Film vs Digital - Dynamic Range

vijay_nebhrajani · December 7, 2008

Bernie: Who cares if it is incidental, accidental, on purpose, or was designed by aliens. The fact is, this IS the way the sensor electronics function, and as the relationship is more-or-less linear, we can infer all sorts of stuff from this.

</blockquote>

We absolutely care. Here is an example of an incidental relationship: Buildings in Manhattan are taller than they are wide. So? This says nothing about buidings in general . In fact, there is no intrinsic relationship between the height of buildings and their width.

So:

<ol>

<li>If all you are saying that there is an incidental relationship, then you are agreeing with me and Ken Rockwell that there is no general or intrinsic relation between dynamic range and bit depth. </li>

<li>If you are saying that there is an intrinsic relationship, then this is immediately disproved by the example of nonlinear ADCs.</li>

</ol>

Which is it?

vijay_nebhrajani · December 8, 2008

Helen: Another question - are all the ADCs used in digital cameras integer-output ADCs, or have ADCs with a floating-point output been implemented in cameras?

</blockquote>

Doesn't matter. A set of bits is a set of bits; it can be converted and interpreted in any manner you like. See the IEEE 754 standard - here's the <a title="IEEE 754" href="http://en.wikipedia.org/wiki/IEEE_754" title="IEEE 754">Wikipedia link</a> . This standard defines how to interpret bits as floating point numbers. Any integer representation can be converted into floating point (which just defines a sign, a mantissa (now significand) and an exponent) and any floating point representation can be converted into an integer representation.

Here's an example: 25 = 0.25 * 10^2; all you need to know is the mantissa (25, and the exponent, 2). In binary, this is represented by 11001 for the mantissa and 10 for the exponent. These are concatenated together, like this: 1100110.

250 is 0.25 * 10^3, so a binary representation is 11001 (for 25) and 11 (for 3) which equals, after concatenation: 1100111.

This is a matter of interpretation of what binary digits stand for. Now for a simpler example, suppose we have an ADC that "measures" voltage - i.e., a 1 volt input reads out 1, a 3 volt input reads out 3 etc., and that we want to make a "voltmeter" out of this that measures upto 230V.

We can use choose the bit depth of our voltmeter ADC to be 8. Then it will read out exactly what the voltage is, in integral whole numbers. It won't be able to handle more than 255 volts; but thats OK since our input voltage will never exceed 230. We can't use a linear 7-bit ADC, because that can only handle a max of 127 volts. But we can use a non-linear 7 bit ADC, and that is an example for another time.

Now, suppose the marketing department comes along and says - you know what, our competitor has a voltmeter that can read to 0.1 volt; so we must too.

As engineers, we say - no problem. We increase the bit depth of our ADC to 12 bits. Now we read out values that are multiplied by 16 for every volt. Thus, for 1 volt we read out 16, for 2 volts we read out 32 and so on.

All we do finally is divide the values by 16, and we get back our original volts, but guess what, we get an extra digit of precision.

Heres how: Previously 13 volts would have read out 13, but so would have 13.5 volts. Now 13.0 volts reads out 13 * 16 or 208, and 13.5 volts reads out 13.5 * 16, which is 216. When we divide this readout by 16, we get 13.0 volts and 13.5 volts respectively, or an extra digit of precision.

Thus, our bit depth increase increased precision, but not dynamic range - we can still measure upto 255 volts, but we can no do so with more precision.

And we happily represented floating point numbers (13.5 volts) with 12-bit integer values, because we knew in advance that the lower four bits were to be used as if they were to the "right" of the "decimal" point (in this case it is a binary point really).

bernardwest · December 8, 2008

Vijay, why won't you just do an example that shows me how what I am saying is wrong? It would be so much less frustrating for both of us.

jamesnjohnson · December 8, 2008

I'm just going to resort to taking photos and if I like what the cameras does cool, if not I move on to the next one.

jamesnjohnson · December 8, 2008

Here is one I took Sunday morning. <img src="http://farm4.static.flickr.com/3271/3089819121_1927c6ee39_b.jpg" alt="Sun" width="1024" height="685" /> It still needs some work. I need to print it out to really geet a feel for it.

It was taking with Fujifilm S5

F16 1/200 ISO 100

alexdi · December 8, 2008

Wow. Hell of a thread, this. Vijay, your knowledge is tremendous, though it seems as if you and a few of the others have been in a bit of a loop for the last couple days. Nonetheless, I though the engineering perspective was fascinating.

I would be weary about quoting too extensively from Wikipedia. The first line you pulled about exposure latitude, I wrote. It's rather oversimplified relative to the extrapolation you've provided in this thread. In fact, you might consider rewriting it with a reference to the length of the linear portion of a tone curve.

vijay_nebhrajani · December 8, 2008

David: Wow. Hell of a thread, this. Vijay, your knowledge is tremendous, though it seems as if you and a few of the others have been in a bit of a loop for the last couple days. Nonetheless, I though the engineering perspective was fascinating.

</blockquote>

Thank you, David. Appreciate the compliment.

Re: Wikipedia - what you wrote is quite correct, but as you said yourself, it could be elaborated upon. Maybe I'll do it someday. But if you wish you could do it yourself - I think it is important to spread knowledge sooner. Of course if you'd rather I did it, I shall; but it'll have to wait for some time.

vijay_nebhrajani · December 8, 2008

James: I'm just going to resort to taking photos and if I like what the cameras does cool, if not I move on to the next one.

</blockquote>

Move on to the next picture or to the next camera? If the latter, you are one lucky guy.

jamesnjohnson · December 8, 2008

Well most likely to the next photo.

I will move on to the next camera but not right away. I arrived at the Fuji S5 after owning D70s/D200 combo. Now I'm a Fuji S2 Pro/ S5 Pro combo. Sold Nikons.

The S5 has what I am looking for in a camera. So there is no need for me to jump on the latest cameras that come out. Unless, unless Fujifilm comes out with a Full Frame SuperCCD with 12mp resolution, 14 stops of capture DR.

mauro_franic · December 8, 2008

James, "Here is one I took Sunday morning."

That is a great shot. Try increasing the contrast only on the land portion.

helenbach · December 8, 2008

Vijay,

I think that in the context of this discussion it matters very much whether the output represents floating point or integer data - and I think that you missed the point again. The thing about a lot of this discussion is whether there is an intrinsic link between bit depth and dynamic range that can be represented by that bit depth. If the ADC algorithm is linear AND the output is integer, then there is a direct relationship between the two. If the ADC algorithm is non-linear and the output is floating point, the dynamic range is not limited by the bit depth. (Of course I realise that binaries used and I know how they can be used - you seem to think that I'm thick. I'm referring to whether they represent integers or floating point numbers.)

Floating point is used in some HDR image formats that use linear luminance data for this very reason. An engineer could choose to use a floating point output to match the range of a high dynamic range sensor with fewer bits than would be necessary if an integer output was used.

Floating point output fits two observations of the results I see in actual image files: halving the luminance halves the (integer rounded) value in the linear Raw output; better shadow resolution than I would expect from linear integer data after a tone curve has been applied (ie tone curve aplied to floating point Raw data before integer rounding). This latter observation is what prompted me to question my original understanding that all cameras used a linear, integer algorithm - the tonal resolution that could be recovered from the shadows at what should have been the lower limit seemed to good to be true. Ultimately I'm only interested in the results I can achieve with the equipment I have, and this is just my curiosity.

Are you happy that the LL page you linked to is not showing Raw data, but data that has been transformed by a tone curve, and thus cannot be used to show what is happening at the ADC, or the extent of the linearity of the response?

Best, Helen

PS You do a little edit of a typo and you lose all the formatting...

vijay_nebhrajani · December 8, 2008

To continue with the voltmeter example - I said something not entirely true: I said "We can't use a linear 7-bit ADC, because that can only handle a max of 127 volts. "

I hould have more properly said, "We can't use a linear 7-bit ADC to measure with the same precision , because it can only handle a max of 127 volts with a precision of 1 volt ."

OK, on to the other side of the example: Say our marketing dept. comes to us and says: "Our voltmeters are really good, but what would really get us some more market share is if we could make a voltmeter that was cheaper, even if it were not as precise. We could live with an error in measurement of +/- 1 volt."

As engineers we say "No problem" and proceed to reduce the bit depth of the ADC to 7 bits and make it read out 1 for any voltage between 0V and 2V; 2 for any voltage between 2V and 4V and so on. If we want to measure 13 volts, the ADC reads out 7. We multiply the ADC readout by 2 and get 14V, which is acceptable, because it is within our specified error margin of +/- 1V.

What did we lose by dropping a bit from the ADC? The ability to precisely measure down to 1 volt; i.e., precision. The dynamic range? Still can do the specified 230V max, i.e., unchanged.

Bernie, this is a general example. Meaning that it covers all cases. Meaning that it also covers the specific case where the application is a camera, where the voltage comes from a photodetector, and the output is fed to image processing algorithms rather than a voltmeter's voltage display. The term 'precision' for imaging simply means number of tones, or tonal range.

I honestly don't have the ability to explain and give examples beyond this - and I have argued these arcane points only so that some future reader isn't misled by false ideas. At this point, I am quite convinced that any future reader can read my arguments and examples and decide for themselves what they want to believe. They can get confirmation of my arguments from engineering textbooks and from any number of other sources. You on the other hand, are a man of unshakeable faith, so I can't really convince you; you too can decide to believe whatever you want.

jamesnjohnson · December 8, 2008

<a style="outline-width: 0px; outline-style: initial; outline-color: initial; font-style: inherit; font-family: inherit; color: #800080; text-decoration: none; padding: 0px; margin: 0px; border: 0px initial initial;" href="http://www.photo.net/photodb/user?user_id=3995956">Mauro Franic</a> , Thanks for the tip. I will continue to work on it.

<a style="outline-width: 0px; outline-style: initial; outline-color: initial; font-style: inherit; font-family: inherit; color: #800080; text-decoration: none; padding: 0px; margin: 0px; border: 0px initial initial;" href="http://www.photo.net/photodb/user?user_id=3995956">Mauro Franic</a> started this thread off with some practical examples that we as photographers could have agreed with or challenged using the best methods i.e. Photographs and technique. Some of us actual took it up and posted photographs to prove the DR range of digital vs film. I thank Mauro Franic for his time and work along with asking folks to contribute.

I agree that the technical aspects are important but gets to a point where all the number throwing becomes pointless. When start talking about what engineers could, should have or what if comments, things start to lose focus, which is the what the final image will look like. Work with what we have and how to make it work.

Mauro reply to my post is a good example from a photographic point of view, technique, a suggestion to improve the contrast of a particular part of the scene which in turn help expand the tone if done right.

Now I know folks will come back at me with the scientific and engineering points of photography is important. If so I say back it up with an image or two explain how it can be improved with scientific scenarios brought up.

vijay_nebhrajani · December 8, 2008

Helen, you are confusing several things, I'm sorry to say.

First, the dynamic range of the numerical range 1 to 255 is fixed for integers: it is 255. That's it. This in no way implies that the dynamic range of the voltage input has to be the same, i.e., 255; this is because each number actually represents a voltage range itself (a quantization step). By making the voltage range represented by each number large or small, I can make two hundred and fiftyfive steps represent a large or small total voltage range.

Dynamic range just has to do with the total voltage range.

255 numbers can represent exactly 255 steps, or tonal values; so once you decide on a bit depth for an ADC, you immediately fix, once and for all - the tonal range (the number of steps of the ADC).

The dynamic range is not affected.

For instance, let us take a photodiode that can capture 10 stops of dynamic range - the max voltage it can generate is 1024 times its minimum voltage. Say its minimum voltage is 1 mV and max is 1.024 V. Lets say I use a linear 4-bit ADC to measure this. I can make the ADC read out 1 for any voltage between 1 mV and 64 mV; 2 for any voltage between 64 mV and 128 mV and so on. You could only read out 16 values, or 16 tones; or said differently the tonal range of the ADC is 16.

You could say "But there are six stops of dynamic range between 1 mV and 64 mV that you represent with only one value so the real dynamic range you are measuring is only 4 stops. 4 stops = 4 bits, so bingo."

You couldn't be more wrong. You lost 6 stops worth of tonal information by compressing it all into one tone. This is a loss of precision, wherein you can't distinguish 1 mV from 64 mV anymore; not a loss of dynamic range - because you can still measure the 1 mV to 64 mV range and represent it as a 1, quite distinct from all the other outputs of the ADC. A loss of dynamic range would occur if any voltage below 64 mV would cause the ADC to readout 0; i.e., if the ADC didn't respond at all.

This is the difference between tonal range and dynamic range. It may be a subtle one, but you do need to appreciate it so that the whole issue isn't convoluted beyond all belief.

For the record, engineers won't put in an ADC that compresses several stops worth of tonal information into one tone; but they might put in an ADC that chops up a stop into more steps; i.e., if an ADC chopped up a stop into 8 parts, the next generation sensor could chop it up into 32 parts. This is done, again - to improve the precision of measurement; not to improve the dynamic range. The dynamic range is fixed by the photodetectors (which are a distinct and different device than ADCs) - if they can't go above or below certain extrema, thats it - no matter what you do you can't change that. That is a physical characteristic of the devices, and you can't change that period.

Let me give you a practical example: most digital cameras have had a dynamic range in the 8-9 stop range, but recent cameras have gone from 12-bit ADCs to 14-bit ADCs, and their dynamic range has remained more or less the same, or even gone down (because pixels have shrunk). What does that tell you vis-a-vis bit depth and dynamic range?

As for the DPreview curves; these are dynamic response curves - the response of the sensor when you change the exposure. I would be very surprised if they applied tone curves to modify this because that would be really poor methodology. Also those curves do look like the response of photodiodes, which do have that kind of S-curve behavior. So I'm not saying that it isn't possible; I'm saying it isn't probable.

And I don't think you are dense at all, sorry if I came across like that. I have often read your posts and mostly have agreed with you. In this specific case though, we are at an arcane enough level that I may be zeroing in on a level of engineering detail that you can't appreciate. I'm trying my best to explain it though. Maybe I'm not that good a communicator; but it is quite rare for me to communicate with nontechnical audiences. Apologies for that.

vijay_nebhrajani · December 8, 2008

James, you are quite correct of course. This level of arcane detail isn't really important to photography. However, there were several misconceptions in this thread (and even generally) about things such as: film has logarithmic response and so on.

How should we have proceeded to correct the misconceptions? Or should we simply have let those misconceptions propagate? Maybe the moderators need to open up a separate technical section for such discussions. But wait, isn't this the "film and processing" forum? I thought this forum was technical, but maybe I'm mistaken.

Anyway, I'm outta here, because I have nothing further to add (this is the second time I've said so, so I really need to go.) Bye guys, have fun; I'll go take some pictures of linear logs.

jamesnjohnson · December 8, 2008

With newer cameras can one shoot 12 or 14bits? I believe some of the Nikons you can. If so this will be the best way to test if more bits gives you more dynamic range. Simple.

helenbach · December 8, 2008

Vijay,

You say that you are being misunderstood. I think that you are consistently misunderstanding what I am trying to say because your replies cover the stuff I already know well, and you say that I am misunderstanding or confusing things that I understand clearly. In the same vein, I have the impression that you are completely missing the point in many cases. As I said earlier, this sort of thing is much easier face-to-face, because such misdirections can be nipped in the bud.

An example. I wrote, carefully: "The thing about a lot of this discussion is whether there is an intrinsic link between bit depth and dynamic range that can be represented by that bit depth." Now that does not require a response that has anything to do with sensor dynamic range, because I specifically restricted it to the dynamic range that can be represented by a particular bit depth. One of the things about the rambling nature of this thread is that too many variables have been in the air at once.

"As for the DPreview curves; these are dynamic response curves - the response of the sensor when you change the exposure. I would be very surprised if they applied tone curves to modify this because that would be really poor methodology."

Look carefully. I think that you will see that they have a tone curve applied. Ask them if you don't believe me. I hope that you don't mind me asking this, but how much experience do you have in handling Raw files and testing digital cameras?

Best, Helen

bernardwest · December 8, 2008

you seem to think that I'm thick

</blockquote>

Helen, don't worry, Vijay treats everyone like they are thick. He's not singling you out :)

Helen, that last post of yours was absolutely spot on the money! I am so sick and tired of being condescended to by Vijay, and having to read through his posts stuffed with fluff that is so simple the punk next door could understand it, and/or is totally irrelevent to the question at hand. I believe this is a tactic that he does to obfuscate from the fact that he might not know an answer to one's question. And your last sentence really hits the nail on the head. From reading what he has said on this subject in this thread, it is clear to me that he has very little experience with linear raw data and digital imaging.

On your point about floating point stuff, it's been a thousand years since I studied that in maths, so I might do some revision and then get into a discussion with it about you. There is clearly a different treatment of data going on when comparing a raw converter with photoshop. If you save a linear raw as a 16 bit tiff and take one copy into photoshop and another copy into say lightroom and severely boost the shadows in both, you will get posterisation very quickly in the photoshop image, but the lightroom image will remain intact. I understood this had something to do with raw converters working in floating point math, but I seem to remember seeing it written somewhere that this isn't the case. So I don't know. Another factor maybe the fact that photoshop isn't really 16 bit, but is actually only 15 bit processing. But I'm not sure what effects that has in reality on an image. I'd be interested to hear your thoughts on this. Cheers.

bernardwest · December 8, 2008

Vijay, you still haven't done a photographic example where you show ALL inputs and outputs. You conveniently forgot to mention the light input to the scene. Exposure is the key component in this argument. When you halve exposure, you halve the output. Absolutely undeniable (and don't forget YOU AGREE WITH THIS STATEMENT. You have stated it above, and you haven't recanted it, so I assume you still believe this), but congniscent of the noise effect, of course. If you continue to believe this (which I recommend you do, as this is what actually happens), then you can't decouple stops from bits. I have showed you numerous examples of this, I have quoted three of the biggest names in the field, and Helen and Rishi (essentially) have agreed with me. You on the other hand have shown one incomplete example, and quoted Ken Rockwell and Wikipedia. If that's what you believe to be a convincing argument, then good luck to you.

robert lee · December 8, 2008

This is amongst the longest thread I've ever come across and perhaps one of the least useful. Unfortunately, half of it seems to have been about ego fluffing and the other half speculations about how many angels can dance on the tip of a pin.

Unless one is participating on the design team of an imager or imaging product, it is entirely misleading to try speaking authoritatively about it. The difference between knowing "about" something, and knowing it is huge.

On the sensor front, a good way to get a feel of how these things work and behave is to go dig up the data sheet and application notes of a real product. For example, go to Dalsa's web site. Look at something like the Dalsa FTF5033C. This is a 36mm x 24mm 17MP color CCD designed specific for still image capture.

For some hands on experience without spending too much money, go Google for machine vision systems. Low end models can be had for around $500. Alternatively, you might try charming an eval kit from a distributor or manufacturer.

On the RAW conversion front, just dig through the source for open source converters like dcraw. No need to speculate - it's all splayed out in the open. A 5 minute glance shows that the internal processing is most likely done in (double precision) float. In fact, I wouldn't be surprised at all if on deeper inspection fixed point is in place only for initial input conversion and TIFF output generation.

rishij · December 8, 2008

Bernie,

Sorry to say this, man, but I can't say I agree with you because I truly no longer have any idea what you're trying to prove.

I already showed you an example of how a non-linear ADC can encode a dynamic range wider than its bit-depth. Vijay confirmed this is possible. You call Vijay out on saying "....A nonlinear ADC can encompass a larger dynamic range with fewer bits," claiming that somehow this indicates that Vijay IS relating ADC bit-depth to DR. But really, Vijay was just backing up my example of the 8-bit ADC encoding a 10-stop DR scene, since an 8-bit linear ADC couldn't encode the 10-stop DR (it would clip, right Vijay?).

You then point out that it would seem that digital cameras DON'T use non-linear ADCs, so the only ADCs relevant to our discussion are linear ADCs.

Fair enough.

But to that I say: do you see any digital cameras where the potential output of the photocells encompass a DR larger than the bit-depth of the included ADCs? I think not.

They include ADCs with bit depths higher than the full range of values that any given photocell can output, thereby NOT making the ADC the limiting factor in the DR represented by the final RAW file.

So, what are you trying to prove? This discussion was over a long time ago. I haven't looked at the floating-point discussion b/c I'm wary of how much more time it'll eat away. But this whole ADC/bit-depth/DR thing... I don't understand why we're still talking about it.

As the DR of photocells get higher and higher, sure we'll need higher and higher bit-depth ADCs to encode that information well . If you start making photocells that can output up to 20,000 different voltages from min to max, I would venture to say that a linear 14-bit ADC would not represent all the information there (some information would be clipped), and a 14-bit non-linear ADC could represent that entire range but not as well as a 16-bit linear ADC.

Vijay correct me if I'm wrong.

Bernie, I still don't understand what you're trying to prove.

Rishi

ryan_drake_stoker · December 8, 2008

Just to clear things up, this is basic maths:

Dynamic range is defined as the ratio of maximum recordable value to the minimum recordable value, obtained by dividing max by min, look it up in the dictionary.

Number of discrete values recorded by a binary number of X bits is 2 to the power of X. Simple scale able to describe every integer in the range has a minimum value of 0 and a maximum value of (2 power X) - 1. Look it up in any basic computer text book.

8bit = 256 values = 0 to 255

255 divided by 0 is infinity.

Largest dynamic range recordable by 8bit binary (or any other number of bits greater than 0) is infinite.

Bit depth and dynamic range have nothing to do with each other.

Dynamic range of an imaging system is defined by the lowest discernible amount of light detected (when black become not black) divided by the highest recordable amount of light (when really bright light gray becomes white). If you want record that digitally in 8bit the bottom value is assigned to 0 and the top 255 irrespective of the ratio between the two.

bernardwest · December 8, 2008

Rishi, thanks for the response. I'm happy to keep nutting this out with you as, like I said, you at least approach these things with an open mind. I'll try and answer some of your quotes directly. Let's go back to one of your previous posts to see what you said:

Just to make sure I understand what you're saying... you claim that:

<ol>

<li>When the bit depth of the ADC is greater than the DR of the recorded scene, then final DR is *not* dependent on the ADC bit depth. </li>

<li>When the bit depth of the ADC is lower than the DR of the recorded scene, the the bit depth of the ADC is DR limiting (i.e. you won't be able to capture more stops than the # of stops the ADC can represent). </li>

</ol>

In which case, I agree with both statements IF the ADC is linear. However, with a non-linear ADC and/or companding, 2. above is not necessarily true.

</blockquote>

Couple of points: You know you are actually relating DR to bit-depth with your choice of words here, don't you? You can't claim on one hand that they are unrelated, but then write statements like "When the bit depth of the ADC is greater than the DR of the recorded scene". Bit-depth and DR are different things. You can't compare them to each other, unless you tacitly agree that there is a relationship between them.

The next point is that I agree 100% with what you wrote, and this has been the crux of my argument the WHOLE time. Where you have probably lost my argument is from trying to follow along with Vijay's lecturing on side topics.

I already showed you an example of how a non-linear ADC can encode a dynamic range wider than its bit-depth.

</blockquote>

Yes, but then I showed you how that is not reflected in the REALITY of digital imaging. Here's what I said:

Dropping one stop (i.e. 1024 -> 512) doesn't lead to a halving of histogram levels (i.e. 256-> 128; instead it leads to a 1/16th drop). As this isn't what we see happening with dslrs, we should be able to confidently say that the ADC doesn't function this way.

</blockquote>

Next point:

You call Vijay out on saying "....A nonlinear ADC can encompass a larger dynamic range with fewer bits," claiming that somehow this indicates that Vijay IS relating ADC bit-depth to DR.

</blockquote>

This is exactly what he was relating, IF you take the quote in context (which you haven't). He was comparing non-linear to linear ADC's. When he talks about "fewer bits" he is referring to fewer bits than a linear ADC. If bits have nothing to do with DR, then you can't "encompass" a "larger" DR with bits. In your and Vijay's proposition, you can encompass ANY dynamic range with ANY number of bits. That implies no link between the two. You guys can't have it both ways.

Vijay was just backing up my example of the 8-bit ADC encoding a 10-stop DR scene, since an 8-bit linear ADC couldn't encode the 10-stop DR (it would clip, right Vijay?).

</blockquote>

Good question Rishi! Let's hear Vijay wiggle his way out of that one. Because Vijay has claimed ad naseum that there is no link between bits and DR. I suspect we'll get some more obfuscation from him.

But to that I say: do you see any digital cameras where the potential output of the photocells encompass a DR larger than the bit-depth of the included ADCs? I think not.

</blockquote>

Rishi, you are implicitly relating bit's to DR here. And again:

They include ADCs with bit depths higher than the full range of values that any given photocell can output, thereby NOT making the ADC the limiting factor in the DR represented by the final RAW file.

</blockquote>

If we put aside the fact that you DO relate bits to DR, I have never claimed that the ADC bit depth is the only factor in DR. In almost all examples I have presented, I have used the case of where the max voltage of the photodiode = the max input votage of the ADC. Where I haven't used this case, I have taken account for it in my examples. And I have also pointed out that it is no different which way you approach it, as the principle is still the same. As long as you account for a particular approach in your results, they are entirely valid (and reflected in reality as well).

But this whole ADC/bit-depth/DR thing... I don't understand why we're still talking about it.

</blockquote>

Because Vijay refuses to accept (like numerous experts, Helen and yourself (even though you're having trouble admitting it)) that dynamic range is related to bit depths in linear ADC's.

</blockquote>

</blockquote>

jlarson6130 · December 8, 2008

You cannot divide a number by Zero.

bernardwest · December 8, 2008

8bit = 256 values = 0 to 255 255 divided by 0 is infinity. Largest dynamic range recordable by 8bit binary (or any other number of bits greater than 0) is infinite.

</blockquote>

Ryan, thanks for chipping in. But there is a problem with what you say. You correctly state that 8 bits represent 256 values. But then you claim that they are respresented by 0-255 therefore DR is infinite. The use of a zero-based index is a computing convention. I could easily refer to them as 2-257, in which case the DR would be about 129. You see the problem with this? 0-255 is only a convention. The REAL values are 1-256. Cheers.

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