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Hi experts,
BR
Frank
Hey Frank,
May I know the reason of why "When the output-enable (OE) input is low, all outputs are placed in the high-impedance state." ?
Its an active High enable pin.
Which part is the +/- 2mA in 8.3.2 driving?
The current required to over drive that inverter in series with the 4kohm to change the state of the I/O.
Is this 4kOhm a real resistor or a equivalent resistance?
It is a real resistor.