Sensitivity of the new SD1 sensor ?

A Foveon sensor uses about 8% of the light that it received.

There's two things going on. First is the photons that simply aren't absorbed in any of the sensitive layers.

This paper from Richard Lyon at Foveon explains some of that.

http://www.foveon.com/files/CIC10_Lyon_Hubel_FINAL.pdf

Have a look at page 3 of 7, figure 6 wavelenght vs. quantum efficiency. That actually looks a lot like the QE plot of a decent Bayer sensor, except we'd expect closer to 45% in the green.

But the big problem with that chart is that it's plotted all pretty in red, green, and blue, and the Foveon outputs aren't really red, green, and blue. Look at the spectral sensitivity curves, figure 7. There's actually wavelengths where a photon can show up equally in the top, middle, or bottom layers of the sensor. So. light of that wavelenght isn't interpreted as a color, it's interpreted as "white". One stimulus that can produce outputs in all the channels is "bad" from a signal processing standpoint. That's called "mutual information", and you have to uncouple the mutual parts and extract the "separable" information before anything makes sense.

If you actually look at the outputs of a Foveon sensor, the three outputs are more like white, yellow, and red. The top layer is all mutual information, you have to make the green and red that "contaminate" it go away before you can see the green.

The diagram on slide 26 of the AeroSense 2003 presentation by David Gilblom (Alternate Vision Corp), Sang Keun Yoo (HanVision), and Peter Ventura (Foveon) explained it pretty well. The link was:
http://www.alt-vision.com/r/documents/AeroSense_2003_Oral.pdf

But that appears dead. I'm not sure about the legalities of hosting the copy that I have.

The way it works is that a red photon can be absorbed by any of the three layers, at random. The red layer actually has the least probability of absorbing the red photon. But let's make the probabilities all equal, for the example.

So, imagine 30 red photons hit a pixel. 10 are absorbed in the first (white) layer. 10 are absorbed in the second (yellow) layer. 10 are absorbed in the bottom (red) layer.

Basically, 2/3 of the photons have turned into "noise" that must be subtracted from the white and yellow layers before they can become blue or green.

Same thing with green photons. They cause as much "noise" in the white layer as "signal" in the yellow layer.

The real number is actually 4.1 of red has to be subtracted away from the white channel. That's 1/4.1 or 24% of stuff that you get to keep.

Multiply that by the 32% QE, and you've only got 7.8% of your light actually used.

A Bayer sensor uses closer to 35% of photons. Virtually no destructive mutual information. About 4x the efficiency of a Foveon.

There's also the issue of differential sensing and uncorrelated noise, making reading low values from the sensor twice as noisy as a non-differential Bayer, so it's really more like 8x (or that the Foveon actually only uses about 4% of the light, in a low light situation).

--
Rahon Klavanian 1912-2008.

Armenian genocide survivor, amazing cook, scrabble master, and loving grandmother. You will be missed.

Ciao! Joseph

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