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# How can convert a 12v power supply to 3v? Answered

I am working on a laser spirograph project, I am using 12v dc fans on the inside but also want to power a 3v laser diode off the same source. Does anyone know how I can allow the full 12v to be used on the fans but drop down the voltage for the laser?

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Here is an instructable using a DC to Dc converter:

https://www.instructables.com/id/93-efficient-DC-to-DC-Converter/

The data sheet says the input voltage can be 4.5 to 14 volts. The output can be adjusted from one half volt to 6 volts. Just make sure you 12 volts does not exceed 14 and this device will work fine. Murata makes several different devices for different current / wattage requirements;

http://www.futureelectronics.com/en/Search.aspx?dsNav=Ntk:PlainTextSearch%7coktr%7c3%7c,Ny:True,Nea:True

PC power supply? Simplify everything...?
12v 5v & 3v.

Steve already gave you a \$3 circuit for a 3V
constant current source..  -.-.  .  -.  --.

+1 Most PC power supplies can support up to about 20A on there 3.3V rail. You can get a cheap PSU for about \$40 that will more then cover your needs.

You can probably pick one up for free. I find computers tossed out at the curb several times a year that still have working power supplies and more.

use a 3v voltage regulator, which will regulate the voltage in 3v, great for your laser diode

There are several ways;

Simplest and least robust is just using a resistor or resistor voltage divider; given a current flow you can calculate the voltage drop of a given resistor across a voltage at a given current - its not ideal because you burn off 9 volts of energy to have 3 usable left over. Benefit is it's ridiculously simple and uses a \$0.01 part.

Next best is using a linear voltage regulator, either variable or fixed (like a 7803) - you input voltage and ground, and it outputs a rock solid 3 volts, so long as you feed it about 4.2 or more volts, up to about (something like) 32. The higher the input voltage, the less efficient it is, because it works just like the resistor - burning off extra voltage as heat.

For both of the above, the AMOUNT of energy it has to dissipate is proportional to how much current your laser diode draws. If just a few mA, then the regulation only has to burn off a few mA at 9 volts (12 down to 3) - but if you're drawing a few amps, the regulation has to burn off a few amps at 9 volts, which makes for a VERY hot regulator.

Lastly, there are switch-mode power supplies. They actively contribute just the electrons required to make the required voltage (or current) and turn off the rest without ever actually resisting their flow - a much more efficient, but often more expensive option (a few dollars instead of a few dimes or a few pennies).

For all of the above, search '3 volt regulator' and you'll come up with lots of options. the 7803 is SUPER common and can be scavenged from many electronics that have a microcontroller or processor on them (sometimes 3.3 volts, but who's counting?)

Some HORRIBLY confused units in there..... "burning off mA" ?????

agreed, 'burning off volts at x mA'. More current = more current has to pass = more wasted heat.

Laser diodes need constant CURRENTS, just like LEDs. Use a circuit like this one.