LSF0204: Threshold voltage of channel and pull-up resistor

For all voltages between GND and Vref_A, the LSF works as an analog switch. For voltages above Vref_A, the switch is open, and the voltages are determined by the pull-up resistors.

1. When translating down from B to A, the output signal is clamped to the Vref_A voltage. However, a pull-up resistor on the A side will speed up rising edges, and makes the output signal more robust in case there are leakage currents that try to pull the high level down.

2. When the B-side pin is used as an input, and if the signal is driven by a push/pull driver, then you do not need a pull-up resistor. Otherwise, you need one, and its value depends on the desired speed for rising edges.

Also see [FAQ] How do the LSF translators work?

Thanks for your answer.

I want to clarify the below question.

Is it right when Vref_B is high, the FET of LSF0204 is high impedance, so the A side's voltage is Vref_A?

The status of the FET switches depends on the voltage at the I/O pins. The FET is off (high impedance) when both pins are at Vref_A or higher.

Thank you for kind answer.