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

[Don Klipstein]

In article <i987sd$v80$1@news.eternal-september.org>, Pete Verdon wrote:

Rich Grise wrote:

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.

http://www.superbrightleds.com/

But I don't want a floodlight, I want one or two plain oldfashioned
LEDs, probably basic red,

<I SNIP from here>

If you wanna see something illuminated by one or more LEDs at night, I
strongly advise choosing white over red.

White ones have roughly double the luminous efficacy of red ones, and
that is according to "photopic vision" / "day vision".
The disadvantage of red for illuminating things from low power worsens
when the application is a nighttime/"dark" one. Human "night vision"
("scotopic vision") peaks at a shorter wavelength than human photopic
vision does.
--
- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In article &lt;i997ut$48j$1@news.eternal-september.org&gt;, baron wrote:

Pete Verdon Inscribed thus:

if you want to save power, you could use a simple PWM circuit.

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

http://www.bigclive.com/joule.htm

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.

If the LED voltage drop is less than the supply voltage by about or more
than one "diode drop" (roughly .6 volt, .3 for less-lossy Schottky diode),
then the circuit topology seriously changes.

If the load voltage is below the supply voltage by more than a "diode
drop", then the favored circuit topology is a "buck regulator" or "buck /
bucking switching regulator". Preferably load current as opposed to load
voltage is what is regulated.
--
- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In article &lt;pan.2010.10.15.22.32.25.601043@example.net&gt;, Rich Grise wrote:

On Fri, 15 Oct 2010 01:39:40 +0100, Pete Verdon wrote:
Rich Grise wrote:

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.

http://www.superbrightleds.com/

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.
...

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,

<I SNIP from here>

At 10 mA, red LEDs tend to have voltage drop of 1.6 to 1.95 volts.
Even longer-wavelength lower-efficiency ones with GaAs substrate tend to
have voltage drop around 1.6 volts, slightly exceeding 1.5 volts, at 10
or even 5 mA.
--
- Don Klipstein (don@misty.com)

 

[Rich Grise]

On Sun, 17 Oct 2010 08:19:14 +0000, Don Klipstein wrote:

In article &lt;pan.2010.10.15.22.32.25.601043@example.net&gt;, Rich Grise wrote:
On Fri, 15 Oct 2010 01:39:40 +0100, Pete Verdon wrote:
Rich Grise wrote:

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.

http://www.superbrightleds.com/

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. ...

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

Wire the circuit like this:

+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,

I SNIP from here

At 10 mA, red LEDs tend to have voltage drop of 1.6 to 1.95 volts.
Even longer-wavelength lower-efficiency ones with GaAs substrate tend to
have voltage drop around 1.6 volts, slightly exceeding 1.5 volts, at 10 or
even 5 mA.

Thanks for the correction - it's been a while since I've actually used
an LED, so my numbers were kind of pulled out of a hat, primarily for
the sake of example.

In either case, the formula still applies.

Thanks,
Rich

 

[Don Klipstein]

In &lt;RvGdnXWVMJztmDHRnZ2dnUVZ_sGdnZ2d@giganews.com&gt;, Dan soldier emiritus say:

On 10/5/2010 11:48 PM, Winston wrote:
Pete Verdon wrote:
Over this coming winter, I'm planning a complete refit of my small
boat's electrical system. One of the things I would like to add is a
tiny light at the masthead, to illuminate at night the flag I fly
there to tell the wind direction.

How about a self-glowing flag made with photo luminescent nylon thread?
http://lightlead.com/Photoluminescent_Products/Glow_Yarns.htm

Shine a little UV light at it and it will glow for hours.
http://www.ultravioletledflashlights.com/NightShark.html

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

I have seen this phenomenon also. I am still wondering whether the
mechanism is stimulated emission (something lasers depend on) or the
usually-less-desired "quenching".
--
- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In &lt;a3hra65s16i89p1j2kmjr3hst2kpbeum1s@4ax.com&gt;, default wrote:

On Tue, 05 Oct 2010 23:23:54 +0100, Pete Verdon
news@verdonet.organisation.unitedkingdom.invalid&gt; wrote:
And I edit for space a bit
Over this coming winter, I'm planning a complete refit of my small
boat's electrical system. One of the things I would like to add is a
tiny light at the masthead, to illuminate at night the flag I fly there
to tell the wind direction. This light needs to be just bright enough to
make out the flag immediately above it - it should be invisible from any
appreciable distance. This is because the anti-collision regulations lay
down a complex (but logical) system of lights for identifying different
types of vessel, and having a random superfluous masthead light would
interfere with that. I haven't tested yet, but I suspect a single
standard LED might be all that's needed for dark-adapted eyes to pick
out the mostly-white flag nearby.

This light would be powered by a feed from the "official" navigation
lights further down the mast. These run at a nominal 12v - perhaps up to
14.5 when the engine is running.

How would I best power a single LED from a 12v source? My electronic
learning stopped when I left school, so I don't really know what I
should be looking for. Are there standard voltage convertor chips which
would be suitable?

Like many sailors without a shore power hookup, I'm twitchy about power
usage, so something that doesn't gratuitously waste energy into a big
heatsink would be good, even if compared to other loads the question is
more psychological than practical.

(For any fellow sailors reading this, who are used to seeing a Windex
via the overspill from a tricolour, note that this is a
traditionally-rigged boat with a plain truck masthead, so that doesn't
apply.)

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 &gt;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.

I think that a 1 watt "potentiometer" ("variable resistor") is
sufficient. If even 11V after LED drop appears across total resistance,
at 20 mA this is .22 watt.

Heck, since I know some super-efficient LEDs, I would suggest 10 or 5-6
milliamps max, which means ~1K to 1.8K ohms fixed series resistor (go for
less here), and the variable one then only needs to pass at most an amount
of current that would produce less than 1/8 watt of heat in full
resistance of its resistive part.

For that matter, I would choose a "potentiometer" (a "pot") of 5K to 10K
ohms, in case a milliamp or 2 is sufficient with a super-efficient white
LED (such as the Digi-Key stocked Cree C535A-WJN-CU0V0231).

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.

6 degrees sounds to me so narrow that the LED needs to be distant from
the flag by 6-8 times the width of the flag.

Bright white "low-power" white LEDs tend to have beam angle (2-theta-1/2)
of at least 15 degrees - good for placing no farther than 4 times the
width of the flag, maybe only 3 times.
Ones with better optical efficiency appear to me to be wider beam ones
40-plus degrees that want to have distance from the flag not greatly more
than the width of the flag.

If the size of the flag is small, then the LED that I named can
illuminate it at nighttime at 1 or 2 milliamps.
--
- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In &lt;b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com&gt;,
Bill Bowden wrote:

On Oct 12, 7:58 pm, ehsjr &lt;eh...@nospamverizon.net&gt; wrote:
Pete Verdon wrote:

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.

It's a current regulator circuit.

<I snip mostly from here to edit for space>

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.

Although this is true, I would suggest *not* worrying about mere
milliwatts of power consumption waste, especially "in light of" LEDs that
are both "widebeam" and capable of putting spots in people's eyes at a few
mA. Put one or 2-3 series-wired puppies somewhat-in-front-of-the-flag by
a distance roughly the width of the flag (with plenty of "give-or-take"),
and you probably have all the light you need from around 2 mA, fair chance
1 mA.

Such as with the "in-stock" "Digi-Key-available in-stock" Cree
C535A-WJN-CU0V0231. That one is a nominally 110 degree model, known to me
to make itself visibly glowing in "direct high-noon sunlight" at 2.5
milliamps.

Nichia makes even better "low-power" LEDs, but it appears to me that you
need to buy those 100 at a time from one of their sales offices, such as
the one close to Detroit.
--
- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In &lt;4cb6d6b2$0$11873$e4fe514c@dreader12.news.xs4all.nl&gt;, petrus bitbyter
wrote (slightly edited for space by me):

"Bill Bowden" &lt;bperryb@bowdenshobbycircuits.info&gt; schreef in bericht
news:b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com...
On Oct 12, 7:58 pm, ehsjr &lt;eh...@nospamverizon.net&gt; wrote:
Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" &lt;a...@a.com

You can try a constant current generator with two transistors :

http://img829.imageshack.us/i/26686413.gif/

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.

Excellent advice... Assuming the OP to be able to understand the
schematic
and build the circuit.

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.

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:

I snip to edit dfor space

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.

| 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 &gt;100mA.
But unless ones interested in the electronics, you'd better buy a 12V LED
(car)lamp with the electronics build in.

Though I could agree with this suggestion to use market-available
products, I also suggest usage of 1-3 LEDs "in series string"
(such as Digi-Key-available C535A-WJN-CU0V0231) to work from ~1 to ~2.5
milliamps) along with a sauitable "dropping resistor" as high as around
1.0 kiilo-ohm, maybe 2.2 kilo-ohm (a common rersistor resistance).

In the likely event the current drawn by LED(s) gets down to ~2.5 or
~1 milliamp "ballpark", then I have doubt about practical improvement
over the "obviously wasteful" "dropping resistors" that "more obviosly"
"waste power". At ~2.5 mA or less, I would re-consider "energy
efficiciency requirement" due to small figures for enegy and power
requirements.
--
- Don Klipstein (don@misty.com)