Kodak new small CMOS sensor

as you seem to know what you're talking about (I'm an absolute antitalent in anything electronics): Is there a chance, that the sensor has a logarithmic response curve? Or any other inherent aspect, that allows for a considerably imroved DR?

BTW, I do wonder why about nobody seems to be interested into this news topic!

Kind regards,
Martin

--
http://www.datzinger.net

scale it up and stuff it in the E4 or E5. If there is really a gain in DR it could be a good thing at any scale.
--
Garry

Typically pfets (p-type field effect transistors, MOSFETs that use holes rather than electrons) have lower noise than nfets but they are also a lot slower.

So, one would guess that Kodak decided to try to reduce transistor noise and are using either a slower active pixel, or resized the active transistor to make it faster.

Sacrificing one of the green filters for clear is a good idea for boosting low light sensitivity. I wonder if they increased the charge handling capacity for that pixel or if it goes "snowblind" under normal lighting conditions for the other pixels.

I am glad to see Kodak throwing a few punches. For a while I thought they were down for the count. Sad, since they were the first image sensor company to have access to the CMOS active pixel technology. But, they aren't down and out yet.

I wonder if Kodak's fab partner has been publicly announced?

-Eric
Pres, ImageSensors, Inc. A California Non-Profit, Public Benefit Corporation
see http://www.imagesensors.org

Eric Fossum said:

Sacrificing one of the green filters for clear is a good idea for
boosting low light sensitivity.

Click to expand...

I read that the reason they kept the 2:1:1 ratio and went with a whole row of clear was to keep the demosaic algorithms same or similar. Something to do with the ASIC chips? But if that weren't the case, RGBC would seem to be a better compromise approach.

Eric Fossum said:

I wonder if they increased the
charge handling capacity for that pixel or if it goes "snowblind"
under normal lighting conditions for the other pixels.

Click to expand...

That seemed to be the statement in comments regarding the partial-panchromatic designs. The clear pixel ends up being "shifted" by 2 EV or something. So yeah, it would blow out sooner than the other pixels.

But then, you'd think "so what" on the properly exposed segments...you don't need that data anyways. It would help out on the underexposed pixels to bring up detail.

My ignorant brain is imagining chroma-defiicient detail in the shadows, rather than full extended DR. But B/W shadows would be a cool effect.

Greg

Inspectoreleven said:

So if this sensor measures the absence of light instead of the usual
way...would this mean clean shadows & noise in the highlights?

Yeah, I doubt it too, but I don't know enough of the theory to be
able to back up an answer either way.

Click to expand...

Nope, you still get noise in the shadows. Think of it this way:

This sensor will start out "white", and as photons fall onto it, it start painting it black (or making pixels darker). Really dark pixels mean they're saturated with a lot of light, and have less noise, and the really bright pixels (closer to white) have little light, more variation, and thus lots of noise.

So the sensor has a light of noise in the "bright" areas and little in the dark areas. BUT, you have to remember that what the sensor has recorded is a NEGATIVE of the image (just like negative film) - you still have the invert it to get the real image, and thus the noisy bright areas become noisy shadow areas, as usual.

I would think, based on slide film vs. negative film, that this will reverse the trend of digital having very good shadow performance but clipping out highlights fast - instead we'll find that we're going to have lots of highlight headroom, but less play in the shadow areas if we try to adjust curves and things.

Martin Datzinger said:

as you seem to know what you're talking about (I'm an absolute
antitalent in anything electronics): Is there a chance, that the
sensor has a logarithmic response curve? Or any other inherent
aspect, that allows for a considerably imroved DR?

Click to expand...

Not likely. The devices typically used to absorb photons and convert them into charge are linear in nature until the charge-storing wells are near capacity when they can become nonlinear (but not a controlled, logarithmic relationship).

David

Eric Fossum said:

Typically pfets (p-type field effect transistors, MOSFETs that use
holes rather than electrons) have lower noise than nfets but they are
also a lot slower.

So, one would guess that Kodak decided to try to reduce transistor
noise and are using either a slower active pixel, or resized the
active transistor to make it faster.

Sacrificing one of the green filters for clear is a good idea for
boosting low light sensitivity. I wonder if they increased the
charge handling capacity for that pixel or if it goes "snowblind"
under normal lighting conditions for the other pixels.

Click to expand...

It's always nice to read your posts, Dr. Fossum. I was actually a student of yours long ago.

David

If it works 'opposite'...

George Sears

Yes, smaller CMOS sensors do not compare favorably to smaller CCD sensors, though Canon is also working on this.

George Sears said:

If it works 'opposite'...

Click to expand...

It "darkens" in the sense that it records holes instead of electrons. Just think of it as negative film vs. positive slide film.

Just a bit curious about this: since we're moving back to negatives and recording holes, is there a chance that reciprocity failure will be creeping back?

Wow, that's great! There are a few David G's that come to mind. Why don't you find my email address from imagesensors.org and write me a note! I'd love to find out which DG you are, and what you have been up to the last, uh, + - 20 yrs.
cheers,
Eric

no no no, you guys. Cut it out. No negative stuff going on here, no reciprocity failure etc etc.

When a photon strikes the silicon crystal, it breaks one of the Si-Si bonds. This lets loose and electron to wander in the lattice. But, it also leaves behind a broken bond, aka, hole. Since the bond was initially neutral, with the electron gone it is now positively charged. Furthermore, it is easy for an electron in an adjacent "whole" bond, to move into the broken bond and heal it. However, when that electron moves over, it leaves behind yet another broken bond. It is like the broken bond (hole) is moving in the opposite direction.

So, if you want to count photons, you can either count the electrons that are let loose, or you can count the number of broken bonds (holes) and from that deduce the number of incident photons. The more electrons, the more photons. The more holes, the more photons. You can even count electrons and holes, but you are still going to get the answer.

Hope this helps.
Eric

Do you mean can we make a p-channel CCD? Absolutely. Many early CCDs were p-channel. But it is easier to make a fast n-channel CCD since electrons can be moved about much more quickly in silicon than holes.

As far as the filter design goes, yes, it would work perfectly well with a CCD too, as long as it had anti-blooming control.

see my other response. It still detects the presence of light (photons). It just counts the holes instead of the electrons, when the photon generates and electron-hole pair.

Eric Fossum said:

no no no, you guys. Cut it out. No negative stuff going on here, no
reciprocity failure etc etc.

When a photon strikes the silicon crystal, it breaks one of the Si-Si
bonds. This lets loose and electron to wander in the lattice. But,
it also leaves behind a broken bond, aka, hole.

Click to expand...

Question: in a regular CMOS or CCD sensor, doesn't this hole get "refilled"? Otherwise another photon striking that same crystal wouldn't get detected, since there are no more valence electrons to knock loose.

bearcat said:

Yes, smaller CMOS sensors do not compare favorably to smaller CCD
sensors, though Canon is also working on this.

Click to expand...

Omnivision would strongly disagree, in fact, I've been testing an 2.2mm x 2.1mm sensor from them which performs extremely well!

--
Simon Taylor
http://www.phooto.co.uk

'The belief that something is NOT impossible is the absolute essence of discovery'

Nathan Yan said:

Nathan Yan said:

When a photon strikes the silicon crystal, it breaks one of the Si-Si
bonds. This lets loose and electron to wander in the lattice. But,
it also leaves behind a broken bond, aka, hole.

Click to expand...

Question: in a regular CMOS or CCD sensor, doesn't this hole get
"refilled"? Otherwise another photon striking that same crystal
wouldn't get detected, since there are no more valence electrons to
knock loose.

Click to expand...

There are a ziillion bonds per cubic micron of silicon material so no shortage of bonds to break or valence electrons to set loose.

The holes generally flow to some part of the silicon where they are annihilated by electrons (so-called recombination). This completes the "circuit".

There is some interesting detail on this new sensor at:

http://www.adorama.com/catalog.tpl?op=NewsDesk_Internal&article_num=061407-4

From the interview with John Compton and John Hamilton of Kodak, these comments caught my attention:

JH: The Bayer filter pattern has a very tight 2x2 repeat pattern. So, for a red pixel, you're never more than two pixels away from another red pixel. One of the new patterns uses a pan checkerboard and on the complement of that checkerboard, there is a pair of reds, a pair of greens, another pair of greens and a pair of blues.

Finding the right color edge of something can be a challenge. You've got to tie the color edge to the pan image, which gives you a good idea of where that edge is. What you'd like to do is bring the color out to the edge, but keep it from going any further. If you hold to these edges, it's hard to do the noise cleaning because that is done by averaging pixels that you expect to have about the same value. If you're not careful, you'll be averaging pixels on either side of the edge and you'll ge what we call "color bleed." For instance, if you have skin next to blue jeans, you'll see a cyan halo on the hand.

And if you overclean the image, it looks like plastic, because it is just too smooth. So, it's hard to get the right amount of cleaning ensuring that you reduce the noise, and at the same time, keep the edge definition reasonably good. And so there's been a lot of work done on finding the best way to do that.

--
Darrell
http://members.aol.com/pixbydg/still/life.html
http://members.aol.com/pixbydg/New/Gallery.html

Eric Fossum said:

Eric Fossum said:

Eric Fossum said:

When a photon strikes the silicon crystal, it breaks one of the Si-Si
bonds. This lets loose and electron to wander in the lattice. But,
it also leaves behind a broken bond, aka, hole.

Click to expand...

Question: in a regular CMOS or CCD sensor, doesn't this hole get
"refilled"? Otherwise another photon striking that same crystal
wouldn't get detected, since there are no more valence electrons to
knock loose.

Click to expand...

There are a ziillion bonds per cubic micron of silicon material so no
shortage of bonds to break or valence electrons to set loose.

The holes generally flow to some part of the silicon where they are
annihilated by electrons (so-called recombination). This completes
the "circuit".

Click to expand...

^This is what I thought. So doesn't recombination make it impossible to count the holes?