Electronic – Connector pins protection from overvoltage

Why don't you just design (or buy) a small voltage regulator that feeds your bank of LEDs. Then there are no complex issues like fuses blowing and notifications etc. Think simple. Design a simple small buck regulator that can supply (say) 12 volts from the 24 volt power option.

It looks like this would be OK: -

But there are many other options from Linear technology and Texas Instruments.

if I or anyone else mistakenly sets the outputs at 24V instead of 12V the ringlight is protected

With this idea you set it at 24 volts and it regulates to 12 volts. If you set the input to be 12 volts then the LEDs will be a bit dimmer because the LT3970 won't be able to supply 12 volts from a 12 volt input (it'll be more like 11.75 volts as per graph on pg 6 of data sheet). But it won't fuse!

As an aside, I also want my circuit to be very small, with the fuse fitting in a 150 series inline fuse holder from Littelfuse, catalog # 150274, And the zener diode fitting on this tiny PCB, which I can fit inside the case with existing electronics

I expect the regulator circuit will be smaller.

So if 12V is applied the Zener acts like an open circuit

Of course if you do decide to go down the zener diode route and pick a 5% tolerance one (about as good as they come) it may only be regulating at 11.4 volts and this may blow the fuse under normal circumstances. Go for a 15 volt zener is my advice then there is never any possibility of it blowing the fuse at 12 volts.

Yes, that would work fine, but you can make this even more simple:

^{simulate this circuit – Schematic created using CircuitLab}

Zener diodes behave more or less as a "normal" diode when in forward mode so we can use that to clamp negative voltages.

When using a 12V, 400 mW zener diode the maximum current through the zener diode can be 400mW / 12 V = 33 mA. There will then be 12 V across R1, R1 needs to limit the current to 33 mA so it needs to be: 12 V / 33 mA = 364 ohm or higher. I added some margin and suggest 470 ohms then the actual current will be 12 V / 470 ohm = 26 mA and R1 will dissipate 12 * 26 mA = 0.312 W so you might want to use a 0.5 W resistor (or use two 1 kohm 0.25 W resistors in parallel).

I would make the capacitor C1 10 nF, that should suffice to remove most pulses. Only if you need to measure high frequency signals C1 might need to get a smaller value or must be removed. Then the zener diode might be an issue as well as zeners aren't very fast. But if you stay below 100 kHz or so there should be no issue with this circuit.