Understanding pull down resistor current flow

Question

I'm working on an introduction to Arduino and circuits course and I've run into a stumbling block.

Everything works the way the course says that it should - I just don't understand WHY it works.

Here's the schematic:

The program is easy to understand. When digital input 2 receives input it lights the LED. The whole path with the LED makes sense to me, no need to discuss it. It's the path with the red wire and green wire going into the DIP switch.

My understanding is that the DIP switch is connected to both the digital input and the resistor and the resistor provides a path to ground for any stray voltage that gets induced on the wire by the environment.

What's confusing me is - isn't the digital input also a path to ground? If not, how does the circuit ever get closed? If there is stray voltage the resistor is providing a path to ground, but if the switch is closed then suddenly the digital input is the best path to ground?

When the switch is closed, the 5 volts can either go to the digital input or through the resistor and the resistor path has more impedance and so the 5V goes into the digital input. When the switch is open, why would any stray voltage not take the same path that the 5V takes?

Answer 1

No current flows in or out of the input pin.

It has no path for current to flow, whether the input pin is connected to 5V or 0V (ground).

So no current flows when resistor keeps the voltage at 0V.

When DIP is set ON, only current from 5V flows via resistor to ground.

That is how ideal input pins work. In real life there could be some leakage currents flowing in the order 1 microamps, which for simple beginner circuits can be ignored.

Answer 2

isn't the digital input also a path to ground?

It may be. It may also be a path to Vcc or left floating. Depending on the uC it may support input configuration to be: pull-up (weak or strong), pull-down (weak or strong) or a floating input (which depends on the pin to have a potential defined externally).

This image from (internal pull-down) shows a situation where the external resistor drives the pin up, since R1 << R2, and the button drives it down, since its resistance is almost zero:

Your illustration shows a configuration in which Vcc and GND are switched. If the input pin is configured as floating, you may consider that R2 is so large that practically no current will enter or exit the pin. Hence, any external resistance (R1 or the button) will determine the voltage at the pin.

This post may also be very useful: pull (up/down) resistor

If the input pin is configured as "floating" your diagram would look like this:

This post also has excellent answers on different pin configurations: internal pull-up/down resistors

Yet another example of an input pin configuration with software configurable lower internal resistance: arduino

Answer 3

The key you are missing in the understanding is the fact that the pins have three states. The states are high state, low state and HIGH IMPEDANCE state. When you have the pins configured as digitalWrite(), the pins can be set to either high or low state with digitalWrite(pin,HIGH) and digitalWrite(pin, LOW) respectively. When the pin is set to HIGH mode, the digital pin acts like conncection to Vcc. When the pin is set to LOW mode, the digital pin acts like conncection to ground and this is what you are talking about.

But when you set the pin to read with digitalRead() function, you set the pin in high impedance mode.

^{simulate this circuit – Schematic created using CircuitLab}

The above schematic is the illustration of how the Arduino is actually configured. Here, when the input is configured to INPUT state, the 1 GOhm impedance comes into action. Thus there is no path to ground through the digital pin.

How the circuit works then?

Well, this is pretty simple to understand with the concept of HIGH IMPEDANCE state of input pin. When the dip switch is off, there is no current flowing through the 220Ohm resistor. Thus there is no voltage drop across it meaning the node you are checking at has the same potential as the ground.

When the dip switch is on, there is path for the current to flow throught the resistor. Since the input impedance is very high, the current takes the path of 220 Ohm resistor. Now, the node you are checking is at +5V. (I am assuming your red wire is supplying +5V.)