How to power a single LED from a 12v supply?

I just googled "12V led light" without the quotes and got "About 564,000 results"

Good Luck! Rich

not knowing what type of LED you plan on using... I'll give you some rough numbers.

470 ohm R at .5 watt or better (common value)..

attached it in series to one of the legs, and attach to 12 volt source.

if the LED does not light, reverse connections from the 12 volt source... The + should be leading into the anode side of the LED.

Bye..

And did you notice that all those results were referring to readymade LED light fittings, "bulbs", or other assemblies that are unrelated to my query?

Pete

Hmm, so simple :-)

Presumably, if 2v is being dropped across the LED and the remaining 10v across the resistor, and they're both passing the same current, then five times the energy is being wasted in the resistor. I guess that can't be helped?

Pete

You could use two or three LEDs in series and reduce the current proportionally. That would multiply the electrical efficiency.

For example, one led + 470 ohms uses about 20 mA. Electrical efficiency is 2/12 = 0.16

Two leds + 800 ohms uses 10 mA and makes the same amount of light. Efficiency is 0.33.

But both are pretty small amounts of current. A 40 A-H battery would supply 20 mA for 2000 hours.

John

It's possible to make a very small switching power supply from only a few parts (IC, inductor, two caps) to supply "just enough" voltage for the LED and a small current limiting resistor, if you're really concerned about a few milliamps. However, there are also special ICs designed to drive LEDs - including white LEDs - at a constant current very efficiently.

A quick perusal of Digikey (search for "white led" then choose step-down) found the ZXLD1366 series, for example. 6-60V input,

0-100 mA output, up to 97% efficient.

Sorry. I guess I assumed that by "How to power a single LED from a 12v supply?" that you wanted to power an LED from 12V.

Those LEDs run off 12V, they've got their current regulator built-in.

What's the real problem? Are you looking to light your boat, or are you looking for a construction project?

Thanks, Rich

"Pete Verdon" schreef in bericht news:i8g8hr$kat$ snipped-for-privacy@news.eternal-september.org...

I doubt a simple cheap white low power LED will light much more then just itself. This ones have a forward voltage drop of between 3Vdc and 4Vdc can handle a maximum current of 20mA. For a try out take three of them in series with a 180R or 220R resistor, depending on the forward voltage of the LEDs.

Beware: LEDs can handle only a limited reverse voltage. The LEDs mentioned above usually 5V. So connecting them the wrong way you may blow them. A Schottky diode bridge will prevent this and you can lower the resistor to

180R or 150R.

In the unlikely event that the LEDs produce too much light, replace a LED by an extra series resistor. If you need more light, you'll have to find other (more expensive) LEDs. They draw more current and so produce more light. There are tens or even hundreds of types so you may need some time to find out.

petrus bitbyter

A single LED wants to run at between 1.5V and 2.5V, depending on color (there may be some 3V ones out there, I dunno). This means that if you want to run _one_ LED from 12V you're going to have to either drop a whole lotta volts through a resistor (turning those volts into heat), or you're going to have to run a switching converter (which you probably don't want to undertake).

So:

1: Why not use several LED's in series, enough to make about 6V nominally (check the volts vs. temperature, and go for "works right when it's really, really cold). Use little ones, and run less current -- same light, less current, same voltage, good.

2: Instead of shining a light _up_ to reflect all over from a white flag (or to just plain reflect poorly when things get dirty or wet), why not put a wind vane up there with LEDs along the bottom, shining _down_. That way instead of most of the photons from the light being scattered every which way, most of them will land on deck, meaning that far more of them will land on your retinas for the amount of energy burned off the battery? It'll require a slip ring or some similar cleverness, but it'll be loads more efficient.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

(...)

How about a self-glowing flag made with photo luminescent nylon thread?

Shine a little UV light at it and it will glow for hours.

--Winston

I found out by accident a red laser pointer can darken a piece glowing sew on tape. I have a flashlight that has UV, visible and a pointer. I found the laser makes a dark spot on the tape and I could actually write on it.

It doesn't have anything to do with the topic, just an interesting discovery.

Dan, U.S. Air Force, retired

(...)

Whoa! 'Nothing new under the sun', huh?

An enterprising young engineer could use that to create a demonstration of xerography / laser printing for a deserving outfit like the Children's Discovery Museum.

Cool!

--Winston

Hi,

You can try a constant current generator with two transistors :

In this example, you are sure to get around 20mA between 8 and 15V as input voltage. Of course, this is an example to modify according to your LED power.

Cdlt, F

"Pete Verdon" a écrit dans le message de news: i8g8hr$kat$ snipped-for-privacy@news.eternal-september.org...

----- Original Message ----- From: "haaaTchoum" Newsgroups: sci.electronics.basics Sent: Wednesday, October 06, 2010 11:14 PM Subject: Re: How to power a single LED from a 12v supply?

Excellent advice... Assuming the OP to be able to understand the schematic and build the circuit. Using a 12V battery, you can take two LEDS in series. Twice the light for the same energy cost. Even three LEDs may be possible depending on the forward voltage of the LEDs.

petrus bitbyter

Choose a high efficiency LED for lots of light at little current.

A single led isn't usually a reason to get all sophisticated with drive circuitry since you want it fairly dim... and even a 9 LED pocket torch will run for >20 hours on a few AAA batteries.

I'd mount a single 20 milliamp LED, wire it in series with two resistors, one to limit the current to 20 milliamps and one variable wired as a rheostat to adjust the current between 100% and 10%.

On a 12V battery that would be ~470/500 ohms fixed and 5,000 ohms variable in series with the LED. I'd use a 5 watt wire wound variable.

Choose a color to match the wind telltale or use white.

Use a narrow angle LED to keep the light only on the flag. 6 degree Led aimed up at the flag shouldn't be too visible to others, and you can always shield the light so it only illuminates the flag.

I can certainly build the circuit straight from the diagram, although I don't necessarily understand it well enough to make substitutions (eg a different transistor due to availability, or different resistor values for different LEDs)

This does look like a very interesting possibility.

Pete

That's a nice, logical solution to this specific problem, but not really appropriate to the boat generally. This is a traditional-style gaff-rigged boat with a wooden mast, and a wind-vane would look out of place compared to a traditional burgee on a bamboo staff.

Thanks too for all the other suggestions posted.

Pete

It's a current regulator circuit. Here's a brief explanation: R3 (33 ohms in the circuit) sets the current that will go through the LED, regardless (within reason) of the voltage source.

R2 applies + to the base of Q1 making Q1 conduct and allowing the LED to draw current through R3 and Q1.

When the current through R3 causes a voltage drop across R3 that equals about .6 to .7 volts, Q2 conducts and creates a voltage drop across R2, lowering the drive to the base of Q1, which in turn limits the amount of current Q1 can conduct.

Since we know R3 is 33 ohms, and that Vbe for Q2 is around .6 to .7 volts, we can figure the amount of current that will be allowed by using ohms law: I = E/R so I = .7/33 or about 21 mA. If we figure based on .6 volts, I = .6/33 or about 18 mA

Provided we supply the circuit with a reasonable voltage, the current through R3 (and therefore the LED) will be constant. Reasonable in this case means a minimum voltage high enough to light the LED in the circuit, and a maximum voltage low enough so that the transistors maximum rating is not exceeded. Your 12 volt supply is fine.

As to substituting parts: for a typical LED, you can use any NPN transistors you have on hand. The typical Vbe will be around .6 to .7 volts. There is nothing critical about R2 - it is chosen to keep Q2's collector current well below maximum. R3 is not critical either, but it is chosen so that the LED maximum current is not exceeded.

Now, if you were to use a high power white LED, you need to select components for that higher power - you can't just use whatever you have on hand - and a heat sink for Q1 may be needed.

There is an even simpler circuit using an LM317 regulator IC:

----- +12 -----in|LM317|out---+ ----- | adj [R] | | +---------+ | [LED] | Gnd --------------------+

The value for resistor R is computed by the formula R = 1.25/I where I is the current you want the LED to draw. Say you use a typical LED and you want the current through it to be set about

20 mA. A 62.5 ohm resistor would provide that, and a standard value of 62 ohms, or 56 ohms or 68 ohms would be close enough, yielding currents of ~ 20.16, 22.32 and 18.38 mA, respectively.

You can also use the LM317 or the two transistor circuit with LEDs in series.

Ed

Well, if you want a floodlight illuminating your burgee, why are you making it into such an ordeal? Just get a 12V LED thingie, and slap it up there.

Or any of thousands more.

Or you can get a plain ol' white LED,

and slap a series resistor on it: the resistor value will, of course, be:

(12V - VLED) / (ILED)

But I can't understand what it is that's making you reject all those wonderful other suggestions - if you want to save power, you could use a simple PWM circuit.

But, if none of these suggestions are good enough for you, then do whatever you wanted to do in the first place, and quit sniveling.

"I want a circuit!" "Suggestion A" "No, that's not it." "suggestion b" "no, that's not it." "suggestion c" "no, that's not it."

Just tell me what answer you want, and I'll be happy to tell you what you want to hear. ;-)

Good Luck! Rich