Electronic – Using a high resistance pull down resistor

Firstly, forget the 100 Ω resistor for now. It's not required for the working of the button, it's just there as a protection in case you would make a programming error.

• If the button is pressed P2 will be directly connected to +5 V, so that will be seen as a high level, being "1".
• If the button is released the +5 V doesn't count anymore, there's just the 10 kΩ between the port and ground.

A microcontroller's I/O pin is high impedance when used as input, meaning there flows only a small leakage current, usually much less than the 1 µA, which will be the maximum according to the datasheet. OK, lets' say it's 1 µA. Then according to Ohm's Law this will cause a voltage drop of 1 µA \$\times\$ 10 kΩ = 10 mV across the resistor. So the input will be at 0.01 V. That's a low level, or a "0". A typical 5 V microcontroller will see any level lower than 1.5 V as low.

Now the 100 Ω resistor. If you would accidentally made the pin output and set it low then pressing the button will cause a short-circuit: the microcontroller sets 0 V on the pin, and the switch +5 V on the same pin. The microcontroller doesn't like that, and the IC may be damaged. In those cases the 100 Ω resistor should limit the current to 50 mA. (Which still is a bit too much, a 1 kΩ resistor would be better.)

Since there won't flow current into an input pin (apart from the low leakage) there will hardly be any voltage drop across the resistor.

The 10 kΩ is a typical value for a pull-up or pull-down. A lower value will give you even a lower voltage drop, but 10 mV or 1 mV doesn't make much difference. But there's something else: if the button is pressed there's 5 V across the resistor, so there will flow a current of 5 V/ 10 kΩ = 500 µA. That's low enough not to cause any problems, and you won't be keeping the button pressed for a long time anyway. But you may replace the button with a switch, which may be closed for a long time. Then if you would have chosen a 1 kΩ pull-down you would have 5 mA through the resistor as long as the switch is closed, and that's a bit of a waste. 10 kΩ is a good value.

Note that you can turn this upside down to get a pull-up resistor, and switch to ground when the button is pressed.

This will invert your logic: pressing the button will give you a "0" instead of a "1", but the working is the same: pressing the button will make the input 0 V, if you release the button the resistor will connect the input to the +5 V level (with a negligible voltage drop).

This is the way it's usually done, and microcontroller manufacturers take this into account: most microcontrollers have internal pull-up resistors, which you can activate or deactivate in software. If you use the internal pull-up you only need to connect the button to ground, that's all. (Some microcontrollers also have configurable pull-downs, but these are much less common.)

A GPIO pin, when in INPUT mode, can be thought of as a very very large resistor connected to ground. The GPIO pin is interested in the voltage that is across this resistor. Take the following circuit for example:

^{simulate this circuit – Schematic created using CircuitLab}

A logic HIGH is seen by the Arduino when the voltage at the node labelled GPIO is at, or near, \$V_{CC}\$ (in this case 5V). A LOW is seen when the voltage at GPIO is at or near \$0V\$.

With the switch SW1 open, there are just the two resistors in play - the pull-up, and the internal GPIO port's resistor. So, using simple maths, we can calculate the voltage that would be at GPIO.

First we calculate the ratio of the two resistors, using \$\frac{R2}{R1 + R2}\$, and then multiply it by the voltage, which is \$5V\$. So we have the sum:

$$ \frac{10,000,000}{10,000 + 10,000,000}×5 $$

We can of course simplify that by doing the addition, then cancelling out trailing zeros above and below the line:

$$ \frac{10,000,000}{10,010,000}×5 $$ $$ \frac{1,000}{1,001}×5 $$

And so the answer comes out as \$4.995V\$ - pretty much the full \$5V\$. So the Arduino see that as being HIGH, since it is above its "input logic high threshold", also known as \$V_{IH}\$ in datasheets.

So now what happens when we press the button? Well, basically we create a short circuit across the internal GPIO resistor. So now we can completely ignore that resistor, since we have essentially put a wire across it to short circuit it.

So now our sum gets changed slightly, since \$R2\$ is now \$0\Omega\$ (the resistance of the wire shorting out \$R2\$).

$$ \frac{0}{0 + 10,000}×5 = 0V $$

And of course, \$0V\$ is below the "input logic low threshold", or \$V_{IL}\$.

Another way of looking at it is that the GPIO, when the button is pressed, is directly connected to ground. No amount of tweaking of the resistor \$R1\$ will ever change the fact that the voltage at ground is \$0V\$. The only way you can change that is by short circuiting \$R1\$ so that becomes \$0\Omega\$ as well, and then you have basically short circuited your battery, and all your wires have now melted.

For reference, here is part of Table 28.2 from the ATMega328P data sheet detailing the input voltage thresholds:

We can see there the \$V_{IL}\$ and \$V_{IH}\$ voltages for the \$2.4V - 5.5V\$ \$V_{CC}\$ range listed as \$0.3V_{CC}\$ and \$0.6V_{CC}\$ respectively. Now, this doesn't refer to \$0.3V\$ and \$0.6V\$, but to \$0.3×V_{CC}\$ and \$0.6×V_{CC}\$.

If \$V_{CC}\$ is \$5V\$, then \$V_{IL}\$ is \$0.3 × 5 = 1.5V\$, and \$V_{IH}\$ is \$0.6 × 5 = 3V\$.

So any voltage seen on the GPIO pin that is below \$1.5V\$ is registered as a logic LOW, and any voltage see that is above \$3V\$ is registered as a logic HIGH.