Current vs Voltage Characteristics - Derivation

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

$I=Q_d\;.\;V$

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 VGS -VTH.

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:

$Q_d=WC_{ox}\left(V_{GS}-V_{TH}\right){\color{Blue} }$

Here we are multiplying COX with W to obtain total Capacitance per unit length.

Now make the Drain voltage VDS slightly positive (greater than Zero)

The Channel Voltage varies from VG at the Source end to VG-VDS 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 (QD) in terms of the point at which it is being measured, say x.

$Q_d\;\left(x\right)\;=WC_{ox}\left(V_{GS}\;-\;V\left(x\right)\right)-V_{TH})$

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:

$E\left(x\right)=-\frac{dV}{dX}$

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

$I_{D\;}=\;-WC_{OX\;}\left[\left(V_{GS\;}-\;V\left(X\right)-V_{TH}\right)\right]\mu_nE$

Subsequently replace E with -dV/dx,

$I_D=\;-WC_{ox}\left[V_{GS}-V(x)\;-\;V_{TH}\right]\mu_n\frac{dV(x)}{dx}$

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.

$\int_{x-0}^{x=l}I_Ddx=\int_{V=0}^{V_{DE}}\;WC_{OX}\mu_n\left[V_{GS}-V(x)-V_{TH}\right]dV$

After Integration, the value of ID is:

$I_D=\mu_nC_{ox}\frac WL\left[\left(V_{GS}-V_{TH}\right)V_{DS}-\frac12{V^2}_{DS}\right]$

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

$I_{D,max}=\frac12\mu_nC_{ox}\frac WL\left(V_{GS}-V_{TH}\right)^2$

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

$I_D\approx\mu_nC_{OX}\frac WL\left(V_{GS}-\;V_{TH}\right)V_{DS}$

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.