Re: It is ingenious! Was: Does this mean what I think it means?
Since it does not seem to be generally realized quantum foam is the current version of the unified field theory (any one remember ether?). I recently read a SF where the starship was powered by tapping the charge gradient of the quantum foam. Hence the battery comment. No, I do not think the author had any better idea what that means than I do. Anyone who actually knows me would not mention Bud and my name in the same sentence without including the verb hates. Bud and Coors are the two brews I will not even drink when they are free. -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- David Savage wrote: > At 01:12 AM 19/09/2006, mike wilson wrote: > >>> From: graywolf <[EMAIL PROTECTED]> >>> >>> For those of you who insist that quantum mechanics apply at the real >>> world level, please send me a quart of quantum foam, I want to use it as >>> a battery. Reading too much SF lately . >> That would be a couple of bottles of Bud. Battery acid by any name is >> still battery acid. Foamy or not. > > > Battery acid has more flavour > > > Dave > > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
At 01:12 AM 19/09/2006, mike wilson wrote: > > > > From: graywolf <[EMAIL PROTECTED]> > > > > For those of you who insist that quantum mechanics apply at the real > > world level, please send me a quart of quantum foam, I want to use it as > > a battery. Reading too much SF lately . > >That would be a couple of bottles of Bud. Battery acid by any name is >still battery acid. Foamy or not. Battery acid has more flavour Dave -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
You are missing the fact that the holes in this case is not in molecules but in crystals having energy gaps corresponding to certain energy levels, so they can move around a bit and get excited by both phonons (e.g, thermal) and photons. Now, lets all jump into reciprocal space to look at it from another point of view... .-) DagT Den 18. sep. 2006 kl. 20.24 skrev graywolf: > Oh, lets make it more interesting. A hole is actually a molecule with > one or more missing electrons, hence holes, that can be picked up, > usually from an adjacent molecule. That explains all those moving > holes > I had so much trouble understanding back when transistors came out. > And > to finish off solid state electronics is made postible by having > material that will give off, and receive electrons easily. Hence, > again, > those moving holes. > > There, I have explained in one sentence what whole books have had > trouble getting across , and no fair saying my explaination is > over simplified. > > -- > graywolf > http://www.graywolfphoto.com > http://webpages.charter.net/graywolf > "Idiot Proof" <==> "Expert Proof" > --- > > > Digital Image Studio wrote: >> On 19/09/06, graywolf <[EMAIL PROTECTED]> wrote: >>> Thanks, Toralf, I have skimmed that and bookmarked to read more >>> closely. >>> >>> Getting a bit of an Ah-Ha from that skim, apparently folks are >>> talking >>> about electrons and holes as if they are the same thing. They >>> definately >>> are not. A hole is a space that can accept an electron, and if we >>> are >>> talking about 40K or so holes in the photodiode then that makes >>> sense as >>> that would be the amount of additional free electrons that the >>> diode can >>> accept (note that lots of them will bounce around and never find >>> a hole >>> to fit into). Still in the real world sense the output can not be >>> considered stepped but a fairly smooth analog curve. Sometimes we >>> try to >>> apply too much to our understanding of things. >> >> Yes, but I don't know if here is the place to discuss Solid-state >> technology in detail :-) >> > > -- > PDML Pentax-Discuss Mail List > PDML@pdml.net > http://pdml.net/mailman/listinfo/pdml_pdml.net DagT http://dag.foto.no Beware of internet links. You never know what is on the other side. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Oh, lets make it more interesting. A hole is actually a molecule with one or more missing electrons, hence holes, that can be picked up, usually from an adjacent molecule. That explains all those moving holes I had so much trouble understanding back when transistors came out. And to finish off solid state electronics is made postible by having material that will give off, and receive electrons easily. Hence, again, those moving holes. There, I have explained in one sentence what whole books have had trouble getting across , and no fair saying my explaination is over simplified. -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- Digital Image Studio wrote: > On 19/09/06, graywolf <[EMAIL PROTECTED]> wrote: >> Thanks, Toralf, I have skimmed that and bookmarked to read more closely. >> >> Getting a bit of an Ah-Ha from that skim, apparently folks are talking >> about electrons and holes as if they are the same thing. They definately >> are not. A hole is a space that can accept an electron, and if we are >> talking about 40K or so holes in the photodiode then that makes sense as >> that would be the amount of additional free electrons that the diode can >> accept (note that lots of them will bounce around and never find a hole >> to fit into). Still in the real world sense the output can not be >> considered stepped but a fairly smooth analog curve. Sometimes we try to >> apply too much to our understanding of things. > > Yes, but I don't know if here is the place to discuss Solid-state > technology in detail :-) > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Quoting Digital Image Studio <[EMAIL PROTECTED]>: > On 19/09/06, graywolf <[EMAIL PROTECTED]> wrote: >> Thanks, Toralf, I have skimmed that and bookmarked to read more closely. >> >> Getting a bit of an Ah-Ha from that skim, apparently folks are talking >> about electrons and holes as if they are the same thing. They definately >> are not. A hole is a space that can accept an electron, and if we are >> talking about 40K or so holes in the photodiode then that makes sense as >> that would be the amount of additional free electrons that the diode can >> accept (note that lots of them will bounce around and never find a hole >> to fit into). Still in the real world sense the output can not be >> considered stepped but a fairly smooth analog curve. Sometimes we try to >> apply too much to our understanding of things. > > Yes, but I don't know if here is the place to discuss Solid-state > technology in detail :-) What about tube radio's.:-) Dave > > -- > Rob Studdert > HURSTVILLE AUSTRALIA > Tel +61-2-9554-4110 > UTC(GMT) +10 Hours > [EMAIL PROTECTED] > http://home.swiftdsl.com.au/~distudio//publications/ > Pentax user since 1986, PDMLer since 1998 > > -- > PDML Pentax-Discuss Mail List > PDML@pdml.net > http://pdml.net/mailman/listinfo/pdml_pdml.net > Equine Photography in York Region -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
> > From: graywolf <[EMAIL PROTECTED]> > > For those of you who insist that quantum mechanics apply at the real > world level, please send me a quart of quantum foam, I want to use it as > a battery. Reading too much SF lately . That would be a couple of bottles of Bud. Battery acid by any name is still battery acid. Foamy or not. - Email sent from www.ntlworld.com Virus-checked using McAfee(R) Software Visit www.ntlworld.com/security for more information -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
On 19/09/06, graywolf <[EMAIL PROTECTED]> wrote: > Thanks, Toralf, I have skimmed that and bookmarked to read more closely. > > Getting a bit of an Ah-Ha from that skim, apparently folks are talking > about electrons and holes as if they are the same thing. They definately > are not. A hole is a space that can accept an electron, and if we are > talking about 40K or so holes in the photodiode then that makes sense as > that would be the amount of additional free electrons that the diode can > accept (note that lots of them will bounce around and never find a hole > to fit into). Still in the real world sense the output can not be > considered stepped but a fairly smooth analog curve. Sometimes we try to > apply too much to our understanding of things. Yes, but I don't know if here is the place to discuss Solid-state technology in detail :-) -- Rob Studdert HURSTVILLE AUSTRALIA Tel +61-2-9554-4110 UTC(GMT) +10 Hours [EMAIL PROTECTED] http://home.swiftdsl.com.au/~distudio//publications/ Pentax user since 1986, PDMLer since 1998 -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Thanks, Toralf, I have skimmed that and bookmarked to read more closely. Getting a bit of an Ah-Ha from that skim, apparently folks are talking about electrons and holes as if they are the same thing. They definately are not. A hole is a space that can accept an electron, and if we are talking about 40K or so holes in the photodiode then that makes sense as that would be the amount of additional free electrons that the diode can accept (note that lots of them will bounce around and never find a hole to fit into). Still in the real world sense the output can not be considered stepped but a fairly smooth analog curve. Sometimes we try to apply too much to our understanding of things. For those of you who insist that quantum mechanics apply at the real world level, please send me a quart of quantum foam, I want to use it as a battery. Reading too much SF lately . -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- Toralf Lund wrote: >> So you are saying a 22 bit ADC is overkill. Wonder if Pentax knows that? >> > The marketing people probably don't know, or care; they only know they > can quote a higher number than the competitors. > > The engineers probably deliberately chose to have extra bits, so as to > avoid problems with accumulating round-off errors etc. in the processing > stages. >> BTW, Rob's explanation was clearer, but still not documented for my >> curiosity. >> > I'm not sure this can be considered an authoritative reference, but at > least it contains some actual numbers: > > http://micro.magnet.fsu.edu/primer/digitalimaging/concepts/dynamicrange.html > > - Toralf > > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: It is ingenious! Was: Does this mean what I think it means?
Toralf, I think you got it all just right and I believe we were both talking about the same thing in the first place :-). Sometimes it just takes a lot of writing when trying to describe not so easy things. I have to admit I wasn't looking too deep into the DSLR ccd data sheets when trying to figure out why they chose 22 bits. Plus, it is not very easy to get the bare foot data from the current sheets, since the ccd manufacturers seem to be hiding the basic facts to get more of their sensors sold (?). Astronomical sensors generally have quite big pixel sizes (but there are quite small as well, comparable to DSLR). The smaller the wells, the higher the noise (because of the smaller area to capture photons and smaller space to store electrons. Making better signal processing stages is one way (maybe the only one at this time) to improve the results. Pentax seems to be ahead of the competition with the new (to DSLR world) approach when using better designed signal processing. Let us hope there will also be better sensors in the future :-). Actually the only negative thing I have with the K10D is I hoped they would incorporate a full frame sensor to get more sensitivity and lower noise (more exposure latitude). But maybe it would have competed with their coming 645D too much. BTW I did preorder my K10D today! It will be exciting to see the sample photos of production model K10D. I think what we will see is a little different digital pictures (closer to film kind of images). But I have been proven wrong before... Antti-Pekka Antti-Pekka Virjonen Computec Oy R&D Turku www.computec.fi > -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf > Of Toralf Lund > Sent: Monday, September 18, 2006 3:41 PM > To: Pentax-Discuss Mail List > Subject: Re: It is ingenious! Was: Does this mean what I think it > means? > > Antti-Pekka Virjonen wrote: > >> I think the actual sensors do have the 14-16 bits you mention - [ > ... ] > >> > >> > > > > Hi, > > > > Yes, actually I was talking about usable bits, taking the noise into > > account, thus giving 14-16bits of usable data. > > > > This is so at least with the good quality astronomical ccd sensors > but I > > > > am sure the DSLR camera sensors are coming close nowadays (or am I > just > > hoping?). > > > As has also been mentioned many times, the max. charge or "full-well" > capacity of the sensor depends directly on the pixel area. Apparently > it's approximately 1000 electrons per square micrometer in a normal > CCD > > I'm not sure about astronomical sensors, but in the linear CCDs I've > used at work, the pixels are 12x12 or 14x14 micrometers, which should > give you 2-3 times the data required for 16 bits. That may not be > quite > enough to leave you with 16 "good" bits after you've thrown away the > noise, but 14 should be no problem. > > On a DSLR, on the other hand, the pixel size has been something like > 8x8 > microns for a while now, but on the latest (10MP) cameras, it's down > to > 6x6 or so. In other words, the DSLRs cameras are not "coming close"; > they are actually moving further away in that the pixels keep getting > smaller! > > A different side of this is obviously the noise situation. You can > also > improve the number of usable bits by reducing the noise, and this is > an > area where the DLSRs probably are getting better - but they also have > to, in order to compensate for the smaller pixels. And it wouldn't > surprise me if the astronomical sensors have very low noise, too. > > In any case, when designing measuring systems (using amplifiers and > a/d > > converters) it is a good thing to have plenty of more bits and > precision > > compared to the original signal. This way you will minimize the > added > > noise. When having a 22 bits a/d converter on the K10D you can also > > measure and analyze the "noise" of the low expusure pixels. You will > get > > all the available information out of the sensor. > > > Yes. It definitely seems like I good thing to keep *all* the info for > the early image processing stages. Also, having some extra bits around > will minimise round-off errors that might accumulate across multiple > pixel value adjustment steps. You may think of extra bits as more > decimals in intermediate results. I think this may be the most > important > consideration behind the choice of 22 bits. > > - Toralf > > > > > -- > PDML Pentax-Discuss Mail List > PDML@pdml.net > http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
> So you are saying a 22 bit ADC is overkill. Wonder if Pentax knows that? > The marketing people probably don't know, or care; they only know they can quote a higher number than the competitors. The engineers probably deliberately chose to have extra bits, so as to avoid problems with accumulating round-off errors etc. in the processing stages. > BTW, Rob's explanation was clearer, but still not documented for my > curiosity. > I'm not sure this can be considered an authoritative reference, but at least it contains some actual numbers: http://micro.magnet.fsu.edu/primer/digitalimaging/concepts/dynamicrange.html - Toralf -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Antti-Pekka Virjonen wrote: >> I think the actual sensors do have the 14-16 bits you mention - [ ... ] >> >> > > Hi, > > Yes, actually I was talking about usable bits, taking the noise into > account, thus giving 14-16bits of usable data. > > This is so at least with the good quality astronomical ccd sensors but I > > am sure the DSLR camera sensors are coming close nowadays (or am I just > hoping?). > As has also been mentioned many times, the max. charge or "full-well" capacity of the sensor depends directly on the pixel area. Apparently it's approximately 1000 electrons per square micrometer in a normal CCD I'm not sure about astronomical sensors, but in the linear CCDs I've used at work, the pixels are 12x12 or 14x14 micrometers, which should give you 2-3 times the data required for 16 bits. That may not be quite enough to leave you with 16 "good" bits after you've thrown away the noise, but 14 should be no problem. On a DSLR, on the other hand, the pixel size has been something like 8x8 microns for a while now, but on the latest (10MP) cameras, it's down to 6x6 or so. In other words, the DSLRs cameras are not "coming close"; they are actually moving further away in that the pixels keep getting smaller! A different side of this is obviously the noise situation. You can also improve the number of usable bits by reducing the noise, and this is an area where the DLSRs probably are getting better - but they also have to, in order to compensate for the smaller pixels. And it wouldn't surprise me if the astronomical sensors have very low noise, too. > In any case, when designing measuring systems (using amplifiers and a/d > converters) it is a good thing to have plenty of more bits and precision > compared to the original signal. This way you will minimize the added > noise. When having a 22 bits a/d converter on the K10D you can also > measure and analyze the "noise" of the low expusure pixels. You will get > all the available information out of the sensor. > Yes. It definitely seems like I good thing to keep *all* the info for the early image processing stages. Also, having some extra bits around will minimise round-off errors that might accumulate across multiple pixel value adjustment steps. You may think of extra bits as more decimals in intermediate results. I think this may be the most important consideration behind the choice of 22 bits. - Toralf -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: It is ingenious! Was: Does this mean what I think it means?
> I think the actual sensors do have the 14-16 bits you mention - which > literally means that the max charge is up to some 65000 electrons. > However, they also have a readout noise of at least 10 electrons, > which > is equivalent to 3 or 4 bits. This essentially means that you are left > with a usable dynamic range or latitude equivalent to 12 bits; any > additional bits would just be "measuring the noise", if they contained > any information at all. That's why a 12-bit A/D is traditionally used. > > Furthermore, when you amplify the signal for a higher ISO setting, you > also amplify the noise, thus shifting up the 3 or 4 bits so that > contain > noise, so it enters even the 12 bits you keep. > > But like I said, this has been discussed a lot in the past few weeks. > In > in several rounds before that, too. Enough for now. > > I'm not sure I've read the data sheets of the *exact* sensors used in > the Pentax cameras, either, by the way (I've seen technical data for > various ones of similar type...) > > - Toralf Hi, Yes, actually I was talking about usable bits, taking the noise into account, thus giving 14-16bits of usable data. This is so at least with the good quality astronomical ccd sensors but I am sure the DSLR camera sensors are coming close nowadays (or am I just hoping?). In any case, when designing measuring systems (using amplifiers and a/d converters) it is a good thing to have plenty of more bits and precision compared to the original signal. This way you will minimize the added noise. When having a 22 bits a/d converter on the K10D you can also measure and analyze the "noise" of the low expusure pixels. You will get all the available information out of the sensor. Antti-Pekka Antti-Pekka Virjonen Computec Oy R&D Turku www.computec.fi -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: It is ingenious! Was: Does this mean what I think it means?
> -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf > Of graywolf > Sent: Friday, September 15, 2006 5:37 PM > To: Pentax-Discuss Mail List > Subject: Re: It is ingenious! Was: Does this mean what I think it > means? > > Lets back up a bit here. You have a sensor that has an analog out put. > You send that output to an analog to digital converter. The converter > produces an 8, or 10, or 12, or in this case 22 bit digital > representation of the analog signal. Then that digital representation > is > processed digitally. Note that the analog signal does not change. The > only difference is that the higher bit rates produce a more accurate > representation of the analog signal. There is no magic here, just > technology. > > Rephrasing that, the analog signal is simply broken up into more > digital > pieces, nothing is added to it. > > -- > graywolf > http://www.graywolfphoto.com > http://webpages.charter.net/graywolf > "Idiot Proof" <==> "Expert Proof" Hi, I did not say the digital processing would add anything else than noise! If you look closely at the "if" part: Yes, it has to decide the exposure time and f-stop but *IF* the imaging *AND* image processing has a really wide dynamic range you can still choose the correctly exposed part of that range (choose the most fitting 12 bits space from the full 22 bit space for example). With *imaging* I mean the sensor and with *image processing* I mean the analog amplifier stage and the a/d conversion. Sorry for maybe not being very clear (English not my first language). In another e-mail to this thread I already said this was going into a bit too technical (and theoretical). I know there is no magic in the electronics design (I do it at work) but sometimes I sure feel like a magician... At least in theory (and I am sure we will see that partly in practice with the K10D), using a higher bit depth in the a/d stage is a really good thing to do. Antti-Pekka Antti-Pekka Virjonen Computec Oy R&D Turku www.computec.fi -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
On 17/09/06, graywolf <[EMAIL PROTECTED]> wrote: > So you are saying a 22 bit ADC is overkill. Wonder if Pentax knows that? > BTW, Rob's explanation was clearer, but still not documented for my > curiosity. A theoretically perfect 22 bit ADC should be able to resolve 4,194,304 voltage levels between zero and it the voltage represented by all bits at 1. This is clearly overkill given the noise figures of the source device, if the ADC was 18 bits I'm sure you'd be seeing far less discussion regarding the validity of sampling at such high resolution. I'll try to dredge up some decent links on the subject. Personally I'm sure that the mention of 22 bits is true but purely marketing hype. In in the case of the K10D it's great marketing fodder, regardless of the fact that it's overkill or not. The chip that they've likely used is is marketed as a generic imaging system building block which when coupled with an appropriate sensor (and cooling system) may be entirely justifiable. Let's just hope that in the case of the K10D the pre-processing stages following the ADC live up to the hype and deliver a very usable 12 bit RAW image file. -- Rob Studdert HURSTVILLE AUSTRALIA Tel +61-2-9554-4110 UTC(GMT) +10 Hours [EMAIL PROTECTED] http://home.swiftdsl.com.au/~distudio//publications/ Pentax user since 1986, PDMLer since 1998 -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
So you are saying a 22 bit ADC is overkill. Wonder if Pentax knows that? BTW, Rob's explanation was clearer, but still not documented for my curiosity. -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- John Francis wrote: > On Sat, Sep 16, 2006 at 05:45:09PM -0400, graywolf wrote: >> Can you provide a reference on that? A quick goggle search* finds >> nothing the contradicts my explanation. Your comment may be accurate on >> the quantum level but I do not think we can quite apply it to current >> image sensors, but would be interested in seeing something about where >> you are getting that from. Photons, like all quantum particles, are very >> slippery critters and probably act that way, but I do not know of any >> photo diode that gives out a quantized signal --in a macro world sense. > > The point is that you're not in a macro world. The change in voltage level > that would be indicated by the smallest possible change in a 22-bit number > is well below the change in voltage that would be caused by the presence of > just one additional electron in the sensor site (by a couple of orders of > magnitude). You're trying to measure an "analog" quantity to a precision > that puts you firmly in the quantum domain. > > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
On Sat, Sep 16, 2006 at 05:45:09PM -0400, graywolf wrote: > Can you provide a reference on that? A quick goggle search* finds > nothing the contradicts my explanation. Your comment may be accurate on > the quantum level but I do not think we can quite apply it to current > image sensors, but would be interested in seeing something about where > you are getting that from. Photons, like all quantum particles, are very > slippery critters and probably act that way, but I do not know of any > photo diode that gives out a quantized signal --in a macro world sense. The point is that you're not in a macro world. The change in voltage level that would be indicated by the smallest possible change in a 22-bit number is well below the change in voltage that would be caused by the presence of just one additional electron in the sensor site (by a couple of orders of magnitude). You're trying to measure an "analog" quantity to a precision that puts you firmly in the quantum domain. -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
On 17/09/06, graywolf <[EMAIL PROTECTED]> wrote: > Can you provide a reference on that? A quick goggle search* finds > nothing the contradicts my explanation. Your comment may be accurate on > the quantum level but I do not think we can quite apply it to current > image sensors, but would be interested in seeing something about where > you are getting that from. Photons, like all quantum particles, are very > slippery critters and probably act that way, but I do not know of any > photo diode that gives out a quantized signal --in a macro world sense. > I admit my physics knowledge is not as up to date as it could be. But > I do believe folks are trying to digitalize non-digital phenomena, if > the information was digital we would not need an ADC. It's still an analogue signal it's just not continuous, an ADC is still required in order to produce digital representations of the voltage levels. You have to consider that real size of the photo diode component of the sensor. Its width is only tens of wavelengths (considering active sensitivity to about 700nm) across (6.05 x 6.05um) in the case of the sensor in the K10D and it may only be two or three atoms deep, this is why numbers like 40k electrons for saturation are being bandied about (though I suspect that the sensor in the K10D would be closer to about 32k, unfortunately Sony specs aren't direct comparable with those from the majority of other manufacturers). It's a bit mind contorting but the analogue output of the sensor is governed by the electron count so if the devices were noiseless you would see discrete steps in the analogue output. -- Rob Studdert HURSTVILLE AUSTRALIA Tel +61-2-9554-4110 UTC(GMT) +10 Hours [EMAIL PROTECTED] http://home.swiftdsl.com.au/~distudio//publications/ Pentax user since 1986, PDMLer since 1998 -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Can you provide a reference on that? A quick goggle search* finds nothing the contradicts my explanation. Your comment may be accurate on the quantum level but I do not think we can quite apply it to current image sensors, but would be interested in seeing something about where you are getting that from. Photons, like all quantum particles, are very slippery critters and probably act that way, but I do not know of any photo diode that gives out a quantized signal --in a macro world sense. I admit my physics knowledge is not as up to date as it could be. But I do believe folks are trying to digitalize non-digital phenomena, if the information was digital we would not need an ADC. *Unfortunately, the liberal arts university here in Boone does not have much in the way of science texts in its library (singular). Sigh, I do miss having the U of M libraries at my fingertips, so to speak. -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- John Francis wrote: > On Sat, Sep 16, 2006 at 02:47:56PM -0400, graywolf wrote: >> Still trying to turn things around. The sensor is simply a photocell. >> The brighter the light hitting it the higher the voltage output. >> Visualize analog as curves, digital as steps. Don't get them crossed in >> your mind if you want to understand what is going on. > > Unfortunately, Tom, you're incorrect here. The "analog" output > from that sensor is still quantized into discrete steps (either > the number of electrons in the well, or the number of photons > that hit the sensor - take your pick). Both of those numbers > are below 2^17 (around 128000). So you're never going to get > more than that many different values in your digital output. > (Or in your "analog" input curve, either, if you look closely). > > > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
On Sat, Sep 16, 2006 at 02:47:56PM -0400, graywolf wrote: > Still trying to turn things around. The sensor is simply a photocell. > The brighter the light hitting it the higher the voltage output. > Visualize analog as curves, digital as steps. Don't get them crossed in > your mind if you want to understand what is going on. Unfortunately, Tom, you're incorrect here. The "analog" output from that sensor is still quantized into discrete steps (either the number of electrons in the well, or the number of photons that hit the sensor - take your pick). Both of those numbers are below 2^17 (around 128000). So you're never going to get more than that many different values in your digital output. (Or in your "analog" input curve, either, if you look closely). -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Still trying to turn things around. The sensor is simply a photocell. The brighter the light hitting it the higher the voltage output. Visualize analog as curves, digital as steps. Don't get them crossed in your mind if you want to understand what is going on. All the talk about sensors like they are photon counters is pretty counter-productive also, as they definitely are not. It is not like they have a capacity of x photons, but that at a certain point the out put voltage cannot go any higher even if the light is brighter. If you have ever used an old fashioned light meter with a swinging needle, think of a sensor as an super tiny array of them. The ADC converts the voltage into a digital signal with the voltage as a binary number. At this point we are still dealing with the output of a single pixel, the next step is integrating all those signals that is where it gets complicated. -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- Jim King wrote: >> Lets back up a bit here. You have a sensor that has an analog out put. >> You send that output to an analog to digital converter. The converter >> produces an 8, or 10, or 12, or in this case 22 bit digital >> representation of the analog signal. Then that digital >> representation is >> processed digitally. Note that the analog signal does not change. The >> only difference is that the higher bit rates produce a more accurate >> representation of the analog signal. There is no magic here, just >> technology. >> >> Rephrasing that, the analog signal is simply broken up into more >> digital >> pieces, nothing is added to it. > > But the analog signal output is not a smooth curve; it is a stepped > voltage representing the capture of from 0 to about 50,000 > electrons. Sampling at 22 bits will simply give a number of similar > outputs for each step where the voltage is the same. The benefit > will come from averaging these to reduce noise. At least that is how > I envision it... > > Regards, Jim > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
graywolf wrote on Fri, 15 Sep 2006 07:39:11 -0700 > Lets back up a bit here. You have a sensor that has an analog out put. > You send that output to an analog to digital converter. The converter > produces an 8, or 10, or 12, or in this case 22 bit digital > representation of the analog signal. Then that digital > representation is > processed digitally. Note that the analog signal does not change. The > only difference is that the higher bit rates produce a more accurate > representation of the analog signal. There is no magic here, just > technology. > > Rephrasing that, the analog signal is simply broken up into more > digital > pieces, nothing is added to it. But the analog signal output is not a smooth curve; it is a stepped voltage representing the capture of from 0 to about 50,000 electrons. Sampling at 22 bits will simply give a number of similar outputs for each step where the voltage is the same. The benefit will come from averaging these to reduce noise. At least that is how I envision it... Regards, Jim -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
>> It's been mentioned about two point six zillion times by now that you >> probably don't, though. The sensor itself doesn't have a lot more than >> 12-bits worth of latitude, so producing more bits in the A/D doesn't >> help a lot. >> > > I disagree with that a little. It's still the same sensor reveiving the > photons no matter if you set the ISO to 100 or 1600 in a normal > DSLR. When setting the camera to ISO 1600 you just capture a lot > less photons in one pixel well than you would if you set ISO 100 > (having the same amount of light available on both exposures). This > is because you will have less light entering the sensor to get the > correct exposure. Every single photon generates one electon (not > exactly but to about 70% success rate) and there is a lot of > space for those electrons. A good modern CCD sensor does have 14-16 bits > worth of depth (or latitude). Don't know about these DSLR camera > sensors though, maybe they are just 12 bits :(. I think the actual sensors do have the 14-16 bits you mention - which literally means that the max charge is up to some 65000 electrons. However, they also have a readout noise of at least 10 electrons, which is equivalent to 3 or 4 bits. This essentially means that you are left with a usable dynamic range or latitude equivalent to 12 bits; any additional bits would just be "measuring the noise", if they contained any information at all. That's why a 12-bit A/D is traditionally used. Furthermore, when you amplify the signal for a higher ISO setting, you also amplify the noise, thus shifting up the 3 or 4 bits so that contain noise, so it enters even the 12 bits you keep. But like I said, this has been discussed a lot in the past few weeks. In in several rounds before that, too. Enough for now. I'm not sure I've read the data sheets of the *exact* sensors used in the Pentax cameras, either, by the way (I've seen technical data for various ones of similar type...) - Toralf -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Lets back up a bit here. You have a sensor that has an analog out put. You send that output to an analog to digital converter. The converter produces an 8, or 10, or 12, or in this case 22 bit digital representation of the analog signal. Then that digital representation is processed digitally. Note that the analog signal does not change. The only difference is that the higher bit rates produce a more accurate representation of the analog signal. There is no magic here, just technology. Rephrasing that, the analog signal is simply broken up into more digital pieces, nothing is added to it. -- graywolf http://www.graywolfphoto.com http://webpages.charter.net/graywolf "Idiot Proof" <==> "Expert Proof" --- Antti-Pekka Virjonen wrote: >> -Original Message- >> It's been mentioned about two point six zillion times by now that you >> probably don't, though. The sensor itself doesn't have a lot more than >> 12-bits worth of latitude, so producing more bits in the A/D doesn't >> help a lot. > > I disagree with that a little. It's still the same sensor reveiving the > photons no matter if you set the ISO to 100 or 1600 in a normal > DSLR. When setting the camera to ISO 1600 you just capture a lot > less photons in one pixel well than you would if you set ISO 100 > (having the same amount of light available on both exposures). This > is because you will have less light entering the sensor to get the > correct exposure. Every single photon generates one electon (not > exactly but to about 70% success rate) and there is a lot of > space for those electrons. A good modern CCD sensor does have 14-16 bits > worth of depth (or latitude). Don't know about these DSLR camera > sensors though, maybe they are just 12 bits :(. That would be a little > disappointing, heh. (Ok, I need to do some CCD data sheet reading I > guess). > > If you shoot at 1600, the amplifier gain is set high because > less photon generated electrons are available to be read from the CCD > wells to get the output voltage values for all the pixels in the picture > to closely fill the full dynamic range of the following a/d converter. > The overexposed cells or wells (still far from flowing over) overflow > the amplifier (at the set gain) and the a/d reads full scale value for > those "pixels" (overexposure). > > If you set the same camera to lower ISO (like 100). You still use the > same ccd, just the wells of the sensor get a lot more photons and > a lot more electrons are generated. Then you use a lower gain amplifier > to get the output matched as close as possible to the input range of the > > a/d conversion. > > When using the 22 bits a/d you can have a lower gain amplifier and then > choose the suitable part of the full scale input of the a/d converter to > get the (for example) 12 bits of the wanted exposure and latitude (which > > is a small part of the whole dynamic range of the conversion). > At least, this is how I believe it could work in theory. In real world > the CCD well depth vs. noise level (dynamic range) is of course less > than 22 bits but you still can have an optimized fixed gain amplifier. > It is a lot more easier to design a good fixed gain amp than a variable > one. > >> Not amplifying the signal (in a variable manner) probably does help *a >> bit* as one source of noise has been removed (and the 12 bits are the >> latitude after noise has been chopped off), but it's probably not too >> significant compared to the noise in the actual sensor and/or the >> interface to it. >> >> - Toralf > > Of course, the real world "gain" from all this technology is not > world breaking but I think we may well notice an easily visible > difference between Pentax and the current competition. I don't think > either that we can set the post prosessing ISO value from the whole > 100-1600 range, but I am pretty convinced there will be an advantage > over the current models. > > It would be nice if they would explain exactly why they decided to > have 22 bits of a/d. > > This got a bit too technical... I even did some amplifier design today > and looked at bit too deep into the future *grin*. > > Can't help it, being an electronics design engineer... > > Let's just wait and see, > Antti-Pekka > > > Antti-Pekka Virjonen > > Computec Oy > R&D Turku > > www.computec.fi > > > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: It is ingenious! Was: Does this mean what I think it means?
> -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf > Of Hans Imglueck > Sent: Friday, September 15, 2006 9:30 AM > To: Pentax-Discuss Mail List > Subject: Re: It is ingenious! Was: Does this mean what I think it > means? > > Hi, > > but the camera has to choose the exposure time for the shot and > therefore everything is fixed after the exposure. Setting an ISO in > a digital camera is just another word for under- or overexposure. So > I think what you discribe is just not possible - but a nice > imagination! > > Best regards, Hans. Yes, it has to decide the exposure time and f-stop but if the imaging and image processing has a really wide dynamic range you can still choose the correctly exposed part of that range (choose the most fitting 12 bits space from the full 22 bit space for example). I guess we'll have to wait and see, Antti-Pekka Antti-Pekka Virjonen Computec Oy R&D Turku www.computec.fi -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
RE: It is ingenious! Was: Does this mean what I think it means?
> -Original Message- > It's been mentioned about two point six zillion times by now that you > probably don't, though. The sensor itself doesn't have a lot more than > 12-bits worth of latitude, so producing more bits in the A/D doesn't > help a lot. I disagree with that a little. It's still the same sensor reveiving the photons no matter if you set the ISO to 100 or 1600 in a normal DSLR. When setting the camera to ISO 1600 you just capture a lot less photons in one pixel well than you would if you set ISO 100 (having the same amount of light available on both exposures). This is because you will have less light entering the sensor to get the correct exposure. Every single photon generates one electon (not exactly but to about 70% success rate) and there is a lot of space for those electrons. A good modern CCD sensor does have 14-16 bits worth of depth (or latitude). Don't know about these DSLR camera sensors though, maybe they are just 12 bits :(. That would be a little disappointing, heh. (Ok, I need to do some CCD data sheet reading I guess). If you shoot at 1600, the amplifier gain is set high because less photon generated electrons are available to be read from the CCD wells to get the output voltage values for all the pixels in the picture to closely fill the full dynamic range of the following a/d converter. The overexposed cells or wells (still far from flowing over) overflow the amplifier (at the set gain) and the a/d reads full scale value for those "pixels" (overexposure). If you set the same camera to lower ISO (like 100). You still use the same ccd, just the wells of the sensor get a lot more photons and a lot more electrons are generated. Then you use a lower gain amplifier to get the output matched as close as possible to the input range of the a/d conversion. When using the 22 bits a/d you can have a lower gain amplifier and then choose the suitable part of the full scale input of the a/d converter to get the (for example) 12 bits of the wanted exposure and latitude (which is a small part of the whole dynamic range of the conversion). At least, this is how I believe it could work in theory. In real world the CCD well depth vs. noise level (dynamic range) is of course less than 22 bits but you still can have an optimized fixed gain amplifier. It is a lot more easier to design a good fixed gain amp than a variable one. > Not amplifying the signal (in a variable manner) probably does help *a > bit* as one source of noise has been removed (and the 12 bits are the > latitude after noise has been chopped off), but it's probably not too > significant compared to the noise in the actual sensor and/or the > interface to it. > > - Toralf Of course, the real world "gain" from all this technology is not world breaking but I think we may well notice an easily visible difference between Pentax and the current competition. I don't think either that we can set the post prosessing ISO value from the whole 100-1600 range, but I am pretty convinced there will be an advantage over the current models. It would be nice if they would explain exactly why they decided to have 22 bits of a/d. This got a bit too technical... I even did some amplifier design today and looked at bit too deep into the future *grin*. Can't help it, being an electronics design engineer... Let's just wait and see, Antti-Pekka Antti-Pekka Virjonen Computec Oy R&D Turku www.computec.fi -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
> Hi, > > Wanted to add: > Maybe the raw files contain the 22 bit data. If so, you can set the ISO > to anything you wish in the post prosessing (and as many times as you > wish). > May guess is this: The signal is converted into 22 bits without a variable gain stage (like you say). The ISO setting is then applied in the image processor, along with various other (optional) adjustment steps, and the processed data is output as a 12-bit "raw" image. But the unprocessed input data is also saved temporarily, and may be run through the image processor additional times, if you change the processing parameters on the camera. Whether this "fully unprocessed" data can actually be read out, is a different matter. > > 5) With a relatively low gain amplifier and 22 bit A/D (and both having > low > noise) you will get a really wide exposure latitude. The K10D differs > from the competition in this respect. It is like how negative film > differs from > slide film. > It's been mentioned about two point six zillion times by now that you probably don't, though. The sensor itself doesn't have a lot more than 12-bits worth of latitude, so producing more bits in the A/D doesn't help a lot. Not amplifying the signal (in a variable manner) probably does help *a bit* as one source of noise has been removed (and the 12 bits are the latitude after noise has been chopped off), but it's probably not too significant compared to the noise in the actual sensor and/or the interface to it. - Toralf -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
Re: It is ingenious! Was: Does this mean what I think it means?
Hi, but the camera has to choose the exposure time for the shot and therefore everything is fixed after the exposure. Setting an ISO in a digital camera is just another word for under- or overexposure. So I think what you discribe is just not possible - but a nice imagination! Best regards, Hans. Original-Nachricht Datum: Fri, 15 Sep 2006 08:45:01 +0300 Von: "Antti-Pekka Virjonen" <[EMAIL PROTECTED]> An: "Pentax-Discuss Mail List" Betreff: It is ingenious! Was: Does this mean what I think it means? > Hi, > > I was thinking about the 22 bit A/D converter and about the ISO > setting last night. > > Then it struck me, right into the back of my head... :-) > > It is simply ingenious (if this is what Pentax has done): > > 1) By utilizing a 22 bit A/D you get a wide dynamic range > > 2) When you have a wide dynamic range on the A/D converter, you don't > have to > use a high gain amplifier stage (to get high ISO sensitivity) but you > can use > a constant (lower) gain low noise (optimized) amplifier. > > 3) This way you can "choose" the ISO setting after exposure, before > storing > the image into the raw (or converting to jpg). You can analyze the data > and > then decide "how to actually expose". I think this is what Pentax may be > doing > with the "automatic ISO setting". It's like having all the different ISO > > negative films in your camera the same time. > > (Of course you can have a variable gain amp if you wish to get > "standard" > DSLR behaviour...) > > 5) With a relatively low gain amplifier and 22 bit A/D (and both having > low > noise) you will get a really wide exposure latitude. The K10D differs > >from the competition in this respect. It is like how negative film > differs from > slide film. > > In theory you will get a very nice package. We'll see how it will > perform > in reality, but I am sure it will do quite nicely. Pentax wouldn't have > used a 22 bits A/D if there wasn't a good reason. > > Antti-Pekka > > > > Antti-Pekka Virjonen > > Computec Oy > R&D Turku > Fiskarsinkatu 7 D > FIN-20750 Turku Finland > > Puh. +358 20 7908 300 > GSM +358 500 789 753 > Telefax +358 20 7908 319 > > Y-tunnus 1974184-5 > Kotipaikka Helsinki > > www.computec.fi > > > -- > PDML Pentax-Discuss Mail List > PDML@pdml.net > http://pdml.net/mailman/listinfo/pdml_pdml.net -- "Feel free" - 10 GB Mailbox, 100 FreeSMS/Monat ... Jetzt GMX TopMail testen: http://www.gmx.net/de/go/topmail -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net
