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Can you change laser diode for LED to produce white light? Answered

Hi All,

I have an existing circuit that I'm removing from a donor device that currently uses a class 3R / 5mw laser diode running off of two 357 batteries (1.55v, I believe). The functionality of this module is ideal for my application (has an on/off button positioned exactly where I would like it), but I want to repurpose it to produce white light, not a focused laser beam. I know next to nothing about circuitry, so pardon me in advance if this is either a dumb question or just not feasible: Is it possible to just de-solder the 5mw laser diode and replace it with a white LED, and if so, what specs do I require for the LED? The existing circuit is tiny, so just making a new module is beyond anything I could do, despite simple LED circuits being relatively easy to put together. Thanks in advance to any answer I might get.


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Jack A Lopez
Jack A Lopez

2 years ago

Hey. I just re-read your question, and I realized my previous answer did not really address what you were asking.

You want to use the circuit that drives a (red?) laser diode, to drive a white LED instead, by just swapping the laser diode with a white LED.

I expect that would work, if the Vd, the characteristic forward voltage drop for the LED, is low enough.

I am guessing the existing circuit that drives the laser diode is a constant current driver, and the magnitude of that constant current is only about 2 mA.

How can I guess the magnitude of the current?

Because the product of voltage across, and current through, that diode is only about 5 mW.

e.g. P = V*I = (2.5 V)*(2 mA) = 5 mW, where those numbers are approximate.

Jack A Lopez
Jack A Lopez

2 years ago

Well, there exist white LED modules (i.e white LED plus driver and power converting circuitry), that are probably a good match to two small alkaline cells, like your 357, or LR44 coin cells.

Actually these modules are cleverly disguised in a package that is roughly the same shape as the the old incandescent flashlight bulbs, made for two cell (e.g. 2xAA) flashlights.

It turns out a design with a single white LED, will possibly be more complicated.

You asked about how LEDs are specified.

The usual numbers used to specify an LED, are its approximate forward voltage drop (in volts), Vd, and its maximum rated current (in milliamperes), Imax.

So if you want to drive a single LED, with two alkaline cells in series, giving approximately 1.5+1.5 = 3.0 volts, then Step 1 is to pick a LED with Vd less than 3.0V.

Does such a LED exist? I think so. I think I have seen a white LED, with Vd = 2.8 V,and I saw it in a piece of junk from this place,


that was powered by 2xAA cells. You know, if want some white LEDs to play with, and you have a DollarTree in your town, that might be the place to get them from.

Step 2, is choose a resistor to put in series with the LED, for to limit the current to some value less than, Imax.

By the way, choosing how much current flows through the LED is equivalent to choosing how bright it will be. The amount of light emitted is roughly proportional to current.

The math for calculating the size of the current limiting resistor is simple, but noobs don't grok this math. So to help out the noobs, there exist web-based calculators, most famously this one,


for to do the math for you.

In some no frills designs, particularly those powered by coin cells, the series resistor is mysteriously omitted.

It turns out the reason why that works is because the coin cells actually have significant internal series resistance.

It is like, for coin cells, the resistors are built in.

Actually, I was looking at the Wikipedia page, "List of battery sizes",


and I noticed next to your cells, your 357, or LR44, someone left a note that says "Typical internal resistance: 8 ohms"

So I guess two of those cells in series would have a total series resistance of about 16 ohms, for what that is worth as a current limiting resistor.