​Depletion-Region-Estimate

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

Depletion-Region-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

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