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

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

1. ### genxiumNew 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_tWell-Known MemberMost 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. ### genxiumNew 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

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5. ### moffyMember

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