Question on MOS strong inversion

  • Thread starter htzzz
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  • #1
htzzz
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Hi, I am a bit confused about the MOS strong inversion case.
When Vg>>Vth, is the equation describing gate voltage at threshold still applicable? If not, what is the expression for corresponding surface potential and depletion width. Some books suggest that surface potential and depletion width won't change much, please explain why. If it's not easy to explain by posting, could anyone please send me a link or reference? Thank you!
 

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  • #2
phyzguy
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As you increase the charge on the gate, this charge needs to be compensated by charge in the semiconductor. At first, the charge is compensated by an increasing depletion region, which "uncovers" fixed charge in the semiconductor. However, as the gate voltage increases, you reach a point where it is energetically more favorable for charge on the gate to be compensated by inversion charge at the surface instead of further depletion. This is the threshold voltage. At this point the depletion region stops growing (this depletion region width is usually called xdmax), and the inversion charge starts increasing. The surface potential doesn't change much beyond this because the inversion charge is an exponential function of the surface potential, so it only takes a small change in the surface potential to get a large change in the inversion charge.

The maximum depletion width (xdmax) for a uniformly doped semiconductor is given by the relation:
[tex]X_{dmax}=\sqrt{\frac{K_s\epsilon_0 k_B T}{q^2N_A} log(\frac{N_A}{n_i})}[/tex]
 
  • #3
htzzz
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Thank you phyzguy!

So it means as the gate voltage increases, the voltage across the oxide will increase while the surface potential is constant?

And the assumption that semiconductor bulk charge is much greater than inversion charge (Qb >> Qinv) which was used in deriving the threshold voltage is no longer valid after inversion?

And is there a way to quantify the inversion charge besides using Qg - Qb = Qinv in strong inversion?
 

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