Circuit with Diodes, Capacitors, and Resistors

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Discussion Overview

The discussion revolves around analyzing a circuit containing diodes, capacitors, and resistors, specifically focusing on calculating the voltage difference V3(t) - V4(t) after a voltage source is activated. The problem involves initial conditions, the behavior of capacitors, and the effects of diodes in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant outlines the initial conditions and equations governing the circuit, noting the use of node analysis and KCL to express current through capacitors and resistors.
  • Another participant introduces the concept of the half-wave effect, suggesting that capacitors charge and discharge based on the input waveform.
  • A participant expresses uncertainty about the polarity comparison of the capacitors and the implications of diode states on the voltages V1 and V2.
  • One participant hints at starting with high resistor values to analyze the charging behavior of the capacitors before adjusting the time constant to 0.1s.
  • Another participant emphasizes the need to understand the waveform shape at the diodes rather than focusing solely on algebraic expressions, suggesting sketching the waveforms for clarity.
  • A later reply notes that the frequency of the excitation may significantly affect the results, indicating the importance of this factor in the analysis.

Areas of Agreement / Disagreement

Participants express various viewpoints on the behavior of the circuit, with no consensus reached on the implications of the diode states or the exact nature of the voltage waveforms. The discussion remains unresolved regarding the calculations and interpretations of the circuit behavior.

Contextual Notes

Participants mention the initial conditions and assumptions regarding the diodes' states, but there are unresolved aspects regarding the current from the voltage source and the lack of closed loops for KVL application. The discussion also highlights the dependence on the chosen time constant and the waveform shape.

kawaiiRose
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Homework Statement


Compute V3(t)-V4(t). The capacitors are initially discharged, the voltage source is turned on at t = 0, and the diodes’ forward voltage drop is negligible (Vf = 0). What is the value of V3(10s) – V4(10s)?

Vi=4sin(2∏t)
RC= 0.1s

JhX5b9g.jpg


Homework Equations


i (over capacitor) = CdV/dt
i (over resistor) = V/R
Assume Diode is either on (id>0) or off (id=0)

The Attempt at a Solution



I used the node method: node above the top capacitor has voltage V1 and node above the bottom capacitor has voltage V2. Using KCL at those two nodes, I found id1=(CdV1/dt)+(V3/R) and id2=-(CdV2/dt)-(V4/R). Unknowns are id1, id2, V1, V2, V3, and V4.

Based on whether I assume a diode is on/off I can assume V1=Vi or V2=Vi and that id1=0 or id2=0, so that eliminates two unknowns. I'm still left with 4 variables though and can't see any other equations to solve for them. I was thinking use the node between the two diodes and the voltage source, but I don't know the current coming off the source. And there aren't any closed loops for me to use KVL.
 
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Look at what the half-wave effect has - capacitors are charging for half a cycle and then discharging the other half. Compare the polarity of upper and lower caps when semi-charged.

Have you used the expression for v(t) input that results from the diodes?
 
I'm not too sure what you mean by comparing the polarity.

From the diodes, I only know that if D1 is on (and D2 is off), V1=Vi and V2<Vi. If D2 is on (and D1 is off), V2=Vi and Vi<V1. If both are off, V2<Vi<V1. I can sub in Vi but I don't know any other V(t) input I can use in regards to the other voltages.
 
kawaiiRose said:

Homework Statement


Compute V3(t)-V4(t). The capacitors are initially discharged, the voltage source is turned on at t = 0, and the diodes’ forward voltage drop is negligible (Vf = 0). What is the value of V3(10s) – V4(10s)?

Vi=4sin(2∏t)
RC= 0.1s

JhX5b9g.jpg


Homework Equations


i (over capacitor) = CdV/dt
i (over resistor) = V/R
Assume Diode is either on (id>0) or off (id=0)

The Attempt at a Solution



I used the node method: node above the top capacitor has voltage V1 and node above the bottom capacitor has voltage V2. Using KCL at those two nodes, I found id1=(CdV1/dt)+(V3/R) and id2=-(CdV2/dt)-(V4/R). Unknowns are id1, id2, V1, V2, V3, and V4.

Based on whether I assume a diode is on/off I can assume V1=Vi or V2=Vi and that id1=0 or id2=0, so that eliminates two unknowns. I'm still left with 4 variables though and can't see any other equations to solve for them. I was thinking use the node between the two diodes and the voltage source, but I don't know the current coming off the source. And there aren't any closed loops for me to use KVL.

kawaiiRose said:
I'm not too sure what you mean by comparing the polarity.

From the diodes, I only know that if D1 is on (and D2 is off), V1=Vi and V2<Vi. If D2 is on (and D1 is off), V2=Vi and Vi<V1. If both are off, V2<Vi<V1. I can sub in Vi but I don't know any other V(t) input I can use in regards to the other voltages.

Hint -- start with the resistor values very high (so the discharge time constant is many seconds. What does each capacitor charge to (that is Simon's hint)? Now, change the time constant to 0.1s -- does that change your answer by much?
 
The supply if giveing you a waveform of something like: vi(t)=Vsin(wt)
The sketch of this is a sign wave.
This is the wave at each diode.

The top diode means that the circuit after it will only get a particular shape v1(t) driving it.
That shape is not the same as vi(t). It is not a simple "on" vs "off" state either.
You are thinking too algebraically: look for understanding first.
What is that shape? Sketch it. Don't worry about being exact. What does the sketch look like?

Do the same for the effective driving voltage for the lower circuit - what is that?

...and thank you berkman :)
 
BTW, the frequency of the excitation is pretty low, so your answer may change by enough to be important

"Vi=4sin(2∏t)"
 
Piling on! :smile:
 

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