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Electron-Inversion-Density-Function-Of-Surface-Potential 

In the calculation of depletion region width and depletion capacitance the space charges could not move and the holes were largely already relatively negligible and thus modification of charge density due to Vg was not necessary in the first order.

The inversion channel is formed using mobile electron carriers and therefore charge density is going to be a function of Vg applied. See here for explanation of this: Electric Field and Charge Density Forms

For weak inversion where the depletion charge >> inversion charge

Inversion Charge:

{dQ_I}={q}{(n-n_0)}{dx}

Related rates of differential elements of  {phi} and {dx}

{d}{phi}=-{E}{dx}

-{{d}{phi}}/E={dx}   

substituting and setting up for surface interface values:

{dQ_I}={q}{(n_s-n_0)} {{-d}{phi_s}}/E_s

-{dQ_I}/{{d}{phi_s}}={{q}{(n_s-n_0)}} /E_s

The inversion capacitance is:

{C_I}=-{dQ_I}/{{d}{phi_s}}={{q}{(n_s-n_0)}} /E_s

Substituting the identity for surface electric field:

{C_I}=-{dQ_I}/{{d}{phi_s}}={{q}{(n_s-n_0)}} /{sqrt { {2 phi_s Phi_s} }}  where: {(n-n_0)} {approx} {N_a(e^{u_s-2u_F})}

Breaking down the Es in the phrase:

{C_I}=-{dQ_I}/{{d}{phi_s}}={{q}{(n_s-n_0)}} /{sqrt { {2 phi_s {qN_a / varepsilon_s}} }}

Simplifying:

{C_I}=-{dQ_I}/{{d}{phi_s}}={{sqrt{{q}{varepsilon_s}{N_a}}}(e^{u_s-2u_F})} /{sqrt { {2 phi_s} }} ….this is almost the same as equation 2.1.28 of CMOS Analog Design Using All Region MOSFET Modeling

If you want to include the linear term of depletion region electric field then substitute  {(phi_s - phi_t)}   for  {phi_s}

{C_I}=-{dQ_I}/{{d}{phi_s}}={{sqrt{{q}{varepsilon_s}{N_a}}}(e^{u_s-2u_F})} /{sqrt {{2(phi_s - phi_t)}}} ………And the phrase now exactly matches the book.

Note that the inversion capacitance becomes equal to the depletion capacitance at Vt where surface potential = 2*Fermi level as previously explained.

Observation: This problem when approached using the Xd depletion width left me confused because the inversion charge builds up at the surface not Xd down from the interface.  It can be approached that way as we saw with the depletion capacitance the correct form will be arrived at.

 

Phi = {qN_a / varepsilon_s} E_surface = sqrt { {2 phi_s Phi_s} }
 x_d = sqrt { {2 phi_s } / {Phi_s} }  E_surface = Phi_s x_d 
 
{C_D/C_ox}={varepsilon_S}/{t_D}  {/}  {varepsilon_ox}/{t_ox}   gamma =  {varepsilon_s}/{1} {/} {varepsilon_ox}/{t_ox} {sqrt{2 Phi_s }}
 

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