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I'm building a project that needs both 5V and 12V rails which will draw about 1A on the 12V and 0.5A on the 5V. I was thinking of using a 7805 to produce the required 5V but this could generate some significant heat. I am not so much asking if it is acceptable to use a 7805 (I just need a mini heatsink), but rather: are there any ways of doing it cooler? Any tips would be great.

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3 Answers

up vote 12 down vote accepted

Instead of using a linear regulator, which has relatively low efficiency (~50%), use a switching DC-DC converter (efficiency around 80-90%). A decent one to start with is National's "simple switcher." It looks like the LMZ12001 would do the job-- up to 1 A at 5 V, and it accepts input voltages from 0.8-20 V.

http://www.national.com/pf/LM/LMZ12001.html

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switcher circuit design can be difficult, if your not careful with controlling the switching current loops you can induce noise throughout your design not to mention generate large about of EMI. Also keep in mind that the output of the switcher will have a much higher ripple voltage than the output of LDOs. If you have anything noise sensitive you will need to plan for at least an LC filter on the output. – Mark Apr 23 '10 at 0:50
The cheapest switchers I know of are the old (but still in production) MC34063. Just be sure to use use very fast diodes (preferably Schottky). – jpc Apr 23 '10 at 6:50
Yeah, the National simple switchers aren't cheap, but they're pretty hard to screw up. The reason that they're expensive is that they have some of the passives integrated into the chip package. I think it's a worthwhile tradeoff if you're just trying to get something working. They'd be a bad choice in a production design. – pingswept Apr 24 '10 at 16:20
@jpc - definitely Schottky! Not only because of their speed, but also because of the lower voltage drop. Even more relevant for lower \$V_{OUT}\$. – stevenvh Jun 18 '11 at 10:52
up vote 5 down vote

One of the main issues with regulators like the LM7805 (and linear regulators in general) when you step down from large voltages is that a lot of the energy is dissipated as heat. If you are limited to passive cooling only and you have some significant power draw from your load, you might risk thermal overload. Fear not, because if you are willing to put up with a bit more complexity and cost there is another option: Switching regulators, specifically in this case buck converters, which work off of a completely different principle. Depending on what your other requirements are, like noise, stability, EMI considerations, you might look into some of the highly integrated switching regulator modules offered by TI, like the PTN78000W family. That part in particular is nice because of the adjustable output, so you can produce a variety of supply rails off of one part.

Cheers!

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up vote 4 down vote

Switching regulators (aka DC/DC converters) can step down 12V to 5V step without generating too much heat. Since you only need 2.5W on your 5V, don't think you need a heatsink too.

Lineage has some good DC/DC regulators: http://www.lineagepower.com/category.aspx?NID=5e1812f4-fb02-49f4-ba31-6e312010f40d

I'm assuming here that you need a non-isolated converter ie. your 5V does not need be electrically isolated with a transformer from the 12V. Non-isolated dc/dc are cheaper and smaller.

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