Dear forum team, I am using SN74CBTLV3257.
This IC has ground clamp (max =-1.2V). But Has it power clamp?
Otherwise, when VCC = 0V # OE = 0V A = 3.3V signal, Can damage the signal?
Thank you very much, Regards
For VCC=0V, the protection for the I/O pin when a voltage>0 is present at that pin, is brought by the Ioff feature
I understand that for an I/O pin.
But if the device has two pins (one input and one output), and I apply a voltage higher than 0 (3.3V) at the input while Vcc is shorted to ground, can the driver of this input be damaged or there won't be any current flow due to the diode?
Thanks to everybody.
From the datasheet:
"This device is fully specified for partial-power-down applications using Ioff. The Ioff feature ensures that damaging current will not backflow through the device when it is powered down. The device has isolation during power off.
To ensure the high-impedance state during power up or power down, OE should be tied to VCC through a pullupresistor; the minimum value of the resistor is determined by the current-sinking capability of the driver."
I understand that the internal logic is such that, if you tie OE to VCC, the input signal won't be present at the output during power transition.
Thanks Albert for your answer.
I am going to explain better my problem. I am noy using OE pin. The device is allways connected.
I want to make sure what would happen if the device gets unpowered but I still apply a voltage at the input. From the explanation in the datasheet I understand that current can not flow back from the output to the power rail. But what about the input? Is there any possible path between Vcc and the INPUT that can demand current during power off. I don't want the device driving this input to burn.
Any help from any engineer or TI employee?
Thanks to all in advance.
If Vcc is 0V then the Ioff circuit prevents backflow from the inputs or outputs. It is OK to have a voltage ont the input or output when Vcc = 0V.
I have attached a brief description of Ioff.
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