# Consider an n-channel MOSFET having width W, length L and electron mobility in the channel is µn and capacitance per unit area is C_ox. If gate to source voltage V_GS = 0.7V, drain to sour voltage V_DS = 0.2 V, µ_nC._ox = 120µA/V, threshold voltage V_T = 0.4V and (W/L) = 60. Calculate the transconductance gm in mA/V

_{n}and capacitance per unit area is C_ox. If gate to source voltage V_GS = 0.7V, drain to sour voltage V_DS = 0.2 V, µ_nC._ox = 120µA/V, threshold voltage V_T = 0.4V and (W/L) = 60. Calculate the transconductance gm in mA/V

### Right Answer is:

1.44 mA

#### SOLUTION

Given

V_{GS} =0.7

V_{DS} = 0.2

V_{Th} = 0.4V

V_{DS} < V_{GS} – V_{Th} = 0.7 – 0.4 = 0.3

**Linear (Ohmic) Region ** – with V_{GS} > V_{threshold} and V_{DS} < V_{GS} the transistor is in its constant resistance region behaving as a voltage-controlled resistance whose resistive value is determined by the gate voltage, V_{GS} level.

The magnitude of current increases linearly with increasing drain voltage till a particular drain voltage determined by the following relations –

VGS ≥ Vth

VDS < VGS – Vth

For Linear region

${I_D} = \dfrac{{{\mu _n}{C_{ox}}{W_N}}}{{{L_N}}}\left[ {\left( {{V_{GS}} – {V_{TN}}} \right){V_{DS}} – \dfrac{{V_{DS}^2}}{2}} \right]$

where,

μ_{N} is the mobility of electrons in the drain-source channel in cm^{2}/Vsec

C_{ox} is the gate oxide capacitance in F/cm^{2}

W_{N} is the width of the NMOS transistor

L_{N} is the length of the NMOS transistor

V_{TN} is the threshold voltage of the NMOS transistor

Now for transconductance gm diffrentiate the above equation by V_{GS}

**g _{m} = δI_{D}/δV_{GS} = μ_{N}.C_{ox}.V_{DS}.[W/L]**

=120 × 10^{–6} × 60 × 0.2

**= 1.44 mA**