Current vs Voltage Characteristics - Derivation

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Current vs Voltage Characteristics - Derivation

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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 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:

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

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.

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

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.

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Current vs Voltage Characteristics - Derivation

Field Effect Transistors

Field Effect Transistors

Metal Oxide Semiconductor Field Effect Transistor - Mosfet

Metal Oxide Semiconductor Field Effect Transistor - Mosfet

Amplifiers

Amplifiers

Inverters

Inverters

Current vs Voltage Characteristics - Derivation