Question about Shockley's equation for JFETs

Question

I'm currently studying JFETs (Junction Field Effect Transistors) in Navy school. What I know so far is that in JFETs, \$V_{gs}\$ is reversed biased, creating a depletion zone. What this means in plain English is that the more negative the gate is with respect to the source, the more narrower the channel becomes, leading to more resistance in the drain to the source, so the drain to source acts like a resistor. There is a point when current flows constant and we call this the saturation point and denote this as \$v_{gs(off)}\$

We are introduced to this equation and I have no idea where it comes from:

$$I_d = I_{dss} \left(1 - \frac{V_{gs}}{v_{gs(off)}} \right)^2$$

Here \$I_d\$ is the current of the drain and \$I_{dss}\$ is the drain to source saturation. Can someone shed some light as to how we can get this equation?

Answer 1

OK first a little on notation. Your Vgs(off) is also called the Threshold voltage (Vt) and the pinch-of voltage (Vp), I think they all mean the same thing. The voltage where the device is off.
So I cracked open "Art of electronics" (2nd ed.), They talk about two regions of operation. A linear region where the the drain current is proportional to the gate source voltage.
Id ~ (Vgs-Vt) and a saturation region where the current is proportional to the square of that voltage.
Id ~ (Vgs-Vt)^2.

The details are involved. You might read some here.. and then look up the references.