Third Party Manufacturers / Re: Dynamic Range - Try it for yourself, conclude for yourself: 5D III vs. A7r« on: October 08, 2014, 12:02:14 PM »
The reason ISO 100 on a Canon has roughly the same DR as ISO 400 is because the read noise drops by about a stop with each progressive increase to ISO. Hence the flattening of the DR curve on Canon sensors.
Well, it's not that the read noise drops, it's just that all other data & noise off the sensor is amplified 4x compared to ISO 100, which lowers the impact of the read noise. So even though you're throwing away 2 stops of highlight range at ISO 400, 4x (2 EV) darker signal is now being amplified to be brought up to your SNR = 1 (or what have you) threshold, leaving DR largely the same. Cameras with very little downstream read noise don't need the sensor signal amplified to overcome the 30+ electrons of read noise in a Canon camera.
I think there is more to it than that. There is something else going on that flattens the curve relative to read noise at lower ISO, because once your past ISO 400, then you get that linear fit that matches the ideal DR curve. ISO 100 on a 5D III has 33.1e- RN. That is noise added to the signal AFTER the pixels are read (well, after they are amplified and shipped off the sensor, during read). At ISO 200, the pixels are amplified, then read. Since the RN is added down stream, logically, you should still have 33.1e-...the signal was already amplified, so it's still the same strength at this point as an unamplified ISO 100 signal. Similarly, at ISO 400, the pixels are amplified even more, then read. Since RN is added down stream, logically, you should still have 33.1e-. The information coming off the sensor always effectively represents a signal with the same strength, since it's being amplified there before readout.
Your explanation doesn't really explain why read noise drops with higher ISO settings. I don't know exactly why it drops, ...
Yeah, it does.
Let's say, for sake of argument, that 1V = 10,000e- at ISO 100.
Let's say the analog noise is 10mv. 10mV * (10,000e-/1V) = 100e-
Now, let's say that 2V = 10,000e- at ISO 200.
We still have 10mv of analog noise. 10mV * (10,000e-/2V) = 50e-