I'm using a 7808 voltage regulator in to-220 package. Vin is 12V, Vout is 8V. Expected max current should be in the range of 10mA, let's say 50mA at most. Circuit is to be encased. Do I need to add a heat sink to the 7808, or that power will dissipate correctly without?

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    \$\begingroup\$ That depends on how well the 7808 will be (thermally) isolated from the surroundings. With good isolation you can still cook it to death, even at a lowly 0.05 * 4 = 0.2W. \$\endgroup\$ Apr 30, 2016 at 12:55

2 Answers 2


Do the math. The regulator will drop 4 V with 10 mA thru it. (4 V)(10 mA) = 40 mW. That's not much at all, and even a small package like a SOT-23 can safely dissipate that when mounted normally on a board.

However, you mentioned this will be encased. That changes the thermal resistance from junctions to ambient. Ultimately its about keeping the junctions below some maximum temperature.

A typical junction to ambient resistance for a SOT-23 is about 360 °C/W. At 40 mW that means the junction would be 15°C above ambient. If the part can operate at up to 120°C junction temperature, for example, and ambient is 25°C, then the device could dissipate up to 260 mW.

The calculations are the same with a potted circuit, except that the junction to ambient thermal resistance will be different. It can be both higher or lower, depending on the nature of the potting compound and its depth. This is something you have to look into by carefully examining the potting compound datasheet, and possibly talking to the manufacturer and maybe running your own experiments.

One strategy is to grossly over-spec it. For example, if you get a 7808 in a TO-220 package, then it's hard to imagine there will be a problem except for some extreme potting situations.

  • \$\begingroup\$ Thx for the explanation. The 7808 is in a TO-220 package, so I guess I'm ok without heat sink. \$\endgroup\$
    – Antoine
    Apr 30, 2016 at 13:05
  • \$\begingroup\$ @Ant: Probably, but guessing is no way to do a design. Read the datasheets and do the math. \$\endgroup\$ Apr 30, 2016 at 13:15

My rule of thumb (conservative in most situations) is that 600mW is acceptable for a TO-220 without a heat sink. (at a typically stated \$R_{\theta JA}\$ of 65°C/W, that's a rise of 39°C, so in a 70°C environment the junction would be at 109°C).

Your maximum dissipation is (12-8)*50mA + 5mA*12 = 0.26W (5mA is roughly the current the regulator uses itself).

So it appears it will be just fine. However if you happened to omit that it must operate at an ambient of 100°C, or in a vacuum, or some other important 'detail' it may not last long. It gets a bit more complex with SMT parts that have heat spreaders such as copper patches on the board- the stated \$R_{\theta JA}\$ is only valid for a very specific test PCB situation and attempts to apply it to a very different situation may end in tears.

If the calculated dissipation approached my rule of thumb or if the operating conditions were unusual (eg. vacuum or high ambient) I would do the calculations from scratch.


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