What the heck is an f-stop?

Started 8 months ago | Discussions thread
Great Bustard
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Sensor size, f-stop, exposure, total light, and noise.
In reply to Paul Anderegg, 8 months ago

Paul Anderegg wrote:

So fast forward to today, and you've got mostly f10 @ 2000 HD gear, with a few more expensive f11 and maybe 2 or 3 at f12. Which brings me to my inquiry. I am trying to figure out which of the following two cameras would be best suited for night and low light use. A JVC GY-HM890 (f11@2000 1/3" chips) with a constant f1.4 lens, or a Panasonic HPX600 (f12@2000 2/3" chip) with an f1.8/wide-f2.3/tele lens.

First, some important background to answering your question.

There are two primary forms of noise in a photo: luminance noise, which is a function of the total amount of light falling on the sensor (photon noise) and the sensor efficiency, and the chroma noise, which is a function of the CFA (Color Filter Array).

The total amount of light falling on the sensor is the product of the exposure and area of the sensor. For example, if you quadruple the sensor area then, for a given exposure, 4x as much light will fall on the sensor, and the photon noise will be half as much.

The sensor efficiency is a function of the QE and the read noise. The QE is the Quantum Efficiency, which is the proportion of light falling on the sensor that gets recorded. For example, if a sensor has a QE of 0.5, then half of the light falling on it gets recorded. The read noise is the additional noise added by the sensor and supporting hardware, and is usually small compared to the photon noise except for shadows at low ISO and progressively more and more of the photo as the light lessens and we raise the ISO to compensate for the lower brightness.

OK, let's say we have two sensors, one 1/3" and the other 2/3".  Let's also assume the 2/3" sensor is twice the size (4x the area).

For a given lens and exposure (say, f/1.4 1/30), 4x as much light will fall on the 2/3" sensor as the 1/3" sensor, which will result in half the noise if the sensors were equally efficient.  If we instead used f/2 (half the aperture diameter and thus 1/4 the aperture area) 1/30 for the 2/3" sensor (doubling the ISO to compensate for the lower brightness), then the same total amount of light would fall on each sensor, resulting in the same noise if both sensors were equally efficient.

So, an f/1.4 lens on a 1/3 sensor is equivalent to an f/2 lens on a 2/3" sensor, as it will project the same total amount of light on the sensor for a given shutter speed (given equal transmission), thus resulting in the same noise for equally efficiency sensors.

However, the transmission is not necessarily the same for both lenses wide open.  The f/1.4 lens might have a transmission of 70% (1/2 of a stop light lost) and the f/2 lens might have a transmission of 90% (1/6 of a stop light lost).  Thus, 1/3 of a stop more light would fall on the 2/3" sensor than the 1/3" sensor.

In addition, the efficiency of the microlens covering of the sensor might result in additional light losses reaching the photosensitive portion of the sensor, which will be more pronounced at f/1.4 than it will at f/2 (despite the sensors being otherwise equally efficient), so there could be, say, an additional 1/3 stop more light falling on the 2/3" sensor, resulting in a total gain of 2/3 of a stop more light being recorded on the 2/3 sensor.

So, as we can see, the specifics of the transmissivity of the lens, the efficiency of the microlens covering also play important roles at low f-ratios (the microlens issue quickly dissipates after f/2, however).

Hope this helps!

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