A simple question but nearly make me crazy

When the current of M1 is larger than I1,the mos's gm is decided by the I1, right? It is the equation gm=(2KW/LI)^1/2, not the gm=KW/L(Vgs-Vth).But the pervious equation is derived from the equation I=1/2KW/L(Vgs-Vth)^2 and gm=KW/L(Vgs-Vth),so the precondition of using the equation gm=(2KW/LI)^1/2 is that the I is decided by the mos,not the current supply, why do we still can use it, why does the conflict comes out?

usually, Ids depends on Vds in saturation region. Actually, the resistance Rdson control the Ids. I don't know what you want to know.

I think he means I1 is in conflict with the the current source controlled by Vin. Yes there is conflict. They see who's stronger and then compromise. If the current source I1 is ideal or stronger, Vds of the transistor will change so that channel modulation will play a bigger role. Thus they compromise and live happily together. Put lamda in I=1/2KW/L(Vgs-Vth)^2, then there will be no conflict.

To better visualize this, make a set of curves on paper of the transistor characteristics. This is the usual one with the drain current on the vertical axis and the drain-source voltage on the horizontal axis. There will be a set of curves for various gate-source voltages. Then draw a horizontal line at the drain current of 10 mA. This will intersect the transistor curves at several points. That is each Vgs curve will cross the 10 mA horizontal line at one point which at low values of Vgs will be the drain-source breakdown where the curve shoots up to very high current at high Vds values. Then the coordinates of that intersecting point will give you a set of paired values of Vds and Vgs.

If you want to go further, you can draw another curve of these pairs of points and get the "transfer curve" for the circuit.


Note that this method assumes that the current source is an ideal one.

Thanks to you all, I clear about the problem.

I dont really know what do you mean.but in this circuit, current of mos cant exceed I1, but decided by I1, and always equal to it, because of the ideal current source. If Vin approaches power supply, vds will descend rapidly to meet the current limitation.

I think if the current of M1 is larger than I1, M1 will be forced to operate in the triod region while not the saturation region. Then some formulas will be not valid any more.