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2:01 AM
so, it seems, at least according to Wikipedia
that gain per stop is 6dB
and that seems to work out correctly regardless of what FWC is
so its power gain not aplitude gain
I've run the numbers for FWC's of 100,000, 60,000, 40,000, and 25,000
they all come out the same
it may be power gain
I don't know the amount of current running through any sensor though
w = v*a
if we knew how many amps for a given sensor, we could compute power gain
and since that is 10 log, it probably is 3dB per stop
the problem with Canon sensors is that damnable bias offset
at lower ISO settings, the bias offset allows read noise to intrude into signal
because of how DXO computes S/N
the ratio between maximum signal and the average of read noise
Canon sensor DR flattens out at low ISO
usually, the first two or three ISO settings have exactly the same dynamic range
the next stop tends to have slightly less, but usually not a full stop less
from that point on dynamic range drops off as a linear function
Canon's problem is how they deal with read noise...which these days, is quite ineffectively.
Their CDS design is ancient, and clearly not up to the task.
Canon has been on a 500nm process for over a decade, where as most other sensor manufacturers started moving to the 180nm process several years ago.
Canon seems to have a 180nm process of their own, they used it in both the 50mp and 120mp APS-H sensors. They also have lightpipe technology on a 180nm Cu process.
They don't seem to have ramped up any kind of FF or APS-C fabrication capacity on that process yet, for whatever reason.
yes, 6db per stop
The design of the 120mp APS-H also seems to indicate some kind of CP-ADC technology
and the press releases stated "on die image processing", which sounds veyr much like Sony's digital noise reduction in Exmor's CP-ADC
I just wish Canon would get on the ball and use their advanced process alrady.
I dont think I ever saw a camera with less than -3db
2:11 AM
that would have to be an analog approach to ISO 50 or something
rather than a change to exposure and digital compensation
which makes a true 50 iso pretty hard to make
yeah, true ISO 50 doesn't exist in consumer DSLR's as far as I know
some MFD sensors are based at ISO 80
rather than ISO 100
yes I think I had one
making the next full stop ISO 160, then 320, 640, etc.
a dslr wannabe fixed lens mirrorless
2:13 AM
electronic pixelated viewfinder :)
flickering too
yeah...not an EVF fan here
even the newer ones
they still suck
Red even makes a high end high resolution EVF for like $3800
it is pretty nice
I had it 6 months then I bought the 400D
but still nothing close to an OVF
I think I found teh source of my wrong info about 3db per stop
2:15 AM
it was a datasheet for an adc that had gain in 3db steps
14bit adc
so I assumed that would be a direct mapping to iso
so, that would mean gain is divided up into 1/2 stop adjustments
so it seesm the adcs have half stop increments
2:16 AM
Canon ADC's can do some gain
but as far as I know, it is only applied at high ISO
wich explains why the 1/3 stops are "tricks"
there are actually amplifiers for each pixel that apply most gain
and 1/3rd stops are only "tricks" with Canon cameras
Nikon cameras seem to have real third stops
without the exposure shifting and compensation (and corresponding 1/3rd stop loss to DR) that Canon has
that is, until high ISO
hmm teh read noise is chaotic on nikon graphs:
in Canon, historically, ISO 1600 was the last setting that used per-pixel gain
canons graphs are more like ski jump hills as expected
2:18 AM
above ISO 1600, ADC gain was used to achieve each additional stop, with a 1/3rd stop push or pull post-ADC
nikon is up and down
in the 1D X and 5D III, I think the last native stop is ISO 12800
with ADC gain above that
Well, sensorgen.info is based on DXO data
which is not all that accurate
a professor at the uni working with vision claim that they all use a mix of analogue and digital gain already from low isos
even Canon read noise fluctuates a bit
Bargain ISO uses per-pixel gain
Numpty ISO uses ADC gain
Ive read that before
2:21 AM
Idiot ISO uses digital "gain"
which is rally just a digital exposure push or pull
thats why I mentian "trick" 1.3 stops
yeah, it's idiotic :)
I wish Canon didn't do that....it is incredibly cheap
and its wierd since machine vision cameras have gain in e.g. 0.0359 db increments
I would assume that is because they are working at a much finer grain than "stops"
even 1/3rd of a stop is a pretty significant difference in exposure
I'd expect a machine that can "see" to have a much finer grained ability to tune sensor "sensitivity"
but it is still based on a cmos and an adc, like hte dslr, so tehres no reason for tham to make them such that they cant control 1/3 stops on the sensor gain
actually my camera that has that 0.0359 value is ccd. could it be a ccd vs cmos thing
2:34 AM
I wouldn't think so
it is a FINER gradation of amplification
and why would it be more expensive to make
it seems more expensive to so the humty dumpty smart switching
which means you should be able to achieve third stops (or close enough) by amplifying in multiples of the base gain
well the "measured iso" is often off by up to 50 anyway
so 0.0359 should be fine enough
approx.55-56 is 1/3 stop
and maybe 0.0359 isnt the physical sensor limitation but thats just the grains of the interger gain parameter set in the firmware
better get to bed at 3:42 am
17 hours later…
7:58 PM
so one of my co-workers wants a good low-light camera that is small, smaller than most DSLRs, even
8:18 PM
@Aaron well a few things play into that equation
hej (Swedish)
sensor size, maximum aperture, and high ISO performance(which is usually tied to sensor size).
and even better yet, we have a question on this site that explains all of that detail:
Q: How do I tell which point-and-shoot cameras take good low light photos?

Vinko VrsalovicI have a Canon Powershot A710 and I previously had a previous Powershot A-series model. I love them due to the high amount of control they offer. Sadly, they both suck in low light conditions, their flash is very unresponsive, and the camera takes several seconds preparing itself for the next sho...

I personally really like the Canon S series of compacts. The S90, s95, S100, S110
They are pretty good in low light
I have mailed Canon that they should make a model in the same price range as 5D with ~2 Mpixel
@Aaron If you want a small DSLR, you could get any entry level DSLR with a 50mm f/1.8 lens.
Focus on dynamic range and noise
8:29 PM
@JohanLarsson hahah, good luck!
8:53 PM
@dpollitt yeah, I know but I still find it strange that they stack all models on each other 18-25 Mpixel
@JohanLarsson why strange? isn't that what all of the manufacturers basically do?
@dpollitt but they already have that segment covered x5
I think there might be market for an uncompromising 2 Mpixel fullformat
I guess a large % of all pictures taken 2012 is meant for web/screen
and besides, it is very rare that a pic is sharp/noise free enough to have 18 effective Mpixels
Hmmm... anyone used Magic Lantern firmware?
@dpollitt thanks for the URL, I was already reading through that one
1 hour later…
10:23 PM
@JohanLarsson Noise free is relative. If you take a photo at 18mp and scale it down to 2mp size, it will have far superior noise characteristics than a native 2mp sensor. You have the benefit of nine-fold pixel averaging.
@MichaelNielsen 3db per stop sounds correct. 10^(0.3) is approximately 2.
A lot of high resolution sensors these days also have increased quantum efficiency. If we take an 8mp camera from years ago and compare it to an 18mp camera of today, the 18mp camera will perform better on all levels thanks to improved manufacturing quality and higher quantum efficiency.
@EvanKrall The math seems to come out to 6dB though...
which math?
6db can show up when you've got quadratic terms
Gain = 10 log (Vout/Vin)^2
Gain = 20 log (Vout/Vin)
either one
the 20 log rule just makes it simpler
I don't know about power gain
I am not even sure exactly how gain is done in a CIS
the only numbers I have handy is charge in electrons, so I used voltage gain
I mean, power gain is 20 log(Vout/Vin) when you have an ohms-law load - when you double voltage, you quadruple power
/me reads wikipedia
10:31 PM
so, in that case, it seems that each stop is about 6dB
at least, that's what my math came out to last night
{| class="infobox" style="padding:0;" border="0" cellpadding="0" cellspacing="0" width="1" ! align="right"|dB ! align="center" colspan="2"|power ratio ! align="center" colspan="2"|amplitude ratio |- | align="right"| 100 | align="right"|  10 000 000 000|| | align="right"| 100 000|| |- | align="right"| 90 | align="right"| 1 000 000 000|| | align="right"| 31 620|| |- | align="right"| 80 | align="right"| 100 000&nbs...
(click the link, that summary didn't turn out well)
yeah, that one got botched good :P
although...I guess technically I'm using eV
rather than V
I wonder if that might change things...
eV is logarithmic
V is linear
does it matter if you use electron volts rather than volts?
sorry not EV
10:33 PM
not EV as in stops, electron volts :P
eV is a unit of energy
I mean, charge in electrons is effectively charge in units of electron volts
at a particular voltage, yes.
I guess eV would actually be joules of energy, though....
yeah eV and joules are directly convertable
if you have 1 electron at 1 volt, you've got 1 eV of energy
if you have 2 electrons at 1 volt, you've got 2 eV of energy
or 1 electron at 2 volts
10:35 PM
you wold really need to know the current as well
to be able to actually figure out how many dB per stop
because from the information generally available, we only seem to have electrons at max saturation for each stop
we would need to know voltage and current to be able to figure anything else out
so, I can't say
it may be 3dB
DXO indicates that the signal in dB is in the 40's at ISO 100 for modern sensors
I mean it all depends on the base unit
is that dB of eV, V, photons...
and as far as I know, the unit is simply "e-" (electrons)
that makes sense to me
10:42 PM
but still, what is that?
I mean, pixels are electron counters
photon strikes "usually" convert to an electron
so I think photodetectors basically work by emitting an electron when they get hit by a photon
some convert to heat or reflect
@EvanKrall correct
so it should be proportional to number of photons hitting the sensor.
or rather, that site
That would only be the case with 100% Q.E. for the photodiode
but not every photon strike, for photons that make it through the filter stack, the microlens, the color filter, and down the wiring channel and actually reach the photodiode in the first place
will actually convert...most do, but some convert to heat or reflect
and there is also the conversion response level
I don't think it is always a 1:1 ratio of photons to electrons
I think it is a slight fractional ratio
so, it might take like 1.137 photons per electron or something around there
I think it depends on the exact nature of the silicon
and maybe applied voltage
but in any case it should be proportional, right?
10:45 PM
roughly, yes
so 40 dB S/N ratio means (or should mean) there are 10^4 times as many electrons due to the signal as there are due to noise
due to read noise
read noise explicitly
there is still photon noise in the signal itself
not much you can do about that.
from sensorgen:
1D X read noise: 38.2e-
1D X ISO 100 signal: 90367e-
10:49 PM
wouldn't the limits have to do with Q.E.?
I mean, lets say we DO have a perfect sensor
what's Q.E.?
0e- read noise
quantum efficiency, which is effectively the rate of successful photon to electron conversions
modern sensors have around 50% Q.E.
the 1D X is 47%, the 5D III is 49%
so let's say we had a Q.E of 100%
and 0e- read noise
the D800 is 56%
well, before we get that far...it seems Canon gains about 1 stop of ISO for every 8% gain in Q.E.
(Based on the progression of the 5D line)
If we project that forward through to 100% Q.E.
100-49 = 51
51/8 = 6.375
Assuming the improvements are linear
we could see about 6 additional stops of usable ISO at Q.E. 100%
an 8% gain in Q.E. doesn't seem to me like it'd give you a stop on its own
10:53 PM
technicaly speaking, we would only need 97% Q.E., but that is also more realistic...we'll never actually achieve 100% anyway
it's probably just correlation
and the gains are mostly read noise I'd suspect
the 5D C had 25% Q.E., the 5D II had 33%, the 5D III has 49%
8% between 5DC and II, 16% between 5D II and III
8% per stop so far
Interesting discussion, I'm not very up-to-date with the physics/electrnics
@EvanKrall Read noise doesn't really apply at high ISO
Since canon uses a bias offset
read noise is the sensor itself, yes?
10:54 PM
they effectively clip read noise at higher ISO
I'm not sure what you mean by that
@EvanKrall read noise is dark current, fixed pattern (intrinsic pixel hot spots fixed in layout for a given sensor), banding (both dynamic and fixed, crosshatch pattern caused by signal interference and some other fixed artifact (usually vertical) that is caused by sensor design), and PRNU (pixel response non uniformity...differences in the response rate of each pixel...very small contributor to read noise)
Then there is downstream amplifier, conversion, and quantization noise added by the ADC as well (in a Canon design...Sony Exmor works in a radically different way)
@EvanKrall Well, according to both DXO and Roger Clark
read noise in a canon sensor falls off to around 2e- at high ISO
I have an article bookmarked at home that explaines the exact mechanics...I'll link it when I get back home
yeah I should really learn about this
10:57 PM
either way, by around ISO 800, read noise is a minimal contributor
so you can effectively ignore read noise at high ISO
I mean in terms of electrons it makes sense that it falls as you raise ISO
it is pretty much all photon shot noise, in a Poisson distribution
@EvanKrall It depends on sensor design. Exmor has flat read noise, at around 3e- for ALL ISO settings. An exmor sensor is effectively ISO less...you could take a shot at ISO 100 and amplify it in post and get pretty much the same result as using a higher ISO. You might exhibit slightly more noise in the shadows by boosting ISO 100 in post, but in general you wouldn't notice.
Exmor has the benefit of CP-ADC and digital noise mitigation, though, so it is tough to compare it with a Canon sensor (or even an older gen Nikon sensor, for that matter.)
if we actually COULD achieve 6 more stops of ISO with 100% Q.E. (ignoring the caveats in doing that for now)...that would be a native ISO 3276800
And that ISO 3 million should look similar to ISO 51200 on the 1D X right now
(Assuming 8%/stop Q.E. is actually enough of a photon conversion improvement to actually hold up through six more stops of ISO)
gotta go be back in an hour
I think noise is proportional to photodetector size
And if that is the case it would be beneficial with large sensor and few pixels
There is a border around each detector right? That border would approach 100% if pixel count went to infinity and it would be as leaving the lens cap on
11:05 PM
Size plays a roll...but size is just a way of improving Q.E.
Larger pixel surface area, greater chance of converting a photon to an electron.
However, assuming the same pixel size, a pixel that converts more incident photons will generally have less noise at all ISO settings than one that converts fewer incident photons.
@JohanLarsson As for pixel structure, in a Front-Side Illuminated CIS design, the photodiode is at the back. In front of that is a channel, and around the sides of that channel (around the sides of the photodiode) is readout wiring.
Do you happen to have a number for how many % of the photons that gets detected
@JohanLarsson A Backside Illuminated CIS design flips the whole structure, directly exposing the photodiode, with all the readout wiring underneath, thus eliminating the problem entirely.
@JohanLarsson generally speaking, for cameras released over the last four years, anywhere from 25% to 56%
the 5D II was 25% Q.E. (quantum efficiency, the conversion rate as a percentage of photons)
the 5D III is 49%
the D800 is 56%
There are limits to how far we can push Q.E. in a bayer design
each pixel has a color filter
the dyes in the CFA eliminate a LOT of light per pixel
56% is actually a pretty amazing feat, and probably required some special high transmittance dye formula for the color filters
I think all manufacturers are weakening their CFA's these days to improve Q.E., but with that comes more "color blindness".
Color blindness is probably not a problem at 50-60% Q.E., as we seem more than capable of mathematically correcting that.
I am not even sure if it is possible to achieve 70-90% Q.E. with a bayer type sensor...the color filters have to filter out a certain amount of light at each pixel to achieve color...so we are going to hit a wall at some point, and it is probably not far off.
The only way to achieve near-100% Q.E. would be with a monochrome sensor
many $100,000 scientific grade supercooled sensors are monochrome, and those are achieving 90% or so Q.E. at temperatures of -80°C
layered sensor designs might be interesting, once they go mainstream
I know Canon has at least one, maybe two patents for a layered CIS design
Q.E. per photosite might be able to push higher than 60% then
however per color it would still be lower
so you would still have per-color channel noise differentials
@jrista You happen to know the number for 20D?
You know much more than I do, I'm wasting your time :)
11:24 PM
I don't know for the 20D
I would guess around 20% or so
maybe as low as 15%
it can't be too low, otherwise the camera simply would not work
but older cameras natively amplified their signal more at base ISO, because they had such low Q.E.
that is why older cameras used to have a minimum ISO of 200
rather than 100
11:54 PM
I think I've heard of an idea where instead of filtering per pixel, you split the light with prisms
directing all of the red light for a pixel to one sensor, all the blue light to another, etc.
if you did it right you might be able to direct all the photons into photo sensors
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