# Electronic – Pull Down Resistors

In my quest to understand electrical engineering, I have stumbled across this tutorial:

I have understood the diagrams until I got to switches. I am not sure how switches work on the breadboard or the diagrams. This is the specific one I am thinking of (this is of a pull down resistor): The implementation is: Based on the diagram, what I think is happening is: Power goes to the switch, if the button is up then the circuit is not completed. If the button is pressed then the current takes the path of least resistance to pin2 because it has more pull (100ohm < 10kohm).

The way it is described in the tutorial sounds like when the button is up, the circuit is still complete, but the 10k ohm resistor pulls the power to the ground. I am not positive how or why if both the 10k ohm and 100ohm are receiving equal current, the current would get pulled to the ground through a higher resistance than is open to pin 2.

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.)