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..to be updated I will use a power supply inside an enclosure. But I saw some people are screwing these supplies to heatsinks where the bottom of the supply faces the heatsink.

But nothing is mentioned on the datasheet about it.

My question is if this supply operates at at its rated power 50w continuously, does it require heatsink and is there a way to estimate it? Is there a way to estimate how much heat it dissipates?

...to be updated

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  • \$\begingroup\$ Depends on your enclosure to ambient thermal resistance and the ambient air temperature is. Have you measured it? \$\endgroup\$
    – winny
    Commented Sep 6, 2021 at 13:35
  • \$\begingroup\$ Just an ambient derating 5°C/1000m is known (note 8, page 2) \$\endgroup\$
    – Antonio51
    Commented Sep 6, 2021 at 13:43
  • \$\begingroup\$ What is temperature coefficient? Given as ±℃ ℃ 0.03%/ (0 ~ 50 ) \$\endgroup\$
    – floppy380
    Commented Sep 6, 2021 at 13:45
  • \$\begingroup\$ Ok. I did not see this. Perhaps the "stability" of output voltage vs Temperature. \$\endgroup\$
    – Antonio51
    Commented Sep 6, 2021 at 15:28

3 Answers 3

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Those PSUs have a quoted efficiency of 85% to 90%. Heat dissipated will be \$ \frac 1 {0.85} - 1 = 17\% = 26 \ \text W\$ to \$ \frac 1 {0.90} - 1 = 11\% = 17 \ \text W \$.

Under the "Features" heading on the datasheet it promises "Cooling by free air convection". I've seen these used in many industrial systems and never seen a heatsink. It wouldn't hurt to screw it flat side to the cabinet or panel to dissipate more heat by conduction.

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  • \$\begingroup\$ How can I relate whether 26W dissipation require heatsink if the ambient temperature is lets say 50C? \$\endgroup\$
    – floppy380
    Commented Sep 6, 2021 at 13:41
  • \$\begingroup\$ Temperature coefficient given as ±℃ ℃ 0.03%/ (0 ~ 50 ). Does it tell anything useful? \$\endgroup\$
    – floppy380
    Commented Sep 6, 2021 at 13:46
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This is a matter of how reliable do you want it with Arrhenius Law and using Ohm's Law for thermal resistance of the hottest transformer and your enclosure effective thermal resistance. The linear air hot-surface speed matters far more than the internal ambient and CFM volume air flow. Most PC towers are poorly designed for this.

You must choose max hot spot of Ferrite as the most critical element to prevent thermal runaway. Typically I found from doing this design before in a high volume product. The OEM suppliers design for open air, not enclosed, which they have no control over. They measure internal ambient maybe, but not the hotspot. So I decided to do this and found that 2 1U (38mm?) high 1W fans easily remove fans regulated with thermistor and Amplifier insulated on chassis in air path. My chassis was rack mount, sealed lid ends vented with a custom Mylar spoiler on intake to make turbulent flow then laminar flow in exit. That made the hotspot 20'C cooler than without the spoiler. The fans only needed a little power above about 50% at full load at room temp so it could survive 40'C ambient. YMWV

Up to 90% efficiency means best case of 150 W. which could be some value below max power and might mean 85% at 150W (est.) which means 23W if heat on the ferrite XFMR and FETS shared unequally. I suggest you measure the hotspot and ambient heat rise in a box with a fan and see the temp rise above room on the hottest part. I chose 20'C rise max. Above 40'C internal ambient, initially before I figured out this stuff, it was a 45'C rise.

  • then you have to add the load heat of internal dissipation.

You must also consider fan noise and safety requirements. UL mandated some coke spill requirement and crush test for the lid. also it was tested and passed all conducted emissions and UL with caveats approved for painted welds on ground studs.

More details

BTW I epoxied a 10k thermistor to the XFMR and tested HIPOT with DC earth bonded and found 6 failure exceptions in their production as they test HIPOT easier with DC floating. The XFMR capacitance causes more HIPOT primary stress and failures when secondary is earth bonded .

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Doesn't matter, that power supply is not designed for use with a heat sink, it's designed to cool by air convection through its chassis. To work with a heat sink, it would need to have its heat making parts mounted on a flat thermally conductive plate to which you would attach a heat sink. It's not. It's a PCB mounted in an perforated (holey) metal enclosure with standoff distance between PCB bottom and enclosure (so the metal does not short out the PCB). There's an air gap there; putting a heat sink there would be futile.

Its spec sheet discusses its requirements, and you need to follow them if your assembly needs to be permitted/approved.

If you can't secure a stream of fresh air for it (e.g. robot submarine), then you need to find a way to remove heat from the air. For instance two back to back heat sinks bolted through the metal chassis. Preferably not in the direct sun lol.

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