I know how the 74HC4067 works, but I have difficulty understanding its limits when the Enable pin is high, meaning all outputs are inactive.
In this case the acceptable voltage limit should be between GND and VCC, but what is the current limit? If VCC is 5 V, is it acceptable to apply 5 V, 1 A to each of the inputs, for example?
When /E is low, the selected channel is connected to the common I/O pin, and all other channels are disconnected.
When /E is high, all channels are disconnected.
A closed switch (a connection) behaves somewhat like a resistor (this is specified with RON); the current limit is specified in the absolute maximum ratings as drain current.
An open switch does not allow current to flow (except for a very small leakage current); it is not necessary to specify a limit.
This chip has clamping diodes at all I/O pins, but as long as the voltage stays between GND and VCC, no current flows through those.
Enable has no effect on the inputs, other than disabling their impact on the output. The actual voltage you can apply to the inputs is specified as Vcc +0.5v, at which point the current will be limited by clamping diode/ other to be not more than 20mA.
It says so clearly in the spec sheet.
but what is the current limit
There is no limit because the inputs act as an open circuit when disabled. So if you apply any voltage between GND and VCC, only a very tiny leakage current will flow. The chip’s design limits how much leakage there will be at any given temperature.
This leakage is independent of what is the maximum current rating of the supply that feeds those inputs. There’s almost no load, so almost no current flows (a few µA).
The only way to make any appreciable current flow into or out of the inputs when disabled, the input voltage would need to be outside the GND and VCC limits. If that happens, the current must be externally limited to a few mA per pin.
I imagine you’re mixing up the maximum current a power supply can provide with the actual current that will flow in a circuit. In this case, with the chip disabled, or with the output(s) open, no DC current will flow.
