Hi, Everybody! I was tasked by my lab instructor to design an AC to DC converter using a 4 diodes model. I started doing the simulation in Proteus ISIS7 part by part to get the needed values for components to get a final DC output of 4.3V. Currently, I'm doing a simulation of the first block of the converter, which is the rectifier. I have started simulating with a half-wave rectifier to get familiar of how the diode I'm using (1N 4007) may work, and from there I would use the same type of diode to build a 4 diode rectifier. I connected a 12V peak-to-peak sinusoidal AC voltage source in series with a 1N 4007 diode, and a 1Kohm resistor. I used a built-in Oscilloscope to display the output waveform across the resistor. It has a peak value of 11.25 V, which means there's a 0.75 voltage drop across the diode, but I also found that there's -500mV (-0.5V) in the negative cycle, whereas in reality it should be 0V, because the peak inverse voltage of 1N 4007 is set to be 1KV according to ISIS7 software, so how come there's a leakage in current from the reverse side? is it something I have to worry about in the real life design? or is it just something related to the simulation software? Thanks. Edit1: I used Multisim to run the simulation, and I got a negative value for the output voltage when the diode is in reverse mode, but this time, it was in micro-volts. Does this really reflect the application in reality? also, I did a full-wave rectifier, and it was more precise in calculating the output voltage, with no negative value at all! still curious why there's a leakage in current for the half-wave rectifier.