Electrical Engineers (or enthusiasts)

[nintendofreak]

well, here is the deal.

Ive got this 12v power source, and i wanna use it to power something that uses 3v (2x AA)

Now, how would I do that?

I know V=IR and

V=3
I= ???
R= WHAT I AM SOLVING FOR

So... whats I?

Yah, i know dumb question, but makes me wonder...

Explanations would be appreciated

Thankee

 

[FAST6191]

The back of the supply will usually have a wattage/power rating given in watts.

Power (watts) = current (i in amps) x voltage (v in volts).

of course you can also rearrange v=ir and throw it into p=iv to give p=i^2.r and p= v^2 / R

Be careful if you do step down the voltage as the current that may end up flying through the device may blow it up (which if it is 3v will likely not be that much.

 

[nintendofreak]

Got it, easier than I thought.

3V = I (584R)

I= .0051369863

So...

Need 1,800 kOhm 1/8 watt resistor

Thanks!

 

[Caoimhin]

Maybe some induction transformer? I don't really know ^^"

 

[noamkot]

You need to use a voltage regulator or step down converter.
If you actually use a 3V voltage regulator you must make sure that you don't pass the power limit of the regulator.

 

[nintendofreak]

You need to use a voltage regulator or step down converter.
If you actually use a 3V voltage regulator you must make sure that you don't pass the power limit of the regulator.

Yes, that would be the easy way... But i prefer to home make things, it would cost me like 2 dollars to make it myself

 

[outphase]

Another way is to take the Thevenin equivalent of the load circuit. Then figure out a way to drop the voltage down with a resistor (or other impedances)

 

[Veho]

You can use the 12V to power a transistor amplifier, and set the output voltage of the amplifier to 3 V. That way you get the least losses. Any other voltage divider will have dissipation larger than the power output you need for the 3V device you're using, i.e. you'll still be using the entire power output of the source, instead of one fourth (or less) that you need.