Lighting a .010 green fiber optic with led

Not possible, or not simply and reliably anyway.
A white LED will need over 3V; typically 3.2 to 3.6V dependant on type and current.
Also, a 2032 can give next to no sustained current; try and take more than a few mA and the voltage will just drop off.

You need a higher voltage, ideally around 5 - 6V, or a higher power battery and a switch mode power supply to boost the voltage.

With appropriate power, you could use a single transistor current source (or current sink) circuit.

The circuit below is a simple constant current sink.

To make it adjustable, replace R1 and R2 with a resistor from positive (top connection and two signal diodes such as 1N4148 in series to 0V (bottom connection).
The resistor value depends on the supply voltage; use a value that gives a minimum 10mA through the diodes with a near flat battery.

Add the potentiometer (100 Ohms) across the upper of the two diodes and connect the transistor base to the pot wiper.
That gives a variable bias from roughly 0.6V to 1.2V for the base.

The LED would connect in the "Load" position and make Re a suitable value so around 0.6V across it gives the highest LED current you need; eg 20 Ohms would give roughly 30mA maximum.

What is a green fiber optic? And, why do you need a white LED to illuminate it?

Mike.

Thanks, not as simple as the ones I can purchase (maybe why they are so expensive)
The green fiberoptic is a .010 strand of optic monofilament attached to a target bow sight that collects sunlight directing it to the end facing the archer that is used as a visual aiming point on the target.

OK, so it's intended for someone to look directly at the light source; that's a bit different to what I thought was required, from the original question; I took it to mean fibre end had to be visible at a distance in full sun...

As long as you could use an ultra high efficiency LED, the current could be microamps..
eg. One of these is rated 12.9cd at 20mA (12900 mcd, compared to about 20 - 50mcd for a classic "power on" type LED at 20mA.


Use an ICL7660 or similar voltage inverter IC, running from the button cell. The 7660 is made by various companies with different prefix letters.

That will give 6V between the feed from battery positive to the negative output from the converter IC' plenty to run the white LED and a decent voltage across a current limiting resistor.

Experiment with different series resistor values feeding the LED; start with at least 100K (100,000 ohms) and go from there. Once you find a value to give the brightest illumination, we can work out how best to dim it.

Do not run it like a normal LED at milliamps; the output on those ultra high efficiency ones is enough to cause eye damage, at normal LED currents!

I'd be willing to bet that those sights use a green LED and a common clear fiber optic. The green LED is also better suited to a coin cell battery.

Again, thank you for your time. This belongs to a friend., and is what I am trying to replicate except with more dimming range. It retails for
$90.00.
The switch is a grayhill 56s30-01-1-12n-f with off at each end 10 brightness settings 2 through 11. It seems to have just a conductive strip soldered to 10 positions. And no other components. I just need more adjustment past the low end . Can a potentiometer of ?ohms replace this switch? Please excuse my ignorance.

They are green and the light source is white. When the target is in the shade , sometimes the pin brightness on its lowest setting is too bright giving a starburst effect to these too old eyes.
The green led may be my ticket ! That will be plan "B"

That switch has a number of surface mount resistors arranged around the back; it's emulating a potentiometer but with steps rather than continuous rotation.

As Shortbus says, it must be a green LED than needs a rather lower voltage than a white one. Remember that a green (or any colour) LED can have a clear casing, they are not always coloured plastic.

Slb said:

They are green and the light source is white.

Click to expand...

Ok, I don't think those "Pins" are really a fiber optic. I say that because the light must be comming from the side of the pin. Fiber optics the light only is seen at the end of the fiber. That's why so many signals can be sent down a fiber bundle at one time and not get a cross talk of them. Your pins are probably an acrylic rod, acrylic allows light to escape from the sides.

I'm sorry my description was misleading. It's not signal quality fiber , but colored flexible plastic optic fiber Made by nanoptics specifically for this application.