**Current vs Voltage Characteristics - Derivation:**

**Step 1: **

Consider a Semiconductor bar and it is carrying a current I in it.

Let Qd (Coulombs / meter) be the mobile charge density in the direction of the current flow.

Let v (meters / second) be the velocity of the charge.

Then

**Step 2:**

Let us find out the mobile charge density in a MOSFET.

Connect the Source and Drain terminals of the nMOS to ground.

The inversion starts at a value of VGS=VTH and the Gate Oxide Capacitance produces an inversion charge density which is proportional to V_{GS} -V_{TH}.

Now any Gate Voltage greater than the Threshold Voltage VGS>VTH, the charge on the Gate is mirrored by the charge in the channel.

This results in a uniform charge density ie charge / unit length along the Source - Drain path and is equal to:

Here we are multiplying C_{OX} with W to obtain total Capacitance per unit length.

Now make the Drain voltage V_{DS} slightly positive (greater than Zero)

The Channel Voltage varies from VG at the Source end to VG-V_{DS} at the Drain end, this shows that there is a local voltage difference between the Gate and the channel.

This necessitates the need to define the charge density (Q_{D}) in terms of the point at which it is being measured, say x.

Here V(x) is the voltage measured at a point x.

We know that

ID = QD. v

Substituting the value of QD in ID above, and put a negative sign to indicate that the charge carriers are negative,

As already stated earlier v is the velocity of electrons in the channel.

Also we know that the value of v in semiconductors is given by:

v=𝛍E

Where 𝛍 is the mobility of electrons i.e. charge carriers and E is the Electric Field

The Electric Field E(x) is given by:

To indicate the mobility of charge carriers electrons, let us replace the term 𝛍 with 𝛍n.

Replace the value of v with 𝛍E in the equation of ID, we have

Subsequently replace E with -dV/dx,

**Boundary Conditions:**

The Length of the Channel is indicated by L and it starts from point x=0 at Source to x=L at Drain.

Similarly V is V(0) =0 to V(L) = VDS

To calculate the Drain Current ID multiply both sides by dx and then integrate within the boundary conditions.

After Integration, the value of ID is:

The maximum value of ID will occur when VDS=VGS-VTH

The value VGS-VTH is called OverDrive Voltage

W/L is called the Aspect ratio

If VDS ≤ VGS-VTH , the device operates in the Triode Region.

If VDS<<2(VGS-VTH)

From this equation it can be observed that the Drain current ID is directly proportional to the voltage VDS.

From this equation the resistance R can be derived (ID/VDS)

As discussed earlier, increasing the value of VDS >VGS -VTH will make the transistor operate in the saturation region.

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