How can Nikon remain in business with 12MP FF?

[dtaylor]

In the example below (100% crops from the 5DII, processed with DPP using no NR or other adjustments), the light source illumination was the same. For the shot on the right, ND filters were used reduce illumination by a fixed amount to match exposure, giving (IMO) a more common real-world situation where you might need to use ISO 6400.

Imaging Resource tests don't show as dramatic a difference as shutter speed drops. Are you sure you got the exposure right?

 

[epsiloneri]

In the example below (100% crops from the 5DII, processed with DPP using no NR or other adjustments), the light source illumination was the same.

Wow, that really looks convincing. I would never have guessed and am really at a loss how this can be. I will have to test it myself.

Yes, I think it's the case that it's not the shutter speed, but rather having enough light to satisfy the high sensitivity requirement of that ISO setting. More light = less shot noise

That would be he case if the exposures (the number of photons captured) for the two images were different. But in these exposures the total number of photons captured is the same, the only difference is that the photons arrive at different rates (hence you need to expose longer in the second case to get the same number of photons).

Same number of photons = same shot noise, implying that the noise must come from elsewhere. One suggestion is readout noise, but I can't see how that could be the case - the readout is normally independent of the exposure. Thermal noise (same thing as dark noise) grows with time, so that could be it, but normally dark noise is insignificant at exposure times less than a second. But perhaps it isn't, I really see no other explanation.

 

[Meh]

In the example below (100% crops from the 5DII, processed with DPP using no NR or other adjustments), the light source illumination was the same.

Wow, that really looks convincing. I would never have guessed and am really at a loss how this can be. I will have to test it myself.

Yes, I think it's the case that it's not the shutter speed, but rather having enough light to satisfy the high sensitivity requirement of that ISO setting. More light = less shot noise

That would be he case if the exposures (the number of photons captured) for the two images were different. But in these exposures the total number of photons captured is the same, the only difference is that the photons arrive at different rates (hence you need to expose longer in the second case to get the same number of photons).

Same number of photons = same shot noise, implying that the noise must come from elsewhere. One suggestion is readout noise, but I can't see how that could be the case - the readout is normally independent of the exposure. Thermal noise (same thing as dark noise) grows with time, so that could be it, but normally dark noise is insignificant at exposure times less than a second. But perhaps it isn't, I really see no other explanation.

@epsilonari good points this is what I initially thought when I first read Neuro's comment but struggled with the reason because his later examples do in fact show an obvious difference. From your comments in other threads you seem to know much about this type of thing.... a couple of ideas occur to me that perhaps you can comment on...

1) Since the ISO, aperture, and exposure is the same then we know that the same number of photons are collected. However, they are collected over a longer time as you said. Could the slower arrival of photons cause a greater spatial variation of the photons? But this would be "shot noise" which normally scales with number of photons collected not exposure time and 1/60s shouldn't be long enough to cause any unusual effect.

2) The ND filter (4 stops in Neuro's example) is not a perfect filter... could it be causing the noise by unevenly blocking the incoming photons causing additional spatial variation?

 

[poias]

Okay, I will take most of what I say back.

I did not realize that Nikon has had a 24mpx FF, which is apparently its flagship. It is expensive, but it was there.

So, Nikon had the highest mpx FF and the highest ISO FF at the same time? I think I answered my own question -- that is why they are in business.

 

[TimKaldas]

Canon didn't "hose us" by having the 1Dx with 18 MPs. They didn't hose us because we got a native ISO that reaches 51,200 with virtually no substantive drop in resolution. We got 12 fps with nearly no drop in resolution. We got a range of improvements with basically no sacrifice. I wish my 5D IIs had 12 MPs and d3s ISO performance. I export my images at 12 MPs anyways for my clients because that's plenty for most uses. I use high ISOs much more than I use 21 MPs. I think a lot of us were rather frustrated with Canon's endless pursuit of resolution while Nikon focused on perfecting image quality at a perfectly sufficient resolution. Finally Canon is getting that we're not stupid and understand that going any further in resolution is a wasteful distraction and that pros aren't as easily fooled as the point and shoot crowd who believe that more MPs = better. The 1Dx was exactly the camera I wanted Canon to make and frankly I can't wait to get one...or two.

 

[epsiloneri]

1) Since the ISO, aperture, and exposure is the same then we know that the same number of photons are collected. However, they are collected over a longer time as you said. Could the slower arrival of photons cause a greater spatial variation of the photons? But this would be "shot noise" which normally scales with number of photons collected not exposure time and 1/60s shouldn't be long enough to cause any unusual effect.

No, the path of a photon does not depend on the rate, so it shouldn't make a difference for the spatial distribution (otherwise sharp static objects would look blurrier in long exposures).

2) The ND filter (4 stops in Neuro's example) is not a perfect filter... could it be causing the noise by unevenly blocking the incoming photons causing additional spatial variation?

No, the ND filters are in the (almost) parallel beam, meaning that they are way, way out of focus. So any inhomogeneities of the filter will be spread over large parts of the detector and not give rise to the noise seen in neuroanatomist's images. We have to find the explanation elsewhere...

 

[Meh]

1) Since the ISO, aperture, and exposure is the same then we know that the same number of photons are collected. However, they are collected over a longer time as you said. Could the slower arrival of photons cause a greater spatial variation of the photons? But this would be "shot noise" which normally scales with number of photons collected not exposure time and 1/60s shouldn't be long enough to cause any unusual effect.

No, the path of a photon does not depend on the rate, so it shouldn't make a difference for the spatial distribution (otherwise sharp static objects would look blurrier in long exposures).

Yes, that's right of course shot noise is not spatial variation... it's the statistical variation in the arrival rate of the photons incident on each photosite. That variation causes adjacent/nearby photosites to have variation in the number of photons each should have collected. Now, each exposure was the same so the average/total number of photons collected is the same but there is more variation in the longer shutter speed image. Are we sure that shot noise is the square root of the number of photons collected and does not depend on the time over which they were collected (at least in the range of 1/2000s to 1/60s?

I just can't see how the difference in noise for these two equal exposures is due to read noise... the read noise is low and fairly constant. The exposure was the same therefore approximately the same number of photons were collected at each photosite. Therefore, the contribution of read noise to the SNR has to be about the same.

 

[skitron]

...and not give rise to the noise seen in neuroanatomist's images. We have to find the explanation elsewhere...

I believe the issue is that the longer length of time the analog and AD conversion circuits are in use, the more cumulative noise from these circuits there will be imparted on the final image. Remember, circuit noise is a random deviation from ideal performance (i.e. amplitude in the analog domain) and these deviations also have a random frequency in which they occur.

 

[Meh]

...and not give rise to the noise seen in neuroanatomist's images. We have to find the explanation elsewhere...

I believe the issue is that the longer length of time the analog and AD conversion circuits are in use, the more cumulative noise from these circuits there will be imparted on the final image. Remember, circuit noise is a random deviation from ideal performance (i.e. amplitude in the analog domain) and these deviations also have a random frequency in which they occur.

That's true in general, however, the readout of the image sensor and the A/D conversion takes place after the exposure is completed. I'm fairly certain the readout and A/D process would be identical and take the same amount of time regardless of the exposure time.

 

[Meh]

Maybe it is thermal noise. It's normally considered to be insignificant at exposures less than 1 second but certainly it's continuous and would be greater (although possibly still very small) at 1/60s than it is at 1/2000s. Any thoughts?

 

[skitron]

...and not give rise to the noise seen in neuroanatomist's images. We have to find the explanation elsewhere...

I believe the issue is that the longer length of time the analog and AD conversion circuits are in use, the more cumulative noise from these circuits there will be imparted on the final image. Remember, circuit noise is a random deviation from ideal performance (i.e. amplitude in the analog domain) and these deviations also have a random frequency in which they occur.

That's true in general, however, the readout of the image sensor and the A/D conversion takes place after the exposure is completed. I'm fairly certain the readout and A/D process would be identical and take the same amount of time regardless of the exposure time.

So the sensor is actually a storage device? There is no way to have a fixed length of time for the analog and AD to be used *regardless of exposure* unless the sensor actually is a storage device as well. I don't see how that is possible but confess it's not an area of electronics I know anything about.

 

[jaduffy007]

The funny thing though is the way that Nikon users are looking forward to a 36MP camera extolling the virtues of high MP count sensors while forgetting the arguements they used to defend the 12MP they currently have.

Some maybe, but I would say a majority of Nikon shooters, especially the D700 shooters are not in favor of a 36MP cam.
Of course, the landscape shooters are saying "bring it on!"

 

[jaduffy007]

http://minus.com/mwi8ith2f#6
but I wouldn't call 7D performance hideous.

Sorry,but compared to a D3S? Hideous comes pretty close imo

 

[skitron]

Maybe it is thermal noise. It's normally considered to be insignificant at exposures less than 1 second but certainly it's continuous and would be greater (although possibly still very small) at 1/60s than it is at 1/2000s. Any thoughts?

Wiki has a decent and short writeup of different noise types found in all electronics. They all apply all the time, it's just a matter of when do they become significant. Obviously neuro's example reaches 'significant' since we're spending our time talking about it. And that said, the useful principle seems to be that system S/N is closely correlated to instantaneous intensity of light reaching the sensor and lowering the instantaneous intensity and then collecting the same number of photons over time does in fact reveal this to be true. At least that's what I'm taking away from it.

 

[EYEONE]

http://minus.com/mwi8ith2f#6
but I wouldn't call 7D performance hideous.

Sorry,but compared to a D3S? Hideous comes pretty close imo

Ug, I didn't even say the performance was hideous. Just the look of the noise.
Anyway, moving on...

 

[Meh]

That's true in general, however, the readout of the image sensor and the A/D conversion takes place after the exposure is completed. I'm fairly certain the readout and A/D process would be identical and take the same amount of time regardless of the exposure time.

So the sensor is actually a storage device? There is no way to have a fixed length of time for the analog and AD to be used *regardless of exposure* unless the sensor actually is a storage device as well. I don't see how that is possible but confess it's not an area of electronics I know anything about.

In a way, yes, just during the exposure. During the exposure the photodiodes are electrically isolated... they are an electron well. Each photon that's absorbed frees one electron and that electron gets trapped in the well and stays there. By the time the exposure is finished each photodiode will have somewhere between a few (maybe 10) and a lot (many thousands, depends on the size of the photodiode) electrons that together create a voltage. Then the readout circuits start selecting rows and columns and measure the voltage at each photosite. That voltage (analog value) is then passed through the ADC and assigned a digital value. Even after all that is done the voltage is still present in each photosite so a reset circuit is activated that drains all the electrons out of the photodiodes. And the next exposure can then be made.

I may not have that exactly right and I'm omitting a few details so take it as a general explanation.

That is for a CMOS sensor. A CCD sensor works a little differently.

 

[epsiloneri]

Are we sure that shot noise is the square root of the number of photons collected and does not depend on the time over which they were collected (at least in the range of 1/2000s to 1/60s?

Yes, it's called poisson noise. Technically, in the extremely-few photon case (like, you-can-count-them-on-your-fingers few), the photons do not follow poisson statistics (because of the quantum mechanical fact that photons are bosons), but that's not a case we come remotely close to here.

So the sensor is actually a storage device?

Yes, that's precisely right. A fraction of the photons that strike the detector produce electrons (the fraction being called the quantum efficiency, QE, of the detector), and these electrons are then stored in the pixel "electron well". The depth of the well (how many electrons fit in) gives an upper limit on the dynamic range of a pixel. At the end of the exposure, the well is emptied and the number of electrons in each pixel are counted and given a (digital) number.

Dark current (also called thermal noise) arises because at any given temperature, electrons tend to move around (more the higher the temperature), and some of these thermal electrons accidentally fall into the electron wells of the pixels (electrons may also escape from a well this way). For longer exposure times and higher temperatures, the electrons are more likely o re-assemble like this, which is the reason longer exposures generate more thermal noise.

 

[Meh]

@epsiloneri

Thanks. I didn't think we were anywhere near a regime where the photons were not statistically well behaved. Although, in very dark areas of an image I think there would be very few photons collected. Somewhere between 10 and 100.

I'll add something to your explanation of the sensor read out process. In a CMOS sensor the electrons are not emptied during readout. It has to be reset after the readout. In a CCD they are emptied sequentially as they're read out. I think so anyway

And +1 to you sir!

 

[epsiloneri]

I'll add something to your explanation of the sensor read out process. In a CMOS sensor the electrons are not emptied during readout. It has to be reset after the readout. In a CCD they are emptied sequentially as they're read out.

Yes, you're right. The pixel is first measured, then emptied (reset). A CCD has to physically transfer the charge to the border of the chip where it's read out, that's why CCD readouts are destructive while CMOS's are not.

Since it's trivial to set up, I repeated neuroanatomist's experiment but cannot confirm his result that the noise depends on exposure time. I produced a scene that at f/4 was 75% well exposed at 1/2000s and iso 6400 (with a 5D2). Since we are interested in the detector rather than the lens I set the focus to infinity to blur any details of the scene (just a blank homogeneous wall).

For the second image, I used the same setting but with a Hoya NDX400 filter and 1/5s exposure instead. I chose the time carefully so that the exposure would be the same (judging from the on-screen histogram), and predictably the difference in exposure time was a factor 400 (expected because of the 400x ND filter).

I set noise reduction and sharpness enhancement to 0 in DPP, and white balance to "white fluorescent light" to keep it constant (and since it matches the lighting conditions). The same recipe was used on both exposures.

The attached image shows 100% crops of the identical region, the left being from the 1/2000s image and the right from the 1/5s image. Apart from the obvious colour cast introduced by the filter, there is no discernible difference in the noise. Computing the noise (stdvar) of the image confirms the visual impression; for the left/right part of the (R,G,B) channels, the noise is (3.03/3.06, 2.43/2.40, 3.03/3.07), i.e. the difference in noise between left and right is on the order of 1%, clearly insignificant.

 

[Loswr]

Interesting! I'll have to see if I can replicate my own results...