Digital sensors and the "ping-pong balls in buckets" analogy

[john_g]

I can understand that the analogy of photons accumulating in the photosites of a digital camera sensor being like ping-pong balls in buckets is both appropriate and useful for CCD sensors, but is it still true for CMOS sensors? I've found that it's still widely quoted for both types of sensor, but I'd expect the photosites in CMOS sensors to behave like photo-diodes i.e. to generate a current rather than accumulate a charge. Does anybody have a definitive answer to this please?

Thanks,

[beejaybee]

I can understand that the analogy of photons accumulating in the photosites of a digital camera sensor being like ping-pong balls in buckets is both appropriate and useful for CCD sensors, but is it still true for CMOS sensors?

Exactly the same .... photons arrive, some of them knock electrons off the semiconductor sensor so that the photons are either detected or not ... there is no difference between an electron count (accumulation of a charge) and a current when you integrate the current over time, which is what you are doing during the exposure.

Of course there are complications due to some thermal electrons also being liberated & being registered (dark current), plus inaccuracies in the measurement of the number of the number of electrons detected (readout noise), but that's basically it.

[john_g]

Ok, thanks, that's more or less how I saw it. But I do think there's one difference... With the CCD, saturation occurs when sufficient time has elapsed for the 'bucket' to have been completely filled. With CMOS, the saturation occurs immediately if the incident light causes saturation of the photosite. Under these conditions, integrating the output from the CMOS sensor won't approximate the behaviour of the CCD sensor. Do you agree with this or have I misunderstood something?

[beejaybee]

But I do think there's one difference... With the CCD, saturation occurs when sufficient time has elapsed for the 'bucket' to have been completely filled. With CMOS, the saturation occurs immediately if the incident light causes saturation of the photosite. Under these conditions, integrating the output from the CMOS sensor won't approximate the behaviour of the CCD sensor. Do you agree with this or have I misunderstood something?

I think you've misunderstood ... the only real difference is what happens after the saturation occurs: with a CCD you get "bleeding" of the charge into adjacent pixels (the ones "downstream" of the "overflowing" one) unless an "anti blooming gate" is fitted, which isn't always done because the gate causes non linearity of response.

Whatever the technology, light arrives in discrete packets or quanta, these are called photons. Even with analogue film, where the photons knock electrons from the molecules of silver salt in the emulsion - it's the missing electrons which allow the developer to selectively reduce the grains which have been "hit" to metallic silver, thus converting the latent image into a printable negative. More photons being detected during the exposure (higher light intensity or a longer exposure) results in a higher proportion of the grains losing electrons i.e. a darker spot in the negative at that point after development. (Incidentally this is why films have reciprocity failure - some of the electrons liberated later in the exposure fall into "holes" left by electrons liberated earlier, resulting in reduced efficiency the longer the exposure is. This is the one overwhelming practical difference between film and electronic sensors!)

If your theory about saturation of a CMOS sensor was correct, the output from a CMOS sensor would be an array of pixels which are either completely black (no photons detected) or completely white (at least one photon detected). Such an image would be unbearably noisy; the dynamic range of one unit is impossibly restricted for the purpose of constructing an image; that's why practical sensors have a dynamic range of at least 8 bits (256 levels), 12 or 14 bits (4096 or 16384 levels) are normal in commerical cameras using CMOS sensors whilst CCD sensors used in dedicated astronomical cameras usually have a 16 bit (65536 level) output.

[john_g]

Thanks for the replies. I'm not sure I agree with everything you wrote in your last post, but that's most likely because I didn't explain myself very well... my thoughts about a CMOS photosite saturating wouldn't result in just on/off states - I was thinking in terms of a linear response that, at a certain level of brightness, can no longer increase its output i.e. it has achieved the maximum output it can.

But, either way, your observations have led me to the understanding that had been eluding me... I'd assumed that the sensor was the limiting factor in terms of burnt-out highlights but now I can see that it's most likely to be the associated signal processing that must take the blame. And now that I've got that in my head, I know that I should have realised this earlier. After all, if I set my camera to manual and use an appropriate exposure, the image shouldn't have any burnt-out areas but, if I extend the exposure time, the image will become blown. Clearly the light level falling on the sensor hasn't changed, only the time for which the sensor's output was being integrated. Well, I hope that explains it!