My camera has an engineering Dynamic Range of 14 stops, how many bits do I need to encode that DR? Well, to encode the whole Dynamic Range 1 bit could suffice, depending on the content and the application. The reason is simple, dynamic range is only concerned with the extremes, not with tones in between:
![\[ DR = \frac{Maximum Signal}{Minimum Signal} \]](https://i0.wp.com/www.strollswithmydog.com/wordpress/wp-content/ql-cache/quicklatex.com-f89166dae7cfbfc1a37b1ef0f536dd7e_l3.png?resize=194%2C41&ssl=1 "Rendered by QuickLaTeX.com")
So in theory we only need 1 bit to encode it: zero for minimum signal and one for maximum signal, like so
This image is comprised of four squares, two at maximum white and two at deepest black, made up of pixels encoded at 1 bit depth. Think of it as a fax: 0 for black, 1 for white. What is its DR? The answer depends on the physical characteristics of the capture/display medium, not on the bit depth of the information.
If it were displayed on a typical 24″ consumer monitor, the DR of the resulting image would be independent of the 1-bit encoding depth and entirely due to the physical characteristics of the monitor. The ratio of the brightest to the darkest luminance such monitors are currently able to produce in 
is typically about 250:1 (often referred to as static ‘contrast ratio’), for an 8 stops contrast ratio (
).
If it were printed on paper the DR of the resulting image would be independent of the 1-bit encoding depth of the information and entirely due to the physical characteristics of the paper and the ink. The ratio of brightest to darkest reflected luminance in that can be produced by photographic papers and inks is typically about 100:1, or about a 7 stop contrast ratio (
).
If the image had been captured by a modern full frame Digital Still Camera, chances are maximum signal at full scale would have been around 80,000 photoelectrons (e-) and minimum acceptable signal around 5e-, for an eDR of about 14 stops (
). However, the information to describe this image could have been encoded at 1-bit depth.
Black and white newspaper images are created by more or less densely spaced black dots (ones and zeros), as are grayscale inkjet images, a process known as halftoning. 1-bit Audio ADC/DACs produce dynamic ranges above 100dBs.
All of this to say that Dynamic Range and bit depth are not directly related. The question then becomes: How many bits do I need to properly encode all tones in the scene? Next.
I don’t think the analysis is correct.
What is the limiting factor of the minimum value? Why is it 5e in your example and not 0?
Well, this is due to noise.
Noise is the limiting factor for Dynamic Range and Tonal Range which is basically dividing the Dynamic Range into slices according to the noise level.
I see your point R, however the main concept exposed in the article is that ideally DR and bit depth are unrelated.
And there can be usable signal below read noise, for instance one can easily read text. If that’s acceptable in your application, DR is greater than using just noise in the denominator and it is likely greater than bit depth.
You stated that the DR of a monitor wohld be 250/1 in cd per m^2 – in f-stops ld (250) = 8.
In cameras we have 2^(#bits) -1 as the brightest tone which means that theorteically the ld of that which is the number of bits is the upper border of dynamic range?
Hi Wolfgang,
Yes but the key is the denominator: if you have read the previous article you know that the definition of DR used is the ratio of the largest acceptable signal divided by the smallest acceptable signal, so it depends on the smallest acceptable signal, which is application specific.
There are several that are fine with sub-bit signals (for instance forensics or astro), so in their case DR could be greater than bit depth would suggest. On the other hand the lowest acceptable signal to your typical photographer is quite a bit over read noise, so DR is typically smaller than that.