# MOSFET saturation relation between Vgs and Vds

We know that:

$$V_{in} < V_{out} + V_{t0}$$ and $$V_{in} >= V_{t0}$$

Where $V_{in}$ is the input voltage (say voltage gate source)

$V_{out}$ is the output voltage say $V_{ds}$ (voltage drain source)

$V_{t0}$ is the threshold voltage

Say we are neglecting all the higher order effects and also assume that $V_{sb}$ (voltage between substrate and source) is zero.

Then is my logic correct of interpreting the above equation?

That $V_{gs}$ is proportional to $I_d$ (drain current) i.e. it sets a limit that it can be the maximum allowable current before the drain current saturates and starts to pinch off.

And $V_{ds}$ is also proportional to $I_d$. Hence when we increase the $V_{ds}$ the $I_d$ will increase and after some point it will start to saturate hence at that point $V_{ds} + const$ ($V_{t0}$) must be greater than $V_{gs}$ as $V_{gs}$ did set the upper limit on the amount of allowable $I_d$ before pinchoff.

Please correct me if I am wrong.

$V_{gs}$ defines the thickness of the channel under the Gate (in a 3D mosfet model). Think of this channel as a hallway whose width you can increase as you increase $V_{gs}$. Think of people walking through the hallway as electrons ( $I_{d}$). Think of yourself as a god who can force people to have to run through this hallway (force = applied $V_{ds}$, which creates $I_{d}$)
If the hallway is wider ($V_{gs}$ is higher), you can fit more running people (electrons) before people start getting stuck in a tight trafficy environment (current saturation).
If you keep the hallway constant ($V_{gs}$ is constant), the there is only a finite amount of people (finite $I_{d}$) you can force (increase $V_{ds}$) before people start getting stuck and jammed up ($I_{d}$ saturates). If you try to force more people into the jammed hallway (try to increase $I_{d}$ by increasing $V_{ds}$), you will barely be able to. It will be a lot harder for you because the doorway is already jam-packed ($I_{d}$ = saturated) with people (current).