MOSFET Threshold Voltage Formula Derivation

Published by Fudgy McFarlen on June 21, 2014June 21, 2014

At threshold voltage Vt the surface potential is

[pmath] {phi} = {2 phi_F} [/pmath]

Given this condition threshold voltage is:

[pmath] V_T0 = V_FB + {2 phi_F} + {{varepsilon_s}{E_s}}/C_ox [/pmath] ….the 3rd term is V=Q/C

Reorganizing terms:

[pmath] V_T0 = V_FB + {2 phi_F} +{{varepsilon_s}/{1}{/}{varepsilon_ox}/{t_ox}{E_s}} [/pmath]

Substituting the identity for Es:

[pmath] V_T0= V_FB + {2 phi_F} +{{varepsilon_s}/{1}{/}{varepsilon_ox}/{t_ox}{sqrt {{2 phi_s Phi_s}}}} [/pmath]

[pmath] V_T0= V_FB + {2 phi_F} +{{varepsilon_s}/{1}{/}{varepsilon_ox}/{t_ox}{sqrt {{4{phi_F}{Phi_s}}}}} [/pmath]

[pmath] V_T0= V_FB + {2 phi_F} +{{varepsilon_s}/{1}{/}{varepsilon_ox}/{t_ox}{sqrt{2 Phi_s}sqrt{2 phi_F}}} [/pmath]

Substituting for the body effect factor identity gamma:

[pmath] V_T0= V_FB + {2 phi_F} + {gamma}{sqrt{2 phi_F}}[/pmath]

Equation 2.1.59 can be found in: Tsividis:Operation and Modeling of the MOS Transistor :Page 110

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

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