Thus, there is a narrow strip of n-channel between the two-depletion regions where I D can flow. And as soon as I D goes to 0, there is no voltage drop across the channel which was causing the reverse bias of the channel and p wells in the first place. But if they end up touching each other, there is no n-channel left for the drain current to flow. This is known as the channel pinch off and the voltage V DD ( = V DS ) at which this occurs is known as the pinch off voltage.Īs we increase the voltage V DS further, the depletion regions do approach each other at the top, and at V P they come very close to each other. Soon there will be a point where the depletion regions from either side of the channel expand inwards and touch each other until there is hardly any channel left in between. What happens if you keep increasing V DS? The current will increase and so will the width of the depletion region (especially at the top near to the drain) as it gets more and more reverse biased.

The width of the depletion region can be increased by reverse biasing the pn junction. This side-by-side region of immobile positive and negative ions is known as the depletion region – a region depleted of mobile charge carriers. Negative fixed ions are similarly formed on the p side when the hole and the electron combine. As they do so, they leave behind positive fixed ions on the n side. Referencing image 3 below, you can see that when a p-type semiconductor meets an n-type semiconductor, the electrons from the n side move to the p side and combine with the holes there.

A brief discussion of the same is given in the next section, but it's highly recommended that you can refer to the tutorials linked to get a better grasp of the subject.īoth n- and p-type semiconductors have mobile charge carriers – electrons and holes respectively.

Before understanding why this is done in a JFET, it's beneficial to first review what happens when a p-type and n-type semiconductor meet each other side by side.

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