Is there a technical reason why sensors output goes from 0.5 to 4.5V (analog) or 10% - 90% (digital)?
I understand that for diagnostic purpose one doesn't normally want to use the full range especially when sensor values are critical so that short to GND or short to VCC can be detected but why 0.5 and not 0.2 or 0.3? Is it due to noise or ground shifting or similars?
For a 0-5 V signal, how do you determine if a wire is loose and you measure 0 V or if the sensor has bottomed out? If your sensor is open-collector with a 5 V pull up and you read 5 V output, has your sensor shorted to Vcc or are you at max scale? Undeterminable. See 4-20 mA current loops for an even more industrial fail-safe "protocol". Same issue and solved by disallowing 0-4 mA.
By having 0.5 to 4.5 V allowed range, you can determine that a reading of 0 or 5 V is not only unreliable, you can mark down the sensor as failed and have your customer replace it.
As for why 0.5 and not 0.3 V is chosen, I don't have an answer for you. 10-90 % is common in engineering so that might be enough to make it a popular option, or someone made an IEC standard out of it?
For more consumer oriented goods, it can also be as simple as lack of rail-to-rail opamps on the output, being ever so slightly cheaper.
Side note: effects from lack of fail-safe.
It's impossible to sink or source current and get all the way to the supply rail.
Giving up 0.5 V on either end is a reasonable compromise between reduction of range (20 % total) and demands on the output structure. For example the AD8676 (a precision amplifier with Rail-to-Rail output) can drive 2 kΩ to ground over the entire temperature range with less than 0.25 V drop, and can sink a couple mA with similar drop. Some op-amps are not as good.
Maybe cost and simplicity, or just that sensor output specs may be calibrated to be within tolerance in that range, and even if it could drive from 9% to 91%, the manufacturer just does not guarantee calibration, under all voltage and temperature conditions.
For example, take a cheapest op-amp you can find, and it won't have analog output from 0V to VCC, or the actual voltage may depend on how strong load there is.
So if an analog sensor has an analog buffer in it, it may just not be able to drive the output to 0V, or VCC.
Another reason is to have headroom for calculations. For example analog television was defined to be converted to 8-bit digital with 16-235 digital range, not 0-255, but values within that range is still valid and may be encountered. Later on values 0 and 255 were declared invalid for video to use them for synchronizing serial bit streams, which still allows for enough margin for over/undershoots and processing headroom.
There are a few reasons. To start, most sensors have an analog component, and this is why you generally do not go "rail-to-rail". (note: you can do this, but at cost) The reason for this is that you need to keep your support circuitry in saturation, and as you get closer to the rails, you find that you'll slip into ohmic operation, as I show with experimental/model data in this answer. It is important to keep your current mirrors, and references sources behaving well for your analog component to behave as expected.
There is also the digitizing aspect. Most sensors also have a control loop where you are digitizing some sort of analog signal through integration of charge or some other conversion. Generally, analog systems have at least two stability points, one is at zero, and the other is at some operating point. If you have control loop, you need to make sure you stay within the range of operation that satisfies the control conditions for your conversion to digital, or you get values that are not representative of the sensor. In the case of charge integration, it's easier if you stay away from the rails. I generally design my analog sensors to have 200mV of operation above and below the rails. This makes sure that I stay out of ohmic behavior, and I can make a smaller sensor.
To get "rail-to-rail" operation, you can make charge pumps to give you more headroom on your sensors or use external supplies higher than the sensor needs, but this costs area on the IC, and the world of semiconductors is about using the smallest pieces of sand possible.
