I was studying the JFET. In the books and other material it is mentioned that when the voltage between the drain and source is increased then drain current is also increased. After a certain point (Pinch Off) the drain current becomes constant and will remain constant even if the drain voltage is increased.
If we see the physical structure of the device the depletion region will start to increase when the voltage between the drain and source is increased. After the same certain point (Pinch off Point)depletion region touches and then current cannot move forward.
My question here is that then how the current becomes steady or current will pass when the depletion regions are closed. Doesn't the current become zero ?
If the depletion region is closed by Voltage between drain and source then what is the role of voltage between gate and source. Although I can see the steady current value is changed at different Voltage gate to source and a logic of faster build of depletion region (Through Voltage to Source and Drain to source) also comes in mind but what is the main logic behind it ?
Before replying please consider me as basic learner. Thanks
Best Answer
Here is my short (simplified, but clear and descriptive) answer:
With rising Vds the channel becomes smaller and smaller - and the corresponding channel resistance Rds is increasing continuosly. Therefore, and due to the geometrical properties of the device, the ratio Vds/Rds (which is identical to the current Id) is nearly constant.