Application Design Guide
Power Resistor Heat Sinks
This Design Guide presents the concepts and methods
used to determine heat sink requirements for Riedon
high power resistors. The methods follow normal
practice used to design heat sink systems for power
semiconductors, and are directly applicable to power
resistors.
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Thermal Considerations
Power resistors often need heat sinks to maintain a safe operating temperature.
The design engineer must determine the heat sink required for the application,
balancing power dissipation with heat sink size and cooling resources.
The heat sink maintains the critical internal temperature of the device within design
limits. For power semiconductors, this is usually the maximum junction temperature;
for resistors, it is the maximum temperature of the resistance element. If these limits
are exceeded, the device will not operate to specifications, or will be destroyed. A
properly designed heat sink maintains that critical temperature within device limits.
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Thermal Resistance
Thermal resistance is the relationship between the temperature drop
and the internal power dissipation of a device under steady-state
conditions.
R = T/PD (C/Watt) (equation 1)
R is the sum of the thermal
resistances of the heat flow
path across which the
temperature difference T
exists. T is the temperature
difference or driving potential
which causes the flow of heat.
PD is the power dissipated by
the device in watts.
The temperature drop (T) is measured between the region of heat generation (the center of the
resistance element in a resistor) and some reference point, usually the mounting surface of the device.
Note that thermal resistance is defined for steady-state conditions. Under conditions of intermittent or
switching loads, this definition results in unnecessarily conservative and expensive designs.
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The Power Resistor Thermal Resistance System
In a practical power resistor application, R is normally the sum
of three thermal resistance values:
 RR - The thermal resistance associated with the resistor internal design.
 RJ - The thermal resistance of the interface between the device and its heat sink.
 RSA - The thermal resistance of the heat sink to ambient temperature.
RR - The resistor has a thermal resistance determined by its internal
design. This value is given on the datasheet and it specifies how much
hotter the internal resistor is in relation to it’s case ( or backplate ). So, if
you are putting 15W into a device that has a thermal resistance of
3.5C/Watt. The internal temperature will be 15 * 3.5 = 52.5C hotter than
it’s case. This is independent of the heatsink and cannot be lowered.
RJ - The interface between the device and its heat sink is never perfect,
it always has some thermal resistance. Typical values can be from
0.1C/Watt to 1C/Watt, depending on the surface finish of the heatsink
and the thermal grease or pad that is used.
RSA - Finally, the thermal resistance of the heat sink to the ambient
temperature. This is determined by the surface area of the heatsink and
the rate of airflow across it. Using a fan can significantly decrease the size
of heatsink required, often by a factor of 5.
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TINTERNAL
RQR
TCASE
RQJ
THEATSINK
RQSA
TAMB ENT
Determining the Thermal Resistance and Size
of Heat Sinks
The necessary heat sink thermal resistance can be calculated
(including the interface thermal resistance) for a power resistor
system by transforming Equation 1.
RSA+ RJ = [(TI-TA)/PD]-RR
(Equation 2)
Where:
TI is the internal resistance element temperature (C)
TA is the ambient temperature surrounding the heat sink (C)
PD is the power dissipated by the resistor system (watts)
RR is the resistor's thermal resistance (C/watt)
RJ is the thermal resistance of the mounting interface. (C/watt)
RSA is the thermal resistance of the heat sink to ambient. (C/watt)
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Application Example
Let's use an application example to show how to calculate heat sink requirements:
Riedon's PF2202 is rated at 10 watts when mounted on a heat sink
that keeps the internal temperature of the resistance element less
than 155C. Its thermal resistance (RR) is 5.9 C/W. Assuming an
ambient temperature (TA) of 25C, calculate the thermal resistance
of a heat sink and mounting interface that will allow this resistor to
dissipate 8 watts.
Solving equation 2 for this data determines that RSA+RJ must be 10.35 °C/watt or less
ΔT= 155°C-25°C (TA)= 130°C
8W X 5.9°C/W = 47.2°C
130°C - 47.2°C = 82.8°C / 8W = 10.35°C/W
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Heat Sink Considerations

With convection cooling, a thermal resistance of 10.75C/watt requires an efficient, heat sink
with the resistor properly mounted using thermal grease or pad and proper pressure. (The
mounting interface could contribute from 0.1 to 0.5 C/watt of thermal resistance to the
system.)

Heat sink thermal resistance depends on the surface area in contact with ambient air, and the
amount of air flow. Thermal resistance is lowered by increasing the area or by increasing the
air flow by using a fan.

If the heat sink is too large, there are two options: A fan that provides higher air flow over a
smaller heat sink. Or, a resistor with a lower thermal resistance.
In the previous example the thermal resistance of the PF2202 was
5.9C/watt . This meant there was a 42.7C temperature rise internal to
the part. Using a PF2205 resistor will lower this value to 18.4C because
that device has only a 2.3C/watt thermal resistance. The PF2205 is more
expensive than the PF2202, but chances are the heatsink savings will
more than offset this. Finishing the calculations, a PF2205 would require
a RSA+RJ of only 13.95 vs. 10.35 for the PF2202, a 25% smaller
heatsink!
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