MOSFET Depletion Region Width Xd

Published by Fudgy McFarlen on June 15, 2014June 15, 2014

Gauss’s Law

$int{}{}{E circ dA} = q/ varepsilon_s =Phi$ This is the electric field flux

With a uniform charge density that is much more wide and long than it is thick you can use the following:

E circ A = q_total / varepsilon_s

$E circ A = -{qN_a Ax}/{varepsilon_s}$

$E = -{qN_a / varepsilon_s} x$ So as you progress upwards from 0 the flux increases.

d{phi} / dx = -E

$int{}{}{d{phi}/{dx}} = {qN_a}/{varepsilon_s} {int{0}{x_d}{x}}$

${phi_s} = {{qN_a}/{varepsilon_s}} {{x_d}^2}/2$

${2 phi_s varepsilon_s}/{qN_a} = {x_d}^2$

$x_d = sqrt {{2 phi_s varepsilon_s}/{qN_a}}$

Gathering up terms and using Gauss’s law:

$Phi = {qN_a / varepsilon_s}$

$x_d = sqrt {{2 phi_s}/{Phi_s}}$

Now that we have the estimate for the depletion width Xd:

$E_surface = {qN_a / varepsilon_s} x_d$

$E_surface = {qN_a / varepsilon_s}{sqrt{{2 phi_s varepsilon_s}/{qN_a}}}$

$E_surface = sqrt {{2 phi_s qN_a}/{varepsilon_s}}$

$E_surface = sqrt{{2 phi_s Phi_s}}$

Now the total charge is

$Q_total = {varepsilon_s} E_surface = {varepsilon_s} {sqrt{{{2 phi_s}{Phi_s}}}}$

Summary

$Phi = {qN_a / varepsilon_s}$ Planar Flux Density $E_s = sqrt { {2 phi_s Phi} }$
$x_d = sqrt { {2 phi_s } / {Phi} }$ $E_s = {x_d} {Phi}$

See equation: A2.1.13: CMOS Analog Design Using All Region MOSFET Modeling

Research Links

Gauss’s Law
Introduction to Solid State Devices: Chapter 10 See page 6, Body Effect coefficient page 49

Categories: ElectronicsMOSFET Basics

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