pinMode and Default Output State of Pin

Projects Programming
1

I have an output tied high with a pull-up resistor. When I use the pinMode statement with OUTPUT as the parameter, the output goes low even though I don't explicitly use digitalWrite(pin,LOW) to set the pin low.
Is this the normal default state of the pin after pinMode is called? To maintain the pin at a high level, should I use a digitalWrite(pin,HIGH) before calling pinMode. Presumably behind the scenes this just sets the appropriate bit in the output register for the port?

2

Eugbug:
Is this the normal default state of the pin after pinMode is called?

Yes, unless you have previously written HIGH to that pin.

3

Why do you have the pull up resistor on the output ?

4

UKHeliBob:
Why do you have the pull up resistor on the output ?

That is sometimes required for external circuitry to prevent short turn on during reset and power up sequence. For example many of those Asia E-bay relay boards use a low input on their relay channels to activate each relay channel, so adding an external pull-up resistor will prevent a short turn on of the relays during an initial power up before the sketch code gets a chance to set the pin to output mode and HIGH output.

Lefty

1 Like
5

UKHeliBob:
Why do you have the pull up resistor on the output ?

It's a pull up resistor. I'm communicating with a DS18B20 temperature sensor which has an open drain output stage so a resistor is needed.

6

Hi,

You can use digitalWrite(pin, HIGH) before use pinMode(pin, OUTPUT).

Best Regards.