The diagram above shows a very simple JFET amplifier. Let's make some DC calculations (no input signal). T1 is a BF245A. The datasheet tells us that the y_fs is 3mA/V if -1<V_GS<0. So we choose V_GS = -0.5V. At that voltage, I_D = 2.5mA. We want V_OUT to be 6V, so RD = 6V/2.5mA = 2.4k.

Since V_G = 0V and V_GS = -0.5V, V_S must be 0.5V. So RS = 0.5V/2.5mA = 200Ω.

Next, we connect a 0.1V_t sinus signal to the IN termial. What will be the the amplitude of the output signal? A 0.1V change in V_GS causes a 0.1V∙3mA/V = 0.3mA change in I_D and thus a 0.3mA∙2.4k = 0.8V change in the output signal. So the voltage gain is 8 (or actually -8; it's an inverting amplifier).

Capacitor CS makes sure that V_S remains constant, so that v_GS = v_IN (for AC signals).

RI is usually 1M or so. It guarantees that V_IN = 0V (DC), while the input resistance remains very high.

Just like the current gain (h_FE) of a bipolar transistor may vary over a wide range, so may the forward transfer admittance of a JFET. In case of the BF245A, y_FS may vary between 3 and 6.5mS. Far worse is the fact that the pinch-off voltage of a BF245A ranges from -0.25 to -8V. That means that at V_GS=-0.5V, I_D can be much less or greater than the 2.5mA mentioned above; that was just a typical value. Resistor RS is therefore often replaced by a current source.