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

[Bill Bowden]

On Oct 12, 7:58 pm, ehsjr <eh...@nospamverizon.net> wrote:

Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" <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.

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---+
              -----       |
               adj       [R]
                |         |
                +---------+
                          |
                        [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

 

[petrus bitbyter]

"Bill Bowden" <bperryb@bowdenshobbycircuits.info> schreef in bericht
news:b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com...
On Oct 12, 7:58 pm, ehsjr <eh...@nospamverizon.net> wrote:

Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" <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.

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---+
----- |
adj [R]
| |
+---------+
|
[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

 

[Pete Verdon]

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. 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.

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
thread has kindly tried to answer it.

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

 

[Bill Bowden]

On Oct 14, 3:08 am, "petrus bitbyter" <petrus.bitby...@hotmail.com>
wrote:

"Bill Bowden" <bper...@bowdenshobbycircuits.info> schreef in berichtnews:b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com...
On Oct 12, 7:58 pm, ehsjr <eh...@nospamverizon.net> wrote:



Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" <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.

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---+
----- |
adj [R]
| |
+---------+
|
[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

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

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 14 Oct 2010 17:27:43 -0700 (PDT), Bill Bowden
<bperryb@bowdenshobbycircuits.info> wrote:

On Oct 14, 3:08 am, "petrus bitbyter" <petrus.bitby...@hotmail.com
wrote:
"Bill Bowden" <bper...@bowdenshobbycircuits.info> schreef in berichtnews:b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com...
On Oct 12, 7:58 pm, ehsjr <eh...@nospamverizon.net> wrote:



Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" <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.

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---+
----- |
adj [R]
| |
+---------+
|
[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

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.

 

[Bill Bowden]

On Oct 14, 8:15 pm, "k...@att.bizzzzzzzzzzzz"
<k...@att.bizzzzzzzzzzzz> wrote:

On Thu, 14 Oct 2010 17:27:43 -0700 (PDT), Bill Bowden



bper...@bowdenshobbycircuits.info> wrote:
On Oct 14, 3:08 am, "petrus bitbyter" <petrus.bitby...@hotmail.com
wrote:
"Bill Bowden" <bper...@bowdenshobbycircuits.info> schreef in berichtnews:b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com...
On Oct 12, 7:58 pm, ehsjr <eh...@nospamverizon.net> wrote:

Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" <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.

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---+
----- |
adj [R]
| |
+---------+
|
[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

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.

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

 

[baron]

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.

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

--
Best Regards:
Baron.

 

[Jasen Betts]

On 2010-10-15, Bill Bowden <bperryb@bowdenshobbycircuits.info> wrote:

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.



--
ɹǝpun uʍop ɯoɹɟ sƃuıʇǝǝɹ⅁

 

[petrus bitbyter]

"Bill Bowden" <bperryb@bowdenshobbycircuits.info> schreef in bericht
news:483df1b6-67ac-4c45-93c3-0d6f2f1cbede@t20g2000yqa.googlegroups.com...
On Oct 14, 3:08 am, "petrus bitbyter" <petrus.bitby...@hotmail.com>
wrote:

"Bill Bowden" <bper...@bowdenshobbycircuits.info> schreef in
berichtnews:b1d2c6a6-29e7-44b2-af17-b81aec7912e2@l8g2000yql.googlegroups.com...
On Oct 12, 7:58 pm, ehsjr <eh...@nospamverizon.net> wrote:



Pete Verdon wrote:
petrus bitbyter wrote:

From: "haaaTchoum" <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.

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---+
----- |
adj [R]
| |
+---------+
|
[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
|

| 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

 

[Rich Grise]

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, 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

 

[Rich Grise]

On Fri, 15 Oct 2010 16:20:08 -0700, Bill Bowden wrote:

On Oct 15, 3:32 pm, Rich Grise <richgr...@example.net> 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, 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

 

[Don Klipstein]

In article <i99hir$fge$1@reversiblemaps.ath.cx>, Jasen Betts wrote:

On 2010-10-15, Bill Bowden <bperryb@bowdenshobbycircuits.info> wrote:

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.

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:

http://members.misty.com/don/ledp.html
--
- Don Klipstein (don@misty.com)

 

[Bill Bowden]

On Oct 15, 3:32 pm, Rich Grise <richgr...@example.net> 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, 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.

-Bill

 

[Sjouke Burry]

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.
cut

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

 

[Don Klipstein]

In article <4cb85a5c$0$11333$e4fe514c@dreader25.news.xs4all.nl>, petrus
bitbyter wrote:

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

 

[Don Klipstein]

In article <54ena6d512pl2rbqqbmodiafpqhd4k563s@4ax.com>, John Larkin wrote:

On Wed, 06 Oct 2010 00:38:36 +0100, Pete Verdon
news@verdonet.organisation.unitedkingdom.invalid> wrote:

Jamie wrote:

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

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

 

[Don Klipstein]

In article <JumdnfWxDuaMIzbRnZ2dnUVZ_qmdnZ2d@giganews.com>, DJ Delorie wrote:

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

 

[Bill Bowden]

On Oct 16, 10:25 am, d...@manx.misty.com (Don Klipstein) wrote:

In article <54ena6d512pl2rbqqbmodiafpqhd4k5...@4ax.com>, John Larkin wrote:
On Wed, 06 Oct 2010 00:38:36 +0100, Pete Verdon
n...@verdonet.organisation.unitedkingdom.invalid> wrote:

Jamie wrote:

 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

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.
==================> --
 - Don Klipstein (d...@misty.com)

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

 

[Don Klipstein]

In <4cabc344$0$31414$e4fe514c@dreader12.news.xs4all.nl>, p. bitbyter wrote:
<SNIP previously quoted material>

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

 

[Don Klipstein]

In <6rSdnQlRgYUFWDbRnZ2dnUVZ_jOdnZ2d@web-ster.com>, Tim Wescott wrote:

<SNIP what was already quoted>

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