Diode circuit with AC excitation source

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Discussion Overview

The discussion revolves around analyzing a diode circuit with an AC excitation source, focusing on the output voltage (Vout) in relation to the input voltage (Vin) and the behavior of capacitors in the circuit. Participants explore the implications of diode biasing and capacitor charging in this context.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that when Vin equals V, the configuration of capacitors C1 and C2 leads to no output voltage (Vout).
  • Another participant challenges this claim, stating that under the assumption that V is much greater than the forward voltage of the diode, the right-hand diode becomes reverse biased and can be removed, while the left-hand diode is forward biased and should be replaced with a wire.
  • A further question is raised regarding the conduction of the left-hand diode, with a participant inquiring whether the capacitors affect the voltage and if they charge to the same voltage, suggesting that the diode may not conduct.
  • One participant clarifies that the two capacitors do not charge to the same voltage unless they are of equal value, indicating that they are in series.
  • A participant provides a visual representation of the circuit when Vin transitions from 0 to +V, assuming +V is significantly greater than the diode's forward voltage.

Areas of Agreement / Disagreement

Participants express disagreement regarding the behavior of the diodes and capacitors in the circuit, with multiple competing views on how to analyze the output voltage and the role of the components involved. The discussion remains unresolved as participants continue to challenge each other's claims.

Contextual Notes

There are limitations regarding assumptions about the values of the capacitors and the forward voltage of the diodes, which may affect the analysis. The discussion also reflects uncertainty about the conditions under which the diodes conduct.

Gbox
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Homework Statement


Graph ##V_{out}##
0wGEWB8.png


Homework Equations

The Attempt at a Solution



When ##V_{in}=V## C1 is positive on the left and negative on the right, and C2 is negative on the the down side and positive on the upper side so there is no Vout?
 
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Gbox said:
When Vin=VVin=VV_{in}=V C1 is positive on the left and negative on the right, and C2 is negative on the the down side and positive on the upper side so there is no Vout?

That's not correct.

For the moment let's assume that V is much greater than the forward voltage of the diode...

When Vin = +V the right hand diode is reverse biased (not conducting) so that can be temporarily removed. The left hand diode is forward biased (conducting) so replace it with a wire. Redraw the circuit to show this.

Then... Are you familiar with the "Potential Divider" circuit?
 
CWatters said:
That's not correct.

For the moment let's assume that V is much greater than the forward voltage of the diode...

When Vin = +V the right hand diode is reverse biased (not conducting) so that can be temporarily removed. The left hand diode is forward biased (conducting) so replace it with a wire. Redraw the circuit to show this.

Then... Are you familiar with the "Potential Divider" circuit?
Why is the left hand diode is conducting? the capacitors do not affect the voltage? do not they both charge to the same voltage and the diode is not conducting?
 
No they don't charge to the same voltage (unless they are same value). The two caps are in series.
 
This is what the circuit looks like when Vin transitions from 0 to +V (assuming +V >> Vd)...

Caps.jpg
 

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