I have the following circuit hooked up on a breadboard.
I vary the gate voltage using a potentiometer. Here is what confuses me: according to wikipedia, the MOSFET is in saturation when V(GS) > V(TH) and V(DS) > V(GS) - V(TH).
If I slowly increase the gate voltage starting from 0, the MOSFET remains off. The LED starts conducting a small amount of current when the gate voltage is around 2.5V or so. The brightness stops increasing when the gate voltage reaches around 4V. There is no change in the brightness of the LED when the gate voltage is greater then 4V. Even if I increase the voltage rapidly from 4 to 12, the brightness of the LED remains unchanged.
I also monitor the Drain to Source voltage while I'm increasing the gate voltage. The drain to source voltage drops from 12V to close to 0V when the gate voltage is 4V or so. This is easy to understand: since R1 and R(DS) form a voltage divider and R1 is much larger than R(DS), most of the voltage is dropped on R1. In my measurements, around 10V is being dropped on R1 and the rest on the red LED (2V).
However, since V(DS) is now approximately 0, the condition V(DS) > V(GS) - V(TH) is not satisfied, is the MOSFET not in saturation? If this is the case, how would one design a circuit in which the MOSFET is in saturation?
Note that: R(DS) for IRF840 is 0.8 Ohms. V(TH) is between 2V and 4V. Vcc is 12V.
Here is the load line that I plotted of my circuit.
Now, from what I've gained from the answers here is that in order to operate the MOSFET as a switch, the operating point should be towards the left of the load line. Am I correct in my understanding?
And If one imposes the MOSFET characteristic curves, on the above graph, then the operating point would be in the so called "linear/triode" region. Infact, the switch should reach that region as quickly as possible in order to work efficiently. Do I get it or am I completely wrong?