1) During powerup, what happens to the output of the SN74AUP1G97 when configured to be an inverter in the following example circuit? VCC=3.3V and it ramps slowly (let's say 1ms/V). The input is pulled up to VCC with a 4.7K resistor, the output is connected to a weak pullup (100K) to VCC at one of its loads.
I'd expect the output to be undriven until VCC reaches a valid operating voltage (0.8V) due to the Ioff feature (therefore the output is tracking VCC), at which point the inverter logic turns on and drives the output to 0V, subject to the propagation delay for 0.8V operation. Is this accurate?
2) Let's say VCC=2.5V and the inputs are at 3.3V (in any configuration). Due to the Ioff feature, if the input(s) are active prior to VCC being enabled, no current should flow through the device (in either direction). Does this also imply that no current or voltage will leak onto VCC if it is undriven?
Thank you,
Nick
