Who can explain the relationship.....

Discussion in 'Casual Photo Conversations' started by fotolopithecus, Mar 4, 2018.

  1. A bit OT, but, If I have a sensor A that is 4 times the area of sensor B, will camera A have 4 times the light gathering power of camera B or twice the power?

    I want to compare the Oly tg-5 sensor size ~ 28mm sq with Rx0 sensor ~ 117 mm sq at f/2 and f/4 respectively. the oly is12mp, the sony 15mp. ignore differences due to other factors

    edit: both are ~ 24mm (35mm equiv) focal lengths at their widest
  2. Sensor A will have 4 times the light gathering power. But from your point of view there won't be any speed advantage as the extra-gathering effect goes to areas outside that of the smaller sensor.

    If this doesn't make sense, consider the same situation with a film camera. Imagine that it is a native 4x5" camera, but you have a roll-film back as well. The 4x5 films have (very roughly) 4 times the area of the roll film, so they can actually gather roughly 4 times the light. But obviously you don't change exposure settings, right?

    When you compare two different sized sensors, but make the fields of view match by using the appropriate focal lengths, the lens f-number is the great equaliser. At the same f-number, any focal length lens will deliver the same light energy to a given unit area on the sensor. (The physical aperture sizes will vary relative to the focal length so as to cancel out the image magnification effect.)
  3. If by "light gathering power" you mean baseline sensitivity, then the relative area of the light cell would be the determining factor, not the total size of the sensor. There's not a direct relationship because the net sensitivity is affected by signal processing. Some designers trade sensitivity for low noise and bit depth, medium format sensors, for example. The cell size is not necessarily proportional to the sensor size, since the percent coverage can vary widely, from about 75%to over 85%. Backlit sensors have close to 100% coverage. The cell diameter is often listed in the camera's specifications. Another factor in cell design is it's capacity for electrons (triggered by photons). In order to achieve 16 bit depth, the cell would have to hold at least 65,536 electrons. I've never seen that in specifications, but MF digital often cites 16 bit depth. Without careful noise control, those low bits would be lost in the mud.

    I know that my digital Hasselblad can pull details out of shadows that were completely dark to my eye when I took the shot. The sensitivity is only ISO 50 to 400. The cells are 9 microns in diameter, A Sony A7Sii, 12 MP, has cells approximately 8 microns in diameter, and delivers ISO 100 to 409,600 (200,000 fairly cleanly). Choices!

    Are the Oly and RX0 competing for your dollars? You can attach the RX0 to a mic stand and run it remotely from a smart phone, or on you hat when skiing down hill. I don't think MF would be appropriate in either case. MF would be better for product photography and portraits.
  4. they were. i’ve just looked at this site

    Compare digital camera sensor sizes: 1″-Type, 4/3, APS-C, full frame 35mm

    and it suggests the oly has a crop factor of 6 whereas the sony has a crop factor of 2.7 which suggest they are similar (12 vs 10.8) so i’ll keep my oly. but that sony is so cute.
  5. Thanks, Bill & Ed, for your contributions
  6. You're quite welcome, if it was actually any help. I only spoke of "light collection power," disregarding pixel size. Larger pixels do have more surface area, so can collect proportionally more photons, etc.
  7. I'm sorry, but you said the Oly had a square, 28 mm sensor, which would put it in the far low end of medium format. In fact, it is a 1/2.3" sensor with a 5.5x cropping factor. The RX0 has a 1" sensor with a 3.1x cropping factor. Both sizes reflect an archaic video designation (related to the diameter of a vacuum tub video sensor). 4/3 is also misleading, but in fact is a half-frame camera (16x24 mm) with a crop factor of 2.0. The Olympus has a BSI (back illuminated) sensor, which is more efficient than the Sony, which has a conventional CMOS sensor.
  8. i didn’t. i said it had a 28mm sq sensor meaning 28mm squared as I don’t know how to write “to the power of 2” on the internet :)
  9. I looked this up to see if there were any keyboard shortcuts to accomplish this. I could only find things to do if one is in a specific word processing program like Word. But I did discover something new to me. Turns out, accepted convention among mathematicians and scientists (in other words, PN photographers:rolleyes:) is to use the caret symbol to indicate an exponent, to wit:


    . . . would indicate 28mm squared.
  10. I understand. The Olympus sensor has an AREA of 28 mm^2, or about 4.3 mm x 6.5 mm and 12 MP. That corresponds to a 1/2.3" sensor. Each pixel is approximately 2.3 microns^2. The Sony has a 1.0" sensor, which has an area of 90 mm^2 and 21 MP. Each pixel has an area of 4.2 microns^2. Even considering the benefit of a BSI sensor, the Olympus would have only a little over half the baseline sensitivity of the Sony. All else being equal, the same light exposure (intensity x time) would release nearly twice as many electrons in the Sony, compared to the Olympus. Of course, all else is NEVER equal.

    A sensor cell is basically a capacitor in which the charge is generated by exposure to light. The voltage (or incremental voltage) is equal to the charge (coulombs) divided by the capacitance (farads). Measuring that tiny voltage is where the fun begins.
    Norman likes this.
  11. Thanks, Ed. I appreciate the input.

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