D3 sensor efficiency

That's interesting reading, Emil. Thanks for looking into this and I hope your numbers are accurate (no offense).

Any chance you will be doing something similar for the D300 (v 40D)? It would be interesting to read the raw numbers for these two as the debates on image quality commence after launch.

Please guys/girls.. someone earlier on this thread talked about D200 having a dynamic range around 11 stops.. and that´s just b-s h i t , the MF-digital backs have aprox. 12 steps and the difference in raw-files between MF and Dslr is extreme! i would guess that the D200 have around 7,5 steps and the D40x around 8 steps dynamic range.

I dont for a moment believe that the D3 has dynmic range in the same league as the MF-backs.

Danlo said:

Please guys/girls.. someone earlier on this thread talked about D200
having a dynamic range around 11 stops.. and that´s just b-s h i t ,
the MF-digital backs have aprox. 12 steps and the difference in
raw-files between MF and Dslr is extreme! i would guess that the D200
have around 7,5 steps and the D40x around 8 steps dynamic range.

I dont for a moment believe that the D3 has dynmic range in the same
league as the MF-backs.

Click to expand...

Fine, don't believe it. The Canon 1-series has had 11.5 stops of DR since the Mark II. Realize that what is being discussed is the engineering definition of DR, which is raw saturation level divided by read noise. This is a more liberal definition than what some people call "useable dynamic range", obtained from looking at the output of raw converted images of step wedges etc. Just make sure you are comparing apples to apples.

--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/

Could you explain how this notion of dynamic range is related to the dynamic range we can use in an image, and possibly other useful things like latitude?
--
Jim

JimPearce said:

Could you explain how this notion of dynamic range is related to the
dynamic range we can use in an image, and possibly other useful
things like latitude?
--

Click to expand...

I'll give the example of the 40D, which John Sheehy tested to have 11.1 stops DR according to the engineering definition:
http://www.pbase.com/jps_photo/image/87419422/original.jpg
Imaging-resource ran a step-wedge raw image through Imatest
http://www.imaging-resource.com/PRODS/E40D/E40DIMATEST.HTM

and found the DR according to various "quality" standards -- where the noise reaches various fractions of the luminosity jump between steps; low quality is where the noise is 1 stop, medium half a stop of noise, medium high a quarter stop, and high quality is .1 stop of noise. IR finds that the low quality DR is 11.2 stops (interesting, since the DR in such a test cannot exceed the engineering DR), the medium quality DR is 10.1 stops, medium-high is 9.3 stops, and the high quality DR is 7.7 stops. So as your standards go up as to how small the noise should be relative to the illumination level, the DR goes down.

So the engineering DR corresponds rather closely to what others call "low quality" DR. Knowing the read noise and the quantum efficiency of the sensor, one can calculate the DR according to various other quality standards since these two figures entirely determine the noise characteristic as a function of exposure. I have not bothered to do so however.

--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/

ejmartin said:

I have now also measured the (full well) capacity of the pixels of
the Nikon D3. I found that at the base ISO of 200, the full well
capacity is 53,400 electrons, a new high among DSLR's. The previous
record holder was the Canon 1Dmk3, whose well capacity at ISO 200 is
about 39,500.

Click to expand...

I'm not a scientist, so I don't know, but don't photosites collect photons, not electrons?

--
Jay Philip Williams Photographic Design
http://www.jpwphoto.com

Based upon published data by C. Buil:
http://www.astrosurf.com/buil/nikon_test/test.htm
and R. Clark:

http://www.clarkvision.com/imagedetail/digital.sensor.performance.summary/index.html

camera: 5D,20/30D,40D,300D,400D,350XT,D50,D70, D200, D3
Pixel size(PS): 8.2,6.4,5.7,7.4, 5.7,6.4,7.8,7.8,6.05,8.4
pixel area(PA): 67.2,41,32.5,55,32.5,41,61,61,37,71
relativePixelarea(RPA) 2.1,1.3,1,1.7,1,1.3,1.9,1.9,1.1,2.2
IG 4.1,3.1,3.1,2.7,2.8,2.6,3.7,3,2,8.3
IG/RPA 2,2.46,3.1,2.7,1.6,2.7,2,1.6,1.8,3.8

(Ranking is alphabetical when equal)

Camera rank order by RPA (area per individual pixel)
D3,2.2
5D, 2.1
D50, 1.9
D70, 1.9
300D, 1.7
20/30D, 1.3
350XT, 1.3
200D, 1.1
40D, 1
400D, 1

Camera rank order by IG (Inverse Gain,12bit ADU, electrons to photons/time unit/pixel)

D3, 8.3
5D, 4.1
D50, 3.7
20/30D, 3.1
40D, 3.1
D70, 3
300D,2.8
400D 2.7
350XT,2.6
200D, 2

Camera rank order by IG/RPA:

D3, 3.8
40D, 3.1
400D, 2.7
20/30d, 2.46
350XT, 2
5D, 2
D50, 2
D200, 1.8
300D 1.6
D70 1.6

This is a measure of the camera's QE/pixel (quantum efficiency/area)
area. It shows the D3 ahead, but not by an amount which would require
more than incremental changes in technology, such as fill factor, dye
transmission, etc. We can see in this table, that Canon has made large
gains in QE/RPA, but that Nikon has caught up, and surpassed Canon. We
also can see, for example how the D50, was able to match the 350XT and
5D in IG/RPA. I can predict that the D40X has got a higher IG/RPA,than
the D200 due to sensor level improvements,based upon published DSLR
photography,reviews and if the trend holds true, is probably similar
to the 20/30D at 2.5, with an IG, of 3.2.
Note the strong relationship between pixel area and IG, when comparing
cameras of similar development age.

cheers

Duncan

Buil reports a QE for the D3 that is about 25% higher than I found. I suspect his result is the more accurate, as the 3rd party image I was analyzing was rather flawed for the purpose of the analysis, while presumably since he had the camera in his hands he could take the image he needed in a controlled way. A 12% overestimate of the photon noise in my measurement would account for the difference, and is close to the margin of error of what I had to work with. And in this measurment, any error tends to overestimate the noise, and therefore underestimate the QE.

So this is even better news for the D3 if confirmed
--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/

The article by Christian Buil states, "The ISO sensitivity levels corresponds to an electronic gain extremely different between Canon and Nikon. Assuming that the quantum efficiency of sensors are equivalent (it is highly probable in first approximation), the displayed sensitivity of 6400 ISO for the Nikon D3 #1 corresponds to a sensitivity of 600 ISO for the Canon 40D, while a sensitivity displayed at 6400 ISO for the Nikon D3 #2, corresponds to a sensitivity of 2400 ISO for the Canon 40D. Analysis shows that the 'displayed' ISO announced by Nikon is very over evaluated relative to Canon competitor."

From the presented data, I find this puzzling. He compared the Canon 40D to the Nikon D3, both operating with a bit depth of 14. What he calls inverse gain for both cameras at ISO 400 was 0.78 electron/ADU for the Canon and 2.1 electrons/ADU for the Nikon. The Canon pixel pitch is 5.7 microns and that for the Nikon is 8.2. The ratio of the pixel areas is 1:2.22. If quantum efficiency, fill factor, etc are similar for the two cameras, the Nikon should collect 2.22 times more photoelectrons than the Canon at equivalent integration times, and this should affect the gain proportionally. The actual gain ratio is 1:2.7, roughly as expected.

Can anyone explain Christian's analysis?

Bill

p.s.

For Emil: Christian does take truncation of the data into account when calculating read noise.

Jay Williams said:

Jay Williams said:

I have now also measured the (full well) capacity of the pixels of
the Nikon D3. I found that at the base ISO of 200, the full well
capacity is 53,400 electrons, a new high among DSLR's. The previous
record holder was the Canon 1Dmk3, whose well capacity at ISO 200 is
about 39,500.

Click to expand...

I'm not a scientist, so I don't know, but don't photosites collect
photons, not electrons?

Click to expand...

Photons that "hit" a particular photosite cause a electric charge to build up on that photosite. As more photons hit ,more charge or electrons gather at that point. So they do collect photons but that energy is stored as an electric charge.

Cheers.

Jay Williams said:

--
Jay Philip Williams Photographic Design
http://www.jpwphoto.com

Click to expand...

ejmartin said:

Buil reports a QE for the D3 that is about 25% higher than I found.

Click to expand...

...A full 25% increase is really, really odd. Something else (obviously) is going on. The question is 25% over what baseline, what reference?

PIX

--

TIP: If you do not like this post, simply press the 'COMPLAINT' button. Mommy/Daddy are just one click away.

Bill Janes said:

The actual gain ratio is
1:2.7, roughly as expected.

Click to expand...

...2.7 is A WHOLE LOT more than 2.2. It truly is.

This is all messed up, by any means. Something else is involved on this computations, that is not quite clear, yet. But there is something wrong.

That simple.

PIX

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TIP: If you do not like this post, simply press the 'COMPLAINT' button. Mommy/Daddy are just one click away.

Bill Janes said:

The article by Christian Buil states, "The ISO sensitivity levels
corresponds to an electronic gain extremely different between Canon
and Nikon. Assuming that the quantum efficiency of sensors are
equivalent (it is highly probable in first approximation), the
displayed sensitivity of 6400 ISO for the Nikon D3 #1 corresponds to
a sensitivity of 600 ISO for the Canon 40D, while a sensitivity
displayed at 6400 ISO for the Nikon D3 #2, corresponds to a
sensitivity of 2400 ISO for the Canon 40D. Analysis shows that the
'displayed' ISO announced by Nikon is very over evaluated relative to
Canon competitor."

From the presented data, I find this puzzling. He compared the Canon
40D to the Nikon D3, both operating with a bit depth of 14. What he
calls inverse gain for both cameras at ISO 400 was 0.78 electron/ADU
for the Canon and 2.1 electrons/ADU for the Nikon. The Canon pixel
pitch is 5.7 microns and that for the Nikon is 8.2. The ratio of the
pixel areas is 1:2.22. If quantum efficiency, fill factor, etc are
similar for the two cameras, the Nikon should collect 2.22 times more
photoelectrons than the Canon at equivalent integration times, and
this should affect the gain proportionally. The actual gain ratio is
1:2.7, roughly as expected.

Can anyone explain Christian's analysis?

Bill

p.s.

For Emil: Christian does take truncation of the data into account
when calculating read noise.

Click to expand...

And how can ISO 6400 on the Nikon equate to 600 on the Canon? Surely, exposure / aperture equation is the same - it's not going to be 3-1/2 stops off or someone would have noticed by now ;-)

--
John Walker
http://jhwalker.smugmug.com/

PIXmantra said:

Bill Janes said:

The actual gain ratio is
1:2.7, roughly as expected.

Click to expand...

...2.7 is A WHOLE LOT more than 2.2. It truly is.

Click to expand...

When you take into account there may be differences in quantum efficiency, fill factor, microlens efficiency, color array filters, etc., the difference may not be significant.
--
Bill Janes

I can't figure out what he is talking about, unless the camera he was using was defective in some way.

BTW, my result for electrons at raw saturation are consistent with what John Sheehy is reporting, but I think he is using similar publicly available raw files that are less than ideal for this measurement.

Certainly the result I quoted should be regarded as a lower bound, since the systematic errors tend to suppress the resulting QE measurement.
--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/

jwalker019 said:

And how can ISO 6400 on the Nikon equate to 600 on the Canon?

Click to expand...

...This is purely fictional.

I have ISO samples from 1D3 and D3, and D3 6400 is pretty much the same as 1D3 with Luma Noise reduction set to a modest +15, and chroma to +10, in Lightroom. And the resulting effect looks almost exactly as D3, with NO degradation of detail (albeit SHARPER on the 1D3), though, which is what most likely is happening on D3, but ON-BOARD.

D3's almost total lack of chrominance noise accross its ISO band is SUPREMELY SUSPICIOUS.

We'll see.

PIX

--

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Jay Williams said:

I'm not a scientist, so I don't know, but don't photosites collect
photons, not electrons?

Click to expand...

When a photon strikes the surface of a material, if its energy exceeds a certain threshold (dependent on the material) it will kick out an electron. This is called the photoelectric effect, and was one of the first bits of evidence for the quantum nature of light. Einstein won the 1921 Nobel Prize in Physics for his explanation of this effect (not, as many people assume, for the theory of relativity).

Bottom line is that each electron accumulating in the photosite traces back to a photon in the light signal being detected.

--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/

Bill Janes said:

The article by Christian Buil states, "The ISO sensitivity levels
corresponds to an electronic gain extremely different between Canon
and Nikon. Assuming that the quantum efficiency of sensors are
equivalent (it is highly probable in first approximation), the
displayed sensitivity of 6400 ISO for the Nikon D3 #1 corresponds to
a sensitivity of 600 ISO for the Canon 40D, while a sensitivity
displayed at 6400 ISO for the Nikon D3 #2, corresponds to a
sensitivity of 2400 ISO for the Canon 40D. Analysis shows that the
'displayed' ISO announced by Nikon is very over evaluated relative to
Canon competitor."

From the presented data, I find this puzzling. He compared the Canon
40D to the Nikon D3, both operating with a bit depth of 14. What he
calls inverse gain for both cameras at ISO 400 was 0.78 electron/ADU
for the Canon and 2.1 electrons/ADU for the Nikon. The Canon pixel
pitch is 5.7 microns and that for the Nikon is 8.2. The ratio of the
pixel areas is 1:2.22. If quantum efficiency, fill factor, etc are
similar for the two cameras, the Nikon should collect 2.22 times more
photoelectrons than the Canon at equivalent integration times, and
this should affect the gain proportionally. The actual gain ratio is
1:2.7, roughly as expected.

Can anyone explain Christian's analysis?

Click to expand...

I think he made a mistake and did not factor in the 14bit RAW ADU of the D3, and underestimated the actual sensitivity of the D3 pixels by a factor of 4.

Bill Janes said:

To confuse the situation further, Christian Buil uses the term
"inverse gain". He reports further that Nikon does process the raw
data before writing them to the memory card, and that they are not
true raw files. This apparently applies only to exposures greater
than one second.

Click to expand...

I can confirm this: if you don't have noise reduction off the camera will somehow pre-process raw data before writing it. In my D80 that even happened above ISO-800 with the noise reduction off at any shutter speed. Getting reliable sensor measurements from such a system is unreliable at best...

I don't know about other Nikon models, but we might assume it is so also, probably at different ISOs (not 800).

For testing the D3 I would recommend using a few different ISOs to check how noise behaves... otherwise you might be fooled by the camera.

Just my 2 cents...

OK, I went back and did a somewhat more thorough job of analyzing available raw data. My previous methodology had somewhat overestimated the noise, leading to an underestimate of the efficiency of the D3 sensor. Before I was measuring the std dev of small uniform patches. This time, I used the difference of the two green subarrays of the sensor's Bayer array to calculate the noise, which should be more immune to subtle tonal gradients which were a systematic error in the initial measurement using standard deviations of small patches; this technique is much more robust since overall tonal gradients cancel between the two subarrays. I analyzed raw files found on the net from four different D3 bodies, at least one of which is a production model; they were all remarkably consistent with one another.

The result: the D3 collects 15.4 electrons per 12-bit raw level at ISO 200, or about 63,000 electrons at raw saturation for this ISO (and about 18% better than my initial rough measurement). This is to be compared with the 1Dmk3, which collects 10.2 electrons/12-bit raw level at this ISO, or a little over 39,000 electrons at raw saturation. The D3 collects about 50% more photons per pixel assuming that the calibration of ISO gain between the two cameras is comparable. Since S/N ratio goes as the square root of the photoelectron count, the D3 has about a 1/4 stop advantage over the 1Dmk3 at the pixel level in midtones and highlights at a given ISO; the two are about equal in shadows, since read noise is about the same.

If we compare the efficiency per unit of sensor area, however, the advantage largely disappears: The D3 is only about 5% less noisy for given sensor area (as for instance you would find by shooting the two at the same focal length; the D3 would just have a larger field of view).
--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/