The circuit works and it provides a stable 3.3V output, but it heats up and I'm afraid it can overheat and cause problems.

  • \$\begingroup\$ The problem is not the source. It is the load. How much current are you using in the 3.3V section? \$\endgroup\$ – JRE Apr 23 '17 at 11:19
  • \$\begingroup\$ @JRE I'm using about 700mA \$\endgroup\$ – Semko Toruj Apr 23 '17 at 11:22
  • \$\begingroup\$ @MaxPower Then you can calculate how much power your regulator will have to evacuate, and then find a radiator with low enough thermal resistance. Or better replace that linear regulator with a buck switching regulator, like a LM2576. \$\endgroup\$ – user2233709 Apr 23 '17 at 11:45
  • \$\begingroup\$ @user2233709 That is so cool, I will definitely use switching regulator in future, but for now I already have a circuit ready. I will try calculating, but I'm not exactly sure how :) Thanks for answers \$\endgroup\$ – Semko Toruj Apr 23 '17 at 11:52
  • \$\begingroup\$ @MaxPower Enric Blanco did the math for you. \$\endgroup\$ – user2233709 Apr 23 '17 at 11:56

Check the datasheet and do the math. Short answer: YES, it will overheat A LOT!

With that huge 8.7 V voltage drop, if the load required the full 0.8 A that the LD1117V33 can output then the power dissipation would be approx 7 W, which is well below the 15 W maximum rating.

However, the temperature rise at 7 W dissipation without using a heatsink (50 C/W) would be a mind-boggling 350 C, and the junction temperature would go far above the maximum 150 C rating.

At Tamb = 25 C, the maximum allowable temperature raise would be 125 C. At 50 C/W the maximum allowable power dissipation for that temperature rise would be 2.5 W. This would mean a maximum 0.36 A current draw from the load. And you should derate this to 50% (or 70% at most) for reliability reasons, so let's say your maximum current draw should be 180-250 mA.

If your load draws more current than that, then you'll need a heatsink in order to reduce the temperature rise. The maximum thermal resistance (junction to ambient) for a temperature rise of 125 C at 7 W dissipation is 17.8 C/W. The case itself has 3 C/W (junction to case), so the heatsink should have a maximum of 14.8 C/W (case to ambient). You should derate this too, so you actually want a maximum of 7.4-10.4 C/W for your heatsink.

However, the best solution by far is to use a switching regulator / DC-DC step down converter. A linear regulator is very inefficient power-wise, when used with such huge voltage drops. The switching regulator will not need a heatsink because it doesn't dissipate that much power to begin with.

  • \$\begingroup\$ Thanks! That's a delicious answer and it clears a lot of questions :) \$\endgroup\$ – Semko Toruj Apr 23 '17 at 11:56
  • \$\begingroup\$ Please note that operating at higher ambient temperatures will reduce you maximum allowable temperature rise. This in turn will mean either 1) lower maximum current draw without heatsink, or 2) lower maximum thermal resistance required for the heatsink. You might want to redo the math according to your actual operating temperature range (I just did the math for 25 C as a reference for you to follow). \$\endgroup\$ – Enric Blanco Apr 23 '17 at 12:03

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