Trying to identify the LED in the Coleman Slim-Lantern key-f

[wylbur37]

I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

Since no information appears on the LED itself, does anyone know how I
could go about finding out what the specs are? Better yet, if someone
has one of these flashlights and already knows what the specs are,
please mention them.

By the way, there's an evaluation (with photos) done on this item at
the following URL (but no information about the LED specs) ...
http://ledmuseum.jpoproductions.com/clant.htm

Thanks.

 

[Jerry G.]

I would use any of the standard type high output LED's that are in the
colour white area. I am sure if you venture in to some of the catalogue
from the major electronics parts suppliers, you will find something similar.
Considering that most of these LED's have a 1.6 to 2.2 Volt drop, I would
suspect that the current pull is about 60 to 80 ma. I have a feeling that
if you buy these LED's in small quantity they will be on the expensive side.

A number of years ago, I had one of these myself, but it had a red LED in
it. It looks interesting and more realistic, that they are now using a white
LED in these. I will have to go get one for curiosity. We have some
sporting goods stores around here that sell these.


--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
=========================================


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404140322.bb1f288@posting.google.com...
I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

Since no information appears on the LED itself, does anyone know how I
could go about finding out what the specs are? Better yet, if someone
has one of these flashlights and already knows what the specs are,
please mention them.

By the way, there's an evaluation (with photos) done on this item at
the following URL (but no information about the LED specs) ...
http://ledmuseum.jpoproductions.com/clant.htm

Thanks.

 

[Watson A.Name \"Watt Sun]

wylbur37 wrote:

I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

Since no information appears on the LED itself, does anyone know how I
could go about finding out what the specs are? Better yet, if someone
has one of these flashlights and already knows what the specs are,
please mention them.

By the way, there's an evaluation (with photos) done on this item at
the following URL (but no information about the LED specs) ...
http://ledmuseum.jpoproductions.com/clant.htm

Since you didn't say what color the LED was, I had to go to the website
above to find out. Now that that issue is resolved and the LED is given
as a white LED, you can get a replacement from
http://www.whitelightled.com/ for a dollar plus postage. The white LED
is commonly known as a Nichia NSPW500BS. Go to the above website or
else to www.nichia.com to get specifications.

> Thanks.

 

[Don Klipstein]

In article <8028c236.0404140322.bb1f288@posting.google.com>, wylbur37 wrote:

I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

I can say that the resistor is probably 51 and not 52 ohms since 51 is a
standard value and 52 is not.

As for the LED: Based on the LEDMuseum page, I am guessing that this is
a plain old ordinary common (by today's standards) "super red" or "super
bright red" or "ultrabright red" GaAlAsP LED with a typical forward
voltage drop at 20 mA around 1.85-1.9 volts, maximum rated current of 30
mA, brightness probably 3,000 mcd (possibly less), and peak wavelength
probably around 660 nm.

- Don Klipstein (don@misty.com)

 

[Watson A.Name - \"Watt Su]

"Jerry G." <jerryg50@hotmail.com> wrote in message
news:c5k89t$jpt$1@news.eusc.inter.net...

I would use any of the standard type high output LED's that are in the
colour white area. I am sure if you venture in to some of the
catalogue
from the major electronics parts suppliers, you will find something
similar.
Considering that most of these LED's have a 1.6 to 2.2 Volt drop, I
would

I think you need to read up on current LED technology. White LEDs use a
blue LED with phosphor, and their forward V drop is _not_ 1.5 to 2.2V,
but 3.1 to 3.6V.

suspect that the current pull is about 60 to 80 ma. I have a feeling
that
if you buy these LED's in small quantity they will be on the expensive
side.

I've been running white LEDs at 50 to 60 mA, and I'm finding them dying
in a matter of days to weeks. The 20 mA that they are rated for is
okay, but 30 mA is iffy and and above that will start to overheat them,
and they lose their long lifetime.

A number of years ago, I had one of these myself, but it had a red LED
in
it. It looks interesting and more realistic, that they are now using a
white
LED in these. I will have to go get one for curiosity. We have some
sporting goods stores around here that sell these.

I got one of those Coleman key fob lanterns, but it was not flattened
out, it was just like a regular lantern, with two button cells in the
base and an amber LED in the globe. It sucked, because the LED shined
up into plastic, not out to the location where the light was needed.
It's now stuck away somewhere where it won't be a pain in the butt to
someone.

--

Greetings,

Jerry Greenberg GLG Technologies GLG


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404140322.bb1f288@posting.google.com...
I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

Since no information appears on the LED itself, does anyone know how I
could go about finding out what the specs are? Better yet, if someone
has one of these flashlights and already knows what the specs are,
please mention them.

By the way, there's an evaluation (with photos) done on this item at
the following URL (but no information about the LED specs) ...
http://ledmuseum.jpoproductions.com/clant.htm

Thanks.

 

[wylbur37]

don@manx.misty.com (Don Klipstein) wrote in message news:<slrnc7rpg6.106.don@manx.misty.com>...

In article <8028c236.0404140322.bb1f288@posting.google.com>, wylbur37 wrote:
I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

I can say that the resistor is probably 51 and not 52 ohms since 51 is a
standard value and 52 is not.

As for the LED: Based on the LEDMuseum page, I am guessing that this is
a plain old ordinary common (by today's standards) "super red" or "super
bright red" or "ultrabright red" GaAlAsP LED with a typical forward
voltage drop at 20 mA around 1.85-1.9 volts, maximum rated current of 30
mA, brightness probably 3,000 mcd (possibly less), and peak wavelength
probably around 660 nm.

No, it's not a red LED, it's a *WHITE* LED.
Sorry I neglected to mention that.

 

[wylbur37]

"Watson A.Name - \"Watt Sun, the Dark Remover\"" <NOSPAM@dslextreme.com> wrote in message news:<107s90arisj914@corp.supernews.com>...

I've been running white LEDs at 50 to 60 mA, and I'm finding them dying
in a matter of days to weeks. The 20 mA that they are rated for is
okay, but 30 mA is iffy and and above that will start to overheat them,
and they lose their long lifetime.

How would you reduce the current of the power source to make it compatible
with the rated current of the LED?
After all, the standard formula for the desired resistance is ...

R = (V8 - Vf)/If

R = the resistor
V8 = dc supply voltage
Vf = rated forward voltage of the LED
If = rated forward current of the LED at specified forward voltage

Notice that nowhere in the formula is the current of the power source.
So if the voltage of the source is already equal to the rated voltage
of the LED, how would you reduce the current?

 

[Watson A.Name - \"Watt Su]

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404150300.6cf40740@posting.google.com...

"Watson A.Name - \"Watt Sun, the Dark Remover\""
NOSPAM@dslextreme.com> wrote in message

news:<107s90arisj914@corp.supernews.com>...

I've been running white LEDs at 50 to 60 mA, and I'm finding them
dying
in a matter of days to weeks. The 20 mA that they are rated for is
okay, but 30 mA is iffy and and above that will start to overheat
them,
and they lose their long lifetime.

How would you reduce the current of the power source to make it
compatible
with the rated current of the LED?
After all, the standard formula for the desired resistance is ...

R = (V8 - Vf)/If

R = the resistor
V8 = dc supply voltage
Vf = rated forward voltage of the LED
If = rated forward current of the LED at specified forward voltage

Notice that nowhere in the formula is the current of the power source.
So if the voltage of the source is already equal to the rated voltage
of the LED, how would you reduce the current?

That last paragraph, above, has got me bumfuzzled. I'm not sure what
you mean.

 

[William P.N. Smith]

wylbur37nospam@yahoo.com (wylbur37) wrote:

So if the voltage of the source is already equal to the rated voltage
of the LED, how would you reduce the current?

Well, you wouldn't, but in the real world it's never that simple.
First, the "voltage" of the LED isn't a specific value, it's a range
of values depending on manufacturing tolerances, temperature, and
phase of the moon. Likewise, the voltage of your power source isn't
exactly <so many> volts, it's a battery with a nominal voltage, which
in the real world will start out high and end up low.

Many tiny flashlights don't use a dropping resistor at all, they just
depend on the internal resistance of the battery to limit the current
to one that will not burn out the LED immediately, though it may show
some degredation in light output over time.

There are lots of solutions, from higher voltages and current limiting
resistors to complex switching buck/boost converters. You pretty much
get what you pay for.

--
William Smith
ComputerSmiths Consulting, Inc. www.compusmiths.com

 

[William P.N. Smith]

"Watson A.Name - \"Watt Sun, the Dark Remover\""
<NOSPAM@dslextreme.com> wrote:

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
Notice that nowhere in the formula is the current of the power source.

That last paragraph, above, has got me bumfuzzled. I'm not sure what
you mean.

Wylbur may not understand that the supply current and the LED current
are identical...

--
William Smith
ComputerSmiths Consulting, Inc. www.compusmiths.com

 

[Jim Yanik]

"Watson A.Name - \"Watt Sun, the Dark Remover\"" <NOSPAM@dslextreme.com>
wrote in news:107s90arisj914@corp.supernews.com:

I think you need to read up on current LED technology. White LEDs use a
blue LED with phosphor, and their forward V drop is _not_ 1.5 to 2.2V,
but 3.1 to 3.6V.



I've been running white LEDs at 50 to 60 mA, and I'm finding them dying
in a matter of days to weeks. The 20 mA that they are rated for is
okay, but 30 mA is iffy and and above that will start to overheat them,
and they lose their long lifetime.

But those lithium coil cells have a high internal resistance.I don't think
they can deliver 50 ma.

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404140322.bb1f288@posting.google.com...
I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

--
Jim Yanik
jyanik-at-kua.net

 

[Jim Yanik]

William P.N. Smith <> wrote in
newst1t70th3jaqpaqgoe391l60rac4rprcog@4ax.com:

"Watson A.Name - \"Watt Sun, the Dark Remover\""
NOSPAM@dslextreme.com> wrote:
"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
Notice that nowhere in the formula is the current of the power source.

That last paragraph, above, has got me bumfuzzled. I'm not sure what
you mean.

Wylbur may not understand that the supply current and the LED current
are identical...

What about the coin cell's internal resistance?
--
Jim Yanik
jyanik-at-kua.net

 

[wylbur37]

"Watson A.Name - \"Watt Sun, the Dark Remover\"" <NOSPAM@dslextreme.com> wrote in message news:<107sv0dl29r3730@corp.supernews.com>...

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404150300.6cf40740@posting.google.com...
"Watson A.Name - \"Watt Sun, the Dark Remover\""
NOSPAM@dslextreme.com> wrote in message
news:<107s90arisj914@corp.supernews.com>...

I've been running white LEDs at 50 to 60 mA, and I'm finding them
dying
in a matter of days to weeks. The 20 mA that they are rated for is
okay, but 30 mA is iffy and and above that will start to overheat
them,
and they lose their long lifetime.

How would you reduce the current of the power source to make it
compatible
with the rated current of the LED?
After all, the standard formula for the desired resistance is ...

R = (V8 - Vf)/If

R = the resistor
V8 = dc supply voltage
Vf = rated forward voltage of the LED
If = rated forward current of the LED at specified forward voltage

Notice that nowhere in the formula is the current of the power source.
So if the voltage of the source is already equal to the rated voltage
of the LED, how would you reduce the current?

That last paragraph, above, has got me bumfuzzled. I'm not sure what
you mean.

I'd like to take the LED from the key-fob flashlight and make it into
a booklight (so I can read a book or use a computer in the dark so as
not to disturb others in the room).

However, since I plan to have this LED on continuously for several
hours at a time, it's not practical or economical to run it on the
original lithium disk batteries (because the batteries would quickly
be exhausted).

So I figured on using an AC adapter that puts out the same 6V that the
two batteries in series had put out. But if the AC adapter puts out
100mA of current (as it says on the unit) and if the LED is rated at
20mA (as someone had suggested), then wouldn't I be overloading the
LED? (After all, you said in your original posting that running your
LEDs at a higher current than their rating would prematurely burn
them out).

So, assuming that's true, the idea is to find some way to reduce the
current that the LED is exposed to from 100mA to 20mA when running it
on the AC adapter.

 

[Michael A. Covington]

To reduce the current in an LED, use a larger resistor.

Ohm's Law is:

R = E / I

where R is resistance, E is voltage, and I is current.

In the case of an LED circuit, E is the voltage drop across the resistor.


Example:

A good old-fashioned red LED whose internal voltage drop is 1.8 volts.

A 6-volt power source.

You want 20 mA through the LED.

R = (6 - 1.8) / 0.020 = 210 ohms.



We should also check the power rating the resistor, since in some cases a
1/8-watt resistor will not be big enough. (The power rating of the resistor
is a maximum, so anything larger than the actual power dissipation will be
safe.)

P = E * I = (6 - 1.8) * 0.020 = 0.084 watt

This is well under 1/8 watt, so a common 1/8-watt resistor will do fine.



Do similar calculations for any LED and any voltage source.

Or, if you like to experiment, simply hook up a milliammeter in series with
the power source. Start with a resistor that you know is too large (like
1000 ohms) and measure the current flow. Change resistors until you get
what you want.


Crucially, unlike a light bulb, and LED does not limit its own current. The
voltage drop across an LED is nearly constant regardless of the current.
That's why I treated it as constant in the calculation above. That's also
why an LED requires a resistor. If you connect it directly to a voltage
source, it will generally burn out, and if it doesn't, its current will vary
tremendously with small fluctuations in the voltage source.


This must be one of the most frequently asked questions in all of
electronics. I haven't seen anyone post a good answer recently.



--
----------------------------------------------------------------------------
---------
Michael A. Covington, Associate Director
Artificial Intelligence Center, The University of Georgia
http://www.ai.uga.edu/~mc

 

[Watson A.Name - \"Watt Su]

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404160156.6e06bd1a@posting.google.com...

[snip]

Notice that nowhere in the formula is the current of the power
source.
So if the voltage of the source is already equal to the rated
voltage
of the LED, how would you reduce the current?

That last paragraph, above, has got me bumfuzzled. I'm not sure
what
you mean.

I'd like to take the LED from the key-fob flashlight and make it into
a booklight (so I can read a book or use a computer in the dark so as
not to disturb others in the room).

However, since I plan to have this LED on continuously for several
hours at a time, it's not practical or economical to run it on the
original lithium disk batteries (because the batteries would quickly
be exhausted).

So I figured on using an AC adapter that puts out the same 6V that the
two batteries in series had put out. But if the AC adapter puts out
100mA of current (as it says on the unit) and if the LED is rated at
20mA (as someone had suggested), then wouldn't I be overloading the
LED? (After all, you said in your original posting that running your
LEDs at a higher current than their rating would prematurely burn
them out).

So, assuming that's true, the idea is to find some way to reduce the
current that the LED is exposed to from 100mA to 20mA when running it
on the AC adapter.

I did exactly the same thing, I put several white LEDs in a wall wart AC
adapter to let them run for the last 5 or 6 months. I used a 6VDC, 100
mA adapter, which had a voltage of about 7.7 to 8VDC with no load on it.
I used 150 ohm resistors in series with the LEDs, but if you have only
one LED, then use a 180 or 220 ohm resistor. For a 220 ohm resistor,
you should measure 4.4VDC across it at 20 mA. For a 180 ohm resistor,
you should measure 3.6VDC across it at 20 mA. A bit higher or lower
should be okay. To find the current, just divide the voltage you
meassure by the resistance.

View the following with courier font.

220 ohm
+ o----/\/\/\--------+
|
Wall Wart --- --> white LED
6VDC 100 mA \ / -->
--- Cathode is flattened point
|
= o------------------+

 

[Watson A.Name - \"Watt Su]

"Michael A. Covington" <look@www.covingtoninnovations.com.for.address>
wrote in message news:_dmdnT2ww9y8S-Ld4p2dnA@speedfactory.net...

To reduce the current in an LED, use a larger resistor.

Ohm's Law is:

R = E / I

where R is resistance, E is voltage, and I is current.

In the case of an LED circuit, E is the voltage drop across the
resistor.


Example:

A good old-fashioned red LED whose internal voltage drop is 1.8 volts.

A 6-volt power source.

You want 20 mA through the LED.

R = (6 - 1.8) / 0.020 = 210 ohms.

We should also check the power rating the resistor, since in some
cases a
1/8-watt resistor will not be big enough. (The power rating of the
resistor
is a maximum, so anything larger than the actual power dissipation
will be
safe.)

P = E * I = (6 - 1.8) * 0.020 = 0.084 watt

This is well under 1/8 watt, so a common 1/8-watt resistor will do
fine.

Do similar calculations for any LED and any voltage source.

Or, if you like to experiment, simply hook up a milliammeter in series
with
the power source. Start with a resistor that you know is too large
(like
1000 ohms) and measure the current flow. Change resistors until you
get
what you want.

Or to save having to change resistors, just put a 1000 ohm pot in there,
and adjust to get the correct current. Then measure the pot and use the
next higher standard value.

BTW, I dunno about you, Mike, but most resistors (at least the thru hole
type, anyway) that I see are 1/4W, not 1/8W.

Crucially, unlike a light bulb, and LED does not limit its own
current. The
voltage drop across an LED is nearly constant regardless of the
current.
That's why I treated it as constant in the calculation above. That's
also
why an LED requires a resistor. If you connect it directly to a
voltage
source, it will generally burn out, and if it doesn't, its current
will vary
tremendously with small fluctuations in the voltage source.


This must be one of the most frequently asked questions in all of
electronics. I haven't seen anyone post a good answer recently.

Hey, yours is a good one ;-)

--
---------
Michael A. Covington, Associate Director
Artificial Intelligence Center, The University of Georgia
http://www.ai.uga.edu/~mc

 

[H. R. Bob Hofmann]

wylbur37nospam@yahoo.com (wylbur37) wrote in message news:<8028c236.0404150246.30a3546f@posting.google.com>...

don@manx.misty.com (Don Klipstein) wrote in message news:<slrnc7rpg6.106.don@manx.misty.com>...
In article <8028c236.0404140322.bb1f288@posting.google.com>, wylbur37 wrote:
I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

I can say that the resistor is probably 51 and not 52 ohms since 51 is a
standard value and 52 is not.

As for the LED: Based on the LEDMuseum page, I am guessing that this is
a plain old ordinary common (by today's standards) "super red" or "super
bright red" or "ultrabright red" GaAlAsP LED with a typical forward
voltage drop at 20 mA around 1.85-1.9 volts, maximum rated current of 30
mA, brightness probably 3,000 mcd (possibly less), and peak wavelength
probably around 660 nm.

No, it's not a red LED, it's a *WHITE* LED.
Sorry I neglected to mention that.\

Start with V=IR, R=V/I

The wall wart will put out the voltage stated, at a current up to the
value given (150ma I think you said). The voltage drop across the
series resistor and
the LED will have to add up to the output voltage of the wallwart. The
LED drop is more or less constant voltage, so as you increase the
resistance of the series resistor, the current trhough the total
resistor and LED goes down,. So you need to measure the output
voltage of the wallwart feeding a pure resistance at let's say 30 ma,
that may be a little higher than the rated voltage when the output
current is a full 150 ma (say you get 7V). Then put the LED in series
with the resistor and see how bright it is, it should be fairly dim
because there is less current flowing. Measure the voltage across the
resistor (say 4V) Now, divide the voltage across the resistor (4V) by
the current you want thru the LED, say 0.030 amperes (30 ma), 4/0.03 =
133 ohms.
That is the nominal valuue you want for the resistor.

Remember, voltage is like water pressure, current is like the size of
the pipe used to deliver the water.

H. R. (Bob) Hofmann

 

[Soeren]

wylbur37nospam@yahoo.com (wylbur37) wrote in
news:8028c236.0404160156.6e06bd1a@posting.google.com:

So I figured on using an AC adapter that puts out the same 6V that
the two batteries in series had put out.

Just remember there is a resistor in the circuit (in between the lithium
cells).

(It is *not* the voltage of the batteries you need for calculating the
resistor for use with an adapter, it is the LEDs voltage drop - measure it
or calculate with ~3V6)

But if the AC adapter puts
out 100mA of current (as it says on the unit)

The adapter doesn't "put out", the circuit it feeds draws from it. The
100mA is a rating of the maximum current you can draw (without overloading
the adapter).

and if the LED is rated
at 20mA (as someone had suggested), then wouldn't I be overloading
the LED? (After all, you said in your original posting that running
your LEDs at a higher current than their rating would prematurely
burn them out).

You control the current of the LED by choosing the right size resistor as
others have allready explained.

So, assuming that's true, the idea is to find some way to reduce the
current that the LED is exposed to from 100mA to 20mA when running it
on the AC adapter.

You need to replicate the resistor (51 Ohm) in the lantern, but you just
need a higher value.
In the lantern, when the battery cells is full, the current draw is around
47mA (given a LED voltage drop of 3V6). While this is acceptable in a
lantern used only for a few seconds (or minutes) occasionally, any long
term use at this current will degrade and eventually kill the LED too
fast. go for a max. current of 20..25mA.

The resistor will then be:
(6-3.6)/0.025 = 96 Ohm
(6-3.6) / 0.020 = 120 Ohm
Standard values (E12) are 100 Ohm and 120 Ohm.

The LED will not give the same amount of light as in the lantern, of
course, but it will last longer.


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *

 

[Don Klipstein]

In article <8028c236.0404150246.30a3546f@posting.google.com>, wylbur37 wrote:

don@manx.misty.com (Don Klipstein) wrote in message
news:<slrnc7rpg6.106.don@manx.misty.com>...

As for the LED: Based on the LEDMuseum page, I am guessing that this is
a plain old ordinary common (by today's standards) "super red" or "super
bright red" or "ultrabright red" GaAlAsP LED with a typical forward
voltage drop at 20 mA around 1.85-1.9 volts, maximum rated current of 30
mA, brightness probably 3,000 mcd (possibly less), and peak wavelength
probably around 660 nm.

No, it's not a red LED, it's a *WHITE* LED.
Sorry I neglected to mention that.

Oops, I was going from memory of an LEDMuseum page on those Coleman LED
mini-lanterns...

Typical voltage drop of an InGaN white LED (blue with broadband
yellow-glowing phosphor) at 20 mA is 3.5 volts. Give or take about 1/4
volt. The maximum voltage drop at 20 mA before the LED is considered by
the manufacturer rejectable as defective due to excessive voltage drop at
20 mA is 4 volts (for most white LEDs).

- Don Klipstein (don@misty.com)

 

[wylbur37]

Two questions ...

1. I'm aware that when running an LED on a DC power source, the
polarity must be observed or the LED won't light. But when the rated
voltage of the LED is less than the source voltage and a resistor has
to be used, does it make any difference whether the resistor is
attached to the anode side of the LED or the cathode side?

2. The AC adapter for my "itty-bitty booklight" is rated "4.8V 500mA",
but it's actually 4.8 volts *AC*, not DC. I hooked up an LED (with a
resistor on the "+" side of the LED) to the socket and found that it
"works". But just because it "lights up" doesn't mean that it's
"right" (in other words, if you overload an LED by running it on a
higher voltage, it'll "light up" too, but you'll be damaging it). So
the question is, is it OK to run an LED on an AC power source? and if
"yes", is it sufficient to have one resistor or do you have to have
one on either side of the LED?