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Guest (visitor) (OP)
23 Jun 02 12:35
Hi,
  I am looking for suggestions on how to build a DC voltage converter.  Specs as stated above, the 12V source can draw up to 15A, and the converter must provide at least 2.7A @ 3.3V.  I have found a couple of regulators that take 12 to 3.3 except they usually max out at 1 or so amps.  should i just use several in parallel or where can I get a bigger regulator?
Lowest possible cost is the main concern, not size/appearance/efficiency.  I can add heatsink/fans if required.  Can I use a bunch of zeners instead?  Any benefits to using a switching regulator like this:
http://www.national.com/pf/LM/LM2655.html
Thanks, -Mike
buzz41 (Electrical)
23 Jun 02 15:11
Using a linear regulator, resistors, zeners, means you will have to figure out how you are going to dissipate ~25 watts. That might range from very easy to impossible, depending on the application.

There are 3 amp linear regs around. Using the right sized power resistor to feed it will put most of the heat at the resistor. About 3 ohm and 25+ watt rating. The reg, then, needs to handle 7 watts.

Overall, switching regulation will probably be more practical. I noticed an IC recently designed for your nearly exact application. It was similar to your IC link, needed only the inductor and maybe the schottky.  
 
elektron (Electrical)
24 Jun 02 9:43
Hi mjbw,

It looks like you have a nice, stiff source of current.  I would consider the National Semi LM2676S-3.3 for a direct conversion.  This gives you 3 amperes capability.  Go to the National Semiconductor website and then into Webench, power.  Input your input voltage minimum and maximum, your output voltage and current requirements and Webench will give you all the part values and part numbers of the 7 other components needed.  Run the electrical and thermal simulations and be sure your design will not heat up excessively.  If it appears to do so, look up the recommended heat sink theat Webench gives you on the heat sink manufacturer's website.  Then you will have to choose a sink that has a lower thermal resistance.

Another approach if you need quiet power is to regulate down to 5V with a switcher, then use a linear regulator for the rest.

If you try to use just a linear regulator, your heat sink will have to be enormous compared to the switcher.  This is because of the low efficiency (27.5%).  The switcher has a higher part count, but gives you an efficiency of 86%.  Thus you only have to have a heat sink sufficient to dissipate the 14% loss versus a loss of 72.5%.

National will even send you free samples of the chips, and a PC board with all parts is available for around $30 (including overnight shipping.)

Just remember that you can always go higher in current, just don't get silly with it (10A regulator for 100mA load).
Lewish (Electrical)
24 Jun 02 11:53
Take a look at this:
http://www.linear.com/prod/datasheet?datasheet=248&product_family=power
Because the "switch" is external to the regulator IC, you can size it for any amperage that you need.  They have a version just for 3.3V.  Finding a suitable inductor may be a challenge, but is possible.

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