What compact digicam has the biggest CCD pixels?

Archived from groups: rec.photo.digital (More info?)

Pixel size = CCD size in square microns, divided by total number of pixels.

Nikon D70: CCD = 15 x 22 mm = 330 million sq. microns, divided by 6 MP =
about 50 square microns per pixel, pretty good.

Random compact point and shoot: CCD = 6 x 8 mm = 48 million sq. microns,
divided by 4 mp = 12 square microns per pixel, much noisier.

What I want is a compact P/S with a good 6x8 mm sensor and about 1 MP
-- do they still make those?

Archived from groups: rec.photo.digital (More info?)

You could get lucky, but I'd say that marketing has dictated that
manufacturers need to cram in as many megapixels as possible in order
to be attractive. Ask any consumer salesperson how to improve image
quality and the answer is inevitably going to be "get the model with
more megapixels." A newly released 1 megapixel compact would
effectively be considered obsolete...

Archived from groups: rec.photo.digital (More info?)

paul <paul@not.net> writes:
> The Olympus C3030 has a "half inch" sensor 4:3 ratio (12.7x9.5mm?)
> sensor with effective 3.14MP. It cost $800 in the year 2000.

Nah, that "half inch" has somewhat less sensor area than that, like
half that amount. "Half inch" refers to the diameter of vidicon tube
from the tube era that would have had a similar amount of active
imaging area. Good point about the old Olympus though.

Archived from groups: rec.photo.digital (More info?)

paul <paul@not.net> writes:
> > Nah, that "half inch" has somewhat less sensor area than that, like
> > half that amount. "Half inch" refers to the diameter of vidicon tube
> > from the tube era that would have had a similar amount of active
> > imaging area. Good point about the old Olympus though.
>
> ehhh that's why I put it in quotes <grin>.
> I don't know, I'd like to see a chart.

All you need to do is ask:

http://www.dpreview.com/learn/?/key=Sensor_Sizes

Archived from groups: rec.photo.digital (More info?)

In article <7x1x9334dj.fsf_-_@ruckus.brouhaha.com>,
Paul Rubin <http://phr.cx@NOSPAM.invalid> wrote:

> Pixel size = CCD size in square microns, divided by total number of pixels.
>
> Nikon D70: CCD = 15 x 22 mm = 330 million sq. microns, divided by 6 MP =
> about 50 square microns per pixel, pretty good.
>
> Random compact point and shoot: CCD = 6 x 8 mm = 48 million sq. microns,
> divided by 4 mp = 12 square microns per pixel, much noisier.
>
> What I want is a compact P/S with a good 6x8 mm sensor and about 1 MP
> -- do they still make those?

Forget about the theoretical math. Sensors aren't that good yet.
Factors like production quality, chip design, manufacturing process, and
image processing still dominate the image quality. Read the reviews,
check out some sample pictures, and then try some out.

Your specifications of a good 1MP camera is contradictory. The lack of
resolution would ruin many photos.

Archived from groups: rec.photo.digital (More info?)

Paul Rubin wrote:

> paul <paul@not.net> writes:
>
>>>Nah, that "half inch" has somewhat less sensor area than that, like
>>>half that amount. "Half inch" refers to the diameter of vidicon tube
>>>from the tube era that would have had a similar amount of active
>>>imaging area. Good point about the old Olympus though.
>>
>>ehhh that's why I put it in quotes <grin>.
>>I don't know, I'd like to see a chart.
>
>
> All you need to do is ask:
>
> http://www.dpreview.com/learn/?/key=Sensor_Sizes


OK so the Oly 1/1.8-inch CCD=
7.15x5.32mm = 38sq mm (38 million sq. microns)
/3.14MP = 12 square microns per pixel.

The examples on that page come out to 6.4 to 7.6sq microns/px so that's
pretty good but not the 60-76sq microns/px of a DSLR.

Archived from groups: rec.photo.digital (More info?)

Paul Rubin wrote:
> Pixel size = CCD size in square microns, divided by total number of
> pixels.
>
> Nikon D70: CCD = 15 x 22 mm = 330 million sq. microns, divided by 6
> MP = about 50 square microns per pixel, pretty good.
>
> Random compact point and shoot: CCD = 6 x 8 mm = 48 million sq.
> microns, divided by 4 mp = 12 square microns per pixel, much noisier.
>
> What I want is a compact P/S with a good 6x8 mm sensor and about 1 MP
> -- do they still make those?

You may be asking the wrong question. Sensor size itself may be more
important than pixel size, as you can downsample to reduce the number of
pixels. Bigger sensors can capture more photons, and produce a better
signal-to-noise ratio. However, as the pixels get smaller, they fraction
of their area which captures light may decrease.

Given all of the above, the 2/3 inch sensor (8.8 x 6.6mm) is perhaps the
largest compact camera sensor, and can be found in the Nikon 5700 (5MP)
and Nikon 8400 and 880 (8MP) cameras.

Cheers,
David

Archived from groups: rec.photo.digital (More info?)

paul <paul@not.net> writes:
> > ....you should be able to get good printable monochrome
> > shots with no flash even in the worst indoor lighting like dimly lit
> > bars. That's the total fix to red-eye ;-).
>
> Downsample an underexposed/corrected 6MP image to 1MP in that lighting
> and it might look good?

I've been wondering about that and I think some improvement might
result, but you don't get the whole factor of six. For example,
almost all consumer video cameras these days have megapixel sensors so
they can double as still cameras. As a result, they have worse low
light performance at normal video resolution (720x480) than the
previous generation of camcorders which only had video resolution
sensors.

Archived from groups: rec.photo.digital (More info?)

David J Taylor wrote:
> Paul Rubin wrote:
>
>>Pixel size = CCD size in square microns, divided by total number of
>>pixels.
>>
>>Nikon D70: CCD = 15 x 22 mm = 330 million sq. microns, divided by 6
>>MP = about 50 square microns per pixel, pretty good.
>>
>>Random compact point and shoot: CCD = 6 x 8 mm = 48 million sq.
>>microns, divided by 4 mp = 12 square microns per pixel, much noisier.
>>
>>What I want is a compact P/S with a good 6x8 mm sensor and about 1 MP
>>-- do they still make those?
>
>
> You may be asking the wrong question. Sensor size itself may be more
> important than pixel size, as you can downsample to reduce the number of
> pixels. Bigger sensors can capture more photons, and produce a better
> signal-to-noise ratio. However, as the pixels get smaller, they fraction
> of their area which captures light may decrease.


I'm not sure. I think it's the pixel size that overloads when too small.


>
> Given all of the above, the 2/3 inch sensor (8.8 x 6.6mm) is perhaps the
> largest compact camera sensor, and can be found in the Nikon 5700 (5MP)


OK I started a chart:
<http://www.edgehill.net/1/Misc/photography/sensors/excel/sensor-size.htm>
That shows the Nikon 5700 2/3 @ 5MP as 11.6sq microns/px
It is an excel chart saved as html & can be opened with excel.


> and Nikon 8400 and 880 (8MP) cameras.


7.3sq microns/px

Archived from groups: rec.photo.digital (More info?)

Kevin McMurtrie <mcmurtri@dslextreme.com> wrote:

[snip]

>CCDs are also prone to overheating.
>45 minute exposures with my Canon 350D have less noise than 8 second
>exposures from older CCD sensors.

You can do 45 minute minute exposures? If so, this means these
cameras are like real film cameras. I have a feeling that next years
bonus and tax refund have just been spoken for.

Heading off to dpreview to read the specs.

Wes

--
Reply to:
Whiskey Echo Sierra Sierra AT Alpha Charlie Echo Golf Romeo Oscar Paul dot Charlie Charlie
Lycos address is a spam trap.

Archived from groups: rec.photo.digital (More info?)

paul wrote:

> Paul Rubin wrote:
>
>> paul <paul@not.net> writes:
>>
>>> The Olympus C3030 has a "half inch" sensor 4:3 ratio (12.7x9.5mm?)
>>> sensor with effective 3.14MP. It cost $800 in the year 2000.
>>
>>
>>
>> Nah, that "half inch" has somewhat less sensor area than that, like
>> half that amount. "Half inch" refers to the diameter of vidicon tube
>> from the tube era that would have had a similar amount of active
>> imaging area. Good point about the old Olympus though.
>
>
>
> So what does it mean when they state a sensor's size as
> "1/1.8-inch CCD" ?

It's a scam to fool you from knowing how small it really
is! See:
http://www.dpreview.com/learn/?/Gl [...] zes_01.htm

Roger

Archived from groups: rec.photo.digital (More info?)

Kevin McMurtrie <mcmurtri@dslextreme.com> writes:

> Modern cameras like the Canon 20D, 300D, and 350D have far less
> noise than lower resolution cameras of not long ago. Like I said
> before, it's more about the sensor quality. The older CCD chips
> grabbed lots of light but they leaked electricity rapidly, bloomed,
> and the quality from pixel to pixel was terrible.

Do you have data to back up this claim? If it was as you say, then
they simply would not work at all. Even the tiniest leakage of loss in
readout sends the signal to zero. If you wonder about that, try
calculating 0.999^2000 and see how many 9s you need to get a
reasonable answer.

> My Canon 350D has less noise at ISO 800 than my Oly C2000Z, C3030Z,
> and C4040Z had at ISO 100. When using the higher pixel count for
> noise filtering, the Canon 350D is cleaner at ISO 1600 than the Olys
> were at ISO 100. CCDs are also prone to overheating. 45 minute
> exposures with my Canon 350D have less noise than 8 second exposures
> from older CCD sensors.

How do CCDs get hot? They have no electronicss on the chip, unlike CMOS
chips, and rare static during exposure.

--
Paul Repacholi 1 Crescent Rd.,
+61 (08) 9257-1001 Kalamunda.
West Australia 6076
comp.os.vms,- The Older, Grumpier Slashdot
Raw, Cooked or Well-done, it's all half baked.
EPIC, The Architecture of the future, always has been, always will be.

Archived from groups: rec.photo.digital (More info?)

Paul Rubin wrote:

> Pixel size = CCD size in square microns, divided by total number of pixels.
>
> Nikon D70: CCD = 15 x 22 mm = 330 million sq. microns, divided by 6 MP =
> about 50 square microns per pixel, pretty good.
>
> Random compact point and shoot: CCD = 6 x 8 mm = 48 million sq. microns,
> divided by 4 mp = 12 square microns per pixel, much noisier.
>
> What I want is a compact P/S with a good 6x8 mm sensor and about 1 MP
> -- do they still make those?


Here are some links:
http://nordicgroup.us/digicam/dslr [...] resolution
-table of DSLR's showing "pixel pitch" of around 8 microns. I don't know
how pitch is calculated, your calc giveas around 50 microns for these
cameras.

http://www.outbackphoto.com/dp_ess [...] essay.html
http://www.outbackphoto.com/dp_ess [...] essay.html
Biased by the author's Foveon connection but otherwise very lucid
discussion of sensor/pixel sizes. Interestingly though, he talks about
Foveon downgrading to average out for better SNR at fewer MP like you
discuss. I'm still not clear this can't be done with a simple
downsampling in photoshop but I don't know.

I think sensor size is one part of the formula but there are large pixel
cameras with bad noise & more pixels really does count for more (within
limits). A table showing actual signal to noise ratio (SNR) would answer
this question. I don't know if SNR is as easily graphed as MTF
(resolution) but still MTF like pixel count is more important than
noise. Dynamic range is the other big factor which probably is closely
tied to pixel size. And the quality of the lens is awfully important
too. That's how MTF is usually used though I suppose the sensor quality
effects MTF. Unfortunately none of these critical factors are spelled
out for the smaller MP cameras. I agree it'd be nice to have a super
quality compact 1 to 3MP but I guess the reality is the smaller cameras
are just not taken seriously in terms of quality control so have poorer
lenses, more noise, etc. and the pixel size is targeted as small as
possible for the day's technology. If it has bigger pixels at the low
end, it's because they can't afford to make better smaller pixels.

Archived from groups: rec.photo.digital (More info?)

On Sat, 23 Apr 2005 21:33:55 -0700, paul wrote:

> -table of DSLR's showing "pixel pitch" of around 8 microns. I don't
> know how pitch is calculated, your calc giveas around 50 microns for
> these cameras.

I didn't look at the link, and didn't examine any formula, but
simple "mind math" showed that if the former was a linear
measurement and the latter was area, they'd be very close, and this
was proved by a calculator. 4 * 4 * pi == 50.265. Of course this
may just be a coincidence.

Archived from groups: rec.photo.digital (More info?)

[A complimentary Cc of this posting was sent to

<prep@prep.synonet.com>], who wrote in article <87r7h1qks7.fsf@prep.synonet.com>:
> readout sends the signal to zero. If you wonder about that, try
> calculating 0.999^2000 and see how many 9s you need to get a
> reasonable answer.

???

perl -wle "print 0.999**2000"
0.135199925397499

Comparing the camera of 2000 and cameras of today are comparing days
and night. The sensors of today are within an order of magnitude of
theoretical limits. If you take large sensors (those used in dSLRs)
then they are within 7x or 8x of theoretical limits.

Cameras of 2000 were a pure junk comparing to what is available today
(at least compacts; I did not have my hands on a dSLR then).

Hope this helps,
Ilya

Archived from groups: rec.photo.digital (More info?)

Ilya Zakharevich <nospam-abuse@ilyaz.org> writes:

>You mean "smaller pixels with the same exposition"; but this is not
>necessarily how cameras are used. If you increase exposition to
>compensate for smaller pixels, you get the same contribution of read
>noise. Example:

> a) 8 microns square pixel; you expose it as ISO1600; you get (e.g.)
> 2500 electrons for white; noise of white is 50 electons; read noise
> is (e.g.) 12 electrons; total noise is 51 electron;

> b) 2 microns square pixel; you expose it as ISO100; you get the same
> 2500 electrons for white; noise of white is 50 electons; read noise
> is (e.g.) 12 electrons; total noise is the same 51 electron.

True enough. But the difference is that ISO 1600 is near the top end of
the useful ISO range for the DSLR. You can choose to expose at ISO 800,
400, 200, and perhaps 100. Each halving of ISO doubles the number of
electrons per pixel, improving the signal to noise by sqrt(2), and
giving more dynamic range to boot.

On the other hand, ISO 100 is at or near the *bottom* of the ISO range
available in a small-sensor camera. You can't reduce it much without
saturating the sensor, since you're already close to the full-well
capacity of the tiny sensor elements. You can select higher ISOs, with
even noisier results.

So your example shows that the read noise of a DSLR at just about its
worst ISO (in terms of image noise) is equal to a small-sensor P&S
camera operating at or near its best.

Dave

Archived from groups: rec.photo.digital (More info?)

Ilya Zakharevich wrote:

> Roger N. Clark (change username to rnclark)
> <username@qwest.net>], who wrote in article <426A927B.1070300@qwest.net>:

> You mean "smaller pixels with the same exposition"; but this is not
> necessarily how cameras are used. If you increase exposition to
> compensate for smaller pixels, you get the same contribution of read
> noise.

Yes, if you want to compare iso 1600 images to iso 100 images,
which is hardly a fair comparison. Compare the same ISO and
the smaller pixels then read noise dominates more and more as the
pixel size drops. And the effect is worse in the darker
portions of the image.

> Of course, cheaper cameras come with lousier sensors (even when the
> performance is rescaled for the size); probably, the read noise is
> worse too. E.g., with the same throughput QE a 2/3'' sensor should
> perform as 8 times less sensitive than half-frame sensor; however, in
> my measurements, 50ISO with 2/3'' sensors is only marginally better
> than 800ISO with haolf-frame (with the same QE it should behave as
> 400ISO).

1) Sensor size is irrelevant. It is pixel size that matters.
2) I've shown that modern P&S cameras, like the S60 are photon
noise limited. It is a fundamental property of scaling
pixel size that no matter what the camera throughput and
quantum efficiency of the sensor, larger pixels will
produce higher signal-to-noise images.
3) Your 8 to 10x numbers are based on flawed calculations.
Improvements can come by 2 methods: improving the
transmission of the optics (lens, blur filter, color filters
over the pixels), and sensor quantum efficiency.
Both of these are unlikely to change unless a radical new
detector comes out, and then it is likely only a factor
of 2 better for small as well as large sensors. Transmission
is already high and is unlikely to change. Fill factors
are already effectively 100% from what I'm reading.
Here is a page listing quantum efficiency:
http://www.britastro.org/vss/ccdtable.html
(see spectral response rows). Here is a page
showing fill factor:
http://www.fillfactory.com/htm/tec [...] h_fill.htm

>>Note too the diffraction spot diameter: for f/2.8 it is
>>3.2 microns for blue (4700 angstroms) light, 4.1 microns for
>>red light (6000 angstroms). Pixels smaller than 4 microns
>>have images that are diffraction limited (many high megapixel
>>P&S have smaller pixels than this)!
>
> What planet are you on? Today's 2/3'' can produce "throughput" 75%
> MTF at 150 lp/mm (with Adobe demosaicer and sharpening mild enough to
> not introduce visual artefacts with line art images [no artefacts at
> least when viewed on CRT; I did not check yet on LCD]). See the
> thread on "MTF of monitors" for an example.

I'm in the real world with real physics. You are in the
world of fake math. Look up the diffraction spot size of
a lens.
Diffraction spot diameter = 2.44 * w * f_ratio,
where w = wavelength, and f_ratio is the f/ratio
of the optical system.

Here is a table from:
http://clarkvision.com/imagedetail [...] ize.matter

================================================
red= Green= Blue=
0.6 0.53 0.47
micron micron micron
================================================
f/ratio diffraction spot diameter in microns
================================================
2 2.9 2.6 2.3
2.8 4.1 3.6 3.2
4 5.9 5.2 4.6
5.6 8.2 7.2 6.4
8 11.7 10.3 9.2
11 16.1 14.2 12.6
16 23.4 20.7 18.3
19 27.8 24.6 21.8
22 32.2 28.5 25.2
32 46.8 41.4 36.7
45 65.9 58.2 51.6
64 93.7 82.8 73.4
================================================

For decent photography where you want some control over depth of
field, you want to get to f/8 (and even that is not much depth
of field in many situations), diffraction strongly influences
image quality, and it should be obvious that diffraction is
less of a problem with a larger pixel.

The diffraction limit for an f/2.8 lens at 5000 angstrom (green)
wavelength is: 585 lines/mm = Rayleigh Resolution Criterion,
which is ~9% MTF. That requires pixel spacing if 585*2/1000
= 1.2 microns. But that is only 9% MTF! Yes, you could
see lines on a test target, but that does not make for a
clean sharp image. The 0% MTF is the Dawes limit, which
for a diffraction limited f/2.8 lens is only 550 lines
per mm for 6500 angstrom light (red). So pixels spacing
less than this will show now detail at all. Yet you say
on your web page that the optimum pixel size is less than
a micron.

Do you know of ANY diffraction limited f/2.8 lens on a P&S
camera (diffraction limited over the whole field of view)?
I do not.

>
> Hope this helps,

No it doesn't. You are providing more confusion.
Here is the problem. You cite a number that you pull from
somewhere without consideration of other factors. For example
deriving optimum pixel size that derives quantum efficiencies
10 times less than manufacturer's specifications, and ignore
diffraction. I have honestly tried to help you pointing
out some of these things. You reject them all and then
declare your erroneous result again. This is becoming
no different than the film versus digital religious wars.

Theory is fine when it includes all components in the real world.
Then there is the real world.....

Roger

Archived from groups: rec.photo.digital (More info?)

"David J Taylor" <david-taylor@blueyonder.co.not-this-bit.nor-this-part.uk> writes:
> This is part of how I see the P&S versus DSLR trade-off. Bigger, more
> inconvenient and heavy package but with the advantages of higher
> sensitivity, lower noise and interchangeable lenses. Decide what you want
> to do, pay your money and take your choice.

There have been millions of relatively inexpensive compact full frame
35mm point-and-shoot cameras made with high quality lenses, sometimes
even zooms. I don't see why compact digicams can't be made with large
sensors. It must just be a cost thing.

Archived from groups: rec.photo.digital (More info?)

On Mon, 25 Apr 2005 16:05:11 +0000 (UTC), davem@cs.ubc.ca (Dave
Martindale) wrote:

>The size of sensors is usually expressed as one over some number in
>inches, for example 1/1.8". This is an odd notation, but what it does
>tell you is that the actual image capture area diagonal is about
>16/1.8 = 8.9 mm. From that and the sensor aspect ratio (usually 1.33,
>but sometimes 1.5) you can calculate the horizontal and vertical size.
>In our example, it's about 7.1 x 5.3 mm.

Almost.

From:
http://www.dpreview.com/learn/?/Gl [...] zes_01.htm

"Sensors are often referred to with a "type" designation using
imperial fractions such as 1/1.8" or 2/3" which are larger than the
actual sensor diameters. The type designation harks back to a set of
standard sizes given to TV camera tubes in the 50's. These sizes were
typically 1/2", 2/3" etc. The size designation does not define the
diagonal of the sensor area but rather the outer diameter of the long
glass envelope of the tube. Engineers soon discovered that for various
reasons the usable area of this imaging plane was approximately two
thirds of the designated size. This designation has clearly stuck
(although it should have been thrown out long ago). There appears to
be no specific mathematical relationship between the diameter of the
imaging circle and the sensor size, although it is always roughly two
thirds."


--
Bill Funk
Change "g" to "a"

Archived from groups: rec.photo.digital (More info?)

On 26 Apr 2005 10:58:12 -0700, Paul Rubin
<http://phr.cx@NOSPAM.invalid> wrote:

>Big Bill <bill@pipping.com> writes:
>> It would seem to me that the lenses used on compact, small-CCD cameras
>> are made specifically to fit eh CCD size used. Many are about 1/3"
>> diagonal. The lenses are smaller, because they only need to cover that
>> small CCD.
>
>Correct.
>
>> Putting a full 35mm size CCD in the same camera body will require a
>> much larger lens (and all incidental hardware). This *will* cost more.
>> There's no way around it.
>
>Well, larger, I don't know about "much larger". The lenses in small
>35mm P/S cameras are not that large.
>
>> Plus, the compact cameras aren't empty; there's 'stuff' already in the
>> nooks and crannies that exist. To make room fro the larger CCD and
>> lens, the 'stuff' would need to go somewhere; if miniaturization is
>> used to reduce the size of the 'stuff', that also increases cost. If
>> not, it increases size.
>> At least that's how I see it.
>
>Take a small P/S digicam like a Canon SD200. That clearly contains
>all the stuff in those nooks and crannies. Now take a small 35mm P/S
>like an Olympus Stylus. That clearly contains a full frame "sensor"
>(film frame) plus a lens capable of covering the sensor. (It also
>contains space for the film cartridge and takeup spool). Now imagine
>a digicam the size of the SD200 and Olympus Stylus put together. The
>result is still a pretty small camera, maybe the size of a Canon A70.
>But it clearly has space for a full frame sensor and the associated
>electronics. Yes it would cost a lot, but I'm not even sure that's
>prohibitive. High end 35mm P/S cameras like the Contax T cost more
>tha low end DSLR's of today, so people are willing to pay that much
>for compact cameras.

I'm not so sure that you can compare the innards of a film camera with
that of a digital camera.
The film camera needs the cannister/takeup reels, but the digital
needs the extra electronics to process the image,and of course, a card
to hold the image files (with some pretty amazing things done to the
smaller cards, so that hey are easier to lose! :-) ).
Well, OK, maybe the same volume.
I still wonder why, if it's so possible for the same price, it's not
being done. I think the above has somewhat to do with it.

--
Bill Funk
Change "g" to "a"

Archived from groups: rec.photo.digital (More info?)

On 26 Apr 2005 13:05:25 -0700, in rec.photo.digital , Paul Rubin
<http://phr.cx@NOSPAM.invalid> in <7xis29nw1m.fsf@ruckus.brouhaha.com>
wrote:

>Big Bill <bill@pipping.com> writes:
>> The film camera needs the cannister/takeup reels, but the digital
>> needs the extra electronics to process the image,and of course, a card
>> to hold the image files (with some pretty amazing things done to the
>> smaller cards, so that hey are easier to lose! :-) ).
>> Well, OK, maybe the same volume.
>
>As explained, a camera the size of a small 35mm AND a small digital
>PUT TOGETHER, should be able to hold everything that's in both of them,
>and still be a fairly small camera.
>
>> I still wonder why, if it's so possible for the same price, it's not
>> being done. I think the above has somewhat to do with it.
>
>It couldn't be done at the same price as today's compact digicams.
>But it could probably be done at the price of today's entry level
>DSLR's. It's an open question whether anyone would pay that much for
>a compact digicam with a large sensor. People still seem to want
>gimmick features like extreme pixel counts.

It is not so much gimmick as explainable with a single number. S/N is
a bit hard to understand, but megapixels/wattage (a la
stereos)/horsepower/etc forms a nice single number. For a camera there
are multiple numbers to track. Not a problem for aficionados, but
difficult for the general public. I suspect that big numbers on the
high end stuff actually helps sell the lower end stuff.

--
Matt Silberstein

All in all, if I could be any animal, I would want to be
a duck or a goose. They can fly, walk, and swim. Plus,
there there is a certain satisfaction knowing that at the
end of your life you will taste good with an orange sauce
or, in the case of a goose, a chestnut stuffing.

Archived from groups: rec.photo.digital (More info?)

>On Mon, 25 Apr 2005 16:05:11 +0000 (UTC), davem@cs.ubc.ca (Dave
>Martindale) wrote:
>
>>The size of sensors is usually expressed as one over some number in
>>inches, for example 1/1.8". This is an odd notation, but what it does
>>tell you is that the actual image capture area diagonal is about
>>16/1.8 = 8.9 mm.

Big Bill <bill@pipping.com> writes:
>Almost.

>"Sensors are often referred to with a "type" designation using
>imperial fractions such as 1/1.8" or 2/3" which are larger than the
>actual sensor diameters. The type designation harks back to a set of
>standard sizes given to TV camera tubes in the 50's. These sizes were
>typically 1/2", 2/3" etc. The size designation does not define the
>diagonal of the sensor area but rather the outer diameter of the long
>glass envelope of the tube. Engineers soon discovered that for various
>reasons the usable area of this imaging plane was approximately two
>thirds of the designated size. This designation has clearly stuck
>(although it should have been thrown out long ago). There appears to
>be no specific mathematical relationship between the diameter of the
>imaging circle and the sensor size, although it is always roughly two
>thirds."

I'm afraid I don't see where your "almost" comes from.

The quote above suggests that you take the "type" and multiply by 2/3
to convert from "nominal tube diameter" to the diagonal of the image
area. 2/3 of one inch (25.4 mm) is 16.9 mm. What I wrote suggested
starting with 16 mm instead of 16.9. There's no disagreement between
the two quotes above, because one says "about" 16 and the other says
"roughly" 16.9 mm (from 2/3 inch).

On the other hand, what I wrote gave an example of how to take the magic
number 1/1.8, plus the aspect ratio of the image and the pixel count of
the sensor, and calculate an actual pixel pitch. The result will be
approximate because the 16 mm value is approximate and "1.8" doesn't
have many significant figures. But it works pretty well in practice.

Dave

Archived from groups: rec.photo.digital (More info?)

[A complimentary Cc of this posting was sent to
Roger N. Clark (change username to rnclark)
<username@qwest.net>], who wrote in article <426CF488.5030905@qwest.net>:
> > You mean "smaller pixels with the same exposition"; but this is not
> > necessarily how cameras are used. If you increase exposition to
> > compensate for smaller pixels, you get the same contribution of read
> > noise.

> Yes, if you want to compare iso 1600 images to iso 100 images,

Almost right. Comparing half-frame to 2/3'' sensors (with same pixel
count), it is 8x smaller area; so with the same throughput QE, it is
comparing ISO800 with ISO100. (Although, as I said, the assumption of
the same throughput QE does not currently hold.)

> which is hardly a fair comparison.

Actually, it may be quite fair. Depends on the situation. E.g., you
need 3 f-stops larger aperture to get the same depth of field with
2/3''; this exactly compensates for the difference in sensibility.

[Of course, this assumes that in both situations the lenses perform
similarly good (with goodness measured w.r.t. diffraction-bound
lens of the same f-stop). This is not always true, the zoom lenses used
with current 2/3'' are only 2 f-stops "better in optical quality"
than fixed-focal-length lenses used with film; do not know about
new lenses designed for half-frames.]

In addition, the smaller sensor is easier to motion-compensate; this
may also improve "practical sensitivity" of this sensor.

> 1) Sensor size is irrelevant. It is pixel size that matters.

While correct, this is irrelevant, since current "leaders" have the
same pixel count.

> 2) I've shown that modern P&S cameras, like the S60 are photon
> noise limited.

a) You mean electron noise here, not photon noise; but let keep this
discussion in one thread, not in many;

b) This is very strange data for S60: with 8 times smaller sensor
you quote 2.5 times smaller full well. Are you sure the data is
for 100ISO?

> It is a fundamental property of scaling
> pixel size that no matter what the camera throughput and
> quantum efficiency of the sensor, larger pixels will
> produce higher signal-to-noise images.

This is manifestly wrong. What matters is product of throughput QE
and the area.

Or do you mean "with the same technology for both sensors"? Then of
course it is true. So with (imminent) 6x improvement in throughput
you can get 10000ISO from large sensors vs 1200IS for small ones. The
people who *need* 10000ISO will have a clear choice. But the people
who do not need more than 1000ISO (with practically no noise) will
*also* have a clear choice.

> 3) Your 8 to 10x numbers are based on flawed calculations.

I think 8x is theoretical, with practical about 6x; but let us keep it
in a separate thread.

> >>Note too the diffraction spot diameter: for f/2.8 it is
> >>3.2 microns for blue (4700 angstroms) light, 4.1 microns for
> >>red light (6000 angstroms). Pixels smaller than 4 microns
> >>have images that are diffraction limited (many high megapixel
> >>P&S have smaller pixels than this)!
> >
> > What planet are you on? Today's 2/3'' can produce "throughput" 75%
> > MTF at 150 lp/mm (with Adobe demosaicer and sharpening mild enough to
> > not introduce visual artefacts with line art images [no artefacts at
> > least when viewed on CRT; I did not check yet on LCD]). See the
> > thread on "MTF of monitors" for an example.
>
> I'm in the real world with real physics. You are in the
> world of fake math. Look up the diffraction spot size of
> a lens.

I see that it is you who are in the world of fake math. Take a test
shot, and see by yourselves. Or look at the example image I provided
(actually, it is not the best possible; when I have time, I will
upload the slightly improved one).

BTW, diffraction spot size is absolutely irrelevant in the age of DSP.
The only important thing is the throughput MTF curve vs throughput
narrow-band noise; it shows what CAN'T be extracted from the image
with DSP.

Hope this helps,
Ilya

Archived from groups: rec.photo.digital (More info?)

HvdV wrote:

> Hi Roger & Ilya,
>
>>
>> Diffraction spot diameter = 2.44 * w * f_ratio,
>> where w = wavelength, and f_ratio is the f/ratio
>> of the optical system.
>
> <snip useful table>
>
>> For decent photography where you want some control over depth of
>> field, you want to get to f/8 (and even that is not much depth
>> of field in many situations), diffraction strongly influences
>> image quality, and it should be obvious that diffraction is
>> less of a problem with a larger pixel.
>>
>> The diffraction limit for an f/2.8 lens at 5000 angstrom (green)
>> wavelength is: 585 lines/mm = Rayleigh Resolution Criterion,
>> which is ~9% MTF. That requires pixel spacing if 585*2/1000
>
> On the danger of being pedantic:
> According to my text book on optics the original Rayleigh criterion is
> about
> two-line resolution in spectroscopes: the lines can be said to be 'just
> resolved' when the maximum of the image of one coincides with the
> minimum of
> the other.
> This was extended to two-point resolution by coinciding the maximum of one
> with the minimum of the Airy disk diffraction pattern of the other:
> R = 0.61 lambda/sin(half_aperture) (for aperture << 1, circular aperture,
> incoherent light)
> For small apertures sin(half_aperture) ~ 1/(2f_number) so:
> R = 0.61 * lambda * 2 * f_number
> To detect the minimum between the maxima you need two pixels per Rayleigh
> distance so the 'Rayleigh pixel size' is 2.44 * lambda * f_number.
> So this rediscovers Rayleigh's work from 1879!

HvdV,
Concerning both above and below, it seems your equation differs by a
factor of 2 from that in "Star Testing Astronomical Telescopes,
A Manual for Optical Evaluation and Adjustment" by HR Suiter,
William Bell, Richmond, 1994, page 49, equation 3.5:

S'max = 1/(F*w)

where F = focal ratio, w = wavelength, and S'max is the critical
sampling (0% MTF) in cycles (line pairs) per length (given by the units
of the wavelength). For 0.47 micron wavelength, F=2.8 the
equation gives S'max = 1/(2.8*0.00047 mm) = 760 line pairs/mm.
The Rayleigh limit is at about 82% of the critical limit,
which for 0.47 microns wavelength would be about 620 lp/mm.

In a diffraction limited system, the Rayleigh resolution
criteria occurs when the separation of two objects is precisely
the radius of the theoretical diffraction disk (Suiter, 1994,
page 286). The radius is defined from the maximum to the
first minimum of the diffraction disk. This separation
causes a minimum in the response at the radius, so the
radius is the ideal sampling distance.

The diffraction disk radius, R_airy, is (Suiter, 1994, page 11,
equation 1.1):

R_airy = 1.22*w*f/D

where w = wavelength, f = focal length, and D = aperture
diameter. This reduces to:

R_airy = 1.22*w*F (where F = focal ratio).

For 0.47 micron wavelength, F=2.8 R_airy = 1.22*.47*2.8 = 1.6 microns.

This is the Nyquist sampling interval, meaning one sample at the
peak, one sample at the minimum.

I haven't gone through your equations in detail, but I
suspect you used an equation assuming a diameter and divided
by 2 getting the factor of two smaller sampling, when the
equation was already for radius.

I believe this is correct. It matches examples given in Suiter.
Equation 1.1 is quite well known and far predates Suiter, 1994.
I'm just quoting Suiter as it is an excellent book that brings
a lot of information together into one place.

Roger

>
>> = 1.2 microns. But that is only 9% MTF! Yes, you could
>> see lines on a test target, but that does not make for a
>> clean sharp image. The 0% MTF is the Dawes limit, which
>> for a diffraction limited f/2.8 lens is only 550 lines
>> per mm for 6500 angstrom light (red). So pixels spacing
>> less than this will show now detail at all. Yet you say
>> on your web page that the optimum pixel size is less than
>> a micron.
>
> However, if we take the blue 470 nm light then the critical sampling
> rate is:
> delta_x = lambda/(4 * sin(half_aperture) ~ lambda * f_number/2
> so for 470nm the sampling distance is ~.66 micron. To capture all
> information
> passed by the lens this is the pixel spacing you need. It could very
> well be
> that readout noise problems and the like make it not feasible, but that is
> something else.
>
>>
>> Do you know of ANY diffraction limited f/2.8 lens on a P&S
>> camera (diffraction limited over the whole field of view)?
>
> I get the impression spherical aberration is so strong in these P&S lenses
> that all rays above say f/8 are 'lost'. With that the whole computation
> above
> goes to pieces. Add chromatic aberration and you end up with a 640x480
> image
> at best.
> Quite likely such lenses can be made though, especially for smaller focal
> lengths, and using the space where the mirror is in a (d)SLR. But high
> quality camera makers have a hard time at the moment.
>
> -- Hans

Archived from groups: rec.photo.digital (More info?)

Hi Ilya,
>
>
> I see that it is you who are in the world of fake math. Take a test
Ah, but it is a nice world! Unless you mean the math itself is fake ;-)
>
> BTW, diffraction spot size is absolutely irrelevant in the age of DSP.
> The only important thing is the throughput MTF curve vs throughput
> narrow-band noise; it shows what CAN'T be extracted from the image
> with DSP.
Well, you could say the PSF shape is equivalent to the MTF. Of course the
actual PSF hardly ever looks like a diffraction limited PSF, diffraction
limited in the sense of a aberration-free system. The PSF shape and the noise
statistics together are the main ingredients to the image formation, ok.

-- Hans

Archived from groups: rec.photo.digital (More info?)

HvdV wrote:

> When the two Airy patterns overlap in 'peak on the first dark ring of
> the other' fashion, a local minimum in between the peaks 26.5% lower
> than the peak results. To sample this configuration you need a (point)
> sample at the minimum, so the pseudo Nyquist interval is R_airy/2. In
> this case .8 microns.
> Incidently, the figure of the 26.5% dip is also sometimes referred to as
> 'Rayleigh criterion'. In practice the value is greatly affected by the
> degree of coherency, not mentioning aberrations.
>
>> This is the Nyquist sampling interval, meaning one sample at the
>> peak, one sample at the minimum.
>
> Yes, the minimum between the two maxima.

HvDv,
This is where we disagree. The minimum occurs at R_airy, not R_airy/2.
The distance between samples is R_airy. You have double Nyquist sampling.
Nyquist sampling occurs at the maxima and minima.

> The fact that this value is different than the Rayleigh value above is
> no surprize since the idea behind the Rayleigh criterion is, though
> practical, arbitrary. If the Rayleigh criterion would have demanded a 1%
> dip in an n-line pattern the values would have been closer. The value of
> a bandwidth related criterion is, IMO, that it is independent of
> aberrations and the exact shape of the PSF.

Yes, I agree completely. The shape of the MTF is a major difference
between film and digital cameras and the major factor in the film versus
digital wars.

Roger