CCD exact size?

[Stephen Birdsong]

Does anyone know the exact dimensions of the xl2 ccd? I'm not looking for the answer, the xl2 has 1/3 inch chips, but its exact length/ width. Does it differ from the xl1s, even by a millimeter or so?

stephen

[Luis Caffesse]

No it doesn't differ from the XL1s, but the image area does.

Then again, the size of the CCD itself isn't the only factor.
The chips in the XL2 have significantly more pixels.

If you do a search on this, there was quite a bit of discussion about this when the XL2 first came out.

[Chris Hurd]

What I can tell you is that the printed specs in the Canon XL2 brochure are incorrect. Their brochure states a diagonal measurement of 0.289 inches for the 16:9 target area and a diagonal measurement of 0.236 inches for the 4:3 target area. Both of those numbers are in error, though.

If those measurements were true, it would imply that the full CCD surface has a diagonal measurement of 0.333 inches. However, one-third inch CCD's do NOT actually measure one-third of an inch (or 0.333 inches). They actually measure somewhat smaller than one-third of an inch. The reason for this goes back to the days before charge-coupled devices (CCDs) were employed in video cameras.

Video cameras used to have either orthocon, plumbicon or saticon imaging tubes, and the diameter of the tube was stated as the size of the camera... either two-thirds inch, one-half inch, one-third inch, etc. Of course the actual imaging area was smaller than that... it was a 4:3 rectangular shape to fit *inside* the circular diameter of the tube. When the CCD was first used instead of a tube, its surface area was only as big as that previous 4:3 area (to preserve the properties of existing lenses such as focal length, etc.). In other words, the terms "one-third inch, one half-inch" are holdovers from the tube days. The CCD's are actually smaller than that. Somewhere around here I've got a bookmark or two which references the actual diagonal measurement for a one-third inch 4:3 CCD, I'll have to dig them up, but it is definitely smaller than 0.333 inches.

And Luis is right, the overall *size* of the CCDs are identical between the XL1, XL1S and XL2, but the XL2 has different target areas for 16:9 and 4:3 that are smaller than the full surface area of the sensor. See my XL2 CCD article at http://www.dvinfo.net/canonxl2/articles/article06.php for more info. Hope this helps,

[Luis Caffesse]

Thanks for the concise response Chris,
I was just trying to keep it simple without starting another,
"what do you mean my 1/3" chip isn't 1/3"??" thread.

:)


By the way, I believe the diagonal of a 1/3" chips is actually .236 inches. Which, oddly enough, is the measurement Canon gives for their 4:3 target area (which makes all of this more confusing for people).



By the way Stephen,
Here is a link to the thread I was remembering.
This topic was covered in quite a bit of detail there.

[Chris Hurd]

Thanks Luis,

That link is the all-important discussion we had on this exact topic last year.

By the way, here's an excellent overview of CCD technology from Nikon:

http://www.microscopyu.com/articles/.../ccdintro.html

To partially quote:

"The rectangular geometry and common dimensions of CCDs result from their early competition with vidicon tube cameras, which required the solid-state sensors to produce an electronic signal output that conformed to the prevailing video standards at the time. Note that the "inch" designations do not correspond directly to any of the CCD dimensions, but represent the size of the rectangular area scanned in the corresponding round vidicon tube. A designated "1-inch" CCD has a diagonal of 16 millimeters and sensor dimensions of 9.6 x 12.8 millimeters, derived from the scanned area of a 1-inch vidicon tube with a 25.4-millimeter outside diameter and an input window approximately 18 millimeters in diameter. Unfortunately, this confusing nomenclature has persisted, often used in reference to CCD "type" rather than size, and even includes sensors classified by a combination of fractional and decimal terms, such as the widely used 1/1.8-inch CCD that is intermediate in size between 1/2-inch and 2/3-inch devices."

[Boyd Ostroff]

<<<-- Originally posted by Chris Hurd : Somewhere around here I've got a bookmark or two which references the actual diagonal measurement for a one-third inch 4:3 CCD -->>>

Perhaps it's this? http://www.dpreview.com/news/0210/02...ensorsizes.asp

[Stephen Birdsong]

so, in short, the exact measurements are a mystery, known only to a few canon engineers?

[Chris Hurd]

I suppose it's on a "need to know" basis.

I mean, if you had that information, what could you do with it?

How would knowing that affect usability? For example I don't know what the piston ring clearances are in the cylinder walls of my GMC Sierra's engine, but that doesn't affect how I drive it... see what I mean?

By the way -- great link, Boyd -- thanks for that!

[Stephen Birdsong]

actually, its because I'm designing a relay lens for a diy 35mm adapter, and I'm about to upgrade to an xl2. I need to know that the adapter I build with the xl1s is going to work flawlessy with the xl2, and secondly, I don't even know the specs of the xl1s. Since there are engineers doing the machining for me, this is a very important piece of information. I'm upgrading, so I'm going to build with the xl2 specs in mind. If its information that is not available, then I guess I'll have to deal with estimating it. Chris, you seem to be rather knowledgable about this, what would you say?

stephen

[Chris Hurd]

Ah so. Well then you do need to know, it seems!

For what it's worth, a one-third inch CCD measures 6.000mm on the diagonal; 4.800mm on the width and 3.600mm on the height. And this would cover the entire CCD surface area. The effective image area is just slightly smaller than that for an XL1 or XL1S. For the XL2, the 16:9 target area is smaller than that; and the 4:3 target area is smaller still. I think we could probably figure it out though. Hope this gives you a starting point,

[A. J. deLange]

You could measure it. Put a 35 mm prime on the XL2 and then move the camera closer to or farther from an object of known size until it completely fills the frame in the vertical direction (don't use the viewfinder to determine when the frame is full). Then measure the distance from the lens front to the object. The height of the image is approximately equal to the size of the object times the focal length divided by the distance to the object. This is approximate because you don't really know what the focal length of the lens is nor exactly where to measure to on the lens.

A more accurate result could be obtained by putting the lens on a 35 mm camera, posititioning the camera so that the lens is exactly the same distance from the object as it was when on the XL2 and then taking a picture. The height of the image on the developed film will then be the same as its height on the XL2 CCD when its frame was filled. If you have a digital camera whose sensor size is known you can measure the digital image size as a fraction of the vertical extent of the image and multiply by the sensor height to get the actual image height in the XL2.

[Stephen Birdsong]

hrm.
(still trying to wrap my head around that)

[A. J. deLange]

A particular lens will form an image of a size dependent on its effective focal length and distance from the object regardless of what camera it is mounted on so if you can determine what distance it takes for a particular lens to fill the screen vertically on the XL2 (on which you can't measure the image size) and then switch cameras (keeping distance to the object the same) to one for which you can measure the image size that measurement will give you the image size on the XL2.

If I put a 35 mm Nikkor on my XL2 I have to back off until the front lens flange is 5325.5 mm from a Putora chart to have the points on the chart touch the upper and lower boundaries of the frame. If I take off the XL2 and put on a Nikon D1x keeping the distance of the front flange of the lens the same distance from the chart the points measure 1.415 inches apart on the D1x image in PhotoShop. Photoshop says the picture is 8.167 inches high so the Putora points span 1.415/8.167 = 17.326% of the D1x's sensor height. The manual for the camera says the sensor is 15.6 mm tall so the image of the points from the chart spans 15.6*.17326 = 2.70 mm on the D1x and the same height on the XL2. Thus that's our estimate of the height of the CCD active area in 16:9 mode. The width would be 16*2.7/9 = 4.8mm.

The other scheme uses the thin lens formula hi =ho*f/so where hi = image height, ho object height, f the focal length and so the object distance. The Putora points are 396 mm apart so their image using a 35 mm lens at 5324.5mm is 396*35/5324.5 = 2.6 mm which is the estimated height of the CCDs using this approximation.

[Charlie Wilkinson]

This is crazy. We should never be having to guess about basic issues like how many bits wide the video circuits are or the exact size of the imager. Sombody at Canon (probably a bunch of somebodys) has the exact measurements and no doubt some very nice diagrams as well. Safe to say they didn't build the CCD from a dinner napkin sketch or just holding their fingers up ("make it about... oh... _that_ big").

If a Canon customer has a legitimate need for this information and it's not a "trade secret", they should be able to get it, period. I would suggest working your way up through the marketing / customer relations food chain until you find someone who can make it happen. It's just a matter of persistance and politeness. (Knowing Japanese probably wouldn't hurt either. ;-)

If you run into any undue resistance, make sure they understand you are part of a large community of XL2 owners and prospective owners, many of whom have been frustrated by similar issues and are quite interested to see how this turns out.

My $.02 worth...

-cw-

PS - A.J., I'm not intending to disparage your efforts. You've been quite helpful and obviously know your stuff. I'm just saying it really shouldn't be necessary to go through all that. But if measurements derived in that fashion are good enough for Stephen, who am I to complain?