Fast, efficient IR LEDs?

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Perhaps accurately controllable optical attenuation can be done? =20 Don't know myself but it sounds possible to me.

I'd have to think about Phil's circuit but would need more info for that (and time ...). My case with the single diode, not at liberty to say. But we built dozens and they all cal'd automagically.

Just don't wait until T minus 360 days :-)

Good, so it seems automatic. 14V sound like a white-knuckle ride :-)

Hehe, that's what I used to say as well. No home without a fully certified cardiology ultrasound scanner. And another one for ob/gyn if the freshly married couple is inclined to ...

It can, but it's expensive.

The greenies appear to be the most efficient. Or, at least, visibly most efficient, if not in real life.

John

Just for the heck of it, I asked Jonathan to measure the low-current linearity of some visible and IR led's. I know that some LEDs can make visible light at 1 nA, so it will be interesting to see if there is a linearity knee somewhere. A red LED driving a silicon PIN diode makes a visually perfect straight-line graph plotted linearly from 0 to 55 mA.

I theory, LED voltage is the log of current, so at some very low current there won't be enough voltage across the junction to make a photon of anywhere near the expected wavelength. It could be that materials defects will kill things before that point.

I did some googling on LED behavior at low currents and found nothing useful.

John

Academics have tried to do single-photon generation with LEDs. Pulsed at very low current. IIRC one paper was from Syracuse, NY. But I don't think you get access to this stuff directly on the web, probably needs some paid access like IEEE Explore. Or good connections to a university.

Well, everything is radiating, although at 26meV and no bias, there's damned little all the way out at 2eV. You're looking at the tail end of two decades of exponent there.

There's no reason why, for instance, you can't get 2.000eV photons from a semiconductor with 1.984V across it. You can electrolyze water the same way -- it does proceed below the reaction voltage, it's just endothermic and slow as hell. You don't get something for nothing, so there's still current draw, with an electron plopping through the bandgap for every photon radiated, it's just falling through a slightly lower voltage.

Question for Phil: does the process of photon production cause a blip in the diode voltage? This should be detectable as shot noise on the diode's terminal voltage -and- on a photodiode directly in front of the LED, and there should be perfect correlation between the two effects (minus quantum efficiency, so maybe you'll only detect 1 in 5 events at the photodiode). Is this measurable? I think it should be.

Tim

If the Fed Reserve finds out they'll want the tax that's owed on those

Don't know but generating single photons is done differently, or at least will be some day:

It should be. Macroscopically, voltage noise across the junction must modulate intensity. I don't know if you could observe this at the single photon level.

John

Normally of course you can't put current sources in series, but by applying optical feedback you can make series-connected photodiodes work.

Since the diodes have essentially infinite impedance, the noise of each one splits in half, with half going through each capacitor.

The shot noise current from each diode divides by the ratio of the conductances of the two branches. Thus goes through the other diode's capacitance, and so into the external circuit, but half just circulates round through its own capacitance and hence doesn't contribute to the output noise.

The shot noise currents from the two diodes are uncorrelated, and so the RMS noise current arriving in the external circuit is

/ |2eI_dc| |2eI_dc| \ i_N = sqrt| |------| + |------| | = sqrt(eI_dc) \ | 2 | | 2 | /

which is 3 dB below the shot noise of a primary photocurrent of I_dc. That's the same SNR you'd get by parallelling the two, but (crucially) you don't double the capacitance or the photocurrent by doing so.

That makes it a good trick for low photocurrents, though not one you'd use every day.

Cheers

Phil Hobbs

OK. I read you! Thanks! ...Jim Thompson

Not at all. Most Si photodiodes are good to 30-60V. With many (e.g. the BPW34) the capacitance stops decreasing at 10V or so, but they continue to speed up with increasing bias, because the series resistance keeps going down. That's because it's dominated by the (very thin) diffusion zone, so cranking up the bias until they're really really fully depleted makes the speed go up amazingly. Silvio Donati's book on photodetectors is an excellent read for this sort of stuff.

Cheers

Phil Hobbs

Where do you get that book? Searching on "Silvio Donati" "photodetector" seems to scramble google's brains ;-) ...Jim Thompson

Ok, yes, the big ones can do that. I was thinking about the ones I used, mostly from Japan, fast ones with low capacitance. Their abs max is between 5V and 15V. Not sure by how much you could exceed that before going *phut* but I couldn't allow that kind of system to be ECO'd in that mode anyhow.

Please be so kind as to give a few makers and part numbers for silicon PIN diodes usable that way. Thanks.

..and HOW does one determine / detect a single photon? Tap into nerve of a cat?

Cheers

Phil Hobbs

Thanks! ...Jim Thompson