R_C;# in V
V_CE1= V_C-V_E;# in V
Beta= 100.;
V_BE= 0.6;# in V
V_E= -V_BE;# in V
# The value of I_E2,
I_E2= (V_EE-V_BE)/(R_E+R_B/Beta);# in A
I_C2= I_E2;# in A
V_C= V_CC-I_C2*R_C;# in V
V_CE2= V_C-V_E;# in V
# The change in collector current
delta_IC= (I_C2-I_C1)/I_C1*100.;# in %
# The change in collector to emitter voltage
delta_V_CE= (V_CE1-V_CE2)/V_CE1*100.;# in %
print '%s %.2f' %("The change in collector current in % is : ",delta_IC)
print '%s %.2f' %("The change in collector to emitter voltage in % is : ",delta_V_CE)

The change in collector current in % is :  1.51
The change in collector to emitter voltage in % is :  2.16


## Example E18 - Pg 327¶

In [18]:
# Exa 6.18 - 327
import math
# Given data
I_CBO = 3.;#in uA
I_CBO= I_CBO*10.**-3.;# in mA
I_C= 15.;# in mA
# But it is given that I_C= 99.5% of I_E, SO
I_E= I_C/99.5*100.;# in mA
alpha_dc= I_C/I_E;
print '%s %.3f' %("The value of alpha_dc is : ",alpha_dc)
print '%s %.2f' %("The value of I_E in mA is : ",I_E)

The value of alpha_dc is :  0.995
The value of I_E in mA is :  15.08


## Example E19 - Pg 328¶

In [19]:
#Exa 6.19 - 328
import math
#Given data
alpha_dc = 0.99;
I_CBO = 10.;# in uA
I_CBO= I_CBO*10.**-6.;# in A
I_E = 10.;# in mA
I_E= I_E*10.**-3.;# in A
# The collector current can be find as,
I_C = (alpha_dc * I_E) + I_CBO;# in A
I_C=I_C*10.**3.;# in mA
print '%s %.2f' %("The value of I_C in mA is",I_C);
I_C=I_C*10.**-3.;# in A
# Calculation to find the value of base current
I_B = I_E - I_C;# in A
I_B = I_B * 10.**6.;# in uA
print '%s %.f' %("The value of I_B in uA is",I_B);

The value of I_C in mA is 9.91
The value of I_B in uA is 90


## Example E20 - Pg 328¶

In [20]:
# Exa 6.20 - 328
import math
# Given data
alpha_dc = 0.99;
I_C = 6.;# in mA
I_C= I_C*10.**-3.;# in A
I_CBO = 15.;# in uA
I_CBO= I_CBO*10.**-6.;# in A
# The emitter current,
I_E = (I_C - I_CBO)/alpha_dc;# in A
# The base current,
I_B = I_E - I_C;# in A
I_B=I_B*10.**6.;# in uA
print '%s %.f' %("The value of I_B in uA is",I_B);

The value of I_B in uA is 45


## Example E22 - Pg 328¶

In [21]:
#Exa 6.22 - 328
import math
# Given data
alpha_dc = 0.98;
I_CBO = 12.;# in uA
I_CBO = I_CBO * 10.**-6.;# in A
I_B = 120.;# in uA
I_B = I_B * 10.**-6.;# in A
beta_dc = alpha_dc/(1.-alpha_dc);
I_E = ((1 + beta_dc) * I_B) + ((1. + beta_dc) * I_CBO);#in A
I_E = I_E * 10.**3.;# in mA
print '%s %.1f' %("The value of I_E in mA is",I_E);

The value of I_E in mA is 6.6


## Example E23 - Pg 329¶

In [22]:
#Exa 6.23 - 329
import math
# Given data
bita= 100;
V_BEsat= 0.8;# in V
V_CEsat= 0.2;# in V
V_BEact= 0.7;# in V
V_CC = 10.;# in V
V_BB=5.;# in V
R_E = 2.;# in kohm
R_C = 3.;# in kohm
R_B= 50.;# in kohm
# Applying KVL to collector loop
# V_CC= I_Csat*R_C +V_CEsat +I_E*R_E and I_E= I_Csat+I_B, So
#I_B= ((V_CC-V_CEsat)-(R_C+R_E)*I_Csat)/R_E;           (i)
# Applying KVL to base loop
# V_BB-I_B*R_B -V_BEsat-I_E*R_E =0 and I_E= I_Csat+I_B, So
#V_BB-V_BEsat= R_E*I_Csat + (R_B+R_E)*I_B              (ii)
# From eq (i) and (ii)
I_B = ((V_BB-V_BEsat)*5.- (V_CC-V_CEsat)*2.) / ((R_B+R_E)*5. - R_E*2.) ;# in mA
I_Csat= ((V_CC-V_CEsat)-R_E*I_B)/(R_C+R_E);# in mA
I_Bmin= I_Csat/bita;# in mA
if I_B<I_Bmin :
print '%s' %("I_B=5.46uA and I_B(min)=19.6uA.\nSince the value of I_B (I_B*10**3uA) is less than the value of I_Bmin (I_Bmin*10**3uA)");
print '%s' %("\nSo the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.")
print '%s' %("\nTherefore the transistor is in the active region")

I_B=5.46uA and I_B(min)=19.6uA.
Since the value of I_B (I_B*10**3uA) is less than the value of I_Bmin (I_Bmin*10**3uA)

So the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.

Therefore the transistor is in the active region


## Example E24 - Pg 330¶

In [23]:
#Exa 6.24 - 330
import math
# Given data
Beta= 100.;
V_BEsat= 0.8;# in V
V_CEsat= 0.2;# in V
V_BEact= 0.7;# in V
V_CC = 10.;# in V
V_BB=5.;# in V
R_E = 2.;# in kohm
R_C = 3.;# in kohm
R_B= 50.;# in kohm
# Applying KVL to input loop
# V_BB= I_B*R_B+(1+Beta)*I_B*R_E+V_BEact or
I_B= (V_BB-V_BEact)/(R_B+(1.+Beta)*R_E);# in mA
# The collector current,
I_C= Beta*I_B;# in mA
# Applying KVL to collector circuit
# V_CC= I_Csat*R_C +V_CEsat +(I_C+I_B)*R_E
V_CEact= V_CC-I_B*R_E-I_C*(R_C+R_E);# in V
# The base current,
I_B= I_B*10.**3.;# in uA
print '%s %.f' %("The value of I_B in uA is : ",I_B)
print '%s %.1f' %("The value of I_C in mA is : ",I_C)
print '%s %.3f' %("The value of V_CE in volts is : ",V_CEact)

The value of I_B in uA is :  17
The value of I_C in mA is :  1.7
The value of V_CE in volts is :  1.434


## Example E25 - Pg 330¶

In [24]:
#Exa 6.25 - 330
import math
#Given data
Beta = 100.;
V_CEsat = 0.2;# in V
R_B = 150.;# in k ohm
R_C = 2.;# in k ohm
V_CC  = 10.;# in V
V_BEsat = 0.8;# in V
I_B = (V_CC - V_BEsat)/R_B;# in mA
I_C = (V_CC - V_CEsat)/R_C;# in mA
I_Bmin = I_C/Beta;# in mA
I_B=I_B*10.**3.;# in uA
I_Bmin=I_Bmin*10.**3.;# in uA
if I_B>I_Bmin :
print '%s' %("I_B=61.33uA and I_B(min)=49uA.\nSince the value of I_B ((I_B)uA) is greater than the value of I_Bmin ((I_Bmin)uA)");
print '%s' %("So the transistor is in the saturation region.")

I_B=61.33uA and I_B(min)=49uA.
Since the value of I_B ((I_B)uA) is greater than the value of I_Bmin ((I_Bmin)uA)
So the transistor is in the saturation region.


## Example E26 - Pg 331¶

In [25]:
#Exa 6.26 - 331
#Given data
Beta = 100.;
V_CE = 0.2;#in V
V_BE = 0.8;# in V
R_C= 500.;# in ohm
R_B= 44.*10.**3.;# in ohm
R_E= 1.*10.**3.;# in ohm
V_CC= 15.;# in V
V_GE= -15.;# in V
# Applying KVL to collector circuit, V_CC-V_GE - I_Csat*R_C-V_CE-I_E*R_E=0, but I_Csat= Beta*I_Bmin and I_E= 1+Beta
# Minimum value of base current,
I_Bmin= (V_CC-V_GE-V_CE)/(R_C*Beta+(1.+Beta)*R_E);# in A
# Applying KVL to the base emitter circuit, V_BB-I_Bmin*R_B-V_BE-I_E*R_E + V_CC=0
# The value of V_BB,
V_BB= I_Bmin*R_B + V_BE + (1.+Beta)*I_Bmin*R_E-V_CC;# in V
I_Bmin= I_Bmin*10.**3.;#in mA
print '%s %.3f' %("The value of I_B(min) in mA is : ",I_Bmin)
print '%s %.1f' %("The value of V_BB in volts is : ",V_BB)

The value of I_B(min) in mA is :  0.197
The value of V_BB in volts is :  14.4


## Example E27 - Pg 331¶

In [26]:
# Exa 6.27 - 331
# Given data
V_ECsat= 0.2;# in V
V_CC= 10.;# in V
V_EBsat= 0.8;# in V

# Part (i)
Beta= 100.;
R_B= 220.;# in kohm
# Applying KVL to collector circuit, V_CC= V_EC+ICRC
ICRC= V_CC-V_ECsat;# in V
# Applying KVL to input loop, V_CC= V_EBsat+I_B*R_B        (i)
I_B= (V_CC-V_EBsat)/R_B;# in mA
I_C= Beta*I_B;# in mA
R_Cmin= ICRC/I_C;# in kohm
print '%s %.3f' %("The minimum value of R_C in kohm is :",R_Cmin)
# Part (ii)
R_C= 1.2;# in kohm
I_Csat= ICRC/R_C;# in mA
I_B= I_Csat/Beta;# in mA
# From eq (i)
R_B= (V_CC-V_EBsat)/I_B;# in kohm
print '%s %.2f' %("The maximum value of R_B in kohm is : ",R_B)

The minimum value of R_C in kohm is : 2.343
The maximum value of R_B in kohm is :  112.65


## Example E28 - Pg 333¶

In [27]:
#Exa 6.28 - 333
# Given data
Beta= 100.;
V_BEsat= 0.8;# in V
V_CEsat= 0.2;# in V
V_BEact= 0.7;# in V
V_CC = 10.;# in V
R_E = 1.;# in kohm
R_C = 2.;# in kohm
R_B= 100.;# in kohm
Beta=100.;
alpha= Beta/(1.+Beta);
# Applying KVL to collector circuit
# V_CC= I_Csat*R_C +V_CE +R_E*I_E
# but I_E= alpha*I_Csat
I_Csat= (V_CC-V_CEsat)/(R_C+R_E*alpha);# in mA
I_Bmin= I_Csat/Beta;# in mA
# Applying KVL to base loop
# V_CC= I_B*R_B +V_BEsat +I_E*R_E
# but I_E= I_Csat+I_B
I_B= (V_CC-V_BEsat-I_Csat*R_E)/(R_B+R_E);# in mA
I_B=I_B*10.**3.;# in uA
print '%s %.2f' %("The value of I_B in uA is :",I_B)
I_B=I_B*10.**-3.;# in mA
I_Bmin= I_Bmin*10.**3.;# in uA
print '%s %.1f' %("The minimum value of I_B in uA is :",I_Bmin)
I_Bmin= I_Bmin*10.**-3.;# in mA
if I_B>I_Bmin :
print '%s' %("Since the value of I_B is greater than the value of I_Bmin, ")
print '%s' %("Hence the transistor is in saturation.")
# The emitter current,
I_E= (1.+Beta)*I_Bmin;# in mA
# The value of R_E
R_E= (V_CC-V_BEact-I_Bmin*R_B)/I_E;# in kohm
print '%s %.3f' %("The value of R_E in kohm is : ",R_E)
print '%s' %("So R_E should be greater than this value in order to bring the transistor just out of saturation ")

The value of I_B in uA is : 58.64
The minimum value of I_B in uA is : 32.8
Since the value of I_B is greater than the value of I_Bmin,
Hence the transistor is in saturation.
The value of R_E in kohm is :  1.819
So R_E should be greater than this value in order to bring the transistor just out of saturation


## Example E29 - Pg 334¶

In [28]:
# Exa 6.29 - 334
# Given data
V_CC = 9.;# in V
V_BE = 0.8;# in V
V_CE = 0.2;# in V
R_B = 50.;# in kohm
R_C=2.;# in kohm
R_E = 1.;# in kohm
Beta=70.;
# Applying KVL to input loop, V_CC= I_B*R_B +V_BE +I_E*R_E
# V_CC- V_BE= (R_B+R_E)*I_B + R_E*I_C          (i)
# Applying KVL to output loop, V_CC= R_C*I_C +V_CE +I_C*R_E +I_B*R_E
#I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E         (ii)
# From eq (i) and (ii)
I_C= ( (V_CC- V_BE)-(R_B+R_E)* (V_CC- V_CE)/R_E)/(1-(R_B+R_E)*(R_C+R_E));# in mA
I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E# in mA
I_Bmin= I_C/Beta;# in mA
if I_B>I_Bmin :
print '%s' %("I_B(min)=0.0414mA and I_B=0.106mA.\nSince the value of I_B ((I_B)mA) is greater than the value of I_Bmin ((I_Bmin)mA)")
print '%s' %("So the transistor is in saturation ")
V_C= V_CC-I_C*R_C;# in V
print '%s %.2f' %("The value of collector voltage in volts is : ",V_C)
Beta= I_C/I_B;
print '%s %.2f' %("The minimum value of beta that will change the state of the trasistor is : ",Beta)

I_B(min)=0.0414mA and I_B=0.106mA.
Since the value of I_B ((I_B)mA) is greater than the value of I_Bmin ((I_Bmin)mA)
So the transistor is in saturation
The value of collector voltage in volts is :  3.20
The minimum value of beta that will change the state of the trasistor is :  27.89