LED's on a vehicle

I am trying to determine the optimum arrangment for multiple LED's powered by a 12V battery/alternator system on a motor vehicle.

I could mate each LED with a 500 ohm resistor, which is how they came from the HK eBay vendor. This will waste 2/3 of the energy consumed and might create a heat problem if I pot the electronics in epoxy.

I could put three LED's in series with a 150 ohm resistor, but they would significantly brighten and dim as the vehicle's voltage ranged between 12.5 volts (engine off) and 15 volts (alternator charging).

I am leaning toward using one 7808 regulator feeding multiple clusters, with each cluster containing two LED's in series and a 50 ohm resistor. This would put most of the energy loss in the regulator, which I can put in a protected spot with a heat sink. The constant 8V would solve the variable brightness. The regulator, rated at 2.2 amps, could in this fashion illuminate over 200 LED's, which is more than enough.

Any suggestions?

Ken C

Reply to
Ken C
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Why not pulse the LEDs? Using power FETs, this should yield the best possible efficiency. By pulse, I mean full ON or full OFF. None of the losses associated with series resistors or linear voltage regulators.

Reply to
Charles Schuler

How many FETs would that require for 200 LEDs?

I suppose you could put them all in series and hit them with a constant current pulse at 300 volts or so. But if one burned out, it would hard to find.

-BIll

Reply to
Bill Bowden

As I'm sure you know, you really want to limit the current rather than the voltage. You can do this with an adjustable voltage regulator and a resistor.

See:

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Gareth

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Reply to
Gareth

Why not just wire them up, ten at a time, in series? There will be a 1.2 volt drop across each of them which, as I recall but may be wrong about, is tolerable.

R
Reply to
Roger Dewhurst

Because a car's 12V supply isn't actually 12V. It varies widely.

Graham

Reply to
Pooh Bear

Typical red LEDs drop ~ 1.8v. Put 6 of them in series with your 150 ohm resistor, and the string will draw ~28 ma at

15 V, and ~11 mA at 12.5 volts. This will waste about 118 mW, worst case.

You could run 4 in series with 68 ohms and a 7809 regulator to get ~ 26 mA through the LEDs. It will waste 202 mW per string, worst case. Because the car is a hostile environment, you don't want to run the 7809 anywhere near its maximum rating. You need a heat sink and probably a pass transistor to run a decent number of LEDs. Plus you'll need to account for transients ( at a minimum a transorb and capacitors).

All in all, I think the first approach is best. So what if it's dimmer when the engine is off?

Ed

Reply to
ehsjr

Hadn't thought about that. I remember the LM3909.

These are blue LED's with a 3.3V average drop at 20 mA. Could I put a

50 ohm resistor in series with each LED and pulse them with the full 12V on a 5% duty cycle at maybe 10kHz (or whatever looks continuous).

What chip/circuit/kit would accomplish this?

Each LED would be at the end of a 4 ft pair of wires. Would this arrangement disrupt radio reception?

Ken C

Reply to
Ken C

Right now I am stuck using Google Groups, which doesn't have a nonproportional font for posting ASCII diagrams, so bear with me while I describe a simple constant current circuit. This uses a npn transistor at the low end of your led string. If you want the transistor at the top of the string, just use pnp and flip everything.

Draw a npn transistor. Draw a common red led from transistor base to ground. Draw a resistor from transistor base to battery positive. Draw a resistor from emitter to ground. Draw your diode string between battery positive and the collector. The emitter resistor will have a fixed voltage impressed on it, which is the difference between the led voltage and the base-emitter voltage of the transistor. That will be about one volt, which makes calcs easy. Tempcos of the "base" led and b/e junction will cancel, or nearly so. Collector current, which is the led string current, will be equal to one volt divided by the value of the emitter resistor. Put enough leds in the string to minimize the voltage drop across the transistor. This is the most efficient way I can think of to run led's, if you are willing to wire one transistor for about every six or seven led's you run.

Reply to
kell

3 of the blues in series should work. Set the peak current with a current regulator (check the manufacturer's specs for recommended peak currents) and then set the duty cycle for the desired brightness (the maximum safe duty cycle will depend on the peak current that you choose). If you use a series resistor as your current regulator, the current will be temperature dependent.
Reply to
Charles Schuler

Yes, it does, under "options" or something like that.

And if you draft something in, say, notepad, you can paste it in any old font you want - it's just plain old ascii, so we can view it however we want to. :-)

Good Luck! Rich

Reply to
Rich Grise

I used to post ASCII under Google. Can't do that any more.

Reply to
kell

I made another post before I figured out how to draw this in ASCII (thanks to Rich Grise for the suggestion about notepad). View in a fixed-width font.

enough led's in series here to drop 10 or 11 volts +--->|--->|--->|-------+ | | | |/c B+---+--------R1--+-------| NPN | |\\e | | red led | | R2 | | | | | | +----+----+ | | ground

Size R1 to illuminate the red led. Maybe 680 ohms. This red led just serves as a voltage reference. But it lights up too :) You can use R2, which has a fixed voltage of approximately 1 volt across it, to trim the current in the led string. If you want .02 amps, R2 would be in the 50 ohm ballpark.

Reply to
kell

Thanks for the mention here - This is a very, very good way to start a Monday morning. :-)

Looks good to me!

But then again, as has probably been noted, I'm easy. ;-)

Cheers! Rich

Reply to
Rich Grise

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