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Quick question - what do you think the heat capacity of a LM317 might be? The use case is that I want to ignite e-matches, which requires are short duration high current and would prefer to just use bunch of linear regulators without a heatsink. But I have no idea what the thermal capacity of a LM317 might be, and I don't want to kill the regulators!

I think I would need something like 10-15W of heat dissipation for maybe 100ms.

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  • \$\begingroup\$ Just look it up for a typical package like a T0-220 the "joules per degC" figure should be quoted somewhere. \$\endgroup\$
    – Andy aka
    Dec 27, 2020 at 15:25
  • \$\begingroup\$ So, 1.5J. Weigh one, and estimate what fraction of its weight is copper alloy (the tab and leads). Look up the heat capacity in tables. If not enough, calculate the mass of copper required and bolt to the tab. \$\endgroup\$
    – user16324
    Dec 27, 2020 at 15:27
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    \$\begingroup\$ What is the output current requirement and voltage requirement? What is the input voltage to regulator? \$\endgroup\$
    – Justme
    Dec 27, 2020 at 15:28
  • \$\begingroup\$ Away from pc. || Datasheet usually provides pulse duration versus watts curve. || If you tell us your actual need we can probably provide a better and cheaper solution. \$\endgroup\$
    – Russell McMahon
    Dec 27, 2020 at 19:29
  • \$\begingroup\$ Thank you, very usable and something I might consider another time. I have however abandoned the idea of using linear regulators I think. \$\endgroup\$ Dec 28, 2020 at 20:22

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Over just 100 ms I don't think the thermal capacity of the packaged part matters as much as the thermal capacity of the chip itself. The regulator die may reach a very high temperature, and fail, before the heat can be dissipated to the surrounding package.

You also need to worry about the maximum instantaneous current. If you are passing a "high current" then the average temperature of the part may be the least of your concerns.

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  • \$\begingroup\$ I get your point, the power might not be able to dissipate quickly enough to assume an even temperature distribution. \$\endgroup\$ Dec 28, 2020 at 20:28
  • \$\begingroup\$ @Davidskovgaardhansen - it sounds like you assume that high temperature is the only damaging effect of overcurrent. This is incorrect. High current can directly damage parts, e.g. through electromigration. \$\endgroup\$
    – TLW
    Nov 28, 2021 at 23:28
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Apart from "why do you need a regulator at all if you're just trying to fry a fuse?" one of the claimed features of the LM317 is that it will shut itself down on a variety of excess conditions. Internal shutdown on thermal overload and output short circuit protected are direct quotes from the (vintage) datasheet. The new datasheet has a few changes to the wording but nothing major.

A capacitor bank and a crowbar seem far more suited to the problem at the fancy end. A battery and a switch (or the other kind of crowbar) at the less fancy end.

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  • \$\begingroup\$ Very true, I might just apply the full 12V voltage to the e match. I am however a bit concerned with the very high current I would get. I might be able to connect the e matches in series instead of parallel, but I am uncertain if I want this. I have also considered using a super cap charged through a linear regulator. Or perhaps just apply my 12V supply directly to the e matches though a MOSFET. The current will momentarily be very high, though. \$\endgroup\$ Dec 28, 2020 at 20:27
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Rather than use a "bunch of linear regulators", you might want to use a simple buck circuit. It sounds as though you may not need to actually regulate your output, just step the voltage down. If so, the buck circuit could be a simple square wave generator with a fixed duty cycle, a MOSFET, an inductor and a Schottky diode. You will probably have to dissipate much less heat with a buck circuit, than with a "bunch of linear regulators" running at full capacity.

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  • \$\begingroup\$ This sounds like it should be a comment rather than an answer, since you didn't try to answer the original question. \$\endgroup\$ Dec 27, 2020 at 23:58
  • \$\begingroup\$ Sure, I considered that, it might be the best option. \$\endgroup\$ Dec 28, 2020 at 20:22
  • \$\begingroup\$ I'm not supposed to give you a design in the answer section of your question here, because it really is a different question. However, if you want to ask for a design for a buck converter separately, I will help you. I know you need 10-15W for 500 msec. Where is power coming from? What is it's voltage? What is it's maximum current? How are you switching it, and what is an acceptable start-up delay before the circuit reaches 10-15 W? \$\endgroup\$ Dec 28, 2020 at 20:37
  • \$\begingroup\$ I did consider a buck converter. Ultimately, I think I will go with a linear regulator charging a super capacitor as this seems to be easier to design for my use case. \$\endgroup\$ Dec 30, 2020 at 22:05
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Numbers extracted from datasheet:

TO-220 (NDE) is your best bet: \$R_{θJA}\$ = 23.5°C/W and \$R_{θJC}\$ = 16.2°C/W.

Maximum power:

$$ P_{D_{Max}} = {(T_J - T_A) \over R_{θJA}} = {(150°C - 20°C) \over 23.5°C/W} = 5.53W$$

If ambient temperature is 20°C, the maximum temperature of case will be:

$$ T_C = P_D\ R_{θJC} + T_A = 5.53W \times 16.2°C/W + 20°C = 99.6°C $$

These are the theoretical maximums, so no way you are getting to 10-15W. It definitely won't last 100ms and will a case temperature of 99.6°C achieve what you want.

Ultimately, this is dependent upon speed of thermal shutdown circuit, but odds are good, thermal shutdown will activate.

TI's Understanding Thermal Dissipation and Design of a Heatsink

From LDO Thermal Design Concept:

enter image description here

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  • \$\begingroup\$ Right, I get your point. You are however assuming steady-state and the point is I don't want the system in steady-state - it is only temporary. \$\endgroup\$ Dec 28, 2020 at 20:25
  • \$\begingroup\$ The heat is not instantaneous. \$\endgroup\$ Dec 29, 2020 at 1:16
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Not sure if this should be a separate comment or an edit?

Anyway, for my use case I have decided to go with a design where I charge a super capacitor and discharge it through the electric match through a relay. So I will use a 5F capacitor as I need fairly low ESR. The biggest drawback is the charging time as I would prefer a bit lower capacitance, but I see this design as more fool proof compared to using linear regulators and simpler to implement compared to a buck converter. Thank you for the help.

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  • \$\begingroup\$ Make sure you don't inadvertently spotweld the relay contacts. \$\endgroup\$
    – TLW
    Nov 28, 2021 at 23:31

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