Pe lec-thermal design-1

1. Thermal Design
1Sharif kakar

2. Heat transfer takes place by one or more of
following three fundamental mechanisms.
1-Conduction:
 It is a process by which heat flows from the
heated end of a metal rod towards the
cooler end.
 A physical medium is required for heat
transfer.
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3. 2-Radiation:
Heat radiation is similar to other forms of
radiations.
No physical medium is required between
the heat source and its destination.
Example:
The heat transfer from the sun to the
earth through space.
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4. 3-Convection:
 It occurs when a physical medium that has
been heated by conduction or radiation
moves away from the source of heat.
Example:
 A fan pushes air past a heated surface-
forced convection.
 Heated air rises and is replaced by cooler
air from below or sides-natural convection.
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5. 5
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6. Semiconductor devices are cooled by all
three heat transfer mechanisms.
Heat is generated near the junction and
transferred to the metal case through
semiconductor material by conduction.
Heat is radiated from the surface of the
case to the air.
Heating of the surrounding air creates air
flow around the device resulting
convection.
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7.  Semiconductor devices are usually
enclosed in metal cases.
 In most of the power devices one terminal is
in direct contact with the metal case to
increase conduction.
 Power devices can be mounted on metal
heat-sinks to improve conduction and
radiation.
 Heat-sink conduct heat outward to metal fins
that increase the total surface area from
where the conduction and radiation to air
can take place.
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8. The heat Transfer Equation is
ThAPd =
where
Pd is the rate of heat transfer
h is the heat transfer coefficient
A is the area involved in heat transfer.
T is the temperature difference between heat
transfer points.
(1)
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Heat transfer coefficient ‘h’ depends on the heat transfer
mechanism used and various factors involved in it. It is
difficult to evaluate ‘h’ in real thermal designs.Sharif kakar

9. It is more convenient to evaluate the
thermal design in semiconductors in
terms of thermal resistances.
The thermal resistance is defined as
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hAP
T
R
d
1
=

= (2)
)( jaR
TT
P
Aj
d

−
= (3)
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10.  Ohm’s law analogy is usually used to form the
heat flow models.
 The temperature difference ‘T’ could be
thought of as a voltage drop ‘V’.
 Thermal resistance ‘R’ corresponds to electrical
resistance ‘R’ .
 Power dissipation ‘Pd’ is analogous to current ‘I’.
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Therefore the equation (2) can be rewritten as
dPRT = 
(4)
where
Pd is in watts and R is in degrees per watt and T is
in degrees C. Sharif kakar

11. 11
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12. The total thermal resistance of the
semiconductor device from junction to ambient
is
( )sacscajcja RRRRR  ++= 
The thermal resistance Rca is usually very high
compared to Rcs and Rsa therefore the model
can be simplified as
(5)
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sacsjcja RRRR  ++= (6)
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13. 13
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14. In some low power applications, power devices
can be used without the heatsink.Therefore, the
total thermal resistance will be
cajcja RRR  +=
Rjc and Rca are specified by the manufacturer
and given in the data sheet of the device.
(7)
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15. 15
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16. Case to Sink thermal resistance “Rcs”
depends on package type, interface
flatness and mounting pressure.
It also depends whether silicon grease or
insulation material is used between
device case and the sink.
Typical values of the thermal resistance
“Rcs” for various packages are listed in
the following table.
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17. Package Insulation Washer Rcs
With SG Without SG
TO-03
Non insulated 0.1 0.3
Mica(50-100uM) 0.5 to 0.7 1.25 to 1.45
TO-220
Non insulated 0.3 to 0.5 1.5 to 2.0
Mica(50-100uM) 2.0 to 2.5 4.0 to 6.0
TO-3P
Non insulated 0.1 to 0.2 0.4 to 1.0
Mica(50-100uM) 0.5 to 0.7 1.2 to 1.5
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18.  The thermal resistance for a flat square
plate heat sink may be approximated by
the following equation.
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( ) ffsa C
A
CR
6503.3 25.0
+








=


(8)
where
 is the thermal conductance
 is the thickness of the heat sink in mm
A is the area of the sink in cm2
Cf is the correction factor for the position, surface emissivity and
orientation.
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19.  Above equation is valid for only one power
device mounted in the center of the plate at a
static ambient temperature of 45oC approx.
without other radiators in the near vicinity.
 Thermal resistance of the sink ca be decreased
by employing finned type design instead of flat
metal plate.
 This will increase total surface area of the sink
 The above discussion give you a fair idea that
how complex is to design a heat sink.
 Fortunately, heat sinks of 100’s of shapes are
available in the market and the thermal
resistance is given by the manufacturer.
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20. Example-1:
A certain power transistor dissipate 2
watts.The thermal resistance from
junction to case is 8o
C/W and case to
ambient is 20o
C/W.The free air
temperature is 25o
C. Calculate the
junction temperature and case
temperature.
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21. Solution-1
Ja
aj
d
R
TT
P

−
=
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aJadj TRPT += 
CT o
j 8125)208(2 =++=
Jcdcj RPTT =−
CT o
c 65)82(81 =−=
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22. Example-2:
The maximum permissible junction
temperature of a certain power device is
150o
C. It is desired to operate the device at
15W in an ambient temperature of 40o
C. Rjc
= 0.5o
C/W and Rca = 10o
C/W. Determine
whether a heat sink is required for this
application. If sink is required, determine
the maximum thermal resistance it can have
assuming Rcs nearly zero.
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23. Solution-2
caJcJa RRR  +=
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Jadaj RPTT =−
5.10105.0 =+=JaR
CTT o
aj 5.1575.1015 ==−
CTT o
aj 5.197405.1575.157 =+=+=
197.5 will exceed the maximum limit of 150o
C, therefore,
heat sink is required. Sharif kakar

24. Solution-2
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Jadaj RPTT =−
WCR o
Ja /33.7=
sacsJcJa RRRR  ++=
Set Tj to maximum value of 150o
C.
JaR=− 1540150
saR++= 05.033.7
WCR o
sa /83.6=
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25.  The maximum permissible power dissipation of a
semiconductor device is specified by the
manufacturer at a certain temperature (either
junction or case).
 The decrease in device power dissipation at
elevated temperature is called derating.
 The datasheet for the device also includes
a derating curve which indicates how much a
device will dissipate without getting damaged at
any given case temperature and this must be
taken into account while designing a system.
 Manufacturer also specify a derating factor in
W/o
C.
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26. Example-3:
Suppose a device has maximum power of
20 W at 25o
C and a derating factor of
100mW/o
C at temperature above 25o
C.
The maximum permissible power
dissipation at 100o
C is
Pd=20W-(100o
C-25o
C)X100mW/ o
C
Pd=20-7.5=12.5W
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27. Example-4:
The temperature of a certain
semiconductor device is 100o
C when it is
dissipating 1.2W.The device temperature
can not exceed 100o
C. If the total thermal
resistance from device to ambient is
50o
C/W, above what ambient
temperature should the dissipation be
derated.
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28. Solution-4
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cadac RPTT =−
CT o
a 402.150100 =−=
2.150100 =− aT
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