How do we use the linear regime? What exactly are we calculating when we draw the curve $I_d = f(V_{ds})$?

Well, the $I_d$ vs $V_{ds}$ curve is a property of the FET, for a given $V_{gs}$, and it's some function $I_d = f(V_{ds})$ (with a linear turn-on and then a saturation). For each value of $V_{gs}$ there's a different curve. In today's example, we were trying to figure out what $V_{gs}$ value to use to make a FET switch (like in Eggleston Fig. 5.9) turn on (i.e., pull a large $I_d$) or turn off.

To figure that out, we use Ohm's Law to write $I_d = \frac{V_{dd}-V_{ds}}{R_d}$. That's a straight line with y-intercept $V_{dd}/R_d$ and x-intercept $V_{dd}$ on the $I_d$ vs $V_{ds}$ plot. For a given $V_{gs}=V_g$ (where $V_g$ is the input voltage), the intersection of this line and the corresponding $I_d$ vs $V_{ds}$ curve tells us the value of $I_d$ you get for $V_g$ at the input.

So, let's take the JFET example (see Fig. 5.10). When we want the switch on (large $I_d$), the intersection of the line and top curve gives a large $I_d$. So $V_{gs}=0$ will turn the switch on. For switch off, we want small $I_d$, so we'd pick the bottom curve with $V_{gs}=-5 V$ (or thereabouts) where the intersection gives small $I_d$.

Note that the other kinds of FETs give different ranges of $V_{gs}$ corresponding to the different $I_d$ vs $V_{gs}$ curves. So if we are using different kinds of FETs, different input voltages will turn the FET on and off.