Question regarding the output for AC circuit with capacitors

  • Thread starter karan000
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  • #1
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Homework Statement


df1746a830.jpg

Both diodes are ideal.

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


Homework Equations


Q=CV

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:

Answers and Replies

  • #2
CWatters
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1. For t=0 to Pi:
Cap1 and Cap2 are in series and hence Q1=Q2, therefore Vc1 = Vc2.

Check that. What about the left hand diode?

(I assume you mean Vi is +ve during t=0 to Pi)
 
  • #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.
 
  • #4
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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
 
  • #5
CWatters
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2. For the first negative cycle, Vc2 = Vi + Vc1 = 2Vi. As the polarity of the output terminals reversed, the output should be -(2Vi).

Think about the right hand diode. How does a -ve cycle affect the output at all?
 
  • #6
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Think about the right hand diode. How does a -ve cycle affect the output at all?

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?
 
  • #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
 
  • #8
8
1
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
Many thanks for your help!
 

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