RC circuit as a delay

Question

I'm bad at electronics. I have to create a circuit which, after a normally-closed switch is opened, sends a signal (could be high or low) to a microcontroller and, after approximately 1 second, drives an Enable Not pin of a motor driver (TMC2209) high.

I came out with the following circuit:

The diode is for keeping the delay caused by the capacitor in the DRV portion. I can't change what's in the rectangle, I'm working with a pre-made module. 500ohm resistor is for discharging the capacitor. Pull-up above the switch is for quickly driving MCU input high.

It works in the simulator, but would it work in the real world? From other questions I saw that the RC delay circuits aren't very good and a 555 timer would be better, but it's definitely more complex.

Answer 1

Since you work with small 3v3 voltage the diode has a huge impact how deep the cap is discharged and affects the timing. So I would use a transistor instead.

Here the discharging resistor is not needed. The 22k in base will limit the transistor (the cap discharging) current to few mA only.

Answer 2

If you are sure that the DRV ENN input is a Schmitt trigger, then what you have should work. In the switch-closed state, the DRV ENN signal will be over 0.8 V because of the two 20 K resistors feeding the 500 ohm resistor. This might be too high to assure a logic 0 state at the input. Changing the diode to a small-signal Shottkey should assure a valid low signal, but be sure to verify this with a prototype build before committing to production.

Answer 3

Thank you @AnalogKid and @Michal Podmanický, but I found a different solution. I'm going to use this simple circuit:

I've taken the inspiration from the Art of Electronics. They use buffers, but NOT gates are cheaper. It has the advantage of protecting both the input signal from the capacitor and the capacitor from the output. Without the second NOT I'd have to solve some voltage divider equations. The first NOT gate charges and discharges the capacitor.