DC regulator to 5v from a 32v source?

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Is there such a DC regulator/converter or a way to get a 32v source down to a 5v to power LEDs and another low powered circuit in a simple manner?

I've seen 5v regulators but only went up to as far as 20v source. I'm sure there is a simpler way that I'm just not seeing... maybe using two regulators?

Thanks...

:idea:
 
it's called a buck converter

in telecom systems, they work off a 48v bus which is brought down to 5V -- these are pretty common. A buck switching converter will operate with very good efficiency (compared to a linear regulator which basically just burns watts). Don't forget that in powering LED's you can put several in series (but not in parallel unless each has its own current limiting resistor.)

The best and easiest explanations can be found on National Semi's website -- www.national.com --

if pressed, you can take a linear regulator and convert it into a switching regulator, btw.
 
A simple adjustable regulator will do that (LM317, etc). the voltage limit on the 317 is 35 volts difference between input and output. You are within that, so you are golden.

If you wanted to use a 317 on a higher input voltage, you could use a pre-regulating pass transistor referenced to the 317 output.

This isn't all that efficient though. for every watt you use out the 5 volt side of the 317, you'll be burning up more than 6 watts in heat from the regulator.


Sheldon
 
Yes definately a cheap transformer... I found a 5/12v .75A power supply at a surplus store which I think was used in an external computer drive enclosure which might work! Thanks for all the suggestions!!

.... Come to think of it.... I have an old external drive enclosure somewhere around too.....
 
You could use a zener regulator if you're only needing some tens of mA.

Get a zener that's good for at least twice the current you want to draw and the exact voltage you want to regulate (5V in this case). Connect the zener as the bottom half of a voltage divider with the top half supplying at least the current through the zener you're expecting to draw.

Let's say you've got a 32V minimum input, want a 5V output, and want to draw around 50mA:

Voltage drop across the zener is 5V, so you want to bias it with 50mA for load plus 10mA min for the zener so 60mA across 32-5V=27V, which is 450 ohms.

Assuming maximum voltage of 40V, resistor dissipation is (40V-5V)^2/R = 2.7W, current is 40V-5V/450 = 78mA.
At no load, zener dissipation is maximum at 5V*78mA = 390mW.
 
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