What causes noise floor in an image sensor?

Question

In the book Image Sensors and Signal Processing for Digital Still Cameras, it says that Read noise, or noise floor, is defined as noise that comes from the readout electronics.Noise generated in a detector is not included. In CCD image sensors, the noise floor is determined by the noise generated by the output amplifier, assuming that the charge transfer in the CCD shift registers is complete. In CMOS image sensors, the noise floor is determined by the noise generated by readout electronics, including the amplifier inside a pixel.

I remember that noise floor is dominated by dark electrons. What is the reason behind noise floor? In cameras, the output of sensors is subtracted by black level. So noise floor will not be one problem of sensor because it can be eliminated completely.

For the user of sensors, black level need to be subtracted. Why don't analog output subtract black analog value before ADC, so the digital output of sensors won't include noise floor.

Answer 1

In cameras, the output of sensors is subtracted by black level. So noise floor will not be one problem of sensor because it can be eliminated completely.

I'm going to answer this one by analogy.

Let's say you build a hut in a swamp. The ground is really muddy, so you build a floor several feet off the ground. Does this mean that the mud doesn't exist? Does it mean you can dig a cellar?

The black level cuts off data (noise and signal) below a certain level. Usually, this is set just above the noise floor — like building your floor just above the mud. If you build your floor below the mud, the bottom of your house is less useful. If you build the floor high, you're wasting space (but, uh, I guess protecting against floods).

Answer 2

"Why don't analog output subtract black analog value before ADC, so the digital output of sensors won't include noise floor."

Because the opamp that substracts the black level is noisy, the higher the ISO, the larger the amplification factor, and the more internal noise is being superimposed.

The DC-cancellation circuit is a low-pass filter, but it lets high frequency sensor noise pass through, plus it adds its own high-frequency noise.

See your previous question for more technical details.

Answer 3

Have you ever turned up your home stereo so loud while listening to something quiet that you start to hear a hissing or buzzing noise in the background? Then you have discovered the "noise floor". No amplification circuit is immune to noise, they should just have very little of it that it is not noticed in normal situations. However, when you boost the signal enough, you will start to notice it.

What causes the noise? All electrical circuits emit and receive electromagnetic radiation: ie, radio waves. External radio waves, and radio waves emitted as interference by other components within the same circuit or device, both combine to generate tiny fluctuations in current. This is natural to all electronics.

This is what is happening when your camera is boosting the signal it receives. The higher the ISO, the more it will be boosted, and as you can see the cost of boosting ISO too far is that the noise, which is normally not noticeable, becomes noticeable from a certain point onwards.

It is not possible to cancel out noise floor by just subtracting a constant value, because the noise isn't a constant level but random variations overlaid on your signal - see below illustration.

Top: pure signal without noise, bottom: noise overlaid - as you can see the overlaid noise means it can't be removed with a simple "subtraction" operation.

Separating it from your signal cannot be done mathematically or by an analog circuit without damaging the signal you want to keep - even the most sophisticated digital noise reduction algorithm is unable to remove noise without altering or removing some signal.

Some aspects of this explanation have been simplified.