# Consider an n-channel metal-oxide-semiconductor field-effect transistor (MOSFET) with a gate-to-source voltage of 1.8 V. Assume that $\frac{W}{L}=4$

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Consider an n-channel metal-oxide-semiconductor field-effect transistor (MOSFET) with a gate-to-source voltage of 1.8 V. Assume that $\frac{W}{L}=4$, μnCox = 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 ________.

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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 (gd) 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:

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

gd = 28.35 μ Seimens