Understanding Reverse Bias Diode Output Voltage

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Discussion Overview

The discussion centers around the behavior of reverse bias diodes in a circuit, particularly focusing on the output voltage observed in different scenarios, including those involving an AC source. Participants explore the conditions under which a diode may conduct and the implications for voltage readings across the output.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that a reverse bias diode should block voltage, leading to an expectation of 0V output, while questioning why a specific output voltage of -4.3V is observed.
  • Others propose that the AC source causes the diode to switch between forward and reverse bias, suggesting that the output voltage can become negative during certain phases of the AC cycle.
  • A participant mentions that when the AC source swings negative, it can forward bias the diode, resulting in a calculated output voltage of -4.3V across the resistor.
  • There is a suggestion to derive an equation for output voltage as a function of time to better understand the behavior of the diode during different phases of the AC waveform.
  • Some participants express understanding of the concepts discussed, indicating a grasp of the relationship between the AC source and the diode's behavior.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the behavior of the diode under AC conditions, particularly regarding the transition between forward and reverse bias. However, there remains uncertainty about the implications of these transitions for output voltage, with no consensus reached on the overall understanding of the situation.

Contextual Notes

Participants note that the problem may be oversimplified and that the output voltage is only -4.3V at specific points in the AC waveform, indicating potential limitations in the discussion's scope.

Who May Find This Useful

This discussion may be useful for individuals studying semiconductor physics, electrical engineering, or anyone interested in understanding diode behavior in AC circuits.

Petrucciowns
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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
 
Last edited by a moderator:
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Petrucciowns said:
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

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?
 
Petrucciowns said:
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?
 
Petrucciowns said:
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!
 

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