I have 3 ICs in my circuit that have active-low reset signals. By pressing a tactile button I would like to reset all 3 of them.

two of them operates with 3.3V and one of them operates with 1.8V.

How can I achieve this? Should I put for each RESET signal a pull-up resistor connected to the desired VDD?

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Should I consider a simple-reset supervisor IC?

Thank you very much.

  • \$\begingroup\$ You cannot pullup to 3.3 the chips that work with 1.8 \$\endgroup\$
    – Undertalk
    Oct 20, 2018 at 10:17
  • \$\begingroup\$ @Undertalk what do you suggest me to do? \$\endgroup\$ Oct 20, 2018 at 10:47
  • \$\begingroup\$ Use a level shifter like hnilica.cz/radek/zim/Electronics/Level_Shifting.html and add a small capacitor across the switch to debounce \$\endgroup\$
    – Undertalk
    Oct 20, 2018 at 13:18
  • 1
    \$\begingroup\$ Why not a simple voltage divider before the 1.8V reset? Just R1/(R1+R2) = 1.8 / 3.3. You can either make R1,R2 >> the pullup resistor, or you have to do a bit more calculations taking the pull-up into account. Assume the reset input is very high impedance. \$\endgroup\$
    – Oldfart
    Oct 20, 2018 at 14:31

1 Answer 1


For the two logic voltages, add a level translator from 1.8 to 3.3 or from 3.3 to 1.8. For example, the SN74LVC1T45.

Add a supervisor if you need supervisory functions (such as analog brownout detect and timer). Some chips have a mechanical switch input.

You will still need the level translator.

You may need two supervisor chips (possibly) plus some logic if you have to reliably reset parts in both power domains under all conditions- for example, if the supervisor is running from the 3.3V supply and the 1.8V supply glitches it may not catch it, or vice versa. It may or may not be necessary to reset the parts on the other domain if a reset takes place on the opposite domain. That's a system-level design issue.


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