Best way to protect inputs from over voltages

[Smajdalf]

@killzone_kid

Then go find a topic on current protection. you singled out my reply as if I suggested some nonsense when in fact it is a common way of doing it. Thanks for wasting my time

Common does not mean correct! Usually it does work. But if you have some heavy load on the Vcc rail but Vcc is not powered you can easily damage the pin with low impedance voltage source.

[mb107]

Whats you experience with these rugged units. Will it be able to take 14V on a regular operational basis. All the other solutions presented take up too much real estate for my situation. According to this data from Arduino the 3.3 BLE will die.

[MarkT]

@killzone_kid

And? the question was how do you protect it from over voltage not from backfeeding

Either way the chip could end up fried, and the chip's input absolute maximum voltage ratings are exceeded. That's over-voltage in my book.

[raschemmel]

@Smajdalf

The LED limits the voltage on the Arduino pin to about 2V. A strange design.

(is your comment in reference to Post #7, with the assumption the led has
a V(forward) of 2.1V , resulting in a voltage drop of 3.9V across the two
resistors ? If so, I rather doubt the led is going to light with only 372uA.
(V=I*R => I=V/R => I =3.9V/10470 ohms=3.724e-4=0.0003724A)

How so ?

[Smajdalf]

For reference there is the strange schematic again:

@Idahowalker

Assuming the optocoupler is off the circuit is simple: R3, R2 and LED1 are in series. Assuming forward drop of the LED is 2 V only 3 V are left for R3 + R2. So the current is about 3 V / 10470 Ohm ~ 300 uA and voltage dropped over R2 is then 470 Ohm * 300 uA ~ 150 mV. Arduino pin "sees" 2 + 0.15 ~ 2 V. Too low to be detected as HIGH reliably.

Of course the LED will be dim but it is the least problem in this circuit.

[herbschwarz]

Maybe someone just added the LED because he thought it was a good idea without doing the maths.