# "2/3 inch", etc. sensor???

Discussion in 'Digital Photography' started by cube, Sep 2, 2006.

1. ### cube macrumors G5

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May 10, 2004
#1
Why are sensors called "2/3 inch", etc. if those are not any of their dimensions?

2. ### Chundles macrumors G4

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Jul 4, 2005
#2
So that we guys will buy bigger lenses to compensate.

3. ### oblomow macrumors 68020

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#3
http://hannemyr.com/photo/crop.html

4. ### cube thread starter macrumors G5

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May 10, 2004
#4
Thanks, but that article doesn't say why these sensors are called like that.

5. ### oblomow macrumors 68020

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#5
ok. you're not talking about the aspect ratio of the senors are you?
Normal(analog) camera's have a ratio of 2/3 (24*36mm). dSLR have the same
ratio. P&S camera's generally have a 3/4 ratio.http://en.wikipedia.org/wiki/Four_Thirds_System

6. ### cube thread starter macrumors G5

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May 10, 2004
#6
No, I'm not talking about the aspect ratio. I'm talking about sensors called "1/1.8 inch" and "2/3 inch".

7. ### iMeowbot macrumors G3

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Aug 30, 2003
#7
It's not even that. The sensors are specified in different formats, 1/2-inch, 1/3, 1/5 etc. Nominally this is supposed to be the diagonal measurement of the active imaging area, but in real life the active areas are much smaller than that. The number is really to help manufacturers match up lenses with sensors. There is a little table at Micron with typical nominal and actual sizes.

8. ### cube thread starter macrumors G5

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May 10, 2004
#8
The Micron wording is a bit confusing, but if I understand well they definitely mean that this measure is the size of the image (diameter?) produced by the corresponding lens.

But where does the image end?

9. ### iMeowbot macrumors G3

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Aug 30, 2003
#9
The image ends where it ends The effect is what you are probably thinking: the lenses are designed to cover more area than the sensors will see, so some of the potential angle is wasted.

There are painful little formulas and calculators around the net to figure out what you will actually get.

10. ### oblomow macrumors 68020

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#10
OK, now I understand the original question. And have learned something too!
( I could have asked somebody at work, it work at a company that makes the machines for Micron...)

11. ### jared_kipe macrumors 68030

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#11
I suspect it is one of the dimensions, probably the diagonal measurement.

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May 10, 2004
13. ### jared_kipe macrumors 68030

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#13
Where are you getting your information?

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May 10, 2004
15. ### jared_kipe macrumors 68030

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Dec 8, 2003
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Seattle
#15
Why are you CCD specs from a manufacturer. Maybe it is the sensor package, the part that needs to be soldered to a board.

But everything for 11mm diagonal always says (2/3 type) wich I would take to be the aspect ratio.

16. ### cube thread starter macrumors G5

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May 10, 2004
#16
It's not just in CCD specs. It appears in many photo sites.

And many of those google results say "2/3 inch".

Here you have one photo site.

And listing the diagonal of a package would be useless. What people need to know is the width and height.

17. ### extraextra macrumors 68000

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California
#17
Does this help?

Or has the answer already been found? I got lost somewhere in the middle of the thread.

18. ### hayduke macrumors 65816

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#18
19. ### MobiusStrip macrumors regular

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Dec 11, 2009
#19
That link just goes to a general area.

Here's the real explanation; it's more outdated crap bogging us down (like interlace and 29.97 FPS):

"Sensors are often referred to with a "type" designation using imperial fractions such as 1/1.8" or 2/3" which are larger than the actual sensor diameters. The type designation harks back to a set of standard sizes given to TV camera tubes in the 50's. These sizes were typically 1/2", 2/3" etc. The size designation does not define the diagonal of the sensor area but rather the outer diameter of the long glass envelope of the tube. Engineers soon discovered that for various reasons the usable area of this imaging plane was approximately two thirds of the designated size. This designation has clearly stuck (although it should have been thrown out long ago). There appears to be no specific mathematical relationship between the diameter of the imaging circle and the sensor size, although it is always roughly two thirds."

20. ### jackerin macrumors 6502a

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Finland
#20
This thread is from 4 years ago. In case you didn't notice.