Do full frame sensors have a higher exposure? [duplicate]

Question

I was reading a piece on the comparison of APS-C to full frame sensors. In this article, it mentioned that full frame sensors can have larger individual "photo sites", and can therefore capture more light.

Here is the quote...

It is not the number of pixels that really counts - APS-C models such as Canon's EOS 7D and EOS 550D have almost as many photosites, thanks to their 18-million pixel count - but the size of the photosites is crucial to image quality.

Bigger individual light sensors capture more light - and this means that less electronic noise is created. You notice this most as you increase the ISO setting - with this noise creating a coloured mosaic pattern that is particularly noticeable in shadow areas.

Is this true? If I control for variables like aperture relative to the sensor, would a full frame sensor get higher exposure?

Answer 1

The full frame sensor will not be brighter under the same exposure conditions (Same light in scene, same focal length and f-number, same exposure time, etc.). It will collect more light, but it will also spread that light over an equally proportionally larger area. The brightness, which is defined as the amount of light energy per unit area, will be the same. The advantage of the larger pixels will not be in increased brightness, but in reduced noise (due to the averaging of the random nature of light - what we call shot noise - over a larger area) and increased dynamic range if the pixels are larger on the FF sensor (due to higher full well capacity for the same thickness silicon wafer).

Answer 2

Exposure is per unit area — see Why does illuminance stay the same for a given f-stop even when focal length changes?. That means that if you measure exposure for a given shutter-speed-and-aperture on one half of the frame, it'll be the same on the other. So, that's what matters for exposure settings.

But, full-frame sensors do have an inherent low-light advantage.

Here's a way to look at it: digital sensors are 36×24mm for full frame, or 24×16mm for APS-C. When you take a correctly-exposed image, each square mm on each sensor gets the same amount of brightness. If you want to print at, say, 12×18" (mixing imperial and metric), you need to enlarge 12.7× from the full-frame sensor — or 19.05× from the APS-C one. A square mm from the full-frame camera becomes 1.27×1.27 centimeters, or 1.61cm². A square mm from the smaller-sensor camera becomes 3.63cm²! That means that in your final print, the same amount of light is spread out over 2.25× the area.

Of course, we don't print the smaller print much more darkly. Instead, we effectively amplify the brightness as we enlarge. Stretching the same amount of light into a larger area inherently gives worse results. When there's plenty of signal — lots of light — this generally doesn't matter, but when it's dim and there's a lot of noise, it does.

This is generally the reason that cameras with full-frame sensors are regarding as having about a one-stop advantage in ISO noise over APS-C. ("One stop" is 2×, of course.) In a digital camera, ISO is amplification, and for a given print size from images at the same ISO, full-frame images are literally amplified only half as much.

The issue with size of photosites is a different, technical one not actually related to sensor size and is largely obsolete with modern technology. See Do megapixels matter with modern sensor technology? for more. Even with infinitely-good technology, you can't beat the physical reality of the above. Bigger is always more light. However, in most practical cases, you can get to the point where smaller is good enough. (Otherwise, we'd all be carrying around 8×10" large-format digital cameras....)

Answer 3

If we consider film photography, to correctly expose any particular film, the same number of photons must be absorbed by the film per unit area for any particular scene. This means that a full-frame 35mm camera will need to receive twice as many photons as a half-frame camera to achieve the same results under identical conditions. Note that this does not mean that the scene your are photographing needs to be twice as bright.

The same must be true for digital sensors having the same sensitivity (which is a direct analogue of film speed). Of course, how many of the photons are actually "used" by the sensor is another matter. So, a lot more light (i.e. photons) will be required for a full-frame sensor than for an APS-C sensor having the same sensitivity. However, the lighting conditions of the scene that you are photographing will be the same.

Answer 4

The job of the photosite is to collect photon hits during the exposure. The photosites contain a photodiode and a storage area to hold the charge as it accumulates during the exposure. The more photon hits, the greater the charge. When the exposure is complete, the charge is moved into storage. Software marches the charges, row by row transferring the charges into a transfer register. Here the charges are read and converted to a voltage. Because the voltages are weak they are amplified to a useful level. Next the voltage is converted to a digital signal.

The amount of amplification that is applied is a key factor. Low charge levels require more amplification. This will be a function of the light level during the exposure and this will be intertwined with the ISO setting. If high amplification is applied, some unwanted static will be induced. This is tantamount to turning up the volume on a radio. In other words, static rides piggyback on the good signal. In digital imaging, the static manifests itself as noise. Noise is a granularity akin to grain in the conventional film imaging process.

You avoid elevated noise by keeping the ISO moderate and by getting the exposure correct. While true, smaller image sensors have smaller photosites, the imaging chip continues to evolve. The differences between the full size sensor FX and the compact sensor APS-C, becomes less noticeable.

P.S. Larger chips have larger photosites thus they collect more photon hits during the exposure. The charge in the photosite is higher thus less amplification is needed thus less noise.