I'm "building" a regulated linear power supply using LM317, Min(1.25V) Max(24V) Max 1.5A, and I have like +/- 30.88V (10% of ripple(3.25v), 32.5V Maximum and 29.25V minimum, medium 30.88V) in input of the LM317(I'm using 4700uF electrolytic cap), but assuming that LM317 have a huge ripple rejection (66dB, that reduces 3.4v to 1.62mv of ripple) I will say that I have like 32.5V getting in my LM317, my power dissipation at wrost case (in LM317) if I am correct, its something like 46.48 watts (P = 32.5V - 1.25V) x 1.5A so my question is, is it 46,87W to much and problably will I need a heat sink to hold it right?

Sorry guys I'm, new in this game :p Ty everyone!

(I'm using a 24V/3A transformer, bridge rectifier made with 4x1N5408)

  • 1
    \$\begingroup\$ I don't think there is a heatsink of any size that will allow you to run nearly 50 watts through an LM317. \$\endgroup\$
    – Ron Beyer
    May 23, 2020 at 16:05
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    \$\begingroup\$ (10%(3.25v) ripple, Max value of ripple is 32.5V Maximum and 29.25V minimum, half of ripple(RMS) is 30.88V) - what is this strange language you speak? Be clear. \$\endgroup\$
    – Andy aka
    May 23, 2020 at 16:17
  • \$\begingroup\$ What is your output voltage and current? Why does the supply actually need to be linear? High current linear supplies are... huge and heavy for a reason. And they are typically extinct except in the most sensitive applications (and conversely some unregulated ones). You might also consider if you could use a linear regulator after a switching one that accomplishes most of the change. \$\endgroup\$ May 23, 2020 at 17:21
  • \$\begingroup\$ Thanks guys, Chris my output voltage and current is 24V/1.5A Max \$\endgroup\$
    – Jeep nN
    May 23, 2020 at 17:34

1 Answer 1


Is 48 W dissipation too much?

Let's look at the thermal resistance between the chip and the tab on the back.

enter image description here

Figure 1. The TI LM317 datasheet shows the minimum thermal resistance between the junction and case is 15.9°C/W for the KCS (TO220) version.

That means that with 48 W the temperature rise would be 48 × 15.9 = 763°C above ambient if we were able to hold the tab at ambient temperature. Add in the thermal resistance of the chip to the heatsink and the heatsink to the air and you will increase the temperature further.

Table 7, the Abolute Maximum Ratings (which you want to be operating well below) states, "[Max] Operating virtual junction temperature 150°C."

Yes, it's too much. Forget this design. I first used that chip > 40 years ago. Things have moved on.

  • \$\begingroup\$ Thank you very much, no doubts now! And sorry if that was a stupid question! \$\endgroup\$
    – Jeep nN
    May 23, 2020 at 17:48
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    \$\begingroup\$ No, not stupid if you don't know. TI have a technical document on the subject. It might be worth a quick scan to learn a few new words and the context where they're used. Understanding might come too! ti.com/lit/an/spra953c/spra953c.pdf?ts=1590250523406 \$\endgroup\$
    – Transistor
    May 23, 2020 at 17:53

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