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I am trying make a variable voltage supply using part of this circuit:

enter image description here

Instead of a transformer and a rectifier, I am using a solar panel, which gives 37 V (open circuit) and about 30 V with a load. I also skipped the capacitors. I used a heat sink on the LM338 (not very big but not very small, either.)

It all works fine and I can regulate the voltage.

The problem is, it gets VERY hot.

I used a load of 0.5 ampere (an LED 12 V lamp,) and regulated voltage to about 12 V. The temperature was 70°C.

Have I missed something? The LM338 is said to be able to give 5 amperes from 1.2 V to 30 V.

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    \$\begingroup\$ Welcome to EE.SE. Please post the schematic in your question rather than a link. (Leave the link to the original article so you can credit the author.) Have you calculated the power dissipation in your LM338 (P = VI where V is the voltage drop across the regulator)? What answer did you get? Put all the info in your question. \$\endgroup\$
    – Transistor
    Mar 5, 2021 at 15:05
  • \$\begingroup\$ What sort of heatsink are you using? \$\endgroup\$
    – user_1818839
    Mar 5, 2021 at 15:53
  • \$\begingroup\$ Thanks, i tried post the link image, but did not work :(. I got the answer from "Justme" and i belive thats the case. I still belive i can use larger hetasink and use 0.5 Amp, the way i want it. My heatsink was not very big. \$\endgroup\$
    – mathias
    Mar 5, 2021 at 17:05
  • \$\begingroup\$ Any way the goal was to use the panel for variable voltage (up to 28 volts), and as much power as possible (even at lower voltage), but i guess that is not possible this way. \$\endgroup\$
    – mathias
    Mar 5, 2021 at 17:10
  • \$\begingroup\$ It is normal for linear regulator, use two secondary winding transformer, ones that gives you 15V DC (for 12V output very common used) and one with 28V DC. Switch between them at input of LM338 according desired output. This switching you can do automaticly with logic and relay. Or use DC/DC. \$\endgroup\$
    – user208862
    Mar 5, 2021 at 20:07

2 Answers 2

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With supply voltage input of 30V, output voltage of 12V, and load current of 0.5A, the regulator has to dissipate 9W as heat.

Which means, it does get very hot, so your circuit works exactly as expected.

While it can provide up to 5A of current, it does not mean it can do 5A, or even 0.5A when it is dropping 30V to 12V.

As the chip has junction-ambient thermal resistance of roughly 23-35 °C/W (depends on the package), it means that at 9W dissipation the temperature should rise beyond maximum operating temperature of 125 °C quite quickly and is most likely saved by overtemperature shutdown circuitry.

So a linear regulator will not work with your needs of regulating 30V to 12V at 0.5A.

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  • \$\begingroup\$ Thanks allot for that answer, yeah that sounds logical, did not think about that. Maybe a power resistor in series with the solar panel, can solve that problem?. Glad for tip on that :) best regards Mathias \$\endgroup\$
    – mathias
    Mar 5, 2021 at 16:20
  • \$\begingroup\$ But one thing i like to point, is that the circuit says "1.25V - 37V 5A Adjustable Power Supply with LM338". That can not be totaly true? cause i guess no one can cool down lm 338 good enough when using about (40 volt) and drop the voltage to say 5 volt, and load it with 5 amps. If i understand all correct?. 37-5=32, 32*5=160 Watt. \$\endgroup\$
    – mathias
    Mar 5, 2021 at 17:34
  • \$\begingroup\$ The statement is true. It can give out any voltage from 1.25V to 37V, and it can provide up to 5A current. No problems there. But like with any other linear regulator, there are limitations that must be taken into account, or the power dissipation limit will become too large. So it can give out 5A but maybe when Vin is only 3V larger than Vout and it has good heat sink to dissipate 15W. And it can bring down 40V to 5V, but it can't do it at 5A at all, and providing 428mA it has to also dissipate same 15W into a heat sink. \$\endgroup\$
    – Justme
    Mar 5, 2021 at 18:27
  • \$\begingroup\$ Thank you. Yes then i under it correct :). Best regards Mathias \$\endgroup\$
    – mathias
    Mar 5, 2021 at 20:24
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Curiousity got the better of me, and I had a look at the datasheet.

Page 6 shows has this chart:

enter image description here

The solid line shows the maximum DC current you can expect given a specific input to output voltage difference.

If you had 30V in, and 12V out, then that's a difference of 18V. You could expect a maximum DC current of about 4 amperes.

That is assuming you keep the LM338 cool. Dropping 18V at 4 amperes means the LM338 will have to dissipate 72 watts of power.

You're looking at a massive chunk of metal if you want to operate it in those conditions for very long.

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  • \$\begingroup\$ Thanks for that answer!. Yeah i guess i need very big metal cooler for it :). \$\endgroup\$
    – mathias
    Mar 5, 2021 at 20:28
  • \$\begingroup\$ Will make some experiments with it , and check what i can get :) \$\endgroup\$
    – mathias
    Mar 5, 2021 at 20:29

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