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I assumed that a Schmitt Trigger inverter would act as an open circuit when unpowered, but discovered today that this is not the case. What does a Schmitt Trigger inverter act as when unpowered? I'm using the 74HC14.

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    \$\begingroup\$ You should put your question in the text part too, not just in the title. \$\endgroup\$ – Wouter van Ooijen Mar 19 '16 at 7:25
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The ST 74HC14 datasheet shows the internal input and output protection circuits as follows:

enter image description here

As can be clearly seen, any input exceeding Vcc (even if Vcc is zero) will cause current to flow through the input-to-Vcc diode and power up the chip.

Whether or how long this will happen depends on what is driving the input (can it provide much current) and what else is connected to the Vcc line (other chips, LEDs, etc.). If the source current is limited the chip should survive. If not then either the 74HC14 or the driving chip may be damaged.

Note that since there are six identical inputs on the chip, any one of them going high will tend to power up the chip as described. If all inputs are low the chip will power down.

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In theory, when a part is operated outside it working conditions (as documented in the datasheet), there are no guarantees on its behaviours. NONE.

In practice, most logic chip have input protection circuits that feed 'excess' input voltage to the power rail. So I would expect that an inverter with its input high has its output (somewhat - the drive characteritics will likely be far from optiomal) low. But with its input low there is no power, so the output could be either high-impedance or low or something inbetween.

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