MOSFET-Substrate-Bias-Effects

  • The source and body may not be at the same potential. The curve represented as EFi is the Fermi level from the p substrate through the reverse biased source–substrate junction to the source contact.
  • The space charge region width under the oxide increases from the original xd @ threshold value when a VSB  > 0 is applied.  Since the maximum depletion region depth Xd was due to 2Ef it is now due to 2Ef + Vsb and the maximum Xd is bigger before Vt is reached.

@Threshold with:   

V_SB = 0

Q_Space = qN_A x_d_th 

With Vsb applied

V_SB = V_SB

Q_Space = qN_A x_d_V_SB 

And thus

Delta Q_Space = qN_A [ x_d_V_SB - x_d_th ] 

Substituting from the identity table below:

Delta Q_Space = {sqrt{2}}qN_A ( sqrt { {2 phi_s + V_SB}/{Phi_s} } - sqrt { {2 phi_s}/{Phi_s} } )

Factoring

Delta Q_Space = {sqrt{2}qN_A} / sqrt { {Phi_s} } ( sqrt { { 2 phi_s + V_SB } } - sqrt { { 2 phi_s } } )

The change in the threshold voltage is

Delta V_T = - {  { Delta }{Q_Space}}/{C_ox}  

Delta V_T = {{sqrt{2}}{qN_A }}/{C_ox circ sqrt { Phi_s }}[sqrt { 2 phi_s + V_SB  } - sqrt { 2 phi_s } ] 

And here we find the body effect factor.  It is defined as:

gamma ={sqrt{2}{qN_A }}/{C_ox circ sqrt { {Phi_s} }} 

 

The following is my work and needs to be verified

gamma ={sqrt{2}{qN_A / varepsilon_s}{varepsilon_s}}/{C_ox circ sqrt { {Phi_s} }} 

Substituting Identity

gamma ={sqrt{2}{{Phi_s}{varepsilon_s} } }/{{C_ox}{sqrt{ {Phi_s}} }} 

Since:

1/t_ox = {{C_ox}/{varepsilon_s}}

The following simple expression results

gamma = t_ox {{varepsilon_s}/{varepsilon_ox}} {sqrt{2 Phi_s }}

 

Identity Table

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 
 

 

 

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1 Comment

Prof. Von Nostrand MonsterID Icon Prof. Von Nostrand · June 17, 2014 at 10:10 pm

You have achieved V-pinch off. 

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