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Let's consider a CMOS sensor with N number of pixels. For the sake of simplicity, let's consider an hypothetical Black and White sensor. Each pixel catches the energy provided by the input luminance seen through entrance pupil of the lens. It generates a charge (Coulomb) per pixel area which is proportional to the input luminance. This charge is read through an external circuitry and eventually amplified. The output signal is a voltage or current which is proportional to such a charge and hence to the input luminance.

So, the image signal for each pixel is a current (or a voltage) proportional to the input luminance. Now, how is such a signal displayed?

More Current provided by a certain pixel --> More white on the screen on that pixel? If it works in that way, the perceived luminance of an image will depend on the pixel area

More Current provided by a certain pixel normalized to the pixel area --> More white on the screen on that pixel? If it works in that way, bigger pixels does not imply brighter images

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    \$\begingroup\$ I have to ask... what photographic problem are you trying to solve? This sounds like an electrical engineering question to me. IMHO one of the negative aspects of this site is how it attracts/tolerates engineering questions, losing focus on actual photography. \$\endgroup\$
    – osullic
    Nov 25, 2022 at 23:11
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    \$\begingroup\$ @osullic my photographic problem is: will a big sensor provide higher exposure with same iso, shutter speed and F number? \$\endgroup\$
    – Kinka-Byo
    Nov 26, 2022 at 1:41
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    \$\begingroup\$ OK, so, ask that. \$\endgroup\$
    – osullic
    Nov 26, 2022 at 11:37

2 Answers 2

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More Current provided by a certain pixel --> More white on the screen on that pixel? If it works in that way, the perceived luminance of an image will depend on the pixel area

More Current provided by a certain pixel normalized to the pixel area --> More white on the screen on that pixel? If it works in that way, bigger pixels does not imply brighter images

Both are essentially correct... everything related to exposure is light/image area and not really light/photosite. I.e. pixels do not stand alone; they combine to create an image. When receiving light, the size of the pixel affects both how much light it receives, and also how much image area it represents (i.e. self normalizing).

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You are ommitting the analog to digital conversion. A fully saturated pixel, no matter the absolute charge, will be assigned the value 255 (or whatever the highest value is according to bit depth). No charge will be assigned 0, and so forth for values between.

The display is from the digital image (after a lot of processing), so 255 -> white. As a simplified explanation.

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  • \$\begingroup\$ I'd reckon that most of the cameras people talk about in this forum capture more than eight bits of intensity. I'm guessing that 12 bits or 14 bits are fairly typical (But, I don't follow the cutting-edge products, so my information may be out of date.) OTOH, If you let the camera process the image internally and save the result in an 8-bit file format instead of saving RAW,... That's a whole 'nother topic. \$\endgroup\$
    – Solomon Slow
    Nov 28, 2022 at 21:34

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