How do pull down and pull up resistors PULL??????

I understand that it keeps the reading from an input at a specific logic level but how does it do the pulling in the first place???? How does it even change anything? I thought that resistors are only something to add extra resistance, hence the name. Of course, it probably has something to do with the way the current flows, but that's where I'm stuck. It's obvious that I'm missing some very important assumption that would allow me to understand this, so please, explain the way you think about it and understand it. I've been through my book's sucky explanation and 10 other websites' similar explanation with nothing, so maybe I need some unconventional way of explaining this now very annoying topic.

Well the pullup or pulldown part could be done with a wire.

Problem is, if you look at the wires in my pic attached, with wires in place of resistors when the switch is closed, the 5V and Gnd would be shorted together and boom.

With a resistor on one side of the switch and a wire on the other, you have a voltage divider with a 0 on one side, and I’ll let you do the numbers ala Wiki.

Of course, it probably has something to do with the way the current flows, but that's where I'm stuck.

Current flows from high potential to low potential. In this case, the resistor is between 5V and the IO pin. The pin only in INPUT mode needs 1uA to make it read High. So a resistor between the pin (internal or external) can be pretty high resistance, like 10K. If an external "device" then wants to make the pin go low, say all the way to Gnd, the resistor limits the current that will flow from 5V: 5V/10,000 = 0.5mA. The 328P with it's 20mA drivers can also make the pin go low if it is set to OUTPUT mode.

Think of it like a gate with a spring on it gate closed your pulled up gate open the spring pulls lightly to pull it closed after you walk in. Same thing resistor pulls little current to pullup you close a switch it pulls to ground you let switch up and just like the spring the resistor pulls up to VDD use a 10K to 22k for most cases.

Pull down is the same way but your holding the input open with just a little current and the switch closes it with a greater force higher current.

Pull down is the same way but your holding the input open with just a little current and the switch closes it with a greater force higher current.

I think in this case the switch would be connected to VDD rather than GND. So perhaps the analogy should be:

"...resistor pulls little current to pulldown you close a switch it pulls to VDD you let switch up and just like the spring the resistor pulls down to GND"

Its fanciful language, the image is that the voltage is pulled around, and lower impedances pull harder. Compare to the analogy between electrical potential and water pressure, where is voltages that "pull" or "push".

Up and down refer to the conventional way circuit diagrams are arranged with +ve power rail at the top and -ve power rail at the bottom.

What actually happens is more precisely (but less intuitively) explained by Ohm's and Kirchoff's laws.

be80be: Think of it like a gate with a spring on it gate closed your pulled up gate open the spring pulls lightly to pull it closed after you walk in.

That is pretty much the perfect analogy in this case.

Forget about thinking of a resistor as "resisting" current. A resistor is used to control current; to adjust the behaviour of each part of a circuit. Always think in terms of Ohm's law (and I am not going to enumerate it here :D).

A pull-up resistor - or device - connects between an input and the supply voltage. The (CMOS) input itself has an extremely high resistance, such that it provides essentially no current to the resistor in either direction. If the other circuit connected to the input is providing no current because it provides no connection to anything at the time, then there can be no current flowing through the pull-up resistor. If there is no current flowing through it, then Ohm's law defines that there can be no voltage across it so the input pin will be held by the resistor, at the supply voltage.

If you provide some current to the resistor, then a voltage will be developed across it. You also need to understand Ohm's law to be commutative, that is you must think of it operating in either direction automatically and simultaneously. So when you now connect your input to ground with a switch, the full supply voltage will be across the resistor and it will be passing a current according (again by Ohm's law) to its resistance. By passing a current, it will be "pulling" the input up, though to no effect since the supply is (hopefully) capable of providing that current and more. Like the spring on the gate pulling it closed and applying a force even though you are holding it open.

I deliberately pose this in terms of pull-up resistors to emphasise that for practical and safety reasons, circuits should by preference be arranged with input switches to ground and pull-ups to supply.