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Why is the impedance of this cross-coupled MOSFET -1/gm?

Discussion in 'Mathematics and Physics' started by genxium, Sep 23, 2011.

  1. genxium

    genxium New Member

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    This is a figure I encountered in this article: LC_Oscillator, I calculated the impedance in this way and I got -2/gm, is there anyone can give me a hand? I think maybe I misunderstood something,some concepts about "impedance for port".

    My calculation

    For ideal NMOS FET,
    , hence denote the voltage difference applied on
    is
    , and the corresponding current is
    , then

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    Hence the equivalent impedance for this component combination is
    .
     

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  2. alec_t

    alec_t Well-Known Member Most Helpful Member

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    How do you get I = -id1 = id2 ? Where is I flowing?
    If I is the current difference id1-id2 (tho' it's not apparent from the drawing why it should be), then I = gm*(vg1 - vg2) = gm*(vd2-vd1) = gm*(-V), so Z = V/I = -V/(gm*V) = -1/gm.
     
    Last edited: Sep 24, 2011
  3. genxium

    genxium New Member

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    Thanks a lot for the reply!

    In my opinion,
    is assumed to be flowing into d2, and, out of d1, to comply with the direction of
    , hence
     
  4. dave

    Dave New Member

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  5. moffy

    moffy Member

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    You got -2/gm because you are evaluating Vd2 - Vd1, instead of say just Vd2. Each MOSFET contributes 1/2.
     

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