From what I understand of digital cameras, they are basically a lens plus a tiny two-dimensional array of millions of photo-diodes. And from what I understand of photo-diodes, they create a voltage when in the light, with higher-intensity light immediately causing a higher voltages.

However, if this were all true, there would be no need for an exposure in digital cameras: the individual voltages could be read and (assuming our voltage-reader is sensitive enough and electrical noise is negligible) we'd get as accurate an image as possible almost instantly.

But, this is not what happens. So where is my understanding incorrect? And are there any digital cameras that work this way?

Sorry if this is a better fit for electronics.SE - but I felt like this question would be more interesting to this audience.

  • \$\begingroup\$ Sounds really boring(the idea of no exposure time, not the way you asked the question) :P \$\endgroup\$
    – dpollitt
    Commented Jan 10, 2013 at 3:53
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    \$\begingroup\$ "assuming our voltage-reader is sensitive enough and electrical noise is negligible" Those are pretty big assumptions to start off with... \$\endgroup\$
    – user
    Commented Jan 11, 2013 at 22:03

4 Answers 4


I'm visiting from Electronics, so I'll add a little bit of electronics/semiconductor physics background to a couple of the answers you've already gotten.

The key misunderstanding I think you have is that a photodiode doesn't create a voltage in response to light, it creates a current. Each photon that hits the photodiode generates a mobile electron inside the device (really an "electron-hole pair", but if you want that level of detail you'd better take the question over to EE.SE). Millions of electrons together constitute a measurable electrical current. Finally when this current is used to charge a capacitor, then you have a measurable voltage which can be sensed or recorded to form a pixel in your image.

This is why, as cmason says, the sensor needs some time to fill each "bucket", and as mattdm says, it takes time for an accumulator to fill to the point it can be measured to form an image.

  • \$\begingroup\$ I'm glad this answer got written, I originally wrote an answer like this, trying to explain the semiconductor physics but decided that I would probably not make it clear enough. \$\endgroup\$
    – Phil
    Commented Jan 11, 2013 at 16:45
  • \$\begingroup\$ @Phil, I think the final para of your answer, where you focus on having to wait for the actual photons to arrive, actually gets to the fundamental limit of why we need non-zero exposure time. I was just trying to hit the key misunderstanding I saw in the OP's premises about the question. \$\endgroup\$
    – The Photon
    Commented Jan 11, 2013 at 16:59
  • \$\begingroup\$ Just to make sure I understand: it's simply because the current that's created by such tiny photo-diodes is too small to instantaneously measure accurately/free of noise, so we have to add a capacitor so we can measure the total charge that flows over a period of time. Is this correct? \$\endgroup\$ Commented Jan 14, 2013 at 20:14
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    \$\begingroup\$ @BlueRaja-DannyPflughoeft, the fundamental reason is what Phil said --- you have to wait for enough photons to arrive to create a low noise image. It basically comes down to the same thing for electrons. On the electronics side, it's also true that our methods for measuring current generally require converting them to voltage first. \$\endgroup\$
    – The Photon
    Commented Jan 14, 2013 at 20:29

Digital cameras attempt to do exactly that, it is only because of noise that they do not. Such as camera could be described as having an arbitrarily high ISO, and consequently correct exposure would be obtained with an arbitrarily short shutter speed.

Making a low resolution large format back out of large photo diodes could be a fun project.

I also think that in the future 'multi exposure' systems will be integrated into sensors- record the sensor values mid exposure but keep the shutter open, to get more detail in the blacks.

The following is a rough calculation of the energy captured by a pixel of a modern DSLR during an exposure in room lighting:

Warren Mars's Photon Behavior site provides a table of the number of photons incident upon the pixels of various size under various lighting conditions for a 1/60th second exposure.

The smallest pixel listed in the chard it a 70µm² pixel, three times larger than those of the D7000; the D7000's imager has a pixel size of 4.78µm

Under 'living room light' this gives a value of about 110000 photons per pixel on a D7000.

A red photon has about 1.6*10E-19 J of energy. It can be seen that the energy per pixel is on the order of 10E-14 J. A very small amount of energy to measure indeed.

Pixel on Sensor

For more information (and source of image): http://www.gyes.eu/photo/sensor_pixel_sizes.htm

It should also be noted that fundamentally a zero second exposure camera is impossible, since it would allow no time for photons to hit the surface. Suppose we create a photon counting camera- that is one that can provide a 100% accurate zero noise count of the photons that hit each pixel. To get a 10 bit image the brightest pixels require that 1024 photons. In room lighting (Using the pixel pitch from the D7000) 2 million photons hit each pixel every second. Dividing the 2 million photons by the number of brightness levels (1024) we get a theoretical maximum frame rate of 1950 frames per second. 1/1950 would be the minimum possible exposure time for a 10bit image under room lighting.

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    \$\begingroup\$ Interestingly, on that last bit: at an aperture of f/1.4 and a shutter speed of ¹⁄₁₉₅₀, ISO 6400 should theoretically get you correct exposure at room lighting. There may not be 10 bits of real dynamic range in there, and it'll be pretty noisy, but I think real world cameras today are in the ballpark of your theoretical maximum. \$\endgroup\$
    – mattdm
    Commented Jan 10, 2013 at 19:54
  • \$\begingroup\$ Digital cameras attempt to do exactly that, it is only because of noise that they do not. - erm, this cannot be correct though. If it were, any light-level that caused voltages not near the noise-level could be read instantaneously; and any voltages around or below the noise level could not be read at all. "Exposing" the diodes for a short time to average the values could help when we're very slightly above the noise level, but in every other case, there would be no need for an exposure at all. \$\endgroup\$ Commented Jan 10, 2013 at 23:00
  • \$\begingroup\$ Also your last paragraph is incorrect; if photo-diodes measure the instantaneous amplitude of the light (which I believe they do), you would not need to leave the image "exposed" for any period of time at all - the values could simply be read instantly. I think you are confusing this for "exposing the diodes to light for 0 time," which is mixing up the way digital cameras work with the way analog cameras work. \$\endgroup\$ Commented Jan 10, 2013 at 23:02
  • \$\begingroup\$ There is a lot more to digital camera sensors that I did not go into here. What I was getting at is that most digital cameras attempt to take a correctly metered photo with as little exposure time as possible- thus increasing the resistance to camera shake and moving subjects. I analysed a low light situation to make it apparent that an exposure time of zero seconds makes no physical sense. \$\endgroup\$
    – Phil
    Commented Jan 10, 2013 at 23:30
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    \$\begingroup\$ I got some time this after noon and decided to research this topic more. I discovered that there are some inaccuracies in my theory and math. I do not have the time right now to fix them, and do not think they impact the ability of my answer to answer the question. Please do not take any of the numbers presented as repeatable fact at this time. I do not want to spread any misinformation. \$\endgroup\$
    – Phil
    Commented Jan 11, 2013 at 7:14

The brighter light immediately causes a higher voltage, but not hugely higher. That's the crucial part. If you want to have an image that looks like the eye expects it to, you either need to amplify the signal (increasing the differences between high and low, both correct and incorrect due to noise) or you need to read for longer, increasing the actual sample. The latter is what the sensors used in digital cameras do.

Each photosite is not just a light-sensitive photodiode, but also contains an accumulator called a "well". As the photodiode continues to produce voltage (as it is exposed to light), the accumulator fills. If the light hitting a particular site is bright, that well fills quickly. If the light is dim, it fills slowly. When the exposure is finished, the level of the well is sampled and converted to a digital value.

Of course, in bright light, there's a lot of data, so a short exposure paints an accurate picture (if you'll pardon the turn of phrase). In low light, though, there's just not much energy to measured. If you just take a quick sampling, noise from reading the sensor and other unavoidable real-world randomness will indroduce variation as strong as the "legitimate" difference between the more full and more empty photosites, and there's no way to tell which is which.

This is what happens when you take an underexposed image and try to crank up the amplification in software: noise, noise, noise, and maybe just blackness. And any instantaneous read (without an accumulator well) would not have enough data to be useful.

Simple as that, really. Turns out that modern sensors are better at this than chemical-process film: it's why we can have seemingly insane ISO values of 25k and above. Those are able to measure finely enough that a large amount of amplification can be applied without noise becoming overwhelming. Fundamentally, though, compared to the magical instant-read device, we're still in the same ballpark.

  • \$\begingroup\$ I don't believe this is correct, please see my first comment above. \$\endgroup\$ Commented Jan 10, 2013 at 23:04
  • \$\begingroup\$ Which part do you not believe to be correct? (I'm not even sure what you mean by "above", since answers can be viewed in different order.) \$\endgroup\$
    – mattdm
    Commented Jan 10, 2013 at 23:43

The simplest answer is that light is particle based, consisting of photons. A digital sensor is not a single photon trigger, but a bucket to be measured. I believe this is where you are confused: a sensor is not binary, nor are they sensitive to a single photon: a photon does not 'turn on' the sensor photo site. Instead, what is measured is how full the bucket is. Enough time must be given to properly fill the bucket, or no image will be recorded.

Brighter scenes emit more and higher energy photons, thus filling the bucket quicker. Overfilling the bucket overexposes the image, losing detail or 'washing out' the image. To prevent this washing out, you simply shorten the time you collect photons.

  • \$\begingroup\$ As I understand it, photo-diodes do not work as you described. Instead, they measure the amplitude of the light (or, if you prefer the particle-explanation, the rate at which photons are entering) instantaneously. If your bucket-version were correct, that would explain the need for an exposure; but unfortunately, I don't believe it is. \$\endgroup\$ Commented Jan 10, 2013 at 23:05
  • \$\begingroup\$ I tried to spell this out more clearly in my answer; a photosite is not just a photodiode. \$\endgroup\$
    – mattdm
    Commented Jan 11, 2013 at 0:22
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    \$\begingroup\$ @BlueRaja-DannyPflughoeft - Believe it or not (as you desire); belief has little to do with it. There is no "light pressure", only a photon rate. (There is also a photon energy, but that's determined by wavelength, not intensity.) That rate is probabilistic, so the minimum case (measuring the time between two successive photon interactions at a site) is not a statistically valid measurement of the average rate of photon emission. Indirect measurement (potential electrical energy caused by the interaction of photons with the sensor) relies on accumulation. \$\endgroup\$
    – user2719
    Commented Jan 11, 2013 at 0:32

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