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I've built a LED PWM controller according to the below schematic and the voltage regulator is getting extremely hot (I can't touch the heatsink).

I've measured input and output voltages and currents and these are:
Vin = ~24 V
Vout = ~4.85 V
Iin = 80-95 mA
Iout = 80-95 mA

If my regulator is indeed L78S05 (because the package says UL75S05, but can't find any datasheet for this) then it should handle up to 2A easily and up to 35 V on input. Also if I understand correctly, it reduces the voltage by dissipating excess power as heat, so how hot would it get if it would be supplied by 35 V and 2A?

Unless there is something wrong with my construction and it shouldn't get so hot right now. If so, could you help me find what is wrong?

wiring diagram

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    \$\begingroup\$ Dropping 24V down to 5V with linear regulator is extremely inefficient. You can feel this inefficiency. It can handle what it says it can handle only if properly cooled. Your heatsink is probably not sufficient. \$\endgroup\$
    – Eugene Sh.
    Jul 26, 2019 at 14:13
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    \$\begingroup\$ 24v - 5v = 19v being dropped by the 7805. 19v * 0.1A = 1.9W... way too much power dissipation for this device. Consider a switcher instead, which is much more efficient and will not get as hot. \$\endgroup\$
    – rdtsc
    Jul 26, 2019 at 14:20
  • \$\begingroup\$ So I assume if I would first drop voltage to 12 V with one regulator and then with second regulator to 5V they would not get so hot, but overall temperature inside the case would be the same, right? \$\endgroup\$
    – Marek
    Jul 26, 2019 at 14:42
  • \$\begingroup\$ @rdtsc I see that these convertes has no way to attach heatsink, can I be sure that they won't get hot and melt? \$\endgroup\$
    – Marek
    Jul 26, 2019 at 14:45
  • \$\begingroup\$ @Marek I would not speculate about temperature in the case, as it has more affecting factors. But yes, this might help to manage the temperature on each regulator. \$\endgroup\$
    – Eugene Sh.
    Jul 26, 2019 at 14:59

1 Answer 1

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Check your heat sink - it should have a thermal resistance value in degrees Celsius per watt. You have a voltage drop of 24-4.85 = 19.15 volts. At 95 mA, you are dissipating 0.095 x 19.15 or 1.8 watts. Your part has a thermal resistance of 50 degrees C/W, so the temperature rise would be 90 degrees C above ambient room temperature, or about 115 C (definitely too hot to touch!). A typical TO-220 heat sink is around 24 degrees C/W, which would reduce the temperature rise to 45 C, so the heat sink temperature would be around 70 C.

The thermal resistance junction-to-case is 5 degrees C/W for your part, so at 1.8 W the junction temperature is 9 degrees higher than your heat sink temperature. The part is rated at 150 degrees operating, so it should work, but you will be depleting your battery much faster than if you used a switcher as @Eugenesh suggests.

If you try to run at 35 volts and 2 amps, your voltage drop is 30.15 volts so you would be dissipating 60W. Since the junction-to-case thermal resistance is 5 degrees per watt, the junction would have to rise by 300 Celsius even if the heat sink was maintained at room temperature. So although the part may be good to 2 amps in one application with a small voltage drop, and may also operate at 35 volts in another application with low current, these can not occur together.

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  • \$\begingroup\$ Thanks! I just realized that NodeMCU can be powered with max 10 V, so I think I will either go with DC-DC converter to 9V or two regulators first to 18 V (or 12 V as it will solve another of my issues) and second to 10 V. \$\endgroup\$
    – Marek
    Jul 26, 2019 at 16:00
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    \$\begingroup\$ In your first paragraph you say the thermal resistance is 50 C/W and then talk about adding a 24 C/W heatsink. You are mixing up the \$\Theta_{JA}\$ and \$\Theta_{JC}\$ without being clear. Then in the second paragraph you state a new value without clarification. I know what you mean but the OP probably does not. \$\endgroup\$
    – Elliot Alderson
    Jul 26, 2019 at 16:25

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