Heatsink Requirements

  31 Aug 14 - Converted to JavaScript.

One thing that needs to be taken into account is ensuring that your regulator or FET doesn't overheat. Regulators are fitted with an internal over temperature cut-out, but you don't want the power dying when you trying to use the output.

The following table gives the thermal and power ratings for devices used on this website.

Explanation:

The following shows the equations to calculate the need and necessary heatsink, with a worked example is at the bottom for a LM317T regulator. The maximum allowable temperature rise (T_Δ) needs to be determined to ensure the regulator will not overheat can be determined using the following equation where T_AMB = the Ambient Air Temperature around the device.

For a voltage regulator,

For a FET,

The thermal coefficient required is given by:

If the value of θ calculated by this equation is lower than the devices own heatsink effect (θ_JA) it will need an external heatsink. The heatsink required can be calculated using the following equation, where θ_CHS is the thermal coefficient between case and heatsink (normally taken as 0.5°C/W when using an insulating washer and thermal grease):

Example:

• The input voltage to a LM317T regulator is 15V so that a stable 12V supply can be produced, but the output can also be at 5V.
• The maximum output current of 1.5A may be drawn so that is used for the calculations.
• The maximum temperature rise for the LM317T is worked out as T_Δ = T_MAX - T_AMB = 125°C - 25°C = 100°C.
• The power dissipated will be (15v - 5v) x 1.5A = 15W
• The required thermal coefficient (θ) is therefore = 100°C / 15W = 6.67 °C/W
• This value is much lower than the devices own heatsink ability (50°C/W) so we will need an external heatsink.
• The required heatsink performance is θ_HS = 6.67°C/W - 5°C/W - 0.5°C/W = 1.17°C/W.
• This would be an extremely large heatsink, but it could be done especially is the case is metal and also forms part of the effective heatsink. Normally the difference between supply voltage and output is not so great and therefore a much small heatsink would be required (if at all).

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