Like a bipolar transistor, a JFET consists of three layers of P and N silicon. However, these layers are connected in a different way. The picture below shows the inside of an N-channel JFET.
The two P layers are connected to each other and form the Gate. The Source and the Drain are both connected to the N layer. That's why this is called an N-channel JFET.
Let's connect an N-channel JFET to some voltage sources:
While discussing the inside of a diode, we saw that the depletion zone between a PN junction increases if the reverse voltage increases. So if VGS becomes more negative, the depletion zone between the P layers and the N layer will become thicker, narrowing down the channel between the Source and the Drain. The voltage at which the channel is closed, is called the pinch-off voltage.
Now let's see what happens if VGS remains constant and VDS increases:
At first ID increases as well, just like the D-S channel were a resistor. However, increasing VDS also makes VGD more negative, narrowing down the channel. When VGD has reached a certain voltage, the pinch-off voltage, ID cannot increase any further. The FET has now become saturated.
So if VGD is between 0 and the pinch-off voltage, the S-D channel looks like a resistor; its resistance can be controlled by VGS. If VGD becoms less than the pinch-off voltage, the S-D channel acts like a current source that can be controlled by VGS.
We see that both VGS and VGD have a "pinch-off value". Since JFETs are symmetrical, the pinch-off voltages are the same. You may even swap the Drain and the Source!