SURVEY - Do FT / mFT users know the difference from "full frame"? Replies wanted!!

Started Apr 26, 2013 | Discussions thread
Great Bustard Forum Pro • Posts: 41,751

JosephScha wrote:

Great Bustard wrote:

JosephScha wrote:

yes, I do know. I used to actually join in equivalence discussion but gave up when told that despite equal light intensity at the sensor f stops were not equivalent on m43 because total light was less... so wrong.

It's a shame that you can't understand that f/2.8 on mFT puts the same amount of light on the sensor as f/5.6 on FF for a given shutter speed, which, in turn, results in the same noise for equally efficient sensors.

Let's see.  f/2.8 on mFT puts the same amount of light on a 1/4 ff size sensor, which means the light per unit area is quadrupled, compared to f/5.6 on FF.  I get that.  What I don't get is why you and many others care about total light instead of light per unit area.  I read your post "DPR mans up and nails it".

Apologies for not having seen your post until now, which is ironic, 'cause I responded at the tail of the subthread (I often read the tail end of subthreads and skip the posts prior).

To answer your question, the photon noise, which is the dominant source of noise in a photo except for the deep shadows, is a function of the total amount of light that makes up the photo.

A 12MP 4/3 sensor is 4000x3000 light sensitive sights.  If I remember correctly, they are about 4 micro inches square.  An "equally efficient" full frame sensor with the same light sensitive element size, if it's 4x larger, would be about 48MP. But no one makes that.  FF cameras have larger sensor sights, I claim THAT is why they have better dynamic range: because they get more light per sensor element.   More light in total over the sensor I claim does not matter.

By "equally efficient" I mean:

  • Same QE
  • Same read noise / µphoto

The QE is the Quantum Efficiency of the sensor -- the proportion of light falling on the sensor that is recorded.  For example, most modern sensors have a QE of around 50%, which means they record half the light falling on them.

The read noise is the additional noise added by the sensor and supporting hardware.  The "/ µphoto" is "per microphoto", where a microphoto is one-millionth of the photo.  So, for example, we don't compare the read noise of one pixel of a 36 MP sensor to one pixel of a 16 MP sensor, but the aggregate read noise of 36 pixels from a 36 MP sensor to 16 pixels from a 16 MP sensor, as they cover the same are of the photo.

As a thought experiment, let's assume that Nikon goes nuts and decides to make a 256MP FF sensor.  The photo sites would be tiny - much smaller than m43. Even at the same f/stop, there is less light falling on each of those small photo sites than on a 16MP m43 sensor's photo slites.   Would you still claim that that FF sensor will have better dynamic range than an m43 sensor because in total it receives more light?  Or will it have less because each photo sensor receives far less light?

Now we'd be comparing 256 of those pixels from the 256 MP sensor to 16 pixels on the 16 MP sensor.  QE doesn't seem to be affected by pixel size at all.  The issue, then, is read noise, which can vary considerably from sensor to sensor, regardless of sensor size or pixel count, as well as the ISO setting.

Obviously, I think the latter (less dynamic range because less light at each photo site).  I suppose in software you could average pixels together, to compensate ... but why would you have to do that if it's total light on the sensor that matters?  I think it's light on each photo site that matters.  I think it is just too simple to say f/2.8 on mFT is "equaivalent" for noise to f/5.6 on FF for a given shutter speed.

We compare the IQ of photos area for area, not pixel for pixel.

But what REALLY bothers me about saying f/2.8 on mFT is "equivalent" to f/5.6 on FF is that is misleads too many people who somehow think this talks about exposure.  I think the more important point is that at the same ISO, and the same shutter speed, f/2.8 on mFT gives the same exposure as f/2.8 on any other sized sensor.  This does not consider dynamic range effects.

The problem is that exposure is irrelevant in cross-format comparisons -- it's the total amount of light that made up the photo that matters.

I used to have a pany FZ7 with a 1/2.5" sensor that had pretty poor dynamic range and egregious noise at is 200.

That's because so little light fell on the sensor compared to larger sensor systems for a given exposure.

Well... you must know I think it's because very little light fell on each of the very tiny photo sites on that 6MP teeny sensor.  Given the same total light, I'll bet a 1MP sensor (which wouldn't have been a sellable product) would have had significantly better dynamic range, because each photo site would collect a lot more "signal" (photons).  Dynamic range is basically signal to noise ratio.

You would lose that bet.  For example, which camera has a greater DR, the D800 or the D600?

I was looking for a similar size and weight camera with better image quality and m43 fit my needs. ...

Makes perfect sense.


I honestly do look forward to your reply.  I want to understand why total light on the sensor means more than light per photo site, which if photo sites are the same size on the sensors being compared means we can compare light per unit area.

Read this:

It answers all of that.

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