2018-05-15 06:10 AM
2018-05-15 06:55 AM
maybe, only because the 2nS is not enough time to overheat anything.
You could divert that energy by using a schottky diode from ground on that pin, clamping to -0.4v max
2018-05-15 06:57 AM
Hello
tomas.bengtsson@prevas.se
,For the Input voltage on FT pins:VSS − 0.3V ; with maximum:VDD+4.0V
For more details, please check therelated
,in the Table Voltage characteristicsNoted that V
IN
maximum value must always be respected. You should refer to Table 16 for the values of the maximum allowed injected current.Best Regards,
Imen.
2018-05-15 09:07 AM
maybe, only because the 2nS is not enough time to overheat anything.
This depends on the actual circuitry 'behind the pin', I would be cautious.
MOSFETs, for example, have a dV/dt limit. Exceeding that limit will kill it, without much energy involved (relative to it's total admissible power dissipation). I've seen FETs dying that way.