Electronic – Turn off LED with Hall effect sensor

Using VpH as an example the current will be limited to around 90mA by the resistor R7. The datasheet says the minimum current guaranteed to blow a fuse is 300mA and even recommends a 0.1uF capacitor to ensure enough current is available.

A simple solution more likely to work would be to scrap the resistor divider and use something like an LM317 adjustable regulator and switch the voltage set resistor shown as R2 in the typical application circuit instead. Using 1% resistors should keep you well within the tolerances given in the datasheet.

Also take note of markrages answer, switch bounce is likely to be a problem so ideally apart from a more robust supply you should be switching voltages from a microcontroller. The LM317 datasheet shows Iadj is limited to less than 100uA so you could achieve that using any form of transistor or analog switch that doesn't introduce too much resistance.

Minimal circuit could be one chip like an analogue asic but that would cost many, many thousands of dollars. Minimal could of course mean cheap or small or both. Stretching it a bit, minimal could mean low power.

Putting all this to one side, amplify the hall sensor's output, feed that into a suitably referenced analogue comparator. That will produce a high signal when the magnet is close. Feed that into the clock input of a D type flip flop configured as a divide by two and drive the LED, via a resistor from the output.