New large sensors limited by the lens in low light?

Started Aug 30, 2011 | Discussions thread
Antisthenes
Forum MemberPosts: 62
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Re: Some thoughts on noise
In reply to cpw, Sep 3, 2011

cpw wrote:

We form the variance, V1 = sum((mi - m)^2) / N = sum((QE*ni - QE*n)^2) / N
so V1 = QE^2*(variance of photon fluctuations) = (QE^2)n = QE*m.
This V1 is sub-poissonian in m !

Your variance calculation uses the same population size N for the mi and ni values.

This is incorrect, as the mi population size must be smaller than the ni 's by a factor QE.

Your calculation thus leads to an underestimation of variance( mi ) relative to variance( ni ), and to an incorrect conclusion that V1 is sub-Poissonian.

The correct calculation is:

Variance(ni) = sum((ni - n)^2) / N

and

V1 = sum((mi - m)^2) / (QE * N) = QE * Variance(ni)

which shows that V1 is, as expected, not sub-Poissonian — its mean and variance are equal.

Where does the rest of the photoelectron shot noise come from to make it to 70.7 e? Well, suppose now we make the light magic again, and turn off its fluctuations.
[snip]

The math for this is found in several places, I found it in the following references:
4. http://www.qis.ex.nii.ac.jp/qis/documents_YY/y3_02chp1_txt.pdf (p. 14)

As explained above, the calculation of your V1 value was incorrect, so your further discussion introducing a "magic light source" and a fictive "binomial variable with variance V2" is quite moot.

You also overlook the fact that your chosen QE is a theoretic construct that is time-invariant . As such, its variance, by definition , is zero. Modelling a QE with a binomial variable, which has a non-zero variance, is therefore incorrect.

The Burgess variance theorem mentioned e.g. on pp. 14 and 15 of your reference URL #4 is typically used to calculate the resulting mean and variance of a chain of stochastic systems with non-null variance.

An example of such a chain would be a photon source — presumably Poissonian, — followed by a slightly noisy photodetector. The detector, being noisy, would obviously introduce a random gain that has a non-null variance, requiring the use of the Burgess theorem. A QE ratio, however, has zero variance and thus has a statistical behavior that is completely different from an amplifier with a random gain component.

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