# Why are larger sensors better at low light?

The top answer of What point and shoots are good in low light conditions? says that (1) a fast lens/wide aperture (2) reasonable ISO 400+ handling and (3) a large sensor when put together are critical in shooting in low light.

The first I understand (it lets in more light), the second I understand (the "film" is more sensitive to light). Sorry I do not understand the third factor.

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Large sensors also mean better noise control as higher ISO <en.wikipedia.org/wiki/Image_noise#Effects_of_sensor_size>; – Anand Sep 15 '11 at 5:41

Its easiest to understand the difference when both the larger and smaller sensor have the same megapixels. If we have a couple hypothetical cameras, one with a smaller APS-C sensor and one with a Full Frame sensor, and assume both have 8 megapixels, the difference boils down to pixel density.

An APS-C sensor is about 24x15mm, while a Full Frame (FF) sensor is 36x24mm. In terms of area, the APS-C sensor is about 360mm^2, and the FF is 864mm^2. Now, calculating the actual area of a sensor that is functional pixels can be rather complex from a real-world standpoint, so we will assume ideal sensors for the time being, wherein the total surface area of the sensor is dedicated to functional pixels, assume that those pixels are used as efficiently as possible, and assume all other factors affecting light (such as focal length, aperture, etc.) are equivalent. Given that, and given that our hypothetical cameras are both 8mp, then its clear that the size of each pixel for the APS-C sensor is smaller than the size of each pixel for the FF sensor. In exact terms:

APS-C:
360mm^2 / 8,000,000px = 0.000045mm^2/px
-> 0.000045 mm^2 * (1000 µm / mm)^2 = 45µm^2 (square microns)
-> sqrt(45µm^2) = 6.7µm

FF:
864mm^2 / 8,000,000px = 0.000108mm^2/px
-> 0.000108 mm^2 * (1000 µm / mm)^2 = 108µm^2 (microns)
-> sqrt(108µm^2) = 10.4µm

In simpler, normalized terms of "pixel size", or the width or height of each pixel (commonly quoted on photo gear web sites), we have:

APS-C Pixel Size = 6.7µm pixel
FF Pixel Size = 10.4µm pixel

In terms of pixel size, a FF 8mp camera has 1.55x larger pixels than an APS-C 8mp camera. A one-dimensional difference in pixel size does not tell the whole story, however. Pixels have two-dimensional area over which they gather light, so taking the difference between the area of each FF pixel vs. each APS-C pixel tells the whole story:

108µm^2 / 45µm^2 = 2.4

An (idealized) FF camera has 2.4x, or about 1 stop worth, the light gathering power of an (idealized) APS-C camera! That is why a larger sensor is more beneficial when shooting in low light...they simply have greater light gathering power over any given timeframe.

In alternative terms, a larger pixel is capable of capturing more photon hits than a smaller pixel in any given timeframe (my meaning of 'sensitivity').

Now, the example and computations above all assume "idealized" sensors, or sensors that are perfectly efficient. Real-world sensors are not idealized, nor are they as easy to compare in an apples-to-apples fashion. Real-world sensors don't utilize every single pixel etched into their surface at maximum efficiency, more expensive sensors tend to have more advanced "technology" built into them, such as microlenses that help gather even more light, smaller non-functional gaps between each pixel, backlit wiring fabrication that moves column/row activate and read wiring below the photo-sensitive elements (while normal designs leave that wiring above (and interfering with) the photo-sensitive elements), etc. Additionally, full-frame sensors often have higher megapixel counts than smaller sensors, complicating matters even more.

A real-world example of two actual sensors might be to compare the Canon 7D APS-C sensor with the Canon 5D Mark II FF sensor. The 7D sensor is 18mp, while the 5D sensor is 21.1mp. Most sensors are rated in rough megapixels, and usually have a bit more than their marketed number, as many border pixels are used for calibration purposes, obstructed by sensor filter mechanics, etc. So we'll assume that 18mp and 21.1mp are real-world pixel counts. The difference in light-gathering power of these two current and modern sensors is:

7D APS-C: 360mm^2 / 18,000,000px * 1,000,000 = 20µm^2/px
5DMII FF: 864mm^2 / 21,100,000px * 1,000,000 = 40.947 ~= 41µm^2/px

41µm^2 / 20µm^2 = 2.05 ~= 2

The Canon 5D MkII Full-Frame camera has about 2x the light gathering power of the 7D APS-C camera. That would translate into about one stops worth of additional native sensitivity. (In reality, the 5DII and 7D both have a maximum native ISO of 6400, however the 7D is quite a bit noisier than the 5DII at both 3200 and 6400, and only really seems to normalize at about ISO 800. See: http://the-digital-picture.com/Reviews/Canon-EOS-7D-Digital-SLR-Camera-Review.aspx) In contrast, an 18mp FF sensor would have about 1.17x the light gathering power of the 21.1mp FF sensor of the 5D MkII, since fewer pixels are spread out over the same (and larger than APS-C) area.

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@jrista: Do the 2-megapixel cameras of 1999 (2.5" sensors?) have large pixels? – William C Sep 15 '11 at 4:49
@William: Regarding the Canon Pro70, don't forget that there have been many other advancements in sensor design since 1998. Even though the pixel size is larger on those cameras, technologically they were extremely primitive compared to todays sensors. For one, the pixel size is probably smaller (4nm?)...pixels had large gaps and no microlenses back then. CCD readout was much noisier, prone to read streaking, charge overflow into neighboring cells, etc. Sensitivity of the Pro70 was a lot lower too, ISO 100-200 in "high res" mode and ISO 400 in "low res" mode. – jrista Sep 15 '11 at 5:23
I should point out that the units on these pixel sizes should be µm (micrometers or microns), not nm (nanometers). a 10nm pixel would be tiny -- transistors in computer processors these days are generally on the order of 45nm wide. I've edited jrista's answer to take that into account. – Evan Krall Sep 15 '11 at 6:00
And, in fact, visible light is 380nm-740nm, so a 10nm pixel would literally be smaller than a single wavelength of light. – Evan Krall Sep 15 '11 at 6:05
It should be made clear that the extra light gathering ability of large sensors assumes the same f-stop. This is not always possible in practice as maintaining the same angle of view means using a lens with a longer focal length, which tend to have smaller max apertures, e.g. when using a 200 f/2.0 on an APS-C body, you'll get a similar amount of light as using a 300 f/2.8 on full frame - as there is no 300 f/2.0 [currently in production]. – Matt Grum Sep 15 '11 at 7:09

Strictly speaking it is NOT the sensor-size that makes it better, it IS the pixel-size.

Larger pixels have more surface areas to capture light and accumulate a higher voltage from the release of electrons when photons (light) hits the surface. The inherent noise being mostly random is therefore relatively lower compared to the higher voltage which increases the signal-to-noise ratio (S/N).

The implied data you were missing is that larger sensors tend to have larger pixels. Just compare a full-frame 12 MP D3S with a cropped 12 MP D300S. Each pixel has 2.25X more surface area which is why the D3S has such a stellar high-ISO performance.

EDIT (2015-11-24):

For the anonymous downvoter non-believer, there is a newer and better example. Sony has two nearly identical full-frame cameras, the A7S II and the A7R II. Their sensors are the same size but the former has 12 MP of resolution, while the latter 42 MP. The low-light performance and ISO range of the A7S II is quite ahead of the A7R II, reaching ISO 409,600 vs 102,400. That is two stops difference only for having the larger pixels.

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When you normalize to print size or resolution, it is sensor size not pixel size that makes a difference. If you take a 24MP APS-C sensor and a 6MP APS-C sensor, the 24MP will have more noise per pixel but if you downsize the image to 6MP then the noise averages out and you have (in theory) the same amount of noise as the 6MP camera images. On the other hand if you print the images at the same size the noise on the 24MP print will be much finer grained and less visible at the same viewing distance as the 6MP print. – Matt Grum Sep 15 '11 at 6:28
@Matt - Sadly, most people get so excited about how many pixels they get these days that they forget about comparing on prints they would actually make! – Itai Sep 15 '11 at 12:49
@Itai - That's because many people won't actually print them. They either post online, use them as computer backgrounds, or display them in a digital frame. Photo printing seems to be getting less and less common, sadly. – John Cavan Sep 15 '11 at 13:19
@John Even if you don't print, resizing a large high res image for the web averages out the noise to the same degree! – Matt Grum Sep 15 '11 at 13:24
It's important to distinguish between per pixel noise and per image noise. Per pixel noise is heavily dependent on the pixel size, per image noise is most dependent on sensor size (pixel size has a very small influence). However since most people view and print images, not pixels, per image noise is a much more relevant measure, therefor it really is the sensor size that matters most. – Matt Grum Nov 24 '15 at 15:11

The size of the single pixel is nearly irrelevant. That is urban legend!

Given two identical cameras with a sensor of the same size but a different pixel count (say 2MP and 8MP) - and therefor a different pixel size. The amount of light that gets on the sensor depends on the diameter of the lens, not of the pixel size. No doubt the 8MP picture will be noisier that the 2MP picture, but if you scale down the 8MP to 2MP you will get nearly the same picture - with nearly the same noise level. That's simple maths. I say nearly because the sensor logic costs size. As you will have 4 times the logic on a 8MP sensor that on a 2MP, you will get less net light-sensitive sensor area. But that won't cost you 1 stop (=50%), maybe a little bit, but not that much!

What actually makes the difference are the lenses. If you shot a picture, you won't be interested in metrics - neither sensor size, pixel size nor focal length. You want to catch a face, a group of people, a building or something else from a given distance. What you are interested in is angle of view. Your focal length will depend on the sensor size and angle of view. If you have a tiny sensor, you will also have a tiny focal length (say some few mm). A lens with a tiny focal length will never catch a lot of light, as it will be limited in diameter. A larger sensor will need a larger focal length, a lens with the same speed will have a greater diameter and therefor catch much more light.

Who needs 10MP or more except for printing posters? Scaled down to a few MP all pictures look ok. Sensor size does not limit your picture quality directly, but your lens will. Although the lens size often depends on the sensor size (must not). But I have seen cameras with small sensors and lots of MP but greats lenses (say greater that 2cm diameter) that shoot great pictures.

I've written an article on that a while ago. It's in German, I hadn't the time to translate it into English - sorry for that. It's more verbose and explains some issues (especially the noise issue) a bit more in detail.

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For completeness - the comparison has to be made between sensors of the same age and technology. Also, to counter the "dead area" of the pixel logic problem, the microlens arrays were introduced. Last - I don't see how the lens' diameter affects the amount of light falling on the sensor (do you mean the aperture??). – ysap Sep 15 '11 at 15:28
To make my point clearer - if the light approaching the lens forms a cone, and the FoV determines the cone's head angle, then the physical size of the lens, being proportional to the size of the sensor, should not change the amount of light falling on the sensor. The aperture, however does affect that. – ysap Sep 15 '11 at 15:32
Of course, diameter = aperture :) So, the greater the aperture, the more light will get on the sensor. But you can't take FoV as cone of light. The relevant cone of light has it's origin at the object, you'r looking straight onto it. The greater your aperture, the bigger that cone. – craesh Sep 15 '11 at 15:37
Yes, but the aperture is given in relative numbers. The light gathering ability of a 50mm f/2 lens on 35mm sensor should be the same as a ~35mm f/2 lens on an APS-C sensor. This is why the actual aperture iris is not necessarily located at the lens front but can be located anywhere on the light path. – ysap Sep 15 '11 at 15:54
What you mean is the f-number or relative aperture, sometimes the numerical aperture. That is the focal length divided by the aperture (or entrance pupil). The aperture is (as I wrote above) the diameter of the lens. Ok, as camera lenses get more and more complex, the diameter of the first lens will not necessarily be the same as the focal length divided by the lowest f-number. But in principle, they should match. The greater the aperture, the more light comes into the camera. That's roughly comparable to buildings with bigger/smaller windows. – craesh Sep 15 '11 at 16:47

The size of an individual pixel is unimportant. Several small pixels can be combined mathematically into one large one, trading detail for sensitivity.

A large sensor camera has, for a given angle of view, a longer focal length lens than a small sensor camera. This longer lens has, for a given f-stop, a large physical aperture (opening in the iris). This results in more light entering the system, and accounts for the better low light performance. It also accounts for the shallower depth of field.

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If nothing else, this answer ignores read noise - several small pixels do perform worse than one large pixel. – Philip Kendall Nov 23 '15 at 21:21
@PhilipKendall Apart from that statement in the first paragraph the rest of the answer is correct, it would suggest just removing that part. – Matt Grum Nov 24 '15 at 8:50
Do you have a reference to support this assertion? I offer clarkvision.com/articles/digital.sensor.performance.summary – Neil P Dec 9 '15 at 14:45

The surface of the digital sensor is covered with photosites. These record the image of the outside world as projected by the lens. During the exposure, image forming rays in the form of photons bombard the surface of the sensor. The photon hits are in proportion to scene brightness. In other words, photosites that receive photon hits that correspond to brightly lit areas of the scene, receive more photon hits than photosites that correspond to dimly lit image areas. When the exposure is complete, the photosites contain an electric charge in proportion to scene brightness. Nevertheless, the degree of charge in all photosites is too feeble to be useful unless amplified. The next step in the image forming process is to amplify the charges.

Amplification is like turning up the volume of a radio or TV. Amplification boots the strength of the image signal but it also induces distortion in the form of static. In digital imaging we don’t call this distortion static; we call it “noise”. The noise induced is actually called fixed pattern noise. This is because each photosite has slightly different characteristics. In other words they each respond to amplification differently. The result is, some photosites that had few photon hits will image as black when they should image as dark gray or gray. This is fixed pattern noise. We mitigate by not upping the amplification (keeping the ISO low) and by software in the camera.

Since fixed pattern noise is generally due to high amplification, it stands to reason that more photon hits at any given photosite generate a higher charge and need less amplification. The bottom line is, larger imaging chips sport larger photosites with larger surface area allows for more photon hits during the exposure. More hits translate to less amplification; thus less distortion due to fixed pattern noise.

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