Laser noise

Indium Gallium Arsenide power wavehead photodetector.

What about using a lens to spread the light?

Wouldn't that effect or distort the noise or influence the reading otherwise?

Well, at the least it would reduce the absolute value of noise read. But if the light were spread evenly, and you knew the area ratio of spread to direct light...

True if you had the ratio just do the math.

Spreading would be fine, as long as it does not spread to my eyes. But thats a good idea.

greg

Like a filter?

Three front-surface reflections from window glass, plus an ordinary photodiode. Make sure the multiple reflections miss each other.

Cheers

Phil Hobbs

So you'd have 4% of 4% of 4%, or about .000064 of the original energy, right?

Use a clear optical glass beam splitter. Pick the incident angle to get as much or less as you want. Terminate the beam into a beam stop (brick?).

All sorts of ways of reducing the power density including lenses to spread it (as someone else mentioned), or a simple beam sampler using a glass plate to pick up 8 or 10 percent. A couple times and you get down to reasonable power levels. Or get a proper beamsplitter that can be obtained to reflect a very low percentage of the beam power. An optical filter or even a mirror that reflects most of the beam and transmits a small percentage can also be used, but probably only after the lens to reduce the power density so as not to damage it.

Then any photodiode will be fine.

Those expensive devices have some means of reducing the power to acceptable levels for their sensors.

Not as long as the noise relative to the total power into the sensor is what's measured.

-- sam | Sci.Electronics.Repair FAQ:

Important: Anything sent to the email address in the message header above is ignored unless my full name AND either lasers or electronics is included in the subject line. Or, you can contact me via the Feedback Form in the FAQs.

tml

is

in the

Pre-apologize for the naive question: Are you measuring amplitude noise and/or spectral noise? Are these valid questions?

Depends on the angle, but it's that order of magnitude. Figuring 5% per bounce, that 6 W comes down to about 750 uW. Keeping the reflections from overlapping is really key--interference between the different beams will make the ratio go all over the place.

Ordinary hardware store black paint, e.g. ultra-flat black Krylon, is a really really good index match to fused quartz, so if you have a quarter-inch quartz windows lying around, you can come in at a fairly oblique angle, let it bounce around inside a few times, and pick the reflection you want. If the angle of incidence is more than, say, 20 degrees, you have to worry about the polarization dependence of the reflection coefficient.

Laser noise is commonly space- and polarization-dependent--it's a surprisingly deep subject.

Cheers

Phil Hobbs

That's true only if the beam isn't vignetted--i.e. if you use a lens to spread the 6W out over a solar cell that can handle the power. Laser noise (especially in gas and diode lasers) often has a pronounced spatial variation. For instance, when there are multiple modes, each mode in itself may be much noisier than the total beam.

Cheers

Phil Hobbs

I like spitting the beam with a diffraction grating or even with a pair of these newer 3D glasses.

I'm mostly concerned with power supply noise or whatever is predominant.

greg

What type of laser are you using (sorry if I forgot!)?

If it's a diode laser, looking at an electrical signal from the power supply that corresponds to the laser diode current may be nearly as good as looking at the laser's output and doesn't require a sensor. :)

Right, but getting to the right point is somewhat of a risk. I don't want to send anymore lasers back to China.

Greg