- Fred Miranda Jr
- Jul 10, 2012
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I spend endless hours in front of the computer perfecting techniques for the digital format.
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List_of_Zeiss_lenses.xhtml
I'd like to point out that the Zeiss ZE lenses are all manual focus. For the 99.9% of us for whom autofocus is a requirement, the Canon L lenses are the crème de la crème.
Yes. It doesn't change the fact it's horrible value.
You're clearly not the target market.
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May 16, 2013 00:21
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- Fred Miranda Jr
- Jul 10, 2012
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I spend endless hours in front of the computer perfecting techniques for the digital format.
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I meant to type “intensity (light per unit area)” instead of “intensity per unit area.” That was my mistake, rushing to type this out on a day when I teach multiple classes. Anyway, now I have time to write more.
BTW, this isn’t directly exposure triangle stuff – this is lower level than that. This is image sensor and lens layout level crap. And I am sure that I am going to use a term inappropriately here and there – if that genuinely causes confusion, let me know.
So, I am going to start by making a bunch of statements, and if you disagree with any of them, just say which one and provide information to support that. Then I’ll combine them in an example to demonstrate what I have been trying to say.
1.) Light intensity on the sensor (illuminance) is a function of the light source (which I will assume is constant, like a perfect white wall with even reflectivity) and f-stop.
2.) Light intensity on the sensor for a given FOV and f-stop is constant regardless of whether the sensor is FF, APS-C, 1”, etc.
3.) Light intensity represents the number of photons per unit area.
4.) The larger the sensor that light falls on with a given intensity, the more total photons the sensor will collect.
5.) The total number of photons collected by a pixel on a sensor is given by the total number of photons hitting the sensor (intensity times surface area) divided by the portion of the sensor taken up by that pixel.
6.) Let’s say that I have a sensor (size doesn’t matter) and I have exposed it such that each pixel receives 100 photons. Now, I cut each pixel in half, and with the same exposure conditions, each pixel now receives half the number of photons (50).
6a.) Say I have a 20MP full frame sensor. I have a light source and lensing setup that sends light to the sensor such that each pixel receives 1,000 photons (let’s say at 100mm and 1/60th – it’s arbitrary and doesn’t matter; this is just to appease people who keep trying to inappropriately bring the exposure triangle back into this). My image processor defines this level of CMOS excitation as matching ISO 100 sensitivity for the intensity of light falling on it (not a clear sentence, I know, but I am tired).
6b.) Now, I have a 40MP FF sensor. I expose it the same as before. Because each pixel is half the area of the pixels in 6a., each pixel now receives 500 photons. But wait, the intensity of the light is the same, so the exposure must be the same too. Thus, I have to boost the signal at each pixel in order to keep the 40MP exposure the same as my 20MP exposure. In other words, one of two things has to happen:
Option 1: For every electron that 1 photon caused the 20MP pixel to send, each photon has to cause the 40MP pixel to send 2 electrons in order to fill its part of the exposure triangle and match the ISO 100 sensitivity standard. (Boost at the pixel level*)
Option 2: The image processor for the 40MP sensor requires half the number of electrons from a pixel than the 20MP sensor’s image processor requires from its pixels in order to match the ISO 100 standard. (Boost at the image processor level*)
*Yes, the image processor can “know” what area is covered by its pixels and use that current*known area math to figure out what illuminance the sensor is getting at that pixel, but each smaller pixel will still have a lower SNR, and noise is why I brought sensor/pixel size up in the first place.
7.) If I have a FF sensor and an APS-C sensor each receiving the same intensity of light, the APS-C sensor, being 43% the area of the FF sensor, will receive 43% of the total number of photons that the FF sensor receives with that exposure.
8.) If my FF and APS-C sensors each have the same number of pixels, each pixel will collect the same percentage of the total photons that fall on the sensor.
9.) If I expose a 20MP FF and a 20MP APS-C sensor to an illuminance as in 6a., each FF pixel will receive 1,000 photons, while each APS-C pixel will receive 430 photons.
9a.) If every photon that hits a FF pixel in 9. excites the CMOS to send the image processor X number of electrons, then every photon that hits an APS-C pixel in 9. will also cause X number of electrons to be sent. (Assume no pixel-level boosting here – this is, of course, the Canon thread, and we don’t need no stinkin’ dynamic range!)
9b.) Thus when exposed to the same illuminance (FOV and f-stop from the same scene/white wall), each pixel on the APS-C sensor will send the image processor 43% the number of electrons that each FF pixel will send. The image processor will have to compensate for this reduced total light per pixel/reduced total electrons per pixel by boosting the signal from the pixels, introducing noise.
Ok, still with me? Are we in agreement that light intensity is independent of the total area that we use to sample that illuminance? If we have a FF sensor, an APS-C sensor, and a 1” sensor, and give them all the same FOV and f-stop, they will all be receiving the same intensity of light. Agreed?
So, let’s say I am a sensor manufacturer. I design a FF sensor with 20 million pixels, so each pixel is 0.043^2 microns in area (Pixel Mk 1). Let’s say that with an arbitrary f-stop (light intensity) and exposure time, each Pixel Mk 1 will receive 1,000 photons. I also want to design an APS-C sensor and a 1” sensor that also use Pixel Mk 1. Once I have done that, I have an 8.6 megapixel APS-C sensor with the same light sensitivity as my FF sensor, and a 2.7 megapixel 1” sensor.
So, I take them to marketing, proud of having made crop sensors with the same light sensitivity as full-frame. I would get told “No one wants to buy 8.6 and 2.7 megapixel sensors! Go give all of them 20MP!”
So, I go back and design Pixel Mk2 for my APS-C sensor, which has an area of 0.0185^2 microns per pixel, and Pixel Mk3 for my 1” sensor, which has an area of 0.0058^2 microns per pixel.
I then expose all of them with the same conditions that I mentioned before. Each Pixel Mk1 on my FF sensor receives 1,000 photons. Each Pixel Mk 2 on my APS-C sensor receives 430 photons, and each Pixel Mk 3 on my 1” sensor receives 135 photons. Remember that each sensor is receiving the same intensity of light. But because the APS-C and 1” sensors’ pixels are smaller than the FF sensor’s pixels, they take in less light when the total sensor is exposed to a given illuminance.
But, I need them all to be showing the same exposure. I have two ways that I could do this. The first is that I could boost the smaller sensors (such that the APS-C sensor reads 430 photons the same as the FF sensor reads 1,000 photons, and the same for 135 photons on my 1” sensor).
The second is that I could build lenses that capture the same amount of light, but lens it down to smaller areas to match the smaller sensors (thus increasing the intensity of the light on the sensors). But wait, if according to that textbook, the f-stop defines the intensity of the light, wouldn’t changing the intensity of the light on the sensor while keeping the FOV change the f-stop?
Let’s do some math. My original FF exposure was at [arbitrary simple to work with values here: 100mm and f/8, for a 12.5mm aperture diameter and 156.25mm^2 aperture area] and it delivered an intensity of 1,000 photons per 0.043^2 microns. To match the FOV, I need focal lengths of 66.7mm and 37mm for the APS-C and 1” lenses, respectively. Aperture diameter for those lenses will be 8.3mm and 4.6mm, with aperture areas of 69.4mm^2 and 21.4mm^2, in order to place the same intensity of light on the sensor as the FF sensor gets (i.e., to be at f/8).
So, in order to place as much light on each pixel as the FF sensor receives at our selected FOV (increasing the intensity), we need to increase the total light on the sensor. We can’t change the focal length without changing the FOV, so we have to change the aperture. If we open the APS-C aperture up to 12.5mm, we will increase the intensity of the light on the sensor by 2.25 times, meaning that each pixel now captures ~1,000 photons. But the f-stop has now become 66.7mm/12.5mm = f/5.3, And if we open the 1” aperture up to 12.5mm, we will increase the intensity of the light on the sensor by 7.3 times, meaning that each pixel now captures ~1,000 photons, but the f-stop has now become 37mm/12.5mm = ~f/3.
TL/DR:
If you have smaller pixels, you need to have a higher intensity of light to excite them to a given level than larger pixels need. Even if they are on different sized sensors, you will need to change the f-stop to do that. If you want to keep the f-stop the same, then you will need to accept that each pixel will get less light, and boost the sensitivity of each pixel to compensate.
Also, I hate you all. Goodnight.
Hello! Have you considered posting on my forums at http://www.fredmiranda.com/forum/ ? You'd fit right in! Come on over.
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May 16, 2013 00:21
