The transconductance of an n-channel MOSFET operating in the satu

SOLUTION

Concept:

For a MOSFET in saturation, the current is given by:

\({I_{D\left( {sat} \right)}} = \frac{{W{μ _x}{C_{ox}}}}{{2L}}{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)

W = Width of the Gate

Cox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

Vth = Threshold voltage

The transconductance of a MOSFET is defined as the change in drain current(ID) with respect to the corresponding change in gate voltage (VGS), i.e. 

\({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\)

\(g_m = \frac{{W{μ _x}{C_{ox}}}}{{L}}{\left( {{V_{GS}} - {V_{th}}} \right)}\)

Application:

Given:

\(g_{m1} = (\mu_x C_{ox})(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)}_1\)

Since the transistor current for both the cases is the same:

I_D(sat)1 = I_D(sat)2

\(\frac{{μ _x}{C_{ox}}}{2}(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)_1^2}=\frac{{μ _x}{C_{ox}}}{2}(\frac{{W}}{{L}})_2{\left( {{V_{GS}} - {V_{th}}} \right)_2^2}\)

Given:

\((\frac{W}{L})_2=2(\frac{W}{L})_1\)   ---(1)

\(\frac{{μ _x}{C_{ox}}}{2}(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)_1^2}=\frac{{μ _x}{C_{ox}}}{2}(\frac{{2W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)_2^2}\)

\((V_{GS}-V_{th})_2=\frac{(V_{GS}-V_{th})_1}{√2}\)

Now, the new transconductance will be:

\(g_{m2} = (\mu_x C_{ox})(\frac{{W}}{{L}})_2{\left( {{V_{GS}} - {V_{th}}} \right)}_2\)

\(g_{m2} = (\mu_x C_{ox})(\frac{2{W}}{{L}})_1\frac{{\left( {{V_{GS}} - {V_{th}}} \right)}_1}{√2}\)   ---(2)

Using Equation (1) and (2), we can write:

\(g_{m2} = \sqrt2(\mu_x C_{ox})(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)}_1\)

∴ g_m2 = √2 g_m1