capacity multiplier

Discussion in 'General Electronics Chat' started by Fabien_the_french_one, Jun 1, 2015.

  1. Fabien_the_french_one

    Thread Starter New Member

    May 26, 2015
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    0
    Hello,
    I'm currently trying to understand this circuit which is supposed to be a basic capacity multiplier :
    [​IMG]
    To compute the equivalent impedance of it, the website from which it's cpoied gives :
    VA = VB = 0
    VE = i/jCω
    VS = – R1⋅i
    hence
    VS = – jR1⋅Cω⋅VE
    moreover,
    VE – VS = R2⋅(I – i)
    thus,
    I = i + (I – i) = jCωVE + (VE – VS)/R2
    I = jCωVE + VE/R2 + jR1⋅CωVE/R2
    I=VE(1/R2+jCω(1+R1R2))
    Then they write that this is equivalent to an impedance (C⋅(1 + R1/R2)) // R2
    (where // means "parallel to", I don't know if it's an international notation for it)

    But, one thing seems illogical to me : if it was equivalent to it, sould not there be I = (Vs-Ve) (1/R2+jCω(1+R1R2)) ?
    I would like to control a capacity by choosing R1, but I don't understand how to use it because it's the (Vs-Ve) voltage which interrests me .

    thank you in advance if you have an answer

    Best regards
     
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