sensor size - myth vs marketing

Started Mar 18, 2012 | Discussions thread
Great Bustard
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Here's a little more than you wanted to hear.
In reply to jubilatu, Mar 18, 2012

jubilatu wrote:

Recently i read an article regarding the Olympus OMD where the tester said that the little MFT sensor "completely surpass" the 5d2 at ISO 1600 and above (noise, DR, etc).

So, if a sensor 4 time smaller could be better than the bigger one, than also the nikon V1 sensor (for example), could be better than MFT. and if MFT is better than FF, than the nikon V1 is better than canon 5d2, isn't it ?

how much is real physics and how much is pure marketing?

We have to begin with a few basics on noise and DR first. The dominant factors in the noise in a photo are:

  1. The total amount of light that falls on the sensor: Total Light = Exposure x Sensor Area

  2. The proportion of that light recorded by the sensor (QE -- Quantum Efficiency)

  3. The additional noise added by the sensor and supporting hardware (Read Noise)

The photon noise is determined by Total Light and QE. The total noise (N) is the quadrature sum of the photon noise (P) and read noise (R): N = sqrt (P² + R²). The apparent noise in the photo is the ratio of the noise to signal in the photo: NSR = N / S.

The read noise is dominant in the shadows, but the photon noise is dominant elsewhere in the photo.

The DR (Dynamic Range) is the range of light levels where detail can be recorded (the number of stops from the noise floor to the saturation limit).

This site:

http://www.sensorgen.info/

has a list of QE, read noise, and DR figures for various sensors. The DR figures for the sensorgen site are pixel-level measures using the read noise for the noise floor. DxOMark uses the 100% NSR (Noise to Signal Ratio) for the noise floor -- the signal where the read noise and photon noise are the same. In addition, DxOMark gives two measures for DR:

  1. Screen -- the DR / pixel

  2. Print -- the DR / area, where the area is one-eight millionth of a photo

Note that two systems with the same DR can have rather different looks to them, because the DR does not account for the photon noise in the photo, but I will skip the specifics for now -- this is already probably more than you wanted to know.

OK, now we can address your question. For a given exposure, four times as much light will fall on a FF sensor than a 4/3 sensor. If the FF sensor and mFT sensor are equally efficient (same QE and read noise), then the FF sensor will have significantly less apparent noise because the signal is four times larger.

In order to trump that, the mFT sensor would need to have four times the QE (impossible, since the max possible QE is 100%, and the 5D2 has a QE of 33%) or significantly less read noise.

In low light, that would be quite a challenge given that the 5D2 has only 3.2 electrons of read noise at ISO 3200. However, at base ISO, the read noise of the 5D2 is quite high, at 28 electrons. On the other hand, the pixel saturation is also quite high at 64600 electrons, which cancels out the high read noise to large extent, leaving the 5D2 with 11.2 stops of DR at base ISO.

So, unless the pixels for the mFT sensor had proportionallyl less read noise, but proportionally the same (or higher) saturation limits, then neither will the DR nor the noise performance be any better.

In other words, it's looking pretty unlikely for this feat to be accomplished unless, of course, there is some very advanced NR (noise reduction) going on in the processing of the mFT photo that is not going on in the processing of the 5D2 photo.

However, I don't see a DR of 12 stops vs the 11.2 stops of the 5D2 as being "reasonable", as well as closing the noise gap by 2/3 of a stop with a sensor that sported a 50% QE.

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