# Is pinch off voltage of JFET constant?

The drain current is related to gate to source voltage by the following equation:

$I_d = I_{dss} (1 - \frac{V_{gs}}{V_p})^2$

Now, with changing $V_{gs}$, I think $V_p$ also changes. I am studying Electronic devices and circuit theory by Boylestad and Nashelsky where they are using $V_p$ defined at $V_{gs}=0$ for calculation of $I_d$ at all levels of $V_{gs}$ though it is nowhere mentioned that $V_p$ is going to be constant with changing $V_{gs}$. Is it the case or am I mistaken somewhere? Please clarify the doubt.

## 2 Answers

If you look at the situation where you hold Vgs constant and then increase Vds: Vds appears as a voltage drop along the channel, increasing as you move from source to drain. Therefore the reverse bias voltage between gate and channel also varies with length and is highest at the drain end. The channel acquires a tapered shape and the Id --> Vds relationship becomes non-linear. At high enough Vds the channel pinches off and you get saturation.

The pinch off conditions are dependant upon not just Vgs but also Vds.

This is from the physics of the situation.

What I am not sure of is if Vp is defined for a set of conditions.

• if that is the case Vp MAY not vary even though you get pinch off as very different Vgs levels depending upon Vds.
• but that would be because it is a defined level.
• once defined you should be able to use it in all situations
• unless the transport mechanism changes -> ballistic transport.

-> note I'm not certain about this last part.

• Suggestion to whoever approved the previous suggested edit by Vasiliy: If the edit seems to be directly contradictory to a specific technical detail, check the facts before approving the edit. Some of us do make incorrect edits, the peer review process is supposed to address this. Commented Aug 18, 2013 at 16:40

I've never read this book, but the usage of $V_p$ in the equation you provided is consistent with the definition of $V_p$ as the static pinch-off voltage. In other words:

$$V_p=V_{gs_{pinch-off}}(V_{ds}=0)$$

The above definition defines $V_p$ unambiguously and independently of Drain-to-Source bias.

In other words - this is just a parameter of the device and it does not change.

BTW, there is simple way to calculate this voltage. Denoting the "height" of the channel as $2a$ and requiring that the combined depletion regions' widths equal exactly $2a$ one gets:

$$V_p=\frac {q}{2\epsilon}N_{ch}a^2-V_{bi}$$

where $V_{bi}$ is the built in Gate-to-Channel voltage.