Dynamic Range of 23 Stops

[David Williams]

John Galt will be no more "working" on that that Jim Jannard works on Mysterium sensors.

He may well have hired someone to design a chip on Panavision's behalf.

 

But this is one of those cherished notions that refuses to die. You are never going to get any more than about 12 stops of useful dynamic range out of any sort of digital video camera, at least one based on silicon. It's a technological brick wall where a whole lot of intractible problems come together to make further progress either impossible or ludicrously expensive.

 

And please do not invoke Moore's Law and other psuedo-technical Haikus. ALL electronic image sensors are Analog devices, with digital post processing. Garbage in - Garbage out. Moore's Law is not a real physical law, and in any case it only applies to digital devices, which have not yet reached the theoretical limits of their development.

 

If they want to try to re-write the laws of Physics, well that will provide more years of free entertainment.

 

I think you miss-understand the concept.

[Keith Walters]

I think you miss-understand the concept.

I understand perfectly well what he's saying.

I also understand why it won't work.

If you've seen some of the tortuous techniques attempted by various camera manufacturers to squeeze a lousy single extra stop out of their sensors, you might wonder THEY didn't think of this simple technique.

The answer is that it doesn't work, except for long exposures for absolutely stationary objects. For movie making it's about as practical as a pinhole camera.

[John Hoffler]

what about uses for "Black Silicon"? You guys ever heard of this stuff??

 

http://www.hno.harvard.edu/gazette/1999/12.09/silicon.html

 

http://www.tomshardware.com/news/black-sil...raphy,6485.html

[John Sprung]

what about uses for "Black Silicon"? You guys ever heard of this stuff??

 

http://www.hno.harvard.edu/gazette/1999/12.09/silicon.html

 

http://www.tomshardware.com/news/black-sil...raphy,6485.html

 

One little oops in the second article:

 

> Apparently light sensors using black silicon are 100 times more sensitive to visible light than conventional silicon sensors,

> allowing a pixel 1-micrometer in size to produce the same signal as a traditional pixel 10-micrometer in size.

 

> For consumers, smaller pixels could mean smaller sensors, which could use smaller lenses, resulting in cameras that

> are smaller and cheaper. Smaller pixels could also mean greater pixel densities, introducing a new generation of

> ultra-high resolution cameras that are capable of taking noise-free photos.

 

The writer isn't aware of diffraction. The smaller the chip, the shallower the stop at which it's diffraction limited. For instance, 2/3" chips hit the diffraction limit for red at f/8, 1/3" chips at f/4. Chips significantly smaller won't be practical because, basically, photons are too big. ;-)

 

 

 

 

-- J.S.

[John Hoffler]

One little oops in the second article:

 

> Apparently light sensors using black silicon are 100 times more sensitive to visible light than conventional silicon sensors,

> allowing a pixel 1-micrometer in size to produce the same signal as a traditional pixel 10-micrometer in size.

 

> For consumers, smaller pixels could mean smaller sensors, which could use smaller lenses, resulting in cameras that

> are smaller and cheaper. Smaller pixels could also mean greater pixel densities, introducing a new generation of

> ultra-high resolution cameras that are capable of taking noise-free photos.

 

The writer isn't aware of diffraction. The smaller the chip, the shallower the stop at which it's diffraction limited. For instance, 2/3" chips hit the diffraction limit for red at f/8, 1/3" chips at f/4. Chips significantly smaller won't be practical because, basically, photons are too big. ;-)

 

 

 

 

-- J.S.

 

i was wondering if you could clarify a little more on how this applies to color. I'm aware of the spectral artifacts side of refraction based on limited study of how CCD's work, but i'm not sure how this applies to color. i'm assuming is has to do with the smaller size of the spaces the light has to hit and the fact that red has a larger wave created by the photons than blue...

 

thanks!

[John Sprung]

i was wondering if you could clarify a little more on how this applies to color.

 

The way diffraction works is that instead of light being focused to a point, it makes a filled-in circle with a radius equal to 1.22 times the wavelength times the f/stop number. So, the longest wavelengths, which are red, make the biggest circles. These circles are called "Airy disks" after the scientist who first figured this out. Also, the deeper the stop, the bigger the disk. That's why lenses have a sweet spot in the middle of their aperture range. Shallow stops have limited DOF and use the less well corrected outer parts of the glass. Deep stops have bigger Airy disks. So, somewhere in the middle is best.

 

When the Airy disk radius gets bigger than the photosites on the chip, it's diffraction that limits your resolution, not the sampling grid of the chip. Because green is the most important color for subjective sharpness, you may want to cheat a little on the far end of red (700 nanometers), but really worry about it when it gets into the greens, say 550 - 600 nm.

 

Doing some arithmetic, suppose we take a typical 2/3" HD chip. It has 1920 x 1080 photosites in a rectangle 9.6mm x 5.4mm.

The photosite pitch is (9.6mm/1920) = (5.4mm/1080) = 0.005 mm = 5 microns = 5000 nanometers. To make the Nyquist limit, we want to filter out detail finer that twice that pitch, or 10 microns. That's what the OLPF does.

 

Now look at the lens. Suppose we shoot at f/8, and there's some red light at 700 nm:

 

Airy Radius = 1.22 x 700nm x 8 = 6382 nm. So, the diameter would be 13,664 nm = 13.664 microns. Red is losing resolution to diffraction here.

 

Try it again with a green at 550 nm:

 

Airy Radius = 1.22 x 550nm x 8 = 5368 nm. The diameter for green would be 10.736 microns, just a little softer than Nyquist requires.

 

So, f/8 is the deepest stop that won't suffer significant resolution loss from diffraction on a 2/3" camera.

 

 

 

 

-- J.S.

[John Sprung]

Bump

[Adrian Sierkowski]

You know John, every time i read one of your posts I am reminded about how wrong I was going into film thinking I'd never need math nor fractions again ;) ! I actually found that explanation very interesting and useful, so wanted to say a thank you for that!

[Chris Keth]

You know John, every time i read one of your posts I am reminded about how wrong I was going into film thinking I'd never need math nor fractions again ;) ! I actually found that explanation very interesting and useful, so wanted to say a thank you for that!

 

That's why I like photography and photographing films, actually. It's a beautiful combination of math, science, art, with a big handful of problem solving thrown in.

[Satsuki Murashige]

You know John, every time i read one of your posts I am reminded about how wrong I was going into film thinking I'd never need math nor fractions again ;) ! I actually found that explanation very interesting and useful, so wanted to say a thank you for that!

:) Agreed, thanks John.

 

Do you think what you're describing (red wavelengths making bigger airy discs) contributes to heavily red-saturated subjects appearing out of focus?

[John Sprung]

Do you think what you're describing (red wavelengths making bigger airy discs) contributes to heavily red-saturated subjects appearing out of focus?

 

Thanks, guys -- Diffraction would be part of the problem. The other part, depending on how well the lens is corrected for it, is chromatic abberation. The different colors focus on different planes. It's more a problem with older lenses.

 

 

 

-- J.S.

[Sasha Riu]

The way diffraction works is that instead of light being focused to a point, it makes a filled-in circle with a radius equal to 1.22 times the wavelength times the f/stop number. So, the longest wavelengths, which are red, make the biggest circles. These circles are called "Airy disks" after the scientist who first figured this out. Also, the deeper the stop, the bigger the disk. That's why lenses have a sweet spot in the middle of their aperture range. Shallow stops have limited DOF and use the less well corrected outer parts of the glass. Deep stops have bigger Airy disks. So, somewhere in the middle is best.

 

When the Airy disk radius gets bigger than the photosites on the chip, it's diffraction that limits your resolution, not the sampling grid of the chip. Because green is the most important color for subjective sharpness, you may want to cheat a little on the far end of red (700 nanometers), but really worry about it when it gets into the greens, say 550 - 600 nm.

 

Doing some arithmetic, suppose we take a typical 2/3" HD chip. It has 1920 x 1080 photosites in a rectangle 9.6mm x 5.4mm.

The photosite pitch is (9.6mm/1920) = (5.4mm/1080) = 0.005 mm = 5 microns = 5000 nanometers. To make the Nyquist limit, we want to filter out detail finer that twice that pitch, or 10 microns. That's what the OLPF does.

 

Now look at the lens. Suppose we shoot at f/8, and there's some red light at 700 nm:

 

Airy Radius = 1.22 x 700nm x 8 = 6382 nm. So, the diameter would be 13,664 nm = 13.664 microns. Red is losing resolution to diffraction here.

 

Try it again with a green at 550 nm:

 

Airy Radius = 1.22 x 550nm x 8 = 5368 nm. The diameter for green would be 10.736 microns, just a little softer than Nyquist requires.

 

So, f/8 is the deepest stop that won't suffer significant resolution loss from diffraction on a 2/3" camera.

 

 

 

 

-- J.S.

 

 

 

WOW!!!

 

:)

 

 

You know John, every time i read one of your posts I am reminded about how wrong I was going into film thinking I'd never need math nor fractions again ! I actually found that explanation very interesting and useful, so wanted to say a thank you for that!

 

 

You red my thoughts!

:)