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.


7 Answers 7


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:

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

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.

  • 2
    \$\begingroup\$ @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. \$\endgroup\$
    – jrista
    Commented Sep 15, 2011 at 5:23
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    \$\begingroup\$ 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. \$\endgroup\$
    – Evan Krall
    Commented Sep 15, 2011 at 6:00
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    \$\begingroup\$ And, in fact, visible light is 380nm-740nm, so a 10nm pixel would literally be smaller than a single wavelength of light. \$\endgroup\$
    – Evan Krall
    Commented Sep 15, 2011 at 6:05
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    \$\begingroup\$ 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]. \$\endgroup\$
    – Matt Grum
    Commented Sep 15, 2011 at 7:09
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    \$\begingroup\$ I'd also like to note that ISO settings on digital sensors are actually a function of amplifiers. CCD/CMOS have a fixed response to light that does not change. There's no additional light being manufactured when bumping up the ISO setting. Instead, the processing converters tell how much the total data should be amplified - including noise. It's just like turning up the volume on a noisy cassette tape - you'll hear the hiss that much louder. Note that amplifiers and software improves over time so the effect of increased noise appears reduced. \$\endgroup\$
    – smigol
    Commented Sep 15, 2011 at 15:05

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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    \$\begingroup\$ 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. \$\endgroup\$
    – Matt Grum
    Commented Sep 15, 2011 at 6:28
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    \$\begingroup\$ @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! \$\endgroup\$
    – Itai
    Commented Sep 15, 2011 at 12:49
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    \$\begingroup\$ @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. \$\endgroup\$
    – Joanne C
    Commented Sep 15, 2011 at 13:19
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    \$\begingroup\$ @John Even if you don't print, resizing a large high res image for the web averages out the noise to the same degree! \$\endgroup\$
    – Matt Grum
    Commented Sep 15, 2011 at 13:24
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    \$\begingroup\$ 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. \$\endgroup\$
    – Matt Grum
    Commented Nov 24, 2015 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.

  • \$\begingroup\$ 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??). \$\endgroup\$
    – ysap
    Commented Sep 15, 2011 at 15:28
  • \$\begingroup\$ 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. \$\endgroup\$
    – ysap
    Commented Sep 15, 2011 at 15:32
  • \$\begingroup\$ 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. \$\endgroup\$
    – craesh
    Commented Sep 15, 2011 at 15:37
  • \$\begingroup\$ 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. \$\endgroup\$
    – ysap
    Commented Sep 15, 2011 at 15:54
  • \$\begingroup\$ 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. \$\endgroup\$
    – craesh
    Commented Sep 15, 2011 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.

  • \$\begingroup\$ If nothing else, this answer ignores read noise - several small pixels do perform worse than one large pixel. \$\endgroup\$
    – Philip Kendall
    Commented Nov 23, 2015 at 21:21
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    \$\begingroup\$ @PhilipKendall Apart from that statement in the first paragraph the rest of the answer is correct, it would suggest just removing that part. \$\endgroup\$
    – Matt Grum
    Commented Nov 24, 2015 at 8:50
  • \$\begingroup\$ Do you have a reference to support this assertion? I offer clarkvision.com/articles/digital.sensor.performance.summary \$\endgroup\$
    – Neil P
    Commented Dec 9, 2015 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.


Larger sensors are generally slightly worse in low light for capturing an image. Larger lenses are generally available for larger sensors, and larger lenses are generally better in low light if you don't mind the reduced depth-of-field.

  • 1
    \$\begingroup\$ Hi QuietOC. Welcome to Photo.SE. I hope you're enjoying the site. I was wondering if, perhaps, you might have mangled something in your answer. It doesn't really make much sense to me as it sounds like you are saying large sensors are worse in low light and then saying that larger sensors have larger lenses which are better in low light. Could you clarify what you are trying to say? \$\endgroup\$
    – AJ Henderson
    Commented Aug 19, 2017 at 16:49

There is a lot on the internet claiming that the amount of light gathered by a sensor is proportional to the size of the sensor. This is incorrect. Given the same angle of view of the lens, the same amount of light will be projected onto the sensor irrespective of the size of the sensor. If a full-frame sensor and a MFT sensor have the same number of pixel elements, then each element will detect the same amount of light, irrespective of their size. Think of this: put a piece of paper in the sun behind a circle of glass—nothing happens. Concentrate the light onto a small area of that paper with a magnifying glass of the same diameter as the aforementioned circle of glass and the paper will heat up because the energy density at the area of focus is so much greater. The same is true of image sensors; small sensor = higher energy density than large sensor = same energy per unit area on both sensors. The reason for greater noise on smaller sensors lies elsewhere; perhaps in radio frequency interference between closely packed image sensing elements.

  • 1
    \$\begingroup\$ I think you need to take your thinking a step farther. Same energy per unit area, yes — but the large sensor has more overall area. Larger sensors don't have more light per area, but for the same framing, more overall light is collected. \$\endgroup\$
    – mattdm
    Commented Jul 21, 2019 at 18:03
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    \$\begingroup\$ Another way to think of it: when we enlarge, either optically or digitally, we keep the exposure constant, right? We expect a 12×18 print to have the same brightness and apparent exposure as a 4×6 print. But, in order do this, we need to keep unit-area brightness the same even as the print is further enlarged. So, the larger print has more "added" light. If you start with a larger original, you have to multiply less — so, less apparent noise (or for that matter, film grain). \$\endgroup\$
    – mattdm
    Commented Jul 21, 2019 at 18:12
  • \$\begingroup\$ Thank you. I have been searching for a forum that presents a sensible view of sensor size and resolution. For sensible read "agrees with me" . Now let me add my own comment. In essence if the photon density from the same scene hits a sensor both the large and small device receive the same number of photons. It may be that the smaller sensor has a better signal to noise at that point because of its lower dynamic range. The availabile dynamic range being better optimised. Larger sensors with bigger photosites given the correct illumination conditions can collect more photons because of their w \$\endgroup\$ Commented Sep 20, 2019 at 8:41
  • 1
    \$\begingroup\$ Full well capacity is important. Given the same level of technology, larger pixels have higher full well capacities. This gives them higher dynamic range which allows them to be exposed "brighter" without blowing the highlights. In the process, the shadows are also exposed brighter and thus can have less read/thermal noise as well. \$\endgroup\$
    – Michael C
    Commented Mar 11, 2020 at 1:38

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