small signal circuit analysis

Discussion in 'Homework Help' started by veerendra, Jan 30, 2011.

  1. veerendra

    Thread Starter New Member

    Jun 18, 2010
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  2. Georacer

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    Nov 25, 2009
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    What is your question? Where are you having trouble exactly?
     
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  3. veerendra

    Thread Starter New Member

    Jun 18, 2010
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    hi, my question is substitute fig B in fig A and find Vout/Vin
     
  4. veerendra

    Thread Starter New Member

    Jun 18, 2010
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    friends please solve this
     
  5. Georacer

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    Nov 25, 2009
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    First of all, we won't do your homework for you. You have to make an initial effort to approach the solution, and we can help you after you are stuck. So post your work and we will comment on it.
     
  6. veerendra

    Thread Starter New Member

    Jun 18, 2010
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    here by applying nodal or mesh analysis

    I observed that no current flows to the second loop..

    then how the second loop will be activated....

    note:here 5V is the dependent source

    thanking you
     
  7. narasimhan

    Member

    Dec 3, 2009
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    Its a simple network. Let the BD line be the reference node.
    First mark the two ends(+ve and -ve ends) of the voltage V as V1 and V2.
    Find V1 in terms of Vin by applying voltage-divider rule.
    Then find V2 in terms of V.
    V=V1-V2 and find V in terms of Vin
    Then find Vout.
    If you still don't understand then I'll post the pic of solution.

    There are other ways to solve the prob using 2 port analysis. But to start with my above solution will do.

    No need to mention. Its implicit when you ask the question
     
  8. veerendra

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    Jun 18, 2010
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    how to find v2 interms of v
     
  9. narasimhan

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    Dec 3, 2009
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    Simple voltage=current*resistance. current=5V resistance=r3+r4
    V2=5V(R3+R4)
     
  10. veerendra

    Thread Starter New Member

    Jun 18, 2010
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    sorry 5v is voltage not current
     
  11. Adjuster

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    Dec 26, 2010
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    The output of the first potential divider is equal to the sum of the controlling potential V, and the dependant source voltage "5V".

    The input to the second divider is the controlled voltage source "5V". How is that expressed as a fraction of the output of the first divider?
     
  12. narasimhan

    Member

    Dec 3, 2009
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    Well the diagram says its current source. Either the question is incorrect or the 5V is the voltage dependent current source.
    If its the latter then the problem has a solution(which I have already suggested).
     
  13. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Right enough. The symbol on the diagram is for a current source ( the label "5V" means "5 times V", and does not tell you that its output is a voltage ).

    If the diagram is right my earlier post was nonsense. The problem is soluble in both cases, but you need to know which.

    @ veerendra: Can you confirm what the problem was really about?
     
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