Consider an NMOS device. When V

_{gs}_{t}the transistor is off, whatever the drain voltage i.e. I

_{ds }= 0. When V

_{gs}>V

_{t}, the device conducts.

If V

_{gs}is constant and V_{ds}is variable then the resulting I_{ds}Vs V_{ds }curve have two region-
a)

**Resistive Region:**
When V

_{ds}_{gs}-V

_{t}, V

_{gs}>V

_{t},

Voltage across SiO

_{2}at source, V_{gs}= V_{g}-V_{s}
Voltage across SiO

_{2}at drain, V_{gd}= V_{g}-V_{d}
But, voltage across SiO

_{2}is not constant. Let’s assumed that channel is linear so voltage along the channel increases from the source to the drain. Since, the voltage across SiO_{2}increases from drain to source. So there are infinite capacitances.
Consider one capacitance of length dx situated at a distance x meters from the drain, as shown in figure below-

The capacitance,

*C = Wεdx/D*
The voltage v in excess of V

_{t}across the capacitor is,*v = V*

_{gd}+ x(V_{ds}- V_{t})/L = V_{gs}– V_{ds}+ xV_{ds}/L - V_{t}
The charge, q, induced on this capacitor is,

*q =Cv = Wdx (V*

_{gs}– V_{ds}+ xV_{ds}/L - V_{t})/D
The total charge,

*Q*, induced in the channel is-Now,

*Q = tI*

_{ds}=> I_{ds}= Q/t
Where, t is time to move across the channel.

And,

t = channel length, L/electron velocity

Again,

μ

_{n}= electron velocity/electric field = electron velocity/(V_{ds}/L)
so,

Since, V

_{ds}< V_{gs}- V_{t}
so,

A linear relationship between I

_{ds}and V_{ds}for a constant V_{gs}.
b)

**Saturation Region:**
When V

At V

_{ds}≥ V_{gs}–V_{t}. As the drain voltage rises, the voltage across SiO_{2}at the drain decreases.At V

_{ds}= V_{gs}- V_{t}
V

_{gd}= V_{gs}– V_{ds}= V_{gs}– V_{gs}+ V_{t}
=>V

_{gd}= V_{t}
This is the voltage necessary to just support inversion, this point on I

_{ds}Vs V_{ds}curve is called “Pinch-off”.
At V

_{ds}>V_{gs}– V_{t}, the point at which inversion ceases moves away from the drain.
The voltage difference along the inversion channel from the source to where it ceases is V

_{gs}– V_{t}and excess potential V_{ds}– V_{gs}+ V_{t}is dropped between the end of the inversion channel and the drain. It’s creates a high electric field across this very short distance and electrons are quickly swept across this area.
At and above pinch-off, the voltage between a and b is constant at V

_{gs}– V_{t}. The channel length also be considered to be constant.
Thus current flow is constant and the device is said to be saturated.

So, V

_{ds }= V_{gs}- V_{t}_{}

But, in practical situation, L decreases when V

_{ds }increases so I

_{ds}slightly increases.

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