Putting Red LED's In Osram Dot it light

[Dave.H]

I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.

 

[Dave.H]

I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.

The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.

 

[Tom2000]

I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.

Dave,

It sounds like the white LEDs are in parallel, drawing (theoretically)
about 20 mA per LED.

If you wanted to put your red LEDs in there at the same current draw
and wired in parallel, you'd use something like a 43 ohm 1/4 watt
resistor. If the LEDs are too bright, use a larger value resistor and
gain increased battery life. (56 ohms, approx. 15 mA. 82 ohms,
approx. 10 mA. 150 ohms, approx. 5 mA. 820 ohms, approx. 1 mA.)

If you find that one LED is much brighter than the other two with your
LEDs wired in parallel, use three individual resistors, one for each
LED, to balance the currents. 120 ohms per LED will give you approx.
20 mA per LED. Larger value resistors will lower the current.

Here's how to calculate the resistors and currents:

R = (Batt voltage - LED voltage) / current

Current = (Batt voltage - LED voltage) / R

You want to limit the current through each LED to a maximum of 20
milliamps. High-brightness LEDs put out a decent amount of light at 5
milliamps, or even 1 milliamp. For an outdoor night light, 1 milliamp
might be enough if all you want to do is look at a chart at close
range, or something like that. Experiment to find the current level
that meets your needs. (You might even find that a single LED
provides adequate brightness.)

If you have room in that housing to add a miniature switch, you could
connect two of the LEDs, biased for fairly high current, switched from
the existing power switch. Use your extra switch for a single
LED/resistor combo, biased for very low current. That would give you
a choice between dim and bright light levels, depending upon your
needs.

Have fun!

Tom

 

[Dave.H]

Dave,

It sounds like the white LEDs are in parallel, drawing (theoretically)
about 20 mA per LED.

If you wanted to put your red LEDs in there at the same current draw
and wired in parallel, you'd use something like a 43 ohm 1/4 watt
resistor. If the LEDs are too bright, use a larger value resistor and
gain increased battery life. (56 ohms, approx. 15 mA. 82 ohms,
approx. 10 mA. 150 ohms, approx. 5 mA. 820 ohms, approx. 1 mA.)

If you find that one LED is much brighter than the other two with your
LEDs wired in parallel, use three individual resistors, one for each
LED, to balance the currents. 120 ohms per LED will give you approx.
20 mA per LED. Larger value resistors will lower the current.

Here's how to calculate the resistors and currents:

R = (Batt voltage - LED voltage) / current

Current = (Batt voltage - LED voltage) / R

You want to limit the current through each LED to a maximum of 20
milliamps. High-brightness LEDs put out a decent amount of light at 5
milliamps, or even 1 milliamp. For an outdoor night light, 1 milliamp
might be enough if all you want to do is look at a chart at close
range, or something like that. Experiment to find the current level
that meets your needs. (You might even find that a single LED
provides adequate brightness.)

If you have room in that housing to add a miniature switch, you could
connect two of the LEDs, biased for fairly high current, switched from
the existing power switch. Use your extra switch for a single
LED/resistor combo, biased for very low current. That would give you
a choice between dim and bright light levels, depending upon your
needs.

Have fun!

Tom

I think I have a 47 ohm resistor stashed somewhere, I'll dig that out
and try it once I get the LED's tomorrow.

 

[John Fields]

The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.

---
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)


4.5V 3.1V
/ /
+--[25R]--+--[LED>]--+
| | |
|+ +--[LED>]--+
[BAT] | |
| +--[LED>]--+
| |
+--------------------+
<-- Iled

which means that the current into the LEDs is:


Vbat - Vled 4.5V - 3.1V
Iled = ------------- = ------------- = 0.056A = 56mA
Rs 25R


Unfortunately, since Vf/If is different for each LED (unless they
were very carefully matched) the current won't split equally, with
the result being that one LED may be hogging current beyond its
rating, resulting in a shortened life.


I suggest that you rewire the flashlight like this:

+--[R1]---[LED>]--+
| |
+--[R2]---[LED>]--+
| |
+--[R3]---[LED>]--+
| |
+-----[+BAT]------+


In order to determine the value of each of the resistors use:


Vbat - Vled
R = -------------
Iled


Where R is the value of the resistor, in ohms,

Vbat is the battery voltage, in volts,

Vled is Vf(min) for the LED, from the data sheet, and

Iled is the nominal forward current, in amperes, also from
the data sheet.


For example, If your LEDs are rated for a Vf(min) of 2V at 20mA,
then you'll have:


Vbat - Vled 4.5V - 2.0V
R = ------------- = ------------- = 125 ohms
Iled 0.02A


The closest standard 5% resistor on the low end is 120 ohms, so the
current it would allow through the LED would be:


Vbat - Vled
Iled = ------------- = 0.0208A ~ 21mA,
R

Which would be fine.

 

[Dave.H]

The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.

---
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)

4.5V 3.1V
/ /
+--[25R]--+--[LED>]--+
| | |
|+ +--[LED>]--+
[BAT] | |
| +--[LED>]--+
| |
+--------------------+
<-- Iled

which means that the current into the LEDs is:

Vbat - Vled 4.5V - 3.1V
Iled = ------------- = ------------- = 0.056A = 56mA
Rs 25R

Unfortunately, since Vf/If is different for each LED (unless they
were very carefully matched) the current won't split equally, with
the result being that one LED may be hogging current beyond its
rating, resulting in a shortened life.

I suggest that you rewire the flashlight like this:

+--[R1]---[LED>]--+
| |
+--[R2]---[LED>]--+
| |
+--[R3]---[LED>]--+
| |
+-----[+BAT]------+

In order to determine the value of each of the resistors use:

Vbat - Vled
R = -------------
Iled

Where R is the value of the resistor, in ohms,

Vbat is the battery voltage, in volts,

Vled is Vf(min) for the LED, from the data sheet, and

Iled is the nominal forward current, in amperes, also from
the data sheet.

For example, If your LEDs are rated for a Vf(min) of 2V at 20mA,
then you'll have:

Vbat - Vled 4.5V - 2.0V
R = ------------- = ------------- = 125 ohms
Iled 0.02A

The closest standard 5% resistor on the low end is 120 ohms, so the
current it would allow through the LED would be:

Vbat - Vled
Iled = ------------- = 0.0208A ~ 21mA,
R

Which would be fine.

I have linked to a photo of the back of the PCB to make it easier to
understand the circuit (I'm not very good at tracing out PCB's), hope
this helps.
http://s222.photobucket.com/albums/dd237/ozguy89/?action=view&current=IMG_6808.jpg

 

[John Fields]

I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.

The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.

---
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)

4.5V 3.1V
/ /
+--[25R]--+--[LED>]--+
| | |
|+ +--[LED>]--+
[BAT] | |
| +--[LED>]--+
| |
+--------------------+
<-- Iled

which means that the current into the LEDs is:

Vbat - Vled 4.5V - 3.1V
Iled = ------------- = ------------- = 0.056A = 56mA
Rs 25R

Unfortunately, since Vf/If is different for each LED (unless they
were very carefully matched) the current won't split equally, with
the result being that one LED may be hogging current beyond its
rating, resulting in a shortened life.

I suggest that you rewire the flashlight like this:

+--[R1]---[LED>]--+
| |
+--[R2]---[LED>]--+
| |
+--[R3]---[LED>]--+
| |
+-----[+BAT]------+

In order to determine the value of each of the resistors use:

Vbat - Vled
R = -------------
Iled

Where R is the value of the resistor, in ohms,

Vbat is the battery voltage, in volts,

Vled is Vf(min) for the LED, from the data sheet, and

Iled is the nominal forward current, in amperes, also from
the data sheet.

For example, If your LEDs are rated for a Vf(min) of 2V at 20mA,
then you'll have:

Vbat - Vled 4.5V - 2.0V
R = ------------- = ------------- = 125 ohms
Iled 0.02A

The closest standard 5% resistor on the low end is 120 ohms, so the
current it would allow through the LED would be:

Vbat - Vled
Iled = ------------- = 0.0208A ~ 21mA,
R

Which would be fine.

I have linked to a photo of the back of the PCB to make it easier to
understand the circuit (I'm not very good at tracing out PCB's), hope
this helps.
http://s222.photobucket.com/albums/dd237/ozguy89/?action=view&current=IMG_6808.jpg

---
Yes, that's what I thought.

Including the switch, it's wired like this:


.. +--[25R]--+--[LED>]--+
.. | | |
.. | +--[LED>]--+
.. [SWITCH] | |
.. | +--[LED>]--+
.. | |
.. +-----[+BATTERY]-----+

Looking at the photo, if you wanted to go with my suggestion it
seems to me the easiest way of getting there would be to rewire the
board, using cuts and jumpers, like this:


.. +--[JUMPER]--+--[LED>]--[R]--+
.. | | |
.. | +--[LED>]--[R]--+
.. [SWITCH] | |
.. | +--[LED>]--[R]--+
.. | |
.. +--------[+BATTERY]----------+

If you have the room on the wiring side of the board it'd be an easy
matter to use surface-mount resistors, but if not it doesn't seem
that it'd be that much more difficult to use 1/8 or 1/4 watters on
the component side of the board.

What do you want to do?

 

[Dave.H]

On Wed, 20 Feb 2008 03:28:14 -0800 (PST), "Dave.H"
I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.
The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.
---
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)
4.5V 3.1V
/ /
+--[25R]--+--[LED>]--+
| | |
|+ +--[LED>]--+
[BAT] | |
| +--[LED>]--+
| |
+--------------------+
<-- Iled
which means that the current into the LEDs is:
Vbat - Vled 4.5V - 3.1V
Iled = ------------- = ------------- = 0.056A = 56mA
Rs 25R
Unfortunately, since Vf/If is different for each LED (unless they
were very carefully matched) the current won't split equally, with
the result being that one LED may be hogging current beyond its
rating, resulting in a shortened life.
I suggest that you rewire the flashlight like this:
+--[R1]---[LED>]--+
| |
+--[R2]---[LED>]--+
| |
+--[R3]---[LED>]--+
| |
+-----[+BAT]------+
In order to determine the value of each of the resistors use:
Vbat - Vled
R = -------------
Iled
Where R is the value of the resistor, in ohms,
Vbat is the battery voltage, in volts,
Vled is Vf(min) for the LED, from the data sheet, and
Iled is the nominal forward current, in amperes, also from
the data sheet.
For example, If your LEDs are rated for a Vf(min) of 2V at 20mA,
then you'll have:
Vbat - Vled 4.5V - 2.0V
R = ------------- = ------------- = 125 ohms
Iled 0.02A
The closest standard 5% resistor on the low end is 120 ohms, so the
current it would allow through the LED would be:
Vbat - Vled
Iled = ------------- = 0.0208A ~ 21mA,
R
Which would be fine.

I have linked to a photo of the back of the PCB to make it easier to
understand the circuit (I'm not very good at tracing out PCB's), hope
this helps.
http://s222.photobucket.com/albums/dd237/ozguy89/?action=view&current...

---
Yes, that's what I thought.

Including the switch, it's wired like this:

. +--[25R]--+--[LED>]--+
. | | |
. | +--[LED>]--+
. [SWITCH] | |
. | +--[LED>]--+
. | |
. +-----[+BATTERY]-----+

Looking at the photo, if you wanted to go with my suggestion it
seems to me the easiest way of getting there would be to rewire the
board, using cuts and jumpers, like this:

. +--[JUMPER]--+--[LED>]--[R]--+
. | | |
. | +--[LED>]--[R]--+
. [SWITCH] | |
. | +--[LED>]--[R]--+
. | |
. +--------[+BATTERY]----------+

If you have the room on the wiring side of the board it'd be an easy
matter to use surface-mount resistors, but if not it doesn't seem
that it'd be that much more difficult to use 1/8 or 1/4 watters on
the component side of the board.

What do you want to do?

I'm not at all confident on working with PC boards like that, so I
might actually make up a whole new unit, point to point construction,
leaving the DOT it intact.

 

[Dave.H]

On Feb 21, 12:23 am, John Fields <[email protected]>
wrote:
On Wed, 20 Feb 2008 03:28:14 -0800 (PST), "Dave.H"
I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.
The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.
---
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)
4.5V 3.1V
/ /
+--[25R]--+--[LED>]--+
| | |
|+ +--[LED>]--+
[BAT] | |
| +--[LED>]--+
| |
+--------------------+
<-- Iled
which means that the current into the LEDs is:
Vbat - Vled 4.5V - 3.1V
Iled = ------------- = ------------- = 0.056A = 56mA
Rs 25R
Unfortunately, since Vf/If is different for each LED (unless they
were very carefully matched) the current won't split equally, with
the result being that one LED may be hogging current beyond its
rating, resulting in a shortened life.
I suggest that you rewire the flashlight like this:
+--[R1]---[LED>]--+
| |
+--[R2]---[LED>]--+
| |
+--[R3]---[LED>]--+
| |
+-----[+BAT]------+
In order to determine the value of each of the resistors use:
Vbat - Vled
R = -------------
Iled
Where R is the value of the resistor, in ohms,
Vbat is the battery voltage, in volts,
Vled is Vf(min) for the LED, from the data sheet, and
Iled is the nominal forward current, in amperes, also from
the data sheet.
For example, If your LEDs are rated for a Vf(min) of 2V at 20mA,
then you'll have:
Vbat - Vled 4.5V - 2.0V
R = ------------- = ------------- = 125 ohms
Iled 0.02A
The closest standard 5% resistor on the low end is 120 ohms, so the
current it would allow through the LED would be:
Vbat - Vled
Iled = ------------- = 0.0208A ~ 21mA,
R
Which would be fine.
--
JF
I have linked to a photo of the back of the PCB to make it easier to
understand the circuit (I'm not very good at tracing out PCB's), hope
this helps.
http://s222.photobucket.com/albums/dd237/ozguy89/?action=view&current...

Including the switch, it's wired like this:

. +--[25R]--+--[LED>]--+
. | | |
. | +--[LED>]--+
. [SWITCH] | |
. | +--[LED>]--+
. | |
. +-----[+BATTERY]-----+

Looking at the photo, if you wanted to go with my suggestion it
seems to me the easiest way of getting there would be to rewire the
board, using cuts and jumpers, like this:

. +--[JUMPER]--+--[LED>]--[R]--+
. | | |
. | +--[LED>]--[R]--+
. [SWITCH] | |
. | +--[LED>]--[R]--+
. | |
. +--------[+BATTERY]----------+

If you have the room on the wiring side of the board it'd be an easy
matter to use surface-mount resistors, but if not it doesn't seem
that it'd be that much more difficult to use 1/8 or 1/4 watters on
the component side of the board.

What do you want to do?

I'm not at all confident on working with PC boards like that, so I
might actually make up a whole new unit, point to point construction,
leaving the DOT it intact.

I wouldn't mind just installing the red LEDs and a new resistor in
place of the old ones, and leave all wiring as-is. I don't mind if
the LED's have a shortened life. Not too hard to replace, IMHO. Would
I still go with the 47 ohm?

 

[John Fields]

I wouldn't mind just installing the red LEDs and a new resistor in
place of the old ones, and leave all wiring as-is. I don't mind if
the LED's have a shortened life. Not too hard to replace, IMHO. Would
I still go with the 47 ohm?

 

[Dave.H]

23 am, John Fields <[email protected]>
wrote:
On Wed, 20 Feb 2008 03:28:14 -0800 (PST), "Dave.H"
I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.
The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold-
brown as I mentioned earlier. I think the difference is only one ohm
or so. The green was light in colour making me think it was yellow,
but it does measure 25 ohms.
---
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)
4.5V 3.1V
/ /
+--[25R]--+--[LED>]--+
| | |
|+ +--[LED>]--+
[BAT] | |
| +--[LED>]--+
| |
+--------------------+
<-- Iled
which means that the current into the LEDs is:
Vbat - Vled 4.5V - 3.1V
Iled = ------------- = ------------- = 0.056A = 56mA
Rs 25R
Unfortunately, since Vf/If is different for each LED (unless they
were very carefully matched) the current won't split equally, with
the result being that one LED may be hogging current beyond its
rating, resulting in a shortened life.
I suggest that you rewire the flashlight like this:
+--[R1]---[LED>]--+
| |
+--[R2]---[LED>]--+
| |
+--[R3]---[LED>]--+
| |
+-----[+BAT]------+
In order to determine the value of each of the resistors use:
Vbat - Vled
R = -------------
Iled
Where R is the value of the resistor, in ohms,
Vbat is the battery voltage, in volts,
Vled is Vf(min) for the LED, from the data sheet, and
Iled is the nominal forward current, in amperes, also from
the data sheet.
For example, If your LEDs are rated for a Vf(min) of 2V at 20mA,
then you'll have:
Vbat - Vled 4.5V - 2.0V
R = ------------- = ------------- = 125 ohms
Iled 0.02A
The closest standard 5% resistor on the low end is 120 ohms, so the
current it would allow through the LED would be:
Vbat - Vled
Iled = ------------- = 0.0208A ~ 21mA,
R
Which would be fine.
--
JF
I have linked to a photo of the back of the PCB to make it easier to
understand the circuit (I'm not very good at tracing out PCB's), hope
this helps.
http://s222.photobucket.com/albums/dd237/ozguy89/?action=view&current...
---
Yes, that's what I thought.
Including the switch, it's wired like this:
. +--[25R]--+--[LED>]--+
. | | |
. | +--[LED>]--+
. [SWITCH] | |
. | +--[LED>]--+
. | |
. +-----[+BATTERY]-----+
Looking at the photo, if you wanted to go with my suggestion it
seems to me the easiest way of getting there would be to rewire the
board, using cuts and jumpers, like this:
. +--[JUMPER]--+--[LED>]--[R]--+
. | | |
. | +--[LED>]--[R]--+
. [SWITCH] | |
. | +--[LED>]--[R]--+
. | |
. +--------[+BATTERY]----------+
If you have the room on the wiring side of the board it'd be an easy
matter to use surface-mount resistors, but if not it doesn't seem
that it'd be that much more difficult to use 1/8 or 1/4 watters on
the component side of the board.
What do you want to do?

I'm not at all confident on working with PC boards like that, so I
might actually make up a whole new unit, point to point construction,
leaving the DOT it intact.

I wouldn't mind just installing the red LEDs and a new resistor in
place of the old ones, and leave all wiring as-is. I don't mind if
the LED's have a shortened life. Not too hard to replace, IMHO. Would
I still go with the 47 ohm?

I did this with a 56k, blew one of the LED's so I think I'll look at
other options than modifying the DOT it. Maybe just getting a cheap
flashlight and installing one red LED in that. What resistor value
would I need if this flashlight runs on two AA or AAA batteries? I
think I measured about 40 mA on one LED before they all went out.
Might have been the one that burnt out.

 

[Ecnerwal]

. +--[JUMPER]--+--[LED>]--[R]--+

. | | |
. | +--[LED>]--[R]--+
. [SWITCH] | |
. | +--[LED>]--[R]--+
. | |
. +--------[+BATTERY]----------+

Maybe just getting a cheap
flashlight and installing one red LED in that. What resistor value
would I need if this flashlight runs on two AA or AAA batteries? I
think I measured about 40 mA on one LED before they all went out.
Might have been the one that burnt out.

At an earlier point, I believe you had the reasonable response that
you'd perhaps just build the thing on perfboard from scratch, following
the correct circuit (the one I left above) rather than the flawed one of
the original light. You then appear not to have done that, but that
makes a great deal of sense, as you might now realize, having tried and
fried with the original.

2AA or AAA are 3V.

Typical red LED is 2V nominal. 20mA is typical safe current for generic
LEDs. So you have 20mA and one volt = 50 ohms. 51 is close enough, 63
would give more margin of safety, but somewhat less light.

 

[Dave.H]

. +--[JUMPER]--+--[LED>]--[R]--+
. | | |
. | +--[LED>]--[R]--+
. [SWITCH] | |
. | +--[LED>]--[R]--+
. | |
. +--------[+BATTERY]----------+

Maybe just getting a cheap
flashlight and installing one red LED in that. What resistor value
would I need if this flashlight runs on two AA or AAA batteries? I
think I measured about 40 mA on one LED before they all went out.
Might have been the one that burnt out.

At an earlier point, I believe you had the reasonable response that
you'd perhaps just build the thing on perfboard from scratch, following
the correct circuit (the one I left above) rather than the flawed one of
the original light. You then appear not to have done that, but that
makes a great deal of sense, as you might now realize, having tried and
fried with the original.

Yes that's right. Much easier to install in an old flashlight lamp
base.

2AA or AAA are 3V.

Typical red LED is 2V nominal. 20mA is typical safe current for generic
LEDs. So you have 20mA and one volt = 50 ohms. 51 is close enough, 63
would give more margin of safety, but somewhat less light.

Thanks, I still have the 56 ohm resistor I used when I fried the LED.
If it tests good I'll use that. I also have the two other red LED's
which work, so obviously, I'll be using one of them. Thanks for all
your help. Greatly appreciated!

 

[maxfoo]

I want to put 16,000 mcd red LEDs in an Osram Dot it light for working
outside when I don't want my night vision affected, the light in
question has 3 white LED's and a resistor (red-yellow-black-gold-
brown) voltage measured from the pins of the LED's is 3.122 VDC. How
would I go about installing these 2.0 volt red LED's? Do I just
change the resistor, if so what value? Unit is powered by 3 AAA
batteries.

just make a smps boost converter and you won't need the resistors, and the
battery will last longer too.

Here's a schematic of a "Joule Thief" circuit ...
Should be plenty of room for that pcb you posted.

VCC
+ -------------|
| | toroid |
-----|---. ,---- |
)|( | |
)|( | |
-------' '---|--|
| |
| |
| |
| |
| | LED LED LED
| --------->|-->|-->|------
| 1K | |
| ___ |/ ===
|---|___|--|2N3904 GND
|>
|
|
|
===
GND

 

[Dave.H]

just make a smps boost converter and you won't need the resistors, and the
battery will last longer too.

Here's a schematic of a "Joule Thief" circuit ...
Should be plenty of room for that pcb you posted.

VCC
+ -------------|
| | toroid |
-----|---. ,---- |
)|( | |
)|( | |
-------' '---|--|
| |
| |
| |
| |
| | LED LED LED
| --------->|-->|-->|------
| 1K | |
| ___ |/ ===
|---|___|--|2N3904 GND
|>
|
|
|
===
GND

I was looking at a circuit similar to that, might give it a try.

 

[maxfoo]

I was looking at a circuit similar to that, might give it a try.

I made one using 4 leds, schottky diode, and BUZ11 mosfet. posted it on flickr
Here's the link also have the ltspice simulation for it there...

http://www.flickr.com/photos/23742190@N08/2288673638/