I'm currently studying FETs (Field Effect Transistors) in Navy school. What I know so far is that in FETs, $V_{gs}$ 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?
Best Answer
The exact equations for I-V characteristics of transistors are derived using quantum-mechanics. Several approximations can be used, one of which is based on the shottky barrier analysis
This reference here derives the I-V linear and quadratic approximation (in saturation) for FET transistors.
Another reference here
UPDATE:
As @QMechanic pointed, Electrical Engineering should be better suited for this question.