# Question about Shockley's equation for JFETs

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?

• I think you're talking about a JFET, not a MOSFET.
– Null
Nov 12, 2014 at 3:06
• Please note that W. Shockley´s equation describes the voltage-current relationship for a pn junction only. Hence, the equation as shown has nothing to do with Shockley.
– LvW
Nov 12, 2014 at 8:19
• Sorry - I have to revise/delete my above comment. Hearing about Shockley I immediately think of the exponential characteristics of a pn junction. However, the equation under discussion also is called "Shockley`s equation".
– LvW
Nov 12, 2014 at 8:41