# Diode Output Voltage

The link provided shows two cases of reverse bias Diodes. From my understanding a reverse bias diode is supposed to block the voltage from reaching the output. So from the following picture wouldn't both of the outputs be 0v. I know that the answers in the diagram are correct because I did them in class, but I don't understand why the voltage across the output on the second picture is - 4.3 volts. I understand the voltage drop across the diode is .7 and the output of a circuit with a forward bias diode would be source-.7 ,but I just don't understand this.

http://img268.imageshack.us/img268/5602/reversebias.jpg [Broken]

Last edited by a moderator:

berkeman
Mentor
The link provided shows two cases of reverse bias Diodes. From my understanding a reverse bias diode is supposed to block the voltage from reaching the output. So from the following picture wouldn't both of the outputs be 0v. I know that the answers in the diagram are correct because I did them in class, but I don't understand why the voltage across the output on the second picture is - 4.3 volts. I understand the voltage drop across the diode is .7 and the output of a circuit with a forward bias diode would be source-.7 ,but I just don't understand this.

http://img268.imageshack.us/img268/5602/reversebias.jpg [Broken]

In the diagram with the AC source, the diode is not always reverse biased, right?

Last edited by a moderator:
I think you're right, because the current flows in both directions does that make the output negative if its in the reverse bias position? If so what's the reasoning for this?

berkeman
Mentor
I think you're right, because the current flows in both directions does that make the output negative if its in the reverse bias position? If so what's the reasoning for this?

When the AC source voltage swings negative, that forward biases the diode for VAC < -0.7V.

Write the equation for the output voltage as a function of time for the AC source, and figure out where the forward and reverse bias regions are for the diode. That will leave some voltage on the resistor.

However, this problem is a bit too simplified. There will only be -4.3V on the output resistor load at what point in the AC source waveform drive?

I see, when the source switches to negative the diode becomes forward bias. The .7 volts across the diode becomes negative because of the negative source, and -5 volts peak - .7 volts peak = -4.3v across the resistor.

Is this correct?

berkeman
Mentor
I see, when the source switches to negative the diode becomes forward bias. The .7 volts across the diode becomes negative because of the negative source, and -5 volts peak - .7 volts peak = -4.3v across the resistor.

Is this correct?

Very good. And you see that is only true at the peak negative excursion of the AC supply, right? So plot the voltage across the resistor as a function of time, versus the output of the AC source. That will help to cement the concepts in your mind.

Yes, I understand that now. I hope to see you on some of my future posts, because I will definitely have more!