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Difference between revisions of "equivalent oxide thickness"

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Where <math>A\epsilon_r</math> is the relative [[dielectric constant]] of SiO<sub>2</sub> in our case. Therefore one calculate the equivalent oxide thickness as,
 
Where <math>A\epsilon_r</math> is the relative [[dielectric constant]] of SiO<sub>2</sub> in our case. Therefore one calculate the equivalent oxide thickness as,
  
−
:: <math>\frac{A\epsilon_{SiO2}\epsilon_0}{t_{ox}} = \frac{A\epsilon_{high-κ}\epsilon_0}{t_{oxe}}</math>
+
:: <math>\frac{A\epsilon_{SiO2}\epsilon_0}{t_{ox}} = \frac{A\epsilon_{high-\kappa}\epsilon_0}{t_{oxe}}</math>
  
−
:: <math>\require{cancel}t_{oxe} = \frac{\cancel{A}\epsilon_{high-κ}\cancel{\epsilon_0}}{\cancel{A}\epsilon_{SiO2}\cancel{\epsilon_0}}t_{ox} = \frac{\epsilon_{high-κ}}{\epsilon_{SiO2}}t_{ox}</math>
+
:: <math>t_{oxe} = \frac{\cancel{A}\epsilon_{high-\kappa}\cancel{\epsilon_0}}{\cancel{A}\epsilon_{SiO2}\cancel{\epsilon_0}}t_{ox} = \frac{\epsilon_{high-\kappa}}{\epsilon_{SiO2}}t_{ox}</math>
  
 
Note that the dielectric constant SiO<sub>2</sub> is 3.9
 
Note that the dielectric constant SiO<sub>2</sub> is 3.9
  
−
:: <math>t_{oxe} = \frac{\epsilon_{high-κ}}{3.9}t_{ox}</math>
+
:: <math>t_{oxe} = \frac{\epsilon_{high-\kappa}}{3.9}t_{ox}</math>
  
 
Where <code>t<sub>oxe</sub></code> is the equivalent oxide thickness, <code>ε<sub>high-κ</sub></code> is the [[dielectric constant]] of the [[high-κ]] material used, and <code>t<sub>ox</sub></code> is the physical oxide layer thickness.
 
Where <code>t<sub>oxe</sub></code> is the equivalent oxide thickness, <code>ε<sub>high-κ</sub></code> is the [[dielectric constant]] of the [[high-κ]] material used, and <code>t<sub>ox</sub></code> is the physical oxide layer thickness.

Revision as of 14:21, 23 May 2017

Not to be confused with Oxide Thickness (tOX).

Equivalent Oxide Thickness (EOT), represented by teq or tOXE, is the gate oxide thickness of the SiO2 layer of a transistor that would be required to achieve similar capacitance density as the high-κ material used.

A gate dielectric with a dielectric constant that is substantially higher than that of SiO2 will initially have a much smaller equivalent electrical thickness. This key feature allowed for the industry to continue on with Moore's Law. As the semiconductor industry began to experiment with transitioning from a SiO2 gate oxide to a high-κ material, EOT can be used to quickly compare those materials using existing SiO2-based models.

Equation

One can treat MOSFET behavior like two parallel plate capacitors,

Equation upper C Subscript o x Baseline equals StartFraction upper A epsilon Subscript r Baseline epsilon 0 Over t Subscript o x Baseline EndFraction

Where Equation upper A epsilon Subscript r is the relative dielectric constant of SiO2 in our case. Therefore one calculate the equivalent oxide thickness as,

Equation StartFraction upper A epsilon Subscript upper S i upper O Baseline 2 Baseline epsilon 0 Over t Subscript o x Baseline EndFraction equals StartFraction upper A epsilon Subscript h i g h minus kappa Baseline epsilon 0 Over t Subscript o x e Baseline EndFraction
Equation t Subscript o x e Baseline equals StartFraction CrossOut upper A EndCrossOut epsilon Subscript h i g h minus kappa Baseline CrossOut epsilon 0 EndCrossOut Over CrossOut upper A EndCrossOut epsilon Subscript upper S i upper O Baseline 2 Baseline CrossOut epsilon 0 EndCrossOut EndFraction t Subscript o x Baseline equals StartFraction epsilon Subscript h i g h minus kappa Baseline Over epsilon Subscript upper S i upper O Baseline 2 Baseline EndFraction t Subscript o x

Note that the dielectric constant SiO2 is 3.9

Equation t Subscript o x e Baseline equals StartFraction epsilon Subscript h i g h minus kappa Baseline Over 3.9 EndFraction t Subscript o x

Where toxe is the equivalent oxide thickness, εhigh-κ is the dielectric constant of the high-κ material used, and tox is the physical oxide layer thickness.

Example

For example, consider Hafnium Dioxide (HfO2) which has an Equation epsilon Subscript r Baseline equals 24 (subject to variations in temperature). A layer of just 1 nm in thickness would result in an equivalent oxide thickness of around Equation t Subscript o x e Baseline equals StartFraction 24 Over 3.9 EndFraction 1 nm equals 6.15 nm . This is indeed the material used by Intel following their transition to high-κ at the 45 nm process node.

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