Colour depth is often referred to as being X bits. What does this mean and how does it effect a photograph? What scale is used, i.e. is it linear, exponential, logarithmic, etc.?


3 Answers 3


What is a bit?

Computers store values as binary numbers. Each digit of a binary number is called a bit. 2^N, where N is the number of bits is the maximum number of things that binary number can represent.

Example Please

A black and white image (no gray here, just black and white) can be represented with a color depth of 1 bit. 2^1 = 2. Those two colors are black and white.

Back on older mac computers you could set the color depth: 16 colors, 256 colors, thousands of colors, millions of colors. These options correspond to different bit depth values: 4, 8, 16, and 24 bits. Bit depth on computer monitors always refers to the sum of the red, green, and blue pixels bit depth. If the sum is not divisible by 3, then usually green gets the extra bit since your eye is most sensitive to green.

What are some real world numbers?

Nikon d7000: 14bits per pixel.

Most computer monitors display color with 8 bits per color for a total of 24 bits per pixel.


Image sensors are linear, which means the half the values represent the brightest stop of light, then the next quarter the next stop and so forth. This means that dark values quickly get compressed into a small number of possible values. The higher the bit depth the better quality dark pixels.

How does it affect a photograph?

More bits means more data. That can't be faked. More bits may also mean more quality to work with when processing the images.

Higher values are not always better though. Designing ADC (analog digital converters) with high bit depth is very difficult. This is because the noise level of the converter must be below (V)/2^N where V is the input signal's voltage and N is the bit depth. This voltage, V/2^N is called the least significant bit voltage (often called 'one LSB'). It is the voltage that each bit represents. If the noise level is greater than one LSB the LSB is not storing useful data and should be removed.

Example: A 5 Volt signal is being digitized by a 10 bit ADC. Under what voltage should noise be kept?

Using the equation for LSB voltage: 5/(2^10) = (5/1024)V, 4.88mV.

  • \$\begingroup\$ In reference to the scale, does this mean that there are more light colours than there are dark? \$\endgroup\$ Commented Sep 3, 2012 at 8:18
  • 3
    \$\begingroup\$ @damned truths You can actually represent many many more midtones than either dark or light colours using the RGB colour space. Here's a very simplified example, at the very dark end you have values 1,0,0 0,1,0 and 0,0,1 and those are the only possible colours that have luminance = 1. Now consider the midtones, you have a huge number of combinations that have the same apparent brightness, 127,0,0 126,5,7 4,128,0 etc. etc. etc. \$\endgroup\$
    – Matt Grum
    Commented Sep 3, 2012 at 8:32
  • \$\begingroup\$ Nice answer, however I do not understand your last sentence starting from "If this criteria...". What do you mean by: "more space than it has to"? \$\endgroup\$ Commented Sep 3, 2012 at 11:16
  • 1
    \$\begingroup\$ I've added a little more to explain that part. Basically if you graph the data coming off the ADC when there is no signal connected to the ADC if there are oscillations in the graph then the data is being sampled at a higher bit depth than necessary. \$\endgroup\$
    – Phil
    Commented Sep 3, 2012 at 16:50
  • \$\begingroup\$ 1 bit / 2 colours can be any 2 thy heart may desire. Aquamarine and turquoise if desired. Octarine gets harder. \$\endgroup\$ Commented Sep 3, 2012 at 17:03

Color or tone levels are stored in "words", with each word containing N "bits".
Each bit can be either off or on (or high or low, or 0 or 1).
So, as a binary "bit" has two states it can store one of two "states" or levels.
Combining bits into a "word" allows the word to store any one of a larger number of states.
2 bits can store one of 2 x 2 = 4 states.
8 bits can store one of 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 states.

The number of levels or tones which are available for N bits is given by
2 x 2 x 2 ... N tones or 2^N levels.

2^16 = 16 bits = 65536
2^14 = 14 bits = 16,384
2^12 = 12 bits = 4,096
2^10 = 10 bits = 1024
2^8 = 8 bits = 256
1 bit = on/off = black and white OR any 2 colours of choice.

Where there are very few tone or colour levls available the "real" colours present must be stored as the nearest available value. So, various shades of green "clump togethjer", as do reds or browns or aquamarines or tangerines or octarines or ... .

16 bits = 65,536 levels (see below) is more than enough to make colours continuously variable according to most human eye-brain systems. 14 bits is good enough for most mere mortals and even 12 bits is good enough for most purposes. Get down to 10 bits and people start to see banding and loss of smooth colour changes.

The image below shows an image at resolutions of "16", 8, 4 and 1 bit.
I say "16" as, while the colour data is stored in 16 bits, I chose the original image to suit and it happened to be taken in 2004 with a Minolta 7Hi "bridge" camera with a 12 bit ADC (Analog to Digital Converter).

At 500mm or more from most monitors the left hand two images probably appear similar - and even more so at 1 metre or more. But a closer look will show a vast difference. Examining the original image will show the differences more clearly. (Right click and open in new tab or in an image viewer). The flower whiter parts mainly appear to defocus but the leaves are broken up into bands of similar colours rather than the nearly continuous changes that occur at 16 bits.

flower image


At its very basic level, the higher the colour depth, the smoother tonal gradations an image can display.

This video compares a Nikon D800 at 14bit and a Hasselblad H4D-40 at 16 bit.


  • \$\begingroup\$ Darkcat - link missing \$\endgroup\$ Commented Sep 3, 2012 at 7:23
  • \$\begingroup\$ just noticed!!! fixed :-) \$\endgroup\$ Commented Sep 3, 2012 at 7:24

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