Modifying laser to have additional white LED light

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Crimson Trace LG-431:
Battery:
Manufacturer Recommended:
1x 1/3N Lithium (3V, 11.6x10.8mm)
Or 2x 357 (SR44) Silver Oxide Battery (3V total, 11.6x5.4mm)
Possible alternatives to alter available voltage:
SR54: 11.6x3.1, 1.55V (4.65V Max)
SR55: 11.6x2.1, 1.55V (7.75V Max)
365/366,S16,608: 11.6x1.65, 1.55v (9.3V Max)

The sky is the limit for available voltage, but the device components have voltage and current limits.

Laser:
Output: ~5mW
Wavelength: ~650nM
Operating Voltage: Unknown, need to check.
Constant Voltage Driver: Unknown, need to check.

Switch:
Pressure Operated, connected to Laser housing via PCB ribbon. Not amenable to modification outside of housing.

Patents:
http://www.crimsontrace.com/company/patents
https://www.google.com/patents/US8256154

LEDs usually are fed from a current source since they operate roughly at constant voltage over a wide range of currents. The color of the LED is proportional to the voltage drop; higher energy photons (blue, and "white" [which are actually just blue LEDs with a phosphor that absorbs blue and phosphoresces yellow] LEDs) means higher voltage drop than low energy (red LEDs). So if you just stick one in parallel, unless it's exactly the same type of LED as the laser (which it won't be) only one of them will light up, the one with the lower voltage drop, as all the current will be shunted through it. Even if you figure out some way to balance it, since the laser diode is very likely fed a constant current, using the same circuit for both diodes means the original current will now be shared between both, so neither will operate correctly. Laser diodes in particular will now "lase" unless they have a minimum amount of current passing through. Best thing to do would be to add some circuitry to power the white LED, which only activates when the laser turns on.

>>1049349
Thanks for the reply. The board does have a potentiometer, which I think may allow me to increase the amount of current available after adding the LED, if that won't fry something else.

But that does still leave the problem of the mismatched voltage drops. It seems that laser diodes often have a voltage drop of ~1.5-2V. The only LEDs that operate with a similar voltage drop are non-high intensity red (~1.7V) and maybe high-efficiency, low-current, high intensity red (~1.9V).

On the upside the operating currents for both the laser and LEDs are both ~15-20mA.

It seems like the project could still work, albeit with a lower-intensity red LED. I just need to decide how to proceed with testing and adjusting the device to avoid frying it.

The safest bet might seem to be to very carefully slice/strip s section of the positive and negative and install a red LED and see if I can get simultaneous operation without any adjustments.

bump for you OP, this looks cool

>>1049517
>bump for you OP, this looks cool

Thanks. I've just about concluded that this will work, with a reasonable degree of safety.

The constant current driver is likely providing 20mA, which is suitable for the Laser Diode and a LED. If I wire the LED in series it will get 20mA of current. I will then need to increase the available voltage by adding additional batteries in series (increasing voltage at the cost of battery life in effect) until both components have sufficient voltage to operate: ~1.5Vdropoff for the driver, ~1.5V for the Laser diode, and ~1.5-5V for the LED depending on color and intensity.

A red LED would require the least additional voltage (~1.5-2V), while a high-intensity white may require more (~3-5V). We can easily achieve up to 9.3V, giving a surplus of 6.2V for the LED.

The biggest risk is overtaxing the driver. The nature of the driver remains unclear to me, but if it is a linear driver, it adjust the current at the cost of generating heat in a resistor, probably that tiny black surface mount component labeled (smaller black magic). Too much heat = ruined driver = need a new LG-431.

However, since the laser and light will not be in continuous operation I suspect it will be fine.

I'll buy the needed components and test it as soon as I get back from vacation in a week.