-1
\$\begingroup\$

This question already has an answer here:

I used the LM337 negative regulator to regulate a +12V from a PC power supply. And connected the GND to input and the +12 to ground of the circuit, and i got it work (i mean it's regulating the voltage) but the problem is it's get very hot even with heat sink connected to it. Is that normal because i didn't keep it on for a long time.
It used to regulate 12 volts to 3.2 volts.

\$\endgroup\$

marked as duplicate by m.Alin, PeterJ, Majenko, Dave Tweed Jun 18 '14 at 11:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ How much current? \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 18 '14 at 5:16
  • \$\begingroup\$ 700 mA that used to derive a RGB LED. \$\endgroup\$ – user45798 Jun 18 '14 at 5:18
  • \$\begingroup\$ LM337 is a linear regulator, so (12V - 3.2V) * 0.7A = 6 watts of power must be dissapated. Did you use heat sink grease for thermal connection between LM337 and its heatsink? (I assume TO-220 package) If heat sink is just bolted to LM337 there is not always enough thermal contact. \$\endgroup\$ – MarkU Jun 18 '14 at 5:33
  • \$\begingroup\$ Thank you for answers. So it is normal to get that much of heat ? \$\endgroup\$ – user45798 Jun 18 '14 at 5:54
  • \$\begingroup\$ You'll need to do the thermal calculations to be certain, but 6W will generate a fair amount of heat regardless. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 18 '14 at 6:12
1
\$\begingroup\$

Let's do the math:

\$(12\text{V} − 3.2\text{V}) ⋅ 700\text{mA} = 6.16 \text{W}\$

That's a lot of power, being wasted directly as heat. Consider using a switching regulator instead. Or an actual LED driver (which will be a switching regulator under the hood).

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.