Good day,

I have built a simple power supply that converts 220VAC to 5V and 12V outputs. I used only 1 heatsink for both regulators (LM338). The power supply works, but I am unsure what the common output (Vout left on the schematic) will have on the regulators. One voltage regulator's output is 12V and the other one's is 5V. Below is the schematic of the LM338 voltage regulator and a picture on how the voltage regulators and heatsink look like in my power supply.

enter image description here enter image description here

  • 3
    \$\begingroup\$ The way to do this is to electrically insulate both regulators from the heatsink. There are insulation kits sold to do this (mica washer and a plastic washer for the mounting screw, without which, you aren't getting enough thermal contact for good heatsinking) \$\endgroup\$
    – user16324
    Commented Sep 3, 2016 at 11:47
  • \$\begingroup\$ You are shorting both output together - connecting the 5 volt output to the 12 volt output. The resulting single output voltage is unpredictable. You will only get one output voltage which will be neither 5 volts nor 12 volts. Follow Brian's advice and electrically insulate the tabs from the heat sink, while maintaining thermal contact between the tabs and the heat-sink. Or saw the heat sink into two pieces. \$\endgroup\$
    – FiddyOhm
    Commented Sep 3, 2016 at 11:52
  • \$\begingroup\$ While you are at it, also consider using insulating nuts and bolts. Personally, I won't go through this route though. I'd rather buy another heat sink than going through this trouble. \$\endgroup\$ Commented Sep 3, 2016 at 12:02
  • 2
    \$\begingroup\$ The 78xx series (fixed output voltage) regulators have ground connected to their backplate so there you could mount both on the same heatsink without isolating them. But the 78xx series has a 1 A max output current and so is less powerfull than the LM338. \$\endgroup\$ Commented Sep 3, 2016 at 12:17

2 Answers 2


You need to insulate electrically one or both voltage regulators from the heatsink:-

enter image description hereenter image description here

If you don't then the resulting output will be 12V on both because a 5V regulator normally uses a series pass transistor to control voltage to its load and when presented with a load voltage higher than 5V it will shut that transistor down.

So, you have to insulate one regulator and probably it's best to choose to insulate the regulator that handles the smallest power dissipation, possibly the 12V device (based solely on least voltage drop across it).

  • 1
    \$\begingroup\$ Old guy in new looks. Nice tshirt :) \$\endgroup\$ Commented Sep 3, 2016 at 12:04
  • \$\begingroup\$ Hehe peeping tom!! \$\endgroup\$
    – Andy aka
    Commented Sep 3, 2016 at 12:06
  • \$\begingroup\$ @Matthias if you are happy with these answers, please consider formally accepting one of them - you probably didn't realize that this is the "small price" for otherwise free help! \$\endgroup\$
    – Andy aka
    Commented Nov 22, 2016 at 21:38

Good photo, you may "luck-out" because you've smeared heat-sink grease, and because you've not clamped regulators to heatsink with a screw and insulating shoulder washer. Bad luck may result in blown 5v regulator, or +12v finding its way to your +5v output terminal - the heatsink electrically connects both outputs together if the insulating heatsink grease fails. I've "lucked-out" doing as you've done (but I've clamped down those regulators to the heatsink carefully).

About heat-sinking...you have a good big heatsink, but those regulators are a bit too close to each other - one can overheat the other. If you can keep a finger on each regulator while they're working hard delivering power to a load, then heatsink is likely adequate. Your +5v regulator has a larger voltage drop from un-regulated input to regulated output, so it may get hotter than the +12v regulator for similar output current. I'm assuming that both regulators have their inputs fed from the same DC voltage source.

  • \$\begingroup\$ I'm interested in your finger /regulator temperature method. Is this a throwaway comment, or can it be adopted as a rule of thumb? I ask (and hijack this thread) as I have s similar situation and am concerned over how hot mine run. \$\endgroup\$
    – Paul Uszak
    Commented Sep 3, 2016 at 20:57
  • \$\begingroup\$ to Paul Uszak, Yes, a rule-of-thumb - a quick and dirty test. Most silicon should be run below 150 degrees(C) - I design for 100 C. A failed finger-test requires a much closer look at thermal design headroom. \$\endgroup\$
    – glen_geek
    Commented Sep 3, 2016 at 21:32

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