pull down resistor's value for low deep sleep current

[obod0002]

I've read that using an external pull down resistor might reduce the deep sleep current of an ESP32 compared to using the internally available resistor.
As one of my ESP32's is in deep sleep almost all the time, this might make sense.

But which value should I use to safe the most energy of my batteries and to not risk unexpected behaviour due to too high resistor values?

As I need wakeup from ext0, RTC is powered.

Thanks for some insights

[ESP_Sprite]

The correct answer is: It Depends. If you have a noisy environment, you may go as low as 1K, if your environment is squeaky clean, 100K would do the job as well. Keep in mind, by the way, that a pulldown will only use power if there's current flowing through it, that is, if an external source 'overpowers' the pulldown resistor. In that case, you can find out the 'burned' power using Ohms law. If you're in a situation where that doesn't happen that often (e.g. because your active high wakeup signal is a pulse of limited length) the pulldown resistor value doesn't really matter.

[obod0002]

so if we take a door sensor and assume that most of the time the door is closed, would it make a difference if a reed relais connects the two wires when the door is closed or if it's connected when the door is opened?

Is the pulse you're talking of 'only' that fragment of a second where the state changes or the entire time the reed is shut which could by weeks?

Thanks for your help

[boarchuz]

Ignore the ESP32 for a moment, and notice that you've got a simple circuit with a resistor and a switch between V+ and V-. When (and only when) that switch is closed, the circuit is complete, and current will flow across the resistor.
So you want the switch to be in an open (disconnected) state most of the time to avoid that.
Since you know the door is usually closed (ie. the magnet is nearby), you're looking for a 'normally-closed' reed switch (which will be open in the presence of a magnet).