Sensor Sensitivity: an ISO question

[xpatUSA]

Over on another forum, there was a bit of a bash-out as to what is meant by "the ISO of a sensor". The discussion was liberally spattered with words like "native", "base" but no actual numbers, units, definitions or whatever appeared that seemed satisfactory. Not talking here about base "camera sensitivity" e.g. 200 for Nikons, 100 for Canons, etc.

I do have a copy of Kodak's application note on the subject and also I have access to ISO:12232-2006. My problem with those documents et al is the statement in just about the first paragraph of every one, where it says:

ISO = 10/H where H is the exposure at the sensor face in lux-seconds for some specified condition.

So, you would expect to do a quick calc and thereby get your sensor's "ISO", right? The trouble, and the basis of my question, is that out comes a seemingly stupid number from what should have been a simple calculation. Here's what I did:

To find H at saturation (one possible condition), I used data for a particular sensor which had the following values: Well capacity 77,000 electrons. Quantum Efficiency 0.45.

From that data came a requirement for 77,000/0.45 = 171,000 photons (on average) to fill, i.e. saturate, the well. That figure lead to an exposure in lux-seconds of 171,000/5580 = 30.67 lux-seconds. (I Googled to find 5580 which is representative of several different values depending on wavelength).

So, applying the formula ISO = 10/H, I get ISO = 10/30.67 = 0.326 !!

The value 0.326 makes no sense to me at all. None. Zip. Nada.

Please, what I am I not understanding? I just don't get it. There's something missing, I'm sure of it. Help!

[Mike Buckley]

Not talking here about base "camera sensitivity" e.g. 200 for Nikons, 100 for Canons, etc.

I don't know about Canon cameras, but some Nikons have a base ISO of 100 and others have a base ISO of 200. I don't know how to understand your fundamental question, much less how to asnwer it.

[Polar01]

The problem here I believe is that some people have too much time on their hands. Really who cares if the sensor can be set to appear to the old film sensitivity of ISO 50 and can be increase to ISO 62,000. If you are worried about this then you have too much time on your hand make better use of your time and take some images. That is just my thinking on the subject.

Cheers:

Allan

[Brocken]

I think these may be the kind of thing you're looking for, Ted:

http://www.hpl.hp.com/techreports/97/HPL-97-54.pdf

http://www.cis.rit.edu/research/mcsl.../PDFs/Quan.pdf

[Colin Southern]

I don't know about Canon cameras, but some Nikons have a base ISO of 100 and others have a base ISO of 200. I don't know how to understand your fundamental question, much less how to asnwer it.

Canon generally have a base ISO of 100.

[Manfred M]

Ted - I suspect that you have identified half of the issue; which is the native ability of the sensor to ennumerate the number of photons that hit the individual photodiodes. I think this is the value you have calculated. But this is only half of what happens. The signal strength is quite low and has to be amplified before additional processing can occur. You have to look at the combination of the detectors and analogue amps when you calculate the base ISO rating.

[jiro]

Maybe, this article will give you a different formula to work on about ISO and digital sensor design.

http://www.dxomark.com/index.php/Abo...SO-sensitivity

[xpatUSA]

The problem here I believe is that some people have too much time on their hands. Really who cares if the sensor can be set to appear to the old film sensitivity of ISO 50 and can be increase to ISO 62,000. If you are worried about this then you have too much time on your hand make better use of your time and take some images.

My oh my, how tolerant we are today! Must be all that GST, PST and those Hydro bills at a guess . . enough to make anyone lash out at the nearest overpaid retired person with an interest in how things work ;-)

Thanks to everybody else for their informative replies!

[xpatUSA]

So, applying the formula ISO = 10/H, I get ISO = 10/30.67 = 0.326 !!

The value 0.326 makes no sense to me at all. None. Zip. Nada.

Please, what I am I not understanding? I just don't get it. There's something missing, I'm sure of it. Help!

Found it. I had failed to account for the active area of the sensor pixel.

At a particular wavelength, the saturated ISO for that sensor works out to be 10 (a coincidence, though) and that could be considered the base ISO of the sensor. It follows that amplification would indeed be required (as Manfred said above) to get a camera ISO of even 50!!

[Mike Buckley]

So, am I understanding correctly that what most people commonly refer to as the camera's base ISO is instead simply its lowest ISO setting and that the true base ISO is a lower value that has to be amplified to achieve the camera's lowest ISO setting?

(Please keep the response in simple lay terms so this dummy can understand it. "Yes" or "No" would be helpful responses. If clarification is needed, I probably wouldn't understand it.)

[Manfred M]

So, am I understanding correctly that what most people commonly refer to as the camera's base ISO is instead simply its lowest ISO setting and that the true base ISO is a lower value that has to be amplified to achieve the camera's lowest ISO setting?

(Please keep the response in simple lay terms so this dummy can understand it. "Yes" or "No" would be helpful responses. If clarification is needed, I probably wouldn't understand it.)

Mike; while I'm by no means an expert, I do have a basic understanding of how sensors work. The simple answer to your question is no. You cannot separate the photo detector part of the sensor from the amplifier part; they are part and parcel of the same piece of hardware. The actual signal from the photodiodes is so low as to be impossible for other circuit elements to use, so amplification is necessary so that other external circuits can actually use it. Circuits don't care about ISO settings, rather they need a specific voltage range to operate. As photographers we do care and this base voltage requirement defines the base ISO of the sensor.

As an analogy; if you enter a room with the lights controlled by a dimmer switch. If the dimmer is turned all the way down so you can just barely see a tiny bit of light from the light fixture, this is pretty useless to anyone trying to do something in the room. Technically the light is on, but... So now, when you turn up the light level using the dimmer control, you are increasing the voltage , you get to a point where the light level is high enough that you can actually see and do things.

[blakemcguire]

I think Manfred nailed it. Amplification is part of the sensor itself, either at the cell itself on a CMOS, or just off of it on a CCD.

Another analogy would be an electric guitar. You could grab a patch cord and plug it directly into a speaker, and it IS producing some current, but it's not enough to hear through the speaker, so we need to amplify it by a fair margin to make it even faintly audible.

[arith]

I do not understand it To me; iso like asa or din is to do with film and speed, size of grain and the rest. But on a sensor it means speed and dynamic range.

So 200iso is twice as fast as 100, 1600iso 16 times faster but probably only a 1/4 the dynamic range.

This is all I think I need to know

[xpatUSA]

The sensor I've been looking at has 3 analog outputs of 550 millivolts at maximum signal level. As far as I can find out, the sensor voltage output is proportional to the well fill level only, i.e. 550mV = well full. From there, it gets converted off-sensor into a digital signal, via a buffer amplifier and a 12-bit digitizer, known to some as an ADC. It is the off-sensor buffer amplifier which sets the camera ISO by having it's gain varied by the camera controls.

All of which is too much information for me - I was only interested in the photo-site performance and how it related to the ISO basic standard formula. As soon as we digress into the areas of camera ISO, scene luminance, signal-to-noise ratio, dynamic range, headroom, gray cards I get very confused. Especially when the ISO knob on your camera sets what the manufacturer thinks you want (quite legally, through the provision of "latitude" in the ISO standard and at least four different ways to define ISO itself).

Over here in the USA there's a series of commercials for a credit card that always end with the question: "what's in YOUR wallet?". To paraphrase:

"What does YOUR ISO 200 set?"

[Mike Buckley]

Thank you to Blake and Manfred for the clear explanations in lay terms. Not that I could possibly think of blaming them if I didn't understand them, but I actually think I do understand them. I think I'll quit while I'm still under such enjoyable delusions.

[dje]

Ted I'm no expert on this subject either but here are a few of my thoughts.

I don't have access to the full ISO standard (12232) but I have read an article in Wikepedia which summarises the thrust of the standard here. This refers to two formulae -

Standard Output Sensitivity S=10/H where H is the value that will lead to 18% saturation

Saturation based Speed S = 78/H where H is the value that gives max saturation.

I would have thought the the second formulae is the one you should be using. This is consistent with the DXO labs article that Willie refers to in his post above.

I would also have thought that the "base" ISO for a camera corresponded to the lowest noise value achievable ie without amplification but I could be wrong.

I look at ISO for digital cameras as follows :

• ISO values are a carry over from the days of film and were a measure of the speed of the film
• The film speed was determined by the chemical make-up of the film. The higher the speed, the more grainy-ness (noise)
• With digital, an attempt to retain the same sensitivity ratings as for film was made by introducing ways of "standardiing" ISO ratings for digital cameras. This allowed the same exposure settings "triangle" to be retained (shutter speed, aperture and ISO).
• With digital, the same compromise applies. With the base ISO, the noise is as low as it gets. To get a higher ISO setting and hence more sensitivity to light, amplication must be added and this adds noise.

That's my simple view of the world !!

Dave

[xpatUSA]

Ted I'm no expert on this subject either but here are a few of my thoughts.

I don't have access to the full ISO standard (12232) but I have read an article in Wikepedia which summarises the thrust of the standard here. This refers to two formulae -

Standard Output Sensitivity S=10/H where H is the value that will lead to 18% saturation

Saturation based Speed S = 78/H where H is the value that gives max saturation.

I would have thought the the second formulae is the one you should be using. This is consistent with the DXO labs article that Willie refers to in his post above.
Dave

Hi Dave,

When researching, I looked at the formula with the constant of 78. ISO says:

:6.2.1 Focal plane measurement
The saturation based speed, Ssat, of an electronic still picture camera is defined as:
Ssat = 78/Hsat ( eqn. 5)
where Hsat is the minimum focal plane exposure, expressed in lux-seconds (lx⋅s), that produces the maximum valid (not clipped or bloomed) camera output signal.


NOTE Equation (5) provides 1/2 “stop" of headroom (41 % additional headroom) for specular highlights above the signal level that would be obtained from a theoretical 100 % reflectance object in the scene, so that a theoretical 141 %
reflectance object in the scene would produce a focal plane exposure of Hsat. Therefore, an 18 % reflectance test card in the scene would produce a focal plane exposure of 128/1 000 Hsat. Thus, the multiplicative constant 78 in Equation (5) is equal to 10 times 1 000/128, where the value 10 is the constant from Equation (1)."

I thought that was the one to use until I read the about headroom. I also read ISO 2721:1982, where the constant 10 came from but it's description there just went on about stops and stuff. It said:

"Exposure in the focal plane
The nominal exposure H in the focal plane for a film of ISO Speed S (arithmetic) and with a luminance range of from 4 to 4 096 cd/m^2 (corresponding approximatively to exposure values EV = 5 to EV = 15 for ISO 100/21° film) should be:


H = Ho/S
with the constant Ho = 10 lx.s"

Not much help there, except that 10 sprouted units of lx.s making S (i.e.) a dimensionless number

BUT from what you're saying, 10 is probably the wrong choice. Unfortunately, the constant 78 includes 1/2 stop of head room so that is wrong too. Because I'm still only talking about the sensor, not the camera, then is it perhaps valid to somehow remove the headroom from the constant?

Your view of ISO for digital cameras makes sense to me, by the way. Indeed, as regards the "base ISO" if that ISO uses the constant 78 the noise is presumably lower with a constant that allows no headroom at all - the equivalent of "exposing to the right" all the way.

So, if the constant 10 refers to 18% saturation and the constant 78 refers to 100% saturation less 1/2 a stop, the constant for a saturated sensor (full well) be 78 plus 1/2 a stop?

[xpatUSA]

Ted I'm no expert on this subject either but here are a few of my thoughts.

I don't have access to the full ISO standard (12232) but I have read an article in Wikepedia which summarises the thrust of the standard here. This refers to two formulae -

Standard Output Sensitivity S=10/H where H is the value that will lead to 18% saturation

Saturation based Speed S = 78/H where H is the value that gives max saturation.

I would have thought the the second formulae is the one you should be using. This is consistent with the DXO labs article that Willie refers to in his post above.
Dave

Hi Dave,

When researching, I looked at the formula with the constant of 78. ISO says:

"Focal plane measurement
The saturation based speed, Ssat, of an electronic still picture camera is defined as:
Ssat = 78/Hsat ( eqn. 5)
where Hsat is the minimum focal plane exposure, expressed in lux-seconds (lx⋅s), that produces the maximum valid (not clipped or bloomed) camera output signal.

NOTE Equation (5) provides 1/2 “stop" of headroom (41 % additional headroom) for specular highlights above the signal level that would be obtained from a theoretical 100 % reflectance object in the scene, so that a theoretical 141 % reflectance object in the scene would produce a focal plane exposure of Hsat. Therefore, an 18 % reflectance test card in the scene would produce a focal plane exposure of 128/1 000 Hsat. Thus, the multiplicative constant 78 in Equation (5) is equal to 10 times 1 000/128, where the value 10 is the constant from Equation (1)."

I thought that was the one to use until I read the note about headroom which, at first glance, implied to me that 10 was the value without headroom. I also read ISO 2721:1982, where the constant 10 came from but it's description there just went on about stops and stuff. It said:

"Exposure in the focal plane
The nominal exposure H in the focal plane for a film of ISO Speed S (arithmetic) and with a luminance range of from 4 to 4 096 cd/m^2 (corresponding approximatively to exposure values EV = 5 to EV = 15 for ISO 100/21° film) should be:

H = Ho/S
with the constant Ho = 10 lx.s"

Not much help there, except that the constant 10 sprouted units of lx.s making S (i.e. ISO) a dimensionless number for what that's worth!

BUT, from what you're saying, 10 is probably the wrong choice. Unfortunately, the constant 78 includes 1/2 stop of head room so 78 is wrong too. Because I'm still only talking about a saturated sensor, not the camera, then is it perhaps valid to somehow remove the headroom from the constant?

Your view of ISO for digital cameras makes sense to me, by the way. Indeed, as regards the "base ISO" if that base ISO uses the constant 78 the noise would be lower with a constant that allows no headroom at all - the equivalent of "exposing to the right" I reckon.

So, if the constant 10 refers to 18% saturation and the constant 78 refers to 100% saturation less 1/2 a stop, should the constant used for a saturated sensor (full well) be 78 lx.s plus 1/2 a stop, i.e. 110 lx.s?

[Colin Southern]

Perhaps more to the point, does any of this make any "Real World" difference?

Basically, lower the ISO the greater the dynamic range, but the slower the shutterspeed or wider the aperture that will be needed (with the resulting consequences that those may bring).

[dje]

Perhaps more to the point, does any of this make any "Real World" difference?

No it doesn't Colin but you've got to allow us old codgers with too much time on our hands and our grasp of technical issues sliiping gradually away a little bit of latitude to "excercise our minds" !!

Dave