Analog saturation (like digital) comes into play when you enter into a situation in which either (a) mechanical limits are exceeded, or (b) the output of a circuit should exceed the supply voltage. An example of the former would be an amp which is trying to drive a speaker to play audio beyond the maximum amplitude which the speaker was designed for - the volume is turned up too high for the speaker and you get distortion.
An example of the latter is an operational amplifier (op amp). Suppose we have an op amp with supply voltages of +5V and -5V. Further, let's suppose it is connected in such a way that it has a gain of -10 (I'm using it in an inverting configuration as the math is simpler). It will have a linear response range for input voltages between -0.5V and +0.5V. For input voltages outside of this range, it saturates at the supply voltage. The voltage at which saturation occurs can be increased (to a point, I think it's +/- 18V) simply by increasing the supply voltage.
I suspect that CMOS/CCD sensors operate in a similar fashion, but having not worked with them I'm not entirely sure. I do know that A-to-D circuits have a similar problem as the op amps, but are further limited by the number of bits in the output of the circuit. Further, some A-to-D circuits are simply a D-to-A circuit with a counter and a voltage comparing circuit in them. Add to this that a CMOS/CCD sensor will probably have some sort of amplifier attached to it to bring the voltages up to something easier to convert to a numeric representation, which will have the same sort of saturation limits as the previously mentioned op amp.
I hope this answered your question.