The Canon EOS 5D Mark V is in the works [CR2]

BillB

EOS 6D MK II
May 11, 2017
1,150
385
Why not? What if the sensor accumulates exactly zero photoelectrons?

You cannot avoid that problem without treating signal quantization as a sort of sensor's nonlinearity by itself.
The practical problem would seem to be constructing the ADC to return a bit value if no photons are detected. Wouldn't a simple "if measured value is less than n photons, then return 0" get you out of the theoretical box?
 

Kit.

EOS 6D MK II
Apr 25, 2011
1,347
740
The practical problem would seem to be constructing the ADC to return a bit value if no photons are detected.
It practically is doable, but then we get a theoretical infinity in our definition of "DR".

Wouldn't a simple "if measured value is less than n photons, then return 0" get you out of the theoretical box?
Then our sensor is not linear anymore, and the chosen value of n will determine its DR.
 

Quarkcharmed

EOS 5DMkIV
Feb 14, 2018
517
369
Australia
www.michaelborisenko.com
Why not? What if the sensor accumulates exactly zero photoelectrons?

You cannot avoid that problem without treating signal quantization as a sort of sensor's nonlinearity by itself.
It's not possible, there's no physical systems without noise. As far as I understand, zero value after applying 7 hi ADC represents the noise level or is close to it. After a 14-bit ADC, the highest value 16383 has to be interpreted as 14 stopsbrighter than the lowest value and all values are linear (raw files are gamma 1.0).
 

Kit.

EOS 6D MK II
Apr 25, 2011
1,347
740
It's not possible, there's no physical systems without noise. As far as I understand, zero value after applying 7 hi ADC represents the noise level or is close to it. After a 14-bit ADC, the highest value 16383 has to be interpreted as 14 stopsbrighter than the lowest value and all values are linear (raw files are gamma 1.0).
That would mean that the DR of your sensor is 7 stops.

However, you are mistaken in the assumption that the level of noise does not depend on the level of the signal. When we are talking about photon shot noise, a signal with full 14 bits of value will be accompanied with ~7 bits of noise; a much lower signal with 8 bits of value will be accompanied with ~4 bits of noise, and a total darkness (a signal with all bits equal 0 on a linear sensor) will have no photon shot noise at all (there will still be dark current noise, of course, but t can be very small and not flip the lowest bit most of the time).
 

Michael Clark

Now we see through a glass, darkly...
Apr 5, 2016
990
484
The number of electrons and therefore the voltage in a pixel is a linear function of number of photons, and the voltage is then converted to a digital value via ADC. Now with the steps, they're normally defined as the smallest detectable level of the signal, effectively the noise, and the DR is defined as max level/noise level, so by definition a linear ADC can't produce a DR larger than its bitness.
There's no "voltage" collected by a pixel, there is an amount of energy properly defined as a charge. Photons vibrating at different wavelengths of light release slightly different amounts of energy when they are absorbed by a photosite. The full well capacity of many cameras is well beyond the maximum number of steps possible with 14 bits. The EOS 1D X, for instance, has FWC of over 90K electrons, which requires 17 bits to allow a unique digital value for each increase of one electron. That's 5.5X more information than can be expressed in 14 bits.
 

Michael Clark

Now we see through a glass, darkly...
Apr 5, 2016
990
484
That would mean that the DR of your sensor is 7 stops.

However, you are mistaken in the assumption that the level of noise does not depend on the level of the signal. When we are talking about photon shot noise, a signal with full 14 bits of value will be accompanied with ~7 bits of noise; a much lower signal with 8 bits of value will be accompanied with ~4 bits of noise, and a total darkness (a signal with all bits equal 0 on a linear sensor) will have no photon shot noise at all (there will still be dark current noise, of course, but t can be very small and not flip the lowest bit most of the time).
Further, in the real world there can be some charges that are negative by the time the information reaches the ADC, because the system has absorbed more energy than the sensor produced for that particular photosite. Other photosites that collected the same amount of charge can still be positive when that information reaches the ADC. That's why we call the variability of the effect the system has on the analog signal between each pixel well and the ADC "noise".
 

Quarkcharmed

EOS 5DMkIV
Feb 14, 2018
517
369
Australia
www.michaelborisenko.com
There's no "voltage" collected by a pixel, there is an amount of energy properly defined as a charge. Photons vibrating at different wavelengths of light release slightly different amounts of energy when they are absorbed by a photosite. The full well capacity of many cameras is well beyond the maximum number of steps possible with 14 bits. The EOS 1D X, for instance, has FWC of over 90K electrons, which requires 17 bits to allow a unique digital value for each increase of one electron. That's 5.5X more information than can be expressed in 14 bits.
Yes, but after ADC the limiting factor is the bitness of the ADC, which was my original point.

 

Michael Clark

Now we see through a glass, darkly...
Apr 5, 2016
990
484
Yes, but after ADC the limiting factor is the bitness of the ADC, which was my original point.

But the bitness only limits the distance between each step for the same total distance between "0" and 2^n - 1. It doesn't require that each doubling of the number of steps means a doubling of the total brightness between "0" and 2^n - 1. A 2:1 or 1:2 slope is still just as linear as a 1:1 slope is.
 

Quarkcharmed

EOS 5DMkIV
Feb 14, 2018
517
369
Australia
www.michaelborisenko.com
But the bitness only limits the distance between each step for the same total distance between "0" and 2^n - 1. It doesn't require that each doubling of the number of steps means a doubling of the total brightness between "0" and 2^n - 1. A 2:1 or 1:2 slope is still just as linear as a 1:1 slope is.
But that paper (and other papers on this matter that I saw) suggests/implies that the slope is 45 degrees, i.e. 1:1. The standard for raw files is to use gamma 1.0, which is a straight line at a 45-degree slope:

I agree it doesn't have to be. But it is in the current sensor implementations, as far as I can see.
 

stevelee

FT-QL
Jul 6, 2017
1,250
281
Davidson, NC
In addition to the features already mentioned for the 5D Mark V the addition of focus bracketing would be very useful for in field macro photography. If it has this feature and some increase in resolution among other improvements I would upgrade from my 5D mark IV.
That could help sometimes. Often for macro, you want the lens left alone and the camera moved on a rail.
 

dtaylor

Canon 5Ds
Jul 26, 2011
1,394
830
All the papers I've read so far on digital sensors point out they're linear. A single pixel can have more than 14 stops of well capacity, but it's then clamped and converted lineary in ADC. So the number of bits in ADC is the upper bound for the resulting dynamic range.
The highest ranked cameras at DxO have sub-14ev DR scores in their screen test. Only the print test scores exceed 14ev. We know their print scores come from downsampling (which is valid methodology). But it would appear that today's best sensors, in terms of base ISO DR, would not exceed 14-bits with a linear ADC. They're not >14 stops at the pixel level.

(And I'm kicking myself because I've debated this very topic before, assuming ADCs were linear, then assuming they couldn't be perfectly linear, then realizing my own mistake and actually checking the scores that weren't downsampled.)
 

dtaylor

Canon 5Ds
Jul 26, 2011
1,394
830
The full well capacity of many cameras is well beyond the maximum number of steps possible with 14 bits. The EOS 1D X, for instance, has FWC of over 90K electrons, which requires 17 bits to allow a unique digital value for each increase of one electron. That's 5.5X more information than can be expressed in 14 bits.
I don't think the ADC spits out a unique digital value for every single-electron increase. ETTR is a thing precisely because it maximizes the tonal steps as there are more steps dividing the highlight stops than the shadow stops. (Luminous Landscape has a page somewhere explaining this.)

Clark reported the read noise as 35.2 and the FWC as 88,600, for a (per pixel) DR of 11.3.
 

Kit.

EOS 6D MK II
Apr 25, 2011
1,347
740
I don't think the ADC spits out a unique digital value for every single-electron increase.
An "ideal" one does.

ETTR is a thing precisely because it maximizes the tonal steps as there are more steps dividing the highlight stops than the shadow stops.
ETTR is a thing mostly because while the absolute shot noise is directly proportional to the square root of luminous energy, the relative shot noise is inversely proportional to the square root of luminous energy.
 

dtaylor

Canon 5Ds
Jul 26, 2011
1,394
830
ETTR is a thing mostly because while the absolute shot noise is directly proportional to the square root of luminous energy, the relative shot noise is inversely proportional to the square root of luminous energy.
When ETTR first hit sites like LL, any benefits related to noise or shadow push were mentioned as a side note if they were mentioned at all. It was all about tonal separation, which is an observable benefit.
 

Kit.

EOS 6D MK II
Apr 25, 2011
1,347
740
When ETTR first hit sites like LL, any benefits related to noise or shadow push were mentioned as a side note if they were mentioned at all. It was all about tonal separation, which is an observable benefit.
Probably because the early non-FF digital cameras were really poor at it?
 

stevelee

FT-QL
Jul 6, 2017
1,250
281
Davidson, NC
No, we were talking about specific DR measurement method from http://www.photonstophotos.net/Charts/PDR.htm
Those charts don't tell you the maximum you could pull out of your raw files. The charts are based on a model where you print the image on 8"x10" paper and view it from a certain distance (I understand it's just the math model - they don't actually print the images).
At best, subtractive colors on a piece of paper will not have a great dynamic range.
 
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tron

EOS 5D SR
Nov 8, 2011
4,059
359
At best, subtractive colors on a piece of paper will not have a great dynamic range.
So it seems that practically these are imaginary numbers (neither in screen nor in paper)! The real DR is less for all camera models.
 
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JohnC

EOS M50
Sep 22, 2019
46
39
Gainesville,GA
So it seems that practically these are imaginary numbers (neither in screen nor in paper)! The real DR is less for all camera models.
If printing yes. Pigment inks also have less range than dye inks but in general they also have more lightfastness. Having said that I have due prints hangin in my office for over a decade that look great.

With monitors you have more range, although that can be limited by the specific shot OR the capabilities of the monitor.
 

Kit.

EOS 6D MK II
Apr 25, 2011
1,347
740
At best, subtractive colors on a piece of paper will not have a great dynamic range.
If you consider them as the origin of the signal, then yes.

Otherwise you can compress the dynamic range of the original signal before printing, although the result will not always look "natural".