Wired as per:
- 5V goes straight to the end stop switch (normally closed) and from there straight to ground with no resistance other than the wire and whatever connectors you have in the wires. This is a short circuit from the 5V rail to ground. It is a problem! The 5V going through the resistor to signal isn’t a problem, however when the circuit is closed, the resistor is also in line with the path to ground, which has the existing pullup resistor acting as a voltage divider with the insert resistor. It still might seem to work… but it’s just bad.
- Remember that there js already a resistor doing exactly this in place on the board. The addition of another resistor lime this acts to reduce the overall reaistance between the 5V rail and the signal pin. The SKR Pro problem (when it exists) is that the resistance is too high, and the LED creates a path to ground that drops the voltage at the signal too low to register as “logic high” even though it lights the LED. This is the correct fix.
- Will only result in the stop constantly registering triggered. Probably the path to ground with the resistor in it will not be enough to lower the voltage at the signal pin to “logic low” on order for the pin to register as open.
The intended circuit is like this: a pull up resistor between the 5V rail and the signal pin forms a voltage divider with the LED and a current resistor. If the voltage at the junction is higher than about 2.5V the pin registers logic high. If it is lower, it registers logic low. If the current resistor for the LED is too low value or the Vf of the LED is too low, it drops the voltage at the pin. If the pull up resistor value is too high, that also drops the voltage at the pin. I don’t know which value is wrong on the SKR boards that have the trouble, but is is one of these three. By putting a second resistor in parallel with the pull up resistor, we lower the resistance between the 5V pin and the signal pin, which increases the voltage at the divider. We want resistance though because when we short that pin tk ground, that is the only thing limiting the flow of current from 5V to ground. (Or, I suppose ground to 5V if you want to get texhnical.) While the signal pin itself does have sk.e flow, we basically ignore it because unless we really overwhelm it, 5V isn’t enough to make any current flow between it and ground. We could put 5V directly to it without harming it. The resistor isn’t to protect the signal pin, it is to protect the 5V source.
Consider the following concept: put 3 wires to the switch. 5V goes to the NO terminal on the switch, G goes to the NC terminal on the switch and S goes to the C terminal on the switch. When the switch is open, it shorts S to G. This pulls the pin to logic low, and the board reports open. Press the switch and it shorts 5V to S. This will pull the pin to logic high, and the board will report triggered. This works. The logic pin will tolerate 5V (though 3.3V is safer.)