Digital camera sensor damage from the sun

I've read so many conflicting things about this subject. For

example; photographing a sunset with the entire sun still above the

horizon in the photo. I've read where this will damage the sensor in

the camera, and read that in older models that is true, but newer

models can handle it. Wondering if anyone has any first hand

experience with this. Thanks

Specifically, Nikon and Fuji pro series cameras.

No first hand experience, other than typical sunset shots and astronomy and solar observing. You will do no damage to the sensor itself, as it won't "see" the sun until you click and lift the reflex mirror out of the way, open the shutter curtain, and expose the sensor. However, you <i>could</i> do damage to the mirror, viewfinder, exposure meter, and most importantly, your eyes if you are using a long telephoto lens that will magnify the intensity of the sunlight. A magnified sun will generate alot of heat, to the point of melting and warping. Generally, when the sun is low on the horizon, you could have enough haze and atmosphere to filter it to make it safe to view through a long telephoto. Witness the recent Venus transit photos taken unfiltered that show the planet perfectly against the sun even at very high magnifications. At low magnification (200mm and down I'd say), you are generally ok. Anything longer, and you run a risk.<p>

In short, play it safe and if in doubt, filter in front of the lens. You run the risk of frying your camera viewfinder and your retina well before you damage your sensor.

D100, 300mm f/2.8 TC14...<div></div>

Many of the documents on the web that discuss CCD sensor damage are not talking about sunset photos. They are talking about astronomical photographs of the sun where the sun takes up 90% to 100% of the frame as magnified by a telescope.<div></div>

Cameras which have an EVF and no mirror and mechanical shutter ARE prone to sensor burn out when pointed at the sun. Cameras (film or digital) with a mechanical shutter but no mirror can have the shutter damaged by the sun, especially using a telephoto lens. In a DSLR, sensors do degrade over time - more hot pixels appear, and colour rendition can be affected. Some of this is due to cosmic radiation.

You can damage a sensor by exposing it to the sun.

However, I don't have any idea how much is necessary.

 

Justification: Spacecraft have these nice doodads on them called star trackers. They're

basically digital camera which constantly look out into space and identify the arrangement

of stars to determine how the spacecraft is positioned. When commanding a spacecraft to

maneuver, you have to make sure that you don't ever park it so that a star tracker has the

sun in its field of view, as that will damage it. However, the star trackers can safely be

pointed at the full moon (30% the brightness of the sun, I think), and the sun-lit Earth

(also pretty bright) without any damage. I think the star trackers can also be slewed across

the sun safely, which takes no more than maybe ten or twenty seconds for the spacecraft

I'm familiar with.

 

The sun is of course, significantly brighter in space. I'm not sure if our trackers have

anything like a neutral density filter on them.

 

Anyways, my unscientific rule of thumb would be: if you' can look through the

viewfinder under the same conditions that you'll be exposing the sensor to, then it'll be

safe.

Can someone provide a physical rationale explaining why image scale (use of a telephoto vs. normal lens) should have anything to do with this? ISTR that the potential for damage depends on the amount of solar energy reaching the sensor and the sensor's ability to dissipate that energy. In which case damage potential on a given sensor is dictated by focal ratio and exposure time and has nothing to do with focal length.

 

Granted, if you cross the damage threshold (which is quite high for CCDs), then the longer the focal length, the more area of the sensor is going to get screwed up....

The sun's disk is about 100,000 times brighter than a sunlit scene, or about 15 stops (you need a solar filter with that amount of reduction to photograph the sun). To put that another way, a straight exposure would be around 600 nanoseconds at f/16 and ISO 50. It's quite easy to verify that this is about the right factor. The sun subtends about half a degree in the sky (we can consider the rest of the sky to be dark in comparison). Its radius is therefore about R x pi/720, and its disk area is R^2 x pi^3/(720^2) where R is the distance to the sun. Compare with a hemishpere of radius R, whose area is 2 x pi x R^2, and we have a ratio of 720^2 x 2 / (pi^2), or 105,049.

In case you hadn't worked it out - the sun's image is a disc with a diameter of approximately 1/110th of the lens focal length. With a wide angle lens, this is quite small, and any heating produced locally can more easily be conducted away, preventing serious damage. Moreover, any fairly minor change of direction that the lens is pointed in will mean a different area is capturing the sun's image. With a telephoto, the area heated increases and may not be so easily conducted away. If you've ever played with a magnifying glass and focussed the sun onto paper to light a fire, you'll know it takes several seconds before it ignites. Obviously faster lenses (or a larger diameter magnifying glass) collect more of the sun's rays and will cause damage sooner at any given focal length. Also, the reflectivity of the surface on which the sun is imaged affects how much of the energy is absorbed - hence you can burn a hole in a black coated shutter although it isn't quite at prime focus.

<< If you've ever played with a magnifying glass and

focussed the sun onto paper to light a fire, you'll know it takes several seconds before it ignites. >>

 

Last time I played with a magnifying glass I had to focus to a small point to get anything to burn. A point much much smaller than the area covered by a 35mm frame.

 

 

<< With a wide angle lens, this is quite small, and any heating produced locally can more easily be conducted away, preventing serious damage. >>

 

Would the 6 or 7mm lenses on a digital point and shoot not be considered wide-angle or is this going to fall into another semantics argument?

<p><i>In case you hadn't worked it out - the sun's image is a disc with a diameter of approximately 1/110th of the lens focal length. With a wide angle lens, this is quite small, and any heating produced locally can more easily be conducted away, preventing serious damage.</i>

 

<p>Conduction is constant regardless of the size of the sun's image, though; it is also isotropic. Most of the heat gets conducted out the back of the chip and into the package regardless of the image size. A big image may make a few percent or so difference, but I have reason to doubt that this would result in crossing some damage threshold.

 

<p><i>Moreover, any fairly minor change of direction that the lens is pointed in will mean a different area is capturing the sun's image. With a telephoto, the area heated increases and may not be so easily conducted away.</i>

 

<p>This makes sense for shutterless cameras. But few if any of them can zoom to such a long focal length that the sun takes up a big percentage of the sensor area. And even if they did, most of these cameras are f/5 to f/9 at the long end.

 

<p>Meanwhile, on the shuttered camera side, nobody is going to be shooting the sun with the camera set to bulb.

 

<p>Finally, any photographer smart enough to be concerned about sensor damage by the sun is also going to be using a tripod, right? ;-)

 

<p>I really doubt the real-life applicability of the shaky-camera principle; as Steven said, I wonder if anyone has first hand experience with this.

 

<p><i>Also, the reflectivity of the surface on which the sun is imaged affects how much of the energy is absorbed - hence you can burn a hole in a black coated shutter although it isn't quite at prime focus.</i>

 

<p>I think another important physical quantity at issue here is the emissivity. For CCD and CMOS sensors this quantity is around 0.88 at the d-Line (wavelength 587.56nm). So it should dump heat rapidly through radiation.

 

<p>The other thing to consider is reactivity and how that scales with temperature. For fabric shutters, reactivity (primarily oxidation) is very high and does not require a large temperature increase to accomplish. For titanium shutters, reactivity is rather lower. But in CCD and CMOS sensors, it is almost non-existent. The materials are relatively inert. The basic mechanism of heat-mitigated damage to CCD and CMOS devices is through recrystallization. Looking at materials tables, I gather that temperatures around 1,000 degrees C, sustained for some minutes or possibly hours, is required to recrystallize these sensors. This is the reason for the skepticism I express above.

 

<p>Real-life experience backs this up. One of the SOHO mission's camera shutters <a href="http://www.qrg.northwestern.edu/projects/vss/docs/mission/3-why-sun-damage-micas.html">was jammed open</a> for <i>several hours</i> while staring at the sun. Despite being exposed not only to light, but also to radiation far more energetic than visible light (such as extreme ultraviolet), this produced no damage except where an active region's image happened to be falling on the sensor. Even then it did not destroy the sensor or render it in any way inoperable.

 

<p>So I am skeptical that anyone has damaged their sensor taking a picture of the sun. Some alternative explanations for instances of apparent sensor damage may be that anti-reflective coatings on the near-IR exclusion filter were burned (or perhaps a plastic IR exclusion filter was melted); that the glue that is found in some CCD and CMOS packages was melted; that other electronics near the sensor were destroyed through conductive overheating; or even that something volatile inside the chamber got heated and subsequently condensed onto near-sensor surfaces. (This latter is the typical form of solar damage to spacecraft star trackers, I am told.) These are fairly mundane repair issues, though, not garbage collector issues like real sensor damage.

Thank you for this discussion. This was a concern of mine.<br><br>-s