I'm a beginner to Arduino and circuits. I've spent the last couple hours hung up on the concept of how current flows through switches and pull-down resistors, having read every thread and watched every YouTube video I could find on the subject.

I understand that pull-down (and pull-up) resistors are necessary in order to prevent pins from floating. What confuses me is that, if the input pins effectively have a built-in 100 megohm resistance (according to this page), then how is it that a 10k ohm pull-down resistor is enough to dissuade current from flowing directly into ground even when the switch is closed?

Here is a diagram of my circuit:

enter image description here

Am I conflating the high-impedance state of input pins with resistance more generally?

Would really appreciate any clarification on this. Thanks!

  • Huh? 5V across 10kohm is 500uA. – Ignacio Vazquez-Abrams Oct 4 '17 at 1:37
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    @IgnacioVazquez-Abrams Sorry, I'm not totally sure what you're getting at. – T. Peppers Oct 4 '17 at 2:10
  • I'm not sure what your question means. Of course some current flows from the positive rail to ground when the button is closed. How much? Well... – Ignacio Vazquez-Abrams Oct 4 '17 at 2:11
  • @IgnacioVazquez-Abrams Ah, I'm wondering why all the current doesn't proceed across the resistor to ground when the switch is closed, given that there's a very high built-in resistance right before the pin. If I'm missing something really obvious, I apologize. You may be overestimating my knowledge here. – T. Peppers Oct 4 '17 at 2:14
  • Ohm's Law – Ignacio Vazquez-Abrams Oct 4 '17 at 2:15

Yes, all of the current that flows goes through that resistor to ground. That way almost all the voltage drops across that one resistor and you are able to read the voltage with the pin on the high side of it. Only a very very small amount of current flows into the pin when you read a voltage. It just needs to charge up a very small capacitor.

  • Thank you! Just to clarify high side vs low side (these terms are new to me) - in the diagram in my question, isn't my pin reading it on the low side, between ground and load? I'm getting my understanding of these concepts from this page. Let me know if I'm missing something. – T. Peppers Oct 4 '17 at 4:51
  • @T.Peppers. In your circuit your input pin is connected to the low side: when your switch is open, you pin read 0V. – user31481 Oct 4 '17 at 7:34
  • But when you press the button, your pin is on the high side of that resistor. The button has almost no resistance compared to the resistor, so all the voltage drops across the resistor. – Delta_G Oct 5 '17 at 2:08

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