0
\$\begingroup\$

I have a XC6220B331MR-G 3.3 V LDO voltage regulator that supplies power to the MCU along with some other components on my board.

The regulator has an enable pin which I connected to VIN so that the regulator is always on. I wanted to add a push switch which I can press to cut the power momentarily, which will reset the MCU and everything connected to it.

This is what I have right now, and I'm not sure how to incorporate the switch in the design.

enter image description here the button in the design.

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

You can easily do that with pull-up resistor and push button.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • \$\begingroup\$ Yea that works, I had a follow up question. Why 100k Resistor and not a different value? \$\endgroup\$ Sep 17, 2022 at 0:43
  • 1
    \$\begingroup\$ @MostafaIbrahim Actually, it may differ according to the project. When you close the switch, you will draw some current from the battery to the ground. A value between 10k and 100k may be reasonable to limit this current. Choosing too large values may cause delayed enable when the switch is opened, depending on the amount of capacity on the enable pin. \$\endgroup\$
    – devscg
    Sep 17, 2022 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.