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According to 3.4.2. section "data precision" on this website, the Canon 450D has 14bit deph. Canon CR2 stores RAW data as JPEG Lossless format.

  • Image size: 4312x2876
  • CFA: 4313x2877 (bayer sensor scheme)

As 14bit depth (2^14, 16,384 variations) are stored as unsigned __int16 (max value 65,535) which has size of 2 bytes. 24,802,624 bytes =~ 24mb per file (IFD#3 only)

We could have data precision in 64 bits:

As 64bit depth (2^64, 18,446,744,073,709,551,615 variations) are stored as unsigned __int64 (max value 18,446,744,073,709,551,615) which has size of 8 bytes. 99,210,496 bytes =~ 100mb per file (IFD#3 only)

I'm not saying JPEG Lossless format stores information in unsigned __int, but I found out the normal JPEG uses a WidthxHeightx3 matrix in YCbCr color space that can be converted using libjpg. I don't know how

Not talking about "you couldn't tell the difference", the point is about we being able to capture (there's much more room available for values) some sort of a ultra high dynamic range image which only blown up thing would be pointing to the sun directly or bulb mode exposures.

Maybe I am misunderstanding something, who knows. The Canon 450D seems to have a 128MB buffer by the way.

  • For the next to browse this, I did test out the linked to site for malware and it looks to be safe to browse. Only Twitter and Google Analytics calls from what I can tell. – dpollitt Dec 26 '15 at 20:19
  • @dpollitt Thanks for formatting my question! I hope it is not off-topic here. – T. Braga Dec 26 '15 at 20:27
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The 14 bit depth is the limit of the physical sensors capabilities, it isn't just that the engineers decided to throw away useful data.

An increasing number of bits available in a sensor reflects an increasingly larger complexity of circuitry and precision needed to resolve those progressively finer and finer details. Complexity and precision don't come for free.

While a 64 bit ADC could be hooked up to today's imaging sensors, the bottom 50 bits would be just noise, which is of no use to anyone. Attempts to improve the Signal to noise ratio would dramatically increase the cost of the sensors, plus getting 64 bits of useful data is likely not even possible with today's semiconductor capabilities. If you had a military-scale budget, you may be able to achieve 24 bits, but even that would be pushing it.

  • That's why I'm saying I must be misunderstanding things and my post is naturally confusing. I think about removing ceiling limit as the measure will add up to a number rather than inspect an infinitesimal change in sensor. We can remove dark current and overall noise in this hypothetical situation. Making various measurements over time and storing current or voltage data, just like a capacitor it will add up but the pixels are not yet mapped to a range. Thanks for your response! – T. Braga Dec 26 '15 at 21:43
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    and it's not just "today's semiconductor capabilities". using so many bits you would be encoding brightness levels far below single photon! (assuming the max level stays similar to current sensors) – szulat Dec 26 '15 at 23:40
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I think there's a fundamental misunderstanding in the value of this:

Not talking about "you couldn't tell the difference", the point is about we being able to capture (there's much more room available for values) some sort of a ultra high dynamic range image which only blown up thing would be pointing to the sun directly or bulb mode exposures.

Adding more bits won't give you more dynamic range, at least beyond a reasonable threshold. The photosite itself is an analog device, and the digital conversion just chops that up into digital levels. If you're really starved for bits, you might decide to chop off the low end (mostly noise anyway), but with 14 bits as demonstrated by real-world devices, that's not really an issue. So, all you get by going to 64 bits is chopping up the existing range more finely, which might be useful in some situations — I can certainly see 16 bits — but is going to mostly just be excess precision.

Now, if you're talking about designing photosites with a lot more inherent dynamic range in the first place... that's a different question.

  • Thanks for the clarification! Your answer also helped a lot! – T. Braga Dec 28 '15 at 13:56

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