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Rezarf
03-03-2011, 09:46 PM
So if I want to run a set of LED lights in my trailer tent that use a battery pack of 3 AA batteries, and if I want to run 2-3 sets of these lights... is it possible to convert the little battery boxes with the AA's in them to 12v through some kind of converter?

I really have no clue on this one, any help is mucho appreciated.

TIMZTOY
03-03-2011, 10:11 PM
i dont think that you can convert the led's but you could possible make or convert the supply.

why not get 12v led's ? and just run a marine battery in the trailer.

Inukshuk
03-03-2011, 10:24 PM
AA are 1.5 volts, so three are 4.5 volts.

Car cell chargers put out 5v. Car USB sources are 5v too. You can also just get a 12 to 5v converter at radio shack. Check the milliamp ratings for your loads and supplies.

leiniesred
03-04-2011, 09:50 AM
http://www.radioshack.com/product/index.jsp?productId=2062599

7805 will give you 5 volts at 1 amp for about a buck and a half. Plus it has all kinds of built in protection. over current, thermal...really hard to kill one of these little regulators.

Might work just fine right there at 5 volts especially with nice long power cables to drop a little more voltage, but it might be a bit too much voltage for your LED array. Maybe put a resistor in line with with the LED array to drop the voltage down to the original 4.5 volts. Can't tell you which resistor until we know how much power the LED array is pullin'.




I like the idea of finding a 12V cell phone charger that puts out 4.5 volts too. Simplest solution.

Here is an off-the-shelf solution, but it is rather expensive at $23 if you ask me:
http://www.radioshack.com/product/index.jsp?productId=3896270


Here is a good paper on how to measure/ calculate the value and wattage of the resistor you'll want to get the LED array back down to 4.5 volts. You can skip the transistor part and focus on the handy, well explained formulas for "R."

http://picprojects.org.uk/projects/inf/drivingLEDs.pdf

DaveInDenver
03-04-2011, 10:23 AM
The linear regulator option is fine, but I suspect all you need to think about how they are wired together and maybe add a resistor.

Do you have a manual or schematic or something for the LED lamp you are planning to use?




LEDs are just diodes and have a specific voltage drop. What you have to do is limit the current.

http://www.kpsec.freeuk.com/images/ledres.gif

The Vs can be just about any value, the LED is just going to drop whatever it's designed to drop. Let's say for sake of argument it's 1V and you are using a 3V supply. They also have a maximum current rating (although you probably will run it below that, they should have a suggested operating current). Let's say here that current is 10mA.

To find the value of the R in this circuit, you take VS minus VL and divide by the operating current. That means in this simple circuit the resistor should be R = (3 - 1) / 0.010 = 200 ohms. It would have to be sized to handle 20mW (a typical carbon resistor from Radio Shack is 1/4 watt, so 250mW). That's P = I^2 x R = (0.010 * 0.010) x 200 = 0.0001 * 200 = 0.020W = 20mW.

When you do multiple LEDs, you ALWAYS connect in series, like this:

http://www.kpsec.freeuk.com/images/ledser.gif

In that case each LED drops it's specified voltage and the current limiting resistor is found by the series total voltage drops, VL.

So assuming that your current assembly is designed with say three 20mA LEDs that each drop 0.7V (ignore the 3 x 2V on the pic) and for 4.5V, it would have a 120 ohm current limiting resistor.

4.5 - ( 3 x 0.7) / 0.020 = 120

It's probably just as simple as redesigning for 12V.

12 - 2.1 / 0.020 = 480 ohms

So you can either just put a 480 - 120 = 360 ohm resistor in series or replace the 120 with a 480.

DaveInDenver
03-04-2011, 10:31 AM
I should mention that my values are ideal, not real. Resistors don't come in infinite range of values and so you'll find values close to the ones you calculate. Also plan for 13.5~14V, not 12V, unless you disconnect them during charging. If they'll /never/ be on under charge, then ~12V is fine. But sure as you assume that you'll find yourself clicking them on looking for a jacket one night pulling into camp late, trying to find a flat spot to park.

More info on the lamps you're using would be helpful... I'm just assuming they are simple lights. There are much more complex ways to drive LEDs designed to maximize illumination, minimize power. My last post assumes they are cheap, thus not fancy, fixtures. If they do have PWM drivers, then adding a resistor (1) may not work or (b) might even be unnecessary (it might work over a wide range of input voltage). Stephen's regulator idea would also be the simplest option if a resistor doesn't work. That's 3 or 4 parts on a perf board, shoved into a little plastic box.

AxleIke
03-04-2011, 11:10 AM
Sure, but you can run a set of resistors in series and parallel to come up with almost any resistance you need.

DaveInDenver
03-04-2011, 11:14 AM
Sure, but you can run a set of resistors in series and parallel to come up with almost any resistance you need.
It's not necessary to get exact values for this. If the calc says 324.67 ohms, pick 330 and the difference in illumination is going to be undetectable to your eyes. The tolerance on a carbon resistor is 5%, so it might be 315 to 345 ohms anyway, 'pecially over temps. About the only time we bother stacking to get exact values is with caps. But mostly multiple parts is avoided because it makes reliability and parts stress de-ratings a bigger pain in the rear.

TIMZTOY
03-04-2011, 11:18 AM
couldnt he also just get more led's in serries to make up to 12v ?

DaveInDenver
03-04-2011, 11:19 AM
couldnt he also just get more led's in serries to make up to 12v ?
Yup, 4.5 x 3 = 13.5, interestingly enough. That's why I wanted to see how they're wired and their specs. It could be just that simple... I was just assuming that he might not want that many LEDs for space or brightness reasons.

Rezarf
03-05-2011, 09:50 AM
Here is what I am trying to wire up into my 12v system...

http://i.ebayimg.com/21/!Bb76WewCGk~$(KGrHqMH-EUEquRH5Vo)BKyRm4RdmQ~~_12.JPG