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I am looking to power an Arduino Due through the 5V pin directly from an external regulator. The Arduino takes 120 mA @ 5 V, and I initially tried a MAX5023LASA+T from a 32 V power source, rated to 5 V and 150 mA. However after about 10 s the chip went into a thermal shutdown loop and was scorching hot to touch. Looking at the Vin and thermal derate, I can see why this is, but the chip seems to be terrible for the job.

Now I am looking at the ZLDO1117G50TA, from a pre-regulated 12 V supply. This chip is rated to a 5 V, 1 A. With a draw of 160 mA on the 12 V supply, the regulator eventually after about a minute, went also into a thermal shutdown loop. When I put the same 12 V into the Vin pin of the Arduino, the power draw from 12 V was about 60 mA, and the onboard regulator ran at about 30 degC, stable.

Am I simply using the wrong regulators? What am I missing in the ZLD01117 that it is struggling to make 5 V, is doing so horrendously inefficiently, and subsequently cooking itself? Is there a better choice? I am using the filter capacitors recommended in the datasheet.

FYI I need the 5V for several other systems, so can't just be done with using the Arduino's onboard regulation.

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    \$\begingroup\$ How are you cooling these regulators? Linear regulators dissipates heat according to P=(Vin-Vout)*I. So without any cooling they will inevitably get hot if you run any significant current through then, with high voltage drop. \$\endgroup\$
    – Klas-Kenny
    Commented Jan 21, 2022 at 14:13
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    \$\begingroup\$ Do you have proper heat sinking on the regulator? \$\endgroup\$
    – GodJihyo
    Commented Jan 21, 2022 at 14:13
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    \$\begingroup\$ @JCollins "If we're saying current in = current out then is a linear regulator just a fancy resistor?" Pretty much, yes. \$\endgroup\$
    – Klas-Kenny
    Commented Jan 21, 2022 at 14:24
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    \$\begingroup\$ @J Collins not really, they’re useful for a lot of things, but if you have a large voltage difference they’re not useful for efficiency. \$\endgroup\$
    – Ryan
    Commented Jan 21, 2022 at 14:48
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    \$\begingroup\$ @JCollins They are cheap, simple and low-noise. Great for some applications, but definitely not for all. \$\endgroup\$
    – Klas-Kenny
    Commented Jan 21, 2022 at 16:05

2 Answers 2

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The first regulator has an absolute max rating of 1.5 W, yet you're dissipating 4 W through it ((32-5) V times 0.15 A).

In the second case it is dissipating (12-5) x 0.16 = 1.12 W, which assuming a thermal resistance of 100°C / W would give a junction temperature of 112°C + ambient temperature, very close to the maximum allowed 150°C.

With such a high difference between input and output voltage, you need either cooling (heat sinks, fans), or switching regulators instead of linear regulators.

If you don't want to make your own, there are plenty of ready-to-use modules, some of which are even pin-compatible with a classic TO220 linear regulator. Search for "potted DC/DC converter" or "DC/DC converter module".

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  • \$\begingroup\$ Great answer, exactly what I'm after. However a switching regulator seems to need a host of extra components, an inductor most notably. Are there any single-chip solutions? \$\endgroup\$
    – J Collins
    Commented Jan 21, 2022 at 14:30
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    \$\begingroup\$ Yes, you can buy a single cube of plastic with leads sticking out, and everything included inside. Or a ready-made PCB. \$\endgroup\$ Commented Jan 21, 2022 at 14:31
  • \$\begingroup\$ @JCollins some brand names: RECOM (Google "dc/dc converter module"), Pololu. \$\endgroup\$ Commented Jan 21, 2022 at 14:33
  • \$\begingroup\$ Very good answer, upvoted. Don't stray into uncharted part recommendation territory, though... Could you edit the module descriptions into your question, can get missed in comments. Thanks. \$\endgroup\$
    – TonyM
    Commented Jan 21, 2022 at 14:40
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    \$\begingroup\$ @TonyM as I am adventurous I'll stray a little bit in the comments, and see what happens ;) I think this will be helpful to the OP. \$\endgroup\$ Commented Jan 21, 2022 at 14:45
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Yes, the regulators are not suitable for the task you need them to perform. You are exceeding their limits.

Example 1: Drop from 32V to 5V at 120mA. The regulator needs to dissipate 3.24 watts. It is rated for max 1.5W and it will overheat.

Example 2: Drop from 12V to 5V at 160mA. Regulator needs to dissipate 1.12 watts. It should handle it with enough cooling, but if you can't transfer enough heat away from it the temperature will rise too much.

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