I found this here: Multiple buttons connected to reset pin which is what I want to do but it doesn't really explain how it works. I copied it to a simulator to try and understand it but even after removing/changing some resistors I'm failing to see why they are there and why it works the same without them. I understand the top right of the circuit and how that resets but the diodes and 47k resistors don't fully make sense. I was also able to remove the middle 10k and the circuit worked so is that needed too?

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  • \$\begingroup\$ Check what happens with the voltage supply rail. Without those resistors, you would be shorting it to ground. \$\endgroup\$
    – Oskar Skog
    Mar 4, 2018 at 12:00
  • 1
    \$\begingroup\$ If the TP3.3 has no internal pullup then 47k is needed to ensure ESD safe. CMOS , should never be floating. Otherwise stray E field may induce false noise with logic 0 \$\endgroup\$ Mar 4, 2018 at 13:36

1 Answer 1


The diodes perform an OR function.

The middle 10K is there to charge the capacitor back to 3.3V. If you leave it out the charge has to come from the 47K pullups and the current has to come from the reverse current of the diodes. That would not be a great idea. I forgot the 47K resistors: as mentioned by Oskar: without you short the power rail. They also make sure the outputs at the RHS are high if no button is pushed.

You mentioned you want to do the same thing. If it is just multiple reset buttons you can do it with several buttons in parallel. in the circuit above the RHS signals are used.

  • \$\begingroup\$ Multiple reset buttons while being able to tell which button was pressed once the microcontroller starts up. The idea I guess is the cap takes long enough to recharge that you can read the io line to see what button was pressed. \$\endgroup\$ Mar 4, 2018 at 13:01
  • \$\begingroup\$ @Ryan Detzal. I used just. The post already says: "This is not the ideal circuit." \$\endgroup\$
    – Oldfart
    Mar 4, 2018 at 13:12

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