Application Design Guide Power Resistor Heat Sinks

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1 Application Design Guide Power Resistor Heat Sinks

2 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.

3 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.

4 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/P D ( 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. P D 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.

5 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. T I N T E R N A L R QR T C A S E R QJ T H E A T S I N K R QSA T A M B E N T 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.

6 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 = [(T I -T A )/P D ]-R R (Equation 2) Where: T I is the internal resistance element temperature ( C) T A is the ambient temperature surrounding the heat sink ( C) P D 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)

7 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 (T A ) 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 C/watt or less ΔT= 155 C-25 C (T A )= 130 C 8W X 5.9 C/W = 47.2 C 130 C C = 82.8 C / 8W = C/W

8 Heat Sink Considerations With convection cooling, a thermal resistance of 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 vs for the PF2202, a 25% smaller heatsink!