While reading posts about photodiodes and counting photons I noticed that the topic is more than interesting, it could be useful.
I have a use to an affordable photodiode which is sensitive at around
By the way, if you are going to answer "yes" to my topic, please tell the name of the diode or its manufacturer too, please.
Leif M
300 nm is within the range of normal blue-enhanced silicon diodes, as long as they have quartz windows. Once you get down to 240 or thereabouts, your options become a lot more limited, and of course 100 nm is vacuum ultraviolet, so you can't do anything very useful in air except at very short ranges, which tends to be unhealthy. So you're looking at probably three orders of magnitude difference in cost for doing measurements across that range.
What are you hoping to do with it?
Cheers
Phil Hobbs
Well, these people do everything but the "cheap" part. I'm using them at 13 nm.
.highlandtechnology.com=A0 jlarkin at highlandtechnology dot com
Phil and John
I read from Wikipedia that every flame emits at these wavelenghts, so what about a flame detector. A few meters or yards would be enough for a distance.
at 13 nm. Heh yes, I understand. But usually price and sales are inversely proportional.
Leif
Incwww.highlandtechnology.com jlarkin at highlandtechnology dot com
The gas-tube flame detectors are fairly cheap. They are breakdown devices, so you power them from current-limited AC and look at the resulting average current. They have only cudely linear response. The ones I've used were sunlight blind but would fire from a lighted match a couple of feet away.
There are semiconductor (silicon carbide?) flame detectors around now.
See fig 3.
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John Larkin Highland Technology Inc
Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
Well, the old vacuum-tube photocells are nearly ideal for this, but why bother? Just put a fluorescent screen next to an IR through visible photodiode/phototransistor/integrated sensor.
This sensor seems to do the job.
jp.hamamatsu.com/resources/products/etd/pdf/R2868_TPT1008E02.pdf
I see you can get UV LEDs down to ~240nm without much trouble. Could you double that up to ~120nm and just about see through metals?
Best regards, Spehro Pefhany
Frequency doubling LEDs is hard, because you can't get enough intensity in the crystal for good efficiency. (There aren't that many crystals that work at 120 nm, either--I can't think of one off-hand, but then I'm a plain vanilla CW guy whenever possible.)
And IIRC it's really only alkali metals that become reasonably transparent in the UV, though there may be a few others. Even that would be pretty cool, though--I've never seen the experiment done.
Cheers
Phil Hobbs
Apparently, multivalent metals and alkali metals turn transparent above the plasma frequency.
There's a lot of variation in the reported numbers on the net- due to the ne (carrier density)?
There's a cool relationship between the plasma frequency and the London depth in superconductors. Reminded me what 1/sqrt(u0 * e0) is equal to!
So *that's* how Scotty did it!
Si Señor, but I don't see the relationship.
There's a kind of inductance due to the kinetic energy of the carriers. It doesn't usually affect room temperature electrical conduction in metals measurably because the carriers never get a chance to get moving very fast- they typically bump into something every 0.1ps or so ("relaxation time" in the Drude model of conductivity).
At really high frequencies (such as ~100nm hard UV) the kinetic reactance becomes important relative to the resistance in typical metals, or if the resistivity goes to zero (as in a superconductor) it also becomes important. In the latter case, it allows the wave to penetrate into the conductor (typically by ~100nm or so).
Inductance? I'd think it was more of a resistance. Energy can be stored?
Transparent aluminum. Huh. Hadn't noticed that phenomenon. ;-)
Sure, kinetic energy, on top of the energy stored in the magnetic field.
You should have sent away for the X-ray glasses from those little ads in the back of the comic books.
Best regards, Spehro Pefhany
Never held up a CD to the light? The layer is a bit thicker than "transparent", but the light which shines through is nonetheless -- blue! :)
Wonder how much is lost to refraction. Obviously, CDs are quite shiny. Could thin layers of metal, separated by dielectric layers, provide a good match to free space while filtering light in a useful way?
Reminds me of something I was reading about recently that was talking about nanometer layers of gold. Can't remember what it was actually about.
Tim
You mean reflection? There should be no significant refraction, the sides are flat and parallel.
You mean like sun glasses? ;-)
Isn't gold used for anti-reflective coatings on glass (gold-colored building glass isn't uncommon)? Partially aluminized surfaces are used for beam splitters and "one-way" mirrors.
I'm pretty sure Spehro is talking about a different phenomenon, though.
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