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Question regarding the output for AC circuit with capacitors

  1. Jun 24, 2016 #1
    1. The problem statement, all variables and given/known data
    df1746a830.jpg
    Both diodes are ideal.

    Answers:
    (a) Vi
    (b) -Vi
    (c) -2Vi
    (d) none of the above


    2. Relevant equations
    Q=CV

    3. The attempt at a solution
    1. For t=0 to Pi:
    Cap1 and Cap2 are in series and hence Q1=Q2, therefore Vc1 = Vc2.

    From this we get Vi = Vc1 + Vc2, hence Vc1 =Vc2 = Vi/2.

    2. For t=Pi to 2Pi,
    95c95fd031.jpg

    The circuit should look like 3 batteries in series as such:
    fc5e62682b.jpg ,

    Therefore Vo = -(Vi + Vi/2 + Vi/2) = -2Vi,
    And so the answer is (c).

    BUT...
    Couldn't the circuit look like this too?

    851ddeac22.png

    And so Vo = - (1.5Vi - 0.5Vi) = -Vi
    Thus the answer is (b) ?

    BUT...
    We are just taking the voltage drop across the resistor, so Vo= Vc2 = 1/2Vi
    So the answer is (d).


    Which is correct and what exactly am I misunderstanding?
     
    Last edited: Jun 24, 2016
  2. jcsd
  3. Jun 24, 2016 #2

    CWatters

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    Check that. What about the left hand diode?

    (I assume you mean Vi is +ve during t=0 to Pi)
     
  4. Jun 24, 2016 #3

    CWatters

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    It's worth spending a moment just looking at the two diodes. What do the diodes mean for the polarity of the output voltage? Think.. What would need to happen to make it +ve? What would need to happen to make it go -ve.
     
  5. Jun 24, 2016 #4
    I hope this is a step in the right direction,

    1. For the first positive cycle, Vc1 = Vi and Vc2 does not charge is there is a short circuit due to the left ideal diode
    2. For the first negative cycle, Vc2 = Vi + Vc1 = 2Vi. As the polarity of the output terminals reversed, the output should be -(2Vi).
    3. For any positive cycle afterwards, the Vo remains -(2Vi) due to the short circuit from left diode.
    4. For any negative cycle afterwards, Vo remains -(2Vi) as verified from 2.

    48d6ce5501.jpg
     
  6. Jun 24, 2016 #5

    CWatters

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    Think about the right hand diode. How does a -ve cycle affect the output at all?
     
  7. Jun 24, 2016 #6
    The diode is reverse biased, so no current flow. Hence Vc2 to remains uncharged also and Vo is 0 for both the positive and negative cycle?
     
  8. Jun 24, 2016 #7

    CWatters

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    That's what I think. To get Vo = -Vi (or -2Vi) the output diode has to be the other way around like this..

    Gzaiv.jpg
     
  9. Jun 24, 2016 #8
    Many thanks for your help!
     
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