I was randomly browsing around some Introduction to circuits websites when I came along this:
The text that went with it is:
Now let us consider what happens when using the output of a chip, such as the 74HC04, to operate an external device. For example, the circuit on the right drives an LED. When the gate is HIGH, then there is no path to GND for cathode of the LED L1. When the gate is LOW, then output pin 2 is connected to ground, and current flows. Since R3 only allows 15mA of current to pass, the gate is safe from being overloaded. Remember that most gates can handle 20mA of current. The same holds true for most microcontrollers.
Can someone explain how this works (I realize this is probably super newbish to most people) I understand how logic gates work, that is an OR gate correct? My "assumption" is that since it's an OR gate....when the Input (or 1) is set to low (or zero) it allows current to pass through?....but if thats the case shouldn't it be like....facing the other way. I guess im confused at why the LED is on the output side with Vcc. I guess I just don't understand why the OR gate isn't facing the other way. Im probably being dumb lol