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I want to buy my forth digital camera, and the final decision is some recent Panasonic Lumix G vs. an Olympus OM-D E-10 (of course the Panasonic is more expensive). As I have had two Panasonic GH models already, I found out that even RAW corrections seem somewhat limited, and I thought it's due to the fact that my models had 10 bits of "color depth" only (8 bits for video). I've had other RAW material, and the impression was that I could get more out of those. Using exiftool I got these values:

Olympus ORF:
Bits Per Sample                 : 16

Olympus JPG:
Bits Per Sample                 : 8

Panasonic JPG:
Bits Per Sample                 : 8

Panasonic RW2:
Bits Per Sample                 : 12

So it seems even the Olympus has much better color resolution than the Panasonic GH-3. Of course I don't have the data for the camera I haven't bought yet (G91 or G9 most likely). Panasonic did not specify.

So basically I wonder: Is there a list of camera models showing their color bit depths for images and videos?

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    \$\begingroup\$ Just because a file format stores values in 16 bits doesn’t mean the sensor resolves 16 bits. \$\endgroup\$
    – Eric S
    Commented Jan 7, 2022 at 3:26
  • \$\begingroup\$ Not all Panasonic RW2 files are equal. For many of their cameras, if you use electronic shutter the actual bit depth at analog-to-digital conversion is only 10-bits, but it's still encoded in a 12-bit or 16-bit scheme. If you use the mechanical shutter, the bit depth is usually 12-bit, though again, it may be encoded in either a 12-bit or 16-bit container. \$\endgroup\$
    – Michael C
    Commented Jan 7, 2022 at 9:04
  • \$\begingroup\$ @EricS Sensors don't resolve any number of bits. They record analog electrical charges. Analog-to-digital conversion of the signal from the sensor is what determines bit depth. But one must also remember that bit-depth does not necessarily equal a specific number of stops of dynamic range or color bit-depth output. One can use very small increments to use 16-bits over a range of one stop. Or one could use very large increments to use 8-bits over a range of more than eight stops. Obviously, if one uses a low bit-depth with a high dynamic range, blocking and banding will be much more noticeable. \$\endgroup\$
    – Michael C
    Commented Jan 7, 2022 at 9:08
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    \$\begingroup\$ @MichaelC I get your point, but sensors do have noise and that noise limits the effective resolution. Digitizing a noisy signal with more bits doesn't necessarily increase the effective resolution. That is what I was getting at. \$\endgroup\$
    – Eric S
    Commented Jan 7, 2022 at 16:15
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    \$\begingroup\$ There is something wrong with your data; the OMD-E10, GH-3, and the GH-9 all have 12bit ADC's. Jpegs will always report 8bit, but there are almost no digital still cameras that have 16bit ADC's that I know of (some CCD monochrome astro sensors do). In either case neither the file depth (raw is typically 16bit, jpeg 8bit), nor the ADC's maximum accuracy (typically 12 or 14 bit), has much to do with the bit depth of the image file data; which to a certain extent depends on the scene recorded. I would worry about other things... \$\endgroup\$
    – Steven Kersting
    Commented Jan 7, 2022 at 18:07

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DXOMARK lists bit depth for many cameras (use the search feature) along with lots of other interesting sensor information in the measurements section. You can also do a side by side comparison of up to 3 bodies. Note, these are lab tests. The scoring sometimes doesn't translate in to real world shooting experience.

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  • \$\begingroup\$ DxO isn't really measuring raw bit depth. They're measuring (normalized for display size) color bit depth within the sRGB color space after converting to JPEG, which is by definition limited to 8 bits per channel. The scores can approach 24 bits because there are three color channels, thus 8 x 3 = 24. \$\endgroup\$
    – Michael C
    Commented Jan 7, 2022 at 8:54
  • \$\begingroup\$ @MichaelC, the DXO measurements (e.g. color sensitivity) are taken from raw images ("screen" data). Only the "print" values are for a normalized jpeg file. dxomark.com/dxomark-camera-sensor-testing-protocol-and-scores \$\endgroup\$
    – Steven Kersting
    Commented Jan 7, 2022 at 17:18
  • \$\begingroup\$ Raw image data has no color. Each photosite outputs a single monochromatic brightness value for all the light that made it down the well. Color filter arrays do NOT have hard cutoff, there's a lot of overlap, just as there is with our retinal cones. In fact, without that overlap there's no way to create color. Light (nor X-ray or UV, etc.) itself has no intrinsic color. Color is a product of the system perceiving electromagnetic radiation. Screens showing "raw" images are showing a processed, jpeg-like version that has been demosaiced, color balanced, gamma corrected, etc. \$\endgroup\$
    – Michael C
    Commented Jan 8, 2022 at 21:09
  • \$\begingroup\$ The fact that the colors of Bayer mask filters are not the same colors as the RGB channels of our output devices makes DxO's claims about using unprocessed raw image data to measure "color sensitivty" dubious at best. They're not true representations of how sensitive the sensor is to 8+8+8 bits in the R+G+B channels when they're measuring monochromatic tonal values measured behind three differently colored filters, none of which are exactly green or blue and particularly one that isn't even remotely "red". \$\endgroup\$
    – Michael C
    Commented Jan 8, 2022 at 21:18
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    \$\begingroup\$ 12 stops of signal-to-noise ratio does not have to be encoded using 12-bits. You can set the white point at one stop higher than the black point and use every single value in 16-bits to finely measure the gradations between the two. Or you can set the black point 14 stops lower than the white point and only use 8-bits to encode that difference. Both cases would be ridiculously extreme. But there's no rule that says you must encode signals using the same number of bits as the number of stops between the highest analog voltage possible (full well value) and the noise floor. \$\endgroup\$
    – Michael C
    Commented Jan 9, 2022 at 0:18

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