# How current is steady after pinch off voltage?

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 ?

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.

To answer this question, it is necessary first to recall the form of the current density in a semiconductor device according to the customary Drift-Diffusion model, then to see what's happen precisely in a JFET: I apologize for not being elementary, but the answer requires some concepts which, I hope, I'll made understandable.

The current density across the channel transversal section of a JFET. Since the JFET, as all Field Effect devices, is a majority carrier device, let's consider a $$\n\$$-channel JFET and see that the biggest contribution to the current density is due only to the motion of electrons (call it $$\J_n\$$), which has the following form $$J_n=\underbrace{q\mu_n nE}_\text{Drift}+ \underbrace{qD_n\nabla n}_\text{Diffusion}\label{1}\tag{1}$$ where

• $$\q \$$ is the elementary electron charge,
• $$\n\$$ is the conduction electron density in the region considered,
• $$\E\$$ is the electric field applied to the device,
• $$\\mu_n\$$ and $$\D_n\$$ can be considered as proportionality constants and their meaning is not useful to this discussion, so we'll not describe it.

The current density consist of two contribution of very different nature. Without giving too many details, we can say that the drift term at the right side of equation \eqref{1} identifies a ohmic (i.e. resistive) contribution since is proportional to the electric field (and thus the the voltage applied to the device) and its magnitude depends on the availability of electrons and thus on the magnitude of $$\n\$$ in a given region. On the other hand, the diffusion term is proportional to the magnitude of variation of the charge density $$\n\$$ trough its gradient $$\\nabla n\$$, and not on the magnitude of $$\n\$$ itself.

What happens in a JFET at Pinch OFF? Let's consider the qualitative behavior of the current conduction by using the following picture, where it is supposed $$\V_{GS}=0\mathrm{V}\$$ for simplicity and without restriction to generality. Let's call $$\V_P\equiv V_{GS_\mathrm{th}}\$$ the thresold or pinch-off or ON gate voltage the voltage that should be applied to between the gate and the source of the JFET in order to bring to (almost) zero its drain current.

• In a) we have $$\V_{DS} and the depleted region near each gate-channel $$\pn\$$ junction increases its width when moving from the source to the drain terminals, where the conduction electron charge density $$\n\$$ is nearly $$\0\$$. However, this region does not yet extend in such a way to fill the whole channel section: there is still a conduction path between the source and the drain, thus the dominating contribution to the drain current is given by the drift term in \eqref{1} and $$\I_D\$$ rises almost proportionally to $$\V_{DS}\$$.
• In b) we have exactly $$\V_{DS}=V_P\$$, thus there exists a full channel section of the JFET where $$\n\simeq 0\$$: the drift contribution to the drain current ceases to be the main one, and is in fact reduced consistently. However $$\I_D\$$ does not ceases because the diffusion term in \eqref{1} becomes very high since, near the depleted section of the channel, there is a very strong variation of $$\n\$$ as it goes to its doping defined level to almost zero very quickly and this implies $$\\nabla n\gg0\$$. Thus $$\I_D\$$, from now on, is almost independent on the rise of $$\V_{DS}\$$ since the main contribution to its magnitude is the diffusion term that does not depend on the applied drain voltage. This is depicted in part c) of the picture which shows that, during the rise of $$\V_{DS}\$$ above the $$\V_P\$$ value, a greater part of the channel region becomes depleted, however influencing little the value of $$\I_D\$$ due to the fact that $$\\nabla n\simeq \text{const.}\$$ and due to the relative magnitude of the two terms in \eqref{1}.