right answer will always be, smaller pixels, **lower noise**

**NOISE**

simple as that

Is it that simple? What about the signal?

now it is motor cycles again

please learn to discuss the subject, noise

small pixel always gives lower noise than bigger, the signal has not been up to discussion

You and Jrista has some problem to know what we discuss or not

Must it be like this?

and are you going to make you funny about me again?

head room, signal/noise , fluorite glass , etc etc what next?

Ok, lets get to the root of this here.

1. Assume we have a hypothetical sensor that introduces ZERO read noise whatsoever...no dark current noise, no high frequency noise...no read noise of any kind from any electronic source in the camera, on the sensor die or anywhere else...just for discussions sake.

2. That sensor captures an image projected by a lens in dim light, at an ISO setting of 1600.

3. Is that image noise free, or is there an intrinsic component of noise that is a very part of the image itself?

This is a test. Your answer does matter. You will be judged upon it. Go!

test? what kind of tests?

I got guests here in my summer house, what kind of answer do you want?

**A pragmatic answer will be**-YOU don't se any difference from middle grey up to high light, (I can call it infinity)

But you are going to se better results in the shadows - then there are a huge among of other answers

BUT first my guests who are staying here 2 days more.

and PS You will be judged upon it. Go!

what in h... is that?

And there you go, peoples! No answer, more obfuscation, and some beating around the bush about "guests".

I think we can safely come to the conclusion that Mikael does not understand the concept of intrinsic noise in an image signal.

**Boys who do you think you are.**

well Neuro and Jrista if you don't understand my answer about signal noise and results below middle grey and above middle grey here will come a quick answer

As I said it is one answer of many.

but here is one

translate it can you do by your self

swedish to english use google

Det enda som egentligen händer är ju att det då är helt och hållet [effektiv QE] som bestämmer bruset...

Brusvärdet är då alltid roten av ljusmängden (+tillskott normalt sett, men inte i detta fallet då). Bruset vid 18% grå eller ljusare kommer knappt påverkas alls av att det inte finns något läsbrus, det är en så liten del av bruset totalt sett här - även på ISO1600.

För t.ex 6D är vitpunkten i råfilen på ISO1600 ca 4400e- (FWC/16, eftersom ISO1600 är 16ggr mer än bas-ISO).

Mellangrått blir då 18% av 4400 = 792e-, och fotonbruset av detta är sqrt(792) = 28,1e- för 18% grå

Eftersom läsbruset i en normal kamera är ca 3e- vid ISO1600 hade detta knappt påverkat signalen. Skulle man lagt till 3e- till 28.1e- fotonbrus hade det blivit:

sqrt(28.1^2 + 3^2) = 28.3e- (+0.2e- inte en märkbar skillnad, knappt mätbar)

Under 18% grå kommer bruset växa saktare när man går neråt mot det mörkare om man INTE har något läsbrus. Detta är skillnaden jmf med läsbrus.

Vid 1% exponering (-6.6Ev) är den infångade ljusmängden bara 44e-, så bruset blir:

sqrt(44) = 6.6e-

Här är läsbruset märkbart, då

sqrt(6.6^2 + 3^2) = 7.3e-.

Skillnaden är då 6.6e- utan läsbrus vs 7.3e- med normalt läsbrus - ingen jättestor skillnad här heller, men antagligen märkbar

6.6e- av 44e- signal = 20*log(44/6.6) = 16.4dB brusavstånd

7.3e- av 44e- signal = 20*log(44/7.3) = 15.6dB med läsbrus.

Den viktiga skillnaden kommer ju när läsbruset hade börjat dominerar fotonbruset, nere i skuggorna. Samma sak som DR-punkten alltså.

På -10Ev har den "läsbrusfria" kameran fortfarande ca 6dB SNR. Kameran med 3e- läsbrus kommer då ha nästan noll i SNR, dvs lika mycket brus som signal. Här syns det mycket skillnad.

Den läsbrusfria pixeln kan helt utan förluster delas upp i hur små pixlar som helst utan att bruset över bildytan (per detalj) ökar, har man läsbrus kommer detta vid en viss pixelmängd börja bli "för många brustillskott" för att det ska finas någon vinst med det.

right answer will always be, smaller pixels, **lower noise**

**NOISE**

simple as that

Is it that simple? What about the signal?

now it is motor cycles again

please learn to discuss the subject, noise

small pixel always gives lower noise than bigger, the signal has not been up to discussion

You and Jrista has some problem to know what we discuss or not

Must it be like this?

and are you going to make you funny about me again?

head room, signal/noise , fluorite glass , etc etc what next?

Ok, lets get to the root of this here.

1. Assume we have a hypothetical sensor that introduces ZERO read noise whatsoever...no dark current noise, no high frequency noise...no read noise of any kind from any electronic source in the camera, on the sensor die or anywhere else...just for discussions sake.

2. That sensor captures an image projected by a lens in dim light, at an ISO setting of 1600.

3. Is that image noise free, or is there an intrinsic component of noise that is a very part of the image itself?

This is a test. Your answer does matter. You will be judged upon it. Go!

test? what kind of tests?

I got guests here in my summer house, what kind of answer do you want?

**A pragmatic answer will be**-YOU don't se any difference from middle grey up to high light, (I can call it infinity)

But you are going to se better results in the shadows - then there are a huge among of other answers

BUT first my guests who are staying here 2 days more.

and PS You will be judged upon it. Go!

what in h... is that?

And there you go, peoples! No answer, more obfuscation, and some beating around the bush about "guests".

I think we can safely come to the conclusion that Mikael does not understand the concept of intrinsic noise in an image signal.

**Boys who do you think you are.**

well Neuro and Jrista if you don't understand my answer about signal noise and results below middle grey and above middle grey here will come a quick answer

As I said it is one answer of many.

but here is one

translate it can you do by your self

swedish to english use google

Det enda som egentligen händer är ju att det då är helt och hållet [effektiv QE] som bestämmer bruset...

Brusvärdet är då alltid roten av ljusmängden (+tillskott normalt sett, men inte i detta fallet då). Bruset vid 18% grå eller ljusare kommer knappt påverkas alls av att det inte finns något läsbrus, det är en så liten del av bruset totalt sett här - även på ISO1600.

För t.ex 6D är vitpunkten i råfilen på ISO1600 ca 4400e- (FWC/16, eftersom ISO1600 är 16ggr mer än bas-ISO).

Mellangrått blir då 18% av 4400 = 792e-, och fotonbruset av detta är sqrt(792) = 28,1e- för 18% grå

Eftersom läsbruset i en normal kamera är ca 3e- vid ISO1600 hade detta knappt påverkat signalen. Skulle man lagt till 3e- till 28.1e- fotonbrus hade det blivit:

sqrt(28.1^2 + 3^2) = 28.3e- (+0.2e- inte en märkbar skillnad, knappt mätbar)

Under 18% grå kommer bruset växa saktare när man går neråt mot det mörkare om man INTE har något läsbrus. Detta är skillnaden jmf med läsbrus.

Vid 1% exponering (-6.6Ev) är den infångade ljusmängden bara 44e-, så bruset blir:

sqrt(44) = 6.6e-

Här är läsbruset märkbart, då

sqrt(6.6^2 + 3^2) = 7.3e-.

Skillnaden är då 6.6e- utan läsbrus vs 7.3e- med normalt läsbrus - ingen jättestor skillnad här heller, men antagligen märkbar

6.6e- av 44e- signal = 20*log(44/6.6) = 16.4dB brusavstånd

7.3e- av 44e- signal = 20*log(44/7.3) = 15.6dB med läsbrus.

Den viktiga skillnaden kommer ju när läsbruset hade börjat dominerar fotonbruset, nere i skuggorna. Samma sak som DR-punkten alltså.

På -10Ev har den "läsbrusfria" kameran fortfarande ca 6dB SNR. Kameran med 3e- läsbrus kommer då ha nästan noll i SNR, dvs lika mycket brus som signal. Här syns det mycket skillnad.

Den läsbrusfria pixeln kan helt utan förluster delas upp i hur små pixlar som helst utan att bruset över bildytan (per detalj) ökar, har man läsbrus kommer detta vid en viss pixelmängd börja bli "för många brustillskott" för att det ska finas någon vinst med det.

What, nobody willing to hit up FreeTranslation?

The only thing that really happens is, of course, the fact that it is completely [effective QE] which determines noise ...

Brusvärdet is always the root of ljusmängden ( addition normally, but not in this case). Noise at 18% gray or lighter will hardly be affected at all of the fact that there is no läsbrus, it is such a small proportion of overall noise here - even at ISO1600.

For example 6D is white spot in råfilen on ISO1600 approximately 4400 e- (attractive FWC/16, since ISO1600 is 16x more than base-ISO).

Mellangrått will be 18% of 4400 = 792e-, and fotonbruset of this is sqrt(792) = 28.1e- for 18% gray

as läsbruset in a normal camera is approximately 3e- at ISO1600 had barely had an impact on the signal. It would have added to the 3e- to 28.1e- fotonbrus had it been:

sqrt(28.1 ^2 3 ^ 2) = 28.3e- ( 0.2e- not a noticeable difference, barely measurable)

under 18% gray noise will grow more slowly when it goes down to the darker if you do not have a läsbrus. This is the difference comparison with läsbrus.

At 1% exposure (-6.6Ev) is the captured ljusmängden only 44e-, so noise becomes:

sqrt(44) = 6.6e-

this is läsbruset noticeably, then

sqrt(6.6) ^2 3 ^ 2) = 7.3e-.

The difference is then 6.6e- without läsbrus vs 7.3e- with normal läsbrus - no enormous difference here either, but probably noticeable

6.6e- of 44e- signal = 20 * log(44/ 6.6) = 16.4dB brusavstånd

7.3e of 44e- signal = 20 * log(44/ 7.3) = 15.6dB with läsbrus.

The important difference, of course, will when läsbruset had begun to dominate fotonbruset, down in the shadows. Same as DR-point then.

-10Ev has the "läsbrusfria" the camera is still approximately 6dB SNR. The camera with 3e- läsbrus will have almost zero in SNR, i.e. , as much noise as a signal. This can be seen very difference.

The läsbrusfria pixel can completely without losses are divided up in how small pixels at any time without the noise of image area (per detail) increases, it has läsbrus this will at a certain pixelmängd start to become "too many brustillskott" for it to appear any profit with it.

Not all of those words translated right, but the just of it looks like the maths. Nothing hurts my brain more than maths. All I know is that I have cameras with sensor sizes ranging from the size of my pinky nail to the size of your average card reader, and every time I print a print outside of that camera's native resolution, the outcome is exponentially better with bigger sensors and bigger pixels. As an example:

http://motionblurdaily.com/2013/06/18/why-we-use-what-we-use-medium-format-vs-full-frame-vs-aps-c/Or, as another example, check out the photos below. One is from a Canon SX40, with a sensor the size of my pinky nail and a pixel size of 1.5 µm, at base ISO. One is from my Canon T2i with an APS-C sensor and a 4.2µm pixel size, and the third is from a medium-format Phase One camera with a huge sensor (37mm x 49mm) and a 6.8µm pixel pitch. Can you guess which is which?