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I'm trying to obtain 5V DC from a 37V DC power supply. The load is a computer fan that at 5V drains 31mA (around 0.15W). I'm using a LM350 voltage regulator (LM350 Datasheet). I get a very precise 5V output but the chip becomes really hot. I measured that the the power supply provides 49mA at 37V. I think that the LM370 use 48mA - 31mA = 17mA, so it dissipates 37V * 17mA = 0.63W. (Please correct me if these calculations are wrong) It bothers me that the voltage regulator requires more energy than the load. It's normal or I'm doing this in a stupid way? There is a better way to supply a computer fan from a 37V DC tension?

Thanks :)

P.s. On the datasheet it says that the maximum input-output voltage differential is 35V. I'm using it with a 32V voltage differential. Is it safe? I need a better voltage regulator?

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  • \$\begingroup\$ 1) Can you use a 12 V fan instead? 2) Perhaps you can find a switch-mode voltage regulator which accepts 37 V as its input. 3) Does the 37 V source have any lower-voltage taps which you could use? \$\endgroup\$ Commented Sep 11, 2016 at 16:41
  • \$\begingroup\$ 1) That's a good idea. Maybe I've already one of it in my old PC. 3) Nope :( I've built the power supply for a gainclone. It has only +-37V DC output \$\endgroup\$ Commented Sep 11, 2016 at 16:47
  • \$\begingroup\$ I think it is misleading to say 'step down' in the title - to me that implies a switcher... \$\endgroup\$ Commented Sep 11, 2016 at 16:52
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    \$\begingroup\$ I once faced the same design problem except it was 48V so I chose two 1.5" 24V fans in series. PWM is a better way to go with a choke. ( ie buck) \$\endgroup\$ Commented Sep 11, 2016 at 17:15
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    \$\begingroup\$ Per its datasheet, the LM350 should be sinking less than 0.1 mA through its ADJ pin, not 17 mA. So I'm not sure where that 0.63 W is going. \$\endgroup\$ Commented Sep 12, 2016 at 3:32

2 Answers 2

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You are correct in that the linear regulator appears to be using 17mA at 37V so that is a power of:

Vin * Iregulator
37V * 0.017A = 0.629W

But you need to also consider that the linear regulator that you are using is also dissipating power of:

(Vin - Vout) * Iload
(37V - 5V) * 0.031A = 0.992W

Thus you have a total power loss in the linear regulator of:

0.629W + 0.992W = 1.621W !!!

You really need to be using a switching buck regulator for this job. If you do not want to build such a thing you can find a plethora of ready made modules on web sites the like of Amazon or eBay. They do not cost a lot either.

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    \$\begingroup\$ Your proposed OP Amp solution is not any better than the linear regulator in terms of power dissipation. Plus that opamp cannot source enough current anyway. You do need the switcher module. Do not take the single example I showed as the only price option. Do your own search. It really is dumb to burn up the Watts like this. If you cannot get the efficient solution for the 37V->5V then redesign your main supply to provide a 5V output (or 12V as there are plenty of 12V fans). \$\endgroup\$ Commented Sep 11, 2016 at 17:16
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    \$\begingroup\$ If you try to stick with the linear power hog solution you will also need a second fan to cool the regulator. \$\endgroup\$ Commented Sep 11, 2016 at 17:32
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    \$\begingroup\$ @EnricoBarberis Could you give some context for 'pretty expensive'? A random search shows (amazon.com/Adjustable-Supply-Module-replace-LM2596s/dp/…) as prefabricated modules these can be purchased for $1.30 each. The LM350 alone (random search on the same site) seems to cost at least $0.50. \$\endgroup\$
    – abligh
    Commented Sep 11, 2016 at 17:38
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    \$\begingroup\$ @abligh - Be careful what modules you recommend. Not all can handle a 37V input. The ones you linked are specified to support an input up to 28V. \$\endgroup\$ Commented Sep 11, 2016 at 17:42
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    \$\begingroup\$ @MichaelKaras fair point. To be clear I was not recommending that particular module, just making it clear that LM2596-based modules are cheap. \$\endgroup\$
    – abligh
    Commented Sep 11, 2016 at 18:44
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It bothers me that the voltage regulator requires more energy than the load. It's normal or I'm doing this in a stupid way?

It's inevitable given that you're using a linear regulator, and your output voltage is less than half of your input voltage. An ideal linear regulator has the same current on its input and its output. That means that the input power and output power are in the same proportion as the input voltage and the output voltage, and the difference between the input power and the output power has to be dissipated in the regulator. In a less-than-ideal regulator, the input current will be larger. In the case of 37V input and 5V output, it follows that at best 5/37 ~= 13.5% of the input power reaches the load, and at least 32/37 ~= 86.5% of the input power is turned into heat.

The mistake in your math is subtracting current instead of power. Your fan is drawing 31mA * 5V = 155mW. Your regulator is drawing 48mA * 37V = 1776mW, and dissipating 1776mW - 155mW = 1621mW. Even if it was perfectly efficient, it could not draw less than 31mA * 37V = 1147mW, and could not dissipate less than 1147mW - 155mW = 992mW. If it did, it wouldn't be a linear regulator.

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