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

P

#### Pete Verdon

Jan 1, 1970
0
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.

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.

Thanks for any advice you can give,

Pete

(For any fellow sailors reading this, who are used to seeing a Windex
via the overspill from a tricolour, note that this is a
apply.)

R

#### Rich Grise

Jan 1, 1970
0
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.

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.

Thanks for any advice you can give,

(For any fellow sailors reading this, who are used to seeing a Windex
via the overspill from a tricolour, note that this is a
apply.)

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

Good Luck!
Rich

J

#### Jamie

Jan 1, 1970
0
Pete said:
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.

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.

Thanks for any advice you can give,

Pete

(For any fellow sailors reading this, who are used to seeing a Windex
via the overspill from a tricolour, note that this is a
apply.)
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..

P

#### Pete Verdon

Jan 1, 1970
0
Rich said:
I just googled "12V led light" without the quotes and got "About 564,000
results"

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

P

#### Pete Verdon

Jan 1, 1970
0
Jamie said:
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.

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

R

#### Rich Grise

Jan 1, 1970
0
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?
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

P

#### petrus bitbyter

Jan 1, 1970
0
Pete Verdon said:
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.

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.

Thanks for any advice you can give,

Pete

(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.)

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

W

#### Winston

Jan 1, 1970
0
Pete said:
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.

(...)

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

--Winston

W

#### Winston

Jan 1, 1970
0
Dan wrote:

(...)
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.

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

H

#### haaaTchoum

Jan 1, 1970
0
Hi,

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.

Cdlt,
F

P

#### petrus bitbyter

Jan 1, 1970
0
----- Original Message -----
From: "haaaTchoum" <[email protected]>
Newsgroups: sci.electronics.basics
Sent: Wednesday, October 06, 2010 11:14 PM
Subject: Re: How to power a single LED from a 12v supply?

Hi,

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.

Cdlt,
F

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

P

#### Pete Verdon

Jan 1, 1970
0
petrus said:
From: "haaaTchoum" <[email protected]>
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.

Pete

P

#### Pete Verdon

Jan 1, 1970
0
Tim said:
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'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

E

#### ehsjr

Jan 1, 1970
0
Pete said:
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

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---+
----- |
| |
+---------+
|
[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

R

#### Rich Grise

Jan 1, 1970
0
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.
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/
http://www.theledlight.com/
http://www.superbrightleds.com/cgi-bin/store/index.cgi?action=DispPage&Page2Disp=/BA9S6_specs.htm
http://www.futurlec.com/LED_Lamps.shtml

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

P

#### petrus bitbyter

Jan 1, 1970
0
"Bill Bowden" <[email protected]> schreef in bericht
Pete said:
petrus bitbyter 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. 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---+
----- |
| |
+---------+
|
[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

| 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

P

#### Pete Verdon

Jan 1, 1970
0
Rich said:
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. 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
replacements and the like, are completely the wrong thing.
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)

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
But I can't understand what it is that's making you reject all those
wonderful other suggestions -

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

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

K

#### [email protected]

Jan 1, 1970
0
"Bill Bowden" <[email protected]> schreef in bericht
Pete Verdon wrote:
petrus bitbyter wrote:
From: "haaaTchoum" <[email protected]>
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
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:
-----
+12 -----in|LM317|out---+
----- |
| |
+---------+
|
[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.

| 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

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?

You can't see the difference between 40W, 60W, 75W, and 100W light bulbs?
There's a legend that controlling a LED with very short pulses
increase the perceived brightness, thus saving power.

Urban legend. The eyes integrate over significant time.
Any truth to that?

No.

B

#### baron

Jan 1, 1970
0
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.

<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.

J

#### Jasen Betts

Jan 1, 1970
0
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.

pulses increase the visibility, but not the brightness.

A
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