Diode circuit with AC excitation source

Click For Summary
SUMMARY

The discussion focuses on analyzing a diode circuit with an AC excitation source, specifically addressing the output voltage (Vout) when the input voltage (Vin) is applied. It is established that when Vin is positive, the left diode becomes forward biased and conducts, while the right diode is reverse biased and does not conduct. The capacitors C1 and C2 do not charge to the same voltage due to their series configuration, which affects the overall circuit behavior. The concept of the "Potential Divider" is also introduced as a relevant principle in understanding the circuit dynamics.

PREREQUISITES
  • Understanding of diode behavior in circuits
  • Familiarity with AC excitation sources
  • Knowledge of capacitor charging in series
  • Basic principles of potential divider circuits
NEXT STEPS
  • Study diode characteristics and their impact on circuit design
  • Explore AC circuit analysis techniques
  • Learn about capacitor behavior in series and parallel configurations
  • Investigate potential divider applications in electronic circuits
USEFUL FOR

Electrical engineering students, circuit designers, and hobbyists interested in understanding diode circuits and their behavior under AC excitation.

Gbox
Messages
54
Reaction score
0

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?
 
Physics news on Phys.org
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
 

Similar threads

Engineering How Does a Capacitor Affect Ripple Voltage in a Diode Circuit?
  • · Replies 15 ·
Replies
15
Views
7K
Engineering DC source rather than a clamper circuit?
  • · Replies 6 ·
Replies
6
Views
2K
Why does amplifier sink power for Vout>0 and sources power for Vout<0?
  • · Replies 1 ·
Replies
1
Views
2K
Engineering Understanding the output waveform and using LTSpice
  • · Replies 1 ·
Replies
1
Views
2K
Engineering What is the output voltage of this ideal diode circuit?
  • · Replies 19 ·
Replies
19
Views
5K
Engineering How to analyze diode circuits? (Find waveform generated)
  • · Replies 2 ·
Replies
2
Views
2K
How to determine drain-source voltage for this MOSFET amplifier.
  • · Replies 0 ·
Replies
0
Views
2K
Engineering Ac Circuit Ideal Diode: Find Vout-t
  • · Replies 2 ·
Replies
2
Views
2K
How do I calculate the voltage of a diode in a simple electronics circuit?
  • · Replies 9 ·
Replies
9
Views
2K
Engineering Constant current source transistor circuit - differences between 2 types?
  • · Replies 5 ·
Replies
5
Views
3K