A simple way to put Canon's sensors on top.

[Mikael Risedal]

1.Canon need to invest in new sensor lines to keep up regarding modern sensor design and resolution

2. Canon need to shorten the analog signal chain as Sony/Panasonic/Toshiba/Aptina etc and have for example raw vise ADC to lower the read out noise

3.Today Canon sensor tech is old, based on solutions from 2004 and modified during the years but with still high read out noise at base iso

--
Member of Swedish Photographers Association since 1984
Canon, Hasselblad, Leica,Nikon, Linhoff, Sinar
Ernest Hemingway to Irving Penn:?“Your photos are really good. What camera do you use?”?Irving Penn to Ernest Hemingway:?“Your novels are excellent. What typewriter do you use?”

 

[Rick Knepper]

At higher ISO settings, Canon sensors are as good as the best of them. So, if Canon were to offer an optional ISOless interface, increase the bit depth of the capture files to 20 bits, and have the cameras shoot permanently at ISO 3200 in the ISOless shooting mode, they'd match or beat Sony's sensors in terms of noise and DR.

By making the ISOless UI optional, it would still allow those who find the current noise and DR levels to be "good enough" and prefer setting the ISO themselves to continue as before without any bother.

 

[rrccad]

1.Canon need to invest in new sensor lines to keep up regarding modern sensor design and resolution

safe to say that the 40M pixel APS-C sensor they've already have out in production wasn't using the 500nm line. They already had a 180nm sensor line in production. even chipworks acknowledged that canon had 180nm fabs.

2. Canon need to shorten the analog signal chain as Sony/Panasonic/Toshiba/Aptina etc and have for example raw vise ADC to lower the read out noise

patented tech though. canon did however come up with a column parallel ADC method and patented it last year i think.

3.Today Canon sensor tech is old, based on solutions from 2004 and modified during the years but with still high read out noise at base iso

see above.

 

[Press Correspondent]

You do fill the wells at any ISO. The amplification is done by applying a higher voltage to the sensor for each photon to produce more than one electron and this fills the wells, because the well capacity limit is in electrons, not in photons. The well capacity in electrons is actually the same at any ISO, but appears reduced at higher ISO when normalized to the number of photons by dividing the output by the amplification ratio.

This also explains why fractional ISO values are digital fakes: e.g. 1 photon can produce 1 or 2 electrons, depending on the voltage (ISO), but not 1.5 electrons.

I think you must study the subject how a sensor works and collecting photons due the time and f-stop

FWC or almost full well capacity do you have only at base iso

Is this so Mikael? Then please educate us. This forum is about learning photography. So please, do tell where my explanation was wrong and explain us how this actually works or, since you struggle with English so much, provide web links that clearly explain your point. Otherwise your unsubstantiated objection is worth nothing and is not true only because you said so.

 

[Press Correspondent]

At higher ISO settings, Canon sensors are as good as the best of them. So, if Canon were to offer an optional ISOless interface, increase the bit depth of the capture files to 20 bits, and have the cameras shoot permanently at ISO 3200 in the ISOless shooting mode, they'd match or beat Sony's sensors in terms of noise and DR.

By making the ISOless UI optional, it would still allow those who find the current noise and DR levels to be "good enough" and prefer setting the ISO themselves to continue as before without any bother.

At low ISO the DR is limited mostly by the sensor read noise. At high ISO the DR is limited by the well capacity and therefore the photon noise. In other words, at high ISO the DR limitation is in the light itself even before it hits the sensor. Therefore increasing the number of bits in the electronics would not increase the DR at high ISO.

not the well capacity but the QE

No Mikael, not QE. QE is measured in %. Even if you have a 100% QE, the DR is still limited by the amount of light due to its random statistical nature. In the absence of the read noise the smallest signal with the SNR > 1 is 2 photons. The fact that many overlook is that even this small signal has its own photon noise of 1.4 plus the photon quantization noise of 0.5 for the total of 1.9 < 2. Therefore the DR is limited by the well capacity normalized to the number of photons. Specifically, in the absence of the read noise, DR = Log2(Well Capacity / 2).

 

[Jack Hogan]

brightcolours wrote: The idea is this:

With ISO 3200, you do not fill the wells. You simply amplify the weaker signal, weaker because you collected less light. The weak signal makes for noise.

Well, before there was a signal there was a lot (in the case of Canons) of Amp/ADC noise at low ISOs. So when you introduce a weak signal, you end up with a poor SNR ratio - what we perceive as a noisy picture.

Now, what happens if you on purpose "over expose" ISO 3200, till you do use the full well capacity?

So next, you say, we increase the ISO to 3200. That amplifies the weak signal and the lot of noise equally and you end up with the same 'noisy picture' SNR ratio.

If all the noise, or the bulk of it, came before the amplification. Clearly, we want to amplify the signal as soon as possible to minimize the noise.

Ok, then you would spec a low noise Amp/ADC system as I suggested earlier, ideally one that has a noise floor lower than the sensor's, similar to Sony's Exmor on-board design. But that's not possible with Canon's existing components simply by raising 'gain'. It requires a redesign, and that comes with compromises elsewhere.

Just to be clear, the massive (compared to the sensor's) noise introduced by Canon's Amp/ADC subsystem is introduced during amplification/conversion. Therefore in practice we are only able to see it after it by determining the standard deviation of the resulting ADUs. So the correct visualization for a photographer of how Total read noise develops as ISO is raised is in ADUs, not e- (it wasn't there yet), and for well behaved Canon cameras (like the 6D and 5dmiii) Total read noise in ADUs develops pretty much like in the graph I showed earlier.

On the other hand we often show results in e- by dividing the ADU value by ISO-related 'gain' for convenience, because it makes operations like adding in quadrature more direct. However if we want to express data in e-, all ADU data needs to be divided by 'gain' and transformed in e-. That's why you see the FWC in e- at Sensorgen.info decrease 1:1 with 'gain' as it is raised. In reality FWC in e- stays always the same, its value at base ISO. But as the noise floor of the Amp/ADC is raised through amplification the end result is 1:1 loss of 'DR': for all intents and purposes we pretend that the e- ceiling has come down, while in practice it's the ADU floor that has gone up (as shown in the graph).

Jack

 

[Press Correspondent]

1.Canon need to invest in new sensor lines to keep up regarding modern sensor design and resolution

2. Canon need to shorten the analog signal chain as Sony/Panasonic/Toshiba/Aptina etc and have for example raw vise ADC to lower the read out noise

3.Today Canon sensor tech is old, based on solutions from 2004 and modified during the years but with still high read out noise at base iso

Canon didn't make film. We bought the best film available from other companies. Recording medium was not Canon's field. Why should Canon make sensors now instead of buying the best available sensors from others? The difference I would pay for 5D3 with a Sony sensor would be more than the cost of this sensor for Canon.

 

[Mikael Risedal]

You do fill the wells at any ISO. The amplification is done by applying a higher voltage to the sensor for each photon to produce more than one electron and this fills the wells, because the well capacity limit is in electrons, not in photons. The well capacity in electrons is actually the same at any ISO, but appears reduced at higher ISO when normalized to the number of photons by dividing the output by the amplification ratio.

This also explains why fractional ISO values are digital fakes: e.g. 1 photon can produce 1 or 2 electrons, depending on the voltage (ISO), but not 1.5 electrons.

I think you must study the subject how a sensor works and collecting photons due the time and f-stop

FWC or almost full well capacity do you have only at base iso

Is this so Mikael? Then please educate us. This forum is about learning photography. So please, do tell where my explanation was wrong and explain us how this actually works or, since you struggle with English so much, provide web links that clearly explain your point. Otherwise your unsubstantiated objection is worth nothing and is not true only because you said so.

You mean , educate you ! first of all so achieves FWC at base iso due the max in falling light time/f-stop and generating electrons until the capacitor can not hold more electrons and it must be an readout, second at high iso FWC has nothing to with the results as you wrote , what is important is QE

Every iso step above base iso the electrons/charge a halving .

This might help you to understand, read the figures from lowest iso and FWC = charge/electrons and then up in "ISO" range

http://www.sensorgen.info/NikonD4.html

--
Member of Swedish Photographers Association since 1984
Canon, Hasselblad, Leica,Nikon, Linhoff, Sinar
Ernest Hemingway to Irving Penn:?“Your photos are really good. What camera do you use?”?Irving Penn to Ernest Hemingway:?“Your novels are excellent. What typewriter do you use?”

 

[Mikael Risedal]

1.Canon need to invest in new sensor lines to keep up regarding modern sensor design and resolution

safe to say that the 40M pixel APS-C sensor they've already have out in production wasn't using the 500nm line. They already had a 180nm sensor line in production. even chipworks acknowledged that canon had 180nm fabs.

2. Canon need to shorten the analog signal chain as Sony/Panasonic/Toshiba/Aptina etc and have for example raw vise ADC to lower the read out noise

patented tech though. canon did however come up with a column parallel ADC method and patented it last year i think.

3.Today Canon sensor tech is old, based on solutions from 2004 and modified during the years but with still high read out noise at base iso

see above.

I have already answered about that patent, nothing new really

 

[Mikael Risedal]

1.Canon need to invest in new sensor lines to keep up regarding modern sensor design and resolution

2. Canon need to shorten the analog signal chain as Sony/Panasonic/Toshiba/Aptina etc and have for example raw vise ADC to lower the read out noise

3.Today Canon sensor tech is old, based on solutions from 2004 and modified during the years but with still high read out noise at base iso

Canon didn't make film. We bought the best film available from other companies. Recording medium was not Canon's field. Why should Canon make sensors now instead of buying the best available sensors from others? The difference I would pay for 5D3 with a Sony sensor would be more than the cost of this sensor for Canon.

now you are wrong again, do you know how much Canon charge internally for its stitched 24x36mm sensors due to a small output compared to others? and with a old tech regarding read out, QE ?

 

[Jack Hogan]

Mikael Risedal wrote: not the well capacity but the QE

No Mikael, not QE. QE is measured in %. Even if you have a 100% QE, the DR is still limited by the amount of light due to its random statistical nature. In the absence of the read noise the smallest signal with the SNR > 1 is 2 photons. The fact that many overlook is that even this small signal has its own photon noise of 1.4 plus the photon quantization noise of 0.5 for the total of 1.9 < 2. Therefore the DR is limited by the well capacity normalized to the number of photons. Specifically, in the absence of the read noise, DR = Log2(Well Capacity / 2).

That would be correct if we only had one photosite and one exposure. But then we would not know the amount of noise. In fact in that case we also would not know what the signal was

Images (in our eyes, in our raw files) are created by sampling Luminance from the scene, the signal, at a high enough sampling rate for the desired level of detail retention and converting the result of the samples to digital units. The sampling rate is given by photosite pitch. We determine properties like signal, noise, SNR and DR by measuring the statistics at the output of a number of these photosites in a patch of uniform signal (the bigger the better).

Luminance from the scene for a given Exposure corresponds to a number of photons per unit area, say about 10 per photosite. But in fact because of the uncertainty inherent in quanta of light some photosites may see 13, some 9, some 7... etc, as you suggest, according to poisson statistics. Ibnformation theory then tells us that the signal is the mean of these values, which could very well turn out to be, say, 10.395. If the sample were large enough that's an accurate representation of the actual physical signal arriving on top of the photosite. Not 10, but 10.395.

Next the photons go through the filters on top of the photosite and hit silicon, producing photoelectrons with, say, an average probability (effective absolute QE) of 15%, also according to poisson statistics. We simplify and say that from a signal of 10.395 photons we get 1.559 photoelectrons (or e-) with shot noise and related SNR equal to the square root of that, or about 1.249.

Next we compute read noise. It also is not an integer, if not by impossible chance.

So as you see QE does enter the equation and in the right conditions the signal effectively comes in floating point notation So we could very well have physically real values with decimal places and less than 1 (photons, electrons, SNRs). The issue is only whether we can measure them precisely with our equipment (ADC), and that depends on the amount of noise present in the system: we need just enough (for dithering) but no more. The sweetspot these days seems to be a noise floor of around 1 ADU.

 

[Mikael Risedal]

1.Canon need to invest in new sensor lines to keep up regarding modern sensor design and resolution

safe to say that the 40M pixel APS-C sensor they've already have out in production wasn't using the 500nm line. They already had a 180nm sensor line in production. even chipworks acknowledged that canon had 180nm fabs.

2. Canon need to shorten the analog signal chain as Sony/Panasonic/Toshiba/Aptina etc and have for example raw vise ADC to lower the read out noise

patented tech though. canon did however come up with a column parallel ADC method and patented it last year i think.

3.Today Canon sensor tech is old, based on solutions from 2004 and modified during the years but with still high read out noise at base iso

see above.

180nm line is mainly for compact and APS sensors, Canon has a old line and a 180nm line, as an example Sony has more than 6 lines and now down to 45nm for stacked sensors

 

[Press Correspondent]

You do fill the wells at any ISO. The amplification is done by applying a higher voltage to the sensor for each photon to produce more than one electron and this fills the wells, because the well capacity limit is in electrons, not in photons. The well capacity in electrons is actually the same at any ISO, but appears reduced at higher ISO when normalized to the number of photons by dividing the output by the amplification ratio.

This also explains why fractional ISO values are digital fakes: e.g. 1 photon can produce 1 or 2 electrons, depending on the voltage (ISO), but not 1.5 electrons.

I think you must study the subject how a sensor works and collecting photons due the time and f-stop

FWC or almost full well capacity do you have only at base iso

Is this so Mikael? Then please educate us. This forum is about learning photography. So please, do tell where my explanation was wrong and explain us how this actually works or, since you struggle with English so much, provide web links that clearly explain your point. Otherwise your unsubstantiated objection is worth nothing and is not true only because you said so.

You mean , educate you ! first of all so achieves FWC at base iso due the max in falling light time/f-stop and generating electrons until the capacitor can not hold more electrons and it must be an readout, second at high iso FWC has nothing to with the results as you wrote , what is important is QE

Every iso step above base iso the electrons/charge a halving .

This might help you to understand, read the figures from lowest iso and FWC = charge/electrons and then up in "ISO" range

http://www.sensorgen.info/NikonD4.html

It is hard to read your English, but it sounds like you agree that the well capacity is lower at higher ISO. And therefore the wells do fill up at any ISO. In other words, an overexposure at ISO 3200 will saturate the sensor, not just the ADC. You seem to disagree earlier. Those here who think that you can overexpose at ISO 3200 by 5 stops without clipping the sensor and then just use a wider ADC to get more DR are wrong.

 

[Press Correspondent]

Mikael Risedal wrote: not the well capacity but the QE

No Mikael, not QE. QE is measured in %. Even if you have a 100% QE, the DR is still limited by the amount of light due to its random statistical nature. In the absence of the read noise the smallest signal with the SNR > 1 is 2 photons. The fact that many overlook is that even this small signal has its own photon noise of 1.4 plus the photon quantization noise of 0.5 for the total of 1.9 < 2. Therefore the DR is limited by the well capacity normalized to the number of photons. Specifically, in the absence of the read noise, DR = Log2(Well Capacity / 2).

That would be correct if we only had one photosite and one exposure. But then we would not know the amount of noise. In fact in that case we also would not know what the signal was

Images (in our eyes, in our raw files) are created by sampling Luminance from the scene, the signal, at a high enough sampling rate for the desired level of detail retention and converting the result of the samples to digital units. The sampling rate is given by photosite pitch. We determine properties like signal, noise, SNR and DR by measuring the statistics at the output of a number of these photosites in a patch of uniform signal (the bigger the better).

Luminance from the scene for a given Exposure corresponds to a number of photons per unit area, say about 10 per photosite. But in fact because of the uncertainty inherent in quanta of light some photosites may see 13, some 9, some 7... etc, as you suggest, according to poisson statistics. Ibnformation theory then tells us that the signal is the mean of these values, which could very well turn out to be, say, 10.395. If the sample were large enough that's an accurate representation of the actual physical signal arriving on top of the photosite. Not 10, but 10.395.

Next the photons go through the filters on top of the photosite and hit silicon, producing photoelectrons with, say, an average probability (effective absolute QE) of 15%, also according to poisson statistics. We simplify and say that from a signal of 10.395 photons we get 1.559 photoelectrons (or e-) with shot noise and related SNR equal to the square root of that, or about 1.249.

Next we compute read noise. It also is not an integer, if not by impossible chance.

So as you see QE does enter the equation and in the right conditions the signal effectively comes in floating point notation So we could very well have physically real values with decimal places and less than 1 (photons, electrons, SNRs). The issue is only whether we can measure them precisely with our equipment (ADC), and that depends on the amount of noise present in the system: we need just enough (for dithering) but no more. The sweetspot these days seems to be a noise floor of around 1 ADU.

I don't disagree with most of this, but it has nothing to do with my argument. Essentially you have two points. The first is that noise is only meaningful either over an area or over time in multiple exposures. You need more than one data point to have any statistics. Yes, so? The sensor does have more than one pixel, what is the problem? Noise still is calculated as a square root of the signal in one pixel, not a square root of the signal averaged over an area. I don't see how this would remove the DR limitation due to the photon noise.

Your second point is that QE reduces the number of photons. Well, this still would not reduce the noise. For example, consider the number of photons is 100 and the photon noise is 10 with a 10% SNR. Then you have a 10% QE and the signal you actually see is 10 times less. Do you think the noise would also be 10 times less and the SNR remain 10%? Or do you think the noise would be a square root of the signal thus reducing the SNR about 3 times? I believe the latter is correct, because the quantum statistics still applies to the signal. And if so, your argument changes nothing, and the DR is still limited by lowest signal that is above its own noise.

 

[Great Bustard]

It is hard to read your English, but it sounds like you agree that the well capacity is lower at higher ISO. And therefore the wells do fill up at any ISO. In other words, an overexposure at ISO 3200 will saturate the sensor, not just the ADC. You seem to disagree earlier. Those here who think that you can overexpose at ISO 3200 by 5 stops without clipping the sensor and then just use a wider ADC to get more DR are wrong.

...the ISO setting on the camera has squat to do with the FWC of the pixels. For example, if a pixel can absorb 80000 photons, then it can do so at any ISO setting, since the ISO setting merely applies a gain to that signal *after* the fact.

Indeed, it doesn't make any sense at all that the ISO setting would affect the pixels' ability to absorb light -- it merely affects what is done with that light after it turns photons into electrons.

 

[Great Bustard]

OK, let's assume only two significant sources of read noise, pixel noise and ADC noise, where the ADC noise comes after the amplification.

The following chart assumes a pixel noise of 2 electrons and an ADC noise of 27 electrons. The ordering is ISO / actual / predicted:

100 / 26.8 / 27.1

200 / 14.6 / 13.6

400 / 7.9 / 7.0

800 / 5.1 / 3.9

1600 / 3.1 / 2.6

3200 / 2.3 / 2.2

6400 / 2.0 / 2.0

I mean, not too shabby an approximation, methinks. So, would not having a 20 bit image file allow the camera to use ISO 3200 or ISO 6400 all the time without blowing highlights?

That's good. But don't forget that all the ADC does is digitize an actual analog signal. In this case the signal at ISO100 is bound on the lower end by read noise (mainly Amp/ADC) of 27e- and at the upper end by FWC of around 76600 e- (sensorgen.info data for the 6D I assume) = a 'DR' of about 11.5 stops by this definition.

With the 6D' Amp/ADC, those two figures would correspond to 4.6 ADUs of noise and 13,235 ADUs saturation at ISO100 - every Raw level equivalent to about 76600e-/13235ADU = 5.88e-/ADU.

OK, I have to admit to cognitive failure. How do you convert electrons to ADUs? I thought one ADU for the 6D would correspond to 76600 electrons / 2^14 = 4.675 electrons so a noise of 27.1 electrons would correspond to 5.8 ADUs.

This, of course, results in the same DR that you computed above DR = log2 (2^14 / 5.8) = 11.5 stops, using the read noise for the noise floor over the area of one pixel.

If you were to use a 20-bit ADC to encode the same range you would have 'gain' of something like 76600/2^20 = 0.073e-/DN. So 27e- would correspond to 367DN and 76600e- to 2^20DN (DN here meaning 20-bit ADUs). DR of course reamains what it is.

My misunderstanding on the computation of ADUs is likely why I don't understand why this would be the case. Does not the noise floor drop at ISO 3200 while the FWC remains the same (with the 20 bits allowing the full FWC to be recorded)?

Amp/ADC read noise in ADU remains unchanged throughout the ISO range (4.6 or 367 ADUs as you prefer), but Sensor noise grows with ISO, becoming the main souce of noise by ISO1600 as per the graph above.

 

[Press Correspondent]

It is hard to read your English, but it sounds like you agree that the well capacity is lower at higher ISO. And therefore the wells do fill up at any ISO. In other words, an overexposure at ISO 3200 will saturate the sensor, not just the ADC. You seem to disagree earlier. Those here who think that you can overexpose at ISO 3200 by 5 stops without clipping the sensor and then just use a wider ADC to get more DR are wrong.

...the ISO setting on the camera has squat to do with the FWC of the pixels. For example, if a pixel can absorb 80000 photons, then it can do so at any ISO setting, since the ISO setting merely applies a gain to that signal *after* the fact.

Indeed, it doesn't make any sense at all that the ISO setting would affect the pixels' ability to absorb light -- it merely affects what is done with that light after it turns photons into electrons.

I understand what you are saying. If the sensor and amplifier were separate consecutive units, you would be correct. I am not a sensor engineer and I may be wrong. If you provide a link to a convincing evidence then I would learn and benefit from this forum. But until then I maintain that the amplification is done by applying a higher voltage to the sensor thus reducing the well capacity. To be clear, I understand that there is an amplifier after the sensor, but it is not where the ISO amplification is done. Prove me wrong by a clear reference. I don't mind to be wrong, my goal here is to learn, not to satisfy my ego.

 

[Great Bustard]

It is hard to read your English, but it sounds like you agree that the well capacity is lower at higher ISO. And therefore the wells do fill up at any ISO. In other words, an overexposure at ISO 3200 will saturate the sensor, not just the ADC. You seem to disagree earlier. Those here who think that you can overexpose at ISO 3200 by 5 stops without clipping the sensor and then just use a wider ADC to get more DR are wrong.

...the ISO setting on the camera has squat to do with the FWC of the pixels. For example, if a pixel can absorb 80000 photons, then it can do so at any ISO setting, since the ISO setting merely applies a gain to that signal *after* the fact.

Indeed, it doesn't make any sense at all that the ISO setting would affect the pixels' ability to absorb light -- it merely affects what is done with that light after it turns photons into electrons.

I understand what you are saying. If the sensor and amplifier were separate consecutive units, you would be correct. I am not a sensor engineer and I may be wrong. If you provide a link to a convincing evidence then I would learn and benefit from this forum. But until then I maintain that the amplification is done by applying a higher voltage to the sensor thus reducing the well capacity. To be clear, I understand that there is an amplifier after the sensor, but it is not where the ISO amplification is done. Prove me wrong by a clear reference. I don't mind to be wrong, my goal here is to learn, not to satisfy my ego.

The proof is beyond my knowledge base, so I will have to defer to those more knowledgeable than myself to give you the proof. In other words, I'll shoot Bob an email. ;-)

 

[j900]

high iso 18% noise and low iso noise floor are not the same thing. Canon currently has high iso 18% noise on par with competition but completely falls apart in low iso noise floor.

 

[Mako2011]

It is hard to read your English, but it sounds like you agree that the well capacity is lower at higher ISO. And therefore the wells do fill up at any ISO. In other words, an overexposure at ISO 3200 will saturate the sensor, not just the ADC. You seem to disagree earlier. Those here who think that you can overexpose at ISO 3200 by 5 stops without clipping the sensor and then just use a wider ADC to get more DR are wrong.

...the ISO setting on the camera has squat to do with the FWC of the pixels. For example, if a pixel can absorb 80000 photons, then it can do so at any ISO setting, since the ISO setting merely applies a gain to that signal *after* the fact.

Indeed, it doesn't make any sense at all that the ISO setting would affect the pixels' ability to absorb light -- it merely affects what is done with that light after it turns photons into electrons.

I understand what you are saying. If the sensor and amplifier were separate consecutive units, you would be correct. I am not a sensor engineer and I may be wrong. If you provide a link to a convincing evidence then I would learn and benefit from this forum. But until then I maintain that the amplification is done by applying a higher voltage to the sensor thus reducing the well capacity. To be clear, I understand that there is an amplifier after the sensor, but it is not where the ISO amplification is done. Prove me wrong by a clear reference. I don't mind to be wrong, my goal here is to learn, not to satisfy my ego.

From Ron's site

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www.ronbigelow.com

"....the photons reach the sensor and excite electrons.... These excited electrons are freed from the molecules to which they are attached. When a voltage is applied, these free electrons create a current and flow into a capacitor. This creates a charge on the capacitor. The charge is then measured to create a voltage measurement. This voltage measurement is processed by the camera to determine how much light reached the pixel during exposure"

"...After the freed electrons flow into a capacitor and the voltage of the capacitor is measured, the voltage is amplified before any further processing is performed"

".....The way that a digital camera increases the ISO is to apply a greater amount of amplification to the voltages that come from the pixels' capacitors"

So if the increase in voltage associated with ISO is applied to the signal from the capacitor, and not the individual pixels prior to capacitor storage, is actual Full Well Capacity not affected by ISO?