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.
Answer
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.
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