Digital BW, The Print

--- In DigitalBlackandWhiteThePrint@yahoogroups.com, Jeff Medkeff <medkeff@g...> 
wrote:
> Roy Harrington wrote:
> 
> 
> 
> > Basically you increase the sensor dynamic range by either: reducing the noise
> > at the low end and/or increasing the clip point at the high end.  The bits of
> > the A/D have to fine enough to take advantage of the reduced noise at the
> > low end and coarse enough to not get clipped at the high end.
> 
> It sounds to me as though we are approaching this from two different 
> perspectives. When you say the bits have to be coarse enough to take 
> advantage of reduced noise, and fine enough not to clip, it sounds to me 
> as though you are saying something about the digital sampling of the 
> signal being read off the sensor. And if I understand correctly you are 
> dead right - there is no point to making the luminous flux density delta 
> per ADU so coarse that the 100 (or whatever) highest ADU values all clip 
> to white. Not only do you throw away values, but you also increase the 
> chances of posterization in the resulting digital image.
> 
> You state correctly, above, that to increase sensor dynamic range, you 
> can reduce noise. You also say you can increase the clip point at the 
> high end. I'm familiar with using the term "clip" in reference to 
> digital sampling, so it appears as though what you are saying is that to 
> increase dynamic range you should take your highest ADU - 4096, say - 
> and calibrate that to what is detected from a higher luminous intensity 
> (luminous flux per unit solid angle) from a scene at a given EV. Maybe 
> that is what you are suggesting, or maybe not - that's what I'm hearing.
> 
> Either way: Doing that won't work.
> 
> Well, I admit it will work in one special case: It will work when your 
> sensor's highest ADU value represents a number of electrons that is less 
> than the full well potential of the sensor. In your terms above, if the 
> "bits of the A/D" are a bit beyond "fine enough," and are in fact too 
> fine, then your digital image won't have the dynamic range it could have 
> had if there were more bits available at the same sampling rate. By 
> sampling rate, I mean electrons per ADU.
> 
> But if your highest ADU value already represents a number of electrons 
> that is equal to the full well potential of the sensor, then adding bits 
> at the *same* sampling rate buys you nothing. This would be too "coarse" 
> sampling, as you put it, and all those added ADU values would be clipped.
> 
> Increasing the sampling rate (reducing electrons per ADU) while 
> increasing bits *can* solve this clipping problem - and I'd like to have 
> this in all my cameras. Unfortunately it will not result in greater 
> dynamic range. The reason is that the maximum luminous intensity and the 
> minimum noise recorded by the sensor are both limited at the analog stage.
> 
> The analog stage is mostly what I've been talking about. We are always, 
> always, always limited by the photosite well. A well can hold between 
> zero and n electrons. If you exceed n, and try to stuff n+x electrons 
> into the well, an average of x electrons will be lost to other parts of 
> the sensor. This is because the well achieves enough voltage to jump the 
> resistant gap between it and another capacitor or some convenient path 
> to ground. In either case, these electrons have overflowed their bucket, 
> and are no longer present in the well when the sensor is read out.
> 
> In this situation the well is said to be "saturated." This sometimes 
> results in blooming, but it does not result in clipping as I understand 
> the term, because this has happened before any sampling of the well is 
> made; there is no chance for the amplifier output to distort or the ADC 
> to clip as these electrons never get there.
> 
> So a well can hold between 0 and n electrons. For a big well, n is a 
> larger number; for a small well, n is a smaller number. Quantum 
> efficiency determines how many electrons end up in the well, as a 
> percentage of quanta of photons that are incident upon the photoelectric 
> surface.
> 
> If the sensor QE is 50%, and 1k photons strike the sensor, then 500 
> electrons (on average) are dumped into the photosite well. The number of 
> photons striking the sensor is determined by the luminous intensity of 
> the scene. So if we double the EV, we will have 2k photons striking the 
> sensor and 1k electrons in the well. If we double it again, we'll have 
> 4k photons and 2k electrons.
> 
> Suppose the well capacity is 1k electrons. We've tried to stuff twice 
> that many into the well. Half those electrons will find some other way 
> out of the well, than through the readout amplifiers. Does adding bits 
> to the sampling do anything to recover the lost 1k electrons?
> 
> The correct answer to this question is "no." The analog limitation of 
> well depth is fundamental and exists because those lost electrons are 
> not amplified and never affect the signal that the ADCs see.
> 
> And zero is a possible number of electrons in the well. This does not 
> lend itself to useful representation with a proportion, which is 
> probably one reason why dynamic range is specified on sensor spec sheets 
> with luminous flux density and electrons of noise, rather than a ratio.
> 
> To increase the dynamic range of the sensor, we have to either reduce 
> noise, or increase the saturation point of the well. The latter means 
> increasing the number of electrons it can hold, because that correlates 
> linearly with the maximum brightness in the scene that the sensor can 
> usefully record.
> 
> What I originally started this discussion with, was a statement that 
> photographers should beware the abuse of bit depth specifications. It 
> does not always tell you anything about the ability to record the 
> dynamic range of a scene, because no matter how many bits you have, the 
> highest n ADUs could all be representing saturated photosite wells that 
> contain no useful information about the scene you've just photographed. 
> I think bit rate is going to be the next great digital camera scam, once 
> everybody gets over the megapixel fetish - it already appears to be 
> happening in the digital back realm.
> 
> --
> Jeff Medkeff
> Eagle River, Alaska

I think we are pretty much in agreement.   My coarse and fine terminology was
probably a little vague.

Like you said the analog well is the main determinant of dynamic range.  The
dynamic range is pretty simple to state at that level.  When I said ratio of the
max signal / min signal, max is just how many electrons fill the well i.e.
the saturation point or what I called the clipping point.  Any more signal (more
light or more electons) don't register in the well.  The min is not just the 
minimum number of electrons in the well.  As you noted 0 doesn't work well in
the denominator.  Min is the minimum number of electrons that you can read as
being the result of light hitting rather than just noise.  In other works it's the minimum
amount of signal that is distinguishable above the noise level.

So for example, if a well can contain 100,000 electrons when it's saturated and the
smallest number of electrons that indicates a signal is 100 electrons then the 
dynamic range = 100,000/100 or 1000-to-1.    If you had a larger well that could
contain 200,000 electrons but the noise level made it such that 200 electrons was
the smallest signal, both wells would have the same dynamic range.

On the digital side, matching the A/D to the first well would mean a 10-bit A/D
such that 0 matched 0 electrons, 1 matched 100 electrons, ... 1000 matched 100,000
electrons.  What I meant by "fine" enough is that you don't want the A/D to first
register a 1 with 200 electrons because you'd miss the 100 electron signal. On the other 
end it must be "coarse" enough not to max out the A/D value 1023 at 50,000 electrons
because you'd miss all the larger signals.  If you had less bits in the A/D you'd be
missing some signals and losing dyn.range and if you had more bits they wouldn't 
buy you any more dyn.range because the analog stage doesn't have any more to offer.  
The bit depth of 10 bits is sufficient and necessary to store the 1000-to-1 dyn.range.

The other aspect that I think tends to be mis-interpreted is that 100 electron
minimum distinguishable signal between 0 and 1 does NOT imply that you can
distinguish 100 electrons throughout the whole scale.  Although there are A/D
possible values 1000, 999, 998, ...  the noise level is probably a lot higher in
the 100,000 to 99,900 electron range.  If noise was 1% then the best you could
distinguish would be 1000,  990,  980, ...   So the fact that 10 bits = 1024 values
and can represent 1000-to-1 DR does not mean there are 1000 distinguishable grays.

Roy