# Understanding p JFET

I would like to make sure that I understood p JFET (https://electronics.stackexchange.com/a/97350/78253) right:

With SWitch1 in its current position, there is voltage at the gate of the J1 p JFET, the depletion layer is large, current through J1 is blocked, D1 is dark.

With SWitch1 closing to the ground, there is no voltage at the gate of the J1 p JFET, the depletion layer is very small or not there at all, current can flow through J1, D1 is shining (correct voltage V1 and resistance R1 provided).

Correct or what did I make wrong?

• The arrow indicates a diode. Throwing the switch will destroy the JFET. Aug 12 '15 at 11:07

Keep in mind a couple of things that are valid for every JFET (both N- and P-channel):

1. For many models source and drain terminals are interchangeable, i.e. there is absolutely no difference between the two (their datasheets will often tell you this upfront).

2. Even when S and D are not interchangeable, they work almost the same, i.e. if you exchange them you'll get the device still working, but with some (maybe also big) performance degradation. So, even in this case, it doesn't make much difference what terminal (D or S) you consider to understand the general operation of the device in a circuit.

3. The current through the channel, i.e. the current from D to S (or S to D, remember the "interchangeability" thing of points 1 and 2) depends on the voltage across gate and source (or drain, ... interchangeability above... from now on I will simply mention S when I could also mention D instead). The larger the voltage (in absolute value) the narrower the channel, hence less current from D to S.

4. JFETs are depletion-mode devices (P- and N-channel alike). This means that with no gate-source voltage applied the channel is already formed and it can conduct current. In this latter situation (Vgs=0) the current is called $I_{DSS}$ in datasheets, and it is (practically) the maximum current your device can handle. In your case, if you want to switch an LED, you must ensure that the current the LED needs is less than the $I_{DSS}$ of the JFET you are using (much less, if you want the JFET to operate in the ohmic region, so that it doesn't dissipate excessive power).

The only (main) difference between N- and P- channel devices is the polarity of Vgs needed to control the DS current. Since a JFET has a PN junction (i.e. a rectifier diode) from gate to channel, it is paramount not to bring this diode into conduction, otherwise the JFET won't work and may also be damaged. Therefore the gate diode must always be reverse biased (or slightly forward biased, but let's not go there for simplicity).

The arrow on the gate tells you the direction of the diode (think of it as the arrow in the diode symbol). Therefore, to reverse bias that diode you have to apply a positive Vgs (gate more positive than source) for P-channel devices, whereas Vgs must be negative (gate more negative than source) for N-channel devices.

In your case, when the switch is like it is drawn, S and G are shorted, hence Vgs=0, so the JFET will conduct (and this would be true also for a N-channel type). If you connect G to ground you will have the source more positive than gate, and this is bad because the gate diode will be brought into conduction, and this will damage the device since there is nothing that limits the current in the gate diode in that case.

Incorrect.

If you connect the gate of a P channel JFET to ground while one of the other pins (JFETs are symmetric, source and drain are interchangeable) is connected to a positive voltage you will get current through the gate.

P-JFET is essentially a bar of P semiconductor wrapped in N semiconductor. This means that you get a diode with its cathode at the gate. You should never forward-bias this diode.

So, if you apply zero volts to the gate, then maximum current flows trough the channel. If you connect the gate to a positive voltage, then depletion layer forms between channel and gate and the current is reduced.

A n-JFET is controlled like a vacuum tube - zero volts on the gate open it, negative voltage shut it off.

The gate voltage here is relative to the drain or source, not ground.

Study this graph: -

Source is connected to positive rail (as per your diagram). VGS in the graph is positive and therefore control voltages are from the positive rail (VGS = 0V) and above (VGS = +X volts)

With VGS at 0V the most current flows for a given DS voltage. As you take VGS more positive less current flows thru the drain.

It's probably easier to visualize this with an N channel JFET: -

Here, source is at 0V and the gate is taken below 0V to gradually switch off the device.