Electrical – Calculating the minimum and maximum \$R_{load}\$ for a zener voltage regulator

Yes, the dropout voltage is the headroom the regulator needs to work with. This is the minimum amount the input voltage must be above the output voltage for the regulator to work properly. So as you say, a 3.3 V regulator with 1.1 V dropout requires at least 4.4 V input.

Stuff happens, so it's usually a good idea to give the regulator a little more room. However, going too far will be less efficient, so there is a tradeoff. The range you quote of 4.75 to 5.25 volts could well be a reasonable tradeoff. Basically, that's a "5 V" supply.

The output current of a linear regulator is its input current (minus the small current on the ground pin, but that is small enough to ignore for the point I'm trying to make). Since the input voltage is higher than the output but the currents are the same, the output has less power. The difference in input power and output power is wasted in the regulator as heat. Another way to look at this is that the regulator will dissipate the difference between the input and output voltages times the current as heat. For example, consider a case where the input is 5.0 V, the output 3.3 V, and the current 500 mA. The voltage accross the regulator is 5.0V - 3.3V = 1.7V. That times 500 mA is 850 mW, which will go into heating the regulator. That would be fine for something like a TO-220 package, but a SOT-23, for example, would burn up at 850 mW.

Assuming you're trying to protect an input (the outside world connects to the LDO input).. it does not provide protection against input voltages below ground. Absolute maximum input voltage is -0.3V.

It might work with a series Schottky diode, say for +/-30V inputs, but you'd have to evaluate whether the few hundred mV loss in voltage was acceptable for normal inputs (what your \$V_{IH}\$ is vs. \$V_{OH}\$) . I don't see where thermal protection would ever come into play, either as an input or as an output. If you exceed the 30V abs. max. input voltage, the part may fail.

You're going to need something like a capacitor in series with a resistor loading the output to keep it from oscillating. That will slow down the response considerably (you may want to add a load resistor to ground as well) and the capacitor could conceivably damage the GPIO under some conditions without a series resistor (if Vdd gets crowbarred).

Once you look at one diode, LDO, 3 resistors and a ceramic capacitor per input, it doesn't look quite as clean.