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When configuring a pin as output on the ESP32, does it set an internal pull-up or pull-down resistor?

I am connecting a bus buffer to an output pin, and I am not sure whether I should use a pull-up or pull-down resistor on the bus.

EDIT - Schematic: ESP32 with bus buffer

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  • \$\begingroup\$ How are you connecting the bus buffer? Can you draw a schematic? \$\endgroup\$
    – anrieff
    Commented Mar 19, 2020 at 10:38
  • \$\begingroup\$ just added the schematic \$\endgroup\$ Commented Mar 19, 2020 at 10:52

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Output pins output a certain voltage. Adding internal pull-up or pull-down resistors to output pins is silly because they won't do anything except for wasting power.

The output pin is equivalent to a 0-ohm pull-up or pull-down resistor already (depending on whether your software makes it high or low).

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  • \$\begingroup\$ is it correct to say that only input pins require pullup/pulldown resistor to avoid floating potential if nothing is driving it? \$\endgroup\$ Commented Mar 19, 2020 at 13:14
  • \$\begingroup\$ @F.Heisenberg Yes. And output pins are always driven - by themselves. \$\endgroup\$ Commented Mar 19, 2020 at 14:06
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    \$\begingroup\$ What if the ouput pin is open drain? \$\endgroup\$ Commented Mar 19, 2020 at 14:08
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    \$\begingroup\$ @F.Heisenberg Well then it's floating instead of high, so you want a pull-up resistor to make it high. Most microcontroller outputs aren't open drain. \$\endgroup\$ Commented Mar 19, 2020 at 14:09
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    \$\begingroup\$ I2C outputs are open-drain, but internal pull-ups are too weak to use with I2C pins, they are in the order of tens of kilo-ohms. I am not saying that it does not work, it will just be extremely slow if it works, and not within rise time specs. \$\endgroup\$
    – Justme
    Commented Mar 19, 2020 at 14:33
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An output pin that is a push-pull output is a strong output - using an internal pull resistor (down or up) makes no sense.

Edit (due to new information): Since the ESP32 pin may not be an output when it boots up, it can be a floating input so the buffer output state is not known. Obviously to keep buffer output state stable and known during ESP32 boot, you need a pull-down (or pull-up) at the buffer input.

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  • \$\begingroup\$ True. But, if you're concerned that the bus might float if nothing is driving it (in the case of a shared bus) then it is considered good design practice to put weak pullups or pulldowns on the bus. You could use the pullups/pulldowns on the buffer, if they are provided by the device manufacturer, to do this, rather than adding a bunch of discrete resistors. \$\endgroup\$
    – SteveSh
    Commented Mar 19, 2020 at 10:47
  • \$\begingroup\$ I thought so too. But should I add a pull-up or pull-down resistor? Can I add both? \$\endgroup\$ Commented Mar 19, 2020 at 10:51
  • \$\begingroup\$ @Justme: can you provide a equivalent circuit of the output? Why is there no internal pull-up for example? \$\endgroup\$ Commented Mar 19, 2020 at 10:52
  • \$\begingroup\$ @SteveSh in that case, it's not an output pin! \$\endgroup\$ Commented Mar 19, 2020 at 12:50
  • \$\begingroup\$ @F.Heisenberg An output pin has a 0 ohm pull-up or pull-down resistor, depending on whether you set it high or low. \$\endgroup\$ Commented Mar 19, 2020 at 12:51

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