Photon Transfer Curve

Hi All,

I've been pulling my hair out over this one .... maybe someone has
some insight.

Any ideas what would cause the shot-noise region of a photon transfer
curve to have a slope significantly different from 1/2? It is plotted
on a log-log graph, as usual. The strange thing is that I can fit a
straight line to this region with low residual error. The problem is
the slope of the line is somewhere around 0.2.

The device under test is a CMOS imager. PRNU removed by frame
differencing. I observe the onset of saturation because the noise
suddening drops.

Regards,
pk

Anyone have any thoughts?

Thanks,
pk

wrote:
Hi All,

I've been pulling my hair out over this one .... maybe someone has
some insight.

Any ideas what would cause the shot-noise region of a photon transfer
curve to have a slope significantly different from 1/2? It is plotted
on a log-log graph, as usual. The strange thing is that I can fit a
straight line to this region with low residual error. The problem is
the slope of the line is somewhere around 0.2.

The device under test is a CMOS imager. PRNU removed by frame
differencing. I observe the onset of saturation because the noise
suddening drops.

Regards,
pk

Hi,

wrote:
....
I've been pulling my hair out over this one .... maybe someone has
some insight.

Any ideas what would cause the shot-noise region of a photon transfer
curve to have a slope significantly different from 1/2? It is plotted
on a log-log graph, as usual. The strange thing is that I can fit a
straight line to this region with low residual error. The problem is
the slope of the line is somewhere around 0.2.

The value of the slope depends on the e/ADU conversion factor, maybe
you have a factor of 5 e/ADU (2.5e/ADU, because of differencing)?

The device under test is a CMOS imager. PRNU removed by frame
differencing.

What results in a double variance, so this is why you expect a
slope of 1/2?

Just guessing,
Jens

The value of the slope depends on the e/ADU conversion factor, maybe
you have a factor of 5 e/ADU (2.5e/ADU, because of differencing)?

For example:
if you have a conversion factor of 5e/ADU and in the graph
is a variance of 20 for a mean value of 100 ADUs, that would
belong to a mean value of 500 electrons, which generates
a variance of also 500e, or -because of differencing- to
1000e (so i made a mistake, differencing would lead to a
higher slope-value, so you have 10e/ADU?).
In the case of 500e variance, the measures value of variance
in ADUs isnt the same, because you get the mean of the
_squares_ of the diffs.
The standard deviation for 500e would be 22.36e, thats a
value of 4,472 ADUs and the square is 20 ADUs.
You get only 20 ADUs for the variance by 100 ADUs mean value,
what is just the conversion factor.

Regards,
Jens

Hi Jens,

Thanks so much for your reply.

The value of the slope depends on the e/ADU conversion factor, maybe
you have a factor of 5 e/ADU (2.5e/ADU, because of differencing)?

I guess this is what's confusing me. On a log-log scale of standard
deviation versus mean, the shot noise region should have a slope of 1/2
regardless of the conversion factor, right? I thought the conversion
factor was where that line intercepts the line standard deviation = 1.

The device under test is a CMOS imager. PRNU removed by frame
differencing.

What results in a double variance, so this is why you expect a
slope of 1/2?

After frame differencing, I take the standard deviation of the
resulting frame and divide by sqrt(2). this is the same as dividing
the varience by 2 to return it to it's proper value.

Thanks again,
Patrick

Hi Jens:

Jens Dierks wrote:
The value of the slope depends on the e/ADU conversion factor, maybe
you have a factor of 5 e/ADU (2.5e/ADU, because of differencing)?

For example:
if you have a conversion factor of 5e/ADU and in the graph
is a variance of 20 for a mean value of 100 ADUs, that would
belong to a mean value of 500 electrons,

understood.
which generates
a variance of also 500e, or -because of differencing- to
1000e (so i made a mistake, differencing would lead to a
higher slope-value, so you have 10e/ADU?).

can you explain why the varience is also 500e for this case?

In the case of 500e variance, the measures value of variance
in ADUs isnt the same, because you get the mean of the
_squares_ of the diffs.
The standard deviation for 500e would be 22.36e, thats a
value of 4,472 ADUs and the square is 20 ADUs.
You get only 20 ADUs for the variance by 100 ADUs mean value,
what is just the conversion factor.

Regards,
Jens

Hi,
Hi Jens,

Thanks so much for your reply.

The value of the slope depends on the e/ADU conversion factor, maybe
you have a factor of 5 e/ADU (2.5e/ADU, because of differencing)?

I guess this is what's confusing me. On a log-log scale of standard
deviation versus mean, the shot noise region should have a slope of 1/2
regardless of the conversion factor, right?

No, the standard deviation goes with the squareroot of electrons.
A problem could be, that the cmos sensor has a nonlinear output,
but im not sure about that.


I thought the conversion
factor was where that line intercepts the line standard deviation = 1.

The conversion factor is the inverse of the slope from the curve:
mean value and variance.


The device under test is a CMOS imager. PRNU removed by frame
differencing.

What results in a double variance, so this is why you expect a
slope of 1/2?

After frame differencing, I take the standard deviation of the
resulting frame and divide by sqrt(2). this is the same as dividing

Yes, but the standard deviation shouldnt give a constant slope.

From other post:
which generates
a variance of also 500e, or -because of differencing- to
1000e (so i made a mistake, differencing would lead to a
higher slope-value, so you have 10e/ADU?).

can you explain why the varience is also 500e for this case?

500 electrons mean value should produce sqrt(500)e standard
deviation and 500e variance.

Maybe this could help:
http://ccavador.free.fr/prism_ccdtest/CCDtest-US.html

Regards,
Jens

Hi again,
wrote:
I guess this is what's confusing me. On a log-log scale of standard
deviation versus mean, the shot noise region should have a slope of 1/2
regardless of the conversion factor, right? I thought the conversion
factor was where that line intercepts the line standard deviation = 1.

Sorry, forget what i was saying, on a log-log scale you should get
a slope of 1/2 dB!
I wasnt counting in dB units.

How did you made the flatfield images and what a camera is it,
is there a kind of AGC or other things to mention (AD-bitdepth)?

Regards,
Jens

Hi Jens,

Thanks for the explanations. Things are making a little more sense
now!

How did you made the flatfield images and what a camera is it,
is there a kind of AGC or other things to mention (AD-bitdepth)?

Flat field images were acquired using an integrating sphere with LED
illumination (not monochromatic, but not too broad spectrum). I
acquire two flat fields under the same conditions and measured the mean
and standard deviation from a 50 x 50 pixel region.

No AGC. It's a CMOS imager and the output is going into a LabView
acquistion board (12 bit).

Regards,
Patrick

Hi Patrick,
Thanks for the explanations. Things are making a little more sense
now!

How did you made the flatfield images and what a camera is it,
is there a kind of AGC or other things to mention (AD-bitdepth)?

Flat field images were acquired using an integrating sphere with LED
illumination (not monochromatic, but not too broad spectrum). I
acquire two flat fields under the same conditions and measured the mean
and standard deviation from a 50 x 50 pixel region.

No AGC. It's a CMOS imager and the output is going into a LabView
acquistion board (12 bit).

Have you made a linearity curve of the sensor, for example by
making different exposure times with the same illumination?
The amount of readout noise could also be very high, but this
should be seperateable from the photon shot noise in the graph?

My ideas are shrinking, so what else can cause the fact that
the shot noise doesnt increase so much as it should (what a
lower slope means)?

Regards,
Jens