Chapter 5: Transistor Bias Circuits

Example 5.1, Page Number: 146

In [1]:
%matplotlib inline

Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].
For more information, type 'help(pylab)'.
In [2]:
# variable declaration
V_BB=10.0;        #voltage in volt
V_CC=20.0;        #voltage in volt
B_DC=200.0;       #B_DC value
R_B=47.0*10**3;   #resistance in ohm
R_C=330.0;        #resistance in ohm
V_BE=0.7;       #voltage in volt

#current
I_B=(V_BB-V_BE)/R_B;    #base current
I_C=B_DC*I_B;           #Q POINT
V_CE=V_CC-I_C*R_C;      #Q POINT
I_C_sat=V_CC/R_C;       #saturation current
I_c_peak=I_C_sat-I_C;   #peak current
I_b_peak=I_c_peak/B_DC; #peak current in ampere

#result
print "Q point of I_C = %.3f amperes" %I_C
print "Q point of V_CE = %.2f volts" %V_CE
print "peak base current = %.4f amperes" %I_b_peak

Q point of I_C = 0.040 amperes
Q point of V_CE = 6.94 volts
peak base current = 0.0001 amperes

Example 5.2, Page Number: 149

In [3]:
# variable declaration
B_DC=125.0;   #DC value
R_E=10.0**3;  #resistance in ohm

#calculation
R_IN_base=B_DC*R_E;  #base resistance

#Result
print "DC input resistance, looking at base of transistor = %d ohm" %R_IN_base

DC input resistance, looking at base of transistor = 125000 ohm

Example 5.3, Page Number: 151

In [4]:
# variable declaration
B_DC=100;         #DC value
R1=10*10**3;       #resistance in ohm
R2=5.6*10**3;      #resistance in ohm
R_C=1*10**3;       #resistance in ohm
R_E=560;          #resistance in ohm
V_CC=10;          #voltage in volt
V_BE=0.7          #voltage in volt

#calculation
R_IN_base=B_DC*R_E;  #calculate base resistance
#We can neglect R_IN_base as it is equal to 10*R2
print "input resistance seen from base = %d ohm" %R_IN_base
print "which can be neglected as it is 10 times R2"

V_B=(R2/(R1+R2))*V_CC;   #base voltage
V_E=V_B-V_BE;            #emitter voltage
I_E=V_E/R_E;             #emitter current
I_C=I_E;                 #currents are equal
V_CE=V_CC-I_C*(R_C+R_E); #voltage in volt

#result
print "V_CE = %.2f volts" %V_CE
print "I_C = %.3f amperes" %I_C
print "Since V_CE>0V, transistor is not in saturation"

input resistance seen from base = 56000 ohm
which can be neglected as it is 10 times R2
V_CE = 1.95 volts
I_C = 0.005 amperes
Since V_CE>0V, transistor is not in saturation

Example 5.4, Page Number: 154

In [5]:
# variable declaration
V_EE=10.0;           #voltage in volt
V_BE=0.7;            #voltage in volt
B_DC=150.0;          #DC value
R1=22.0*10**3;       #resistance in ohm
R2=10.0*10**3;       #resistance in ohm
R_C=2.2*10**3;       #resistance in ohm
R_E=1.0*10**3;       #resistance in ohm

#calculation
R_IN_base=B_DC*R_E;    #R_IN_base>10*R2,so it can be neglected
print "input resistance as seen from base = %d ohm" %R_IN_base
print "it can be neglected as it is greater than 10 times R2"
V_B=(R1/(R1+R2))*V_EE;   #base voltage
V_E=V_B+V_BE;            #emitter voltage
I_E=(V_EE-V_E)/R_E;      #emitter current
I_C=I_E;                 #currents are equal
V_C=I_C*R_C;             #collector voltage
V_EC=V_E-V_C;            #emitter-collector voltage

#result
print "I_C collector current = %.4f amperes" %I_C
print "V_EC emitter-collector voltage = %.2f Volts" %V_EC

input resistance as seen from base = 150000 ohm
it can be neglected as it is greater than 10 times R2
I_C collector current = 0.0024 amperes
V_EC emitter-collector voltage = 2.24 Volts

Example 5.5, PAge Number: 154

In [6]:
# variable declaration
R1=68.0*10**3;      #resistance in ohm
R2=47.0*10**3;      #resistance in ohm
R_C=1.8*10**3;      #resistance in ohm
R_E=2.2*10**3;      #resistance in ohm
V_CC=-6.0;          #voltage in volt
V_BE=0.7;           #voltage in volt
B_DC=75.0;          #DC value

#calculation
R_IN_base=B_DC*R_E;
print "input resistance as seen from base"
print "is not greater than 10 times R2 so it should be taken into account"
#R_IN_base  in parallel with R2
V_B=((R2*R_IN_base)/(R2+R_IN_base)/(R1+(R2*R_IN_base)/(R2+R_IN_base)))*V_CC;
V_E=V_B+V_BE;      #emitter voltage
I_E=V_E/R_E;       #emitter current
I_C=I_E;           #currents are equal
V_C=V_CC-I_C*R_C;  #collector voltage
V_CE=V_C-V_E;      #collector-emitter voltage

#result
print "collector current = %.4f amperes" %I_C
print "collector emitter voltage = %.2f volts" %V_CE

input resistance as seen from base
is not greater than 10 times R2 so it should be taken into account
collector current = -0.0006 amperes
collector emitter voltage = -3.46 volts

Example 5.6, Page Number: 156

In [7]:
# variable declaration
V_CC=12.0;       #voltage in volt
R_B=100.0*10**3; #resistance in ohm
R_C=560.0;       #resistance in ohm
#FOR B_DC=85 AND V_BE=0.7V
B_DC=85.0;       #DC value
V_BE=0.7;        #base-emitter voltage

#calculation
I_C_1=B_DC*(V_CC-V_BE)/R_B;  #collector current
V_CE_1=V_CC-I_C_1*R_C;       #collector-emittor voltage
#FOR B_DC=100 AND V_BE=0.6V
B_DC=100.0;                  #DC value
V_BE=0.6;                    #base emitter voltage
I_C_2=B_DC*(V_CC-V_BE)/R_B;               #collector current
V_CE_2=V_CC-I_C_2*R_C;                    #voltage in volt
p_del_I_C=((I_C_2-I_C_1)/I_C_1)*100;      #percent change in collector current
p_del_V_CE=((V_CE_2-V_CE_1)/V_CE_1)*100;  #percent change in C-E voltage

#result
print "percent change in collector current = %.2f" %p_del_I_C
print "percent change in collector emitter voltage = %.2f" %p_del_V_CE

percent change in collector current = 18.69
percent change in collector emitter voltage = -15.18

Example 5.7, Page Number: 159

In [8]:
# variable declaration
V_CC=20.0;         #voltage in volt
R_C=4.7*10**3;     #resistance in ohm
R_E=10.0*10**3;    #resistance in ohm
V_EE=-20.0;        #voltage in volt
R_B=100*10**3;     #resistance in ohm
#FOR B_DC=85 AND V_BE=0.7V
B_DC=85;           #DC value
V_BE=0.7;          #base-emitter voltage
I_C_1=(-V_EE-V_BE)/(R_E+(R_B/B_DC));
V_C=V_CC-I_C_1*R_C;  #colector voltage
I_E=I_C_1;           #emittor current
V_E=V_EE+I_E*R_E;    #emittor voltage
V_CE_1=V_C-V_E;      #CE voltage
print "I_C_1 = %.3f" %I_C_1
print "V_CE_1 = %.2f" %V_CE_1
#FOR B_DC=100 AND V_BE=0.6V
B_DC=100;          #DC value
V_BE=0.6;          #base-emitter voltage
I_C_2=(-V_EE-V_BE)/(R_E+(R_B/B_DC));
V_C=V_CC-I_C_2*R_C;#colector voltage
I_E=I_C_2;         #emittor current
V_E=V_EE+I_E*R_E;  #emittor voltage
V_CE_2=V_C-V_E;    #CE voltage
print "I_C_2 = %.3f" %I_C_2
print "V_CE_2 = %.2f" %V_CE_2

p_del_I_C=((I_C_2-I_C_1)/I_C_1)*100;
p_del_V_CE=((V_CE_2-V_CE_1)/V_CE_1)*100;
print "percent change in collector currrent = %.2f" %p_del_I_C
print "percent change in collector emitter voltage = %.2f" %p_del_V_CE
print "answers in book are approximated"

I_C_1 = 0.002
V_CE_1 = 14.61
I_C_2 = 0.002
V_CE_2 = 14.07
percent change in collector currrent = 2.13
percent change in collector emitter voltage = -3.69
answers in book are approximated

Example 5.8, Page Number: 161

In [9]:
# variable declaratio
V_CC=10.0;        #voltage in volt
B_DC=100.0;       #Dc value
R_C=10.0*10**3;   #resistance in ohm
R_B=100.0*10**3;  #resistance in ohm
V_BE=0.7;         #base-emittor voltage

#calculation
I_C=(V_CC-V_BE)/(R_C+(R_B/B_DC));  #collector current
V_CE=V_CC-I_C*R_C;                 #CE voltage

#result
print "Q point of collector current %.4f amperes" %I_C
print "Q point of collector-emitter voltage %.3f volts" %V_CE

Q point of collector current 0.0008 amperes
Q point of collector-emitter voltage 1.545 volts