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Does pin 5 or pin 6 use more power if they're just being used as a signal to an external device?

pinMode(5, INPUT_PULLUP);

pinMode(6, OUTPUT);
digitalWrite(6, HIGH);

.

EDIT: Thanks to some links below, I hopped thru a series of threads until I found the website www.gammon.com.au/power which did tests with a 328 and found the following results with the pins disconnected (which is different than my question but seems relevant nonetheless):

In SLEEP_MODE_PWR_DOWN:

  • All pins as outputs, and LOW: 0.35 µA.
  • All pins as outputs, and HIGH: 1.86 µA.
  • All pins as inputs, and LOW (in other words, internal pull-ups disabled): 0.35 µA (same as before).
  • All pins as inputs, and HIGH (in other words, internal pull-ups enabled): 1.25 µA.
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It all depends on the external device. If it has high-impedance inputs, then it makes no difference. The Arduino itself has high-impedance on its pins configured as INPUT. If the device draws current from this signal, then chances are it will draw more current from an OUTOUT HIGH than from an INPUT_PULLUP. How much more? We cannot say without knowing the specifics of that device.

If you model the device as a simple resistor to ground (which is almost certainly overly naive), then the current draw is

I = VCC / (R + Ro)

where R is the value of the resistor and Ro the output resistance of the Arduino: about 25 Ω for an OUTPUT and 32 kΩ for an INPUT_PULLUP.

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  • Thanks, for us noobs is the signal wire to most transistors/MOSFETs considered high-impedance?
    – rfii
    Mar 26 at 16:10
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    @rfii: The input of a MOSFET (the gate) is high impedance (it's actually capacitive, thus high impedance for DC). The input of a BJT (the base) is low impedance, and it needs a current-limiting resistor. Mar 26 at 19:20

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