Chapter 6 - Thermal Stability

Example E1 - Pg 333

In [1]:
#Ex6_1 Pg-333
#calculate The relation for thermal stability
Ta=25. #ambient temperature in degree celsius
tetha=10. #thermal resistance 
Pd=2. #power dessipated in transistor
Tj=Ta+tetha*Pd #collector base junction transistor temperature
print '%s %.0f' %("\nCollector base junction transistor temperature=degree celcius\n",Tj)
print '%s' %("tetha=10 degree celcius/watt means for every watt consumed its temperature will rise by 10 degree celcius")
print '%s' %("\nWhile using a transistor,we must keep in mind that it must not reach a condition called thermal runaway.\nThe heat released at collector base junction must not exceed the rate at which heat can dissipated under steady state. For this,\n ")
print '%s' %("(del_Pd/del_Tj) < (1/tetha)")
print '%s' %("This is the relation for thermal stability.")

Collector base junction transistor temperature=degree celcius
 45
tetha=10 degree celcius/watt means for every watt consumed its temperature will rise by 10 degree celcius

While using a transistor,we must keep in mind that it must not reach a condition called thermal runaway.
The heat released at collector base junction must not exceed the rate at which heat can dissipated under steady state. For this,
 
(del_Pd/del_Tj) < (1/tetha)
This is the relation for thermal stability.

Example E2 - Pg 335

In [2]:
#Ex6_2 Pg-335
#calculate the dissipation capability at 50 degree celcius
print '%s' %("Draw a vertical line from temperature axis at 50 degree celcius to cut the 71 degree celcius line.\nJoin the point of intersection P through a horizontal line at Y-axis.\nThe point where it intersects Y-axis gives the value of permissible dissipation equal to 45 of maximum rating.\n")
per=.45 #permissible dissipation in percentage
max_diss=165. #maximum dissipation
diss_cap=per*max_diss #dissipation capability
print '%s' %("The dissipation capability at 50 degree celcius")
print '%s %.0f' %("= %.0f mW \n ",diss_cap)
print '%s' %("Its value ranges from (0.2) degree celcius/watt to (1000) degree celcius/watt for a transistor  that has an efficient heat sink")
print '%s' %("Tj = Ta + tetha*Pd")
print '%s' %("The above equation reflects that collector-junction temperature Tj of the transistor will be higher than ambient temperature Ta by an amount equal to the product of tetha and Pd.")

Draw a vertical line from temperature axis at 50 degree celcius to cut the 71 degree celcius line.
Join the point of intersection P through a horizontal line at Y-axis.
The point where it intersects Y-axis gives the value of permissible dissipation equal to 45 of maximum rating.

The dissipation capability at 50 degree celcius
= %.0f mW 
  74
Its value ranges from (0.2) degree celcius/watt to (1000) degree celcius/watt for a transistor  that has an efficient heat sink
Tj = Ta + tetha*Pd
The above equation reflects that collector-junction temperature Tj of the transistor will be higher than ambient temperature Ta by an amount equal to the product of tetha and Pd.

Example E3 - Pg 335

In [3]:
#Ex6_3 Pg-335
#calculate The operating point coordinates
Rb=200.*10.**(3.) #base resistance in ohm
Vcc=10. #supply voltage in V
Vbe=0.7 #voltage drop in V
Rl=2.*10.**(3.) #load resistor in ohm
Beta=50. #transistor gain
print '%s' %("If Beta=50")
Ib=(Vcc-Vbe)/Rb #base current in A
Ic=Beta*Ib #collector current
Vce=Vcc-Ic*Rl #collector emitter voltage
print '%s %.2f %.2f' %("The operating point coordinates are [V,mA]\n",Vce, Ic*10**3)
print '%s' %("\nIf Beta=60")
Beta2=60 #second transistor gain 
Ic2=Beta2*Ib #collector current
Vce2=Vcc-Ic2*Rl #collector emitter voltage
print '%s %.2f %.2f' %("The operating point coordinates are [V,mA]\n",Vce2, Ic2*10**3)

If Beta=50
The operating point coordinates are [V,mA]
 5.35 2.33

If Beta=60
The operating point coordinates are [V,mA]
 4.42 2.79

Example E4 - Pg 335

In [4]:
#Ex6_4 Pg-335
#calculate The collector emitter voltage
Rb=330.*10.**(3.) #base resistance in ohm
Vcc=15. #supply voltage in V
Vbe=0.7 #voltage drop in V
Rl=3.3*10.**(3.) #load resistor in ohm
Beta=60. #transistor gain
Ib=(Vcc-Vbe)/Rb #base current in A
Ic=Beta*Ib #collector current (value in textbook is wrong)
Vce=Vcc-(Ic+Ib)*Rl 
print '%s %.2f' %("The collector emitter voltage=V",Vce)
#collector emitter voltage (value in textbook is wrong) 

The collector emitter voltage=V 6.28