# Is there a difference between connecting the input of a logic gate to the anode or cathode of LED

I am still learning about basic circuit analysis and I was wondering if there is a difference between the two circuits in real life. Technically the polarity at the anode of the LED will be higher than the cathode to allow current to flow, so i'm which is the more "correct" way to connect the logic inputs.

simulate this circuit – Schematic created using CircuitLab

• What are you trying to do with this circuit? It doesn't look to me like it does anything useful at all. Dec 2, 2019 at 14:32

You have made "life hard on yourself" by not simplifying the circuits such that it is easier to see what their difference is.

I have re-drawn your circuits, they are exactly the same circuits as yours. But don't take my word for it, check that yourself!

simulate this circuit – Schematic created using CircuitLab

Now note how the only difference is in the way that the input of the AND gate is connected.

From the LEDs point of view, the circuits are the same. Each LED is directly in parallel with the input voltage source.

Do realize that this circuit doesn't "do" much. LEDs always need a series resistor. Also LEDs need more than 1 V to light up.

Your question doesn't make much sense either, don't think of connecting the input of the logic gate to anode or cathode, think of it in terms of connecting to supply voltage or ground.

There is a difference, not really regarding polarity afaik, however, a LED is a diode. And a diode has a forward voltage. Depending on the LED color (among others) this can be typically between 1.8 and 2 or more volts.

This means the input voltage of AND1 logic gates is lower than the output of V1/V2.

What even is more important, you need to add a resistor to prevent the LED from burning. A LED needs a resistor to limit the current. If you add the resistor, and you don't want the resistor to be inside the circuit from V1/V2 to AND1/AND2, than you have to add the resistor according the left scheme where the resistor is just before or after the LED (and not in the path of V1 to AND1).