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Active-Low enable input driven by high voltage signal

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mspider65:
Due to the chip shortage I am trying to adapt a circuit to use what is available.
In particular, I have to replace a Step-Down which, starting from the voltage of an e-bike battery, should generate 5V.
The doubt concerns the Enable signal of the new DC-DC which is now active low.
What I have is an output (PSU_ON) which in case of ON is at battery voltage while in case of OFF it is at GND or floating.
The EN input must be less than 0,8V to enable the chip and must be above 1,8V to disable the chip.
Max Voltage at enable pin is 7V.
I thought of using the circuit you see below but I ask those who have more experience than me for an opinion if it can be good or if there is a better way to go.
In particular i have a very small footprint to accommodate everything and any suggestion to lower the number of nedded component is welcome.

gcewing:
At VIN = 60V, the high level at EN will be about 6.9V, which is getting rather close to the absolute maximum of 7V. You might want to decrease R4 a little, although that would reduce your high level margin at lower VIN.

It might be better to replace R4 with a zener to clamp the high level to 4 or 5 volts.

mspider65:
You are right, a Zener is much better to clamp the EN voltage regardless of the VIN value.
As for R1, how far can its value be safety increased to reduce the current when the circuit is off?

iMo:
So we waste our time?
https://www.eevblog.com/forum/beginners/active-low-enable-input-driven-by-high-voltage-signal/msg3699757/#msg3699757

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