Hi,
(If i need negative voltage i`d turn the battery sign upside down and set voltage to 12.0V...but the function is the same..)
Let´s go step by step.
it is a depletion mode FET and it is slightely conductive, so there is a I-DS of about 7.1mA
Now from the bottom:
* There is the GND symbol. This is the voltage reference. It is 0V fixed.
* Then there is a battery. The "+" sign at the top, but with the voltage setut to "-12V" you get a "-12V" reading at the top.
* then 100R. Gate current through R3 is 0.0A. Current through FET is about 7.1mA. Causing a voltage drop of 0.71V: -12V-(-0.71V) = -11.29V
* then R3 to gate. No current, no voltage drop: therfore the same -11.29V
* then R1: again 7.1mA giving a voltage drop of 0.71V: -11.29V - (-0.71V) = -10.58V
* FET: Voltage at Gate: -11.29V, Voltage at Source: -10.58V giving a V_GS of -0.71V
What else do you want to know?
Klaus