7812 heat sink sizing using graphs


I'm trying to determine if my heat sink is large enough for my LM 7812. I realize there are several posts and YouTube videos on computing heat sink size, but my data sheet does not include the typical Theta data, so i don't see how to compute the heat sink the traditional way. However, there is a maximum average power dissipation graph (figure 4). Here's the data sheet:


Operating parameters:

  1. Input voltage is 28 VDC (comming from a battery)
  2. Current load is 200 mA
  3. Devices being powered: Uno, 20x4 LCD, Hall effect current sensor (15 mA)
  4. ambient will be a summer day (32 deg c).
  5. operating time 2 hours before unit will be switched off
  6. Heatsink 26deg C/W http://www.rapidonline.com/electronic-components/fischer-elektronik-fk-242sa220v-to220-vertical-mount-clip-on-heat-sink-26-c-w-36-0522


Pd = (28 - 12) x 0.2 = 3.2 watts

Based on figure 4, I think I should be OK.

My heatsink is getting pretty hot, hence my question if my heatsink is large enough

Thanks in advance.


Change to a switching regulator instead. 85-96% efficient or more vs 42% efficient.

...but my data sheet does not include the typical Theta data,

It is there, it is in note (3) under the Absolute Maximum Ratings table.

(3) Thermal resistance of the TO-3 package is typically 4°C/W junction to case and 35°C/W case to ambient. Thermal resistance of the TO- 220 package (NDE) is typically 4°C/W junction to case and 50°C/W case to ambient.

It's such a simple calculation that the datasheets never show it.

3.2 W looks right, if 200mA really is your total load. Then multiply by the heatsink degrees/Watt to get the temperature rise. 83.2 degrees. This is the rise above ambient, so add ambient temperature to find the actual temperature: 115.2

Then you need to work out how hot the 'junction' inside the case is getting. The junction-to-case thermal resistance is buried in note(3) in that datasheet: 4C/W. So add another 12.8 degrees to get 128 degrees. The maximum junction temperature is 150C so you are within its ratings, but only just.

How much of this 200mA is going through the Arduino's 5V regulator? You need to do the same calculation for the Arduino regulator as it will overheat and die at moderate loads with 12V input.