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.


3 Answers 3


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).

  • To be measurable, the signal should be larger that noise floor. Why? The noise floor is always present, and signal will be added to noise floor to obtain measured value. The signal can be got by subtracting measured value by noise floor. Oct 29, 2015 at 1:11
  • Noise floor is the sum of all random noise. It is not a number. It is a sequence of random values what you don't know, so you cannot subtract that from the measured value. You can subtract the estimator of the noise floor, that is the moving average of noise, but that does not cancel out high frequency noise. You are missing out on some fundamental of random processes and noise.
    – TFuto
    Nov 4, 2015 at 15:36
  • E.g. pick a number between 1 and 10, and then add a random number between 1 and 100. Repeat this. You will have something like: 23 51 103 75 31 5 41, etc... Although I chose the numbers completely randomly between 1 and 100, you cannot tell what numbers I chose between 1 and 10. The random number between 1 and 100 is the noise floor, the number you chose between 1 and 10 is the signal you would like to get.
    – TFuto
    Nov 4, 2015 at 15:38
  • Noise floor is not a number and obey probability distribution such as Gaussian noise. However we usually need some indicator to compare noise floor of different sensors. From the link en.wikipedia.org/wiki/Noise_floor, The noise floor limits the smallest measurement that can be taken with certainty since any measured amplitude can on average be no less than the noise floor. Nov 6, 2015 at 11:52
  • Now you are talking about spectrum and SNR. You cannot get the signal back by substracting the noise spectrum from the signal or the signal spectrum, unless the signal is a time-invariant signal, like a carrier (e.g. a sine). If the signal is time variant, spectrum is only an averaging estimator, so you cannot get the signal data back with the above method.
    – TFuto
    Nov 9, 2015 at 13:46

"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.

  • Thanks. I read your reply to my previous question. Opamp noise is related to ISO setting. Does it means that black level calibration should be performed for different ISO setting? Nov 9, 2015 at 1:13
  • Yes, definitely.
    – TFuto
    Nov 9, 2015 at 13:46

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.

  • Thanks. From my understanding, noise floor may obey some probability distribution such as Gaussian distribution. Gaussian distribution is described by two parameters: means and standard deviation. Here subtraction operation means that noise is subtracted by its means and obeys N(0, sigma). Boosting one signal doesn't improve SNR(signal-noise ratio). I don't know why it can affect image quality? Nov 9, 2015 at 1:06
  • You boost the signal so that you can use the entire dynamic range of the ADC. (You are normalizing the input maximum voltage level to let's say 1 V, the max. input voltage for the ADC). If you don't use the entire range, you lose resolution (less effective bits in the digital data). With that boost, you also increase the noise power. If you have a lot of light, there is no need to amplify the signal, so the noise is not amplified as well. If there is little light, you have to boost ISO, amplify the signal, and you also amplify noise. The SNR stays the same after analog amplification.
    – TFuto
    Nov 9, 2015 at 13:48
  • However, with correctly using the ADC's full dynamic range, you actually get more effective resolution, and with each bit, you get SNR improvement after the ADC.
    – TFuto
    Nov 9, 2015 at 13:52

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