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

| The problem with that regulator is it doesn't do much for efficiency. | You still waste the same power using the regulator or just a single | resistor. If he uses a couple white LEDs in series at 3.5 volts and a | 270 ohm resistor, he gets 18mA at 12 volts and 28 mA at 14.5 volts. | Should be in spec for small 20000 mcd LEDs. | | -Bill | The advantage of the resistor only solution is it's symplicity. The disadvantage is the current variation of a 30%, which will vary the light yield accordingly.

Using the proposed current source you will have a constant current, so constant light. You can at least use two LEDs in series for more light. There is not much to earn on the power efficiency side. Even if you go using a switcher, you will need an extremely efficient one to compensate for the power used by that switcher itself.

That picture will change however when ordinary 20-50mA LEDs do not produce enough light and you need to use more powerfull LEDs that require >100mA. But unless ones interested in the electronics, you'd better buy a 12V LED (car)lamp with the electronics build in.

petrus bitbyter

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I don't think I can notice a 30% change in light intensity. The eyes are sort of logarithmic aren't they? Maybe 2X to notice anything? There's a legend that controlling a LED with very short pulses increase the perceived brightness, thus saving power.

Any truth to that?

-Bill

But I don't want a floodlight, I want one or two plain oldfashioned LEDs, probably basic red, the kind of thing that gets used as panel indicators. Most of the time I can see the flag by the light of the moon and stars, but if it's overcast and completely inky black, as it was the night I came up with the idea, then I just need the faintest glow to see the flag without affecting my night vision or claiming to be a motorboat (which I would be by showing a big white light at the masthead). The ready-made 12v units you keep pointing to, intended as car bulb replacements and the like, are completely the wrong thing.

You say "of course" - to me it seemed likely but not blindingly obvious, what if there was some factor I hadn't considered, so I posted on sci.electronic.**BASICS** to ask the question. Everyone else in this thread has kindly tried to answer it.

I have "rejected" precisely two suggestions - your 12v floodlight and inspection-lamp products, because they don't come anywhere near the goal of providing a discreet glimmer of light from a tiny unit, and Tim's idea of a self illuminating vane. The vane idea was reasonable given the requirements, even if not quite what I was looking for, so I hope I turned that idea down politely.

The other suggestions in this thread have been useful and interesting, and I will no doubt be using one (or a combination) of them when I come to build the thing. I'm grateful for them.

Sounds like exactly what I need. But simple to you is not simple to me. If you can point me to a diagram of such a circuit, I could probably solder it. That way this part of the conversation might even have been useful :-)

Pete

berichtnews: snipped-for-privacy@l8g2000yql.googlegroups.com...

You can't see the difference between 40W, 60W, 75W, and 100W light bulbs?

Urban legend. The eyes integrate over significant time.

No.

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Yes, I can tell the difference from 40 watts to 75 watts, but not from

60 watts to 75. I use both in my table lamp and keep forgetting which is installed. I have to study the markings to figure out which is which. But my eyes aren't as good nowadays.

-Bill

Pete Verdon Inscribed thus:

What about the "Joule Thief" circuit ! Originally used to extract the last dregs of energy from a 1.5v dry cell.

One I made produced a reasonable amount of light drawing less than 5ma. I'm sure that it could easily be made to run from a 12v source with a couple of extra components.

--
Best Regards:
                Baron.

pulses increase the visibility, but not the brightness.

--
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| | I don't think I can notice a 30% change in light intensity. The eyes | are sort of logarithmic aren't they? Maybe 2X to notice anything? | There's a legend that controlling a LED with very short pulses | increase the perceived brightness, thus saving power. | |Any truth to that? | |-Bill | I don't know about the eyes exactly but the light yield of a LED is not proportional to the current through it. There is some optimum between light yield and life time. A, let's say 10%, rise of the current above the normal operating current will provide only little more light - though your eyes will register the difference - but will shorten the LEDs lifetime considerable. So for serious design you'll need to know the manufacturer and obtain the datasheet for the LED you want to use.

If you want to dim a LED, PWM is told to do a better job then linear current control. More light for the same average (and effective) current and better control.

petrus bitbyter

If the pulse rate is fast enough for the LED to appear continuously on, then puling does not increase the efficiency of human vision.

There is a legend that it does - but when pulsing too fast to visibly flicker increases visual efficiency, then the efficiency gain is in the LED due to a nonlinearity of the LED.

Meanwhile, most white low power LEDs have peak efficiency at 1.6-6 milliamps.

I discuss this at more length in:

--
 - Don Klipstein (don@misty.com)

OK. Get two red LEDs. Note their forward voltage and current rating.

Wire the circuit like this:

k k +12V -----[R]------[LED]------[LED]-----12V. Return (i.e., the negative of your battery.

Now, if you want 10 mA through your LEDs (which is half their typical rated current, a nice compromise), and each LED has a 1.2V forward drop (for this example) then there's 12 - 2.4 volts left that you need to drop through your resistor. 12 - 2.4 = 9.6.

R = E/I, so 9.6 / 0.010 = 9600 ohms.

So, if you've got bog-standard Rat Shack-type red LEDs, typically

1.2V forward at 10 ma, you'd use a 1K resistor; 9.6V * 0.010A = .096 watts, so a 1 watt resistor should be fine.

Two LEDs; you'll have to figure out how to mount them; one 1K, 1W resistor, and wires, and you're done.

I've marked the LEDs in my schematic with a "K" for the cathode - usually that's the lead next to the flat on the flange.

If you can figure out how to lash this all up, then you're done.

But why did it take so long to extract from you what your actual goal is?

Why didn't you just say that you wanted two red LEDs in the first place?

Have Fun! Rich

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-Bill

Well, at least I rounded it right, to 1K. :-)

Thanks! Rich

PWM does improve control, but it does not always improve efficiency.

Most white low power LEDs have luminous efficiency maximized by having instantaneous current somewhere in/near the range of 1.6 to 6 mA. Most low power blue and green LEDs have luminous efficiency maximized by having instantaneous current near/in the range of 1.5-4.5 mA.

(The difference is because the LED chip emission has wavelength shifting to longer more-visible wavelength as current decreases, while much of the output of usual white LEDs is from a phosphor whose spectrum does not shift as instantaneous current varies.)

--
 - Don Klipstein (don@misty.com)

cut ??? =960 ohms...........

Given the application, I could suggest that many white LEDs are blazing bright at 5 mA. For example, either the Cree C513A-WSN-CV0Y0151 (nominal beamwidth of 55 degrees, nominally 5.185 candela at 20 mA) or the Cree C535A-WJN-CU0V0231 (nominal beamwidth 110 degrees, nominally 2.26 candela at 20 mA). Both of these critters are available from Digi-Key.

Expect about 30% of the nominal 20mA brightness at 5 mA.

Heck, expect about 15-16% of their nominal 20mA brightness at 2.5 mA.

Voltage drop of these at a couple to a few mA is close enough to 2.8 volts. (I just tried one of the 110 degree ones, which I have a few of.) Put 3 of these in series, and voltage drop is close enough to 8.4 volts. That makes efficiency ~70%.

When total LED voltage drop is 8.4 volts and supply voltage is 12 volts, a dropping resistor would have ~3.6 volts across it. To pass 2.5 mA, its value would be 1440 ohms. The nearest standard value is 1500 ohms, usually referred to as 1.5K.

The power wasted in such a dropping resistor is 2.5 mA times 3.6 volts, or 9 milliwatts. (Give-or-take due to tolerances in LED voltage drop, resistor value, and supply voltage.) The total power consumption is ~30 milliwatts.

Three of these LEDs and a 1.5K 1/4 watt resistor is tolerant of much higher voltages without exceeding the ratings of any parts. And if the

12V battery in question is being charged (voltage closer to 14V), there is little need to conserve every possible milliwatt of power, so LED current of 3.75 mA is OK in that case.

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One bit of caution: White LEDs are usually static-sensitive. If the LED leads or anything connected to them (other than a direct power supply connection) is subject to being touched by humans or otherwise subject to static electricity after installation, it might be a good idea to add a

1N4148 diode in parallel with the LED gang or possibly even in parallel with each LED. The 1N4148 diodes would be connected to the LEDs in "antiparallel" fashion, cathode-to-anode and anode-to-cathode.

Also, handle white LEDs carefully. Mainly, avoid touching either lead (with human fingers or an ungrounded soldering iron) when the other lead is connecting to ground or anything large or a soldering iron. ===================

--
 - Don Klipstein (don@misty.com)

Zetex ZXLD1366...

The 97% efficiency refers to 3% loss in this LED driver IC at favorable conditions including higher LED voltage drop near or over 30 volts (9-plus white LEDs in series with each other). Supply voltage must significantly exceed the load voltage. At 12 volts with 1 white LED, the efficiency is

75%, loss is 25%.

(According to bottom graph of page 8 of

I suspect this loss does not include losses in the inductor.

Also, this IC draws 1.6 mA on its own, even if forced into shutdown mode. At 12V, that is waste of 19.2 milliwatts. I recently posted elsewhere in this thread what I consider to be a workable arrangement where a mere dropping resistor wastes only 9 milliwatts. That is achieved by using only 2.5 mA through three white LEDs, Cree C513A-WSN-CV0Y0151 or C535A-WJN-CU0V0231.

This IC is useful for much higher LED currents of around .1 to 1 amp. Meanwhile, the originally posted application (illuminating a small flag on a boat to see wind direction) sounds to me like one with a requirement for a much smaller amount of light than good white LEDs produce with hundreds of milliamps.

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 - Don Klipstein (don@misty.com)

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I think the OP wants to use RED Leds for night vision reasons. There was a LED driver using a buck converter posted by Win Hill a couple years ago, but I can't find it. Seems like it just used an inductor in series with LED and transistor switch and a diode to keep the current flowing while the switch was off, and some sort of duty cycle to match things up. Maybe a couple CMOS inverters wired as an oscillator with the right duty cycle controlling the transistor switch would do the trick?

-Bill

My experience is otherwise. I know of ones that visibly glow in direct sunlight, put spots in my eyes when viewing directly in dimmish room light, and are too bright for my comfort for usage of one of them as a bedroom nightlight, at a mere 2.5 milliamps! One of these is even a Digi-Key-available one with nominal beam width of 110 degrees! (Cree C535A-WJN-CU0V0231)

(Plan on forward voltage drop close to 3 volts.)

--
 - Don Klipstein (don@misty.com)

I do know - blue, violet, purple, UV, white, non-yellowish green, and pink LEDs tend to have voltage drop around 3 volts - mostly more at "full power".

--
 - Don Klipstein (don@misty.com)