Choosing an Appropriate Pull-up/Pull-downResistor for. The first thing to recognize when dealing with an open drain output is whether a pull-upor a pull-down. When I first got involved in digital electronics, it took me awhile to understand the concepts of pull-up and pull-down resistors and when to use up or dow.

I'm trying to understand exactly how a resistor can 'pull up' on voltage? For that matter, a 'pull down' resistor only underscores the fact that neither pull-up or pull-down resistors do anything. They are just their to prevent a short. Why are they called 'Pull-up' and 'Pull-down' resistors, and how can Ohm's law prove they are useful for anything other than preventing a short to Ground? I've looked high and low for a real answer on this, bought books, and more.

Will someone *please* explain this so my new-to-electronics mind can understand this. I appreciate all the instantaneous responses. Contour Shuttle Pro 2 Driver Mac more. I looked at wiki, and the problem is, you are all way over my head on this. Ayreon Actual Fantasy Rar.

You all have told me what it does, but not really how it does it at the electron and voltage level. A resistor in series drops voltage, not current. A resistor in parallel drops current, not voltage. I want to be able to really understand this, not just spout what I'm told. I need a real example with real values for everything and a 'walk through' from Vcc to Ground on each leg to understand when the resistor is doing something and conversely when it is not. You can apply all the voltage you like across a resistor that connects to an open (floating) pin. There will be no current flow and hence no voltage across the resistor, no matter what it's value is.

This effectively eliminates the resistor. The open end of the floating pin will go high (to full voltage) simply because there is maximum attraction from the power-source, trying to make current move. So again, the pull-up resistor is only there to prevent a short-circuit. The fact that voltage goes high on an open pin (as described above) is in direct contrast to Ohm's Law, unless this is a result of the resistances within the power-source interfering: E = I * R E = 0 * (Near Infinity + Pull-Up Resistor) E = 0 Help me get my mind in line with yours, please, because I keep coming back to one perspective, which means I'm not being given enough discriminatory information (new information). Thank you very much for your patience and help. If I'm wrong, don't be afraid to say so, and explain WHY. My Qwest for understanding far outweighs my pride.