ltspicemosfetnmossaturation
I have a question regarding a mosfet simulation I am trying to perform.
Here is the circuit in question.
We can see that Vd=1kV , Vs=0V and Vg=10V(it is a pulsed source)
Now to my understanding I should be seeing the 1kV at the output when the we have the saturation equation satisfied. 
When I simulate it on the other hand I do not see that response.
Here is the response I see: 
First of all, "saturation" in mosfets means that change in VDS will not produce significant change in the Id (drain current). You can think about MOSFET in saturation as a current source. That is regardless of the voltage across VDS (with limits of course) the current through the device will be (almost) constant.
Now going back to the question:
According to wikipedia, the MOSFET is in saturation when V(GS) > V(TH) and V(DS) > V(GS) - V(TH).
That is correct.
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.
You increased The Vgs above Vth of the NMOS so the channel was formed and device started to conduct.
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.
You increased the Vgs making the device conducting more current. At Vgs = 4V the thing that is limiting amount of current is no longer transistor but resistor that you have in series with transistor.
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).
Everything looks in order here.
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?
No it is not. It is in linear or triode region. It behaves as resistor in that region. That is increasing Vds will increase Id.
If this is the case, how would one design a circuit in which the MOSFET is in saturation?
You already have. You just to need take care for operating point (make sure that conditions that you have mention are met).
A) In linear region you can observe following: -> when increasing the SUPPLY voltage, the LED will get brighter as the current across resistor and transistor will rise and thus more will be flowing through the LED.
B) In saturation region something different will happen -> when increasing SUPPLY voltage, the LED brightness will not change. The extra voltage that you apply on the SUPPLY will not translate to bigger current. Instead it will be across MOSFET, so the DRAIN volage will rise together with supply voltage (so increase supply by 2V will mean increasing drain volage by almost 2V)
For a given gate-source voltage and above a certain drain-source voltage, the MOSFET behaves like a constant current sink. Say you have 3 volts applied to the gate with respect to the source - if if the drain-source voltage rises to 2 volts, the drain current reaches a plateau of about 5 milliamps. If you increase the drain-source voltage (whilst keeping gate voltage at 3 volts), the drain current remains largely at 5 mA.
This is broadly referred to as "constant current" mode of operation.
Is the current drawn determined by the transistor or the load itself?
The current is determined by the transistor mainly but, because there is a slight slope there is a small change in current with a big change in drain load resistance.
for example, at saturation is this the max current it can draw or the amount of current it will draw?
This is the typical current it will draw for a given gate-source voltage (3 volts in my example).
Best Answer
Your circuit is a source follower. This means that the voltage at the source "follows" the gate. In fact it lags behind by a couple of volts so, if the gate is at (say) 10 volts you might see 6 to 8 volts (typically) at the source.
A source follower has no voltage amplification just like an emitter follower has no voltage amplification. If you want either to act as a switch then, the gate needs to be driven with respect to the source by 10 volts.
Image from this question and answer. Or from this answer: -
The answer explains that the saturation region for a BJT is not the same region for a MOSFET.