Convert Brinkman LED Flashlight to IR?

[DougC]

I have a Brinkman LED flashlite I want to convert to an IR led.
A schematic of the Brinkman driver is here:
http://edusite10.tripod.com/led3/brink/brinkmann.gif
The IR LED is the GaAlAs type noted halfway down the page here:
http://ledmuseum.home.att.net/ledir.htm
The second page notes that the IR LED requires 50 ma max, but the
Brinkman circuit peaks at 300 ma and averages 85 ma. What resistor would
need to be changed to lower the Brinkman circuit? If I can find a
trimmer I would like to start it out low and adjust it up as needed--it
is for use with NV goggles, and so the IR LED may not need to be run at
its maximum power.....
-end-

 

[Don Klipstein]

In article <102iiejnfo5804e@corp.supernews.com>, DougC wrote:

I have a Brinkman LED flashlite I want to convert to an IR led.
A schematic of the Brinkman driver is here:
http://edusite10.tripod.com/led3/brink/brinkmann.gif
The IR LED is the GaAlAs type noted halfway down the page here:
http://ledmuseum.home.att.net/ledir.htm
The second page notes that the IR LED requires 50 ma max, but the
Brinkman circuit peaks at 300 ma and averages 85 ma. What resistor would
need to be changed to lower the Brinkman circuit? If I can find a
trimmer I would like to start it out low and adjust it up as needed--it
is for use with NV goggles, and so the IR LED may not need to be run at
its maximum power.....

Fine the power supply rail-to-rail voltage with the LEDs in use
operating.
(Just for reference, so many of the "usual" / "regular sizes" white LEDs
have a rated maximum continuous current of 30 mA and peak of 100 mA.)

Infrared LEDs that only tolerate 50 mA at max. current drop approx.
1.3-1.5 volts at max. current. Subtract 1.3 volts from the rail-to-rail
DC supply voltage (unlikely to exceed the battery voltage if the battery
voltage is normally near or over 4.5 volts). Divide this difference by
..05 amp and you get a moderately alarmist required resistor in ohms. Use
the next higher value among those available at your favorite parts
supplier.

- Don Klipstein (don@misty.com)

 

[DougC]

Thank you--you are describing just using a dropping transistor, which I
did not find info on until after I posted:
http://edusite10.tripod.com/led3/bulb/index.html
It is much easier to do--I could not figure out how to modify the
pulsing circuit of the Brinkman....??? but who cares. There is plenty of
room in the front end of the Brinkman for a trimmer pot if I remove the
pulsing circuit components.

Don Klipstein wrote:

Fine the power supply rail-to-rail voltage with the LEDs in use
operating.
(Just for reference, so many of the "usual" / "regular sizes" white LEDs
have a rated maximum continuous current of 30 mA and peak of 100 mA.)

Infrared LEDs that only tolerate 50 mA at max. current drop approx.
1.3-1.5 volts at max. current. Subtract 1.3 volts from the rail-to-rail
DC supply voltage (unlikely to exceed the battery voltage if the battery
voltage is normally near or over 4.5 volts). Divide this difference by
.05 amp and you get a moderately alarmist required resistor in ohms. Use
the next higher value among those available at your favorite parts
supplier.

- Don Klipstein (don@misty.com)

 

[Watson A.Name - \"Watt Su]

"DougC" <dcimperg@norcom2000.com> wrote in message
news:102iiejnfo5804e@corp.supernews.com...

I have a Brinkman LED flashlite I want to convert to an IR led.
A schematic of the Brinkman driver is here:
http://edusite10.tripod.com/led3/brink/brinkmann.gif
The IR LED is the GaAlAs type noted halfway down the page here:
http://ledmuseum.home.att.net/ledir.htm
The second page notes that the IR LED requires 50 ma max, but the
Brinkman circuit peaks at 300 ma and averages 85 ma. What resistor
would
need to be changed to lower the Brinkman circuit? If I can find a
trimmer I would like to start it out low and adjust it up as
needed--it
is for use with NV goggles, and so the IR LED may not need to be run
at
its maximum power.....
-end-

It looks like there's a mistake in that circular wiring diagram.

You don't need the circuit if you use IR LEDs because IR LEDs need only
1.5V or so to conduct that much current, compared to well over 3V for
white LEDs. So you can just remove the circuit and replace it with a
single resistor. The resistor should be 1.5V / .05A or 30 ohms.