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I have this cheap voltage regulator (DIY kit). After purchase I have seen that it requires a huge heat-sink AND a cooler.

The input is 24V AC and the output will be around 12V DC and I will be used to power a 20W audio amplifier.

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My question is: how this regulator compares with a modern "switching" regulator which requires a much lower heatsink and no cooler.
Do I want to use the above regulator or buy a "switching" one?

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  • \$\begingroup\$ This answer has related information. (skip the first few paragraphs unless you're interested in voltage dividers) :) \$\endgroup\$ – bitsmack Jul 27 '17 at 21:18
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    \$\begingroup\$ Possible duplicate of Best design choice: linear regulator or switch converter \$\endgroup\$ – Dmitry Grigoryev Aug 4 '17 at 11:21
  • \$\begingroup\$ You should mention your application, otherwise no one can correctly recommend a solution. \$\endgroup\$ – horta Aug 4 '17 at 13:50
  • \$\begingroup\$ Once again, either may be used for general purpose which is why this question has so many close votes and also why both types are still in existence. At least you've got some good answers below. \$\endgroup\$ – horta Aug 5 '17 at 19:11
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Use of the linear regulator versus a switching regulator will entail a number of considerations:

  1. The linear will produce a cleaner (less noise) output voltage.
  2. The switcher will be much smaller for the same output capabilities.
  3. The linear will generate lots of heat to keep you warm in winter.
  4. The switcher will save you money in saved mains electrical energy usage.
  5. The linear will typically provide better response time to stepped output load change.
  6. The linear with a fan will help drown out other background noise.
  7. ...
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  • \$\begingroup\$ OK. It looks like I want a "switching" regulator. Many thanks! \$\endgroup\$ – Ultralisk Jul 27 '17 at 21:21
  • \$\begingroup\$ Don't be too sure you need a switching regulator! The heat generation is rarely at peak values, heatsink size doesn't imply that using 25 mA to trickle-charge a battery is going to warm your room. \$\endgroup\$ – Whit3rd Jul 29 '17 at 20:19
  • \$\begingroup\$ The linear regulator (particularly with NPN pass elements as above) will very probably be much easier to compensate. \$\endgroup\$ – Peter Smith Aug 4 '17 at 12:50
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This is a linear regulator, which means that the output current is almost equal to the input current, so the efficiency is roughly equal to output voltage / input voltage. The power loss is roughly the voltage drop times the load current, and all of this goes into heat.

Modern switching regulators in general have an efficiency that is relatively independent of the difference between input and output.

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  • \$\begingroup\$ "The power loss is roughly the voltage drop times the load current" --- That's my question. This seems to be a huge disadvantage over switching. So, I have any reason to use this linear regulator, or better go buy directly a switching regulator and ditch this one? \$\endgroup\$ – Ultralisk Jul 27 '17 at 20:25
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    \$\begingroup\$ Linear regulators still have their place because of their simplicity, in low power applications, and because they generate no switching noise. Otherwise you are right on target. So you would not use a high power linear unless you are trying to be very quiet. \$\endgroup\$ – John Birckhead Jul 27 '17 at 20:27
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    \$\begingroup\$ Linear supplies typically provide a "cleaner" output with much less ripple. They're great for sensitive analog circuitry, where efficiency is not the driving factor. So, as always, it depends... \$\endgroup\$ – Chris Knudsen Jul 27 '17 at 20:28
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    \$\begingroup\$ if you want low noise and low heat, you can use both; the switching to drop most of the surplus voltage, and the linear after that to provide a nice clean output; almost as good as pure-linear for most apps \$\endgroup\$ – dandavis Jul 27 '17 at 22:06

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