The bit width of the A/D conversion

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Note that the bit width of the A/D converter needs to be sufficiently large to avoid restricting the sensor’s dynamic range. How large it must be can be derived from the ratio between full well capacity and readout noise. Let’s say the system has a storage capacity of 1,00,000 electrons and a readout noise of 100 electrons. The ratio is 1,00,000/100 = 1,000. In this scenario, 10-bit digitization is the best option as it provides 1,024 stages (210 x 1,024). 8-bit, on the other hand, has only 256 levels.

What happens if we digitize the dynamic range with an insufficient bit width? Let’s imagine the following situation: We shoot a subject with an exposure range of exactly 10 exposure stops. So the brightest details in the highlights are 1,024x as bright as the darkest details in the shadows. We control the exposure exactly so that the brightest part just reaches the saturation value of the sensor. To further calculate with favorable numbers, we assume the recording system has a signal-to-noise ratio of 60 dB, so the noise level is a little less than 1/1,000 of the maximum output value. Since the system is linear, the output voltage value of a pixel in the shadows is exactly 1/1,024 of a pixel in the highlights. We adjust the amplifier to ensure that the brightest pixel’s output value is 1.023 volts, while the darkest pixel’s output value is almost exactly 1 millivolt.

If we digitize the data output in this way with an 8-bit A/D converter, the output value of 1 mV becomes 0.The A/D converter truncates the dynamic range to 8 exposure levels, resulting in the loss of information in the shadow area.  We cannot undo this loss at any point in further image processing!

However, if the A/D converter has a 10-bit width, the output value drops to 1, preserving the information. With 12-bit, it would even be 4, and we would have preserved gradation at this lowest exposure level. However, we would also be quantizing the noise more finely, and that can be a disadvantage.

It’s hard to understand at first, isn’t it? Because at first glance, you might think that the lightest or darkest parts would be preserved after digitization, and the tonal values in between would only be graded finer or coarser according to the bit width. But the relationship is as follows: We digitize linearly, and so the image location of the value 255 is also 255x brighter than that of the value 1. Since an exposure level represents the increase or decrease of the amount of light by a factor of 2, we also have only a dynamic range of 8 exposure levels after the linear 8-bit digitization (we remember the section „The minimum size of brightness differences“ in which we stated that due to the properties of our visual system, only a tone value scale appears to us as „correct“ if the levels differ by a stated facto):

1. 255/2=128, 
2. 128/2=64, 
3. 64/2 = 32, 
4. 32/2 = 16, 
5. 16/2 = 8, 
6. 8/2 = 4, 
7. 4/2 = 2, 
8. 2/2 = 1

Strictly speaking, there are only 7 exposure levels, as the lowest value is either 1 or 0, meaning there are no further gradations. However, this issue pertains to the definition of the dynamic range.

We will continue with this example in the section „Gamma Correction“ to illustrate how the further processing has a practical effect.

Next Practical consideration of the variables dynamic range, sensitivity, full well capacity and pixel area

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