LM317T Heatsinking
LM317T Heatsinking
Choosing a heatsink for use with an LM317T voltage regulator
electric circuit | generalThe
LM317T is an adjustable voltage regulator chip in a TO-220 package (as pictured below). This chip can be used to supply a load current of up to
1.5 Amps at 1.2 to 37 Volts (from an input of 3 to 40 Volts).
The larger the output current, and the larger the difference between the input and desired output voltage, the greater the
heat which will be dissipated by the chip during the
voltage regulation process. This can cause the chip to heat up, and so a
heatsink is often used to speed up heat removal and prevent
overheating.
Calculate Power Dissipated by an LM317T Regulator
The power (heat) dissipated by an LM317T regulator can be calculated using the following simple formula:
Power Dissipated (Watts) = (Vin - Vout) * IL
where Vin is the input voltage, Vout is the output voltage,
and IL is the load current (measured in Amps).
For example, if an
LED bulb rated at 3.3 Volts (and a current of 30mA) is to be powered from a 9V battery via an LM317T voltage regulator, the power dissipated will be (9V - 3.3V) * 0.030
= 0.171 Watts.
Heatsinking an LM317T
An
LM317T can safely be used to dissipate up to
0.25 Watts without any external heatsinking. Therefore in the example above with the LED powered by a 9V battery, the 0.171 Watts of heat dissipated would not require a
heatsink.
However, if five such LED bulbs were to be powered (in parallel), the total current requirement would be 0.15A (=5 * 0.030A), and so the heat dissipated in the regulator would be 0.855 Watts. A heatsink would
definitely be required or the regulator would overheat.
Heatsinks (such as the one pictured above, and sold here:
bolt-on heatsink) have their
thermal resistance specified - measured in degrees Celcius of heat rise per Watt of power dissipated. For example, the heatsink above has a thermal resistance of
21 degrees Celcius per Watt - i.e. if the regulator attached to it is dissipating a constant 1 Watt, the regulator will rise 21 degrees above ambient temperature. The better (larger and more efficient) a heatsink is, the
lower the
thermal resistance will be.
Typically you do not want a regulator chip to run at over
60 degrees Celcius. Therefore a heat sink should be chosen such that its thermal resistance will keep the temperature of the attached regulator below that maximum tolerable temperature. Therefore the maximum thermal resistance acceptable can be calculated according to the following formula:
Max. Heatsink Thermal Resistance = (60 - AmbientTemp) / Power
For example in a room at a temperature of 20 degrees, a regulator dissipating 9 Watts of power would require a heatsink with a thermal resistance of
less than (60 - 20) / 9 = 4.5 degrees per Watt.
Note that the absolute maximum dissipation rating for an LM317T is
15 Watts. If you need a voltage regulator which will dissipate more than that, consider the
LM338 as an alternative.
Article Last Modified: 12:40, 2nd Mar 2009Comment on this Article
If you have any comments on this article, please email them to
neil@reuk.co.uk.
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