I've designed a simple PWM RGB LED slow fader to be used as a garden lighting effect. My circuit works great, but I under-estimated the amount of heat that is generated by the 7805 linear regulator.

It's mains powered, with a 6Vac transformer and a 5V 1A linear regulator. If all the LED's are at full brightness then it draws around 600mA.

I've mounted my circuit board in a plastic enclosure with a transparent lid, and it's rated at IP67 (and I want to keep it that way!).

I've put quite a small heatsink on the regulator, and it takes around 3 or 4 hours of continuous use to get to a temperature that is just about too hot to touch, I'm guessing around 70-80°C.

My plan is to give this to my Dad for him to use in his garden, but obviously I don't want it to melt or catch fire.

My questions are:

  1. Is this an acceptable temperature for it to operate at?
  2. Is it likely to get any hotter if left on for longer? I didn't want to test this as I didn't want to damage it, but the datasheet says the operating temp is max 125°C so I guess it would be ok.
  3. What can I do to make it run cooler, given that I don't want to drill vents into the enclosure and ruin it's IP67 rating?
  4. If it does happily operate at a high temperature, do I need to be worried about heat conduction through the PCB tracks into other components that may be damaged? Will it melt the solder?

3 Answers 3


The regulator should have thermal limiting (check the data sheet for details), and will shut down if the temperature gets too high. It won't damage any other parts, and definitely won't melt any solder!

Using a larger heatsink will reduce the temperature, of course. A switching regulator will be a lot more efficient and will just get slightly warm.

  • \$\begingroup\$ I've never used a switching regulator before, is it just a single component? Can I use it as a direct replacement for the linear reg? \$\endgroup\$
    – BG100
    Mar 17, 2011 at 22:12
  • 1
    \$\begingroup\$ You can get them as single components, as a drop-in replacement for a linear regulator, made by TI. They are usually built from a chip, an inductor and a few other parts. Nat Semi makes a large range. \$\endgroup\$ Mar 17, 2011 at 23:01
  • \$\begingroup\$ National Semi's SIMPLE SWITCHER modules are switching regulators packaged with an inductor, so all you need are 6 additional components, including 2 caps for input/output, and 2 resistors for setting the output voltage. They're somewhat pricey compared to a plain controller and associated passives, but they have a niche. \$\endgroup\$
    – Nick T
    Mar 18, 2011 at 2:52
  • \$\begingroup\$ Thanks for the answer... I don't really want to change anything at this stage as the circuit is built and working. I think I'm just going to stick a bigger heat sink on it and let it heat up... you've put my mind at rest about the thermal limiting. I guess if it gets too hot it will just switch off for a bit, which is not a big problem for this application. \$\endgroup\$
    – BG100
    Mar 19, 2011 at 0:20

You probably have too high of an unregulated voltage coming out of your transformer.

Ideally, you want to have the VDrop across the linear regulator be pretty much just the regulators drop-out at full load. You can also use a low-drop-out regulator to reduce the dissipated power (the 7805 is not LDO).

What is the voltage on the input to the linear reg?

Also, as Leon Heller was mentioning above, there are some pretty easy-to-implement switching voltage regulators, particularly Nation Semiconductor's Simple Switcher Regulators.
However, for any switching regulator like these, to make them perform properly, you really need either some really tight hand wiring, or a custom PCB. You can not really breadboard them successfully.

OTOH, if you're willing to spend a little bit more, you can get drop-in switching replacements for something like the 7805. A number of manufacturers sell little 7805-sized switching regulator modules, which even have the same footprint. However, they're generally about $10-$15 dollars.

Edit - Does anyone know why my dollar sign symbols are vanishing?

  • \$\begingroup\$ meta.electronics.stackexchange.com/questions/440 \$\endgroup\$
    – markrages
    Mar 18, 2011 at 3:34
  • \$\begingroup\$ Thanks for the answer. I don't think I'll use a switching regulator in this case, but I'll certainly keep it in mind for my next project. \$\endgroup\$
    – BG100
    Mar 19, 2011 at 0:21

I fixed my overheating linear regulator just by adding a big chunky heatsink. I lifted the regulator out of the PCB, attached hook up wire to the leads, and mounted it on a big chunk of metal I cut from an old power supply.

I've run it for about 4 hours and it's only slightly warm... problem solved!

circuit with big heatsink

  • \$\begingroup\$ Looking very neat too! \$\endgroup\$
    – Linker3000
    Mar 22, 2011 at 23:10
  • \$\begingroup\$ The problem here is that your IP67 case will work like a backing oven. As there is essentially no airflow at all the inside will get hot over time. The heatsink does not reduce the thermal energy, it just dissipates it better. But it can only spread it into itself or into the air. Making a thermally good connection between the heatsink and your case (by using some kind of thermal pad) you might be able to ease these problems. Also the long wires between the 7805 and your board might pose a problem. At least add a small capacitor (100-470nF ceramic) to the 7805s input and output pins to ground. \$\endgroup\$ Jul 23, 2011 at 12:31
  • \$\begingroup\$ Also consider adding a diode between the in and output pins of the 7805. Otherwise it might get fried if the box gets unplugged. \$\endgroup\$ Jul 23, 2011 at 12:32

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