Let's say I plug the voltage and ground rails of a breadboard into an arduino's 5 volt and ground pins. Then I add one LED to the breadboard and power it up. *Let's ignore resistors for right now to try and hopfully keep things simple.Let's say the LED is this: https://www.sparkfun.com/products/5313.4V forward dropMax current 20mAWhat do those specs mean? 3.4V forward drop. Does that mean if I apply 5V to the anode it will drop to 1.6V out the cathode? So if that were true then if I put a second LED in series it should be less than half as bright, no?
Max current makes sense... That's the max amount of current you can give it before it fries. But again, I don't know how you figure current.
If I wanted to light 10 LED's how would I figure out what I need?
Also, I've seen people but resistors on the anode side (which makes sense to me, if that's the way current flows) and then I've seen them put them on the cathode side - whats going on here?
I get the whole water through a pipe analogy but for some reason it doesn't feel like it's clicked with me yet.I think my biggest question is where does the current come from? Or how do you find it?
Quote from: bwoogie on May 22, 2013, 08:54 pmAlso, I've seen people but resistors on the anode side (which makes sense to me, if that's the way current flows) and then I've seen them put them on the cathode side - whats going on here?Nothing is going on. The resistor will restrict the amount of current flowing through it, and therefore also through the LED, whether it's placed before or after the LED: the LED is still getting the same (restricted) amount of current flowing through it. It will make a difference if you have two LEDs and one resistor for both, but you don't need to delve into that right now.
Basically the whole circuit is going to be the same the whole way around right? It doesnt have more current in one spot and less in the other?
I keep a printout of different Ohm's Law formulas taped to my workshop wall, I suggest you do the same