# JFET Depletion Regions

Discussion in 'General Electronics Discussion' started by vivensub, Oct 11, 2015.

1. ### vivensub

8
0
Sep 18, 2015
So this is an extract from "Electronic Devices and Circuits", by David A Bell, 5th Edition, Oxford University Press, about an n-channel JFET.

"When a gate-source voltage(VGS) is applied with the gate negative with respect to the source, the gate-channel pn junctions are reverse biased. The channel is more lightly doped than the gate material, so the depletion regions penetrate deep into the channel."

Now, I can understand what the term "Depletion Region Penetration" means, because I read that concept before in BJTs, but how do we decide, given a particular biasing, where the depletion region penetrates furthest?(n-side or p-side)

If someone could clear up this doubt without too much electronics jargon, it'd be helpful.
(I've only done Diodes, Rectifiers, and BJTs before this)
Thanks.
-Vivek

2. ### Martaine2005

2,967
817
May 12, 2015
Hi Vivek,
Probably 3 or 4 members will be here soon to explain all. But be warned! It cannot be explained properly without the jargon.
Grab a large drink and sit back. It's gonna be a bumpy ride!!!

Edit: I just read this HERE and have it wrong too.

Martin

Last edited: Oct 11, 2015
3. ### Ratch

1,088
331
Mar 10, 2013
Evidently, you have not studied depletion regions (DR) very much. If you did, you would realize that it takes more lightly doped (concentration) material to balance the charge of the higher doped material on the other side of the DR. Therefore, you can expect the width of the lightly doped part of the DR to extend out farther than the opposite side.

http://physics.stackexchange.com/questions/88573/change-in-the-width-of-depletion-layer-with-doping

Ratch

4. ### Ratch

1,088
331
Mar 10, 2013
Why can't it be explained simply? It is a straightforward principle that does not need a no holds barred brawl like the one we had discussing the concept of whether a current or voltage controls the current of a BJT.

A BJT controls current by passing charge through the depletion region (DR). A FET controls current by passing charge along the DR, i.e. orthogonally (90°) to what a BJT does.

Ratch

5,164
1,081
Dec 18, 2013
If we take a N channel JFET for example. When you connect a voltage source across the drain and source you will get a current through the device. As the voltage across the drain and source increases the depletion region nearest the drain elongates its self up towards the drain. Sort of making a triangle shape in the channel. But this will only elongate to a maximum, so the current will sort of level out and any further increase of drain source voltage will not greatly increase the current through the device.

Now the only way to shrink the channel further to turn off the device is to bring out the depletion region in the channel at the bottom of the device. This is done by reducing the gate source voltage. So just as the drain elongated the depletion region when it had a positive potential with respect to the source. The source does the same with the source having a positive potential with respect to the gate. This then pushes the depletion region out and closes the channel. This is called pinch off.

Thanks