LM7812 Voltage Regulator Heat Sink Sizing


I'm planning to use the following LM7812 voltage regulator for a project:


Vin will be an 18VDC laptop type power supply rated at over 3A. The 12VDC output will power my Arduino. I need to use the 18V power supply because I want to run a number of series LEDs off of the 18V, but I need the Arduino to run the code that controls the MOSFETs that the LEDs will be connected to.

Anyway, given the fact the the Arduino will not be needing to source a substantial amount of current, can someone tell me:

A) How to determine the size of heat-sink needed (heat dissipating capability)

B) How to determine the physical size of the heat sink to buy, and links to possible heat sinks that will work for my given application and chosen LM7812.


You first need to determine how much current you are going to draw. This will determine the amount of power dissipated in the TO-220.

For example if the output of your supply is 12V and the regulator is 5V you have a 7V drop. If you draw 100mA the dissipation is 7 * 100mA or 0.7W. The thermal resistance (Rja junction to air) of a TO-220 package is around 65 degC per W (from the datasheet). The maximum junction temperature of the Fairchild part is 125DegC. In this example 0.7 * 65 = 45 degC rise. For room temperature operation you have a junction temperature of 70 which is well below the 125DegC maximum.

With a heatsink the thermal resistance is determined by summing the Rjc + Rcs + Rsa = Rja

Rjc is junction to case thermal resistance of the part (5degC/W for the 7812) Rcs is the case to sink thermal resistance of the thermal grease (typically 1degC/W) Rsa is the sink to air thermal resistance of the heatsink

(* jcl *)

www: http://www.wiblocks.com twitter: http://twitter.com/wiblocks blog: http://luciani.org

Thanks for the quick reply...

I assume you mean that I'll need to determine the amount of current required from the output of the LM7812. With that assumption, I guess I'll go with the max for the Arduino input... (which I don't even know)... Would that be the max the current that the Arduino can source???

So, say, I source 500mA from the LM7812 output. That would give 18V(input)-12V(output) = 6V. 6V*500mA = 3W. 3W*65C/W = 195C. Since 195C is above the Topr of 125C for the LM7812, I assume that definitely would mean I need a heat sink.

Can you please possibly give me a dumbed down definition of what Rjc, Rcs, Rsa, and Rja are? And (assuming my previous paragraph was correct) how would my power requirements fit into the equation you provided. That would really help..

Thanks again.

Almost correct -- the 195DegC is the temperature rise. The temperature would be 220DegC (195DegC + 25DegC) which is still well above the maximum operating temperature.

The j subscript refers to the semiconductor junction. This is the point that is referenced in the datasheets. c is case of the component, s is the heatsink and a is air or ambient. Since the resistance to heat flow (thermal resistance) is between two points you have Rjc, Rcs, etc.

The sum of all the series between the junction and the air is what determines the temperature rise (Rja). When the device sits in free air the thermal resistance (Rja) is the combination of Rjc + Rca. This number is given my the manufacturer.

When you mount the device on a heatsink you have the junction connected to the case connected to grease connected to sink connected to air or Rja = Rjc + Rcs + Rsa

The Rjc and Rcs are in the 7812 datasheet. The Rsa is the heatsink datasheet.

(* jcl *)

www: http://www.wiblocks.com twitter: http://twitter.com/wiblocks blog: http://luciani.org

Take a look at:- http://www.thebox.myzen.co.uk/Tutorial/Power.html and http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html