A camera's dynamic range

[Abitconfused]

A camera's "dynamic range" must more accurately be described as "potential dynamic range" because many images have a rather limited tonal span. Yet a camera boasting a greater dynamic range suggests greater color accuracy in any pixel regardless of tones present in a scene. True?

[DanK]

I don't think so. AFAIK. Dynamic range refers solely to the range of luminance the sensor can capture. I don't believe it has any bearing on color accuracy as long as the DR isn't exceeded.

[Abitconfused]

Then if we increase ISO and acquire digital noise in pixels (poorer color accuracy) is it because we have exceeded the dynamic range of the sensor?

[george013]

I don't think so. AFAIK. Dynamic range refers solely to the range of luminance the sensor can capture. I don't believe it has any bearing on color accuracy as long as the DR isn't exceeded.

It's all in the A/D conversion as I showed before.


For a A/D conversion you need a range and an amount of steps, or bitdepth
Normally, at base iso, the maximum range is used and the minimum and maximum of that range, in 8 bit 0 and 255, corresponds to the maximum and minimum of the monitor. A value half way that sensor range will have a value of 128 and so on the monitor.
When you increase the iso with a factor of two, you divide that 'sensor range' by two. The steps don't change. So in figure 3 the value on the left side half way the base range will get the value 255 and that correspondes to the maximum value of your monitor. By making the range smaller you're multiplying the output. And also the noise.

George

[xpatUSA]

There is a Standard for the measurement of a Digital Camera's Dynamic Range: ISO 15739-2003

To save you having to buy it, Doug Kerr discusses it at length here:

http://dougkerr.net/Pumpkin/articles...amic_range.pdf

In this thread, any speculative statement made without reference to the Standard or, at least to a credible reference like the above, is likely to be a waste of Forum time ...

[xpatUSA]

Then if we increase ISO and acquire digital noise in pixels (poorer color accuracy) is it because we have exceeded the dynamic range of the sensor?

No. Choosing to under-expose the sensor by cranking up the ISO uses less of the sensor's dynamic range.

Digital noise is not "acquired" by increasing ISO - it stays the same, more or less:

http://info.adimec.com/blogposts/rea...does-it-matter

See 'read noise'.

[Manfred M]

Dynamic range changes with ISO. The best dynamic range is always found at the "base" ISO of the camera. The moment you change "ISO", all you are doing is amplifying the signal from your sensor, and dynamic range decrease.

Head over to the DxOMark website and look up your camera. You will see how the dynamic range decreases as you turn up the gain.

https://www.dxomark.com/Cameras/Niko...--Measurements

[Abitconfused]

So are we amplifying the signal or underexposing the sensor? It seems underexposing the sensor would be more effective as amplifying the signal would also amplfy stronger signals.

[Manfred M]

So are we amplifying the signal or underexposing the sensor? It seems underexposing the sensor would be more effective as amplifying the signal would also amplfy stronger signals.

The sensor sensitivity capabilities are fixed. What you are describing is exactly how our cameras work. "Base" ISO requires no additional gain. When we increase gain, we are amplifying the signal from what you refer to as "underexposing the sensor".

[george013]

So are we amplifying the signal or underexposing the sensor? It seems underexposing the sensor would be more effective as amplifying the signal would also amplfy stronger signals.

It's a digital game. By reducing the range using for the A/D conversion at the input site but not at the output site you're in matter of fact amplifying the signal. Focusing on that "digital game" will help you also to understand issues with colors.
Looking at the diagram of dynamic range in the same DxO site you'll see that dynamic range and iso are related with each other. A change of 1 EV is roughly a doubling or halve of the iso.
But you don't see it darker for the output is unchanged.

George

[xpatUSA]

So are we amplifying the signal or underexposing the sensor?

We are amplifying the signal plus noise from all sources, not just "the signal". A double-whammy because "noise" now includes 'shot' (Poisson) noise from the captured photons themselves.

It seems [that] under-exposing the sensor would be more effective as amplifying the signal would also amplify stronger signals.

What?

Let's talk in signal electrons (e- = captured photons). Let's choose 40,000e- as close to sensor saturation. Let's say the sensor produces 40e- of noise even with the lens cap on. Let's say the camera electronics add 10e- equivalent read noise. Let's say that the 'shot' noise is sqrt(signal).

1) let's use 100 ISO and capture 40,000e-. Noise (rms) = sqrt(40^2+10^2+40,000) = 204e- r.m.s.

2) now let's use 800 ISO and capture 5,000e-. Noise (rms) = sqrt(40^2+10^2+5,000) = 82e- r.m.s.

The Signal to Noise Ratio (SNR) for (1) is 40,000/204 = 7.6EV. The SNR for (2) is 5,000/82 = 5.9EV.

Quite a difference, eh?

For SNR, more is better ...

[Manfred M]

We are amplifying the signal plus noise from all sources, not just "the signal". A double-whammy because "noise" now includes 'shot' (Poisson) noise from the captured photons themselves.



What?

Let's talk in signal electrons (e- = captured photons). Let's choose 40,000e- as close to sensor saturation. Let's say the sensor produces 40e- of noise even with the lens cap on. Let's say the camera electronics add 10e- equivalent read noise. Let's say that the 'shot' noise is sqrt(signal).

1) let's use 100 ISO and capture 40,000e-. Noise (rms) = sqrt(40+10+sqrt(40,000)) = 16e- r.m.s.

2) now let's use 800 ISO and capture 5,000e-. Noise (rms) = sqrt(40+10+sqrt(5,000)) = 11e- r.m.s.

The Signal to Noise Ratio (SNR) for (1) is 40,000/16 = 11.3EV. The SNR for (2) is 5,000/11 = 8.8EV.

Quite a difference, eh?

For SNR, more is better ...

Interesting Ted. That explains the noise issues, but how does this link to the reduced dynamic range and reduced colour bit depth as ISO settings increase, as per the DxO Mark graphs? Can all these effects be explained by the signal to noise issues?

My guess is yes, but would like to know why. I assume that parts of the data contain so much noise as to render them less useful, so these are excluded from the data used to generate the image. I assume this is the same issues are in play when discussing the "ISO-less sensor".

[xpatUSA]

Interesting Ted. That explains the noise issues, but how does this link to the reduced dynamic range and reduced colour bit depth as ISO settings increase, as per the DxO Mark graphs? Can all these effects be explained by the signal to noise issues? My guess is yes, but would like to know why.

Thank you, Manfred. Your guess is as good as mine. Unfortunately, I don't subscribe to DxO Mark which means that I have no idea how they derive their graphs and therefore can not help with that. I usually try to bootleg standards (tsk) or refer to respected authors in the genre, of whom Doug Kerr is one.

I assume that parts of the data contain so much noise as to render them less useful, so these are excluded from the data used to generate the image. I assume [that the] the same issues are in play when discussing the "ISO-less sensor".

[xpatUSA]

Interesting Ted. That explains the noise issues, but how does this link to the reduced dynamic range and reduced colour bit depth as ISO settings increase, as per the DxO Mark graphs? Can all these effects be explained by the signal to noise issues?

My guess is yes, but would like to know why.

Manfred, taking a second look and avoiding DxO's products:

The very basis of the ISO Standard mentioned is that, at the high end, the sensor is saturated (just) and, at the low end, the signal level is equal to the "noise" level, i.e. SNR = 0 dB.

The Standard does not mention ISO or any other form of under-exposure AFAIK (Kerr has a copy, I do not).

That is to say, messing with the ISO setting of a camera does not affect the sensor's dynamic range at all, but the SNR of a resulting image is obviously affected by so doing.

As to "color bit depth" our meanings must vary - because every image I show on my screen has a color bit depth of 8 bits per channel, no matter what ISO was used for the shot.

https://wolfcrow.com/blog/what-is-co...a-color-model/
.

[Abitconfused]

Thank you, I get it as never before!

[dje]

Interesting Ted. That explains the noise issues, but how does this link to the reduced dynamic range and reduced colour bit depth as ISO settings increase, as per the DxO Mark graphs? Can all these effects be explained by the signal to noise issues?

Here's an example for Dynamic Range for my D610

ISO100

Full Well Capacity 75000 electrons (this causes the A/D to reach it's maximum output eg DN=16383 for 14 bit)
Read noise 8 electrons

DR= FWC/RN = 75000/8=9375 which is 13.2 stops

ISO600

A gain of 6 times is applied to the output of the sensor thus a charge of 75000/6=12,500 causes the A/D converter to reach it's maximum output.

DR = 12,500/RN=12500/8 =1562 which is 10.6 stops.

Dave

[george013]

Here's an example for Dynamic Range for my D610

ISO100

Full Well Capacity 75000 electrons (this causes the A/D to reach it's maximum output eg DN=16383 for 14 bit)
Read noise 8 electrons

DR= FWC/RN = 75000/8=9375 which is 13.2 stops

ISO600

A gain of 6 times is applied to the output of the sensor thus a charge of 75000/6=12,500 causes the A/D converter to reach it's maximum output.

DR = 12,500/RN=12500/8 =1562 which is 10.6 stops.

Dave

Can you explain me how you come to 6 times with iso 600?

George

[dje]

Can you explain me how you come to 6 times with iso 600?

George

There's a linear relationship between sensor exposure in lux secs and the analogue output from a pixel. ISO sensitivity is defined as 78/Exposure required to reach saturation. Increasing the gain by 6 means you need 6 times less exposure to reach saturation and hence the "effective ISO sensitivity" increases by 6 times.

Dave

PS I'm talking 6 times, not 6 stops.

[xpatUSA]

Can you explain me how you come to 6 times with [ISO] 600?

George

Because the amplifier is before the ADC, the "sensitivity" of the sensor is effectively increased.

The simple ISO formula for saturation-based ISO, a.k.a. 'ISO speed', is:

ISO = 78/Hsat.

Now let's say 75,000e- causes saturation (well full), as Dave stated, H would be 0.78 lux-sec making ISO = 78/0.78 = 100 ISO.

If the gain changes to 6, H"sat" becomes 0.78/6 = 0.13. ISO now = 78/0.13 = 600 ISO.

Ipso facto, the factor of 6 is correct, QED.

[edit] Pardon me, Dave, we 'crossed in the post' . . sorry.

[george013]

Thanks both of you Dave and Ted. I never knew how the iso was calculated.
I just was wondering for the result was what I would expect: an increase of the sensitivity with some less as 3 stops results in an decrease of dr with some less as 3 stops. Now the why is getting more clear to me.

A gain of 6 times is applied to the output of the sensor thus a charge of 75000/6=12,500 causes the A/D converter to reach it's maximum output.

Just to be sure, this has the same effect as what I wrote. In an A/D conversion we need a range and an amount of steps. Multiplying that analogue signal by 6 and keeping the maximal at 75000 will give the same result as setting the maximal at 12500. In a 8 bit conversion both maxima will get the digital number of 255. And will be shown as the whitest part of your output device.


George