Jack DuMoulin
Forum Enthusiast
In a recent thread several folks took exception to Michael Reichman’s claim that the highlights are represented by more data than the shadow areas and that in order to maximize your exposure you should expose to the right of the histogram. Since it appeared that some posters were starting to roll up their sleeves I decided not to participate. I am simply not interested in arguments and flames. I would rather go for a walk with my 10D and take some photos.
MR and TK are correct. The objection appears to be based on the premise that Michael (and Thomas Knoll) as well as their followers assumed that a CCD or CMOS sensor is nonlinear. This is not true. CCDs and CMOS sensors are linear devices. So how does this support MR’s claim? Well, the issue is based on the way a light source relates to image illumination and the associated coding into digital data points. I’ll try and explain this in simple terms so that anyone can dissect and follow it.
Fact #1 Sensors are linear devices. In other words if you double the light input, the sensor voltage will double.
Fact #2 Image illumination varies with the square of the light source intensity. To put it in simple language if you double the intensity of the light source (or double the lens opening while holding shutter speed constant), the image illumination will increase four fold. We can restate this by saying that if we want to double the image illumination we will have to increase the aperture by √2 or 1.414. This is what our f stops are based on. They vary in steps of X1.4 and for every f stop increase or decrease the image illumination and by corollary the sensor voltage will be doubled or halved.
Anyone doubting fact #2 can easily verify it by soliciting any good technical photographic text.
So let’s see what happens if we look at a series of f stops and compare them with sensor voltage, the binary coded value of this voltage and the number of bits required to represent this value:
F Stop/Sensor voltage/Binary Data/Bits
2 5 4096/ 12
2.8/ 2.5/ 2048/ 11
4 1.25 1024/ 10
5.6/ .625/ 512/ 9
6.8/ .3125/ 256/ 8
As we see from the table for every reduction in f stop the voltage of the sensor is cut in half in keeping with our Fact #2. I arbitrarily picked 5V as the maximum value of the sensor representing the highlight clipping level. The actual voltage may vary but the principal is the same. We also see from the above that f 2 is represented by 4096 binary data points while f 6.8 contains only 256 data points. The table also illustrates that if you underexpose by one f stop you leave half your total data on the table, just as MR stated in his article. You are effectively only using 11 bits of the 12-bit A/D converter. That’s all there is to it. Let me hasten to add that there is absolutely nothing new about this; it has been understood and used by digital engineers for decades. I have designed some digital equipment that is based on the above principle, so if I am wrong, I am in deep trouble. However, since these controls have been in operation in several OSB plants thoughout North America for several years and are still opreating, I am not too worried.
))
I hope I explained it plainly enough for everyone to understand and perhaps we can put this issue to rest.
Jack
MR and TK are correct. The objection appears to be based on the premise that Michael (and Thomas Knoll) as well as their followers assumed that a CCD or CMOS sensor is nonlinear. This is not true. CCDs and CMOS sensors are linear devices. So how does this support MR’s claim? Well, the issue is based on the way a light source relates to image illumination and the associated coding into digital data points. I’ll try and explain this in simple terms so that anyone can dissect and follow it.
Fact #1 Sensors are linear devices. In other words if you double the light input, the sensor voltage will double.
Fact #2 Image illumination varies with the square of the light source intensity. To put it in simple language if you double the intensity of the light source (or double the lens opening while holding shutter speed constant), the image illumination will increase four fold. We can restate this by saying that if we want to double the image illumination we will have to increase the aperture by √2 or 1.414. This is what our f stops are based on. They vary in steps of X1.4 and for every f stop increase or decrease the image illumination and by corollary the sensor voltage will be doubled or halved.
Anyone doubting fact #2 can easily verify it by soliciting any good technical photographic text.
So let’s see what happens if we look at a series of f stops and compare them with sensor voltage, the binary coded value of this voltage and the number of bits required to represent this value:
F Stop/Sensor voltage/Binary Data/Bits
2 5 4096/ 12
2.8/ 2.5/ 2048/ 11
4 1.25 1024/ 10
5.6/ .625/ 512/ 9
6.8/ .3125/ 256/ 8
As we see from the table for every reduction in f stop the voltage of the sensor is cut in half in keeping with our Fact #2. I arbitrarily picked 5V as the maximum value of the sensor representing the highlight clipping level. The actual voltage may vary but the principal is the same. We also see from the above that f 2 is represented by 4096 binary data points while f 6.8 contains only 256 data points. The table also illustrates that if you underexpose by one f stop you leave half your total data on the table, just as MR stated in his article. You are effectively only using 11 bits of the 12-bit A/D converter. That’s all there is to it. Let me hasten to add that there is absolutely nothing new about this; it has been understood and used by digital engineers for decades. I have designed some digital equipment that is based on the above principle, so if I am wrong, I am in deep trouble. However, since these controls have been in operation in several OSB plants thoughout North America for several years and are still opreating, I am not too worried.
))I hope I explained it plainly enough for everyone to understand and perhaps we can put this issue to rest.
Jack
