capacity multiplier

Thread Starter

Fabien_the_french_one

Joined May 26, 2015
5
Hello,
I'm currently trying to understand this circuit which is supposed to be a basic capacity multiplier :

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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