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

P

petrus bitbyter

Jan 1, 1970
0
"Bill Bowden" <[email protected]> schreef in bericht
"Bill Bowden" <[email protected]> schreef in
bericht
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?
| 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

R

Rich Grise

Jan 1, 1970
0
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.

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

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

960 ohms, not 9600.
Well, at least I rounded it right, to 1K.

Thanks!
Rich

D

Don Klipstein

Jan 1, 1970
0
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.

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.)

S

Sjouke Burry

Jan 1, 1970
0
Rich said:
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

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

D

Don Klipstein

Jan 1, 1970
0
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.

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.

========================

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.
===================

D

Don Klipstein

Jan 1, 1970
0
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.

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
http://www.diodes.com/datasheets/ZXLD1366.pdf)

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.

D

Don Klipstein

Jan 1, 1970
0
p. bitbyter wrote: said:
I doubt a simple cheap white low power LED will light much more then just
itself.

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.)

<SNIP from here stuff that is mostly good>

D

Don Klipstein

Jan 1, 1970
0
Tim Wescott wrote: said:
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).

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".

<I SNIP from here stuff that I at least mostly agree with>

D

Don Klipstein

Jan 1, 1970
0
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.

D

Don Klipstein

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.

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 /
voltage is what is regulated.

D

Don Klipstein

Jan 1, 1970
0
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

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.

R

Rich Grise

Jan 1, 1970
0
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

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

D

Don Klipstein

Jan 1, 1970
0
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".

D

Don Klipstein

Jan 1, 1970
0
On Tue, 05 Oct 2010 23:23:54 +0100, Pete Verdon

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.

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.

D

Don Klipstein

Jan 1, 1970
0
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.

D

Don Klipstein

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

| 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.

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.

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