I have a guide stating that the following diagram represents a DRL AND gate:
Suppose that both A and B have 5-V inputs with a corresponding switch. I tried to simulate it for learning purposes and placed the ammeter right before the LED. The problem is that the current does not change at all. I suspect that the diodes prevent the current from moving through since they are pointing to the left.
"Pointing to the left" is correct, the idea is that if one of your inputs is low, that diode is open and the R + D of the branch take the rest of the 5V voltage drop, giving L on your output. Only when both are high, you get an H output.
Your issue is that when your inputs are not high, they are floating instead of drawing the current through the diodes. Try connecting the R and S junction to the ground with a large resistance.
This AND gate circuit takes either high (5V) or low (0V) inputs and generates an output.
Due to this circuits construction, an unconnected input behaves as a high, not as a low (like one might think).
So in your simulation you are inputting either high (unconnected) or high (5V) but never inputting a low (0V).
If your inputs are always high you aren't going to ever get anything but high on the output.
Just connect the switches to ground and your simulation will be fixed.
In general the diode needs to point to a lower voltage to get any current flow across it. And the voltage drop needs to be in the 0.6V~0.7V range to get any significant current.
Assuming 2V across the LED and no current in the didoes, the voltage at the left side of R4 is...
2V + (5V - 2V) * 1K / 11K = 2.27V.
Therefore the inputs must be connected to a voltage below 1.57V (2.27 - 0.7V) before you will start to see much change in current at either the diodes or the LED.
