Sensor performance, quantum efficiency, sensorgen and DXOmark

Started Feb 5, 2014 | Discussions thread
Jack Hogan Veteran Member • Posts: 6,487
What Peak is being measured?

Chrisk98 wrote:

Nearly 4 years ago I presented here a way to estimate the quantum efficiency of camera sensors by analizing SNR data from DXOmark. ...The paper can be found here.

I had a chance to re-read your fine paper. There are a couple of passages that I am having difficulty following:

1) What Peak QE is being measured? Not green, right?

I get the feeling that the green channel is implied, since you mention (correctly I think) that that's the channel that most likely saturates first when DxO takes its measurements.

However the photoelectron Signal that the QE estimates are based on is computed from a mix of the three channels: DxO averages the channels' measured SNRs to compute a 'grayscale' Full SNR curve - from which we derive the Signal.

As an example, here is how a normalized photon and photoelectron count could look like under the constant spectral radiance assumption of the paper. The peak strength of the color filters in the example is 0.5, 0.35 and 0.25 for G,B, and R resp. The shapes are the photopic eye response's as assumed in the paper.

Note that constant spectral radiance does not mean constant photons/wavelenght throughout the range, which here is taken to be 400-700nm (the paper says 500-700 but I assume that's a typo).

We can figure out peak QE for this example as discussed in the paper by determining the Signal/lx-s/Area (FWC) in e- as the sum of the photoelectrons under each curve collected by the four photosites of an RGGB quartet, divided by four. One can compute the denominator of equation (14) (/lx-s/Area) at a pivot wavelength of 530nm as the paper suggests. The result is a peak QE of 39.2%, obviously a weighted average of the four peak QEs - not 50%. Am I correct?

2) Assuming that all three channels' CFA curves have the same shape (we know this is a stretch and never true in practice) peak QE as discussed in the paper can be calculated as follows from the peak QE's of the individual channels:

QEpeak = (­sqrt(QEpk_G)/2+sqrt(QEpk_B)/4+sqrt(QEpk_R)/4)^2

Plugging in the peak QE values for the CFA channels of the example in 1), yields a QEpeak of 39.2% as calculated earlier.

Taking a more practical example, Christian Buil's Figure 3 for the 5DII - with QEpk_G=33.8%, QEpk_B=29.3% and QEpk_R=16.8% - the same formula would result in a peak QE of 27.9%.

Buil states that "QE is the conversion rate between the incident photons and photo-electrons generated in the detector."

If one actually integrates the curves, effectively counting the photoelectrons hitting the four RBBG photosites and then applies formula (14) to the quartet one gets a Peak QE of 24.8%, indicating that the shapes of the three color filters are dissimilar, as we know. Your table shows a 5DII of 33%.

So I am still unclear: can you better define what QE the method in your paper is estimating?

Thanks for your input.

Jack

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