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Why am I getting 5V at point A?
The microcontroller IO is 3.3V.

enter image description here

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    \$\begingroup\$ (point A left bottom at the net including pin 12/signal 6Y - signal flow is right-to-left) \$\endgroup\$
    – greybeard
    Commented May 6, 2023 at 7:52
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    \$\begingroup\$ What are "5V switches"? It seems to me you have just normal switches with pull-up resistors. Why can't you just "pull up" to +3.3V (instead of +5V), and power the 74HC14 from +3.3V too? And if that's possible, why do you even need the 74HC14 in the middle? \$\endgroup\$ Commented May 6, 2023 at 7:54

1 Answer 1

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An open drain/collector inverter will not drive its outputs high and the pullups voltage will set Vout.

An inverter that is not open-drain will drive its outputs to near its V+ supply voltage - in this case = 5V.

The MC74ACT05DR2 open drain inverter will do what you want.

The 74AC05 also will. Datasheet for both here
It's intended to emulate an LS type open drain part.

An alternative is to use a 3V3 supply inverter and 5v to 3V3 input divider.


Q: Please specify the inverter type that you are using.

A: That's a possible but not certain solution. It will work IF the inverters are tolerant of 5V input.

As I said, ideally you need either open drain inverters or - not mentioned, inverters designed for level shifting.

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  • \$\begingroup\$ Thanks, you mean to say i should connect IC VCC pin to 3.3V right? \$\endgroup\$ Commented May 6, 2023 at 6:58
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    \$\begingroup\$ That's a possible but not certain solution. It will work IF the inverters are tolerant of 5V input.|| As I said, ideally you need either open drain inverters or - not mentioned, inverters designed for level shifting. \$\endgroup\$
    – Russell McMahon
    Commented May 6, 2023 at 7:03
  • \$\begingroup\$ @greybeard Thanks. Yes :-( . Corrected in both cases. \$\endgroup\$
    – Russell McMahon
    Commented May 7, 2023 at 11:37

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