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

[genxium]

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, [LATEX]i_g=0[/LATEX], hence denote the voltage difference applied on [LATEX]d_1,d_2[/LATEX] is [LATEX]V[/LATEX], and the corresponding current is [LATEX]I[/LATEX], then

[LATEX]
\left \{ \begin{array}{c} v_{g_1}=v_{d_2}&v_{g_2}=v_{d_1}&v_{s_1}=v_{s_2}\\i_{d_1}\approx g_m \cdot (v_{g_1}-v_{s_1}-v_{th})\\i_{d_2} \approx g_m \cdot (v_{g_2}-v_{s_2}-v_{th})\\I=-i_{d_1}=i_{d_2}\\V=v_{d_2}-v_{d_1} \end{array} \right
[/LATEX]

Hence the equivalent impedance for this component combination is [LATEX]Z=\frac{V}{I}=-\frac{2}{g_m}[/LATEX].

 

[alec_t]

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.

 

[genxium]

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.

Thanks a lot for the reply!

In my opinion,[LATEX] I [/LATEX] is assumed to be flowing into d2, and, out of d1, to comply with the direction of [LATEX]V[/LATEX], hence [LATEX]2I=i_{d_2}-i_{d_1}[/LATEX]

 

[moffy]

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