Question

Consider an n-channel metal-oxide-semiconductor field-effect transistor (MOSFET) with a gate-to-source voltage of 1.8 V. Assume that \(\frac{W}{L}=4\), μ_nC_ox = 70 × 10^-6 AV^-2, the threshold voltage is 0.3 V, and the channel length modulation parameter is 0.09 V⁻¹. In the saturation region, the drain conductance (in micro seimens) is ________.

Answer 1

Concept:

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

\({{I}_{D}}=\frac{1}{2}{{μ }_{n}}{{C}_{ox}}\left( \frac{W}{L} \right){{\left( {{V}_{Gs}}-{{V}_{T}} \right)}^{2}}\left( 1+\lambda {{V}_{DS}} \right)\)

W = Width of the Gate

Cox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

Vth = Threshold voltage

Calculation:

The drain conductance (g_d) is calculated as the rate of change of drain current with respect to the Drain to source voltage, i.e.

\(g_d= \frac{{\partial {{\rm{I}}_{\rm{D}}}}}{{\partial {{\rm{V}}_{{\rm{DS}}}}}} \)

\({{g}_{d}}=\frac{1}{2}{{μ }_{n}}{{C}_{ox}}\left( \frac{W}{L} \right){{\left( {{V}_{Gs}}-{{V}_{T}} \right)}^{2}}\lambda \)

Putting on the respective values, we get:

\({{g}_{d}}=\frac{1}{2}{\times 70\times 10^{-6}}\left( 4\right){{\left( {1.8-0.3} \right)}^{2}}\times0.09\)

g_d = 28.35 μ Seimens