Archived posting to the Leica Users Group, 1999/10/14
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]From: Alexey Merz <alexey@webcom.com> Sent: Thursday, October 14, 1999 05:47 Subject: Re: [Leica] Re: do your homework, please! > Nope. The electon signal accumulated in the CCD potential > well has to be amplified prior to "counting" in the A/D > converter. No matter how much you amplify it, you still get discrete levels. Every additional electronic adds one indivisible unit charge. > Digital implies the lack of a physical model as > you've said. Yes. > Pixel intensity is NOT transferred as a straight > count of electrons but as an analog signal (roughly) > proportional to the electron count, to the A/D > converter. That's because it is an analog signal, not digital data. > Thus a CCD camera is NOT a purely digital device. No camera is a purely digital device. The endpoints of all image-processing systems are always analog. > Again you use the term "best" in a nonsensical fashion. > The "best" charge ratio is the one most suited to the > specific application at hand. The more resolution (depth of modulation) you want, the higher the ratio has to be. > The relationship between photons entering a well > and electrons stored in that well is relatively > linear (one of the best characteristics of CCDs) ... > ... but that relationship does NOT correspond to > "F-stops" because the SLOPE of the line defining > this relationship varies depending on the CCD and > how it is operated. I wasn't talking about the rest of the chain. For that matter, I wasn't bringing noise or other implementation details into consideration. -- Anthony