I recently came across this question in a series of articles about cross coupled pairs by Razavi:
In Figure 6, M1 and M2 are biased
and balanced by I1 and I2 ^ h I I 1 2 = .
At t = 0, Iin jumps from zero
to a small positive value, I0. We
intuitively expect that VX rises and
VY falls. However, viewing the XCP
as a resistance equal to -2/ , gm we
obtain V g XY = -( /2 m) ( I u0 t), concluding
that VX should descend
and VY should ascend! How do we
explain the discrepancy between
these two results?
based on the figure in the image.
Now, intuitively I expected that Vx would fall and Vy would rise as a positive current is flowing into Vx, doesn't that imply Vy has a more positive potential? Or is this a sign convention thing?
Best Answer
It's just a matter of stability, this circuit exhibits negative resistance and is hence unstable.
Any static quiescent point is a possible equilibrium, so if you force \$I_\text{in}\$ current and \$V_\text{Y}=-V_\text{X}={I_\text{in}}/{g_\text{m}}\$ circuit will (only ideally) remain in this condition if non perturbed and exhibit the expected negative resitance.
But if you, from any such an unstable equilibrium point, try apply any disturbance, system output will diverge.
You could try a different experiment: replace current disturbance \$I_\text{in}\$ with a voltage one.
In this case intuition (VD1 up, VG2 up, ID2 up, some current is sunk into Vin) matches math analisys.
Now system has become stable, why? What is the boundary condition?
The key point is negative resistance:
When you drive by an ideal current source, as anticipated, total equivalent resistance seen at X,Y nodes is negative \$r_\text{o}=-2/g_\text{m}\$
On the other hand driving by ideal voltage source you actually shunt ro with 0Ω internal resistance depleting negative resistance.